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Unformatted text preview: MATH 220: PDES AND BOUNDARIES We have used the Fourier transform and other tools (factoring the PDE) to solve PDEs on R n . We now study how we can use these results to solve problems on the half space, or indeed on intervals, cubes, etc. As you have shown on your problem set, the solution of the wave equation on R n (so u is a function on R n x × R t ) is even, resp. odd, in x n if the initial conditions and the inhomogeneity are even, resp. odd in x n . That is, write x = ( x ′ , x n ) where x ′ = ( x 1 , . . . , x n − 1 ). The wave equation is u tt − c 2 Δ u = f, u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) . If f ( x ′ , x n , t ) = ± f ( x ′ , − x n , t ) , φ ( x ′ , x n ) = ± φ ( x ′ , − x n ) , ψ ( x ′ , x n ) = ± ψ ( x ′ , − x n ) for all x and t , i.e. if f, φ, ψ are all even (+), resp. odd ( − ), functions of x n , then u is an even, resp. odd function of x n as well, i.e. u ( x ′ , x n , t ) = ± u ( x ′ , − x n , t ) . Recall that this was based on considering u ( x ′ , x n , t ) ∓ u ( x ′ , − x n , t ), and showing that it solves the homogeneous wave equation with 0 initial conditions. You also showed that if u is continuous, and is an odd function of x n , then u ( x ′ , , t ) = 0 for all x ′ and t , while if u is a C 1 and is an even function of x n , show that ∂ x n u ( x ′ , , t ) = 0 for all x ′ and t . These observations reduce the solution of the wave equation in x n > 0 with either Dirichlet or Neumann boundary condition to solving the PDE on all of R n . For the sake of definiteness, suppose we want to solve the Dirichlet problem: u tt − c 2 Δ u = f, x n ≥ , u ( x ′ , , t ) = 0 (DBC) , u ( x, 0) = φ ( x ) , u t ( x, 0) = ψ ( x ) , x n ≥ 0 (IC) . Here f , φ and ψ are given functions, defined in x n ≥ 0 only. To solve the PDE, we consider the odd extensions of f, φ, ψ in x n , i.e. define f odd ( x ′ , x n , t ) = braceleftbigg f ( x ′ , x n , t ) , x n ≥ , − f ( x ′ , − x n , t ) , x n < , , and analogously φ odd ( x ′ , x n ) = braceleftbigg φ ( x ′ , x n ) , x n ≥ , − φ ( x ′ , − x n ) , x n < , , with a similar definition for ψ . The resulting function is odd and continuous, provided that f ( x ′ , , t ) = 0 = φ ( x ′ , 0) = ψ ( x ′ , 0), i.e. if the data are compatible with the boundary condition. Indeed, for x n < 0 then φ odd ( x ′ , x n ) = − φ ( x ′ , − x n ) = − φ odd ( x ′ , − x n ) , and similarly in all other cases. 1 2 Now let v be the solution of the wave equation on R n with these odd data: v tt − c 2 Δ v = f odd , v ( x, 0) = φ odd ( x ) , v t ( x, 0) = ψ odd ( x ) (IC) . As we have seen, v is an odd function of x n , hence v ( x ′ , , t ) = 0 for all x ′ and t ....
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This note was uploaded on 03/16/2010 for the course CME 303 taught by Professor Vasy during the Fall '10 term at Stanford.
 Fall '10
 Vasy

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