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## Main Question or Discussion Point

For circular region, why is [tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex] ?

Where [itex]\; \hat{n} \:[/itex] is the outward unit normal of [itex]C_R[/itex].

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

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

[tex] 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)}] [/tex]

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

[tex]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[/tex]

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

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

[tex]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[/tex]

I don't understant how:

[tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex]

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

Thanks

Alan

Where [itex]\; \hat{n} \:[/itex] is the outward unit normal of [itex]C_R[/itex].

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

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

[tex] 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)}] [/tex]

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

[tex]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[/tex]

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

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

[tex]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[/tex]

I don't understant how:

[tex]\frac{\partial}{\partial n}G(r,\theta,r_0,\phi)= \frac{\partial}{\partial r_0}G(r,\theta,r_0,\phi) [/tex]

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

Thanks

Alan

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