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Unformatted text preview: MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 6 Rik Sengupta [email protected] April 7, 2010 1. Stein and Shakarchi, p. 213, problem 3 We observed that the solution u ( x,t ) of the Cauchy problem for the wave equation given by formula (3) depends only on the initial data on the base on the backward light cone. It is natural to ask if this property is shared by any solution of the wave equation. An affirmative answer would imply uniqueness of the solution. Let B ( x ,r ) denote the closed ball in the hyperplane t = 0 centered at x and of radius r . The backward light cone with base B ( x ,r ) is defined by L B ( x ,r ) = { ( x,t ) ∈ R d × R :  x x  ≤ r t, ≤ t ≤ r } . Theorem. Suppose that u ( x,t ) is a C 2 function on the closed upper halfplane { ( x,t ) : x ∈ R d ,t ≥ } that solves the wave equation ∂ 2 u ∂t 2 = Δ u. If u ( x, 0) = ∂u ∂t ( x, 0) = 0 for all x ∈ B ( x ,r ), then u ( x,t ) = 0 for all ( x,t ) ∈ L B ( x ,r ) . In words, if the initial data of the Cauchy problem for the wave equation vanishes on a ball B , then any solution u of the problem vanishes in the backward light cone with base B . The following steps outline a proof of the theorem. (a) Assume that u is real. For 0 ≤ t ≤ r let B t ( x ,r ) = { x :  x x  ≤ r t } , and also define ∇ u ( x,t ) = ∂u ∂x 1 ,..., ∂u ∂x d , ∂u ∂t . Now consider the energy integral E ( t ) = 1 2 Z B t ( x ,r ) ∇ u  2 dx = 1 2 Z B t ( x ,r ) ∂u ∂t 2 + d X j =1 ∂u ∂x j 2 dx. Observe that E ( t ) ≥ 0 and E (0) = 0. Prove that 1 E ( t ) = Z B t ( x ,r ) ∂u ∂t ∂ 2 u ∂t 2 + d X j =1 ∂u ∂x j ∂ 2 u ∂x j ∂t dx 1 2 Z ∂B t ( x ,r ) ∇ u  2 dσ ( γ ) . (b) Show that ∂ ∂x j ∂u ∂x j ∂u ∂t = ∂u ∂x j ∂ 2 u ∂x j ∂t + ∂ 2 u ∂x 2 j ∂u ∂t . (c) Use the last identity, the divergence theorem, and the fact that u solves the wave equation to prove that E ( t ) = Z ∂B t ( x ,r ) d X j =1 ∂u ∂x j ∂u ∂t ν j dσ ( γ ) 1 2 Z ∂B t ( x ,r ) ∇ u  2 dσ ( γ ) , where ν j denotes the j th coordinate of the outward normal to B t ( x ,r ). (d) Use the CauchySchwarz inequality to conclude that d X j =1 ∂u ∂x j ∂u ∂t ν j ≤ 1 2 ∇ u  2 , and as a result, E ( t ) ≤ 0. Deduce from this that E ( t ) = 0 and u = 0. Solution. We solve the parts one by one. (a) We simply differentiate the given expression for E ( t ), taking care to use the extended form of Leibniz’s rule for integration. Recall that this rule states that when the limits of integration and the integrand are all functions of the parameter α , then d dα Z b ( α ) a ( α ) f ( x,α ) dx = db ( α ) dα f ( b ( α ) ,α ) da ( α ) dα f ( a ( α ) ,α ) + Z b ( α ) a ( α ) ∂ ∂α f ( x,α ) dx....
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This note was uploaded on 05/07/2010 for the course MAT 330 taught by Professor Sergiuklainermann during the Spring '10 term at Princeton.
 Spring '10
 SERGIUKLAINERMANN
 Differential Equations, Equations, Partial Differential Equations, Fourier Series

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