# Can I simplify a PDE with Mathematica?

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 $$\bar{f}_{z\bar{z}}=(\bar{f}_z)_{\bar{z}}$$.

• 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. May 16 '20 at 12:15
• Thank you for your comments @flinty. I changed the coefficients to be more clear. I appreciate any help. May 16 '20 at 16:46
• 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? May 16 '20 at 17:30
• I added on the post some notation, thank you for you patience, @flinty. May 16 '20 at 17:43

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... *)
$$$$
`