If $u\in C^2(B_{R}(0))\cap C^0(B_R(0)))$ is harmonic, then
$\large u(x)=\frac{R^2-|x|^2}{nW_nR}\int_{\partial B_R(0)}\frac{u(y)}{|x-y|^n}\,ds(y)=\int_{\partial B_R(0)}k(x,y)u(y)\,ds(y)$
by the equation from Poisson Integral Formula.
let $\large x_0\in\partial{B_R(0)}$, for $x\in B_R(0)$
When proving $ u(x) = \left\{ \begin{array}{l l} \int_{\partial B_R(0)}k(x,y)\varphi(y)\,ds(y) & \quad \text{if $x\in B_R(0)$ }\\ \varphi(x） & \quad \text{if $x\in \partial B_R(0)$ } \end{array} \right. $ is a solution to Dirichlet Problem where $u|\partial B(R)=\varphi(x)$.
My note from the PDE class says $\large u(x_0)=\int_{\partial B_R}k(x,y)\varphi(x_0)\,ds(y)$. I don't understand why this is true?
This is a part of proving the equation above is actually a solution to the Dirichlet problem, so it is important. But I just cannot figure out why. Can anyone give me some hint? Thanks in advance!
I will certainly upvote your answer if anyone helps me!
Well, that identity is true because of how you defined $u$: Remember that, for any $x\in B_R(0)$ we have that $$ \int_{\partial B_R(0)} k(x,y)ds(y)=1 $$ so that, fixing $x_0\in \partial B_R(0)$, we get $$ u(x_0)=\varphi(x_0) = \int_{\partial B_R(0)} k(x,y) \varphi(x_0) ds(y). $$