Constant Coefficient PDE

Constant Coefficient PDE - A NALYSIS TOOLS W ITH...

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ANALYSIS TOOLS WITH APPLICATIONS 399 21. Constant Coefficient partial differential equations Suppose that p ( ξ )= P | α | k a α ξ α with a α C and (21.1) L = p ( D x ):= Σ | α | N a α D α x = Σ | α | N a α μ 1 i x α . Then for f S c Lf ( ξ )= p ( ξ ) ˆ f ( ξ ) , that is to say the Fourier transform takes a constant coe cient partial di f erential operator to multiplication by a polynomial. This fact can often be used to solve constant coe cient partial di f erential equation. For example suppose g : R n C is a given function and we want to f ndaso lut iontotheequat ion Lf = g. Taking the Fourier transform of both sides of the equation Lf = g would imply p ( ξ ) ˆ f ( ξ )=ˆ g ( ξ ) and therefore ˆ f ( ξ )=ˆ g ( ξ ) /p ( ξ ) provided p ( ξ ) is never zero. (We will discuss what happens when p ( ξ ) has zeros a bit more later on.) So we should expect f ( x )= F 1 μ 1 p ( ξ ) ˆ g ( ξ ) ( x )= F 1 μ 1 p ( ξ ) F g ( x ) . De f nition 21.1. Let L = p ( D x ) as in Eq. (21.1). Then we let σ ( L ):= Ran ( p ) C and call σ ( L ) the spectrum of L. Given a measurable function G : σ ( L ) C , we de f ne (a possibly unbounded operator) G ( L ): L 2 ( R n ,m ) L 2 ( R n ,m ) by G ( L ) f := F 1 M G p F where M G p denotes the operation on L 2 ( R n ,m ) of multiplication by G p, i.e. M G p f =( G p ) f with domain given by those f L 2 such that ( G p ) f L 2 . At a formal level we expect G ( L ) f = F 1 ( G p ) F g. 21.0.3. Elliptic examples. As a speci f c example consider the equation (21.2) ¡ + m 2 ¢ f = g where f,g : R n C and = P n i =1 2 /∂x 2 i is the usual Laplacian on R n . By Corollary 20.16 (i.e. taking the Fourier transform of this equation), solving Eq. (21.2) with f,g L 2 is equivalent to solving (21.3) ¡ | ξ | 2 + m 2 ¢ ˆ f ( ξ )=ˆ g ( ξ ) . The unique solution to this latter equation is ˆ f ( ξ )= ¡ | ξ | 2 + m 2 ¢ 1 ˆ g ( ξ ) and therefore, f ( x )= F 1 ³ ¡ | ξ | 2 + m 2 ¢ 1 ˆ g ( ξ ) ´ ( x )= : ¡ + m 2 ¢ 1 g ( x ) . We expect F 1 ³ ¡ | ξ | 2 + m 2 ¢ 1 ˆ g ( ξ ) ´ ( x )= G m F g ( x )= Z R n G m ( x y ) g (
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400 BRUCE K. DRIVER where G m ( x ):= F 1 ¡ | ξ | 2 + m 2 ¢ 1 ( x )= Z R n 1 m 2 + | ξ | 2 e · x d ξ. At the moment F 1 ¡ | ξ | 2 + m 2 ¢ 1 only makes sense when n =1 , 2 , or 3 because only then is ¡ | ξ | 2 + m 2 ¢ 1 L 2 ( R n ) . For now we will restrict our attention to the one dimensional case, n =1 , in which case (21.4) G m ( x )= 1 2 π Z R 1 ( ξ + mi )( ξ mi ) e iξx dξ.
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Constant Coefficient PDE - A NALYSIS TOOLS W ITH...

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