# Coupled semilinear PDEs

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!

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How is this semilinear? Ignoring the integrability constraint it looks rather linear to me. – Willie Wong Oct 2 '10 at 1:35
Using the integrability constraint, your equation reduces to $\partial_x^2 z + a^2 \partial_y^2 z = b (\partial_y z + a \partial_x z)$, which is elliptic type. So I assume you are prescribing some sort of boundary data? – Willie Wong Oct 2 '10 at 1:42
Given the integrability condition, this is an overdetermined system. If you substitute for $q_x$ in PDE2 using the integrability condition, you get an ODE in $y$ only. So you can specify initial data along the curve $x = 0$ that satisfies the ODE and then use PDE1 and PDE2 to propagate the solution in $x$. However, you need to verify that the integratibility condition is preserved by PDE1 and PDE2. – Deane Yang Oct 2 '10 at 1:56
Without the integrability condition this is just an ode along the vector field $\partial_x + a\partial_y$ – Deane Yang Oct 2 '10 at 2:20
There is a systematic theory known as Cartan-Kahler theory that is useful for analyzing a system like this. But for a system like this you can probably do it by hand. Basically, you should differentiate all of the equations and try to find all possible linear combinations of these second order equations that cause all of the second derivatives to disappear. By doing this, you will derive additional first order equations. You will find that either there are no solutions at all, or you can find a solution as I describe above. – Deane Yang Oct 2 '10 at 3:43

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 $f_y=g_x$.
Thanks for the answer, Denis and everyone else. I can know a couple of additional things at a closed boundary contour. Along that contour, both $p$ and $q$ tend to infinity. But the ratio $q/p$ is finite and known. So, in your solution, I can specify $\theta$ at the boundary, but $\rho$ is infinity. I have also posed this as a second order PDE by substituting $p_y = q_x$ in PDE1. I have asked it here: mathoverflow.net/questions/41050/… For this second-order PDE, we may ignore the fact that it arises from a coupled first order system. – user9728 Oct 4 '10 at 19:25