# A "simple" coupled system of first order PDEs

I'm trying to solve/ understand the the following system of PDEs: $$\begin{cases} \partial_xf(x,y)+\partial_yg(x,y) &= 0,\\ \partial_xg(x,y)-\partial_yf(x,y) &= 0, \end{cases}$$ for the functions $$f$$ and $$g$$. Here's what I saw so far:

• by deriving each line by $$x$$ or $$y$$ and substituing in the other equations, we can see that de system becomes $$\begin{cases} (\partial^2_x+\partial^2_y)f(x,y) &= 0,\\ (\partial^2_x+\partial^2_y)g(x,y) &= 0, \end{cases}$$ so each function must satisfy the 2-dimensional Laplace equation, for which I know the solutions. But, if I'm not mistaken, this manipulation results in finding a particular case form of solutions, i.e. a subset of solutions in the bigger set of solutions that the initial system admits. Is that right ?
• these equations are the Cauchy-Riemann equations, thus this system is exactly equivalent to asking that the complexe map $$F(x+iy)=g(x,y)+if(x,y)$$ is holomorphic (or $$\mathbb{C}$$-differentiable) but, up to my knowledge, there exist no generic explicit expression for such maps. Does this means that this system of PDEs is just a constraint on the form of $$f$$ and $$g$$ but that there is no generic explicit solution ?

In general, how to solve this system ?

I'm also very interested in a way (coordinate change for example) to obtain two decoupled equations.

• Your question is too general to give a complete answer. For PDEs it is very rare that you can just "solve" an equation without specifying more information. For example, are you looking for a solution in a bounded domain $\Omega \subset \mathbb{R}^2$? If so then you need to specify boundary conditions. If you are looking for solutions in $\mathbb{R}^2$ then, as you mentioned, the real and imaginary parts of every entire function will satisfy your equations, so there will be no general formula for the solution. Mar 14, 2021 at 4:01
• @Globemaster17 this my problem, since there is an simple and explicit solution to the 2d Laplace equation, this would give us a general form for F, a generic entire function, no ? I know that that if both component are harmonic it does not means that it's holomorphic, it's just a necessary condition, but this still means that an entre function must have "sub-form" (a particular case of this form), doesn't it ?
– xpsf
Mar 15, 2021 at 10:13

## 1 Answer

The answer to the first question is no. It is a theorem that if $$g$$ and $$f$$ have the indicated first partial derivatives in an open connected set, then $$g+if$$ is analytic. In particular $$g$$ and $$f$$ have second derivatives. Therefore every solution to CR is harmonic, not just a subset of them.

For the second question I suppose you could say that a convergent power series is a generic explicit solution.