TITLE: Wave equation $u_{xx}+u_{xt}- u_{tt}=0$
QUESTION [7 upvotes]: 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.

REPLY [4 votes]: 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?