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Let $ (M,g) $ be a smooth Riemann manifold without boundary, $ S^{n-1} $ is an $ n $-dimensional sphere, and $ T>0 $. Consider a weak solution $ u:M\times[0,T]\to S^{n+1} $ of $$ \partial_tu-\Delta u=|\nabla u|^2u\text{ in }M\times[0,T],\text{ and }u(x,0)=u_0(x)\text{ on }M $$ for $ u_0\in H^1(M) $ such that $ \partial_tu\in L^2(0,T,L^2(M)) $, $ \nabla u\in L^{\infty}(0,T,L^2(M)) $. Precisely, $ u $ satisfies $$ \int_0^T\int_{M}\partial_tu\varphi\mathrm{d}v_g\mathrm{d}t+\int_0^T\int_{M}\nabla u\cdot\nabla\varphi\mathrm{d}v_g\mathrm{d}t=\int_0^T\int_M|\nabla u|^2u\varphi\mathrm{d}v_g\mathrm{d}t $$ for any $ \varphi\in C_c^{\infty}(M\times(0,T)) $. Moreover, we assume that $ u $ satisfies the energy inequality $$ \int_0^T\int_M|\partial_tu|^2\mathrm{d}v_g\mathrm{d}t+\sup_{0\leqq t\leqq T}\int_{M}|\nabla u|^2\mathrm{d}v_g\leqq\int_{M}|\nabla u_0|^2\mathrm{d}v_g. $$ From the definition, we have that for $ \theta\in C_0^{\infty}(M\times(0,T)) $, $ \theta \partial_tu\in L^2(M\times(0,T)) $ is the legal test function. However it seems that $ \Delta u(\theta\partial_tu) $ doesn't make sense. On the other hand, if $ u $ is smooth, one can use integration by parts to obtain $$ \int_0^T\int_M\left(\theta|\partial_tu|^2-\frac{1}{2}|\nabla u|^2\partial_t\theta+\partial_tu\nabla u\cdot\nabla\theta\right)\mathrm{d}v_g\mathrm{d}t=0, $$ here all the terms in the equality is valid, is that mean such formal results can be done rigorously?

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