A How Do You Construct the Chain Rule for Green's Function in Half-Space?

[Shackleford]

This is from Evans page 37. I seem to be missing a basic but perhaps subtle point.

Definition. Green's function for the half-space $\mathbb{R}^n_+,$ is
\begin{gather*}
G(x,y) = \Phi(y-x) - \Phi(y-\tilde{x}) \qquad x,y \in \mathbb{R}^n_+, \quad x \neq y.
\end{gather*}
What's the proper way to construct the chain rule here for $\partial y_n$ given that $\Phi$ is a function of $y-x$? Instead $(y-x)_n$? I think I setup my $u$ substitution correctly, but I'm unclear on $\frac{\partial u}{\partial y_n}$.

I seem to be missing a subtle point here.
\begin{gather*}
\begin{split}
\Phi(y-x) & = \frac{1}{n(n-2)\alpha (n)} \cdot \frac{1}{|y-x|^{n-2}} = \frac{1}{n(n-2)\alpha (n)} \cdot |y-x|^{-(n-2)}, \\
x & =(x_1,...,x_{n-1},x_n) \qquad \tilde{x}=(x_1,...,x_{n-1},-x_n), \\
y-x & = (y_1-x_1,...,y_{n-1}-x_{n-1},y_n-x_n), \\
y-\tilde{x} & =(y_1-x_1,...,y_{n-1}-x_{n-1},y_n+x_n), \\
\frac{\partial \Phi}{\partial y_n}(y-x) & = \frac{\partial \Phi}{\partial u} \frac{\partial u}{\partial y_n} \qquad u = |y-x|, \; \frac{du}{dy_n} =\frac{y_n-x_n}{|y-x|}, \\
& = \frac{-(n-2)}{n(n-2)\alpha (n)} |y-x|^{-(n-2)-1} \frac{y_n-x_n}{|y-x|} \\
& = \frac{-1}{n\alpha (n)} \frac{|y-x|}{|y-x|^n} \frac{y_n-x_n}{|y-x|} = \frac{-1}{n\alpha (n)} \frac{y_n-x_n}{|y-x|^n} \\
\end{split}
\end{gather*}

 

[Shackleford]

I think I figured it out.
\begin{gather}
\begin{split}
u & = |y-x| = \left( \sum_{i=1}^{n} (y_i-x_i)^2 \right)^\frac{1}{2} \\
& = \frac{du}{dy_n} = \frac{1}{2}\left( \sum_{i=1}^{n} (y_i-x_i)^2 \right)^\frac{-1}{2} 2\sum_{i=1}^{n}(y_i-x_i) = \frac{y_n-x_n}{|y-x|^n}. \\
\frac{\partial \Phi}{\partial y_n}(y-x) & = \frac{\partial \Phi}{\partial u} \frac{\partial u}{\partial y_n} = \frac{-(n-2)}{n(n-2)\alpha (n)} |y-x|^{-(n-2)-1} \frac{y_n-x_n}{|y-x|} \\
\end{split}
\end{gather}