Hyperbolic Equations and Systems

6B.3 The Wave Equation solved the wave equation with constant coefficients by explicit formulas. This chapter does without formulas. Waves in real media travel through materials whose properties vary from place to place, and many physical systems (elasticity, electromagnetism, acoustics, general relativity) are systems of equations rather than a single one. For these, existence, uniqueness and finite speed of propagation are proved by energy estimates, the method of 6B.3 The Wave Equation's energy proofs in general form, and existence by Galerkin's method or by adding a small viscosity and removing it.

Two structures carry the theory. For second-order equations $u_{tt} + Lu = f$ with $L$ elliptic, the energy $\frac12\|u_t\|^2 + \frac12B[u, u]$ is nearly conserved, and Galerkin's method goes through as for parabolic equations (6A.7 Parabolic Regularity), except that nothing is gained in regularity: data in $H^1\times L^2$ give solutions in $H^1\times L^2$ for all time, no better. For first-order systems $\mathbf u_t + \sum B_j\partial_j\mathbf u = \mathbf f$, the key property is that the matrices $B_j$ be symmetric: then the $L^2$ norm is nearly conserved, and the system is well posed. Symmetric hyperbolic systems, introduced by Kurt Friedrichs in 1954, include Maxwell's equations, linear elasticity, acoustics and, in suitable coordinates, the Einstein equations.

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

• define weak solutions of second-order hyperbolic equations, and prove their existence by Galerkin's method and uniqueness with a time-integrated test function;
• explain why hyperbolic equations do not smooth, and state the regularity and finite-propagation results;
• prove the energy estimate for symmetric hyperbolic systems, and recognise the wave equation, Maxwell's equations and elasticity as such systems;
• decide whether a constant-coefficient system is hyperbolic from the eigenvalues of its symbol;
• compute P- and S-wave speeds in an elastic solid, and explain the gauge structure of Einstein's equations and its relation to DeTurck's trick.

14.1Why the Earth's outer core is liquid

14.2Second-order hyperbolic equations

Let $U$ be bounded with smooth boundary, and consider

where $Lu = -\sum\partial_j(a^{ij}(x, t)\partial_iu) + \sum b^i(x, t)\partial_iu + c(x, t)u$ is uniformly elliptic with smooth coefficients, $a^{ij} = a^{ji}$. Write $B[u, v; t]$ for its bilinear form, and assume for simplicity that $B[u, u; t] \geq \beta\|u\|_{H^1_0}^2$ (true if $b = 0$ and $c \geq 0$; the general case adds a lower-order term, as Gårding's inequality does in 6A.6 Weak Solutions and Elliptic Regularity).

(As in 6B.11 Linear Parabolic Equations in Full, the regularity in time makes $u$ continuous into $L^2$ and $u'$ continuous into $H^{-1}$, so the initial conditions make sense.)

The trick in the uniqueness proof recurs: when the "natural" test function lacks regularity, integrate in time to gain a derivative, at the cost of handling a time integral. The energy method with the test function $u'$ is still the idea; Ladyzhenskaya's version makes it rigorous for weak solutions.

No smoothing, but no loss. Regularity for hyperbolic equations is "conserved, not created". If $g \in H^2\cap H^1_0$, $h \in H^1_0$ and $f' \in L^2(L^2)$, the solution has $u \in L^\infty(H^2)$, $u' \in L^\infty(H^1_0)$, $u'' \in L^\infty(L^2)$, by differentiating the equation in time and applying the energy estimate to $u'$, then elliptic regularity at each time (Evans, section 7.2.3). With smooth data satisfying compatibility conditions at the boundary, the solution is smooth. But rough data give rough solutions forever: unlike 6B.11 Linear Parabolic Equations in Full, there is no instant smoothing, because the energy is conserved rather than dissipated.

Finite propagation speed. With variable coefficients, disturbances travel at most at the speed $\sqrt{\Lambda}$ determined by the largest eigenvalue of $(a^{ij})$, and the proof is 6B.3 The Wave Equation's shrinking-ball energy argument with balls replaced by the sets swept out by the characteristic cones of the operator (Evans, section 7.2.4). Taylor I, §2.6, proves it on Riemannian and Lorentzian manifolds in the language of energy–momentum tensors.

14.3Symmetric hyperbolic systems

Many wave phenomena are naturally first-order systems: Maxwell's equations relate first derivatives of the electric and magnetic fields, and the equations of acoustics relate first derivatives of pressure and velocity. Consider

for $\mathbf u : \mathbb{R}^n\times[0, T] \to \mathbb{R}^m$, with $m\times m$ matrices $B_j$ (we take the whole space, which avoids boundary conditions, a subtle matter for hyperbolic systems). The system is symmetric hyperbolic if each $B_j$ is a symmetric matrix, smooth and bounded with bounded derivatives.

Symmetry is the whole point. Without it, the term $\int\mathbf u\cdot B_j\partial_j\mathbf u$ contains derivatives of $\mathbf u$ that cannot be integrated away, and no $L^2$ estimate holds; with it, the first-order terms contribute only through the derivatives of the coefficients. Applying the same computation to $\partial^\alpha\mathbf u$, which solves a system of the same form with extra lower-order terms, gives $H^k$ estimates for every $k$.

Evans (section 7.3.2) proves existence by vanishing viscosity: solve the parabolic system $\mathbf u^\varepsilon_t + \sum B_j\partial_j\mathbf u^\varepsilon = \varepsilon\Delta\mathbf u^\varepsilon + \mathbf f$, which has smooth solutions by the theory of 6B.11 Linear Parabolic Equations in Full; the energy estimate holds for $\mathbf u^\varepsilon$ uniformly in $\varepsilon$ (the viscous term only helps: it contributes $-\varepsilon\|\nabla\mathbf u\|^2 \leq 0$), so a subsequence converges as $\varepsilon \to 0$ to a solution of the hyperbolic system. Uniqueness is the energy estimate applied to a difference. This is 6B.2 Conservation Laws and Shocks's vanishing-viscosity idea, now for linear systems where it causes no shocks.

14.3.1Examples

The wave equation. If $u_{tt} = \Delta u$, then $\mathbf u = (u_t, \partial_1u, \ldots, \partial_nu)$ satisfies $\partial_t(u_t) = \sum_j\partial_j(\partial_ju)$ and $\partial_t(\partial_ju) = \partial_j(u_t)$, a first-order system $\mathbf u_t = \sum_jB_j\partial_j\mathbf u$ in which $B_j$ has ones in the positions $(0, j)$ and $(j, 0)$ and zeros elsewhere: symmetric. The $L^2$ norm of $\mathbf u$ is twice the energy of 6B.3 The Wave Equation.

Maxwell's equations. In a vacuum, in units where the speed of light is $1$, the electric and magnetic fields satisfy

(together with the constraints $\operatorname{div}\mathbf E = \operatorname{div}\mathbf B = 0$, which are preserved by the evolution). The curl is $\operatorname{curl}\mathbf F = \sum_jC_j\partial_j\mathbf F$ with antisymmetric $3\times3$ matrices $C_j$, so for $\mathbf u = (\mathbf E, \mathbf B)$ the system has $B_j = \begin{pmatrix}0 & -C_j\\ C_j & 0\end{pmatrix}$, which is symmetric because $C_j^{\mathsf T} = -C_j$ (Exercise 14.7). The energy estimate is conservation of electromagnetic energy $\frac12\int(|\mathbf E|^2 + |\mathbf B|^2)$. In matter with variable permittivity and permeability, the system becomes symmetrizable: symmetric after multiplying by a positive definite matrix, which is enough (Taylor III, §16.2).

Acoustics and elasticity. Linearised gas dynamics around a fluid at rest, with pressure $p$ and velocity $\mathbf v$, is the symmetric system $p_t + \rho c^2\operatorname{div}\mathbf v = 0$, $\rho\mathbf v_t + \nabla p = 0$ after scaling $p$ by $\frac{1}{\sqrt{\rho}c}$. Linear elasticity, written for the velocity and the stress, is symmetric hyperbolic too.

14.4Constant coefficients and hyperbolicity

For constant matrices $B_j$, the Fourier transform turns $\mathbf u_t + \sum B_j\partial_j\mathbf u = 0$ into the ODE system $\hat{\mathbf u}_t = -2\pi i\,B(\xi)\hat{\mathbf u}$, where $B(\xi) = \sum_j\xi_jB_j$ is the symbol. So $\hat{\mathbf u}(\xi, t) = e^{-2\pi itB(\xi)}\hat{\mathbf g}(\xi)$, and by Plancherel the solution is bounded in $L^2$, uniformly in $t \in [0, T]$, exactly when $\sup_\xi\|e^{-2\pi itB(\xi)}\| < \infty$. If $B(\xi)$ has a non-real eigenvalue $\mu$ for some $\xi$, then scaling $\xi \mapsto s\xi$ scales $\mu$ by $s$, and the corresponding mode grows like $e^{2\pi s|\operatorname{Im}\mu|t}$, unbounded as $s \to \infty$: Hadamard's ill-posedness (6A.1 What a PDE Is). So a necessary condition is that $B(\xi)$ have real eigenvalues for every $\xi$. It is not quite sufficient: one also needs $B(\xi)$ to be diagonalisable with uniformly bounded condition number of the eigenvector matrix (strong hyperbolicity), and real symmetric matrices are the simplest case, where the eigenvectors are orthonormal (Evans, section 7.3.3; Exercise 14.8). The eigenvalues of $B(\xi)$ for unit $\xi$ are the propagation speeds in the direction $\xi$; for Maxwell they are $0$ and $\pm1$, for the Lamé system $v_P$, $v_S$ and $0$.

14.5History

Augustin Cauchy and Siméon Poisson developed the theory of elasticity in the 1820s, and Poisson showed in 1829 that an elastic solid carries two kinds of waves with different speeds; George Stokes analysed them further in 1849. Kurt Friedrichs introduced symmetric hyperbolic systems in 1954, after earlier work by Jean Leray and Ivan Petrovsky on hyperbolic equations and systems in the 1930s and 1950s. Olga Ladyzhenskaya developed the Galerkin and energy theory of hyperbolic and parabolic equations in the 1950s, and the time-integrated test function used for uniqueness here is in the spirit of her work (Evans, section 7.2.2, gives the argument). Richard Oldham (1906), Beno Gutenberg (1913), Harold Jeffreys (1926) and Inge Lehmann (1936) worked out the structure of the Earth's core from seismic waves. Yvonne Choquet-Bruhat proved local well-posedness of Einstein's equations in 1952.

14.6Exercises

Solution

(a) $\operatorname{div}\Delta\mathbf u = \Delta\operatorname{div}\mathbf u$ and $\operatorname{div}\nabla\theta = \Delta\theta$, so $\rho\theta_{tt} = \mu\Delta\theta + (\lambda + \mu)\Delta\theta$. (b) $\operatorname{curl}\nabla = 0$ and $\operatorname{curl}\Delta = \Delta\operatorname{curl}$. (c) $\frac{v_P^2}{v_S^2} = \frac{\lambda + 2\mu}{\mu} = 3$. (d) $\frac{d}{3.5} - \frac d6 = 20$, so $d\big(\frac{6 - 3.5}{21}\big) = 20$ and $d = 168$ km.

Solution

$d_k'' + k^2d_k = 0$, with $d_k(0) = (g, w_k)$, $d_k'(0) = (h, w_k)$; so $d_k = (g, w_k)\cos kt + \frac{(h, w_k)}{k}\sin kt$, and $u_m = \sum_{k\leq m}d_kw_k$, the partial sum. Energy: $\sum_{k\leq m}(d_k'^2 + k^2d_k^2) = \sum_{k\leq m}\big(k^2(g, w_k)^2 + (h, w_k)^2\big)$, constant in $t$.

Solution

(a) $(\operatorname{curl}\mathbf F)_1 = \partial_2F_3 - \partial_3F_2$, etc. So $C_1$ has $(C_1)_{23} = -1$, $(C_1)_{32} = 1$; $C_2$ has $(C_2)_{13} = 1$, $(C_2)_{31} = -1$; $C_3$ has $(C_3)_{12} = -1$, $(C_3)_{21} = 1$: each antisymmetric. (b) $\frac{d}{dt}\frac12\int(|\mathbf E|^2 + |\mathbf B|^2) = \int\mathbf E\cdot\operatorname{curl}\mathbf B - \int\mathbf B\cdot\operatorname{curl}\mathbf E = 0$, the identity being the integrated form of $\operatorname{div}(\mathbf B\times\mathbf E) = \mathbf E\cdot\operatorname{curl}\mathbf B - \mathbf B\cdot\operatorname{curl}\mathbf E$. (c) $C(\xi) = \sum\xi_jC_j$ is the matrix of $\mathbf v \mapsto \xi\times\mathbf v$ (up to sign), with eigenvalues $0, \pm i$; the symbol $\begin{pmatrix}0 & -C(\xi)\\ C(\xi) & 0\end{pmatrix}$ has eigenvalues $0$ (twice) and $\pm1$ (each twice): light travels at speed $1$ in every direction, and the zero eigenvalues correspond to the static (longitudinal) parts removed by the constraints.

Solution

$B^2 = 0$, so the exponential series stops after two terms. For $\hat{\mathbf g} = (0, 1)$, $\hat{\mathbf u} = (-2\pi i\xi t, 1)$, of size about $2\pi|\xi|t$, unbounded in $\xi$; so no estimate $\|\mathbf u(t)\|_{L^2} \leq C\|\mathbf g\|_{L^2}$. But $|\xi|\,|\hat{\mathbf g}|$ is controlled by the $H^1$ norm, so $\|\mathbf u(t)\|_{L^2} \leq C(1 + t)\|\mathbf g\|_{H^1}$.

Solution

$u = \sum\frac{h_k}{k}\sin kt\sin kx$, $\|\partial_xu\|^2 + \|u_t\|^2 = \frac\pi2\sum h_k^2(\sin^2kt + \cos^2kt) = \|h\|^2$. For $u(t) \in H^2$ we need $\sum k^4\frac{h_k^2}{k^2}\sin^2kt = \sum k^2h_k^2\sin^2kt < \infty$, and also $u_t(t) \in H^1$ needs $\sum k^2h_k^2\cos^2kt < \infty$; adding, both together need $\sum k^2h_k^2 < \infty$, that is, $h \in H^1_0$. So at no time are both $u$ and $u_t$ one degree smoother than the data. For the heat equation the coefficients carry $e^{-k^2t}$, and every Sobolev norm is finite for $t > 0$.

Solution

(a) $\partial_iV_j + \partial_jV_i = \partial_i\partial_kh_{jk} + \partial_j\partial_kh_{ik} - \partial_i\partial_j\operatorname{tr}h$, which is exactly the part of $2\dot R_{ij}$ other than $-\Delta h_{ij}$. (b) For $h_{ij} = \partial_iX_j + \partial_jX_i$: $\operatorname{tr}h = 2\operatorname{div}X$, $\partial_kh_{jk} = \partial_j\operatorname{div}X + \Delta X_j$, so $V_j = \Delta X_j$; then $-\frac12\Delta h_{ij} = -\frac12(\partial_i\Delta X_j + \partial_j\Delta X_i)$ and $\frac12(\partial_iV_j + \partial_jV_i) = \frac12(\partial_i\Delta X_j + \partial_j\Delta X_i)$: they cancel. The symbol of $\dot R$ vanishes on such $h$, and in the decomposition the cancellation is between the Laplacian term and the gauge term. (c) With $V = 0$, $-2\dot R_{ij} = \Delta h_{ij}$. In Lorentzian signature $\Delta$ is replaced by the wave operator, giving $\square h_{ij} = 0$.