Potential theory

Laplace and Poisson equations and separation of variables.

In this chapter 8

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Laplace and Poisson equations

Laplace and Poisson equations

Theorem

Harmonic functions

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Shows that a harmonic function equals its average over every sphere and every ball lying in its domain, that the converse holds for $C^2$ functions, that consequently it attains its maximum and minimum only on the boundary unless it is constant, and that the Dirichlet problem for Poisson's equation therefore has at most one solution, which depends continuously on its data.

Prerequisites: Green's identities

Definition(Harmonic and Poisson)AdaptedAdapted from Strauss (2008), §6.1, p. 152; Ahlfors (1979), §4.6.1, p. 162; Evans (2010), §2.2, p. 25

Let $U \subseteq \mathbb{R}^n$ be open. A function $u : U \to \mathbb{R}$ is harmonic in $U$ if $u \in C^2(U)$ and

\nabla^2 u \;=\; \sum_{i=1}^n \frac{\partial^2 u}{\partial x_i^2} \;=\; 0 \qquad \text{at every point of } U .

The equation $\nabla^2 u = 0$ is Laplace's equation; the inhomogeneous equation $\nabla^2 u = f$ with $f$ given is Poisson's equation. Ahlfors, working in the plane, builds the continuity of $u$ and of its first and second partial derivatives into the definition and remarks that the regularity demanded can be relaxed; Strauss calls a solution of Laplace's equation a harmonic function without naming a smoothness class.

Two pieces of notation are used throughout. $B(x,r)$ is the open ball of radius $r$ about $x$ and $\partial B(x,r)$ its bounding sphere. If $\alpha(n)$ denotes the volume of the unit ball of $\mathbb{R}^n$ , then

\lvert B(x,r)\rvert = \alpha(n)\, r^n, \qquad \lvert \partial B(x,r)\rvert = n\,\alpha(n)\, r^{n-1} ,

so that $\alpha(2) = \pi$ , $\alpha(3) = \tfrac{4}{3}\pi$ , and the two sphere areas are $2\pi r$ and $4\pi r^2$ . We write $\partial u/\partial n = \hat{\mathbf{n}}\cdot\nabla u$ for the outward normal derivative, as in Green's identities.

Signs. Evans writes Poisson's equation as $-\Delta u = f$ , so that the operator $-\Delta$ is the non-negative one; Strauss and Ahlfors write $\Delta u = f$ and $\Delta u = 0$ . We use $\nabla^2$ for the Laplacian and put the sign where the source we are following puts it, saying each time which it is.

Theorem(Mean value property)AdaptedAdapted from Evans (2010), §2.2.2, Thm 2, p. 25; Strauss (2008), §7.1, Eq. (6), p. 181

A harmonic function carries no information at the centre of a ball that is not already carried, on average, by its values on the surrounding sphere. Precisely: let $U \subseteq \mathbb{R}^n$ be open, let $u$ be harmonic in $U$ , and let $x$ and $r > 0$ be such that the ball $B(x,r)$ and its boundary sphere lie in $U$ . Then

u(x) \;=\; \frac{1}{\lvert \partial B(x,r)\rvert}\int_{\partial B(x,r)} u \; dS \;=\; \frac{1}{\lvert B(x,r)\rvert}\int_{B(x,r)} u \; dV . \tag{MV}

In particular, the right-hand sides do not depend on $r$ . Evans states the two equalities together in $\mathbb{R}^n$ for $u \in C^2(U)$ and "each ball $B(x,r) \subseteq U$ "; we read that as including the boundary sphere, which the surface integral needs. Strauss derives the spherical form in three dimensions only, and remarks that the same idea works in $n$ dimensions.

Theorem(Converse)AdaptedAdapted from Evans (2010), §2.2.2, Thm 3, p. 26

Let $U \subseteq \mathbb{R}^n$ be open and let $u \in C^2(U)$ satisfy the first equality of (MV) for every ball $B(x,r) \subseteq U$ . Then $u$ is harmonic in $U$ .

Idea of the proofOriginal

Write $\phi(r)$ for the average of $u$ over the sphere $\partial B(x,r)$ and watch it as $r$ grows. Rescaling the integral to the fixed unit sphere lets us differentiate in $r$ under the integral sign, and the derivative that comes out is exactly the average of the normal derivative $\partial u/\partial n$ over the sphere. By Green's first identity with $v \equiv 1$ , that integral equals the integral of $\nabla^2 u$ over the ball, which is zero. So $\phi$ is constant in $r$ ; and as $r \to 0^+$ the average over a shrinking sphere tends to $u(x)$ by continuity. That gives the sphere form of (MV), and the ball form follows by integrating the sphere form in $r$ .

The converse is the same computation read backwards. The identity $\phi'(r) = \frac{r}{n}\times(\text{average of } \nabla^2 u \text{ over } B(x,r))$ holds for every $C^2$ function, harmonic or not. If the averages are constant in $r$ then $\phi' \equiv 0$ , so the average of $\nabla^2 u$ over every ball vanishes; a continuous function whose average over every ball vanishes is identically zero.

ProofAdaptedAdapted from Evans (2010), §2.2.2, pp. 25–26; Strauss (2008), §7.1, pp. 180–181

Step 1 (the spherical average, rescaled). Fix $x$ with $B(x,R) \subseteq U$ and for $0 < r < R$ put

\phi(r) \;=\; \frac{1}{\lvert\partial B(x,r)\rvert}\int_{\partial B(x,r)} u(y)\; dS(y) .

Substituting $y = x + rz$ maps $\partial B(x,r)$ onto the unit sphere $\partial B(0,1)$ and scales the surface element by $r^{n-1}$ , so, since $\lvert\partial B(x,r)\rvert = n\alpha(n)r^{n-1}$ ,

\phi(r) \;=\; \frac{1}{n\alpha(n)}\int_{\partial B(0,1)} u(x + rz)\; dS(z) . \tag{1}

The domain of integration in (1) no longer depends on $r$ .

Step 2 (differentiate). The integrand of (1) is $C^1$ in $(r,z)$ on $[0,R)\times\partial B(0,1)$ , because $u \in C^2$ , and the unit sphere is compact; so we may differentiate under the integral sign:

\phi'(r) \;=\; \frac{1}{n\alpha(n)}\int_{\partial B(0,1)} \nabla u(x+rz)\cdot z\; dS(z) .

Undoing the substitution, $z = (y-x)/r$ is the outward unit normal to $\partial B(x,r)$ at $y$ , and $dS(z) = r^{-(n-1)}dS(y)$ , so

\phi'(r) \;=\; \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} \nabla u(y)\cdot\frac{y-x}{r}\; dS(y) \;=\; \frac{1}{\lvert\partial B(x,r)\rvert}\int_{\partial B(x,r)} \frac{\partial u}{\partial n}\; dS . \tag{2}

The derivative of the spherical average is the average of the normal derivative.

Step 3 (use Green's identity). The ball $B(x,r)$ has a smooth boundary and $u \in C^2$ on its closure, so part 2 of Green's identities — the case $v \equiv 1$ — applies. That unit displays its identities for a bounded region of $\mathbb{R}^3$ ; what we need here is the $n$ -dimensional version, which is the form recorded in the closing paragraph of the same statement block (Evans's hypotheses: $U \subset \mathbb{R}^n$ bounded and open with $C^1$ boundary, $u \in C^2(\overline U)$ ), and the ball $B(x,r)$ with $\overline{B(x,r)} \subseteq U$ meets them. It turns (2) into

\phi'(r) \;=\; \frac{1}{n\alpha(n)r^{n-1}}\int_{B(x,r)} \nabla^2 u\; dV \;=\; \frac{r}{n}\cdot\frac{1}{\lvert B(x,r)\rvert}\int_{B(x,r)} \nabla^2 u\; dV , \tag{3}

using $\lvert B(x,r)\rvert = \alpha(n)r^n$ in the second form. Note that (3) has used nothing about $u$ beyond $u \in C^2$ .

Step 4 (harmonicity). Now let $u$ be harmonic. Then the integrand in (3) vanishes, so $\phi'(r) = 0$ for all $0 < r < R$ and $\phi$ is constant there.

Step 5 (the limit at the centre). Let $\varepsilon > 0$ . Since $u$ is continuous at $x$ , there is $\delta > 0$ with $\lvert u(y) - u(x)\rvert < \varepsilon$ for $\lvert y - x\rvert < \delta$ . For $r < \delta$ , averaging this inequality over $\partial B(x,r)$ gives $\lvert \phi(r) - u(x)\rvert \le \varepsilon$ , because an average of $u(x)$ is $u(x)$ . Hence $\phi(r) \to u(x)$ as $r \to 0^+$ , and by Step 4 the constant value of $\phi$ is $u(x)$ . This is the first equality of (MV).

Step 6 (from spheres to the ball). Integrating in polar coordinates about $x$ and inserting the sphere result,

\int_{B(x,r)} u\; dV = \int_0^r\left(\int_{\partial B(x,s)} u\; dS\right) ds = \int_0^r u(x)\, n\alpha(n)s^{n-1}\, ds = u(x)\,\alpha(n)r^n = u(x)\,\lvert B(x,r)\rvert ,

which is the second equality of (MV).

Step 7 (the converse). Let $u \in C^2(U)$ satisfy the sphere equality for every ball in $U$ . Then $\phi$ is constant on $(0,R)$ for every admissible $R$ , so $\phi' \equiv 0$ , and (3) — which needed only $u \in C^2$ — says that

\int_{B(x,r)} \nabla^2 u\; dV = 0 \qquad\text{for every } B(x,r)\subseteq U .

Suppose $\nabla^2u(x_0) \ne 0$ for some $x_0 \in U$ ; replacing $u$ by $-u$ if necessary, assume $\nabla^2 u(x_0) > 0$ . Since $\nabla^2 u$ is continuous, it stays above $\tfrac12\nabla^2u(x_0) > 0$ on some ball $B(x_0,\rho) \subseteq U$ , whose integral is then strictly positive. That contradicts the display. Hence $\nabla^2 u \equiv 0$ on $U$ . $\square$

RemarkAdaptedAdapted from Ahlfors (1979), §4.6.1, Thm 19 and Eq. (60), p. 164, and §4.6.2, pp. 164–165

Ahlfors reaches the plane case of (MV) by a route with no divergence theorem in it. Write ${*}du = -u_y\,dx + u_x\,dy$ for the conjugate differential. His Theorem 19 (Eq. (60), p. 164) is that $\int_\gamma (u_1\,{*}du_2 - u_2\,{*}du_1) = 0$ whenever $u_1$ and $u_2$ are harmonic in a region $\Omega$ and the cycle $\gamma$ is homologous to zero in $\Omega$ . He proves it by reducing the claim to the boundary of a rectangle, where single-valued conjugate functions exist and the integrand is the imaginary part of an analytic differential, so Cauchy's theorem applies.

Two specialisations give the plane mean value property for a $u$ harmonic on the disc $\lvert z\rvert < \rho$ . On a circle $\lvert z\rvert = r$ one has ${*}du = r(\partial u/\partial r)\,d\theta$ .

With $\Omega$ the punctured disc $0 < \lvert z\rvert < \rho$ , $u_1 = \log r$ , $u_2 = u$ , and $\gamma$ the difference of two concentric circles — a cycle homologous to zero there — Theorem 19 gives the same value on both circles for $\int_{\lvert z\rvert = r} u \; d\theta - \log r \int_{\lvert z\rvert = r} r\,\frac{\partial u}{\partial r}\; d\theta ,$ so this combination is independent of $r$ . Harmonicity in the annulus between the two circles is all that is used.
With $u_1 = 1$ and $u_2 = u$ , Theorem 19 collapses to his earlier Eq. (58), $\int_\gamma {*}du = 0$ . On the punctured disc that only repeats that $\int_{\lvert z\rvert = r} r(\partial u/\partial r)\,d\theta$ is independent of $r$ ; a single circle is not homologous to zero there. To get the stronger conclusion that this integral vanishes, one needs $u$ harmonic on the whole disc, for then the circle $\lvert z\rvert = r$ is by itself homologous to zero and (58) applies to it alone.

Subtracting the second from the first leaves $\int_{\lvert z\rvert = r} u\,d\theta$ independent of $r$ , which is the mean value property; in the annulus, where only the first bullet is available, Ahlfors gets instead a circular mean of the form $a\log r + \beta$ (his Theorem 20, Eq. (61), p. 165). The conclusion is the same as Step 5's; the hypotheses differ in that Ahlfors's argument is about a region in the plane and is sensitive to the topology of $\gamma$ , whereas the argument above is dimension-free.

Theorem(Weak and strong maximum principle)AdaptedAdapted from Evans (2010), §2.2.3a, Thm 4, p. 26; Strauss (2008), §6.1, pp. 154–155; Stakgold & Holst (2011), §1.3, p. 73 and §8.3, pp. 495–496; Ahlfors (1979), §4.6.2, Thm 21, p. 166

Let $U \subseteq \mathbb{R}^n$ be open and bounded, and let $u \in C^2(U)\cap C(\overline U)$ be harmonic in $U$ . Then:

Weak maximum principle. $\max_{\overline U} u = \max_{\partial U} u, \qquad \min_{\overline U} u = \min_{\partial U} u .$
Strong maximum principle. If in addition $U$ is connected and $u(x_0) = \max_{\overline U} u$ for some $x_0 \in U$ , then $u$ is constant on $U$ .

Part 2 applied to $-u$ gives the same statement for an interior minimum. Equivalently: a nonconstant harmonic function on a connected bounded $U$ attains its maximum and its minimum on $\partial U$ and at no interior point.

The sources state this with different scaffolding. Evans assumes $U$ open and bounded and proves part 2 first, deducing part 1 from it. Strauss states it for a connected bounded open set in two or three dimensions, with the interior conclusion included, and proves the boundary bound of part 1 by the device below. Stakgold and Holst give the weak version for a bounded domain $\Omega$ with boundary $\Gamma$ and a continuous $u$ , in the form: if $-\nabla^2 u \le 0$ in $\Omega$ then $\max u$ is on $\Gamma$ , if $-\nabla^2 u \ge 0$ then $\min u$ is on $\Gamma$ , and if $\nabla^2 u = 0$ then both are — so for them the harmonic case is the overlap of two one-sided principles for subsolutions and supersolutions. Ahlfors states it as Theorem 21 without a boundedness hypothesis, for a harmonic function on a plane region: if it is not constant it attains no maximum value and no minimum value at any point where it is harmonic, and the extrema over a closed bounded subset are therefore attained on that subset's boundary.

Idea of the proofOriginal

There are two proofs, and they are worth having both because they need different things.

The first proves only part 1, and needs no mean value property. At an interior maximum of a $C^2$ function each second pure derivative is $\le 0$ , so the Laplacian is $\le 0$ there. That nearly contradicts $\nabla^2 u = 0$ , but not quite, since $\le 0$ allows $= 0$ . The repair is to tilt $u$ by a small multiple of $\lvert x\rvert^2$ , whose Laplacian is the positive constant $2n$ : the tilted function has strictly positive Laplacian, so it has no interior maximum at all, and letting the tilt go to zero transfers the boundary bound to $u$ .

The second proves part 2 from the ball form of (MV). If $u$ reaches its global maximum $M$ at an interior point, then $M$ is also the average of $u$ over a small ball around that point; but $u \le M$ there, and a function $\le M$ whose average is $M$ must equal $M$ throughout. So the set where $u = M$ swallows a whole ball around each of its points: it is open. It is also closed in $U$ , being a level set of a continuous function, and connectedness then forces it to be all of $U$ .

Proof(Part 1, by perturbing with

\varepsilon\lvert x\rvert^2

)AdaptedAdapted from Strauss (2008), §6.1, p. 155; Stakgold & Holst (2011), §1.3, pp. 72–73

Step 1 (no interior maximum for a strict supersolution). Let $w \in C^2(U)$ satisfy $\nabla^2 w > 0$ at every point of $U$ . If $w$ had a local maximum at $x_1 \in U$ , then each function $t \mapsto w(x_1 + t e_i)$ would have a local maximum at $t = 0$ , so $\partial^2 w/\partial x_i^2 (x_1) \le 0$ by the second-derivative test, for each $i$ . Summing over $i$ gives $\nabla^2 w(x_1) \le 0$ , contradicting the hypothesis. So $w$ has no local maximum in $U$ .

Step 2 (the tilted function). For $\varepsilon > 0$ set $v_\varepsilon(x) = u(x) + \varepsilon\lvert x\rvert^2$ . Since $\nabla^2\lvert x\rvert^2 = 2n$ and $u$ is harmonic,

\nabla^2 v_\varepsilon = 0 + 2n\varepsilon > 0 \qquad\text{in } U ,

so Step 1 applies to $v_\varepsilon$ : it has no local maximum in the open set $U$ .

Step 3 (where the maximum is). Here boundedness of $U$ and continuity of $u$ on $\overline U$ are used: $\overline U$ is compact and $v_\varepsilon$ is continuous on it, so $v_\varepsilon$ attains a maximum on $\overline U$ . By Step 2 that maximum is not attained in $U$ , hence it is attained at some point of $\partial U$ .

Step 4 (remove the tilt). Let $L = \max_{y\in\partial U}\lvert y\rvert$ , finite because $U$ is bounded. For every $x \in \overline U$ ,

u(x) \;\le\; v_\varepsilon(x) \;\le\; \max_{\partial U} v_\varepsilon \;=\; \max_{y \in \partial U}\bigl(u(y) + \varepsilon\lvert y\rvert^2\bigr) \;\le\; \max_{\partial U} u \;+\; \varepsilon L^2 ,

the first inequality because $\varepsilon\lvert x\rvert^2 \ge 0$ . The left and right ends do not depend on $\varepsilon$ except through the last term, so letting $\varepsilon \to 0^+$ gives $u(x) \le \max_{\partial U} u$ for all $x \in \overline U$ . Since $\partial U \subseteq \overline U$ , the reverse inequality is automatic, and the two maxima are equal.

Step 5 (the minimum). Apply Steps 1–4 to $-u$ , which is harmonic as well, and use $\min_{\partial U} u = -\max_{\partial U}(-u)$ . $\square$

Proof(Part 2, from the ball mean value property)AdaptedAdapted from Evans (2010), §2.2.3a, p. 26; Stakgold & Holst (2011), §8.3, pp. 495–496

Step 1 (set-up). Let $M = \max_{\overline U} u$ , which exists because $\overline U$ is compact and $u \in C(\overline U)$ , and let

E = \{\, x \in U : u(x) = M \,\} ,

assumed nonempty: $x_0 \in E$ .

Step 2 ( $E$ is open). Let $x \in E$ and let $r$ be any radius with $0 < r < \operatorname{dist}(x,\partial U)$ , so that $B(x,r) \subseteq U$ . The ball form of (MV) gives

M = u(x) = \frac{1}{\lvert B(x,r)\rvert}\int_{B(x,r)} u\; dV .

Hence $\int_{B(x,r)} (M - u)\, dV = 0$ . The integrand is continuous and non-negative, since $M$ is the maximum of $u$ over all of $\overline U$ . A continuous non-negative function with vanishing integral vanishes identically: if $M - u(y) > 0$ at some $y \in B(x,r)$ , it would exceed half that value on a small ball about $y$ and the integral would be positive. So $u \equiv M$ on $B(x,r)$ , and therefore $B(x,r) \subseteq E$ . Thus $E$ is open.

Step 3 ( $E$ is closed in $U$ ). $E = u^{-1}(\{M\}) \cap U$ and $u$ is continuous, so $E$ is the intersection of $U$ with a closed set: it is relatively closed in $U$ .

Step 4 (connectedness). $U$ is connected, and $E$ is a nonempty subset of $U$ that is both open and relatively closed. The only such subset is $U$ itself, for otherwise $U = E \cup (U\setminus E)$ would split $U$ into two disjoint nonempty relatively open pieces. Hence $E = U$ and $u \equiv M$ on $U$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Boundedness of $U$ , in Steps 3 and 4 of the first proof. It is what makes $\overline U$ compact and $L$ finite. Both conclusions genuinely fail on an unbounded domain: on the half-plane $U = \{(x,y) : y > 0\}$ the function $u = y$ is harmonic and vanishes on $\partial U$ , so $\max_{\partial U} u = 0$ while $u$ is unbounded above on $U$ . Ahlfors's Theorem 21 avoids the issue by asserting only the local statement — no interior extremum — and then drawing the boundary conclusion on a closed bounded set.
Continuity on $\overline U$ . Without it there is nothing to maximize on the boundary. On the punctured disc $U = \{0 < \lvert \mathbf{x}\rvert < 1\} \subset \mathbb{R}^2$ the function $u = \log(1/\lvert\mathbf{x}\rvert)$ is harmonic in $U$ and zero on the outer boundary circle, but it has no continuous extension to the boundary point $\mathbf{0}$ , and its supremum over $U$ is $+\infty$ .
Connectedness, in Step 4 of the second proof only. If $U$ is the union of two disjoint open balls, $u = 0$ on one and $u = 1$ on the other is harmonic on $U$ , continuous on $\overline U$ , and attains its maximum at every interior point of the second ball without being constant on $U$ . Part 1 survives: each component has its own boundary. This is why Evans attaches connectedness to part 2 and not to part 1.
$u \in C^2$ , in Step 1 of the first proof. The second-derivative test needs the pure second derivatives to exist. (By Evans's smoothness theorem, a continuous function with the mean value property is automatically $C^\infty$ inside $U$ , so for harmonic functions this costs nothing in the interior; see Poisson's integral formula.)
Which proof proves what. The first proof never uses the mean value property and gives only part 1; it is the one Strauss and Stakgold–Holst give, and it extends at once to subsolutions, $\nabla^2 u \ge 0$ , since Step 2 only needs $\nabla^2 v_\varepsilon > 0$ . The second proof uses (MV) and gives the sharper part 2 but says nothing about subsolutions directly. Strauss states the interior conclusion with the weak proof and defers the interior argument to his §6.3; Stakgold and Holst likewise prove the weak version in their §1.3 and the strong version only in §8.3, after the mean value theorem is available.

Theorem(Uniqueness, comparison and stability for the Dirichlet problem)AdaptedAdapted from Evans (2010), §2.2.3a, Thm 5, p. 27; Strauss (2008), §6.1, pp. 155–156; Stakgold & Holst (2011), §1.3, pp. 73–74

Let $\Omega \subset \mathbb{R}^n$ be a bounded domain with boundary $\Gamma$ , let $f \in C(\Omega)$ and $h \in C(\Gamma)$ , and consider

-\nabla^2 u = f \ \text{ in } \Omega, \qquad u = h \ \text{ on } \Gamma . \tag{D}

Solutions are required to lie in $C^2(\Omega)\cap C(\overline\Omega)$ .

Uniqueness. Problem (D) has at most one solution.
Comparison. If $u_1$ solves (D) with data $(f_1,h_1)$ , $u_2$ solves it with data $(f_2,h_2)$ , and $f_1 \ge f_2$ in $\Omega$ together with $h_1 \ge h_2$ on $\Gamma$ , then $u_1 \ge u_2$ in $\Omega$ .
Continuous dependence on the data. Let $v$ solve $-\nabla^2 v = 1$ in $\Omega$ , $v = 0$ on $\Gamma$ . Then every solution of (D) obeys $\lVert u\rVert_\infty \;\le\; \lVert v\rVert_\infty \,\lVert f\rVert_\infty \;+\; \lVert h\rVert_{\infty,\Gamma} ,$ where $\lVert\cdot\rVert_\infty$ is the supremum over $\overline\Omega$ and $\lVert h\rVert_{\infty,\Gamma} = \sup_\Gamma\lvert h\rvert$ . In particular the solution map is continuous from the supremum norm on the data to the supremum norm on the solution.

Stakgold and Holst prove 2 and 3 from the weak maximum principle alone, state 3 first for $h = 0$ (with the general case as their Exercise 1.3.2(a)), and observe that no constant smaller than $\lVert v\rVert_\infty$ works in 3, because taking $f \equiv 1$ makes it an equality. Evans states 1 for $f \in C(U)$ , $g \in C(\partial U)$ and solutions in $C^2(U)\cap C(\overline U)$ ; Strauss states and proves 1 for the form $\nabla^2 u = f$ .

Idea of the proofOriginal

All three parts are the weak maximum principle applied to a difference. Subtracting two solutions of (D) kills the source and the boundary data, leaving a harmonic function with zero boundary values; the maximum principle squeezes it between $0$ and $0$ . For comparison, the difference is not harmonic but a subsolution, $-\nabla^2 w \le 0$ , with $w \ge 0$ on $\Gamma$ , and the one-sided principle still puts its maximum on the boundary. For the bound, the function $v$ is used as a barrier: $\lVert f\rVert_\infty v \pm$ (the boundary bound) dominates $u$ by comparison, and evaluating the barrier gives the estimate.

ProofAdaptedAdapted from Evans (2010), §2.2.3a, p. 27; Strauss (2008), §6.1, pp. 155–156; Stakgold & Holst (2011), §1.3, pp. 73–74

Step 1 (uniqueness). Let $u_1, u_2$ both solve (D) and put $w = u_1 - u_2$ . By linearity $-\nabla^2 w = f - f = 0$ in $\Omega$ and $w = h - h = 0$ on $\Gamma$ , and $w \in C^2(\Omega)\cap C(\overline\Omega)$ . Part 1 of the maximum principle, applied to the bounded domain $\Omega$ , gives

\max_{\overline\Omega} w = \max_\Gamma w = 0, \qquad \min_{\overline\Omega} w = \min_\Gamma w = 0 ,

so $0 \le w \le 0$ : $w \equiv 0$ and $u_1 = u_2$ .

Step 2 (a one-sided principle for subsolutions). Let $w \in C^2(\Omega)\cap C(\overline\Omega)$ satisfy $-\nabla^2 w \le 0$ in $\Omega$ , that is $\nabla^2 w \ge 0$ . The proof of part 1 of the maximum principle goes through unchanged: $v_\varepsilon = w + \varepsilon\lvert x\rvert^2$ has $\nabla^2 v_\varepsilon \ge 2n\varepsilon > 0$ , so Steps 1–4 there give $\max_{\overline\Omega} w = \max_\Gamma w$ . (Only the upper bound survives; a subsolution may well take its minimum inside.)

Step 3 (comparison). Put $w = u_1 - u_2$ . Then $-\nabla^2 w = f_1 - f_2 \ge 0$ in $\Omega$ . Careful with the sign: this says $\nabla^2 w \le 0$ , so it is $-w$ that is a subsolution in the sense of Step 2. Applying Step 2 to $-w$ ,

\max_{\overline\Omega}(-w) = \max_\Gamma (-w) = \max_\Gamma (h_2 - h_1) \le 0 ,

because $h_1 \ge h_2$ on $\Gamma$ . Hence $-w \le 0$ , i.e. $u_1 \ge u_2$ throughout $\overline\Omega$ .

Step 4 (the barrier). Let $u$ solve (D), set $F = \lVert f\rVert_\infty$ and $H = \lVert h\rVert_{\infty,\Gamma}$ , and let $v$ be as in part 3. Consider

w_\pm = F\,v \pm u + H .

Then $-\nabla^2 w_\pm = F \pm f$ in $\Omega$ , which is $\ge 0$ since $\lvert f\rvert \le F$ ; and on $\Gamma$ , $w_\pm = 0 \pm h + H \ge 0$ since $\lvert h\rvert \le H$ . So $w_\pm$ is a nonnegative-data supersolution, and comparison against the zero solution — Step 3 with $(f_2,h_2) = (0,0)$ , $u_2 \equiv 0$ — gives $w_\pm \ge 0$ in $\Omega$ . Reading the two signs,

-\,F v - H \;\le\; u \;\le\; F v + H \qquad\text{in } \overline\Omega .

Step 5 (evaluate). Note that $v \ge 0$ by Step 3 (its data $1 \ge 0$ and $0 \ge 0$ dominate the zero data), so $\lvert F v + H\rvert \le F\lVert v\rVert_\infty + H$ . Taking the supremum of Step 4 over $\overline\Omega$ ,

\lVert u\rVert_\infty \;\le\; \lVert v\rVert_\infty\,\lVert f\rVert_\infty + \lVert h\rVert_{\infty,\Gamma} ,

which is part 3. If two data sets $(f_1,h_1)$ and $(f_2,h_2)$ are given, applying this to the difference of the solutions — which solves (D) with data $(f_1-f_2, h_1-h_2)$ — bounds $\lVert u_1-u_2\rVert_\infty$ by $\lVert v\rVert_\infty\lVert f_1-f_2\rVert_\infty + \lVert h_1-h_2\rVert_{\infty,\Gamma}$ : the solution depends continuously on the data. $\square$

RemarkOriginal

Uniqueness has two independent proofs in this handbook, and they assume different things. The one above uses the maximum principle and needs only $u \in C^2(\Omega)\cap C(\overline\Omega)$ , with no regularity of $\Gamma$ and no derivatives of $u$ on the boundary. The energy proof in Green's identities puts $v = u$ in Green's first identity and needs $u \in C^2(\overline\Omega)$ and a piecewise $C^1$ boundary, because it integrates by parts; in exchange it handles the Neumann and Robin conditions, which the maximum principle does not. Strauss gives both, the maximum-principle proof in his §6.1 and the energy proof in his §7.1; Stakgold and Holst give both as well, in their §1.3 and §8.3.

Example(The mean value property tested in two and three dimensions)Original

(a) A plane polynomial. The function

u(x,y) = x^3 - 3xy^2

is harmonic on $\mathbb{R}^2$ : $u_{xx} = 6x$ and $u_{yy} = -6x$ . (It is $\operatorname{Re}(x+iy)^3$ , so harmonicity is also Strauss's third example.) Take the centre $(x_0,y_0) = (1,2)$ , where $u = 1 - 3\cdot 1\cdot 4 = -11$ . Parametrize the circle of radius $r$ about it by $(1 + r\cos\theta,\; 2 + r\sin\theta)$ and expand:

\frac{1}{2\pi}\int_0^{2\pi} u\,d\theta = \frac{1}{2\pi}\int_0^{2\pi}\Bigl[(1+r\cos\theta)^3 - 3(1+r\cos\theta)(2+r\sin\theta)^2\Bigr] d\theta .

Every term with an odd total power of $\cos\theta$ or $\sin\theta$ averages to zero, and $\langle\cos^2\theta\rangle = \langle\sin^2\theta\rangle = \tfrac12$ , $\langle\cos\theta\sin\theta\rangle = 0$ . The surviving terms are

1 + 3\cdot 1\cdot r^2\langle\cos^2\theta\rangle \;-\;3\bigl[\,1\cdot 4 + 1\cdot r^2\langle\sin^2\theta\rangle + 0\,\bigr] = 1 + \tfrac32 r^2 - 12 - \tfrac32 r^2 = -11 ,

independent of $r$ , as (MV) requires. The two $r^2$ terms cancel because $u_{xx} + u_{yy} = 0$ : indeed for any $C^2$ function the $O(r^2)$ term of the circular average is $\tfrac14 r^2\,\nabla^2u(x_0,y_0)$ , which is the two-dimensional case of (3). Numerically the average is $-11.000000$ for $r = 0.5$ , $1$ and $3$ , and the average over the solid disc of radius $2$ is also $-11.000000$ .

(b) Newton's theorem in three dimensions. Let $\mathbf{a} = (0,0,2)$ and

u(\mathbf{x}) = \frac{1}{\lvert\mathbf{x}-\mathbf{a}\rvert} ,

the Coulomb potential of a unit point source at $\mathbf{a}$ . It is harmonic on $\mathbb{R}^3\setminus\{\mathbf{a}\}$ (see the fundamental solution), hence on a neighbourhood of the closed unit ball, which does not contain $\mathbf{a}$ . So (MV) applies with $x = \mathbf{0}$ , $r = 1$ and predicts that the average of $u$ over the unit sphere is $u(\mathbf{0}) = 1/\lvert\mathbf{a}\rvert = \tfrac12$ . Checking directly, with $\theta$ the angle from the $z$ -axis and $dS = \sin\theta\,d\theta\,d\varphi$ on the unit sphere,

\frac{1}{4\pi}\oint_{\partial B(0,1)}\frac{dS}{\lvert\mathbf{x}-\mathbf{a}\rvert} = \frac12\int_0^\pi \frac{\sin\theta\, d\theta}{\sqrt{1 + 4 - 4\cos\theta}} = \frac12\int_{-1}^{1}\frac{dt}{\sqrt{5-4t}} = \frac12\left[-\tfrac12\sqrt{5-4t}\right]_{-1}^{1} = \frac12\cdot\frac{3-1}{2} = \frac12 ,

using $t = \cos\theta$ . This is Newton's theorem: a sphere of uniform surface density attracts an external point exactly as if its mass were concentrated at its centre, read here in the reciprocal direction. Note that the mean value property says nothing when the singularity is inside the sphere, and indeed for $\lvert\mathbf{a}\rvert < 1$ the same integral gives $1$ , not $1/\lvert\mathbf{a}\rvert$ .

Example(The stability constant for the unit disc is sharp)Original

Take $\Omega$ to be the unit disc in $\mathbb{R}^2$ and compute the barrier $v$ of part 3 explicitly. We need $-\nabla^2 v = 1$ with $v = 0$ on $r = 1$ . Looking for a radial solution and using $\nabla^2 v = v'' + v'/r$ for radial functions in the plane,

v(r) = \frac{1 - r^2}{4} \quad\Longrightarrow\quad v'' + \frac{v'}{r} = -\frac12 - \frac12 = -1 ,

and $v(1) = 0$ . So $\lVert v\rVert_\infty = v(0) = \tfrac14$ , and part 3 reads

\lVert u\rVert_\infty \;\le\; \tfrac14\lVert f\rVert_\infty + \lVert h\rVert_{\infty,\Gamma}

for every solution of $-\nabla^2 u = f$ in the unit disc with $u = h$ on its boundary.

The constant $\tfrac14$ cannot be reduced. Take $f \equiv 8$ and $h \equiv 0$ . Then $u = 2(1-r^2)$ solves the problem, since $\nabla^2 u = -8$ and $u(1) = 0$ , and $\lVert u\rVert_\infty = u(0) = 2$ , while the bound gives $\tfrac14\cdot 8 + 0 = 2$ exactly. This is the sharpness remark of Stakgold and Holst, which they make for $f \equiv 1$ .

A case with both terms active. Take $f \equiv 8$ and $h = 3\cos\theta$ on the boundary circle, i.e. $h = 3x$ . Since $3x$ is harmonic and equals $3\cos\theta$ on $r=1$ , the solution is

u = 2(1-r^2) + 3x ,

whose supremum over the closed disc is attained where $2(1-x^2-y^2)+3x$ is largest: setting the gradient to zero gives $y = 0$ and $-4x + 3 = 0$ , so $x = \tfrac34$ and $\lVert u\rVert_\infty = 2(1 - \tfrac9{16}) + \tfrac94 = \tfrac78 + \tfrac94 = \tfrac{25}{8} = 3.125$ . The bound gives $\tfrac14\cdot 8 + 3 = 5$ . The estimate holds with room to spare, as it must: the two extreme cases that make each term sharp cannot be arranged at the same place.

ApplicationOriginal

Electrostatics and the shielding of fields. Strauss's first example is Maxwell's equations in a charge-free region: $\nabla\times\mathbf{E} = 0$ gives $\mathbf{E} = -\nabla\phi$ , and $\nabla\cdot\mathbf{E} = 4\pi\rho$ (Gaussian units) then gives $\nabla^2\phi = -4\pi\rho$ , so the potential is harmonic wherever the charge density vanishes. The maximum principle is then a physical statement with teeth: inside a charge-free cavity the potential can have no interior maximum or minimum, so a charged particle cannot sit in stable electrostatic equilibrium in free space. That is Earnshaw's theorem, and it is why electrostatic traps for ions must be time-dependent. The same argument shows that a conductor held at a single potential encloses a cavity at exactly that potential — a Faraday cage — since the boundary data are constant and uniqueness forces the constant solution.

Steady flow and steady temperature. For an incompressible irrotational flow, Strauss's second example, the velocity potential is harmonic, and the maximum principle says the extremes of the potential sit on the boundary. For steady heat conduction with no sources the temperature is harmonic, and the weak maximum principle says that the hottest and coldest points of a body in steady state are on its surface: a result that is obvious physically and would be hard to prove without it. The comparison principle is the formal version of the statement that raising the temperature anywhere on the boundary, or adding heat anywhere inside, raises the temperature everywhere.

Error control for numerical solvers. Part 3 of the uniqueness statement is what makes a discretized Poisson solver trustworthy. Suitable discretizations of $-\nabla^2$ satisfy a discrete maximum principle — the five-point finite-difference stencil on a uniform grid, and piecewise-linear finite elements on a triangulation with no obtuse angle, the condition that keeps the off-diagonal stiffness entries non-positive — and the discrete analogue of the barrier estimate — with $\lVert v\rVert_\infty = \tfrac14$ on the unit disc, as computed above — converts a bound on the residual $f$ into a bound on the error in the computed solution, in the supremum norm and with no reference to derivatives of the exact solution. The mean value property is the statement that the exact solution is the fixed point of the averaging operator, which is precisely what the five-point stencil on a uniform grid discretizes, each value being replaced by the average of its four neighbours; Strauss's fourth example makes the probabilistic version of this explicit, with $u$ the probability that a Brownian particle started at a point leaves the region through a chosen piece of its boundary, and that interpretation is the basis of Monte Carlo (walk-on-spheres) solvers for the Dirichlet problem.

Harmonic functions as the real parts of analytic functions. In the plane, Strauss's and Ahlfors's third setting, an analytic function splits into a harmonic real part and a harmonic imaginary part, so the maximum principle for harmonic functions and the maximum modulus principle for analytic functions are the same theorem seen twice. Ahlfors notes the direction that is less often used: the maximum principle for an analytic $f$ with no zeros follows from the harmonic one applied to $\log\lvert f\rvert$ .

Theorem

Fundamental solution of Laplace's equation

Adapted

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Finds all radial harmonic functions, normalizes them into the fundamental solution $\Phi$ with $-\nabla^2\Phi = \delta_0$ — the logarithm in the plane and $1/(4\pi r)$ in space — and proves that for $f \in C_c^2(\mathbb{R}^n)$ the Newtonian potential $u = \Phi * f$ is $C^2$ and solves Poisson's equation $-\nabla^2 u = f$ in all of $\mathbb{R}^n$ .

Prerequisites: The Dirac delta in calculations, Green's identities, Cylindrical and spherical coordinates

Theorem(The radial harmonic functions)AdaptedAdapted from Evans (2010), §2.2.1a, Eq. (5), p. 21; Strauss (2008), §6.1, pp. 158, 160

Laplace's equation is unchanged by rotations about the origin, so it is natural to look for solutions that depend only on $r = \lvert x\rvert$ . Let $v$ be twice differentiable on $(0,\infty)$ and put $u(x) = v(\lvert x\rvert)$ on $\mathbb{R}^n\setminus\{0\}$ . Then

\nabla^2 u \;=\; v''(r) + \frac{n-1}{r}\,v'(r) , \tag{1}

so $u$ is harmonic away from the origin if and only if

v'' + \frac{n-1}{r}v' = 0 \qquad (r>0) . \tag{2}

Every solution of (2) is

v(r) = \begin{cases} b\log r + c, & n = 2, \\[2pt] \dfrac{b}{r^{n-2}} + c, & n \ge 3, \end{cases} \qquad b, c \text{ constants.} \tag{3}

Strauss obtains the two cases $n = 2$ and $n = 3$ separately from the polar and spherical forms of the Laplacian, writing (2) as $(r u_r)_r = 0$ and $(r^2u_r)_r = 0$ and integrating; Evans does the general $n$ at once, as above. For $n = 1$ the equation reads $v'' = 0$ , and the harmonic functions are the affine ones; no singular solution appears, which is why the one-dimensional case gives no hint of what follows.

Idea of the proofOriginal

Writing out $\nabla^2$ for a radial function is a chain-rule computation, and the only thing worth noticing is where the $n-1$ comes from: each of the $n$ coordinate directions contributes a term $v''x_i^2/r^2$ plus a term $v'(1/r - x_i^2/r^3)$ , and summing $x_i^2 = r^2$ collapses the first group to $v''$ and the second to $(n-1)v'/r$ . Equation (2) is then first order in $v'$ : it says that $r^{n-1}v'$ is constant, which is the statement that the flux of $\nabla u$ through the sphere of radius $r$ does not depend on $r$ — exactly what Green's identity with $v\equiv1$ predicts for a harmonic function. Integrating once more gives (3), and the split at $n = 2$ is the familiar split between $\int r^{-1}dr$ and $\int r^{-p}dr$ .

ProofAdaptedAdapted from Evans (2010), §2.2.1a, p. 21

Step 1 (derivatives of $r$ ). For $x \ne 0$ and each $i$ , differentiating $r = (x_1^2+\cdots+x_n^2)^{1/2}$ gives $\partial r/\partial x_i = x_i/r$ .

Step 2 (the Laplacian of a radial function). By the chain rule, $u_{x_i} = v'(r)\,x_i/r$ , and differentiating again, using the product rule and Step 1,

u_{x_ix_i} = v''(r)\frac{x_i^2}{r^2} + v'(r)\left(\frac{1}{r} - \frac{x_i^2}{r^3}\right) .

Summing over $i = 1,\dots,n$ and using $\sum_i x_i^2 = r^2$ ,

\nabla^2 u = v''(r)\frac{r^2}{r^2} + v'(r)\left(\frac{n}{r} - \frac{r^2}{r^3}\right) = v'' + \frac{n-1}{r}v' ,

which is (1). Hence (2) is equivalent to harmonicity off the origin.

Step 3 (solve (2)). Suppose first that $v'$ never vanishes on an interval. Then (2) may be divided by $v'$ and reads $\bigl(\log\lvert v'\rvert\bigr)' = (1-n)/r$ , whose solutions are $\log\lvert v'\rvert = (1-n)\log r + \text{const}$ , that is

v'(r) = \frac{a}{r^{n-1}} \tag{4}

for a constant $a \neq 0$ . Equation (4) also holds, with $a = 0$ , when $v' \equiv 0$ ; and these are the only possibilities, since (2) is a linear first-order equation for $v'$ and (4) is its general solution. (Equivalently, multiply (2) by $r^{n-1}$ to get $(r^{n-1}v')' = 0$ directly, which is Strauss's form.)

Step 4 (integrate (4)). For $n = 2$ , $v' = a/r$ integrates to $a\log r + c$ . For $n \ge 3$ , $v' = a r^{1-n}$ integrates to $\frac{a}{2-n}r^{2-n} + c$ . Renaming the constant $b$ in each case gives (3). $\square$

Definition(The fundamental solution)AdaptedAdapted from Evans (2010), §2.2.1a, Eq. (6), p. 22; Stakgold & Holst (2011), §2.5, pp. 175–176 and §8.3, Eqs. (8.3.16)–(8.3.17), p. 496

Among the radial harmonic functions (3) we pick out one, normalized so that it is the potential of a unit source. Following Stakgold and Holst, a fundamental solution for a linear operator $L$ with pole at $\xi$ is a distributional solution of

L E = \delta(x - \xi) ,

which by their criterion means $(E, L^*\phi) = \phi(\xi)$ for every test function $\phi \in C_0^\infty(\mathbb{R}^n)$ . Such an $E$ is never unique — any solution of the homogeneous equation may be added — and for an isotropic problem one singles out a particular one by requiring spherical symmetry about the pole and prescribed behaviour at infinity. For a constant-coefficient $L$ it is enough to find the solution with pole at the origin and translate: $E(x,\xi) = E(x-\xi,0)$ .

For $L = -\nabla^2$ this selection gives Evans's fundamental solution of Laplace's equation

\Phi(x) \;=\; \begin{cases} \displaystyle -\frac{1}{2\pi}\log\lvert x\rvert, & n = 2, \\[8pt] \displaystyle \frac{1}{n(n-2)\alpha(n)}\,\frac{1}{\lvert x\rvert^{\,n-2}}, & n \ge 3, \end{cases} \tag{5}

defined for $x \ne 0$ . Here, as in harmonic functions, $\alpha(n)$ denotes the volume of the $n$ -dimensional unit ball, so that a ball and a sphere of radius $r$ measure $\lvert B(x,r)\rvert = \alpha(n)r^n$ and $\lvert\partial B(x,r)\rvert = n\alpha(n)r^{n-1}$ . In three dimensions $n(n-2)\alpha(n) = 3\cdot1\cdot\tfrac43\pi = 4\pi$ , so

\Phi(x) = \frac{1}{4\pi\lvert x\rvert} \qquad (n = 3) ,

which is Stakgold and Holst's $E(x,\xi) = 1/(4\pi\lvert x - \xi\rvert)$ . By construction $\Phi$ is harmonic on $\mathbb{R}^n\setminus\{0\}$ , and so is $x \mapsto \Phi(x-y)$ away from $x = y$ , since Laplace's equation is translation invariant. Evans records the size of its derivatives: there is a constant $C$ with

\lvert D\Phi(x)\rvert \le \frac{C}{\lvert x\rvert^{\,n-1}}, \qquad \lvert D^2\Phi(x)\rvert \le \frac{C}{\lvert x\rvert^{\,n}} \qquad (x\ne 0) . \tag{6}

Sign conventions, which differ between our sources. Evans and Stakgold–Holst normalize against $-\nabla^2$ , so that $-\nabla^2\Phi = \delta_0$ and $\Phi > 0$ for $n \ge 3$ : the potential of a positive point source is positive. Riley, Hobson and Bence normalize against $\nabla^2$ itself, defining $F$ by $\nabla^2 F(\mathbf{r},\mathbf{r}_0) = \delta(\mathbf{r}-\mathbf{r}_0)$ , and therefore get the opposite sign,

F(\mathbf{r},\mathbf{r}_0) = -\frac{1}{4\pi\lvert\mathbf{r}-\mathbf{r}_0\rvert} \quad (n=3), \qquad F(\mathbf{r},\mathbf{r}_0) = \frac{1}{2\pi}\ln\lvert\mathbf{r}-\mathbf{r}_0\rvert + \text{const} \quad (n=2) .

So $F = -\Phi$ . We follow Evans and Stakgold–Holst: in this unit $-\nabla^2\Phi = \delta_0$ . Arfken writes the same content as the identity $\nabla^2(1/r) = -4\pi\delta(\mathbf{r})$ , proved in the Dirac delta in calculations, which is $-\nabla^2\Phi = \delta_0$ divided by $4\pi$ .

Theorem(The normalizing constant is a flux)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.5.3, Eqs. (21.87)–(21.90), p. 757

Let $n \ge 2$ and let $\Phi$ be as in (5). Then for every $r > 0$

\oint_{\partial B(0,r)} \frac{\partial\Phi}{\partial n}\; dS \;=\; -1 , \tag{7}

where $\partial/\partial n$ is the derivative along the outward normal of the ball. This single equation is what fixes the constants in (5), and it is how Riley, Hobson and Bence derive them: integrate $\nabla^2 F = \delta$ over a ball centred on the pole, get $1$ on the right, and convert the left side by the divergence theorem into the flux of $\nabla F$ through the sphere; radial symmetry then turns that flux into $(\text{area})\times F'(r)$ and determines $F$ .

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §21.5.3, p. 757

Step 1 (the normal derivative). $\Phi$ is radial, $\Phi(x) = v(r)$ , so on $\partial B(0,r)$ the outward normal is radial and $\partial\Phi/\partial n = v'(r)$ , a constant on the sphere.

Step 2 (evaluate $v'$ ). For $n \ge 3$ , $v(r) = \frac{1}{n(n-2)\alpha(n)}r^{2-n}$ , so

v'(r) = \frac{2-n}{n(n-2)\alpha(n)}\,r^{1-n} = -\frac{1}{n\alpha(n)}\,r^{1-n} .

For $n = 2$ , $v(r) = -\frac{1}{2\pi}\log r$ gives $v'(r) = -\frac{1}{2\pi r}$ , which is the same formula since $n\alpha(n) = 2\pi$ when $n = 2$ .

Step 3 (multiply by the area). The sphere $\partial B(0,r)$ has area $n\alpha(n)r^{n-1}$ , so

\oint_{\partial B(0,r)} \frac{\partial\Phi}{\partial n}\; dS = n\alpha(n)r^{n-1}\cdot\left(-\frac{1}{n\alpha(n)}r^{1-n}\right) = -1 ,

independently of $r$ , which is (7). $\square$

RemarkOriginal

Equation (7) is the whole content of the normalization, and it also explains why the flux is independent of $r$ : by part 2 of Green's identities, the flux of $\nabla\Phi$ through $\partial B(0,r)$ equals $\int_{B(0,r)}\nabla^2\Phi$ , which gets no contribution from the region where $\Phi$ is harmonic. All of the $-1$ sits at the origin. Put the other way round: no classical function can have $\nabla^2\Phi = 0$ everywhere and a nonzero flux through a closed surface, so the singularity is not a blemish on $\Phi$ but the source itself.

Theorem(The Newtonian potential solves Poisson's equation)AdaptedAdapted from Evans (2010), §2.2.1b, Thm 1, pp. 22–23; Stakgold & Holst (2011), §8.3, Eq. (8.3.18), p. 496

Let $f \in C_c^2(\mathbb{R}^n)$ : twice continuously differentiable with compact support. Define the Newtonian potential of $f$ as the convolution

u(x) \;=\; \int_{\mathbb{R}^n} \Phi(x-y)\,f(y)\; dy \;=\; \begin{cases} \displaystyle -\frac{1}{2\pi}\int_{\mathbb{R}^2}\log\lvert x-y\rvert\, f(y)\; dy, & n = 2,\\[10pt] \displaystyle \frac{1}{n(n-2)\alpha(n)}\int_{\mathbb{R}^n}\frac{f(y)}{\lvert x-y\rvert^{\,n-2}}\; dy, & n \ge 3 . \end{cases} \tag{8}

Then

$u \in C^2(\mathbb{R}^n)$ , and
$-\nabla^2 u = f$ at every point of $\mathbb{R}^n$ .

In three dimensions (8) reads $u(x) = \int f(y)\,dy/(4\pi\lvert x-y\rvert)$ , which is Stakgold and Holst's superposition of point-source potentials: the volume element at $y$ carries source strength $f(y)\,dy$ and contributes $f(y)\,dy/(4\pi\lvert x-y\rvert)$ to the potential at $x$ . Evans notes that the smoothness demanded of $f$ is far more than is needed, and refers to Gilbarg and Trudinger for weaker hypotheses.

Idea of the proofOriginal

The formula is the superposition one would write down without thinking, but the obvious proof of it is wrong, and seeing why is the point. One cannot simply move two derivatives inside the integral in (8): by the Hessian estimate (6), $\lvert D^2\Phi\rvert$ is of size $\lvert x-y\rvert^{-n}$ near the singularity, which is not integrable in $n$ dimensions, so the differentiated integral does not even converge.

The repair has two moves. First, write the convolution the other way round, $u(x) = \int\Phi(y)f(x-y)\,dy$ , so that the derivatives fall on $f$ , which is smooth and compactly supported; difference quotients then converge uniformly and give $u \in C^2$ with $\nabla^2u(x) = \int\Phi(y)\,\nabla^2_x f(x-y)\,dy$ . Second, cut a ball $B(0,\varepsilon)$ out around the singularity. On the small ball, $\Phi$ is integrable — that is the whole difference between $\Phi$ and $D^2\Phi$ — so that piece is $O(\varepsilon^2)$ or $O(\varepsilon^2\lvert\log\varepsilon\rvert)$ and disappears. Outside the ball, $\Phi$ is smooth and harmonic, so Green's second identity moves both derivatives from $f$ onto $\Phi$ , killing the volume term and leaving only boundary terms on the sphere $\partial B(0,\varepsilon)$ . One of those terms carries a factor $\Phi$ and vanishes with $\varepsilon$ ; the other carries $\partial\Phi/\partial\nu$ , which by the normalization (7) is exactly $1/\lvert\partial B(0,\varepsilon)\rvert$ , so that term is an average of $f$ over a shrinking sphere, and it tends to $-f(x)$ . That limit is the theorem.

ProofAdaptedAdapted from Evans (2010), §2.2.1b, pp. 23–24

Throughout, $C$ denotes a constant that may change from line to line but does not depend on $\varepsilon$ .

Step 1 ( $\Phi$ is locally integrable). Near the origin $\lvert\Phi(y)\rvert \le C\lvert y\rvert^{2-n}$ for $n\ge3$ and $C\lvert\log\lvert y\rvert\rvert$ for $n=2$ . In polar coordinates $dy = n\alpha(n)\rho^{n-1}d\rho$ , so

\int_{B(0,\varepsilon)}\lvert\Phi\rvert\,dy \le C\int_0^\varepsilon \rho^{2-n}\rho^{n-1}d\rho = C\varepsilon^2 \quad (n\ge3), \qquad \int_{B(0,\varepsilon)}\lvert\Phi\rvert\,dy \le C\varepsilon^2\lvert\log\varepsilon\rvert \quad (n=2). \tag{9}

Both are finite, and both tend to $0$ with $\varepsilon$ . The same computation with one power of $\rho$ fewer bounds the surface integral,

\oint_{\partial B(0,\varepsilon)}\lvert\Phi\rvert\,dS \le C\varepsilon \quad (n\ge3), \qquad \oint_{\partial B(0,\varepsilon)}\lvert\Phi\rvert\,dS \le C\varepsilon\lvert\log\varepsilon\rvert \quad (n=2). \tag{10}

In particular the integral (8) converges absolutely for every $x$ : this is Stakgold and Holst's remark that the singularity of the kernel is mild enough to be cancelled by the volume element.

Step 2 (move the derivatives onto $f$ ). Substituting $y \mapsto x - y$ in (8),

u(x) = \int_{\mathbb{R}^n}\Phi(y)\,f(x-y)\; dy . \tag{11}

Fix $i$ and $h \ne 0$ . Subtracting two copies of (11),

\frac{u(x+he_i) - u(x)}{h} = \int_{\mathbb{R}^n}\Phi(y)\,\frac{f(x+he_i-y) - f(x-y)}{h}\; dy .

Because $f \in C^2_c$ , its first derivatives are uniformly continuous, so the difference quotient of $f$ converges to $f_{x_i}(x-y)$ uniformly in $y$ as $h \to 0$ ; and $\Phi$ is integrable on the (bounded) region where the quotients are not zero, by Step 1. Hence we may pass to the limit under the integral:

u_{x_i}(x) = \int_{\mathbb{R}^n}\Phi(y)\,f_{x_i}(x-y)\; dy .

Repeating the argument with $f_{x_i}$ in place of $f$ , which is legitimate because $f_{x_i} \in C^1_c$ ,

u_{x_ix_j}(x) = \int_{\mathbb{R}^n}\Phi(y)\,f_{x_ix_j}(x-y)\; dy \qquad (i,j = 1,\dots,n) . \tag{12}

The right side of (12) is continuous in $x$ , again by uniform continuity of $f_{x_ix_j}$ and integrability of $\Phi$ , so $u \in C^2(\mathbb{R}^n)$ . This is part 1. Note that no derivative has been taken of $\Phi$ so far.

Step 3 (split off the singularity). Summing (12) over $i = j$ and splitting the domain at radius $\varepsilon$ ,

\nabla^2 u(x) = \underbrace{\int_{B(0,\varepsilon)}\Phi(y)\,\nabla^2_x f(x-y)\,dy}_{I_\varepsilon} \;+\;\underbrace{\int_{\mathbb{R}^n\setminus B(0,\varepsilon)}\Phi(y)\,\nabla^2_x f(x-y)\,dy}_{J_\varepsilon} .

By (9), $\lvert I_\varepsilon\rvert \le \lVert D^2f\rVert_\infty\int_{B(0,\varepsilon)}\lvert\Phi\rvert \le C\varepsilon^2$ for $n\ge3$ and $C\varepsilon^2\lvert\log\varepsilon\rvert$ for $n=2$ ; either way $I_\varepsilon \to 0$ .

Step 4 (integrate by parts once). In $J_\varepsilon$ the $x$ -Laplacian of $f(x-y)$ equals its $y$ -Laplacian, because each differentiation in $y$ brings one factor $-1$ and they appear in pairs. On the region $\Omega_\varepsilon = \mathbb{R}^n\setminus B(0,\varepsilon)$ — intersected with a large ball containing the support of $y \mapsto f(x-y)$ , outside which everything vanishes — the function $\Phi$ is smooth and $f(x-\cdot)$ is $C^2$ , so Green's first identity applies:

J_\varepsilon = \int_{\Omega_\varepsilon}\Phi\,\nabla^2_y f(x-y)\,dy = -\int_{\Omega_\varepsilon}\nabla\Phi(y)\cdot\nabla_y f(x-y)\,dy + \oint_{\partial B(0,\varepsilon)}\Phi(y)\,\frac{\partial f}{\partial\nu}(x-y)\,dS(y) ,

where $\boldsymbol\nu$ is the unit normal on $\partial B(0,\varepsilon)$ pointing into the ball, i.e. outward from $\Omega_\varepsilon$ (the sign in Green's identity is fixed by that choice, and there is no contribution from the large sphere, where $f$ vanishes). Call the two terms $K_\varepsilon$ and $L_\varepsilon$ . By (10), $\lvert L_\varepsilon\rvert \le \lVert Df\rVert_\infty\oint_{\partial B(0,\varepsilon)}\lvert\Phi\rvert\,dS \le C\varepsilon$ for $n\ge3$ and $C\varepsilon\lvert\log\varepsilon\rvert$ for $n = 2$ , so $L_\varepsilon \to 0$ .

Step 5 (integrate by parts again; the volume term dies). Applying Green's first identity to $K_\varepsilon$ in the other direction,

K_\varepsilon = \int_{\Omega_\varepsilon}\nabla^2\Phi(y)\,f(x-y)\,dy \;-\; \oint_{\partial B(0,\varepsilon)}\frac{\partial\Phi}{\partial\nu}(y)\,f(x-y)\,dS(y) .

The volume term is zero, because $\Phi$ is harmonic on $\Omega_\varepsilon$ — the singularity has been excised, and this is the step that needed the excision.

Step 6 (the surviving boundary term). On $\partial B(0,\varepsilon)$ the normal $\boldsymbol\nu$ points towards the origin, $\boldsymbol\nu = -y/\lvert y\rvert$ , so by Step 2 of the previous proof

\frac{\partial\Phi}{\partial\nu}(y) = -v'(\varepsilon) = \frac{1}{n\alpha(n)\varepsilon^{n-1}} = \frac{1}{\lvert\partial B(0,\varepsilon)\rvert} .

It is constant on the sphere, and its value is exactly the reciprocal of the sphere's area — this is where the normalization (7) is used. Therefore

K_\varepsilon = -\frac{1}{\lvert\partial B(0,\varepsilon)\rvert}\oint_{\partial B(0,\varepsilon)}f(x-y)\, dS(y) = -\frac{1}{\lvert\partial B(x,\varepsilon)\rvert}\oint_{\partial B(x,\varepsilon)} f(z)\, dS(z) ,

substituting $z = x - y$ , which maps the sphere of radius $\varepsilon$ about $0$ onto the sphere of radius $\varepsilon$ about $x$ . The right side is minus the average of $f$ over a sphere shrinking to $x$ , and $f$ is continuous, so by the argument of Step 5 of harmonic functions it tends to $-f(x)$ .

Step 7 (collect). $\nabla^2 u(x) = I_\varepsilon + K_\varepsilon + L_\varepsilon$ for every $\varepsilon > 0$ , and the left side does not depend on $\varepsilon$ . Letting $\varepsilon \to 0^+$ and using Steps 3, 4 and 6,

\nabla^2 u(x) = 0 + (-f(x)) + 0 = -f(x) ,

that is $-\nabla^2 u = f$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Compact support of $f$ , in Steps 2 and 4. It makes (8) converge at infinity, makes the difference quotients vanish outside a bounded set, and kills the outer boundary term in Step 4. Without decay the convolution need not exist at all: in three dimensions $f \equiv 1$ gives a divergent integral.
$f \in C^2$ , in Steps 2, 3 and 4. Two derivatives are moved onto $f$ , and $\lVert D^2f\rVert_\infty$ is used to bound $I_\varepsilon$ . Evans states that much less suffices, and the formula remains correct for $f$ merely Hölder continuous with compact support, but that is not what is proved here.
The excision of $B(0,\varepsilon)$ , in Step 5. This is the hinge. The term $\int\nabla^2\Phi\,f$ vanishes only because the origin has been removed; if one keeps it and writes $\nabla^2\Phi = -\delta_0$ in the sense of distributions, the same term produces $-f(x)$ directly, which is Evans's "formal computation" and Arfken's way of using $\nabla^2(1/r) = -4\pi\delta(\mathbf{r})$ . The proof above is that formal computation made legitimate.
$n \ge 2$ . The constants in (5) are singular at $n=2$ (where $r^{2-n}$ is constant) and the logarithm replaces the power. In $n = 1$ there is no singular radial solution at all, and the fundamental solution of $-d^2/dx^2$ is $-\tfrac12\lvert x\rvert$ , which grows rather than decays.
Behaviour at infinity, and why two dimensions is different. For $n\ge3$ , $\Phi \to 0$ at infinity, and this is the extra condition that picks $\Phi$ out of the family (3): Riley, Hobson and Bence impose $F \to 0$ as $\lvert\mathbf{r}\rvert\to\infty$ and conclude that the additive constant is zero. For $n = 2$ they point out that this condition cannot be imposed, because $\log\lvert x\rvert$ is unbounded in both directions, and the additive constant stays undetermined. The non-uniqueness is exactly Stakgold and Holst's remark that a fundamental solution is determined only up to a solution of the homogeneous equation; adding a constant, or any entire harmonic function, gives another fundamental solution.
What the theorem does not say. It gives a solution of $-\nabla^2u = f$ on all of $\mathbb{R}^n$ , not the only one: $u + h$ solves the same equation for any harmonic $h$ . Uniqueness needs a condition at infinity (for $n\ge3$ , decay, by the maximum principle applied on large balls), or boundary data on a bounded region, which is the subject of Green's functions for Laplace and Poisson problems.

Example(A smooth source in three dimensions, solved twice)Original

Take $n = 3$ and the radial source

f(x) = \begin{cases}(1 - r^2)^3, & r = \lvert x\rvert \le 1,\\ 0, & r > 1,\end{cases}

which is in $C_c^2(\mathbb{R}^3)$ but not $C^3$ : with $s = r^2$ the profile $(1-s)^3$ and its first two derivatives in $s$ vanish at $s = 1$ , while the third does not. So Evans's hypotheses hold exactly and no more.

Route 1: solve the ODE. For a radial $u$ , $-\nabla^2 u = -\frac{1}{r^2}(r^2u')'$ , so $-\nabla^2u = f$ says $(r^2u')' = -r^2f$ and hence $r^2u'(r) = -M(r)$ with

M(r) = \int_0^r s^2 f(s)\, ds = \frac{r^3}{3} - \frac{3r^5}{5} + \frac{3r^7}{7} - \frac{r^9}{9} \qquad (r\le1), \qquad M(1) = \frac{16}{315} ,

the last value from $\tfrac13 - \tfrac35 + \tfrac37 - \tfrac19 = \tfrac{105 - 189 + 135 - 35}{315}$ . Outside the support $M$ is constant, so $u' = -M(1)/r^2$ and the decaying solution is

u(r) = \frac{16}{315\,r} \qquad (r \ge 1) .

Integrating $u' = -M(r)/r^2$ inwards from $r=1$ gives, for $r \le 1$ ,

u(r) = \frac{1}{8} - \frac{r^2}{6} + \frac{3r^4}{20} - \frac{r^6}{14} + \frac{r^8}{72} .

(Check the match at $r=1$ : $\tfrac18 - \tfrac16 + \tfrac3{20} - \tfrac1{14} + \tfrac1{72} = \tfrac{16}{315} = 0.0507937$ , and the two expressions for $u'$ both give $-\tfrac{16}{315}$ there. Check the equation: $u'' + 2u'/r = -1 + 3r^2 - 3r^4 + r^6 = -(1-r^2)^3$ , so $-\nabla^2 u = f$ inside, as required.)

Route 2: evaluate the convolution at the centre. Formula (8) at $x = 0$ with $\Phi = 1/(4\pi\lvert y\rvert)$ and $dy = 4\pi s^2\,ds$ gives

u(0) = \int_{\mathbb{R}^3}\frac{f(y)}{4\pi\lvert y\rvert}\,dy = \int_0^1 \frac{(1-s^2)^3}{4\pi s}\,4\pi s^2\, ds = \int_0^1 s(1-s^2)^3\, ds = \left[-\frac{(1-s^2)^4}{8}\right]_0^1 = \frac18 .

The two routes agree: $u(0) = \tfrac18$ exactly. Notice that the integrand $s(1-s^2)^3$ is bounded even though $\Phi$ blows up at the origin — the factor $s^2$ from the volume element beats the factor $s^{-1}$ from the kernel, which is Step 1 of the proof in one line.

The far field. The total source strength is $Q = \int f\,dy = 4\pi M(1) = 64\pi/315$ , and outside the support $u = 16/(315r) = Q/(4\pi r) = Q\,\Phi(x)$ . A localized source looks like a point source of the same strength from far away; the systematic version of this statement is the multipole expansion.

Example(The same problem in the plane: the potential does not decay)Original

Take $n = 2$ and a uniform source on the disc of radius $a$ : $f = \rho_0$ for $r \le a$ and $f = 0$ for $r > a$ . (This $f$ is discontinuous, so it falls outside Evans's hypotheses; we solve the ODE directly and verify the result, and the point of the example is the behaviour at infinity, not the regularity.) With $-\nabla^2 u = -\frac1r(ru')'$ ,

u(r) = \frac{\rho_0}{4}\bigl(a^2 - r^2\bigr) \quad (r \le a), \qquad u(r) = -\frac{\rho_0a^2}{2}\log\frac{r}{a} \quad (r \ge a) .

Check. Inside, $u' = -\rho_0r/2$ and $u'' = -\rho_0/2$ , so $u'' + u'/r = -\rho_0$ and $-\nabla^2u = \rho_0$ . Outside, $u$ is a multiple of $\log r$ , which is harmonic by (3), so $-\nabla^2u = 0$ . At $r = a$ both expressions give $u = 0$ , and both give $u' = -\rho_0a/2$ : the solution is $C^1$ across the edge, as it must be, and its second derivative jumps there by exactly the jump in $f$ .

What this shows. The total source strength is $Q = \pi a^2\rho_0$ , and the exterior potential is $-\frac{Q}{2\pi}\log(r/a)$ , which tends to $-\infty$ as $r\to\infty$ rather than to $0$ . In two dimensions the Newtonian potential of a nonzero net source cannot be made to vanish at infinity, which is why Riley, Hobson and Bence keep an undetermined constant in the planar fundamental solution and impose no condition at infinity. The field $-u'(r) = Q/(2\pi r)$ does decay, and it is the field, not the potential, that is physical here: this is the potential of a uniformly charged infinite cylinder of radius $a$ , read in the plane transverse to its axis.

ApplicationOriginal

Electrostatics and gravitation. Arfken's route to the fundamental solution is the physicist's: Gauss's law $\nabla\cdot\mathbf{E} = \rho/\varepsilon_0$ with $\mathbf{E} = -\nabla\varphi$ gives Poisson's equation $\nabla^2\varphi = -\rho/\varepsilon_0$ , and a point charge $q$ at the origin means $\rho = q\,\delta(\mathbf{r})$ . Comparing with the Coulomb potential $\varphi = q/(4\pi\varepsilon_0 r)$ produces $\nabla^2(1/r) = -4\pi\delta(\mathbf{r})$ , which Arfken emphasises is meaningful only under an integral sign. In the normalization of this unit, $\varphi = \Phi * (\rho/\varepsilon_0)$ , which is Riley, Hobson and Bence's Eq. (21.91),

\varphi(\mathbf{x}) = \int \frac{\rho(\mathbf{y})}{4\pi\varepsilon_0\lvert\mathbf{x}-\mathbf{y}\rvert}\; dV(\mathbf{y}) ,

the integral over all space. The same formula with $-4\pi G$ in place of $1/\varepsilon_0$ is the Newtonian gravitational potential of a mass distribution, which is where the name "Newtonian potential" comes from. The theorem above is the statement that this universally used formula really does solve the field equation, and the proof is where one sees that the $1/r$ singularity of each point contribution is harmless after integration but fatal if one differentiates first.

Boundary-value problems. Riley, Hobson and Bence introduce the fundamental solution not for its own sake but as the singular part of a Green's function: for a region $V$ with boundary $S$ one writes $G = F + H$ with $F$ the fundamental solution, carrying the delta, and $H$ harmonic in $V$ , chosen to make $G$ satisfy the boundary condition. Every construction in Green's functions for Laplace and Poisson problems and the method of images begins this way, and the representation formula they derive from Green's second identity reduces to (8) when the boundary is pushed to infinity and the data vanish there.

Screened potentials and other operators. The pattern of the definition — solve the homogeneous equation away from the pole, then fix the strength of the singularity by the integrated form of the equation — is Stakgold and Holst's general recipe, and it is not special to the Laplacian. For $-\nabla^2 + q^2$ , which governs steady diffusion in an absorbing medium (and, in particle physics, a massive scalar field), the same recipe gives $e^{-qr}/(4\pi r)$ in three dimensions — the Yukawa or Debye-screened potential — and $K_0(q\lvert x\rvert)/(2\pi)$ in two, as Stakgold and Holst set as their Exercise 2.5.3. The Laplacian is the $q \to 0$ limit, and the comparison shows what the long range of the Coulomb and Newtonian potentials is really due to: the absence of a mass or screening term.

Numerics: boundary integral and fast multipole methods. Because (8) is a convolution with an explicit kernel, a charge distribution given on a mesh of $N$ points produces a potential computable at all $N$ points in $O(N\log N)$ or $O(N)$ operations instead of $O(N^2)$ , by grouping distant sources — the fast multipole method, and more broadly the boundary-element method, in which an elliptic problem on a region is converted into an integral equation on its boundary using $\Phi$ as the kernel. The local integrability established in Step 1 is what makes those singular boundary integrals convergent, and the inverse-power bounds (6) are what make the far-field grouping accurate.

Theorem

Poisson's integral formula

Adapted

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Solves the Dirichlet problem for Laplace's equation in a disk by separating variables and summing the resulting Fourier series into the Poisson kernel, proves from the kernel's three properties that the integral has the prescribed continuous boundary values, records the corresponding formula for a ball in three dimensions, and deduces that every harmonic function is infinitely differentiable inside its domain.

Prerequisites: Harmonic functions, Cylindrical and spherical coordinates, Fourier series

Definition(The Dirichlet problem for a disk, and the Poisson kernel)AdaptedAdapted from Strauss (2008), §6.3, Eqs. (1)–(2), p. 165 and Eq. (17), p. 170; Stein & Shakarchi (2003), §2.5.4, Lemma 5.5, p. 55; Stakgold & Holst (2011), §8.3, Eq. (8.3.2), p. 491

Let $D = \{\,\mathbf{x}\in\mathbb{R}^2 : \lvert\mathbf{x}\rvert < a\,\}$ be the open disk of radius $a$ and $C = \partial D$ its boundary circle, with polar coordinates $(r,\theta)$ on $D$ . Given a function $h$ on $C$ , the Dirichlet problem for the disk is

u_{xx} + u_{yy} = 0 \quad\text{in } D, \qquad u = h \quad\text{on } C . \tag{1}

The Poisson kernel of the disk of radius $a$ is

P(r,\theta) \;=\; \frac{a^2 - r^2}{a^2 - 2ar\cos\theta + r^2} \;=\; 1 + 2\sum_{n=1}^\infty \left(\frac{r}{a}\right)^{\!n}\cos n\theta \qquad (0 \le r < a) . \tag{2}

For $a = 1$ and with the angle written as a single variable, Stein and Shakarchi's normalization is

P_r(\theta) = \sum_{n=-\infty}^{\infty} r^{\lvert n\rvert}e^{in\theta} = \frac{1-r^2}{1-2r\cos\theta+r^2} ,

the same function: they arrive at it as the kernel whose convolution with $f$ produces the Abel means $A_r(f)(\theta) = \sum_n r^{\lvert n\rvert}a_n e^{in\theta}$ of the Fourier series of $f$ . So the Poisson kernel of the disk and the Abel-summation kernel of Fourier series are one object, and Poisson's formula below is Abel summation of the boundary data's Fourier series.

Stakgold and Holst write the interior Dirichlet problem for $-\nabla^2$ and warn about an artefact of polar coordinates: the lines $\theta = \pi$ and $\theta = -\pi$ are the same ray and $r = 0$ is an interior point, so a candidate solution must be checked for good behaviour at both — which is where the periodic conditions and the boundedness requirement at the origin come from, rather than from any physical boundary.

Theorem(Poisson's integral formula for a disk)AdaptedAdapted from Strauss (2008), §6.3, Thm 1 with Eqs. (13)–(15), pp. 168–169; Ahlfors (1979), §4.6.3, Thm 22, Eq. (64), p. 168 and §4.6.4, Thm 23, p. 169

Let $h$ be continuous on the circle $C = \{r = a\}$ and define, for $r < a$ ,

u(r,\theta) \;=\; \bigl(a^2 - r^2\bigr)\int_0^{2\pi} \frac{h(\varphi)}{a^2 - 2ar\cos(\theta-\varphi) + r^2}\;\frac{d\varphi}{2\pi} \;=\;\int_0^{2\pi} P(r,\theta-\varphi)\,h(\varphi)\,\frac{d\varphi}{2\pi} . \tag{3}

Then $u$ is harmonic in $D$ , it is infinitely differentiable there, and

\lim_{\mathbf{x}\to\mathbf{x}_0} u(\mathbf{x}) = h(\mathbf{x}_0) \qquad\text{for every } \mathbf{x}_0 \in C , \tag{4}

so that $u$ extends to a continuous function on $\overline D$ with boundary values $h$ . It is the only harmonic function in $D$ with this property.

Writing $\mathbf{x}$ for the point with polar coordinates $(r,\theta)$ and $\mathbf{x}'$ for the boundary point with polar coordinates $(a,\varphi)$ , the law of cosines gives $\lvert\mathbf{x}-\mathbf{x}'\rvert^2 = a^2 + r^2 - 2ar\cos(\theta-\varphi)$ , and since arc length on $C$ is $ds = a\,d\varphi$ , formula (3) takes the geometric form

u(\mathbf{x}) \;=\; \frac{a^2 - \lvert\mathbf{x}\rvert^2}{2\pi a}\oint_{\lvert\mathbf{x}'\rvert = a} \frac{u(\mathbf{x}')}{\lvert\mathbf{x}-\mathbf{x}'\rvert^2}\; ds . \tag{5}

Ahlfors states (5) the other way round — as a representation of a given harmonic function through its boundary values, for $u$ harmonic in $\lvert z\rvert < R$ and continuous on $\lvert z\rvert\le R$ — and then proves separately (his Theorem 23, due to Schwarz) that the integral on the right has the right boundary limits whenever the data are merely piecewise continuous and continuous at the point in question. Stein and Shakarchi prove the same for any integrable $f$ on the circle, with the boundary limit holding at every point of continuity of $f$ and uniformly if $f$ is continuous.

Idea of the proofOriginal

Two different things have to happen, and it is worth keeping them apart.

Finding the formula. Separating variables in polar coordinates gives $\Theta'' + \lambda\Theta = 0$ with periodic conditions, hence $\lambda = n^2$ , and an Euler equation for the radial factor, hence $r^{\pm n}$ (and $1, \log r$ for $n = 0$ ). Half of these blow up at the centre and are discarded — the regularity requirement at $r = 0$ is doing the work that a boundary condition would do. What survives is $r^n\cos n\theta$ and $r^n\sin n\theta$ , and matching at $r = a$ identifies the coefficients as the Fourier coefficients of $h$ divided by $a^n$ . Substituting them back and exchanging the sum with the integral produces an explicit series in $(r/a)^n\cos n(\theta-\varphi)$ , which is the real part of two geometric series and therefore sums in closed form. The result is the kernel (2).

Proving that the formula works. Harmonicity is easy, because each term of the series is harmonic and one may differentiate the integral (3) under the integral sign: the integrand is smooth in $\mathbf{x}$ for $\mathbf{x}$ inside and $\mathbf{x}'$ on the circle, so they never collide. The boundary limit is the real content, and it follows from three properties of the kernel: it is positive, it has total mass one, and for angles bounded away from zero it becomes uniformly small as $r \to a$ . A kernel with those three properties concentrates all of its mass near $\theta = 0$ , so the integral (3) is a weighted average of values of $h$ taken almost entirely from a small arc around the point being approached; continuity of $h$ then pins the average to $h$ at that point. Uniqueness is immediate from the maximum principle.

Proof(Derivation of the formula)AdaptedAdapted from Strauss (2008), §6.3, pp. 165–167; Stakgold & Holst (2011), §8.3, pp. 492–493

Step 1 (separate). In polar coordinates $\nabla^2 u = u_{rr} + r^{-1}u_r + r^{-2}u_{\theta\theta}$ (see cylindrical and spherical coordinates). Substituting $u = R(r)\Theta(\theta)$ , dividing by $R\Theta$ and multiplying by $r^2$ gives

\frac{r^2R'' + rR'}{R} = -\frac{\Theta''}{\Theta} .

The two sides depend on different variables, so each is a constant $\lambda$ :

\Theta'' + \lambda\Theta = 0, \qquad r^2R'' + rR' - \lambda R = 0 . \tag{6}

Step 2 (the angular factor). The coordinate $\theta$ is an angle, so $u$ must be unchanged by $\theta\mapsto\theta+2\pi$ ; the natural conditions on $\Theta$ are therefore the periodic ones, $\Theta(\theta+2\pi) = \Theta(\theta)$ for all $\theta$ (Stakgold and Holst impose $\Theta(-\pi) = \Theta(\pi)$ and $\Theta'(-\pi) = \Theta'(\pi)$ , which is the same requirement). The solutions of $\Theta''+\lambda\Theta = 0$ are periodic with period $2\pi$ exactly when $\lambda = n^2$ with $n = 0, 1, 2, \dots$ , and then

\Theta(\theta) = A\cos n\theta + B\sin n\theta \quad (n \ge 1), \qquad \Theta = A \quad (n = 0) .

Step 3 (the radial factor). With $\lambda = n^2$ the radial equation in (6) is of Euler type, so we try $R = r^\alpha$ : substituting gives $\alpha(\alpha-1) + \alpha - n^2 = 0$ , that is $\alpha^2 = n^2$ and $\alpha = \pm n$ . For $n \ge 1$ this gives $R = Cr^n + Dr^{-n}$ ; for $n = 0$ the two roots coincide and the second solution is $\log r$ , so $R = C + D\log r$ .

Step 4 (reject the singular solutions). We want $u$ harmonic on all of $D$ , including the centre. Both $r^{-n}$ and $\log r$ are unbounded as $r \to 0$ , so $D = 0$ in each case. Strauss calls this the "boundary condition" at $r = 0$ ; it is not a boundary condition at all, but the demand that the solution be a genuine function on the disk. Superposing what is left,

u(r,\theta) = \tfrac12 A_0 + \sum_{n=1}^\infty r^n\bigl(A_n\cos n\theta + B_n\sin n\theta\bigr) . \tag{7}

Step 5 (match the boundary data). Setting $r = a$ in (7), the requirement $u = h$ on $C$ reads

h(\theta) = \tfrac12 A_0 + \sum_{n=1}^\infty a^n\bigl(A_n\cos n\theta + B_n\sin n\theta\bigr) ,

which is the Fourier series of $h$ with coefficients $a^nA_n$ and $a^nB_n$ . Hence (Fourier series)

A_n = \frac{1}{\pi a^n}\int_0^{2\pi}h(\varphi)\cos n\varphi\; d\varphi, \qquad B_n = \frac{1}{\pi a^n}\int_0^{2\pi}h(\varphi)\sin n\varphi\; d\varphi . \tag{8}

Step 6 (substitute and sum). Putting (8) into (7) and using $\cos n\varphi\cos n\theta + \sin n\varphi\sin n\theta = \cos n(\theta-\varphi)$ ,

u(r,\theta) = \int_0^{2\pi} h(\varphi)\left[1 + 2\sum_{n=1}^\infty\left(\frac ra\right)^{\!n}\cos n(\theta-\varphi)\right]\frac{d\varphi}{2\pi} .

The interchange of sum and integral is legitimate because $r/a < 1$ is fixed and the series converges uniformly in $\varphi$ ( $h$ is bounded, being continuous on a circle). The bracket is the second expression in (2); to sum it, write $\cos n\psi = \tfrac12(e^{in\psi}+e^{-in\psi})$ with $\psi = \theta-\varphi$ and use the geometric series with ratio $(r/a)e^{\pm i\psi}$ , of modulus $r/a < 1$ :

1 + \sum_{n=1}^\infty\left(\frac ra\right)^{\!n}e^{in\psi} + \sum_{n=1}^\infty\left(\frac ra\right)^{\!n}e^{-in\psi} = 1 + \frac{re^{i\psi}}{a - re^{i\psi}} + \frac{re^{-i\psi}}{a-re^{-i\psi}} .

Putting the three terms over the common denominator $(a-re^{i\psi})(a-re^{-i\psi}) = a^2 - 2ar\cos\psi + r^2$ , the numerator is

(a^2 - 2ar\cos\psi + r^2) + re^{i\psi}(a - re^{-i\psi}) + re^{-i\psi}(a-re^{i\psi}) = a^2 - r^2 ,

since the two added terms contribute $2ar\cos\psi - 2r^2$ . So the bracket equals $(a^2-r^2)/(a^2-2ar\cos\psi+r^2) = P(r,\psi)$ , which turns the display into (3). $\square$

Proof(The three kernel properties, and the boundary limit)AdaptedAdapted from Strauss (2008), §6.3, properties (i)–(iii) and Eqs. (18)–(20), pp. 170–171; Stein & Shakarchi (2003), §2.5.4, Lemma 5.5, p. 56

Step 1 (positivity). For $0\le r<a$ ,

a^2 - 2ar\cos\theta + r^2 = (a-r)^2 + 2ar(1-\cos\theta) = (a-r)^2 + 4ar\sin^2(\theta/2) \;\ge\; (a-r)^2 > 0 , \tag{9}

so the denominator of $P$ is positive and bounded below, and since $a^2 - r^2 > 0$ ,

P(r,\theta) > 0 \qquad (r < a) .

The middle expression in (9) is Stein and Shakarchi's split; the right-hand one is Strauss's, and it is the sharper of the two for what follows.

Step 2 (unit mass). Integrating the series in (2) term by term — legitimate for fixed $r<a$ , where it converges uniformly — and using $\int_0^{2\pi}\cos n\theta\,d\theta = 0$ for $n\ge1$ ,

\int_0^{2\pi} P(r,\theta)\,\frac{d\theta}{2\pi} = 1 \qquad (0\le r<a) . \tag{10}

(Ahlfors obtains the same identity as the special case $u\equiv1$ of the representation formula.)

Step 3 (concentration). Let $\delta > 0$ . For $\delta \le \lvert\theta\rvert \le \pi$ we have $\sin^2(\theta/2) \ge \sin^2(\delta/2)$ , so by (9)

0 < P(r,\theta) \;\le\; \frac{a^2-r^2}{4ar\sin^2(\delta/2)} \qquad (\delta\le\lvert\theta\rvert\le\pi) . \tag{11}

The right-hand side tends to $0$ as $r \to a^-$ , uniformly in such $\theta$ . So: given $\delta>0$ and $\eta>0$ , there is $r_0 < a$ with $P(r,\theta) < \eta$ for all $r_0 < r < a$ and all $\delta\le\lvert\theta\rvert\le\pi$ .

Step 4 (harmonicity). Each term $r^n\cos n\theta$ , $r^n\sin n\theta$ of (7) is harmonic — they are the real and imaginary parts of $z^n$ , or directly: $\nabla^2(r^n\cos n\theta) = n(n-1)r^{n-2}\cos n\theta + nr^{n-2}\cos n\theta - n^2r^{n-2}\cos n\theta = 0$ . On any disk $r \le \rho < a$ the series (7) and all of its term-by-term derivatives converge uniformly, because the coefficients $a^nA_n, a^nB_n$ are bounded (they are Fourier coefficients of a bounded function) and $(\rho/a)^n$ decays geometrically. Hence $u$ may be differentiated term by term any number of times, and $\nabla^2u = 0$ on $r<\rho$ for every $\rho<a$ , so on $D$ . The same argument applied to the series in (2) shows that $P(r,\theta)$ is itself harmonic in the disk, which is Strauss's third property of the kernel: each term $(r/a)^n\cos n\theta$ is harmonic, and so is the sum.

Step 5 (the boundary limit: set-up). Fix $\theta_0$ . By (10),

u(r,\theta_0) - h(\theta_0) = \int_0^{2\pi} P(r,\theta_0-\varphi)\bigl[h(\varphi) - h(\theta_0)\bigr]\frac{d\varphi}{2\pi} . \tag{12}

Let $\varepsilon > 0$ and let $H = \max_C\lvert h\rvert$ , finite since $h$ is continuous on a circle. By continuity of $h$ at $\theta_0$ choose $\delta>0$ with $\lvert h(\varphi)-h(\theta_0)\rvert < \varepsilon$ whenever $\lvert\varphi-\theta_0\rvert<\delta$ .

Step 6 (split the integral). Split (12) at $\lvert\varphi-\theta_0\rvert = \delta$ and bound each piece.

Near part. On $\lvert\varphi-\theta_0\rvert<\delta$ the bracket is smaller than $\varepsilon$ in absolute value, and $P>0$ , so this piece is at most $\varepsilon\int_0^{2\pi}P(r,\theta_0-\varphi)\frac{d\varphi}{2\pi} = \varepsilon$ by (10). (Positivity, Step 1, is what allows the bracket to be bounded and the kernel left alone.)
Far part. On $\lvert\varphi-\theta_0\rvert\ge\delta$ the bracket is at most $2H$ , and by Step 3 the kernel is less than $\eta$ once $r$ is close enough to $a$ ; so this piece is at most $2H\eta$ .

Choosing $\eta = \varepsilon$ gives $\lvert u(r,\theta_0) - h(\theta_0)\rvert \le \varepsilon(1 + 2H)$ for all $r$ sufficiently close to $a$ . Since $\varepsilon$ was arbitrary, $u(r,\theta_0)\to h(\theta_0)$ as $r\to a^-$ .

Step 7 (from radial to unrestricted limits). The estimate in Step 6 is uniform in $\theta_0$ if $\delta$ is chosen by uniform continuity of $h$ , which holds since $C$ is compact. Hence $u(r,\theta) \to h(\theta)$ uniformly in $\theta$ , and therefore for any sequence $\mathbf{x}_k\to\mathbf{x}_0\in C$ with $\mathbf{x}_k = (r_k,\theta_k)$ ,

\lvert u(\mathbf{x}_k) - h(\mathbf{x}_0)\rvert \le \sup_\theta\lvert u(r_k,\theta)-h(\theta)\rvert + \lvert h(\theta_k)-h(\theta_0)\rvert \to 0 ,

the first term by uniformity and the second by continuity of $h$ . This is (4), and it also shows $u$ extends continuously to $\overline D$ .

Step 8 (uniqueness). If $u_1,u_2$ are harmonic in $D$ and both satisfy (4) with the same $h$ , then each extends continuously to $\overline D$ with the same boundary values, so $w = u_1-u_2$ is harmonic in $D$ , continuous on $\overline D$ , and zero on $C$ . Since $D$ is bounded, the weak maximum principle of harmonic functions gives $\max_{\overline D}w = \min_{\overline D}w = 0$ , hence $w \equiv 0$ . $\square$

RemarkOriginal

Ahlfors's derivation of (5) is shorter than the one above and uses no Fourier series at all, at the price of assuming that a solution exists. The mean value property gives the value of a harmonic function at the centre of the disk; a Möbius transformation of the disk onto itself carries any interior point to the centre, and transplanting the mean value identity through it produces (5), with the kernel appearing as the Jacobian $d\theta^*/d\theta$ of the induced map of the boundary circle. Ahlfors even gives the resulting picture: to compute $P_U(z)$ , replace each boundary value $U(\theta)$ by the value at the point diametrically opposite with respect to $z$ — the second intersection with the circle of the line through $e^{i\theta}$ and $z$ — and take the ordinary average. Our Step 6 is instead his and Schwarz's arc-splitting argument in the language of kernels.

Theorem(Poisson's formula for a ball in three dimensions)AdaptedAdapted from Strauss (2008), §7.4, Eqs. (16)–(17), p. 195

Let $D = \{\,\mathbf{x}\in\mathbb{R}^3 : \lvert\mathbf{x}\rvert < a\,\}$ and let $h$ be continuous on the sphere $\lvert\mathbf{x}\rvert = a$ . Then the solution of $\nabla^2u = 0$ in $D$ , $u = h$ on $\partial D$ , is

u(\mathbf{x}_0) \;=\; \frac{a^2 - \lvert\mathbf{x}_0\rvert^2}{4\pi a}\oint_{\lvert\mathbf{x}\rvert=a}\frac{h(\mathbf{x})}{\lvert\mathbf{x}-\mathbf{x}_0\rvert^3}\; dS \qquad (\lvert\mathbf{x}_0\rvert < a) . \tag{13}

In spherical coordinates, with $r_0 = \lvert\mathbf{x}_0\rvert$ and $\psi$ the angle between $\mathbf{x}$ and $\mathbf{x}_0$ , this reads

u(r_0,\theta_0,\phi_0) = \frac{a\,(a^2-r_0^2)}{4\pi}\int_0^{2\pi}\!\!\int_0^\pi \frac{h(\theta,\phi)}{\bigl(a^2+r_0^2-2ar_0\cos\psi\bigr)^{3/2}}\;\sin\theta\, d\theta\, d\phi . \tag{14}

Compare the disk, formula (5): the kernel is $\lvert\mathbf{x}-\mathbf{x}_0\rvert^{-2}$ over $2\pi a$ in two dimensions and $\lvert\mathbf{x}-\mathbf{x}_0\rvert^{-3}$ over $4\pi a$ in three, in each case multiplied by $a^2 - \lvert\mathbf{x}_0\rvert^2$ . At the centre, $\mathbf{x}_0 = \mathbf{0}$ , formula (13) collapses to $u(\mathbf{0}) = \frac{1}{4\pi a^2}\oint h\, dS$ , the mean value property.

NoteOriginal

Strauss obtains (13) not by separating variables but from the Green's function of the ball, which he constructs by reflecting the source point $\mathbf{x}_0$ across the sphere to $\mathbf{x}_0^* = a^2\mathbf{x}_0/\lvert\mathbf{x}_0\rvert^2$ and taking $G = -\frac{1}{4\pi\rho} + \frac{a}{\lvert\mathbf{x}_0\rvert}\frac{1}{4\pi\rho^*}$ with $\rho = \lvert\mathbf{x}-\mathbf{x}_0\rvert$ and $\rho^* = \lvert\mathbf{x}-\mathbf{x}_0^*\rvert$ ; formula (13) is then $\oint h\,\partial G/\partial n\,dS$ . That construction belongs to the method of images, and the general representation formula it uses to Green's functions for Laplace and Poisson problems, so we state (13) here and do not reprove it. Strauss notes that the same reflection carried out in two dimensions, starting from $\frac{1}{2\pi}\log r$ , reproduces formula (5) — the disk formula obtained in this unit by a completely different route.

Theorem(Harmonic functions are infinitely differentiable)AdaptedAdapted from Strauss (2008), §6.3, 'Differentiability', p. 170; Evans (2010), §2.2.3b, Thm 6, p. 27

Let $U$ be any open subset of the plane and let $u$ be harmonic in $U$ . Then $u$ has partial derivatives of every order at every point of $U$ ; that is, $u \in C^\infty(U)$ .

Evans proves the corresponding statement in $\mathbb{R}^n$ under a weaker hypothesis — $u \in C(U)$ satisfying the mean value property for every ball in $U$ — and warns that nothing follows about the boundary: $u$ may fail to extend smoothly to $\partial U$ , and may fail even to extend continuously. The conclusion is interior and only interior.

Idea of the proofOriginal

Smoothness is not visible in Laplace's equation, which mentions only two derivatives; it has to be imported from a formula. Poisson's formula is that formula. Its integrand, as a function of the interior point $\mathbf{x}$ , is a smooth function of $\mathbf{x}$ — the only thing that could spoil it is the vanishing of $\lvert\mathbf{x}-\mathbf{x}'\rvert$ , and that cannot happen because $\mathbf{x}$ is inside and $\mathbf{x}'$ is on the circle. So one may differentiate under the integral sign as often as one likes. Since every point of an open set is the centre of some small disk contained in it, and a function differentiable on each of a family of disks covering $U$ is differentiable on $U$ , the local statement gives the global one.

ProofAdaptedAdapted from Strauss (2008), §6.3, p. 170

Step 1 (a disk about the point). Let $\mathbf{x}_0 \in U$ . Since $U$ is open, choose $\rho>0$ with the closed disk $\overline B(\mathbf{x}_0,\rho) \subset U$ , and let $B = B(\mathbf{x}_0,\rho)$ .

Step 2 ( $u$ on $B$ is a Poisson integral). $u$ is harmonic on $B$ and continuous on $\overline B$ (it is $C^2$ on $U \supseteq \overline B$ ). Let $v$ be the Poisson integral (5) on $B$ built from the boundary values $u|_{\partial B}$ . By Step 8 of the earlier proof — uniqueness — $v = u$ on $B$ . So on $B$ ,

u(\mathbf{x}) = \frac{\rho^2 - \lvert\mathbf{x}-\mathbf{x}_0\rvert^2}{2\pi\rho}\oint_{\partial B}\frac{u(\mathbf{x}')}{\lvert\mathbf{x}-\mathbf{x}'\rvert^2}\; ds(\mathbf{x}') . \tag{15}

Step 3 (differentiate under the integral sign). Fix a closed disk $\overline B(\mathbf{x}_0,\sigma)$ with $\sigma<\rho$ . For $\mathbf{x}$ in it and $\mathbf{x}'\in\partial B$ we have $\lvert\mathbf{x}-\mathbf{x}'\rvert \ge \rho-\sigma > 0$ , so the integrand of (15) is a $C^\infty$ function of $\mathbf{x}$ there, and every $\mathbf{x}$ -derivative of it is bounded on $\overline B(\mathbf{x}_0,\sigma)\times\partial B$ (a product of compact sets) and continuous. The hypotheses of the theorem on differentiating an integral with respect to a parameter are therefore met, for derivatives of every order, and

D^\alpha u(\mathbf{x}) = \oint_{\partial B} D^\alpha_{\mathbf{x}}\!\left[\frac{\rho^2-\lvert\mathbf{x}-\mathbf{x}_0\rvert^2}{2\pi\rho\,\lvert\mathbf{x}-\mathbf{x}'\rvert^2}\right] u(\mathbf{x}')\; ds(\mathbf{x}')

for every multi-index $\alpha$ , each such derivative being continuous in $\mathbf{x}$ . Hence $u \in C^\infty\bigl(B(\mathbf{x}_0,\sigma)\bigr)$ , and in particular all derivatives of $u$ exist at $\mathbf{x}_0$ .

Step 4 (globalize). $\mathbf{x}_0$ was an arbitrary point of $U$ , and differentiability of all orders is a pointwise property, so $u \in C^\infty(U)$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Continuity of $h$ , in Steps 5–6 of the kernel proof. It is used only at the point $\theta_0$ being approached, which is why Ahlfors's Theorem 23 and Stein and Shakarchi's Theorem 5.7 can take $h$ merely piecewise continuous or integrable and still conclude the limit at every point of continuity. The example below has a two-point jump and illustrates exactly this.
Unrestricted versus radial limits, in Step 7. Condition (4) demands $u(\mathbf{x})\to h(\mathbf{x}_0)$ as $\mathbf{x}\to\mathbf{x}_0$ in any manner, and that is what makes the solution unique. If one asks only for the radial limit $\lim_{r\to1}u(r,\theta) = h(\theta)$ for each fixed $\theta$ , uniqueness fails: Stein and Shakarchi's Exercise 18 points out that $u(r,\theta) = \frac{\partial P_r}{\partial\theta}(\theta) = \frac{-2r(1-r^2)\sin\theta}{(1-2r\cos\theta+r^2)^2}$ is harmonic in the unit disc (a derivative of a harmonic function with respect to the rotation parameter), satisfies $\lim_{r\to1}u(r,\theta) = 0$ for every $\theta$ , and is not identically zero. It is unbounded near $z = 1$ , so it has no continuous extension to $\overline D$ and (4) fails for it at that one boundary point. One boundary point is enough.
Boundedness of the domain, in Step 8 only. Uniqueness is the maximum principle's contribution and inherits its need for a bounded domain.
Existence can fail for other domains. Poisson's formula settles the disk and the ball completely, and it is tempting to assume a continuous $h$ on a nice-looking boundary always has a solution. Stakgold and Holst give three warnings. For the punctured unit disk $0<\lvert\mathbf{x}\rvert<1$ , whose boundary consists of the circle $\lvert\mathbf{x}\rvert=1$ plus the one extra point $\mathbf{0}$ , the data $h=0$ on the circle and $h=1$ at the puncture admit no solution at all. For the unit disk with suitable continuous $h$ , $\lvert\nabla u\rvert$ can blow up at the boundary, so the solution lies in $C^2(\Omega)\cap C(\overline\Omega)$ but not in $C^2(\overline\Omega)$ . And for a three-dimensional domain with a sharp inward spike — in the extreme case, the unit ball with a radial segment removed — there are continuous data with no solution; they interpret this physically as charge leaking out along the spike. Their sufficient condition is the exterior sphere condition: if at every boundary point there is an open ball outside $\Omega$ whose boundary touches $\Gamma$ at that point and nowhere else, then continuous data on $\Gamma$ always admit a harmonic solution lying in $C^2(\Omega)\cap C(\overline\Omega)$ .
The kernel's two-sided bound gives more than the boundary limit. Ahlfors notes the elementary inequalities, in our notation with $\lvert\mathbf{x}\rvert = r$ and radius $a$ , $\frac{a-r}{a+r} \;\le\; \frac{a^2-r^2}{\lvert\mathbf{x}-\mathbf{x}'\rvert^2} \;\le\; \frac{a+r}{a-r} ,$ which follow from $a - r \le \lvert\mathbf{x}-\mathbf{x}'\rvert \le a+r$ . For $u \ge 0$ they turn (5) into Harnack's inequality $\frac{a-r}{a+r}\,u(\mathbf{0}) \;\le\; u(\mathbf{x}) \;\le\; \frac{a+r}{a-r}\,u(\mathbf{0}) ,$ since the average of $u$ over the circle is $u(\mathbf{0})$ . Ahlfors emphasises that this holds only for positive harmonic functions, and uses it to prove Harnack's principle: a monotone sequence of harmonic functions either tends to $+\infty$ uniformly on compact sets or converges to a harmonic limit, uniformly on compact sets.
Conventions. Strauss writes the kernel as $P(r,\theta)$ for a disk of radius $a$ and divides by $2\pi$ inside the integral; Stein and Shakarchi write $P_r(\theta)$ for the unit disc and define convolution with the same $\frac{1}{2\pi}$ ; Ahlfors writes the kernel as $(R^2-\lvert a\rvert^2)/\lvert z-a\rvert^2$ with the interior point called $a$ and the radius $R$ . Only names differ.

Example(A jump on the boundary circle: harmonic measure of a semicircle)Original

Take $a = 1$ and the discontinuous data

h(\varphi) = \begin{cases} 1, & 0 < \varphi < \pi,\\ 0, & -\pi < \varphi < 0,\end{cases}

so $h$ is continuous except at the two points $\varphi = 0$ and $\varphi = \pi$ . Poisson's formula still makes sense — $h$ is bounded and integrable — and gives

u(r,\theta) = \frac{1}{2\pi}\int_0^\pi P(r,\theta-\varphi)\; d\varphi .

Evaluation. Write $z = re^{i\theta}$ . Then

\frac{1+z}{1-z} = \frac{(1+z)(1-\bar z)}{\lvert 1-z\rvert^2} = \frac{(1 - r^2) + 2i\,r\sin\theta}{\lvert 1-z\rvert^2} ,

whose real part is positive for $\lvert z\rvert<1$ ; so $w = \log\frac{1+z}{1-z}$ is analytic on the unit disc with the principal branch, and

\operatorname{Im}w = \arg\frac{1+z}{1-z} = \arctan\frac{2r\sin\theta}{1-r^2} .

Set

U(r,\theta) = \frac12 + \frac1\pi\arctan\frac{2r\sin\theta}{1-r^2} .

$U$ is harmonic: it is an affine function of the imaginary part of an analytic function. $U$ has the right boundary values: at $z = e^{i\varphi}$ with $\sin\varphi \neq 0$ , the real part of $(1+z)/(1-z)$ is $(1-r^2)/\lvert1-z\rvert^2 = 0$ while the imaginary part has the sign of $\sin\varphi$ , so $\arg = \pm\pi/2$ and $U \to \tfrac12\pm\tfrac12$ , which is $1$ on the upper semicircle and $0$ on the lower. $U$ is bounded, by $0 \le U \le 1$ . By the uniqueness argument of Step 8, applied on the disc minus two boundary points (or simply by checking against the integral), $u = U$ .

Numerical check of the identification. Evaluating the integral $\frac{1}{2\pi}\int_0^\pi P(r,\theta-\varphi)d\varphi$ numerically against $U$ :

$r$	$\theta$	integral	$U(r,\theta)$
$0$	$0.3$	$0.500000$	$0.500000$
$0.5$	$\pi/2$	$0.7951672$	$0.7951672$
$0.5$	$0.3$	$0.6194765$	$0.6194765$
$0.9$	$-1.0$	$0.0397219$	$0.0397219$

At the centre, $u(0,\theta) = \tfrac12$ , which is the mean value property: half the boundary circle carries the value $1$ . At $r = \tfrac12$ , $\theta = \tfrac\pi2$ the exact value is

u = \frac12 + \frac1\pi\arctan\frac{2\cdot\frac12\cdot1}{1-\frac14} = \frac12 + \frac{1}{\pi}\arctan\frac43 = 0.7951672 .

What the example illustrates. The function $u$ is $C^\infty$ everywhere inside the disc, as the smoothness theorem demands, even though its boundary data have two jumps: smoothness is an interior statement and knows nothing about the boundary. The boundary limit (4) holds at every point of the circle except $z = \pm1$ , in agreement with Ahlfors's and Stein–Shakarchi's versions of the theorem, which ask for continuity of the data only at the point approached. At $z = 1$ the limit genuinely fails: approaching along the upper semicircle gives $1$ , along the lower gives $0$ , and approaching along the real axis gives $\tfrac12$ — the average of the two one-sided values, which is the general rule for a jump (Stein and Shakarchi's Exercise 17). The level sets $u = c$ are the circular arcs through $\pm1$ , so $u$ is the harmonic measure of the upper semicircle.

Example(A ball in three dimensions, checked against an explicit harmonic polynomial)Original

Take $a = 1$ in (13) and the axially symmetric data $h = \cos^2\theta$ on the unit sphere, i.e. $h = z^2$ in Cartesian coordinates.

The answer, found by inspection. Write $\cos^2\theta = \tfrac13\bigl(2P_2(\cos\theta) + 1\bigr)$ , using $P_2(t) = \tfrac12(3t^2-1)$ . The function

u = \frac13 + \frac{2}{3}\,r^2P_2(\cos\theta) = \frac13 + \frac{r^2(3\cos^2\theta - 1)}{3} = \frac13 + \frac{2z^2 - x^2 - y^2}{3}

is a polynomial, harmonic because $\nabla^2(2z^2-x^2-y^2) = 4-2-2 = 0$ , and equals $\tfrac13 + \tfrac13(3\cos^2\theta-1) = \cos^2\theta$ on $r=1$ . By uniqueness it is the solution.

Checking Poisson's formula. Put $\mathbf{x}_0 = (0,0,\tfrac12)$ , so $r_0 = \tfrac12$ and $\psi = \theta$ . With $\lvert\mathbf{x}-\mathbf{x}_0\rvert^2 = 1 + \tfrac14 - \cos\theta$ and $dS = \sin\theta\,d\theta\,d\phi$ on the unit sphere, (14) gives

u(\mathbf{x}_0) = \frac{1 - \tfrac14}{4\pi}\cdot 2\pi\int_{-1}^{1}\frac{t^2\, dt}{\bigl(\tfrac54 - t\bigr)^{3/2}} = \frac{3}{8}\int_{-1}^1\frac{t^2\,dt}{(\tfrac54-t)^{3/2}} ,

substituting $t = \cos\theta$ . The integral evaluates to exactly $\tfrac43$ (numerically $1.3333333$ ), so $u(\mathbf{x}_0) = \tfrac38\cdot\tfrac43 = \tfrac12$ ; and the polynomial gives $\tfrac13 + \tfrac14(3-1)/3 = \tfrac13+\tfrac16 = \tfrac12$ . The two agree.

The centre, and the mean value property. Setting $\mathbf{x}_0 = \mathbf{0}$ in (13) gives $u(\mathbf{0}) = \frac{1}{4\pi}\oint\cos^2\theta\,dS = \frac12\int_{-1}^1t^2\,dt = \tfrac13$ , which is the constant term of the polynomial, as it must be: the degree- $2$ solid harmonic $r^2P_2(\cos\theta)$ vanishes at the origin and averages to zero on every sphere about it.

ApplicationOriginal

Heat, and why the Dirichlet problem is the natural one. Stein and Shakarchi reach Laplace's equation in the disc as the long-time limit of the heat equation: hold a prescribed temperature distribution $f$ on the rim of a disc, wait, and the interior settles into the steady state $u$ with $\nabla^2u = 0$ and $u = f$ on the rim. Poisson's formula says what the equilibrium temperature at an interior point is — a weighted average of the rim temperatures, weighted by $(a^2-r^2)/\lvert\mathbf{x}-\mathbf{x}'\rvert^2$ , which favours the part of the rim nearest the point and does so more and more sharply as the point approaches the rim. The same formula with the same reading describes the electrostatic potential inside a cylindrical capacitor whose surface potential is prescribed and which is long enough to be treated as two-dimensional, which is Strauss's reading of his formula (14).

Probability: harmonic measure and exit distributions. Because the kernel is positive and integrates to one, Poisson's formula exhibits $u(\mathbf{x})$ as the expectation of $h$ against a probability measure on the boundary — the harmonic measure seen from $\mathbf{x}$ . That measure is exactly the exit distribution of a Brownian particle started at $\mathbf{x}$ : the probability that it first touches the circle in the arc $A$ is $\frac{1}{2\pi a}\int_A \frac{a^2-\lvert\mathbf{x}\rvert^2}{\lvert\mathbf{x}-\mathbf{x}'\rvert^2}ds$ . The worked example computes exactly this for $A$ a semicircle, and the answer $\tfrac12 + \tfrac1\pi\arctan\frac{2r\sin\theta}{1-r^2}$ is the probability that a Brownian particle started at $re^{i\theta}$ leaves the disc through the upper half of the circle. This is the probabilistic reading of Strauss's fourth motivating example for Laplace's equation, and it underlies walk-on-spheres Monte Carlo solvers, which sample exit points of successive spheres instead of discretizing the equation.

Summability of Fourier series. Read with the roles reversed, (3) is a statement about Fourier series, and this is Stein and Shakarchi's reason for introducing it. A continuous function's Fourier series may diverge at some points; but $u(r,\theta)$ , which is its Abel mean $\sum_n r^{\lvert n\rvert}a_ne^{in\theta}$ , converges to $f$ at every point of continuity and uniformly for continuous $f$ . So "evaluate the Fourier series inside the disc and then let $r\to1$ " always works where termwise summation on the circle may not. Damping the $n$ th coefficient by $r^{\lvert n\rvert}$ is the oldest and simplest of the summability methods, and solving a Dirichlet problem is what it amounts to geometrically.

Interior regularity as a tool. The smoothness theorem is used constantly without being mentioned: it is what licenses differentiating a harmonic function as many times as an argument needs, and what makes the term-by-term manipulations of multipole and spherical-harmonic expansions legitimate inside the region of harmonicity (multipole expansions). It also marks the difference between elliptic and hyperbolic equations: Strauss points out that the analogous smoothing holds for the diffusion equation but is simply false for the wave equation, where a kink in the initial data travels along a characteristic for ever (the one-dimensional wave equation). In numerical analysis, interior smoothness is why spectral methods converge so fast for elliptic problems away from the boundary, and why the loss of accuracy in such a computation is almost always a boundary effect.

Method

Separation of variables in standard coordinates

Adapted

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Gives the recipe for reducing Laplace's and Helmholtz's equations to ordinary differential equations by seeking product solutions, proves the lemma that makes the separation constants constant, carries the spherical case through in full, and catalogues which special functions - trigonometric, hyperbolic, Bessel, Legendre and spherical harmonic - each of the Cartesian, cylindrical and spherical systems produces.

Prerequisites: Cylindrical and spherical coordinates, Spherical harmonics, Bessel functions of the first kind, Neumann and Hankel functions, Orthogonality and Fourier-Bessel series, Spherical Bessel functions, Modified Bessel functions, Associated Legendre functions, Sturm-Liouville eigenvalue problems, Fourier series

Definition(Separable solutions and separation constants)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.1, Eq. (21.1), pp. 713–714; Hassani (2013), §13.1, pp. 396–397

Let a linear partial differential equation be given in independent variables $x_1,\dots,x_N$ . A solution of the product form

u(x_1,\dots,x_N) = X_1(x_1)\,X_2(x_2)\cdots X_N(x_N) \tag{1}

is called separable in those variables, and looking for solutions of the form (1) is the method of separation of variables. The requirement is only that each factor depend on one variable alone; the factors may, and usually do, share numerical parameters. Riley, Hobson and Bence illustrate the distinction with three functions of $(x,y,z,t)$ : $xyz^2\sin bt$ separates completely; $(x^2+y^2)z\cos\omega t$ separates in the last two variables only, since the first two occur in a sum that cannot be factored out; and $xy + zt$ separates in none of them.

The parameters that appear when (1) is substituted into the equation are the separation constants. In the examples below they turn out to be the eigenvalues of the ordinary differential operators that the method produces, so the special functions of mathematical physics enter as the eigenfunctions.

Throughout we use the Laplacian in the three standard systems (cylindrical and spherical coordinates): with $\rho$ the distance from the $z$ -axis, $\varphi$ the azimuth and $r$ the distance from the origin,

\nabla^2 = \frac1\rho\frac{\partial}{\partial\rho}\!\left(\rho\frac{\partial}{\partial\rho}\right) + \frac{1}{\rho^2}\frac{\partial^2}{\partial\varphi^2} \qquad\text{(plane polars)},

\nabla^2 = \frac1\rho\frac{\partial}{\partial\rho}\!\left(\rho\frac{\partial}{\partial\rho}\right) + \frac{1}{\rho^2}\frac{\partial^2}{\partial\varphi^2} + \frac{\partial^2}{\partial z^2} \qquad\text{(cylindrical polars)},

\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\!\left(\sin\theta\frac{\partial}{\partial\theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2}{\partial\varphi^2} \qquad\text{(spherical polars)} .

The two equations to be separated are Laplace's equation $\nabla^2u = 0$ and the Helmholtz equation $\nabla^2 u + k^2u = 0$ , which reduces to Laplace's when $k = 0$ .

Method(The separation lemma)Original

Let $A \subseteq \mathbb{R}^p$ and $B\subseteq\mathbb{R}^q$ be nonempty sets, let $F : A\to\mathbb{R}$ and $G : B\to\mathbb{R}$ , and suppose

F(\mathbf{s}) + G(\mathbf{t}) = 0 \qquad\text{for all } \mathbf{s}\in A,\ \mathbf{t}\in B . \tag{2}

Then $F$ and $G$ are each constant, with $G \equiv -\lambda$ and $F\equiv\lambda$ for a single number $\lambda$ .

Idea of the proofOriginal

This is the only genuine piece of logic in the method, and it is a one-line argument: in (2) the variables $\mathbf{s}$ and $\mathbf{t}$ may be moved independently, so freezing one of them exhibits the other's function as a constant. Everything else in separation of variables is calculus and bookkeeping. It is worth stating the lemma separately because the textbook phrase "a function of $x$ alone equals a function of $y$ alone, so both are constant" is used dozens of times in what follows, and because it makes visible what the method actually needs: that the variables range over a product set, which is exactly the condition that fails on a domain not bounded by coordinate surfaces.

ProofOriginal

Step 1. Fix any $\mathbf{t}_0 \in B$ and set $\lambda = -G(\mathbf{t}_0)$ . Putting $\mathbf{t} = \mathbf{t}_0$ in (2) gives $F(\mathbf{s}) = -G(\mathbf{t}_0) = \lambda$ for every $\mathbf{s}\in A$ . So $F$ is the constant $\lambda$ .

Step 2. Substituting $F \equiv \lambda$ back into (2) gives $G(\mathbf{t}) = -\lambda$ for every $\mathbf{t}\in B$ . So $G$ is the constant $-\lambda$ . $\square$

Method(Separation of variables)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.1, pp. 714–715 and §21.3, pp. 725–734; Hassani (2013), §13.1, pp. 396–397; Arfken, Weber & Harris (2013), §9.4, pp. 414–415

Choose coordinates that fit the boundary. Pick a coordinate system in which every piece of the boundary is a level surface of one coordinate: a box in Cartesians, a disk or annulus or wedge in plane polars, a cylinder in cylindrical polars, a ball or spherical shell in spherical polars. This is the step that decides whether the method will work at all.
Separate the time, if there is any. Write $\Psi(\mathbf{r},t) = u(\mathbf{r})T(t)$ and let $S$ denote the whole spatial operator, so that the equation reads $S\Psi = \partial\Psi/\partial t$ (heat, Schrödinger) or $S\Psi = \partial^2\Psi/\partial t^2$ (wave, Klein–Gordon). Dividing by $uT$ puts a function of $\mathbf{r}$ on one side and a function of $t$ on the other; by the separation lemma both equal a constant, Hassani's $\alpha$ , and one is left with an ODE in $t$ and a homogeneous equation in space, $\nabla^2u(\mathbf{r}) + f(\mathbf{r})u(\mathbf{r}) = 0 , \tag{3}$ which is Hassani's master equation. (Hassani writes the product as $\Psi = RT$ and then lets the letter $\Psi$ stand for the spatial factor as well in his master equation; we reserve $\Psi$ for the full solution and call the spatial factor $u$ , which is the letter steps 3–7 and the rest of this unit use, so that $u$ in (3) and $u$ in step 3 are the same object.) For the heat and wave equations $f$ is the constant $k^2$ and (3) is Helmholtz's equation; for the Schrödinger equation $f(\mathbf{r}) = 2m(E - V(\mathbf{r}))/\hbar^2$ . Hassani notes that $\alpha$ is conventionally taken real except for the Schrödinger equation, where a purely imaginary $\alpha$ is more convenient because it cancels the $i$ in the equation, and that in every case the admissible $\alpha$ are fixed by the boundary conditions, not by the separation. This time-separation constant is used nowhere else in the unit; the letter $\alpha$ is reused below, in the cylindrical catalogue and the cylinder example, for the unrelated axial separation constant.
Posit a product and divide by it. Substitute $u = X_1X_2\cdots X_N$ into (3) and divide the result by $u$ . Then multiply by whatever factor (typically $r^2$ or $\rho^2$ ) is needed to make one variable appear in one group of terms only.
Introduce separation constants. Apply the separation lemma to each split, naming the constants in whatever algebraic form will be convenient later — $-n^2$ , $k^2$ , $\ell(\ell+1)$ . With $N$ variables there are $N-1$ independent constants, and in a Cartesian separation of the wave equation they satisfy one algebraic relation among themselves, as in Riley, Hobson and Bence's example where $\mu^2 = l^2+m^2+n^2$ .
Impose the conditions the coordinates hide. These are not boundary conditions in the physical sense, and forgetting them is the commonest error:
- Single-valuedness in an angle. $\Phi(\varphi+2\pi) = \Phi(\varphi)$ forces the azimuthal separation constant to be $m^2$ with $m$ an integer. (The exception is a genuinely multivalued potential, such as the magnetic scalar potential of a current, where Riley, Hobson and Bence keep the term $\Phi = C\varphi + D$ with $C = I/2\pi$ .)
- Regularity at the origin or on the axis. If the region contains $\rho = 0$ or $r = 0$ , discard $\rho^{-m}$ , $\ln\rho$ , $r^{-(\ell+1)}$ , $Y_m(k\rho)$ , $y_\ell(kr)$ . If it contains the polar axis $\theta = 0,\pi$ , discard $Q_\ell^m(\cos\theta)$ .
- Behaviour at infinity. On an unbounded exterior region, discard the solutions that grow. What is left after these deletions is an eigenvalue problem of Sturm–Liouville type in each variable, and the admissible separation constants are its eigenvalues.
Read off the special functions. Each separated ODE is one of a short list, and its solutions are the classical special functions; the catalogue is in the note below.
Superpose, and fit the inhomogeneous data by orthogonality. A single product rarely satisfies the given data. Sum over all admissible separation constants and determine the coefficients from the one inhomogeneous condition by the orthogonality of the eigenfunctions: Fourier coefficients for $\cos m\varphi,\sin m\varphi$ , Fourier–Bessel coefficients for $J_m(k_n\rho)$ (orthogonality and Fourier–Bessel series), Legendre coefficients for $P_\ell(\cos\theta)$ , and the full double sum over $\ell$ and $m$ in spherical harmonics (spherical harmonics) when the data are not axially symmetric.

Note(Catalogue of separated equations)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.1, Eqs. (21.6)–(21.7), p. 715, §21.3.1, Eqs. (21.27)–(21.49), pp. 726–734, and §21.3.2, Eqs. (21.52)–(21.57), pp. 737–741; Arfken, Weber & Harris (2013), §9.4, Eqs. (9.29)–(9.41), pp. 415–417, Eqs. (9.58)–(9.63), pp. 421–422, and Eqs. (9.74)–(9.84), pp. 424–427

For the Helmholtz equation $\nabla^2u + k^2u = 0$ ; Laplace's equation is the case $k = 0$ throughout. In each system the product is substituted and the equation divided by it, which is legitimate only where no factor vanishes; as in the proof below, the separated ODEs listed here then hold on the whole of each coordinate interval because both sides of each are continuous.

Cartesian, $u = X(x)Y(y)Z(z)$ . With constants $-l^2, -m^2, -n^2$ ,

X'' = -l^2X, \qquad Y'' = -m^2Y, \qquad Z'' = -n^2Z, \qquad l^2+m^2+n^2 = k^2 .

Each factor is $Ae^{i l x} + Be^{-ilx}$ : a sine and cosine pair if $l^2>0$ ; a pair of exponentials $e^{\pm\beta x}$ , equivalently $\sinh\beta x$ and $\cosh\beta x$ , if $l^2<0$ , where $\beta = \sqrt{-l^2}$ is real and $l = \mathrm{i}\beta$ is imaginary (the exponent is the real number $\sqrt{-l^{2}}$ , not $\sqrt{l^{2}} = \pm l$ , which is imaginary here); and the linear function $A + Bx$ if $l^2 = 0$ . For Laplace's equation ( $k=0$ ) the three constants sum to zero, so they cannot all share a sign: either one of them vanishes, or one has the opposite sign to another. So Laplace's equation in a box never produces an oscillating factor in every direction; some direction carries hyperbolic or linear behaviour instead, which is why a Dirichlet datum on one face decays into the box.

Plane polars, $u = P(\rho)\Phi(\varphi)$ . With constant $n^2$ ,

\Phi'' + n^2\Phi = 0, \qquad \rho^2P'' + \rho P' + \bigl(k^2\rho^2 - n^2\bigr)P = 0 .

Single-valuedness makes $n$ an integer. For $k = 0$ the radial equation is the Euler equation $\rho^2P''+\rho P' - n^2P = 0$ with solutions $\rho^{\,n},\rho^{-n}$ (and $1,\ln\rho$ when $n = 0$ ); for $k\ne0$ the substitution $\mu = k\rho$ turns it into Bessel's equation of order $n$ , with solutions $J_n(k\rho)$ and $Y_n(k\rho)$ . Riley, Hobson and Bence warn against confusing $Y_n$ , the Bessel function of the second kind, with a spherical harmonic $Y_\ell^m$ ; the latter carries two indices.

Cylindrical polars, $u = P(\rho)\Phi(\varphi)Z(z)$ . With constants $m^2$ (azimuthal) and $\alpha^2$ (axial),

\Phi'' + m^2\Phi = 0, \qquad Z'' = \alpha^2 Z, \qquad \rho^2P'' + \rho P' + \bigl[(k^2+\alpha^2)\rho^2 - m^2\bigr]P = 0 .

So the solutions are

u = \bigl[A J_m\bigl(\sqrt{k^2+\alpha^2}\,\rho\bigr) + BY_m\bigl(\sqrt{k^2+\alpha^2}\,\rho\bigr)\bigr]\,\bigl[C\cos m\varphi + D\sin m\varphi\bigr]\,\bigl[Ee^{\alpha z}+Fe^{-\alpha z}\bigr] .

The sign convention in the axial constant. We write the axial equation as $Z'' = \alpha^2Z$ , so that a real $\alpha$ gives exponential behaviour in $z$ and an imaginary $\alpha$ oscillatory behaviour; substituting back into $\nabla^2u + k^2u = 0$ then puts $k^2 + \alpha^2$ , not $k^2-\alpha^2$ , in the radial equation. Riley, Hobson and Bence write the axial factor as $E\mathrm{e}^{\mathrm{i}\alpha z} + F\mathrm{e}^{-\mathrm{i}\alpha z}$ (p. 740), that is $Z'' = -\alpha^2Z$ , and their radial argument is correspondingly $\sqrt{k^2-\alpha^2}$ . The two conventions differ only by $\alpha\mapsto\mathrm{i}\alpha$ ; everything here and in the cylinder example below is in ours.

For Laplace's equation ( $k = 0$ ) the radial argument is $\alpha$ and the solution is $[AJ_m(\alpha\rho)+BY_m(\alpha\rho)][C\cos m\varphi + D\sin m\varphi][Ee^{\alpha z}+Fe^{-\alpha z}]$ : Bessel functions radially, trigonometric functions azimuthally, exponentials axially. (Riley, Hobson and Bence call the axial constant $k$ in this case and write the same solution as $[AJ_m(k\rho)+BY_m(k\rho)][C\cos m\varphi+D\sin m\varphi][Ee^{-kz}+Fe^{kz}]$ , Eq. (21.35).) When $k^2 + \alpha^2 < 0$ the radial argument becomes imaginary and the Bessel functions are replaced by the modified Bessel functions $I_m, K_m$ .

Spherical polars, $u = R(r)\Theta(\theta)\Phi(\varphi)$ . With constants $\ell(\ell+1)$ and $m^2$ ,

\Phi'' + m^2\Phi = 0, \qquad \frac{1}{\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \left[\ell(\ell+1) - \frac{m^2}{\sin^2\theta}\right]\Theta = 0 ,

r^2R'' + 2rR' + \bigl[k^2r^2 - \ell(\ell+1)\bigr]R = 0 .

The polar equation becomes the associated Legendre equation under $\mu = \cos\theta$ , with solutions $P_\ell^m(\mu)$ and $Q_\ell^m(\mu)$ (associated Legendre functions), finite polynomial solutions requiring $\ell$ a non-negative integer and $0\le\lvert m\rvert\le\ell$ ; the product $\Theta\Phi$ regular on the axis is the spherical harmonic $Y_\ell^m(\theta,\varphi)$ . The radial equation for $k = 0$ is the Euler equation with solutions $r^\ell$ and $r^{-(\ell+1)}$ ; for $k\ne0$ the substitution $R = r^{-1/2}S$ with $\mu = kr$ reduces it to Bessel's equation of order $\ell+\tfrac12$ , so that $R = r^{-1/2}J_{\ell+1/2}(kr)$ or $r^{-1/2}Y_{\ell+1/2}(kr)$ , which after normalization are the spherical Bessel functions $j_\ell(kr)$ and $y_\ell(kr)$ .

Method(The separated solutions in spherical polars)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.1, Eqs. (21.42)–(21.49), pp. 732–734 and §21.3.2, Eqs. (21.56)–(21.57), pp. 740–741

Let $u = R(r)\Theta(\theta)\Phi(\varphi)$ be a solution of Laplace's equation $\nabla^2u = 0$ of product form on a region that is a product of coordinate intervals in $r \in (0,\infty)$ , $\theta\in(0,\pi)$ and $\varphi$ , where the spherical coordinates are regular. Let the three factors be twice continuously differentiable, and let none of them vanish identically. (Note what this last hypothesis does not say: it does not make $u$ itself nonvanishing, and indeed $u$ vanishes somewhere inside the region in all but the simplest cases. Step 1 of the proof shows that this weaker hypothesis is nevertheless all the separation needs.)

1. What the separation alone gives. There are constants $m^2$ and $\lambda = \ell(\ell+1)$ — complex numbers, subject to no further restriction — such that $\Phi'' = -m^2\Phi$ , such that $M(\mu) = \Theta(\theta)$ with $\mu = \cos\theta$ solves the associated Legendre equation with those parameters, and such that $R$ solves the Euler equation $r^2R'' + 2rR' - \ell(\ell+1)R = 0$ . Those three ODEs are the entire content of the separation.

Naming their solutions takes a little care, because the parameters are so far unrestricted. Here $P_\ell^m$ and $Q_\ell^m$ denote that equation’s two standard solutions — Legendre’s function of first kind, and the one of second kind — taken at complex degree $\ell$ and order $m$ ; associated Legendre functions defines them only for an integer $\ell\ge0$ with $\lvert m\rvert\le\ell$ , which is the only case used from part 2 onwards. Generically — that is, for parameters with $\ell\ne-\tfrac12$ and $m\ne0$ at which $P_\ell^m$ and $Q_\ell^m$ are moreover linearly independent —

u(r,\theta,\varphi) = \bigl(Ar^\ell + Br^{-(\ell+1)}\bigr)\bigl(C\cos m\varphi + D\sin m\varphi\bigr)\bigl[E P_\ell^m(\cos\theta) + FQ_\ell^m(\cos\theta)\bigr] , \tag{4}

Each of the three provisos is a real restriction, and the third is not claimed to be a complete list. When $\ell = -\tfrac12$ the two radial exponents $\ell$ and $-(\ell+1)$ coincide and the radial factor is $r^{-1/2}(A + B\ln r)$ instead. When $m = 0$ the azimuthal factor is $C\varphi + D$ . And $P_\ell^m, Q_\ell^m$ really can fail to be an independent pair: if $\ell$ is a non-negative integer and $m$ an integer with $m>\ell$ , then $P_\ell^m\equiv0$ and the second solution has to be found some other way. No such parameter pair arises in part 2, where $\ell$ and $m$ are integers with $\lvert m\rvert\le\ell$ .

2. With the two regularity conditions. Suppose in addition that

$u$ is single-valued in the azimuth, $\Phi(\varphi+2\pi) = \Phi(\varphi)$ , and
$u$ is finite on the polar axis $\theta = 0,\pi$ .

Then $m$ is an integer, $\ell$ is a non-negative integer with $0\le\lvert m\rvert\le\ell$ , and $F = 0$ , so that

u(r,\theta,\varphi) = \bigl(Ar^\ell + Br^{-(\ell+1)}\bigr)\bigl(C\cos m\varphi + D\sin m\varphi\bigr) P_\ell^m(\cos\theta) . \tag{4a}

The three factors are then linked only through $\ell$ and $m$ . Finiteness at the origin sets $B = 0$ as well, and decay at infinity sets $A = 0$ .

Both extra conditions are needed. Without single-valuedness, $u = \varphi$ is harmonic and of product form but is not of the form (4a) — it is the $m=0$ , $C\ne0$ case, and it is exactly the magnetic scalar potential kept in step 5 of the method. Without finiteness on the axis, $u = r^{1/2}P_{1/2}(\cos\theta)$ is a product solution with $\ell = \tfrac12$ , so the indices need not be integers at all. This does not contradict part 2, because the second condition fails: $P_{1/2}$ is perfectly finite at the north pole, $P_{1/2}(1) = 1$ , but diverges logarithmically at the south pole $\theta\to\pi$ , where $\cos\theta\to-1$ . Numerically $P_{1/2}(-0.9) = -0.50127$ , $P_{1/2}(-0.99999) = -3.4946$ and $P_{1/2}(-1+10^{-7}) = -4.9605$ — slow divergence, but divergence.

The Helmholtz equation. For $\nabla^2u + k^2u = 0$ the angular factors and the two conditions on them are unchanged, and only the radial factor differs:

u(r,\theta,\varphi) = r^{-1/2}\bigl[AJ_{\ell+1/2}(kr) + BY_{\ell+1/2}(kr)\bigr]\bigl(C\cos m\varphi+D\sin m\varphi\bigr)\bigl[EP_\ell^m(\cos\theta)+FQ_\ell^m(\cos\theta)\bigr] . \tag{5}

Under the same two regularity conditions, $\ell$ and $m$ are integers with $0\le\lvert m\rvert\le\ell$ and the coefficient of $Q_\ell^m$ vanishes; finiteness at the origin sets $B = 0$ in addition. (Riley, Hobson and Bence call the solution $F$ in their Eq. (21.57), where it collides with their coefficient $F$ of $Q_\ell^m$ ; we write the solution as $u$ , as in (4), so that $F$ here is the coefficient only.)

Riley, Hobson and Bence are careful to point out that nothing in the derivation of (4) forces $\ell$ to be an integer; integrality is imposed afterwards, by the requirement that the polar factor be finite on the axis, and is then inherited by the radial factor.

Idea of the proofOriginal

The separation is done in two cuts, and the order matters. Multiplying the equation by $r^2$ after dividing by $u$ isolates all the $r$ -dependence in one group: that is the first cut, and its separation constant is written $\ell(\ell+1)$ because the radial equation it leaves is an Euler equation whose indices are then the tidy pair $\ell, -(\ell+1)$ . The remaining equation involves $\theta$ and $\varphi$ only; multiplying it by $\sin^2\theta$ isolates the $\varphi$ -dependence, which is the second cut, with constant $m^2$ . Single-valuedness in $\varphi$ makes $m$ an integer, and what is left for $\theta$ is the associated Legendre equation once $\cos\theta$ is used as the variable in place of $\theta$ .

The one subtlety is that the two conditions fixing the integers come from different places. The integrality of $m$ comes from the geometry of the azimuth — $\varphi$ and $\varphi+2\pi$ are the same direction. The integrality of $\ell$ comes from the behaviour of the Legendre equation at its singular points $\cos\theta = \pm1$ , which are the poles of the sphere: only for integer $\ell$ does that equation have a solution bounded there.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.1, pp. 731–734

Step 1 (first cut, and where the division is legal). Substitute $u = R(r)\Theta(\theta)\Phi(\varphi)$ into $\nabla^2u = 0$ in spherical polars. One wants to divide by $u$ , and $u$ is not nonvanishing: in all but the simplest cases it is zero on a surface inside the region, because $P_\ell^m(\cos\theta)$ and $\cos m\varphi$ have interior zeros and $Ar^\ell + Br^{-(\ell+1)}$ has one whenever $AB<0$ . So divide only where the division is legal. Put

I_R = \{r : R(r)\ne 0\}, \qquad I_\Theta = \{\theta : \Theta(\theta)\ne 0\}, \qquad I_\Phi = \{\varphi : \Phi(\varphi)\ne 0\} .

Each is open in its own coordinate interval, the three factors being continuous, and each is nonempty — this, and only this, is what the hypothesis that no factor vanishes identically gives, and it is all that is needed. Their product $I_R\times I_\Theta\times I_\Phi$ is exactly the subset of the region where $u \ne 0$ , and it is again a product set, which is what the separation lemma requires. On it, dividing by $u$ and multiplying by $r^2$ gives

\underbrace{\frac{1}{R}\frac{d}{dr}\!\left(r^2\frac{dR}{dr}\right)}_{\text{depends on } r \text{ only}} \;+\; \underbrace{\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2\Phi}{d\varphi^2}}_{\text{depends on }(\theta,\varphi)\text{ only}} \;=\; 0 ,

the factor $\sin^2\theta$ being nonzero because $0<\theta<\pi$ on the region. By the separation lemma, with $\mathbf{s} = r$ ranging over $I_R$ and $\mathbf{t} = (\theta,\varphi)$ over $I_\Theta\times I_\Phi$ , each group is a constant; call the first $\lambda$ , so the second is $-\lambda$ .

Step 1a (an identity true where the division was legal is true everywhere). Each separated equation below is first obtained only on the set where its own factor is nonzero, and then holds on the whole coordinate interval, by the following argument, which we use without further comment. Take the radial equation $r^2R''+2rR'-\lambda R = 0$ , which multiplying the first group of Step 1 by $R$ gives on $I_R$ . Let $Z$ be the complement of $I_R$ in the $r$ -interval, a closed set. A point of $\partial Z$ is a limit of points of $I_R$ , and both sides of the equation are continuous in $r$ ( $R$ being $C^2$ ), so the equation passes to the limit and holds there. On the interior of $Z$ the function $R$ vanishes on a neighbourhood, so $R = R' = R'' = 0$ and both sides are $0$ . Since the interval is $I_R\cup\partial Z\cup\operatorname{int}Z$ , the equation holds throughout it. The same argument applies verbatim to $\Theta$ on its interval and to $\Phi$ on its own.

Step 2 (the radial equation). By Steps 1 and 1a, on the whole $r$ -interval

r^2R'' + 2rR' - \lambda R = 0 ,

an Euler equation. Substituting $r = e^t$ and $R(r) = S(t)$ , so that $r\,d/dr = d/dt$ and $r^2d^2/dr^2 = d^2/dt^2 - d/dt$ , it becomes the constant-coefficient equation $S'' + S' - \lambda S = 0$ , with solutions $e^{\lambda_1t}, e^{\lambda_2t}$ where $\lambda_1+\lambda_2 = -1$ and $\lambda_1\lambda_2 = -\lambda$ . Hence

R(r) = Ar^{\lambda_1} + Br^{\lambda_2}, \qquad \lambda_1 + \lambda_2 = -1 .

Writing $\lambda_1 = \ell$ , the relation $\lambda_1+\lambda_2 = -1$ gives $\lambda_2 = -(\ell+1)$ and $\lambda = -\lambda_1\lambda_2 = \ell(\ell+1)$ . Nothing so far requires $\ell$ to be an integer, or even real: it is a name for the first root.

The two roots coincide exactly when $\lambda_1 = \lambda_2 = -\tfrac12$ , that is when $\ell = -\tfrac12$ and $\lambda = -\tfrac14$ . There $S''+S'-\lambda S = 0$ has the repeated root $-\tfrac12$ , so its general solution is $S(t) = (A + Bt)\mathrm{e}^{-t/2}$ and

R(r) = r^{-1/2}\bigl(A + B\ln r\bigr) ,

which is the exceptional radial factor recorded in part 1 of the statement. Every other value of $\ell$ gives two distinct powers.

Step 3 (second cut). The second group of Step 1, with $\lambda = \ell(\ell+1)$ , reads, on $I_\Theta\times I_\Phi$ ,

\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \frac{1}{\Phi\sin^2\theta}\Phi'' + \ell(\ell+1) = 0 .

Multiply by $\sin^2\theta$ :

\underbrace{\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \ell(\ell+1)\sin^2\theta}_{\theta \text{ only}} \;+\; \underbrace{\frac{\Phi''}{\Phi}}_{\varphi\text{ only}} \;=\; 0 .

By the separation lemma again, applied on the product set $I_\Theta\times I_\Phi$ , each of the two groups is a constant: $\Phi''/\Phi \equiv -m^2$ on $I_\Phi$ , and the $\theta$ -group $\equiv m^2$ on $I_\Theta$ .

Step 4 (the azimuthal factor). Multiplying by $\Phi$ gives $\Phi'' + m^2\Phi = 0$ on $I_\Phi$ , hence on the whole $\varphi$ -interval by Step 1a. It gives $\Phi = C\cos m\varphi + D\sin m\varphi$ for $m \ne 0$ and $\Phi = C\varphi + D$ for $m = 0$ . Since $\varphi$ and $\varphi + 2\pi$ label the same half-plane, a single-valued $u$ requires $\Phi(\varphi+2\pi) = \Phi(\varphi)$ : hence $C = 0$ in the $m = 0$ case, and $m$ must be an integer in the other.

Step 5 (the polar factor). With the constant $m^2$ inserted, the $\theta$ -group of Step 3 reads, on $I_\Theta$ ,

\frac{\sin\theta}{\Theta}\frac{d}{d\theta}\!\left(\sin\theta\frac{d\Theta}{d\theta}\right) + \ell(\ell+1)\sin^2\theta = m^2 .

Change the variable to $\mu = \cos\theta$ . Then $d\mu/d\theta = -\sin\theta$ , so $d/d\theta = -\sin\theta\,d/d\mu = -(1-\mu^2)^{1/2}d/d\mu$ , and $\sin\theta\,d\Theta/d\theta = -(1-\mu^2)\,dM/d\mu$ where $M(\mu) = \Theta(\theta)$ . Substituting, multiplying through by $M$ and dividing by $\sin^2\theta = 1-\mu^2$ ,

\frac{d}{d\mu}\!\left[(1-\mu^2)\frac{dM}{d\mu}\right] + \left[\ell(\ell+1) - \frac{m^2}{1-\mu^2}\right]M = 0 , \tag{6}

which is the associated Legendre equation — on the image of $I_\Theta$ under $\theta\mapsto\cos\theta$ to begin with, and then, by Step 1a, on the whole interval. Its solutions are $M = EP_\ell^m(\mu) + FQ_\ell^m(\mu)$ .

Step 6 (integrality of $\ell$ , and the axis). Equation (6) is singular at $\mu = \pm1$ , which are the two poles $\theta = 0,\pi$ of the sphere. The function $Q_\ell^m$ diverges there, so a solution finite on the polar axis has $F = 0$ ; and $P_\ell^m$ is finite there only when $\ell$ is a non-negative integer with $0\le\lvert m\rvert\le\ell$ , which is the condition established in associated Legendre functions. This is where integrality of $\ell$ enters, and through $\lambda = \ell(\ell+1)$ it feeds back into the radial exponents of Step 2.

Step 7 (assemble). Multiplying the three factors gives (4), which is part 1 of the statement; imposing the single-valuedness of Step 4 and the finiteness on the axis of Step 6 gives part 2 and the form (4a).

Step 8 (the Helmholtz case). Repeating Steps 1–6 for $\nabla^2u + k^2u = 0$ changes only Step 1, where the extra term contributes $k^2r^2$ to the $r$ -group. So Steps 3–6 are untouched — the angular factors are identical — and the radial equation becomes

r^2R'' + 2rR' + \bigl[k^2r^2 - \ell(\ell+1)\bigr]R = 0 . \tag{7}

Put $R(r) = r^{-1/2}S(r)$ . Then $R' = r^{-1/2}S' - \tfrac12 r^{-3/2}S$ and $R'' = r^{-1/2}S'' - r^{-3/2}S' + \tfrac34r^{-5/2}S$ , so

r^2R'' + 2rR' = r^{3/2}S'' + r^{1/2}S' - \tfrac14 r^{-1/2}S ,

and multiplying (7) by $r^{1/2}$ gives

r^2S'' + rS' + \left[k^2r^2 - \left(\ell+\tfrac12\right)^2\right]S = 0 ,

using $\ell(\ell+1) + \tfrac14 = (\ell+\tfrac12)^2$ . With $\mu = kr$ this is Bessel's equation of order $\ell+\tfrac12$ , whose solutions are $J_{\ell+1/2}(kr)$ and $Y_{\ell+1/2}(kr)$ . Hence (5). $\square$

Remark(Where the hypotheses are used: nonvanishing factors, and a product domain for the separation lemma)Original

Two hypotheses of the statement are consumed in Step 1, and nowhere else. The first is that no factor vanishes identically. It is far too weak to make $u\ne0$ on the region — no hypothesis short of assuming that outright would — but it is exactly enough to make the set $I_R\times I_\Theta\times I_\Phi$ on which one may divide by $u$ nonempty, and continuity then carries each separated equation across the zeros of its own factor. The second is the product form of the region: Step 1 applies the lemma with $\mathbf{t} = (\theta,\varphi)$ ranging over its angular factor independently of $r$ . If the region were, say, a ball dented on one side, the admissible $(\theta,\varphi)$ would depend on $r$ , the two groups would not be free to vary independently, and the conclusion that each is constant would be unjustified. (The restriction to $I_R\times I_\Theta\times I_\Phi$ does no damage here, because cutting a product set down factor by factor leaves a product set.) This is the precise reason the method demands a region bounded by coordinate surfaces, and why step 1 of the recipe is the decisive one.

Remark(Separation finds solutions; it does not prove they are all of them)AdaptedAdapted from Evans (2010), §4.1.1, pp. 159–161

A single product (4) is one solution; a sum of products is another. That a superposition can meet arbitrary data is a completeness statement about the eigenfunctions, proved separately — Fourier series for the trigonometric factors, Fourier–Bessel series for the radial ones, the completeness of spherical harmonics for the angular ones. Uniqueness then comes from the maximum principle or an energy argument (harmonic functions), not from the separation. Evans makes the same point about his separated heat equation: only the single product $u = \mathrm{e}^{-\lambda t}w$ is produced by the separation itself, and the finite sums and infinite series built from it rest on the linearity of the equation and on a completeness statement he defers to his later chapters.

Remark(What the sources add: angular factors, time factors, and the Schrödinger exception)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, pp. 737–741; Hassani (2013), §13.2, Eqs. (13.11)–(13.14), pp. 398–400

Why every equation built from $\nabla^2$ has the same angular factors. Riley, Hobson and Bence make this the general point of their section: provided $\theta$ and $\varphi$ enter the equation only through $\nabla^2$ , the second cut of the proof is untouched, so the angular factors are spherical harmonics whatever the radial term. Hassani's operator version makes it structural: $\nabla^2 = -r^{-2}L^2 + r^{-2}(\mathbf{r}\cdot\nabla)(\mathbf{r}\cdot\nabla) + r^{-2}\mathbf{r}\cdot\nabla$ with $L = \mathbf{r}\times(-i\nabla)$ depending only on the angles, so that for any $f = f(r)$ the master equation (3) splits into the eigenvalue problem $L^2Y = \lambda Y$ on the sphere and a radial ODE containing $f$ and $\lambda$ . (Hassani writes this angular eigenvalue $\alpha$ ; we write $\lambda$ , which is the $\lambda = \ell(\ell+1)$ of the proof above, to keep $\alpha$ for the axial constant.) In quantum mechanics this is the statement that angular momentum is conserved in any central potential, and it is why the hydrogen atom, the three-dimensional oscillator and a free particle all have the same angular wavefunctions and differ only radially.
Which equations separate the time cleanly. Step 2 works because the equation is linear with coefficients independent of $t$ . Riley, Hobson and Bence note that for the wave equation the resulting time factor $e^{\pm i\omega t}$ with $\omega = kc$ accompanies the spatial Helmholtz solution of the same $k$ , so that a separated solution has a definite frequency; but in spherical polars the zeros of $j_\ell(kr)$ are not equally spaced in $r$ except when $\ell = 0$ , so such a solution has no definite wavelength. For the diffusion equation the time factor is the decaying $e^{-k^2\kappa t}$ instead.
The Schrödinger equation is close to Helmholtz but not equal to it. Riley, Hobson and Bence point this out for the hydrogen atom: the stationary-state equation is $\nabla^2u + \frac{2m}{\hbar^2}\bigl(E + \frac{e^2}{4\pi\varepsilon_0 r}\bigr)u = 0$ , which has an $r$ -dependent coefficient rather than the constant $k^2$ , so the radial equation is not Bessel's. The angular factors are nevertheless spherical harmonics, by the previous point. See the Schrödinger equation.

Remark(Not every separable system is one of these three)Original

The method also separates in elliptic, parabolic, prolate and oblate spheroidal, and other coordinate systems, each producing its own family of special functions (Mathieu, Lamé, spheroidal wave functions). The three systems treated here cover the geometries that occur most often, and they are the ones whose special functions this handbook develops.

Example(Cartesian: Laplace's equation in a cube)Original

Solve $\nabla^2u = 0$ in the unit cube $0<x,y,z<1$ with $u = 0$ on the five faces $x=0$ , $x=1$ , $y=0$ , $y=1$ , $z=0$ , and

u(x,y,1) = \sin\pi x\,\sin\pi y

on the top face.

Separate. With $u = X(x)Y(y)Z(z)$ and the constants of the catalogue, $X'' = -l^2X$ , $Y''=-m^2Y$ , $Z''=-n^2Z$ and $l^2+m^2+n^2 = 0$ (Laplace, so $k=0$ ). The four homogeneous side faces force $X(0)=X(1)=0$ and $Y(0)=Y(1)=0$ , so $X = \sin p\pi x$ and $Y = \sin q\pi y$ with $p,q$ positive integers, and $l^2 = p^2\pi^2$ , $m^2 = q^2\pi^2$ . Then $n^2 = -(p^2+q^2)\pi^2 < 0$ , so the $z$ -factor is hyperbolic, as the catalogue predicts:

Z(z) = E\sinh\gamma_{pq}z + F\cosh\gamma_{pq}z, \qquad \gamma_{pq} = \pi\sqrt{p^2+q^2} .

The face $z=0$ forces $F = 0$ .

Fit the data. The datum is already a single eigenfunction, $p = q = 1$ , so only one term survives and

u(x,y,z) = \sin\pi x\,\sin\pi y\,\frac{\sinh(\gamma z)}{\sinh\gamma}, \qquad \gamma = \pi\sqrt2 = 4.4428829 .

Check. $\nabla^2 u = (-\pi^2 - \pi^2 + \gamma^2)u = (\gamma^2 - 2\pi^2)u = 0$ since $\gamma^2 = 2\pi^2$ . All six boundary conditions hold by construction.

A number. At the centre of the cube,

u\left(\tfrac12,\tfrac12,\tfrac12\right) = \frac{\sinh(\gamma/2)}{\sinh\gamma} = \frac{1}{2\cosh(\pi/\sqrt2)} = \frac{1}{2\cdot 4.6645326} = 0.1071919 .

The datum has decayed by a factor of about nine over half the height of the cube — the exponential decay of a Dirichlet datum into the interior at the rate set by $\gamma$ , which is the Cartesian signature of harmonicity. Had the datum been $\sin 3\pi x\sin 4\pi y$ instead, $\gamma$ would be $5\pi$ and the centre value $1/(2\cosh(5\pi/2)) = 3.9\times10^{-4}$ : finer structure on the boundary dies faster, which is the mechanism behind interior smoothing.

Example(Cylindrical: Dirichlet eigenvalues of a cylinder)Original

Find the lowest frequencies at which a scalar field can resonate inside a cylinder of radius $a = 1$ and height $L = 2$ while vanishing on the whole boundary; that is, find the smallest $k$ with a nonzero solution of the Helmholtz equation $\nabla^2u + k^2u = 0$ in the cylinder and $u = 0$ on its curved surface, bottom and top.

Separate. From the catalogue, with the region containing the axis $\rho = 0$ we drop $Y_m$ , and with $u$ vanishing at $z = 0$ and $z = L$ the axial factor must be $\sin(p\pi z/L)$ , so $\alpha^2 = -(p\pi/L)^2$ with $p$ a positive integer. The radial factor is then $J_m(\kappa\rho)$ with the catalogue's radial argument $\kappa = \sqrt{k^2+\alpha^2}$ , that is

\kappa^2 = k^2 + \alpha^2 = k^2 - \frac{p^2\pi^2}{L^2} ,

and the condition $u = 0$ on $\rho = a$ requires $J_m(\kappa a) = 0$ , i.e. $\kappa a = j_{m,n}$ , the $n$ th positive zero of $J_m$ . Hence the admissible values are

k^2_{m,n,p} = \left(\frac{j_{m,n}}{a}\right)^{\!2} + \left(\frac{p\pi}{L}\right)^{\!2} , \qquad u = J_m\!\left(\frac{j_{m,n}\rho}{a}\right)\bigl(C\cos m\varphi + D\sin m\varphi\bigr)\sin\frac{p\pi z}{L} .

Numbers. With $a = 1$ and $L = 2$ , the relevant Bessel zeros are $j_{0,1} = 2.4048256$ , $j_{1,1} = 3.8317060$ , $j_{2,1} = 5.1356223$ , $j_{0,2} = 5.5200781$ , and $(p\pi/L)^2 = 2.4674011\,p^2$ . The seven lowest eigenvalues are

$(m,n,p)$	$j_{m,n}^2$	$(p\pi/L)^2$	$k^2$	$k$	multiplicity
$(0,1,1)$	$5.783186$	$2.467401$	$8.250587$	$2.872384$	1
$(0,1,2)$	$5.783186$	$9.869604$	$15.652790$	$3.956361$	1
$(1,1,1)$	$14.681971$	$2.467401$	$17.149372$	$4.141180$	2
$(1,1,2)$	$14.681971$	$9.869604$	$24.551575$	$4.954955$	2
$(0,1,3)$	$5.783186$	$22.206610$	$27.989796$	$5.290538$	1
$(2,1,1)$	$26.374616$	$2.467401$	$28.842018$	$5.370476$	2
$(0,2,1)$	$30.471262$	$2.467401$	$32.938663$	$5.739222$	1

The modes with $m \ne 0$ come in pairs, $\cos m\varphi$ and $\sin m\varphi$ , so they are doubly degenerate; those with $m = 0$ are simple. The lowest mode is $J_0(2.4048256\,\rho)\sin(\pi z/2)$ , nodeless inside, with $k = 2.8723835$ .

A check that the separation constants are doing what they should. Each $k^2$ is a sum of a radial and an axial piece, and the two pieces are the eigenvalues of the two separated operators; the algebraic relation among the separation constants in step 4 of the recipe is here simply $\kappa^2 + (p\pi/L)^2 = k^2$ . Lengthening the cylinder, $L\to\infty$ , drives the axial piece to zero and the spectrum down onto the two-dimensional values $(j_{m,n}/a)^2$ — the drumskin eigenvalues — as it must.

Example(Spherical: a sphere held at potential

V_0\cos^3\theta

)Original

A sphere of radius $a$ carries the surface potential $h(\theta) = V_0\cos^3\theta$ . Find the potential inside and outside, with $u$ bounded at the origin and $u\to0$ at infinity.

Expand the datum in Legendre polynomials. From $P_1(\mu) = \mu$ and $P_3(\mu) = \tfrac12(5\mu^3-3\mu)$ we get $\mu^3 = \tfrac25P_3(\mu) + \tfrac35P_1(\mu)$ . So with $\mu = \cos\theta$ ,

h = V_0\left[\tfrac35P_1(\mu) + \tfrac25P_3(\mu)\right] .

Choose the surviving terms of (4). The datum is independent of $\varphi$ , so $m = 0$ and the polar factors are the Legendre polynomials $P_\ell$ ; the solution must be finite on the axis, so $F = 0$ . Inside the sphere finiteness at the origin kills $r^{-(\ell+1)}$ ; outside, decay at infinity kills $r^\ell$ . Hence

u_{\text{in}} = \sum_{\ell\ge0}A_\ell r^\ell P_\ell(\mu), \qquad u_{\text{out}} = \sum_{\ell\ge0}B_\ell r^{-(\ell+1)}P_\ell(\mu) .

Match at $r = a$ . By the orthogonality of the $P_\ell$ , only $\ell = 1$ and $\ell = 3$ appear, and $A_1a = \tfrac35V_0$ , $A_3a^3 = \tfrac25V_0$ , $B_1a^{-2} = \tfrac35V_0$ , $B_3a^{-4} = \tfrac25V_0$ . So

u_{\text{in}} = V_0\left[\frac35\frac ra P_1(\mu) + \frac25\frac{r^3}{a^3}P_3(\mu)\right], \qquad u_{\text{out}} = V_0\left[\frac35\frac{a^2}{r^2}P_1(\mu) + \frac25\frac{a^4}{r^4}P_3(\mu)\right] .

Checks. Both are harmonic, being superpositions of the solid harmonics $r^\ell P_\ell$ and $r^{-(\ell+1)}P_\ell$ of (4). At $r = a$ both reduce to $V_0[\tfrac35P_1+\tfrac25P_3] = V_0\cos^3\theta$ , so the potential is continuous across the surface. Numerically, on the polar axis $\theta = 0$ (where $P_\ell(1) = 1$ for all $\ell$ ):

\frac{u_{\text{in}}(a/2,0)}{V_0} = \frac35\cdot\frac12 + \frac25\cdot\frac18 = 0.35 , \qquad \frac{u_{\text{out}}(2a,0)}{V_0} = \frac35\cdot\frac14 + \frac25\cdot\frac1{16} = 0.175 .

At the centre, $u_{\text{in}}(0) = 0$ , which is the mean value property: $\cos^3\theta$ is odd under $\theta\mapsto\pi-\theta$ and so averages to zero over the sphere. Far away, $u_{\text{out}} \approx \tfrac35V_0a^2\cos\theta/r^2$ : the leading behaviour is a dipole, because the datum has no $\ell = 0$ component and hence no net charge — the systematic version of this reading is multipole expansions.

The discontinuity in the normal derivative. Differentiating the two expressions and setting $r = a$ ,

\left.\frac{\partial u_{\text{in}}}{\partial r}\right|_{r=a} = \frac{V_0}{a}\left[\tfrac35P_1 + \tfrac65P_3\right], \qquad \left.\frac{\partial u_{\text{out}}}{\partial r}\right|_{r=a} = -\frac{V_0}{a}\left[\tfrac65P_1 + \tfrac85P_3\right] .

The radial derivative therefore jumps by $-\frac{V_0}{a}\bigl[\tfrac95P_1 + \tfrac{14}{5}P_3\bigr]$ across the surface, which in electrostatics equals $-\sigma/\varepsilon_0$ for the surface charge density $\sigma$ that holds the sphere at the prescribed potential. The potential itself is continuous and its tangential derivative is continuous; only the normal derivative jumps, and the jump measures the source sitting on the surface.

Application(Electrostatics of conductors)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.1, pp. 734, 735–737

Riley, Hobson and Bence's two spherical examples are the standard pair. An uncharged conducting sphere in a uniform field $E$ must have a potential that approaches $-Er\cos\theta$ at large $r$ and is constant on $r = a$ ; since $\cos\theta = P_1(\cos\theta)$ , only the $\ell = 1$ terms of (4) can appear, and the two conditions fix them completely, giving $v = -Er(1-a^3/r^3)\cos\theta$ — the sphere behaves as a dipole of moment $4\pi\varepsilon_0a^3E$ . A hollow split sphere with its two hemispheres held at $v_0$ and $0$ needs the full Legendre series instead, and the coefficients come out as $v_0/2, 3v_0/4a, 0, -7v_0/16a^3,\dots$ ; on the equator the series sums to $v_0/2$ , the average of the two values, because a Legendre series behaves at a jump exactly as a Fourier series does.

Application(Normal modes and resonant cavities)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, p. 739; Arfken, Weber & Harris (2013), §9.4, Example 9.4.3, pp. 422–424

The cylindrical example above is a cavity calculation for a single scalar field required to vanish on the whole boundary: an acoustic cavity with pressure-release walls, or — Arfken's reading of the same eigenvalue problem — a particle in a cylindrical box, with $k^2 = 2mE/\hbar^2$ . With $\omega = ck$ and $c = 2.998\times10^{10}\,\mathrm{cm\,s^{-1}}$ , a cylinder of radius $a = 1$ cm and height $L = 2$ cm has its lowest such mode at $k = 2.8723835\,\mathrm{cm}^{-1}$ , that is $\omega/2\pi = ck/2\pi = 13.71$ GHz.

An electromagnetic cavity of the same dimensions is not this problem, and resonates lower. A conducting wall forces only the tangential electric field to vanish on it, not every field component, so for a TM mode the axial field $E_z\propto J_m(j_{m,n}\rho/a)\cos(p\pi z/L)$ need have no axial node: $k^2 = (j_{m,n}/a)^2+(p\pi/L)^2$ with $p = 0,1,2,\dots$ , whereas the TE modes are built from the zeros $j'_{m,n}$ of $J'_m$ and need $p\ge1$ . The lowest mode of this cylinder is therefore TM $_{010}$ , with $k = j_{0,1}/a = 2.4048256\,\mathrm{cm}^{-1}$ and $ck/2\pi = 11.47$ GHz, independent of the height $L$ ; the lowest TE mode, TE $_{111}$ , has $k = [(j'_{1,1}/a)^2+(\pi/L)^2]^{1/2} = 2.4202\,\mathrm{cm}^{-1}$ with $j'_{1,1} = 1.8411838$ , giving $11.55$ GHz. The separation itself is word for word the one above; only the admissible axial index changes.

The same separation with the $z$ -dependence dropped gives the modes of a circular drumskin, whose frequencies are $\omega = j_{m,n}c/a$ with $c^2 = T/\sigma$ ; Riley, Hobson and Bence compute the four lowest and note that the second and third are two-fold degenerate. The degeneracy is not an accident of the numbers but a consequence of the rotational symmetry of the disk, which pairs $\cos m\varphi$ with $\sin m\varphi$ ; breaking the symmetry — an elliptical drum — splits each pair. See normal modes and standing waves and normal modes.

Application(Steady heat conduction with mixed data)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.1, pp. 729–731

Riley, Hobson and Bence's semi-infinite cylinder, curved surface at $0^\circ$ and base at $T_0$ , is the model problem for which a Fourier–Bessel series rather than a Fourier series is required: the base datum is expanded in $J_0(k_n\rho)$ with $k_na$ the zeros of $J_0$ , and the coefficients come out as $2T_0/(k_naJ_1(k_na))$ using the recurrence $\frac{d}{dz}[zJ_1(z)] = zJ_0(z)$ . The same calculation with a non-axisymmetric base temperature needs the double series over $m$ and $n$ , with coefficients extracted by the orthogonality of both the Bessel and the trigonometric factors — step 7 of the recipe in its general form.

ApplicationOriginal

Quantum mechanics of central potentials. The reason the periodic table has the shape it has is step 5 and step 6 of this recipe. For any potential $V(r)$ depending on $r$ alone, the angular factors of the stationary Schrödinger equation are the spherical harmonics $Y_\ell^m$ , with $\ell$ a non-negative integer and $2\ell+1$ values of $m$ ; the potential enters only the radial equation and so fixes the energies but not the angular degeneracy. The $s,p,d,f$ labels are the values $\ell = 0,1,2,3$ , and the $2(2\ell+1)$ electrons per subshell are the $2\ell+1$ values of $m$ times two spin states. The operator derivation of Hassani (Hassani (2013), §13.2, pp. 398–400) shows why: $L^2$ commutes with any $\nabla^2 + f(r)$ , so angular momentum and energy can be diagonalized together. See the Schrödinger equation and the Helmholtz equation.

Numerical methods. Separation is also the basis of the fast direct solvers. On a box, a rectangular grid, or a disk, the discrete Laplacian is diagonalized by the discrete analogues of the separated eigenfunctions, so a Poisson solve becomes a forward transform, a division by the eigenvalues $l^2+m^2+n^2$ , and an inverse transform — $O(N\log N)$ work instead of the $O(N^{3/2})$ or worse of an iterative solve (the discrete Fourier transform). The same spectral idea, applied with Chebyshev or Legendre expansions in a non-periodic direction, is what makes spectral methods for elliptic problems on tensor-product domains converge faster than any fixed power of the mesh size.

Application

Multipole expansions

Adapted

Add to PDF

Expands the potential of a localized source outside a sphere containing it, first in Legendre polynomials from the generating function and then in spherical harmonics by the addition theorem, defines the monopole, dipole and quadrupole moments that the terms carry, and shows that the lowest nonvanishing moment is independent of the choice of origin while all higher ones are not.

Prerequisites: Spherical harmonics, Fundamental solution of Laplace's equation, Legendre polynomials, The Dirac delta in calculations

Definition(Point multipoles)AdaptedAdapted from Stakgold & Holst (2011), §2.1, Example 6, pp. 99–100 and §2.2, Eq. (2.2.9), p. 118; Arfken, Weber & Harris (2013), §15.3, Eq. (15.66), p. 738

A point dipole at $\boldsymbol\xi$ with axis the unit vector $\mathbf{l}$ is the limit, as $\varepsilon\to0$ , of a source of strength $+1/\varepsilon$ at $\boldsymbol\xi + \tfrac{\varepsilon}{2}\mathbf{l}$ together with one of strength $-1/\varepsilon$ at $\boldsymbol\xi - \tfrac{\varepsilon}{2}\mathbf{l}$ . Stakgold and Holst take the limit in the sense of distributions: for a test function $\phi$ ,

\frac{1}{\varepsilon}\left[\phi\!\left(\boldsymbol\xi + \tfrac{\varepsilon}{2}\mathbf{l}\right) - \phi\!\left(\boldsymbol\xi - \tfrac{\varepsilon}{2}\mathbf{l}\right)\right] \;\longrightarrow\; \frac{\partial\phi}{\partial l}(\boldsymbol\xi) ,

so the point dipole is the distribution $\phi\mapsto \partial\phi/\partial l\,(\boldsymbol\xi)$ . Equivalently, its volume source density is a derivative of the delta: $(\partial/\partial\xi_i)\,\delta(\mathbf{x}-\boldsymbol\xi)$ is the density of a unit dipole at $\boldsymbol\xi$ with axis along $x_i$ . Note that the charges must be made stronger as they are brought together, or nothing survives.

Arfken takes the same limit at the level of the potential: the two charges $\pm q$ at $z = \pm a$ produce a potential whose $a\to0$ limit at fixed $\mu = 2qa$ is

\psi = \frac{\mu}{4\pi\varepsilon_0}\frac{P_1(\cos\theta)}{r^2} ,

the potential of a point dipole of moment $\mu$ at the origin, oriented along the polar axis. A linear $2^n$ -pole is obtained by iterating the construction $n$ times, and contributes $P_n(\cos\theta)/r^{n+1}$ ; $n = 0,1,2$ are the monopole, dipole and quadrupole.

Throughout, $r_>$ and $r_<$ denote the larger and the smaller of two radii, and $\Omega = (\theta,\varphi)$ a direction. We use the physicists' spherical harmonics $Y_\ell^m$ of spherical harmonics, in which the addition theorem reads as in (3) below.

Application(The Laplace expansion of the inverse distance)AdaptedAdapted from Arfken, Weber & Harris (2013), §15.3, Eqs. (15.60)–(15.63), pp. 736–737 and §16.3, Eq. (16.66), p. 800

Let $\mathbf{r}_1,\mathbf{r}_2$ be measured from a common origin, with magnitudes $r_1,r_2$ and angular coordinates $\Omega_1,\Omega_2$ , and let $\chi$ be the angle between them. Then

\frac{1}{\lvert\mathbf{r}_1-\mathbf{r}_2\rvert} \;=\; \frac{1}{r_>}\sum_{\ell=0}^{\infty}\left(\frac{r_<}{r_>}\right)^{\!\ell} P_\ell(\cos\chi) \;=\;\sum_{\ell=0}^{\infty}\frac{r_<^{\,\ell}}{r_>^{\,\ell+1}}P_\ell(\cos\chi) , \tag{1}

the series converging except when $r_1 = r_2$ . Inserting the spherical harmonic addition theorem turns (1) into

\frac{1}{\lvert\mathbf{r}_1-\mathbf{r}_2\rvert} \;=\; \sum_{\ell=0}^{\infty}\frac{4\pi}{2\ell+1}\frac{r_<^{\,\ell}}{r_>^{\,\ell+1}}\sum_{m=-\ell}^{\ell}Y_\ell^m(\Omega_1)^*\,Y_\ell^m(\Omega_2) , \tag{2}

which holds in a coordinate system of arbitrary orientation, the angle $\chi$ having been traded for the two directions separately. The addition theorem used here is

P_\ell(\cos\chi) = \frac{4\pi}{2\ell+1}\sum_{m=-\ell}^{\ell}Y_\ell^m(\Omega_1)^*\,Y_\ell^m(\Omega_2) . \tag{3}

Arfken notes that the rate of convergence of (1) improves as $r_>/r_<$ grows, and that the expansion of a point charge's potential inside the sphere through the charge — the case $r_< = r$ — is obtained from the same generating function by rearranging the square root the other way.

Idea of the proofOriginal

Everything follows from the Legendre generating function

\bigl(1 - 2ht + h^2\bigr)^{-1/2} = \sum_{\ell=0}^\infty P_\ell(t)\,h^\ell \qquad (\lvert h\rvert<1) ,

because the law of cosines makes $\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ exactly of that shape. Writing $\lvert\mathbf{r}_1-\mathbf{r}_2\rvert^2 = r_1^2 + r_2^2 - 2r_1r_2\cos\chi$ and pulling out whichever of $r_1,r_2$ is larger leaves $(1-2ht+h^2)^{-1/2}$ with $t = \cos\chi$ and $h = r_</r_>$ , which is less than one. Pulling out the smaller radius would give $h>1$ and a divergent series, which is why the expansion must be organized around $r_>$ and why a multipole expansion is a statement about the field outside the source, never inside it.

The passage from (1) to (2) is a change of bookkeeping with one real gain. In (1) the single function $P_\ell(\cos\chi)$ couples the two directions, so the expansion is only useful when one of them is fixed — Arfken's charge on the polar axis. The addition theorem (3) factorizes that coupling into a sum of products, each factor depending on one direction alone. Once that is done, the source variables and the field variables appear in separate factors, and the sum over the source can be carried out once and for all. The numbers it produces are the multipole moments.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §15.3, Eqs. (15.58)–(15.63), pp. 736–737

Step 1 (the law of cosines). With $\chi$ the angle between $\mathbf{r}_1$ and $\mathbf{r}_2$ ,

\lvert\mathbf{r}_1-\mathbf{r}_2\rvert = \bigl(r_1^2 + r_2^2 - 2r_1r_2\cos\chi\bigr)^{1/2} .

Step 2 (pull out the larger radius). Let $r_> = \max(r_1,r_2)$ , $r_< = \min(r_1,r_2)$ and $h = r_</r_> \in [0,1)$ . The expression inside the square root is symmetric in $r_1,r_2$ , so

\lvert\mathbf{r}_1-\mathbf{r}_2\rvert = r_>\bigl(1 - 2h\cos\chi + h^2\bigr)^{1/2} ,

and therefore

\frac{1}{\lvert\mathbf{r}_1-\mathbf{r}_2\rvert} = \frac{1}{r_>}\bigl(1-2h\cos\chi+h^2\bigr)^{-1/2} .

Step 3 (insert the generating function). Since $\lvert h\rvert<1$ and $\lvert\cos\chi\rvert\le1$ , the generating function of the Legendre polynomials applies with $t = \cos\chi$ and gives

\frac{1}{\lvert\mathbf{r}_1-\mathbf{r}_2\rvert} = \frac{1}{r_>}\sum_{\ell=0}^\infty P_\ell(\cos\chi)\,h^\ell ,

which is (1). The generating-function expansion is guaranteed only for $\lvert h\rvert < 1$ ; at $h = 1$ , that is at $r_1 = r_2$ , absolute convergence is lost, and Arfken excludes that case. Note that $\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ itself is perfectly finite there unless also $\chi = 0$ , so what the exclusion limits is the expansion rather than the function.

Step 4 (factorize the angle). Substituting the addition theorem (3) into (1) and exchanging the two finite inner sums with the outer one gives (2) term by term. $\square$

Application(The multipole expansion)AdaptedAdapted from Arfken, Weber & Harris (2013), §16.3, Eqs. (16.72)–(16.74), p. 801 and §15.3, Eq. (15.67), p. 739

Let charges $q_i$ sit at points $\mathbf{r}_i$ , all inside the sphere $\lvert\mathbf{r}\rvert = a$ centred at the origin. For a field point with $r > a$ ,

\psi(\mathbf{r}) \;=\; \frac{1}{4\pi\varepsilon_0}\sum_{\ell=0}^\infty\sum_{m=-\ell}^{\ell} M_\ell^m\,\frac{Y_\ell^m(\theta,\varphi)}{r^{\ell+1}} , \qquad M_\ell^m \;=\; \frac{4\pi}{2\ell+1}\sum_i q_i\,r_i^{\,\ell}\,Y_\ell^m(\theta_i,\varphi_i)^* . \tag{4}

The numbers $M_\ell^m$ are the multipole moments of the distribution. For a continuous density $\rho$ the sums over $i$ become integrals, $M_\ell^m = \frac{4\pi}{2\ell+1}\int \rho(\mathbf{r}')\,r'^{\,\ell}\,Y_\ell^m(\Omega')^*\,dV'$ .

If the charges lie on the polar axis, at $z = a_i$ , the sum collapses to $m = 0$ and (4) becomes Arfken's linear multipole expansion

\psi = \frac{1}{4\pi\varepsilon_0 r}\left[\mu_0 + \frac{\mu_1}{r}P_1(\cos\theta) + \frac{\mu_2}{r^2}P_2(\cos\theta) + \cdots\right], \qquad \mu_n = \sum_i q_i a_i^{\,n} , \tag{5}

with $\mu_0$ the monopole moment (the total charge), $\mu_1$ the dipole moment and $\mu_2$ the quadrupole moment. Both expansions converge for $r$ larger than every $r_i$ , that is at field points farther from the origin than any part of the source.

Normalizations differ between authors. Arfken warns of this explicitly: his $M_\ell^m$ of (4) is the choice that makes the expansion formula simplest, and it does not reproduce the traditional scalings of the low-order moments. With his convention

M_0^0 = \sqrt{4\pi}\sum_i q_i , \qquad M_1^0 = \sqrt{\tfrac{4\pi}{3}}\sum_i q_i z_i ,

so the monopole moment is $\sqrt{4\pi}$ times the total charge and $M_1^0$ is $\sqrt{4\pi/3}$ times the $z$ -component of the usual dipole vector. We follow Arfken's convention and state the factors wherever a traditional moment is wanted.

Idea of the proofOriginal

Take the Laplace expansion (2), put $\mathbf{r}_1 = \mathbf{r}_i$ (a source point) and $\mathbf{r}_2 = \mathbf{r}$ (the field point), and note that the hypothesis $r > a$ decides once and for all which radius is the larger: $r_> = r$ and $r_< = r_i$ for every $i$ . The $\ell$ th term then factors into $r_i^{\ell}Y_\ell^m(\Omega_i)^*$ , which involves the source only, times $Y_\ell^m(\Omega)/r^{\ell+1}$ , which involves the field point only. Summing over the charges touches only the first factor, and what comes out is a finite list of numbers that is all the far field ever needs to know about the source. That is the whole content of the expansion: the exterior field of a localized source is determined by countably many numbers, and the first few of them dominate.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §16.3, p. 801

Step 1 (superpose). By linearity, the potential of the charges is the sum of their individual Coulomb potentials,

\psi(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{\lvert\mathbf{r}-\mathbf{r}_i\rvert} ,

each term being $q_i/\varepsilon_0$ times the fundamental solution $\Phi(\mathbf{x}) = 1/(4\pi\lvert\mathbf{x}\rvert)$ of Laplace's equation centred at $\mathbf{r}_i$ , since $(q_i/\varepsilon_0)\Phi(\mathbf{r}-\mathbf{r}_i) = q_i/(4\pi\varepsilon_0\lvert\mathbf{r}-\mathbf{r}_i\rvert)$ .

Step 2 (expand each term). For each $i$ we have $r_i \le a < r$ , so in (2) the roles are fixed: $r_< = r_i$ , $r_> = r$ . Hence

\frac{1}{\lvert\mathbf{r}-\mathbf{r}_i\rvert} = \sum_{\ell=0}^\infty\frac{4\pi}{2\ell+1}\frac{r_i^{\,\ell}}{r^{\,\ell+1}}\sum_{m=-\ell}^{\ell}Y_\ell^m(\Omega_i)^*\,Y_\ell^m(\Omega) .

Step 3 (exchange the sums). The sum over $i$ is finite, and each expansion converges absolutely for $r > a$ . To see that, bound the inner sum by the Cauchy–Schwarz inequality and then use $\sum_{m=-\ell}^{\ell}\lvert Y_\ell^m(\Omega)\rvert^2 = (2\ell+1)/4\pi$ , which is the addition theorem (3) at $\chi = 0$ :

\sum_{m=-\ell}^{\ell}\bigl\lvert Y_\ell^m(\Omega_i)^*\,Y_\ell^m(\Omega)\bigr\rvert \;\le\; \Bigl(\sum_m \lvert Y_\ell^m(\Omega_i)\rvert^2\Bigr)^{1/2}\Bigl(\sum_m \lvert Y_\ell^m(\Omega)\rvert^2\Bigr)^{1/2} \;=\; \frac{2\ell+1}{4\pi} .

The $\ell$ th term of the expansion of $1/\lvert\mathbf{r}-\mathbf{r}_i\rvert$ is therefore at most $\frac{4\pi}{2\ell+1}\cdot\frac{r_i^{\ell}}{r^{\ell+1}}\cdot\frac{2\ell+1}{4\pi} = r_i^{\ell}/r^{\ell+1} \le r^{-1}(a/r)^{\ell}$ in modulus, a convergent geometric series since $a < r$ . (Applying the pointwise bound $\lvert Y_\ell^m\rvert \le \sqrt{(2\ell+1)/4\pi}$ to each of the $2\ell+1$ terms separately would leave a factor $2\ell+1$ , which Cauchy–Schwarz removes.) So the order of summation may be changed:

\psi(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_{\ell=0}^\infty\sum_{m=-\ell}^{\ell}\left[\frac{4\pi}{2\ell+1}\sum_i q_ir_i^{\,\ell}Y_\ell^m(\Omega_i)^*\right]\frac{Y_\ell^m(\Omega)}{r^{\ell+1}} ,

and the bracket is $M_\ell^m$ , which is (4).

Step 4 (the axial case, for $a_i$ of either sign). Let every charge sit on the polar axis, at $z = a_i$ with $a_i$ positive, negative or zero. A charge with $a_i > 0$ has $r_i = a_i$ and $\theta_i = 0$ ; one with $a_i < 0$ has $r_i = \lvert a_i\rvert = -a_i$ and $\theta_i = \pi$ . By part 3 of spherical harmonics the two axis values are

Y_\ell^m(0,\varphi) = \sqrt{\frac{2\ell+1}{4\pi}}\,\delta_{m0}, \qquad Y_\ell^m(\pi,\varphi) = (-1)^\ell\sqrt{\frac{2\ell+1}{4\pi}}\,\delta_{m0} ,

both real, so in either case only $m = 0$ survives in $M_\ell^m$ , and the two cases combine into the single formula

r_i^{\,\ell}\,Y_\ell^0(\theta_i,\varphi_i)^* = \lvert a_i\rvert^{\,\ell}\,(\operatorname{sgn} a_i)^{\ell}\sqrt{\frac{2\ell+1}{4\pi}} = a_i^{\,\ell}\sqrt{\frac{2\ell+1}{4\pi}} ,

because $(-1)^\ell = (\operatorname{sgn} a_i)^\ell$ when $a_i < 0$ . (A charge at the origin, $a_i = 0$ , contributes $0$ for $\ell \ge 1$ and $q_i$ for $\ell = 0$ , which is what $a_i^{\,\ell}$ gives with $0^0 = 1$ .) At the field point $Y_\ell^0(\theta,\varphi) = \sqrt{(2\ell+1)/4\pi}\,P_\ell(\cos\theta)$ . Substituting both into (4),

\psi = \frac{1}{4\pi\varepsilon_0}\sum_\ell \frac{4\pi}{2\ell+1}\cdot\frac{2\ell+1}{4\pi}\left(\sum_i q_ia_i^{\,\ell}\right)\frac{P_\ell(\cos\theta)}{r^{\ell+1}} = \frac{1}{4\pi\varepsilon_0}\sum_\ell \frac{\mu_\ell\,P_\ell(\cos\theta)}{r^{\ell+1}} ,

which is (5). $\square$

Application(Only the lowest nonvanishing moment is independent of the origin)AdaptedAdapted from Arfken, Weber & Harris (2013), §15.3, p. 739

Move the expansion centre from $\mathbf{0}$ to $\mathbf{r}_p$ , so that each $\mathbf{r}_i$ is replaced by $\mathbf{r}_i - \mathbf{r}_p$ and the moments change accordingly. Then the lowest nonvanishing moment of the distribution is unchanged, while every higher moment is in general changed. In particular:

the monopole moment — the total charge — is always independent of the choice of centre;
the dipole moment is independent of it only when the total charge is zero;
the quadrupole moment is independent of it only when the total charge and the dipole moment both vanish.

Idea of the proofOriginal

A moment is a weighted sum of a solid harmonic over the charges, and a solid harmonic of degree $\ell$ is a homogeneous harmonic polynomial of that degree. Translating such a polynomial keeps it harmonic and keeps its degree- $\ell$ part exactly where it was; everything the translation creates is of lower degree. So moving the centre adds to each moment a combination of moments of strictly lower order and nothing else. If the lower ones are all zero there is nothing to add, which is the theorem; if they are not, the added terms are visible, which gives the converse in the cases named.

ProofOriginal

Step 1 (moments as a linear functional on harmonic polynomials). For $\ell \ge 0$ and $\lvert m\rvert \le \ell$ put

S_\ell^m(\mathbf{x}) \;=\; r^{\ell}\,Y_\ell^m(\Omega)^* , \qquad \mathbf{x} = r\,\Omega .

By part 7 of spherical harmonics, $r^\ell Y_\ell^m$ is a homogeneous polynomial of degree $\ell$ in $x,y,z$ which is harmonic on $\mathbb{R}^3$ , and by the conjugation relation of part 2 there, $Y_\ell^m{}^* = (-1)^m Y_\ell^{-m}$ , so $S_\ell^m = (-1)^m r^\ell Y_\ell^{-m}$ is one too. Write $H_\ell$ for the complex span of $S_\ell^{-\ell},\dots,S_\ell^{\ell}$ ; relabelling $m \mapsto -m$ shows this is also the span of the solid harmonics $r^\ell Y_\ell^m$ .

$H_\ell$ is in fact all of the homogeneous harmonic polynomials of degree $\ell$ . For if $P$ is one, then $P$ is harmonic on the whole of $\mathbb{R}^3$ and regular at the origin, so the second sentence of part 7 writes it as $P = \sum_{l,m}A_{lm}r^lY_l^m$ ; each $r^lY_l^m$ is homogeneous of degree $l$ , and homogeneous polynomials of distinct degrees are linearly independent, so matching degrees against the homogeneous $P$ leaves only $l = \ell$ and puts $P$ in $H_\ell$ .

Define the linear functional $\Lambda(P) = \sum_i q_i\,P(\mathbf{r}_i)$ on polynomials. Then (4) reads $M_\ell^m = \frac{4\pi}{2\ell+1}\Lambda(S_\ell^m)$ , and the moments about the new centre $\mathbf{r}_p$ — got by replacing each $\mathbf{r}_i$ by $\mathbf{r}_i - \mathbf{r}_p$ — are

\tilde M_\ell^m \;=\; \frac{4\pi}{2\ell+1}\sum_i q_i\,S_\ell^m(\mathbf{r}_i - \mathbf{r}_p) \;=\; \frac{4\pi}{2\ell+1}\,\Lambda\bigl(S_\ell^m(\,\cdot\, - \mathbf{r}_p)\bigr) .

In this language, all the moments of order $\ell$ vanish precisely when $\Lambda$ vanishes on $H_\ell$ .

Step 2 (what a translation does to a harmonic polynomial). Let $P$ be a homogeneous harmonic polynomial of degree $N$ and set $Q(\mathbf{x}) = P(\mathbf{x}-\mathbf{r}_p)$ . Then $Q$ is a polynomial of degree $N$ , and it is harmonic, because $\nabla^2$ is translation invariant. Split $Q$ into its homogeneous parts, $Q = \sum_{d=0}^{N} Q_d$ . Each $\nabla^2 Q_d$ is homogeneous of degree $d-2$ , and homogeneous polynomials of distinct degrees are linearly independent, so $\nabla^2 Q = 0$ forces $\nabla^2 Q_d = 0$ for every $d$ separately: each homogeneous part of a harmonic polynomial is itself harmonic. Finally, expanding $P(\mathbf{x}-\mathbf{r}_p)$ and collecting terms of top degree gives $Q_N = P$ , since $P$ is homogeneous of degree $N$ . Hence

P(\mathbf{x}-\mathbf{r}_p) \;=\; P(\mathbf{x}) \;+\; \sum_{d<N} Q_d(\mathbf{x}) , \qquad Q_d \in H_d . \tag{7}

This is the step that the axial moments cannot supply on their own: a shift off the axis destroys the axial symmetry, so the scalar binomial expansion of $(a_i - z_p)^n$ in Step 5 below settles only the axial case.

Step 3 (the lowest nonvanishing moment is unchanged). Let $N$ be the smallest order at which some moment is nonzero, so by Step 1 $\Lambda$ vanishes on $H_d$ for every $d < N$ . Apply (7) with $P = S_N^m$ , whose lower homogeneous parts $Q_d$ lie in $H_d$ with $d<N$ :

\tilde M_N^m = \frac{4\pi}{2N+1}\Bigl[\Lambda(S_N^m) + \sum_{d<N}\Lambda(Q_d)\Bigr] = \frac{4\pi}{2N+1}\Lambda(S_N^m) = M_N^m

for every $m$ and every $\mathbf{r}_p$ . So the whole set of order- $N$ moments survives the shift unchanged; in particular it is still not all zero, so the order $N$ itself does not depend on the centre either.

Step 4 (the three named cases). The converses in the three bullets of the statement come out of (7) at low order, and need no choice of phase convention.

Monopole. $H_0$ consists of the constants, so $P(\mathbf{x}-\mathbf{r}_p) = P(\mathbf{x})$ and $\tilde M_0^0 = M_0^0$ for every $\mathbf{r}_p$ . This is Step 3 with $N = 0$ : the total charge is always centre-independent.
Dipole. Every linear form is harmonic, so $H_1$ is the space of all $P(\mathbf{x}) = \alpha x+\beta y+\gamma z$ , and for such a $P$ , $P(\mathbf{x}-\mathbf{r}_p) = P(\mathbf{x}) - P(\mathbf{r}_p)$ . Hence $\Lambda(P(\,\cdot\,-\mathbf{r}_p)) = \Lambda(P) - P(\mathbf{r}_p)\sum_i q_i$ . The three order-1 moments are the values of $\Lambda$ on the basis $S_1^{-1},S_1^0,S_1^1$ of $H_1$ , so they are all unchanged exactly when $\Lambda(P(\,\cdot\,-\mathbf{r}_p)) = \Lambda(P)$ for every $P\in H_1$ ; by the displayed identity that happens for every $\mathbf{r}_p$ if and only if the total charge $\sum_i q_i$ vanishes, since otherwise a $P$ and an $\mathbf{r}_p$ with $P(\mathbf{r}_p)\ne0$ move one of them.
Quadrupole. Take the real element $S_2^0(\mathbf{x}) = \sqrt{5/16\pi}\,(2z^2-x^2-y^2)$ of $H_2$ and $\mathbf{r}_p = (x_p,y_p,z_p)$ . Since $2(z-z_p)^2 - (x-x_p)^2 - (y-y_p)^2 = (2z^2-x^2-y^2) - 4z_pz + 2x_px + 2y_py + \bigl(2z_p^2-x_p^2-y_p^2\bigr) ,$ we get, with $c = \frac{4\pi}{5}\sqrt{5/16\pi} = \tfrac12\sqrt{4\pi/5}$ , $\tilde M_2^0 - M_2^0 = c\Bigl[-4z_p\sum_i q_iz_i + 2x_p\sum_i q_ix_i + 2y_p\sum_i q_iy_i + \bigl(2z_p^2-x_p^2-y_p^2\bigr)\sum_i q_i\Bigr] .$ The right-hand side is a polynomial in $\mathbf{r}_p$ , so it vanishes for every shift only if its linear part does, which kills all three components of the dipole vector $\sum_i q_i\mathbf{r}_i$ , and its quadratic part does, which kills the total charge. Together with Step 3 this is the third bullet.

Step 5 (the axial moments, and how they change). On the polar axis the general moments collapse to the $\mu_n$ of (5), and the shift is a scalar computation. With the centre moved to $z = z_p$ and all charges at $z = a_i$ , the binomial theorem gives

\tilde\mu_n = \sum_i q_i\,(a_i - z_p)^n = \sum_{j=0}^{n}\binom{n}{j}(-z_p)^{\,n-j}\sum_i q_ia_i^{\,j} = \sum_{j=0}^n\binom nj (-z_p)^{n-j}\mu_j ,

which is (7) read in the one-dimensional case: $\tilde\mu_n$ is $\mu_n$ plus a combination of the strictly lower moments. For $n > N$ it gives $\tilde\mu_n = \mu_n + \binom{n}{N}(-z_p)^{n-N}\mu_N + (\text{terms with }\mu_j,\ N<j<n)$ , whose second term is nonzero for every $z_p \ne 0$ . So moments above the lowest really do move, and the "in general" in the statement cannot be dropped. Arfken reaches the same conclusion from the shape of the binomial expansion of $\lvert\mathbf{r}-\mathbf{r}_p\rvert^{-n}$ , which feeds each term of the expansion into the later ones and never into the earlier ones. $\square$

Remark(How many components a moment has, and why)AdaptedAdapted from Arfken, Weber & Harris (2013), §16.3, pp. 801–802

For a unit charge at $(x,y,z)$ the moments $M_\ell^m$ of (4) are, in Cartesian form, homogeneous polynomials of degree $\ell$ . Arfken tabulates them for $\ell \le 2$ ; for example $M_0^0 = (4\pi)^{1/2}$ , $M_1^0 = (4\pi/3)^{1/2}z$ , $M_1^{\pm1} = \mp(2\pi/3)^{1/2}(x\pm iy)$ , and $M_2^0 = (4\pi/5)^{1/2}\tfrac12(2z^2-x^2-y^2)$ . There are $2\ell+1$ values of $m$ , but the space of homogeneous polynomials of degree $\ell$ in three variables has dimension $\binom{\ell+2}{2}$ , which is larger from $\ell = 2$ on: six monomials $x^2,y^2,z^2,xy,xz,yz$ against five values of $m$ . Arfken resolves the discrepancy by identifying the missing direction: the combination $r^2 = x^2+y^2+z^2$ is rotationally invariant and so belongs with $M_0^0$ , but it carries the wrong power of $r$ to contribute to a solution of Laplace's equation, and therefore cannot appear in a potential. The same happens at $\ell = 3$ , where ten monomials split into seven that span the $M_3^m$ and three with the symmetry of $M_1^m$ but again the wrong radial behaviour. In Cartesian language this is the statement that the quadrupole tensor may be taken traceless.

Remark(Symmetry makes moments vanish, and it is worth checking before computing)AdaptedAdapted from Riley, Hobson & Bence (2006), §29.1, pp. 1077–1078

Riley, Hobson and Bence open their chapter on representation theory with this, using three molecules. Hydrogen chloride has no symmetry operation to forbid a dipole moment, and has one. Carbon dioxide is linear and symmetric: a rotation about the long axis kills any perpendicular component, and a rotation by $\pi$ about a perpendicular axis through the carbon, or a reflection in the plane through the carbon, reverses the parallel component while carrying the molecule to an indistinguishable copy — so that component is zero too, and CO $_2$ has no permanent dipole moment. Ozone is bent, and only the component perpendicular to its two-fold axis is forbidden. The moral to carry away is one-sided: a symmetry operation can rule a component out, but no symmetry argument ever produces a nonzero moment. What survives the symmetry test is merely permitted, and whether it is actually there has to be settled by computation or by experiment.

Remark(What the expansion is not)AdaptedAdapted from Arfken, Weber & Harris (2013), §15.3, Eq. (15.62), p. 736 and p. 739

It is an expansion of the field outside a sphere containing the whole source. Arfken's Eq. (15.62) is the companion statement for a field point closer to the origin than the charge: one single charge $q$ at $z = a$ , seen from $r < a$ , with the roles of $r$ and $a$ in (1) exchanged and the series running in powers of $r/a$ . No shell appears in that statement, and none could: inside a uniform spherical shell of radius $a$ carrying total charge $q$ the potential is the constant $q/4\pi\varepsilon_0 a$ , with no $P_n$ series at all. A field point in the middle of a distribution needs both expansions, split at $r$ , as in the Green's-function expansion of Green's functions for Laplace and Poisson problems. Nor is the expansion restricted to electrostatics: Arfken's third closing observation is that it applies to any inverse-square force, so that the shape of the Earth is described by mass multipoles, and the leading term of gravitational radiation is set by the time dependence of the mass quadrupole.

Example(A planar quadrupole: four charges, no monopole, no dipole)Original

Take $a>0$ and place charges

+q \text{ at } (\pm a,0,0), \qquad -q \text{ at } (0,\pm a,0) ,

four equal magnitudes at the corners of a square in the equatorial plane, alternating in sign. Work in units with $4\pi\varepsilon_0 = 1$ .

The low moments vanish. The total charge is $q+q-q-q = 0$ , so $M_0^0 = 0$ . The dipole vector is $q(a,0,0)+q(-a,0,0)-q(0,a,0)-q(0,-a,0) = \mathbf{0}$ , so $M_1^m = 0$ for all $m$ . Both vanishings are forced by symmetry: the configuration is carried to itself by $\mathbf{r}\mapsto-\mathbf{r}$ with the charges unchanged, so every odd- $\ell$ moment is zero, and by the rotation through $\pi/2$ about the $z$ -axis combined with $q\mapsto-q$ , which kills the $m = 0$ terms.

The quadrupole moments. All four charges have $r_i = a$ and $\theta_i = \pi/2$ , with $\varphi_i = 0,\pi$ (charge $+q$ ) and $\pi/2, 3\pi/2$ (charge $-q$ ). Since $Y_2^{\pm1}\propto\sin\theta\cos\theta\,e^{\pm i\varphi}$ vanishes at $\theta = \pi/2$ , $M_2^{\pm1} = 0$ . For $m = 0$ , $Y_2^0(\pi/2,\varphi)$ is the same constant for all four charges, and the charges sum to zero, so $M_2^0 = 0$ . Only $m = \pm2$ survives. With $Y_2^{\pm2}(\theta,\varphi) = \sqrt{15/32\pi}\,\sin^2\theta\,e^{\pm2i\varphi}$ ,

\sum_i q_i Y_2^{2}(\Omega_i)^* = \sqrt{\frac{15}{32\pi}}\Bigl[q e^{0} + qe^{-2\pi i} - qe^{-\pi i} - qe^{-3\pi i}\Bigr] = \sqrt{\frac{15}{32\pi}}\,\bigl[q+q+q+q\bigr] = 4q\sqrt{\frac{15}{32\pi}} ,

M_2^{2} = M_2^{-2} = \frac{4\pi}{5}\,a^2\cdot 4q\sqrt{\frac{15}{32\pi}} = \frac{2\sqrt{30\pi}}{5}\,qa^2 = 3.8832518\,qa^2 .

The leading far field. Substituting into (4), and using $e^{2i\varphi}+e^{-2i\varphi} = 2\cos2\varphi$ ,

\psi \;\approx\; \frac{1}{r^3}\bigl[M_2^2Y_2^2 + M_2^{-2}Y_2^{-2}\bigr] = \frac{4\pi}{5}\cdot 4q a^2\cdot\frac{15}{32\pi}\cdot\frac{2\sin^2\theta\cos2\varphi}{r^3} = \frac{3qa^2\sin^2\theta\cos2\varphi}{r^3} . \tag{6}

Cross-check against the Cartesian form. The traceless quadrupole tensor $Q_{jk} = \sum_i q_i(3x_jx_k - r_i^2\delta_{jk})$ of this configuration is diagonal with $Q_{xx} = 6qa^2$ , $Q_{yy} = -6qa^2$ , $Q_{zz} = 0$ (each $+q$ contributes $(2,-1,-1)qa^2$ and each $-q$ contributes $(1,-2,1)qa^2$ to the diagonal). The standard Cartesian quadrupole potential $\tfrac12 r^{-3}Q_{jk}\hat n_j\hat n_k$ is then $\tfrac12 r^{-3}\cdot6qa^2(\hat n_x^2-\hat n_y^2) = 3qa^2\sin^2\theta\cos2\varphi/r^3$ , which is (6).

Numerical check and the rate of convergence. Set $a = 1$ , $q = 1$ . At the point $r = 5$ , $\theta = \pi/2$ , $\varphi = 0$ — that is, $(5,0,0)$ — the exact potential is

\frac{1}{4} + \frac{1}{6} - \frac{2}{\sqrt{26}} = 0.02443440 ,

while (6) gives $3/125 = 0.02400000$ . Comparing at several points:

$r$	$\theta$	$\varphi$	exact	from (6)	error
$5$	$\pi/2$	$0$	$0.02443440$	$0.02400000$	$4.34\times10^{-4}$
$10$	$\pi/2$	$0$	$0.00301276$	$0.00300000$	$1.28\times10^{-5}$
$20$	$\pi/2$	$0$	$0.00037539$	$0.00037500$	$3.93\times10^{-7}$
$5$	$\pi/3$	$0.3$	$0.01465879$	$0.01485604$	$-1.97\times10^{-4}$
$10$	$\pi/3$	$0.3$	$0.00185111$	$0.00185701$	$-5.89\times10^{-6}$

Each doubling of $r$ divides the error by about $32 = 2^5$ , so the error decays like $r^{-5}$ — exactly the size of the next surviving term, $\ell = 4$ . The odd- $\ell$ terms are absent by the inversion symmetry noted above, so the first correction to a quadrupole here is a hexadecapole, not an octupole.

The moments are origin-independent here. Because the monopole and dipole moments both vanish, the previous statement says the quadrupole moments cannot change if the expansion centre is moved. Checking directly: shifting all four charges by a vector $\mathbf{d}$ changes $\sum q_i(3x_jx_k - r^2\delta_{jk})$ by terms proportional to $\sum q_i x_j$ (the dipole, zero) and $\sum q_i$ (the total charge, zero). So $Q_{jk}$ is unchanged, as claimed.

Example(When the dipole moment depends on where you put the origin)Original

Place $+2q$ at $z = a$ and $+q$ at $z = -a$ — two charges of the same sign, so the monopole moment does not vanish. Using the axial moments (5) with the origin at $z = 0$ :

\mu_0 = 2q + q = 3q, \qquad \mu_1 = 2q\cdot a + q\cdot(-a) = qa, \qquad \mu_2 = 2qa^2 + qa^2 = 3qa^2 .

Now move the centre to $z = z_p$ . By Step 1 of the proof,

\tilde\mu_0 = 3q, \qquad \tilde\mu_1 = \mu_1 - z_p\mu_0 = qa - 3qz_p, \qquad \tilde\mu_2 = \mu_2 - 2z_p\mu_1 + z_p^2\mu_0 = 3qa^2 - 2qaz_p + 3qz_p^2 .

The monopole moment is unmoved, as it must be. The dipole moment is not: it is $qa$ about the origin, zero about $z_p = a/3$ — the charge-weighted centre — and $-2qa$ about $z_p = a$ . So the sentence "this configuration has dipole moment $qa$ " is meaningless without saying about which point, whereas "this configuration has total charge $3q$ " is not.

The field does not care. At $r = 10a$ on the positive axis, the exact potential (units $4\pi\varepsilon_0=1$ , $q=a=1$ ) is $2/9 + 1/11 = 0.31313131$ . From the origin, (5) gives $\mu_0/r + \mu_1/r^2 + \mu_2/r^3 = 0.3 + 0.01 + 0.003 = 0.31300000$ . From the centre $z_p = 1/3$ , the field point is at distance $r' = 10 - 1/3 = 9.6666667$ and the moments are $\tilde\mu_0 = 3$ , $\tilde\mu_1 = 0$ , $\tilde\mu_2 = 3 - 2/3 + 1/3 = 2.6666667$ , giving $3/9.6666667 + 0 + 2.6666667/9.6666667^3 = 0.31034483 + 0.00295215 = 0.31329698$ . The two truncations miss the exact value by $-1.31\times10^{-4}$ and $+1.66\times10^{-4}$ : both are accurate to four digits, and they bracket the answer. Including the next ( $\ell = 3$ ) term improves each to $0.3131000$ and $0.3130934$ respectively, against the exact $0.3131313$ . Moving the origin reshuffles the moments but is bound to leave the sum of the series alone; the choice $z_p = a/3$ buys a vanishing dipole term at the price of a changed radial variable and a changed quadrupole.

Application(Induced dipoles and polarizability)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.1, p. 734

Riley, Hobson and Bence's conducting sphere is the cleanest instance of a multipole moment being produced rather than measured. An uncharged conducting sphere of radius $a$ in a uniform field $E$ must have a potential that tends to $-Er\cos\theta$ far away and is constant on the surface; since $\cos\theta = P_1(\cos\theta)$ , only the $\ell=1$ terms of the separated solution can appear (separation of variables), and the two conditions determine them, giving $v = -Er(1-a^3/r^3)\cos\theta$ . Comparing the second term with the point-dipole potential $\mu P_1(\cos\theta)/4\pi\varepsilon_0r^2$ identifies an induced dipole moment $4\pi\varepsilon_0a^3E$ : the polarizability of a conducting sphere is $4\pi\varepsilon_0a^3$ , three times its volume times $\varepsilon_0$ . The same calculation with a dielectric instead of a conductor gives the Clausius–Mossotti factor, and that is how the refractive index of a dilute suspension is computed.

ApplicationOriginal

Nuclear and molecular moments as experimental quantities. The $\ell = 2$ moment of a nucleus is measurable through its interaction with the electric field gradient of the surrounding electrons, and a nonzero value is direct evidence that the nucleus is not spherical — prolate for a positive quadrupole moment, oblate for a negative one. The origin-independence statement is what makes such a number well defined: a nucleus has zero dipole moment (by parity), so its quadrupole moment is its lowest nonvanishing one and does not depend on where the coordinate origin is put. By the same token a quoted molecular dipole moment is meaningful only for a neutral molecule, which is why one reads of the dipole moment of water but of the quadrupole moment of an ion.

Gravity: geodesy and gravitational waves. The expansion applies unchanged to Newtonian gravity, with $\rho$ the mass density and the sign reversed. Because mass density is positive, the monopole moment never vanishes, so the gravitational dipole moment of a body is always origin-dependent — and the origin that makes it vanish is the centre of mass. This is why the gravitational field of the Earth is quoted as a series of $J_\ell$ coefficients about the centre of mass, with the flattening appearing as $J_2 \approx 1.08\times10^{-3}$ and everything beyond it some three orders of magnitude smaller ( $J_3 \approx -2.5\times10^{-6}$ , $J_4 \approx -1.6\times10^{-6}$ , each about $1/430$ of $J_2$ ); satellite orbits are fitted to those coefficients, and the measured drift of the coefficients tracks the melting of ice sheets. In general relativity the leading radiative term is the second time derivative of the mass quadrupole, as Arfken notes; the mass monopole and dipole cannot radiate because they are conserved.

Fast summation. The expansion is the mathematical content of the fast multipole method. To evaluate the potential of $N$ sources at $N$ targets, group the sources into cells; for a target far from a cell, replace the sources in the cell by the first few of its multipole moments, computed once for the whole cell. The error is controlled by the convergence rate in (1), which is governed by $r_</r_>$ — the ratio of the cell radius to the distance — and the example above shows how fast that is in practice: truncating at the quadrupole gave three-digit accuracy at five cell radii. This turns an $O(N^2)$ computation into $O(N)$ , and it is the standard tool in astrophysical $N$ -body simulation, molecular dynamics with Coulomb interactions, and boundary-element solvers built on the fundamental solution of Laplace's equation.

Application(Partial waves in scattering)AdaptedAdapted from Arfken, Weber & Harris (2013), §16.3, Eq. (16.61), pp. 798–799

The same factorization of a two-point function into products of one-point functions is what makes partial-wave analysis possible. Arfken's Eq. (16.61) is the companion expansion $e^{i\mathbf{k}\cdot\mathbf{r}} = 4\pi\sum_{\ell m}i^\ell j_\ell(kr)Y_\ell^m(\Omega_k)^*Y_\ell^m(\Omega_r)$ , reached by expanding $e^{ikr\cos\chi}$ in Legendre polynomials, recognising the coefficient integral as the integral representation of $j_\ell$ , and then applying the same addition theorem. Spherical Bessel functions take the place of the powers of $r$ , and the identity converts an incident plane wave into a superposition of spherical waves, each of which is scattered independently by a central potential. See the Helmholtz equation and spherical Bessel functions.

Method

The Helmholtz equation

Adapted

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Reduces the wave and diffusion equations to the Helmholtz equation by separating a time-harmonic factor, collects its elementary solutions - plane waves, cylindrical waves and the spherical waves built from spherical Bessel and Hankel functions - and states and proves the Sommerfeld radiation condition that picks out the outgoing solution and forces a field that is source-free on all of space to vanish.

Prerequisites: Separation of variables in standard coordinates, Spherical Bessel functions, Neumann and Hankel functions, Green's identities, Fundamental solution of Laplace's equation

Definition(Time-harmonic reduction and the Helmholtz equation)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, Eqs. (21.52)–(21.54), p. 737; Hassani (2013), §19.7.2, Eq. (19.38), p. 593; Evans (2010), §6.6, Problem 16, p. 347

A field oscillating at a single angular frequency $\omega$ is time-harmonic: it has the form

u(\mathbf{x},t) = e^{-i\omega t}\,w(\mathbf{x}) , \tag{1}

the physical field being $\operatorname{Re}u$ . Substituting (1) into the wave equation $\nabla^2u = c^{-2}u_{tt}$ gives

\nabla^2 w + k^2 w = 0 , \qquad k = \frac{\omega}{c} , \tag{2}

the Helmholtz equation, with $k$ the wavenumber. Equation (2) reduces to Laplace's equation when $k = 0$ . We also meet the modified Helmholtz equation $\nabla^2 w - k^2w = 0$ , which is (2) with $k^2$ replaced by $-k^2$ : its radial solutions grow or decay exponentially instead of oscillating, and its only bounded solution on all of $\mathbb{R}^n$ is $w \equiv 0$ . (It does have solutions that oscillate in some directions while decaying in others — $\cos(ax)e^{-\sqrt{k^2+a^2}\,z}$ is one — and those evanescent fields are what the waveguide and tunnelling paragraphs below are about; what it has not got is a bounded oscillation filling space.)

Sign and time conventions, which differ between our sources. Riley, Hobson and Bence separate time without choosing a sign, writing $T(t) = Ae^{i\omega t}+Be^{-i\omega t}$ with $\omega = kc$ , and get (2) either way, since only $\omega^2$ enters. Evans fixes $u = e^{-i\sigma t}w$ and writes the equation as $-\Delta w = \lambda w$ with $\lambda = \sigma^2$ , so his $\sigma$ is our $k$ in units with $c = 1$ . Arfken fixes the convention at the other end, by declaring an outgoing wave to be one whose asymptotic radial dependence is $\exp(+ikr)$ . We use $e^{-i\omega t}$ throughout, which makes $e^{i(kr-\omega t)}$ travel outwards and so agrees with Arfken: with the opposite time convention $e^{+i\omega t}$ , the outgoing solution would be $e^{-ikr}$ instead, and every sign in the radiation condition below would flip.

Hassani records why (2) is the equation to study: separating time out of the diffusion equation, the wave equation, or the Schrödinger equation of a free particle leaves (2) in each case, the only difference being the time factor one has discarded. For the wave equation it is $e^{\pm i\omega t}$ with $\omega = kc$ ; for the diffusion equation $\kappa\nabla^2u = u_t$ it is the decaying $e^{-k^2\kappa t}$ , so that there $k^2$ is a decay rate rather than a frequency.

Method(Solving a time-harmonic problem)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, pp. 737–741; Arfken, Weber & Harris (2013), §10.2, Table 10.1 and its footnote, p. 463; Evans (2010), §6.6, Problem 16(c), p. 347

Strip the time. Write the field as $e^{-i\omega t}w(\mathbf{x})$ and reduce the equation to (2) with $k = \omega/c$ . Everything from here on is a problem for a complex-valued function of position; the physical field is recovered at the end as $\operatorname{Re}\bigl(e^{-i\omega t}w\bigr)$ , and amplitudes and phases are the modulus and argument of $w$ .
Choose coordinates and separate. Apply the recipe of separation of variables. The special functions are those of the catalogue there, now with $k \ne 0$ : $J_m(k\rho), Y_m(k\rho)$ in plane polars, $J_m(\sqrt{k^2-\alpha^2}\rho)$ in cylindrical polars, and the spherical Bessel functions $j_\ell(kr), y_\ell(kr)$ in spherical polars.
Interior problem: discard what is singular, then quantize. If the region contains the origin, drop $Y_m$ and $y_\ell$ . A homogeneous boundary condition on a closed surface then admits a solution only for special values of $k$ — the eigenvalues — and the frequencies $\omega = ck$ are the resonances of the cavity.
Exterior problem: impose a condition at infinity. The Helmholtz equation in an unbounded region does not have enough boundary data to be determined, because its solutions oscillate rather than decay and the usual "vanishes at infinity" is too weak to rule anything out. The missing condition is the Sommerfeld radiation condition, which in Evans's form reads $\lim_{r\to\infty} r\left(\frac{\partial w}{\partial r} - ikw\right) = 0 \qquad (n = 3), \tag{3}$ and which Arfken states in the equivalent working form that the asymptotic radial dependence must be $e^{+ikr}$ : energy flows outwards from the source and none comes in from infinity. Statement (3) is what makes the exterior problem well posed; see the uniqueness theorem below.
For the modified Helmholtz equation, use decay instead. With $\nabla^2 - k^2$ the radial solutions are exponentially growing or decaying rather than oscillatory, and the right condition at infinity is simply that $w \to 0$ — Arfken's choice in his Table 10.1, where the three-dimensional modified-Helmholtz Green's function is $-e^{-k\lvert\mathbf{r}_1-\mathbf{r}_2\rvert}/4\pi\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ , the screened potential.
Build the solution from the elementary waves. Plane waves $e^{i\mathbf{k}\cdot\mathbf{x}}$ are the natural basis for a problem with planar symmetry, outgoing spherical waves $h_\ell^{(1)}(kr)Y_\ell^m$ for a radiating or scattering problem, and $j_\ell(kr)Y_\ell^m$ for the field inside a sphere. The plane-wave expansion of spherical Bessel functions converts between the first two.

Method(The elementary solutions)AdaptedAdapted from Evans (2010), §6.6, Problem 16(a),(b), p. 347; Hassani (2013), §19.7.2, Eqs. (19.39)–(19.42) and §19.7.4, pp. 593–594; Riley, Hobson & Bence (2006), §21.3.2, Eqs. (21.55)–(21.57), pp. 738–741; Arfken, Weber & Harris (2013), §10.2, Table 10.1, p. 463

Plane waves. For a constant vector $\mathbf{k}$ , the function $w(\mathbf{x}) = e^{i\mathbf{k}\cdot\mathbf{x}}$ solves (2) if and only if $\mathbf{k}\cdot\mathbf{k} = k^2$ . Writing $\mathbf{k} = k\boldsymbol\omega$ with $\lvert\boldsymbol\omega\rvert = 1$ , the corresponding time-harmonic field is the travelling wave $u = e^{ik(\boldsymbol\omega\cdot\mathbf{x} - ct)}$ , moving with speed $c$ in the direction $\boldsymbol\omega$ .
Spherical waves. In $\mathbb{R}^3$ the functions
$w^\pm(\mathbf{x}) = \frac{e^{\pm ik r}}{r}, \qquad r = \lvert\mathbf{x}\rvert ,$
solve (2) at every point with $r>0$ ; $w^+$ is outgoing and $w^-$ incoming. More precisely, Evans records that
$\Phi(\mathbf{x}) = \frac{e^{ikr}}{4\pi r} \quad\text{satisfies}\quad -\nabla^2\Phi = k^2\Phi + \delta_0 \ \text{ in } \mathbb{R}^3 ,$
so $\Phi$ is the outgoing fundamental solution of the Helmholtz operator: it reduces to the fundamental solution $1/4\pi r$ of Laplace's equation as $k\to0$ . Arfken's Table 10.1 lists the same object as $-e^{ik\lvert\mathbf{r}_1-\mathbf{r}_2\rvert}/4\pi\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ , the sign difference being his convention $\nabla^2G + k^2G = \delta$ against Evans's $-\nabla^2\Phi - k^2\Phi = \delta$ .
The separated solutions in spherical polars. The radial equation is
$R'' + \frac2rR' + \left[k^2 - \frac{\ell(\ell+1)}{r^2}\right]R = 0 ,$
whose solutions are the spherical Bessel functions $R = Aj_\ell(kr) + By_\ell(kr)$ , which Hassani builds out of the half-integer-order Bessel functions through the normalization $z_\ell(x) = \sqrt{\pi/2}\;Z_{\ell+1/2}(x)/\sqrt x$ . So
$w = \sum_{\ell=0}^\infty\sum_{m=-\ell}^{\ell}\bigl[A_{\ell m}j_\ell(kr) + B_{\ell m}y_\ell(kr)\bigr]Y_\ell^m(\theta,\varphi) ,$
with every $B_{\ell m} = 0$ if the region contains the origin, since $y_\ell$ is singular there. The outgoing combination is the spherical Hankel function $h_\ell^{(1)} = j_\ell + iy_\ell$ , and $h_0^{(1)}(kr) = -ie^{ikr}/kr$ recovers part 2.
Cylindrical waves. In cylindrical polars $(\rho,\varphi,z)$ the separated solutions are
$w = Z_m(\kappa\rho)\,e^{im\varphi}\,e^{i\alpha z} , \qquad \kappa = \sqrt{k^2-\alpha^2} , \qquad m \in \mathbb{Z} ,$
where $Z_m$ stands, as in Hassani's normalization above, for either of $J_m$ and $Y_m$ — only $J_m$ if the region contains the axis, since $Y_m$ is singular there — and $\alpha$ is the separation constant of the $z$ direction. The radial argument $\kappa\rho = \sqrt{k^2-\alpha^2}\,\rho$ is Riley, Hobson and Bence's. The three wavenumbers are tied by $\kappa^2 + \alpha^2 = k^2$ , which is the dispersion relation of a waveguide. For $\alpha^2 < k^2$ the factor $Z_m(\kappa\rho)$ oscillates in $\rho$ and the wave propagates along the axis; for $\alpha^2 > k^2$ the argument is imaginary and the radial factor is no longer oscillatory. In the plane, $\alpha = 0$ and the solutions are $J_m(k\rho)e^{im\varphi}$ and $Y_m(k\rho)e^{im\varphi}$ .
Lower dimensions. In two dimensions the role of $e^{ikr}/4\pi r$ is played by $-\tfrac i4 H_0^{(1)}(k\rho)$ , with a Hankel function in place of the exponential; in one dimension by $-\tfrac{i}{2k}e^{ik\lvert x\rvert}$ . Arfken tabulates all three. He also warns that the step from two dimensions to three is a genuine change and not a change of notation — a remark he makes on the previous page about the Laplace Green's functions, where $\frac{1}{2\pi}\ln\lvert\boldsymbol\rho_1-\boldsymbol\rho_2\rvert$ grows at infinity while $-1/4\pi\lvert\mathbf{r}_1-\mathbf{r}_2\rvert$ decays, and which applies just as well to the entries here.

Idea of the proofOriginal

Parts 1 and 2 are two-line verifications, and the only thing worth watching is where each one fails. A plane wave is bounded but does not decay in any direction, so it is a legitimate solution of (2) on all of $\mathbb{R}^3$ with no source anywhere: the Helmholtz equation on all space has a huge supply of global solutions, which is exactly why a condition at infinity is needed to make an exterior problem well posed. A spherical wave does decay, like $1/r$ , but it has a singularity at the origin, and the singularity is the source. Part 2 says that the singularity is precisely a unit point source: near $r = 0$ the factor $e^{ikr}$ is $1 + O(r)$ , so $\Phi$ looks like the Laplace fundamental solution $1/4\pi r$ , whose Laplacian is $-\delta_0$ , and the correction accounts for the term $k^2\Phi$ .

Part 3 is the separation already carried out in spherical polars, read with $k\ne0$ : nothing changes in the angular factors, and the radial factor becomes a Bessel function of half-integer order.

ProofAdaptedAdapted from Evans (2010), §6.6, Problem 16(a),(b), p. 347; Hassani (2013), §19.7.4, p. 594

Step 1 (plane waves). For $w = e^{i\mathbf{k}\cdot\mathbf{x}}$ we have $\partial w/\partial x_j = ik_jw$ and hence $\partial^2w/\partial x_j^2 = -k_j^2w$ . Summing over $j$ ,

\nabla^2 w = -\lvert\mathbf{k}\rvert^2 w ,

so $\nabla^2w + k^2w = (k^2 - \lvert\mathbf{k}\rvert^2)w$ , which vanishes identically if and only if $\lvert\mathbf{k}\rvert = k$ (as $w$ is nowhere zero). Writing $\mathbf{k} = k\boldsymbol\omega$ and restoring the time factor, $u = e^{-i\omega t}e^{ik\boldsymbol\omega\cdot\mathbf{x}} = e^{ik(\boldsymbol\omega\cdot\mathbf{x}-ct)}$ using $\omega = kc$ , which is constant on the planes $\boldsymbol\omega\cdot\mathbf{x} - ct = \text{const}$ moving with speed $c$ .

Step 2 (spherical waves, away from the origin). For a radial $w = v(r)$ , the Laplacian is $v'' + 2v'/r$ (the case $n = 3$ of the formula in the fundamental solution), which can be written $\frac1r(rv)''$ . With $v = e^{\pm ikr}/r$ we get $rv = e^{\pm ikr}$ and $(rv)'' = -k^2e^{\pm ikr}$ , so

\nabla^2 v = \frac1r\bigl(-k^2e^{\pm ikr}\bigr) = -k^2 v ,

and $\nabla^2v + k^2v = 0$ for $r>0$ .

Step 3 (the source at the origin). Put $\Phi = e^{ikr}/4\pi r$ and let $\phi$ be a test function. Write $\Phi = \frac{1}{4\pi r} + \Psi$ with

\Psi = \frac{e^{ikr}-1}{4\pi r} ,

which is bounded near the origin, since $e^{ikr}-1 = O(r)$ ; indeed $\Psi \to ik/4\pi$ as $r\to0$ . Then:

$\frac{1}{4\pi r}$ is the Laplace fundamental solution, so $-\nabla^2\bigl(\frac1{4\pi r}\bigr) = \delta_0$ in the sense of distributions (the Dirac delta in calculations).
By Step 2, $-\nabla^2\Phi = k^2\Phi$ pointwise for $r>0$ ; subtracting the previous item, $-\nabla^2\Psi = k^2\Phi$ for $r>0$ . This also holds in the sense of distributions on all of $\mathbb{R}^3$ , with no term supported at $\mathbf{0}$ , and the reason is that $\Psi$ is too mild a function to carry one. Indeed, work outside a small ball, where $\Psi$ is smooth: Green's second identity on $\{r>\varepsilon\}$ turns $\int\Psi\nabla^2\phi$ into $\int\phi\nabla^2\Psi$ plus a boundary term on $r=\varepsilon$ , giving
$-\int_{r>\varepsilon}\Psi\,\nabla^2\phi\;dV \;=\; k^2\!\int_{r>\varepsilon}\Phi\,\phi\;dV \;+\; \oint_{r=\varepsilon}\bigl(\Psi\,\phi_r - \phi\,\Psi_r\bigr)dS .$
Expanding the exponential, $\Psi = \frac{ik}{4\pi} - \frac{k^2r}{8\pi} + O(r^2)$ , so $\Psi$ and $\Psi_r$ are both bounded near the origin while the sphere $r = \varepsilon$ has area $4\pi\varepsilon^2$ : the boundary term is $O(\varepsilon^2)$ and drops out as $\varepsilon\to0$ . Both volume integrals converge, the first because $\Psi$ is bounded and the second because $\lvert\Phi\rvert\le1/4\pi r$ is locally integrable. So $-\nabla^2\Psi = k^2\Phi$ as distributions, and the only singular contribution in $\Phi$ sits in $\frac{1}{4\pi r}$ .

Adding the two statements gives $-\nabla^2\Phi = k^2\Phi + \delta_0$ in $\mathbb{R}^3$ , which is Evans's part (b).

Step 4 (the radial equation and the spherical Bessel functions). This is Step 8 of the proof in separation of variables: the substitution $R = r^{-1/2}S$ turns the radial equation into Bessel's equation of order $\ell+\tfrac12$ in the variable $kr$ . Hassani's normalization $z_\ell(x) = \sqrt{\pi/2}\,Z_{\ell+1/2}(x)/\sqrt x$ is exactly the factor $r^{-1/2}$ together with a constant chosen so that $j_\ell(x)\sim x^\ell/(2\ell+1)!!$ at small $x$ ; with it, $R = Aj_\ell(kr)+By_\ell(kr)$ . $\square$

Method(The Sommerfeld radiation condition and uniqueness)AdaptedAdapted from Evans (2010), §6.6, Problems 16(c) and 17, pp. 347–348

Throughout, $k$ is real and nonzero, $w$ is complex-valued and defined at least outside some ball, and $w_r = \nabla w\cdot\mathbf{x}/\lvert\mathbf{x}\rvert$ is its radial derivative. The Sommerfeld radiation condition on $w$ is

\lim_{r\to\infty} r\bigl(w_r - ikw\bigr) = 0 , \tag{3}

the limit being uniform in the direction $\mathbf{x}/r$ : for every $\varepsilon>0$ there is an $R_0$ with $\lvert r(w_r-ikw)\rvert \le \varepsilon$ at every point with $r\ge R_0$ . (This is Evans's reading, and it is what makes the surface estimate of Step 6 below work.) Then:

Of the three solutions of (2) found above — all of them solutions away from the origin, where the first two are singular — the outgoing fundamental solution $\Phi = e^{ikr}/4\pi r$ satisfies (3), while the incoming one $e^{-ikr}/4\pi r$ and the plane wave $e^{ik\boldsymbol\omega\cdot\mathbf{x}}$ both fail it.
Uniqueness. If $w$ solves $-\nabla^2w = k^2w$ on all of $\mathbb{R}^3$ and satisfies (3), then $w \equiv 0$ .

Part 2 is the statement that makes the condition useful: it says that the exterior Helmholtz problem — the equation outside an obstacle, data on the obstacle, and (3) at infinity — has at most one solution, because the difference of two solutions would solve the source-free problem everywhere and hence vanish. Evans sets both parts as problems, with the hints reproduced in the proof below, and refers to Colton for a fuller treatment.

Idea of the proofOriginal

Condition (3) is a statement about which way the energy goes. For a field behaving like $A(\theta,\varphi)e^{ikr}/r$ at large $r$ , the radial derivative is $ikA e^{ikr}/r$ plus a term of order $r^{-2}$ , so $w_r - ikw$ is $O(r^{-2})$ and (3) holds; for $A e^{-ikr}/r$ the two terms add instead of cancelling, leaving $-2ikw$ , which is $O(r^{-1})$ , and $r(w_r-ikw)$ does not go to zero. So (3) selects $e^{+ikr}$ over $e^{-ikr}$ , which with the time convention $e^{-i\omega t}$ means outgoing over incoming. A plane wave fails for a different reason: its radial derivative is $ik(\boldsymbol\omega\cdot\hat{\mathbf{x}})w$ , so $w_r - ikw$ is $ik(\boldsymbol\omega\cdot\hat{\mathbf{x}} - 1)w$ , which does not decay at all in any direction other than $\boldsymbol\omega$ — the plane wave is coming in from infinity on one side as much as it is going out on the other.

The uniqueness proof has two halves. First, Green's second identity applied to $w$ and $\bar w$ on a large ball gives a conserved flux: because $\bar w\Delta w - w\Delta\bar w = 0$ for a solution with real $k^2$ , the boundary integral $\int(\bar ww_r - w\bar w_r)\,dS$ vanishes for every $R$ . That integral is $2i\operatorname{Im}\int\bar ww_r$ , so the radiation condition — which controls $w_r - ikw$ — combines with it to force $\int_{\partial B(0,R)}(\lvert w_r\rvert^2 + k^2\lvert w\rvert^2)dS \to 0$ : the field carries no energy at all through large spheres, and both $w$ and $w_r$ die faster than $1/r$ in the mean-square sense. Second, Green's identity against the fundamental solution represents $w$ at any point as an integral over $\partial B(0,R)$ ; the first half shows that integral tends to zero, so $w$ vanishes at that point.

ProofAdaptedAdapted from Evans (2010), §6.6, Problems 16(c) and 17, pp. 347–348

Step 1 ( $\Phi$ satisfies (3), exactly). With $\Phi = e^{ikr}/4\pi r$ ,

\Phi_r = \frac{d}{dr}\frac{e^{ikr}}{4\pi r} = \frac{ike^{ikr}}{4\pi r} - \frac{e^{ikr}}{4\pi r^2} = ik\Phi - \frac{\Phi}{r} ,

so $r(\Phi_r - ik\Phi) = -\Phi = -e^{ikr}/4\pi r$ , whose modulus is $1/4\pi r \to 0$ . Condition (3) holds, with the quantity in it of size exactly $1/4\pi r$ .

Step 2 (the incoming wave fails). With $\Psi = e^{-ikr}/4\pi r$ the same computation gives $\Psi_r = -ik\Psi - \Psi/r$ , so

r(\Psi_r - ik\Psi) = r\left(-2ik\Psi - \frac{\Psi}{r}\right) = -2ikr\Psi - \Psi ,

of modulus $\bigl\lvert{-2ik r\Psi - \Psi}\bigr\rvert \to k/2\pi$ as $r\to\infty$ , since $\lvert r\Psi\rvert = 1/4\pi$ and $\Psi\to0$ . The limit is nonzero because $k \ne 0$ .

Step 3 (the plane wave fails). For $w = e^{ik\boldsymbol\omega\cdot\mathbf{x}}$ , $\nabla w = ik\boldsymbol\omega w$ and so $w_r = ik(\boldsymbol\omega\cdot\hat{\mathbf{x}})w$ with $\hat{\mathbf{x}} = \mathbf{x}/r$ . Hence

r\bigl(w_r - ikw\bigr) = ikr\bigl(\boldsymbol\omega\cdot\hat{\mathbf{x}} - 1\bigr)w ,

and $\lvert w\rvert = 1$ . Along any ray with $\boldsymbol\omega\cdot\hat{\mathbf{x}} \ne 1$ the modulus is $k r\lvert\boldsymbol\omega\cdot\hat{\mathbf{x}}-1\rvert$ , which grows without bound. So (3) fails, and badly.

Step 4 (a vanishing flux). Now let $w$ solve $-\nabla^2w = k^2w$ on $\mathbb{R}^3$ and satisfy (3). Then $\Delta w = -k^2w$ and, taking complex conjugates and using that $k^2$ is real, $\Delta\bar w = -k^2\bar w$ . Therefore

\bar w\,\Delta w - w\,\Delta\bar w = -k^2\bar ww + k^2w\bar w = 0 \qquad\text{everywhere},

and Green's second identity on the ball $B(0,R)$ gives

0 = \int_{B(0,R)}\bigl(\bar w\Delta w - w\Delta\bar w\bigr)\,dV = \oint_{\partial B(0,R)}\bigl(\bar w\,w_r - w\,\bar w_r\bigr)\,dS \qquad\text{for every } R>0 . \tag{4}

Since $w\bar w_r = \overline{\bar ww_r}$ , the integrand of (4) is $2i\operatorname{Im}(\bar ww_r)$ , so (4) says $\operatorname{Im}\oint_{\partial B(0,R)}\bar w\,w_r\,dS = 0$ .

Step 5 (expand the radiation condition). Expanding the square,

\lvert w_r - ikw\rvert^2 = \lvert w_r\rvert^2 + k^2\lvert w\rvert^2 - 2\operatorname{Re}\bigl(\overline{w_r}\,ikw\bigr) = \lvert w_r\rvert^2 + k^2\lvert w\rvert^2 - 2k\operatorname{Im}\bigl(\bar w\,w_r\bigr) ,

using $\operatorname{Re}(ik\,\overline{w_r}w) = -k\operatorname{Im}(\overline{w_r}w) = k\operatorname{Im}(\bar w w_r)$ for real $k$ . Integrating over $\partial B(0,R)$ and inserting Step 4, the last term drops out:

\oint_{\partial B(0,R)}\lvert w_r - ikw\rvert^2\,dS = \oint_{\partial B(0,R)}\bigl(\lvert w_r\rvert^2 + k^2\lvert w\rvert^2\bigr)\,dS . \tag{5}

Step 6 (both sides tend to zero). By (3), whose limit is uniform in direction, there is for each $\varepsilon>0$ an $R_0$ with $\lvert w_r - ikw\rvert \le \varepsilon/r$ at every point with $r \ge R_0$ . The sphere $\partial B(0,R)$ has area $4\pi R^2$ , so the left side of (5) is at most $4\pi R^2\cdot\varepsilon^2/R^2 = 4\pi\varepsilon^2$ for $R\ge R_0$ . Hence

\oint_{\partial B(0,R)}\bigl(\lvert w_r\rvert^2 + k^2\lvert w\rvert^2\bigr)\,dS \;\longrightarrow\; 0 \qquad (R\to\infty) . \tag{6}

Both terms are non-negative, so each tends to zero separately; and since $k\ne0$ , (6) controls $\lvert w\rvert$ as well as $\lvert w_r\rvert$ .

Step 7 (represent $w$ on a large sphere). Fix $\mathbf{x}_0$ and let $R > \lvert\mathbf{x}_0\rvert$ . Apply Green's second identity on $B(0,R)$ to the pair $w$ and $\Phi(\cdot-\mathbf{x}_0)$ , excising a small ball about $\mathbf{x}_0$ exactly as in the proof of the representation formula for Laplace's equation. The volume terms cancel, because both functions satisfy the same equation $\Delta v = -k^2v$ away from $\mathbf{x}_0$ ; the excised sphere contributes $w(\mathbf{x}_0)$ in the limit, because $\Phi$ has the same singularity as the Laplace fundamental solution (Step 3 of the previous proof). What remains is Evans's representation

w(\mathbf{x}_0) = \oint_{\partial B(0,R)}\bigl(\Phi\,w_r - w\,\Phi_r\bigr)\,dS , \qquad \Phi = \Phi(\mathbf{x}-\mathbf{x}_0) . \tag{7}

Step 8 (let $R\to\infty$ ). On $\partial B(0,R)$ with $R > 2\lvert\mathbf{x}_0\rvert$ we have $\lvert\mathbf{x}-\mathbf{x}_0\rvert \ge R/2$ , so $\lvert\Phi\rvert \le 1/(2\pi R)$ and, from the identity $\nabla\Phi = \bigl(ik - 1/\rho\bigr)\Phi\,\hat{\boldsymbol\rho}$ of Step 1 with $\rho = \lvert\mathbf{x}-\mathbf{x}_0\rvert$ , also $\lvert\Phi_r\rvert \le \lvert\nabla\Phi\rvert \le k\lvert\Phi\rvert + \lvert\Phi\rvert/\rho \le k/(2\pi R) + 1/(\pi R^2) \le C/R$ for large $R$ . By the Cauchy–Schwarz inequality on the sphere,

\left\lvert\oint_{\partial B(0,R)}\Phi\,w_r\,dS\right\rvert \le \left(\oint\lvert\Phi\rvert^2dS\right)^{1/2}\left(\oint\lvert w_r\rvert^2dS\right)^{1/2} \le \left(\frac{4\pi R^2}{4\pi^2R^2}\right)^{1/2}\left(\oint\lvert w_r\rvert^2dS\right)^{1/2} ,

whose first factor is bounded and whose second tends to $0$ by (6). The same estimate applies to $\oint w\,\Phi_r\,dS$ , using the $\lvert w\rvert^2$ half of (6). Hence the whole right-hand side of (7) vanishes in the limit, while its left-hand side is a fixed number independent of $R$ . Hence $w(\mathbf{x}_0) = 0$ , and since $\mathbf{x}_0$ was arbitrary, $w \equiv 0$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

$k$ real and nonzero, in Steps 2, 4 and 6. Step 4 needs $k^2 = \overline{k^2}$ , which is where "real" is used; a complex $k$ (an absorbing medium) makes the flux identity (4) false, and then the right condition at infinity is plain decay, as for the modified Helmholtz equation. Step 2 needs $k \ne 0$ to make the incoming wave fail, and Step 6 needs it to extract $\lvert w\rvert$ from (6). At $k = 0$ the statement collapses: the outgoing and incoming fundamental solutions $e^{+ikr}/4\pi r$ and $e^{-ikr}/4\pi r$ both become the single function $1/4\pi r$ , so there is nothing left to choose between, and the condition at infinity reverts to $w\to0$ .
Three dimensions, in Steps 6 and 8. The factor $r$ in (3) and the area $4\pi R^2$ are matched to $n = 3$ . In $n$ dimensions the condition carries $r^{(n-1)/2}$ instead, matching the $r^{-(n-1)/2}$ decay of an outgoing wave; in two dimensions the outgoing solution is $H_0^{(1)}(k\rho)$ , which decays like $\rho^{-1/2}$ , and the condition reads $\lim\rho^{1/2}(w_\rho - ikw) = 0$ .
The condition is not implied by decay. This is the whole point. Both $e^{ikr}/r$ and $e^{-ikr}/r$ tend to zero at infinity, and so does any combination of them; " $w\to0$ " therefore fails to distinguish a radiating source from one absorbing energy from infinity, and fails to make the exterior problem uniquely solvable. Nor is boundedness enough, as the plane wave shows: it is bounded, source-free and not identically zero, so part 2 of the statement would be false with (3) weakened to boundedness.
What part 2 does and does not say. It is a statement about solutions on all of $\mathbb{R}^3$ with no source. It does not say that an exterior problem has no solution — only that it has at most one, which is what one wants. The existence of that one solution is proved by constructing it from the outgoing Green's function, which is the subject of Green's functions for the Helmholtz equation.
Interior resonances are the other side of the same coin. For a bounded region with homogeneous boundary data, (2) has nonzero solutions for a discrete set of $k$ and only for those; for an exterior region with the radiation condition, it has none. That contrast is why cavity problems are eigenvalue problems and radiation problems are not.
A separated solution has a frequency but not a wavelength. Riley, Hobson and Bence make this observation about (2) in spherical polars: the time factor $e^{-i\omega t}$ with $\omega = kc$ is a pure frequency, but the zeros of $j_\ell(kr)$ are not equally spaced in $r$ except for $\ell = 0$ , so the radial profile has no single wavelength. It acquires one only asymptotically, where $j_\ell(kr) \approx \sin(kr - \ell\pi/2)/kr$ .

Example(The radiation condition, tested exactly)Original

Take $k = 2$ and compare three solutions of $\nabla^2w + 4w = 0$ in $\mathbb{R}^3$ (away from the origin where necessary), computing $Q(r) = \sup_{\lvert\mathbf{x}\rvert = r}\lvert r(w_r - 2iw)\rvert$ exactly.

$w$	$w_r$	$r(w_r - ikw)$	$\lvert\,\cdot\,\rvert$ at $r = 10$	$r\to\infty$
$\dfrac{e^{2ir}}{4\pi r}$	$2iw - \dfrac wr$	$-w$	$\dfrac{1}{40\pi} = 0.0079577$	$\to 0$
$\dfrac{e^{-2ir}}{4\pi r}$	$-2iw - \dfrac wr$	$-4irw - w$	$\dfrac{\sqrt{16r^2+1}}{4\pi r}\Big\vert_{r=10} = 0.3184093$	$\to \dfrac{k}{2\pi} = 0.3183099$
$e^{2iz}$ (i.e. $\boldsymbol\omega = \hat{\mathbf{z}}$ )	$2i\cos\theta\,w$	$2ir(\cos\theta - 1)w$	$0$ at $\theta=0$ , $40$ at $\theta=\pi$	unbounded

The first row is Step 1 of the proof with the numbers put in: the quantity in (3) has modulus exactly $1/4\pi r$ , so it decays like $1/r$ and the limit is zero. The second row is Step 2: the modulus is $\sqrt{16r^2+1}/4\pi r$ , which tends to $k/2\pi = 1/\pi = 0.3183099$ and at $r=10$ is already $0.3184093$ , within $0.04\%$ of its limit. The third row is Step 3: along the $-z$ axis, $\cos\theta = -1$ and the modulus is $4r$ , which at $r = 10$ is $40$ and grows without bound; only along the single ray $\theta = 0$ does it vanish.

A superposition makes the selectivity vivid. The standing wave

w = \frac{\sin 2r}{2r} = \frac{1}{4i}\left(\frac{e^{2ir}}{r} - \frac{e^{-2ir}}{r}\right) = j_0(2r)

is a solution of $\nabla^2w+4w = 0$ on all of $\mathbb{R}^3$ , with no singularity at the origin (where $w \to 1$ ), and it decays like $1/r$ . It is a counterexample to any uniqueness statement based on decay alone; and indeed it is half incoming and half outgoing, so it fails (3). Explicitly, combining the first two rows of the table linearly,

r\bigl(w_r - ikw\bigr) = e^{-ikr} - \frac{\sin kr}{kr} ,

whose modulus tends to $1$ rather than to $0$ — numerically $0.98226$ at $r=10$ , $1.00437$ at $r=50$ and $1.00213$ at $r=100$ . Part 2 of the statement is consistent with the existence of this solution precisely because the solution fails (3).

Example(Resonances of a spherical shell cavity)Original

Find the values of $k$ for which $\nabla^2w + k^2w = 0$ has a nonzero solution in the shell $1 \le r \le 2$ with $w = 0$ on both bounding spheres.

Separate. The region excludes the origin, so both spherical Bessel functions survive and the radial factor for each $\ell$ is $R(r) = Aj_\ell(kr) + By_\ell(kr)$ . The two boundary conditions give a homogeneous $2\times2$ system for $(A,B)$ , which has a nonzero solution exactly when

D_\ell(k) \;=\; j_\ell(k)\,y_\ell(2k) - y_\ell(k)\,j_\ell(2k) \;=\; 0 . \tag{8}

The case $\ell = 0$ in closed form. Here $j_0(x) = \sin x/x$ and $y_0(x) = -\cos x/x$ , so

D_0(k) = \frac{\sin k}{k}\cdot\frac{-\cos 2k}{2k} + \frac{\cos k}{k}\cdot\frac{\sin 2k}{2k} = \frac{\sin 2k\cos k - \cos 2k\sin k}{2k^2} = \frac{\sin k}{2k^2} .

So $D_0(k) = 0$ exactly when $k = n\pi$ , $n = 1,2,\dots$ . This is no accident: for $\ell = 0$ the substitution $R = \chi/r$ turns the radial equation into $\chi'' + k^2\chi = 0$ with $\chi(1) = \chi(2) = 0$ , which is a string of length $1$ — the shell behaves, in its spherically symmetric modes, exactly like a one-dimensional box of width $1$ , the thickness of the shell.

Higher $\ell$ , numerically. Solving (8) for the first few roots:

$\ell$	$k_{\ell,1}$	$k_{\ell,2}$	$k_{\ell,3}$	degeneracy $2\ell+1$
$0$	$3.141593$	$6.283185$	$9.424778$	1
$1$	$3.286007$	$6.360678$	$9.477196$	3
$2$	$3.555788$	$6.513068$	$9.581270$	5
$3$	$3.922520$	$6.735565$	$9.735534$	7

The $\ell = 0$ roots are $\pi, 2\pi, 3\pi$ to six figures, confirming the closed form. The lowest resonance of the cavity is $k = \pi$ , simple; the next is $k = 3.286007$ , threefold degenerate. Reading down any column, the roots rise with $\ell$ , as they must: a larger $\ell$ adds the repulsive term $\ell(\ell+1)/r^2$ to the radial equation, which pushes the eigenvalue up. Reading along a row instead — successive roots at fixed $\ell$ — the spacing tends to $\pi$ , the $\ell=0$ value, because at large $k$ the centrifugal term becomes negligible beside $k^2$ : at $\ell=3$ the two gaps printed are $2.813045$ and $2.999969$ .

Example(A pulsating sphere: the radiation condition doing work)Original

A sphere of radius $a$ in air oscillates radially at angular frequency $\omega$ , its surface moving with radial velocity amplitude $U_0$ . Find the radiated sound field and the acoustic power.

Set up. The acoustic pressure satisfies the wave equation, so $p = \operatorname{Re}\bigl(e^{-i\omega t}w\bigr)$ with $\nabla^2w+k^2w=0$ , $k = \omega/c$ , outside the sphere. The problem is spherically symmetric, so only $\ell = 0$ appears; of the two $\ell=0$ solutions $e^{\pm ikr}/r$ , the radiation condition keeps only the outgoing one. This is the step that makes the problem solvable at all: without (3), any combination of the two would do, and the answer would not be determined by the data on $r = a$ . So

w(r) = A\,\frac{e^{ikr}}{r} .

Apply the surface condition. Linearized momentum balance, $\rho_0\partial\mathbf{v}/\partial t = -\nabla p$ , gives for the time-harmonic parts $-i\omega\rho_0 v_r = -w'(r)$ , so $v_r = w'(r)/(i\omega\rho_0)$ . Using $w' = ikw - w/r$ and imposing $v_r = U_0$ at $r = a$ ,

U_0 = \frac{1}{i\omega\rho_0}\,A\,\frac{e^{ika}}{a}\left(ik - \frac1a\right) \quad\Longrightarrow\quad A = \frac{i\omega\rho_0 a^2 U_0\,e^{-ika}}{ika - 1}, \qquad \lvert A\rvert = \frac{\omega\rho_0 a^2U_0}{\sqrt{1 + k^2a^2}} .

Radiated power. Far from the sphere, $v_r \to w/(\rho_0c)$ , so the time-averaged intensity is $\lvert w\rvert^2/2\rho_0c = \lvert A\rvert^2/2\rho_0cr^2$ and the power through a large sphere is

P = 4\pi r^2\cdot\frac{\lvert A\rvert^2}{2\rho_0cr^2} = \frac{2\pi\lvert A\rvert^2}{\rho_0 c} = \frac{2\pi\rho_0c\,k^2a^4U_0^2}{1+k^2a^2} .

Numbers. Air at room conditions: $\rho_0 = 1.2\ \mathrm{kg\,m^{-3}}$ , $c = 343\ \mathrm{m\,s^{-1}}$ . Take $a = 0.05\ \mathrm m$ , $f = 100\ \mathrm{Hz}$ and $U_0 = 0.01\ \mathrm{m\,s^{-1}}$ . Then

k = \frac{2\pi f}{c} = 1.831832\ \mathrm{m^{-1}}, \qquad ka = 0.0915916 ,

\lvert A\rvert = 0.0187710\ \mathrm{Pa\,m}, \qquad P = 5.37872\times10^{-6}\ \mathrm W .

So the pressure amplitude at $1$ m is $0.0188$ Pa — root-mean-square $0.01327$ Pa, which against the reference $2\times10^{-5}$ Pa is $56.4$ dB SPL — and the sphere radiates about $5.4$ microwatts.

The two limits, and why small loudspeakers are inefficient. For $ka \gg 1$ the formula gives $P \to 2\pi\rho_0ca^2U_0^2 = (4\pi a^2)\cdot\tfrac12\rho_0cU_0^2$ : the whole surface radiates as if it were a piece of plane wave, at the best possible rate. For $ka\ll1$ ,

P \approx 2\pi\rho_0c\,k^2a^4U_0^2 ,

smaller by the factor $k^2a^2/(1+k^2a^2)$ , which here is $0.00832$ . A source much smaller than a wavelength radiates with efficiency proportional to $(ka)^2$ , and since $k\propto f$ the power at fixed surface velocity falls off as $f^2$ — which is why a small driver needs a large enclosure, a horn, or a great deal of excursion to produce bass.

ApplicationOriginal

Acoustics, antennas and scattering. Every steady-state radiation calculation is an exterior Helmholtz problem closed by the radiation condition: a loudspeaker in a baffle, a dipole antenna, an ultrasound transducer. For scattering, one splits the field into a prescribed incident part and an unknown scattered part, $w = w_{\text{inc}} + w_{\text{sc}}$ , and imposes (3) on $w_{\text{sc}}$ alone — not on $w$ , which contains the incident plane wave and therefore cannot satisfy it, as Step 3 of the proof shows. That split is exactly what the uniqueness theorem makes legitimate: the scattered field is the unique outgoing solution with the boundary data that the incident field forces on the obstacle. The far-field pattern $A(\theta,\varphi)$ in $w_{\text{sc}}\sim A(\theta,\varphi)e^{ikr}/r$ is the scattering amplitude, and $\lvert A\rvert^2$ integrated over directions is the scattering cross-section; see scattering and the Born approximation.

Waveguides and optical fibres. In a cylinder, step 2 of the recipe with the catalogue of separation of variables gives modes $J_m(\kappa\rho)e^{im\varphi}e^{i\beta z}$ with $\kappa^2 + \beta^2 = k^2$ . A mode propagates only when $\beta^2 > 0$ , that is when $k$ exceeds the cutoff $\kappa$ fixed by the boundary condition; below cutoff, $\beta$ is imaginary and the mode is evanescent. The relation $\kappa^2+\beta^2 = k^2$ is the algebraic relation among separation constants of step 4 of that recipe, and here it is the dispersion relation of the guide — which is where waveguide dispersion, and hence pulse spreading in a fibre, comes from.

Cavity resonances and the density of states. The interior problem of step 3 gives the resonant frequencies of a cavity: a microwave oven, an organ pipe, a laser cavity. Counting eigenvalues below $k$ in a box of volume $V$ gives Weyl's asymptotic $N(k)\approx Vk^3/6\pi^2$ , which multiplied by two polarizations is the mode count behind the Rayleigh–Jeans and then the Planck radiation law. The shell example above shows the structure in miniature: a discrete set of $k$ , with multiplicity $2\ell+1$ forced by the rotational symmetry.

Quantum mechanics of a free particle, and the modified equation in a barrier. The time-independent Schrödinger equation $-\frac{\hbar^2}{2m}\nabla^2\psi = E\psi$ is the Helmholtz equation with $k^2 = 2mE/\hbar^2$ ; Hassani reaches it exactly this way, and the solutions regular at the origin are $j_\ell(kr)Y_\ell^m$ . Where the potential exceeds the energy, $k^2$ becomes negative and the equation turns into the modified Helmholtz equation, whose solutions decay exponentially — quantum tunnelling, in the form in which it is actually computed. The same substitution turns the Coulomb-screening problem into the modified equation, whose three-dimensional Green's function Arfken tabulates as the Yukawa potential $e^{-kr}/4\pi r$ . See the Schrödinger equation.

Numerical methods, and why the radiation condition is a nuisance. A computational domain is finite, so the condition at infinity has to be faked: by a non-reflecting or absorbing boundary condition, which enforces (3) approximately on an artificial outer boundary; by a perfectly matched layer, which continues the solution into complex coordinates so that outgoing waves decay and incoming ones would grow; or by reformulating the problem as a boundary integral equation on the obstacle using the outgoing Green's function, which satisfies (3) exactly and so needs no outer boundary at all. The last route is the fully faithful one, and it is why the boundary-element method dominates exterior acoustics and electromagnetic scattering. It also inherits the one difficulty of the Helmholtz equation that the Laplace equation does not have: the interior resonances of step 3 appear as frequencies at which the integral operator becomes singular, even though the exterior problem itself remains uniquely solvable.

Application

The Schrodinger equation

Adapted

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States the time-dependent and time-independent Schrodinger equations, shows that separating the time turns the first into an eigenvalue problem whose solutions are the stationary states, works out the free particle, the infinite well and the harmonic oscillator with their normalizations and energy levels, and derives the probability current and the continuity equation that makes the normalization persist.

Prerequisites: Separation of variables in standard coordinates, The Helmholtz equation, Hermite polynomials, Eigenfunction expansions, The divergence theorem

Definition(The two Schrodinger equations)AdaptedAdapted from Hassani (2013), §13.1, Eq. (13.5), p. 396; Kreyszig (1978), §11.3, Eqs. (5)–(6), p. 585; Riley, Hobson & Bence (2006), §21.3.2, p. 741; Reed & Simon (1980), §VIII.11, pp. 303–304

A non-relativistic particle of mass $m$ moving in a potential $V(\mathbf{r})$ is described by a complex-valued wave function $\Psi(\mathbf{r},t)$ satisfying the time-dependent Schrodinger equation

-\frac{\hbar^2}{2m}\nabla^2\Psi + V(\mathbf{r})\,\Psi \;=\; i\hbar\,\frac{\partial\Psi}{\partial t} , \tag{1}

where $\hbar = h/2\pi$ and $h$ is Planck's constant. The quantity $\lvert\Psi(\mathbf{r},t)\rvert^2$ is the probability density for finding the particle at $\mathbf{r}$ at time $t$ , so a physically admissible $\Psi$ must satisfy the normalization

\int \lvert\Psi(\mathbf{r},t)\rvert^2\,dV = 1 . \tag{2}

The time-independent Schrodinger equation is the eigenvalue problem

-\frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf{r})\,\psi \;=\; E\,\psi , \qquad\text{that is}\qquad \hat H\psi = E\psi , \tag{3}

with $\hat H = -\frac{\hbar^2}{2m}\nabla^2 + V$ the Hamiltonian operator. Kreyszig notes at once what (3) is for, and he puts the claim carefully: the operator form suggests that which energies are possible depends on the spectrum of $\hat H$ , and a solution counts as physically meaningful only if it stays finite and tends to zero at infinity — it is that restriction, not the differential equation alone, that quantizes the energy. The identification of the admissible energies with the spectrum of $\hat H$ is the stronger statement, and it is Reed and Simon's rather than his: it presupposes a domain on which $\hat H$ is self-adjoint, and the remark below returns to it. Kreyszig's §11.4 records how $\hat H$ is written down: take the classical Hamiltonian $H = E_{\text{kin}} + V$ in terms of positions and momenta, and replace each momentum $p_j$ by the operator $\frac{h}{2\pi i}\partial/\partial q_j = -i\hbar\,\partial/\partial q_j$ (his §11.2, Eq. (2)) and each coordinate $q_j$ by multiplication by $q_j$ . Note the sign: $1/i = -i$ , so Kreyszig's $\frac{h}{2\pi i}\partial/\partial q_j$ , written without a leading minus, is the physicists' $-i\hbar\,\partial/\partial q_j$ .

Units and conventions. Kreyszig writes $h$ where we write $2\pi\hbar$ , so his equation (5) reads $\Delta\psi + \frac{8\pi^2m}{h^2}(E-V)\psi = 0$ , which is (3). Hassani and Riley, Hobson and Bence use $\hbar$ as above. Reed and Simon work in units with $\hbar = 1$ and write the dynamics as $\frac{d}{dt}[U(t)\varphi] = iH[U(t)\varphi]$ , which is (1) with the opposite sign of $i$ — their $U(t) = e^{itH}$ against the physicists' $e^{-i\hat Ht/\hbar}$ . We follow the physicists' convention (1), so that a stationary state carries the time factor $e^{-iEt/\hbar}$ , as in Riley, Hobson and Bence. Nothing physical depends on the choice, since the two are exchanged by complex conjugation, but the sign of the probability current below does.

Application(Separation into stationary states)AdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, p. 741; Kreyszig (1978), §11.3, pp. 583–585; Hassani (2013), §13.1, pp. 396–397

Separable solutions of (1) of the form $\Psi(\mathbf{r},t) = \psi(\mathbf{r})T(t)$ exist exactly when $\psi$ solves (3) for some constant $E \in \mathbb{C}$ , and then

\Psi(\mathbf{r},t) = \psi(\mathbf{r})\,e^{-iEt/\hbar} . \tag{4}

If in addition $E$ is real, such a $\Psi$ is a stationary state: its probability density $\lvert\Psi\rvert^2 = \lvert\psi\rvert^2$ does not depend on $t$ , because the time factor then has modulus one.

That hypothesis has to be carried, because (3) as written is a differential equation with no normalization and no boundary condition attached, and so does not by itself force $E$ to be real: for complex $E$ the time factor has modulus $e^{(\operatorname{Im}E)t/\hbar}$ and the density grows or decays, and the remark below exhibits such a separable solution for the free particle. Reality is automatic at the one place it matters. If $\psi$ is an eigenfunction of $\hat H$ as a self-adjoint operator — in particular if $V$ is real and $\psi$ is normalizable — then $E$ is real, as Step 2 of the proof shows in a line.

If the eigenfunctions $\psi_n$ of (3) with eigenvalues $E_n$ form a complete orthonormal set, the general solution of (1) with initial datum $\Psi(\mathbf{r},0) = f(\mathbf{r})$ is

\Psi(\mathbf{r},t) = \sum_n c_n\,\psi_n(\mathbf{r})\,e^{-iE_nt/\hbar} , \qquad c_n = \int \overline{\psi_n}\,f\;dV , \tag{5}

and $\sum_n\lvert c_n\rvert^2 = 1$ when $f$ is normalized. A superposition of two or more stationary states with different energies is not stationary: its density oscillates at the beat frequencies $(E_n - E_{n'})/\hbar$ .

Kreyszig arrives at (3) along a different road, which is worth recording because it explains the equation's shape. He starts from the wave equation for a scalar field, imposes the time-harmonic factor to get the Helmholtz equation $\Delta\psi + k^2\psi = 0$ with $k = 2\pi/\lambda$ , and then substitutes de Broglie's $\lambda = h/mv$ for the wavelength of matter waves; since $mv^2/2 = E - V$ , the wavenumber becomes $k^2 = 8\pi^2m(E-V)/h^2$ , and the Helmholtz equation becomes (3). So the time-independent Schrodinger equation is the Helmholtz equation with a position-dependent wavenumber. Riley, Hobson and Bence make the same comparison for hydrogen, where the coefficient involves $1/r$ , and record that what one gets there resembles the Helmholtz equation without being it.

Idea of the proofOriginal

Equation (1) is first order in $t$ and second order in space, like the heat equation, but with an $i$ in front of the time derivative — and that single factor of $i$ changes everything. Separating variables produces $T'/T = -iE/\hbar$ instead of the heat equation's real decay rate, so for real $E$ the time factor has modulus one and oscillates instead of decaying. The result is that a Schrodinger eigenmode persists for ever rather than dying out, and an initial state does not relax to anything; it merely has its modes rotate at different rates, which is why quantum evolution is reversible while diffusion is not.

The separation itself is the ordinary argument: divide (1) by $\psi T$ , note that one side depends on $\mathbf{r}$ alone and the other on $t$ alone, and apply the separation lemma. The content is in which values of the separation constant are admissible, and that is decided by the requirement that $\psi$ be normalizable — the three examples below show it deciding differently in three cases.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §21.3.2, p. 741; Hassani (2013), §13.1, pp. 396–397

Step 1 (separate). Put $\Psi = \psi(\mathbf{r})T(t)$ into (1) and divide by $\psi T$ at the points where neither factor vanishes — $\psi$ may well have interior nodes, as the well's own $\psi_2$ does at $x = L/2$ :

\frac{1}{\psi}\left[-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi\right] = i\hbar\,\frac{T'(t)}{T(t)} .

The left side depends only on $\mathbf{r}$ and the right only on $t$ , so by the separation lemma of separation of variables both equal a constant, which we call $E$ . This gives $-\frac{\hbar^2}{2m}\nabla^2\psi + V\psi = E\psi$ and $i\hbar T' = ET$ wherever the division was legitimate; both sides of each identity are continuous, so each extends to the whole domain, and (3) holds there.

Step 2 (the time factor). The equation $T' = -iET/\hbar$ has the general solution $T = Ae^{-iEt/\hbar}$ , for any $E\in\mathbb{C}$ ; absorbing $A$ into $\psi$ gives (4). Now impose the hypothesis that $E$ is real. Then $\lvert T\rvert = 1$ , so $\lvert\Psi\rvert^2 = \lvert\psi\rvert^2$ does not depend on $t$ and $\Psi$ is stationary.

Reality is a genuine extra hypothesis and not a consequence of (3) — the remark below exhibits a separable solution with $E$ purely imaginary — but it is automatic in the case the rest of the unit works in. If $V$ is real and $\psi$ is a normalizable eigenfunction of $\hat H$ , then, with the inner product linear in its first slot and $\hat H$ symmetric on its domain,

E\lVert\psi\rVert^2 = (\hat H\psi,\psi) = (\psi,\hat H\psi) = \overline E\,\lVert\psi\rVert^2 ,

and $\lVert\psi\rVert \ne 0$ , so $E = \overline E$ . Finding a domain on which $\hat H$ really is self-adjoint, rather than merely symmetric, is Reed and Simon's first mathematical problem.

Step 3 (superposition). Each term of (5) solves (1), by Step 2, and (1) is linear, so any finite sum does; for an infinite sum, the convergence of $\sum\lvert c_n\rvert^2$ and the completeness of $\{\psi_n\}$ give convergence in the mean-square sense (eigenfunction expansions). At $t=0$ the sum is $\sum c_n\psi_n$ , which is the expansion of $f$ , so the initial condition holds; the coefficients are $c_n = (f, \psi_n)$ by orthonormality, which is the integral in (5) with our convention that an inner product is linear in its first slot.

Step 4 (normalization is preserved, and a superposition is not stationary). By orthonormality,

\int\lvert\Psi\rvert^2 dV = \sum_{n,n'}\overline{c_n}c_{n'}e^{i(E_n-E_{n'})t/\hbar}\int\overline{\psi_n}\psi_{n'}dV = \sum_n\lvert c_n\rvert^2 ,

independent of $t$ : the normalization (2) holds at all times if it holds at one. But the density itself is

\lvert\Psi(\mathbf{r},t)\rvert^2 = \sum_{n,n'}\overline{c_n}c_{n'}\,\overline{\psi_n(\mathbf{r})}\psi_{n'}(\mathbf{r})\,e^{i(E_n-E_{n'})t/\hbar} ,

in which every off-diagonal pair with $E_n\ne E_{n'}$ contributes a term oscillating at the frequency $\lvert E_n-E_{n'}\rvert/\hbar$ . Only if a single energy is present does the density stand still. $\square$

Application(The probability current and the continuity equation)AdaptedAdapted from Hassani (2013), §14.8.1, Eqs. (14.49)–(14.52), pp. 452–453

For a solution $\Psi$ of (1) with real $V$ , define the probability density and probability current density

\rho = \lvert\Psi\rvert^2 , \qquad \mathbf{J} = \frac{\hbar}{m}\operatorname{Im}\bigl(\overline\Psi\,\nabla\Psi\bigr) = \frac{\hbar}{2mi}\bigl(\overline\Psi\nabla\Psi - \Psi\nabla\overline\Psi\bigr) . \tag{6}

Then

\frac{\partial\rho}{\partial t} + \nabla\cdot\mathbf{J} = 0 . \tag{7}

Consequently $\frac{d}{dt}\int_\Omega\rho\,dV = -\oint_{\partial\Omega}\mathbf{J}\cdot\hat{\mathbf{n}}\,dS$ for any fixed region $\Omega$ , and if $\Psi$ and $\nabla\Psi$ decay suitably at infinity then $\int\rho\,dV$ is constant in time: normalization, once imposed, is permanent.

Hassani obtains (6) and (7) in polar form. Writing $\Psi = A\exp(iS/\hbar)$ with $A$ and $S$ real, equation (1) is exactly equivalent to the pair

\frac{\partial S}{\partial t} + \frac{\nabla S\cdot\nabla S}{2m} + V = \frac{\hbar^2}{2m}\frac{\nabla^2A}{A} , \qquad m\frac{\partial A}{\partial t} + \nabla S\cdot\nabla A + \frac A2\nabla^2S = 0 ,

and with $\rho = A^2$ and $\mathbf{J} = A^2\nabla S/m = \rho\mathbf{v}$ , where $\mathbf{v} = \nabla S/m$ , the second of these is (7). (The italicized claim is Hassani's (Hassani (2013), §14.8.1, after Eq. (14.50), p. 452), who calls the pair "completely equivalent" to the original equation, and we report it as he makes it. It should be read on a region where $\Psi$ has no zero: at a node $A = 0$ , so the phase $S$ is undefined there and the first equation divides by $A$ ; and in any case the polar form fixes $S$ only modulo $2\pi\hbar$ .) His Proposition 14.8.2 then reads the first equation in the limit $\hbar\to0$ : the right side disappears, $S$ satisfies the classical Hamilton–Jacobi equation, and the pair describes a fluid of non-interacting classical particles of mass $m$ in the potential $V$ , with density $\rho$ and current $\mathbf{J}$ .

Idea of the proofOriginal

Equation (7) is Green's second identity in disguise. Multiply (1) by $\overline\Psi$ , multiply the conjugate equation by $\Psi$ , and subtract: the potential term cancels, because $V$ is real, and what is left is $i\hbar\,\partial_t\lvert\Psi\rvert^2$ on one side and $-\frac{\hbar^2}{2m}(\overline\Psi\nabla^2\Psi - \Psi\nabla^2\overline\Psi)$ on the other. The bracket is exactly the integrand of Green's second identity, and the product rule turns it into the divergence of $\overline\Psi\nabla\Psi - \Psi\nabla\overline\Psi$ . So the cancellation of $V$ is where reality of the potential is used, and the identity $\overline\Psi\nabla^2\Psi - \Psi\nabla^2\overline\Psi = \nabla\cdot(\overline\Psi\nabla\Psi - \Psi\nabla\overline\Psi)$ is where the second-order operator becomes a divergence. A complex potential — an absorbing medium, or a decaying state — breaks the first step and probability is then not conserved, which is exactly what one wants it to mean.

ProofAdaptedAdapted from Hassani (2013), §14.8.1, pp. 452–453

Step 1 (two equations). Write (1) as

i\hbar\,\partial_t\Psi = -\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi ,

and take complex conjugates, using $\overline V = V$ :

-i\hbar\,\partial_t\overline\Psi = -\frac{\hbar^2}{2m}\nabla^2\overline\Psi + V\overline\Psi .

Step 2 (combine). Multiply the first by $\overline\Psi$ and the second by $\Psi$ and subtract the second from the first. The terms $V\lvert\Psi\rvert^2$ cancel — this is where the reality of $V$ enters — leaving

i\hbar\bigl(\overline\Psi\,\partial_t\Psi + \Psi\,\partial_t\overline\Psi\bigr) = -\frac{\hbar^2}{2m}\bigl(\overline\Psi\nabla^2\Psi - \Psi\nabla^2\overline\Psi\bigr) .

The left bracket is $\partial_t(\overline\Psi\Psi) = \partial_t\rho$ .

Step 3 (make the right side a divergence). By the product rule for the divergence,

\nabla\cdot\bigl(\overline\Psi\nabla\Psi\bigr) = \nabla\overline\Psi\cdot\nabla\Psi + \overline\Psi\nabla^2\Psi ,

and the same with $\Psi$ and $\overline\Psi$ exchanged. Subtracting, the symmetric first terms cancel:

\overline\Psi\nabla^2\Psi - \Psi\nabla^2\overline\Psi = \nabla\cdot\bigl(\overline\Psi\nabla\Psi - \Psi\nabla\overline\Psi\bigr) .

Step 4 (assemble). Substituting into Step 2 and dividing by $i\hbar$ ,

\partial_t\rho = -\frac{\hbar}{2mi}\nabla\cdot\bigl(\overline\Psi\nabla\Psi - \Psi\nabla\overline\Psi\bigr) = -\nabla\cdot\mathbf{J} ,

which is (7), with $\mathbf{J}$ as in (6). The two expressions for $\mathbf{J}$ in (6) agree because $\Psi\nabla\overline\Psi = \overline{\overline\Psi\nabla\Psi}$ , so their difference is $2i\operatorname{Im}(\overline\Psi\nabla\Psi)$ .

Step 5 (global conservation). Integrating (7) over a fixed region $\Omega$ with piecewise smooth boundary and applying the divergence theorem,

\frac{d}{dt}\int_\Omega\rho\,dV = -\int_\Omega\nabla\cdot\mathbf{J}\,dV = -\oint_{\partial\Omega}\mathbf{J}\cdot\hat{\mathbf{n}}\,dS .

Taking $\Omega$ to be a large ball and letting its radius grow, the surface term vanishes provided $\lvert\overline\Psi\nabla\Psi\rvert = o(r^{-2})$ , and $\int\rho\,dV$ is constant. $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Reality of $V$ , in Step 2. Without it the potential terms do not cancel and (7) acquires a source $\frac{2}{\hbar}\operatorname{Im}(V)\rho$ : probability leaks away where $\operatorname{Im}V < 0$ . Optical-potential models of nuclear reactions use precisely this to represent absorption.
Reality of the separation constant $E$ , in Step 2 of the separation proof. The constancy of $\lvert\Psi\rvert^2$ for a separable solution is a consequence of self-adjointness and not of the differential equation, and (3) on its own does not deliver it. Take $V\equiv0$ in one dimension and $E = i\varepsilon$ with $\varepsilon>0$ real. Then $\psi = e^{\alpha x}$ with $\alpha^2 = -2mi\varepsilon/\hbar^2$ , that is $\alpha = e^{-i\pi/4}\sqrt{2m\varepsilon}/\hbar$ , solves (3), and $\Psi = e^{\alpha x}e^{-iEt/\hbar} = e^{\alpha x}e^{\varepsilon t/\hbar}$ solves (1) with density $\lvert\Psi\rvert^2 = e^{2(\operatorname{Re}\alpha)x}e^{2\varepsilon t/\hbar}$ , which grows without bound in $t$ . It is a separable solution and it is not stationary. What excludes it is not the equation but the normalization (2): $\operatorname{Re}\alpha = \sqrt{m\varepsilon/\hbar^2}>0$ , so $\psi\notin L^2(\mathbb{R})$ and $\psi$ is not an eigenfunction of the self-adjoint $\hat H$ at all. Where a suitable domain is to be found is the first of the three problems recorded in the next remark.
Normalizability as the quantization condition. Equation (3) is a differential equation with a solution for every $E$ ; what rules out most $E$ is the demand that the solution be in $L^2$ , or satisfy a boundary condition. Kreyszig makes the criterion explicit — a solution counts as physically meaningful only if it stays bounded everywhere and dies away at large distance — and observes that, once a potential field is fixed, (3) possesses such solutions for certain values of $E$ and no others. The three examples below show the three outcomes: a continuous spectrum with no normalizable eigenfunctions at all (free particle), a discrete spectrum from a boundary condition (infinite well), and a discrete spectrum from decay at infinity (oscillator).
Completeness, in Step 3 of the separation proof. Formula (5) presumes that the eigenfunctions span the space of initial data. For the well and the oscillator this is the completeness of a Fourier sine series and of the Hermite functions respectively; for a potential with both bound and scattering states, the sum must be supplemented by an integral over the continuous spectrum, and (5) as written is incomplete.
The separation of angular variables survives any central potential. Riley, Hobson and Bence's general remark applies verbatim to (3): if $V = V(r)$ , then $\theta$ and $\varphi$ enter only through $\nabla^2$ , so the angular factors are spherical harmonics $Y_\ell^m$ and only the radial equation feels the potential. That is why hydrogen, the three-dimensional oscillator and a free particle share their angular wavefunctions, and why the orbital labels $s,p,d,f$ mean the same thing in every atom.
What the time-dependent equation is not. It is not a wave equation: it is first order in time, so one initial condition determines the evolution, not two, and there is no finite propagation speed. It is not the heat equation either, despite the shape, because the factor $i$ makes the evolution unitary rather than dissipative: $\lvert c_n\rvert$ is constant in (5), where a heat equation would have $e^{-E_nt}$ and lose the high modes.

Remark(What the operator theory adds: three problems the formal equation leaves open)AdaptedAdapted from Reed & Simon (1980), §VIII.11, pp. 302–304

Reed and Simon set the model up in Hilbert-space terms: a separable Hilbert space, physical states as unit vectors determined up to a phase, observables as self-adjoint operators whose measurement probabilities come from a projection-valued measure, and the dynamics as a continuous one-parameter group of unitary operators $U(t)$ whose self-adjoint generator $H$ is the Hamiltonian. Written out for $\varphi$ in the domain of $H$ , their dynamical law is $\frac{d}{dt}[U(t)\varphi] = iH[U(t)\varphi]$ , which is (1) in their units $\hbar = 1$ and with their sign of $i$ . Three mathematical problems are then left open, and they are the three this unit keeps running into.

Self-adjointness. The physics supplies a formal differential expression for $\hat H$ and no domain. One must find a domain on which it is essentially self-adjoint; where there is none, its self-adjoint extensions have to be surveyed and one of them chosen as the observable. The free-particle solution with imaginary $E$ above is the elementary face of this: the differential equation admits a solution that the operator does not.
Spectral analysis. The point spectrum of $H$ is where the stationary states sit, and a system disturbed from one of them moves to another, radiating at a frequency set by the gap between the two energies. This is the sense in which (3) is an experiment.
Scattering theory. The behaviour of $U(t)\varphi$ for large $\lvert t\rvert$ .

Their formulation also makes the reversibility of (1) structural rather than observed: a one-parameter group of unitary operators has an inverse at every $t$ , and the norm it preserves is exactly the normalization (2).

Example(The free particle: a continuous spectrum and no normalizable eigenstates)Original

Take $V \equiv 0$ in one dimension. Then (3) is $-\frac{\hbar^2}{2m}\psi'' = E\psi$ , whose solutions for $E>0$ are $\psi = e^{\pm ikx}$ with

E = \frac{\hbar^2k^2}{2m} ,

and for $E<0$ are the real exponentials $e^{\pm\kappa x}$ with $E = -\hbar^2\kappa^2/2m$ .

No eigenfunction is normalizable. $\int_{-\infty}^{\infty}\lvert e^{ikx}\rvert^2dx = \int 1\,dx = \infty$ , and $e^{\kappa x}$ blows up at one end. So (3) has no solution satisfying (2) for any $E$ : the free particle has no bound states, and its spectrum is the continuous half-line $E\ge0$ . Kreyszig's criterion — finite and tending to zero at infinity — excludes every $E$ , and the resolution is that the physical states are superpositions,

\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \hat f(k)\,e^{i(kx - \hbar k^2t/2m)}\,dk ,

a wave packet, normalizable when $\hat f\in L^2$ . The exponent shows the dispersion relation $\omega = \hbar k^2/2m$ . Unlike the wave equation, where $\omega = ck$ and every component of the packet travels at the same speed, here the group velocity

\frac{d\omega}{dk} = \frac{\hbar k}{m} = \frac{p}{m}

— the classical velocity of a particle of momentum $p = \hbar k$ — depends on $k$ , and the packet therefore spreads: its envelope is broadened at the rate set by $\omega'' = \hbar/m \ne 0$ , and a packet of initial width $\Delta x_0$ doubles in width after a time of order $m(\Delta x_0)^2/\hbar$ . It is $\omega''\ne0$ that does this and not the $k$ -dependence of the phase velocity $\omega/k = \hbar k/2m$ , which is exactly half the group velocity here: a medium with $\omega = ck + a$ has phase velocity $c + a/k$ , varying with $k$ , and spreads nothing at all, because every component of its envelope still moves at $c$ .

The current of a plane wave. For $\Psi = Ae^{i(kx-\omega t)}$ , formula (6) gives

J = \frac{\hbar}{m}\operatorname{Im}\bigl(\overline A e^{-ikx}\cdot ikAe^{ikx}\bigr) = \frac{\hbar k}{m}\lvert A\rvert^2 = v\,\rho ,

with $v = \hbar k/m = p/m$ the classical velocity. So the current is density times velocity, exactly as in fluid mechanics, and the sign of $k$ decides the direction of flow. Here $\rho$ and $J$ are both constant, so (7) holds trivially, $0 = 0$ ; the example is still worth doing because it fixes the interpretation of $\mathbf{J}$ and the normalization of (6).

Example(The infinite well: energies, a beating superposition, and the current)Original

Take $V = 0$ for $0<x<L$ and $V = +\infty$ outside, which forces $\psi(0) = \psi(L) = 0$ .

The stationary states. Solving (3) as in the previous example and imposing the two boundary conditions gives $\psi = \sin(k x)$ with $kL = n\pi$ , so

\psi_n(x) = \sqrt{\frac2L}\,\sin\frac{n\pi x}{L} , \qquad E_n = \frac{n^2\pi^2\hbar^2}{2mL^2} \qquad (n = 1,2,\dots) ,

the normalization constant coming from $\int_0^L\sin^2(n\pi x/L)dx = L/2$ . Here the quantization comes from the boundary condition rather than from decay at infinity, and the eigenfunctions are an orthonormal basis of $L^2(0,L)$ — the Fourier sine basis — so (5) applies with no continuous part.

Numbers. For an electron ( $m = 9.1093837\times10^{-31}$ kg) in a well of width $L = 1$ nm, with $\hbar = 1.0545718\times10^{-34}$ J s,

E_1 = \frac{\pi^2\hbar^2}{2mL^2} = 6.024667\times10^{-20}\ \mathrm J = 0.376030\ \mathrm{eV} ,

and $E_n = n^2E_1$ , so the first gap is $E_2 - E_1 = 3E_1 = 1.128090$ eV. A photon of that energy has wavelength $hc/(E_2-E_1) = (1239.842\ \mathrm{eV\,nm})/(1.128090\ \mathrm{eV}) = 1099.06$ nm, which is the near infrared, not the visible. The gap falls as $L^{-2}$ , so a narrower well is needed to reach visible light: the band from $1.77$ eV ( $700$ nm) to $3.10$ eV ( $400$ nm) corresponds to $L$ between $0.798$ nm and $0.603$ nm. That $L^{-2}$ dependence of the level spacing on size is the mechanism behind the colour of a quantum dot, which is set by its diameter — though a real dot is a well in a semiconductor, so one must add the bulk band gap to the confinement energy and use the carrier's effective mass in place of $m$ .

A superposition, and its beat. Take the normalized initial state $f = \frac{1}{\sqrt2}(\psi_1+\psi_2)$ , so $c_1 = c_2 = 1/\sqrt2$ in (5):

\Psi(x,t) = \frac{1}{\sqrt2}\Bigl[\psi_1(x)e^{-iE_1t/\hbar} + \psi_2(x)e^{-iE_2t/\hbar}\Bigr] .

Its density is $\rho = \tfrac12\bigl[\psi_1^2+\psi_2^2 + 2\psi_1\psi_2\cos\omega_{21}t\bigr]$ with

\omega_{21} = \frac{E_2-E_1}{\hbar} = \frac{3\pi^2\hbar}{2mL^2} = 1.713871\times10^{15}\ \mathrm{s^{-1}} ,

a period of $2\pi/\omega_{21} = 3.666078$ fs for the electron above. The mean position oscillates: since $\int_0^L x\psi_n^2dx = L/2$ for every $n$ and

\int_0^L x\,\psi_1\psi_2\,dx = 2L\int_0^1 u\sin\pi u\sin2\pi u\;du = 2L\cdot\left(-\frac{8}{9\pi^2}\right) = -\frac{16L}{9\pi^2} ,

(using $\sin\pi u\sin2\pi u = \tfrac12[\cos\pi u - \cos3\pi u]$ and $\int_0^1u\cos a u\,du = (\cos a - 1)/a^2 + (\sin a)/a$ ), we get

\langle x\rangle(t) = \frac L2 - \frac{16L}{9\pi^2}\cos\omega_{21}t = \frac L2 - 0.1801265\,L\cos\omega_{21}t .

The particle sloshes from side to side with amplitude $0.18L$ — a classical-looking motion built from two stationary states, and the clearest demonstration that a superposition of different energies is not stationary.

The current, and an exact check of the continuity equation. The probability of being in the left half is

P_{\text{left}}(t) = \int_0^{L/2}\rho\,dx = \frac12\left[\frac12 + \frac12 + 2\cdot\frac{4}{3\pi}\cos\omega_{21}t\right] = \frac12 + \frac{4}{3\pi}\cos\omega_{21}t ,

using $\int_0^{L/2}\psi_n^2dx = \tfrac12$ and $\int_0^{L/2}\psi_1\psi_2\,dx = 2\int_0^{1/2}\sin\pi u\sin2\pi u\,du = \frac{4}{3\pi} = 0.4244132$ . Differentiating,

\dot P_{\text{left}} = -\frac{4\omega_{21}}{3\pi}\sin\omega_{21}t = -\frac{4}{3\pi}\cdot\frac{3\pi^2\hbar}{2mL^2}\sin\omega_{21}t = -\frac{2\pi\hbar}{mL^2}\sin\omega_{21}t .

Now compute the current at the midpoint from (6). At $x = L/2$ we have $\psi_1 = \sqrt{2/L}$ , $\psi_2 = 0$ , $\psi_1' = 0$ and $\psi_2' = -\frac{2\pi}{L}\sqrt{2/L}$ , so

\overline\Psi\,\partial_x\Psi\Big\vert_{x=L/2} = \frac12\cdot\sqrt{\tfrac2L}\,e^{iE_1t/\hbar}\cdot\left(-\frac{2\pi}{L}\sqrt{\tfrac2L}\right)e^{-iE_2t/\hbar} = -\frac{2\pi}{L^2}\,e^{-i\omega_{21}t} ,

whose imaginary part is $\frac{2\pi}{L^2}\sin\omega_{21}t$ . Hence

J\!\left(\tfrac L2,t\right) = \frac{\hbar}{m}\cdot\frac{2\pi}{L^2}\sin\omega_{21}t = \frac{2\pi\hbar}{mL^2}\sin\omega_{21}t = -\dot P_{\text{left}} ,

which is exactly $\frac{d}{dt}\int_0^{L/2}\rho\,dx = -J(L/2,t)$ , the one-dimensional case of Step 5 — the boundary $x=0$ contributing nothing because $\psi_n(0) = 0$ . The two sides agree identically in $t$ , not just numerically, so this is a complete check of (6) and (7) on a nontrivial solution. Note also that a single stationary state here has $J \equiv 0$ , because $\psi_n$ is real and $\operatorname{Im}(\psi_n\psi_n') = 0$ : nothing flows in a standing wave.

Example(The harmonic oscillator: zero-point energy from the Hermite equation)AdaptedAdapted from Kreyszig (1978), §11.3, Example 11.3-1, Eqs. (7)–(12), pp. 585–588; Kreyszig (1978), §11.3, Prob. 5, p. 588; Kreyszig (1978), §11.3, Probs. 9–10, pp. 589–590

Take $V = \tfrac12 m\omega_0^2x^2$ in one dimension, Kreyszig's Example 11.3-1 in our notation. Equation (3) is

-\frac{\hbar^2}{2m}\psi'' + \tfrac12 m\omega_0^2x^2\psi = E\psi . \tag{8}

Non-dimensionalize. Set $b = \sqrt{\hbar/m\omega_0}$ (Kreyszig's $b^2 = h/2\pi m\omega_0$ , the same quantity), $s = x/b$ and $\Lambda = 2E/\hbar\omega_0$ (his $\Lambda = 4\pi E/\omega_0h$ ). Multiplying (8) by $2/\hbar\omega_0$ turns it into

\frac{d^2\tilde\psi}{ds^2} + \bigl(\Lambda - s^2\bigr)\tilde\psi = 0 . \tag{9}

Strip the Gaussian. Substituting $\tilde\psi = v(s)e^{-s^2/2}$ into (9): $\tilde\psi'' = (v'' - 2sv' + (s^2-1)v)e^{-s^2/2}$ , so (9) becomes

v'' - 2sv' + (\Lambda-1)v = 0 ,

which is Hermite's equation. It has a polynomial solution — and therefore $\tilde\psi$ decays at infinity — exactly when

\Lambda - 1 = 2n , \quad\text{i.e.}\quad \Lambda = 2n+1, \qquad n = 0,1,2,\dots ,

in which case $v = H_n(s)$ , the Hermite polynomial (Hermite polynomials).

No other $\Lambda$ is admissible. The displayed condition is what produces a polynomial; what makes these the only energies is that every other solution is thrown out by the normalization (2), and that needs one further fact. Substituting $v = \sum_k a_ks^k$ into Hermite's equation gives the two-term recursion

(k+2)(k+1)\,a_{k+2} = (2k+1-\Lambda)\,a_k ,

which couples $a_k$ to $a_{k+2}$ and so splits the solution space into an even series $E(s)$ and an odd series $O(s)$ , with $v = \alpha E + \beta O$ . A series terminates only when $\Lambda = 2k+1$ for some $k$ of its own parity. Two things then have to be shown, and Kreyszig leaves the first of them as his Problem 5, p. 588.

A non-terminating series grows too fast. Suppose the even series, say, never terminates, so $2k+1-\Lambda \ne 0$ for every even $k$ and no $a_k$ of that parity vanishes beyond the first. Fix $c$ with $\tfrac12 < c < 1$ . Comparing the recursion's ratio with the ratio of successive non-zero coefficients of $\mathrm{e}^{cs^2} = \sum_j c^js^{2j}/j!$ , which is $2c/(k+2)$ at $k = 2j$ ,

\frac{a_{k+2}}{a_k} = \frac{2k+1-\Lambda}{(k+2)(k+1)} \;\ge\; \frac{2c}{k+2} \qquad\Longleftrightarrow\qquad 2k+1-\Lambda \;\ge\; 2c(k+1) ,

and since $c<1$ the right-hand inequality holds for all $k$ beyond some $k_0$ . From $k_0$ on the ratios are positive, so the coefficients keep one sign, and term by term they dominate a positive multiple of those of $\mathrm{e}^{cs^2}$ . Hence for large $s>0$ the series exceeds $C\mathrm{e}^{cs^2}$ minus the finitely many terms below $k_0$ , and

\lvert\tilde\psi(s)\rvert = \lvert v(s)\rvert\,\mathrm{e}^{-s^2/2} \;\gtrsim\; \mathrm{e}^{(c-1/2)s^2} \longrightarrow \infty ,

because $c > \tfrac12$ . Such a $\tilde\psi$ is not in $L^2(\mathbb{R})$ and fails Kreyszig's criterion of staying finite and dying away at infinity.

The two series cannot rescue each other. One might hope that a growing even series and a growing odd series could cancel. They cannot, and parity is why: if $\tilde\psi = v\mathrm{e}^{-s^2/2}$ were in $L^2(\mathbb{R})$ , then so would be its even and odd parts $\tfrac12[\tilde\psi(s)\pm\tilde\psi(-s)]$ , which are exactly $\alpha E(s)\mathrm{e}^{-s^2/2}$ and $\beta O(s)\mathrm{e}^{-s^2/2}$ . So each series would have to be admissible on its own, and by the previous paragraph a non-terminating one is not.

Hence an admissible $\Lambda$ must terminate one of the two series, which is $\Lambda = 2n+1$ ; and at such a $\Lambda$ the series of the other parity still does not terminate, since $2k+1 = \Lambda$ only at $k = n$ , so its coefficient must be zero and the admissible solution is the polynomial one alone, up to a constant multiple.

Translating back, $E = \tfrac12\hbar\omega_0\Lambda$ gives

E_n = \left(n+\tfrac12\right)\hbar\omega_0 = \left(n+\tfrac12\right)h\nu , \qquad \nu = \frac{\omega_0}{2\pi} ,

which is Kreyszig's formula (12). The levels are equally spaced, and the lowest is not zero: the zero-point energy is $\tfrac12 h\nu$ . Kreyszig remarks that this half-integral offset is characteristic of the oscillator and was not what Planck assumed in 1900, and he calls $n$ the principal quantum number of the oscillator.

Check the ground state exactly. For $n = 0$ , $H_0 = 1$ , so $\psi_0 = C e^{-x^2/2b^2}$ . Then

\psi_0'' = \left(\frac{x^2}{b^4} - \frac{1}{b^2}\right)\psi_0 ,

and with $b^2 = \hbar/m\omega_0$ ,

-\frac{\hbar^2}{2m}\psi_0'' + \tfrac12m\omega_0^2x^2\psi_0 = \left(-\tfrac12 m\omega_0^2x^2 + \tfrac12\hbar\omega_0\right)\psi_0 + \tfrac12m\omega_0^2x^2\psi_0 = \tfrac12\hbar\omega_0\,\psi_0 ,

so $E_0 = \tfrac12\hbar\omega_0$ , confirming the formula at $n = 0$ with no appeal to Hermite's equation. Normalizing, $\int e^{-x^2/b^2}dx = b\sqrt\pi$ , so $C = (\pi b^2)^{-1/4}$ and

\psi_0(x) = \frac{1}{(\pi b^2)^{1/4}}e^{-x^2/2b^2} .

A number. For a diatomic vibration with $\nu = 10^{14}$ Hz, the spacing is $\hbar\omega_0 = h\nu = 6.62607\times10^{-20}$ J $= 0.413567$ eV, and the zero-point energy is half of that, $0.206783$ eV — comparable with a chemical bond energy, which is why zero-point energy matters in reaction rates and in isotope effects.

Three dimensions. Kreyszig's Problems 9 and 10 take $V = \tfrac12 m(\omega_1^2x_1^2+\omega_2^2x_2^2+\omega_3^2x_3^2)$ and separate (3) in Cartesian coordinates, giving $E = \sum_j(n_j+\tfrac12)\hbar\omega_j$ with a product of Hermite functions. For the isotropic case $\omega_1=\omega_2=\omega_3=\omega_0$ he notes that the ground level $E_0 = \tfrac32 h\nu$ is non-degenerate while every higher level is degenerate: $E = (N+\tfrac32)\hbar\omega_0$ is attained by every triple with $n_1+n_2+n_3 = N$ , of which there are $\binom{N+2}{2}$ . So the multiplicities of $N = 0,1,2,3,4$ are $1, 3, 6, 10, 15$ .

ApplicationOriginal

Spectroscopy, which is where the energies are read off. Problem 2 of the previous remark is what makes (3) an experimental question: every line in an atomic or molecular spectrum is a difference $E_n - E_{n'}$ , so the eigenvalues are read straight off a photographic plate or a detector. The equally spaced ladder of the oscillator gives the vibrational bands of a diatomic molecule and, through $\hbar\omega_0 = \hbar\sqrt{k/\mu}$ , measures the bond stiffness $k$ ; the $n^2$ ladder of the well gives the absorption edge of a quantum dot, which is why the colour of a cadmium-selenide nanocrystal is set by its diameter.

Hydrogen and the periodic table. Riley, Hobson and Bence's hydrogen equation is (3) with $V = -e^2/4\pi\varepsilon_0r$ . Because the potential is central, the general point of step 5 in separation of variables applies: the angular factors are the spherical harmonics $Y_\ell^m$ , so the quantum numbers $\ell$ and $m$ and the $2\ell+1$ -fold degeneracy are fixed by the Laplacian alone and are the same for every central potential. Only the radial equation knows about the Coulomb tail, and solving it gives the Laguerre functions and the energies $E_n = -13.6\ \mathrm{eV}/n^2$ . Everything in the shell structure of the elements that does not depend on the particular potential — the $s,p,d,f$ labels, the counts $2,6,10,14$ — comes from the separation, not from the physics of the Coulomb force.

Tunnelling, and the sign of $E-V$ . Read Kreyszig's derivation backwards: the local wavenumber is $k(x)^2 = 2m(E-V(x))/\hbar^2$ , so (3) is a Helmholtz equation where $E>V$ and a modified Helmholtz equation where $E<V$ . In the first region the solution oscillates; in the second it grows or decays exponentially. A particle therefore has a nonzero amplitude inside a barrier it could not classically enter, and the transmission through a barrier of width $d$ goes like $e^{-2\kappa d}$ with $\hbar\kappa = \sqrt{2m(V-E)}$ . That exponential is the operating principle of the scanning tunnelling microscope — with a work function of about $4$ eV, $\kappa = 1.02\times10^{10}\ \mathrm{m^{-1}}$ and a change of one angstrom in the tip height changes the current by about a factor of ten, since $\ln 10/2\kappa = 1.12$ Å — and of the flash memory cell.

The classical limit, and semiclassical computation. Hassani's Proposition 14.8.2 is the precise statement of the correspondence principle in this setting: substituting $\Psi = Ae^{iS/\hbar}$ turns (1) into a pair of real equations, one of which is the continuity equation (7) for $\rho = A^2$ and $\mathbf{J} = \rho\nabla S/m$ , and the other of which becomes the Hamilton–Jacobi equation for $S$ when the term $\frac{\hbar^2}{2m}\nabla^2A/A$ is dropped. So as $\hbar\to0$ a wave function behaves like a fluid of non-interacting classical particles following the characteristics of the Hamilton–Jacobi equation, with $\nabla S/m$ as the velocity field. Keeping one order further in $\hbar$ gives the WKB approximation, and matching WKB solutions across a turning point produces the Bohr–Sommerfeld condition $\int_a^b\sqrt{2m(E-V)}\,dx = (n+\tfrac12)\pi\hbar$ , which Hassani derives from the connection formulas and which reproduces the oscillator's $(n+\tfrac12)$ exactly.

Numerical quantum mechanics. Formula (5) is the design of a spectral time-stepper: expand the initial state in eigenfunctions, multiply each coefficient by the unit-modulus factor $e^{-iE_nt/\hbar}$ , and sum. Because the factors have modulus one, the scheme is unconditionally stable and exactly norm-preserving, which an explicit finite-difference scheme for (1) is not — the standard alternative, the Crank–Nicolson or Cayley scheme, is chosen precisely because its propagator $(1 - i\hat H\Delta t/2\hbar)/(1+i\hat H\Delta t/2\hbar)$ is unitary and so inherits the conservation law (7). The split-operator method does the same with the kinetic part diagonalized by a fast Fourier transform (the discrete Fourier transform) and the potential part diagonal in position, alternating between the two.

Application

The Klein-Gordon equation

Adapted

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Introduces the Klein-Gordon equation as the wave equation with a mass term, derives its dispersion relation and reads the relativistic energy-momentum relation off it, verifies its one-dimensional source function across the light cone and records the three-dimensional one, shows that the massless limit gives back the wave equation and the non-relativistic limit the Schrodinger equation, and computes the conserved density that fails to be positive.

Prerequisites: Classification of second-order equations, The one-dimensional wave equation, The Helmholtz equation, The Schrodinger equation, Bessel functions of the first kind, Several independent variables, The Dirac delta in calculations, Test functions and distributions, Delta-function identities

Definition(The Klein-Gordon equation)AdaptedAdapted from Strauss (2008), §13.5, Eq. (1), pp. 373–374; Hassani (2013), §13.1, Eq. (13.6), p. 396; Gelfand & Fomin (1963), §38.1, Example 3, Eqs. (119)–(120), p. 181

The Klein-Gordon equation is the wave equation with a zeroth-order term,

u_{tt} - c^2\nabla^2 u + \mu^2 u = 0 , \tag{1}

in which $c$ is the propagation speed and $\mu$ is a positive constant. It is Strauss's equation (1) of §13.5, which he introduces as one of the simplest of the hyperbolic models of elementary particles and whose only difference from the wave equation, as he puts it, is the last term. Gelfand and Fomin identify the field it describes: the scalar field of uncharged spin-zero particles of mass $M$ , the $\pi^0$ meson being their example.

The constant in the mass term, and the three conventions in our sources. Strauss writes the last term as $m^2u$ with $m$ called the mass of the particle. For the three terms of (1) to have the same dimensions, the coefficient must have the dimensions of an inverse time; in physical units it is

\mu = \frac{Mc^2}{\hbar} ,

$M$ being the mass, and $c/\mu = \hbar/Mc$ is the reduced Compton wavelength of the particle (the Compton wavelength itself being $2\pi$ times that). Hassani avoids the issue by working in natural units $\hbar = c = 1$ , in which his equation (13.6) reads $\nabla^2\phi - m^2\phi = \partial^2\phi/\partial t^2$ — that is (1) with $c = 1$ and $\mu = m$ . We write $\mu$ for the coefficient, state the relation $\mu = Mc^2/\hbar$ once, and keep $c$ explicit.

A sign to be careful with. Gelfand and Fomin display the equation as $(\Box - M^2)u = 0$ and give the Lagrangian density

\mathcal{L} = \tfrac12\bigl(u_{x_0}^2 - u_{x_1}^2 - u_{x_2}^2 - u_{x_3}^2 - M^2u^2\bigr) , \tag{2}

of which they say (1) is the Euler equation for a field (their Eq. (118), the multiple-integral Euler equation of several independent variables). Working out that Euler equation from (2) gives $u_{x_0x_0} - \nabla^2u + M^2u = 0$ , which is (1) with $c = 1$ ; so the sign of the mass term implied by their Lagrangian is the one Strauss and Hassani use, and their displayed $(\Box - M^2)u = 0$ matches it only for the metric signature in which $\Box = \nabla^2 - \partial_{x_0}^2$ . In our scanned copy the signs inside their d'Alembertian are not legible. We follow Strauss and Hassani, equation (1), whose mass term has the sign that makes the solutions oscillate rather than grow.

Application(Plane waves and the dispersion relation)AdaptedAdapted from Hassani (2013), §27.1.1, pp. 832–833; Strauss (2008), §13.5, Eq. (13), p. 376

The function $u = Ae^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ with amplitude $A \ne 0$ solves (1) if and only if

\omega^2 = c^2\lvert\mathbf{k}\rvert^2 + \mu^2 . \tag{3}

Equation (3) is the dispersion relation of the Klein-Gordon equation. Writing $k = \lvert\mathbf{k}\rvert$ , it has the following consequences.

Phase and group velocity. For $k = \lvert\mathbf{k}\rvert > 0$ , $v_{\text{ph}} = \frac{\omega}{k} = c\sqrt{1 + \frac{\mu^2}{c^2k^2}} \;>\; c , \qquad v_{\text{gr}} = \frac{d\omega}{dk} = \frac{c^2k}{\omega} \;<\; c , \qquad v_{\text{ph}}\,v_{\text{gr}} = c^2 .$ The phase velocity exceeds $c$ and the group velocity falls short of it; their product is exactly $c^2$ . At $k = 0$ the phase velocity is undefined, since $\omega/k$ diverges, and the group velocity is $0$ , so the product identity needs $k>0$ . For $\mu = 0$ both equal $c$ and the medium is dispersionless.
The relativistic energy-momentum relation. Substituting $E = \hbar\omega$ , $\mathbf{p} = \hbar\mathbf{k}$ and $\mu = Mc^2/\hbar$ turns (3) into $E^2 = p^2c^2 + M^2c^4 ,$ and then $v_{\text{gr}} = c^2k/\omega = pc^2/E$ , which is the velocity of a relativistic particle of momentum $p$ and energy $E$ .
Both signs of $\omega$ occur. For each $\mathbf{k}$ there are two roots $\omega = \pm\sqrt{c^2k^2+\mu^2}$ , and both give solutions of (1). There is no $\mathbf{k}$ for which $\omega$ lies between $-\mu$ and $\mu$ : the spectrum has a gap of width $2\mu$ about zero.

Hassani reads the correspondence in the other direction, and this is how the equation was found: replacing $E$ by $i\hbar\partial/\partial t$ and $\mathbf{p}$ by $-i\hbar\nabla$ turns the non-relativistic relation $E = p^2/2M$ into the Schrodinger equation and the relativistic relation $E^2 - p^2c^2 = M^2c^4$ into (1).

Idea of the proofOriginal

Substituting an exponential into a constant-coefficient equation turns each derivative into a multiplication, so the whole equation collapses to one algebraic relation among $\omega$ and $\mathbf{k}$ — the dispersion relation. Everything qualitative about the equation can be read off it. Here the mass term contributes a constant $\mu^2$ that is added to $c^2k^2$ , which has three visible effects: long waves ( $ck \ll \mu$ ) are strongly dispersive and hardly propagate, short waves ( $ck\gg\mu$ ) behave as if the mass were absent, and there is a lowest frequency $\mu$ below which no travelling wave exists at all. The last of these is why a massive field cannot carry a static long-range force, and the first is why a Klein-Gordon pulse spreads where a wave-equation pulse does not.

ProofAdaptedAdapted from Hassani (2013), §27.1.1, p. 832

Step 1 (substitute). For $u = Ae^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ with $A \ne 0$ ,

u_{tt} = -\omega^2u , \qquad \nabla^2 u = -\lvert\mathbf{k}\rvert^2u ,

so (1) becomes $\bigl(-\omega^2 + c^2\lvert\mathbf{k}\rvert^2 + \mu^2\bigr)u = 0$ . Since $u$ vanishes nowhere, this holds identically if and only if (3) does.

Step 2 (the velocities). From (3), $\omega = \sqrt{c^2k^2+\mu^2}$ for the positive root, so

\frac\omega k = \frac{\sqrt{c^2k^2+\mu^2}}{k} = c\sqrt{1+\frac{\mu^2}{c^2k^2}} ,

which exceeds $c$ for every $k>0$ when $\mu>0$ . Differentiating (3) in $k$ gives $2\omega\,d\omega/dk = 2c^2k$ , hence $v_{\text{gr}} = c^2k/\omega$ , and multiplying the two,

v_{\text{ph}}v_{\text{gr}} = \frac\omega k\cdot\frac{c^2k}{\omega} = c^2 .

Since $\omega > ck$ by (3), $v_{\text{gr}} = c^2k/\omega < c$ .

Step 3 (the gap). By (3), $\omega^2 \ge \mu^2$ for every real $\mathbf{k}$ , with equality only at $\mathbf{k} = \mathbf{0}$ . So $\lvert\omega\rvert \ge \mu$ , and the two branches $\omega = \pm\sqrt{c^2k^2+\mu^2}$ are separated by the interval $(-\mu,\mu)$ , which no plane wave occupies. $\square$

Application(The one-dimensional source function, and the light cone)AdaptedAdapted from Strauss (2008), §13.5, Eqs. (12)–(14) and (20), pp. 376–378

Consider the Cauchy problem for (1) on the line,

u_{tt} - c^2u_{xx} + \mu^2u = 0 \quad (-\infty<x<\infty), \qquad u(x,0) = \varphi(x), \quad u_t(x,0) = \psi(x) .

Its source function — the solution with $\varphi = 0$ , $\psi = \delta$ — is

S(x,t) \;=\; \begin{cases} \dfrac{1}{2c}\,J_0\!\left(\mu\sqrt{t^2 - \dfrac{x^2}{c^2}}\right), & \lvert x\rvert < ct,\\[10pt] 0, & \lvert x\rvert > ct \ge 0, \end{cases} \tag{4}

where $J_0$ is the first-kind Bessel function of order zero. Strauss obtains it by Fourier transform: the transform satisfies $\hat S_{tt} = -(c^2k^2+\mu^2)\hat S$ with $\hat S(k,0) = 0$ and $\hat S_t(k,0) = 1$ , hence

\hat S(k,t) = \frac{\sin\bigl(t\sqrt{c^2k^2+\mu^2}\bigr)}{\sqrt{c^2k^2+\mu^2}} , \tag{5}

and inverting (5) gives (4). Two features are worth naming.

The support is the same light cone as for the wave equation. $S$ vanishes outside $\lvert x\rvert \le ct$ , and Strauss notes that it has the same jump discontinuity on the cone as the wave equation's source function: the mass term changes the solution inside the cone but moves no signal faster, and the domain of dependence is unchanged from d'Alembert's formula.
The massless limit is the wave equation. As $\mu\to0$ , $J_0 \to J_0(0) = 1$ and (4) becomes $S = \frac{1}{2c}$ on $\lvert x\rvert<ct$ and $0$ outside, which is the source function of the one-dimensional wave equation.

Carrying the same Fourier calculation out in three dimensions, Strauss arrives at

S(\mathbf{x},t) = \frac{1}{2\pi c}\,\delta\bigl(c^2t^2 - r^2\bigr) \;-\; \mu\,H\bigl(c^2t^2-r^2\bigr)\,\frac{J_1\bigl[(\mu/c)\sqrt{c^2t^2-r^2}\bigr]}{4\pi c^2\sqrt{c^2t^2-r^2}} , \tag{6}

where $r = \lvert\mathbf{x}\rvert$ . The first term is an impulse carried on the cone itself; the second is spread through the cone's interior and is built from $J_1$ . At $\mu = 0$ the second term disappears and (6) is the wave equation's source function, concentrated on the cone alone. So Huygens' principle — a sharp pulse staying sharp — holds for the wave equation in three dimensions and fails for Klein-Gordon: a mass term makes the cone leak into its interior, and an impulse is followed by a decaying tail.

Idea of the proofOriginal

The Fourier transform diagonalizes (1), as it does any constant-coefficient equation: in $k$ the problem is the oscillator $\hat S_{tt} + \omega(k)^2\hat S = 0$ with $\omega(k)^2 = c^2k^2+\mu^2$ , whose solution with the stated data is $\sin(\omega t)/\omega$ . All the information about the mass is in the single function $\omega(k)$ , and the only question left is what the inverse transform of $\sin(t\omega(k))/\omega(k)$ is.

That question has a pretty answer because of a coincidence of integral representations: $J_0$ has one whose integrand is exactly $\sin(t\sqrt{c^2k^2+\mu^2})/\sqrt{c^2k^2+\mu^2}$ . Rather than reproduce Strauss's table lookup, the proof below verifies (4) directly — it is a short computation, and it shows where the Bessel equation comes from. The key is that for a function of the invariant combination $s = \sqrt{t^2-x^2/c^2}$ , the wave operator reduces to an ordinary differential operator in $s$ which is precisely Bessel's operator of order zero.

Two things then remain that a pointwise substitution cannot give. The first is the cone itself: (4) is discontinuous there, and a jump across a surface normally leaves $\delta$ and $\delta'$ terms in the wave operator, so solving on each side separately proves nothing about the union. They cancel here for two reasons worth separating — the cone is characteristic, which kills the $\delta'$ term, and the Bessel profile is a function of $t^2-x^2/c^2$ , which vanishes on the cone to exactly the order that kills the $\delta$ term. The second is the initial data, and those follow from the shrinking support alone: $S$ lives on $[-ct,ct]$ and is bounded by $1/2c$ , so it tends to $0$ , while its integral comes out as $t + O(t^3)$ , which is what makes its $t$ -derivative at $0$ a unit point mass.

ProofOriginal

We verify (4) by substitution. Set $c = 1$ throughout; the general case follows by rescaling $x\mapsto x/c$ , which leaves (1) in the form $u_{tt}-u_{xx}+\mu^2u=0$ .

Step 1 (the invariant variable). Inside the forward cone $\lvert x\rvert<t$ put

s = \sqrt{t^2-x^2} \;>\;0 ,

so that $s_t = t/s$ and $s_x = -x/s$ . Differentiating once more,

s_{tt} = \frac1s - \frac{t^2}{s^3} , \qquad s_{xx} = -\frac1s - \frac{x^2}{s^3} .

Step 2 (the wave operator on a function of $s$ ). For $u = f(s)$ , the chain rule gives $u_{tt} = f''s_t^2 + f's_{tt}$ and $u_{xx} = f''s_x^2 + f's_{xx}$ , so

u_{tt} - u_{xx} = f''\bigl(s_t^2 - s_x^2\bigr) + f'\bigl(s_{tt}-s_{xx}\bigr) .

Now $s_t^2 - s_x^2 = (t^2-x^2)/s^2 = 1$ , and

s_{tt}-s_{xx} = \left(\frac1s - \frac{t^2}{s^3}\right) - \left(-\frac1s - \frac{x^2}{s^3}\right) = \frac2s - \frac{t^2-x^2}{s^3} = \frac1s .

Hence, remarkably,

u_{tt}-u_{xx} = f''(s) + \frac{f'(s)}{s} , \tag{7}

an ordinary differential operator in $s$ alone.

Step 3 (Bessel's equation). Take $f(s) = J_0(\mu s)$ . Then $f' = \mu J_0'(\mu s)$ and $f'' = \mu^2J_0''(\mu s)$ , so by (7) and Bessel's equation of order zero, $J_0''(z) + J_0'(z)/z + J_0(z) = 0$ with $z = \mu s$ ,

u_{tt}-u_{xx} = \mu^2\left[J_0''(\mu s) + \frac{J_0'(\mu s)}{\mu s}\right] = -\mu^2 J_0(\mu s) = -\mu^2 u .

Therefore $u_{tt}-u_{xx}+\mu^2u = 0$ : the function $\tfrac12 J_0(\mu\sqrt{t^2-x^2})$ solves the Klein-Gordon equation at every point strictly inside the cone. (With $c$ restored, $s = \sqrt{t^2-x^2/c^2}$ and the constant is $1/2c$ , which is (4).)

Step 4 (across the cone). Outside the cone $S\equiv0$ , which solves (1) trivially. The two pieces meet on $\lvert x\rvert = t$ , where the interior limit is $\tfrac12 J_0(0) = \tfrac12$ , so $S$ jumps by $\tfrac12$ across the cone (by $1/2c$ once $c$ is restored). Solving on each of the two open sets separately is not enough, because a jump across a surface in general leaves $\delta$ and $\delta'$ terms in the wave operator, and what the statement asserts is that $S$ solves the Cauchy problem on the whole line. We show those terms cancel.

Two preliminaries. First, $J_0$ is even and entire, so

J_0(\mu\sqrt w) = \sum_{j\ge0}\frac{(-1)^j}{(j!)^2}\left(\frac{\mu^2 w}{4}\right)^{j}

is an entire function of $w$ with no branch point at $w = 0$ . Hence, putting $G(w) = \tfrac12 J_0(\mu\sqrt w)$ and

g(x,t) = G(t^2-x^2) , \qquad \phi(x,t) = t - \lvert x\rvert ,

the profile $g$ is smooth on the whole plane — not merely inside the cone — and $S = H(\phi)\,g$ with $H$ the Heaviside step. Second, Step 3 says exactly that $g_{tt}-g_{xx} = -\mu^2g$ wherever $\phi>0$ .

Work on the open half-plane $t>0$ . Near the axis $x=0$ , the only place where $\phi$ fails to be differentiable, $\phi>0$ , so $H(\phi)\equiv1$ and $S = g$ on a neighbourhood; there the equation holds by Step 3. It therefore suffices to work where $x \ne 0$ , and there $\phi$ is smooth with

\phi_t = 1 , \qquad \phi_x = -\operatorname{sgn}x , \qquad \phi_{tt} = \phi_{xx} = 0 .

Differentiating the product $S = H(\phi)g$ twice in the sense of distributions, $\partial_t S = \delta(\phi)\phi_tg + H(\phi)g_t$ and

\partial_t^2 S = \delta'(\phi)\phi_t^2\,g + \delta(\phi)\phi_{tt}\,g + 2\delta(\phi)\phi_t\,g_t + H(\phi)\,g_{tt} ,

with the same formula in $x$ . Subtracting,

S_{tt}-S_{xx} = \delta'(\phi)\bigl(\phi_t^2-\phi_x^2\bigr)g \;+\; \delta(\phi)\bigl(\phi_{tt}-\phi_{xx}\bigr)g \;+\; 2\delta(\phi)\bigl(\phi_tg_t-\phi_xg_x\bigr) \;+\; H(\phi)\bigl(g_{tt}-g_{xx}\bigr) .

Take the four terms in turn.

The $\delta'$ term vanishes because $\phi_t^2-\phi_x^2 = 1 - 1 = 0$ : the level sets of $\phi$ are characteristic for the wave operator. This is where the geometry of the light cone enters, and it is the precise content of the statement that the mass term moves no signal faster.
The second term vanishes because $\phi$ is affine on each of the two half-planes $x>0$ and $x<0$ .
The third vanishes because $g$ depends on $x$ and $t$ only through $t^2-x^2$ . Indeed $g_t = 2tG'(t^2-x^2)$ and $g_x = -2xG'(t^2-x^2)$ , so $\phi_tg_t - \phi_xg_x = g_t + (\operatorname{sgn}x)\,g_x = 2\bigl(t-\lvert x\rvert\bigr)G'(t^2-x^2) = 2\phi\,G'(t^2-x^2) ,$ and $\delta(\phi)\,\phi = 0$ . So the Bessel profile vanishes on the cone to exactly the order needed.
The last term is $-\mu^2H(\phi)g = -\mu^2S$ , by Step 3.

Hence $S_{tt}-S_{xx}+\mu^2S = 0$ in the sense of distributions on $\{t>0\}$ , across the cone and not only on either side of it. The jump $1/2c$ is the same one the wave equation's source function has, which is Strauss's observation; the two computations above are why the mass term leaves it alone.

Step 5 (the initial data). It remains to show that $S(\cdot,t)\to0$ and $S_t(\cdot,0)=\delta$ as distributions in $x$ when $t\downarrow0$ . Everything needed is the shrinking support $\operatorname{supp}S(\cdot,t)\subseteq[-t,t]$ and the bound $\lvert S\rvert\le\tfrac12$ , which holds because $\lvert J_0\rvert\le1$ .

First integrate (4) over its support. The series above converges uniformly on compact sets, so it may be integrated term by term:

\int_{-t}^{t}S(x,t)\,dx = \frac12\sum_{j\ge0}\frac{(-1)^j\mu^{2j}}{4^j(j!)^2}\int_{-t}^{t}\bigl(t^2-x^2\bigr)^j dx .

Substituting $x = tu$ and using the beta integral $\int_{-1}^{1}(1-u^2)^j du = 2\cdot4^j(j!)^2/(2j+1)!$ , the inner integral is $t^{2j+1}\cdot2\cdot4^j(j!)^2/(2j+1)!$ , and every factorial cancels:

\int_{-t}^{t}S(x,t)\,dx = \frac1\mu\sum_{j\ge0}\frac{(-1)^j(\mu t)^{2j+1}}{(2j+1)!} = \frac{\sin\mu t}{\mu} = t + O(t^3) .

Now let $\varphi$ be a test function. Splitting $\varphi(x) = \varphi(0) + [\varphi(x)-\varphi(0)]$ and bounding the second piece by $\lvert x\rvert\sup\lvert\varphi'\rvert$ ,

\left\lvert\int S(x,t)\varphi(x)\,dx - \varphi(0)\,\frac{\sin\mu t}{\mu}\right\rvert \;\le\; \sup\lvert\varphi'\rvert\int_{-t}^{t}\lvert x\rvert\cdot\tfrac12\,dx \;=\; \tfrac12 t^2\sup\lvert\varphi'\rvert .

So $\langle S(\cdot,t),\varphi\rangle = t\,\varphi(0) + O(t^2)$ for every $\varphi$ . Letting $t\downarrow0$ gives $S(\cdot,0)=0$ ; dividing by $t$ first and using that value,

\langle S_t(\cdot,0),\varphi\rangle = \lim_{t\downarrow0}\frac1t\bigl\langle S(\cdot,t)-S(\cdot,0),\varphi\bigr\rangle = \varphi(0) ,

which says $S_t(\cdot,0)=\delta$ . Neither step uses (5), which is the point: the Fourier route is not needed anywhere in this proof. (The two do agree: the integral just computed is $\hat S(0,t)$ , which is the value of (5) at $k=0$ .)

Step 6 (the massless limit). $J_0$ is continuous with $J_0(0) = 1$ , and $\mu\sqrt{t^2-x^2/c^2}\to0$ uniformly on compact subsets of the cone as $\mu\to0$ . So $S \to \frac{1}{2c}$ pointwise inside the cone and $0$ outside, which is the d'Alembert source function. $\square$

Application(The conserved density, and the non-relativistic limit)Original

Let $u$ be a complex-valued solution of (1). Define

\rho = \frac{i\hbar}{2Mc^2}\bigl(\bar u\,u_t - u\,\bar u_t\bigr) , \qquad \mathbf{J} = \frac{\hbar}{M}\operatorname{Im}\bigl(\bar u\,\nabla u\bigr) . \tag{8}

Then $\partial_t\rho + \nabla\cdot\mathbf{J} = 0$ . Moreover:

$\rho$ is not positive. For the plane wave $u = Ae^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ , $\rho = \frac{\hbar\omega}{Mc^2}\lvert A\rvert^2 , \qquad \mathbf{J} = \frac{\hbar\mathbf{k}}{M}\lvert A\rvert^2 , \qquad \frac{\mathbf{J}}{\rho} = \frac{c^2\mathbf{k}}{\omega} = v_{\text{gr}}\hat{\mathbf{k}} ,$ so $\rho$ has the sign of $\omega$ and is negative on the branch $\omega<0$ of the dispersion relation.
Non-relativistic limit. Write $u = e^{-i\omega_0t}\psi$ with $\omega_0 = Mc^2/\hbar = \mu$ . Then $\psi$ satisfies $i\hbar\,\frac{\partial\psi}{\partial t} = -\frac{\hbar^2}{2M}\nabla^2\psi + \frac{\hbar^2}{2Mc^2}\frac{\partial^2\psi}{\partial t^2} , \tag{9}$ exactly. Dropping the last term — legitimate when $\lvert\psi_{tt}\rvert \ll \omega_0\lvert\psi_t\rvert$ , that is when the residual frequencies are small beside $Mc^2/\hbar$ — gives the free Schrodinger equation, and in the same limit $\hbar\omega\to Mc^2$ makes $\rho\to\lvert\psi\rvert^2$ , the Schrodinger probability density.

Hassani records the historical consequence of part 1: that the equation's second time derivative made the resulting probabilities capable of coming out negative, that it was set aside at first for that reason, and that it was nevertheless readmitted in the end as a proper equation for relativistic particles; and the wish for a first-order equation is what led Dirac to factor $E^2-p^2-M^2$ and so to the Dirac equation. In Reed and Simon's framework the point is structural: their model demands a one-parameter unitary group generated by a self-adjoint $H$ , with $\lVert\Psi\rVert = 1$ fixed; (1) is second order in $t$ , so its natural phase space is the pair $(u,u_t)$ , and the conserved quantity that does come out of it — the energy — is quadratic in the derivatives rather than in $u$ alone.

Idea of the proofOriginal

The computation is the same two lines as for the Schrodinger current, run on a second-order equation. Multiply (1) by $\bar u$ , multiply its conjugate by $u$ , subtract: the terms in $\mu^2\lvert u\rvert^2$ cancel because $\mu^2$ is real, and the second time derivatives assemble into $\partial_t(\bar uu_t - u\bar u_t)$ while the Laplacians assemble into a divergence. So one gets a conservation law for free. The difference from the Schrodinger case is where the time derivatives sit: there the equation was first order, so the conserved density came out as $\lvert\Psi\rvert^2$ , manifestly non-negative; here it involves $u_t$ , and nothing forces its sign.

For the non-relativistic limit, the trick is to peel off the rest-energy oscillation $e^{-iMc^2t/\hbar}$ , which is the fastest thing in the solution. Doing so cancels the mass term against the $-\omega_0^2$ coming from differentiating the exponential twice — that cancellation is exactly the statement $\mu = \omega_0$ — and leaves a first-order equation in $t$ plus one leftover second derivative, suppressed by $1/c^2$ .

ProofOriginal

Step 1 (the conservation law). Write (1) as $u_{tt} = c^2\nabla^2u - \mu^2u$ and its conjugate as $\bar u_{tt} = c^2\nabla^2\bar u - \mu^2\bar u$ , using that $c$ and $\mu$ are real. Put

W = \bar u\,u_t - u\,\bar u_t .

Differentiating, the cross terms $\bar u_tu_t$ cancel:

\partial_t W = \bar u\,u_{tt} - u\,\bar u_{tt} = c^2\bigl(\bar u\nabla^2u - u\nabla^2\bar u\bigr) - \mu^2\bigl(\bar uu - u\bar u\bigr) = c^2\bigl(\bar u\nabla^2u - u\nabla^2\bar u\bigr) ,

the mass term dropping out. By the identity of Step 3 of the Schrodinger current proof,

\bar u\nabla^2u - u\nabla^2\bar u = \nabla\cdot\bigl(\bar u\nabla u - u\nabla\bar u\bigr) ,

so $\partial_t W = c^2\nabla\cdot(\bar u\nabla u - u\nabla\bar u)$ . Multiplying by $\dfrac{i\hbar}{2Mc^2}$ and noting that $\dfrac{i\hbar}{2M}(\bar u\nabla u - u\nabla\bar u) = -\dfrac{\hbar}{M}\operatorname{Im}(\bar u\nabla u) = -\mathbf{J}$ ,

\partial_t\rho = -\nabla\cdot\mathbf{J} ,

which is the conservation law, with $\rho$ and $\mathbf{J}$ as in (8).

Step 2 (the plane wave). For $u = Ae^{i(\mathbf{k}\cdot\mathbf{x}-\omega t)}$ we have $u_t = -i\omega u$ and $\nabla u = i\mathbf{k}u$ , so $\bar uu_t = -i\omega\lvert A\rvert^2$ and $\bar u\nabla u = i\mathbf{k}\lvert A\rvert^2$ . Then

\rho = \frac{i\hbar}{2Mc^2}\bigl(-i\omega - \overline{(-i\omega)}\bigr)\lvert A\rvert^2 = \frac{i\hbar}{2Mc^2}(-2i\omega)\lvert A\rvert^2 = \frac{\hbar\omega}{Mc^2}\lvert A\rvert^2 ,

and $\mathbf{J} = \frac{\hbar}{M}\operatorname{Im}(i\mathbf{k}\lvert A\rvert^2) = \frac{\hbar\mathbf{k}}{M}\lvert A\rvert^2$ . Dividing, $\mathbf{J}/\rho = c^2\mathbf{k}/\omega$ , whose magnitude is $v_{\text{gr}}$ by part 1 of the dispersion statement. Since both roots $\omega = \pm\sqrt{c^2k^2+\mu^2}$ are admissible, $\rho$ takes both signs.

Step 3 (peel off the rest energy). Put $u = e^{-i\omega_0t}\psi$ with $\omega_0 = \mu = Mc^2/\hbar$ . Then

u_{tt} = e^{-i\omega_0t}\bigl(\psi_{tt} - 2i\omega_0\psi_t - \omega_0^2\psi\bigr) ,

and substituting into (1), after dividing by $e^{-i\omega_0t}$ ,

\psi_{tt} - 2i\omega_0\psi_t - \omega_0^2\psi - c^2\nabla^2\psi + \mu^2\psi = 0 .

Because $\mu = \omega_0$ , the two terms $-\omega_0^2\psi$ and $+\mu^2\psi$ cancel — this is the whole point of the substitution — leaving

\psi_{tt} - 2i\omega_0\psi_t - c^2\nabla^2\psi = 0 .

Step 4 (rearrange). Multiply by $\dfrac{\hbar}{2\omega_0} = \dfrac{\hbar^2}{2Mc^2}$ and move the $\psi_t$ term to the left:

i\hbar\,\psi_t = \frac{\hbar^2}{2Mc^2}\psi_{tt} - \frac{\hbar^2}{2M}\nabla^2\psi ,

which is (9). If $\psi$ varies on a time scale $T$ with $\omega_0T \gg 1$ , the retained term $i\hbar\psi_t$ is of size $\hbar\lvert\psi\rvert/T$ while the dropped term is of size $\hbar^2\lvert\psi\rvert/(2Mc^2T^2) = \hbar\lvert\psi\rvert/(2\omega_0T^2)$ , smaller by the factor $1/(2\omega_0T)$ . Dropping it gives the free Schrodinger equation.

Step 5 (the density in the limit). In the same regime $u_t = e^{-i\omega_0t}(\psi_t - i\omega_0\psi) \approx -i\omega_0e^{-i\omega_0t}\psi$ , so

\rho \approx \frac{i\hbar}{2Mc^2}\bigl(-i\omega_0 - \overline{(-i\omega_0)}\bigr)\lvert\psi\rvert^2 = \frac{\hbar\omega_0}{Mc^2}\lvert\psi\rvert^2 = \lvert\psi\rvert^2 ,

using $\hbar\omega_0 = Mc^2$ . So the Klein-Gordon density reduces to the Schrodinger probability density, and in particular becomes positive. $\square$

RemarkOriginal

Where the hypotheses are used, and what changes without them.

$\mu$ real, in Step 1 of the last proof and in Step 3 of the dispersion proof. Both the cancellation of the mass term in the conservation law and the reality of $\omega$ need it. Strauss's Exercise 9 shows what an imaginary mass does: substituting $u = e^{-t/2}v$ into the telegraph equation $u_{tt} - c^2\Delta u + \nu u_t = 0$ with $\nu = 1$ gives $v_{tt} - c^2\Delta v - \tfrac14 v = 0$ , which is (1) with $\mu = i/2$ . (His exercise prints the conclusion in $u$ where $v$ is meant; it is the equation satisfied by $v$ .) So a dissipative wave equation is a Klein-Gordon equation with imaginary mass — but the imaginary mass is not itself the damping. With $\mu = i/2$ the dispersion relation (3) reads $\omega^2 = c^2k^2 - \tfrac14$ , whose roots are real for $ck>\tfrac12$ and purely imaginary for $ck<\tfrac12$ , never genuinely complex: an imaginary mass closes the frequency gap of part 3 of the dispersion statement and replaces oscillation by the pair $e^{\pm\lambda t}$ , $\lambda = \sqrt{\tfrac14 - c^2k^2}$ , in the long-wave band. One of those branches grows — the tachyonic instability — so (1) with imaginary $\mu$ is unstable, not damped. The telegraph equation's damping is the prefactor $e^{-t/2}$ that the substitution strips off, and it is uniform in $k$ ; the two fit together because $\lambda \le \tfrac12$ for every $k$ , with equality only at $k=0$ , so $u = e^{-t/2}v$ grows at no wavenumber, and decays strictly at every $k\ne0$ . At $k=0$ exactly the two factors cancel and $u$ is constant, as it must be: the $k=0$ telegraph equation is $u_{tt}+u_t=0$ , whose solutions are $a + be^{-t}$ , and the constant $a$ never dies. Reading the correspondence the other way: a real mass term makes waves disperse but not decay, while an imaginary one destroys the oscillation of the long waves altogether.
The classification is unchanged by the mass term. Strauss's index lists Klein-Gordon under hyperbolic equations, and his point in §13.5 is that the term $\mu^2u$ is of order zero, so the principal part of the operator — the part that decides the type, the characteristics and the domain of dependence (classification of second-order equations) — is that of the wave equation. That is why (4) and (6) are supported in the light cone and why Strauss's remark about the jump on the cone holds: the mass changes the solution inside the cone and nothing about the cone itself.
Huygens' principle fails. The second term of (6) is the price of the mass. In three dimensions the wave equation's source function is a pure delta on the cone, so a flash is seen as a flash; with $\mu>0$ a flash is followed by a reverberating tail $\propto J_1$ , decaying but never ending. The same contrast appears in the wave equation in three dimensions, where the two-dimensional wave equation also has a tail.
What the "negative probability" means and what it does not. Part 1 of the last statement is a theorem: the conserved density (8) changes sign. Hassani reports the historical reading — the equation was initially abandoned because of it — together with the resolution, that the equation was in the end admitted as a proper description of relativistic particles after all. The modern reading is that $\rho$ is a charge density and not a probability density, so its sign is a sign of charge; the equation is then a field equation rather than a one-particle wave equation, and its quantization describes particles and antiparticles. Gelfand and Fomin take the field view from the start: (1) is the Euler equation of the Lagrangian density (2), and §38.2 obtains its conservation laws from Noether's theorem, integrating the divergence identities over the boundary of a space-time cylinder.
The square-root equation. Strauss records the precise sense in which Dirac's equation is a square root of (1). Write $\mathcal{D}$ for the first-order operator built from his gamma matrices, $\mathcal{D} = \frac1c\gamma^0\frac{\partial}{\partial t} + \gamma^1\frac{\partial}{\partial x} + \gamma^2\frac{\partial}{\partial y} + \gamma^3\frac{\partial}{\partial z} ,$ whose square is $\mathcal{D}^2 = c^{-2}\partial_t^2 - \Delta$ , the gamma-matrix relations killing every cross term. Strauss's mass symbol here is a lowercase $m$ of inverse-length dimension — it is $\mu/c$ in our notation, which is what lets $m^2$ be added to $-\Delta$ , and it is not the $M$ of $\mu = Mc^2/\hbar$ — and with it $\bigl(\mathcal{D} + im\bigr)\bigl(\mathcal{D} - im\bigr) = \mathcal{D}^2 + m^2 = \frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \Delta + m^2 ,$ which is $c^{-2}$ times the operator in (1). Strauss displays this as the square $(\mathcal{D}+im)^2$ ; read as a square it would leave the cross term $2im\mathcal{D}$ and reverse the sign of the mass, so it is the product of the two first-order factors, as displayed above, that is the Klein-Gordon operator. Hassani gives the algebraic version in exactly that product form: Dirac looked for $\beta,\alpha_i$ making $(\beta E + \sum\alpha_ip_i - m)(\beta E + \sum\alpha_ip_i + m)$ reduce to $E^2 - \sum p_i^2 - m^2$ , which is what forces the coefficients to be matrices satisfying a Clifford algebra.
The massless limit is not merely a limit of solutions. At $\mu = 0$ the dispersion relation (3) loses its gap, the phase and group velocities both become $c$ , and the source functions (4) and (6) become the wave equation's. But the character of the field also changes qualitatively: only a massless field has a static solution falling off as $1/r$ and so a long-range force, which is why the photon is massless and the force carried by a massive field is screened at the length $c/\mu = \hbar/Mc$ — the same screening as in the modified Helmholtz equation of the Helmholtz equation, with $e^{-\mu r/c}/r$ in place of $1/r$ .

Example(Dispersion for a pion, and the exact relation

v_{\text{ph}}v_{\text{gr}} = c^2

)Original

Take the charged pion $\pi^\pm$ , with rest energy $Mc^2 = 139.57039$ MeV. (A charged spin-zero particle is described by a complex-valued solution of (1), which is the case the conserved density (8) applies to, with $\rho$ read as a charge density. Gelfand and Fomin's uncharged $\pi^0$ is a real field and is lighter, $Mc^2 = 134.9768$ MeV, giving $c/\mu = 1.461933$ fm in place of the $1.413817$ fm below, and an $\hbar\omega$ column smaller by the factor $134.9768/139.57039$ ; the velocity and product columns are the same for either particle, since they depend only on the ratio $ck/\mu$ .) Using $\hbar c = 197.3269804$ MeV fm, the mass constant and length scale of (1) are

\mu = \frac{Mc^2}{\hbar} , \qquad \frac{c}{\mu} = \frac{\hbar c}{Mc^2} = \frac{197.3269804}{139.57039} = 1.413817\ \mathrm{fm} ,

the reduced Compton wavelength. Evaluate (3) at the three wavenumbers $ck = \tfrac12\mu$ , $ck=\mu$ and $ck = 2\mu$ , writing $\hbar\omega$ in MeV:

$ck/\mu$	$\omega/\mu = \sqrt{1+(ck/\mu)^2}$	$\hbar\omega$ (MeV)	$v_{\text{gr}}/c = ck/\omega$	$v_{\text{ph}}/c = \omega/(ck)$	product
$0.5$	$1.1180340$	$156.0444$	$0.4472136$	$2.2360680$	$1$
$1$	$1.4142136$	$197.3823$	$0.7071068$	$1.4142136$	$1$
$2$	$2.2360680$	$312.0889$	$0.8944272$	$1.1180340$	$1$

The last column is exactly $1$ in every row, which is the identity $v_{\text{ph}}v_{\text{gr}} = c^2$ of Step 2 — an algebraic identity, not a numerical coincidence. At $ck = \mu$ , that is at $k = Mc/\hbar$ and so at the Compton wavelength $2\pi\hbar/Mc$ , the group velocity is $c/\sqrt2 = 0.7071068\,c$ and the phase velocity $\sqrt2\,c$ .

Check against relativistic kinematics. The row $ck = \mu$ corresponds to $pc = Mc^2$ , hence $E = \sqrt{(pc)^2+(Mc^2)^2} = \sqrt2\,Mc^2 = 197.3823$ MeV, which is the third column; the Lorentz factor is $\gamma = E/Mc^2 = \sqrt2$ and the particle speed $v = c\sqrt{1-\gamma^{-2}} = c\sqrt{1-\tfrac12} = c/\sqrt2$ , which is the fourth column. So the group velocity of the Klein-Gordon wave packet is the velocity of the relativistic particle, as part 2 of the statement says, and the superluminal phase velocity carries no signal.

The gap. No solution of (1) oscillates more slowly than $\mu$ : the lowest frequency, at $k=0$ , corresponds to $\hbar\omega = Mc^2 = 139.57039$ MeV. For the wave equation ( $\mu=0$ ) there is no such floor, and arbitrarily slow oscillations exist.

Example(The source function solves the equation, and its tail)Original

Verify (4) and watch the massless limit numerically. Put $c = 1$ and $\mu = 3$ , and take the candidate

S(x,t) = \tfrac12 J_0\!\left(3\sqrt{t^2-x^2}\right) \qquad (\lvert x\rvert<t) .

Exact verification. By Step 3 of the proof, $S_{tt}-S_{xx}+9S = 0$ identically inside the cone; symbolic differentiation of $J_0(3\sqrt{t^2-x^2})$ and substitution confirms that the residual is exactly zero. Concretely at $t = 2$ , $x = 1$ , so $s = \sqrt3 = 1.7320508$ and $\mu s = 5.1961524$ :

J_0(5.1961524) = -0.1116107 , \qquad S = -0.0558053 .

The sign is worth noting: the source function is negative there, which cannot happen for the wave equation, where $S \equiv 1/2c$ inside the cone. The Bessel oscillation is the mass term at work.

The tail. Along the ray $x = 0$ the source function is $S(0,t) = \tfrac12J_0(3t)$ , which oscillates and decays like $t^{-1/2}$ :

$t$	$\mu t$	$S(0,t) = \tfrac12 J_0(3t)$
$0.1$	$0.3$	$0.4888131$
$0.5$	$1.5$	$0.2559138$
$1$	$3$	$-0.1300260$
$2$	$6$	$0.0753226$
$5$	$15$	$-0.0071122$

At small $t$ the value is near $1/2c = 0.5$ , the wave-equation answer, because $J_0(z)\approx1-z^2/4$ for small $z$ ; at large $t$ the solution rings. This is the one-dimensional face of the tail inside the light cone that formula (6) exhibits in three dimensions, and it is why Huygens' principle fails for a massive field.

The massless limit, quantitatively. At the fixed point $t = 1$ , $x = 0$ , so $s = 1$ :

$\mu$	$S = \tfrac12 J_0(\mu)$	relative departure from $\tfrac12$
$1$	$0.3825988$	$23.480\%$
$0.3$	$0.4888131$	$2.2374\%$
$0.1$	$0.4987508$	$0.24984\%$
$0.01$	$0.4999875$	$0.0024999\%$

The departure is $\tfrac14\mu^2$ to leading order — $0.25$ , $0.0225$ , $0.0025$ , $0.000025$ against the table's $0.23480$ , $0.022374$ , $0.0024984$ , $0.000024999$ — so the solution approaches the d'Alembert source function quadratically in the mass, as Strauss's remark that (4) converges to the wave equation's source function at $m\to0$ requires.

ApplicationOriginal

Why the range of a force is set by the mass of its carrier. Strip the time dependence from (1) with $\mu>0$ and one gets the modified Helmholtz equation $\nabla^2u - (\mu/c)^2u = 0$ , whose decaying radial solution is $e^{-\mu r/c}/r$ rather than $1/r$ . So a static source of a massive field produces a potential screened at the length $c/\mu = \hbar/Mc$ . This is Yukawa's argument: from the observed range of the nuclear force, about $1.4$ fm, one predicts a carrier of rest energy about $140$ MeV, which is the pion — the number computed in the first example above. The massless limit removes the screening and restores the $1/r$ Coulomb and Newtonian potentials of the fundamental solution of Laplace's equation, which is why electromagnetism and gravity are long-ranged.

Waves in a plasma, and in a waveguide. The relation (3) is not special to particle physics. Transverse electromagnetic waves in a cold plasma obey exactly it, with $\mu$ the plasma frequency $\omega_p$ : below $\omega_p$ no wave propagates and the plasma reflects, which is how shortwave radio reaches over the horizon and why radio below a few megahertz cannot escape the ionosphere. A mode in a hollow waveguide obeys it too, with $\mu = c\kappa$ the cutoff set by the transverse eigenvalue of the Helmholtz equation; the superluminal phase velocity and subluminal group velocity of the table above are measured facts about waveguides. In each case the Klein-Gordon equation is the equation of the field in a medium with a gap, and the "mass" is whatever sets the gap.

Dispersion, and the spreading of pulses. Because $v_{\text{gr}}$ depends on $k$ , a Klein-Gordon pulse spreads, unlike a wave-equation pulse in one dimension. Expanding (3) about a carrier wavenumber $k_0$ gives $\omega \approx \omega_0 + v_{\text{gr}}(k-k_0) + \tfrac12\omega''(k_0)(k-k_0)^2$ , and the quadratic term $\omega''= c^2\mu^2/\omega^3$ is the group-velocity dispersion that broadens the envelope — the same quantity that limits the bit rate of an optical fibre. The exact source function (4) exhibits the extreme case: an impulse does not stay an impulse but develops the ringing tail tabulated above.

The non-relativistic limit in practice. Part 2 of the conserved-density statement is the formal content of the statement that quantum mechanics is the low-energy limit of relativistic field theory, and equation (9) is used as written. Keeping the term $\frac{\hbar^2}{2Mc^2}\psi_{tt}$ to first order and eliminating $\psi_{tt}$ with the leading-order equation produces the relativistic kinetic-energy correction $-p^4/8M^3c^2$ , the first of the fine-structure corrections to atomic energy levels and one of the standard tests of the Dirac theory against the Schrodinger theory of the Schrodinger equation.

Numerics: why the two-sign spectrum matters. A time-stepping scheme for (1) must handle both branches of the dispersion relation, and because $\omega$ grows only linearly in $k$ the stability constraint is the usual Courant condition $c\Delta t \le \Delta x$ of a hyperbolic solver, with the mass term contributing a harmless $O(\mu^2\Delta t^2)$ . The genuinely awkward regime is $\mu\Delta t \gg 1$ , where the carrier oscillation $e^{-i\mu t}$ must be resolved even though nothing physical happens on that timescale; the standard remedy is the substitution of Step 3 — factor out $e^{-i\mu t}$ and integrate (9), whose time scale is the slow one. This is the same device that underlies the derivation of envelope equations such as the nonlinear Schrodinger equation from the nonlinear Klein-Gordon equation that Strauss writes down as his meson model.