Prove equation in Green's function.

[yungman]

Homework Statement

Green's function $G(x_0,y_0,x,y) =v(x_0,y_0,x,y) + h(x_0,y_0,x,y)$ in a region $\Omega \hbox { with boundary } \Gamma$. Also $v(x_0,y_0,x,y) = -h(x_0,y_0,x,y)$ on boundary $\Gamma$ and both $v(x_0,y_0,x,y) \hbox { and }h(x_0,y_0,x,y)$ are harmonic function in $\Omega$

$$v=\frac{1}{2}ln[(x-x_0)^2 + (y-y_0)^2]$$




Let u be continuous and h is harmonic on an open disk around $(x_0,y_0)$ in $\Omega$. Show that

$$_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = 0$$

Hint from the book: Both |u| and $|\frac{\partial h}{\partial n}|$ are bounded near $(x_0,y_0)$, say by M. If $I_r$ denotes the integral in question, then $|I_r| \leq 2\pi M r \rightarrow 0 \hbox { as } r\rightarrow 0$

Homework Equations

Green's 1st identity:

$$\int _{\Omega} ( u\nabla^2 h + \nabla u \cdot \nabla h ) dx dy = \int_{\Gamma} u \frac{\partial h}{\partial n} ds$$

The Attempt at a Solution

$$h=-v \hbox { on } \Gamma \Rightarrow\; \int _{\Omega} ( u\nabla^2 h + \nabla u \cdot \nabla h ) dx dy = -\int_{\Gamma} u \frac{\partial v}{\partial n} ds = -\int_{\Gamma} u \frac{1}{r} ds = -\int^{2\pi}_{0} u d\theta$$

h is harmonic in $\Omega\;\Rightarrow \nabla^2 h = 0$

$$\Rightarrow\; \int _{\Omega} \nabla u \cdot \nabla h \; dx dy = -\int^{2\pi}_{0} u d\theta$$
I really don't know how to continue, please help.

 

[adriank]

h = -v on Γ. You're not integrating over Γ; you're integrating around a small circle C_r around (x₀, y₀). You haven't said what h is, but by the hint, presumably ∂h/∂n is bounded on some disk centered at (x₀, y₀). This is certainly not true if h = -v on this disk, as ∂v/∂n = 1/r is not bounded. If h has no singularity at (x₀, y₀), then certainly ∂h/∂n is bounded on such a disk, as h has a continuous derivative (since by assumption h is harmonic).

Apply the hint.

 

[yungman]

h = -v on Γ. You're not integrating over Γ; you're integrating around a small circle C_r around (x₀, y₀). You haven't said what h is, but by the hint, presumably ∂h/∂n is bounded on some disk centered at (x₀, y₀). This is certainly not true if h = -v on this disk, as ∂v/∂n = 1/r is not bounded. If h has no singularity at (x₀, y₀), then certainly ∂h/∂n is bounded on such a disk, as h has a continuous derivative (since by assumption h is harmonic).

Apply the hint.

Thanks for you help, let me try this:

Let u be continuous and h is harmonic on an open disk around $(x_0,y_0)$ in $\Omega$. Show that

$$_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = 0$$

Hint from the book: Both |u| and $|\frac{\partial h}{\partial n}|$ are bounded near $(x_0,y_0)$, say by M. If $I_r$ denotes the integral in question, then $|I_r| \leq 2\pi M r \rightarrow 0 \hbox { as } r\rightarrow 0$

So $|u| \hbox { and } |\frac{\partial h}{\partial n}|\; \leq\; M$. As $(x,y)\rightarrow\;(x_0,y_0)$

$$_r\stackrel{lim}{\rightarrow}_0 \int_{C_r} u\frac{\partial h}{\partial n} ds = \;^+_- M^2\; \int^{2\pi}_{0} r\;d\theta \;=\; _r\stackrel{lim}{\rightarrow}_0 \; (2\pi M^2r) \;=\; 0$$

I have $M^2 \hbox { because } u\frac{\partial h}{\partial n} \leq M^2$

 

[adriank]

Yes. Now you have to prove that both |u| and |∂h/∂n| are bounded near that point.

 

[yungman]

Yes. Now you have to prove that both |u| and |∂h/∂n| are bounded near that point.

Thanks for your time. But I got M^2 instead of M. It does not matter because r goes to zero.