测试
\[\begin{aligned} &\int_\Omega u_{tt}(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x\\ =&\frac{{\mathop{}\!\mathrm{d}}}{{\mathop{}\!\mathrm{d}} t}\int_\Omega u_t(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x-\int_\Omega u_t (h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x\\ =&\frac{{\mathop{}\!\mathrm{d}}}{{\mathop{}\!\mathrm{d}} t}\int_\Omega u_t(h(x)\cdot \nabla u){\mathop{}\!\mathrm{d}} x-\frac12 \int_\Omega h(x)\cdot \nabla |u_t|^2 {\mathop{}\!\mathrm{d}} x. \end{aligned}\] 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{