<s>[INST] 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!
 [/INST] 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.
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