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The Fokker-Planck equation for several variables is :

$$\frac{\partial W}{\partial t} = L_{FP}W\qquad(1)$$


$$L_{FP} = -\frac{\partial}{\partial x_i}D_i(\{x\})+\frac{\partial^2}{\partial x_i \partial x_j}D_{ij}(\{x\}).\qquad(2)$$

The summation convention for Latin indices is used here. The drift vector $D_i$ and the diffusion tensor $D_{ij}$ generally depend on the N variables $x_1,...,x_N = \{x\}$. The Fokker-Planck equation is an equation for the distribution function $W(\{x\},t)$.

According to [Risken 1989 ch6], If drift & diffusion coefficients do not depend on time & $D_{ij}$ is positive definite everywhere & if the drift coefficient has no singularities, a stationary solution $W_{st}$

$$L_{FP} W_{st} = 0,\qquad(3)$$

may exist.

If one solves the above equation, a possible stationary solution can be

$$W_{st} =\frac{a}{D_{ij}}\exp\left(\int^{x_j}_0 \frac{D_i}{D_{ij}}\mathrm dt_j\right)\qquad (4)$$

where a is a normalization constant. Now I want to expand this probability distribution for $i=1,2$. If I use the Einstein summation convention, it becomes

$$\begin{split}W_{st} =\left\{\frac{a}{D_{11}}\exp\left(\int^{x_1}_0 \frac{D_1}{D_{11}}\mathrm dt_1\right)+\frac{a}{D_{12}}\exp\left(\int^{x_2}_0 \frac{D_1}{D_{12}}\mathrm dt_2\right)+\right.\\\left.\frac{a}{D_{21}}\exp\left(\int^{x_1}_0 \frac{D_2}{D_{21}}\mathrm dt_1\right)+\frac{a}{D_{22}}\exp\left(\int^{x_2}_0 \frac{D_2}{D_{22}}\mathrm dt_2\right)\right\}\end{split}\qquad (5)$$

It seems very strange to me. Is it a really correct probability distribution or I made a mistake somewhere? And if it is correct how can I normalize it? Can anyone help?

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If this is Einstein summation convention, something is not right. I spot three i's..., and four j's. – Thomas Rot Mar 3 '11 at 11:42
@SMH: I guess you should know what the original notation means... Did you derive it yourself? Did you get it from some source? It would be helpful to know which source. Just looking at the equation nobody can now what it means (unless you explicitly tell us that Einstein summation convention was used). – Fabian Mar 3 '11 at 14:53
I have somehow the feeling that $\frac{1}{D_{ij}}$ in the exponent should be $(D^{-1})_{ij}$. Maybe you did a mistake in the derivation of your expression (especially because you don't know what the expression means)? Is the original problem simple enough to be posed in this forum? – Fabian Mar 3 '11 at 15:54
@Fabian: I edited my question and wrote it with complete details. I asked it in mathoverflow too. – SMH Mar 3 '11 at 16:00
I guess the right hand side should be $L_{FP} W$ in your first equation. – Fabian Mar 3 '11 at 16:02
up vote 3 down vote accepted

(I will use the Einstein summation convention and $\partial_i = \frac{\partial}{\partial{x_i}}$)

The stationary solutions fulfill $L_{FP} W_{st}=0$ as explained. This can be rewritten as $\partial_i J_i =0$ where we have introduced the current $$J_i = D_i W_{st} - \partial_j D_{ij} W_{st}.$$ Many FP equations allow for a stationary solution where the current vanishes $J_i=0$ (and not only the divergence is equal to 0). Assuming that, we obtain $D_i W_{st} = \partial_j D_{ij} W_{st} = (\partial_j D_{ij}) W_{st} + D_{ij} \partial_j W_{st} $ which is equivalent to $$ \partial_j \log W_{st} = (D^{-1})_{ji} (D_i - \partial_k D_{ik}).$$ Integrating this equation, we obtain $$ W_{st}(x) = c \exp\left[\int^x dx'_j (D^{-1})_{ji} (D_i -\partial_k D_{ik}) \right]$$ with $c$ some normalization constant. So the last two equations of the questions are in fact wrong.

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@Fabian: But the probability current should be $J_i = D_i W_{st} - \partial_j D_{ij} W_{st}$. Again according to [Risken 1989]. You can see by putting $\frac{\partial W_{st}}{\partial t} = L_{FP} W_{st}=-\partial_i J_i =0$ – SMH Mar 3 '11 at 16:46
I thought $D$ does not depend on position? – Fabian Mar 3 '11 at 17:32
@Willie Wong: ok, I will change it to allow for position dependent $D$'s. – Fabian Mar 3 '11 at 18:04
@Fabin:Can you interpret the expression $(D^{-1})_{ij}$ please? I think I must take the inverse of the $D_{ij}$ matrix and then use the ji th component of it. Is that right? – SMH Mar 3 '11 at 18:38
@SMH: yes, that is correct. – Fabian Mar 3 '11 at 18:39

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