Weak solutions

The variational and Sobolev-space approach to PDE.

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Weak formulations

Weak formulations

Definition · advanced

Weak derivatives and Sobolev spaces

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Defines the weak derivative of a locally integrable function, the Sobolev spaces $H^1$ and $H^1_0$ built from it, and the weak formulation of the Poisson problem, and shows that a smooth weak solution is a classical one.

Prerequisites: L^p spaces, Test functions and distributions, Hilbert spaces, Green's identities

Definition(Weak derivative)AdaptedAdapted from Evans (2010), §5.2.1, Eq. (3), p. 242; Stakgold & Holst (2011), §2.6, Eq. (2.6.2), p. 183

Let $\Omega \subseteq \mathbb{R}^n$ be open. Write

L^1_{\mathrm{loc}}(\Omega) = \{\, u : \Omega \to \mathbb{R} \ \text{measurable} \ : \ u \in L^1(A) \ \text{for every open } A \text{ with } \overline{A} \subset \Omega \text{ compact} \,\}

for the locally integrable functions, and $C_c^\infty(\Omega)$ for the smooth functions whose support is a compact subset of $\Omega$ — the test functions of test functions and distributions.

Let $\alpha = (\alpha_1,\dots,\alpha_n)$ be a multi-index, $|\alpha| = \alpha_1 + \dots + \alpha_n$ , and $D^\alpha = \partial_1^{\alpha_1}\cdots\partial_n^{\alpha_n}$ . Given $u \in L^1_{\mathrm{loc}}(\Omega)$ , a function $v \in L^1_{\mathrm{loc}}(\Omega)$ is the $\alpha$ th weak derivative of $u$ , written $D^\alpha u = v$ , if

\int_\Omega u \, D^\alpha \varphi \, dx \;=\; (-1)^{|\alpha|} \int_\Omega v\, \varphi \, dx \qquad \text{for every } \varphi \in C_c^\infty(\Omega) . \tag{W}

If no such $v$ exists, $u$ has no $\alpha$ th weak derivative. The motivation is that (W) is the identity that repeated integration by parts gives when $u \in C^{|\alpha|}(\Omega)$ : the compact support of $\varphi$ kills every boundary term, so the formula survives with the derivatives on $\varphi$ alone, and that side of it asks nothing of $u$ beyond local integrability.

Stakgold and Holst write the same definition, and both they and Evans call the resulting $v$ the weak or distributional derivative. The two words are justified: (W) says exactly that the distribution $D^\alpha T_u$ is the distribution $T_v$ , where $T_w(\varphi) = \int_\Omega w\varphi\,dx$ . The weak derivative is therefore the special case of the distributional derivative in which the answer happens again to be a function. The delta is the standard example where it is not.

Definition(The weak derivative is well defined, and extends the classical one)AdaptedAdapted from Evans (2010), §5.2.1, Lemma (Uniqueness of weak derivatives), p. 242; Stakgold & Holst (2011), §2.6, the remarks following Eq. (2.6.2), p. 183

Let $u \in L^1_{\mathrm{loc}}(\Omega)$ .

Uniqueness. If $v$ and $\tilde v$ both satisfy (W) for the same $\alpha$ , then $v = \tilde v$ almost everywhere. So $D^\alpha u$ is determined as an element of $L^1_{\mathrm{loc}}(\Omega)$ , that is, up to a null set.
Consistency. If $u \in C^m(\Omega)$ and $|\alpha| \le m$ , then the classical $D^\alpha u$ is also the weak one. Hence the notation is unambiguous.

Both are needed before the symbol $D^\alpha u$ may be used at all; Stakgold and Holst record them in the same breath as the definition, and set the first as their Exercise 2.6.1.

Idea of the proofOriginal

Part 2 is the integration by parts that motivated the definition. Part 1 reduces to a single fact: a locally integrable function that is orthogonal to every test function vanishes almost everywhere. That fact is itself proved by smoothing — a test function concentrated near a point reports the local average of $w$ there, and averages at all small scales determine $w$ off a null set.

ProofOriginal

Step 1 (the vanishing lemma). Let $w \in L^1_{\mathrm{loc}}(\Omega)$ satisfy $\int_\Omega w\varphi\,dx = 0$ for every $\varphi \in C_c^\infty(\Omega)$ . Then $w = 0$ almost everywhere.

Fix a mollifier: a function $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\eta \ge 0$ , support in the unit ball and $\int \eta = 1$ , and set $\eta_\varepsilon(x) = \varepsilon^{-n}\eta(x/\varepsilon)$ . Let $A$ be open with $\overline A \subset \Omega$ compact, and let $\varepsilon < \operatorname{dist}(\overline A, \partial\Omega)$ . For $x \in A$ the function $y \mapsto \eta_\varepsilon(x-y)$ lies in $C_c^\infty(\Omega)$ , because its support is the closed ball of radius $\varepsilon$ about $x$ , which is a compact subset of $\Omega$ by the choice of $\varepsilon$ . The hypothesis, applied to that test function, gives

(w * \eta_\varepsilon)(x) \;=\; \int_\Omega w(y)\,\eta_\varepsilon(x-y)\,dy \;=\; 0 \qquad (x \in A).

On the other hand $w \in L^1(A')$ for a slightly larger $A'$ , and mollification converges in $L^1$ : $w * \eta_\varepsilon \to w$ in $L^1(A)$ as $\varepsilon \to 0$ . The limit of the zero function is zero, so $w = 0$ almost everywhere on $A$ . Exhausting $\Omega$ by countably many such $A$ — for instance $A_k = \{x \in \Omega : |x| < k, \ \operatorname{dist}(x,\partial\Omega) > 1/k\}$ — gives $w = 0$ almost everywhere on $\Omega$ .

Step 2 (uniqueness). Subtract the two instances of (W). The left-hand sides are identical, so

\int_\Omega (v - \tilde v)\,\varphi\,dx = 0 \qquad \text{for every } \varphi \in C_c^\infty(\Omega),

after dividing by $(-1)^{|\alpha|}$ . Step 1 with $w = v - \tilde v$ gives $v = \tilde v$ almost everywhere.

Step 3 (consistency). Let $u \in C^m(\Omega)$ , $|\alpha| \le m$ and $\varphi \in C_c^\infty(\Omega)$ . Choose an open $A$ with smooth boundary such that $\operatorname{supp}\varphi \subset A$ and $\overline A \subset \Omega$ compact. On $A$ both $u$ and $\varphi$ are $C^m$ , so one integration by parts in the variable $x_i$ gives

\int_A (\partial_i u)\,\psi \, dx = -\int_A u\,\partial_i\psi\,dx + \oint_{\partial A} u\,\psi\,\nu_i\,dS

for any $C^1$ function $\psi$ , and the boundary integral vanishes whenever $\psi$ vanishes near $\partial A$ . Starting from $\psi = D^{\alpha - e_i}\varphi$ and iterating $|\alpha|$ times — at each stage the function playing the role of $\psi$ is a derivative of $\varphi$ , hence still supported in $\operatorname{supp}\varphi$ , so every boundary term is zero — produces

\int_\Omega u\,D^\alpha\varphi\,dx = (-1)^{|\alpha|}\int_\Omega (D^\alpha u)\,\varphi\,dx ,

which is (W) with $v$ the classical derivative. Both $u$ and $D^\alpha u$ are continuous, hence locally integrable, so $v$ is an admissible candidate, and by Step 2 it is the only one. $\square$

Definition(Sobolev spaces;

H^1

and

H^1_0

)AdaptedAdapted from Evans (2010), §5.2.2, the definitions of $W^{k,p}(U)$, its norm, $H^k(U)$ and $W^{k,p}_0(U)$, pp. 244–245; Stakgold & Holst (2011), §2.6, the definition of $W^{m,p}(\\Omega)$ and of $H^m(\\Omega)$, p. 183

Fix an integer $k \ge 0$ and $1 \le p < \infty$ . The Sobolev space $W^{k,p}(\Omega)$ consists of those $u \in L^1_{\mathrm{loc}}(\Omega)$ whose weak derivative $D^\alpha u$ exists and lies in $L^p(\Omega)$ for every $\alpha$ with $|\alpha| \le k$ ; this includes $\alpha = 0$ , so $u \in L^p(\Omega)$ as well. Its norm is

\|u\|_{W^{k,p}(\Omega)} \;=\; \Bigl( \sum_{|\alpha| \le k} \int_\Omega |D^\alpha u|^p\,dx \Bigr)^{1/p} .

As always in $L^p$ spaces, functions equal almost everywhere are identified. Stakgold and Holst define the same space by requiring $u \in L^p(\Omega)$ and $D^\alpha u \in L^p(\Omega)$ for $0 \le |\alpha| \le m$ , which is the same condition written differently.

For $p = 2$ one writes $H^k(\Omega) = W^{k,2}(\Omega)$ , the letter recording that this one is a Hilbert space. The case of interest here is $k=1$ :

H^1(\Omega) = \{\, u \in L^2(\Omega) : \nabla u \in L^2(\Omega;\mathbb{R}^n) \,\}, \qquad (u,v)_{H^1} = \int_\Omega \bigl( uv + \nabla u \cdot \nabla v \bigr)\,dx ,

where $\nabla u = (D^{e_1}u,\dots,D^{e_n}u)$ is the vector of first weak derivatives, so that $\|u\|_{H^1}^2 = \|u\|_{L^2}^2 + \|\nabla u\|_{L^2}^2$ . The second piece alone,

|u|_{H^1(\Omega)} = \|\nabla u\|_{L^2(\Omega)} ,

is the Dirichlet seminorm; it is only a seminorm, since it vanishes on constants.

Finally, $H^1_0(\Omega)$ is the closure of $C_c^\infty(\Omega)$ in the norm of $H^1(\Omega)$ : a function $u$ lies in $H^1_0(\Omega)$ exactly when there are $\varphi_m \in C_c^\infty(\Omega)$ with $\|u - \varphi_m\|_{H^1} \to 0$ . Evans defines $W^{k,p}_0(U)$ this way for all $k,p$ and notes that it is to be read as the condition " $D^\alpha u = 0$ on $\partial U$ for $|\alpha| \le k-1$ ", a reading made precise only by the theory of traces, which is not developed here. For our purposes $H^1_0(\Omega)$ is the definition of " $u \in H^1$ and $u = 0$ on $\partial\Omega$ "; no pointwise meaning is attached to the boundary values of an $L^2$ function.

Definition(

H^1(\Omega)

and

H^1_0(\Omega)

are Hilbert spaces)AdaptedAdapted from Evans (2010), §5.2.2, Remark (i), p. 244; Stakgold & Holst (2011), §2.6, the completeness of $W^{m,p}(\\Omega)$ and the Hilbert structure of $H^m(\\Omega)$, p. 183

$(u,v)_{H^1}$ is an inner product on $H^1(\Omega)$ , and $H^1(\Omega)$ is complete in the induced norm. Hence $H^1(\Omega)$ is a Hilbert space, and so is the closed subspace $H^1_0(\Omega)$ .

Idea of the proofOriginal

Completeness is inherited from $L^2$ , one derivative at a time. A Cauchy sequence in $H^1$ is Cauchy in $L^2$ in each of its $n+1$ slots, so each slot has a limit; the only thing to check is that the $n$ limits of the derivatives really are the derivatives of the limit, and that is read off by passing to the limit in the defining identity (W), whose two sides are continuous in the $L^2$ norm.

ProofAdaptedAdapted from Evans (2010), §5.2.3, Thm 2 and its proof, pp. 248–249

Bilinearity and symmetry are immediate, and $(u,u)_{H^1} = \|u\|_{L^2}^2 + \|\nabla u\|^2_{L^2} = 0$ forces $u = 0$ in $L^2$ , so the form is an inner product.

For completeness, let $(u_m)$ be Cauchy in $H^1(\Omega)$ . Then $(u_m)$ and each $(\partial_i u_m)$ are Cauchy in $L^2(\Omega)$ , because each of those norms is dominated by the $H^1$ norm. By the completeness of $L^2$ (the Riesz–Fischer theorem of $L^p$ spaces) there are $u, v_1,\dots,v_n \in L^2(\Omega)$ with $u_m \to u$ and $\partial_i u_m \to v_i$ in $L^2(\Omega)$ .

It remains to identify $v_i$ as the weak derivative of $u$ . Fix $\varphi \in C_c^\infty(\Omega)$ . Each $u_m$ lies in $H^1(\Omega)$ , so

\int_\Omega u_m\,\partial_i\varphi\,dx = -\int_\Omega (\partial_i u_m)\,\varphi\,dx .

Both sides converge: by the Cauchy–Schwarz inequality, $\bigl|\int (u_m - u)\partial_i\varphi\bigr| \le \|u_m-u\|_{L^2}\|\partial_i\varphi\|_{L^2} \to 0$ , and similarly on the right with $\varphi$ in place of $\partial_i\varphi$ . (Here $\varphi$ and $\partial_i\varphi$ are in $L^2(\Omega)$ because they are bounded with compact support.) In the limit,

\int_\Omega u\,\partial_i\varphi\,dx = -\int_\Omega v_i\,\varphi\,dx ,

which is (W) for $\alpha = e_i$ . So $\partial_i u = v_i \in L^2(\Omega)$ , hence $u \in H^1(\Omega)$ and $\|u_m - u\|_{H^1} \to 0$ . (This is the route of Evans's proof that $W^{k,p}$ is complete, specialised to $k=1$ , $p=2$ and written out for the identification step.)

A closed subspace of a complete space is complete, and $H^1_0(\Omega)$ is closed by construction, being a closure. $\square$

Definition(Weak formulation of the Poisson problem)AdaptedAdapted from Evans (2010), §6.1.2, Eq. (7) and the Definitions (8)–(9), pp. 295–296

Let $\Omega \subseteq \mathbb{R}^n$ be open and bounded and let $f \in L^2(\Omega)$ . Define the bilinear form and the linear functional

a(u,v) = \int_\Omega \nabla u \cdot \nabla v \, dx , \qquad F(v) = \int_\Omega f v \, dx \qquad \bigl(u,v \in H^1_0(\Omega)\bigr) .

A function $u$ is a weak solution of the Dirichlet problem

-\Delta u = f \ \text{ in } \Omega, \qquad u = 0 \ \text{ on } \partial\Omega \tag{P}

if $u \in H^1_0(\Omega)$ and

a(u,v) = F(v) \qquad \text{for every } v \in H^1_0(\Omega) . \tag{P$'$}

Three remarks on the formulation.

Both integrals in (P $'$ ) converge absolutely for $u,v \in H^1_0(\Omega)$ and $f \in L^2(\Omega)$ , by the Cauchy–Schwarz inequality; so the definition asks nothing that is not finite.
It is enough to test against $v \in C_c^\infty(\Omega)$ : that set is dense in $H^1_0(\Omega)$ by definition, and both sides of (P $'$ ) are continuous in $v$ for the $H^1$ norm. For $u \in H^1(\Omega)$ the resulting identity $\int\nabla u\cdot\nabla\varphi = \int f\varphi$ says precisely that $-\Delta u = f$ in the sense of distributions, which is the formulation Stakgold and Holst take as primary.
The membership $u \in H^1_0(\Omega)$ carries the boundary condition, and (P $'$ ) carries the equation. Evans makes exactly this split, and chooses $H^1_0$ for that reason.

This is Evans's definition with $a^{ij} = \delta^{ij}$ and $b^i = c = 0$ ; the general second-order divergence-form operator is treated in the Lax–Milgram theorem and the Dirichlet problem, which is also where (P $'$ ) is shown to have exactly one solution.

Definition(Weak and classical solutions agree where both make sense)AdaptedAdapted from Stakgold & Holst (2011), §2.5, the Theorem on p. 172 and its proof; Evans (2010), §6.1.2, the motivation preceding Eq. (7), p. 295

Let $\Omega$ be open and bounded and let $f \in L^2(\Omega)$ .

If $u \in C^2(\Omega) \cap H^1_0(\Omega)$ satisfies $-\Delta u = f$ at every point of $\Omega$ , then $u$ is a weak solution of (P).
Conversely, if $f$ is in addition continuous on $\Omega$ , and if $u$ is a weak solution of (P) with $u \in C^2(\Omega)$ , then $-\Delta u = f$ at every point of $\Omega$ .

Only what the proof uses is assumed. Both halves are interior arguments that never approach $\partial\Omega$ , so nothing is asked of the boundary and no pointwise boundary values appear: the boundary condition enters solely through the membership $u \in H^1_0(\Omega)$ , which is how (P $'$ ) carries it. Continuity of $f$ is needed only in part 2, where a pointwise equation has to be recovered from an integral one.

Stakgold and Holst prove both halves for a general linear operator $L$ of order $p$ with smooth coefficients and a continuous right-hand side: a classical solution is weak, and a weak solution with $p$ continuous derivatives is classical. The statement above is that theorem for $L = -\Delta$ , with their hypothesis " $u$ has $p$ continuous derivatives" read as $u \in C^2(\Omega)$ , and with the continuity of the right-hand side dropped from part 1, which does not use it; (P $'$ ) asks only $f \in L^2(\Omega)$ . The caveat in part 1 — that $u$ must be assumed to lie in $H^1_0(\Omega)$ , rather than merely to vanish on $\partial\Omega$ — is also ours; the distributional formulation that Stakgold and Holst work with does not mention the space, whereas the formulation (P $'$ ) does.

Idea of the proofOriginal

Both directions are the same integration by parts, read in opposite senses. Forwards, multiply the equation by a test function and move one derivative across; the boundary term dies because the test function has compact support. Backwards, move the derivative back, which is allowed because $u$ is now assumed twice differentiable; what is left is the statement that the continuous function $-\Delta u - f$ integrates to zero against every test function, and the only continuous function that does that is zero — otherwise, put a bump where it is positive.

ProofAdaptedAdapted from Stakgold & Holst (2011), §2.5, proof of the Theorem, p. 172

Part 1. Let $\varphi \in C_c^\infty(\Omega)$ and let $A$ be open with $\operatorname{supp}\varphi \subset A$ and $\overline A \subset \Omega$ compact. On $A$ the function $u$ is $C^2$ , so Green's first identity (Green's identities) gives

\int_A \nabla u \cdot \nabla \varphi\,dx = -\int_A (\Delta u)\,\varphi\,dx + \oint_{\partial A} \varphi\,\frac{\partial u}{\partial n}\,dS ,

and the boundary integral vanishes because $\varphi$ is identically zero near $\partial A$ . Since $-\Delta u = f$ pointwise and both integrands vanish off $\operatorname{supp}\varphi$ , this reads $a(u,\varphi) = F(\varphi)$ . Both $a(u,\cdot)$ and $F$ are bounded linear functionals on $H^1_0(\Omega)$ — again by Cauchy–Schwarz, $|a(u,v)| \le \|\nabla u\|_{L^2}\|v\|_{H^1}$ and $|F(v)| \le \|f\|_{L^2}\|v\|_{H^1}$ — and they agree on the dense set $C_c^\infty(\Omega)$ , so they agree on all of $H^1_0(\Omega)$ . With $u \in H^1_0(\Omega)$ by hypothesis, $u$ is a weak solution.

Part 2. Let $u$ be a weak solution lying in $C^2(\Omega)$ , and let $\varphi \in C_c^\infty(\Omega)$ . Running the display above in reverse — the identity holds for any $C^2$ function on $A$ , whatever equation it satisfies — turns $a(u,\varphi) = F(\varphi)$ into

\int_\Omega \bigl(-\Delta u - f\bigr)\,\varphi\,dx = 0 \qquad \text{for every } \varphi \in C_c^\infty(\Omega) .

Put $q = -\Delta u - f$ , a continuous function on $\Omega$ . Suppose $q(x_0) \neq 0$ for some $x_0 \in \Omega$ ; replacing $q$ by $-q$ if necessary, assume $q(x_0) > 0$ . By continuity there is a ball $B(x_0,r)$ with $\overline{B(x_0,r)} \subset \Omega$ on which $q > q(x_0)/2 > 0$ . Take

\varphi(x) = \begin{cases} \exp\!\bigl(-\bigl(r^2 - |x-x_0|^2\bigr)^{-1}\bigr), & |x - x_0| < r,\\[2pt] 0, & \text{otherwise,}\end{cases}

the standard bump: it is smooth, supported in $\overline{B(x_0,r)}$ , and strictly positive on $B(x_0,r)$ . Then $\int q\varphi\,dx \ge \tfrac12 q(x_0)\int_{B(x_0,r)}\varphi\,dx > 0$ , contradicting the display. Hence $q \equiv 0$ , that is, $-\Delta u = f$ throughout $\Omega$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

Local integrability of $v$ in (W). Without it, the right-hand side of (W) need not be defined, and the definition would silently become the distributional one. Many ordinary functions have no weak derivative at all: the Heaviside step on $\Omega = (-1,1)$ satisfies $\int H\varphi' = -\varphi(0)$ , so (W) would force $\int_{-1}^{1} v\varphi\,dx = \varphi(0)$ for every $\varphi$ , which no locally integrable $v$ can do — take bumps shrinking to the origin with $\varphi(0)=1$ and $0\le\varphi\le1$ , so that $\int v\varphi \to 0$ by dominated convergence while $\varphi(0)$ stays at $1$ . Its distributional derivative exists and is $\delta$ . So $H \notin W^{1,p}$ for any $p$ , and the jump is exactly the obstruction.
Compact support of the test functions. This is what kills every boundary term in Step 3 of the first proof, and it is why (W) says nothing whatever about the behaviour of $u$ near $\partial\Omega$ . The boundary condition must therefore be imposed separately, and $H^1_0(\Omega)$ is how it is done.
$C^2$ regularity in Part 2 above. Only with two classical derivatives may the integration by parts be run backwards; without them the symbol $\Delta u(x)$ has no meaning and the conclusion cannot even be stated. The hypothesis is a genuine restriction and not a formality, because a weak solution need not have two derivatives. On $\Omega = (-1,1)$ take $f = \operatorname{sgn}$ , which lies in $L^2(\Omega)$ , and

u(x) = \tfrac12\,x\bigl(1-|x|\bigr), \qquad u'(x) = \tfrac12\bigl(1-2|x|\bigr) .

This $u$ is Lipschitz and vanishes at $\pm1$ , so its truncations $(u-\delta)^+ - (-u-\delta)^+$ have compact support in $\Omega$ and converge to it in $H^1$ , putting $u \in H^1_0(\Omega)$ ; and $u'' = -\operatorname{sgn}$ away from the origin. Integrating by parts separately on $(-1,0)$ and $(0,1)$ therefore gives $\int_{-1}^1 u'\varphi'\,dx = \int_{-1}^1 f\varphi\,dx$ for every $\varphi \in C_c^\infty(\Omega)$ : the two boundary terms at the origin are $\pm\tfrac12\varphi(0)$ and cancel, because $u'$ is continuous there. By the density remark under (P $'$ ), $u$ is a weak solution of (P). But $u'$ has a corner at $0$ , so $u''(0)$ does not exist and there is no pointwise equation at the origin. Such a witness must also have a discontinuous $f$ : in one dimension a weak solution with $f$ continuous is automatically $C^2$ , since then $u'$ has weak derivative $-f$ and so agrees almost everywhere with the $C^1$ function $x \mapsto c - \int_0^x f$ . That a weak solution of (P) is in fact smooth when $f$ and $\partial\Omega$ are — elliptic regularity — is a separate theorem, and it is false for hyperbolic equations, which is the deeper reason the weak theory is worth having.

Boundedness of $\Omega$ . None of the arguments above uses it; it stands in the weak formulation only because that is where it starts to matter. In particular it is not needed for the inclusion $L^2(\Omega)\subset L^1_{\mathrm{loc}}(\Omega)$ , which holds for every open $\Omega$ , by Cauchy–Schwarz on compactly contained sets, these having finite measure. Boundedness is essential in the Lax–Milgram theorem and the Dirichlet problem: the Poincaré inequality, and with it the coercivity of $a$ , needs $\Omega$ to be confined in at least one direction. On $\Omega = \mathbb{R}^n$ the form $a$ is not coercive on $H^1$ , and (P $'$ ) with $f=0$ has only the zero solution for a different reason.
Weak derivatives are not local limits. A function can lie in $H^1$ and still be unbounded on every open set, as Evans shows by superposing the singularities $2^{-k}|x - r_k|^{-\alpha}$ over a sequence $(r_k)$ dense in the unit ball (his Example 4, p. 246). His construction needs $0 < \alpha < (n-p)/p$ , so for $p = 2$ it is available only in dimension $n \ge 3$ . Membership of a Sobolev space buys integrability of derivatives, not pointwise good behaviour.

Method(Putting a boundary-value problem into weak form)AdaptedAdapted from Evans (2010), §6.1.2, the motivation for the definition of a weak solution, p. 295

Multiply and integrate. Multiply the equation by a test function $v$ and integrate over $\Omega$ .
Integrate by parts to move derivatives off the unknown and onto $v$ , usually until both carry the same number. For a second-order equation in divergence form, $-\nabla\cdot(a\nabla u)$ becomes $\int a\nabla u\cdot\nabla v$ minus a boundary integral.
Sort the boundary data. A condition that appears in the surviving boundary integral (a prescribed flux, a Neumann or Robin condition) is natural: substitute it and keep the integral. A condition on $u$ itself (Dirichlet) is essential: it cannot be substituted, and must instead be built into the space of competitors.
Choose the space $V$ . Take the largest space in which every remaining integral is finite — for a second-order problem, $H^1(\Omega)$ — and cut it down by the essential conditions. Homogeneous Dirichlet data on all of $\partial\Omega$ give $V = H^1_0(\Omega)$ .
Name the pieces. Write the result as $a(u,v) = F(v)$ for all $v \in V$ , with $a$ bilinear and $F$ linear.
Check the two bounds. Verify $|a(u,v)| \le M\|u\|_V\|v\|_V$ and $|F(v)| \le L\|v\|_V$ . Together with coercivity, these are exactly the hypotheses of the Lax–Milgram theorem, which then supplies a unique solution.
Inhomogeneous Dirichlet data are reduced to homogeneous ones: find any $w \in H^1(\Omega)$ taking the prescribed boundary values and solve for $u - w \in H^1_0(\Omega)$ , moving $-\Delta w$ to the right-hand side. Evans describes this reduction on p. 297.

Note(Natural boundary conditions)AdaptedAdapted from Stakgold & Holst (2011), §10.4, the weak form (10.4.4)–(10.4.9) of the mixed problem, pp. 711–712; Evans (2010), §6.6, Problem 4, p. 344

Step 3 explains an asymmetry that often puzzles a first reader. Stakgold and Holst carry it out for the mixed problem $-\nabla\cdot(a\nabla u) + bu = f$ with $u = g_D$ on $\Gamma_D$ and $(a\nabla u)\cdot\nu + cu = g_N$ on the complementary piece $\Gamma_N$ . After the integration by parts the surface integral $\int_{\partial\Omega} v\,(a\nabla u)\cdot\nu\,dS$ survives; the competitors vanish on $\Gamma_D$ , which deletes that part of it, and on the rest the boundary condition is substituted, sending $\int_{\Gamma_N} g_N v\,dS$ to the right-hand side and $\int_{\Gamma_N} c\,uv\,dS$ into the form. Nothing whatever is imposed on the competitors over $\Gamma_N$ .

For (P) with Dirichlet data replaced by Neumann data — $a \equiv 1$ , $b = c = 0$ , $\Gamma_D$ empty, $\partial u/\partial n = g$ on all of $\partial\Omega$ — this reads

\int_\Omega \nabla u\cdot\nabla v\,dx = \int_\Omega fv\,dx + \oint_{\partial\Omega} g v\,dS \qquad \text{for all } v \in H^1(\Omega),

with the whole of $H^1(\Omega)$ as the space: the boundary data have been absorbed into the right-hand side. Evans poses the homogeneous case $g = 0$ of exactly this formulation, with the same test space $H^1(U)$ , as his Problem 4 (p. 344). The price is visible in the display: taking $v \equiv 1$ gives the compatibility condition $\int_\Omega f\,dx + \oint_{\partial\Omega} g\,dS = 0$ , necessary for a solution to exist; and a solution is never unique, since adding a constant to $u$ changes neither side of the display. The form is accordingly not coercive on $H^1(\Omega)$ — constants have zero Dirichlet energy — so the Neumann problem is not merely the Dirichlet problem with a larger space; it is the absence of an essential condition, not the surface term, that makes it harder. This absorption of the data is the same phenomenon as the natural boundary conditions of the calculus of variations (natural boundary conditions) — and it is the same calculation, since $a(u,v) = F(v)$ is the first variation of a quadratic functional.

Example(A corner singularity: in

H^1

, not in

H^2

)Original

Let $B$ be the open unit disc in $\mathbb{R}^2$ , and let $S = \{(x_1,0) : 0 \le x_1 \le 1\}$ be the radial slit along the positive $x_1$ -axis. Using polar coordinates with the angle taken in the range $0 < \theta < 2\pi$ (so that $\theta$ is smooth on $B \setminus S$ ), define

u(r,\theta) = r^{1/2}\,\sin\frac{\theta}{2} .

Equivalently $u = \operatorname{Im}\sqrt{z}$ , with the branch cut taken along $S$ . This is the model singularity of a crack tip or a re-entrant corner.

It is continuous, and vanishes on the slit. As $\theta \downarrow 0$ and as $\theta \uparrow 2\pi$ , $\sin(\theta/2) \to \sin 0 = 0$ and $\sin \pi = 0$ respectively, so $u$ extends continuously by $0$ across $S$ and by $0$ at the origin. It is smooth on $B \setminus S$ .
Its classical gradient. In the orthonormal polar frame,

\nabla u = \frac{\partial u}{\partial r}\,\hat{\mathbf{e}}_r + \frac1r\frac{\partial u}{\partial \theta}\,\hat{\mathbf{e}}_\theta = \frac{1}{2}r^{-1/2}\Bigl(\sin\tfrac\theta2\,\hat{\mathbf{e}}_r + \cos\tfrac\theta2\,\hat{\mathbf{e}}_\theta\Bigr), \qquad\text{so}\qquad |\nabla u|^2 = \frac{1}{4r} .

That gradient is the weak gradient on all of $B$ . Call the field in step 2 $g$ ; it is defined off $S \cup\{0\}$ , a null set. Let $\varphi \in C_c^\infty(B)$ and write $\Omega_\varepsilon = B \setminus (\overline{B(0,\varepsilon)} \cup S)$ , on which $u$ is smooth up to the boundary. The divergence theorem gives

\int_{\Omega_\varepsilon}\bigl(u\,\partial_i\varphi + g_i\,\varphi\bigr)dx = \oint_{\partial\Omega_\varepsilon} u\,\varphi\,\nu_i\,dS .

The boundary $\partial\Omega_\varepsilon$ has three parts. On the outer circle $\varphi = 0$ . On the two faces of the slit $u = 0$ , by step 1 — this is the crucial point, and it is why no jump term appears. On the circle $|x| = \varepsilon$ , the integrand is bounded by $\varepsilon^{1/2}\|\varphi\|_\infty$ and the length is $2\pi\varepsilon$ , so that piece is $O(\varepsilon^{3/2}) \to 0$ . Meanwhile $u \in L^2(B)$ and $g \in L^2(B;\mathbb{R}^2)$ (step 4), so the integrals over $B\setminus\Omega_\varepsilon$ tend to zero. Letting $\varepsilon \to 0$ gives (W) for $\alpha = e_i$ : the weak gradient of $u$ on $B$ is $g$ .

The norms, in closed form. With $dx = r\,dr\,d\theta$ ,

\int_B u^2\,dx = \int_0^{2\pi}\!\!\int_0^1 r\sin^2\tfrac\theta2 \cdot r\,dr\,d\theta = \frac13\cdot\pi = \frac{\pi}{3}, \qquad \int_B |\nabla u|^2 dx = \int_0^{2\pi}\!\!\int_0^1 \frac{1}{4r}\cdot r\,dr\,d\theta = \frac{\pi}{2} .

The second integral is the one worth pausing over: the integrand blows up like $1/r$ , but the area element contributes a factor $r$ , and $\int_0^1 dr$ converges. Hence

\|u\|_{H^1(B)}^2 = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6} = 2.61799\ldots, \qquad \|u\|_{H^1(B)} = 1.618022\ldots

(The second number agrees with the golden ratio $1.618034\ldots$ to five figures and no further; the coincidence is empty.)

But $u \notin H^2(B)$ . Writing $u = \operatorname{Im} f$ with $f(z) = z^{1/2}$ , the Cauchy–Riemann equations give $\partial_1 u = \operatorname{Im}f'$ and $\partial_1\partial_1 u = \operatorname{Im}f''$ . Since $f''(z) = -\tfrac14 z^{-3/2}$ ,

\partial_1\partial_1 u = \tfrac14 r^{-3/2}\sin\tfrac{3\theta}{2}, \qquad \int_B (\partial_1\partial_1 u)^2 dx = \frac{1}{16}\int_0^{2\pi}\!\!\sin^2\tfrac{3\theta}{2}\,d\theta \int_0^1 r^{-3}\,r\,dr = \frac{\pi}{16}\int_0^1 \frac{dr}{r^2} = \infty .

One power of $r$ has been lost and the integral now diverges. So $u$ has first weak derivatives in $L^2$ and no second ones.

What equation it solves. $u$ is harmonic on $B\setminus S$ , being the imaginary part of a holomorphic function there, so it is a weak solution of Laplace's equation on the slit disc. On the whole disc it is not: repeating the computation of step 3 with $\nabla u\cdot\nabla\varphi$ in place of $u\,\partial_i\varphi$ , the two faces of the slit now contribute, because the outward normal derivative equals $-\tfrac12 r^{-1/2}$ on each face, and

\int_B \nabla u\cdot\nabla\varphi\,dx = -\int_0^1 r^{-1/2}\,\varphi(r,0)\,dr \neq 0 \quad\text{in general}.

So $-\Delta u$ , as a distribution on $B$ , is a layer supported on the slit. Changing the domain changed the problem — which is the content of step 3 of the method above.

The moral for computation: the solution of a Dirichlet problem on a domain with a re-entrant corner of interior angle $\omega$ behaves like $r^{\pi/\omega}$ there, here $\omega = 2\pi$ and the exponent $\tfrac12$ . It is in the energy space, so the weak theory applies without apology, but a finite element method that assumes extra smoothness will converge slowly unless the mesh is refined at the corner.

ApplicationOriginal

The finite element method. Equation (P $'$ ) is the starting point of every finite element code. One replaces $H^1_0(\Omega)$ by a finite-dimensional subspace $V_h$ — typically continuous functions that are polynomials on each triangle of a mesh, which lie in $H^1$ precisely because continuity across the faces is enough to prevent a jump term in the weak derivative, exactly as in step 3 of the example — and solves $a(u_h,v_h) = F(v_h)$ for all $v_h \in V_h$ . That is a linear system in the coefficients of $u_h$ , with a sparse, symmetric, positive definite matrix. See the finite element method and, for the variational ancestor of the same idea, the Rayleigh–Ritz method.

Why engineering problems are posed weakly in the first place. A composite material has a discontinuous conductivity or permittivity; across the interface $u$ is continuous but $\nabla u$ is not, so no classical solution exists even for perfectly smooth data. The weak formulation $\int a\nabla u\cdot\nabla v = \int fv$ never differentiates $a$ , accepts $a \in L^\infty$ , and recovers the interface conditions — continuity of $u$ and of the normal component of $a\nabla u$ — as consequences rather than hypotheses.

Energy. For the Poisson problem $a(u,u)$ is twice the Dirichlet energy, and $H^1$ is the space on which that energy is finite. That is not a coincidence of notation: the conserved quantity of energy and uniqueness for the wave equation is the squared $H^1$ -type norm of the solution, and the estimates of energy estimates and well-posedness are statements about curves in these spaces. Physically, admitting every finite-energy configuration — rather than only the smooth ones — is what makes the variational principle have a minimiser.

Quantum mechanics. The expectation of the kinetic energy of a state $\psi$ is $\tfrac{\hbar^2}{2m}\int|\nabla\psi|^2$ , finite exactly when $\psi \in H^1(\mathbb{R}^3)$ . This is the form domain of the Hamiltonian, the natural home for the variational estimates of ground-state energies, and the reason that Sobolev inequalities control how singular an attractive potential may be before a system collapses (the framework of quantum mechanics).

Theorem · advanced

The Lax-Milgram theorem and the Dirichlet problem

Adapted

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Proves that a bounded coercive bilinear form on a Hilbert space represents every bounded linear functional exactly once, proves the Poincare inequality that makes the Dirichlet form coercive, and combines the two into existence, uniqueness and an a priori bound for weak solutions of second-order elliptic Dirichlet problems.

Prerequisites: Weak derivatives and Sobolev spaces, Hilbert spaces, The Riesz representation theorem, The contraction mapping theorem

Definition(Bounded and coercive forms)AdaptedAdapted from Stakgold & Holst (2011), §8.5, the hypotheses (1)–(3) in the proof of Thm 8.5.2, p. 551; Evans (2010), §6.2.1, Thm 1 (i)–(ii), p. 297

Throughout, $H$ denotes a Hilbert space over $\mathbb{R}$ , its inner product written $(\cdot,\cdot)$ and the norm it induces $\|\cdot\|$ . Let $a : H \times H \to \mathbb{R}$ be bilinear. Say that

$a$ is bounded if there is $M > 0$ with $|a(u,v)| \le M\|u\|\,\|v\|$ for all $u,v \in H$ ;
$a$ is coercive if there is $m > 0$ with $a(u,u) \ge m\|u\|^2$ for all $u \in H$ .

A linear $F : H \to \mathbb{R}$ is bounded if $|F(v)| \le L\|v\|$ for some $L>0$ and all $v$ . No symmetry is required of $a$ .

Boundedness is continuity: it says the form cannot grow faster than the product of the norms. Coercivity is the quantitative statement that $a(u,u)$ controls $\|u\|^2$ from below, so that the form is "uniformly positive definite" and not merely positive. Evans writes the two conditions with the constants $\alpha$ and $\beta$ ; Stakgold and Holst list the same three inequalities as (1)–(3) of their proof, writing them without modulus signs on the left.

Theorem(Poincaré inequality)AdaptedAdapted from Stakgold & Holst (2011), §10.3, Thm 10.3.5, p. 704, and Thm 10.3.6, p. 705

Let $\Omega \subseteq \mathbb{R}^n$ be a domain of finite width: $\Omega$ lies between two parallel hyperplanes. Let $\Gamma \subseteq \partial\Omega$ have non-zero $(n-1)$ -dimensional measure, possibly $\Gamma = \partial\Omega$ , and put $W^{1,p}_{0,\Gamma}(\Omega) = \{u \in W^{1,p}(\Omega) : u = 0 \text{ on } \Gamma\}$ , where the boundary condition is read in the sense of traces: for a Lipschitz domain, Stakgold and Holst's Trace Theorem 10.3.5 (p. 704) supplies a bounded linear trace operator $T$ , written $Tu = u|_\Gamma$ , carrying $W^{1,p}(\Omega)$ into $W^{1-1/p,p}(\Gamma)$ with $\|Tu\|_{W^{1-1/p,p}(\Gamma)} \le C_{\mathrm{tr}}\|u\|_{W^{1,p}(\Omega)}$ , and " $u=0$ on $\Gamma$ " means $Tu=0$ . It is not a pointwise statement; weak derivatives and Sobolev spaces attaches no pointwise meaning to the boundary values of an $L^2$ function, and none is attached here. Then for $1 \le p < \infty$ there is a constant $\rho = \rho(\Omega,p,\Gamma)$ with

\|u\|_{L^p(\Omega)} \;\le\; \rho\,|u|_{W^{1,p}(\Omega)} \qquad \text{for every } u \in W^{1,p}_{0,\Gamma}(\Omega),

where $|u|_{W^{1,p}} = \bigl(\sum_{|\alpha|=1}\|D^\alpha u\|^p_{L^p}\bigr)^{1/p}$ is the first-order seminorm. Stakgold and Holst refer the proof to Adams's monograph; this is the form of the inequality they use, and it is the one followed here.

In particular the seminorm $|\cdot|_{W^{1,p}}$ , which vanishes on constants and so is not a norm on $W^{1,p}(\Omega)$ , is a norm on $W^{1,p}_{0,\Gamma}(\Omega)$ , and is equivalent there to the full Sobolev norm. For $p=2$ and $\Gamma = \partial\Omega$ : on $H^1_0(\Omega)$ the Dirichlet seminorm $|u|_{H^1} = \|\nabla u\|_{L^2}$ and the norm $\|u\|_{H^1}$ are equivalent. That is the only case used below, and for it the trace operator can be dispensed with altogether: the lemma that follows is stated and proved on $H^1_0(\Omega)$ , which is defined as a closure of $C_c^\infty(\Omega)$ and so needs no boundary values at all.

Lemma(The case needed here, with an explicit constant)Original

Let $\Omega \subseteq \mathbb{R}^n$ be open and contained in a slab of width $d$ , say $\Omega \subseteq \{x : \beta < x_1 < \beta + d\}$ . Then

\|u\|_{L^2(\Omega)} \;\le\; d\,\|\partial_1 u\|_{L^2(\Omega)} \;\le\; d\,|u|_{H^1(\Omega)} \qquad \text{for every } u \in H^1_0(\Omega) .

So the constant may be taken to be the width of the slab.

Idea of the proofOriginal

A function that vanishes at one wall of the slab is the integral of its own derivative along the short direction, so its size at a point is controlled by the derivative over a segment of length at most $d$ . Cauchy–Schwarz converts that into a bound on $|u|^2$ by $d$ times the integral of $|\partial_1u|^2$ along the segment, and integrating over the slab costs one more factor of $d$ . The result is proved for test functions, where the fundamental theorem of calculus is available, and then carried to all of $H^1_0$ by density.

ProofOriginal

Step 1 (test functions). Let $\varphi \in C_c^\infty(\Omega)$ and extend it by zero to the whole slab; the extension is smooth, since $\varphi$ vanishes near $\partial\Omega$ . Write $x = (x_1, x')$ . For each $x_1 \in (\beta,\beta+d)$ , the fundamental theorem of calculus along the $x_1$ -line gives

\varphi(x_1,x') = \int_\beta^{x_1} \partial_1\varphi(t,x')\,dt ,

because $\varphi$ vanishes at $t = \beta$ (it is zero outside a compact subset of $\Omega$ , and $\Omega$ does not reach that hyperplane).

Step 2 (Cauchy–Schwarz). Apply the Cauchy–Schwarz inequality to that integral, enlarging the range of $t$ to the whole slab:

|\varphi(x_1,x')|^2 \;\le\; (x_1 - \beta)\int_\beta^{\beta+d}|\partial_1\varphi(t,x')|^2\,dt \;\le\; d\int_\beta^{\beta+d}|\partial_1\varphi(t,x')|^2\,dt .

The right-hand side no longer depends on $x_1$ . The finite width is used twice in all: here, to bound $x_1-\beta$ by $d$ , and again in Step 3, where the $x_1$ -integration runs over an interval of length $d$ .

Step 3 (integrate). Integrate in $x_1$ over the interval of length $d$ , which multiplies the right-hand side by $d$ , and then in $x'$ over $\mathbb{R}^{n-1}$ :

\int_\Omega |\varphi|^2\,dx \;\le\; d^2 \int_\Omega |\partial_1\varphi|^2\,dx .

Step 4 (density). Let $u \in H^1_0(\Omega)$ and take $\varphi_k \in C_c^\infty(\Omega)$ with $\varphi_k \to u$ in $H^1(\Omega)$ , which exist by the definition of $H^1_0$ in weak derivatives and Sobolev spaces. Then $\|\varphi_k\|_{L^2} \to \|u\|_{L^2}$ and $\|\partial_1\varphi_k\|_{L^2} \to \|\partial_1 u\|_{L^2}$ , so the inequality passes to the limit. Finally $\|\partial_1 u\|_{L^2} \le \|\nabla u\|_{L^2} = |u|_{H^1}$ . $\square$

Remark(The two sources state Poincaré differently)AdaptedAdapted from Stakgold & Holst (2011), §10.3, Thm 10.3.6, p. 705; Evans (2010), §5.6.1, Thm 3, p. 264

Evans states the inequality for $U$ bounded and $u \in W^{1,p}_0(U)$ , as $\|u\|_{L^p(U)} \le C\|Du\|_{L^p(U)}$ , and notes the same consequence: on $W^{1,p}_0(U)$ the norms $\|Du\|_{L^p}$ and $\|u\|_{W^{1,p}}$ are equivalent. Stakgold and Holst assume only finite width — an infinite strip or slab qualifies — and allow the function to vanish on part of the boundary only. Neither statement contains the other verbatim, and neither is wrong: they are hypotheses chosen for different later uses. This unit follows Stakgold and Holst, because the elliptic theorem below is theirs and uses their constant $\rho$ ; the lemma proved above is the special case $p=2$ , $\Gamma = \partial\Omega$ , with $\rho = d$ made explicit, which is what the worked example needs.

The width bound is not sharp. On $\Omega = (0,\pi)$ the lemma gives $\rho = \pi$ , while the best constant is $1$ : expanding $u \in H^1_0(0,\pi)$ in the sine basis, $u = \sum_{k\ge1} b_k\sin kx$ , gives $\|u\|_{L^2}^2 = \tfrac\pi2\sum b_k^2$ and $\|u'\|_{L^2}^2 = \tfrac\pi2\sum k^2b_k^2 \ge \|u\|_{L^2}^2$ , with equality for $u = \sin x$ . In general the best constant is $\lambda_1^{-1/2}$ , where $\lambda_1$ is the lowest Dirichlet eigenvalue of $-\Delta$ on $\Omega$ .

Theorem(Lax–Milgram theorem)AdaptedAdapted from Stakgold & Holst (2011), §8.5, Thm 8.5.2, p. 551

Take $H$ as above. Assume the bilinear form $a$ meets both conditions of the definition, with constants $M$ (boundedness) and $m$ (coercivity), and that the linear functional $F$ obeys $|F(v)| \le L\|v\|$ . Then exactly one vector of $H$ solves the variational problem

u \in H, \qquad a(u,v) = F(v) \quad\text{for every } v \in H . \tag{V}

Evans names the two constants $\alpha$ and $\beta$ instead; the content is the same. Stakgold and Holst place the theorem immediately after their Bounded Operator Theorem and describe both as extending the Riesz representation theorem to the case where the form need not be symmetric.

Idea of the proofOriginal

Riesz representation turns both sides of (V) into inner products: the functional $F$ becomes $(\Phi,\cdot)$ for a single vector $\Phi$ , and, for each fixed $u$ , the functional $a(u,\cdot)$ becomes $(Au,\cdot)$ for a vector $Au$ depending linearly and boundedly on $u$ . Testing against $v = Au - \Phi$ then shows that (V) is nothing but the equation $Au = \Phi$ . Coercivity says that $A$ moves vectors at least a fixed fraction of their length in their own direction, so a small step against the residual $Au - \Phi$ is a contraction; the Banach fixed-point theorem converts that into existence and uniqueness in one stroke.

ProofAdaptedAdapted from Stakgold & Holst (2011), §8.5, Thm 8.5.1 and the proof of Thm 8.5.2, pp. 551–552

Let $M$ , $m$ and $L$ be the constants of boundedness, coercivity and boundedness of $F$ .

Step 1 (the form is an operator). Fix $u \in H$ . Then $v \mapsto a(u,v)$ is linear, and bounded by $M\|u\|$ ; so by the Riesz representation theorem there is a unique vector, call it $Au$ , with

a(u,v) = (Au, v) \qquad \text{for all } v \in H .

The map $u \mapsto Au$ is linear, because $a$ is linear in its first slot and the Riesz vector is unique: for scalars $\mu_1,\mu_2$ and $u_1, u_2 \in H$ , the vector $\mu_1Au_1 + \mu_2Au_2$ represents $a(\mu_1u_1+\mu_2u_2,\cdot)$ , so it must equal $A(\mu_1u_1+\mu_2u_2)$ . It is bounded: putting $v = Au$ gives $\|Au\|^2 = a(u,Au) \le M\|u\|\,\|Au\|$ , whence $\|Au\| \le M\|u\|$ (trivially so if $Au = 0$ ). This is Stakgold and Holst's Bounded Operator Theorem.

Step 2 (the functional is a vector). By Riesz again there is $\Phi \in H$ with $F(v) = (\Phi,v)$ for all $v$ , and $\|\Phi\| = \|F\| \le L$ .

Step 3 ((V) is an equation). Problem (V) now reads $(Au - \Phi, v) = 0$ for all $v \in H$ . Choosing $v = Au - \Phi$ gives $\|Au - \Phi\|^2 = 0$ , hence $Au = \Phi$ ; conversely, if $Au = \Phi$ then $a(u,v) = (Au,v) = (\Phi,v) = F(v)$ for every $v$ , which is (V). So

\text{(V)} \iff Au = \Phi .

Step 4 (a contraction). For a parameter $\sigma > 0$ , the equation $Au = \Phi$ is equivalent to the fixed-point equation $u = T(u)$ with

T(u) = u - \sigma\bigl(Au - \Phi\bigr) = (I - \sigma A)u + \sigma\Phi .

Since $T(u) - T(v) = (I-\sigma A)(u-v)$ , we need $\|(I - \sigma A)w\| \le \gamma\|w\|$ with $\gamma<1$ . Expand, using $(Aw,w) = a(w,w) \ge m\|w\|^2$ and $\|Aw\| \le M\|w\|$ :

\|(I-\sigma A)w\|^2 = \|w\|^2 - 2\sigma (Aw,w) + \sigma^2\|Aw\|^2 \le \bigl(1 - 2\sigma m + \sigma^2M^2\bigr)\|w\|^2 .

The bracket is less than $1$ exactly when $\sigma^2M^2 < 2\sigma m$ , that is, when $0 < \sigma < 2m/M^2$ — a non-empty range, since $m$ and $M$ are positive. It is also non-negative, since it dominates $\|(I-\sigma A)w\|^2/\|w\|^2$ for $w \neq 0$ , so its square root $\gamma$ exists and $\gamma < 1$ . Fix such a $\sigma$ .

Step 5 (conclude). $H$ is complete, and $T$ is a contraction on it, so by the Banach fixed-point theorem $T$ has exactly one fixed point $u$ . By Steps 3 and 4 that $u$ is the unique solution of (V). $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

Coercivity is used only in Step 4, and it is indispensable. Without it the operator $A$ can fail to be onto. On $H = \ell^2$ take $a(u,v) = \sum_k k^{-1}u_kv_k$ : this is bounded ( $M=1$ ) and positive, $a(u,u)>0$ for $u\neq0$ , but not coercive, and the corresponding $A$ — multiplication by $k^{-1}$ — has dense, non-closed range. The functional $F(v) = \sum_k k^{-1}v_k$ is bounded, yet $a(u,v)=F(v)$ for all $v$ would force $u_k = 1$ for every $k$ , which is not in $\ell^2$ . So no solution exists: positivity is not enough.
Boundedness of $a$ is used in Step 1, to get an operator at all, and in Step 4, to control $\|Aw\|$ . An unbounded form has no $A$ defined on all of $H$ , and the whole argument must be recast with form domains.
Boundedness of $F$ is used in Step 2. It is exactly the hypothesis the Riesz theorem needs, and it is what the "check the two bounds" step of the recipe in weak derivatives and Sobolev spaces verifies in applications.
Completeness of $H$ is used in Steps 1, 2 and 5. On an incomplete inner product space the fixed-point iteration converges to nothing, which is precisely the gap that forces one to work in $H^1_0(\Omega)$ rather than in $C_c^\infty(\Omega)$ .
Symmetry is not assumed, and that is the point. If $a$ were symmetric, then coercivity and boundedness would make $a$ itself an equivalent inner product on $H$ , and a single application of the Riesz theorem to the Hilbert space $(H,a)$ would end the proof — a remark Evans makes explicitly (p. 299). The theorem earns its name on non-symmetric forms, which is where first-order transport terms $\sum b^i u_{x_i}v$ put us.
An a priori bound comes free. Putting $v = u$ in (V) gives $m\|u\|^2 \le a(u,u) = F(u) \le L\|u\|$ , hence $\|u\| \le L/m$ (trivially if $u=0$ ). Stakgold and Holst call this the energy estimate and derive it before knowing that a solution exists; it is also a second, one-line proof of uniqueness, since the difference of two solutions satisfies (V) with $F=0$ and so has norm $\le 0$ .

Theorem(Existence, uniqueness and stability for the elliptic Dirichlet problem)AdaptedAdapted from Stakgold & Holst (2011), §10.4, Assumption 10.4.1, Thm 10.4.1 and Thm 10.4.2, pp. 713–715; Stakgold & Holst (2011), §8.5, Exercise 8.5.4, p. 553

Let $\Omega \subseteq \mathbb{R}^n$ be an open set of finite width, contained in a slab of width $d$ . Assume

$a_{ij} \in L^\infty(\Omega)$ with $|a_{ij}| \le c_1$ almost everywhere, for $i,j = 1,\dots,n$ ;
$b \in L^\infty(\Omega)$ with $0 \le b \le c_2$ almost everywhere;
strong ellipticity: there is $\lambda > 0$ with $\sum_{i,j} a_{ij}(x)\eta_i\eta_j \ge \lambda|\eta|^2$ for almost every $x\in\Omega$ and every $\eta\in\mathbb{R}^n$ ;
$f \in L^2(\Omega)$ .

Set, for $u,v \in H^1_0(\Omega)$ ,

a(u,v) = \int_\Omega \Bigl(\sum_{i,j=1}^n a_{ij}\,\partial_i u\,\partial_j v \;+\; b\,uv\Bigr) dx , \qquad F(v) = \int_\Omega f v \, dx .

Then there is exactly one $u \in H^1_0(\Omega)$ with $a(u,v) = F(v)$ for all $v \in H^1_0(\Omega)$ — the weak solution of

-\nabla\cdot\bigl(A\nabla u\bigr) + bu = f \ \text{ in } \Omega, \qquad u = 0 \ \text{ on } \partial\Omega, \qquad A = (a_{ij}) ,

and it obeys the a priori bound

\|u\|_{H^1(\Omega)} \;\le\; \frac{\|f\|_{L^2(\Omega)}}{m}, \qquad m = \min\Bigl\{\frac{\lambda}{2d^2},\ \frac{\lambda}{2}\Bigr\} .

In particular the solution map $f \mapsto u$ is a bounded linear map $L^2(\Omega) \to H^1_0(\Omega)$ , so the problem is well posed: a small change of the source produces a small change of the solution, in the energy norm.

This is the pure Dirichlet case of Stakgold and Holst's Theorems 10.4.1 and 10.4.2, which treat a mixed Dirichlet–Neumann problem with inhomogeneous data: setting their Neumann portion $\Gamma_N$ of the boundary and their boundary data $g$ to nothing removes the trace constant $C_{\mathrm{tr}}$ , the coefficient $c$ on $\Gamma_N$ and the lift $w$ from their constants, leaving $m$ as above, $M = n^2c_1 + c_2$ and $L = \|f\|_{L^2}$ . Their Poincaré constant $\rho$ is written here as the width $d$ , by the lemma above. Their data hypothesis is also weaker than ours: Assumption 10.4.1(5) asks only $f \in H^{-1}(\Omega)$ , the dual of $H^1_0(\Omega)$ , and their $L$ carries $\|f\|_{H^{-1}}$ . Taking $f \in L^2(\Omega)$ is the special case in which the pairing is the integral $\int_\Omega fv$ , and then $\|f\|_{H^{-1}} \le \|f\|_{L^2}$ by Cauchy–Schwarz, so $L = \|f\|_{L^2}$ is admissible, if not optimal. The special case $a_{ij} = a\delta_{ij}$ with $a \ge 1$ and $b \ge 1$ on a smooth bounded plane domain is their Exercise 8.5.4.

Idea of the proofOriginal

Everything is a matter of checking the three constants. Boundedness of $a$ is the Cauchy–Schwarz inequality term by term. Boundedness of $F$ is Cauchy–Schwarz once. Coercivity is where the structure enters: ellipticity bounds $a(u,u)$ below by $\lambda$ times the Dirichlet seminorm alone, which is not the $H^1$ norm — the $L^2$ part is missing, and the hypothesis $b\ge0$ contributes nothing to it. Poincaré supplies the missing part: spend half of the seminorm on buying back the $L^2$ norm, and keep the other half.

ProofAdaptedAdapted from Stakgold & Holst (2011), §10.4, proof of Thm 10.4.1, pp. 713–714, and Thm 10.4.2, p. 715

Write $\|\cdot\|$ for $\|\cdot\|_{L^2(\Omega)}$ and recall $\|u\|_{H^1}^2 = \|u\|^2 + |u|_{H^1}^2$ with $|u|_{H^1} = \|\nabla u\|$ .

Step 1 ( $a$ and $F$ are well defined and bilinear/linear.) Every integrand is a product of two $L^2$ functions and an $L^\infty$ coefficient, so each integral converges absolutely; bilinearity and linearity are then immediate from linearity of the integral.

Step 2 (boundedness of $a$ ). By the triangle inequality and Cauchy–Schwarz applied to each term,

|a(u,v)| \le \sum_{i,j}\|a_{ij}\|_\infty\,\|\partial_i u\|\,\|\partial_j v\| + \|b\|_\infty\|u\|\,\|v\| .

Each factor $\|\partial_i u\|$ , $\|u\|$ is at most $\|u\|_{H^1}$ , and there are $n^2$ terms in the first sum. Hence $|a(u,v)| \le M\|u\|_{H^1}\|v\|_{H^1}$ with $M = n^2c_1 + c_2$ .

Step 3 (boundedness of $F$ ). $|F(v)| \le \|f\|\,\|v\| \le \|f\|\,\|v\|_{H^1}$ , so $L = \|f\|_{L^2}$ will do.

Step 4 (coercivity). Apply strong ellipticity pointwise with $\eta = \nabla u(x)$ , and drop the term $\int bu^2$ , which is non-negative because $b \ge 0$ — this is the only use of that hypothesis:

a(u,u) = \int_\Omega \sum_{i,j} a_{ij}\,\partial_i u\,\partial_j u \,dx + \int_\Omega b u^2 dx \;\ge\; \lambda\int_\Omega |\nabla u|^2 dx = \lambda\,|u|_{H^1}^2 .

Now split the right-hand side in half and use the lemma, $\|u\|^2 \le d^2|u|^2_{H^1}$ , on one half:

\lambda|u|_{H^1}^2 = \frac{\lambda}{2}|u|^2_{H^1} + \frac{\lambda}{2}|u|^2_{H^1} \;\ge\; \frac{\lambda}{2d^2}\|u\|^2 + \frac{\lambda}{2}|u|^2_{H^1} \;\ge\; m\bigl(\|u\|^2 + |u|^2_{H^1}\bigr) = m\|u\|_{H^1}^2 ,

with $m = \min\{\lambda/(2d^2),\lambda/2\}$ . This is the computation of Stakgold and Holst's Theorem 10.4.1 with their Poincaré constant $\rho$ replaced by $d$ .

Step 5 (apply the theorem). $H^1_0(\Omega)$ is a Hilbert space, $a$ is bounded and coercive on it and $F$ is bounded, so Lax–Milgram gives a unique $u \in H^1_0(\Omega)$ with $a(u,v) = F(v)$ for all $v$ . The a priori bound is the remark above, $\|u\|_{H^1} \le L/m$ . $\square$

Note(When the coefficients have the wrong sign)AdaptedAdapted from Evans (2010), §6.2.2, Thm 2 and Thm 3, pp. 299–301

The hypothesis $b \ge 0$ is doing real work, and so is the absence of first-order terms. For the general divergence-form operator $Lu = -\sum_{i,j}(a^{ij}u_{x_i})_{x_j} + \sum_i b^iu_{x_i} + cu$ with $a^{ij},b^i,c \in L^\infty$ , Evans proves only the weaker pair of bounds

|B[u,v]| \le \alpha\|u\|_{H^1_0}\|v\|_{H^1_0}, \qquad \beta\|u\|_{H^1_0}^2 \le B[u,u] + \gamma\|u\|_{L^2}^2

for some $\alpha,\beta>0$ and $\gamma\ge0$ — the first-order term is absorbed with Cauchy's inequality with $\varepsilon$ , at the cost of the $\gamma$ term. When $\gamma>0$ the form is not coercive and Lax–Milgram does not apply. Evans's First Existence Theorem recovers what can be recovered: for every $\mu\ge\gamma$ the shifted problem $Lu + \mu u = f$ in $U$ , $u=0$ on $\partial U$ , has a unique weak solution. That the unshifted problem need not be solvable is no defect of the method: if $-\mu$ is a Dirichlet eigenvalue of $L$ , uniqueness genuinely fails, and the right statement is the Fredholm alternative (the Fredholm alternative).

Example(A one-dimensional problem, with every constant computed)Original

Take $\Omega = (0,\pi)$ , $n=1$ , $a_{11} \equiv 1$ , $b \equiv 0$ and $f(x) = \sin x$ . The weak problem is

\text{find } u \in H^1_0(0,\pi) \text{ with } \int_0^\pi u'v'\,dx = \int_0^\pi (\sin x)\,v\,dx \quad \text{for all } v \in H^1_0(0,\pi),

the weak form of $-u'' = \sin x$ , $u(0)=u(\pi)=0$ .

The constants. Here $c_1 = 1$ , $c_2 = 0$ , $\lambda = 1$ and the width is $d = \pi$ . So

M = n^2c_1 + c_2 = 1, \qquad m = \min\Bigl\{\frac{1}{2\pi^2},\ \frac12\Bigr\} = \frac{1}{2\pi^2} = 0.0506606, \qquad L = \|f\|_{L^2} = \Bigl(\int_0^\pi\sin^2x\,dx\Bigr)^{1/2} = \sqrt{\tfrac\pi2} = 1.2533141 .

The theorem therefore guarantees a unique solution with $\|u\|_{H^1} \le L/m = 2\pi^2\sqrt{\pi/2} = 24.7394$ .

The solution. Try $u = \sin x$ . It lies in $H^1_0(0,\pi)$ : take smooth cut-offs $\chi_k$ equal to $1$ on $[1/k,\pi-1/k]$ , vanishing outside $(1/2k,\pi-1/2k)$ and with $|\chi_k'| \le Ck$ ; then $\chi_k u \in C_c^\infty(0,\pi)$ , and $\|(\chi_k-1)u\|_{H^1} \to 0$ because $|u(x)| \le x$ near $0$ makes $\|u\chi_k'\|^2_{L^2} \le (Ck\cdot k^{-1})^2\cdot k^{-1} \to 0$ , and symmetrically at $\pi$ . Since $-u'' = \sin x$ pointwise, the consistency theorem of weak derivatives and Sobolev spaces makes it the weak solution. Its norm is

\|u\|^2_{H^1} = \int_0^\pi \sin^2 x\,dx + \int_0^\pi\cos^2x\,dx = \frac\pi2+\frac\pi2 = \pi, \qquad \|u\|_{H^1} = \sqrt\pi = 1.7724539 .

How good is the bound? The guaranteed bound $24.7394$ overshoots the truth $1.7724539$ by a factor of $13.96$ . Almost all of the loss is in the Poincaré constant. Put the sharp value $\rho = 1$ for this interval (see the remark above) in place of the lemma's $\rho = d = \pi$ ; the coercivity constant becomes $m = \min\{1/2,1/2\} = 1/2$ and the bound becomes

\frac{L}{m} = 2\sqrt{\tfrac\pi2} = \sqrt{2\pi} = 2.5066283,

which still exceeds the true norm by exactly $\sqrt2$ . That residual factor is not the cost of splitting the seminorm in half in Step 4: with the sharp constant, $a(u,u) = |u|^2_{H^1} = \tfrac12\|u\|^2_{H^1}$ for $u = \sin x$ , so coercivity with $m = 1/2$ is an equality here and the split gives away nothing. The whole $\sqrt2$ is given away in Step 3, which bounds $F$ by discarding the difference between the $L^2$ and $H^1$ norms of its argument: the true value is $F(u) = \int_0^\pi\sin^2x\,dx = \pi/2 = 1.5708$ , while the bound used is $L\|u\|_{H^1} = \sqrt{\pi/2}\cdot\sqrt\pi = \pi/\sqrt2 = 2.2214$ , and $2.2214/1.5708 = \sqrt2$ exactly.

A consistency check. The a priori chain $m\|u\|^2_{H^1} \le a(u,u) = F(u) \le L\|u\|_{H^1}$ reads, with the sharp $m$ ,

\tfrac12\cdot\pi = 1.5708 \;\le\; \int_0^\pi\cos^2x\,dx = \tfrac\pi2 = 1.5708 \;=\; \int_0^\pi \sin^2 x\,dx \;\le\; 1.2533141 \times 1.7724539 = 2.2214 ,

the first inequality being an equality, as just explained, and the last being Cauchy–Schwarz, strict because $f = \sin x$ and $u = \sin x$ are proportional in $L^2$ but $\|u\|_{L^2} < \|u\|_{H^1}$ .

The contraction parameter. With $m=1/2$ and $M=1$ the proof's iteration is guaranteed to converge for every $\sigma \in (0, 2m/M^2) = (0,1)$ , with contraction factor at most $(1-\sigma+\sigma^2)^{1/2}$ , smallest at $\sigma = \tfrac12$ where it equals $\sqrt3/2 = 0.8660$ . The guarantee is pessimistic here. In the orthogonal basis $e_k = \sin kx$ one has $(e_k,e_j)_{H^1} = \delta_{kj}\tfrac\pi2(1+k^2)$ and $a(e_k,e_j) = \delta_{kj}\tfrac\pi2 k^2$ , so the operator $A$ of Step 1 is multiplication by the numbers $c_k = k^2/(1+k^2)$ , which lie in $[\tfrac12,1)$ with least value $c_1 = \tfrac12$ and supremum $1$ . Since $c \mapsto |1-\sigma c|$ is convex, its supremum over those numbers is reached at those two extremes:

\|I-\sigma A\| = \sup_{k\ge1}\bigl|1-\sigma c_k\bigr| = \max\bigl\{\,\bigl|1-\tfrac\sigma2\bigr|,\ |1-\sigma|\,\bigr\} \qquad (0<\sigma<2).

This equals $3/4$ at $\sigma=\tfrac12$ (against the certified $0.8660$ ), $1/2$ at $\sigma=1$ , and $0.4$ at $\sigma=1.2$ ; it is least where the two entries meet, at $\sigma = 4/3$ , where it is $1/3$ . So the true optimum lies strictly outside the certified range $(0,1)$ , and it beats everything inside it: within that range the true factor decreases towards $1/2$ as $\sigma\uparrow1$ without reaching it, whereas $\sigma = 4/3$ gives $1/3$ . The certified range is a safe region, not a search region.

ApplicationOriginal

Finite elements, and why they converge. Replacing $H^1_0(\Omega)$ by a finite-dimensional subspace $V_h$ in (V) gives the Galerkin equations $a(u_h,v_h) = F(v_h)$ for all $v_h \in V_h$ . Lax–Milgram applies verbatim to $V_h$ — a finite-dimensional subspace of a Hilbert space is closed, hence complete, and $a$ is still bounded and coercive on it — so the discrete problem has a unique solution, with the same constants $m$ and $M$ and so the same a priori bound. The constants also control the error: subtracting the two problems gives $a(u-u_h, v_h) = 0$ , and from there $\|u-u_h\|_{H^1} \le (M/m)\inf_{v_h\in V_h}\|u-v_h\|_{H^1}$ — Céa's lemma, which Stakgold and Holst state as near-best approximation (Theorem 10.4.3, p. 716). A finite element method is therefore as accurate as the best approximation the mesh can provide, up to the factor $M/m$ . See the finite element method.

Electrostatics in a composite. The potential in a region filled with materials of permittivity $\epsilon(x)$ , grounded on the boundary, is the weak solution of $-\nabla\cdot(\epsilon\nabla\phi) = \varrho$ with $\phi \in H^1_0$ . The theorem applies with $a_{ij} = \epsilon\delta_{ij}$ as soon as $\epsilon$ is measurable, bounded and bounded below by a positive constant. No continuity of $\epsilon$ is needed, which is exactly the situation at a material interface, and the a priori bound is the statement that the stored energy is controlled by the charge.

Steady diffusion with drift. Transport terms $\mathbf{v}\cdot\nabla u$ make the form non-symmetric, and this is where Lax–Milgram is not replaceable by Riesz. If the drift is strong relative to the diffusion, the $\gamma$ of the note above is positive, coercivity is lost, and the numerical symptom is the spurious oscillation that afflicts centred differencing of advection-dominated problems (the advection-diffusion equation).

Where the method stops. Lax–Milgram is a statement about elliptic problems: coercivity of $a$ on a single space is a form of positivity in all directions at once. Evolution equations have no such form, and their well-posedness rests instead on estimates that control a solution in terms of its past (energy estimates and well-posedness), or on semigroup theory (evolution equations and operator semigroups).

Theorem · advanced

Energy estimates and well-posedness

Adapted

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Proves the basic energy estimates for the forced heat and wave equations on a bounded domain, deduces uniqueness and continuous dependence on the data from them, and shows by the backward heat equation and Hadamard's Cauchy problem for Laplace's equation what the estimates exclude.

Prerequisites: Energy and uniqueness for the wave equation, Maximum principle for the heat equation, L^p spaces, Green's identities, Weak derivatives and Sobolev spaces, The Lax-Milgram theorem and the Dirichlet problem

Definition(Well-posedness)AdaptedAdapted from Evans (2010), §1.3.1, conditions (i)–(iii), p. 6; Strauss (2008), §1.5, pp. 25–26; Stakgold & Holst (2011), §1.2, p. 66

A problem consisting of a partial differential equation on a domain together with initial and boundary conditions is well posed when all three of the following hold.

Existence. At least one solution satisfies the equation and all the auxiliary conditions.
Uniqueness. At most one does.
Stability, also called continuous dependence on the data: a small change in the data produces a small change in the solution.

Otherwise the problem is ill posed. All three sources give this list; Stakgold and Holst put it as three questions and note that an affirmative answer to all three is what the term means.

Clause 3 has no content until "small" is given a meaning, and the meaning is a choice: one must fix a norm on the data and a norm on the solutions. Strauss is emphatic that the choice must be made and makes it twice for the same problem, obtaining stability in the "square integral" sense from the energy method and stability in the "uniform" sense from the maximum principle. A problem can be well posed for one pair of norms and ill posed for another, and in the end the physics decides: the data are measurements, known only to finitely many decimal places, and a model whose output is not controlled by the accuracy of its input predicts nothing. Stakgold and Holst add the modern caveat that ill-posed problems are no longer rejected out of hand, since inverse problems are ill posed by nature and are solved anyway, with regularisation and with care.

What an energy estimate is. An inequality bounding a norm of the solution at time $t$ by the same norm at time $0$ and a norm of the source, with constants depending only on the equation and on $t$ . Such an inequality settles clauses 2 and 3 at one stroke, by linearity: apply it to the difference of two solutions. It says nothing about clause 1.

Theorem(Energy estimate for the heat equation)AdaptedAdapted from Evans (2010), §2.3.4a, Thm 10 and its proof, p. 60; Strauss (2008), §2.3, the energy method and Eq. (4)–(5), pp. 44–45

Let $\Omega \subset \mathbb{R}^n$ be a bounded open set with piecewise $C^1$ boundary, let $T>0$ and $k>0$ , and let $u \in C^2(\overline\Omega\times[0,T])$ satisfy

u_t = k\,\Delta u + f \ \text{ in } \Omega\times(0,T), \qquad u = 0 \ \text{ on } \partial\Omega\times[0,T], \qquad u(\cdot,0) = \varphi ,

with $f$ continuous on $\overline\Omega\times[0,T]$ . Write $N(t) = \|u(\cdot,t)\|_{L^2(\Omega)}$ and $\|f(t)\| = \|f(\cdot,t)\|_{L^2(\Omega)}$ . Then:

(a) The identity.

\frac{d}{dt}N(t)^2 \;=\; -\,2k\int_\Omega |\nabla u|^2\,dx \;+\; 2\int_\Omega f\,u\,dx \qquad (0<t<T).

(b) The estimate.

N(t) \;\le\; N(0) + \int_0^t \|f(s)\|\,ds \qquad (0\le t\le T).

(c) The dissipation identity. If $f \equiv 0$ ,

N(T)^2 + 2k\int_0^T\!\!\int_\Omega |\nabla u|^2\,dx\,dt \;=\; N(0)^2 ,

so in particular $N$ is non-increasing — this is Strauss's inequality (4) — and the total dissipation is exactly the energy lost.

(d) The sharpened estimate. If moreover $\|\nabla v\|^2_{L^2} \ge \lambda\|v\|^2_{L^2}$ for all $v \in H^1_0(\Omega)$ — a Poincaré inequality, with $\lambda = d^{-2}$ available whenever $\Omega$ lies in a slab of width $d$ , by the Lax–Milgram theorem and the Dirichlet problem — then

N(t) \;\le\; e^{-k\lambda t}N(0) \;+\; \int_0^t e^{-k\lambda(t-s)}\|f(s)\|\,ds .

Consequences. Solutions are unique (the estimates bound whatever solutions exist; they do not produce one): if two solutions have the same $\varphi$ , the same $f$ and the same boundary values, their difference $w$ satisfies the hypotheses with zero data, so $N_w \equiv 0$ and $w \equiv 0$ . This is Evans's Theorem 10 and Strauss's second proof of uniqueness. And it is stable in $L^2$ : two solutions with sources $f_1,f_2$ and initial data $\varphi_1,\varphi_2$ obey

\|u_1(\cdot,t) - u_2(\cdot,t)\|_{L^2} \;\le\; \|\varphi_1-\varphi_2\|_{L^2} + \int_0^t\|f_1(s)-f_2(s)\|_{L^2}\,ds ,

which with $f_1 = f_2$ is Strauss's stability statement (5). Parts (b) and (d), and the source term throughout, go beyond what either source states: Evans and Strauss both run the argument with $f$ removed by subtraction.

Idea of the proofOriginal

Multiply the equation by the solution itself. On the left that produces $\tfrac12\partial_t(u^2)$ , and on the right, after one integration by parts, $-k|\nabla u|^2$ together with the work $fu$ done by the source. The boundary term is killed by the boundary condition. So the $L^2$ norm has a derivative with one term that is always negative — the dissipation, which is what distinguishes diffusion from wave propagation — and one term bounded by $\|f\|$ times the norm itself. Throwing the dissipation away gives (b); keeping it and feeding it back through Poincaré's inequality gives the exponential decay in (d). Grönwall's lemma is not needed in a separate form: both follow from integrating a differential inequality for $N$ rather than for $N^2$ .

ProofAdaptedAdapted from Evans (2010), §2.3.4a, the computation of $\\dot e(t)$ in the proof of Thm 10, p. 60; Strauss (2008), §2.3, the pointwise identity and its integration, p. 44

Step 1 (differentiate). The functions $u^2$ and $\partial_t(u^2) = 2uu_t$ are continuous on the compact set $\overline\Omega\times[0,T]$ , and $\Omega$ has finite volume, so one may differentiate under the integral sign exactly as in energy and uniqueness for the wave equation:

\frac{d}{dt}N(t)^2 = \frac{d}{dt}\int_\Omega u^2\,dx = 2\int_\Omega u\,u_t\,dx .

Step 2 (use the equation and integrate by parts). Substituting $u_t = k\Delta u + f$ ,

\frac{d}{dt}N^2 = 2k\int_\Omega u\,\Delta u\,dx + 2\int_\Omega fu\,dx .

By Green's first identity (Green's identities), which applies because $u \in C^2(\overline\Omega)$ and $\partial\Omega$ is piecewise $C^1$ ,

\int_\Omega u\,\Delta u\,dx = -\int_\Omega|\nabla u|^2dx + \oint_{\partial\Omega} u\,\frac{\partial u}{\partial n}\,dS ,

and the boundary integral vanishes because $u = 0$ at every point of $\partial\Omega$ . This proves (a). (A homogeneous Neumann condition $\partial u/\partial n = 0$ kills the same integral, so parts (a), (b) and (c) hold verbatim for it. Part (d) does not: the boundary condition is used a second time, in Step 5, and under Neumann data part (d) is false — see the remark below.)

Step 3 (the differential inequality). Discard the dissipation term, which is $\le 0$ , and apply the Cauchy–Schwarz inequality to the other:

\frac{d}{dt}N^2 \le 2\int_\Omega fu\,dx \le 2\|f(t)\|\,N(t) .

To divide by $N$ without worrying about zeros, put $N_\varepsilon = (N^2+\varepsilon^2)^{1/2}$ for $\varepsilon>0$ . Then $N_\varepsilon \ge N$ and $N_\varepsilon \ge \varepsilon > 0$ , and

N_\varepsilon' = \frac{1}{2N_\varepsilon}\frac{d}{dt}N^2 \le \frac{\|f(t)\|\,N}{N_\varepsilon} \le \|f(t)\| .

Integrating from $0$ to $t$ and letting $\varepsilon \to 0$ (both $N_\varepsilon(t) \to N(t)$ and $N_\varepsilon(0)\to N(0)$ ) gives (b).

Step 4 (dissipation). With $f \equiv 0$ , (a) reads $\frac{d}{dt}N^2 = -2k\int_\Omega|\nabla u|^2dx$ . Both sides are continuous in $t$ ; integrate from $0$ to $T$ and rearrange to get (c). Since the integral on the left of (c) is non-negative, $N(T) \le N(0)$ , and the same holds with $T$ replaced by any $t$ .

Step 5 (the sharpened estimate). Now keep the dissipation. Fix $t$ and write $w = u(\cdot,t)$ , which is $C^1$ on $\overline\Omega$ and vanishes on $\partial\Omega$ . The Poincaré hypothesis is a statement about $H^1_0(\Omega)$ , which is by definition the closure of $C_c^\infty(\Omega)$ in the $H^1$ norm (weak derivatives and Sobolev spaces), so before using it we must exhibit test functions converging to $w$ . This takes three sub-steps: truncate, mollify, and let the truncation level go to zero.

(5a) Truncate. For $\delta>0$ put $w_\delta = (w-\delta)^+ - (-w-\delta)^+$ , which equals $w - \delta\operatorname{sgn}(w)$ where $|w|>\delta$ and $0$ where $|w|\le\delta$ . It is Lipschitz, being a composition of $w$ with a piecewise linear function of slope $0$ or $1$ , and almost everywhere

\nabla w_\delta = \nabla w\,\mathbf{1}_{\{|w|>\delta\}} .

Indeed, on each of the two open sets $\{w>\delta\}$ and $\{w<-\delta\}$ the two functions differ by a constant, on the open set $\{|w|<\delta\}$ the left side is $0$ , and on $\{|w|=\delta\}$ both sides vanish almost everywhere, by the level-set argument of (5c) below applied to $\{w=\delta\}$ and $\{w=-\delta\}$ . Moreover $w_\delta$ vanishes off $\{|w|\ge\delta\}$ , which is a closed subset of the compact set $\overline\Omega$ on which $|w|>0$ ; since $w$ vanishes on $\partial\Omega$ , that set misses the boundary, so it is a compact subset $K_\delta \subset \Omega$ , and $w_\delta$ has compact support in $\Omega$ .

(5b) Mollify. Fix $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\eta\ge0$ , support in the unit ball and $\int\eta = 1$ , and put $\eta_\sigma(x) = \sigma^{-n}\eta(x/\sigma)$ . Extend $w_\delta$ by zero to $\mathbb{R}^n$ . For $\sigma < \operatorname{dist}(K_\delta,\partial\Omega)$ the convolution $\eta_\sigma * w_\delta$ is smooth and supported in the closed $\sigma$ -neighbourhood of $K_\delta$ , so it lies in $C_c^\infty(\Omega)$ . Differentiating under the integral sign and then moving the derivative onto $w_\delta$ ,

\partial_i(\eta_\sigma * w_\delta)(x) = \int \partial_{x_i}\eta_\sigma(x-y)\,w_\delta(y)\,dy = -\int \partial_{y_i}\bigl[\eta_\sigma(x-y)\bigr] w_\delta(y)\,dy = \bigl(\eta_\sigma * \partial_i w_\delta\bigr)(x) ,

the last equality being the definition of the weak derivative, applied to the test function $y\mapsto\eta_\sigma(x-y)$ , which lies in $C_c^\infty(\Omega)$ by the choice of $\sigma$ . Both $w_\delta$ and $\nabla w_\delta$ lie in $L^2(\mathbb{R}^n)$ , being bounded with compact support, and mollification converges in $L^2$ : by Minkowski's integral inequality,

\|\eta_\sigma * h - h\|_{L^2} \le \int \eta(z)\,\|h(\cdot-\sigma z) - h\|_{L^2}\,dz ,

and translation is continuous on $L^2$ — for a continuous $g$ of compact support because such a $g$ is uniformly continuous, and then for every $h \in L^2$ by the usual three-term estimate, splitting $h$ into such a $g$ plus a remainder of small $L^2$ norm, which is possible because $C_c$ is dense in $L^2$ (the $L^p$ spaces). Applying this to $h = w_\delta$ and to $h = \partial_i w_\delta$ gives $\eta_\sigma * w_\delta \to w_\delta$ in $H^1$ , so $w_\delta \in H^1_0(\Omega)$ .

(5c) Let $\delta\to0$ . The two halves of the $H^1$ norm must be estimated separately, because $w_\delta - w$ is not supported in $\{|w|\le\delta\}$ : on $\{|w|>\delta\}$ , where the truncation bites, it equals $-\delta\operatorname{sgn}(w)$ . What is true is that $|w_\delta - w| \le \delta$ at every point — it is $|w|\le\delta$ on $\{|w|\le\delta\}$ and exactly $\delta$ elsewhere — so, $\Omega$ being bounded,

\|w_\delta - w\|^2_{L^2} \;\le\; \delta^2|\Omega| \;\xrightarrow[\delta\to0]{}\; 0 ,

while the gradients, by the formula of (5a), differ only on $\{|w|\le\delta\}$ :

\|\nabla w_\delta - \nabla w\|^2_{L^2} = \int_{\{|w|\le\delta\}}|\nabla w|^2\,dx \;\xrightarrow[\delta\to0]{}\; \int_{\{w=0\}}|\nabla w|^2\,dx

by dominated convergence, the dominating function being $|\nabla w|^2 \in L^1(\Omega)$ . That last integral is zero: at a point where $w$ vanishes with $\nabla w\neq0$ the implicit function theorem makes $\{w=0\}$ a $C^1$ hypersurface near it, and $\Omega$ is covered by countably many such neighbourhoods, so the points of $\{w=0\}$ with $\nabla w\neq0$ form a null set. Hence $w_\delta \to w$ in $H^1(\Omega)$ , and $u(\cdot,t) = w$ lies in $H^1_0(\Omega)$ , being a limit of its elements in a closed subspace. The Poincaré hypothesis therefore applies and $-2k\|\nabla u\|^2 \le -2k\lambda N^2$ . Hence

\frac{d}{dt}N^2 \le -2k\lambda N^2 + 2\|f(t)\|N , \qquad\text{so}\qquad N_\varepsilon' \le -k\lambda\frac{N^2}{N_\varepsilon} + \|f(t)\| \le -k\lambda N_\varepsilon + k\lambda\varepsilon + \|f(t)\| ,

using $N^2 = N_\varepsilon^2 - \varepsilon^2$ and $\varepsilon^2/N_\varepsilon \le \varepsilon$ . Multiplying by the integrating factor $e^{k\lambda t}$ makes the left side $\bigl(e^{k\lambda t}N_\varepsilon\bigr)'$ ; integrating from $0$ to $t$ and letting $\varepsilon\to0$ gives (d). $\square$

Theorem(Energy estimate for the wave equation with a source)AdaptedAdapted from Strauss (2008), §9.1, the pointwise energy identity Eq. (4), p. 230; Evans (2010), §2.4.3a, the computation of $\\dot E$ in the proof of Thm 5, p. 79

Let $\Omega\subset\mathbb{R}^n$ be a bounded open set with piecewise $C^1$ boundary, $T>0$ , $c>0$ , and let $u \in C^2(\overline\Omega\times[0,T])$ satisfy

u_{tt} = c^2\Delta u + f \ \text{ in } \Omega\times(0,T), \qquad u = 0 \ \text{ on } \partial\Omega\times[0,T], \qquad u(\cdot,0) = \varphi, \quad u_t(\cdot,0) = \psi ,

with $f$ continuous on $\overline\Omega\times[0,T]$ , so that the initial displacement $\varphi$ and the initial velocity $\psi$ are the two pieces of initial data. Put

\mathcal{E}(t) = \int_\Omega\bigl(u_t^2 + c^2|\nabla u|^2\bigr)dx ,

twice the energy of energy and uniqueness for the wave equation. Then

\frac{d\mathcal{E}}{dt} = 2\int_\Omega f\,u_t\,dx \qquad\text{and}\qquad \sqrt{\mathcal{E}(t)} \;\le\; \sqrt{\mathcal{E}(0)} + \int_0^t\|f(s)\|_{L^2(\Omega)}\,ds .

Consequences. With $f\equiv0$ this is conservation of energy, the theorem of energy and uniqueness for the wave equation. With $f$ present it gives uniqueness for the forced problem, and stability in the energy norm: if $u_1,u_2$ solve the problem with sources $f_1,f_2$ and initial data $(\varphi_1,\psi_1)$ , $(\varphi_2,\psi_2)$ , then $w = u_1-u_2$ obeys

\sqrt{\mathcal{E}_w(t)} \le \Bigl(\|\psi_1-\psi_2\|^2_{L^2} + c^2\|\nabla\varphi_1-\nabla\varphi_2\|^2_{L^2}\Bigr)^{1/2} + \int_0^t\|f_1(s)-f_2(s)\|_{L^2}\,ds .

Note which norms appear: the initial velocity in $L^2$ , the initial displacement in $H^1$ , the source in $L^1$ in time and $L^2$ in space. Strauss proves a stability statement of the same shape for the forced wave equation on the line, but in the uniform norm and with the explicitly $T$ -dependent constants $1$ , $T$ and $T^2/2$ , read off from the solution formula; the estimate here uses no formula and so survives variable coefficients and bounded domains.

Idea of the proofOriginal

Multiply by the velocity rather than by the solution, because for a second-order equation in time it is $u_t$ , not $u$ , that turns the principal part into a time derivative plus a divergence. The identity is the one already proved for the unforced equation; the source simply appears on the right as the power it delivers. Since power is at most force times speed, and speed is controlled by the energy itself, the growth of $\sqrt{\mathcal{E}}$ is at most $\|f\|$ — and nothing is dissipated, so there is no decaying factor to help, as there was for the heat equation.

ProofAdaptedAdapted from Strauss (2008), §9.1, the pointwise identity Eq. (4), p. 230; Evans (2010), §2.4.3a, the bounded-domain computation of $\\dot E$ in the proof of Thm 5, p. 79

Step 1 (the pointwise identity). By the local energy identity of energy and uniqueness for the wave equation, valid for any $C^2$ function,

\bigl(u_{tt}-c^2\Delta u\bigr)u_t = \frac{\partial}{\partial t}\Bigl[\tfrac12\bigl(u_t^2+c^2|\nabla u|^2\bigr)\Bigr] + \nabla\cdot\bigl(-c^2u_t\nabla u\bigr) .

The left-hand side is $f u_t$ , by the equation.

Step 2 (integrate). Integrate over $\Omega$ . Differentiation under the integral sign is licensed as before, and the divergence theorem turns the last term into a boundary integral of $-c^2u_t\,\partial u/\partial n$ . Since $u(\cdot,t) = 0$ on $\partial\Omega$ for every $t$ , differentiating in $t$ at a fixed boundary point gives $u_t = 0$ there, so that integral vanishes. This disposal of the flux is not Strauss's: he integrates his identity over all of $\mathbb{R}^3$ and drops the term because the derivatives of $u$ decay at infinity. On a bounded domain it is the Dirichlet condition that kills it, which is the route Evans takes on p. 79, and the route needed here. Hence

\frac{1}{2}\frac{d\mathcal{E}}{dt} = \int_\Omega f\,u_t\,dx ,

which is the stated identity.

Step 3 (bound the right-hand side by the energy). By Cauchy–Schwarz and then $\|u_t\|^2_{L^2}\le\mathcal{E}$ , which holds because the other term of $\mathcal{E}$ is non-negative,

\frac{d\mathcal{E}}{dt} \le 2\|f(t)\|\,\|u_t(\cdot,t)\|_{L^2} \le 2\|f(t)\|\sqrt{\mathcal{E}(t)} .

Step 4 (integrate the differential inequality). As in the parabolic case, work with $\mathcal{E}_\varepsilon = (\mathcal{E}+\varepsilon^2)^{1/2}$ to avoid dividing by zero: $\mathcal{E}_\varepsilon' = \mathcal{E}'/(2\mathcal{E}_\varepsilon) \le \|f(t)\|\sqrt{\mathcal{E}}/\mathcal{E}_\varepsilon \le \|f(t)\|$ . Integrate from $0$ to $t$ and let $\varepsilon\to0$ . $\square$

Remark(Where the hypotheses are used, and what fails without them)Original

The sign of $k$ . Everything above needs $k>0$ ; with $k<0$ the dissipation term changes sign and becomes growth. The backward heat equation is genuinely ill posed, not merely beyond the reach of this proof. On $\Omega=(0,\pi)$ with $k=1$ , the data $g_m(x) = m^{-1}\sin mx$ prescribed at time $T$ are reached from $u_m(x,0) = m^{-1}e^{m^2T}\sin mx$ . As $m\to\infty$ , $\|g_m\|_{L^2} = m^{-1}\sqrt{\pi/2}\to0$ while $\|u_m(\cdot,0)\|_{L^2} = m^{-1}e^{m^2T}\sqrt{\pi/2}\to\infty$ . With $T=1$ and $m=5$ : data of norm $0.2507$ , and a reconstructed initial state of norm $1.80\times10^{10}$ . Clause 3 fails in any pair of norms in which the data tend to zero. Uniqueness, by contrast, does hold backwards: that is Evans's Theorem 11, proved by showing $\log\int w^2$ convex in $t$ , and it illustrates that uniqueness and stability are genuinely different clauses.
The boundary condition. Step 2 of each proof needs the boundary term to vanish or to have the right sign. Dirichlet and Neumann conditions do it; so does a Robin condition $\partial u/\partial n + bu_t = 0$ with $b>0$ , which makes the wave boundary term $-c^2b\oint u_t^2dS \le 0$ and turns the estimate into decay. With $b<0$ the wall feeds the system and the estimate is lost. Dirichlet data are used a second time in Step 5 of the heat proof, where Poincaré's inequality needs $u(\cdot,t)\in H^1_0(\Omega)$ , and there Neumann is not a substitute: the inequality fails on $H^1(\Omega)$ because constants have zero gradient, and so does part (d). Take $f\equiv0$ and $u\equiv1$ , which satisfies the heat equation and $\partial u/\partial n = 0$ ; then $N(t) = |\Omega|^{1/2}$ for every $t$ , contradicting $N(t)\le e^{-k\lambda t}N(0)$ for any $\lambda>0$ . Parts (a), (b) and (c) are indifferent between the two conditions; part (d) is not.
Ellipticity of the spatial operator. Replace $\Delta$ by an operator that is not elliptic and the dissipation term is no longer sign-definite. The extreme case is Laplace's equation treated as an evolution in $x$ : Hassani notes that the two-dimensional Laplace equation has no characteristic curves in the plane, so one might expect its Cauchy problem to be well posed, and that it is not (his Problem 21.4). Explicitly, the data $u(0,y)=0$ , $u_x(0,y) = \epsilon\sin ky$ are met by

u(x,y) = \frac{\epsilon}{k}\,\sinh(kx)\,\sin(ky),

as one checks: $u_{xx}+u_{yy} = \epsilon k\sinh(kx)\sin(ky) - \epsilon k\sinh(kx)\sin(ky) = 0$ , $u(0,y)=0$ and $u_x(0,y) = \epsilon\cosh 0\sin ky = \epsilon\sin ky$ . The data have sup norm $\epsilon$ however large $k$ is, while at any fixed $x>0$ the solution reaches $(\epsilon/k)\sinh(kx)$ , which grows like $\epsilon e^{kx}/2k$ . This is Hadamard's example, and it is why initial-value problems are posed for hyperbolic and parabolic equations and boundary-value problems for elliptic ones (classification of second-order equations).

$u \in C^2$ up to the boundary. Used to differentiate under the integral sign and to integrate by parts. The estimates are nevertheless the right ones for rough solutions too: one proves them first for smooth approximations and passes to the limit, which is exactly how weak solutions in $L^2(0,T;H^1_0(\Omega))$ are constructed (weak derivatives and Sobolev spaces). An a priori estimate is called a priori because it is derived before any solution is known to exist.
What the estimates do not give. Existence. An energy estimate bounds whatever solutions there are; producing one needs a construction — a formula, a series, a Galerkin limit, or a semigroup (evolution equations and operator semigroups).
Which norm. The heat equation is stable in $L^2$ by the argument above, and also in the uniform norm by the maximum principle (maximum principle for the heat equation); Strauss presents the two side by side. The wave equation has no maximum principle, and its natural statement of stability is in the energy norm, where the displacement is measured in $H^1$ and the velocity in $L^2$ . The choice is not cosmetic: the wave equation is not stable in the uniform norm of the initial displacement alone in three dimensions, because the solution involves derivatives of the data (the wave equation in three dimensions).

Method(The energy recipe)Original

Choose the multiplier. Multiply the equation by whatever makes the principal part a time derivative plus a divergence: by $u$ for an equation first order in time, by $u_t$ for one second order in time. For a general system, the multiplier is whatever the conserved quantity is quadratic in.
Integrate over the spatial domain and integrate by parts, so that the equation becomes $\tfrac{d}{dt}(\text{energy}) + (\text{boundary flux}) = (\text{everything else})$ .
Dispose of the boundary term. Check that the boundary condition makes it vanish, or gives it a sign. If it has the wrong sign, the method fails and the failure is usually real.
Keep the sign-definite interior terms. A negative one is dissipation; keep it if you want decay, discard it if you only want a bound.
Bound the rest by the energy. Cauchy–Schwarz for source terms, Poincaré to convert a gradient into a function, and Young's inequality $ab \le \varepsilon a^2 + b^2/4\varepsilon$ to absorb a cross term into a dissipation term you have kept.
Integrate the differential inequality. For $E' \le 2\alpha E + \beta$ , Grönwall gives $E(t) \le e^{2\alpha t}E(0) + \int_0^t e^{2\alpha(t-s)}\beta(s)ds$ ; when $\beta$ is a multiple of $\sqrt E$ , work with $\sqrt E$ instead, as above.
Read off the two clauses. Zero data gives zero energy, hence uniqueness. Applying the estimate to the difference of two solutions gives stability, in the norms the energy happens to be built from — and those norms are the honest answer to "with respect to what?" in clause 3.

Example(Both estimates, on the same interval)Original

Take $\Omega = (0,\pi)$ with zero Dirichlet conditions at both ends, so that the Poincaré constant is $\lambda = 1$ : expanding in the sine basis, $\|v'\|^2_{L^2} = \tfrac\pi2\sum k^2b_k^2 \ge \tfrac\pi2\sum b_k^2 = \|v\|^2_{L^2}$ for $v=\sum b_k\sin kx$ in $H^1_0$ , with equality for $v=\sin x$ . Note also $\|\sin x\|_{L^2} = \sqrt{\pi/2} = 1.2533141$ .

A. Heat, with a decaying source. Let $k=1$ , $\varphi = 0$ and $f(x,t) = e^{-t}\sin x$ . Writing $u = A(t)\sin x$ turns the equation into $A' = -A + e^{-t}$ with $A(0)=0$ , that is $(e^tA)' = 1$ , so

u(x,t) = t\,e^{-t}\sin x , \qquad N(t) = \|u(\cdot,t)\|_{L^2} = t e^{-t}\sqrt{\tfrac\pi2} .

The crude estimate (b) gives $N(t) \le 0 + \int_0^t e^{-s}\sqrt{\pi/2}\,ds = \sqrt{\pi/2}\,(1-e^{-t})$ , which at $t=1$ is $0.7922456$ against the true $e^{-1}\sqrt{\pi/2} = 0.4610685$ . It overshoots by a factor $1.718$ , and it is monotone increasing, so it never sees that $N$ peaks at $t=1$ and decays afterwards.
The sharpened estimate (d) with $\lambda=1$ gives

N(t) \le \int_0^t e^{-(t-s)}e^{-s}\sqrt{\tfrac\pi2}\,ds = \sqrt{\tfrac\pi2}\;t\,e^{-t},

which is the exact value, at every $t$ . Equality was inevitable: the data are a single eigenfunction, so both the Poincaré inequality and the Cauchy–Schwarz step in the proof are equalities, and the differential inequality for $N$ is the differential equation $N' = -N + \|f\|$ .

Check of the identity (a) at a general $t$ : the left side is $\frac{d}{dt}\bigl[\tfrac\pi2 t^2e^{-2t}\bigr] = \pi e^{-2t}(t-t^2)$ ; the right side is $-2\int_0^\pi u_x^2dx + 2\int_0^\pi fu\,dx = -\pi t^2e^{-2t} + \pi te^{-2t}$ . They agree.

B. Wave, with a steady source. Let $c=1$ , $u = u_t = 0$ at $t=0$ , and $f(x) = \sin x$ , independent of time. Now $A'' = -A + 1$ with $A(0)=A'(0)=0$ , so $A = 1-\cos t$ and

u(x,t) = (1-\cos t)\sin x , \qquad \mathcal{E}(t) = \int_0^\pi\bigl(u_t^2+u_x^2\bigr)dx = \frac\pi2\bigl(\sin^2t + (1-\cos t)^2\bigr) = \pi(1-\cos t) .

The estimate gives $\sqrt{\mathcal{E}(t)} \le 0 + t\sqrt{\pi/2}$ . At $t=\pi$ the true value is $\sqrt{2\pi} = 2.5066283$ and the bound is $\pi\sqrt{\pi/2} = 3.9374025$ , larger by a factor $\pi/2$ .
The contrast with A is the point. $\mathcal{E}$ does not decay; it oscillates between $0$ and $2\pi$ forever, returning exactly to zero at $t = 2\pi, 4\pi,\dots$ . There is no dissipation term to keep, so no exponential factor is available, and the bound on $\sqrt{\mathcal{E}}$ grows linearly in $t$ — which is sharp in general, as the resonant source $f = \sin t\,\sin x$ shows, for which $A = \tfrac12(\sin t - t\cos t)$ grows without bound.
Where the slack is. The bound uses $\|u_t\| \le \sqrt{\mathcal{E}}$ , an equality only when $\nabla u = 0$ , which here happens only at $t=0,2\pi,\dots$ . For this solution the loss is exactly computable. From $\|u_t(\cdot,t)\|^2 = \tfrac\pi2\sin^2t$ and $\mathcal{E}(t) = \pi(1-\cos t)$ ,

\frac{\|u_t(\cdot,t)\|}{\sqrt{\mathcal{E}(t)}} = \sqrt{\frac{\sin^2 t}{2(1-\cos t)}} = \cos\frac t2 \qquad (0 \le t \le \pi) ,

and the mean of that factor over $[0,\pi]$ is $\frac1\pi\int_0^\pi\cos(t/2)\,dt = 2/\pi$ . The overshoot $\pi/2$ at $t=\pi$ is its reciprocal, as it must be: the estimate integrates $\|f(s)\|$ where the identity integrates $\|f(s)\|$ times this factor, and here $\|f(s)\|$ is constant. The two halves of $\mathcal{E}$ are not in balance, as they would be for a freely oscillating standing wave: the steady source holds the displacement off centre, and the averages over a period are $\pi/4$ for the kinetic part against $3\pi/4$ for the potential one. What is exactly $1/2$ on average is the squared ratio $\|u_t\|^2/\mathcal{E} = (1+\cos t)/2$ ; taking the square root before averaging is what produces $2/\pi$ instead.

ApplicationOriginal

Deciding what a numerical scheme must preserve. A discretisation inherits well-posedness only if it inherits the estimate. The standard design rule is to build a discrete energy whose evolution mimics the continuous one: a scheme for the wave equation whose discrete $\mathcal{E}$ is conserved cannot blow up, and a scheme for the heat equation whose discrete $N$ is non-increasing cannot oscillate. This is the energy-method route to stability, equivalent in the constant-coefficient case to the Fourier route of stability, the CFL condition and von Neumann analysis but applicable where Fourier analysis is not — variable coefficients, irregular domains, nonlinearities.

Inverse problems. Reconstructing a past temperature field from present measurements, deblurring an image, or inverting a diffusive tracer transport are all the backward heat equation in disguise, and the remark above says they are ill posed: the factor $e^{m^2T}$ amplifies measurement noise at wavenumber $m$ out of all recognition. The response is not to abandon the problem but to regularise it — truncate the high modes, add a penalty on the reconstructed field's norm, or stop an iteration early — each of which replaces the ill-posed problem by a nearby well-posed one and accepts a bias in return for stability.

Continuum mechanics and relativity. The energy identity for the wave equation is the same calculation as the local conservation law of energy and uniqueness for the wave equation, and it generalises wherever a Lagrangian does: to elastodynamics, to Maxwell's equations with the Poynting flux as $\mathbf{S}$ , and to the Einstein equations in harmonic gauge, where energy estimates are the technical heart of the local existence theory. Because the argument never uses a solution formula, it is essentially the only tool that survives into the nonlinear regime, where it gives existence up to the first time the energy ceases to be finite — and the question of whether that time is finite is, for the incompressible Navier–Stokes equations, still open.

Climate and ocean modelling. Long integrations of geophysical fluid models drift if the discrete energy budget is wrong: a scheme that loses a fixed fraction of energy per step damps the eddies the model exists to resolve, and one that gains energy eventually blows up. Energy- and enstrophy-conserving discretisations are chosen precisely so that the continuous estimate has a discrete counterpart, and so that the only dissipation in the run is the one that was put in deliberately.

Theorem · advanced

Evolution equations and operator semigroups

Adapted

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Reads a linear evolution equation as an ordinary differential equation in a Banach space, proves that the solution operators of such an equation form a contraction semigroup whose generator is densely defined and closed and whose resolvent is the Laplace transform of the semigroup, states the Hille-Yosida characterisation of the operators that generate one, and identifies the heat semigroup and its generator explicitly.

Prerequisites: Unbounded and closed operators, Spectrum and resolvent, The heat kernel, Plancherel's theorem, L^p spaces

Definition(Semigroups, generators and resolvents)AdaptedAdapted from Evans (2010), §7.4.1a, Eq. (1)–(2) and the definitions (4)–(7), pp. 409–410; Evans (2010), §7.4.1b, Eq. (8)–(9), p. 410; Evans (2010), §7.4.1c, the definitions of $\rho(A)$ and $R_\lambda$, pp. 412–413

Following Evans (Evans (2010), §7.4.1a, p. 409), let $X$ be a real Banach space and $A : D(A)\to X$ a linear, possibly unbounded, operator whose domain $D(A)$ is a linear subspace of $X$ . The abstract Cauchy problem is

u'(t) = A\,u(t) \quad (t\ge 0), \qquad u(0) = u_0 ,

for an unknown $u : [0,\infty)\to X$ , with $'=d/dt$ meaning the limit of difference quotients in the norm of $X$ . The interesting case is $X$ a space of functions of the space variables and $A$ a differential operator in those variables, when $A$ is necessarily unbounded.

A family $\{S(t)\}_{t\ge0}$ of bounded linear operators $X\to X$ is a semigroup if

$S(0)u = u$ for all $u\in X$ ;
$S(t+s)u = S(t)S(s)u = S(s)S(t)u$ for all $s,t\ge0$ and $u\in X$ ;
$t\mapsto S(t)u$ is continuous from $[0,\infty)$ into $X$ , for each $u\in X$ .

It is a contraction semigroup if in addition $\lVert S(t)\rVert\le1$ for all $t\ge0$ , the operator norm, so that $\lVert S(t)u\rVert\le\lVert u\rVert$ . Condition 3 is strong continuity, and elsewhere the object just defined is usually called a strongly continuous or $C_0$ semigroup; Evans builds strong continuity into the word "semigroup", and we follow him. Think of $S(t)u_0$ as "run the equation for time $t$ starting from $u_0$ ": condition 2 says that running for $s$ and then for $t$ is the same as running for $s+t$ . As Evans (Evans (2010), §7.4.1a, p. 410) puts it, that condition is just the assumption of existence and uniqueness for the Cauchy problem from every starting point.

The (infinitesimal) generator of a contraction semigroup is the operator $A$ with

D(A) = \Bigl\{\,u\in X \;:\; \lim_{t\to0^+}\frac{S(t)u-u}{t}\ \text{exists in } X \,\Bigr\}, \qquad Au = \lim_{t\to0^+}\frac{S(t)u-u}{t}\ \ (u\in D(A)) .

Finally, for a closed operator $A$ , a real number $\lambda$ lies in the resolvent set $\rho(A)$ if $\lambda I - A : D(A)\to X$ is one-to-one and onto, and then the resolvent operator is $R_\lambda = (\lambda I - A)^{-1} : X\to X$ , which is bounded by the closed graph theorem (Uniform boundedness and open mapping theorems). Compare Spectrum and resolvent, which defines $\rho(A)$ for bounded operators over $\mathbb{C}$ ; Evans takes $X$ real and $\lambda$ real throughout §7.4, and so does everything below.

Theorem(Generators, their resolvents, and the Hille-Yosida theorem)AdaptedAdapted from Evans (2010), §7.4.1b, Thm 1, p. 410, and Thm 2, p. 411; Evans (2010), §7.4.1c, Thm 3(ii) and Eq. (14), p. 413; Evans (2010), §7.4.2, Thm 4 and Eq. (17), p. 414, and the Remark with Eq. (23), p. 416

Let $\{S(t)\}_{t\ge0}$ be a contraction semigroup on a real Banach space $X$ , with generator $A$ .

Differential property. If $u\in D(A)$ then $S(t)u\in D(A)$ for every $t\ge0$ , the map $t\mapsto S(t)u$ is differentiable for $t>0$ , and $\frac{d}{dt}S(t)u = A\,S(t)u = S(t)\,Au \qquad (t>0).$ Since $t\mapsto S(t)Au$ is continuous, $t\mapsto S(t)u$ is $C^1$ on $(0,\infty)$ . So $u(t)=S(t)u_0$ solves the abstract Cauchy problem whenever $u_0\in D(A)$ .
Properties of the generator. $D(A)$ is dense in $X$ , and $A$ is a closed operator: if $u_k\in D(A)$ with $u_k\to u$ and $Au_k\to v$ in $X$ , then $u\in D(A)$ and $Au=v$ .
The resolvent is the Laplace transform of the semigroup. Every $\lambda>0$ lies in $\rho(A)$ , and $R_\lambda u = \int_0^{\infty} e^{-\lambda t}\,S(t)u\;dt \qquad (u\in X) , \tag{L}$ the integral converging as a Banach-space-valued improper Riemann integral. Consequently $\lVert R_\lambda\rVert \le 1/\lambda$ for every $\lambda>0$ .
Hille-Yosida theorem. Suppose $A$ is closed and densely defined on $X$ . Then some contraction semigroup has $A$ for its generator exactly when the resolvent condition $(0,\infty)\subseteq\rho(A) \qquad\text{and}\qquad \lVert R_\lambda\rVert \le \frac{1}{\lambda} \quad\text{for all }\lambda>0 \tag{HY}$ is satisfied. Evans states the equivalence; the semigroup is in addition unique, which is not part of his statement but follows from part 3, since the generator fixes $R_\lambda u$ for every $\lambda>0$ and $R_\lambda u$ is the Laplace transform of the bounded continuous function $t\mapsto S(t)u$ , which determines it.
The $\omega$ -contractive variant. Call $\{S(t)\}_{t\ge0}$ $\omega$ -contractive, for $\omega\in\mathbb{R}$ , if $\lVert S(t)\rVert\le e^{\omega t}$ for all $t\ge0$ . For closed, densely defined $A$ the criterion is again a bound on the resolvent, now on a shifted half-line: $A$ generates such a semigroup precisely when $(\omega,\infty)\subseteq\rho(A)$ , with $\lVert R_\lambda\rVert\le(\lambda-\omega)^{-1}$ at every $\lambda$ of that half-line. Part 4 is the case $\omega=0$ . Evans (Evans (2010), §7.4.2, Remark and Eq. (23), p. 416) records this as an easy variant of part 4 and needs it for parabolic equations, where $\omega>0$ .

Parts 1, 2 and 3 are proved below; part 4 is proved here only in the direction "generator $\Rightarrow$ (HY)", which is part 3, and part 5 is stated without proof, on the authority of Evans's remark.

Idea of the proofOriginal

Three short arguments carry parts 1 to 3, and each is the same trick: move an operator past a limit using the commutation rule 2 and the uniform bound $\lVert S(t)\rVert\le1$ .

For part 1, the difference quotient of $t\mapsto S(t)u$ at $t$ is $S(t)$ applied to the difference quotient at $0$ ; boundedness lets $S(t)$ pass through the limit, which is $Au$ . For part 2, the obstacle is that $D(A)$ might be small, and the cure is averaging: $t^{-1}\int_0^tS(s)u\,ds$ tends to $u$ by continuity and lies in $D(A)$ for every $u$ , because applying $S(r)$ to the average just slides the window of integration, and the slid window differs from the original by two short pieces whose difference quotient converges. Part 3 is the observation that the Laplace transform of a semigroup inverts $\lambda I - A$ : applying $S(h)$ to $\int_0^\infty e^{-\lambda t}S(t)u\,dt$ shifts the integration variable, and the mismatch between the shifted and the unshifted integral is a short initial piece whose average over $[0,h]$ tends to $u$ — which is where the $-u$ in $(\lambda I-A)R_\lambda u=u$ comes from.

The hard half of Hille-Yosida, that (HY) is enough, is a different matter and is not proved here; the note below says what it does.

ProofAdaptedAdapted from Evans (2010), §7.4.1b, proof of Thm 1, p. 411, and proof of Thm 2, pp. 411–412; Evans (2010), §7.4.1c, proof of Thm 3(ii), pp. 413–414

Throughout, $u\in D(A)$ where required, and $\lVert S(t)\rVert\le1$ is used freely.

Step 1: $S(t)$ maps $D(A)$ into itself and commutes with $A$ . For $t\ge0$ and $s>0$ , rule 2 gives $S(s)S(t)u = S(t)S(s)u$ , so

\frac{S(s)\bigl(S(t)u\bigr) - S(t)u}{s} \;=\; S(t)\left(\frac{S(s)u-u}{s}\right) .

As $s\to0^+$ the bracket tends to $Au$ and $S(t)$ is bounded, hence continuous, so the right-hand side tends to $S(t)Au$ . By the definition of the generator the left-hand side therefore has a limit, so $S(t)u\in D(A)$ and $A S(t)u = S(t)Au$ .

Step 2: the derivative (part 1). The same display, read as a forward difference quotient at $t$ , shows

\lim_{h\to0^+}\frac{S(t+h)u-S(t)u}{h} = S(t)Au .

For the backward quotient take $0<h<t$ and split

\frac{S(t)u - S(t-h)u}{h} - S(t)Au = S(t-h)\left(\frac{S(h)u-u}{h} - Au\right) + \bigl(S(t-h)-S(t)\bigr)Au .

The first term has norm at most $\lVert h^{-1}(S(h)u-u)-Au\rVert$ , because $\lVert S(t-h)\rVert\le1$ , and this tends to $0$ by the definition of $Au$ ; the second tends to $0$ by strong continuity (rule 3) applied to the vector $Au$ . So the two one-sided derivatives agree and $\frac{d}{dt}S(t)u = S(t)Au = AS(t)u$ , the last equality by Step 1. Both expressions are continuous in $t$ , so $t\mapsto S(t)u$ is $C^1$ on $(0,\infty)$ .

Step 3: $D(A)$ is dense (part 2, first half). Fix $u\in X$ and $t>0$ and set $u^t = \int_0^t S(s)u\,ds$ , a Banach-space-valued Riemann integral of a continuous integrand. Since $s\mapsto S(s)u$ is continuous at $0$ with value $u$ , we have $u^t/t\to u$ as $t\to0^+$ . It remains to show $u^t\in D(A)$ . For $r>0$ , rule 2 gives $S(r)S(s)u = S(r+s)u$ , so

\frac{S(r)u^t - u^t}{r} = \frac1r\left(\int_0^t S(r+s)u\,ds - \int_0^t S(s)u\,ds\right) = \frac1r\int_t^{t+r} S(s)u\,ds \;-\; \frac1r\int_0^{r} S(s)u\,ds ,

after substituting $s\mapsto s-r$ in the first integral and cancelling the overlap $[r,t]$ . As $r\to0^+$ the two averages tend to $S(t)u$ and to $u$ respectively, by continuity of $s\mapsto S(s)u$ . Hence $u^t\in D(A)$ with $Au^t = S(t)u - u$ , and the vectors $u^t/t\in D(A)$ converge to $u$ . As $u$ was arbitrary, $D(A)$ is dense.

Step 4: $A$ is closed (part 2, second half). For $u\in D(A)$ , integrating the identity of Step 2 from $0$ to $t$ (legitimate because $s\mapsto S(s)u$ is continuous on $[0,t]$ and $C^1$ on $(0,t)$ with continuous derivative $S(s)Au$ ) gives

S(t)u - u = \int_0^t S(s)\,Au\;ds . \tag{$\ast$}

Let $u_k\in D(A)$ with $u_k\to u$ and $Au_k\to v$ . Then $\lVert S(s)Au_k - S(s)v\rVert\le\lVert Au_k-v\rVert$ uniformly in $s$ , so passing to the limit in $(\ast)$ written for $u_k$ gives $S(t)u-u = \int_0^t S(s)v\,ds$ . Divide by $t$ and let $t\to0^+$ : the right-hand side tends to $v$ by continuity of $s\mapsto S(s)v$ at $0$ . So the defining limit for $u$ exists and equals $v$ , that is $u\in D(A)$ and $Au=v$ .

Step 5: the resolvent formula (part 3). Fix $\lambda>0$ and set $\tilde R_\lambda u = \int_0^\infty e^{-\lambda t}S(t)u\,dt$ . The integrand has norm at most $e^{-\lambda t}\lVert u\rVert$ , so the improper integral converges absolutely and

\lVert \tilde R_\lambda u\rVert \le \int_0^\infty e^{-\lambda t}\lVert u\rVert\,dt = \frac{\lVert u\rVert}{\lambda} . \tag{$\dagger$}

Apply $S(h)$ for $h>0$ and substitute $\tau = t+h$ :

S(h)\tilde R_\lambda u = \int_0^\infty e^{-\lambda t}S(t+h)u\,dt = e^{\lambda h}\int_h^\infty e^{-\lambda\tau}S(\tau)u\,d\tau = e^{\lambda h}\left(\tilde R_\lambda u - \int_0^h e^{-\lambda\tau}S(\tau)u\,d\tau\right) .

Therefore

\frac{S(h)\tilde R_\lambda u - \tilde R_\lambda u}{h} = \frac{e^{\lambda h}-1}{h}\,\tilde R_\lambda u \;-\; e^{\lambda h}\cdot\frac1h\int_0^h e^{-\lambda\tau}S(\tau)u\,d\tau \;\xrightarrow[h\to0^+]{}\; \lambda\,\tilde R_\lambda u - u ,

the first term because $(e^{\lambda h}-1)/h\to\lambda$ , the second because the average of a continuous integrand over $[0,h]$ tends to its value at $0$ , which is $u$ . Hence $\tilde R_\lambda u\in D(A)$ and

(\lambda I - A)\tilde R_\lambda u = u \qquad\text{for every } u\in X ,

so $\lambda I - A$ is onto. For injectivity, let $u\in D(A)$ . Each Riemann sum $\sum_i e^{-\lambda t_i}S(t_i)u\,\Delta t_i$ lies in $D(A)$ by Step 1, and applying $A$ to it gives $\sum_i e^{-\lambda t_i}S(t_i)Au\,\Delta t_i$ by Step 1 again; letting the mesh and then the truncation point go to their limits, both sums converge, so closedness of $A$ (Step 4) gives $\tilde R_\lambda u\in D(A)$ with $A\tilde R_\lambda u = \tilde R_\lambda Au$ . Hence

\tilde R_\lambda(\lambda I - A)u = \lambda\tilde R_\lambda u - \tilde R_\lambda A u = \lambda \tilde R_\lambda u - A\tilde R_\lambda u = u ,

using the previous display at the last step. So $\lambda I-A$ is one-to-one and onto, $\lambda\in\rho(A)$ , $R_\lambda = \tilde R_\lambda$ is given by (L), and $\lVert R_\lambda\rVert\le1/\lambda$ by $(\dagger)$ . This is the "only if" half of part 4. $\square$

Remark(Where the hypotheses are used)Original

Contractivity is used exactly once in the proof, in the backward difference quotient of Step 2, to bound $\lVert S(t-h)\rVert$ uniformly as $h\to0^+$ ; and once more in Step 5, to make the Laplace integral converge for every $\lambda>0$ . Replacing it by $\lVert S(t)\rVert\le e^{\omega t}$ changes only the range of $\lambda$ for which the integral converges, which is why part 5 of the statement has the same shape with $\lambda-\omega$ in place of $\lambda$ .
Closedness of $A$ is what makes the resolvent bounded (closed graph theorem) and is used in Step 5 to pull $A$ through the integral. It cannot be dropped from Hille-Yosida: if $A$ is not closed, $(\lambda I-A)^{-1}$ need not be a bounded operator and (HY) does not even make sense as stated.
Density of $D(A)$ does not appear in the proof above, because part 2 produces it. It is a hypothesis in part 4 because there one starts from a bare operator $A$ and must know in advance that it has the property every generator has.
The resolvent condition is where the content is. Take $X = L^2(0,1)$ , $Au=u'$ with $D(A)=H^1(0,1)$ , the Sobolev space of Weak derivatives and Sobolev spaces on the interval. This $A$ is densely defined and closed, because the graph norm is the $H^1$ norm and $H^1(0,1)$ is complete. But for every real $\lambda$ the function $u_\lambda(x)=e^{\lambda x}$ lies in $D(A)$ and satisfies $(\lambda I-A)u_\lambda = \lambda e^{\lambda x}-\lambda e^{\lambda x}=0$ , so $\lambda I-A$ is never one-to-one: $\rho(A)=\varnothing$ , (HY) fails as badly as it can, and $A$ generates no semigroup. Shrinking the domain to $D(A) = \{u\in H^1(0,1) : u(1)=0\}$ — which is meaningful because functions in $H^1(0,1)$ are absolutely continuous — repairs it. Now $(\lambda I-A)u=f$ with $u(1)=0$ has the unique solution $u(x)=\int_x^1 e^{-\lambda(y-x)}f(y)\,dy$ , and extending $f$ by zero to $\mathbb{R}$ exhibits $u$ as $f$ integrated against the kernel $z\mapsto e^{-\lambda z}\mathbf{1}_{z>0}$ , whose $L^1$ norm is $\lambda^{-1}$ ; so $\lVert u\rVert_2\le\lambda^{-1}\lVert f\rVert_2$ by Young's inequality and (HY) holds. The semigroup is the truncated left translation $S(t)u(x)=u(x+t)$ for $x+t<1$ and $0$ otherwise, which satisfies $\lVert S(t)u\rVert_2^2 = \int_t^1\lvert u\rvert^2\le\lVert u\rVert_2^2$ and vanishes identically for $t\ge1$ : formula (L) applied to it returns $\int_0^{1-x}e^{-\lambda t}u(x+t)\,dt$ , which is the solution just written, so by the uniqueness in part 4 this is the semigroup generated by $A$ . One boundary condition is the whole difference between an operator that generates nothing and one that generates a contraction semigroup.

Note(What the hard half of Hille-Yosida does)AdaptedAdapted from Evans (2010), §7.4.2, proof of Thm 4, steps 2–6, pp. 414–416

We proved only that a generator satisfies (HY). For the converse Evans (Evans (2010), §7.4.2, Eq. (18)–(21), pp. 414–415) replaces the unbounded $A$ by the bounded Yosida approximation $A_\lambda = -\lambda I + \lambda^2 R_\lambda = \lambda A R_\lambda$ , for which the exponential $S_\lambda(t) = e^{tA_\lambda}$ is an honest convergent power series and is a contraction because $\lVert R_\lambda\rVert\le\lambda^{-1}$ forces $\lVert S_\lambda(t)\rVert\le e^{-\lambda t}e^{\lambda^2 t\lVert R_\lambda\rVert}\le1$ . He then shows $A_\lambda u\to Au$ for $u\in D(A)$ — here the density of $D(A)$ enters, through $\lambda R_\lambda u\to u$ for all $u\in X$ — and that $\lVert S_\lambda(t)u - S_\mu(t)u\rVert\le t\lVert A_\lambda u - A_\mu u\rVert$ , so that $S(t)u=\lim_{\lambda\to\infty}S_\lambda(t)u$ exists, first on $D(A)$ and then on $X$ by density and the uniform bound. The limit is a contraction semigroup, and a last argument identifies its generator as $A$ . The point of recording the route is that it is constructive: it is the same idea as replacing an unbounded operator by a bounded approximation and passing to the limit, which is how one proves the spectral theorem for unbounded operators in The spectral theorem.

Method(Showing that a differential operator generates a semigroup)AdaptedAdapted from Evans (2010), §7.4.3a, Thm 5 and its proof, pp. 416–418

To put an evolution equation $u_t + Lu = 0$ into semigroup form, where $L$ is an elliptic operator in the space variables with coefficients that do not depend on $t$ :

Choose the space and the domain. Take $X$ to be the space in which you want the solution to live, typically $L^2$ of the spatial region, and let $D(A)$ be the subspace cut out by the boundary conditions and by the regularity you expect — for the Dirichlet problem on a bounded domain $U$ with smooth boundary, $X=L^2(U)$ and $D(A)=H_0^1(U)\cap H^2(U)$ . Set $Au=-Lu$ .
Check that $D(A)$ is dense in $X$ . For the domains above this is immediate, since they contain $C_c^\infty(U)$ .
Check that $A$ is closed. This is where an a priori estimate is needed: a bound of the form $\lVert u\rVert_{H^2}\le C(\lVert Au\rVert_{L^2}+\lVert u\rVert_{L^2})$ turns $L^2$ -convergence of $u_k$ and $Au_k$ into $H^2$ -convergence of $u_k$ , which puts the limit back in $D(A)$ .
Solve the resolvent equation. Show that $\lambda u - Au = f$ , that is $Lu+\lambda u=f$ with the boundary conditions, has exactly one solution in $D(A)$ for every $f\in X$ and every $\lambda$ beyond some $\omega$ . This is the elliptic existence theory: see The Lax-Milgram theorem and the Dirichlet problem, and then elliptic regularity to lift the weak solution into $H^2$ .
Estimate the resolvent. Test the equation against $u$ itself and use the coercivity estimate $\beta\lVert u\rVert_{H_0^1}^2\le B[u,u]+\gamma\lVert u\rVert_{L^2}^2$ for the bilinear form $B$ of $L$ , with constants $\beta>0$ and $\gamma\ge0$ ; this is Evans's (Evans (2010), §7.4.3a, Eq. (27), p. 417) energy estimate (27). This gives $(\lambda-\gamma)\lVert u\rVert_{L^2}^2\le\lVert f\rVert_{L^2}\lVert u\rVert_{L^2}$ , hence $\lVert R_\lambda\rVert\le(\lambda-\gamma)^{-1}$ for $\lambda>\gamma$ . See Energy estimates and well-posedness for estimates of this kind.
Conclude. By part 5 of the statement, $A$ generates a $\gamma$ -contractive semigroup, and $u(t)=S(t)u_0$ solves the evolution problem, with $t\mapsto u(t)$ a $C^1$ map into $X$ whenever $u_0\in D(A)$ .

Evans (Evans (2010), §7.4.3a, p. 418) notes two trade-offs. The method needs the coefficients of $L$ to be independent of $t$ , which the Galerkin construction of weak solutions does not; but it delivers a more regular solution from the start, since $u(t)$ lies in $D(A)$ rather than merely in $H_0^1$ .

Example(The heat semigroup on the line)Original

Take $X = L^2(\mathbb{R})$ and let $S(t)$ be convolution with the heat kernel of diffusivity $1$ ,

S(t)f = G_t * f, \qquad G_t(x) = \frac{1}{\sqrt{4\pi t}}e^{-x^2/4t}\ \ (t>0), \qquad S(0)=I .

Write $\hat g(k)=\int g(x)e^{-ikx}dx$ as in The Fourier transform, so that $\lVert g\rVert_2^2 = (2\pi)^{-1}\lVert\hat g\rVert_2^2$ by Plancherel. (The computations below are carried out in $L^2(\mathbb{R};\mathbb{C})$ , in which the real space $L^2(\mathbb{R};\mathbb{R})$ of the abstract theory sits isometrically; all the operators here are real, so no norm is affected.) Since $\hat G_t(k)=e^{-tk^2}$ , the convolution theorem gives

\widehat{S(t)f}(k) = e^{-tk^2}\hat f(k) .

It is a contraction semigroup. $e^{-sk^2}e^{-tk^2}=e^{-(s+t)k^2}$ gives rule 2; $\lvert e^{-tk^2}\rvert\le1$ and Plancherel give $\lVert S(t)f\rVert_2\le\lVert f\rVert_2$ ; and $\lVert S(t)f-f\rVert_2^2 = (2\pi)^{-1}\int\lvert e^{-tk^2}-1\rvert^2\lvert\hat f\rvert^2dk\to0$ as $t\to0^+$ by dominated convergence, with dominating function $4\lvert\hat f\rvert^2$ , which is rule 3.

Its generator is $d^2/dx^2$ on $H^2(\mathbb{R})$ . Let $H^2(\mathbb{R}) = \{f\in L^2 : k^2\hat f\in L^2\}$ , equivalently the $f$ whose first two weak derivatives are in $L^2$ . For $t>0$ the multiplier of $t^{-1}(S(t)-I)$ is $m_t(k) = (e^{-tk^2}-1)/t$ , which satisfies $\lvert m_t(k)\rvert\le k^2$ (because $\lvert e^{-x}-1\rvert\le x$ for $x\ge0$ ) and $m_t(k)\to-k^2$ pointwise. If $f\in H^2$ then $\lvert m_t\hat f + k^2\hat f\rvert^2\le 4k^4\lvert\hat f\rvert^2\in L^1$ , so dominated convergence gives $t^{-1}(S(t)f-f)\to f''$ in $L^2$ . Conversely, if $t^{-1}(S(t)f-f)\to g$ in $L^2$ then $m_t\hat f\to\hat g$ in $L^2$ , hence almost everywhere along a sequence $t_j\to0^+$ ; the pointwise limit is $-k^2\hat f$ , so $\hat g=-k^2\hat f$ and $f\in H^2$ . Thus $D(A)=H^2(\mathbb{R})$ and $Af=f''$ .

A worked case. Let $f(x)=e^{-x^2}$ . Convolving two Gaussians adds their variances: $f$ is $\sqrt{\pi}$ times the density of variance $\tfrac12$ and $G_t$ is the density of variance $2t$ , so

S(t)f(x) = \frac{1}{\sqrt{1+4t}}\,e^{-x^2/(1+4t)} .

Three checks on the abstract statements.

Semigroup law. Replacing $t$ by $s$ and then applying $S(t)$ replaces $1+4s$ by $1+4s+4t$ , which is the value of $1+4\tau$ at $\tau=s+t$ . So $S(t)S(s)f=S(s+t)f$ .
Generator. Differentiating the displayed formula at $t=0$ gives $-2e^{-x^2}+4x^2e^{-x^2}=(4x^2-2)e^{-x^2}$ , and indeed $f''(x)=(4x^2-2)e^{-x^2}$ .
Contraction, with a number. $\lVert S(t)f\rVert_2^2 = (1+4t)^{-1}\int e^{-2x^2/(1+4t)}dx = \sqrt{\pi/2}\,(1+4t)^{-1/2}$ , so $\frac{\lVert S(t)f\rVert_2}{\lVert f\rVert_2} = (1+4t)^{-1/4} .$ At $t=2$ this is $9^{-1/4}=3^{-1/2}=0.577350269\ldots$ , and since $\lVert f\rVert_2=(\pi/2)^{1/4}=1.119515135\ldots$ , we get $\lVert S(2)f\rVert_2 = 0.646352365\ldots\le\lVert f\rVert_2$ , as part 3 of the statement requires.

The resolvent, two ways. Formula (L) with $\lambda=1$ and $x=0$ gives

R_1f(0) = \int_0^\infty e^{-t}\,\frac{1}{\sqrt{1+4t}}\,dt = \frac{e^{1/4}}{2}\int_{1/4}^{\infty} v^{-1/2}e^{-v}\,dv = \frac{\sqrt{\pi}}{2}\,e^{1/4}\operatorname{erfc}(\tfrac12) = 0.545641361\ldots ,

substituting $1+4t = 4v$ and recognising the incomplete gamma function $\Gamma(\tfrac12,\tfrac14)=\sqrt{\pi}\operatorname{erfc}(\tfrac12)$ (see Error function and incomplete Gamma function). Independently, $R_1 = (I - d^2/dx^2)^{-1}$ is convolution with $\tfrac12e^{-\lvert x\rvert}$ , so

R_1f(0) = \frac12\int_{-\infty}^{\infty} e^{-\lvert y\rvert}e^{-y^2}dy = \int_0^\infty e^{-y-y^2}dy = e^{1/4}\int_0^\infty e^{-(y+1/2)^2}dy = \frac{\sqrt{\pi}}{2}e^{1/4}\operatorname{erfc}(\tfrac12) ,

the same number. The bound of part 3 is comfortably satisfied: $\lVert R_1\rVert\le1$ , and indeed the multiplier of $R_1$ is $(1+k^2)^{-1}$ , whose supremum is $1$ .

Application(Markov chains, diffusions and quantum evolution)AdaptedAdapted from Grimmett & Stirzaker (2020), §6.9, Thm (4), p. 289, and the generator (15), p. 293; Grimmett & Stirzaker (2020), §6.10(A), Eq. (1)–(4), pp. 297–298

Continuous-time Markov chains. Let $X$ be a continuous-time Markov chain on a countable state space $S$ , homogeneous, with transition probabilities $p_{ij}(t)=\mathbb{P}(X(t)=j\mid X(0)=i)$ and $P_t$ the matrix of these. Grimmett and Stirzaker (Grimmett & Stirzaker (2020), §6.9, Thm (4), p. 289) prove that $P_0=I$ , that each $P_t$ is stochastic, and that the Chapman-Kolmogorov equations $P_{s+t}=P_sP_t$ hold; they call $\{P_t : t\ge0\}$ the transition semigroup, and they note explicitly that the Chapman-Kolmogorov relation is the semigroup property. That is rules 1 and 2 of the definition above, so probability theory arrives at the same structure from the other direction. The chain's generator $G=(g_{ij})$ has non-negative off-diagonal entries and rows summing to zero, which is their (Grimmett & Stirzaker (2020), §6.9, Eq. (15), p. 293) condition (15), and the Kolmogorov forward and backward equations

P_t' = P_tG, \qquad P_t' = GP_t ,

derived by conditioning on $X(t)$ and on $X(h)$ respectively, are exactly the two halves of part 1 of the statement, $\frac{d}{dt}S(t)u = S(t)Au = AS(t)u$ . When $S$ is finite the generator is a bounded operator, $D(A)$ is everything, and their (Grimmett & Stirzaker (2020), §6.10(A), Eq. (4), p. 298) Eq. (4) solves the equations as the matrix exponential $P_t=\sum_n t^nG^n/n!$ . The abstract theory is what survives when $S$ is infinite and $G$ is unbounded, and Continuous-time Markov chains takes the probabilistic side further.

For a two-state chain with rates $2$ out of state $1$ and $3$ out of state $2$ ,

G = \begin{pmatrix}-2 & 2\\ 3 & -3\end{pmatrix}, \qquad P_t = e^{tG} = \frac15\begin{pmatrix}3&2\\3&2\end{pmatrix} + \frac{e^{-5t}}{5}\begin{pmatrix}2&-2\\-3&3\end{pmatrix} ,

since $G$ has eigenvalues $0$ and $-5$ . At $t=0$ this is $I$ ; at $t=0.2$ it gives $p_{11}=(3+2e^{-1})/5 = 0.747151776\ldots$ , and as $t\to\infty$ every row tends to the stationary distribution $(0.6,0.4)$ , at the rate $e^{-5t}$ set by the non-zero eigenvalue. The row sums are $1$ for every $t$ , as the zero row sums of $G$ require.

Diffusion. The example above is the semigroup side of the heat equation: $S(t)$ is the transition semigroup of Brownian motion run at twice the usual speed, since the law of standard Brownian motion at time $t$ has variance $t$ while $G_t$ has variance $2t$ . Taking $A=\tfrac12 d^2/dx^2$ instead gives standard Brownian motion exactly, and the resolvent formula (L) becomes the statement that the Green's function of $\lambda - \tfrac12d^2/dx^2$ is the expected occupation density of a path killed at rate $\lambda$ ; see Brownian motion. On a bounded domain the same construction with Dirichlet conditions gives the semigroup of Diffusion in bounded domains, and the method block is the route to it.

Quantum mechanics. The unitary counterpart is Stone's theorem: strongly continuous one-parameter groups of unitaries on a complex Hilbert space correspond to self-adjoint generators, $U(t)=e^{-itH}$ , which is how the Schrödinger equation is solved for a general Hamiltonian. The two theories are siblings: contraction semigroups run one way in time and dissipate, unitary groups run both ways and conserve, and in both the content is a condition on the resolvent of the generator.

Method · advanced

The Radon transform

Adapted

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Defines the integral of a function over every hyperplane of $\mathbb{R}^n$ , proves the projection-slice theorem that identifies its one-dimensional Fourier transform with a radial slice of the $n$ -dimensional one, inverts it by filtering and back-projecting, and applies the inversion to computed tomography and to the plane-wave solution of the wave equation.

Prerequisites: Fourier transform in several dimensions, Fourier inversion, Fubini and Tonelli theorems, The wave equation in three dimensions

Definition(Hyperplanes, the Radon transform and back-projection)AdaptedAdapted from Evans (2010), §4.3.2a, the notation and Eq. (30), p. 186; Stein & Shakarchi (2003), Ch. 6, §5.2, Eq. (12), p. 202, and Eq. (13), p. 205

Write $S^{n-1}$ for the unit sphere of $\mathbb{R}^n$ and, for $\omega\in S^{n-1}$ and $s\in\mathbb{R}$ ,

\Pi(s,\omega) \;=\; \{\,y\in\mathbb{R}^n \;:\; y\cdot\omega = s\,\} ,

the hyperplane with unit normal $\omega$ at signed distance $s$ from the origin; $s$ is allowed to be negative, and $\Pi(s,\omega)=\Pi(-s,-\omega)$ , so every hyperplane is named exactly twice.

To integrate over $\Pi(s,\omega)$ , pick an orthonormal basis $b_1,\dots,b_{n-1}$ of the subspace $\omega^{\perp}$ . Every point of $\Pi(s,\omega)$ is $s\omega + y_1b_1+\cdots+y_{n-1}b_{n-1}$ for exactly one $y\in\mathbb{R}^{n-1}$ , and the Radon transform of $f\in C_c^{\infty}(\mathbb{R}^n)$ is

\mathcal{R}f(s,\omega) \;=\; \int_{\Pi(s,\omega)} f\, dS \;=\; \int_{\mathbb{R}^{n-1}} f\Bigl(s\omega + \sum_{j=1}^{n-1} y_j b_j\Bigr)\,dy , \qquad s\in\mathbb{R},\ \omega\in S^{n-1},

a function on $\mathbb{R}\times S^{n-1}$ satisfying $\mathcal{R}f(-s,-\omega)=\mathcal{R}f(s,\omega)$ . The value does not depend on which orthonormal basis of $\omega^{\perp}$ is used: two such bases differ by an orthogonal map of $\mathbb{R}^{n-1}$ , and Lebesgue measure on $\mathbb{R}^{n-1}$ is invariant under orthogonal maps.

Going the other way, the back-projection (or dual Radon transform) of a function $F$ on $\mathbb{R}\times S^{n-1}$ is

\mathcal{R}^{*}F(x) \;=\; \int_{S^{n-1}} F(x\cdot\omega,\omega)\, dS(\omega) , \qquad x\in\mathbb{R}^n .

Because $x\in\Pi(s,\omega)$ exactly when $s=x\cdot\omega$ , this averages $F$ over the hyperplanes through $x$ : it smears each measurement back along the hyperplane it came from. Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 6, §5.2, p. 206) call it dual because $(\mathcal{R}f,F) = (f,\mathcal{R}^{*}F)$ for the natural inner products on $\mathbb{R}\times S^{n-1}$ and on $\mathbb{R}^n$ .

Fourier convention. Throughout, $\tilde f$ is the $n$ -dimensional Fourier transform in the symmetric convention of Fourier transform in several dimensions,

\tilde f(k) = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n} f(x)\,e^{-ik\cdot x}\,d^nx ,

and $\mathcal{F}_1$ is the same convention with $n=1$ , applied to the variable $s$ with $\omega$ held fixed:

\mathcal{F}_1 g(\sigma) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} g(s)\,e^{-i\sigma s}\,ds .

(The one-variable unit The Fourier transform puts the whole $2\pi$ on the inverse instead, so its $\hat g$ is $\sqrt{2\pi}\,\mathcal{F}_1g$ ; we use $\mathcal{F}_1$ here so that the two transforms in play are normalised alike.)

Method(Properties, projection-slice, and inversion by filtered back-projection)AdaptedAdapted from Evans (2010), §4.3.2a, Thm 3, pp. 186–187, Thm 4 and Eq. (31), p. 187, and Thm 5 and Eq. (32)–(34), p. 188; Stein & Shakarchi (2003), Ch. 6, Lemma 5.2, p. 204, Cor. 5.3, p. 205, and Thm 5.4, p. 206

Let $f\in C_c^{\infty}(\mathbb{R}^n)$ with $n\ge 2$ .

Support. If $f = 0$ outside the ball $B(0,R)$ then $\mathcal{R}f(s,\omega)=0$ for $\lvert s\rvert\ge R$ . In particular $\mathcal{R}f(\cdot,\omega)\in C_c^{\infty}(\mathbb{R})$ for each $\omega$ .
Derivatives. $\mathcal{R}\bigl(\partial_{i}f\bigr)(s,\omega) = \omega_i\,\partial_s\,\mathcal{R}f(s,\omega)$ for $i = 1,\dots,n$ , and hence $\mathcal{R}(\Delta f) = \partial_s^2\,\mathcal{R}f$ .
Projection-slice theorem. For every $\sigma\in\mathbb{R}$ and $\omega\in S^{n-1}$ , $\mathcal{F}_1\bigl[\mathcal{R}f(\cdot,\omega)\bigr](\sigma) \;=\; (2\pi)^{(n-1)/2}\,\tilde f(\sigma\omega) . \tag{PS}$ The one-dimensional transform of one projection is the $n$ -dimensional transform restricted to the line $\mathbb{R}\omega$ through the origin.
Uniqueness. If $\mathcal{R}f = \mathcal{R}g$ for $f,g\in C_c^{\infty}(\mathbb{R}^n)$ , then $f=g$ .
Filtered back-projection. Let $\Lambda_n$ be the one-dimensional operator with Fourier multiplier $\lvert\sigma\rvert^{\,n-1}$ , that is $\mathcal{F}_1[\Lambda_n g](\sigma) = \lvert\sigma\rvert^{\,n-1}\mathcal{F}_1 g(\sigma)$ . Then $f \;=\; \frac{1}{2(2\pi)^{\,n-1}}\;\mathcal{R}^{*}\bigl[\Lambda_n\,\mathcal{R}f\bigr] . \tag{FBP}$
Odd dimensions: a local filter. If $n = 2k+1$ is odd then $\lvert\sigma\rvert^{n-1}=\sigma^{2k}$ , so $\Lambda_n = (-1)^k\partial_s^{2k}$ and (FBP) becomes the decomposition of $f$ into plane waves $f(x) \;=\; \int_{S^{n-1}} \gamma(x\cdot\omega,\omega)\,dS(\omega), \qquad \gamma(s,\omega) \;=\; \frac{(-1)^k}{2(2\pi)^{2k}}\,\frac{\partial^{2k}}{\partial s^{2k}}\,\mathcal{R}f(s,\omega) . \tag{PW}$ For $n=3$ this reads $f = -\dfrac{1}{8\pi^2}\,\Delta\bigl(\mathcal{R}^{*}\mathcal{R}f\bigr)$ .

Idea of the proofOriginal

A hyperplane integral kills all the oscillation of $f$ transverse to $\omega$ and keeps only the oscillation along $\omega$ . That one sentence is the projection-slice theorem: the exponential $e^{-i\sigma s}$ in the one-dimensional transform of $\mathcal{R}f(\cdot,\omega)$ is constant on each hyperplane $\Pi(s,\omega)$ , so it may be pulled inside the hyperplane integral, where it becomes $e^{-i(\sigma\omega)\cdot x}$ ; collapsing the iterated integral back to one over $\mathbb{R}^n$ then gives $\tilde f(\sigma\omega)$ . Everything else follows from it. Uniqueness, because the slices $\sigma\mapsto\tilde f(\sigma\omega)$ over all $\omega$ cover $\mathbb{R}^n$ . Inversion, because Fourier inversion is an integral over $\mathbb{R}^n$ , and writing it in polar coordinates $k=\sigma\omega$ splits it into a radial integral — which is a one-dimensional inverse transform of the projection, weighted by the Jacobian $\lvert\sigma\rvert^{n-1}$ — and an angular integral, which is exactly back-projection. The Jacobian is the filter: back-projection alone over-counts low frequencies, because many hyperplanes pass near the origin and few pass far from it, and $\lvert\sigma\rvert^{n-1}$ is the precise correction.

ProofAdaptedAdapted from Evans (2010), §4.3.2a, proofs of Thm 3, p. 187, Thm 4, p. 188, and Thm 5, p. 189

Fix $\omega$ and an orthonormal basis $b_1,\dots,b_{n-1}$ of $\omega^{\perp}$ . The map $T_\omega(s,y)= s\omega+\sum_j y_jb_j$ is an orthogonal linear map of $\mathbb{R}\times\mathbb{R}^{n-1}$ onto $\mathbb{R}^n$ , so it has Jacobian determinant of modulus $1$ and $\int_{\mathbb{R}^n}h\,dx = \int_{\mathbb{R}}\int_{\mathbb{R}^{n-1}} h(T_\omega(s,y))\,dy\,ds$ for every integrable $h$ .

Step 1 (support, part 1). If $\lvert s\rvert\ge R$ then every point of $\Pi(s,\omega)$ has $\lvert y\rvert\ge\lvert y\cdot\omega\rvert = \lvert s\rvert \ge R$ , so the hyperplane misses $B(0,R)$ and the integrand vanishes identically. Hence $\mathcal{R}f(s,\omega)=0$ . Smoothness in $s$ holds because $f$ is smooth and compactly supported, so differentiation under the integral sign is legitimate.

Step 2 (derivatives, part 2). Differentiating under the integral sign in the direction $\omega$ ,

\partial_s\,\mathcal{R}f(s,\omega) = \int_{\mathbb{R}^{n-1}} \nabla f\bigl(T_\omega(s,y)\bigr)\cdot\omega \;dy = \int_{\Pi(s,\omega)} \nabla f\cdot\omega\; dS .

For a tangential direction the corresponding integral vanishes: for fixed $j$ and fixed values of the other coordinates, $t\mapsto f(T_\omega(s,y)+tb_j)$ is a smooth compactly supported function of one variable, so $\int_{\mathbb{R}} \nabla f\cdot b_j\,dt = 0$ by the fundamental theorem of calculus, and integrating over the remaining $n-2$ coordinates gives $\int_{\Pi(s,\omega)}\nabla f\cdot b_j\,dS = 0$ . Here the compact support of $f$ is used. Expanding $\nabla f$ in the orthonormal basis $b_1,\dots,b_{n-1},\omega$ of $\mathbb{R}^n$ and taking the $i$ -th component,

\mathcal{R}(\partial_i f)(s,\omega) = \sum_{j=1}^{n-1} (b_j\cdot e_i)\int_{\Pi(s,\omega)}\nabla f\cdot b_j\,dS \;+\; \omega_i\int_{\Pi(s,\omega)}\nabla f\cdot\omega\,dS = \omega_i\,\partial_s\,\mathcal{R}f(s,\omega) .

Applying this twice gives $\mathcal{R}(\partial_i^2f)=\omega_i^2\partial_s^2\mathcal{R}f$ , and summing over $i$ gives $\mathcal{R}(\Delta f)=\partial_s^2\mathcal{R}f$ because $\sum_i\omega_i^2=\lvert\omega\rvert^2=1$ .

Step 3 (projection-slice, part 3). Every $x=T_\omega(s,y)$ has $x\cdot\omega = s$ , since $\omega\cdot\omega=1$ and $b_j\cdot\omega=0$ . So the factor $e^{-i\sigma s}$ is the restriction to $\Pi(s,\omega)$ of the function $x\mapsto e^{-i(\sigma\omega)\cdot x}$ , and may be carried inside the hyperplane integral:

\mathcal{F}_1\bigl[\mathcal{R}f(\cdot,\omega)\bigr](\sigma) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\int_{\mathbb{R}^{n-1}} f\bigl(T_\omega(s,y)\bigr)\,e^{-i(\sigma\omega)\cdot T_\omega(s,y)}\,dy\,ds = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}^n} f(x)\,e^{-i(\sigma\omega)\cdot x}\,dx ,

the last equality by the change of variables above. The right-hand side is $(2\pi)^{n/2}(2\pi)^{-1/2}\tilde f(\sigma\omega)$ , which is (PS). The double integral converges absolutely because $f$ is continuous with compact support, so Fubini's theorem (Fubini and Tonelli theorems) licenses the rearrangement.

Step 4 (uniqueness, part 4). If $\mathcal{R}f=\mathcal{R}g$ then by (PS) $\tilde f(\sigma\omega)=\tilde g(\sigma\omega)$ for all $\sigma\in\mathbb{R}$ , $\omega\in S^{n-1}$ . Every $k\in\mathbb{R}^n$ is of the form $\sigma\omega$ , so $\tilde f=\tilde g$ , and $f=g$ by Fourier inversion.

Step 5 (inversion, part 5). Since $f\in C_c^\infty(\mathbb{R}^n)$ , $\tilde f$ is a Schwartz function, so $\lvert k\rvert^{\,m}\lvert\tilde f(k)\rvert$ is integrable for every $m$ and all the integrals below converge absolutely. Fourier inversion in polar coordinates $k=\sigma\omega$ , $\sigma>0$ , with $d^nk = \sigma^{n-1}d\sigma\,dS(\omega)$ , gives

f(x) = \frac{1}{(2\pi)^{n/2}}\int_{0}^{\infty}\!\!\int_{S^{n-1}} \tilde f(\sigma\omega)\,e^{i\sigma(\omega\cdot x)}\,\sigma^{n-1}\,dS(\omega)\,d\sigma .

The integrand of

J \;=\; \int_{\mathbb{R}}\int_{S^{n-1}} \tilde f(\sigma\omega)\,e^{i\sigma(\omega\cdot x)}\,\lvert\sigma\rvert^{\,n-1}\,dS(\omega)\,d\sigma

is unchanged by $(\sigma,\omega)\mapsto(-\sigma,-\omega)$ , which also preserves $dS$ ; hence the part of $J$ with $\sigma<0$ equals the part with $\sigma>0$ , and $f(x) = J/\bigl(2(2\pi)^{n/2}\bigr)$ . Here the evenness built into the pair $(s,\omega)$ is used, and it is what makes $\lvert\sigma\rvert^{n-1}$ , rather than $\sigma^{n-1}$ , the right weight in even dimensions.

Now substitute $\tilde f(\sigma\omega) = (2\pi)^{-(n-1)/2}\mathcal{F}_1[\mathcal{R}f(\cdot,\omega)](\sigma)$ from (PS) and exchange the order of integration:

f(x) = \frac{1}{2(2\pi)^{n/2}(2\pi)^{(n-1)/2}}\int_{S^{n-1}}\left[\int_{\mathbb{R}} \lvert\sigma\rvert^{\,n-1}\,\mathcal{F}_1[\mathcal{R}f(\cdot,\omega)](\sigma)\,e^{i\sigma(\omega\cdot x)}\,d\sigma\right] dS(\omega) .

The inner bracket is $\sqrt{2\pi}$ times the inverse one-dimensional transform of $\lvert\sigma\rvert^{n-1}\mathcal{F}_1[\mathcal{R}f(\cdot,\omega)]$ evaluated at $s=\omega\cdot x$ , that is $\sqrt{2\pi}\,(\Lambda_n\mathcal{R}f)(\omega\cdot x,\omega)$ . Since $(2\pi)^{n/2}(2\pi)^{(n-1)/2}(2\pi)^{-1/2} = (2\pi)^{\,n-1}$ , we get

f(x) = \frac{1}{2(2\pi)^{\,n-1}}\int_{S^{n-1}} (\Lambda_n\mathcal{R}f)(x\cdot\omega,\omega)\,dS(\omega) = \frac{1}{2(2\pi)^{\,n-1}}\,\mathcal{R}^{*}\bigl[\Lambda_n\mathcal{R}f\bigr](x) ,

which is (FBP).

Step 6 (odd dimensions, part 6). Let $n=2k+1$ . By the derivative rule for $\mathcal{F}_1$ , $\mathcal{F}_1[\partial_s^{2k}g](\sigma) = (i\sigma)^{2k}\mathcal{F}_1g(\sigma) = (-1)^k\sigma^{2k}\mathcal{F}_1g(\sigma)$ , and $\sigma^{2k}=\lvert\sigma\rvert^{\,n-1}$ ; so $\Lambda_n g = (-1)^k\partial_s^{2k}g$ , and (FBP) becomes (PW) with the stated $\gamma$ . For $n=3$ , $k=1$ , the factor is $-1/(2(2\pi)^2) = -1/(8\pi^2)$ , and $\partial_s^{2}\mathcal{R}f(x\cdot\omega,\omega) = \Delta_x\bigl[\mathcal{R}f(x\cdot\omega,\omega)\bigr]$ because $\lvert\omega\rvert=1$ ; moving $\Delta_x$ outside the $\omega$ -integral gives $f = -(8\pi^2)^{-1}\Delta(\mathcal{R}^{*}\mathcal{R}f)$ . $\square$

Remark(Where the hypotheses are used)Original

Compact support enters only once, in Step 2, where the tangential integrals $\int_{\Pi}\nabla f\cdot b_j\,dS$ are made to vanish. Without decay in the tangential directions the derivative rule fails, and so does the whole scheme: a plane wave $f(x)=F(x\cdot\omega_0)$ is constant along the hyperplanes $\Pi(s,\omega_0)$ , so $\mathcal{R}f(s,\omega_0)=\pm\infty$ and the transform is not even defined. The transform needs decay in $n-1$ directions at once, which is more than the integrability that defines the Fourier transform.
Smoothness enters in Step 1 (differentiating under the integral sign) and in Step 5, where rapid decay of $\tilde f$ makes $\lvert\sigma\rvert^{n-1}\tilde f(\sigma\omega)$ integrable. Merely integrable $f$ has $\tilde f$ bounded but with no decay at all, and the filtered integral need not converge. This is the mathematical face of the practical fact that filtering amplifies high frequencies, so the ramp filter must be tapered before it is applied to noisy data.
$\lvert\sigma\rvert$ , not $\sigma$ , in part 5: with $\sigma^{n-1}$ in place of $\lvert\sigma\rvert^{n-1}$ the symmetry step would produce a cancellation rather than a doubling whenever $n$ is even, and $J$ would be $0$ . The two agree exactly when $n$ is odd, which is the case treated in (PW).
Odd versus even $n$ . The filter $\Lambda_n$ is a differential operator, hence local, only when $n-1$ is even. For even $n$ — the case of planar computed tomography, $n=2$ , where $\Lambda_2$ has multiplier $\lvert\sigma\rvert$ — it is a non-local operator, and reconstructing $f$ near a point needs the projections of the whole object, not just those passing nearby. Evans (Evans (2010), §4.3.2a, (35), pp. 189–190) draws the same contrast from the support side: in odd dimensions $\mathcal{R}f(s,\omega)=0$ for all $\lvert s\rvert\le R$ forces $f\equiv0$ on $B(0,R)$ , which (PW) gives at once and which has no such one-line proof in even dimensions.

Note(Conventions, and a difference between the two sources)AdaptedAdapted from Evans (2010), §4.3.2a, Eq. (31)–(32), pp. 187–188; Stein & Shakarchi (2003), Ch. 6, Lemma 5.2, p. 204, and Thm 5.4, p. 206

The Radon transform itself involves no Fourier convention, so $\mathcal{R}$ is the same object in both sources; only the statements that mention $\tilde f$ change shape.

Evans (Evans (2010), §4.3.1, pp. 177–178) uses the same symmetric transform as this handbook, but states (PS) with the unnormalised one-dimensional transform $\bar u(r,\omega)=\int\tilde u(s,\omega)e^{-irs}ds$ , in the form $\bar u(r,\omega)=(2\pi)^{n/2}\hat u(r\omega)$ . Since $\bar u = \sqrt{2\pi}\,\mathcal{F}_1[\mathcal{R}u]$ , this is (PS). He writes $\tilde u$ for what we write $\mathcal{R}u$ ; we avoid the tilde for the Radon transform because this handbook already uses it for the Fourier transform on $\mathbb{R}^n$ .
Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 6, Lemma 5.2, p. 204) work in $\mathbb{R}^3$ with the $2\pi$ -in-the-exponent transform, in which (PS) carries no constant at all: $\widehat{R(f)}(s,\gamma) = \hat f(s\gamma)$ . They also prove the $n=3$ results for all $f$ in the Schwartz class $\mathcal{S}(\mathbb{R}^3)$ , a larger class than the $C_c^{\infty}(\mathbb{R}^n)$ of Evans and of the statement above.
In the inversion formula Evans writes $r^{n-1}$ where part 5 above has $\lvert\sigma\rvert^{\,n-1}$ . He applies the formula only for odd $n$ , where the two coincide, and the $(-1)^k$ in his (34) matches ours; readers transporting his (32) to even $n$ should use the modulus, for the reason given in the remark.
The two constants can be checked against each other. Stein and Shakarchi's (Stein & Shakarchi (2003), Ch. 6, Thm 5.4, p. 206) Theorem 5.4 says $\triangle(R^{*}R(f)) = -8\pi^2 f$ , which is exactly the $n=3$ case of (PW) rearranged. Their result is derived in a different Fourier convention from ours, and the agreement is a genuine check: the identity contains no transform, so it must come out the same either way.

Method(Reconstructing an object from its projections)Original

Given measured projections $p(s,\omega)=\mathcal{R}f(s,\omega)$ for $\omega$ in a set of directions and $s$ on a grid, recover $f$ as follows. Steps 2–4 are filtered back-projection; steps 1 and 5 are what practice adds to the theorem.

Fix the geometry. Decide $n$ , the sampling in $s$ , and the directions $\omega$ . Hyperplanes are named twice, $\Pi(s,\omega)=\Pi(-s,-\omega)$ , so for $n=2$ the angles need only cover a half-turn.
Transform each projection. For each fixed $\omega$ , compute $\mathcal{F}_1[p(\cdot,\omega)]$ , in practice by the FFT.
Filter. Multiply by $\lvert\sigma\rvert^{\,n-1}$ and transform back, giving $q(\cdot,\omega) = \Lambda_n p(\cdot,\omega)$ . In odd dimensions the same thing is done in the $s$ variable directly, as $(-1)^k\partial_s^{2k}$ with $n=2k+1$ .
Back-project and scale. Form $\mathcal{R}^{*}q(x)=\int_{S^{n-1}}q(x\cdot\omega,\omega)\,dS(\omega)$ by summing, for each reconstruction point $x$ , the filtered value of the hyperplane through it in each measured direction; then multiply by $1/\bigl(2(2\pi)^{n-1}\bigr)$ . By (FBP) the result is $f$ .
Temper the filter. $\lvert\sigma\rvert^{n-1}$ grows without bound, so with noisy data replace it by $\lvert\sigma\rvert^{n-1}w(\sigma)$ for a window $w$ that cuts off above the frequency the sampling supports. This is a deliberate departure from the exact formula: it reconstructs a smoothed version of $f$ in exchange for stability.

The order matters only for cost, not for the answer: filtering and back-projection commute with each other in the sense that one may instead back-project first and apply the corresponding $n$ -dimensional filter afterwards, but filtering the one-dimensional projections is far cheaper.

Example(Exact reconstruction of a Gaussian in three dimensions)Original

Take $n=3$ and $f(x)=e^{-\lvert x\rvert^{2}}$ on $\mathbb{R}^3$ . (This is Schwartz rather than compactly supported; every integral below converges absolutely, and the reconstruction formula holds, as Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 6, Thm 5.4, p. 206) prove it for the whole Schwartz class.)

Step 1: the projections. Fix $\omega$ and write $x = s\omega+u$ with $u\in\omega^{\perp}$ . Then $\lvert x\rvert^2 = s^2+\lvert u\rvert^2$ by orthogonality, so the integrand factorises and

\mathcal{R}f(s,\omega) = e^{-s^2}\int_{\mathbb{R}^{2}} e^{-\lvert u\rvert^{2}}\,du = \pi\,e^{-s^{2}} ,

using $\int_{\mathbb{R}^2}e^{-|u|^2}du = (\sqrt{\pi})^2=\pi$ . The projections do not depend on $\omega$ , as they must not, since $f$ is radial.

Step 2: the filter. Here $k=1$ , so $\Lambda_3 = -\partial_s^2$ . Differentiating twice,

\partial_s^2\bigl(\pi e^{-s^2}\bigr) = \pi\,(4s^{2}-2)\,e^{-s^{2}}, \qquad\text{so}\qquad \gamma(s,\omega) = -\frac{1}{8\pi^{2}}\,\pi(4s^{2}-2)e^{-s^{2}} = \frac{(1-2s^{2})\,e^{-s^{2}}}{4\pi} .

Step 3: back-projection. Let $\rho=\lvert x\rvert>0$ . For a function $h$ of one variable, writing $t=\cos\theta$ for the angle between $\omega$ and $x$ ,

\int_{S^{2}} h(x\cdot\omega)\,dS(\omega) = 2\pi\int_{-1}^{1} h(\rho t)\,dt = \frac{2\pi}{\rho}\int_{-\rho}^{\rho} h(v)\,dv ,

the first equality because the area element of $S^2$ is $2\pi\,dt$ in the variable $t=\cos\theta$ . With $h=\gamma$ and the antiderivative $\int (1-2v^2)e^{-v^2}dv = v\,e^{-v^{2}}$ — check by differentiating: $\frac{d}{dv}\bigl(ve^{-v^2}\bigr) = (1-2v^2)e^{-v^2}$ — we get

\int_{S^{2}}\gamma(x\cdot\omega,\omega)\,dS(\omega) = \frac{2\pi}{\rho}\cdot\frac{1}{4\pi}\Bigl[v\,e^{-v^{2}}\Bigr]_{-\rho}^{\rho} = \frac{1}{2\rho}\cdot 2\rho\,e^{-\rho^{2}} = e^{-\rho^{2}} ,

which is $f(x)$ . The reconstruction is exact at every point.

A number to check. At $\lvert x\rvert = 1$ the back-projected filtered data is

-\frac{1}{8\pi^{2}}\int_{S^{2}}\partial_s^{2}\mathcal{R}f(x\cdot\omega,\omega)\,dS(\omega) = -\frac{1}{4}\int_{-1}^{1}(4t^{2}-2)e^{-t^{2}}\,dt = -\frac{1}{4}\Bigl[-2t\,e^{-t^{2}}\Bigr]_{-1}^{1} = -\frac{1}{4}\bigl(-4e^{-1}\bigr) = e^{-1} = 0.367879441\ldots,

and $f(x) = e^{-1}$ as well.

What back-projection alone would give. Omitting the filter and the constant, $\mathcal{R}^{*}\mathcal{R}f(x) = \pi\int_{S^2}e^{-(x\cdot\omega)^2}dS(\omega) = (2\pi^{2}/\rho)\int_{-\rho}^{\rho}e^{-v^{2}}dv = 2\pi^{5/2}\operatorname{erf}(\rho)/\rho$ . At $\rho=1$ this is $2\pi^{5/2}\operatorname{erf}(1) = 2(17.493418\ldots)(0.8427008\ldots) = 29.483435\ldots$ , and it decays like $1/\rho$ rather than like a Gaussian: back-projection smears the object out, and the filter is what undoes the smearing.

Application(Computed tomography, and plane waves for the wave equation)AdaptedAdapted from Stein & Shakarchi (2003), Ch. 6, §5.1, pp. 199–200, and §5.3, p. 207; Evans (2010), §4.3.2b, Example 6, Eq. (36)–(38), p. 190

X-ray tomography. A narrow X-ray beam of incident intensity $I_0$ traversing an object of attenuation coefficient $\rho$ along a line $L$ emerges with intensity $I = I_0\exp\bigl(-\int_L\rho\bigr)$ ; this is the law Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 6, §5.1, pp. 199–200) start from. A scanner measures $I$ and $I_0$ , so each beam returns the line integral $\int_L\rho = \log(I_0/I)$ , and a scan returns these for a large family of lines through one planar slice. In the plane that family of line integrals is the Radon transform, $n=2$ , and the method above with the ramp filter $\lvert\sigma\rvert$ is the reconstruction algorithm that turns the measurements into the image. The degrees of freedom match: a function on $\mathbb{R}^2$ and a function of lines in $\mathbb{R}^2$ each depend on two parameters. In $\mathbb{R}^3$ the line data is over-determined — four parameters for a line against three for a point — and Stein and Shakarchi (Stein & Shakarchi (2003), Ch. 6, §5.2, p. 203) note that integrals over all lines determine integrals over all planes, since a plane integral is an iterated line integral; this is why the clean three-dimensional theory above is stated for hyperplanes.

As an illustration of forward data, a uniform ball $B(0,R)\subset\mathbb{R}^3$ of unit density has $\mathcal{R}\mathbf{1}_{B(0,R)}(s,\omega) = \pi(R^{2}-s^{2})$ for $\lvert s\rvert\le R$ and $0$ otherwise, because $\Pi(s,\omega)\cap B(0,R)$ is a disc of radius $\sqrt{R^2-s^2}$ . For $R=2$ and $s=1$ this is $3\pi = 9.42477796\ldots$ — a parabolic profile in $s$ , which is the signature of a sphere in a projection and is used to calibrate scanners.

Plane-wave solution of the wave equation. For a unit vector $\omega$ and a $C^2$ function $\gamma$ of one variable, the plane wave $u(x,t)=\gamma(x\cdot\omega-t)$ solves $u_{tt}=\Delta u$ : both sides equal $\gamma''$ , the right-hand side because $\lvert\omega\rvert=1$ . Superposing over all directions, Evans (Evans (2010), §4.3.2b, Example 6, Eq. (36)–(38), p. 190) observes that for odd $n$ the function

u(x,t) = \int_{S^{n-1}} \gamma(x\cdot\omega - t,\omega)\, dS(\omega) , \qquad \gamma = \frac{(-1)^k}{2(2\pi)^{2k}}\,\partial_s^{2k}\mathcal{R}g,\quad n=2k+1,

solves $u_{tt}=\Delta u$ with $u=g$ and $u_t=0$ at $t=0$ . The initial value is right by (PW); the initial velocity vanishes by an antipodal cancellation. Indeed part 2 of the statement gives $\partial_s^{2k}\mathcal{R}g = \mathcal{R}(\Delta^kg)$ , so $\gamma = \mathcal{R}l$ for $l = (-1)^k(2(2\pi)^{2k})^{-1}\Delta^kg$ , and Step 2 of the proof then gives $\partial_s\gamma(s,\omega)=\int_{\Pi(s,\omega)}\nabla l\cdot\omega\,dS$ . Replacing $\omega$ by $-\omega$ leaves the hyperplane $\Pi(x\cdot\omega,\omega)$ unchanged and reverses the sign of $\nabla l\cdot\omega$ , so $u_t(x,0) = -\int_{S^{n-1}}\partial_s\gamma(x\cdot\omega,\omega)\,dS(\omega) = 0$ . Where Kirchhoff's formula builds the solution out of spherical means, this builds it out of plane waves; for a problem whose data are known through their projections, the second is the one to use.

A check on the two. Take $n=3$ and $g(x)=e^{-\lvert x\rvert^2}$ , so that $\gamma(s,\omega) = (1-2s^2)e^{-s^2}/(4\pi)$ from the example above. At the origin all directions give the same value, so

u(0,t) = 4\pi\,\gamma(-t,\omega) = (1-2t^{2})\,e^{-t^{2}} .

Kirchhoff's formula with $c=1$ , $\psi=0$ gives $u(0,t)=\partial_t\bigl[t\,\mathcal{M}_t g(0)\bigr]$ , and the spherical mean of $e^{-|y|^2}$ over $\lvert y\rvert=t$ is $e^{-t^2}$ , so $u(0,t)=\partial_t\bigl(te^{-t^{2}}\bigr) = (1-2t^{2})e^{-t^{2}}$ — the same function. At $t=1$ both give $u(0,1) = -e^{-1} = -0.367879441\ldots$ : the wave has passed and the signal at the origin has gone negative.