• $d\in\mathbb N$
  • $\lambda$ denote the Lebesgue measure on $\mathbb R^d$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with Lipschitz boundary
  • $G:=\left\{\nabla p:p\in H^1(\Lambda)\right\}$
  • $H:=\left\{u_0\in L^2(\Lambda,\mathbb R^d):\nabla\cdot u_0=0\text{ and }\gamma_\nu u_0=0\right\}$$^1$

It's easy to see that $G$ is a closed subspace of $L^2(\Lambda,\mathbb R^d)$ with $H=G^\perp$ and $$L^2(\Lambda,\mathbb R^d)=G\oplus H\tag 1\;.$$ Now, let $u\in L^2(\Lambda,\mathbb R^d)$. By $(1)$, $$u=\nabla p+u_0\tag 2$$ for some unique $(\nabla p,u_0)\in G\times H$.

Given $u$, how can we formulate a variational problem whose unique solution is $p$?

My idea is the following: Let $W:=\left\{w\in H^1(\Lambda,\mathbb R^d):\int_\Lambda w\:{\rm d}\lambda=0\right\}$, $$a(p,q):=\langle\nabla p,\nabla q\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }p,q\in W$$ and $$\ell(q):=\langle u,\nabla q\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }q\in W\;.$$ By the Lax-Milgram theorem, there is a unique $p\in W$ with $$a(p,q)=\ell(q)\;\;\;\text{for all }q\in W\tag 3$$ and it's easy to see that the $p$ in $(2)$ satisfies $(3)$.

So, we should be done, shouldn't we? However, I've read that the variational formulation for the problem of finding the $p$ in $(2)$ needs to incorporate the Neumann boundary condition $\gamma_\nu\nabla p=\nabla u$. So, my question is: Does the solution of $(3)$ satisfy this condition or do we need to find an other variational formulation which directly incorporates this condition?

$^1$ It's well-known that there is a unique bounded linear operator $\gamma_0$ from $H^1(\Lambda,\mathbb R^d)$ to $L^2(\partial\Lambda,\mathbb R^d)$ with $$\gamma_0\left.u\right|_\Lambda=\left.u\right|_{\partial\Lambda}\;\;\;\text{for all }u\in C^1(\overline\Lambda)\tag 4$$ and a unique bounded linear operator $\gamma_\nu$ from $E$ to $H^{1/2}(\partial\Lambda)$ with $$\langle\gamma_0w,\gamma_\nu u\rangle_{H^{1/2}(\partial\Lambda)}=\langle v,\nabla\cdot u\rangle_{L^2(\Lambda)}+\langle\nabla v,u\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u\in E\text{ and }v\in H^1(\Lambda)\tag 5$$ and $$\gamma_\nu\left.u\right|_{\Lambda}=\left.u\right|_{\partial\Lambda}\cdot\nu\;\;\;\text{for all }v\in C^1(\overline\Lambda,\mathbb R^d)\tag 6\;.$$


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