7
$\begingroup$

Let $G:\mathbb{R}^d\to\mathbb{R}^{d\times d}$ be a matrix-valued smooth function. Let us define a quantity by $$ \begin{align*} \nabla^2\cdot G(x) &=\sum\limits_{i=1}^{d}\sum\limits_{j=1}^{d}\frac{\partial^2}{\partial x_i\partial x_j}G_{ij}(x) \end{align*} $$ Is there an intuitive explanation interpretation for that quantity?

Some background. I stumbled upon this question trying to understand the Fokker-Planck equation. If we have an SDE defined by $$\begin{align*} dX_t = f(X_t,t)dt + g(X_t,t)dW_t \end{align*}$$ Then the Fokker-Planck equation is given by: $$ \begin{align*} \frac{d}{dt}p_t(x) =& -\nabla\cdot [fp_t]+\frac{1}{2}[\nabla^2\cdot [gg^Tp_t]]\\ \end{align*} $$ The first term is the negative divergence whereas the second term is defined by the quantity above. The divergence has an intuitive interpretation as the infinitesimal net outflow from a point (see here).

$\endgroup$

2 Answers 2

6
$\begingroup$

To help the interpretation you may want to rewrite$^\ast$ the Fokker-Planck equation as

\begin{align*} \frac{d}{dt}p(x) =& -\nabla\cdot [\tilde{f}p -D\cdot\nabla p],\\ \end{align*} with $D=\tfrac{1}{2}gg^\top$ and $\tilde{f}=f-\nabla\cdot D$. Then this has the form of a convection-diffusion equation, with a conserved current that is the divergence of the sum of two currents: a convection current $\tilde{f}p$ driven by a force $\tilde{f}$, and a diffusion current $-D\cdot \nabla p$ driven by a density gradient $\nabla p$, with $D$ the diffusion tensor.

$^\ast$ In the form written in the OP a part of the convection current is hidden in the second term, which complicates the interpretation of the Fokker-Planck equation as a convection-diffusion equation.

$\endgroup$
5
$\begingroup$

This is by no means a complete or authoritative answer, just what I've personally managed to figure out. Any corrections or refinements would be greatly appreciated. In fact, I am posting this partially so that I can check my own intuition as well.

The second term you mention is the "diffusion/heat flux" term - any probability mass at a point will spread in an infinitesimal area around the point due to the noise term in the SDE. It is similar to the divergence term in that it measures net infinitesimal outflow from a point, however instead of going in the direction $f$, it disperses around the point, with dispersion rate controlled by $g$.

One may ask why you get a second order term due to the noise, instead of first order. This is essentially the same reason why the heat equation is of second order, not first. Lets restrict to one dimension for simplicity (in the multidimensional case we need to talk about how to interpret the mixed partials).

Fixing some $t$, the net outflow of probability mass out of a point $x$ due to noise is proportional to $p_t \cdot g(x)$, while the net inflow is proportional to the average of $p_t \cdot g$ in a small neighbourhood of $x$. Now the key fact is that - to first order, there is no difference between the value of a function at a point and its average surrounding values. More precisely, if we Taylor expand $p_t \cdot g$ around $x$ to first order, we have $p_t \cdot g(y) - p_t \cdot g (x) \sim \nabla (p_t \cdot g)(x) (y - x)$, and integrating this gives us $0$ due to the oddness of the linear function $\nabla (p_t \cdot g)(x)$. Thus there is no difference between net outflow and inflow due to first order terms.

However, the second order terms are crucially different. The function $y \to \nabla^2 (p_t \cdot g)(x) (y-x)^2$ is even, meaning that there is a genuine second order difference between a function and its nearby values, and so we will see an actual net change in mass, controlled by the noise intensity $g$ and the net difference in probability mass between nearby points. The fact that Brownian motion is rough - its increments over a small timeframe $t$ are proportional to $\sqrt t$ - means that these second order terms do provide a meaningful first order change in time in the probability density.

Perhaps the clearest case is when $X$ is simple Brownian motion - i.e. $f = 0$, $g = \text{Id}$. In this case the Fokker Planck equation is just the heat equation. To understand this case more thoroughly, I can recommend the book Random Walk and the Heat Equation by Lawler, which explains in both discrete and continuous time how the Laplacian term appears.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.