I want to solve $- \Delta u = f$ in $\Omega$ with $u = \phi$ on $\partial \Omega$. But if I have the solutions of (1) and (2) below : $$- \Delta u_1 = f \; \text{in } \Omega , \; u_1 = 0 \; \text{on } \partial \Omega \tag{1}$$ $$- \Delta u_2 = 0 \; \text{in } \Omega , \; u_2 = \phi \; \text{on } \partial \Omega \tag{2}$$ Then how can I solve the problem $- \Delta u = f$ in $\Omega$ with $u = \phi$ on $\partial \Omega$ by using $u_1 , u_2$ ?
Hint: The Laplacian $\Delta$ is a linear operator. –  Rahul Jul 27 '12 at 22:34
The solution is $v=u_1+u_2$. In fact, we have $$-\Delta v= -\Delta u_1 - \Delta u_2 = f \quad \mbox{in} \Omega$$ and $$v= 0 +\phi = \phi\quad \mbox{on} \quad \partial \Omega.$$