Heat integro - differential equation 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.
 A: 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 $n\ne 0$,
$$
  u_n(t)=e^{-4\pi^2n^2D_0t}u_n(0).
$$
A: Apart from your question, let me say that a heat equation with non-constant conductivity would rather be
$\frac{\partial u}{\partial t} = \frac{\partial }{\partial x}(D \frac{\partial u}{\partial x})$.
A: 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 $D(t)$.
A: I wonder if this can be solved using a Feynman-Kac 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 1-dimensional 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 straight-forward to incorporate boundaries. Of course, you can choose $\sigma$ (with some technical conditions).  I hope this helps.
