Time dependent parameter for heat equation

Question

:\begin{cases} u_{t}=ku_{xx} & (x, t) \in [0, \infty) \times (0, \infty) \\ u(x,0)=g(x) & IC \\ u(0,t)=0 & BC \end{cases}

What is the solution for $k$ as a function of $t$. I suspect it requires having to integrate $k(t)$ over the time to find the mean value

Answer 1

By an odd extension $g(x):=-g(-x)$ for $x<0$, you can solve the heat equation on the whole real line (which I assume you know how to do?). Then restrict this solution to the half line.

Answer 2

Of course use separation of variables:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=k(t)X''(x)T(t)$

$\dfrac{T'(t)}{k(t)T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2k(t)\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-s^2\int_0^tk(r)~dr}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-s^2\int_0^tk(r)~dr}\sin xs~ds+\int_0^\infty C_2(s)e^{-s^2\int_0^tk(r)~dr}\cos xs~ds$

$u(0,t)=0$ :

$\int_0^\infty C_2(s)e^{-s^2\int_0^tk(r)~dr}~ds=0$

$C_2(s)=0$

$\therefore u(x,t)=\int_0^\infty C_1(s)e^{-s^2\int_0^tk(r)~dr}\sin xs~ds$

$u(x,0)=g(x)$ :

$\int_0^\infty C_1(s)\sin xs~ds=g(x)$

$\mathcal{F}_{s,s\to x}\{C_1(s)\}=g(x)$

$C_1(s)=\mathcal{F}^{-1}_{s,x\to s}\{g(x)\}$

$\therefore u(x,t)=\int_0^\infty\mathcal{F}^{-1}_{s,x\to s}\{g(x)\}e^{-s^2\int_0^tk(r)~dr}\sin xs~ds$