The Wave Equation

The heat equation and the wave equation look almost alike, $u_t = \Delta u$ and $u_{tt} = \Delta u$, and they behave in opposite ways. Heat smooths instantly, spreads at infinite speed, and forgets its past; waves keep every detail of their initial shape, travel at a finite speed, and can be run backward as easily as forward. Both behaviours are encoded in the symbol (6A.1 What a PDE Is): the heat equation damps a Fourier mode $e^{2\pi ix\cdot\xi}$ at the rate $4\pi^2|\xi|^2$, while the wave equation makes it oscillate forever at the frequency $2\pi|\xi|$.

This chapter solves the wave equation completely in one, two and three space dimensions, and proves the two facts that matter most: energy is conserved, and disturbances travel at speed at most $1$. Along the way it finds a surprise that has to do with the dimension of space: in three dimensions a sharp sound stays sharp (Huygens' principle), while in two dimensions it leaves a lingering tail.

The wave equation is not on the main route to the Poincaré conjecture, because the Ricci flow is parabolic. But it is the model hyperbolic equation, and the contrast with the heat equation explains why the Ricci flow had to be a heat equation to be useful: only a parabolic flow smooths out the metric as it goes (Exercise 3.14). The wave equation is also the equation of general relativity: Einstein's equations, in suitable coordinates, are a system of wave equations for the metric (6B.14 Hyperbolic Equations and Systems).

By the end of this chapter you will be able to:

• derive the wave equation for a vibrating string, and compute the frequencies of a string and a drum;
• prove d'Alembert's formula, solve problems on a half-line and a finite string by reflection, and solve the inhomogeneous equation by Duhamel's principle;
• derive the Euler–Poisson–Darboux equation for spherical means, and from it Kirchhoff's formula in three dimensions and Poisson's formula in two;
• explain Huygens' principle and why it fails in even dimensions;
• prove conservation of energy and finite speed of propagation by energy methods, and solve the wave equation on a bounded domain by eigenfunctions.

3.1Strings and their pitch

Deriving the string equation. Let a string under tension $T$, with mass $\mu$ per unit length, be displaced vertically by a small amount $u(x, t)$. The piece between $x$ and $x + dx$ has mass $\mu\,dx$, and the tension pulls along the string at each end. The vertical component of the tension at $x$ is $T\sin\theta \approx T\tan\theta = Tu_x$ for small slopes, pulling down at the left end and up at the right. Newton's law for the vertical motion is

so $u_{tt} = c^2u_{xx}$ with $c^2 = T/\mu$. The approximation $\sin\theta \approx \tan\theta$ is the small-amplitude assumption; for large vibrations the equation becomes nonlinear. Rescaling time, we take $c = 1$ from now on.

3.2One dimension: d'Alembert

6A.1 What a PDE Is solved $u_{tt} = u_{xx}$ by changing to characteristic coordinates $\xi = x + t$, $\eta = x - t$, in which the equation is $u_{\xi\eta} = 0$. Every $C^2$ solution is

a wave moving left plus a wave moving right, each keeping its shape. Fitting the initial conditions gives:

The formula exhibits the two defining features of waves.

No smoothing. If $g$ has a corner at $x_0$, the solution has corners along the two lines $x \pm t = x_0$ for all time. The regularity of $u(\cdot, t)$ is exactly that of the data: $u \in C^k$ needs $g \in C^k$ and $h \in C^{k-1}$, and no better. The singularities travel along the characteristics $x \pm t = \text{const}$.

Finite speed and the domain of dependence. The value $u(x_0, t_0)$ depends only on $g$ at the two points $x_0 \pm t_0$ and on $h$ on the interval between them. This interval is the domain of dependence of $(x_0, t_0)$. Conversely, data near a point $y$ at time $0$ influence only the range of influence $|x - y| \leq t$. Information travels at speed exactly $1$.

Reflection. On the half-line $x > 0$ with a fixed end, $u(0, t) = 0$, extend $g$ and $h$ to odd functions and apply d'Alembert: the solution of the extended problem is odd in $x$, hence vanishes at $0$ (6A.3 The Heat Equation on ℝⁿ used the same trick for heat). A pulse travelling towards the fixed end comes back inverted. With a free end, $u_x(0, t) = 0$, extend evenly, and the pulse comes back upright. A string of length $L$ fixed at both ends is solved by extending the data to be odd about $0$ and $2L$-periodic: the solution is periodic in time with period $2L$, and its Fourier sine series is the sum of standing waves $\sin\frac{k\pi x}{L}\big(a_k\cos\frac{k\pi t}{L} + b_k\sin\frac{k\pi t}{L}\big)$, the harmonics of Mersenne's laws (Exercise 3.8).

Sources: Duhamel. For $u_{tt} - u_{xx} = f(x, t)$ with zero initial data, the impulse $f(\cdot, s)\,ds$ delivered in the time interval $[s, s + ds]$ gives the string a velocity $f(\cdot, s)\,ds$, which then evolves freely. So, by d'Alembert with $g = 0$ and $h = f(\cdot, s)$, started at time $s$,

half the integral of $f$ over the backward characteristic triangle with apex $(x, t)$. Differentiating under the integral confirms it (Exercise 3.9). This is Duhamel's principle of 6A.3 The Heat Equation on ℝⁿ once more, with the wave equation's "solution operator for velocity data" in place of the heat semigroup.

3.3Spherical means

In higher dimensions there is no factorisation into left- and right-moving waves. The trick, due to Poisson, is to average over spheres, reducing to a one-dimensional problem in the radius.

For a function $u(x, t)$ on $\mathbb{R}^n\times[0, \infty)$, let

be its average over the sphere of radius $r > 0$ about $x$, and similarly $G(x; r)$ and $H(x; r)$ for the initial data $g$ and $h$. As $r \to 0$, $U(x; r, t) \to u(x, t)$.

The operator $U_{rr} + \frac{n-1}{r}U_r$ is the radial part of the Laplacian (6A.2 Harmonic Functions, exercise on the radial Laplacian). The lemma says that spherical averages of a wave obey the radial wave equation, whether or not the wave itself is radial.

3.4Three dimensions: Kirchhoff's formula

For $n = 3$ the Euler–Poisson–Darboux equation has a small miracle: multiplying by $r$ turns it into the one-dimensional wave equation. Let $\tilde U = rU$. Then

So $\tilde U$ solves $\tilde U_{tt} = \tilde U_{rr}$ on the half-line $r > 0$, with $\tilde U(0, t) = 0$ (since $U$ stays bounded as $r \to 0$), and initial data $\tilde G = rG$, $\tilde H = rH$.

Notice the loss of derivatives: to get a $C^2$ solution we needed $g \in C^3$ and $h \in C^2$. In three dimensions, a singularity of the data can be focused: data with a singularity on a sphere of radius $t_0$ produce a stronger singularity at the centre at time $t_0$ (Exercise 3.12). No such loss happens in one dimension, and the energy estimates below avoid it by measuring regularity in $L^2$ rather than in $C^k$.

3.5Two dimensions: Poisson's formula by descent

To solve in two dimensions, pretend there is a third: a solution $u(x_1, x_2, t)$ in the plane is a solution $\bar u(x_1, x_2, x_3, t) = u(x_1, x_2, t)$ in space that happens not to depend on $x_3$. Apply Kirchhoff to $\bar u$, with data independent of $x_3$, and evaluate the spherical averages as integrals over the disc below: this is Hadamard's method of descent.

Parametrise the upper hemisphere of $\partial B(\bar x, t)$ (with $\bar x = (x, 0)$) as the graph of $\gamma(y) = (t^2 - |y - x|^2)^{1/2}$ over the disc $B(x, t)$ in the plane. Its area element is $\sqrt{1 + |\nabla\gamma|^2}\,dy = \frac{t}{(t^2 - |y - x|^2)^{1/2}}dy$. Both hemispheres contribute equally, so for a function $\bar g(y_1, y_2, y_3) = g(y_1, y_2)$,

since $|B(x, t)| = \pi t^2$. With the formula $u = \partial_t(tG) + tH$ from the proof of Kirchhoff, this gives Poisson's formula:

(Evans, section 2.4.1; the derivative of $\frac{t^2}{2}\mathop{\rlap{\,\,-}\int}\nolimits_{B(x,t)}\frac{g}{\sqrt{\cdots}}$ is computed by rescaling to the unit disc first, Exercise 3.10).

3.6Huygens' principle, and why it fails in the plane

Compare the two formulas. In three dimensions, $u(x, t)$ depends only on the data on the sphere $\partial B(x, t)$: a disturbance at a point $y$ at time $0$ is felt at $x$ exactly at the time $t = |x - y|$ and at no other time. A sharp pulse, a hand clap, stays sharp. This is the strong Huygens' principle, and it is why sound in our three-dimensional world carries speech and music cleanly: each sound arrives once and is gone.

In two dimensions, $u(x, t)$ depends on the data in the whole disc $B(x, t)$. A disturbance at $y$ arrives at $x$ at time $|x - y|$, but then lingers, decaying like $t^{-1}$ but never switching off (Exercise 3.11). A clap in Flatland would sound like a clap followed by a fading hiss. The method of descent explains why: a point source in the plane is a line source in space, and the line's more distant parts send their signals later.

In general, the strong Huygens principle holds for the wave equation in odd dimensions $n \geq 3$ and fails in even dimensions; in one dimension the $h$ term spreads over the whole interval, so it fails too (Evans, section 2.4.1, gives the formulas for all $n$). Jacques Hadamard asked in 1923 which second-order hyperbolic equations satisfy Huygens' principle; the question, "Hadamard's problem", turned out to be subtle, with exotic examples found much later.

3.7Energy and finite speed

The formulas are special to the constant-coefficient wave equation in $\mathbb{R}^n$. The two most important properties of waves follow from a method that works for any hyperbolic equation: energy.

The two terms are the kinetic energy $\frac12\int u_t^2$ and the potential (strain) energy $\frac12\int|\nabla u|^2$; energy moves between them but its total is fixed. Applied to the difference of two solutions with the same data, whose energy is $0$ at time $0$ and hence always, it proves uniqueness: $u_t = \nabla u = 0$, so the difference is constant, and it is $0$ initially. Unlike the heat equation's energy, which only decreases (6A.3 The Heat Equation on ℝⁿ), this one is conserved: waves are reversible.

The inequality $|u_t\,\partial_\nu u| \leq \frac12(u_t^2 + |\nabla u|^2)$ is where the speed $1$ enters: the energy flux through a surface is at most the energy density times the speed of light. If the ball shrank faster than at unit speed, the boundary term would still win; if it shrank slower, it would not, and energy could leak in from outside. The cone in the theorem is the domain of dependence in every dimension. Taylor I, §2.6, proves the same result on Riemannian manifolds, where the cone is replaced by the set of points within geodesic distance $t_0 - t$.

3.8The wave equation on a bounded domain

On a bounded region $U$ with Dirichlet boundary conditions, the eigenfunctions $\phi_k$ of $-\Delta$, with eigenvalues $0 < \lambda_1 \leq \lambda_2 \leq \cdots$, form an orthonormal basis of $L^2(U)$ (4A.8 Compact Operators and Spectra). Expand $u(\cdot, t) = \sum_ka_k(t)\phi_k$. The equation $u_{tt} = \Delta u$ becomes $a_k'' = -\lambda_ka_k$ for each $k$, so

where $g_k = \int g\phi_k$ and $h_k = \int h\phi_k$. Each mode oscillates at its own frequency $\sqrt{\lambda_k}$ forever, without decay; compare the heat equation, where the same expansion has $e^{-\lambda_kt}$ and every mode dies (6A.3 The Heat Equation on ℝⁿ). Energy is $\frac12\sum_k\big(\lambda_k|g_k|^2 + |h_k|^2\big)$, constant, by Parseval (Exercise 3.13). The series converges in the energy norm when $g \in H^1_0$ and $h \in L^2$, and defines a weak solution; 6B.14 Hyperbolic Equations and Systems constructs it in general by Galerkin's method, exactly as 6A.7 Parabolic Regularity did for parabolic equations.

	heat $u_t = \Delta u$	wave $u_{tt} = \Delta u$
Fourier mode $\xi$	decays like $e^{-4\pi^2\lvert\xi\rvert^2t}$	oscillates as $\cos(2\pi\lvert\xi\rvert t)$
speed of propagation	infinite	at most $1$
smoothing	instant: $C^\infty$ for $t > 0$	none: singularities travel on characteristics
energy $\int\lvert\nabla u\rvert^2$	decreases	conserved (with $u_t^2$)
backward in time	ill-posed	well-posed (reversible)
natural scaling	$(\lambda x, \lambda^2t)$	$(\lambda x, \lambda t)$

3.9History

Jean le Rond d'Alembert derived and solved the equation of the vibrating string in 1747. Leonhard Euler and Daniel Bernoulli disputed what solutions are allowed: Bernoulli proposed sums of sines (1753), and Euler objected that a plucked string has a corner, which a sum of sines seemed unable to represent. The dispute was settled only with Fourier's work and the theory of convergence of Fourier series (2B.7 Fourier Series and the First Heat Equation). Siméon Denis Poisson solved the three-dimensional wave equation by spherical means in 1819, and Gustav Kirchhoff gave his formula in 1882 in work on the theory of light. Christiaan Huygens stated his principle of wavefronts in his Traité de la lumière (1690). Hadamard's method of descent and his problem date from his 1923 lectures on Cauchy's problem. Mersenne's Harmonie universelle appeared in 1636. Choquet-Bruhat's theorem on the Einstein equations dates from 1952.

3.10Exercises

Solution

(a) $c = 2Lf_1 = 2(0.65)(110) = 143$ m/s. (b) $T = \mu c^2 = 0.004\times143^2 \approx 82$ N. (c) Frequency is inversely proportional to length: $0.325$ m for the octave (half the string, the twelfth fret), and $0.65\cdot2^{-1/12} \approx 0.6135$ m for the semitone, so the first fret sits about $3.65$ cm from the nut.

Solution

$1$, $1.593$, $2.136$, $2.295$: not integers, and not close to them, so the overtones of a drum are inharmonic. Orchestral timpani are tuned so that several modes come out close to harmonic ratios (with the help of the air enclosed in the kettle), which is why they can play definite notes.

Solution

(a) $\frac2\pi\int_0^\pi g\sin kx\,dx = \frac{8a}{\pi^2}\frac{\sin(k\pi/2)}{k^2}$ (integrate by parts twice; $g'' = -\frac{4a}{\pi}\delta_{\pi/2}$), which vanishes for even $k$ and is $\frac{8a}{\pi^2}\frac{(-1)^{(k-1)/2}}{k^2}$ for odd $k$. (b) $u = \frac{8a}{\pi^2}\sum_{k\text{ odd}}\frac{(-1)^{(k-1)/2}}{k^2}\sin kx\cos kt$. Even harmonics have a node at the midpoint, where the pluck is, so they are not excited. (c) $u(x, \frac\pi2) = \frac12(\tilde g(x + \frac\pi2) + \tilde g(x - \frac\pi2))$ with $\tilde g$ the odd $2\pi$-periodic extension; for $0 \leq x \leq \frac\pi2$ the two terms are $\frac{2a}{\pi}(\frac\pi2 - x)$ and $-\frac{2a}{\pi}(\frac\pi2 - x)$, so $u = 0$, and likewise on $[\frac\pi2, \pi]$. At $t = \frac\pi2$ the string is momentarily flat (and moving), and at $t = \pi$ it is the inverted triangle.

Solution

Let $w(x, t; s) = \frac12\int_{x-(t-s)}^{x+(t-s)}f(y, s)\,dy$, so $u = \int_0^tw(x, t; s)\,ds$ and $w(x, s; s) = 0$. Then $u_t = \int_0^tw_t\,ds$ (the boundary term is $w(x, t; t) = 0$), with $w_t = \frac12(f(x + t - s, s) + f(x - t + s, s))$; and $u_{tt} = w_t(x, t; t) + \int_0^tw_{tt}\,ds = f(x, t) + \int_0^tw_{tt}\,ds$. Since $w$ solves the wave equation in $(x, t)$ for each $s$, $\int_0^tw_{tt} = \int_0^tw_{xx} = u_{xx}$.

Solution

With $y = x + tz$, $dy = t^2dz$ and $(t^2 - |y - x|^2)^{1/2} = t(1 - |z|^2)^{1/2}$, so $\frac t2\cdot\frac{1}{\pi t^2}\int_{B(x,t)}\frac{g\,dy}{\sqrt{\cdots}} = \frac{1}{2\pi}\int_{B(0,1)}\frac{g(x + tz)}{(1 - |z|^2)^{1/2}}dz$, and $tG = \frac{t}{2\pi}\int_{B(0,1)}\frac{g(x + tz)}{(1 - |z|^2)^{1/2}}dz$. Then $\partial_t(tG) = \frac{1}{2\pi}\int\frac{g(x + tz) + t\nabla g(x + tz)\cdot z}{(1 - |z|^2)^{1/2}}dz$; changing back to $y$ gives $\frac12\mathop{\rlap{\,\,-}\int}\nolimits_{B(x,t)}\frac{t\,g + t\nabla g\cdot(y - x)}{(t^2 - |y - x|^2)^{1/2}}dy$. The $h$ term is $tH = \frac12\mathop{\rlap{\,\,-}\int}\nolimits_{B(x,t)}\frac{t^2h}{\sqrt{\cdots}}$.

Solution

Poisson's formula with $g = 0$ gives $u = \frac12\cdot\frac{1}{\pi t^2}\int_{B(x,t)}\frac{t^2h}{\sqrt{t^2 - |y - x|^2}}dy = \frac{1}{2\pi}\int_{B(0,a)}\frac{dy}{\sqrt{t^2 - |y - x|^2}}$ once $B(0, a) \subset B(x, t)$. As $t \to \infty$ the integrand is $\approx\frac1t$, and the area is $\pi a^2$. In three dimensions Kirchhoff gives $u = t\mathop{\rlap{\,\,-}\int}\nolimits_{\partial B(x,t)}h$, and for $t > R + a$ the sphere $\partial B(x, t)$ does not meet $B(0, a)$.

Solution

(a) $\partial B(0, t)$ is the sphere $|y| = t$, where $h = \psi(t)$, so the average is $\psi(t)$ and $u(0, t) = t\psi(t)$. (b) At the origin the average is over exactly the sphere $|y| = t$, on which $h$ is constant, so the jump of $\psi$ at $1$ appears undamped in $u(0, t)$ at $t = 1$. At a point $x \neq 0$, the sphere $\partial B(x, t)$ crosses the sphere $|y| = 1$ transversally, and the average of a function with a jump across a transverse surface varies continuously with $t$. The centre is a focus: every ray from the singular sphere arrives there at once.

Solution

With $a_k = g_k\cos\omega_kt + \frac{h_k}{\omega_k}\sin\omega_kt$, $\omega_k = \sqrt{\lambda_k}$: $\int u_t^2 = \sum a_k'^2$ and $\int|\nabla u|^2 = \sum\lambda_ka_k^2$. For each $k$, $a_k'^2 + \lambda_ka_k^2 = (-g_k\omega_k\sin + h_k\cos)^2 + \omega_k^2(g_k\cos + \frac{h_k}{\omega_k}\sin)^2 = \omega_k^2g_k^2 + h_k^2$, the cross terms cancelling.

Solution

(a) $\partial_xu_k = \cos(k(x - t))$, which has sup $1$ at every time, while $\sup|u_k(\cdot, 0)| = \frac1k$. (b) The heat kernel damps each frequency $k$ by $e^{-k^2t}$, which beats the factor $k$ that a derivative costs. (c) A hyperbolic equation transports irregularities of the initial data without damping them, so a rough initial metric stays rough; any singularity in the data persists along characteristics. Only a parabolic equation damps high frequencies and so makes the metric smoother and more uniform as time goes on, which is what a proof of geometrisation needs.