Solutions of PDE I am looking for solutions $g \colon \mathbb R^2 \times \mathbb R^2 \to \mathbb C$ of the PDE
$$\partial_{x_1} g(x,y) + \partial_{y_1} g(x,y) = a(y_1+x_1+i(x_2-y_2))g(x,y), \\ \partial_{x_2} g(x,y) + \partial_{y_2} g(x,y) = a(y_2+x_2+i(y_1-x_1))g(x,y),$$
where $a>0$ and $(x,y) = (x_1,x_2,y_1,y_2) \in \mathbb R^2 \times \mathbb R^2$. I already know that $f(x,y) := \exp(a(x\cdot y - i x \wedge y))$ with $x\wedge y := x_1 y_2 - x_2 y_1$ is a solution but I was so far not able to find any other (smooth) solution. Does anyone have an answer to this? General approaches to solve this kind of PDE are also appreciated!

As the system of PDEs strongly decouples, with one equation only dependent on the $(x_{1}, y_{1})$ coordinates while the other is only dependent on the $(x_{2}, y_{2})$ coordinates, it makes sense to solve the problems separately.
For the first PDE, the method of characteristics implies (labelling $g \to g_{1}$)
$$\frac{d x_{1}}{1} = \frac{d y_{1}}{1} = \frac{d g_{1}}{a (y_{1} + x_{1} + i (x_{2} - y_{2})) g_{1}}$$
The first equality gives
$$x_{1} - y_{1} = C_{1}$$
Using componendo-divindendo on the first equality and setting it equal to the last ratio gives
\begin{align} 
\frac{d g_{1}}{a (y_{1} + x_{1} + i (x_{2} - y_{2})) g_{1}} &= \frac{d (x_{1} + y_{1})}{2} \\
\implies \frac{dg_{1}}{g_{1}} &= \frac{a (y_{1} + x_{1} + i (x_{2} - y_{2})) d(x_{1} + y_{1})}{2} \\
\implies \ln g_{1} &= \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} + C_{2} \\
\implies g_{1} &= C_{2} \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right) \\
&= f_{1}(x_{1} - y_{1}) \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right)
\end{align}
which you can check satisfies the first equation. A similar calculation for the second PDE (by symmetry, let $x_{1} \leftrightarrow x_{2}, y_{1} \leftrightarrow y_{2}$ and relabel $g \to g_{2}$) then shows
\begin{align}
\implies g_{2} &= f_{2}(x_{2} - y_{2}) \exp \left( \frac{a}{4} (x_{2} + y_{2})^{2} + \frac{i a (x_{1} - y_{1})(x_{2} + y_{2})}{2} \right)
\end{align}
The final solution is then a superposition of the two
\begin{align}
g = \ &g_{1} + g_{2} \\
= \ &f_{1}(x_{1} - y_{1}) \exp \left( \frac{a}{4} (x_{1} + y_{1})^{2} + \frac{i a (x_{2} - y_{2})(x_{1} + y_{1})}{2} \right) \\
+ \ &f_{2}(x_{2} - y_{2}) \exp \left( \frac{a}{4} (x_{2} + y_{2})^{2} + \frac{i a (x_{1} - y_{1})(x_{2} + y_{2})}{2} \right)
\end{align}
Note that we get the result you found by choosing $f_{1}, f_{2}$ appropriately.
