# Canonical form for hyperbolic PDE?

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!

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

• Why is the coefficient of the second term in the character function -10? Why wouldn't it be +10 or +5? – user114014 Jan 28 '14 at 14:21
• Also, what do you mean by "compute it to find the new equation" - what did you do there? Thanks! – user114014 Jan 28 '14 at 14:27
• @user114014 The first question is... I really don not know how to answer it and I admit that I didn't learn PDE very well. I just know that if you meet a PDE like $a_{11}u_{xx}+2a_{12}u_{xy}+a_{22}u_{yy}=0$, the first step is to compute the character function $a_{11}(y_{x})^2-2a_{12}y_x+a_{22}=0$. And the solution can be seen as a variable change. And also the variable change will make the PDE simple. But I don not know the really meaning behind it. – gaoxinge Jan 29 '14 at 2:16
• @user114014 The second question is "compute it to find new equation". We have a variable change, so the $u$ will be represented by $\zeta,\eta$ instead of $x,y$. Also, the PDE will change. And I will improve the answer. – gaoxinge Jan 29 '14 at 2:19