Prove that $\lim_{\epsilon \rightarrow 0}\int_{\partial B_\epsilon} (φ∇g · n − g∇φ · n) ds = 2πφ(0, 0)$

Question

Suppose $φ : \mathbb{R^2}\rightarrow \mathbb{R}$ is any $C^1$ function and let $g:\mathbb{R^2}-\{(0,0)\}\rightarrow \mathbb{R}$ given by $g(x, y) := \ln\sqrt{x^2+y^2}$

Prove that $\lim_{\epsilon \rightarrow 0}\int_{\partial B_\epsilon} (φ∇g · n − g∇φ · n) ds = 2πφ(0, 0)$,

where $B_\epsilon$ denotes the disk centered at $(0, 0)$ of radius $\epsilon$ and $n$ denotes the outer unit normal to the circle $\partial B_\epsilon.$

I tried to use Green's Theorem, but things got more complicated since I had terms like $\frac{\partial φ(\nabla g)}{\partial x}$. Is there a simple method to show it? Any help is appreciated.

Answer 1

Change into polar coordinates: (There $n$ is the same as $\hat r$)

$$ \lim_{\epsilon \rightarrow 0} \int_{\partial B_\epsilon} (φ∇g · n − g∇φ · n) ds \\ = \lim_{\epsilon \rightarrow 0} \int_{\partial B_\epsilon} (φ(\frac{\partial g}{\partial r} \hat r + \frac{\partial g}{r \partial \theta} \hat \theta) · \hat r − g(\frac{\partial \phi}{\partial r} \hat r + \frac{\partial \phi}{r \partial \theta} \hat \theta) · \hat r) ds \\ = \lim_{\epsilon \rightarrow 0} \int_{\partial B_\epsilon} (φ \frac{\partial g}{\partial r} − g \frac{\partial \phi}{\partial r} ) ds\\ = \lim_{\epsilon \rightarrow 0} \int_0^{2\pi} (φ \frac{\partial (\ln r)}{\partial r} − (\ln r) \frac{\partial \phi}{\partial r} ) \epsilon d\theta\\ = \lim_{\epsilon \rightarrow 0} \int_0^{2\pi} ( \frac{φ}{\epsilon} − (\ln \epsilon) \frac{\partial φ}{\partial r} ) \epsilon d\theta \\ = \lim_{\epsilon \rightarrow 0} \int_0^{2\pi} ( φ − (\epsilon \ln \epsilon) \frac{\partial φ}{\partial r} ) d\theta \\ = \lim_{\epsilon \rightarrow 0} \int_0^{2\pi} φ|_{r=\epsilon} d\theta \\ =2\pi φ(0,0) $$

The one trick part is to show why $\epsilon \ln \epsilon$ goes to zero. The last line is valid when that is shown, and we know that $φ$ is continuous.