• $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be bounded and open with Lipschitz boundary $\partial\Lambda$
  • $\nu$ denote the outer unit normal vector of $\partial\Lambda$
  • $E:=\left\{u\in L^2(\Lambda,\mathbb R^d):\nabla\cdot u\in L^2(\Lambda^)\right\}$
  • $G:=\left\{\nabla p:p\in H^1(\Lambda)\right\}$

It's known that there are a unique bounded linear Operator $\gamma_0$ from $H^1(\Lambda,\mathbb R^d)$ to $L^2(\partial\Lambda,\mathbb R^d)$ with $$\gamma_0\left.u\right|_\Lambda=\left.u\right|_{\partial\Lambda}\;\;\;\text{for all }u\in C^1(\overline\Lambda)\tag 1$$ and a unique bounded linear operator $\gamma_\nu$ from $E$ to $H^{1/2}(\partial\Lambda)$ with $$\langle\gamma_0w,\gamma_\nu u\rangle_{H^{1/2}(\partial\Lambda)}=\langle v,\nabla\cdot u\rangle_{L^2(\Lambda)}+\langle\nabla v,u\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }u\in E\text{ and }v\in H^1(\Lambda)\tag 2$$ and $$\gamma_\nu\left.u\right|_{\Lambda}=\left.u\right|_{\partial\Lambda}\cdot\nu\;\;\;\text{for all }v\in C^1(\overline\Lambda,\mathbb R^d)\tag 3\;.$$ Moreover, it's known that $$L^2(\Lambda,\mathbb R^d)=G\oplus H\tag 4$$ with $$H:=\left\{w\in H^2(\Lambda,\mathbb R^d):\nabla\cdot w=0\text{ and }\gamma_\nu w=0\right\}\;.$$

Now, let $u\in L^2(\Lambda,\mathbb R^d)$. Using $(4)$, we obtain $$u=\nabla p+w\tag 5$$ for some unique $(\nabla p,w)\in G\times H$. Given $u$, I want to numerically compute $w$. In order to do that, I guess that I need to first compute $p$ and then obtain $w$ by $(5)$.

Since I plan to compute $p$ by a Galerkin method, I need to find a variational formulation for that problem. How can we do that?

Sure, we should be in the well-known case of a Poisson equation: Since $\nabla\cdot w=0$, we obtain $w\in E$ and $$\nabla\cdot u=\nabla\cdot\nabla p\;,\tag 6$$ i.e. $$\langle\nabla\phi,u\rangle_{L^2(\Lambda,\:\mathbb R^d)}=\langle\nabla\phi,\nabla p\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for all }\phi\in C_c^\infty(\Lambda)\;.\tag 7$$ However, this shouldn't be sufficient to come up with a variational formulation of the problem. What I mean is: I guess that we need to specify a boundary condition for $p$ using $\gamma_0$ and $\gamma_\nu$.

How should we do that and what are the correct test function and solution spaces?

  • $\begingroup$ $ \mathbf u\boldsymbol\nu = \frac{\partial p}{\partial \nu}$. Were all those spaces and operators really that necessary? $\endgroup$ – uranix Dec 20 '16 at 14:34
  • $\begingroup$ @uranix What is $\mathbf u\boldsymbol\nu$? And in which sense is this an answer to my questions? $\endgroup$ – 0xbadf00d Dec 20 '16 at 17:56
  • $\begingroup$ This is the missing boundary condition for the Poisson's equation. $\endgroup$ – uranix Dec 20 '16 at 20:59
  • $\begingroup$ @uranix My problem is the derivation of a variational formulation. Look: Assume the $u$ in the question is not only in $L^2(Λ,\mathbb R^d)$, but even in $E$ such that we can apply $γ_\nu$ to it. Then, taking the divergence on both sides of $(5)$ we see that $$Δp=∇⋅∇ p=∇⋅u\in L^2(Λ)\tag 8$$ (since $∇⋅w=0$). Moreover, $$γ_\nu∇ p=γ_\nu u\tag 9$$ (since $γ_νw=0$). Thus, we obtain $$\langle γ_0 v,γ_\nu u\rangle_{H^{1/2}(∂Λ)}=\langle v,∇⋅u\rangle_{L^2(Λ)}+\langle∇ v,∇ p\rangle_{L^2(Λ,\:ℝ^d)}\tag{10}$$ (by multiplying $(8)$ with $v$, integrate and applying the Stokes formula) for all $v\in H^1(Λ)$. $\endgroup$ – 0xbadf00d Dec 21 '16 at 13:04
  • $\begingroup$ @uranix Now, the problem is that I don't know what I need to do with the term $\langle v,∇⋅u\rangle_{L^2(Λ)}$ (to be precise, I want to eliminate the divergence such that the term only contains $u$ itself). It would be clear what I can do with it, if $v$ would belong to $C_c^\infty(\Lambda)$ (or $H_0^1(\Lambda)$), by the Definition of the weak divergence. But we cannot choose $v$ from those spaces, since they vanish at the boundary and hence "don't see the boundary condition". $\endgroup$ – 0xbadf00d Dec 21 '16 at 13:06

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