Consider the following temperature problem: $$u_t(x,t)=ku_{xx}(x,t), \;0\leq x \leq \pi,\;\; t,k >0$$ with boundary conditions: $$u_x(0,t)=u(0,t)$$ $$u_x(\pi,t)=u(\pi,t)$$ $$u(x,0)=f(x)$$ I know the function $u$ will be in the form fn an infinite series $$u(x,t)=\sum_{n=0}^\infty c_n y_n(x)T_n(t)$$
But I am quite confused on what the functions $y_n$ and $T_n$ are. I also wish to know how I can compute the coefficients $c_n$ in terms of the function $f$.
I know that I should, to solve this, consider the following Sturm-Liouville problem:
$\text{Given parameters }c > 0 \text{ and } \beta > 0, \text{ let } y=y(x) \text{ for } 0 \leq x \leq c. \text{ We have }$
$$y''+ \lambda y=0$$
$\text{ with boundary conditions }$
$$y'(0)=\beta y(0),$$
$$y'(c)=\beta y(c),$$
I am aware that if $c=\pi, \beta=1$, this SL problem describes the temperature problem above. How is this so? I would love an explanation on how temperature problems are associated with SL problems in general.
As for the functions $y_n$ my best guess is that they are the eigenfunctions associated with the SL problem. If this is the case, since $\lambda=0$ is not an eigenvalue for the SL problem, is the $n=0$ term for $u(x,t)$ simply 0?
Explanation into any of these would be helpful, I don't quite understand how to solve this temperature problem.
This is a problem that can be solved using separation of variables, which is accomplished by looking for separated solutions $u(x,t)=X(x)T(t)$: $$ u_t = ku_{xx} \\ X(x)T'(t)=kX''(x)T(t) \\ \frac{T'(t)}{T(t)}=k \frac{X''(x)}{X(x)} $$ Because the left side depends only on $t$ and the right side depends only on $x$, there must be a separation constant $\lambda$ such that $$ \frac{T'(t)}{T(t)}=-\lambda,\; -\lambda=k \frac{X''(x)}{X(x)}. $$ This leads to solutions $T,X$, in terms of $\lambda,k$ and new constants $A,B,C$: $$ T'(t)=-\lambda T(t),\;\; X''(x)+\frac{\lambda}{k}X(x)=0 \\ T(t)=Ae^{-\lambda t},\\ X(x)=B\sin\left(\sqrt{\frac{\lambda}{k}}x\right)+C\cos\left(\sqrt{\frac{\lambda}{k}}x\right) $$ The boundary conditions $X'(0)=X(0)$ and $X'(\pi)=X(\pi)$ become $$ B\sqrt{\frac{\lambda}{k}}=C,\\ \;\;B\sqrt{\frac{\lambda}{k}}\cos\left(\sqrt{\frac{\lambda}{k}}\pi\right)-C\sqrt{\frac{\lambda}{k}}\sin\left(\sqrt{\frac{\lambda} {k}}\pi\right) \\ = B\sin\left(\sqrt{\frac{\lambda}{k}}\pi\right)+C\cos\left(\sqrt{\frac{\lambda}{k}}\pi\right) $$ The $\cos$ terms in the second equation cancel because of the first equation. Therefore, the eigenvalue equation is the following equation in $\lambda$: $$ B\left(\frac{\lambda}{k}+1\right)\sin\left(\sqrt{\frac{\lambda}{k}}\pi\right)=0, $$ which has solutions $\lambda=-k$ and $\lambda=kn^2$ for $n=0,1,2,\cdots$.