In the heat equation: $$\partial u(x,t)=D\partial_{xx}u(x,t)$$ the diffusion coefficient $D$ is in general a constant or a given function of $u(x,t)$ in the nonlinear equation. Suppose I have a diffusion coefficient depending on the integral of $u(x,t)$. In this case I have: $$\partial_t u(x,t)=\left[\int_{L}^L u(x,t)dx\right]\partial_{xx}u(x,t)$$ If the $IC$ and $BC$ are: $$u(0,x)=u_0(x)$$ $$u(L,t)=u_L$$ $$\partial_x u(x,t)=f(t)\mid_{x=L}$$ how can I solve this equation? Thanks.
There is a simple way to manage this equation using a Fourier series. We assume a boundary at $0$ and $L$ and that exists the Fourier series for the solution $$ u(x,t)=\sum_{n=\infty}^{\infty}u_n(t)e^{i\frac{2\pi n}{L}x} $$ then you note that $$ D(t)=\int_{L}^L u(s,t)ds=\int_{L}^L\sum_{n=\infty}^{\infty}u_n(t)e^{i\frac{2\pi n}{L}x}=2L u_0(t) $$ and you are left with the following set of ordinary equations $$ \partial_t u_n(t)=4\pi^2n^2u_0(t)u_n(t). $$ This yields $\partial u_0(t)=0$ and so, $u_0=constant=D_0$ and so for $n\ne 0$, $$ u_n(t)=e^{4\pi^2n^2D_0t}u_n(0). $$
Apart from your question, let me say that a heat equation with nonconstant conductivity would rather be $\frac{\partial u}{\partial t} = \frac{\partial }{\partial x}(D \frac{\partial u}{\partial x})$.

$\begingroup$ @Thibaut Demaerel: Yes. It's true. How can this help to solve this equation? $\endgroup$ – Riccardo.Alestra May 15 '13 at 9:31

$\begingroup$ Which equation do you want to solve, yours or the more realistic heat equation given in this answer? They don't produce the same solution in general (given the same IC and BC). $\endgroup$ – Thibaut Demaerel May 15 '13 at 9:45

Well, I think that $$ D(t)=\int_{L}^L u(s,t)ds, $$ thus, both equations are the same ($\partial_x (D(t) \partial_x u)=D(t) \partial_x ^2 u$).
I have some questions:
1) ¿Both BC concern the same point x=L?
2) You will need some hypotheses on the initial data. I mean, positiveness, positive mean or something that ensures that $D(t)>0$ at least for short times $t<\tilde{t}$. If these hypotheses does not exist then you can take an odd initial data and the appropriate BC to have a steady solution. For instance, $u(x,t)=sin(x)$ with $L=\pi$ would be a steady solution corresponding to $u_L=0$ and $f(t)=1$, isn't it?.
Anyway, I think that the usual framework should work as long as you have the appropriate hypotheses to close the problem and the rights estimates for $D(t)$.

$\begingroup$ 1) Both BC concern the same point L 2) Because we are speaking about temperature, all the initial data are definite positive. $\endgroup$ – Riccardo.Alestra May 15 '13 at 10:45

$\begingroup$ Mmm I'm not sure whether you can solve the classical heat equation with both BC at the same point... $\endgroup$ – guacho May 15 '13 at 11:55
I wonder if this can be solved using a FeynmanKac type argument. Let $X_{t,s}(a)$ be a path labeled at the time $t$ (i.e. $X_{t,t}(a)=a$) going back to $s$. Let $X_{t,s}(a)$ satisfy the following stochastic equation: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(X_{t,s}(a),s))}\ \hat{{\rm{d}}} W_s $$ where $W_s$ is a 1dimensional Wiener process and $\hat{{\rm{d}}}$ indicates that this is a backwards Ito differential $\sigma$ is some arbitrary smooth function of $u$. Then, the following is true:
$\textbf{Claim:}$ A function $u(x,t)$ is a solution to the equation $$ \partial_t u = \sigma(u) \triangle u $$ if and only if the pair $(u,X)$ satisfies the following stochastic system: $$ {\rm{d}}X_{t,s}(a) =\sqrt{2\sigma(u(X_{t,s}(a),s))}\ \hat{{\rm{d}}} W_s $$ $$ u(x,t)= \mathbb{E}\left[u_0(X_{t,0}(a))\right] $$ where the expectation is taken over Brownian motions.
Here I am assuming we are solving the heat equation on the real line with no boundary conditions but it is straightforward to incorporate boundaries. Of course, you can choose $\sigma$ (with some technical conditions). I hope this helps.