0
$\begingroup$

I'm a beginner in Mathematica and I search for help with a calculation problem with some derivatives and basic operations.

I need to simplify a PDE that depends of a smooth function $f:U\subset\mathbb{C}\to\mathbb{C}$, something like, $$Af^2+2fB+C=0\tag{1}$$ where $A$, $B$ and $C$ depends of $f$ and $\bar{f}$ and their complex derivatives, of order until 3.

Of course I can compute this by hand, but in my problem it will be a very long computation and I thought about to try Mathematica for this computation.

I would like to know if I can to define a symbolic complex function on Mathematica and to do basic operations as to compute derivatives in $z$ and $\bar{z}$. For example, if

$A=(f_{z\bar{z}}+|f_{z}|^2+f_zf_{\bar{z}})_z$

$B=((\bar{f}_{z\bar{z}})^2+|\bar{f}_{z}|^4+(\bar{f}_z\bar{f}_{\bar{z}})^2)_z$

$C=((f_{z\bar{z}})^3+|f_{z}|^6+(f_zf_{\bar{z}})^3)_z,$

Can I simplify equation $(1)$ using Mathematica?

Thank for your attention. I appreciate any help.

Notation:

The derivatives on $z$ and $\bar{z}$ are the Wirtinger derivatives. Then, given a smooth function $f:U\subset\mathbb{C}\to\mathbb{C}$, we have

$f_z = \frac{\partial f}{\partial z}, \bar{f}_z = \frac{\partial \bar{f}}{\partial z}, f_{\bar{z}} = \frac{\partial f}{\partial {\bar{z}}},\bar{f}_{\bar{z}} = \frac{\partial \bar{f}}{\partial {\bar{z}}}, |f_z|^2=f_z\bar{f}_{\bar{z}}, |\bar{f}_z|^2=\bar{f}_z f_{\bar{z}}, f_{z\bar{z}}=(f_z)_{\bar{z}}$ and $\bar{f}_{z\bar{z}}=(\bar{f}_z)_{\bar{z}}$.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ Could you explain the notation for me? What does $B=(f^2+\bar{f}_{z}^2f_{z\bar{z}}^2)_z$ mean for instance? I'm confused by the subscripts and exponents. $\endgroup$
    – flinty
    May 16, 2020 at 12:15
  • $\begingroup$ Thank you for your comments @flinty. I changed the coefficients to be more clear. I appreciate any help. $\endgroup$
    – Irddo
    May 16, 2020 at 16:46
  • 1
    $\begingroup$ What does $f_{z\bar{z}}$ mean? Shorthands make it hard to decipher and convert for Mathematica. I'm confused as to whether you mean a product, or a derivative D[f[Conjugate[z],z] times D[f[z],z]. Also what have you tried so far? $\endgroup$
    – flinty
    May 16, 2020 at 17:30
  • $\begingroup$ I added on the post some notation, thank you for you patience, @flinty. $\endgroup$
    – Irddo
    May 16, 2020 at 17:43

1 Answer 1

Reset to default
1
$\begingroup$

This stuff doesn't want to simplify. My best effort so far:

fz[g_] := (Dt[Re[g], z] - I Dt[Im[g], z])/2  (*Wirt df/dz*)
fw[g_] := (Dt[Re[g], z] + I Dt[Im[g], z])/2  (*Wirt df/dOverscript[z, _]*)
cz[g_] := fz[Conjugate[g]]  (*dOverscript[f, _]/dz*)
cw[g_] := fw[Conjugate[g]]  (*dOverscript[f, _]/dOverscript[z, _]*)
fzz[g_] := (fz@*fz)[g]
fzw[g_] := (fw@*fz)[g]
fwz[g_] := (fz@*fw)[g]
fww[g_] := (fw@*fw)[g]
czz[g_] := (cz@*cz)[g]
czw[g_] := (cw@*cz)[g]
cwz[g_] := (cz@*cw)[g]
cww[g_] := (cw@*cw)[g]
n1[g_] := fz[g]*cw[g]
n2[g_] := cz[g]*fw[g]

{a, b, c} = {
   fz[fzw[f[z]] + n1[f[z]] + fz[f[z]] fw[f[z]]],
   fz[czw[f[z]]^2 + n2[f[z]]^4 + (cz[f[z]] cw[f[z]])^2],
   fz[fzw[f[z]]^3 + n1[f[z]]^6 + (fz[f[z]] fw[f[z]])^3]
};
eqn = a f[z]^2 + 2 f[z] b + c == 0;

Simplify[eqn] (* will take forever... *)
```
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.