Maximum principle for the Laplace's equation:
$$\nabla^2 u= \Delta u=u_{xx}+u_{yy}$$ $$\Delta u=f(x,y) \text{ Poisson }$$
Problem with boundary values of the form Dirichlet:
$$\left.\begin{matrix} \Delta u=f(x,y), (x,y) \in D\\ u=h(x,y), (x,y) \in \partial D \end{matrix}\right\}:(0)$$ The equation is linear so it reduces into the solution of two problems:
$$\left.\begin{matrix} u=u_1(x,y)+u_2(x,y) \\ \Delta u_1=f(x,y) \text{ in } D\\ u_1=0 \text{ in } \partial D \end{matrix}\right\}:(1)$$ and $$\left.\begin{matrix} \Delta u_2=0 \text{ in } D\\ u_2=h(x,y) \text{ in } \partial D \end{matrix}\right\}:(2)$$
We suppose that $(1)$ stands and we will prove the $(2)$
$$(0) \Rightarrow \Delta u_1+\Delta u_2=f(x,y) \text{ in } D, (1) \Rightarrow \Delta u_1 = f(x,y) \text{ in } D$$ So $\Delta u_2 =0 \text{ in } D$
$$u=u_1+u_2=h \text{ in } \partial D, (1) \Rightarrow u_1=0 \text{ in } \partial D$$ So $u_2=h \text{ in } \partial D$
Therefore $(0)$ is reduced at solving the problems:
$$\left.\begin{matrix} \Delta u_1=f(x,y) \text{ in } \overline{D} \end{matrix}\right\}:(1')$$
$$\left.\begin{matrix} \Delta u_2=0 \text{ in } D\\ u_2=h(x,y)-u_1(x,y)=h(x,y)-f(x,y) \text{ in } \partial D \end{matrix}\right\}:(2')$$
($\overline{D}=D \cup \partial D$) $$$$ Could you explain me why it is: $$\left.\begin{matrix} \Delta u_2=0 \text{ in } D\\ u_2=h(x,y)-u_1(x,y)=h(x,y)-f(x,y) \text{ in } \partial D \end{matrix}\right\}:(2')$$ and not: $$\left.\begin{matrix} \Delta u_2=0 \text{ in } D\\ u_2=h(x,y)-u_1(x,y)=h(x,y) \text{ in } \partial D \end{matrix}\right\}:(2')$$ ??
Isn't $u_1(x,y)=0$ at $\partial D$???