Solution to Anisotropic Heat Equation I am trying to find the solution to a 1-D anisotropic heat equation. The domain is a line segment of length L (i.e., it's a line segment extending from $x = 0$ to $x = L$).
The form of the equation is:
$ \frac{\partial \phi(x,t)}{\partial t} = \frac{\partial f(x)}{\partial x} \frac{\partial \phi(x,t)}{\partial x}  $
where $\phi(x,t)$ is the solution to the equation and $f(x)$ is a function which we incorporate into the equation to make it anisotropic. In this problem, the boundary values are $\phi(0,t) = 0$ and $\phi(L,t) = T$ (T is some arbitrary value), and we also have the constraint that $0 < f(x) < 1$. Once, we solve for $\phi(x,t)$, we take $lim_{t \to \infty}$ (i.e., we find the steady state value).
Can someone explain step-by-step how to derive the solution to this equation? I don't have a strong background in solving PDEs, so I would appreciate any help. Thank you.
 A: To find the limit as $t \to \infty$, it is equivalent to find the steady state solution.  So
$$ \frac\partial{\partial x}\left(f(x)\frac{\partial\phi_\infty}{\partial x}\right) = 0,$$
which implies
$$ f(x)\frac{\partial\phi_\infty}{\partial x} = C, $$
where $C$ is a constant to be determined, which implies
$$ \frac{\partial\phi_\infty}{\partial x} = \frac C{f(x)}, $$
which implies
$$ \phi_\infty(x) = D + C \int_0^x \frac{dy}{f(y)} ,$$
where $D$ is another constant.  Now put in the boundary conditions, to obtain $D = 0$ and $C = T \left(\int_0^L \frac{dy}{f(y)} \right)^{-1}$,
and we get
$$ \phi_\infty(x) = T \frac{\int_0^x \frac{dy}{f(y)}}{\int_0^L \frac{dy}{f(y)}} .$$
Now to "prove" that $\phi(t,x) \to \phi_\infty(x)$ as $t \to \infty$, we apply the following standard argument.  (I put "prove" in quotes because this proof isn't totally rigorous.  To make it totally rigorous you can use a so called "maximal principle," but that is too much to explain here.)
Consider $\psi(t,x) = \phi(t,x) - \phi_\infty(x)$.  This satisfies the equation
$$ \frac{\partial\psi}{\partial t} = \frac\partial{\partial x}\left(f(x)\frac{\partial\psi}{\partial x}\right) ,\tag 1$$
with $\psi(t,0) = \psi(t,L) = 0$.  Multiply both sides of equation (1) by $\psi(t,x)$, and then integrate with respect to $x$ from $x=0$ to $x=L$, and then integrate by parts to obtain
$$ \frac12 \frac\partial{\partial t}\int_0^L |\psi(t,y)|^2 \, dy = - \int_0^L f(y) |\psi(t,y)|^2 \, dy .$$
Write $N(t) = \int_0^L |\psi(t,y)|^2 \, dy$, and $F = \min_{0 \le y \le L} f(y)$.  Then
$$ \int_0^L f(y) |\psi(t,y)|^2 \, dy \ge F N(t) ,$$
that is,
$$ \frac12 \frac{\partial N(t)}{\partial t} \le -F N(t) .$$
Since $N(t) \ge 0$, we see that $N(t) \le N(0) e^{-2Ft}$, that is, $N(t) \to 0$.
The non-rigorous part is to conclude, from $N(t) \to 0$, that $\psi(t,x) \to 0$.  But this at least plausible.
