Chem Differential Eq HW Solutions Fall 2011 158

Chem Differential Eq HW Solutions Fall 2011 158 -...

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158 Chapter 12 Green’s Functions and Conformal Mappings -3 -2 -1 1 2 3 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 -3 -2 -1 1 2 3 -6 -4 -2 2 4 6 -6 -4 -2 2 4 6 41. If u ( x, y ) does not depend on y , then u is a function of x alone. We have u y = 0 and so u yy = 0. If u is also harmonic, then u xx + u yy = 0. But u yy = 0, so u xx = 0. Integrating with respect to x twice, we Fnd u ( x, y )= ax + b . 45. We reason as in Example 4 and try for a solution a function of the form u ( r, θ )= a Arg z + b, where z = x + iy and a and b are constant to be determined. Using the boundary conditions in ±igure 18, we Fnd u ( r 9 π 10 )=60 a 9 π 10 + b = 60; u ( r, 3 π 5 )=0 a 3 π 5 + b =0; a ± 9 π 10 - 6 π 10 ² = 60 or a = 200 π ; b = - 120; u ( r, θ )= 200 π Arg z - 120 . In terms of ( x, y ), we can use (10) and conclude that, for y> 0, u ( x, y )= 200 π cot - 1 ± x y ² - 120 . 49. As in Exercise 47, we think of the given problem as the sum of two simpler
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Unformatted text preview: subproblems. Let u 1 be harmonic in the upper half-plane and equal to 100 on the x-axis for 0 < x < 1 and 0 for all other values of x . Let u 2 be harmonic in the upper half-plane and equal to 20 on the x-axis for-1 < x < 0 and 0 for all other values of x . Let u = u 1 + u 2 . Then u is harmonic in the upper half-plane and equal to 20 on the x-axis for-1 < x < 0; 100 for 0 < x < 1; and 0 otherwise. (Just add the boundary values of u 1 and u 2 .) Thus u 1 + u 2 is the solution to our problem....
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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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