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Unformatted text preview: Partial Differential Equations Math 442 C13/C14 Fall 2009 Homework 2 due September 18 1. (Strauss 2.1.1.) Solve u tt = c 2 u xx , u ( x, 0) = e x , u t ( x, 0) = sin x . Solution: We use dAlemberts formula with ( x ) = e x , ( x ) = sin x. Then we have integraldisplay x + ct x ct sin s ds = cos( x + ct ) cos( x ct ) , and so we have u ( x,t ) = 1 2 ( e x + ct + e x ct ) + 1 2 c (cos( x + ct ) cos( x ct )) . 2. (Strauss 2.1.7.) We define an odd function to be any function f such that f ( x ) = f ( x ) for all x . Prove that if that the initial conditions , are odd functions, then so is the solution u ( x,t ) for any fixed time t . Solution: There are two completely different solutions: Again use dAlembert. We know that the solution is u ( x,t ) = 1 2 ( ( x + ct ) + ( x ct )) + 1 2 c integraldisplay x + ct x ct sin s ds. Now, we compute u ( x,t ) = 1 2 ( ( x + ct ) + ( x ct )) + 1 2 c integraldisplay x + ct x ct ( s ) ds Using the fact that is odd, we have ( x ct ) = ( ( x + ct )) = ( x + ct ) , ( x + ct ) = ( ( x ct )) = ( x ct ) . Now consider the integral, and make a change of variables r = s , giving integraldisplay x + ct x ct ( s ) ds = integraldisplay x ct x + ct ( r ) dr = integraldisplay x + ct x ct ( r ) dr. (Note in the last equality there are actually three minus signs, since we use the fact that is odd, and we flip the domain of integration.) Putting all of this together gives u ( x,t ) = u ( x,t ) and thus u is odd. We showed that reversing time gives a solution to the wave equation, and so does reversing space: if we choose v ( x,t ) = u ( x,t ), then we see that v tt ( x,t ) = u tt ( x,t...
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 Spring '09
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 Equations

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