# Question on normal derivative of Green's function.

• Aug 23rd 2010, 11:52 AM
yungman
Question on normal derivative of Green's function.
For circular region, why is $\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi)$ ?
Where $\; \hat{n} \:$ is the outward unit normal of $C_R$.
Let circular region $D_R$ with radius $R \hbox { and possitive oriented boundary }\; C_R$. Let $u(r_0,\theta)$ be harmonic function in $D_R$.

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

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

Where $\; \theta \;$ is the angle of $\; u(r_0,\theta_0) \;$ and $\; \phi \;$ 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 $u(r_0,\theta_0) in D_R$. The standard formular for Dirichlet problem is:

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

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

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

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

I don't understant how:

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

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

Thanks

Alan
• Aug 23rd 2010, 04:07 PM
yungman
Did I put this question in the wrong section? I don't know where to put this as this is beyond ODE or maybe PDE. Please move this to the correct sub forum.
• Aug 31st 2010, 08:38 AM
yungman
Anyone please? Even if you don't have the answer, point me where to look. I am really out of ideas. I have five PDE book and I can't find any help!!!
• Sep 11th 2010, 08:44 AM
Rebesques
I believe $(r,\theta)$ are polar coordinates. So $r_0=r/|r|$ is a unit vector. Now, since the region is a circle, the normal $n$ is exactly $r_0$, and we get
$\frac{\partial}{\partial n}G=\langle \nabla_{(r,\theta)} G,n\rangle=\langle \nabla_{(r,\theta)} G,r_0\rangle= \frac{\partial}{\partial r_0}G$.