For a surface $z(x,y)$, let $p = z\_x$ and $q = z\_y$. I have a pair of coupled semilinear PDEs in p and q.   

PDE1: $p\_x - a p\_y = b q$.   

PDE2: $q\_x - a q\_y = -b p$.


Note that $a$ and $b$ are functions of $(x,y)$. Is there a general way to solve such coupled systems of semilinear PDEs? (We may or may not assume any relationship between $p$ and $q$. Since they come from a surface, we know $p\_y = q\_x$, but I'll be happy if the coupled system can be solved even without that integrability constraint.)


Thanks!



Coupled semilinear PDEs
First, disregard the constraint $p\_y=q\_x$. Consider the integral curves of the vector field $V:=(1,-a)$, namely the solutions of the ODE
$$\frac{dy}{dx}=-a(x,y).$$
They are parametrized by $x$. Your equations are ODEs along these curves,
$$\dot p=bq,\qquad\dot q=-bp.$$
Set $q+ip=\rho e^{i\theta}$. Then the ODEs become $\dot\rho=0$, $\dot\theta=b$. You see that you need an initial data over a curve transversal to $V$, for instance along a curve $x={\rm cst}$.


If we now take in account the constraint, the system becomes *over-determined*. Such systems usually do not have a solution, unless the data (here $a,b$ and the initial data) are very special. To see how it begins, let us set $h:=p-aq$. Then we have $h\_x=(b-a\_x)q$ and $h\_y=-bp-a\_yq$. By Schwarz, we obtain a new equation
$$(a\_xq-bq)\_y=(a\_yq+bp)\_x.$$
This is a fourth differential equation. With the three others, you may solve 
$$(p\_x,p\_y,q\_x,q\_y)=F(p,q,x,y),$$
unless a $4\times4$ determinant vanishes. And the tale does not end there. You can continue and find other first-order PDEs, which yield incompatible in just one more step. Again, unless you are very lucky.


**edit after a few hours.** Once you have the first derivatives in terms of $p$ and $q$, you may apply Schwarz, $(p\_x)\_y=(p\_y)\_x$ and $(q\_x)\_y=(q\_y)\_x$. This gives $\partial\_yF\_1(p,q,x,y)=\partial\_xF\_2(p,q,x,y)$ and $\partial\_yF\_3(p,q,x,y)=\partial\_xF\_4(p,q,x,y)$. Eliminating first derivatives, there remain two equations $p=P(x,y)$ and $q=Q(x,y)$. Remark that you may not any more impose initial data, because you obtain explicit form of $p,q$ without really solving a differential equation. You have only use elimination and Schwarz identity. Finally, it happens in general that $P$ and $Q$ do not solve at all the differential equation.


In theory, you could explicit a necessary and sufficient condition in terms of $a$ and $b$ in order that your overdetermined system admit a solution. A very classical and simple situation is the system $u\_x=f$, $u\_y=g$, where the condition for having a solution is $f\_y=g\_x$.


