Context: Consider the advection-diffusion equation with periodic boundary conditions (PBC) over a flat square domain $L \times L$. The scalar density $\rho $ is transported by a prescribed field $\mathbf{v}=-\nabla U$, where $U(\mathbf{x})$ is a scalar potential that has the periodicity imposed by the PBC. The density $\rho$ evolves as

$$ \partial_t \rho(\mathbf{x},t) = -\nabla \cdot [ \mathbf{v}(\mathbf{x}) \rho(\mathbf{x},t) - \nabla \rho(\mathbf{x},t) ] = 0 $$

The steady-state solution is found by imposing $\partial_t \rho(\mathbf{x},t) =0$ and has the usual Gibbs form:

$$ \rho(\mathbf{x}) \, \propto \, e^{-U(\mathbf{x}) } $$

The problem: I am wondering how to find the steady-state the case in a slightly more general case, where

$$\mathbf{v} = -\nabla U + \mathbf{q}$$

The potential $U$ has the periodicity imposed by the PBC and $\mathbf{q} =(q_x,q_y)$ is a constant vector field. Hence, the equation we have to solve is

$$ \nabla \cdot [ \, \rho(x,y) \, \mathbf{q} - \rho(x,y) \nabla U(x,y) - \nabla \rho(x,y) \, ] = 0 $$

with the periodic conditions $\rho(0,y) = \rho(L,y)$, $\rho(x,0) = \rho(x,L)$, $U(0,y) = U(L,y)$, $U(x,0) = U(x,L)$. For simplicity, I tried to consider the case $\mathbf{q}=(q,0)$, but the problem still seems non-trivial.

Question: Any idea or reference about the diffusion-advection equation in periodic boundary conditions (in particular about the steady state)? Which is the "Gibbs-like solution" in this case?

Further considerations: I have the feeling that finding a solution is not easy because the potential that generates the constant field $\mathbf{q}$ is $-\mathbf{x}\cdot \mathbf{q}$, that is not periodic (i.e. it does not satisfy the PBC contitions).

Moreover, define the total current in the steady state as

$$ \mathbf{J}(x,y) = \rho(x,y) \, [\mathbf{q} - \nabla U(x,y)] - \nabla \rho(x,y) \, , $$

so that we have to find the $\mathbf{J}$ such that

$$ \nabla \cdot \mathbf{J} = 0 \quad \Rightarrow \quad \mathbf{J} = R \nabla g $$

where $R$ is a 90-degrees rotation and $g$ an unknown scalar potential. Note that $g$ does not have to respect the PBC, but $\mathbf{J}$ does: (probably) the most general form of $g$ is

$$ g(x,y) = G(x,y) + a x + b y $$

where $G$ respects the PBC and $a$ and $b$ are constants. Even though this problem is more likely to be studied by physicists, I have the feeling that the problem is intimately related to the topology of the 2D torus, so I posted also a similar question on math SE.

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    $\begingroup$ What happens when you try a frame transformation, i.e. rewriting the problem in terms of $x' = x - qt$? $\endgroup$ – Daniel Jul 24 '20 at 17:27
  • $\begingroup$ Hi @Daniel, sorry for late reply. Yes I tried but unfortunately it does not seem useful. $\endgroup$ – Quillo Jul 26 '20 at 15:10
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    $\begingroup$ An observation: If $U$ is constant, no solution exists (unless $q = 0$ also). And I think I can show that solutions "blow up" (have energy pushed to arbitrarily high frequencies) as $\nabla U \rightarrow \mathbf{0}$, if they exist. Have you tried studying the 1D version of this problem? $\endgroup$ – Daniel Jul 29 '20 at 3:34
  • $\begingroup$ Yes, the 1D version is basically the same. You end up with a formal solution $\rho(x) \propto e^{q x -U(x)}$ that cannot be periodic. Maybe I found something interesting here: sciencedirect.com/science/article/pii/… ..it seems the "correct" thing to do is to consider the adjoint version of the equation. $\endgroup$ – Quillo Jul 29 '20 at 10:26
  • $\begingroup$ @ Quillo just updated my answer to finish the proof. $\endgroup$ – Daniel Sep 16 '20 at 14:35

Any $\rho$ which solves the equation on the whole torus must also be a solution locally on every subset. In particular, it must be solution on the (non-toroidal) open $L \times L $ square. Since solutions on the torus are a subset of the solutions on the square, the question becomes: Do there exist solutions on the square which happen to match at the boundaries?

On this square, we can define $V = U - \mathbb{x} \cdot \mathbb{q}$, and we have an ordinary advection-diffusion equation. We know there exist solutions of the form $\alpha e^{-V(\mathbb{x})}$. We also know that $U$ is periodic, so $V$ can only be periodic if $\mathbb{q} = \mathbb{0}$. However $e^{-V}$ could still be periodic if $\mathbb{q}$ is imaginary. Specifically, we have periodic solutions for $\mathbb{q} = \frac{2\pi i}{L}\mathbb{n}, \mathbb{n} \in \mathbb{Z}^2$.

For other $\mathbb{q}$, solutions proportional to $ e^{-V(\mathbb{x})} $ cannot extend to solutions on the whole torus. The remaining question: Are such solutions the whole solution space?

Now, Matthew Kvalheim points to Zeeman, 1988. Theorem 3 reads

Let $U$ be a vector field on a compact manifold $X$ without boundary, and let $\epsilon$ > 0. Then the Fokker-Planck equation for $U$ with $\epsilon$-diffusion has a unique steady state, and all solutions tend to that steady state.

The torus is a compact manifold without boundary, Zeeman's $U$ is our $-\nabla V$, and we have $\epsilon = 1$, so the theorem tells us a solution $\rho$ must exist and is unique (up to an overall scalar). Unfortunately, this proof is not constructive.

In one dimension, variation of parameters gives the solution $$\rho = C_1 e^{-V}\left(C_2 + \int_0^x e^V\right)$$ and the requirement $\rho(0) = \rho(L)$ fixes $C_2$. We can try to extend this to two dimensions as follows: Assume $\rho$ is of the form $\alpha(x)e^{-V}$. Then the equation becomes $$ \nabla \cdot [-\nabla V \alpha(x)e^{-V} - \nabla (\alpha(x) e^{-V})] = 0 $$ which simplifies to $$ \nabla \cdot (\nabla\alpha(x) e^{-V}) = 0 $$ Solutions are $$ \nabla\alpha(x) e^{-V} = \nabla \times \mathbf{\psi} $$ for $\mathbf{\psi} = \mathbf{e}_z $ and $g$ some scalar function. Then $$ \nabla\alpha(x) = e^V(\nabla \times \mathbf{\psi}) $$ If $$ \nabla \times (e^V(\nabla \times \mathbf{\psi})) = 0 $$ then this has solution $$ \alpha(x,y) = C + \left(\int_0^x -e^V g_y dx\right) + \left(\int_0^y e^V g_x dy\right) $$ The requirement of periodic boundary conditions picks out some unique $g$, $C$ up to an overall constant. We need $$ \alpha(x,0) = \alpha(x,L)e^{-Lq_y} $$ or $$ C + \left(\int_0^x -e^V g_y dx\right) = Ce^{-Lq_y} + \left(\int_0^x -e^V g_y dx\right)e^{-Lq_y} + \left(\int_0^L e^V g_x dy\right)e^{-Lq_y} $$ At $x = 0$ this simplifies to $$ C = \frac{1}{e^{Lq_y} - 1}\int_0^L e^V g_x(0,y) dy $$ It remains to find $g$.

I'm not sure that there is a nice expression for the solution in general. Some miscellaneous thoughts:

  • When $U = 0$, $\rho = C$ is a solution, which corresponds to $\alpha = e^V, g = xq_y-yq_x$. This shows that $g$ may be defined only on the square, not on the torus.
  • When $\nabla U \gg q$ or $q \gg \nabla U $, we can start with the nearby known solution and series expand.
  • $\begingroup$ Why would it be true that $\nabla \cdot \psi = 0$? $\endgroup$ – Matthew Kvalheim Sep 19 '20 at 6:31
  • $\begingroup$ $\psi = \psi_z \mathbb{e}_z$, and because this is a 2D problem $\frac{\partial}{\partial z} = 0$. $\endgroup$ – Daniel Sep 19 '20 at 14:58
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    $\begingroup$ I think your conclusion $\alpha =$ const. would imply that there are no solutions on the torus since $V$ is not periodic. But this contradicts the fact that this PDE always has solutions on the torus (see e.g. Theorem 3 of "Stability of Dynamical Systems", Zeeman, 1988). So far I think I spot one error in your reasoning: on $\mathbb{R}^2$ I don't think it is true that $\nabla \cdot \mathbf{J} = 0$ $\implies$ that $\mathbf{J}$ is the curl of something. Rather, as Quillo wrote, it implies $\mathbf{J} = R \nabla g$ for some $g$. Hence I think you should have $\nabla \alpha e^{-V} = R \nabla g$. $\endgroup$ – Matthew Kvalheim Sep 19 '20 at 17:18
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    $\begingroup$ I found the mistake, I had lost a $\nabla \times $ in the third-to-last line. $\endgroup$ – Daniel Sep 19 '20 at 20:31
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    $\begingroup$ Oh, I see now that you're right that the rotation/curl are equivalent - thanks for pointing this out. Yes, I think you're right that $\rho = C$ is a solution in the $U = 0$ case, and by Zeeman's Theorem 3 this is the unique solution up to a constant multiplier as you say. $\endgroup$ – Matthew Kvalheim Sep 19 '20 at 21:26

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