So, the final answer would be 0. Is that correct?

[evinda]

Hello! (Wave)

I want to find the value of $\lim_{\epsilon \to 0}\int_{|x|=\epsilon} \left( \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}}- \frac{\phi}{|x|}\right) dS$, where $\phi$ is a test function and $|x|=\sqrt{x_1^2+ x_2^2}$.

Does it hold that $\int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS=0$ ? How can we justify it?

 

[evinda]

I have thought the following:

It holds that $ \int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS= \int_{|x|=\epsilon} \ln{\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS= \ln{\epsilon }\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$ and

$\int_{|x|=\epsilon} \frac{\phi}{|x|} dS= \frac{1}{\epsilon} \int_{|x|=\epsilon} \phi dS$

So we have that $\lim_{\epsilon \to 0}\int_{|x|=\epsilon} \left( \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}}- \frac{\phi}{|x|}\right) dS= \lim_{\epsilon \to 0} \left( \ln{\epsilon }\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS - \frac{1}{\epsilon} \int_{|x|=\epsilon} \phi dS \right)$Is it right so far? How could we continue? (Thinking)

 

[GJA]

Hi evinda!

Does it hold that $\int_{|x|=\epsilon} \ln{|x|} \frac{\partial{\phi}}{\partial{\eta}} dS=0$ ? How can we justify it?

Yes this can be shown. This will be a consequence of Green's First Identity (see https://en.wikipedia.org/wiki/Green's_identities) with $\psi\equiv 1$. (Note: Green's First Identity is a consequence of the integration by parts formula, so some may say integration by parts is the true identity being applied.) Take a look at that formula to see if you can get what you're looking for. Let me know if anything is still unclear. Good luck!

 

[evinda]

How can we show that $\int_{|x|=\epsilon} \frac{\partial{\phi}}{\partial{\eta}} dS$ is bounded ?

 

[evinda]

I want to calculate the limit $\lim_{\epsilon \to 0} \int_{|x|=\epsilon} \frac{\phi}{|x|} dS$.I have thought the following:$\lim_{\epsilon \to 0} \int_{|x|=\epsilon} \frac{\phi}{|x|} dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \cdot \lim_{\epsilon \to 0}\int_{|x|=\epsilon} \phi dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \int_{|x|=0} \phi dS=\lim_{\epsilon \to 0} \frac{1}{\epsilon} \phi(0) \int_{|x|=0} 1 dS $.

But $\int_{|x|=0} 1 dS$ is equal to 0, isn't it?