$p(\mathbf{x},t\vert\mathbf{y},t)=\delta\left(\mathbf{z}-\mathbf{y}\right)\tag{3.5.7}$ For a small $\delta t$, the solution of the Fokker-Planck equation will still be on the whole sharply peaked and hence derivatives of $A_j(\mathbf{z},t)$ and $B_{ij}(\mathbf{z},t)$ will be negligible compared to those of $p$. We are thus reduced to solving, approximately $$\frac{\partial p(\mathbf{z},t\vert\mathbf{y},t')}{\partial t}=-\sum_iA_i\left(\mathbf{y},t\right)\frac{\partial p(\mathbf{z},t\vert\mathbf{y},t')}{\partial z_i}+\sum_{i,j}B_{ij}(\mathbf{y},t)\frac{\partial^2 p(\mathbf{z},t\vert\mathbf{y},t')}{\partial z_i\partial z_j}\tag{3.5.8}$$ where we have also neglected the time dependent of $A_i$ and $B_{ij}$ for small $t-t'$. Equation (3.5.8) can now be solved, subject to the initial condition(3.5.7), and we get \begin{align} P(\mathbf{z},t+\Delta t\vert\mathbf{y},t) &=\left(2\pi\right)^{-N/2}\left\{\operatorname{det}\left[\underline{B}(\mathbf{y},t)\right]\right\}^{1/2}[\Delta t]^{-1/2} \\ &\times\exp\left\{-\frac{1}{2}\frac{\left[\mathbf{z}-\mathbf{y}-A(\mathbf{y},t)\Delta t\right]^T[\underline{B}(y,t)]^{-1}\left[\mathbf{z}-\mathbf{y}-A(\mathbf{y},t)\Delta t\right]}{\Delta t}\right\},\tag{3.5.9} \end{align} that is, a Gaussian distribution with variance matrix $\underline{B}(\mathbf{y},t)$ and mean $\mathbf{y}+A(\mathbf{y},t)\Delta t$.

I need help with the solution. How is equation 3.5.8 solved to obtain 3.5.9; Equation 3.5.7 is the initial condition here. This a page from the famous book Handbook of stochastic methods by CW Gardiner.

  • $\begingroup$ We strongly discourage images of text and mathematics. Please use a combination of text and Mathjax instead. $\endgroup$ Commented Apr 1, 2023 at 19:56
  • $\begingroup$ hint: the solution of $\frac{\partial u}{\partial t}=\frac{1}{2}\frac{\partial^2 u}{\partial x^2}$ is $u(x,t)=\frac{1}{\sqrt{2\pi t}}exp[-\frac{x^2}{2t}]$ $\endgroup$
    – hyportnex
    Commented Apr 1, 2023 at 20:23
  • $\begingroup$ It looks like a change of coordinates are in order to reduce it to the diffusion equation hyportnex indicates. $\endgroup$
    – Kyle Kanos
    Commented Apr 2, 2023 at 1:36


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