# Constructing a coupled Fokker-Planck equation

Consider the $$d$$-dimensional FPE with constant diffusion coefficient, $$\partial_t \rho = -\sum_{i=1}^d \partial_{\theta_i} (b_i(\theta) \rho) + \sum_{i,j}\partial_{\theta_i}\partial_{\theta_j}\rho \;\;\;\;\;\;\;\;\;\; (1)$$ which describes the evolving probability density $$\rho(\theta,t)$$ of a diffusing particle $$\theta_t$$ that follows the SDE $$d\theta_t = b(\theta_t)dt + dW_t.$$

where the drift is $$b(\theta) = (b_1(\theta), \dots, b_d(\theta))$$.

Now, I'm wondering how we can reverse-engineer a coupled equation of this form from $$d$$ one-dimensional equations having dependent drifts, and then combine those as the marginals of $$\rho$$?

Suppose in each dimension $$i=1,\dots,d$$ we have the one-dimensional FPE where the drift in that dimension is the $$i$$th component of the drift, $$b_i(\theta)$$ (note that $$b_i$$ depends on the $$d$$-dimensional $$\theta$$). $$\partial_t \rho_i = -\partial_{\theta_i}(b_i(\theta) \rho) + \partial^2_{\theta_i} \rho$$

If we assume independent marginals (i.e. $$\rho = \prod_i \rho_i(\theta_i)$$), the diffusion term $$\sum_{i,j}\partial_{\theta_i}\partial_{\theta_j} \rho$$ becomes $$\sum_i \partial_{\theta_i}^2 \rho = \Delta \rho$$ which seems fine, but the drift term seems to need some kind of product/chain rule for multiple dimensions?

If these equations can't be combined into the form (1), then this would imply that $$d$$ 1-dimensional FPEs with drifts in the same variable do not necessarily combine to one coupled $$d$$-dimensional FPE. Is this true?

Edit: the work I've done Let $$\vec{\rho} = (\rho_1(\theta_1), \dots, \rho_d(\theta_d))$$. Let $$\partial_i$$ denote $$\partial_{\theta_i}$$, and consider equation (1).

LHS: $$\partial_t \vec{\rho} = (\partial_t \rho_1, \dots, \partial_t\rho_d)$$. Fine.

RHS term 2: $$\sum_{i,j} \partial_{i,j} \rho = \sum_i \partial_i^2 \rho_i$$ since $$\partial_{i,j} \rho = \partial_i (\partial_j \rho_j(\theta_j)) = 0$$. Fine.

RHS term 1: \begin{align} -\sum_i \partial_i (b_i(\theta) \vec{\rho}) &= -\sum_i \Big(\partial_i b_i(\theta) \vec{\rho} - b_i(\theta) \partial_i \vec{\rho}\Big) \\ &=-(\rho_1, \dots, \rho_d) \Big( \sum_i \partial_i b_i(\theta) \Big) - (b_1(\theta) \partial_1\rho_1, \dots, b_d(\theta)\partial_d\rho_d) \end{align} now, the trouble comes with the term $$\sum_i \partial_i b_i(\theta)$$. Since $$b_i(\theta) \in \mathbb{R}$$, each $$\partial_i b_i(\theta)$$ is a constant. In the 1-dimensional FPE, each term should show up as $$\rho_i\partial_i b_i(\theta)$$ rather than $$\rho_i\sum_i\partial_i b_i(\theta)$$. Am I doing something wrong?

• I think what you mean is $b(X) = (b_{1}(X_{1},\dots,b_{d}(X_{d}))$; $\theta$ is a scalar, no? Also, what do you mean by "since $b_{i}(\theta) \in \mathbb{R}$, each $\partial_{i} b_{i}(\theta)$ is a constant"?
– user711689
Commented Jul 18, 2021 at 3:00

Okay, so it seems that my mistake was to consider each component of the $$d$$-dimensional equation, i.e. treating $$\rho$$ as $$\rho(\theta) = (\rho_1, \dots, \rho_d)$$ rather than treating it as $$\rho = \prod_{i=1}^d \rho_i$$. It turns out the 1D and $$d$$D coupled are equivalent. However I'm still not sure if this just happens to work out in this case, or if it's possible that we have $$d$$ 1D FPEs that don't work out to be a $$d$$D coupled FPE.
Anyways, recall that the 1D equation is $$\partial_t \rho_i = -\rho_i \partial_i b_i(\theta) - b_i \partial_i \rho_i + \partial_{\theta_i}^2 \rho_i$$
while the $$d$$D equation is \begin{align*} \partial_t \rho &= -\sum_{i=1}^d \partial_{\theta_i} (b_i(\theta) \rho) + \Delta \rho \\ &= -\sum_i \left(\partial_i b_i(\theta) \rho + b_i(\theta) \partial_i \rho \right) + \Delta \rho \\ &= -\rho \sum_i \partial_i b_i(\theta) - \sum_i b_i(\theta) \partial_i \rho + \sum_i \partial_i^2 \rho \end{align*}
Multiplying the 1D equation on both sides by $$\prod_{j\neq i} \rho_j$$ and summing over all $$i$$, we get \begin{align*} \partial_t\rho = \sum_i \big(\partial_t \rho_i \prod_{j\neq i} \rho_j\big) &= -\rho_i \partial_i b_i(\theta)\prod_{j\neq i} \rho_j - b_i \partial_i \rho_i \prod_{j\neq i} \rho_j + \partial_{\theta_i}^2 \rho_i \prod_{j\neq i} \rho_j \\ &= -\rho \sum_i \partial_i b_i(\theta) - \sum_i b_i(\theta) \partial_i \rho + \sum_i \partial_i^2 \rho \end{align*}
which is exactly equal to the $$d$$D equation.