# F10HW02 - Problem 1.14 Consider the electrostatic Green...

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Unformatted text preview: Problem 1.14 Consider the electrostatic Green functions of Section 1.10 for Dirichlet and Neumann bound- ary conditions on the surface S bounding the volume V . Apply Green's theorem (1.35) with integration variable ~ y and ' = G ( ~x; ~ y ), = G ( ~x ; ~ y ), with r 2 y G ( ~ z; ~ y ) = 4 ( ~ y ~ z ). Find an expression for the di erence [ G ( ~x; ~x ) G ( ~x ; ~x )] in terms of an integral over the boundary surface S . Starting with equation 1.35: Z V ' r 2 r 2 ' ¡ d 3 y = I S ' @ @n @' @n da Z V G ( ~x; ~ y ) r 2 y G ( ~x ; ~ y ) G ( ~x ; ~ y ) r 2 y G ( ~x; ~ y ) ¡ d 3 y = I S ' @ @n @' @n da Z V G ( ~x; ~ y ) [ 4 ( ~ y ~x )] d 3 y Z V G ( ~x ; ~ y ) [ 4 ( ~ y ~x )] d 3 y = I S ' @ @n @' @n da 4 G ( ~x; ~x ) + 4 G ( ~x ; ~x ) = I S ' @ @n @' @n da G ( ~x; ~x ) G ( ~x ; ~x ) = 1 4 I S G ( ~x; ~ y ) @G ( ~x ; ~ y ) @n G ( ~x ; ~ y ) @G ( ~x; ~ y ) @n da 1.14.a For Dirichlet boundary conditions on the potential and the associated boundary condition on the Green function, show that G D ( ~x; ~x ) must be symmetric in ~x and ~x . Note that G D ( ~x; ~ y ) = 0 and G D ( ~x ; ~ y ) = 0 on the bounding surface. Therefore: G D ( ~x; ~x ) G D ( ~x ; ~x ) = 1 4 I S : G D ( ~x; ~ y ) @G D ( ~x ; ~ y ) @n : G D ( ~x ; ~ y ) @G D ( ~x; ~ y ) @n da G D ( ~x; ~x ) G D ( ~x ; ~x ) = 0 ) G D ( ~x; ~x ) = G D ( ~x ; ~x ) 1 1.14.b For Neumann boundary conditions, use the boundary conditions (1.45) for G N ( ~x; ~x ) to show that G N ( ~x; ~x ) is not symmetric in general but that G N ( ~x; ~x ) F ( ~x ) is symmetric in ~x and ~x , where F ( ~x ) = 1 S I S G N ( ~x; ~x ) da y G N ( ~x; ~x ) G N ( ~x ; ~x...
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## This note was uploaded on 12/08/2011 for the course PHYSICS 505 taught by Professor Liu during the Winter '08 term at University of Michigan.

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F10HW02 - Problem 1.14 Consider the electrostatic Green...

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