To find the solution of dirichlet problem for poisson equation on $B_r(0)\subseteq \mathbb{R}^d$, I have to take the normal derivative of green's function, where:
$$\Gamma(x,y):=\Gamma(|x-y|) = \begin{cases}\frac{1}{2\pi}\ln(|x-y|) \; \; if \; d=2
\\ \frac{1}{d(2-d)\omega_d}|x-y|^{2-d} \; \; if \; d>2
\end{cases}$$
$$ G(x,y):= \begin{cases} \Gamma(|x-y|)-\Gamma\Big(\frac{|y|}{r}\Big|x-\frac{r^2}{|y|^2}y\Big|\Big) \; \; \text{for} \;y\not=0 \\
\Gamma(|x|) – \Gamma(r) \; \; \text{for} \; y=0
\end{cases}$$
I know I can compute this derivative in the "standard" way:
$$\frac{\partial G}{\partial \nu_x}(x,y)=\frac{1}{r}\sum_{i=1}^{d}x_i\frac{\partial G}{\partial x_i}(x,y) \qquad x\in \partial B_r(0)$$
However my professor calculates it in another way, using the fact that:
$$ G(x,y)=\Gamma\left((|x|^2+|y|^2-2x\cdot y)^\frac{1}{2} \right) – \Gamma\left( \left(\frac{|x|^2|y|^2}{r^2}+r^2-2x\cdot y \right)^\frac{1}{2}\right) $$
he deduces that:
$$ \frac{\partial}{\partial \nu_x}G(x,y)= \frac{\partial}{\partial {|x|}}G(x,y) = \frac{1}{d\omega_d}\frac{|x|}{|x-y|^d}-\frac{1}{d\omega_d}\frac{|x|}{|x-y|^d}\frac{|y|^2}{r^2}$$
I can't understand the calculations, in particular how can I handle the dot product part when I do the derivative with respect to $|x|$?
Thank you very much
Best Answer
Use $x \cdot y = |x||y|\cos(\theta)$, where $\theta$ is the angle between $x$ and $y$, to calculate the derivative of the dot product.
For example, $$ \begin{aligned} \frac{\partial}{\partial |x|}\Gamma\left((|x|^2+|y|^2-2x\cdot y)^\frac{1}{2} \right) &= \frac{1}{d w_d}|x-y|^{1-d} [\frac{\partial}{\partial |x|}\left((|x|^2+|y|^2-2x\cdot y)^\frac{1}{2} \right)]\\ &= \frac{1}{d w_d} |x-y|^{-d}(|x|-|y|\cos\theta). \end{aligned} $$ Similarly, $$ \begin{aligned} \frac{\partial}{\partial |x|}\Gamma\left( \left(\frac{|x|^2|y|^2}{r^2}+r^2-2x\cdot y \right)^\frac{1}{2}\right) &= \frac{1}{d w_d} |\frac{|y|}{R}x-\frac{R^2}{|y|^2}y||^{-d}(\frac{|x||y|^2}{R^2}-|y|\cos\theta)\\ &\text{( because $x$ is on $\partial B_r(0)$ )}\\ &=\frac{1}{d w_d} |x-y|^{-d}(\frac{|x||y|^2}{R^2}-|y|\cos\theta). \end{aligned} $$
Putting them together one obtains the result.