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This is concerning Poisson's equation with oblique boundary condition (Gilbarg Trudinger p121)

We let $\Gamma(|x-y|)$ denote the fundamental solution to Laplace's equation. Also, let $x-y^{*} = (x_1-y_1, \cdots ,x_{n-1}-y_{n-1},x_n+y_n)$. Finally, let $\zeta = \frac{(x-y^{*})}{|x-y^{*}|}$

I don't understand the computations to get from this

$$ \Theta = -2b_n\int_0^{\infty}{e^{as}D_n\Gamma(x-y^{*}+\textbf{b}s)ds}$$ where $a\leq 0$

to this

$$ \Theta = -|x-y^{*}|^{2-n}\left( (\frac{2 b_n}{n\omega_n})\int_0^{\infty}e^{a|x-y^{*}|s}\frac{\zeta_n+b_ns}{(1+2({\textbf{$\zeta$}\cdot\textbf{b})s+s^2)^{\frac{n}{2}}}}ds\right)$$

where $\omega_n$ is the volume of the n-ball. I guess I'm getting stuck in one regard because we only define the fundamental solution for $|x-y|$, and I've never seen something where you are adding a vector inside. Also, how do we get the additional term in the exponent? If someone could point me in the right direction, I would appreciate it. Thanks.

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Please don't edit this again. It's quite difficult to read unless the type is large with all the superscripts going on. – Euler....IS_ALIVE Sep 19 '12 at 22:05
Note that you can get parentheses (or any opening and closing delimiters like braces, brackets, absolute value bars, ...) of appropriate size by preceding them with \left and \right. – joriki Sep 22 '12 at 8:16
up vote 1 down vote accepted

The part about adding a vector inside the argument of $\Gamma$ is not a problem, because $\Gamma$ is a function of a vector, not of a scalar. That is, the definition is $$ \Gamma(x) = c_n |x|^{2-n}, \qquad x\in\mathbb{R}^{n}\setminus\{0\} $$ as opposed to $$ \Gamma(r) = c_nr^{2-n}, \qquad r>0. $$

The transformation of the integral is the result of the change of variable where you replace $s$ by $|x-y^*|s$. Let us compute the normal derivative of $\Gamma$ first. We have $$ \Gamma(x-y^*+bs) = \frac{|x-y^*+bs|^{2-n}}{n(2-n)\omega_n}, $$ and so

$$ D_n\Gamma(x-y^*+bs) = \frac{|x-y^*+bs|^{-n}}{2n\omega_n} \cdot D_n|x-y^*+bs|^2 = \frac{|x-y^*+bs|^{-n}}{2n\omega_n} \cdot 2(x_n-y^*_n+b_ns). $$ Now we introduce the new variable $r$ by $s=|x-y^*|r$. Using $r$ and $\zeta$, we can rewrite $D_n\Gamma$ as $$ D_n\Gamma(x-y^*+bs) = \frac{(\zeta_n+b_nr)|x-y^*|}{n\omega_n|\zeta+br|^n|x-y^*|^n} = \frac{|x-y^*|^{1-n}}{n\omega_n}\cdot\frac{\zeta_n+b_nr}{|\zeta+br|^n}. $$ We are done once we take into account $ds=|x-y^*|dr$, and $$ |\zeta+br|^2 = |\zeta|^2 + 2r(\zeta\cdot b) + |b|^2r^2 = 1 + 2r(\zeta\cdot b) + r^2. $$

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@Euler....IS_ALIVE: Have a look at the update. – timur Sep 23 '12 at 4:13

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