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I'm having trouble reducing this elliptic equation to canonical form.

$$\frac{\partial^2 u}{\partial x^2} + 2\frac{\partial^2 u}{\partial x \partial y} + 5\frac{\partial^2 u}{\partial y^2} + 3\frac{\partial u}{\partial x} + u = 0$$

I know it's elliptic because I checked: $B^2 - AC < 0$, $$\begin{align} A = 1,\\ B = 1,\\ C = 5,\\ B^2 - AC = 1 - (1)(5) = -4 < 0 \end{align}$$ so it's elliptic.

I think the characteristics are to be found from the equation:

$$\xi_x^2 + 2\xi_x \xi_y + 5\xi_y^2 = 0$$

And then I tried to solve for $\xi_x/\xi_y$ as follows:

$$\frac{\xi_x}{\xi_y} = -1 ± \frac{\sqrt{(1 - (1)(5)}}{2} =-\frac{1}{2} ± i$$

$$\frac{\xi_x}{\xi_y} = -\frac{1}{2} ± i =-\frac{dy}{dx},$$

Trying to solve, I obtained:

$$\xi = \phi_+x + y$$ where $$\phi_+x = -\frac{1}{2} + i$$ and $$\eta = \phi_–x + y$$ where $$\phi_–x = -\frac{1}{2} + i$$

But I'm really not sure where to go from here, or if I'm even on the right track. I'm finding reductions to canonical form really difficult, and now that imaginary numbers are in the mix I'm completely stuck. I appreciate any help. Thanks in advance!

user114014
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2 Answers2

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The equation $$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}=\Phi(x,y,u,u_x,u_y)\qquad(1)$$ is

$\mathbf {hyperbolic }\quad {\mathrm if}\quad b^2-ac>0$,

$\mathbf {parabolic }\quad {\mathrm if}\quad b^2-ac=0$,

$\mathbf {elliptic }\quad {\mathrm if}\quad b^2-ac<0$.

The charateristic equation $$a\,dy^2-2b\,dx\,dy+c\,dx^2=0\qquad(2)$$ splits into two equations $$a\,dy-(b+\sqrt{b^2-ac})\,dx=0,\qquad(3)$$ $$a\,dy-(b-\sqrt{b^2-ac})\,dx=0.\qquad(4)$$ Elliptic case. Let $\quad\phi(x,y) + i\psi(x,y)=c\quad$ solution of $(3)$ or $(4)$. Then change variables $$\xi=\phi(x,y),\quad\eta=\psi(x,y)$$ reduces equation $(1)$ to canonical form $$u_{\xi\xi}+u_{\eta\eta}=\Phi_1(x,y,u,u_\xi,u_\eta).$$

In our case $a=1,\;b=1,\;c=5,\; b^2-ac=-4<0$. From $(4)$ we get $$\mathit{dy}-\left( 1-2 i\right) \, \mathit{dx}=0,$$ $$y-x+2x\,i=c,$$ $$\xi=y-x,\quad\eta=2x.\qquad(5)$$ Finaly canonical form of equation $u_{xx}+2u_{xy}+5u_{yy}+3u_x+u=0\quad$ is $$4u_{\xi\xi}+4u_{\eta\eta}-3u_\xi+6u_\eta+u=0$$ or $$u_{\xi\xi}+u_{\eta\eta}=\frac{3 u_\xi }{4}-\frac{3 u_\eta}{2}-\frac{u}{4}.$$

Aleksas Domarkas
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The characteristic equation congruent to the general partial equation:
$$A\frac{\partial^2}{\partial x^2}u +B\frac{\partial^2}{\partial x \partial y}u +C\frac{\partial^2}{\partial y^2}u+D\frac{\partial}{\partial x}u+E\frac{\partial}{\partial y}u+Fu=G $$

is:

$$Ay'^2-By'+C=0 \text{ or }A\gamma^2-B\gamma+C=0$$

This will lead to the below equation if you choose $\mathbf s=-\gamma_1 x + y$ and $\mathbf r = -\gamma_2 x + y$

(Where $\eta = \gamma_1$ and $\xi = \gamma_2$)

$$\left(\frac{4AC-B^2}{A}\right)\ \frac{\partial ^2 u}{\partial s \partial r} +(\eta D+E)\frac{\partial u}{\partial s} +(\xi D+E)\frac{\partial u}{\partial r}+Fu(s, r) = G(s,r)$$ If the $(4ac -b^2)$ is positive , same as yours, the answer is the elliptic canonical form of the main equation. (Wikipedia elliptical resource)

Which the two complex answers are conjugate of each other which means: $$\xi = \eta^*$$ We have to change our variables once more(to cancel the unreal part): $$\alpha= \xi+\eta = 2\Re\{\eta\}$$ $$\beta= j(\xi-\eta) =2\Im\{\eta\}$$

Ali Farbooodi
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