# Trace theorem, Sobolev space

Let $\Omega$ be a domain of $\mathbb{R}^d,$ and let $\Omega_1$ and $\Omega_2$ be a partition of $\Omega$, and let $\Gamma=\partial \Omega_1 \cap \partial \Omega_2 \subset \Omega$. Let $u_i=u/\Omega_i, i=1,2$. We suppose that $u\in H^1_0(\Omega)$ solution for the variational problem $$\int_{\Omega} k(x)\nabla u \cdot \nabla vdx=\int_{\Omega} fvdx, \forall v \in H^1_0(\Omega)$$. where $k$ is is a constant piecewise function and $f\in L^2(\Omega).$ The question is: prove that $u_1=u_2$ on $\Gamma.$

For this, I use the trace theorem. We know that $\gamma_0(u_1)=u_1/\partial \Omega_1$ is continuous on $H^1(\Omega_1)$ and $\gamma_0(u_2)=u_2/\partial \Omega_2$ is continuous on $H^1(\Omega_2)$. How do we justify that $u_1=u_2$ on $\Gamma$? (in two cases $u\in H^2$ and $u$ not in $H^2$? The initial question is to prouve that, if $u \in H^1_0(\Omega)$ is the solution of the variational proble $$\int_{\Omega} k \nabla u \cdot \nabla v dx = \int_{\Omega} fv dx,\forall v \in H^1_0(\Omega)$$ then $u$ is the solution of the problem $$\begin{cases} &-_i\Delta u_i=f,\quad \mbox{dans} \Omega_i,i=1,2\\ &u_i=0,\quad \mbox{sur} \partial \Omega\\ &u_1=u_2\quad \mbox{sur} \Gamma\\ & k_1 \nabla u_1 \cdot n = k_2 \nabla u_2 \cdot n \quad \mbox{on} \Gamma \end{cases}$$ so we have to suppose two cases toprouve that $u_1=u_2$ on $\Gamma:$ $u\in H^2$ and $u\in H^1_0$ how we can do this proof?

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In what sense are you defining the restriction of $u_i$ on $\Gamma$? –  detnvvp Aug 23 '13 at 22:10
in the strong sens. –  jijii Aug 23 '13 at 22:11
What do you mean by that? –  detnvvp Aug 23 '13 at 22:11
beacause, the initial question is to prouve that, if $u \in H^1_0(\Omega)$ is the solution of the variational problem $\int_{\Omega}k(x)\nabla u\cdot \nabla v dx=\int_{\Omega} fvdx, \forall v \in H^1_0$, then $u$ is the solution of the problem $$-k_i \Delta u_i=f \mbox{in}\Omega_i,i=1,2;u_i=0,u_1=u_2on \Gamma, k_1\nabla u_1\cdot n=k_2\nabla u_2\cdot n on \Gamma$$ and my problem is to prouve that $u_1=u_2$ on $\Gamma$. –  jijii Aug 23 '13 at 22:15
In order to be able to talk about traces, you should have a regularity assumption on $\Gamma$; if there is no such assumption, the trace cannot be defined. –  detnvvp Aug 23 '13 at 22:16