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Unformatted text preview: Physics 505 Fall 2007 Midterm Solutions This midterm is a two hour open book, open notes exam. Do all three problems. [30 pts] 1. Consider a twodimensional problem defined in the region between concentric circles of radii a and b . b a [10] a ) Using polar coordinates, the Dirichlet Greens function may be expanded as G ( , ; , ) = X m = g m ( , ) e im (  ) Write down the appropriate differential equation for g m ( , ). In two dimensions, the Greens function satisfies 2 G ( ~x,~x ) = 4 (2) ( ~x ~x ) Using polar coordinates, we note that 2 = 1 + 1 2 2 2 and (2) ( ~x ~x ) = 1 (  ) (  ) As a result, we have 1 + 1 2 2 2 G ( , ; , ) = 4 (  ) (  ) = X m 2 (  ) e im (  ) where we have used the completeness relation X m e im (  ) = 2 (  ) Inserting the expansion G ( , ; , ) = X m g m ( , ) e im (  ) into the above and matching powers of e i (  ) then gives the differential equation 1  m 2 2 g m ( , ) = 2 (  ) (1) [20] b ) Solve the Greens function equation for g m ( , ) subject to Dirichlet boundary conditions and write down the result for G ( , ; , ). Note that the m = 0 case may need to be treated separately. We start with the m 6 = 0 case. The homogeneous equation corresponding to the Greens function equation (1) is 1  m 2 2 g m ( , ) = 0 This is easy to solve as it is equidimensional in . The two independent solutions are of the form m and 0 m . Because of the deltafunction source in (1), we break up the interval into a and b . Hence we write g m ( , ) = Au ( ) a Bv ( ) b where u ( ) = a m a m , v ( ) = b m b m (2) are appropriately chosen to satisfy the Dirichlet boundary conditions g m ( ,a ) = 0 and g m ( ,b ) = 0. Note that these expressions are valid for both positive and negative m . From (1), we must now satisfy the matching and jump conditions g < = g > , g < = g > + 2 (3) where g < and g > are the values of g m ( , ) for immediately to the left and right of the delta function at , respectively. These conditions give rise to a set of two equations which may be solved to determine the two unknowns A and B . Alternatively, by symmetry of the Greens function, we may write g m ( , ) = Au ( < ) v ( > ) where < = min( , ) and > = max( , ), and where A is a and independent constant. In this case, the first condition of (3) is automatically satisfied, while the second one gives Au ( ) v ( ) = Au ( ) v ( ) + 2 or equivalently A = 2 u ( ) v ( ) u ( ) v ( ) 1 Note that the determinant is simply the Wronskian of...
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 Fall '08
 Stephens,P
 Physics, Magnetism

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