# Solving simple system of linear PDEs

Can the system $$\cases{\frac{\partial g_1}{\partial x}-\frac{\partial g_2}{\partial y}=f_1\\ \frac{\partial g_2}{\partial x}+\frac{\partial g_1}{\partial y}=f_2}$$ (all the functions are on a region in $$\mathbb{R}^2$$)
be solved explicitly using the "anti-derivatives" of $$f_1$$ and $$f_2$$?
(I need only one solution, and not necessarily the general one)

I'm aware that the system can be transformed to two Poisson equation of the functions separately, but the expression for the solution (as far as I know. maybe I'm wrong) is far from explicit.

• You could try using the method of characteristics for systems of PDEs; see here and here for more. Commented Dec 27, 2022 at 12:58
• @Matthew Cassell : If I understand correctly, the method of characteristics is used to solve homogeneous equation/system, so given a particular solution of an inhomogeneous system, you can find the general one. Since I don't have any solutions and I want just one (there aren't boundary conditions), this doesn't seem to address my problem. Commented Dec 28, 2022 at 23:56
• You didn't understand correctly. The method of characteristics can solve inhomogeneous problems. Your problem can be written as $$I\vec{g}_{x}+A\vec{g}_{y}=\vec{f}$$ with $I$ the identity matrix, $$A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$$ and $\vec{g}=(g_{1},g_{2}),\vec{f}=(f_{1},f_{2})$. You can diagonalise $A$ so that the problem becomes $$I\vec{g}_{x}+PDP^{-1}\vec{g}_{y}=\vec{f}$$ Multiply through by $P^{-1}$ and define $P^{-1}\vec{g}=\vec{G},P^{-1}\vec{f}=\vec{F}$ to get $$\vec{G}_{x}+D\vec{G}_{y}=\vec{F}$$ The system is now decoupled and immediately solvable. Commented Dec 29, 2022 at 2:28

## 1 Answer

Not really, but it depends on what you consider as "explicit". For example, you could take the 2D Fourier transform of your problem and the solution will depend explicitly on the Fourier transforms of $$f_1$$ and $$f_2$$.

Nonetheless, I see a more direct trick. I don't know in which context your system of PDEs arises, but it can be seen as inhomogeneous Cauchy-Riemann equations for the complex variable $$z = x+iy$$ and the function $$g(z,\bar{z}) = g_1(x,y) + ig_2(x,y)$$, with the source term $$f(z,\bar{z}) = \frac{1}{2}(f_1(x,y) + if_2(x,y))$$, such that $$\partial_{\bar{z}}g = f$$, where $$\partial_{\bar{z}} = \frac{1}{2}(\partial_x + i\partial_y)$$ is the Wirtinger derivative. N.B. : $$f$$ and $$g$$ are not holomorphic in that case, that is why they also depend on the conjugate variable $$\bar{z}$$.

The solution is then given by $$g(z,\bar{z}) = A(z) + \int_D \frac{f(z',\bar{z}')}{z'-z} \frac{\mathrm{d}z'\wedge\mathrm{d}\bar{z}'}{2\pi i},$$ where $$D\subset\mathbb{R}^2$$ is your said region and $$A(z)$$ an holomorphic function on $$D$$. So, the antiderivatives of $$f_1,f_2$$ do not appear in the solution, because it is basically made of the convolution of $$f$$ with the kernel $$\frac{1}{2\pi iz}$$.