# Does the solution of my variational problem satisfy a specific Neumann boundary condition?

Let

• $d\in\mathbb N$
• $\lambda$ 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\;.$\$