I'm trying to find an explicit formula for heat equation in the interval $(0,L)$, with $L>0$. I have the next boundary-initial value problem. \begin{equation} \begin{cases} u_t-u_{xx}=0, &\text{in } (0,L)\times (0,\infty), \\ u(0,t)=u(L,0)=0 & \\ u(x,0)=f \end{cases} \end{equation} with $f$ a smooth function such that $f(0)=f(L)=0$.
I have the next suggestion: Extend a function $f$ to $\mathbb{R}$ such that $f(x)$ and $f(L-x)$ are odd.
My question is: how do I extend this function?
After that, I think that I just have to apply the fundamental solution for Heat equation (Evans PDE book).
Thanks a lot and sorry for my bad english.
The solution is $$ u(x,t) = \sum_{n=1}^{\infty}a_n\sin(n\pi x/L)e^{-n^2\pi^2 t/L^2} $$ where the constants $a_n$ are chosen so that $$ f(x)=u(x,0)=\sum_{n=1}^{\infty}a_n\sin(n\pi x/L). $$ Therefore, $$ \int_0^{L}f(x)\sin(n\pi x/L)dx =a_n\int_0^L\sin^2(n\pi x/L)dx \\ \implies a_n = \frac{\int_{0}^{L}f(x)\sin(n\pi x/L)dx}{\int_0^L\sin^2(n\pi x/L)dx} $$