How to solve these two second-order coupled PDE? I have two second-order equations governing the behaviour of two spatial function which are coupled:
$$
0 = A f(x,y) + B \frac{\partial^2 f(x,y)}{\partial x^2} + C \frac{\partial^2 f(x,y)}{\partial y^2} + D \frac{\partial^2 g(x,y)}{\partial x \partial y} \\
0 = A g(x,y) + B \frac{\partial^2 g(x,y)}{\partial y^2} + C \frac{\partial^2 g(x,y)}{\partial x^2} + D \frac{\partial^2 f(x,y)}{\partial x \partial y}
$$
I plan to solve this numerically. Can I at least uncoupled them?
Thanks!
 A: Note Edited In:  I must apologize, but in my haste I mis-read the equations as
$0 = A f(x,y) + B \dfrac{\partial^2 f(x,y)}{\partial x^2} + C \dfrac{\partial^2 f(x,y)}{\partial y^2} + D \dfrac{\partial^2 g(x,y)}{\partial x \partial y}, \tag{1}$
$0 = A g(x,y) + B \dfrac{\partial^2 g(x,y)}{\partial x^2} + C \dfrac{\partial^2 g(x,y)}{\partial y^2} + D \dfrac{\partial^2 f(x,y)}{\partial x \partial y}; \tag{2}$
I must be not quite awake yet this morning; late night gigging does is not always miscible with early morning math, like oil and water.  Nevertheless, I will let my solution stand, respecting the prospect in might be helpful.  End of Note.
Having said the above, (1) and (2) can be uncoupled as follows:
set
$U(x, y) = f(x, y) + g(x, y), \tag{3}$
$V(x, y) = f(x, y) - g(x, y); \tag{4}$
if we now add (1) and (2) we obtain
$0 = A (f(x,y) + g(x, y)) + B \dfrac{\partial^2 (f(x,y) + g(x, y))}{\partial x^2} + C \dfrac{\partial^2 (f(x,y) + g(x, y))}{\partial y^2} + D \dfrac{\partial^2 (f(x,y) + g(x, y))}{\partial x \partial y}$
$= A U(x, y) +  B \dfrac{\partial^2 U(x, y)  }{\partial x^2} + C \dfrac{\partial^2 U(x, y)}{\partial y^2} + D \dfrac{\partial^2 U(x, y)}{\partial x \partial y}, \tag{5}$
that is,
$A U(x, y) +  B \dfrac{\partial^2 U(x, y)  }{\partial x^2} + C \dfrac{\partial^2 U(x, y)}{\partial y^2} + D \dfrac{\partial^2 U(x, y)}{\partial x \partial y} = 0, \tag{6}$
and similarly, subtracting yields
$A V(x, y) +  B \dfrac{\partial^2 V(x, y)  }{\partial x^2} + C \dfrac{\partial^2 V(x, y)}{\partial y^2} - D \dfrac{\partial^2 V(x, y)}{\partial x \partial y}=0, \tag{7}$
where the sign of $D$ is negative to accomodate the fact that subtraction is "asymmetric":  $g - f = -(f - g)$.  (6) and (7) are decoupled, and may be solved seperately, and then $f(x, y)$ and $g(x, y)$ may be recovered from
$f(x, y) = \dfrac{1}{2}(U(x, y) + V(x, y)), \tag{8}$
$g(x, y) = \dfrac{1}{2}(U(x, y) - V(x, y)). \tag{9}$
Of course the above does not address the issues of boundary conditions, well-posedness, etc., but I think the boundary conditions for $U(x, y)$, $V(x, y)$ may follow a pattern similar to (3), (4).  Worth looking at, though.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: You have two equations
$$
L_1 (f(x,y)) = L_2(g(x,y))\\
L_1 (g(x,y)) = L_2(f(x,y))
$$
where the operators $L_i$ are given as
$$
L_1 = A + D\frac{\partial^2}{\partial x\partial y}\\
L_2 = -B\frac{\partial^2}{\partial x^2}-C\frac{\partial^2}{\partial y^2}
$$
Since 
$$
[L_1,L_2] = 0
$$
i.e. they commute we can have (this is not true if the coefficients were not constant)
$$
L_2\left[L_1 (f(x,y)\right] = L_1\left[L_2 (f(x,y)\right] = L_1\left[L_1(g(x,y))\right] = L_2\left[L_2 (g(x,y)\right] 
$$
thus we have
$$
L_1^2 g(x,y) = L_2^2g(x,y) 
$$
