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The diffusion equation with constant diffusion $D$ can be represented as: \begin{equation} \frac{\partial \phi(r, t)}{\partial t}=D \Delta \phi(r, t) \end{equation} where

  • $\Delta$ is the Laplace operator and
  • $\phi(r,t)$ represents a concentration at a point $r\in\mathbb{R}^n$ at time $t$.

When the diffusion is on a network, the Laplacian operator can be discretized and take the form of a matrix representation. The diffusion equation then takes the form: \begin{equation} \frac{d \phi_{i}(t)}{d t}=D \sum_{j} A_{i j}\left(\phi_{j}(t)-\phi_{i}(t)\right) \end{equation} where now

  • $\phi_i(t)$ represents a concentration on the vertex $i$ at time $t$ and
  • $A_{ij}=1$ if there exists an edge between $i$ and $j$.

Consider now the case where the diffusion is not constant but is now a function depending on space and time: $D\to D(r,t)$. The diffusion equation simply is:

\begin{equation} \frac{\partial \phi(r, t)}{\partial t}=\nabla \left[D(r,t) \nabla\phi(r, t)\right] \end{equation}

What happens to the network case now? Writing out the discrete version of the Laplacian gives me: \begin{equation} \frac{d \phi_{i}(t)}{d t}=\sum_{j} A_{i j}D_{i}(t)\left(\phi_{j}(t)-\phi_{i}(t)\right)+"(\nabla D)(\nabla\phi)" \end{equation}

But I have no idea how to discretize $\nabla$ and it feels wrong anyway. Intuitively I would expect something like: \begin{equation} \frac{d \phi_{i}(t)}{d t}=-\phi_{i}(t)+f\left(\sum_{j} A_{i j}\left(\phi_{j}(t)(t)\right)\right) \end{equation} Where $f$ is some function related to $D$ so that we recover the non-linear behaviour of the continuous case.

What am I missing? These notes follow the approach I took, but are limited to constant diffusion. I was not able to find any lecture notes that cover non-linear diffusion on networks.

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1 Answer 1

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$\newcommand{\R}{\mathbb R}$You do not need to "discretize $\nabla$". Also, you wrote the diffusion equation incorrectly. The correct version is this: \begin{equation} \frac{\partial f(r,t)}{\partial t}=\nabla\cdot[B(r,t)\,\nabla f(r,t)], \end{equation} where $f:=\phi$, $B:=D$, and $\cdot$ denotes the dot product. In the coordinate form, this equations is \begin{equation} \frac{\partial f(r,t)}{\partial t}=\sum_{j=1}^n [B(r,t)\,(D_j^2 f)(r,t)+(D_j B)(r,t)\, (D_j f)(r,t)], \end{equation} where $D_j$ is the operator of the partial differentiation with respect to the $j$th coordinate of $r\in\R^n$.

Now discretization becomes straightforward, by replacing the partial derivatives by the corresponding differences: \begin{equation} \frac{df_i(r,t)}{dt}= \sum_j A_{i,j}[B_i(t)\,(f_j(t)-f_i(t))+ (B_j(t)-B_i(t))(f_j(t)-f_i(t))] \end{equation} or, simply, \begin{equation} \frac{df_i(r,t)}{dt}= \sum_j A_{i,j}B_j(t)(f_j(t)-f_i(t)). \end{equation}

More generally, we can write \begin{equation} \frac{df_i(r,t)}{dt}= \sum_j A_{i,j}B_{i,j}(t)(f_j(t)-f_i(t)), \end{equation} where the $B_{i,j}$'s are nonnegative functions. This will describe a general continuous-time random walk on the network. One may want to recall at this point that the diffusion equation describes an approximation of jump processes (which are continuous-time random walks) by processes continuous in space.

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