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Unformatted text preview: MATH 251-3, Fall 2007 Simon Fraser University Assignment 7: Solutions Additional Question Due: 4:30pm, Monday 5 November 2007 1. In this problem, you will show that if f ( x, y ) is harmonic (that is, satisfies Laplaces equation), then z ( x, y ) = f ( x 2- y 2 , 2 xy ) is harmonic. That is: suppose f ( x, y ) is any solution of Laplaces equation f xx + f yy = 0, or equivalently, f ( u, v ) satisfies f uu + f vv = 0; this is the meaning of f is harmonic. By substituting u = x 2- y 2 and v = 2 xy , we obtain a new function of x and y , which we call z ( x, y ) . The objective is to show that we also always have 2 z/x 2 + 2 z/y 2 = 0. You may assume that all partial derivatives are contin- uous. [As an example: if f is the function f ( x, y ) = 3 + 2 y- xy , or f ( u, v ) = 3 + 2 v- uv , then z ( x, y ) = f ( x 2- y 2 , 2 xy ) = 3+2(2 xy )- ( x 2- y 2 )(2 xy ) = 3+4 xy- 2 x 3 y +2 xy 3 , a new function of x and y .] To approach this problem, let z = f ( u, v ), where u = x 2- y 2 and v = 2 xy . (a) Use the chain rule to show that z x = 2 x f u + 2 y f v (notation) = 2 xf u + 2 yf v ....
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