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I have been trying to solve the following PDE:

$Au_{xx} + Bu_{xy} + Cu_{yy} = 0$

In my case, I know that this is an elliptical PDE, that is $B^2 - 4AC < 0$. Does there exists an analytic solution for such problem ? If not, then can you suggest any numerical technique ?

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Don't accept the answers too fast, you may miss the better answer or influence the ambitions of the SE users. I give you a downvote for a leasson. –  doraemonpaul Dec 10 '12 at 0:02

2 Answers 2

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You can always do a linear coordinate change after which the equation is $u_{xx} + u_{yy} = 0$, whose solutions are called harmonic functions. So if $L$ denotes the linear transformation in question, the solutions will be of the form $h(L^{-1}(x,y))$, where $h$ is any harmonic function.

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I would like to add that if you are working on a nice enough domain (say with $C^1$ boundary) then you can get explicit solutions for this PDE by specifying boundary conditions and using Green's functions. –  Alex Youcis Dec 9 '12 at 17:14

Since all terms of the PDE are in same order and constant coefficient, you can use the "factorization method" to solve it:

The auxiliary polynomial of this PDE can be $A\lambda^2+B\lambda+C$ or $C\lambda^2+B\lambda+A$

By factorizing them in $\mathbb{C}$ :

$A\lambda^2+B\lambda+C=A\left(\lambda+\dfrac{B+\sqrt{B^2-4AC}}{2A}\right)\left(\lambda+\dfrac{B-\sqrt{B^2-4AC}}{2A}\right)$ or $C\lambda^2+B\lambda+A=C\left(\lambda+\dfrac{B+\sqrt{B^2-4AC}}{2C}\right)\left(\lambda+\dfrac{B-\sqrt{B^2-4AC}}{2C}\right)$

$\therefore$ the general solution of $Au_{xx}+Bu_{xy}+Cu_{yy}=0$ is $u(x,y)=f_1\left(2Ay+\left(B+\sqrt{B^2-4AC}\right)x\right)+f_2\left(2Ay-\left(B+\sqrt{B^2-4AC}\right)x\right)$ or $u(x,y)=f_1\left(2Cx+\left(B+\sqrt{B^2-4AC}\right)y\right)+f_2\left(2Cx-\left(B+\sqrt{B^2-4AC}\right)y\right)$

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