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

Solution 1:

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.