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 $ ?
About Poisson Equation.
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2Hint: The Laplacian $\Delta$ is a *linear* operator. – 2012-07-27
1 Answers
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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 $$ v= 0 +\phi = \phi\quad \mbox{on} \quad \partial \Omega.$$