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I was wondering about how to solve a simple linear Fokker-Planck equation using space-time Laplace transform on space interval $[0,+ \ \infty)$,

$$\frac{\partial f(x,t)}{\partial t}= k_1 \frac{\partial f(x,t)}{\partial x} + k_2 \frac{\partial^2 f(x,t)}{\partial x^2}.$$

The usual method is to do Laplace transform in time and then solve the spatial differential equation by transforming it into a Sturm-Liouville problem.

But I feel that for a special case of semi-infinite, i.e., $[0, \ \infty)$, one can solve it easily if used Laplace transform for space-time instead of only time.

EDIT: The OP didn't specify boundary conditions, making the problem ill-defined. Among the many possible boundary conditions confining the process to the interval $[0,+\infty)$, one of the most used ones are the reflecting boundary conditions, $$\left[k_1 f(x,t)+k_2\frac{\partial}{\partial x}f(x,t)\right]_{x=0}=0,\quad \forall t.$$

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    $\begingroup$ How about the boundary conditions? $\endgroup$
    – Cesareo
    Commented Mar 11 at 0:07
  • $\begingroup$ @Cesareo The question was edited adding boundary conditions. $\endgroup$
    – Javi
    Commented Mar 13 at 15:06

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This is the farthest I could reach, which does not solve the question, but can be used for others to give a definitive answer.

Supposing the initial condition for the process is a well-located state, $$ f(x,t=0)=\delta (x-x_0)$$ with $x_0>0$. We can compute the temporal Laplace transform: $$ \tilde{f} (x,s)=\int _0^{+\infty} dt \, f(x,t)\,e^{-st}, $$ and $$ -\delta(x-x_0) +s\tilde{f} =k_1 \,\partial_x \tilde{f}+k_2\, \partial^2_x \tilde{f}$$ Now introducing the spatial Laplace transform together with the reflecting boundary, $$ \tilde{g} (z,s)=\int _0^{+\infty} dx \, \tilde{f}(x,s)\,e^{-zx}, $$ and $$ -e^{z\,x_0} +s\tilde{g} =k_1 \,z \,\tilde{g}+k_2\, z^2 \tilde{g}-z \tilde{f}(s,x=0).$$ Therefore, $$ \tilde{g}(z,s)=\frac{-e^{z\,x_0}-a\,z}{k_1 z+k_2 z^2-s}.$$ To solve the problem, one should do the spatio-temporal inverse Laplace transform of $ \tilde{g}(z,s)$ bearing in mind the consistence relation $$ a= \tilde{f}(s,x=0),$$ which does not seem easy to do.

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