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Unformatted text preview: 1 PHYS809 Class 11 Notes Construction of a Green function To illustrate how the completeness relation is used in constructing a Green function, consider the two dimensional Dirichlet Green function inside a rectangle of sides a and b . The Green function is the solution of ( ) ( ) ( ) 2 , , , 4 , G x y x y x x y y πδ δ ′ ′ ′ ′ ∇ =  (1.1) with boundary condition ( ) , , , 0 on 0, , 0, and . G x y x y x x a y y b ′ ′ = = = = = (1.2) For the homogeneous equation, the solution involves the set of orthonormal functions 2 sin , n x a a π (1.3) for which the completeness relation is ( ) 1 2 sin sin . n n x n x x x a a a π π δ ∞ = ′ ′ = ∑ (1.4) Hence we look for a Green function of form ( ) ( ) 1 2 , , , , sin sin . n n n x n x G x y x y g y y a a a π π ∞ = ′ ′ ′ ′ = ∑ (1.5) Substituting this into the Poisson equation, we get ( ) 2 2 2 2 2 1 1 2 2 sin sin 4 sin sin . n n n n g n x n x n n x n x g y y a a a y a a a a π π π π π πδ ∞ ∞ = = ′ ′ ∂ ′ =  ∂ ∑ ∑ (1.6) Since the sin n x a π functions are linearly independent, we must have ( ) 2 2 2 2 2 4 . n n g n g y y y a π πδ ∂ ′ =  ∂ (1.7) 2 This has to be solved subject to the boundary conditions ( ) , n g y y ′ = on y = and . y b = To do this we split the range into two intervals y y ′ ≤ < and . y y b ′ < ≤ The solution g n is continuous at . y y ′ = Integrating the above equation over the interval [ ] , y y ε ε ′ ′ + and letting ε → gives the jump condition 4 , n n y y g g y y π ′ ′ + ∂ ∂ =  ∂ ∂ (1.8) which shows that there is a change in gradient of g n at . y y ′ = In y y ′ ≤ < the solution that satisfies the boundary condition at y = is ( ) , sinh , n n y g y y A a π ′ = (1.9) where A will depend on . y ′ Similarly, in y y b ′ < ≤ the solution that satisfies the boundary condition at y b = is ( ) ( ) , sinh , n n b y g y y B a π ′ = (1.10) where B will also depend on . y ′ The continuity and jump conditions at y y ′ = give ( ) ( ) sinh sinh , cosh cosh 4 . n b y n y A B a a n b y n n n y B A a a a a π π π π π π π ′ ′ = ′ ′ =  (1.11) Solving for A and B , we get ( ) sinh sinh 4 , 4 . sinh sinh n b y n y a a a a A B n b n b n n a a π π π π ′ ′ = = (1.12) The solution for ( ) , n g y y ′ is 3 ( ) ( ) ( ) sinh sinh , in 0 , 4 , sinh sinh sinh , in ....
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 Fall '11
 MacDonald
 Sin, nπ b

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