2
$\begingroup$

On the one hand, we have the diffusion equation: \begin{align} \frac{\partial\rho}{\partial t}&=D \nabla^2 \rho \end{align} and on the other hand, we have Fick's first law: \begin{align} \vec J = - D \nabla \rho \, . \end{align} If we apply $\nabla$ to Fick's law: \begin{align} \nabla \vec J = - D \nabla^2 \rho \end{align} and insert this into the diffusion equation, we find \begin{align} \frac{\partial\rho}{\partial t}&=\nabla \vec J \, . \end{align} If we now assume that the current $\vec J$ can be described in terms of a velocity field $\vec u$: $$ \vec J \equiv \rho \vec u,$$ this yields exactly the continuity equation: \begin{align} \frac{\partial\rho}{\partial t}&=D \nabla (\rho \vec u) \, . \end{align} Is there any error in the steps above? I'm somewhat puzzled by the result because the continuity equation is typically associated with advection and not with diffusion.

$\endgroup$
1
  • $\begingroup$ Shouldn't your 3rd equation be applying a divergence, not a gradient? Then the 4th equation should be a negative on the RHS making it the continuity equation (i.e., time variation of a density plus the divergence of a flux equals zero in the absence of sources or sinks). $\endgroup$ – honeste_vivere Mar 12 at 23:22
2
$\begingroup$

Yes, indeed, Diffusion equation is essentially a continuity equation. More general Fokker-PLanck type equation (i.e. a diffusion equation with a drift term), $$\partial_t \rho(\mathbf{r},t) = \nabla \cdot[\mathbf{f}(\mathbf{r})\rho(\mathbf{r},t)] + D\nabla^2\rho(\mathbf{r},t)$$ can be written as a continuity equation $$\partial_t \rho(\mathbf{r},t) = -\nabla\mathbf{J}(\mathbf{r},t),$$ where the current is defined as $$\mathbf{J}(\mathbf{r},t) = -\mathbf{f}(\mathbf{r})\rho(\mathbf{r},t) - D\nabla\rho(\mathbf{r},t).$$ Thus, converting a diffusion-like equation to a continuity equation is a qquestion of correctly defining the current.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.