Chem Differential Eq HW Solutions Fall 2011 165

Chem Differential Eq HW Solutions Fall 2011 165 - ± = 1 2...

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Section 12.7 Green’s Functions and Conformal Mappings 165 Solutions to Exercises 12.6 1. The function φ ( z )= z - 1 is a conformal mapping of Ω, one-to-one, onto the unit disk. Apply Theorem 3; then for z = x + iy and z 0 = x 0 + iy 0 in Ω, we have G ( x, y, x 0 ,y 0 )=l n ± ± ± ± ± x + iy - 1 - x 0 - iy 0 +1 1 - ( x 0 + iy 0 - 1) ( x + iy - 1) ± ± ± ± ± =l n ± ± ± ± ( x - x 0 )+ i ( y - y 0 ) 1 - ( x 0 - 1 - iy 0 )( x - 1+ iy ) ± ± ± ± . = 1 2 ln ² ( x - x 0 ) 2 +( y - y 0 ) 2 ³ - 1 2 ln [1 - ( x 0 - 1 - iy 0 )( x - 1+ iy )] . = 1 2 ln ² ( x - x 0 ) 2 +( y - y 0 ) 2 ³ - 1 2 ln ² ( - x 0 x + x 0 + x - y 0 y ) 2 +( y 0 x + yx 0 - y 0 - y ) 2 ) ³ . 5. The function φ ( z )= e z maps Ω, one-to-one, onto the upper half-plane. Apply Theorem 4; then for z = x + iy and z 0 = x 0 + iy 0 in Ω, we have G ( x, y, x 0 ,y 0 )=l n ± ± ± ± e z - e z 0 e z - e z 0 ± ± ± ± (Note that e z 0 = e z 0 . ) =l n ± ± ± ± e x cos y + ie x sin y - e x 0 cos y 0 - ie x 0 sin y 0 e x cos y + ie x sin y - e x 0 cos y 0 + ie x 0 sin y 0 ± ± ± ± = 1 2 ln ± ± ± ± ± ( e x cos y - e x 0 cos y 0 ) 2 + ( e x sin y - e x 0 sin y 0 ) 2 ( e x cos y - e x 0 cos y 0 ) 2 + ( e x sin y + e x 0 sin y 0 ) 2 ± ± ±
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Unformatted text preview: ± = 1 2 ln e 2 x + e 2 x-2 e x + x ( cos y cos y + sin y sin y ) e 2 x + e 2 x-2 e x + x ( cos y cos y-sin y sin y ) = 1 2 ln e 2 x + e 2 x-2 e x + x cos( y-y ) e 2 x + e 2 x-2 e x + x cos( y + y ) . 9. We use the result of Exercise 5 and apply Theorem 2, Section 12.3. Accordingly, u ( x , y ) = 1 2 π ´ ∞-∞ g ( x ) ∂G ∂y ± ± ± y = π dx....
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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