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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?

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  • $\begingroup$ 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"? $\endgroup$
    – user711689
    Commented Jul 18, 2021 at 3:00

1 Answer 1

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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.

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