Solution to a Sturm–Liouville problem

Question

Suppose we want to find a solution to the following Sturm-Liouville problem: \begin{align} &u_t=u_{xx},\quad 0\leq x\leq\pi,\ t>0\\\ &u(x,0)=x\\ &u(0,t)=u(\pi,t)=0 \end{align} I thought about just using separation of variables to compute the series solution, that is, look for solutions of the form: \begin{equation} u(x,t)=X(x)T(t) \end{equation} But i've not been able to quite get the fourier coefficients right, and therefore I can't seem to find the solution. Summarizing, I want to know what is the actual solution to the heat equation.

Answer 1

When you set $u(x,t)=X(x)T(t)$, then you can use separation of variables to solve the resulting equations: $$ \frac{T'(t)}{T(t)} = \lambda = \frac{X''(x)}{X(x)} \\ X(0)=X(\pi)=0. $$ This gives $X_n(x)=\sin(nx)$ for $n=1,2,3,\cdots,$ with corresponding $\lambda_n=-n^2$ and $T_n(t)=e^{-n^2 t}$. The general solution is $$ u(x,t)=\sum_{n=1}^{\infty}C_ne^{-n^2 t}\sin(nx) $$ where the constants $C_n$ are determined by $$ x= u(x,0)=\sum_{n=1}^{\infty}C_n\sin(n x). $$ Therefore, $$ \int_0^{\pi}x\sin(n x)dx = C_n\int_0^{\pi}\sin^2(n x)dx \\ C_n = \frac{\int_0^{\pi}x\sin(n x)dx}{\int_0^{\pi}\sin^2(n x)dx} $$ The full solution is $$ u(x,t)=\sum_{n=1}^{\infty}\frac{\int_0^{\pi}x\sin(n x)dx}{\int_0^{\pi}\sin^2(n x)dx}e^{-n^2 t}\sin(n x). $$ The denominator is easily evaluated: $$ \int_0^{\pi}\sin^2(nx)dx = \frac{1}{2}\int_0^{2\pi}\sin^2(nx)dx \\ = \frac{1}{4}\int_0^{2\pi}\sin^2(nx)+\cos^2(nx) dx = \frac{\pi}{2} $$