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Let $\Omega \subset \mathbb R^n$ and $\Gamma \subset \partial \Omega$ and $\Gamma_n=\partial \Omega \setminus \Gamma$, $V_0=\{v \in H^1(\Omega) : v=0 \text{ on } \Gamma\}$.

Given the variational statements $$a(u,v)=\int_\Omega a(x)\nabla u(x) \cdot \nabla v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx}$$ or $$a(u,v)=\int_\Omega a(x)\nabla u(x) \cdot \nabla v(x)+\bigl [\nabla \cdot (b(x) v(x)) \bigr]u(x) \, \mathrm{dx}$$ for $b\in \mathbb R^n$ where we seek $u\in V_0$ such that $a(u,v)=f(v):=\int_{\Omega} f(x)v(x)\, \mathrm{dx}+ \int_{\Gamma} g(x)v(x)\, \mathrm{ds}$ for some $f \in L^2(\Omega)$ and all $v \in V_0$, how do I find the corresponding elliptic PDE and boundary conditions asssuming extra regularity? For "simple" weak forms such as $$a(u,v)=\int_\Omega a(x)\nabla u(x) \cdot \nabla v(x)+c(x)u(x)v(x) \, \mathrm{dx}$$ it is just $-a(x)\Delta u(x)+c(x)u(x) =f(x) $ in $\Omega$ and $u=0$ on $\partial \Omega$ if I got this right.

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    $\begingroup$ Do the terms with $a$ really contain only $v$ or should it be $\nabla v$? $\endgroup$
    – gerw
    Commented Jan 17, 2020 at 11:39
  • $\begingroup$ @gerw thanks, forgot it and then copy pasted it into other integrals $\endgroup$
    – Tesla
    Commented Jan 17, 2020 at 12:06
  • $\begingroup$ @Tesla How do you define $f(v)$? $\endgroup$
    – ahdahmani
    Commented Jan 18, 2020 at 12:48
  • $\begingroup$ @ahdahmanii sorry added details to it $\endgroup$
    – Tesla
    Commented Jan 18, 2020 at 13:19

1 Answer 1

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Assuming that $\Omega $ is Lipschitz- continuous subset of $\mathbb{R}^n$ and making use of the identity I.2.17 and theorem 2.4 and 2.5 in [V. Girault and P. A. Raviart, Finite Element Methods for Navier-Stokes Equations. Theory and Algorithms, Springer Series in Computational Mathematics, 5 Springer, Berlin, 1986.] Which state that\

  • $\mathcal{D}(\bar{\Omega})^n$ is dense in $ H(\operatorname{div},\Omega)$\

and

  • the map $\gamma: v \rightarrow v\cdot\eta|_{\Gamma_n}$ defined on $\mathcal{D}(\bar{\Omega})^n$ can be extended by continuity to a linear and continuous mapping from $H(\operatorname{div},\Omega)$ into $H^{-\frac{1}{2}}(\Gamma_n)$\ So, we have the following Green formula

$$(v,\nabla \phi)+(\operatorname{div} v,\phi)=(v\cdot \eta,\phi)_{\Gamma_n} \quad \forall v \in H(\operatorname{div},\Omega), \forall \phi \in H^1(\Omega)$$

Therefore, we have for every $v \in V_0$

\begin{align} a(u,v)=\int_\Omega a(x)\nabla u(x) \cdot \nabla v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx} \\ = \int_\Omega -\operatorname{div} \big(a(x)\nabla u(x)\big)v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx}\\+\int_{\Gamma_n} a(x)\nabla u(x) \eta v(x) \mathrm{ds} \end{align}

while $\eta$ is the outward unit normal vector. Now, taking $v \in \mathcal{D}(\Omega)$ we have $$ \int_\Omega -\operatorname{div} \big(a(x)\nabla u(x)\big)v(x)+\bigl (b(x) \cdot \nabla u(x) \bigr)v(x) \, \mathrm{dx}=\int_{\Omega} f(x)v(x)\, \mathrm{dx}$$ So, $$ -\operatorname{div} \big(a(x)\nabla u(x)\big)+\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \text{a.e. in } \Omega$$ Which can be extended into all $\Omega$

Making use of this, we find $$ a(v) \frac{\partial u}{\partial \eta}=g(x), \quad \forall x \in \Gamma_n$$

Finely, the equation is

$$ \left\{\begin{aligned} -\operatorname{div} \big(a(x)\nabla u(x)\big)+\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \forall x \in \Omega\\ a(x) \frac{\partial u}{\partial \eta}=g(x), \quad \forall x \in \Gamma_n \end{aligned}\right. $$

The second one can be treated the same way to get

$$ \left\{\begin{aligned} -\operatorname{div} \big(a(x)\nabla u(x)\big)-\bigl (b(x) \cdot \nabla u(x) \bigr)=f(x) \quad \forall x \in \Omega\\ a(x) \frac{\partial u}{\partial \eta}+b(x)u(x)\eta=g(x), \quad \forall x \in \Gamma_n \end{aligned}\right. $$

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  • $\begingroup$ Thanks a lot, will have a look at the identities you used. Do you happen to know the equation to the second weak form I stated? $\endgroup$
    – Tesla
    Commented Jan 18, 2020 at 14:24
  • $\begingroup$ And do the identies 2.17 and theorem 2.3, 2.5 have names so I can look them up on the internet? $\endgroup$
    – Tesla
    Commented Jan 18, 2020 at 14:26
  • $\begingroup$ @Tesla, yes, I'm sorry, I have fixed it $\endgroup$
    – ahdahmani
    Commented Jan 18, 2020 at 15:27
  • $\begingroup$ thanks a lot. can you just state the theorems very shortly that you used so that i can understand the first steps? $\endgroup$
    – Tesla
    Commented Jan 18, 2020 at 15:29
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    $\begingroup$ @Tesla further details added. look at it now $\endgroup$
    – ahdahmani
    Commented Jan 18, 2020 at 15:58

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