aligass2004yi

2022-06-30

How to solve these two second-order coupled PDE?
$0=Af\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y}\phantom{\rule{0ex}{0ex}}0=Ag\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+C\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+D\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y}$

mar1nerne

$\begin{array}{}\text{(1)}& 0=Af\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y},\end{array}$
$\begin{array}{}\text{(2)}& 0=Ag\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}g\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}f\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y};\end{array}$
Having said the above, (1) and (2) can be uncoupled as follows set:
$\begin{array}{}\text{(3)}& U\left(x,y\right)=f\left(x,y\right)+g\left(x,y\right),\end{array}$
$\begin{array}{}\text{(4)}& V\left(x,y\right)=f\left(x,y\right)-g\left(x,y\right);\end{array}$
if we now add (1) and (2) we obtain
$0=A\left(f\left(x,y\right)+g\left(x,y\right)\right)+B\frac{{\mathrm{\partial }}^{2}\left(f\left(x,y\right)+g\left(x,y\right)\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}\left(f\left(x,y\right)+g\left(x,y\right)\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}\left(f\left(x,y\right)+g\left(x,y\right)\right)}{\mathrm{\partial }x\mathrm{\partial }y}$
$\begin{array}{}\text{(5)}& =AU\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y},\end{array}$
that is,
$\begin{array}{}\text{(6)}& AU\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }{y}^{2}}+D\frac{{\mathrm{\partial }}^{2}U\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y}=0,\end{array}$
and similarly, subtracting yields
$\begin{array}{}\text{(7)}& AV\left(x,y\right)+B\frac{{\mathrm{\partial }}^{2}V\left(x,y\right)}{\mathrm{\partial }{x}^{2}}+C\frac{{\mathrm{\partial }}^{2}V\left(x,y\right)}{\mathrm{\partial }{y}^{2}}-D\frac{{\mathrm{\partial }}^{2}V\left(x,y\right)}{\mathrm{\partial }x\mathrm{\partial }y}=0,\end{array}$
where the sign of $D$ is negative to accomodate the fact that subtraction is "asymmetric": $g-f=-\left(f-g\right)$. (6) and (7) are decoupled, and may be solved seperately, and then $f\left(x,y\right)$ and $g\left(x,y\right)$ may be recovered from
$\begin{array}{}\text{(8)}& f\left(x,y\right)=\frac{1}{2}\left(U\left(x,y\right)+V\left(x,y\right)\right),\end{array}$
$\begin{array}{}\text{(9)}& g\left(x,y\right)=\frac{1}{2}\left(U\left(x,y\right)-V\left(x,y\right)\right).\end{array}$
Of course the above does not address the issues of boundary conditions, well-posedness, etc., but I think the boundary conditions for $U\left(x,y\right)$, $V\left(x,y\right)$ may follow a pattern similar to (3), (4). Worth looking at, though.

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