Wave equation $u_{xx}+u_{xt}- u_{tt}=0$ Does anybody know how we can solve the equation $u_{xx}+  u_{xt}-  u_{tt}=0$ with $u(x,0):=g(x)$ and $u_t(x,0):=h(x)?$ I mean it is known how to do this for the wave equation see here but I don't know how to do this in the more general case with the mixed term in it.
 A: I would try a solution of type $F(z)=F(x-ct)$, or $G(w)=G(x+ct)$. For $F$ this gives
$$
u_{xx}=F''(z)\\
u_{xt}=-cF''(z)\\
u_{tt}=c^2F''(z)
$$
so
$$
(1-c-c^2)F''(z)=0\implies c_{1,2}=\frac{1\pm\sqrt{5}}{2}.
$$
So $F(x-c_it)$ is a solution for every choice of $F$.
Same reasoning for $G$.
A: We are dealing with the wave equation $u_{xx}+u_{tt}-u_{tt}=0$. Notice that it has a mixed term. So we cannot use d'Alembert's formula here. (The formula assumes a difference of squares when "factoring" the derivatives.) Here, we have to "factor" the derivatives manually.
Note that we can rewrite our wave equation as $$\left( \frac{\partial^2}{\partial x^2} + \frac{\partial}{\partial x}\frac{\partial}{\partial t} - \frac{\partial^2}{\partial t^2}\right)u=0.$$ And "factoring" out the derivatives (using the quadratic formula to help) gives $$\left(\frac{\partial}{\partial x}+\frac{1-\sqrt{5}}2 \frac{\partial}{\partial t} \right)\left(\frac{\partial}{\partial x}+ \frac{1+\sqrt{5}}2 \frac{\partial}{\partial t} \right)u=0.$$
Now set $\frac{\partial}{\partial \xi}:=\frac{\partial}{\partial x}+\frac{1-\sqrt{5}}2 \frac{\partial}{\partial t}$ and $\frac{\partial}{\partial \eta}:=\frac{\partial}{\partial x}+\frac{1-\sqrt{5}}2 \frac{\partial}{\partial t}$. Then the wave equation becomes the Canonical form $$\frac{\partial}{\partial \xi}\frac{\partial}{\partial \eta}u=0$$ or $u_{\xi \eta}=0$. Furthermore, the chain rules of $\frac{\partial}{\partial \xi}=\frac{\partial}{\partial x}\frac{dx}{d\xi}+\frac{\partial}{\partial t}\frac{dt}{d\xi}$ and $\frac{\partial}{\partial \eta}=\frac{\partial}{\partial x}\frac{dx}{d\eta}+\frac{\partial}{\partial t}\frac{dt}{d\eta}$ imply $$\frac{dx}{d\xi}=1, \quad \frac{dx}{d\eta}=1, \quad \frac{dt}{d\xi}=\frac{1-\sqrt{5}}2, \quad \frac{dt}{d\eta}=\frac{1+\sqrt{5}}2.$$ Thus, we  obtain $$x=\xi+\eta, \quad t=\frac{1-\sqrt{5}}2\xi+\frac{1+\sqrt{5}}2\eta,$$ which means we can use these to perform change of variables.
Can you finish the rest from here?
A: Hint: note that your equation can be expressed as follows:
$$(D^2_x + D_x D_t  - D^2_t)\, u = (D_x- \varphi^- D_t)  (D_x - \varphi^+ D_t) u  = 0, $$ where $\varphi^{\pm} = (-1 \pm \sqrt{5})/2$ are the two solutions of $s^2+s-1=0$. What if we now define $v := (D_x - (-1+\sqrt{5})D_t/2) u$? Can you solve 1st order PDEs?
Cheers!

Of course, you will have noticed that $-\varphi^- = \varphi$ is the so-called golden ratio, satisfying $\varphi = 1 + \varphi^{-1}$.
