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I have the following problem: $$ u_{xx}(x,t)=cu_{tt}(x,t),\\ \ u(x,0)=0, \\ \ u_t(x,0)=0, \\ \ u(0,t)=0, \\ \ u_x(L,t)=e^{i\omega t}. $$

This problem represents a bar of lenght L with a periodic excitation at the end. I asked for help to a professor and he told me I could transform it into another problem with homogeneous boundary conditions but with a inomogeneous equation. He told me the problem would turn into something like that

$$ u_{xx}(x,t)=cu_{tt}(x,t)+\delta(x-L)e^{i\omega t},\\ \ u(x,0)=0, \\ \ u_t(x,0)=0, \\ \ u(0,t)=0, \\ \ u(L,t)=0. $$

I understand both problems are physically equivalent. But mathematically it is not clear to me how did the nonhomogeneous boundary condition turn into the term with the dirac delta. Do you know how can this be done?

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Im not quite sure on this one, but I would try to put: $$v(x,t) = u(x,t) - \phi(x)$$ And then try to determine $\phi(x)$ such that, $$ v_{xx}(x,t)=cv_{tt}(x,t)+\delta(x-L)e^{i\omega t},\\ \ v(x,0)=0, \\ (2) \ v'(x,0)=0, \\(1) \ v(0,t)=0, \\ \ v(L,t)=0. $$

For instance, if you want $(1)$ above to hold, then you want $$u(0,t) - \phi(0)=0$$Where you know that $u(0,t)=0$. This is just an idea. I think you should edit $(2)$ above, which is copied from you, which derivative do you mean?

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  • $\begingroup$ I meant the time derivative. I'm not sure you can just try to find v like you did. The reason is one should find the second form for this problem without previous knowledge about its form. $\endgroup$ – AMAS Feb 28 '14 at 20:43
  • $\begingroup$ I dont understand what you mean, what kind of knowledge do you have when you want to find the second form? $\endgroup$ – user117449 Feb 28 '14 at 20:57
  • $\begingroup$ Nothing is known. Just the first form of the equation and the hint to findo another form with a nonhomogeneous equation with homogeneous boundary. When you wrote "try to determine \phi(x) such that..." you were already using information from the answer, i.e., using information of how the new nonhomogeneus term in the equation is (\delta times exp). $\endgroup$ – AMAS Feb 28 '14 at 21:01

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