Consider the diffusion equation,

$\frac{\partial n(x,t)}{\partial t}=D\frac{\partial ^2n(x,t)}{\partial x^2}$,

inside a box from $x=0$ to $x=L$ subject to the boundary conditions

$n(x=0)=0$, $n(x=L)=1$.

One can show that


is a steady-state solution that satisfies the boundary conditions. Since this is a steady-state solution, I wouldn't think there is any average current of particles. Rather if we start the system in this state, it will remain in this state forever, without any need to invoke particle creation or absorption at the boundaries. However, Fick's first law of diffusion says there is a diffusive flux at each of the boundaries, and a current runs from high density to low density over the whole region. Can someone help break this apparent paradox?


Fick's First Law:

It postulates that the flux goes from regions of high concentration to regions of low concentration, with a magnitude that is proportional to the concentration gradient (spatial derivative), or [...] the concept that a solute will move from a region of high concentration to a region of low concentration across a concentration gradient.

So, despite steady state, there is movement of solute from right ($x=L$) to left ($x=0$) and the diffusion flux is:


The only way to stop all diffusion is to specify:


In that steady state case there's no gradient at all and $n(x)=n_0$ where $n_0$ is some average concentration. It could be achieved by 'stoppering' the tube at both ends.

So there's no paradox here.

The situation is comparable to a heat conductive bar clamped at the ends a two different temperatures. At steady state all temperatures are time-indifferent and depend only on the spatial coordinate but there's plenty of heat flowing from the hot end to the cold end.

Incidentally, your initial equation is wrong. It should be:

$$\frac{\partial n(x,t)}{\partial t}=D\frac{\partial^2 n(x,t)}{\partial x^2}$$

In the general case $n$ is a function of both $x$ and $t$. The differential equation then requires partial differentials $\partial$.

Also written in shorthand as:


  • $\begingroup$ Thanks! The conductive bar analogy is very useful. $\endgroup$
    – Ian
    Jun 8 '20 at 2:14
  • $\begingroup$ Yes, the heat equation and the diffusion equation produce very analogous results. Thanks for the upvote! $\endgroup$
    – Gert
    Jun 8 '20 at 5:39

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