hw 02 solution

B a 0 ie ex ay z x bx z y by z z bz ez 0 we try to

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Unformatted text preview: ey such that Ay z0 ey Ax Ay y Ax x , y being an arbitrary function of x and y ; z0 is an arbitrary constant Ay Bx z Bx z z x, y , By z x, y Similarly then x Ax y By Ax Ax z0 z x z0 z Ay Bx y By z x y z0 z Bz z x y x Bz x, y x, y Bz x , y , z0 y x y The functions x , y and x , y have to satisfy the constraint : Bz x , y , z0 x, y x x0 0 x gy, One can find that many such pairs actually exist. For example one can choose g still being an arbitrary function PROBLEM 3 fx x a for any test function f we have x f x x x f x x x f 0 f0 0 and Bz x , y , z0 so x x b integrating by parts fx ' x x fx f x 'x 'x x x f' x f x x x 'x f' 0 x f' x x x f' 0 c fxx x x 0f 0 0 d fxx 'x x fxx' x x f' x x fx x x f' 0 0 f0 f0 a e fx ax x fx ax x a f y a y y a f y a y y a f 0 a a f0 a more generally, if h x is a one to one function in an neighborhood V of 0 and h has no roots in V , we can change the variable of integration from x to x h x in that neighborhood fx y 0 iff x hxx 0 so that x fx fx hxx hxx x x f fx hxx hx hxx hxx hxx x h hxx h f y hx f0 h0 y hx y x h' x x f 0 h0 sgn h 0 h 0 0 h' 0 a a f h1 x fx x x2 a2 x a fx a fa h1 a h1 x x a x a fx h2 x x a x , apply the previous result with a, h2 x x2 a2 x x fx fa h2 a f a fa 2a g let f a be a test function, then fa it results that x a x a x...
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This document was uploaded on 03/17/2014 for the course PHYSICS 600 at Purdue.

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