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Chem Differential Eq HW Solutions Fall 2011 162

# Chem Differential Eq HW Solutions Fall 2011 162 - 162...

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162 Chapter 12 Green’s Functions and Conformal Mappings 17. We map the region onto the upper half-plane using the mapping f ( z ) = z 2 (see Example 1). The transformed problem in the uv -plane is 2 U = 0 with boundary values on the u -axis given by U ( u, 0) = u if 0 <u< 1 and 0 otherwise. To solve the problem in the uv -plane, we apply the Poisson integral formula ((5), Section 7.5). We have U ( u,v ) = v π -∞ U ( s, 0) ( u - s ) 2 + v 2 ds = v π 1 0 s ( u - s ) 2 + v 2 ds. Now use your calculus skills to compute this integral. We have v π 1 0 s ( u - s ) 2 + v 2 ds = v π 1 0 ( s - u ) ( s - u ) 2 + v 2 ds + v π 1 0 u ( s - u ) 2 + v 2 ds = v π 1 2 ln[( s - u ) 2 + v 2 ] 1 0 + 1 - u - u u t 2 + v 2 dt ( s - u = t ) = v 2 π ln (1 - u ) 2 + v 2 u 2 + v 2 + u π tan - 1 t v 1 - u - u = v 2 π ln (1 - u ) 2 + v 2 u 2 + v 2 + u π tan - 1 1 - u v - tan - 1 - u v = v 2 π ln (1 - u ) 2 + v 2 u 2 + v 2 + u π tan - 1 1 - u v + tan - 1 u v . The solution in the xy -plane is φ ( x,y ) = U f ( z ). To find the formula in terms of ( x,y ), we write z = x + iy , f ( z ) = z 2 = x 2 - y 2 + 2 ixy = ( u,v ). Thus u = x 2 - y 2 and v = 2 xy and so φ ( x,y ) = U ( x 2 - y 2 , 2 xy ) = xy π ln (1 - ( x 2 - y 2 )) 2 + (2 xy ) 2 ( x 2 - y 2 ) 2 + (2 xy ) 2 + x 2 - y 2 π tan - 1 1 - x 2 + y 2 2 xy + tan - 1 x 2 - y 2 2 xy = xy π ln (1 - ( x 2 - y 2 )) 2 + (2 xy ) 2 ( x 2 + y 2 ) 2 + x 2 - y 2 π tan - 1 1 - x 2 + y 2 2 xy + tan - 1 x 2 - y 2 2 xy . (I verified this solution on Mathematica and it works! It is harmonic and has the right boundary values.)
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