Domain decomposition Could any body help me to understand that how 
\begin{align*}
-\Delta u=f \hspace{0.2cm}\text{in}\hspace{0.2cm}\Omega\\
u=0\hspace{0.2cm}\text{on}\hspace{0.2cm}\Gamma
\end{align*}
equivalent to solving 
\begin{align*}
-\Delta u_{1}=f \hspace{0.2cm}\text{in}\hspace{0.2cm}\Omega_{1}\\
u_{1}=0\hspace{0.2cm}\text{on}\hspace{0.2cm}\Gamma\cap\partial\Omega_{1}
\end{align*} 
\begin{align*}
-\Delta u_{2}=f \hspace{0.2cm}\text{in}\hspace{0.2cm}\Omega_{2}\\
u_{2}=0\hspace{0.2cm}\text{on}\hspace{0.2cm}\Gamma\cap\partial\Omega_{2}
\end{align*}
with 
\begin{align*}
u_{1}=u_{2}\hspace{0.2cm}\text{on}\hspace{0.2cm} \partial\Omega_{1}\cap\partial\Omega_{2}\\
\frac{\partial u_{1}}{\partial\eta_{1}}+\frac{\partial u_{2}}{\partial\eta_{2}}=0\hspace{0.2cm}\text{on}\hspace{0.2cm}\partial\Omega_{1}\cap\partial\Omega_{2}
\end{align*}
where $\overline{\Omega}=\bigcup^{2}_{i=1}\overline{\Omega}_{i}$ and $u_{i}=u|_{\Omega_{i}}$, $\Gamma$ is boundary of $\Omega$, $\partial\Omega_{i}$ are boundary of $\Omega_{i}$($\Omega$ is open  with Lipschitz boundary and same with $\Omega_{i}$, $f\in L^{2}(\Omega)$)? If any body can give any reference regarding this will be nice. There is a proposition given in the book of M. Fortin but without proof. I will be grateful. 
 A: Your information states that the original domain $\Omega$ is replaced with two domains $\Omega_1$, $\Omega_2$ where $\bar{\Omega} = \bar{\Omega_1} \cup \bar{\Omega_2}$. That is quite general, and might allow some funny selections of the sub domains to yield $\Omega$, so I wonder if some of the other conditions ensure that only reasonable domains are chosen.
The first two new Poisson equations make sense, they require to solve the Poisson equation
$$
-\Delta u_i = f
$$ 
on each domain $\Omega_i$ and require to respect the boundary condition
$$
u_i = 0
$$ on the shared boundary $\Gamma \cap \partial \Omega_i$ of original and new domain. 
The last two equations deal with stitching the solutions together the right way: 
They ask that on the shared boundary of the new domains both solutions have the same values 
$$
u_1 = u_2
$$
and that their first derivatives (along the normal of the boundary from their viewpoint) are the same.
$$
\partial_n u_1 = - \partial_n u_2
$$ 
That seems reasonable.
One interesting bit is that nothing is asked for the second order partial derivatives, so the requirements on orders zero and one seem enough.
I have yet to find an argument why this is sufficient, sorry.
At least the second order derivatives should be continous that way, if I am not wrong. 
The other interesting bit is that the identification only is required for the shared boundary, so it must enforce the identity on the whole shared area $\Omega_1 \cap \Omega_2$.
I believe this uniqueness property can be explained by looking at the difference function $v$ of the two (maybe different) solutions $u_1$ and $u_2$:
$$
v = u_1 - u_2
$$
It satisfies
$$
\Delta v = \Delta u_1 - \Delta u_2 = -f + f = 0
$$
so $v$ is a solution of a Laplace equation, thus it is a harmonic function which has the interesting mean value property. 
From that one can infer that the solution on a domain is bounded by its minimum and maximum on the boundary (Maximum-Minimum Principle). 
See for example P. Duchateau and D.W. Zachmann: Partial Differential Equations, Schaum's Outline.
The difference function vanishes on the boundary thanks to the boundary condition and the above implies that the difference vanishes within the shared domain as well. So the condition just on the boundary is sufficient already.
