# THE WAVE EQUATION. d = 1: D Alembert s formula We begin with the initial value problem in 1 space dimension { u = utt u xx = 0, in (0, ) R, (2)

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1 THE WAVE EQUATION () The free wave equation takes the form u := ( t x )u = 0, u : R t R d x R In the literature, the operator := t x is called the D Alembertian on R +d. Later we shall also consider the inhomogeneous or forced equation u = f for some given function f (t, x). Since the equation is second order in t, an initial value problem involves initial conditions on both u and t u, and takes the form { ( t x )u = f in (0, ) t R d x u = g, t u = h on {t = 0} R d, x. where g, h : R d R and f : [0, ) R d R are prescribed functions. In contrast with the heat equation, the free wave equation is reversible in time: if u(t, x) is a solution, then so is ũ(t, x) := u( t, x). d = : D Alembert s formula We begin with the initial value problem in space dimension { u = utt u xx = 0, in (0, ) R, () u(0, x) = g(x), t u(0, x) = h(x), Theorem 0. (D Alembert s formula). If u is a smooth solution to (), then u(t, x) = [g(x + t) + g(x t)] + d = 3: Spherical means x+t x t h(y) dy. In three space dimensions, the solution to the initial value problem { u = 0, in (0, ) R 3, (3) u(0, x) = g(x), t u(0, x) = h(x), is expressed in terms of averages of the initial data over spheres. If Σ R 3 is a surface, write u ds := u ds A(Σ) Σ Σ Theorem 0. (Kirchhoff s formula). If u is a smooth solution to (3), then u(t, x) = t h(y) ds y + g(y), y x ds y + g(y) ds y t = h(y) ds y + g(y), ˆn y ds y + g(y) ds 4πt 4πt 4πt y.

2 THE WAVE EQUATION Proof. Fix x R 3. For t, r > 0, introduce the spherical averages U(t, r; x) := u(t, y) ds y = u(t, y) ds 4πr y G(r; x) := g(y) ds y, H(r, x) := H(y) ds y. Lemma 0.3. U is smooth on [0, ) t [0, ) r and satisfy the Euler-Darboux-Poisson equation { Utt U rr r U r = 0 in (0, ) t (0, ) r, U(0, r) = G(r), U t (0, r) = H(r) Further, U can be extended as a smooth even function of r to R. Proof. We have U(t, r) = u(t, y) ds y = B (0) u(t, x + rω) ds ω. The second expression is well-defined and smooth for all r R. Thus if we use it to define U for r 0, then U is smooth for all (t, r) (0, ) t R r, and satisfies U(t, r) = U(t, r), U(t, 0) = lim U(t, r) = u(t, x). r 0 + Differentiating and using the divergence theorem and polar coordinates, we have U r = ω, ( u)(t, x + rω) ds ω = u(t, y), ˆn 4π B (0) 4πr y ds y = u(t, y) dy = r [ ] u(t, y) ds 4πr 4πr y ρ dρ, Bρ(x) and so B r (x) r (r U r ) r = u(t, y) dy = ( t u)(t, y) ds y = U tt. This is the Euler-Darboux-Poisson equation. The initial conditions U(0, r) = G(r), U t (0, r) = H(r) are immediate. Now set Ũ(t, r) := ru(t, r), and similarly G := rg, H := rh. Recalling that U(t, ), G, H can be extended as smooth even functions on R, we see that Ũ(t, ), G, and H are odd in r. For t, r > 0, the Euler-Darboux-Poisson equation implies that Ũ tt Ũ rr = ru tt ru rr U r = 0. In view of the identity Ũ(t, r) = Ũ(t, r), this equation holds for r 0 as well (at r = 0, we have Ũ tt = 0 U tt = 0). By the D Alembert formula, when 0 r t so Ũ(t, r) = [ G(r + t) + G(r t)] + = [ G(t + r) G(t r)] + U(t, r) = r [ G(t + r) G(t r)] + r 0 r+t r t t+r t r t+r t r H(y) dy H(y) dy, H(y) dy,

3 THE WAVE EQUATION 3 Taking r 0, we conclude that u(t, x) = lim U(t, r; x) = G (t) + H(t) r 0 + = g(y) ds 4πt y + 4πt g(y), y x ds y + h(y) ds y. t 4πt d = by descending from d = 3 Now consider the equation in two space dimensions { u = ( t x )u = 0, in (0, ) R, (4) u(0, x) = g(x), t u(0, x) = h(x), x R Theorem 0.4 (Poisson s formula). Assume that u is a smooth solution. Then u(t, x) = h(y) π t y x dy + g(y), y x + g(y) πt t y x B t (x) Proof. We simply view u as a solution to the wave equation in R R 3 which does not depend on x 3, and apply Kirchoff s formula. B t (x) dy. ũ(t, x, x x, x 3 ) := u(t, x, x ), g(x, x, x 3 ) := g(x, x ), h(x, x, x 3 ) := h(x, x ). Writing x = (x, x, x 3 ), we then have { ( t x )u = 0, in (0, ) R 3, ũ(0, x) = g( x), t ũ(0, x) = h( x), x R 3. For (t, x) (0, ) R, we apply Kirchoff s formula at the point (t, x, 0) (0, ) R 3 to obtain u(t, x) = ũ(t, x, 0) = h(ỹ) ds ỹ + g(ỹ), ỹ x + g(ỹ) ds ỹ, 4πt ỹ x =t 4πt ỹ x =t = h(y) ds ỹ + g(y), y x) + g(y) ds 4πt ỹ x =t 4πt ỹ ỹ x =t = h(y) ds ỹ + g(y), y x) + g(y) ds ỹ, πt πt ỹ x =t, y 3 0 ỹ x =t, y 3 0 where ỹ := (y, y 3 ), x := (x, 0). For the last line we used symmetry to write the integral over the whole sphere ỹ x = t as twice the integral over the upper hemisphere. Parametrizing the upper hemisphere by y 3 = t y x, y x t, we have ds ỹ = + y 3 dy = u(t, x) = π y x t t t y x dy, and therefore h(y) t y x dy + πt y x t g(y), y x + g(y) t y x dy.

4 4 THE WAVE EQUATION Remark (Finite speed of propagation/huygen s principle). From the representation formulae, we see that the solution at a point (t, x) depends only on the values of the initial data in the ball {y : y x t}; moreover, in odd dimensions, only the values on the sphere {y : y x = t} matter. Here is a slightly different point of view. For each (t, x), consider the backwards light cone = {(s, y) : s t, y x = t s}. C (t,x) Then the value of u(t, x) at the tip (t, x) depends only on the values of u on the cone and in the interior region; it is not influenced by what happens in the exterior of the cone. In particular, if we know that the initial data is zero outside some ball x R, then the solution will be zero in the exterior region { x R + t}. The forced wave equation Using Duhamel s principle, we can solve { u = ( t x )u = f in (0, ) t R d x (5) u = 0, t u = 0 on {t = 0} R d x. For simplicity assume f is a smooth function. Theorem 0.5. For each s, let u( ; s) be the solution to { u(, s) = 0, in (s, ) R d x u(s, x; s) = 0, t u(s, x; s) = f (s, x). Then u(t, x) := is smooth on [0, ) R d, and satisfies (5). t 0 u(t, x; s) ds Proof. Straight computation using the Leibniz integral rule. For instance, in d = 3 we use Kirchhoff s formula to write u(t, x) = t 0 s)[ ] f (s, y) ds y ds. 4π(t y x =t s Remark. Finite speed of propagation for the free wave equation implies that the solution u(t, x) depends only on the values of the forcing term in the backward light cone {(s, y) : y x t s}. Energy methods One can in fact deduce many properties of the wave equation without appealing to the explicit representation formula above. The following technique, based on considering the energy of solutions, is very robust. Let Ω R d be an open bounded set with smooth boundary, and let Ω T = (0, T] Ω and Γ T = Ω T \ Ω T denote the usual cylinder and parabolic boundary, respectively. Proposition 0.6. Suppose u is a C (Ω T ) solution to { u = 0 in ΩT u = 0 on [0, T] Ω.

5 THE WAVE EQUATION 5 Define the energy where (6) E(t) := Ω t u(t, x) + x u(t, x) dx = e(u)(t, x) := tu(t, x) + xu(t, x) Ω e(u)(t, x) dx, is the energy density of u. Then E is independent of t; thus, the energy is conserved. Proof. Differentiate in t and apply the divergence theorem together with the zero boundary condition to obtain E (t) t u t u t u x u dx = 0. Ω Corollary 0.7 (Uniqueness of solutions). There is at most one solution u C (Ω T ) to the initial-boundary value problem u = f in Ω T u = g on Γ T t u = h on {t = 0} Ω. Proof. If u and u are two solutions, their difference u u solves the free wave equation with zero initial and boundary conditions. By energy conservation and the FTOC, u u is constant, and since u = u when t = 0, equality holds everywhere. Proposition 0.8. Let u be a smooth solution to u = 0 such that u(0, x) = 0 and t u(0, x) = 0 for x R. Then u(t, x) = 0 for {(t, x) : x R t, t 0}. Proof. For t < t, let Σ t,t denote the spacetime region Σ t,t = {(t, x) : x R t, t [t, t ]}. The boundary of this region (see Figure below) consists of the t time slice, the t time slice, and the lateral boundary, i.e. the portion of Σ t,t on the light cone: Σ t,t = {(t, x) : x R t } {(t, x) : x R t } {(t, x) : x = R t, t [t, t ]} = S t S t S cone. Now multiply the equation u = 0 by t u integrate over Σ t,t, using the identity t u u = ( t u u) t u, u, to see that 0 = t u t u t x u dxdt = Σ t,t Σ t,t t( t u) + t x u x ( t u u) dxdt = ( t, x ) (e(u), t u u ) dxdt, Σ t,t

6 6 THE WAVE EQUATION Figure. The boundary of the region Σ t,t. where e(u) is the energy density (6). Applying the divergence theorem to the last integral, we discover 0 = (e(u), t u u), (ˆn t, ˆn x ) ds Σ t,t = e(u)(t, x) dx e(u)(t, x) dx + ˆn t e(u) ˆn x, t u u ds, S t S t S cone where ˆn = (ˆn t, ˆn x ) is the outward unit normal to Σ t,t ; thus ˆn = (, 0) on S t, ˆn = (, 0) on S t, ˆn t > 0 and ˆn t = ˆn x on S cone. The last relations ˆn t > 0, ˆn t = ˆn x are key to obtaining a useful estimate, and follows from the fact that the light cone S cone is adapted to light rays that propagate at unit speed. Indeed, note now that by the Cauchy-Schwarz and arithmetic mean-geometric mean inequalities, t u + x u ˆn x, t u x u ˆn x t u x u ˆn t = ˆn t e(u). Therefore the integral ˆn t e(u) ˆn x, t u x u ds S cone is nonnegative, and we derive e(u)(t, x) dx = x R t (7) e(u)(t, x) dx + ˆn t e(u) ˆn x, t u x u ds x R t S cone e(u) dx. x R t The integral over the truncated light cone S cone can be interpreted as the energy flux through the light cone x = R t between times t and t ; this flux is nonnegative provided that ˆn t / ˆn x, i.e. energy can t propagate faster than speed.. Taking t = 0 in this inequality and recalling that e(u) is nonnegative, we have 0 e(u) dx 0 x R t for all 0 t R; therefore, e(u)(t, x) = 0 in the solid cone Σ 0,R := {(t, x) : x R t, t [0, R]}. As u(0, x) = 0 when x R, we integrate t u forward in time to conclude that u(t, x) = 0 in Σ 0,R as well.

7 THE WAVE EQUATION 7 Question. How would you modify this proof for the wave equation t u c x u = 0, where c > 0 is a constant?

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