Question on normal derivative of Green's function.

For circular region, why is $\displaystyle \frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) $ ?

Where $\displaystyle \; \hat{n} \:$ is the outward unit normal of $\displaystyle C_R$.

Let circular region $\displaystyle D_R$ with radius $\displaystyle R \hbox { and possitive oriented boundary }\; C_R$. Let $\displaystyle u(r_0,\theta)$ be harmonic function in $\displaystyle D_R$.

The Green's function for Polar coordinate is found to be:

$\displaystyle G(r,\theta,r_0,\phi) = \frac{1}{2} ln[R^2 \frac{r^2+r_0^2 -2rr_0 cos(\theta-\phi)}{r^2r_0^2 + R^4 - 2rr_0R^2 cos(\theta-\phi)}] $

Where $\displaystyle \; \theta \;$ is the angle of $\displaystyle \; u(r_0,\theta_0) \;$ and $\displaystyle \; \phi \;$ is the angle of the two points used in Steiner Invertion.

Next I want to solve the Dirichlet problem using Green's function. For any value of a hamonic function $\displaystyle u(r_0,\theta_0) in D_R$. The standard formular for Dirichlet problem is:

$\displaystyle u(r_0,\theta_0) = \frac{1}{2}\int_{C_R} u(r,\theta) \frac{\partial}{\partial n}G(r,\theta,r_0,\phi) ds$

Where $\displaystyle \frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \nabla G(r,\theta,r_0,\phi) \;\cdot \widehat{n} $

But the book just simply use $\displaystyle \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) $ Which is only a simple derivative of G respect to $\displaystyle \; r_0 \;$ where in this case $\displaystyle \; r_0 = R \;$ !!!

$\displaystyle u(r_0,\theta_0) = \frac{1}{2}\int_{C_R} u(r,\theta) \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) ds$

I don't understant how:

$\displaystyle \frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) $

How can a normal derivative become and simple derivative respect to $\displaystyle \; r_0 \;$ only? I know $\displaystyle \widehat{r}_0 \;\hbox { is parallel to outward normal of }\;\; C_R \;$ but the magnitude is not unity like the unit normal. Can anyone explain to me?

Thanks

Alan