Entropy Dissipation Semi-Discretization Schemes for Fokker–Planck Equations

Abstract

We propose a new semi-discretization scheme to approximate nonlinear Fokker–Planck equations, by exploiting the gradient flow structures with respect to the 2-Wasserstein metric in the space of probability densities. We discretize the underlying state by a finite graph and define a discrete 2-Wasserstein metric in the discrete probability space. Based on such metric, we introduce a gradient flow of the discrete free energy as semi discretization scheme. We prove that the scheme maintains dissipativity of the free energy and converges to a discrete Gibbs measure at exponential dissipation rate. We exhibit these properties on several numerical examples.

Appendix

Generally, to obtain $\lambda _{\mathcal {F}}(\rho )$ in Definition 6 is not easy. Below, we give simple 1-d model example to illustrate situations in which $\lambda _{\mathcal {F}}(\rho )$ can be explicitly obtained, and its dependence on the graph structure (the boundary conditions of the PDE).

A 1-d model problem Suppose that the free energy contains only the linear entropy term, so that the gradient flow is the heat equation:

$$\begin{aligned} \frac{\partial \rho }{\partial t}=\Delta \rho , \quad x\in (a,b). \end{aligned}$$

(16)

Here, we consider either (i) Neumann boundary conditions (zero flux) $\frac{\partial \rho }{\partial x}|_{x=a}=\frac{\partial \rho }{\partial x}|_{x=b}=0$, or (ii) periodic boundary conditions $\rho (t,a)=\rho (t,b)$.

We approximate the solution of (16) by (3), with a uniform discretization $\Delta x=\frac{b-a}{n-1}$:

$$\begin{aligned} \frac{d\rho _i}{dt}=\frac{1}{\Delta x^2}\left\{ \sum _{j\in N(i)}\rho _j(\log \rho _j-\log \rho _i)_+ -\sum _{j\in N(i)}\rho _i(\log \rho _i-\log \rho _j)_+\right\} . \end{aligned}$$

(17)

The above two types of boundary conditions lead to distinct graph structures.

(i) A lattice graph $L_{n}$:

(ii) A cycle graph $C_n$:

In both cases, (17) is the gradient flow of the discrete linear entropy

$$\begin{aligned} \mathcal {H}(\rho )=\sum _{i=1}^n\rho _i\log \rho _i, \end{aligned}$$

and the unique Gibbs measure is $\rho ^{\infty }=(\frac{1}{n}, \ldots , \frac{1}{n})$. We are going to estimate how fast the solution $\rho (t)$ of the semi-discretization scheme (17) converges to the equilibrium $\rho ^{\infty }$.

As we have seen in Theorem 7, the asymptotic convergence rates are determined by $\lambda _{\mathcal {F}}(\rho )$:

$$\begin{aligned} \begin{aligned}&\lambda _{\mathcal {H}}(\rho ^{\infty })=\min _{\Phi \in \mathbb {R}^n} \left\{ \frac{1}{\Delta x^4}\sum _{(i,j)\in E}\sum _{(k,l)\in E}h_{ij, kl}(\Phi _i-\Phi _j)_+(\Phi _k-\Phi _l)_+~:\right. \\&\quad \left. \sum _{(i,j)\in E}\left( \frac{\Phi _i-\Phi _j}{\Delta x}\right) ^2_+\rho _i=1\right\} , \end{aligned} \end{aligned}$$

(18)

where

$$\begin{aligned} h_{ij, kl}=f_{ik}+f_{jl}-f_{il}-f_{jk}, \quad \text {and} \quad f_{ij}(\rho ^\infty )=\frac{\partial ^2}{\partial \rho _i\partial \rho _j} \mathcal {H}(\rho )|_{\rho =\rho ^{\infty }}={\left\{ \begin{array}{ll} \frac{1}{\rho _i^{\infty }} \quad &{}\text {if}~ i=j;\\ 0\quad &{}\text {if}~i\ne j.\quad \end{array}\right. } \end{aligned}$$

For the present model, we can find exact values of (18) for the above two graphs.

Theorem 13

We have

$$\begin{aligned} \begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })=\frac{\pi ^2}{(b-a)^2}+o(1), \quad (L_n) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })=\frac{4\pi ^2}{(b-a)^2}+o(1).\quad (C_n) \end{aligned} \end{aligned}$$

Proof

First, consider the lattice graph $L_n$. Without loss of generality, let $(\Phi _i)_{i=1}^n$ in (18) satisfy the relation

$$\begin{aligned} \Phi _1\ge \Phi _2 \ge \cdots \ge \Phi _n. \end{aligned}$$

(19)

Denote $\xi :=(\xi _i)_{i=1}^{n-1}\in \mathbb {R}^{n-1}_+$ by

$$\begin{aligned} \xi _i:=\frac{\Phi _{i+1}-\Phi _{i}}{\sqrt{n}\Delta x}, \quad 1\le i\le n, \end{aligned}$$

(20)

and substitute $\rho ^{\infty }$ into (18), to obtain

$$\begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })= \min _{\xi \in \mathbb {R}^{n-1}_+}\left\{ \frac{1}{\Delta x^2}\xi ^TA\xi \quad :\quad \xi ^T\xi =1\right\} , \end{aligned}$$

where

$$\begin{aligned} A=\begin{pmatrix} 2 &{} -\,1 &{} &{} \\ -\,1 &{} 2 &{} -\,1 &{} \\ \\ &{}\ddots &{} \ddots &{}\ddots &{}\\ \\ &{} &{} -\,1&{} 2 &{} -\,1\\ &{} &{} &{}-\,1 &{} 2 \end{pmatrix}\in \mathbb {R}^{(n-1)\times (n-1)}. \end{aligned}$$

It is simple to observe that A is positive definite and that¹

$$\begin{aligned} \begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })=&\frac{1}{\Delta x^2}\times (\text {the smallest eigenvalue of}~ A)=\frac{1}{\frac{(b-a)^2}{(n-1)^2}}\left[ 2-2\cos \left( \frac{\pi }{n-1}\right) \right] \\ =&\frac{\pi ^2}{(b-a)^2}+o(1). \end{aligned} \end{aligned}$$

Next, we analyze the convergence rate for the cycle graph $C_n$. Again we assume the relation (19) and let $\xi$ as in (20). Since $C_n$ has one more edge than $L_n$, we let $\eta \in \mathbb {R}$:

$$\begin{aligned} \eta :=\frac{\Phi _{1}-\Phi _{n}}{\sqrt{n}\Delta x}=\sum _{i=1}^{n-1} \xi _i. \end{aligned}$$

Substituting $\rho ^{\infty }$ into (18), we have

$$\begin{aligned} \begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })=&\min _{(\xi , \eta )\in \mathbb {R}^{n}_+} \left\{ \frac{1}{\Delta x^2} [\xi ^TA\xi +2\xi _1\eta +2\xi _{n-1}\eta +2\eta ^2]\, :\right. \\&\quad \left. \xi ^T\xi +\eta ^2=1, ~\eta =\sum _{i=1}^{n-1}\xi _i\right\} . \end{aligned} \end{aligned}$$

(21)

The following transformations reduce (21) to a simpler eigenvalue problem. Let

$$\begin{aligned} \begin{pmatrix}\xi \\ \eta \end{pmatrix}=P\xi ,\quad \text {where}\quad P=\begin{pmatrix} I \\ \mathbf 1 \end{pmatrix}\in \mathbb {R}^{n\times (n-1)} \end{aligned}$$

with the identity matrix $I\in \mathbb {R}^{(n-1)\times (n-1)}$ and $\mathbf 1 \in \mathbb {R}^{n-1}$ being the vector of all 1’s. Then, (21) becomes

$$\begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })= \min _{\xi \in \mathbb {R}^{n-1}_+}\left\{ \frac{1}{\Delta x^2}(P\xi )^T B (P\xi ){:}(P\xi )^T(P\xi )=1\right\} , \end{aligned}$$

(22)

where

$$\begin{aligned} B=\begin{pmatrix} A &{} b^{\text {T}} \\ b &{} 2 \end{pmatrix}\in \mathbb {R}^{n\times n} \quad \text {with}\quad b^T\in \mathbb {R}^{n-1},\ b=(1,0,\ldots , 0, 1), \end{aligned}$$

and A is as above.

Below, we compute (22). First, we give explicit formulas for the eigenvalues and eigenvectors of B. $\square$

Lemma 14

Let $n\ge 3$. For each $k=0,1,\ldots , n-1$, the eigenvalues of B are

$$\begin{aligned} \lambda _k=2- 2\cos \left( \frac{2k\pi }{n}\right) . \end{aligned}$$

For $k=0, 1,\ldots , n-1$, the associated eigenvectors in un-normalized form are:

$$\begin{aligned} v_k=(v_k(j))_{j=1}^{n}, \quad w_k=(w_k(j))_{j=1}^{n}, \end{aligned}$$

where, for $j=1,\ldots , n-1$,

$$\begin{aligned} v_k(j)=\sin \left( \frac{2\pi k j}{n}\right) ,\quad w_k(j)=\cos \left( \frac{2\pi k j}{n}\right) ; \end{aligned}$$

and when $j=n$,

$$\begin{aligned} v_k(n)=-\sin \left( \frac{2\pi k j}{n}\right) ,\quad w_k(j)=-\cos \left( \frac{2\pi k j}{n}\right) . \end{aligned}$$

Proof

The proof is by direct computation. We just show the details for the case of $j=1$. We have

$$\begin{aligned} \begin{aligned} (Bv_k)(1)&=2v_k(1)-v_{k}(2)+v_{k}(n)\\&=2\sin \left( \frac{2\pi k }{n}\right) -\sin \left( \frac{2\cdot 2\pi k}{n}\right) -0\quad \text {By double angle formula}\\&=\left( 1-2\cos \frac{2k\pi }{n}\right) v_k(1). \end{aligned} \end{aligned}$$

And

$$\begin{aligned} \begin{aligned} (Bw_k)(1)&= 2w_k(1)-w_{k}(2)+w_{k}(n)\\&=2\cos \left( \frac{2\pi k }{n}\right) -\cos \left( \frac{2\cdot 2\pi k}{n}\right) +1\quad \text {By double angle formula}\\&=\left( 1-2\cos \frac{2k\pi }{n}\right) w_k(1). \end{aligned} \end{aligned}$$

$\square$

Note that in Lemma 14, many eigenvalues are repeated. As a consequence, obviously there are only two eigenvectors associated to each repeated eigenvalues, and not four; the repeating eigenvalues, in fact, have identical pairs $v_k$, $w_k$, up to sign. However, the eigenvalue equal to 0 is simple, with associated eigenvector $w_0=(1,\ldots , 1, -\,1)^T$. Moreover, aside from this 0 eigenvalue, all other eigenvalues are positive.

Now, observe that $P^Tw_0=0$, and therefore the matrix $V=[w_0,P]$ is invertible and

$$\begin{aligned} BV=V\begin{bmatrix} 0&0 \\ 0&C\end{bmatrix}, \end{aligned}$$

where $C\in \mathbb {R}^{n-1,n-1}$. Further, notice that $P^TP$ is positive definite and thus it has a unique positive definite square root $(P^TP)^{1/2}$. Thus, $\xi ^TP^TBP\xi$, subject to $(P\xi )^TP\xi =1$, can be rewritten as

$$\begin{aligned} \xi ^TP^TBP\xi =\xi ^TP^TPC\xi =\xi ^T(P^TP)^{1/2}(P^TP)^{1/2}C(P^TP)^{-1/2}(P^TP)^{1/2}\xi \end{aligned}$$

and thus, with $x=(P^TP)^{1/2}\xi$, we end up with the problem

$$\begin{aligned} \min _{x: \ x^Tx=1} x^T\left[ (P^TP)^{1/2}C(P^TP)^{-1/2}\right] x\,. \end{aligned}$$

Finally, we notice that the matrix $\left[ (P^TP)^{1/2}C(P^TP)^{-1/2}\right]$ is symmetric, and it is obviously similar to C, so that indeed

$$\begin{aligned} \begin{aligned}&\min _{x: \ x^Tx=1} x^T\left[ (P^TP)^{1/2}C(P^TP)^{-1/2}\right] x \\&\quad =\min _{\xi \in \mathbb {R}^{n-1}_+}\left\{ (P\xi )^T B (P\xi )~:~ (P\xi )^T(P\xi )=1\right\} =\text {The second smallest eigenvalue of}~ B. \end{aligned} \end{aligned}$$

(23)

Putting it all together, (22) gives

$$\begin{aligned} \begin{aligned} \lambda _{\mathcal {H}}(\rho ^{\infty })=&\frac{1}{\Delta x^2}(\text {the second smallest eigenvalue of}~ B)\\ =&\frac{1}{\frac{(b-a)^2}{(n-1)^2}}\left[ 2-2\cos \left( \frac{2\pi }{n}\right) \right] =\frac{4\pi ^2}{(b-a)^2}+o(1), \end{aligned} \end{aligned}$$

and the proof of Theorem 13 is completed. $\square$