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Unformatted text preview: 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 Greens function may be expanded as G ( x, y, z ; x , y , z ) = X m =1 X n =1 g mn ( z, z ) sin mx a sin mx a sin ny b sin ny b Write down the appropriate differential equation that g mn ( z, z ) must satisfy. Note that sin kx satisfies the completeness relation X m =1 sin mx a sin mx a = a 2 ( x- x ) Hence the Greens function equation 2 x G ( ~x,~x ) =- 4 3 ( ~x- ~x ) has an expansion X m,n 2 x g mn ( z, z ) sin mx a sin mx a sin ny b sin ny b =- 4 ( z- z ) 4 ab X m,n sin mx a sin mx a sin ny b sin ny b Working out the x and y derivatives on the left-hand side yields X m,n d 2 dz 2- m a 2- n b 2 g mn ( z, z ) sin mx a sin mx a sin ny b sin ny b =- 16 ab ( z- z ) X m,n sin mx a sin mx a sin ny b sin ny b 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- 2 mn g mn ( z, z ) =- 16 ab ( z- z ) (1) where mn = p ( 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 Greens 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 Greens function equation (1) by first noting that the homoge- neous equation is of the form g 00 ( z )- 2 mn g ( z ) = 0 This is a second-order linear equation with constant coefficients admitting the familiar solution g ( z ) = Ae mn z + Be- 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 mn z < z < z v ( z ) = sinh[ mn ( c- z )] z < z < c The Greens 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 ab then give the system A sinh mn z = B sinh[ mn ( c- z )] , A cosh mn z =- B cosh[ mn ( c- z )] + 16 ab mn which may be solved to yield...
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