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.
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.
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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} $$ \pars{\nabla^{2} - s^{2}}\tilde{\rm u}\pars{\vec{r},s} = -s{\rm u}\pars{\vec{r},0} $$
