# Advection-diffusion with periodic boundary conditions and tilt

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 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 (the constant \mathbf{q} defines the so-called "tilt" of the potential $$U$$). 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 "tilt" potential that generates the constant field $$\mathbf{q}$$ is $$-\mathbf{x}\cdot \mathbf{q}$$. This "tilt" contribution males the total potential $$U-\mathbf{x}\cdot \mathbf{q}$$ not periodic (i.e. it does not satisfy the PBC).

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 $$g$$ is 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.

• What happens when you try a frame transformation, i.e. rewriting the problem in terms of $x' = x - qt$? Jul 24, 2020 at 17:27
• Hi @Daniel, sorry for late reply. Yes I tried but unfortunately it does not seem useful. Jul 26, 2020 at 15:10
• 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? Jul 29, 2020 at 3:34
• 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. Jul 29, 2020 at 10:26
• This doesn’t quite answer the question, but in case it is of interest: I wrote a paper studying systems of this type in some detail (hence my interest in Quillo and @Daniel ‘s discussion). arxiv.org/abs/2108.06431. Theorem 3 gives (small diffusion) asymptotic information about $\rho$ (cf. Remark 4). However the paper is more focused on approximating the flux of $\mathbf{J}$ through suitable hypersurfaces (Theorems 1, 2, 4; Prop. 3). Aug 23, 2021 at 2:36

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.
• Why would it be true that $\nabla \cdot \psi = 0$? Sep 19, 2020 at 6:31
• $\psi = \psi_z \mathbb{e}_z$, and because this is a 2D problem $\frac{\partial}{\partial z} = 0$. Sep 19, 2020 at 14:58
• 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 $\nabla \alpha e^{-V} = R \nabla g$. Sep 19, 2020 at 17:18
• Wait, aren't those equivalent? $$R\nabla g = (-g_y, g_x)$$ while $$\nabla \times (g \mathbb{e}_z) = (-g_y, g_x, 0)$$ (where $g$ is a scalar function and subscripts denote partial derivatives) Sep 19, 2020 at 19:34
• Regarding Zeeman, however, I agree that this seems to contradict. Right now I'm thinking about the $U = 0$ case - shouldn't $\rho = C$ then be a solution? Sep 19, 2020 at 19:46