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So this has been bothering me for a little while. When you consider scalar diffusion like $\frac{dc}{dt} = D\nabla^2 c$, where $c=c(x,y)$, most people would say that the scalar will move downhill. Now it is certainly true that the scalar is going to move from regions of high concentration to regions of low concentration. However, most of the time it is said that the diffusion is in the direction of the gradient. (In other words, the level curves move perpendicular to themselves and not at a slight angle)

This assumption is mentioned off-hand in a lot of places like: http://www.math.umn.edu/~olver/ln_/vc2.pdf (see paragraphs near Proposition 5.3). I however cannot find a proof that it is always true for any scalar concentration field.

This seems like such a simple thing, but the more I look into it the more I don't know if it's true. When thinking about this, don't just assume a circularly symmetric distribution, consider something like a wedge shape for instance. Shouldn't the point of the wedge grow faster down the opening than along it, thus bending its angle away from the gradient?

I'd appreciate an explanation why diffusion follows the gradient, or a counter-example.

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From a global perspective we can think of the diffusion equation as the minimization problem $$ \frac{dc}{dt} = - \frac{\delta \mathcal{F}}{\delta c} $$ Where we have $ \mathcal{F} = \int \frac{1}{2}|\nabla c |^{2} dV $. If nothing else this at least shows that the diffusion equation is driven by minimizing the gradient of the field.

From a local perspective we can define a normal coordinate($n$) and a coordinate along the level curve$(s)$ of $c$ and expand the laplacian as $$ \nabla^{2} c = \partial_{nn} c + \frac{\kappa \partial_{n} c }{1 + n\kappa} + \frac{\partial_{ss}c}{(1 + n \kappa)^{2}} - \frac{n\partial_{s}\kappa\partial_{s}c }{(1+n\kappa)^{3}} $$ Where $\kappa$ is the curvature of the curve. Since we expanded along a level curve of $c$ the partial derivative along the curve vanish and we find that the evolution of the curve is only dependent on the normal coordinate and the curvature of the curve.

Unfortunately I am not entirely sure this answers your question.

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  • $\begingroup$ Just out of curiosity, how did you get to that minimization problem..? $\endgroup$ – JalfredP Nov 5 '16 at 15:14
  • $\begingroup$ What do you mean? Just take the variational derivative and you get the diffusion equation. Formulations in terms of the variation of a "free energy" functional is quite usual when you motivate the Allen-Cahn or Cahn-Hilliard equation ( or for that matter more exotice phase field equations). $\endgroup$ – Strum Nov 5 '16 at 15:30
  • $\begingroup$ I am just not used to it. If you have some reference I'd like to see it (: $\endgroup$ – JalfredP Nov 5 '16 at 15:37
  • $\begingroup$ I suppose you can view it as (one of) the simplest possible limit of the traditional phase field equations given here en.wikipedia.org/wiki/… $\endgroup$ – Strum Nov 5 '16 at 15:51
  • $\begingroup$ I will look into where you got that from, and I appreciate the recognition of curvature dependence, but I believe this may still be short of the real proof. It only shows that diffusion tries to minimize the gradient of the field. However, I'm more concerned with what path it takes to minimize the gradient of the field. Could it conceivably have level curves grow in a direction non-perpendicular to the level curves at any moment. $\endgroup$ – tnevins Dec 5 '16 at 14:15
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The diffusion equation

$${dc\over dt} = D\Delta c$$

is the result of two equations:

1- continuity equation (conservation of the particles) $${dc\over dt}=-\nabla\cdot\textbf{j}$$ 2 - where $\textbf{j}$ is the flux and is defined as $\textbf{j} =-D\nabla c$ so that combining the two you get the diffusion equation.

As you can see from (2), the flux of particles is anti-parallel to the gradient of $c$ so that particles really move uphill.

That has a mainly a statistical reason: less particles comes from where the concentration is low with respect to where it is high, so you have a net flux towards smaller concentration.

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  • $\begingroup$ Considering that this is working from the definition it seems pretty straightforward. I think there is a slight nuance that makes me not sure if this is a complete proof though. While the flux vector is certainly anti-parallel to the level curves, I am not just concerned with what direction the particles move. Rather I am concerned with how the overall field moves (how the level curves move). When two areas of the level curve can diffuse to the same place (think an initial wedge of particles), can this local interpretation work? This is part of the picture I think though. $\endgroup$ – tnevins Dec 5 '16 at 14:11

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