# Solving the Hyperbolic PDE $u_{xx}-2u_{xy}-3u_{yy}=0$

I'm trying to show that the general solution of the hyperbolic PDE, $$u_{xx}-2u_{xy}-3u_{yy}=0,$$ is $$u(x,t)=F(3x+y)+G(x-y)$$.

I thought I could reduce the given PDE to form two ODEs. So far I have computed,

\begin{align} u_{xx}+u_{xy}-3u_{xy}-3u_{yy}=0 \\ (u_x+u_y)_x-3(u_x+u_y)_y=0 \end{align} Letting $$v=u_x+u_y$$ gives the first order PDE, $$v_x-3v_y=0.$$ I'm unsure if I am on the right track and how to continue further. Is there a more systematic way of solving this problem?

Thank you very much.

• You can just differentiate the given solution and show that it satisfies the PDE – Dylan Feb 25 at 10:42
• I see your point. You would be correct to do so. Though, I wish to derive this result. – user557493 Feb 25 at 10:44

There is a way to derive the solution, by reducing the PDE to its canonical form. For hyperbolic equations, use a substitution $$(x,y)\to (\xi,\eta)$$ such that

\begin{align} a\xi_x + \Big(b+\sqrt{b^2-ac}\Big)\xi_y &= \xi_x + \xi_y = 0 \\ a\eta_x + \Big(b - \sqrt{b^2-ac}\Big)\eta_y &= \eta_x - 3\eta_y = 0 \end{align}

where $$a=1$$, $$b=-1$$, $$c=-3$$.

Solving the above gives $$\xi = x-y$$ and $$\eta = 3x+y$$ and the original PDE reduces to

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

which has a general solution

$$u = F(\eta) + G(\xi) = F(3x+y) + G(x-y)$$