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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!

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2 Answers 2

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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!!!

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  • $\begingroup$ @big_gie: I'll try to think a little more about your real question and see if I come up with anything truly useful! My apologies once again! $\endgroup$ Commented Mar 6, 2014 at 19:53
  • $\begingroup$ Hi Robert, thanks for your reply! I must say the error was mine; I mistakenly entered your version. After playing with it myself, I did what you suggested (before seeing your answer) and realized I entered the equation wrong... So I edited the question. Hopefully a similar approach can decoupled the "fixed" equations... $\endgroup$
    – big_gie
    Commented Mar 6, 2014 at 21:57
  • $\begingroup$ @big_gie: well, it will be tougher! I didn't realize I was dealing with your pre-edited version; that brings me a little relief! $\endgroup$ Commented Mar 6, 2014 at 22:24
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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 $$ L_1^2 g(x,y) = L_2^2g(x,y) $$

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