The wave equation

Solutions of the wave equation in one and three dimensions.

In this chapter 4

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Waves

Waves

Theorem

The one-dimensional wave equation

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Gives the general solution $F(x-ct)+G(x+ct)$ of $u_{tt}=c^2u_{xx}$ and d'Alembert's formula for the initial-value problem, proves that the formula solves it and is the only solution, reads off the domains of dependence and influence, and extends both to a half-line or an interval by odd and even reflection.

Prerequisites: Classification of second-order equations, Partial derivatives and the chain rule

Definition(The problem, and its characteristics)AdaptedAdapted from Strauss (2008), §2.1, Eq. (1)–(2), p. 33, and the characteristic coordinates on p. 34; Arfken, Weber & Harris (2013), §9.6, Eq. (9.89), p. 435; Evans (2010), §2.4, the two initial conditions of the wave equation, p. 64; Stein & Shakarchi (2003), Ch. 1 §1, the change of variables on p. 10

Fix a constant $c > 0$ , the wave speed. The one-dimensional wave equation is

u_{tt} = c^2 u_{xx} ,

for a real-valued $u(x,t)$ . For a stretched string of tension $\tau$ and line density $\rho$ it is the small-displacement equation of motion with $c = \sqrt{\tau/\rho}$ ; Strauss writes the tension as $T$ .

Two features of the equation are used throughout. First, the operator factors: writing $\partial_t = \partial/\partial t$ and $\partial_x = \partial/\partial x$ ,

\partial_t^2 - c^2\partial_x^2 = (\partial_t - c\,\partial_x)(\partial_t + c\,\partial_x) = (\partial_t + c\,\partial_x)(\partial_t - c\,\partial_x),

the two orders agreeing because $c$ is constant. Second, the lines

x - ct = \text{const} \qquad\text{and}\qquad x + ct = \text{const}

in the $(x,t)$ plane are the characteristics of the equation (classification); the equation is hyperbolic, with two real families of them. The corresponding characteristic coordinates are

\xi = x - ct, \qquad \eta = x + ct .

The initial-value problem (or Cauchy problem) on the whole line is: given $\varphi$ and $\psi$ on $\mathbb{R}$ , find $u$ on $\mathbb{R}\times[0,\infty)$ with

u_{tt} = c^2u_{xx} \quad\text{in } \mathbb{R}\times(0,\infty), \qquad u(x,0) = \varphi(x), \quad u_t(x,0) = \psi(x) \quad (x \in \mathbb{R}).

The datum $\varphi$ prescribes where the string starts and $\psi$ how fast each point of it is moving at that instant. Two initial conditions are the right number because the equation is second order in $t$ ; the physical reading, which Evans gives, is that for a vibrating body one may prescribe the displacement and the velocity independently at the start.

Conventions. We write the general solution with the right-moving part first, $u = F(x-ct) + G(x+ct)$ , as Riley, Hobson and Bence and Arfken do. Strauss and Evans write the two terms in the opposite order, $f(x+ct)+g(x-ct)$ ; nothing but the naming of the two functions differs. Evans and Stein and Shakarchi take $c = 1$ and call the data $g,h$ and $f,g$ respectively. The naming of the characteristic coordinates above is also ours and is the reverse of the sources': Strauss (p. 34) sets $\xi = x+ct$ , $\eta = x-ct$ , and Stein and Shakarchi (p. 10) do the same with $c=1$ , so their $\xi$ is our $\eta$ . Since the equation $v_{\xi\eta}=0$ is symmetric in the two, the exchange affects nothing but which of $F$ and $G$ is called first.

Theorem(General solution)AdaptedAdapted from Strauss (2008), §2.1, Eq. (3) and the two proofs on p. 34; Evans (2010), §2.4.1a, Remark (i), p. 66; Riley, Hobson & Bence (2006), §20.4, Eq. (20.26), p. 693

Let $u$ be a $C^2$ function with $u_{tt} = c^2u_{xx}$ on $\mathbb{R}^2$ , or on a characteristic rectangle $\{(x,t) : a < x - ct < b,\ a' < x+ct < b'\}$ , or on the half-plane $\{(x,t) : t > 0\}$ . Then there are $C^2$ functions $F$ and $G$ of one variable with

u(x,t) = F(x - ct) + G(x + ct) .

Conversely every such sum, with $F$ and $G$ twice differentiable, solves the equation. Strauss states it on the whole plane, requiring of his two profiles nothing beyond two derivatives apiece; the rectangle and the half-plane are the same argument on a smaller domain (Step 2 below), and the half-plane is the case the uniqueness argument of Step 3 needs. Evans adds the reading that the general solution of $u_{tt}-u_{xx}=0$ is the sum of the general solutions of the two first-order equations $u_t \mp u_x = 0$ , which is what the factorisation says.

The two terms have a direct meaning, which Riley, Hobson and Bence put in terms of an observer: an observer moving with velocity $+c$ keeps $x - ct$ fixed and therefore sees the value $F(x-ct)$ never change, so $F$ is a profile of permanent shape sliding to the right at speed $c$ . Likewise $G(x+ct)$ slides to the left at speed $c$ . Neither profile need be sinusoidal or periodic, and the two are unrelated to each other (Arfken, p. 435).

Theorem(d'Alembert's formula)AdaptedAdapted from Strauss (2008), §2.1, Eq. (8) and the paragraph after it, p. 36; Evans (2010), §2.4.1a, Eq. (8) and Theorem 1, p. 65; Riley, Hobson & Bence (2006), §20.4, Eq. (20.31), p. 694

Assume $\varphi \in C^2(\mathbb{R})$ and $\psi \in C^1(\mathbb{R})$ , and define

u(x,t) = \frac{1}{2}\bigl[\varphi(x - ct) + \varphi(x + ct)\bigr] + \frac{1}{2c}\int_{x-ct}^{x+ct} \psi(s)\, ds . \tag{D}

Then $u \in C^2(\mathbb{R}\times[0,\infty))$ , it satisfies $u_{tt} = c^2u_{xx}$ in $\mathbb{R}\times(0,\infty)$ , and

\lim_{\substack{(x,t)\to(x_0,0)\\ t>0}} u(x,t) = \varphi(x_0), \qquad \lim_{\substack{(x,t)\to(x_0,0)\\ t>0}} u_t(x,t) = \psi(x_0) \qquad\text{for every } x_0 \in \mathbb{R}.

Moreover no other $C^2$ function solves the problem. The existence half is Evans's Theorem 1, stated there for $c = 1$ with data $g \in C^2(\mathbb{R})$ , $h \in C^1(\mathbb{R})$ ; Strauss records the same smoothness bookkeeping ( $\varphi \in C^2$ and $\psi \in C^1$ give $u \in C^2$ ) and attributes the formula to d'Alembert in 1746.

Formula (D) is a sum of averages. The first term is the mean of $\varphi$ over the two endpoints of the interval $[x-ct, x+ct]$ ; the second is $t$ times the mean of $\psi$ over that interval, since $\frac{1}{2c}\int_{x-ct}^{x+ct}\psi = t \cdot \frac{1}{2ct}\int_{x-ct}^{x+ct}\psi$ . Stein and Shakarchi point out that this is the reading which generalises to higher dimensions (the wave equation in three dimensions).

Idea of the proofOriginal

In the characteristic coordinates $\xi = x-ct$ , $\eta = x+ct$ the equation becomes $u_{\xi\eta} = 0$ : the operator was already a product of two first-order factors, and each factor is a derivative along one family of characteristics. An equation saying that one mixed derivative vanishes is solved by inspection: if $\partial_\eta(\partial_\xi u) = 0$ then $\partial_\xi u$ does not depend on $\eta$ , and integrating once in $\xi$ leaves a function of $\xi$ plus a constant of integration that may depend on $\eta$ . That is the general solution. The initial conditions then become two ordinary equations for the two unknown profiles $F$ and $G$ : the displacement fixes $F+G$ and the velocity fixes $-cF'+cG'$ . Differentiating the first and combining gives $F'$ and $G'$ separately, and one integration finishes the job. Uniqueness is free, because the first step produced every solution, not merely some.

ProofAdaptedAdapted from Stein & Shakarchi (2003), Ch. 1 §1, p. 10; Strauss (2008), §2.1, pp. 34–36; Evans (2010), §2.4.1a, pp. 64–65

Step 1 (the equation in characteristic coordinates). The map $(x,t) \mapsto (\xi,\eta) = (x-ct,\, x+ct)$ is a linear bijection of $\mathbb{R}^2$ , with inverse $x = (\xi+\eta)/2$ , $t = (\eta-\xi)/(2c)$ . Let $v(\xi,\eta) = u(x,t)$ . By the chain rule, $\partial_x = \partial_\xi + \partial_\eta$ and $\partial_t = -c\,\partial_\xi + c\,\partial_\eta$ , hence

\partial_t - c\,\partial_x = -2c\,\partial_\xi, \qquad \partial_t + c\,\partial_x = 2c\,\partial_\eta .

Applying the factorisation of the definition,

0 = (\partial_t - c\partial_x)(\partial_t + c\partial_x)u = (-2c\,\partial_\xi)(2c\,\partial_\eta) v = -4c^2 v_{\xi\eta},

and $c \neq 0$ , so $v_{\xi\eta} = 0$ .

Step 2 (solving $v_{\xi\eta}=0$ ). Let $\Omega$ be the image of the domain of $u$ under $(x,t)\mapsto(\xi,\eta)$ . The argument needs only that $\Omega$ is open, connected, and sliced by intervals: for each fixed $\eta$ the set of admissible $\xi$ is an interval, and for each fixed $\xi$ the set of admissible $\eta$ is an interval. All three domains of the statement qualify. The whole plane and the rectangle $(a,b)\times(a',b')$ do so by inspection; the half-plane $t>0$ is $\{\eta > \xi\}$ , whose slices are the intervals $(-\infty,\eta)$ and $(\xi,\infty)$ . Being connected, $\Omega$ also projects onto intervals $I$ and $J$ of the two axes.

Now $v_{\xi\eta} = \partial_\eta(v_\xi) = 0$ says that $v_\xi$ is constant along each $\eta$ -slice, and that slice is an interval, so $v_\xi$ depends on $\xi$ alone: write $v_\xi(\xi,\eta) = F'(\xi)$ for $\xi \in I$ , a function that is $C^1$ because $v$ is $C^2$ . Fix $\xi_0 \in I$ and put $F(\xi) = \int_{\xi_0}^{\xi}F'$ , which is legitimate because $I$ is an interval; $F$ is $C^2$ . Then $\partial_\xi(v - F) = v_\xi - F' = 0$ on $\Omega$ , and each $\xi$ -slice is an interval, so $v - F$ is constant along it and depends on $\eta$ alone:

v(\xi,\eta) = F(\xi) + G(\eta), \qquad G(\eta) := v(\xi,\eta) - F(\xi) \ \ (\eta \in J).

$G$ is $C^2$ because $v$ and $F$ are. Undoing the substitution gives $u(x,t) = F(x-ct) + G(x+ct)$ , which proves the general-solution statement; the converse direction is the chain rule, since $\partial_t^2 F(x-ct) = c^2F''$ and $\partial_x^2F(x-ct) = F''$ , and likewise for $G$ .

In the half-plane case $I = J = \mathbb{R}$ , so $F$ and $G$ are defined on all of $\mathbb{R}$ . If $u$ is in addition continuous up to $t=0$ , letting $t\to0^+$ in the identity gives $u(x,0) = F(x)+G(x)$ , so the decomposition holds on the closed half-plane as well.

Step 3 (fitting the data). Let $u$ be any $C^2$ solution of the initial-value problem on $\mathbb{R}\times[0,\infty)$ and write it as in Step 2, in the half-plane case, whose last paragraph carries the decomposition down to $t=0$ . For $t>0$ , differentiating the decomposition gives $u_t = -cF'(x-ct) + cG'(x+ct)$ ; both sides are continuous up to $t=0$ , since $u \in C^2$ and $F', G'$ are $C^1$ , so the identity holds at $t=0$ too. Setting $t=0$ in it and in the decomposition itself gives, for all $x$ ,

F(x) + G(x) = \varphi(x), \qquad -cF'(x) + cG'(x) = \psi(x) .

Differentiate the first equation and divide the second by $c$ :

F' + G' = \varphi', \qquad -F' + G' = \tfrac{1}{c}\psi .

Subtracting and adding,

F'(s) = \tfrac12\Bigl(\varphi'(s) - \tfrac1c \psi(s)\Bigr), \qquad G'(s) = \tfrac12\Bigl(\varphi'(s) + \tfrac1c \psi(s)\Bigr) .

Step 4 (integrating, and the constants cancelling). Integrating from $0$ to $s$ ,

F(s) = \tfrac12\varphi(s) - \frac{1}{2c}\int_0^s \psi + A, \qquad G(s) = \tfrac12\varphi(s) + \frac{1}{2c}\int_0^s \psi + B,

with constants $A, B$ . The first equation of Step 3 at $s$ forces $A + B = 0$ . Substituting $s = x-ct$ in $F$ and $s = x+ct$ in $G$ and adding, the constants cancel, the two $\tfrac12\varphi$ terms give the first half of (D), and the two integrals combine as $\int_0^{x+ct} - \int_0^{x-ct} = \int_{x-ct}^{x+ct}$ , giving the second half. So every $C^2$ solution equals the right-hand side of (D): the solution is unique, and no solution other than (D) can exist.

Step 5 (the formula really is a solution). Steps 1–4 assumed a solution existed. Now take $\varphi \in C^2$ , $\psi \in C^1$ and define $u$ by (D). Write $\Psi(s) = \int_0^s\psi$ , which is $C^2$ with $\Psi' = \psi$ by the fundamental theorem of calculus. Then

u(x,t) = \underbrace{\tfrac12\varphi(x-ct) - \tfrac{1}{2c}\Psi(x-ct)}_{F(x-ct)} + \underbrace{\tfrac12\varphi(x+ct) + \tfrac{1}{2c}\Psi(x+ct)}_{G(x+ct)},

a sum of a $C^2$ function of $x-ct$ and a $C^2$ function of $x+ct$ . By the converse half of Step 2 it is $C^2$ on $\mathbb{R}\times[0,\infty)$ and solves the equation. Setting $t=0$ gives $u(x,0) = \varphi(x)$ since $\Psi$ cancels, and

u_t(x,t) = \frac{c}{2}\bigl[\varphi'(x+ct) - \varphi'(x-ct)\bigr] + \frac12\bigl[\psi(x+ct) + \psi(x-ct)\bigr]

equals $\psi(x)$ at $t=0$ . Both $u$ and $u_t$ are continuous up to $t=0$ , which gives the two limits in the statement. $\square$

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

$c \neq 0$ . Step 1 divides by $c^2$ , and Step 3 divides by $c$ . For $c = 0$ the equation is $u_{tt}=0$ , whose solution is $\varphi(x) + t\psi(x)$ : no propagation at all, and the characteristics have collapsed onto one family.
$\varphi \in C^2$ and $\psi \in C^1$ . These are exactly what Step 5 needs, and no more: the formula differentiates $\varphi$ twice and $\psi$ once. Strauss and Evans both state them. If $\varphi \in C^k$ and $\psi \in C^{k-1}$ then $u \in C^k$ and in general no better; Evans notes the consequence that the wave equation, unlike the heat equation, does not smooth its data. A corner in $\varphi$ stays a corner, and travels: Strauss's plucked string, whose $\varphi$ is the "three-finger" tent $b - b|x|/a$ on $|x|<a$ and $0$ outside, has kinks at all three of $x=0$ and $x=\pm a$ , so the resulting $u$ fails to be twice differentiable along the six characteristics issuing from them, and he records that it is nonetheless a weak solution. The formula (D) still makes sense for merely continuous $\varphi$ and locally integrable $\psi$ , and this is the usual way to define the solution in that case (weak derivatives).
Where smoothness is not needed. Nothing in Steps 1–4 uses more than $C^2$ , and no hypothesis of decay, boundedness or integrability appears anywhere: (D) holds for data that grow arbitrarily fast, because the integral is over a bounded interval.
Time reversal. The equation is unchanged by $t\mapsto -t$ : if $u$ solves it, so does $(x,t)\mapsto u(x,-t)$ , which Stein and Shakarchi call the time-reversal property and contrast with the heat equation, which cannot be run backwards. Formula (D) is not itself invariant under $t\mapsto-t$ — replacing $t$ by $-t$ reverses the orientation of the integral and so flips the sign of the $\psi$ term — but it needs no change of form: read with the oriented convention $\int_a^b = -\int_b^a$ and with $t$ allowed to be any real number, the right-hand side of (D) is $F(x-ct)+G(x+ct)$ with the $F,G$ of Step 5, hence a $C^2$ solution on all of $\mathbb{R}^2$ taking the data $\varphi,\psi$ at $t=0$ . In the geometric statements below $ct$ must then be read as $c|t|$ . One consequence is that the initial-value problem is equally well posed forwards and backwards in time.
Degenerate limits of the formula. At $t=0$ the interval of integration collapses and (D) returns $\varphi$ ; this is why no compatibility condition between $\varphi$ and $\psi$ is required on the whole line. On a half-line with a fixed end, by contrast, compatibility conditions do appear — see the method below.

Definition(Domains of dependence and influence)AdaptedAdapted from Strauss (2008), §2.2 (Causality), p. 39, with Figures 1–2

Fix a point $(x,t)$ with $t > 0$ . Strauss's terminology, which we follow, is:

The interval of dependence of $(x,t)$ is the interval $(x-ct,\, x+ct)$ on the initial line; the domain of dependence (or past history) is the closed triangle with apex $(x,t)$ bounded below by that interval and on the sides by the two backward characteristics through $(x,t)$ .
The domain of influence of an initial point $(x_0,0)$ is the set of later points whose solution it can affect, namely the sector $\{(y,s) : s \ge 0,\ |y - x_0| \le cs\}$ bounded by the two forward characteristics through $(x_0,0)$ .

The two notions are the same relation read in opposite directions: $(x_0,0)$ lies in the domain of dependence of $(x,t)$ exactly when $(x,t)$ lies in the domain of influence of $(x_0,0)$ .

Theorem(Causality: finite speed of propagation)AdaptedAdapted from Strauss (2008), §2.2 (Causality), p. 39

Let $\varphi \in C^2(\mathbb{R})$ , $\psi \in C^1(\mathbb{R})$ and let $u$ be given by (D). Then:

$u(x,t)$ depends on $\varphi$ only through its values at the two points $x \pm ct$ , and on $\psi$ only through its values on $[x-ct,\, x+ct]$ . Equivalently, changing the data outside the closed interval $[x-ct,\,x+ct]$ — the closure of the interval of dependence — does not change $u(x,t)$ . The two endpoints may not be excluded: $\varphi(x\pm ct)$ enters (D) directly, so Strauss's open interval of dependence $(x-ct,x+ct)$ names the interval across which $\psi$ is averaged, not the whole set of data that $u(x,t)$ reads.
If $\varphi$ and $\psi$ vanish for $|x| > R$ , then $u(x,t) = 0$ for $|x| > R + ct$ . In Strauss's words, the domain of influence of the interval $\{|x| \le R\}$ is the sector $\{|x| \le R + ct\}$ .

So no part of the solution travels faster than $c$ . Parts of it may lag behind: an initial velocity spreads out at speeds up to $c$ , not exactly $c$ (Strauss, p. 39).

ProofOriginal

Both parts are read off (D), whose right-hand side involves only $\varphi(x\pm ct)$ and the values of $\psi$ on $[x-ct,x+ct]$ . That is assertion 1, and it is a consequence of the uniqueness in Step 4: since (D) is the only solution, there is nothing else for $u(x,t)$ to depend on.

For assertion 2, suppose $\varphi$ and $\psi$ vanish outside $[-R,R]$ , and let $|x| > R + ct$ with $t \ge 0$ . Say $x > R + ct$ (the case $x < -(R+ct)$ is the mirror image). Then

x - ct > R \qquad\text{and}\qquad x + ct \ge x - ct > R ,

the second because $ct \ge 0$ . Hence the whole interval $[x-ct,\, x+ct]$ lies in $(R,\infty)$ , where $\varphi$ and $\psi$ vanish. Every term of (D) is therefore zero, so $u(x,t) = 0$ . $\square$

Method(Reflection at a boundary)AdaptedAdapted from Strauss (2008), §3.2, Eq. (2)–(3), pp. 61–63, and Eq. (5), pp. 63–64; Evans (2010), §2.4.1a (A reflection method), problem (9) and Eq. (10), pp. 66–67; Stein & Shakarchi (2003), Ch. 1 §1, p. 10

To solve the wave equation on a half-line or an interval with homogeneous boundary conditions, extend the data to the whole line so that the boundary condition is enforced by symmetry, apply (D), and restrict.

Half-line $x>0$ , fixed end (Dirichlet). Problem: $v_{tt} = c^2v_{xx}$ for $x>0$ , $v(x,0)=\varphi(x)$ , $v_t(x,0)=\psi(x)$ for $x > 0$ , and $v(0,t)=0$ .

Check compatibility. For a $C^2$ solution the data must satisfy $\varphi(0) = \psi(0) = 0$ (Evans assumes exactly this), since $v(0,t) \equiv 0$ forces $v$ and $v_t$ to vanish at $x=0$ when $t=0$ .
Extend oddly. Put $\varphi_{\mathrm{odd}}(x) = \varphi(x)$ for $x>0$ , $-\varphi(-x)$ for $x<0$ , $0$ at $x=0$ , and similarly for $\psi$ .
Solve on the line. Let $u$ be given by (D) with data $\varphi_{\mathrm{odd}}, \psi_{\mathrm{odd}}$ . Because the equation and the formula are invariant under $x \mapsto -x$ , $u$ is odd in $x$ for every $t$ ; hence $u(0,t) = 0$ automatically.
Restrict. $v = u|_{x>0}$ solves the half-line problem. Unwinding the extension, Strauss's two cases are

v(x,t) = \begin{cases} \dfrac12\bigl[\varphi(x-ct)+\varphi(x+ct)\bigr] + \dfrac{1}{2c}\displaystyle\int_{x-ct}^{x+ct}\psi, & x > c|t| ,\\[2ex] \dfrac12\bigl[\varphi(ct+x)-\varphi(ct-x)\bigr] + \dfrac{1}{2c}\displaystyle\int_{ct-x}^{ct+x}\psi, & 0 < x < c|t| . \end{cases}

Evans's formula (10) is the same pair with $c=1$ . Geometrically: draw the two backward characteristics from $(x,t)$ ; if one of them meets the boundary $x=0$ before the initial line, the reflection flips the sign of the $\varphi$ -contribution carried along it, and the $\psi$ -contributions outside $[ct-x, ct+x]$ cancel in pairs. 5. Free end (Neumann), $v_x(0,t)=0$ . Use the even extension instead, $\varphi_{\mathrm{even}}(x)=\varphi(|x|)$ and $\psi_{\mathrm{even}}(x) = \psi(|x|)$ . An even function has odd derivative, so $u_x(0,t)=0$ automatically, and no compatibility condition on $\varphi(0),\psi(0)$ is needed. For $x > c|t|$ the formula is unchanged, the first case above. For $0<x<c|t|$ the $\varphi$ -terms now add rather than cancel, and so do the two halves of the $\psi$ -integral: substituting $y \mapsto -y$ in $\int_{x-ct}^{0}\psi_{\mathrm{even}}$ turns it into $\int_0^{ct-x}\psi$ , which has the same sign as $\int_0^{ct+x}\psi$ , so

v(x,t) = \frac12\bigl[\varphi(ct+x) + \varphi(ct-x)\bigr] + \frac{1}{2c}\left[\int_0^{ct+x}\psi + \int_0^{ct-x}\psi\right], \qquad 0 < x < c|t| .

It is only when $\psi \equiv 0$ that this is the Dirichlet formula with the minus sign turned into a plus; in general the $\psi$ -term is different, as the simplest example shows: for $c=1$ , $\varphi\equiv0$ and $\psi\equiv1$ on $x>0$ the free end gives $v = \tfrac12[(t+x)+(t-x)] = t$ , which is right, whereas replacing the minus sign by a plus in the Dirichlet $\psi$ -integral $\tfrac{1}{2}\int_{t-x}^{t+x}1$ would give $x$ . Strauss carries out the even-extension version of the reflection method for the diffusion equation on pp. 59–60 and leaves the Neumann wave problem as an exercise, so we state it as the parallel calculation that it is. The physical content of the sign in the $\varphi$ -terms is that a pulse arriving at a fixed end returns inverted, while at a free end it returns upright.

Interval $0<x<L$ , both ends fixed. Extend $\varphi$ and $\psi$ to be odd about both $x=0$ and $x=L$ , equivalently odd on $(-L,L)$ and then $2L$ -periodic, and apply (D) to the extension. This is Strauss's Eq. (5) and the device Stein and Shakarchi use (with $L=\pi$ ) to keep both ends of a string fixed. Unwound, it describes a wave reflecting off the two ends infinitely often, changing sign each time; the compact alternative is to expand in the standing waves of standing waves and normal modes.

Example(A velocity pulse on the whole line)Original

Take $c = 2$ , $\varphi \equiv 0$ and

\psi(x) = x\,e^{-x^2},

which is $C^\infty$ , so the hypotheses hold. Since $\int \psi = -\tfrac12 e^{-x^2} + \text{const}$ , formula (D) gives

u(x,t) = \frac{1}{2c}\int_{x-ct}^{x+ct} s\,e^{-s^2}\,ds = \frac{1}{4c}\Bigl[e^{-(x-ct)^2} - e^{-(x+ct)^2}\Bigr].

Checks. Differentiating, $u_{tt} - c^2u_{xx} = 0$ identically; $u(x,0) = 0$ ; and $u_t(x,0) = \tfrac12[\psi(x)+\psi(x)] = xe^{-x^2}$ by the formula for $u_t$ in Step 5. A number: at $x = 1$ , $t = \tfrac12$ , so $ct = 1$ ,

u\bigl(1,\tfrac12\bigr) = \frac18\bigl[e^{0} - e^{-4}\bigr] = \frac{1 - e^{-4}}{8} = 0.1227105\ldots

Reading the answer. The solution is the difference of two bumps of height $1/(4c) = 1/8$ , one centred on the characteristic $x = ct$ and one, with the opposite sign, on $x = -ct$ . So an initial velocity that is odd in $x$ sends a positive pulse to the right at speed $2$ and a negative one to the left, and at any fixed $x>0$ the displacement rises and falls once as the right-moving bump passes. Note that $u$ does not return to zero behind the pulse here only because the data are not compactly supported. The residual is small very quickly all the same: at a fixed $x$ the dominant term is $\tfrac{1}{4c}e^{-(ct-x)^2}$ , a Gaussian in $t$ , at $x=1$ , $c=2$ the displacement peaks at $u(1,\tfrac12) = 0.1227$ and has fallen to $u(1,2) = 1.54\times10^{-5}$ , about $1.2\times10^{-4}$ of that, by $t=2$ .

Bonus: a half-line problem solved for free. This $u$ is odd in $x$ (swapping $x \to -x$ exchanges the two exponentials and flips the sign), so $u(0,t) = 0$ for all $t$ . By step 3 of the method, its restriction to $x > 0$ is the solution of the Dirichlet half-line problem with $\varphi \equiv 0$ and $\psi(x) = xe^{-x^2}$ for $x>0$ — data which do satisfy the compatibility condition $\psi(0)=0$ .

Example(A pulse reflected from a fixed end)Original

A string occupies $x > 0$ with its end held at $u = 0$ , and $c = 3$ . Release it from rest with the shape

\varphi(x) = x^3 e^{-x}, \qquad \psi \equiv 0 .

This is a single bump with its maximum at $x = 3$ , of height $27e^{-3} = 1.344251$ . Compatibility holds, and more: $\varphi(0) = \varphi''(0) = 0$ , so the odd extension is $C^3$ across the origin. (The odd extension of a function is $C^k$ when its even-order derivatives up to order $k$ vanish at $0$ : the odd-order derivatives of an odd extension are even and match automatically, while the even-order ones are odd and match only if they vanish. From $\varphi(x) = x^3 - x^4 + \tfrac12x^5 - \cdots$ we read $\varphi(0)=\varphi''(0)=0$ , which is what gives $C^3$ , and $\varphi''''(0) = -24 \neq 0$ , which is the obstruction to $C^4$ . The value $\varphi'''(0)=6$ is of odd order and no obstruction at all.) The solution is therefore a genuine $C^2$ solution.

With $\psi \equiv 0$ the reflection formulas reduce to

v(x,t) = \tfrac12\bigl[\varphi(x+ct) + \varphi(x-ct)\bigr] \ \ (x>ct), \qquad v(x,t) = \tfrac12\bigl[\varphi(ct+x) - \varphi(ct-x)\bigr] \ \ (0<x<ct).

Watch the point $x = 1$ , where the two half-bumps of height $\tfrac12\varphi$ pass in turn.

Before reflection, $t = \tfrac{1}{12}$ , so $ct = \tfrac14 < x$ : $v = \tfrac12[\varphi(1.25) + \varphi(0.75)] = \tfrac12[0.559580 + 0.199280] = 0.379430$ .
The left-moving half has hit the wall and come back; $t=1$ , $ct = 3$ : $v(1,1) = \tfrac12[\varphi(4) - \varphi(2)] = \tfrac12[1.172201 - 1.082682] = 0.044759$ . Without the wall the same data would give $\tfrac12[\varphi(4)+\varphi(2)] = 1.127442$ at that point: the reflected pulse has cancelled nearly all of it.
The inverted pulse arrives in full; $t = \tfrac43$ , $ct = 4$ : $v\bigl(1,\tfrac43\bigr) = \tfrac12[\varphi(5) - \varphi(3)] = \tfrac12[0.842243 - 1.344251] = -0.251004$ . The displacement is now negative. The second term, $-\tfrac12\varphi(3) = -0.672125$ , is the crest of the inverted reflected pulse passing $x=1$ ; the first, $+\tfrac12\varphi(5) = 0.421122$ , is the tail of the right-moving half-bump, and the two partly cancel.

Checks. $v(0,t) = \tfrac12[\varphi(ct)-\varphi(ct)] = 0$ for every $t$ , as required. The crest of the inverted reflected wave sits where $ct - x = 3$ , that is on the line $x = 3t - 3$ , which is the characteristic $x + ct = 3$ reflected in the boundary — exactly the picture Strauss draws (his Figure 2, p. 62). This sign inversion is the familiar behaviour of a pulse sent down a rope whose far end is clamped.

ApplicationAdaptedAdapted from Strauss (2008), §2.2, pp. 40–41; Riley, Hobson & Bence (2006), §20.4, pp. 694–695

Strings and pulses. For a string of tension $\tau$ and density $\rho$ the formula says that a pluck ( $\psi = 0$ ) splits into two half-amplitude copies of the initial shape moving apart at $c = \sqrt{\tau/\rho}$ , while a hammer blow ( $\varphi = 0$ , $\psi$ concentrated) produces a plateau spreading at speed $c$ . Riley, Hobson and Bence read the three terms of (D) in just this way: each half of the initial displacement travels towards $x$ from a distance $ct$ on either side, and the integral accumulates the transverse displacement delivered by every part of the initial motion that has had time to reach $x$ .

Time of arrival. Causality turns into an engineering statement: a disturbance at one end of a $0.65\,\mathrm{m}$ steel string with $c = 126.5\,\mathrm{m\,s^{-1}}$ cannot be felt at the other end before $0.65/126.5 = 5.14\,\mathrm{ms}$ , whatever the data are. The same arithmetic gives arrival times in seismology (P and S waves have different $c$ , hence the delay used to locate an earthquake) and in transmission lines, where $u$ is voltage and reflections at an unmatched termination are exactly the sign-flipping reflections of the method above.

Electromagnetism and relativity. Each Cartesian component of $\mathbf{E}$ and $\mathbf{B}$ in vacuum satisfies the wave equation with $c$ the speed of light, so causality says that a signal at $(x_0,t_0)$ cannot outrun light. Strauss identifies this as the cornerstone of relativity, and notes that in three dimensions the principle is sharper still — signals travel at exactly $c$ , which is Huygens's principle (the wave equation in three dimensions).

Where it is used next. The two-case reflection formula is the engine of the three-dimensional solution, where the spherical mean of $u$ satisfies a half-line problem in the radius; the causality statement is proved again, without any formula and in any dimension, by the energy method of energy and uniqueness for the wave equation; and the characteristics that carry $F$ and $G$ are the model case for the method of characteristics and for the shock-forming equations of conservation laws and shocks.

Theorem

Energy and uniqueness for the wave equation

Adapted

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Defines the energy of a solution of the wave equation, proves it is conserved on the line, in all of space and on a bounded domain with fixed or free boundary, and uses that conservation to prove that the initial/boundary-value problem has at most one solution.

Prerequisites: The one-dimensional wave equation, The divergence theorem, Differentiation under the integral sign

Definition(Energy, energy density and energy flux)AdaptedAdapted from Strauss (2008), §2.2, Eq. (1), p. 40; Strauss (2008), §9.1, Eq. (6), p. 230; Stein & Shakarchi (2003), Ch. 3, Exercise 10, p. 90

The string. For a string of tension $\tau$ and line density $\rho$ , obeying $\rho u_{tt} = \tau u_{xx}$ with $c = \sqrt{\tau/\rho}$ , put

\mathrm{KE}(t) = \frac{\rho}{2}\int u_t^2\, dx, \qquad \mathrm{PE}(t) = \frac{\tau}{2}\int u_x^2\, dx, \qquad E(t) = \mathrm{KE}(t) + \mathrm{PE}(t) ,

the integrals being over $\mathbb{R}$ for the infinite string and over $[0,L]$ for a string of length $L$ . The name of the first is forced by $\tfrac12 mv^2$ ; the second is the potential energy stored in the stretching, and Strauss notes that for the mathematics only the total $E$ matters, so that a different constant multiple of $E$ would do as well. Stein and Shakarchi define exactly these two integrals on $[0,L]$ for a string with $c^2 = \tau/\rho$ . Observe that $E \ge 0$ , and $E = 0$ only if $u_t \equiv u_x \equiv 0$ .

General dimension. For $u_{tt} = c^2\Delta u$ on a region of $\mathbb{R}^n$ , the energy density and energy flux are

e = \frac12\bigl(u_t^2 + c^2|\nabla u|^2\bigr), \qquad \mathbf{S} = -c^2 u_t\,\nabla u ,

and the total energy over a region $V$ is $E(t) = \int_V e\, dx$ , which for $V = \mathbb{R}^3$ is Strauss's Eq. (6). Dividing the string's $E$ by $\rho$ turns it into this $E$ with $n=1$ , since $\tau/\rho = c^2$ ; the physical energy of the string is $\rho$ times the quantity used here. For the string the flux is $S = -\tau u_x u_t$ (after multiplying by $\rho$ ), which is exactly the rate at which the piece of string to the left of $x$ does work on the piece to the right: the transverse force it exerts is $-\tau u_x$ and the velocity of the point of application is $u_t$ .

Theorem(Local energy identity)AdaptedAdapted from Strauss (2008), §9.1, Eq. (4) and Eq. (7), p. 230

Let $u$ be of class $C^2$ on an open set of $\mathbb{R}^n\times\mathbb{R}$ . Then, identically,

\bigl(u_{tt} - c^2\Delta u\bigr)\, u_t \;=\; \frac{\partial e}{\partial t} + \nabla\cdot\mathbf{S} \qquad\text{with } e = \tfrac12\bigl(u_t^2 + c^2|\nabla u|^2\bigr),\ \ \mathbf{S} = -c^2u_t\nabla u . \tag{E}

In particular, if $u$ solves the wave equation then

\frac{\partial e}{\partial t} + \nabla\cdot \mathbf{S} = 0 ,

a conservation law of exactly the form of the continuity equation: energy is neither created nor destroyed, only transported, at the rate $\mathbf{S}$ through each surface. Strauss writes the same identity as $\partial_t e = c^2\nabla\cdot(u_t\nabla u)$ and, in three dimensions, as the statement that a certain four-component field has vanishing four-dimensional divergence.

ProofOriginal

Expand both terms on the right. By the product rule,

\frac{\partial e}{\partial t} = u_t u_{tt} + c^2\,\nabla u\cdot\nabla u_t ,

the second term because $\partial_t\tfrac12|\nabla u|^2 = \nabla u\cdot\partial_t\nabla u$ and the mixed partials of a $C^2$ function may be taken in either order. By the product rule for the divergence,

\nabla\cdot\mathbf{S} = -c^2\bigl(\nabla u_t\cdot\nabla u + u_t\,\Delta u\bigr) .

Adding, the two $c^2\nabla u\cdot\nabla u_t$ terms cancel and what is left is $u_t(u_{tt} - c^2\Delta u)$ . $\square$

Theorem(Conservation of energy on the line)AdaptedAdapted from Strauss (2008), §2.2, Eq. (1) and the derivation above it, pp. 40–41

Let the string have constant tension $\tau>0$ and line density $\rho>0$ , so that $c = \sqrt{\tau/\rho}$ and the equation $u_{tt}=c^2u_{xx}$ is the same as $\rho u_{tt} = \tau u_{xx}$ . Let $\varphi \in C^2(\mathbb{R})$ and $\psi \in C^1(\mathbb{R})$ vanish outside an interval $\{|x| \le R\}$ , and let $u$ be what d'Alembert's formula delivers from that data. Then

E(t) = \frac12\int_{-\infty}^{\infty}\bigl(\rho\, u_t^2 + \tau\, u_x^2\bigr)\, dx

is finite and independent of $t$ . Its value is $E(0) = \tfrac12\int(\rho\psi^2 + \tau\varphi'^2)\,dx$ , the same expression in the initial data that Stein and Shakarchi give for the conserved energy of a string on $[0,L]$ .

Idea of the proofOriginal

Multiply the equation by the velocity $u_t$ . On the left that is force times velocity, which is power; on the right, by the identity (E), it is the time derivative of the energy density plus the divergence of a flux. So integrating over all of space leaves only the flux through the boundary, and compact support kills it. The one technical point is the licence to differentiate under the integral sign, which the finite propagation speed of d'Alembert's formula supplies: for times in a bounded interval the integrand lives inside one fixed bounded interval of $x$ , so the integral is really over a compact set.

ProofAdaptedAdapted from Strauss (2008), §2.2, pp. 40–41

Fix $T > 0$ and work for $0 \le t \le T$ .

Step 1 (the integrand has fixed compact support). By the causality statement of d'Alembert's formula, $u$ vanishes identically on the open set $\{|x| > R + ct,\ t>0\}$ , hence so do all its derivatives there. Set $R_T = R + cT$ . If $|x| \ge R_T$ and $0 \le t < T$ then $|x| \ge R + cT > R + ct$ , so $(x,t)$ lies in that open set and $u_t = u_x = 0$ there; at $t = T$ the same holds by continuity of $u_t$ and $u_x$ . So $u_t = u_x = 0$ on the whole of $\{|x| \ge R_T,\ 0 \le t \le T\}$ , endpoints included, and

E(t) = \frac12\int_{-R_T}^{R_T}\bigl(\rho u_t^2 + \tau u_x^2\bigr)\, dx ,

a finite integral over a fixed compact interval.

Step 2 (differentiating). The integrand and its $t$ -derivative are continuous on the compact rectangle $[-R_T,R_T]\times[0,T]$ , because $u \in C^2$ . So differentiation under the integral sign applies:

\frac{dE}{dt} = \int_{-R_T}^{R_T}\bigl(\rho\, u_t u_{tt} + \tau\, u_x u_{xt}\bigr)\, dx .

Step 3 (moving one derivative). Integrate the second term by parts, which is legitimate since $u_x$ and $u_{xt}$ are continuous:

\int_{-R_T}^{R_T} \tau\, u_x u_{xt}\, dx = \Bigl[\tau\, u_x u_t\Bigr]_{-R_T}^{R_T} - \int_{-R_T}^{R_T} \tau\, u_{xx} u_t\, dx .

The bracket vanishes: by Step 1 both $u_x$ and $u_t$ are zero at the two endpoints $x = \pm R_T$ . (This is the only place where the compact support of the data is used.)

Step 4 (the equation). Therefore

\frac{dE}{dt} = \int_{-R_T}^{R_T} u_t\bigl(\rho\, u_{tt} - \tau\, u_{xx}\bigr)\, dx = 0 ,

since $\rho u_{tt} = \tau u_{xx}$ pointwise, which is the equation $u_{tt}=c^2u_{xx}$ multiplied by $\rho$ , using $c^2 = \tau/\rho$ . As $T$ was arbitrary, $E$ is constant on $[0,\infty)$ .

For $t<0$ , apply what has just been proved to $\check u(x,t) = u(x,-t)$ . By the time-reversal remark of d'Alembert's formula, $\check u$ solves the same equation with the data $(\varphi, -\psi)$ , which again vanish outside $\{|x|\le R\}$ ; and the energy of $\check u$ at time $t$ is the energy of $u$ at time $-t$ , because $E$ depends on $u$ only through $u_t^2$ and $u_x^2$ and the sign that $t\mapsto -t$ puts on $u_t$ is squared away. So $E$ is constant on $(-\infty,0]$ too, and the two constants agree at $t=0$ . Evaluating there, where $u_t = \psi$ and $u_x = \varphi'$ , gives the stated value. $\square$

Theorem(Conservation on a bounded domain)AdaptedAdapted from Strauss (2008), §9.1, Eq. (4)–(6), p. 230, and §9.1 Exercises 6–7, p. 234; Evans (2010), §2.4.3a, the computation of $\dot E$ in the proof of Thm 5, p. 79

Let $V \subseteq \mathbb{R}^n$ be a bounded open region with piecewise $C^1$ boundary $\partial V$ and outward unit normal $\hat{\mathbf{n}}$ , let $T>0$ , and let $u \in C^2(\overline V\times[0,T])$ satisfy $u_{tt} = c^2\Delta u$ in $V\times(0,T)$ . Then, with $E(t) = \int_V e\,dx$ ,

\frac{dE}{dt} = c^2\oint_{\partial V} u_t\,\frac{\partial u}{\partial n}\; dS \qquad (0 < t < T),

where $\partial u/\partial n = \hat{\mathbf{n}}\cdot\nabla u$ . Consequently $E$ is constant in each of the following cases:

fixed boundary (Dirichlet) $u = 0$ on $\partial V\times[0,T]$ , because then $u_t = 0$ there;
free boundary (Neumann) $\partial u/\partial n = 0$ on $\partial V\times[0,T]$ ;
more generally whenever $u_t\,\partial u/\partial n$ integrates to zero over $\partial V$ at each time.

For an absorbing (Robin) condition $\partial u/\partial n + b\, u_t = 0$ with a constant $b > 0$ , the same formula gives $dE/dt = -c^2 b\oint_{\partial V} u_t^2\, dS \le 0$ : the energy decreases, and the boundary is a sink. Strauss sets both of these as exercises (§9.1, Exercises 6 and 7, p. 234) after deriving the conservation law (4)–(6) on p. 230; the computation below is the one his derivation indicates.

ProofOriginal

Since $u \in C^2(\overline V\times[0,T])$ and $V$ is bounded, $e$ and $\partial e/\partial t$ are continuous on the compact set $\overline V\times[0,T]$ . That licenses differentiating under the integral sign. The fixed-limit theorem of differentiation under the integral sign is stated there for an integral over a real interval $[a,b]$ , so for $n\ge2$ we use not the statement but its proof, which transfers verbatim: by the mean value theorem each difference quotient $\bigl(e(x,t+h)-e(x,t)\bigr)/h$ equals $\partial e/\partial t$ at some intermediate time; uniform continuity of $\partial e/\partial t$ on the compact set $\overline V\times[0,T]$ makes those values converge to $\partial e/\partial t(x,t)$ uniformly in $x$ ; and integrating an error of size $\varepsilon$ over a set of finite volume $|V|$ costs at most $|V|\varepsilon$ . Nothing else about $[a,b]$ was used. Hence

\frac{dE}{dt} = \int_V \frac{\partial e}{\partial t}\, dx = \int_V \bigl(u_t u_{tt} + c^2\nabla u\cdot\nabla u_t\bigr)\, dx .

By the identity (E) the integrand equals $u_t(u_{tt}-c^2\Delta u) - \nabla\cdot\mathbf{S} = -\nabla\cdot\mathbf{S}$ , the first term vanishing because $u$ solves the equation in $V$ . The field $\mathbf{S} = -c^2u_t\nabla u$ is $C^1$ on $\overline V$ , so the divergence theorem applies:

\frac{dE}{dt} = -\int_V \nabla\cdot\mathbf{S}\, dx = -\oint_{\partial V}\mathbf{S}\cdot\hat{\mathbf{n}}\, dS = c^2\oint_{\partial V} u_t\,\frac{\partial u}{\partial n}\, dS .

For the Dirichlet case, $u(\cdot,t) = 0$ on $\partial V$ for every $t$ , so differentiating in $t$ at a fixed boundary point gives $u_t = 0$ there and the integrand vanishes. For the Neumann case $\partial u/\partial n = 0$ and the integrand vanishes again. For the Robin case substitute $\partial u/\partial n = -b u_t$ . $\square$

Theorem(Uniqueness for the initial/boundary-value problem)AdaptedAdapted from Evans (2010), §2.4.3a, problem (45) and Theorem 5, p. 79

Let $V \subseteq \mathbb{R}^n$ be a bounded open set with smooth boundary, let $T>0$ , and write

V_T = V\times(0,T], \qquad \Gamma_T = \overline{V_T}\setminus V_T ,

so $\Gamma_T$ consists of the bottom $\overline V\times\{t=0\}$ together with the side $\partial V\times[0,T]$ . Given $f$ on $V_T$ , $g$ on $\Gamma_T$ and $h$ on $V\times\{t=0\}$ , consider

u_{tt} - c^2\Delta u = f \ \text{ in } V_T, \qquad u = g \ \text{ on } \Gamma_T, \qquad u_t = h \ \text{ on } V\times\{t=0\} .

There is at most one $u \in C^2(\overline{V_T})$ solving this problem.

This is Evans's Theorem 5, which he states for $c=1$ ; the substitution $s = ct$ turns $u_{tt} = c^2\Delta u$ into $u_{ss} = \Delta u$ and carries one version into the other. Note how Evans packages the data: the single function $g$ on $\Gamma_T$ prescribes both the initial displacement (on the bottom) and the boundary values (on the side), so Dirichlet boundary conditions are built into the formulation, and $h$ prescribes the initial velocity. Nothing is assumed about $f$ beyond its being the same for the two solutions compared.

Idea of the proofOriginal

Uniqueness for a linear problem means that the only solution with zero data is zero. The difference of two solutions has zero source, vanishes on the whole of $\Gamma_T$ — so in particular on the side, where it also has zero time derivative — and starts with zero velocity. Its energy is therefore zero at $t=0$ and conserved, and an energy of zero forces the velocity and the gradient to vanish everywhere. A function whose gradient vanishes at all times and which starts at zero is identically zero.

ProofAdaptedAdapted from Evans (2010), §2.4.3a, proof of Thm 5, p. 79

Let $u$ and $\tilde u$ both solve the problem and put $w = u - \tilde u$ . Subtracting the two problems, and using that the equation and the boundary conditions are linear and the data identical,

w_{tt} - c^2\Delta w = 0 \ \text{ in } V_T, \qquad w = 0 \ \text{ on } \Gamma_T, \qquad w_t = 0 \ \text{ on } V\times\{t=0\} .

Step 1 (the energy is conserved). Let $E(t) = \tfrac12\int_V\bigl(w_t^2 + c^2|\nabla w|^2\bigr)dx$ . Because $w = 0$ at every point of the side $\partial V\times[0,T]$ , we have $w_t = 0$ there as well. Since $w \in C^2(\overline{V_T})$ and $\partial V$ is smooth, the previous statement applies with the Dirichlet alternative, so $E$ is constant on $[0,T]$ .

Step 2 (it starts at zero). On the bottom, $w(\cdot,0) = 0$ throughout $V$ , so $\nabla w(\cdot,0) = 0$ there, and $w_t(\cdot,0) = 0$ by hypothesis. Hence $E(0) = 0$ , and by Step 1, $E(t) = 0$ for all $0 \le t \le T$ .

Step 3 (from zero energy to zero function). The integrand of $E$ is continuous and non-negative, and its integral vanishes, so it vanishes identically: a continuous non-negative function with zero integral is zero, since otherwise it would be bounded below by a positive constant on some ball. Therefore

w_t \equiv 0 \quad\text{and}\quad \nabla w \equiv 0 \qquad\text{in } V_T .

Step 4 (integrating). Fix $x \in V$ . Then $t\mapsto w(x,t)$ has zero derivative on $[0,T]$ and $w(x,0)=0$ , so $w(x,t) = 0$ for all $t$ . Hence $w \equiv 0$ in $V_T$ , that is $u = \tilde u$ . $\square$

Note(The same method gives finite propagation speed)AdaptedAdapted from Evans (2010), §2.4.3b, Thm 6 and its proof, pp. 80–81; Strauss (2008), §9.1, pp. 230–233

Energy also reproves causality, in any dimension and with no solution formula. Fix $x_0 \in \mathbb{R}^n$ , $t_0 > 0$ and let $K(x_0,t_0) = \{(x,t) : 0\le t\le t_0,\ |x-x_0|\le c(t_0 - t)\}$ be the backward wave cone with apex $(x_0,t_0)$ . Evans's Theorem 6 (stated for $c=1$ ) says: if $u \in C^2$ solves $u_{tt} = c^2\Delta u$ in $\mathbb{R}^n\times(0,\infty)$ and $u = u_t = 0$ on the ball $B(x_0,ct_0)$ at $t=0$ , then $u \equiv 0$ in $K(x_0,t_0)$ . The proof applies the energy argument to the local energy $e(t) = \tfrac12\int_{B(x_0,c(t_0-t))}(u_t^2 + c^2|\nabla u|^2)dx$ over the shrinking ball; differentiating produces a boundary term from the moving boundary, and the Cauchy–Schwarz inequality shows that it cannot be positive, so $e$ is non-increasing from $e(0)=0$ . Strauss reaches the same conclusion by integrating the identity (E) over a cone frustum in four-dimensional space-time and checking that the integral over the slanted side has a sign (his Eq. (7)–(12) and Figures 3–5). We state the result and do not prove it here; both arguments are short, and both are in the cited pages. In one dimension it is the causality statement proved from the formula in d'Alembert's formula.

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

$u \in C^2$ up to the boundary. Used three times: to differentiate $E$ under the integral sign, to integrate by parts or apply the divergence theorem, and to interchange mixed partials in the identity (E). Nothing weaker will do for this argument, though the conclusions survive for weak solutions by other means (energy estimates).
Compact support, or a boundary condition. On an unbounded domain some control at infinity is indispensable, since otherwise $E$ need not even be finite: for the travelling wave $u = \sin(x-ct)$ on $\mathbb{R}$ the energy density is $\tfrac12(c^2\cos^2 + c^2\cos^2)$ , whose integral diverges. Strauss assumes the data vanish outside a bounded interval for exactly this reason in one dimension, and in three dimensions asks that the derivatives of $u$ tend to zero as $|x|\to\infty$ "in an appropriate sense"; the hypothesis used above, that the data have compact support, is one concrete way to secure that, and finite propagation speed then keeps the support compact at every later time.
The boundary condition must not pump energy in. The boundary term $c^2\oint u_t\,\partial u/\partial n\, dS$ is the power delivered through $\partial V$ . Dirichlet and Neumann conditions make it vanish; the Robin condition with $b>0$ makes it negative. With $b<0$ the energy can grow, and the proof of uniqueness collapses — not because uniqueness is then false, but because this argument no longer reaches it.
Energy controls derivatives, not the function. $E = 0$ gives $w_t = \nabla w = 0$ , and only Step 4 turns that into $w = 0$ . The step uses the initial condition, which is why the Neumann problem for the wave equation is still uniquely solvable even though $\nabla w \equiv 0$ leaves a constant undetermined at each fixed time: the constant is pinned at $t=0$ . By contrast, in the Neumann problem for Laplace's equation, where there is no time variable to pin it down, an additive constant genuinely survives.
Why not a maximum principle? For the heat equation, uniqueness follows from the maximum principle (maximum principle for the heat equation). The wave equation has no maximum principle: on $V=(0,L)$ with $\varphi\equiv 0$ , $u=0$ at both ends and $\psi>0$ inside, the solution is positive for small $t>0$ , so its maximum over $\overline{V_T}$ is attained at a point off $\Gamma_T$ , while $\max_{\Gamma_T}u=0$ . So energy is the natural substitute, and it is what generalises to variable coefficients and to nonlinear wave equations.
Conserved, not merely bounded. The wave equation's energy is exactly constant, whereas a diffusion's decays. That is the quantitative form of time reversibility, and it is why wave solutions do not forget their initial data.

Example(A vibrating string, energy mode by mode)Original

Take a string of length $L$ with both ends fixed and write a finite superposition of standing waves (the normal modes, derived in standing waves and normal modes; here we simply write them down and check them),

u(x,t) = \sum_{n=1}^{N} \sin\frac{n\pi x}{L}\Bigl(A_n\cos\omega_n t + B_n\sin\omega_n t\Bigr), \qquad \omega_n = \frac{n\pi c}{L},

which satisfies $\rho u_{tt} = \tau u_{xx}$ and $u(0,t)=u(L,t)=0$ : each term is a product $X_n(x)T_n(t)$ with $X_n = \sin(n\pi x/L)$ vanishing at $x=0,L$ , and $T_n'' = -\omega_n^2 T_n$ matches $c^2X_n''/X_n = -c^2(n\pi/L)^2 = -\omega_n^2$ , so $u_{tt}=c^2u_{xx}$ term by term.

The energy, computed. Insert this into $E = \tfrac12\int_0^L(\rho u_t^2 + \tau u_x^2)dx$ . Both $\{\sin(n\pi x/L)\}$ and $\{\cos(n\pi x/L)\}$ are orthogonal on $[0,L]$ , with $\int_0^L\sin^2(n\pi x/L)dx = \int_0^L\cos^2(n\pi x/L)dx = L/2$ , so all cross terms in $n \neq m$ drop out and

E = \frac{L}{4}\sum_{n=1}^N\Bigl[\rho\,\omega_n^2\bigl(A_n\sin\omega_nt - B_n\cos\omega_nt\bigr)^2 + \tau\Bigl(\frac{n\pi}{L}\Bigr)^2\bigl(A_n\cos\omega_nt + B_n\sin\omega_nt\bigr)^2\Bigr].

Now $\rho\omega_n^2 = \rho(n\pi c/L)^2 = \tau(n\pi/L)^2$ , because $\rho c^2 = \tau$ . The two square brackets therefore carry the same coefficient, and $(\,\cdot\,)^2 + (\,\cdot\,)^2 = A_n^2 + B_n^2$ by the Pythagorean identity. Hence

E = \frac{\tau\pi^2}{4L}\sum_{n=1}^{N} n^2\bigl(A_n^2 + B_n^2\bigr),

independent of $t$ , as conservation requires. (Checked symbolically for $N=3$ with general $A_n, B_n$ .) 2. Equipartition. For a single mode with $B_n = 0$ , $\mathrm{KE} = E\sin^2\omega_n t$ and $\mathrm{PE} = E\cos^2\omega_n t$ : the energy sloshes completely between the two forms twice per period, and each has time-average $E/2$ . 3. A number. A steel string with $L = 0.650\,\mathrm{m}$ , $\tau = 80\,\mathrm{N}$ and $\rho = 5.0\times10^{-3}\,\mathrm{kg\,m^{-1}}$ has $c = \sqrt{80/0.005} = 126.49\,\mathrm{m\,s^{-1}}$ and fundamental frequency $c/2L = 97.30\,\mathrm{Hz}$ . Pluck it into the shape $u(x,0) = h\sin^3(\pi x/L)$ with $h = 3\,\mathrm{mm}$ and release it from rest. The identity $\sin^3\theta = \tfrac14(3\sin\theta - \sin3\theta)$ puts that shape in the form above at once, with $N = 3$ ,

A_1 = \tfrac34 h = 2.25\,\mathrm{mm}, \qquad A_3 = -\tfrac14 h = -0.75\,\mathrm{mm},

$A_2 = 0$ and every $B_n = 0$ (released from rest). So

E = \frac{80\pi^2}{4(0.650)}\bigl[(2.25\times10^{-3})^2 + 9(0.75\times10^{-3})^2\bigr] = 3.075\times10^{-3}\,\mathrm{J} .

A millijoule is the right order of magnitude for a plucked guitar string, and the $n^2$ weighting shows why energy put into high modes is expensive: the third harmonic here carries the same energy as the fundamental although its amplitude is three times smaller. 4. A travelling wave, for contrast. On the infinite line let $u = F(x - ct)$ with $F$ a $C^2$ bump. Then $u_t = -cF'$ and $u_x = F'$ , so $\rho u_t^2 = \rho c^2F'^2 = \tau F'^2 = \tau u_x^2$ : kinetic and potential densities are equal at every point and every instant, and $E = \tau\int F'^2$ . The flux is $S = -\tau u_xu_t = \tau cF'^2 = c\,e$ (in the normalisation where $e$ carries the factor $\rho$ ), so the energy of a travelling wave moves bodily at speed $c$ , which is the statement that the wave transports its energy and nothing else.

ApplicationAdaptedAdapted from Strauss (2008), §2.2, p. 40; Evans (2010), §2.4.3, p. 79

Uniqueness is what makes a model a model. A boundary-value problem for a drum, a cavity or a waveguide is only predictive if its solution is determined by the data. The theorem above says precisely that for the wave equation: give the shape of the region, the initial displacement and velocity, and a fixed or free condition on the boundary, and the motion is determined.

Electromagnetism, with a caution. Each Cartesian component $u$ of $\mathbf{E}$ or $\mathbf{B}$ in vacuum satisfies the wave equation, so the identity (E) applies to it and has exactly the form of Poynting's theorem, $\partial_t e + \nabla\cdot\mathbf{S} = 0$ . The quantities are not the electromagnetic ones, however: $\tfrac12(u_t^2+c^2|\nabla u|^2)$ built from one component is not the field energy density, and $-c^2u_t\nabla u$ is not $\mathbf{E}\times\mathbf{H}$ ; summing over the six components does not repair either, and Poynting's theorem is a consequence of Maxwell's equations themselves, not of the wave equation each component happens to satisfy. Nor does the uniqueness theorem transfer to a cavity component by component, because a perfect conductor imposes $\hat{\mathbf{n}}\times\mathbf{E} = 0$ and $\hat{\mathbf{n}}\cdot\mathbf{B} = 0$ on the wall, which is not a Dirichlet condition on each Cartesian component separately. Uniqueness for the cavity field is proved by running the same energy argument on Maxwell's equations directly, with the true field energy and the true Poynting flux; what carries over is the method, not the theorem.

Numerical analysis. Energy is the standard tool for proving that a discretisation of a wave problem is stable: a scheme with a discrete energy that does not grow cannot blow up, and symplectic or energy-conserving integrators are built to inherit the conservation law exactly rather than approximately. A scheme that violates it shows the damage as amplitude drift over long runs. The stability condition on the time step (von Neumann stability analysis) is in essence the requirement that the numerical domain of dependence contain the true one.

Where it leads. Because the energy argument never uses a solution formula, it survives variable coefficients, lower-order terms, non-constant wave speeds and many nonlinearities. It is the prototype of the a priori estimate on which the modern well-posedness theory of evolution equations rests (energy estimates and well-posedness), and the quantity $E$ is the squared norm in the Sobolev space natural to the problem (weak derivatives and Sobolev spaces). For a physicist the same computation, done with a Lagrangian density instead of by hand, is Noether's theorem applied to invariance under time translation.

Theorem

The wave equation in three dimensions

Adapted

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Shows that the spherical means of a solution of the wave equation satisfy the Euler-Poisson-Darboux equation, turns that into a half-line problem in one dimension to derive Kirchhoff's formula for the Cauchy problem in three dimensions, reads off Huygens' principle, and descends to the two-dimensional Poisson formula, where Huygens' principle fails.

Prerequisites: The one-dimensional wave equation, The divergence theorem, Surface integrals and flux, Differentiation under the integral sign

Definition(Spherical means)AdaptedAdapted from Evans (2010), §2.4.1b, Eq. (12)–(13), p. 67; Strauss (2008), §9.2, Eq. (4), p. 236; Stein & Shakarchi (2003), Ch. 6 §3.2, Eq. (5), p. 189

Let $\omega_n$ be the surface area of the unit sphere of $\mathbb{R}^n$ , so $\omega_2 = 2\pi$ and $\omega_3 = 4\pi$ . For a continuous $f$ on $\mathbb{R}^n$ , $x \in \mathbb{R}^n$ and $r > 0$ , the spherical mean of $f$ on the sphere of centre $x$ and radius $r$ is

\mathcal{M}_r f(x) \;=\; \frac{1}{\omega_n r^{\,n-1}}\int_{|y-x| = r} f(y)\, dS(y) \;=\; \frac{1}{\omega_n}\int_{|z|=1} f(x + rz)\, dS(z),

the second expression following by the substitution $y = x+rz$ . The two are equal because $\omega_n r^{n-1}$ is the area of the sphere $\partial B(x,r)$ , so $\mathcal{M}_rf(x)$ is the average value of $f$ there; $\mathcal{M}_rf(x)\to f(x)$ as $r \to 0^+$ by continuity.

For a solution $u$ of the wave equation and for data $\varphi, \psi$ we write

U(x; r,t) = \mathcal{M}_r u(\cdot,t)(x), \qquad G(x;r) = \mathcal{M}_r\varphi(x), \qquad H(x;r) = \mathcal{M}_r\psi(x) ,

and regard $x$ as fixed, so that $U$ is a function of the radius $r$ and the time $t$ . Evans uses this notation, writing the average with a barred integral sign; Strauss writes $\bar u(r,t)$ for the mean of $u$ about the origin; Stein and Shakarchi write $M_t(f)(x)$ , taking the radius equal to the time (they set $c=1$ ).

The point of the definition is that $u(x,t) = \lim_{r\to 0^+} U(x;r,t)$ : if we can find $U$ , we can recover $u$ .

Theorem(The Euler-Poisson-Darboux equation)AdaptedAdapted from Evans (2010), §2.4.1b, Lemma 1 and Eq. (14), p. 67

Let $n \ge 2$ and let $u \in C^2(\mathbb{R}^n\times[0,\infty))$ solve

u_{tt} = c^2\Delta u \ \text{ in } \mathbb{R}^n\times(0,\infty), \qquad u = \varphi,\ u_t = \psi \ \text{ on } \mathbb{R}^n\times\{t=0\} .

Fix $x$ . Then for $r>0$ and $t>0$ ,

U_{tt} = c^2\Bigl(U_{rr} + \frac{n-1}{r}\,U_r\Bigr), \qquad U(x;r,0) = G(x;r), \quad U_t(x;r,0) = H(x;r) . \tag{EPD}

The differential operator on the right is the radial part of the Laplacian, as Evans notes: a function of $|x|$ alone has $\Delta = \partial_r^2 + \frac{n-1}{r}\partial_r$ . Evans states the lemma for $c=1$ and for $u \in C^m$ with $m \ge 2$ , concluding in addition that $U \in C^m(\overline{\mathbb{R}_+}\times[0,\infty))$ — a regularity statement up to $r=0$ . We use it below in two ways: through the limits $U \to u$ and $rU \to 0$ as $r\to0^+$ , and through the smoothness of $\widetilde G(r) = rG(x;r)$ and $\widetilde H(r)=rH(x;r)$ up to $r=0$ , since the half-line formula applied in Step 3 of the Kirchhoff proof needs $\widetilde G \in C^2$ and $\widetilde H \in C^1$ on $[0,\infty)$ .

Idea of the proofOriginal

Averaging a function over a sphere destroys all of its angular structure, so after averaging the Laplacian can only act radially. The cleanest way to see what is left is not to compute $\Delta$ in polar coordinates but to integrate the equation over the solid ball and use the divergence theorem: the integral of $\Delta u$ over $B(x,r)$ equals the flux of $\nabla u$ out of $\partial B(x,r)$ , and that flux is nothing but $r$ -derivative of the mean, up to the area factor. One differentiation in $r$ then converts the integrated identity into a differential equation. The whole calculation is one application of the divergence theorem and two of differentiation under the integral sign.

ProofAdaptedAdapted from Strauss (2008), §9.2, the alternative proof of Eq. (5), p. 236; Evans (2010), §2.4.1b, proof of Lemma 1, p. 68

Fix $x$ and write $U(r,t)$ for $U(x;r,t)$ .

Step 1 (differentiating in $r$ averages the normal derivative). Write $U(r,t) = \omega_n^{-1}\int_{|z|=1}u(x+rz,t)\,dS(z)$ . The integrand and its first and second derivatives in the parameters $(r,t)$ are continuous on a compact neighbourhood of any $(r,t)$ with $r>0$ , times the unit sphere, because $u \in C^2$ . That licenses differentiating under the integral sign. The theorem of differentiation under the integral sign is stated for an integral over a real interval $[a,b]$ ; here the integration is over the unit sphere, so we use its proof rather than its statement. The proof transfers verbatim, since all it uses about $[a,b]$ is that the integration set has finite total measure (here $\omega_n$ ) and that the parameter derivative is uniformly continuous on a compact product; we appeal to it in this form here and again in Step 3. It gives

U_r(r,t) = \frac{1}{\omega_n}\int_{|z|=1}\nabla u(x+rz,t)\cdot z\; dS(z) = \mathcal{M}_r\Bigl(\frac{\partial u}{\partial n}\Bigr)(x),

because at the point $y = x+rz$ of $\partial B(x,r)$ the outward unit normal is exactly $z = (y-x)/r$ . Multiplying by the area,

\oint_{\partial B(x,r)} \frac{\partial u}{\partial n}\; dS = \omega_n r^{\,n-1}\, U_r(r,t) . \tag{1}

Step 2 (the divergence theorem). $\nabla u(\cdot,t)$ is $C^1$ on the closed ball, so the divergence theorem applies to it:

\int_{B(x,r)} \Delta u\; dy = \oint_{\partial B(x,r)}\frac{\partial u}{\partial n}\; dS = \omega_n r^{\,n-1} U_r . \tag{2}

Step 3 (the left-hand side in terms of $U$ ). Since $u$ solves the equation, $\Delta u = c^{-2}u_{tt}$ . Slicing the ball into spheres and using the same interchange of differentiation and integration as in Step 1, now in the parameter $t$ (so that the mean of $u_{tt}$ is $U_{tt}$ ),

\int_{B(x,r)}\Delta u\; dy = \frac{1}{c^2}\int_0^r\Bigl(\oint_{\partial B(x,\varrho)} u_{tt}\; dS\Bigr) d\varrho = \frac{\omega_n}{c^2}\int_0^r \varrho^{\,n-1}\,U_{tt}(\varrho,t)\; d\varrho . \tag{3}

Step 4 (differentiate in $r$ ). Comparing (2) and (3) and cancelling $\omega_n$ ,

r^{\,n-1}U_r(r,t) = \frac{1}{c^2}\int_0^r \varrho^{\,n-1}U_{tt}(\varrho,t)\, d\varrho .

The integrand is continuous in $\varrho$ , so the right-hand side is differentiable in $r$ with derivative $c^{-2}r^{n-1}U_{tt}$ . Differentiating the left-hand side,

r^{\,n-1}U_{rr} + (n-1)r^{\,n-2}U_r = \frac{1}{c^2}r^{\,n-1}U_{tt} .

Dividing by $r^{n-1}$ , which is legitimate for $r>0$ , gives (EPD). The initial conditions are the means of the initial conditions, by the definition of $G$ and $H$ . $\square$

Theorem(Kirchhoff's formula)AdaptedAdapted from Strauss (2008), §9.2, Eq. (3), p. 234; Evans (2010), §2.4.1c, Eq. (21)–(22), p. 69, and Thm 2 with $n=3$, p. 73; Stein & Shakarchi (2003), Ch. 6, Thm 3.6, p. 191

Let $n = 3$ . If $u \in C^2(\mathbb{R}^3\times[0,\infty))$ solves $u_{tt} = c^2\Delta u$ with $u = \varphi$ , $u_t = \psi$ at $t=0$ , then for $t>0$

u(x,t) \;=\; \frac{\partial}{\partial t}\Bigl[\,t\,\mathcal{M}_{ct}\varphi(x)\,\Bigr] \;+\; t\,\mathcal{M}_{ct}\psi(x) \;=\; \frac{\partial}{\partial t}\left[\frac{1}{4\pi c^2 t}\oint_{S}\varphi\, dS\right] + \frac{1}{4\pi c^2 t}\oint_{S}\psi\, dS , \tag{K}

where $S$ is the sphere of centre $x$ and radius $ct$ . This is Strauss's Eq. (3), which he attributes to Poisson although it is known as Kirchhoff's formula. Stein and Shakarchi write it as $\partial_t(tM_t\varphi) + tM_t\psi$ with $c=1$ . Carrying out the $t$ -derivative gives Evans's equivalent form, an average over the single sphere $\partial B(x,ct)$ :

u(x,t) = \mathcal{M}_{ct}\Bigl[\, t\,\psi(y) + \varphi(y) + \nabla\varphi(y)\cdot(y-x)\,\Bigr](x) . \tag{K$'$}

Existence. Conversely, if $\varphi \in C^3(\mathbb{R}^3)$ and $\psi \in C^2(\mathbb{R}^3)$ and $u$ is defined by (K), then $u \in C^2(\mathbb{R}^3\times[0,\infty))$ , it solves $u_{tt}=c^2\Delta u$ for $t>0$ , and $u \to \varphi$ , $u_t \to \psi$ as $t\to 0^+$ . These are the hypotheses of Evans's Theorem 2 specialised to $n=3$ (he requires $g \in C^{m+1}$ , $h \in C^m$ with $m = (n+1)/2$ , so $m=2$ here), and they are more than the $\varphi\in C^2$ , $\psi\in C^1$ that suffice in one dimension. The two halves together say that (K) is the solution: the first half is already the uniqueness assertion, since it shows that any $C^2$ solution of the Cauchy problem must equal the right-hand side of (K); the second says that under the stronger hypotheses on the data that right-hand side is a solution.

Idea of the proofOriginal

In three dimensions, and only in odd dimensions, the Euler-Poisson-Darboux equation can be turned back into the one-dimensional wave equation. The trick is to look not at the mean $U$ but at $rU$ : the factor $r$ absorbs the awkward first-order term $\frac{2}{r}U_r$ exactly. Since $rU \to 0$ as $r\to 0$ , the new unknown satisfies the one-dimensional wave equation on the half-line $r>0$ with a homogeneous Dirichlet condition at the origin — the reflection problem already solved in d'Alembert's formula. Writing down its solution and then dividing by $r$ and letting $r\to0$ recovers $u(x,t)$ itself, because the limit of $rU/r$ is $U$ . Only the behaviour of the half-line formula near the boundary $r=0$ is used, which is why the answer involves a single sphere of radius $ct$ and no interior data at all.

ProofAdaptedAdapted from Strauss (2008), §9.2, Eq. (8)–(14), pp. 237–238; Evans (2010), §2.4.1c, Eq. (17)–(22), p. 69

Fix $x$ and set

\widetilde U(r,t) = r\,U(x;r,t), \qquad \widetilde G(r) = r\,G(x;r), \qquad \widetilde H(r) = r\,H(x;r) .

Step 1 ( $\widetilde U$ solves the one-dimensional wave equation). For $r > 0$ , using (EPD) with $n=3$ ,

\widetilde U_{tt} = r\,U_{tt} = c^2\Bigl(r\,U_{rr} + 2U_r\Bigr) = c^2\bigl(U + rU_r\bigr)_r = c^2\,\widetilde U_{rr} ,

the middle step because $(U + rU_r)_r = U_r + U_r + rU_{rr}$ . At $t=0$ , $\widetilde U = \widetilde G$ and $\widetilde U_t = \widetilde H$ .

Step 2 (the boundary condition). $U$ is bounded near $r = 0$ , since it is an average of the continuous function $u(\cdot,t)$ over a small sphere and tends to $u(x,t)$ . Hence $\widetilde U(0,t) = \lim_{r\to0^+} rU = 0$ for every $t$ . So $\widetilde U$ solves the Dirichlet problem for the one-dimensional wave equation on the half-line $r>0$ , with data $\widetilde G, \widetilde H$ — which satisfy the compatibility conditions $\widetilde G(0)=\widetilde H(0)=0$ for the same reason.

Step 3 (the half-line formula). The reflection method of d'Alembert's formula constructs a solution of the half-line Dirichlet problem with data $\widetilde G, \widetilde H$ and gives it by the two-case formula; what we need is that our $\widetilde U$ is that solution, which requires knowing the half-line problem has only one $C^2$ solution. (Evans, p. 69, and Strauss, p. 237, pass over this point.) It does, and the reason is short: reflect the uniqueness we already have on the whole line. If $w$ is a $C^2$ function on $\{r\ge0\}\times[0,\infty)$ with $w_{tt}=c^2w_{rr}$ for $r>0$ , $w(0,t)=0$ , and zero initial data, let $W$ be its odd extension in $r$ . Then $W$ is $C^2$ across $r=0$ : among $W$ and its derivatives of order at most two, the odd functions of $r$ are $W$ , $W_t$ , $W_{rr}$ and $W_{tt}$ , and all four vanish at $r=0$ — $w(0,t)\equiv 0$ gives $w = w_t = w_{tt} = 0$ there, and the equation then gives $c^2w_{rr}(0,t) = w_{tt}(0,t) = 0$ — so they are continuous across $r=0$ , while the even ones, $W_r$ and $W_{rt}$ , match from the two sides automatically. So $W$ is a $C^2$ solution on the whole line with zero data, hence $W\equiv0$ by the uniqueness half of d'Alembert's theorem, and $w\equiv0$ . Applying this to the difference of two solutions gives uniqueness.

Therefore, in the region $0 < r < ct$ where the backward characteristic from $(r,t)$ meets the boundary,

\widetilde U(r,t) = \frac12\Bigl[\widetilde G(ct+r) - \widetilde G(ct-r)\Bigr] + \frac{1}{2c}\int_{ct-r}^{ct+r}\widetilde H(s)\, ds .

Step 4 (let $r\to0$ ). Since $\widetilde U(0,t)=0$ ,

u(x,t) = \lim_{r\to0^+} U(x;r,t) = \lim_{r\to0^+}\frac{\widetilde U(r,t) - \widetilde U(0,t)}{r} = \widetilde U_r(0,t) .

Differentiating the two terms of Step 3 in $r$ and setting $r=0$ : the first gives $\tfrac12[\widetilde G'(ct+r) + \widetilde G'(ct-r)] \to \widetilde G'(ct)$ , and the second, differentiating the integral with respect to each of its two endpoints, gives $\tfrac{1}{2c}[\widetilde H(ct+r)+\widetilde H(ct-r)] \to \tfrac1c\widetilde H(ct)$ . Hence

u(x,t) = \widetilde G'(ct) + \frac1c \widetilde H(ct) .

Step 5 (unwinding the tildes). By definition $\tfrac1c\widetilde H(ct) = \tfrac1c\, ct\, H(x;ct) = t\,\mathcal{M}_{ct}\psi(x)$ , the second term of (K). For the first, $\widetilde G'(s) = \frac{d}{ds}[sG(x;s)]$ evaluated at $s = ct$ , and by the chain rule $\frac{\partial}{\partial t}\bigl[t\,G(x;ct)\bigr] = G(x;ct) + ct\,G_s(x;ct) = \bigl[sG(x;s)\bigr]'\big|_{s=ct}$ , which is the same number. This is the first term of (K). Writing the means as integrals over $S = \partial B(x,ct)$ , whose area is $4\pi c^2t^2$ , gives the second display of (K).

Step 6 ((K $'$ ) from (K)). In the form $\mathcal{M}_{ct}\varphi(x) = \omega_3^{-1}\int_{|z|=1}\varphi(x+ctz)dS(z)$ ,

\frac{\partial}{\partial t}\bigl[t\,\mathcal{M}_{ct}\varphi(x)\bigr] = \mathcal{M}_{ct}\varphi(x) + \frac{t}{\omega_3}\int_{|z|=1} c\,\nabla\varphi(x+ctz)\cdot z\; dS(z),

and in the second term $ct\,z = y - x$ at the point $y = x+ctz$ , so the integrand is $\nabla\varphi(y)\cdot(y-x)$ averaged over $\partial B(x,ct)$ . Adding the $\psi$ term gives (K $'$ ). The existence half of the statement is Evans's Theorem 2, whose proof we do not repeat. $\square$

Note(Huygens' principle)AdaptedAdapted from Strauss (2008), §9.2, p. 235, and §2.2, p. 41; Evans (2010), §2.4.1d, Remark (i), p. 74; Stein & Shakarchi (2003), Ch. 6 §3.2, pp. 192–194

Read (K $'$ ) backwards. The number $u(x,t)$ is built from the values of $\psi$ , $\varphi$ and $\nabla\varphi$ on the sphere $|y-x| = ct$ alone — Evans makes this remark explicitly — and not from the ball it bounds. Equivalently, the data at a point $x_1$ of the initial plane influence the solution only on the surface of the light cone $|x - x_1| = ct$ emanating from it. This is Huygens' principle, and Strauss draws the two consequences:

a signal in three dimensions travels at exactly the speed $c$ , neither faster nor slower. Sharp images and undistorted sound follow. Separate an ear from a source by a distance $d$ : the single delay $d/c$ then relates what arrives to what was emitted, and nothing emitted at any other moment is superposed on it;
the causality principle of d'Alembert's formula, which in one dimension only bounds the speed by $c$ , is in three dimensions an exact statement about arrival times.

Stein and Shakarchi phrase the same thing with backward and forward light cones, and add the finite-speed corollary in the form we use below: when both data vanish outside $|y - x_0| \le \delta$ , so does the solution at time $t$ outside the concentric ball whose radius is $\delta$ enlarged by $|t|$ (they take $c=1$ ). Huygens' principle is a feature of odd dimensions: Strauss records that signals propagate sharply exactly in dimensions $n = 3, 5, 7,\dots$ , so that three is the smallest dimension in which they do.

Theorem(Poisson's formula in two dimensions)AdaptedAdapted from Strauss (2008), §9.2, Eq. (19), p. 239; Evans (2010), §2.4.1c, Eq. (27), p. 71; Stein & Shakarchi (2003), Ch. 6, Thm 3.7, p. 194

Let $n=2$ , let $D = \{y \in \mathbb{R}^2 : |y - x| \le ct\}$ be the closed disc of radius $ct$ about $x$ , and let $u \in C^2(\mathbb{R}^2\times[0,\infty))$ solve $u_{tt}=c^2\Delta u$ in $\mathbb{R}^2\times(0,\infty)$ with $u = \varphi$ , $u_t = \psi$ at $t=0$ . (The regularity class is the one Evans states, p. 70; the proof needs it in order to apply (K) to the lifted function $\bar u$ , which inherits it.) Then

u(x,t) = \frac{\partial}{\partial t}\left[\frac{1}{2\pi c}\int_{D} \frac{\varphi(y)}{\sqrt{c^2t^2 - |y-x|^2}}\; dy\right] + \frac{1}{2\pi c}\int_{D}\frac{\psi(y)}{\sqrt{c^2t^2 - |y-x|^2}}\; dy . \tag{P}

This is Strauss's Eq. (19). Evans's Eq. (27), with $c=1$ , is the same formula with the $t$ -derivative carried out,

u(x,t) = \frac{1}{2}\,\frac{1}{\pi t^2}\int_{B(x,t)}\frac{t\varphi(y) + t^2\psi(y) + t\,\nabla\varphi(y)\cdot(y-x)}{\bigl(t^2 - |y-x|^2\bigr)^{1/2}}\; dy ,

and Stein and Shakarchi write it as $\partial_t\bigl(t\widetilde M_t\varphi\bigr) + t\widetilde M_t\psi$ with their disc means $\widetilde M_t$ . The singularity of the integrand at the rim $|y-x| = ct$ is integrable, since $\int_0^{R}\varrho\,d\varrho/\sqrt{R^2-\varrho^2} = R$ is finite.

Huygens' principle fails here. The value $u(x,t)$ depends on the data over the whole disc, not just its rim. Strauss's image is a pebble dropped on a pond: a water bug at distance $\delta$ first feels the wave at $t = \delta/c$ , and then keeps feeling ripples for ever, which die away like $t^{-1}$ .

Idea of the proofOriginal

A function of two variables is also a function of three that happens not to depend on the third. So a two-dimensional Cauchy problem is a three-dimensional one with data constant along the $y_3$ -axis, and Kirchhoff's formula already solves it. All that is left is bookkeeping: an average over a sphere of radius $ct$ in $\mathbb{R}^3$ , of a function that does not depend on $y_3$ , is an integral over the disc of radius $ct$ in $\mathbb{R}^2$ , counted twice because the sphere has two hemispheres over that disc, and weighted by the surface element of a hemisphere, which is what produces the factor $(c^2t^2 - |y-x|^2)^{-1/2}$ . Evans calls this the method of descent.

ProofAdaptedAdapted from Strauss (2008), §9.2, Eq. (15)–(19), pp. 238–239; Evans (2010), §2.4.1c, Eq. (23)–(27), pp. 70–71

Step 1 (lift to three dimensions). Given $\varphi,\psi$ on $\mathbb{R}^2$ define $\bar\varphi(y_1,y_2,y_3) = \varphi(y_1,y_2)$ and $\bar\psi$ likewise, and $\bar u(y_1,y_2,y_3,t) = u(y_1,y_2,t)$ . Since $\partial^2_{y_3}\bar u = 0$ , the three-dimensional Laplacian of $\bar u$ equals the two-dimensional Laplacian of $u$ , so $\bar u$ solves the three-dimensional problem with data $\bar\varphi, \bar\psi$ . Writing $\bar x = (x_1,x_2,0)$ , Kirchhoff's formula (K) gives $u(x,t) = \bar u(\bar x, t)$ in terms of means over the sphere $\partial\bar B(\bar x, ct) \subset \mathbb{R}^3$ .

Step 2 (a sphere average becomes a disc integral). The part of $\partial \bar B(\bar x, ct)$ with $y_3 \ge 0$ is the graph $y_3 = \gamma(y) = \sqrt{c^2t^2 - |y-x|^2}$ over the disc $D$ , and the part with $y_3 \le 0$ is the graph of $-\gamma$ . By the graph formula for the surface element (surface integrals), $dS = \sqrt{1 + |\nabla\gamma|^2}\, dy$ , and

\nabla\gamma = \frac{-(y-x)}{\sqrt{c^2t^2-|y-x|^2}} \quad\Longrightarrow\quad \sqrt{1+|\nabla\gamma|^2} = \sqrt{\frac{c^2t^2 - |y-x|^2 + |y-x|^2}{c^2t^2-|y-x|^2}} = \frac{ct}{\sqrt{c^2t^2-|y-x|^2}} .

Since $\bar\varphi$ takes the same value at $(y,\gamma(y))$ and $(y,-\gamma(y))$ , the two hemispheres contribute equally, and the sphere has area $4\pi c^2t^2$ , so

\mathcal{M}_{ct}\bar\varphi(\bar x) = \frac{1}{4\pi c^2t^2}\cdot 2\int_D \varphi(y)\,\frac{ct\; dy}{\sqrt{c^2t^2-|y-x|^2}} = \frac{1}{2\pi c t}\int_D \frac{\varphi(y)}{\sqrt{c^2t^2-|y-x|^2}}\; dy .

Step 3 (substitute). The same identity holds with $\psi$ in place of $\varphi$ . Putting both into (K),

u(x,t) = \frac{\partial}{\partial t}\Bigl[t\cdot\frac{1}{2\pi ct}\int_D\frac{\varphi}{\sqrt{c^2t^2-|y-x|^2}}dy\Bigr] + t\cdot\frac{1}{2\pi ct}\int_D\frac{\psi}{\sqrt{c^2t^2-|y-x|^2}}dy ,

and the factors of $t$ cancel inside both brackets, giving (P). $\square$

Remark(Where the hypotheses are used, and how the dimensions differ)Original

$u \in C^2$ and the divergence theorem. The derivation of (EPD) differentiates under the integral sign twice and uses the divergence theorem on a ball; both need $u(\cdot,t)$ to be $C^2$ up to the sphere, which is why the lemma is stated for $C^2$ (Evans, for $C^m$ with $m\ge2$ ).
Loss of derivatives. Compare the hypotheses: $\varphi \in C^2$ , $\psi\in C^1$ in one dimension, but $\varphi\in C^3$ , $\psi\in C^2$ for $n=3$ , and more still as $n$ grows — Evans's Theorem 2 (p. 73) asks for $\varphi \in C^{m+1}$ , $\psi \in C^m$ with $m=(n+1)/2$ in odd dimensions $n\ge3$ , and his descent for even $n$ (p. 75) needs $m = (n+2)/2$ . The reason is visible in (K $'$ ): the formula contains $\nabla\varphi$ , which d'Alembert's does not. Evans draws the moral that for $n>1$ a solution need not stay as smooth as its initial displacement, because irregularities in $\varphi$ can focus at a later time; he adds that the energy norm of energy and uniqueness does not deteriorate, which is one reason that norm, and not the supremum norm, is the right measure of size for the wave equation.
Where $n=3$ is special. Step 1 of the Kirchhoff proof works only because $n-1 = 2$ , which is exactly what the single factor $r$ absorbs. For general odd $n = 2k+1$ the same job is done by the operator $\bigl(r^{-1}\partial_r\bigr)^{k-1}r^{2k-1}$ , which is Evans's route to the formula in all odd dimensions; for even $n$ no such transformation exists, and one descends from $n+1$ instead.
The two formulas compared. In (K $'$ ) the data enter through a sphere of radius $ct$ , a set of measure zero in space; in (P) through a solid disc. That single difference is Huygens' principle and its failure. It has a sharp consequence. Suppose $\varphi \equiv 0$ and $\psi \ge 0$ is continuous, not identically zero, and supported in the ball $|y - x_0| \le a$ . Fix $x$ with $|x - x_0| = d > a$ . In three dimensions, (K) gives $u(x,t) = t\,\mathcal{M}_{ct}\psi(x)$ , which vanishes for every $t$ with $ct > d + a$ , because the sphere of radius $ct$ about $x$ then lies entirely outside the support of $\psi$ : the signal passes and is over. In two dimensions, (P) gives a positive integrand on a set of positive measure as soon as $ct > d+a$ , so $u(x,t) > 0$ for all later times; and since $ct\big/\sqrt{c^2t^2 - |y-x|^2} \to 1$ uniformly for $y$ in the support as $t\to\infty$ (the quotient is $\bigl(1 - |y-x|^2/c^2t^2\bigr)^{-1/2}$ and $|y-x|$ is bounded there),

\lim_{t\to\infty} t\,u(x,t) = \frac{1}{2\pi c^2}\int_{\mathbb{R}^2}\psi(y)\, dy ,

the $t^{-1}$ decay that Strauss reports. The limit is positive, so the tail never ends.

Backwards in time. The equation is reversible, and (K) and (P) continue to hold for $t<0$ unaltered, provided the sphere $S$ of (K) and the disc $D$ of (P) are read at radius $c|t|$ , which is what they mean for negative $t$ ; the explicit factors of $t$ and the derivative $\partial/\partial t$ stay exactly as written. It is not correct to substitute $|t|$ for every $t$ : that returns $u(x,|t|)$ rather than $u(x,t)$ , and the two differ whenever $u$ is not even in $t$ . The second example below is the counterexample: with $\varphi\equiv0$ , $\psi(y)=|y|^2$ and $c=1$ the solution is $u(x,t)=t|x|^2+t^3$ , so $u(x,-1) = -|x|^2-1$ , whereas replacing $t$ by $|t|$ throughout gives $+|x|^2+1$ . Stein and Shakarchi accordingly state the support property with $|t|$ : the radius, not the formula, is what carries the absolute value.

Example(Kirchhoff's formula on polynomial data)Original

Take $n=3$ , any $c>0$ , and

\varphi(y) = y_1y_2, \qquad \psi(y) = y_3^2 ,

both polynomials, hence $C^\infty$ , so the existence hypotheses hold.

The means. Over $\partial B(x,r)$ , write $y = x + rz$ with $|z|=1$ . The average of $z_i$ is $0$ by symmetry, and the average of $z_iz_j$ is $\tfrac13\delta_{ij}$ : it vanishes for $i\neq j$ because $z_i z_j$ is odd under $z_i \mapsto -z_i$ , and the three diagonal averages are equal and sum to the average of $|z|^2 = 1$ . Hence

\mathcal{M}_r\varphi(x) = x_1x_2 + \tfrac{r^2}{3}\delta_{12} = x_1x_2, \qquad \mathcal{M}_r\psi(x) = x_3^2 + \tfrac{r^2}{3}.

Apply (K) with $r = ct$ :

u(x,t) = \frac{\partial}{\partial t}\bigl[t\,x_1x_2\bigr] + t\Bigl(x_3^2 + \frac{c^2t^2}{3}\Bigr) = x_1x_2 + t\,x_3^2 + \frac{c^2t^3}{3}.

Check it. $u_{tt} = 2c^2 t$ , while $\Delta u = \partial_1^2(x_1x_2) + \partial_2^2(x_1x_2) + \partial_3^2(tx_3^2) = 0 + 0 + 2t$ , so $u_{tt} = c^2\Delta u$ . At $t=0$ , $u = x_1x_2 = \varphi$ and $u_t = x_3^2 + c^2t^2\big|_{t=0} = \psi$ . All three conditions hold exactly.
Check (K $'$ ) too. The extra term there is the mean of $\nabla\varphi(y)\cdot(y-x) = y_2(y_1-x_1) + y_1(y_2-x_2)$ , whose average is $2x_1x_2 - x_1x_2 - x_2x_1 = 0$ by part 1, so (K $'$ ) returns the same $u$ — as it must.
A number. At $x = (0.4,\,-1.1,\,2.3)$ , $c = 1.7$ , $t = 0.9$ : $u = (0.4)(-1.1) + 0.9(2.3)^2 + (1.7)^2(0.9)^3/3 = -0.44 + 4.761 + 0.70227 = 5.02327$ . Evaluating the two spherical means of (K) by Gauss-Legendre quadrature on the sphere of radius $ct = 1.53$ reproduces $5.0232700$ .

Example(The same data in three and in two dimensions)Original

Now take $\varphi \equiv 0$ and $\psi(y) = |y|^2$ , and compute in both dimensions.

Three dimensions. $\mathcal{M}_r\psi(x) = |x|^2 + r^2$ , since $|x+rz|^2 = |x|^2 + 2r\,x\cdot z + r^2$ and the middle term averages to zero. So (K) gives

u_3(x,t) = t\bigl(|x|^2 + c^2t^2\bigr) = t|x|^2 + c^2t^3 .

Check: $u_{3,tt} = 6c^2t$ and $\Delta_3(t|x|^2) = 6t$ , so $u_{3,tt} = c^2\Delta u_3$ . 2. Two dimensions. By (P), with $z = y - x$ and $R = ct$ ,

u_2(x,t) = \frac{1}{2\pi c}\int_{|z|\le R}\frac{|x|^2 + 2x\cdot z + |z|^2}{\sqrt{R^2 - |z|^2}}\, dz .

The middle term integrates to zero by the symmetry $z \mapsto -z$ . For the other two, in plane polars with $|z| = \varrho = R\sin\theta$ ,

\int_{|z|\le R}\frac{dz}{\sqrt{R^2-\varrho^2}} = 2\pi\!\int_0^R\!\frac{\varrho\,d\varrho}{\sqrt{R^2-\varrho^2}} = 2\pi R, \qquad \int_{|z|\le R}\frac{|z|^2\,dz}{\sqrt{R^2-\varrho^2}} = 2\pi\!\int_0^{\pi/2}\! R^3\sin^3\theta\, d\theta = \frac{4\pi R^3}{3},

using $\int_0^{\pi/2}\sin^3 = 2/3$ . Hence

u_2(x,t) = \frac{1}{2\pi c}\Bigl(2\pi R\,|x|^2 + \frac{4\pi R^3}{3}\Bigr) = t|x|^2 + \frac{2}{3}c^2t^3 ,

using $R = ct$ . Check: $u_{2,tt} = 4c^2t$ and $\Delta_2(t|x|^2) = 4t$ , so this too solves its own equation with the right data. 3. Compare. Identical data, and the same leading term $t|x|^2$ , but the correction is $c^2t^3$ in three dimensions and $\tfrac23 c^2t^3$ in two. The formulas are genuinely different operators, not two notations for one. 4. A number, checked by quadrature. At $x = (0.7,-0.5)$ , $c = 1.3$ , $t = 0.8$ : $u_2 = 0.8(0.74) + \tfrac23(1.69)(0.512) = 0.592 + 0.576853 = 1.168853$ , and numerical integration of (P) gives $1.1688533$ . The three-dimensional value at $x = (0.7,-0.5,0)$ with the same $c,t$ is $0.592 + 1.69(0.512) = 1.457280$ , reproduced by quadrature on the sphere to seven figures. 5. The tail, with numbers. Let $c=1$ , $\varphi\equiv0$ and $\psi(y) = (1-|y|^2)^3$ for $|y|\le1$ , zero outside — a $C^2$ bump, non-negative, supported in the unit ball. Observe the point $|x| = 5$ . The front arrives at $t = 4$ in both dimensions. In three dimensions $u(x,t) = t\,\mathcal{M}_t\psi(x) = 0$ for every $t > 6$ : the sphere of radius $t$ about $x$ has left the support, and nothing is left to hear. In two dimensions $\int_{\mathbb{R}^2}\psi = 2\pi\int_0^1(1-\varrho^2)^3\varrho\, d\varrho = \pi/4$ , so

t\,u(x,t) \longrightarrow \frac{1}{2\pi}\cdot\frac{\pi}{4} = \frac18 = 0.125 .

Numerically, $u(x,20) = 6.4569\times10^{-3}$ (so $tu = 0.12914$ ) and $u(x,1000) = 1.25002\times10^{-4}$ (so $tu = 0.125002$ ): the signal is still arriving at $t = 1000$ , when the three-dimensional one has been silent for a thousand time units.

ApplicationAdaptedAdapted from Strauss (2008), §9.2, p. 235 and p. 239; Evans (2010), §2.4.2, Eq. (44), p. 79

Why we can see and hear. Huygens' principle is the mathematical reason that optical images are sharp and that recorded sound is intelligible: in three dimensions every point of a source contributes to the field at a receiver at exactly one instant, so the signal arrives undistorted apart from the $1/r$ spreading of the amplitude. Strauss's counterfactual is worth keeping in mind: in two dimensions every view and every sound would be mixed with the echoes of all earlier ones.

Retarded potentials and radiation. For the inhomogeneous equation $u_{tt}-\Delta u = f$ in three dimensions, the same spherical-mean machinery produces Evans's retarded-potential solution $u(x,t) = \frac{1}{4\pi}\int_{B(x,t)} f(y, t-|y-x|)\,|y-x|^{-1}dy$ : each source element acts at the receiver after the travel-time delay $|y-x|/c$ . That formula, with $c$ restored, is the retarded solution of Maxwell's equations used throughout antenna theory and radiation physics; see Green's functions for the wave and heat equations.

Where two dimensions really occur. The two-dimensional formula is the right one for genuinely planar waves: ripples on a shallow pond, flexural waves in a plate, and shallow-water tsunami models. It predicts the long decaying tail behind the front, which is observed, and which makes the deconvolution of a seismic or sonar record a harder problem than in three dimensions.

Imaging. Kirchhoff's formula is also run backwards. If one knows $u$ on a surface for all times, the integrals over spheres of radius $ct$ can be inverted to recover the initial data — this is the mathematics of photoacoustic and thermoacoustic tomography, and it is the wave-equation cousin of the Radon transform.

Method

Standing waves and normal modes

Adapted

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Separates the variables in the wave equation for a string with fixed ends, shows that the only separated solutions are the standing waves $\sin(n\pi x/L)$ oscillating at frequency $nc/2L$ , gives the recipe for fitting arbitrary initial data by a Fourier sine series, and says when the resulting series is a genuine solution.

Prerequisites: The one-dimensional wave equation, Energy and uniqueness for the wave equation, Fourier series

Definition(Separated solutions and normal modes)AdaptedAdapted from Strauss (2008), §4.1, Eq. (1)–(4), p. 84, and p. 88; Stein & Shakarchi (2003), Ch. 1 §1, p. 11

Let $L>0$ and $c = \sqrt{\tau/\rho}>0$ . The vibrating-string problem is

\begin{aligned} &u_{tt} = c^2u_{xx} && 0<x<L,\ t>0, \\ &u(0,t) = u(L,t) = 0 && t>0, \\ &u(x,0) = \varphi(x), \quad u_t(x,0) = \psi(x) && 0<x<L . \end{aligned} \tag{S}

The boundary condition says the two ends are clamped. Strauss writes the length as $l$ .

A separated solution is a solution of the equation and the boundary conditions of the product form

u(x,t) = X(x)\,T(t), \qquad X \not\equiv 0,\ T\not\equiv 0 ,

with $X$ depending on $x$ alone and $T$ on $t$ alone. A separated solution of (S) is also called a standing wave: its shape in $x$ is fixed once and for all, and only its amplitude moves, so the zeros of $X$ in $0<x<L$ never move. Those zeros are the nodes, and the points of greatest amplitude the anti-nodes (Stein and Shakarchi).

In the eigenvalue language that Strauss introduces on p. 88, the spatial factor solves

-X'' = \lambda X \ \text{ on } (0,L), \qquad X(0)=X(L)=0 ,

an ordinary differential equation with conditions at two points. A number $\lambda$ for which a solution $X \not\equiv 0$ exists is an eigenvalue and such an $X$ an eigenfunction of the operator $A = -d^2/dx^2$ acting on functions that satisfy the Dirichlet conditions; Normal mode is the name physicists and engineers give to such an eigenfunction; Strauss explains the word by observing that these shapes, alone among all shapes the string could take, go on unaltered as time passes. This is the simplest Sturm-Liouville eigenvalue problem, and, as Strauss puts it, the situation is the infinite-dimensional analogue of the eigenvectors of a matrix.

Method(The modes of a fixed string, and the solution)AdaptedAdapted from Strauss (2008), §4.1, Eq. (5)–(12), pp. 85–87, and the eigenvalue proof on p. 88; Stein & Shakarchi (2003), Ch. 1 §1, Eq. (4) and (6), pp. 14–15

The eigenvalues and modes. The eigenvalues of $-X''=\lambda X$ with $X(0)=X(L)=0$ are exactly

\lambda_n = \Bigl(\frac{n\pi}{L}\Bigr)^2, \qquad X_n(x) = \sin\frac{n\pi x}{L}, \qquad n = 1,2,3,\dots,

each simple, each eigenfunction determined up to a constant multiple. No $\lambda \le 0$ and no non-real $\lambda$ is an eigenvalue.

The separated solutions. Consequently the separated solutions of (S) are exactly the functions

u_n(x,t) = \Bigl(A_n\cos\omega_n t + B_n\sin\omega_n t\Bigr)\sin\frac{n\pi x}{L}, \qquad \omega_n = \frac{n\pi c}{L} = n\omega_1 , \tag{M}

for some $n \in \mathbb{N}$ and some real constants $A_n, B_n$ that are not both zero. The $n$ -th mode has $n-1$ interior nodes, at $x = L/n, 2L/n, \dots, (n-1)L/n$ .

Frequencies. Strauss calls the numbers $\omega_n = n\pi c/L$ the frequencies, and records that some texts instead call $nc/2L$ the frequency; the two differ by the factor $2\pi$ between angular frequency and cycles per second. In terms of the string's constants,

\omega_n = \frac{n\pi}{L}\sqrt{\frac{\tau}{\rho}} \qquad (n=1,2,\dots),

which is Strauss's Eq. (12). The fundamental is $n=1$ and the overtones are its exact integer multiples — the fact, as Strauss notes, that Euler discovered in 1749 and that creates a sensation still.

General solution. Any finite sum $\sum_n u_n$ solves the equation and the boundary conditions, by linearity. Such a sum meets the initial conditions of (S) precisely when

\varphi(x) = \sum_n A_n \sin\frac{n\pi x}{L} \qquad\text{and}\qquad \psi(x) = \sum_n \omega_n B_n\sin\frac{n\pi x}{L} ,

Strauss's Eq. (10)–(11). For general data one therefore takes an infinite series and reads these two identities as Fourier sine expansions, which gives

A_n = \frac{2}{L}\int_0^L \varphi(x)\sin\frac{n\pi x}{L}\, dx, \qquad B_n = \frac{2}{n\pi c}\int_0^L \psi(x)\sin\frac{n\pi x}{L}\, dx . \tag{C}

Stein and Shakarchi obtain the first of these (their Eq. (6), with $L=\pi$ ) from the orthogonality relation below, and both they and Strauss are explicit that at this stage the infinite series is formal; see the note on convergence.

Method(Separation of variables for the vibrating string)AdaptedAdapted from Strauss (2008), §4.1, pp. 84–86; Riley, Hobson & Bence (2006), §21.1–21.2, pp. 714–719; Arfken, Weber & Harris (2013), §9.4, pp. 414–416

Try a product. Substitute $u = X(x)T(t)$ into $u_{tt} = c^2u_{xx}$ and divide by $-c^2XT$ :

-\frac{T''(t)}{c^2T(t)} = -\frac{X''(x)}{X(x)} .

Arfken's criterion for expecting this to work is that the differential operators in the different variables occur additively, with no products of derivatives belonging to different variables, and that the boundary conditions, or at any rate part of them, split into conditions on the individual factors; Hassani makes the same separation of $t$ from $\mathbf{x}$ for the heat, Schrodinger and Klein-Gordon equations, which is why one mode decomposition serves all of them. 2. Name the constant. The left side depends only on $t$ and the right only on $x$ , and they are equal, so each is a constant $\lambda$ — the separation constant. (Strauss's proof: the common value has zero partial derivative in both $x$ and $t$ .) In more variables there is one such constant per variable, subject to one relation: Riley, Hobson and Bence separate the three-dimensional wave equation into four factors with constants $-l^2,-m^2,-n^2,-\mu^2$ tied by $\mu^2 = l^2+m^2+n^2$ , and Arfken counts the independent ones as $n-1$ for a problem in $n$ variables. The result is a pair of ordinary equations,

X'' + \lambda X = 0, \qquad T'' + c^2\lambda T = 0 .

Impose the boundary conditions on the spatial factor. $u(0,t)=u(L,t)=0$ for all $t$ with $T \not\equiv 0$ forces $X(0)=X(L)=0$ . This is the step that selects the allowed $\lambda$ ; it is Arfken's second case, a genuine eigenvalue problem with homogeneous boundary conditions.
Find the eigenvalues. Solve the two-point problem. For the fixed string the answer is $\lambda_n = (n\pi/L)^2$ with $X_n = \sin(n\pi x/L)$ , derived in the proof below.
Solve for the time factor. With $\lambda = \lambda_n$ , $T'' + \omega_n^2T = 0$ with $\omega_n = c\sqrt{\lambda_n} = n\pi c/L$ , so $T$ is $A_n\cos\omega_nt + B_n\sin\omega_nt$ . Writing the factors as exponentials instead is often more convenient in other geometries (Riley, Hobson and Bence give both forms); for a bounded interval the trigonometric form is the natural one.
Superpose. Form $u = \sum_{n\ge1}u_n$ . This solves the equation and the boundary conditions for any coefficients for which the series behaves; the step is legitimate only because the problem is linear and the boundary conditions homogeneous.
Fit the initial data. Expand $\varphi$ and $\psi$ in the eigenfunctions and match coefficients, which gives (C). As Riley, Hobson and Bence emphasise, this is possible in general only because the $X_n$ form a complete set — here, the Fourier sine system (Fourier series). They also point out what the expansion silently assumes about the data: a sine series represents the odd, $2L$ -periodic continuation of $\varphi$ , so if $\varphi(0)$ or $\varphi(L)$ is non-zero the continuation has jumps there and the series converges to the midpoints of the jumps.
Check convergence if an exact solution is wanted, by the criterion in the note below.

Idea of the proofOriginal

Two observations drive everything. The first is that an equation in which $x$ -derivatives and $t$ -derivatives are not mixed can be satisfied by a product, because after dividing by that product the $x$ and $t$ parts of the equation sit on opposite sides of the equals sign; two functions of independent variables can agree only by being constant. The second is that the clamped ends turn the resulting ordinary equation for $X$ into a two-point problem, which almost never has a solution: only for a discrete list of $\lambda$ does a non-zero $X$ fit. The list is found by writing down the general solution of $X''=-\lambda X$ in each of the three cases $\lambda<0$ , $\lambda=0$ , $\lambda>0$ and seeing which can vanish at both ends. Only $\lambda>0$ survives, and then only when $\sqrt\lambda L$ is a multiple of $\pi$ . Finally, because the problem is linear, modes can be added; and because the mode shapes are complete and orthogonal, any initial shape can be written as a combination of them, with coefficients given by integrals. The sum is then a solution with the right data.

ProofAdaptedAdapted from Strauss (2008), §4.1, pp. 85–86 and the eigenvalue argument on p. 88; Stein & Shakarchi (2003), Ch. 1 §1, pp. 11–12 and p. 15

Step 1 (the separation constant is a constant, and the two ODEs hold everywhere). Let $u = XT$ solve $u_{tt}=c^2u_{xx}$ on $(0,L)\times(0,\infty)$ with $X,T$ twice differentiable and neither identically zero. Then

X(x)\,T''(t) = c^2 X''(x)\, T(t) \qquad \text{for all } x \in (0,L),\ t>0 . \tag{$*$}

Let $P = \{x : X(x)\neq0\}$ and $Q = \{t: T(t)\neq0\}$ , both non-empty since $X\not\equiv0$ and $T\not\equiv0$ . Stein and Shakarchi note that dividing by $XT$ is legitimate only off the zeros, and their remedy — check the candidate product directly — proves the converse but not the "exactly" claimed above. The gap is short to close, and closing it matters here, because $x=0$ and $x=L$ are themselves zeros of $X$ .

The constant. For $(x,t) \in P\times Q$ , dividing ( $*$ ) by $c^2X(x)T(t)$ gives $T''(t)/c^2T(t) = X''(x)/X(x)$ . Fix one $t_0 \in Q$ . The right-hand side then equals the fixed number $T''(t_0)/c^2T(t_0)$ for every $x \in P$ , so $X''/X$ is a single constant on $P$ ; call it $-\lambda$ . Feeding that back, $T''/c^2T = -\lambda$ on all of $Q$ as well. One constant serves both, on all of $P\times Q$ at once — no connectedness of $P$ or $Q$ is needed, because the two sides are decoupled.

The zeros. Hence $X''+\lambda X = 0$ on $P$ and $T'' + c^2\lambda T = 0$ on $Q$ . At a zero of $X$ the equation ( $*$ ) itself finishes the job: if $X(x_1) = 0$ then ( $*$ ) reads $0 = c^2X''(x_1)T(t)$ for all $t>0$ , and choosing $t \in Q$ gives $X''(x_1) = 0 = -\lambda X(x_1)$ . Symmetrically, if $T(t_1)=0$ then $X(x)T''(t_1) = 0$ for all $x$ , and choosing $x \in P$ gives $T''(t_1) = 0 = -c^2\lambda T(t_1)$ . So

X'' + \lambda X = 0 \ \text{ on } (0,L), \qquad T'' + c^2\lambda T = 0 \ \text{ on } (0,\infty),

with no exceptional points. The equation for $X$ is needed only on the open interval, where the partial differential equation is posed; $X$ is continuous on the closed interval $[0,L]$ , being twice differentiable there, so the boundary values $X(0)$ and $X(L)$ used in the next step are the limits at the two ends of the solution found on $(0,L)$ .

Step 2 (the boundary conditions pass to $X$ ). $X(0)T(t) = 0$ for all $t$ and $T\not\equiv0$ give $X(0)=0$ , and likewise $X(L)=0$ .

Step 3 (no eigenvalue is zero). If $\lambda = 0$ then $X''=0$ , so $X(x) = C + Dx$ . From $X(0)=0$ , $C=0$ ; from $X(L)=0$ , $DL=0$ , so $D=0$ and $X\equiv0$ . Excluded.

Step 4 (no eigenvalue is negative). If $\lambda<0$ , write $\lambda = -\gamma^2$ with $\gamma>0$ . Then $X''=\gamma^2X$ , whose general solution is $X = C\cosh\gamma x + D\sinh\gamma x$ . Now $X(0)=0$ gives $C=0$ , and $X(L) = D\sinh\gamma L = 0$ gives $D=0$ , because $\sinh\gamma L \ne 0$ for $\gamma L>0$ . Excluded.

Step 5 (no eigenvalue is non-real). This is Strauss's argument, and it is needed because nothing so far rules out complex $\lambda$ . Let $\lambda \in \mathbb{C}$ and let $\gamma$ be either square root of $-\lambda$ ; if $\gamma = 0$ we are in Step 3, so take $\gamma \neq 0$ . The general solution is $X = Ce^{\gamma x} + De^{-\gamma x}$ . From $X(0)=0$ , $D = -C$ , and $C \neq 0$ or $X \equiv 0$ ; from $X(L)=0$ , $Ce^{\gamma L} = Ce^{-\gamma L}$ , hence

e^{2\gamma L} = 1 .

Writing $\gamma = \sigma + i\varsigma$ , $|e^{2\gamma L}| = e^{2\sigma L} = 1$ forces $\sigma = 0$ , and then $e^{2i\varsigma L}=1$ forces $2\varsigma L = 2\pi n$ for some integer $n$ . So $\gamma = in\pi/L$ and $\lambda = -\gamma^2 = n^2\pi^2/L^2$ , which is real and positive. Hence every eigenvalue is of this form, and $n=0$ is excluded by Step 3.

Step 6 (the eigenfunctions). Conversely, for $\lambda = \beta^2$ with $\beta>0$ the general solution of $X''+\beta^2X=0$ is $X = C\cos\beta x + D\sin\beta x$ . Then $X(0)=0$ gives $C=0$ , and $X(L) = D\sin\beta L = 0$ with $D\neq0$ gives $\sin\beta L = 0$ , that is $\beta L = n\pi$ for some $n \in \mathbb{N}$ . Each such $\beta$ gives the one-dimensional space of eigenfunctions spanned by $\sin(n\pi x/L)$ . Combining with Steps 3–5, the eigenvalue list is exactly $\lambda_n=(n\pi/L)^2$ for $n\in\mathbb{N}$ . For each of them the time equation of Step 1 reads $T'' + \omega_n^2T = 0$ with $\omega_n = c\sqrt{\lambda_n} = n\pi c/L$ , whose general solution is $A_n\cos\omega_nt + B_n\sin\omega_nt$ . Multiplying the two factors gives the modes (M), and these are all of them.

Step 7 (orthogonality and the coefficients). For $m,n\in\mathbb{N}$ ,

\int_0^L \sin\frac{m\pi x}{L}\sin\frac{n\pi x}{L}\; dx = \frac{L}{2}\,\delta_{mn},

by the product-to-sum identity $2\sin a\sin b = \cos(a-b) - \cos(a+b)$ and the fact that $\int_0^L\cos(k\pi x/L)dx = 0$ for every non-zero integer $k$ . Suppose now $\varphi(x) = \sum_m A_m\sin(m\pi x/L)$ with the series converging well enough to integrate term by term. Multiplying by $\sin(n\pi x/L)$ and integrating over $[0,L]$ leaves one surviving term, $A_n L/2$ , which gives the first formula in (C). Differentiating the series for $u$ once in $t$ and setting $t=0$ gives $\psi = \sum_m \omega_m B_m\sin(m\pi x/L)$ , and the same computation gives $\omega_nB_n = \frac{2}{L}\int_0^L\psi\sin(n\pi x/L)dx$ , that is the second formula in (C). $\square$

Note(When is the series a solution?)AdaptedAdapted from Strauss (2008), §4.1, p. 86; Stein & Shakarchi (2003), Ch. 1 §1, p. 14; Riley, Hobson & Bence (2006), §21.2, p. 719

Steps 6 and 7 are exact for a finite sum of modes, which is Strauss's Eq. (9). For an infinite series both Strauss and Stein and Shakarchi say plainly that the argument is formal at this stage: Strauss postpones the convergence and differentiability questions to his chapter on Fourier series, remarking only that practically any function on $(0,L)$ can be expanded as in (C); Stein and Shakarchi write that questions of convergence arise and that they will not worry about them yet. We do not prove a convergence theorem here either, and record one sufficient condition that follows from the standard theorem on differentiating a uniformly convergent series:

Let $\varphi$ and $\psi$ be continuous on $[0,L]$ , let $A_n$ and $B_n$ be their coefficients (C), and suppose that
$\sum_{n\ge1} n^2\bigl(|A_n| + |B_n|\bigr) < \infty .$
Then the series $u = \sum_n u_n$ and all of its first and second derivative series converge absolutely and uniformly on $[0,L]\times\mathbb{R}$ , so $u \in C^2$ , may be differentiated term by term, and satisfies the equation and the boundary conditions of (S). Its data are $u(x,0) = \sum_n A_n\sin(n\pi x/L)$ and $u_t(x,0)=\sum_n\omega_nB_n\sin(n\pi x/L)$ : two uniformly convergent sine series whose coefficients are by construction the sine coefficients of $\varphi$ and of $\psi$ . Each therefore sums to its own function, because a continuous function on $[0,L]$ all of whose sine coefficients vanish is zero (completeness of the sine system, Fourier series) and the difference between $\varphi$ and the sum of its series is such a function. Hence $u(x,0)=\varphi(x)$ , $u_t(x,0)=\psi(x)$ , and $u$ is a classical solution of (S).

The hypothesis holds as soon as the sine coefficients of $\varphi$ and $\psi$ are $O(n^{-3-\varepsilon})$ for some $\varepsilon>0$ , which smoothness of the data (together with the compatibility of the data with the odd $2L$ -periodic continuation that a sine series represents) delivers. Decay merely faster than $n^{-3}$ is not enough: $A_n = 1/(n^3\log n)$ is $o(n^{-3})$ , yet $\sum n^2|A_n| = \sum 1/(n\log n)$ diverges. See smoothness and decay of Fourier coefficients and the convergence theorems of Fourier series. When it fails — a plucked string with a corner is the standard case — the series still converges to the function that d'Alembert's formula produces from the periodically extended data, which is a weak and not a classical solution; the uniqueness theorem of energy and uniqueness for the wave equation guarantees that whenever the series is a classical solution, it is the solution.

Remark(Reading the hypotheses, and the other boundary conditions)Original

Where the boundary conditions are used. Only in Steps 2–6, and they do all the work: without them every $\lambda \in \mathbb{C}$ is admissible and there are no discrete frequencies. This is why an infinite string has a continuum of modes ( $e^{i(kx-\omega t)}$ for every real $k$ , with $\omega = ck$ ) and a finite one has a discrete spectrum. The quantisation is a consequence of confinement, not of the equation.
Why $\lambda>0$ matters physically. In the convention fixed above, $X''+\lambda X=0$ and $T''+c^2\lambda T=0$ , a negative or complex $\lambda$ would make $T$ grow exponentially. This is Strauss's convention, and the conclusion reads $\lambda>0$ . Beware the sign. Stein and Shakarchi write the separated equations with the opposite sign, $\phi''-\lambda\phi=0$ and $\psi''-\lambda\psi=0$ (p. 12), so their $\lambda$ is the negative of the $\lambda$ used throughout this unit. In their convention the admissible case is $\lambda<0$ , written $\lambda=-m^2$ , and what they reject is $\lambda\ge0$ , on the ground that the time factor would then not oscillate. Translated into our sign that is the same conclusion, $\lambda>0$ , reached by a different route; only Strauss's route also rules out complex $\lambda$ , which is what Step 5 above does.
Standing waves are pairs of travelling waves. By $2\sin a\cos b = \sin(a-b)+\sin(a+b)$ ,

\sin\frac{n\pi x}{L}\cos\omega_n t = \frac12\sin\frac{n\pi(x-ct)}{L} + \frac12\sin\frac{n\pi(x+ct)}{L} ,

a right-moving and a left-moving wave of half the amplitude — exactly the decomposition of d'Alembert's formula. Arfken makes the converse remark that the totality of the separated solutions carries the same content as the travelling-wave solutions, and the difference between a standing and a travelling wave is that the nodes of the former do not move. Strauss records that it took half a century to settle the relation between the series solution and d'Alembert's.

Free ends. With Neumann conditions $u_x(0,t)=u_x(L,t)=0$ the same calculation gives $X_n = \cos(n\pi x/L)$ and $\lambda_n = (n\pi/L)^2$ — but now, as Strauss stresses, $n = 0$ is an eigenvalue, with the constant eigenfunction, and then $T''=0$ gives $T = A + Bt$ . So the free-ended string has an extra solution $\tfrac12A_0 + \tfrac12B_0t$ describing uniform translation at constant speed, which the clamped string does not. Omitting it is the commonest error in applying the method to Neumann problems.
How the frequencies scale with $L$ and $\tau$ . Replacing $L$ by $L/2$ (clamping the midpoint) doubles every frequency, raising the note by an octave; increasing the tension $\tau$ raises all frequencies as $\sqrt\tau$ . Both are immediate from $\omega_n = (n\pi/L)\sqrt{\tau/\rho}$ and both are what a player does.

Example(A plucked string, exactly)Original

Take a steel string with

L = 0.650\ \mathrm{m}, \qquad \tau = 80\ \mathrm{N}, \qquad \rho = 5.0\times10^{-3}\ \mathrm{kg\,m^{-1}},

so $c = \sqrt{80/0.005} = 126.491\ \mathrm{m\,s^{-1}}$ , $\omega_1 = \pi c/L = 611.36\ \mathrm{rad\,s^{-1}}$ , and the fundamental is $f_1 = c/2L = 97.301\ \mathrm{Hz}$ — a little below G ${}_2$ , which in equal temperament with A ${}_4 = 440\ \mathrm{Hz}$ is $97.999\ \mathrm{Hz}$ , so the string is flat by $12.4$ cents. The period is $T_1 = 2L/c = 10.277\ \mathrm{ms}$ .

Draw the string into the shape

\varphi(x) = h\sin^3\frac{\pi x}{L}, \qquad h = 3.00\ \mathrm{mm},

and release it from rest, so $\psi \equiv 0$ . (This profile vanishes at both ends together with its first two derivatives, so it is compatible with the clamped ends to high order.)

The coefficients, in closed form. The identity $\sin^3\theta = \tfrac14\bigl(3\sin\theta - \sin3\theta\bigr)$ gives

\varphi(x) = \frac{3h}{4}\sin\frac{\pi x}{L} - \frac{h}{4}\sin\frac{3\pi x}{L},

already a finite sine series. By the orthogonality of Step 7 the coefficients are read off:

A_1 = \frac{3h}{4} = 2.25\ \mathrm{mm}, \qquad A_3 = -\frac{h}{4} = -0.75\ \mathrm{mm},

all other $A_n = 0$ and all $B_n = 0$ since $\psi\equiv0$ . (Evaluating the integral in (C) directly for $n=1,\dots,6$ confirms $3h/4,\,0,\,-h/4,\,0,\,0,\,0$ .) 2. The solution. Because the series is finite, no convergence question arises and

u(x,t) = \frac{3h}{4}\sin\frac{\pi x}{L}\cos\omega_1 t \;-\; \frac{h}{4}\sin\frac{3\pi x}{L}\cos 3\omega_1 t

is an exact classical solution of (S). 3. Checks. At $t=0$ and $x=L/2$ : $\sin(\pi/2)=1$ and $\sin(3\pi/2)=-1$ , so $u = \tfrac{3h}{4} + \tfrac{h}{4} = h = 3.00\ \mathrm{mm}$ , the full initial displacement. At $t = T_1/4$ both $\cos\omega_1t = \cos(\pi/2)$ and $\cos3\omega_1t = \cos(3\pi/2)$ vanish, so the string is instantaneously straight and all of its energy is kinetic. At $t = T_1/6 = 1.7129\ \mathrm{ms}$ we have $\omega_1t = \pi/3$ and $3\omega_1t=\pi$ , so

u\Bigl(\frac{L}{2},\frac{T_1}{6}\Bigr) = \frac{3h}{4}\cdot\frac12 - \frac{h}{4}\cdot(-1)\cdot(-1) = \frac{3h}{8} - \frac{h}{4} = \frac{h}{8} = 0.375\ \mathrm{mm}.

Which harmonics sound. Only $n=1$ and $n=3$ are excited: the timbre is a fundamental at $97.30\ \mathrm{Hz}$ plus a third harmonic at $291.9\ \mathrm{Hz}$ at one third of the amplitude. Every even harmonic is missing, and that is no accident: $\varphi$ is symmetric about the midpoint $x=L/2$ , while $\sin(n\pi x/L)$ for even $n$ is antisymmetric about it, so the integral in (C) cancels in pairs. Plucking at the centre always kills the even harmonics, by this symmetry. The familiar triangular pluck, with apex height $h$ at $x=d$ , shows the same effect from the other side: its coefficients work out to $A_n = 2hL^2\sin(n\pi d/L)\big/\bigl(\pi^2n^2d(L-d)\bigr)$ , so plucking at $x = L/m$ silences every mode whose index is a multiple of $m$ .
Energy. By the mode-by-mode formula of energy and uniqueness for the wave equation,

E = \frac{\tau\pi^2}{4L}\sum_n n^2A_n^2 = \frac{80\pi^2}{4(0.650)}\Bigl[(2.25\times10^{-3})^2 + 9(0.75\times10^{-3})^2\Bigr] = 3.075\times10^{-3}\ \mathrm{J},

and this is constant in time although the shape is not.

ApplicationAdaptedAdapted from Strauss (2008), §4.1, Eq. (12) and the fundamental and overtones, p. 87, and §4.1 Exercise 1(a), p. 89; Stein & Shakarchi (2003), Ch. 1 §1, pp. 13–14; Hassani (2013), §13.1, pp. 396–397

Musical acoustics. The whole theory of pitched instruments is in (M): the frequencies of a clamped string are exact integer multiples of the fundamental, which is why a string sounds a definite note rather than a noise, and the relative sizes of the $A_n$ and $B_n$ are the timbre. Where the string is plucked or struck decides which modes are excited, as the example shows; a bridge or a finger placed at a node of a mode leaves that mode untouched, which is the physics of harmonics on a guitar. Strauss's exercise about the violin string clamped at its midpoint (§4.1, Exercise 1(a), p. 89) — the note rises by exactly one octave — is the statement that halving $L$ doubles every $\omega_n$ .

Everything else that is confined. The same three steps — separate, find the eigenvalues forced by the boundary conditions, superpose — solve the heat equation in a rod (with $e^{-\lambda_n kt}$ in place of the oscillating factor), the Schrodinger equation for a particle in a box (whose energies $E_n \propto n^2$ are Strauss's $\lambda_n$ in disguise), and the Klein-Gordon equation; Hassani separates all four in one calculation. Modes of a drum, a cavity, an optical fibre and a planet all come from the same recipe in other coordinate systems (separation of variables in standard coordinates), where the eigenfunctions are Bessel functions or spherical harmonics instead of sines.

Numerics and signal processing. Expanding in modes is the discrete sine transform, so a numerical spectral method for (S) advances each coefficient by an exact rotation at frequency $\omega_n$ rather than by a time-stepping rule, at the price of needing the eigenfunctions in closed form (spectral methods). Conversely, measuring the frequencies of a vibrating object and comparing them with $\omega_n = (n\pi/L)\sqrt{\tau/\rho}$ is how tension, stiffness or a hidden crack is inferred in non-destructive testing: the forward problem here is the one solved above, and the inverse problem is to deduce the operator from its spectrum.

Where it leads. The orthogonality and completeness used in Step 7 are not accidents of the sine function: they hold for every regular Sturm-Liouville problem (Sturm-Liouville eigenvalue problems, eigenfunction expansions), which is the general reason the method works, and the mode expansion of a Green's function in eigenfunction expansion of Green's functions is the same decomposition applied to a forced problem.