# Linear Partial Differential Equation

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.

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).
• 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. Feb 10, 2018 at 8:24
• 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. Feb 10, 2018 at 8:35