Hi I am looking for a complete solution to the pde given below, it is a hyperbolic pde and I specify the initial conditions and boundary conditions (periodic). Thanks for your help.
I show what I do below, but I get stuck and must be making a mistake somewhere. The problem is defined by
\begin{eqnarray} \partial^2_t u(x,t)=\partial^2_x u(x,t)\\ u(0,t)=u(L,t) \\ \partial_x u(0,t)=\partial_x u(L,t)\\ u(x,0)=\sin(\pi x)\\ \partial_t u(x,0)=0 \end{eqnarray} where $x\in[0,L], t\geq 0$. I am looking for an analytical solution, here is my attempt so far (I try separation of variables) \begin{eqnarray} u(x,t)=\chi(x)T(t)\implies \frac{T''(t)}{T}=\frac{\chi''(x)}{\chi}=-\alpha^2 \end{eqnarray} where $-\alpha^2$ is an arbitrary constant, since the function of x and t are equal, they must equal a constant. Now I obtain \begin{eqnarray} T''(t)+\alpha^2 T(t)=0\\ \chi''(x)+\alpha^2 \chi(x)=0 \end{eqnarray} So I realize I have oscillatory solutions \begin{eqnarray} u(x,t)=\chi(x)T(t)=\left[A\cos(\alpha x)+B\sin(\alpha x)\right] \cdot \left[C\cos(\alpha t)+D\sin(\alpha t)\right] \end{eqnarray} where $A,B,C,D$ are to be determined from the initial and boundary conditions. Applying the initial conditions first, I then have \begin{eqnarray} u(x,0)= C [A\cos(\alpha x)+B\sin(\alpha x)]=\sin(\pi x)\\ \implies A=0, \ CB=\alpha =1 \end{eqnarray} Now imposing the other initial condition I get \begin{eqnarray} \partial_t u(x,0)=0= D [B\sin( x)]\implies DB=0 \end{eqnarray} So I have $A=B=D=0$, so then shouldn't $C=0$ also, which then yields $u(x,t)=0$ which is not right. I have not yet imposed periodic boundary conditions. How can I find the solution to this problem? Thanks!
The solutions $\chi$ should satisfy $$ \chi''+\alpha^2\chi = 0\\ \chi(0)=\chi(L) \\ \chi'(0)=\chi'(L). $$ The solutions $\chi(x)=Ae^{i\alpha x}+Be^{-i\alpha x}$ satisfy the periodic conditions iff $\alpha = 2\pi n$ for $n=0,1,2,3,\cdots$. So, $\alpha^2=(2\pi n/L)^2$, and the corresponding solutions are $$ \chi_{n}(x) = e^{2\pi inx/L},\;\;\; n=0,\pm 1,\pm 2,\cdots. $$ The corresponding solutions $T_n(t)$ are $\cos(2n\pi t/L)$ because of the requirement that $T_n'(0)=0$. The general solution is $$ u(x,t) = \sum_{n=-\infty}^{\infty}A_n\cos(2\pi nt/L)e^{2\pi inx/L}. $$ The constants $A_n$ are determined by the initial condition: $$ \sin(\pi x)=u(x,0) = \sum_{n=-\infty}^{\infty}A_n e^{2\pi inx/L},\;\;\; 0 < x < L. $$ Multiplying by $e^{-2\pi inx}$ and integrating over $[0,L]$ gives $$ \int_{0}^{L}\sin(\pi x)e^{-2\pi inx/L}dx = A_n L. $$ The solution: $$ u(x,t) = \sum_{n=-\infty}^{\infty}\left(\frac{1}{L}\int_{0}^{L}\sin(\pi y)e^{-2\pi iny/L}dy\right)e^{2\pi inx/L}\cos(2\pi nt/L). $$