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Let $\Omega \subset \mathbb{R}$ in a bounded polygon domain and $f:\Omega \to \mathbb{R}$ known function.We split the boundary into two parts $\partial \Omega_{1}$ and $\partial \Omega_{2}$ such that $\partial \Omega =\partial \Omega_{1} \cup \partial \Omega_{2}.$ The elliptic problem of Dirichlet & Neumann conditions is :

\begin{align*} -\Delta u +u &=f \quad in \quad \Omega \subset\mathbb{R}^2 \\ u&=0 \quad in \quad \partial \Omega_{1} \\ \frac{\partial u}{\partial \mathbf{n}}&=1 \quad in \quad \partial \Omega_{2} \end{align*}

It was asked to write the problem in a weak form in a suitable solution space $\mathcal{V}$ and the bilinear form $a(\cdot,\cdot):\mathcal{V} \times \mathcal{V} \to \mathbb{R}$ as long as a norm $||\cdot||_{\mathcal{V}}$ and show that is indeed a norm.

My beginning thought :

\begin{align*} -\Delta u+u &=f \\ -\Delta u v+uv &=fv \\ -\int_{\Omega}\Delta u v dV+ \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \int_{\Omega}\nabla u \cdot \nabla v dV- \int_{\partial \Omega_{2}}v \nabla u \cdot \vec{n} dS + \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \int_{\Omega}\nabla u \cdot \nabla v dV- 1 + \int_{\Omega}uv dV&=\int_{\Omega}fv dV \\ \end{align*}

So $$a(u,v) = \int_{\Omega} u'(x) \cdot v'(x) dx- 1 + \int_{\Omega}u (x)\cdot v(x) dx=\int_{\Omega}f(x) \cdot v(x) dx \quad \text{(1)}$$

in $\mathcal{V}_{h}$ subspace of $\mathcal{V}$

$$a(u,v) = \int_{\Omega} u_{h}'(x) \cdot v_{h}'(x) dx- 1 + \int_{\Omega}u(x) \cdot v_{h(x)} dx=\int_{\Omega}f(x) \cdot v_{h}(x) dx \quad \text{(2)} $$

Subtracting (1) from (2) we obtain : $$\int_{\Omega} \left(u'(x)- u'_{h}(x)\right) v'_{h}dx = 0$$

but $$||u ||_{\mathcal{V}} = \left( \int_{\Omega}(v'(x))^2 dx\right)^{1/2}$$ is indeed a norm for the space $\mathcal{V}$

\begin{align*} ||u - u_{h} ||_{\mathcal{V}} &= \inf_{v_{h} \in \mathcal{V}_{h}} \\ ||u - u_{h} ||_{\mathcal{V}}^2&= \int_{\Omega} \left( u'(x) -v'_{h}(x) \right)\left( u'(x) -v'_{h}(x) \right) dx \\ &= \int_{\Omega} \left( u'(x) -v'_{h}(x) \right) u'(x) dx - \underbrace{\int_{\Omega} \left( u'(x) -v'_{h}(x) \right) u'_{h}(x) dx}_{G.O = 0 }\\ &= \int_{\Omega} \left( u'(x) -v'_{h}(x) \right) u'(x) dx - \underbrace{ \int_{\Omega} \left( u'(x) -v'_{h}(x) \right) v'_{h}(x) dx}_{G.O = 0 } \end{align*}

Therefore $$||u - u_{h} ||_{\mathcal{V}}^2 =\int_{\Omega} \left( u'(x) -v'_{h}(x) \right)\left( u'(x) -v'_{h}(x) \right) dx \leq ||u-u_{h}||_{\mathcal{V}}||v-v_{h}||_{\mathcal{V}} $$

and by the definition of inf, we must also have:

$$\inf_{v_{h} \in \mathcal{V}_{h}} ||u-u_{h}||_{\mathcal{V}} \leq ||v-v_{h}||_{\mathcal{V}} $$

right calculation (post)

\begin{align*} \int_{\partial \Omega}v \nabla u \cdot \vec{n} dS &=\int_{\partial\Omega_{1}} v \cdot \frac{\partial u}{\partial n} - \int_{\partial \Omega_{2}} v \cdot \frac{\partial u}{\partial n} dV\\ &=0 + \int_{\partial\Omega_{2}} v \cdot 1 dV \\ &= \int_{\partial\Omega_{2}} v dV \end{align*}

$$\int_{ \Omega}\nabla u \cdot \nabla v dV +\int_{\partial \Omega_{2}} v dV+ \int_{ \Omega}uv dV=\int_{\Omega}fv dV$$

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  • $\begingroup$ I doubt that your form $a$ represents an inner product (though I have not proven it). I think that you want to multiply by $v$ your differential equation and then integrate by parts. $\endgroup$ – nicoguaro May 13 at 21:21
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    $\begingroup$ @nicoguaro, it does not since $a(u,v)\neq a(v,u)$. Tsiantakis, you need to pick appropriate test and trial spaces. They can be the same or different, but for your task, it is easier to pick them the same. Then the weak problem can be obtained by constraining $u$ to be in the trial space, multiplying the PDE by test functions, and if necessary, applying "integration by parts". That gives you the bilinear and linear forms you need. You can handle the Dirichlet b.c.s immediately by choosing the spaces properly, and Neumann b.c.s will be important after integration by parts. $\endgroup$ – Abdullah Ali Sivas May 14 at 0:29
  • $\begingroup$ I think that you answered your question. So, I suggest that you revert your edition and create an answer with it. $\endgroup$ – nicoguaro May 14 at 15:40
  • $\begingroup$ so it is right? @nicoguaro $\endgroup$ – user38211 May 14 at 15:46
  • $\begingroup$ You still need to write the space down. Also where does $-1$ come from? $\endgroup$ – Abdullah Ali Sivas May 14 at 16:15

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