In Evan's textbook PDE: 2.2.4, for a half-space $\mathbb{R}^n_{+}$, if $x=(x_1, \dots, x_n)\in \mathbb{R}^n_{+}$, its reflcetion $\bar{x}=(x_1, \dots, -x_n)$. If we know that $\Phi(y-x)$ it the fundamental solution of the Laplace equation. Let $\phi^{x}(y):=\Phi(y-\bar{x})$. We try to use the reflection tricks to solve following equation:
$$
\begin{cases}
\Delta \phi^{x}=0 \mbox{ in } \mathbb{R}^n_{+} \\
\phi^{x}= \Phi(y-x) \qquad \mbox{ on } \partial\mathbb{R}^n_{+}
\end{cases}
$$
Why can we note that $\phi^{x}=\Phi(y-\bar{x})=\Phi(y-x)$ for $y\in \partial \mathbb{R}^n_{+}$ for the half-space?
My idea: For $y\in\partial \mathbb{R}^n_{+}=\{(y_1,\dots, y_{n-1}, 0): y_i\in \mathbb{R}\}$, we know that $$\Phi(y-\bar{x})=\Phi(y_1-x_1, \dots, y_{n-1}-x_{n-1}, y_n+x_n)=\Phi(y_1-x_1, \dots, y_{n-1}-x_{n-1}, x_n)$$. But $$\Phi(y-x)=\Phi(y_1-x_1, \dots, y_{n-1}-x_{n-1}, -x_n)$$. Why are these two functions equal on the boundary?
Best Answer
Recall that the fundamental solution for the Laplace operator is radially symmetric, i.e., $\Phi(x) = f(|x|)$. It is then enough to show that for $y\in\partial\mathbb{R}_+^n$, $|y-x| = |y-\bar{x}|$ for all $x\in\mathbb{R}_+^n$. If $y$ is on the boundary, then we have $y = (y_1,\dots,y_{n-1},0)$, $x = (x_1,\dots,x_n)$, and $\bar{x} = (x_1,\dots,-x_n)$. We then have $$|x-y|^2 = (x_1-y_1)^2+\dots+(x_{n-1}-y_{n-1})^2+x_n^2 = (x_1-y_1)^2+\dots+(x_{n-1}-y_{n-1})^2+(-x_n)^2 = |\bar{x}-y|^2.$$ Therefore, for and $y$ on the boundary, $x$ and $\bar{x}$ are the same distance from $y$. Since $\Phi$ depends only on the length of its argument, we then have that $$\Phi(x-y) = \Phi(\bar{x}-y), \ \forall y\in\partial\mathbb{R}_+^n, \ \forall x\in\mathbb{R}_+^n.$$