How can we prove $\langle(Ψ\cdot∇)Φ,Φ\rangle_{L^2(Λ,\:ℝ^d)}+\frac12\langle(∇\cdotΨ)Φ,Φ\rangle_{L^2(Λ,\:ℝ^d)}=0$ for $Φ,Ψ\in H_0^1(Λ,ℝ^d)$?

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be nonempty and open
• $\Phi,\Psi\in H_0^1(\Lambda,\mathbb R^d)$

I've read that $$\langle(\Psi\cdot\nabla)\Phi,\Phi\rangle_{L^2(\Lambda,\:\mathbb R^d)}+\frac12\langle(\nabla\cdot\Psi)\Phi,\Phi\rangle_{L^2(\Lambda,\:\mathbb R^d)}=0\tag1\;,$$ but how can we prove it? Note that $$(\Psi\cdot\nabla)\Phi=\sum_{i=1}^d(\Psi\cdot\nabla)\Phi_i\tag2\;.$$

Answer 1

Let $\lambda$ denote the Lebesgue measure on $\mathbb R^d$ and $$b(u,v,w):=\int_\Lambda(u\cdot\nabla)v\cdot w\:{\rm d}\lambda\;\;\;\text{for }u,v,w\in H^1(\Lambda,\mathbb R^d)\;.$$ Note that

\begin{equation} \begin{split} b(u,v,w)&=\sum_{i=1}^d\sum_{j=1}^d\langle u_jw_i,\frac{\partial v_i}{\partial x_j}\rangle_{L^2(\Lambda)}\\&=-\sum_{i=1}^d\sum_{j=1}^d\langle\frac\partial{\partial x_j}u_jw_i,v_i\rangle_{L^2(\Lambda)}\\&=-\sum_{i=1}^d\sum_{j=1}^d\langle\frac{\partial u_j}{\partial x_j}w_i+\frac{\partial w_i}{\partial x_j}u_j,v_i\rangle_{L^2(\Lambda)}\\&=-\langle(\nabla\cdot u)v,w\rangle_{L^2(\Lambda,\:\mathbb R^d)}-b(u,w,v) \end{split}\tag3 \end{equation}

for all $v\in H^1(\Lambda,\mathbb R^d)$ and $u,w\in C_c^\infty(\Lambda,\mathbb R^d)$. Now, if $\Lambda$ is bounded and $d\le 4$, then $b$ is a bounded trilinear form on $H_0^1(\Lambda,\mathbb R^d)\times H_0^1(\Lambda,\mathbb R^d)\times H_0^1(\Lambda,\mathbb R^d)$ and hence we obtain $(1)$ from $(3)$ by a density argument.