Use Lax-Milgram theorem to prove the existence of weak solution for an elliptic equation

Let $\Omega$ an open bounded and regular domain to $\mathbb{R^n}$ and let $\{\overline{\Omega_1},\overline{\Omega_2}\}$ a partition of $\Omega.$ $\bar{\Omega} = \bar{\Omega_1} \cup \bar{\Omega_2}.$ We put $\Gamma = \partial \Omega_1 \cap \partial \Omega_2$ the interface between $\Omega_1$ and $\Omega_2$ such that $\Gamma \subset \Omega.$

We put $u_1 = u/\Omega_1$ (restriction $u$ to $\Omega_1$) and we condider the problem:

$$-k_i \Delta u_i = f , x \in \Omega_1 , i=1,2$$ $$u_1=0 , x \in \partial \Omega$$ $$u_1 = u_2 , x \in \Gamma$$ $$k_1 \nabla u_1 n = k_2 \nabla u_2 n , x \in \Gamma$$

$k(x)$ is an piecewise constant , with $k(x) = k_i > 0, i = 1,2$ and $f \in L^2(\Omega)$

My questions are, how:

1-prouve that this problem admit a unique solution in an adequat Hilbert space $V.$

2- Prouve that $\exists c > 0, ||u||_V \leq c$ and prouve that $u$ verfies the minimum of energy.

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Hi, @jijiii, did you see my comment in math.stackexchange.com/questions/402922/… ? – Shuhao Cao May 31 '13 at 16:55
For the second condition on $\Gamma$, what is $\eta$, do you mean $k_1\nabla u\cdot n = k_2\nabla u\cdot n$? where $n$ is the unit vector normal to the $\Gamma$. – Shuhao Cao May 31 '13 at 16:57
yes, n is the unit vector normal to $\Gamma.$ ($n$ and not $\eta$) – jijiii May 31 '13 at 17:02

The big picture is: If $u$ satisfies those two continuity conditions on $\Gamma$ together with the equation in each subdomain, then $u$ solves the weak problem for the following diffusion equation on the whole domain: \left\{\begin{aligned} -\nabla \cdot (k\nabla u )&= f \quad\text{in } \Omega, \\ u&=0 \quad\text{on } \partial \Omega. \end{aligned}\right.

Multiply both equations by a same $H^1_0(\Omega)$ test function $v$: $$-\int_{\Omega_1}k_1 \Delta u_1\,v -\int_{\Omega_2}k_2 \Delta u_2\,v= \int_{\Omega}f\,v.$$ Integrating by parts using Green's identity on each domain: $$\int_{\Omega_1}k_1 \nabla u_1\cdot\nabla v - \int_{\partial \Omega_1}k_1 (\nabla u_1\cdot n_1)v + \int_{\Omega_2}k_2 \nabla u_2\cdot\nabla v - \int_{\partial \Omega_2}k_2 (\nabla u_2\cdot n_2)v = \int_{\Omega}f\,v,$$ where $n_1$ and $n_2$ are the unit vector normal to the boundaries of $\Omega_1$ and $\Omega_2$ respectively. Now we focus on the boundary part, for $\partial \Omega_1\cap \partial \Omega_2 = \Gamma$: \begin{aligned} &\int_{\partial \Omega_1}k_1 (\nabla u_1\cdot n_1)v + \int_{\partial \Omega_2}k_2 (\nabla u_2\cdot n_2)v \\ =& \int_{\partial \Omega_1\backslash\Gamma}k_1 (\nabla u_1\cdot n_1)v + \int_{\partial \Omega_2\backslash\Gamma}k_2 (\nabla u_2\cdot n_2)v + \int_{\Gamma} (k_1 \nabla u_1\cdot n_1 + k_2\nabla u_2\cdot n_2)v \\ =& \int_{\partial \Omega}k (\nabla u\cdot n)v + \int_{\Gamma} (k_1 \nabla u_1\cdot n_1 + k_2\nabla u_2\cdot n_2)v. \end{aligned} The first term vanishes because $v=0$ on $\partial \Omega$. Second term vanishes because $$k_1 \nabla u_1\cdot n - k_2\nabla u_2\cdot n = k_1 \nabla u_1\cdot n_1 + k_2\nabla u_2\cdot n_2 = 0,$$ by $n_1$ and $n_2$ share the same magnitude but point the opposite direction pointwisely. By the other interface condition $u_1 = u_2$ on $\Gamma$, we know $u\in H^1_0(\Omega)$, therefore $$\int_{\Omega} k\nabla u\cdot \nabla v = \int_{\Omega}f\,v,\quad \forall v\in H^1_0.\tag{1}$$

To prove the existence of a solution in $V = H^1_0$, we need to fulfill the conditions of Lax-Milgram theorem:

• $\displaystyle \int_{\Omega} k\nabla u\cdot \nabla v \leq C\|u\|_{H^1_0(\Omega)} \|v\|_{H^1_0(\Omega)}$, this is trivial by Cauchy-Schwarz inequality.

• Coercivity relies on Poincaré inequality for $v\in H^1_0$: $\|v\|_{L^2(\Omega)} \leq c\|\nabla v\|_{L^2(\Omega)}$, so we have $$\int_{\Omega} k|\nabla v|^2 \geq \alpha (\|v\|_{L^2(\Omega)}^2 + \|\nabla v\|_{L^2(\Omega)}^2)$$ for some constant $\alpha>0$ (You need to work out this $\alpha$, it depends on $k_1$, $k_2$ and $c$ in Poincaré inequality).

The second question is then trivial, let $v=u$ in (1), using Cauchy-Schwarz and Poincaré inequality on the right hand side: $$\alpha \|u\|_{H^1_0(\Omega)}^2\leq \int_{\Omega} k|\nabla u|^2 = \int_{\Omega}f\,u\leq \|f\|_{L^2(\Omega)} \|u\|_{L^2(\Omega)} \leq c\|f\|_{L^2(\Omega)} \|u\|_{H^1_0(\Omega)},$$ hence the Céa's lemma holds: $$\|u\|_{H^1_0(\Omega)} \leq \frac{c}{\alpha}\|f\|_{L^2(\Omega)}.$$

And if $u$ solves (1), $u$ also minimize the energy functional: $$\mathcal{F}(u) = \frac{1}{2}\int_{\Omega} k|\nabla u|^2 - \int_{\Omega}f\,u.$$

There are several typos in your question, please double check your notes:

1. It should be $u_1 = u|_{\Omega_1}$.
2. It should be $u= 0$ on $\partial \Omega$, not just $u_1$.
3. There should be a dot product between the gradient and the normal, $k_1 \nabla u_1\cdot n= k_2\nabla u_2\cdot n$ because they are vectors.
4. "prove", not "prouve".
5. Céa's lemma has some norm of $f$ on the right hand side, because $\|u\|_V$ depends on that, you can't find a single $c$ satisfying that property.
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$u=u_1 + u_2$??and $k=k_1 + k_2?$ i have problem to prove the first condition of Lax - Milgram – jijiii May 31 '13 at 23:57
@jijiii No, $u|_{\Omega_1} = u_1$ and $u|_{\Omega_2} = u_2$, $k|_{\Omega_1} = k_1$ and $k|_{\Omega_2} = k_2$. – Shuhao Cao Jun 1 '13 at 0:01
so how we can found that $$|a(u,v)| \leq C ||\nabla u|| ||\nabla v||$$? – jijiii Jun 1 '13 at 0:14
@jijiii The first condition of Lax-Milgram can be proved by Cauchy Schwarz like I said in the answer: $$\int_{\Omega} k\nabla u\cdot \nabla v\leq \max\{k\} \|\nabla u\|_{L^2} \|\nabla v\|_{L^2},$$ then $\|\nabla u\|_{L^2} \leq (\|\nabla u\|_{L^2}^2 + \|u\|_{L^2}^2)^{1/2} = \|u\|_{H^1}$ is trivial, if you square both sides you will find that the left is a part of the right. – Shuhao Cao Jun 1 '13 at 0:16
Thank you so much! Please, can you help me in math.stackexchange.com/questions/408421/finite-elements – jijiii Jun 1 '13 at 17:17