Problem with solving PDE I'm trying to solve this equation:
$u_{tt} = u_{x_1x_1} + u_{x_2x_2} + u_{x_3x_3}$
$u(x,0) = x_1^2\sin(x_2+x_3)$
$u_t(x,0) = 0$
In what form to find a solution? I tried in form $u = \alpha(t)x_1^2\sin(x_2+x_3) + \beta(t)\sin(x_2+x_3)$,but this way gives only the trivial solution.
 A: This is the wave equation in $3$ dimensions. 
For initial condition $u(x, 0) = \sin( \alpha x_1 + x_2 + x_3)$, $u_t(x,0) = 0$ we would have the solution $u = (\sin(\alpha x_1 + x_2  + x_3  + \sqrt{2+\alpha^2} t) + \sin(\alpha x_1 + x_2 + x_3 - \sqrt{2+\alpha^2} t))/2$.
Take $-$ the second derivative of this with respect to $\alpha$, and evaluate at $\alpha = 0$: we get 
$$ \dfrac{x_1^2}{2} \left(\sin(x_2 + x_3 - \sqrt{2} t) + \sin(x_2 + x_3 + \sqrt{2} t)\right)+ \dfrac{\sqrt{2}t}{4} \left(\cos(x_2 + x_3 - \sqrt{2}t)  
- \cos(x_2 + x_3 + \sqrt{2} t)\right) $$
which is a solution that satisfies your initial conditions.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\int_{0}^{\infty}\nabla^{2}{\rm u}\pars{\vec{r},t}\expo{-st}\,\dd t
=
\int_{0}^{\infty}{\rm u}_{tt}\pars{\vec{r},t}\expo{-st}\,\dd t
=
-{\rm u}_{t}\pars{\vec{r},0} + s\int_{0}^{\infty}{\rm u}_{t}\pars{\vec{r},t}
\expo{-st}\,\dd t
\\[3mm]&=
-{\rm u}_{t}\pars{\vec{r},0} - s{\rm u}\pars{\vec{r},0}
+
s^{2}\int_{0}^{\infty}{\rm u}_{t}\pars{\vec{r},t}\expo{-st}\,\dd t
\end{align}
$$
\pars{\nabla^{2} - s^{2}}\tilde{\rm u}\pars{\vec{r},s} = -s{\rm u}\pars{\vec{r},0}
$$
