An equality that holds with $v_t \in L^2(0,T;L^2(\Omega))$ but its proof requires $v_t \in L^2(0,T;H^1(\Omega))$

Question

Let $Q=(0,T)\times \Omega$. For all $\varphi \in C_c^\infty(Q)$ such that $0 \leq \varphi \leq 1$, the following holds

$$\int_Q \varphi^2 (\Delta v)v_t = \int_Q |\nabla v|^2 \varphi \varphi_t - 2\int_Q (\nabla v \cdot \nabla \varphi) (\varphi v_t)$$ when $v \in L^2(0,T;H^2(\Omega))$ such that $v_t \in L^2(0,T;H^1(\Omega))$.

It has been claimed that this identity holds for all $v \in L^2(0,T;H^2(\Omega))$ such that $v_t \in L^2(0,T;L^2(\Omega))$ (notice the weaker space the time derivative is in). Of course all terms in the identity make sense with $v$ in this weaker space, BUT the proof (see this answer) uses integrating by parts and one of the intermediary calculations includes the term $\nabla v_t$. So I don't see why the equality is true in the weaker case of $v_t \in L^2(0,T;L^2(\Omega))$ because the proof requires a stronger time derivative.

I am guessing a density argument is used to prove this for all functions in the weaker space. But I can't find a sufficiently smooth dense subset that will do the job because we need to pass to the limit in the Laplacian term on the left hand side. Can anyone tell me how this works? Is there a different way to get this identity maybe with a different proof to the one linked?

• This is related to a Mathoverflow question I posted.

• The source of the claim is Variational methods in the stefan problem by José-Francisco Rodrigues, in the proof of Proposition 4.7.

Answer 1

Edited after a quite lengthly comment discussion.

Let $X:=\left\{v\in L^{2}\left(0,T,H^{2}\right):v_{t}\in L^{2}\left(0,T,L^{2}\right)\right\}$ and let $\left\|v\right\|_{X}:=\left\|v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|\nabla v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|\nabla^{2} v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|v_{t}\right\|_{L^{2}\left(0,T,L^{2}\right)}$.

Let $Y:=\left\{v\in L^{2}\left(0,T,H^{2}\right):v_{t}\in L^{2}\left(0,T,H^{1}\right)\right\}$ and let $\left\|v\right\|_{Y}:=\left\|v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|\nabla v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|\nabla^{2} v\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|v_{t}\right\|_{L^{2}\left(0,T,L^{2}\right)}+\left\|\nabla v_{t}\right\|_{L^{2}\left(0,T,L^{2}\right)}$.

Your assumption is that for $v\in Y$ the following equation holds:

$\int_{Q}\varphi^{2}\left(\Delta v\right)v_{t}=\int_{Q}\left|\nabla v\right|^{2}\varphi\varphi_{t}-2\int_{q}\left(\nabla v\cdot \nabla\varphi\right)\varphi v_{t}$

Now this shall be generalised to $X$ by density. First note that $Y$ is dense in $X$:

$C^{\infty}$ is dense in $X$ respective $\left\|\cdot\right\|_{X}$, $C^{\infty}\subset Y$ and (obviously) $Y\subset X$. Thus we get $Y$ is dense in $X$ respective $\left\|\cdot\right\|_{X}$.

Now fix $v\in X$ and let $\left(v_{n}\right)_{n\in\mathbb{N}}\in Y$ such that $\left\|v-v_{n}\right\|_{X}\rightarrow 0$. This convergence implies:

• $\left\|\Delta v_{n}-\Delta v\right\|_{L^{2}\left(0,T,L^{2}\right)}\rightarrow 0$
• $\left\|v_{nt}-v_{t}\right\|_{L^{2}\left(0,T,L^{2}\right)}\rightarrow 0$
• $\left\|\nabla v_{n}-\nabla v\right\|_{L^{2}\left(0,T,L^{2}\right)}\rightarrow 0$

since $\varphi\in C^{\infty}$ is bounded this gives us, that all the integrants (all are products of $L^{2}$ functions converging strongly in $X$) are converging strongly in $L^{1}\left(0,T,L^{1}\right)$. The integral operator is continuous on $L^{1}$, so we can pass to the limit.