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!
Best Answer
First, we consider the character function: $$3(\frac{dy}{dx})^2-10\frac{dy}{dx}+3=0$$
The solution is $3y-x=c$ and $y-3x=c$
Second, make a variable change: $$(\zeta,\eta)=(3y-x,y-3x)$$
Third, compute it to find the new equation:
$$u_{\zeta\eta}=0$$
Forth, also make a variable change: $$(s,t)=(\zeta+\eta,\zeta-\eta)$$
Finally, $$u_{ss}-u_{tt}=0$$