# About the solutions of a PDE

I would like to know how the following statement is proved. I do not have any reference, buy I want to know where this come from because it is used in a proof that I am working with and I am not familiar with PDE's. Maybe is an extension of a real PDE's result, but I do not know.

Let $g,h:U\subset\mathbb{C}^2\to \mathbb{C}$ be two analytic functions, where $U$ is an open subset of $\mathbb{C}^2$ that contains $z_0:=(x_0,y_0)\in\mathbb{R}^2\subset\mathbb{C}^2$.

Then, the solutions of the equation \begin{equation} g(x,y)\frac{\partial f}{\partial x} + h(x,y)\frac{\partial f}{\partial y} = 0\end{equation} are the functions $f$ that are constant on the solutions of \begin{equation} \frac{dy}{dx} = \frac{h}{g}. \end{equation}

Moreover, such functions are determined by their values on any transversal to the solutions through $z_0 := (x_0,y_0)\in \mathbb{R}^2\subset\mathbb{C}^2$.

I hope someone can help me. Thank you.

• Looks like the method of characteristics. Have you read up on this? – Theoretical Economist Apr 17 '18 at 21:18
• I skimmed the method of characteristics and it seems to be quite analogous, but I can not imagine the geometry on $\mathbb{C}^2$ and this method is based on geometrical facts (the surface where the solutions lie, etc). Is there an extension of the method of characteristics for $\mathbb{C}^2$? – dudas Apr 17 '18 at 21:47

## 1 Answer

If $x(z)$, $y(z)$ are the solutions of the complex ODE system $$\frac{dx(z)}{dz}=g(x(z),y(z))\\ \frac{dy(z)}{dz}=h(x(z),y(z))$$ then you get as in the real case that along these characteristic curves $$\frac{d}{dz}f(x(z),y(z))=\frac{∂f}{∂x}(x(z),y(z))g(x,y)+\frac{∂f}{∂y}(x(z),y(z))h(x,y)=0,$$ so that $f$ is constant along these curves.

With a parameter change for these characteristic curves you can also invert the relation of $x$ and $z$ so that $y$ is then a function of $x$.