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 $\frac{\partial u}{\partial t} = \frac{\partial }{\partial x}(D \frac{\partial u}{\partial x})$. 


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 $D(t)$. 


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. 

