I was trying to solve $$\frac{{\partial}^4\phi}{\partial{z^2}\partial\bar{z}^2}=0$$

Using the Wirtinger Derivatives $$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial }{\partial x} - i\frac{\partial }{\partial y})$$

$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial }{\partial x} + i\frac{\partial }{\partial y})$$

I used the following by multiplying the two wirtinger derivatives by itself, $$\frac{\partial^2}{\partial {z}^2} = \frac{1}{4}(\frac{\partial^2 }{\partial x^2} - 2i\frac{\partial^2 }{\partial y \partial x}-\frac{\partial^2 }{\partial y^2})$$

$$\frac{\partial^2}{\partial \bar{z}^2} = \frac{1}{4}(\frac{\partial^2 }{\partial x^2} + 2i\frac{\partial^2 }{\partial y \partial x}-\frac{\partial^2 }{\partial y^2})$$

And then I fit into the previous equation I wanted to solve and then got two linear partial differential equations by solving the real part and the imaginary part.

I'm having trouble how to solve them simultaneously.

Please help.


1 Answer 1


Well, the whole point of using Wirtinger derivatives is to make this kind of problem easy. One can "trivially" see that $\phi = (\overline{z}+a)p(z) + (z+b)q(\overline{z})$ is a solution, for arbitrary functions p and q. Why is this "trivial"? Because $z$ and $\overline{z}$ behave as if they were (linearly) independent variables. Posing the question you did is very similar to (the same as?) asking for solutions to $$\frac{\partial^4\phi}{\partial^2s\partial^2t}=0$$ for two linearly independent variables $s$ and $t$. (And oh, by the way, you'd get a mess if you tried to solve the above, using $s=u+v$ and $t=u-v$ - the same mess that you are currently getting).

  • $\begingroup$ Could you please refer me some text which contains the wirtinger derivatives and its properties $\endgroup$ Feb 10, 2018 at 7:51
  • $\begingroup$ Too far in my distant past to know of such a text. But you are missing the point - its really just a trick of linear algebra. Solve the s,t problem I give. Make the change to u,v - its just linear algebra. Right? Make another change of variable: $v=iw$ which just multiplies by a constant called $i$. Harmless, right? So $s=u+iw$ and $t=u-iw$ after some "elementary" linear algebra. Finally, when you draw the letter $t$ make it twistier and make the cross-bar of the t miss the top, so it accidentally looks like $\overline s$ when you write it down. Bingo. $\endgroup$
    – Linas
    Feb 10, 2018 at 8:24
  • $\begingroup$ The point here is that $z$ and $\overline z$ really are linearly-independent variables. The overline notation is confusing you: given a value for $z$, its is impossible to obtain a value for $\overline z$ -- there is no (holomorphic) function of $z$ that gives $\overline z$ The only way to obtain $\overline z$ is to know of a very magic number called $i$ and some of its properties, and then to define a whacked idea called "complex conjugation". Instead, just pretend that $i$ is some constant, and you have no clue what it is. Then, all of a sudden, $z$ and $\overline z$ really are independent. $\endgroup$
    – Linas
    Feb 10, 2018 at 8:35
  • $\begingroup$ I understand now. But could I bother you a bit more asking cite me the name of book please? For my future references. $\endgroup$ Feb 10, 2018 at 8:53
  • $\begingroup$ try google? en.wikipedia.org/wiki/Wirtinger_derivatives is what I get when I type it in. $\endgroup$
    – Linas
    Feb 10, 2018 at 9:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .