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

$$fracpartial^2 upartial x^2 + 2fracpartial^2 upartial x partial y + 5fracpartial^2 upartial y^2 + 3fracpartial upartial x + u = 0$$

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

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

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

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

$$fracxi_xxi_y = -1 ± fracsqrt(1 - (1)(5)2 =-frac12 ± i$$

$$fracxi_xxi_y = -frac12 ± i =-fracdydx,$$

Trying to solve, I obtained:

$$xi = phi_+x + y$$ where $$phi_+x = -frac12 + i$$
and
$$eta = phi_–x + y$$ where $$phi_–x = -frac12 + i$$

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!

The characteristic equation congruent to the general partial equation:


$$Afracpartial^2partial x^2u
+Bfracpartial^2partial x partial yu
+Cfracpartial^2partial y^2u+Dfracpartialpartial xu+Efracpartialpartial yu+Fu=G $$

is:

$$Ay'^2-By'+C=0 text or Agamma^2-Bgamma+C=0$$

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 $xi = gamma_2$)

$$left(frac4AC-B^2Aright) fracpartial ^2 upartial s partial r +(eta D+E)fracpartial upartial s +(xi D+E)fracpartial upartial r+Fu(s, r) = G(s,r)$$
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):
$$alpha= xi+eta = 2Reeta$$
$$beta= j(xi-eta) =2Imeta$$

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

$mathbf hyperbolic quad mathrm ifquad b^2-ac>0$,

$mathbf parabolic quad mathrm ifquad b^2-ac=0$,

$mathbf elliptic quad mathrm ifquad b^2-ac<0$.

The charateristic equation
$$a,dy^2-2b,dx,dy+c,dx^2=0qquad(2)$$
splits into two equations
$$a,dy-(b+sqrtb^2-ac),dx=0,qquad(3)$$
$$a,dy-(b-sqrtb^2-ac),dx=0.qquad(4)$$
Elliptic case. Let $quadphi(x,y) + ipsi(x,y)=cquad$ solution of $(3)$ or $(4)$.
Then change variables
$$xi=phi(x,y),quadeta=psi(x,y)$$
reduces equation $(1)$ to canonical form
$$u_xixi+u_etaeta=Phi_1(x,y,u,u_xi,u_eta).$$

In our case $a=1,;b=1,;c=5,; b^2-ac=-4<0$. From $(4)$ we get
$$mathitdy-left( 1-2 iright) , mathitdx=0,$$
$$y-x+2x,i=c,$$
$$xi=y-x,quadeta=2x.qquad(5)$$
Finaly canonical form of equation $u_xx+2u_xy+5u_yy+3u_x+u=0quad$ is
$$4u_xixi+4u_etaeta-3u_xi+6u_eta+u=0$$
or
$$u_xixi+u_etaeta=frac3 u_xi 4-frac3 u_eta2-fracu4.$$