How to solve the heat equation with time-dependent coefficient on a finite interval?

Question

I am trying to solve the following PDE for $x\in[0,L]$:

$$\partial_t h(x,t) = \partial_x^2 h(x,t) + f(t) \partial_x h(x,t) $$

with boundary conditions

$$ h(0,t) = 0 = h(L,t) $$

The initial condition $h(x,0)$ is arbitrary.

I know how to solve this on the real line (a transformation $z\equiv x + \int f(t)$ recovers the standard heat equation in $z$ and $t$) but not on the finite interval where I need to respect the Dirichlet boundary conditions. Is there a general solution for arbitrary $f(t)$?

Edit 1: removed comment that I cared mainly about the limit $h(x,\infty)$

Answer 1

If you don't care about solving it exactly, and only care about limiting behavior, you can carry on in the spirit of this question and define $$E(t)=\frac{1}{2}\int_0^L {u(t,x)}^2\mathrm dx$$ And find $$\dot{E}(t)=\int_{0}^Lu(t,x)\partial_tu(t,x)\mathrm dx \\ =\int_0^Lu(t,x)\big(\partial_x^2 u(t,x)+f(t)\partial_xu(t,x)\big)\mathrm dx$$ Using integration by parts on the first term and straightforward integration of a derivative on the second, we get $$\dot{E}(t)=\big(u(t,x)\partial_xu(t,x)\big)\bigg|_{x=0}^{x=L}-\int_{0}^L{\partial_xu(t,x)}^2\mathrm dx+f(t)\big(u(t,x)\big)\bigg|^{x=L}_{x=0}$$ Using the boundary conditions $u(t,0)=0~,~u(t,L)=0$, this gives us $$\dot{E}(t)=-\int_{0}^L{\partial_xu(t,x)}^2\mathrm dx\leq 0$$ As detailed in the answer of my other question, a mixture of Poincaré's inequality and Grönwall's inequality implies that $E\to 0$ as $t\to\infty$, in other words, the thermal energy inside the rod goes to zero in the long-time limit. In the case of the classical heat equation $f\equiv 0$, this necessarily implies that $u(t,x)\to 0$ as $t\to\infty$ for all $x\in(0,L)$, not just the 2-norm $\Vert u(t,\cdot)\Vert_2\to 0 $ as $t\to\infty$. (I should add that showing this is not trivial, hence the length of the discussion on the other question.)

In this case showing the convergence to $0$ of the 2-norm is not difficult (as I have just demonstrated) but showing the stronger pointwise convergence (i.e $u(t,x)\to 0$ as $t\to\infty$ for all $x\in(0,L)$) would essentially require one to repeat the detailed analysis of my linked answer in the case $f\neq 0$. I suspect that one probably requires certain "nice" properties of $f$, though I don't know what those properties are yet - probably boundedness and a few other things, I don't know.

Hope this helps, for now.

Answer 2

Your boundary conditions imply that for each fixed $t$, the function $h(x,t)$ can be extended as a periodic function of $x$ with period $L$, e.g. $L=2\pi$. Consider then the expansion of $h(x,t)$ as a Fourier series in $x$, whose coefficients depend on $t$.

Write $h(x,t)=\sum_k c_k(t) e^{i k x}$.

and insert into the heat equation, which leads to the requirement $\dot c_k= (-k^2 + i k f(t)) c_k$. Thus set $c_k(t)= c_k(0) e^{- k^2t + i k F(t)}$ where $F(t)$ is the antiderivative of $f(t)$.

This produces a formal periodic solution to the heat equation that can be matched to the initial datum $h(x,0)=f(x)= \sum_k c_k(0) e^{i kx}$.

Now let's try to satisfy the boundary conditions.

Consider the time-varying left-endpoint value $C(t)= h(0,t)= \sum_k c_k(t)$.

Note that the original heat equation has homogeneous solutions of the type $h(x,t)= C=$ constant in $x$ AND $t$. Since you have the freedom to subtract such expressions you can modify the periodic function constructed above so that its endpoint values at $x=0, 2\pi$ are both 0 at time $t=0$. Unfortunately $h(0,t)=C(t)$ will generally not vanish at later times $t$. So I think your boundary conditions are generally impossible to satisfy.