TITLE: Canonical form for hyperbolic PDE?
QUESTION [1 upvotes]: I'm having trouble reducing this hyperbolic equation to canonical form.
$$3\frac{\partial^2 u}{\partial x^2} + 10\frac{\partial^2 u}{\partial x \partial y} + 3\frac{\partial^2 u}{\partial y^2} = 0$$
I know it's hyperbolic because I checked: $B^2 - AC > 0$,
$$\begin{align}
A = 3,\\
B = 5,\\
C = 3,\\
B^2 - AC = 25 - (3)(3) = 16 > 0 \end{align}$$ so it's hyperbolic.
I think the characteristics are to be found from the equation:
$$3\xi_x^2 + 5\xi_x \xi_y + 3\xi_y^2 = 0$$
And then I tried to solve for $\xi_x/\xi_y$ as follows:
$$\frac{\xi_x}{\xi_y} = -5 ± \frac{\sqrt{(25 - (3)(3)}}{6} =-\frac{5 ± 4}{6} =-\frac{dy}{dx} $$
$$\frac{\xi_x}{\xi_y} = -\frac{1}{2} ± i =-\frac{dy}{dx},$$
Trying to solve, I obtained:
$$-\frac{1}{6} = -\frac{dy}{dx}$$ where $$ x = 6y + c $$
and
$$-\frac{3}{2} = -\frac{dy}{dx}$$ where $$ 3x = 2y + c $$
But I'm really not sure where to go from here, or if I'm even on the right track.  I've been trying to use answers to similar questions but I'm pretty stuck.  I appreciate any help.  Thanks in advance!

REPLY [1 votes]: First, we consider the character function:
$$3(\frac{dy}{dx})^2-10\frac{dy}{dx}+3=0$$

And there is something wrong in your statement because you actually consider the character function like this:
  $$3(\frac{dy}{dx})^2+5\frac{dy}{dx}+3=0$$ 

The solution is $3y-x=c$ and $y-3x=c$

Second, make a variable change:
$$(\zeta,\eta)=(3y-x,y-3x)$$

$$u_{xx}=u_{\zeta\zeta}+6u_{\zeta\eta}+9u_{\eta\eta}$$
  $$u_{xy}=-3u_{\zeta\zeta}-10u_{\zeta\eta}-3u_{\eta\eta}$$
  $$u_{yy}=9u_{\zeta\zeta}+6u_{\zeta\eta}+u_{\eta\eta}$$


Third, compute it to find the new equation:  

$$0$$
  $$=3u_{xx}+10u_{xy}+3u_{yy}$$
  $$=3(u_{\zeta\zeta}+6u_{\zeta\eta}+9u_{\eta\eta})$$
  $$+10(-3u_{\zeta\zeta}-10u_{\zeta\eta}-3u_{\eta\eta})$$
  $$+3(9u_{\zeta\zeta}+6u_{\zeta\eta}+u_{\eta\eta})$$
  $$=64u_{\zeta\eta}$$

$$u_{\zeta\eta}=0$$

Forth, also make a variable change:
$$(s,t)=(\zeta+\eta,\zeta-\eta)$$

Finally,
$$u_{ss}-u_{tt}=0$$
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