F07_Midterm_Solutions

# F07_Midterm_Solutions - Physics 505 Fall 2007 Midterm —...

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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 two-dimensional problem defined in the region between concentric circles of radii a and b . b a [10] a ) Using polar coordinates, the Dirichlet Green’s function may be expanded as G ( ρ,φ ; ρ ,φ ) = ∞ X m =-∞ g m ( ρ,ρ ) e im ( φ- φ ) Write down the appropriate differential equation for g m ( ρ,ρ ). In two dimensions, the Green’s function satisfies ∇ 2 G ( ~x,~x ) =- 4 πδ (2) ( ~x- ~x ) Using polar coordinates, we note that ∇ 2 = 1 ρ ∂ ∂ρ ρ ∂ ∂ρ + 1 ρ 2 ∂ 2 ∂φ 2 and δ (2) ( ~x- ~x ) = 1 ρ δ ( ρ- ρ ) δ ( φ- φ ) As a result, we have 1 ρ ∂ ∂ρ ρ ∂ ∂ρ + 1 ρ 2 ∂ 2 ∂φ 2 G ( ρ,φ ; ρ ,φ ) =- 4 π ρ δ ( ρ- ρ ) δ ( φ- φ ) =- X m 2 ρ δ ( ρ- ρ ) e im ( φ- φ ) where we have used the completeness relation X m e im ( φ- φ ) = 2 πδ ( φ- φ ) Inserting the expansion G ( ρ,φ ; ρ ,φ ) = X m g m ( ρ,ρ ) e im ( φ- φ ) into the above and matching powers of e i ( φ- φ ) then gives the differential equation 1 ρ ∂ ∂ρ ρ ∂ ∂ρ- m 2 ρ 2 g m ( ρ,ρ ) =- 2 ρ δ ( ρ- ρ ) (1) [20] b ) Solve the Green’s function equation for g m ( ρ,ρ ) subject to Dirichlet boundary conditions and write down the result for G ( ρ,φ ; ρ ,φ ). 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 Green’s function equation (1) is 1 ρ ∂ ∂ρ ρ ∂ ∂ρ- m 2 ρ 2 g m ( ρ,ρ ) = 0 This is easy to solve as it is equidimensional in ρ . The two independent solutions are of the form ρ m and ρ 0- m . Because of the delta-function source in (1), we break up the ρ interval into a ≤ ρ ≤ ρ and ρ ≤ ρ ≤ b . Hence we write g m ( ρ,ρ ) = Au ( ρ ) a ≤ ρ ≤ ρ Bv ( ρ ) ρ ≤ ρ ≤ b where u ( ρ ) = ρ a m- a ρ m , v ( ρ ) = ρ b m- b ρ m (2) are appropriately chosen to satisfy the Dirichlet boundary conditions g m ( ρ,a ) = 0 and g m ( ρ,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 > , ∂ ∂ρ g < = ∂ ∂ρ g > + 2 ρ (3) where g < and g > are the values of g m ( ρ,ρ ) for ρ immediately to the left and right of the delta function at ρ , 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 Green’s function, we may write g m ( ρ,ρ ) = Au ( ρ < ) v ( ρ > ) where ρ < = min( ρ,ρ ) and ρ > = max( ρ,ρ ), and where A is a ρ and ρ independent constant. In this case, the first condition of (3) is automatically satisfied, while the second one gives Au ( ρ ) v ( ρ ) = Au ( ρ ) v ( ρ ) + 2 ρ or equivalently A =- 2 ρ u ( ρ ) v ( ρ ) u ( ρ ) v ( ρ )- 1 Note that the determinant is simply the Wronskian of...
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## This note was uploaded on 11/21/2011 for the course PHY 505 taught by Professor Stephens,p during the Fall '08 term at SUNY Stony Brook.

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F07_Midterm_Solutions - Physics 505 Fall 2007 Midterm —...

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