Chem Differential Eq HW Solutions Fall 2011 166

Chem Differential Eq HW Solutions Fall 2011 166 - x ↔ x-x...

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166 Chapter 12 Green’s Functions and Conformal Mappings We have G ( x, y, x 0 ,y 0 )= 1 2 ln e 2 x + e 2 x 0 - 2 e x + x 0 cos( y - y 0 ) e 2 x + e 2 x 0 - 2 e x + x 0 cos( y + y 0 ) = 1 2 ln ( e 2 x + e 2 x 0 - 2 e x + x 0 cos( y - y 0 ) ) - 1 2 ln ( e 2 x + e 2 x 0 - 2 e x + x 0 cos( y + y 0 ) ) ∂G ∂y ± ± ± y = π = 1 2 2 e x + x 0 sin( y - y 0 ) e 2 x + e 2 x 0 - 2 e x + x 0 cos( y - y 0 ) ± ± ± y = π - 1 2 2 e x + x 0 cos( y + y 0 ) e 2 x + e 2 x 0 - 2 e x + x 0 sin( y + y 0 ) ± ± ± y = π = e x + x 0 sin y 0 e 2 x + e 2 x 0 - 2 e x + x 0 cos y 0 = sin y 0 e x - x 0 + e x 0 - x - 2 cos y 0 . Thus the Poisson integral formula in this case is u ( x 0 ,y 0 )= sin y 0 2 π ² -∞ g ( x ) e x - x 0 + e x 0 - x - 2 cos y 0 dx. Let us test this formula in a case where we know the solution. Take g ( x )=1for all x . Then, we know that the solution is a linear function of y (see Exercise , Section 12.); in fact, it is easy to verify that the solution is u ( x 0 ,y 0 )= y 0 π . Take g ( x ) = 1 in the Poisson formula and ask: Do we have y 0 π = sin y 0 2 π ² -∞ 1 e x - x 0 + e x 0 - x - 2 cos y 0 dx ? Change variables in the integral:
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Unformatted text preview: x ↔ x-x . Then the last equation becomes y π = sin y 2 π ² ∞-∞ 1 e x + e-x-2 cos y dx. Evaluate the right side in the case y = π 2 . The answer should be 1 / 2. Then try y = π/ 4. The answer should be 1 / 4. (I tried it on Mathematica. It works.) Out of this exercise, you can get the interesting integral formula 2 y sin y = ² ∞-∞ 1 e x + e-x-2 cos y dx (0 < y < π ) ....
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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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