# One dimensional wave equation with a nonhomogeneous boundary condition

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?

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?