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F07_Practice_Midterm_Solutions

# F07_Practice_Midterm_Solutions - Physics 505 Practice...

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Physics 505 Fall 2007 Practice Midterm — Solutions This midterm will be a two hour open book, open notes exam. Do all three problems. 1. A rectangular box has sides of lengths a , b and c c x y z a b a ) For the Dirichlet problem in the interior of the box, the Green’s function may be expanded as G ( x, y, z ; x , y , z ) = m =1 n =1 g mn ( z, z ) sin mπx a sin mπx a sin nπy b sin nπy b Write down the appropriate differential equation that g mn ( z, z ) must satisfy. Note that sin kx satisfies the completeness relation m =1 sin mπx a sin mπx a = a 2 δ ( x - x ) Hence the Green’s function equation 2 x G ( x, x ) = - 4 πδ 3 ( x - x ) has an expansion m,n 2 x g mn ( z, z ) sin mπx a sin mπx a sin nπy b sin nπy b = - 4 πδ ( z - z ) 4 ab m,n sin mπx a sin mπx a sin nπy b sin nπy b Working out the x and y derivatives on the left-hand side yields m,n d 2 dz 2 - a 2 - b 2 g mn ( z, z ) sin mπx a sin mπx a sin nπy b sin nπy b = - 16 π ab δ ( z - z ) m,n sin mπx a sin mπx a sin nπy b sin nπy b

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However, since sin kx forms an orthogonal basis, each term in this sum must vanish by itself. This results in the differential equation d 2 dz 2 - γ 2 mn g mn ( z, z ) = - 16 π ab δ ( z - z ) (1) where γ mn = π ( m/a ) 2 + ( n/b ) 2 is given in part c ). Note that the Fourier sine expansion automatically satisfies Dirichlet boundary conditions for x and y . The remaining boundary condition is that g mn ( z, z ) vanishes whenever z or z is equal to 0 or c . b ) Solve the Green’s function equation for g mn ( z, z ) subject to Dirichlet boundary conditions and write down the result for G ( x, y, z ; x , y , z ). We may solve the Green’s function equation (1) by first noting that the homoge- neous equation is of the form g ( z ) - γ 2 mn g ( z ) = 0 This is a second-order linear equation with constant coefficients admitting the familiar solution g ( z ) = Ae γ mn z + Be - γ mn z However, we want g ( z ) = 0 when z = 0 or z = c . This motivates us to write out the solutions u ( z ) = sinh γ mn z 0 < z < z v ( z ) = sinh[ γ mn ( c - z )] z < z < c The Green’s function solution is then given by g mn ( z, z ) = Au ( z ) z < z Bv ( z ) z > z The matching conditions g < = g > , d dz g < = d dz g > + 16 π ab then give the system A sinh γ mn z = B sinh[ γ mn ( c - z )] , A cosh γ mn z = - B cosh[ γ mn ( c - z )] + 16 π abγ mn which may be solved to yield A = 16 π sinh[ γ mn ( c - z )] abγ mn (cosh γ mn z sinh[ γ mn ( c - z )] + sinh γ mn z cosh[ γ mn ( c - z )]) = 16 π sinh[ γ mn ( c - z )] abγ mn sinh γ mn c = 16 π abγ mn sinh γ mn c
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F07_Practice_Midterm_Solutions - Physics 505 Practice...

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