\[\begin{aligned}
&\int_\Omega u_{tt}(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x\\
=&\frac{{\mathop{}\!\mathrm{d}}}{{\mathop{}\!\mathrm{d}} t}\int_\Omega u_t(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x-\int_\Omega u_t (h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x\\
=&\frac{{\mathop{}\!\mathrm{d}}}{{\mathop{}\!\mathrm{d}} t}\int_\Omega u_t(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x-\frac12 \int_\Omega h(x)\cdot \nabla |u_t|^2 {\mathop{}\!\mathrm{d}} x.
\end{aligned}\]
由于\(\int_a^b f(x) {\mathop{}\!\mathrm{d}} x=F(b)-F(a)\), 于是
\[\begin{aligned}
&\int_\Omega h(x)\cdot \nabla |u_t|^2{\mathop{}\!\mathrm{d}} x=\sum_i^N\int_\Omega h_i\frac{\partial}{\partial x_i}|u_t|^2 {\mathop{}\!\mathrm{d}} x\\
=&\sum_i^N\int_{\partial\Omega}h_i |u_t|^2 \nu_i{\mathop{}\!\mathrm{d}} S-\sum_i^N\int_\Omega |u_t|^2 \frac{\partial h_i}{\partial x_i}{\mathop{}\!\mathrm{d}} x\\
=&0-\int_\Omega (\nabla \cdot h)|u_t|^2 {\mathop{}\!\mathrm{d}} x,
\end{aligned}\]
所以最终有
\[\begin{aligned}
&\int_\Omega u_{tt}(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x\\
=&\frac{{\mathop{}\!\mathrm{d}}}{{\mathop{}\!\mathrm{d}} t}\int_\Omega u_t(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x+\frac12 \int_\Omega (\nabla \cdot h)|u_t|^2 {\mathop{}\!\mathrm{d}} x.
\end{aligned}\]
来源：https://www.cnblogs.com/sunfenglong/p/11333777.html