# Convert partial differential equation (Fokker-Planck equation) from polar to cartesian coordinates

I am trying to convert the following equation in polar form to the cartesian coordinates using Mathematica (Phys. Rev. A 78, 013810 (2008), https://journals.aps.org/pra/abstract/10.1103/PhysRevA.78.013810). I am new in Mathematica, so if somebody can help me it would be great. $$\begin{eqnarray} \partial_{\tau} W(\tau, r, \varphi) &=& \left\{ (r^2-1)\partial_{\varphi} - \frac{1}{16}\left(\frac{1}{r}\partial_r\partial_{\varphi} + \partial^2_r\partial_{\varphi} + \frac{1}{r^2}\partial^3_{\varphi} \right) \right. \nonumber\\ &+& \left. \xi + \frac{\xi}{2} \left(r +\frac{1}{2}\left(\frac{1}{2} + N\right) \frac{1}{r}\right) \partial_r \right. \nonumber\\ &+& \left.\frac{\xi}{4}\left(\frac{1}{2} + N\right) \left(\partial^2_r + \frac{1}{r^2}\partial^2_{\varphi}\right) \right\}W(\tau, r, \varphi), \end{eqnarray}$$

where $$\gamma = re^{i\varphi}$$, $$\tau = -\kappa t$$ with initial condition $$W(0, \gamma) = \frac{2}{\pi} e^{-2|\alpha-\gamma|^2}$$.

• polar coordinates uses $r,\theta$ only. But you have 3 coordinates. Spherical uses $r,\theta,\phi$. Is this what you meant? and what does $\partial_r^{2}$ mean? Is this $w_{rr}$ or $w_r^2$? Or may be you mean cylinderical? Who knows. Notation is very confusing to reader. Try to enter the pde in Mathematica code, that will help clear things. – Nasser Jul 12 '20 at 13:31
• @Nasser The notation is standard. $\partial^2_r W$ means $W_{rr}$. (But I understand your confusion.) The first coordinate is "time" and the second and third are "spatial". It is the second and third OP wants to transform. – Natas Jul 12 '20 at 13:43
• Kuba wrote a function that allows you to perform a change in coordinates of differential equations. You can find it here: SE Answer (Github) – Natas Jul 12 '20 at 13:44
• @Natas I never like this notation and it is not used much. I prefer $w_r$ and $w_{rr}$ notation instead as I find it much more clear. The notation used in the post seems more common in European eastern block countries. – Nasser Jul 12 '20 at 13:53
• @Nasser The notation used in the post is prevalent in the mathematical/theoretical physics community. (And indeed many confusions arise because people don't appreciate that the derivative maps a function to another function.) – Natas Jul 12 '20 at 15:19

Clear["Global*"];
pde = D[w[t, r, phi], r] == (r^2 - 1)*D[w[t, r, phi], phi] -
1/16 (1/r*D[D[w[t, r, phi], r], phi] +
D[D[w[t, r, phi], {r, 2}], phi] +
1/r^2*D[w[t, r, phi], {phi, 3}]) + zeta +
zeta/2*(r + 1/2 (1/2*n)*1/r)*D[w[t, r, phi], r] +
zeta/4*(1/2 + n)*(D[w[t, r, phi], {r, 2}] +
1/r^2*D[w[t, r, phi], {phi, 2}]);


$$w^{(0,1,0)}(t,r,\phi )=\frac{1}{4} \left(n+\frac{1}{2}\right) \zeta \left(\frac{w^{(0,0,2)}(t,r,\phi )}{r^2}+w^{(0,2,0)}(t,r,\phi )\right)+\frac{1}{2} \zeta \left(\frac{n}{4 r}+r\right) w^{(0,1,0)}(t,r,\phi )+\left(r^2-1\right) w^{(0,0,1)}(t,r,\phi )+\frac{1}{16} \left(-\frac{w^{(0,0,3)}(t,r,\phi )}{r^2}-\frac{w^{(0,1,1)}(t,r,\phi )}{r}-w^{(0,2,1)}(t,r,\phi )\right)+\zeta$$

 << MoreCalculus

DChange[pde, "Polar" -> "Cartesian", {r, phi}, {x, y}, w[t, r, phi],
Assumptions -> {r > 0 && -Pi < phi <= Pi}]


gives

{-((y*Derivative[0, 0, 1][w][t, x, y] +
x*Derivative[0, 1, 0][w][t, x, y])/Sqrt[x^2 + y^2]) ==
(1/(16*(x^2 + y^2)))*(16*x^2*zeta + 16*y^2*zeta +
2*(8*x^5 + 8*x*y^2*(-1 + y^2) + 8*x^3*(-1 + 2*y^2) +
4*x^2*y*zeta +
y*(-1 - n + 4*y^2)*zeta)*
Derivative[0, 0, 1][w][t, x, y] +
2*(1 + 2*n)*(x^2 + y^2)*zeta*Derivative[0, 0, 2][w][t, x, y] -
x^3*Derivative[0, 0, 3][w][t, x, y] -
x*y^2*Derivative[0, 0, 3][w][t, x, y] +
16*x^2*y*Derivative[0, 1, 0][w][t, x, y] -
16*x^4*y*Derivative[0, 1, 0][w][t, x, y] +
16*y^3*Derivative[0, 1, 0][w][t, x, y] -
32*x^2*y^3*Derivative[0, 1, 0][w][t, x, y] -
16*y^5*Derivative[0, 1, 0][w][t, x, y] -
2*x*zeta*Derivative[0, 1, 0][w][t, x, y] -
2*n*x*zeta*Derivative[0, 1, 0][w][t, x, y] +
8*x^3*zeta*Derivative[0, 1, 0][w][t, x, y] +
8*x*y^2*zeta*Derivative[0, 1, 0][w][t, x, y] +
x^2*y*Derivative[0, 1, 2][w][t, x, y] +
y^3*Derivative[0, 1, 2][w][t, x, y] +
2*x^2*zeta*Derivative[0, 2, 0][w][t, x, y] +
4*n*x^2*zeta*Derivative[0, 2, 0][w][t, x, y] +
2*y^2*zeta*Derivative[0, 2, 0][w][t, x, y] +
4*n*y^2*zeta*Derivative[0, 2, 0][w][t, x, y] -
x^3*Derivative[0, 2, 1][w][t, x, y] -
x*y^2*Derivative[0, 2, 1][w][t, x, y] +
x^2*y*Derivative[0, 3, 0][w][t, x, y] +
y^3*Derivative[0, 3, 0][w][t, x, y]),
Association["Mapping" -> {x == r*Cos[phi], y == r*Sin[phi]},
"Assumptions" -> {}]}


But DChange complained about Transformation rule is ambiguous which I am not sure why.

You can download DChange from Analogue for Maple's dchange - change of variables in differential expressions written by Kubba.

• Thanks. The same can be achieved by using the TransformedField module. However, the problem is that the derivatives are still in the form of polar coordinates and not in Cartesian. – iknownothing Jul 13 '20 at 8:29