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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 $W(0, \gamma) = \frac{2}{\pi} e^{-2|\alpha-\gamma|^2}$.

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  • $\begingroup$ 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. $\endgroup$
    – Nasser
    Jul 12, 2020 at 13:31
  • $\begingroup$ @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. $\endgroup$
    – Natas
    Jul 12, 2020 at 13:43
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    $\begingroup$ Kuba wrote a function that allows you to perform a change in coordinates of differential equations. You can find it here: SE Answer (Github) $\endgroup$
    – Natas
    Jul 12, 2020 at 13:44
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    $\begingroup$ @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. $\endgroup$
    – Nasser
    Jul 12, 2020 at 13:53
  • $\begingroup$ @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.) $\endgroup$
    – Natas
    Jul 12, 2020 at 15:19

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – iknownothing
    Jul 13, 2020 at 8:29

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