How to show $\int_Q \varphi^2 (\Delta v)v_t = \int_Q |\nabla v|^2 \varphi \varphi_t - 2\int_Q (\nabla v \cdot \nabla \varphi) (\varphi v_t)$

Question

Let $Q=\bigcup_{t \in (0,T)}\Omega \times \{t\}$. I have seen this identity for all $\varphi \in C_c^\infty(Q)$ such that $0 \leq \varphi \leq 1$:

$$\int_Q \varphi^2 (\Delta v)v_t = \int_Q |\nabla v|^2 \varphi \varphi_t - 2\int_Q (\nabla v \cdot \nabla \varphi) (\varphi v_t)$$

How do I prove it?

I get the left hand sides equals $$2\int_Q |\nabla v|^2 \varphi \varphi_t$$ but no idea what to do next.

Assume whatever smoothness of $v$ necessary.

Answer 1

We have, using integration by parts $$ \int_{Q}\varphi^2(\Delta v)v_t= -\int_{Q}\nabla (\varphi^2 v_t)\cdot\nabla v= -2\int_{Q}(\nabla\varphi\cdot\nabla v)\varphi v_t-\int_{Q}(\nabla v_t\cdot\nabla v)\varphi^2. $$ But $$ \int_{Q}(\nabla v_t\cdot\nabla v)\varphi^2=\frac{1}{2}\int_{Q}\frac{\partial}{\partial t}|\nabla v|^2\varphi^2=-\frac{1}{2}\int_{Q}\frac{\partial}{\partial t}(\varphi^2)|\nabla v|^2 =-\int_{Q}|\nabla v|^2\varphi\varphi_t. $$ Note that there are not boundary terms, as $\varphi$ is compactly supported.

(Edited some typos)