Solve the following periodic problem Sturm-Liouville

Question

Solve the following periodic problem $$u_{t}-u_{xx}=0, \quad -\pi<x<\pi,\quad t>0$$ $$u(-\pi,t)=u(\pi,t), \quad u_{x}(-\pi,t)=u_{x}(\pi,t), \quad t\geq 0 $$ $$u(x,0)= \left\{ \begin{array}{lcc} 1 & -\pi\leq x\leq 0 \\ \\0 & \quad 0 \leq x\leq \pi \\ \end{array} \right. $$ I'm trying to solve this Sturm-Liouville problem. But I haven't gotten a solution yet. How can I solve this problem? I need some help or a way to solve this problem. I would like to know a solution. Any help is appreciated

Answer 1

suppose $u(x,t)=X(x)T(t)$, then \begin{equation} \begin{cases} u_t=X(x)T'(t)\\ u_{xx}=X''(x)T(t) \end{cases} \end{equation}

isolating X' with X and T' with T we get $\dfrac{T'(t)}{T(t)}=\dfrac{X''(x)}{X(x)}= -\lambda$

$X''(x)= -\lambda X(x)$
$X''(x)+\lambda X(x)=0$

Since we know from the theory of ODEs that in this type of equation $X(x)=e^{rx}$, applying it on the equation we get

$r^2= -\lambda$
$r=\pm \sqrt{-\lambda}$
supposing $\lambda=n^2$
$r=\pm in$

$X(x)=b e^{inx} + c e^{-inx}$
$X(x)=b (cos(nx) +i sin(nx))+c(cos(-nx)+i sin(-nx))$
$X(x)=b (cos(nx) +i sin(nx))+c_2os(nx)-i sin(nx))$
$X(x)=(b +c)cos(nx) +i(b-c) sin(nx)$
$X(x)=B cos(nx) +C sin(nx)$

then we get $X_n(x)=B_ncos(nx) +C_n sin(nx)$

For $T$ we have

$\dfrac{d T(t)}{dt}= -n^2 T(t)$
$\dfrac{dT}{T}= -n^2 t$
$ln T(t) = -n^2 t + a$
$T(t)=e^{-n^2 t + a}$
$T(t)=Ae^{ -n^2 t}$

and $T_n(t)=A_ne^{ -n^2 t}$

Then

$u(x,t)= \sum_{n} X_n(x)T_n(t)$
$u(x,t)= \sum_{n} (B_ncos(nx) +C_n sin(nx))(A_ne^{ -n^2 t})$

We now have to apply the initial conditions to get $A_n$, $B_n$ and $C_n$