While working on a physical diffusion process, I encountered the following Fokker-Planck equation

$$ \frac{\partial F}{\partial t} = D (x) \frac{\partial^2 F}{\partial x^2} \tag1$$

where $D(x) > 0$ is an anistropic diffusion coefficient. I introduced a temporal ansatz, so that I was looking for solutions of the form $F(x , t) = e^{- \lambda^{2} t} \, F_{\lambda} (x) $. (It corresponds to solutions temporally decreasing). Using a physical argument, allowing me to write $\frac{\partial }{\partial x} \left[ D (x) F_{\lambda}(x)\right] \simeq D(x) \frac{\partial F_{\lambda}}{\partial x}$, the diffusion equation becomes

$$ \left[ i \lambda - \sqrt{D} \frac{\partial }{\partial x}\right] \left[ i \lambda + \sqrt{D} \frac{\partial}{\partial x}\right] F_{\lambda} = 0 \tag2$$

We can now introduce the primitive $V (x)$ so that $\frac{d V}{d x} = \frac{1}{\sqrt{D(x)}}$. (We can assume it exists and is as smooth as needed.) Looking only for the solution of the right operator from (2) (One should note that because of my previous assumption, these two operators commute), I get that $F_{\lambda}$ takes the form

$$ F_{\lambda}(x) = A (\lambda) \, e^{- i \lambda V (x)}$$

The last step is to release the $\textit{lock}$ on $\lambda$ and look for the contributions from all $\lambda \in \mathbb{R}^{*}$, so that under my assumptions, the general solutions of the diffusion equation would take the form

$$ F (t,x) = \int_{- \infty}^{\infty} d \lambda \, e^{- \lambda^{2} t} \,A(\lambda) \, e^{i \lambda V(J)} \tag3$$

Now comes my question.

In order to study the temporal evolution of my solution and its temporal decay rate, I now need to include my initial conditions (for $t = 0$), which verify

$$ F_{0} (x) = \int_{- \infty}^{\infty} d \lambda \, A(\lambda) \, e^{i \lambda V(J)} \tag4 $$

How could I $\textit{invert}$ (4) to obtain the expression of the function $\lambda \mapsto A (\lambda)$, as a function of my initial conditions $x \mapsto F(0,x) = F_{0} (x)$ ?

For example, in the case of the homogeneous heat equation, one has that $V(x) = x$, so that using an inverse Fourier Transform, one can show that $A (\lambda) \propto \int_{-\infty}^{+\infty} dx\, F_0(x) \, e^{- i \lambda x}$

Should I use a kind of $\textit{tuned}$ Fourier Transform ? Would you have any ideas on a way to take into account the initial conditions ? Or references which deal with the same issues ?


1 Answer 1


I got the answer to my own question, so here it is. The key remark is to note that since $D(x) > 0$, the function $x \mapsto V(x)$ is strictly monotonic, which means that the change of variables $y = V(x)$ is a valid one.

Finally, I need to assume that $\lim\limits_{x \to - \infty} V(x) = - \infty$ and $\lim\limits_{x \to + \infty} V (x) = + \infty$.

The initial condition (4) becomes

$$ F_{0} (V^{-1} (y)) = \int_{- \infty}^{+ \infty} d \lambda' \, A (\lambda') \,e^{i \lambda' y} \tag{*}$$

Multiplying (*) by $e^{- i \lambda y}$ and integrating for $y \in ] - \infty ; + \infty[$, one can see that the left-hand side when integrated on $y$ gives a term equal to $2 \pi \, \delta_{\rm D} (\lambda - \lambda')$, so that we obtain

$$ A (\lambda) = \frac{1}{2 \pi } \int_{- \infty}^{+ \infty} d y \, F_{0} (V^{-1}(y)) \, e^{- i \lambda y}$$

Using once again the same change of variables $y = V(x)$, we obtain the final $\textit{inverted}$ expression

$$ A (\lambda) = \frac{1}{2 \pi } \int_{- \infty}^{+ \infty} d x \, \frac{1}{\sqrt{D(x)}} \, F_{0} (x) \, e^{- i \lambda V(x)}$$


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