Normal derivative of Green's function on a disk.

[yungman]

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 formula 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

 

[yungman]

Anyone?

 

[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!

 

[psholtz]

I'm not sure I completely understand the problem statement, but in general on a circle, the radius is normal to the curve.

So if you're interested in "normal" derivatives, you usually only have to consider how things change w/ respect to radius. This is something along the lines of the argument of why (in E&M) electrostatic fields are conservative, why Gauss' law goes like 1/r^2, etc... you can "ignore" the angular components b/c they don't contribute to the overall integral.

However, you're using a function that reads like:

G(r,\theta,r_0,\phi)

rather than simply as:

G(r,\theta,\phi)

and it seems like r_0 is a constant.

Are you asking why a derivative is being taken with respect to something "constant" like r_0?

If so, it could perhaps be a typo.

It would make more sense to take the derivative wrt "r"..

 

[SallyGreen]

Hi Everyone,

please can you tell me why for a circle we have the normal derivative is equal to the tangential derivative, adn what is the antiderivative of a normal derivative?

wishes,
Sally

 

[jambaugh]

youngman,
I think your using "normal derivative" in the context of a directional derivative in the normal direction.

In general a directional derivative will be the gradient dotted with a unit normal vector. Consider then the gradient operator in polar coordinates:
\nabla u(r,\theta) = \hat{r}\frac{\partial u}{\partial r} + \hat{\theta} \frac{1}{r} \frac{\partial u}{\partial \theta}
Since \hat{n}=\hat{r}
\nabla_\hat{n} u = \hat{r}\bullet \nabla u = \partial_r u.

 

[SallyGreen]

Thanks, and what about the tangential derivative. How can I convence my self that is equal to the normal derivative.

I need these info to be able to do the following integral by part::
\int dq G_0(q,r:k)\frac{\partial G(q,r')}{\partial n}
so this where I want to take the antiderivative of the normal derivative.

 

[jambaugh]

Thanks, and what about the tangential derivative. How can I convence my self that is equal to the normal derivative.

I need these info to be able to do the following integral by part::
\int dq G_0(q,r:k)\frac{\partial G(q,r')}{\partial n}
so this where I want to take the antiderivative of the normal derivative.

I'm confused too. In general the tangential derivative and normal derivative of a function relative to a curve will not be the same. In particular for a circle centered at the origin and the function f(r,theta) = f(r) the tangential derivative will be zero and normal derivative will be f'.

 

[SallyGreen]

Thanks, for the moment let us forget the tangential derivative. I only need to compute this integral by parts
∫dqG0(q,r:k)∂G(q,r′)∂n

choosing u=G0(q,r:k) and v=∂G(q,r′)\∂n
so du=∂G0(q,r:k)\∂q dv=?

where ∂G(q,r′)\∂n= n.\grad(G(q,r′))

so what is the antiderivative of the normal derivative?please help me in doing this??