Q:

PDE with mixed derivative

We condsider the linear PDE :$${u_{tt}} + {u_{tx}} - {u_{xx}} = 0$$
with boundary conditions:$$u(t,0) = u(t,0) = 0$$
and initial data:$$u(0,x) = {u_0}(x),{u_t}(0,x) = {u_1}(x)$$.
I tried to apply the separation of variabes methode but it didn't work...Is there any method to solve this kind of equation explicitly ? thanks.

A:

The equation can be solved with the variable change:
$$\begin{cases}
 \xi=t+ax\\
 \eta=t+bx
\end{cases}$$
to transform the equation into $u_{\xi\eta}=0$ with general solution $u=f(\xi)+g(\eta)$ with $f$ and $g$ some single variable, differentiable functions depending on the boundary and initial conditions.
We can compute the partials expressed in terms of the new variables:
$$u_t=u_\xi+u_\eta$$
$$u_x=au_\xi+bu_\eta$$
We need the second partials:
$$u_{tt}=u_{\xi\xi}+2u_{\xi\eta}+u_{\eta\eta}$$
$$u_{xx}=a^2u_{\xi\xi}+2abu_{\xi\eta}+b^2u_{\eta\eta}$$
$$u_{tx}=au_{\xi\xi}+(a+b)u_{\xi\eta}+bu_{\eta\eta}$$
So,
$${u_{tt}} + {u_{tx}} - {u_{xx}} =(1-a^2+a)u_{\xi\xi}+(2-2ab+a+b)u_{\xi\eta}+(1-b^2+b)u_{\eta\eta}=0$$
We can impose to $a$ and $b$ to make zero the coefficients for the first and third terms, so is,
$$1-a^2+a=0\implies a=\dfrac{1\pm\sqrt{5}}{2}$$
$$1-b^2+b=0\implies b=\dfrac{1\pm\sqrt{5}}{2}$$
We can choose $a=\dfrac{1+\sqrt{5}}{2}$ and $b=\dfrac{1-\sqrt{5}}{2}$, leading to,
$$2u_{\xi\eta}=0$$
The solution in terms of $t$ and $x$ is:
$$u=f(t+\dfrac{1+\sqrt{5}}{2}x)+g(t+\dfrac{1-\sqrt{5}}{2}x)$$

