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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 $ ?

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Hint: The Laplacian $\Delta$ is a linear operator. – Rahul Jul 27 '12 at 22:34
up vote 0 down vote accepted

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 $$ v= 0 +\phi = \phi\quad \mbox{on} \quad \partial \Omega.$$

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