I have been trying to solve the 2D PDE $$\frac{\partial p(x,y,t)}{\partial t}=-\frac{\partial }{\partial x}p(x,y,t)+\frac{\partial ^2}{\partial x^2}p(x,y,t)+\frac{\partial ^2}{\partial y^2}p(x,y,t)$$

This is close to the diffusion equation $$-\frac{1}{D}\frac{\partial p(x,y,t)}{\partial t}+\frac{\partial^2}{\partial x^2}p(x,y,t)+\frac{\partial ^2}{\partial y^2}p(x,y,t)=0$$ where D is a constant. This has a solution of $$p(x,y,t)=\frac{1}{4\pi D t}\exp\frac{-(x+y)^2}{(4Dt)}$$ (Source: http://rpdata.caltech.edu/courses/aph162/2006/Protocols/diffusion.pdf)

However, there is an additional drift term. I am hoping someone can help me out with the analytical solution to the first equation. Thank you :)


1 Answer 1


The solution of the diffusion equation that you are referring to here is called the fundamental solution, i.e. the solution starting at $t=0$ at a Dirac centered at $(0,0)$. When a PDE is linear with constant coefficients the fundamental solution may be computed thanks to the Fourier transform.

In your case, if we take the Fourier transform of the equation, we get $$ \frac{d}{dt} \hat{p}(t,\xi_x,\xi_y) = (-i\xi_x-\xi_x^2-\xi_y^2)\hat{p}(t,\xi_x,\xi_y), $$ where $\hat{p}(t,\xi_x,\xi_y) = \int_{\mathbb{R}^2}e^{-i\xi_xx-i\xi_yy}p(t,x,y)d\xi_x d\xi_y$.

The fundamental solution corresponds to the solution for which the initial data in Fourier is identically $1$, which corresponds to a Dirac at $0$ in the physical space. By solving the ODE on $\hat{p}$ we get $$ \hat{p}(t,\xi_x,\xi_y) = e^{-t(i\xi_x+\xi_x^2+\xi_y^2)}. $$

The solution that you are seeking is the inverse Fourier transform of that, namely $$ p(t,x,y) = \frac{1}{(2\pi)^2}\int_{\mathbb{R}^2} e^{i\xi_xx+i\xi_yy-t(i\xi_x+\xi_x^2+\xi_y^2)}d\xi_x d\xi_y. $$ A change of variable and Fubini give you $$ p(t,\sqrt{t}x+t,\sqrt{t}y) = \frac{1}{4\pi^2t}f(x)f(y) $$ where $f(x) =\int_{\mathbb R}e^{i\xi x-\xi^2}d\xi$. Differentiation under the integral sign and integrations by parts will show you that $f'(x)= -f(x)x/2$. Since $f(0) = \sqrt{\pi}$, one has $f(x) = \sqrt{\pi}e^{-x^2/4}$ and putting all together $$ p(t,x,y) = \frac{1}{4\pi t}\exp\left(\frac{-(x-t)^2-y^2}{4t}\right). $$

  • $\begingroup$ Thank you so much. This is really helpful and I appreciate your time. :) $\endgroup$
    – MathIsHard
    Nov 27, 2020 at 18:12

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