Author Topic: Problem 1  (Read 8846 times)

Kun Guo

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Re: Problem 1
« Reply #15 on: October 15, 2012, 09:33:18 PM »
Build on DW's solution,part c).  For -2t < x < 2t, x+3t> 0, Ï• can be solved the same as in part a). x+2t<0, ψ is the same as x< -2t, which is a constant C, since Ï•+ψ.  The u(x,t)= 1/4exp(-x-2t)+C for -2t < x < 2t. Will this be a correct answer to finish up the question?

Victor Ivrii

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Re: Problem 1
« Reply #16 on: October 16, 2012, 02:35:55 AM »
Build on DW's solution,part c).  For -2t < x < 2t, x+3t> 0, Ï• can be solved the same as in part a). x+2t<0, ψ is the same as x< -2t, which is a constant C, since Ï•+ψ.  The u(x,t)= 1/4exp(-x-2t)+C for -2t < x < 2t. Will this be a correct answer to finish up the question?

Probably, it was obtained already (with lost $C$)