If $\Omega$ circle with radius $\frac{1}{2}$ centred at the origin, for what values $\alpha\in \Bbb{R}$ for function $$v_{\alpha}(x,y)=|log(x^2+y^2)|^{\alpha}$$ we have $v_{\alpha}\in H^{1}(\Omega)$?
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Welcome to MSE! Do you have any thoughts and have tried anything you can share? Regards – Amzoti May 04 '13 at 01:23
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Things to consider when deciding if something is a Sobolev function or not:
Is it integrable with the desired power?
Look at its restrictions to lines $x=b$ contained in $\Omega$. Are they locally absolutely continuous, for almost every value of $b$? Same for the lines $y=b$.
If the answer is yes, then first order partial derivatives exist at a.e. point. Find them. Are they integrable to the desired power ($2$ in this case)?
If all answers are yes, this is a Sobolev function. Otherwise it's not.
In the specific example, 1 is easy: for $b\ne 0$ the restriction is continuously differentiable, hence locally Lipschitz, hence locally absolutely continuous.
To find the partial derivatives, you need nothing more complicated than the chain rule.
You may find it convenient to integrate $|\nabla v|^2$ in polar coordinates.
75064
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Thanks, I need the prove that 1) $\int_{\Omega}v_{\alpha}^{2}dxdy <\infty$ 2) $\int_{\Omega}|\nabla v_{\alpha}|^2 dxdy<\infty$ – Jack Clorado May 04 '13 at 03:06
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@JackClorado As I suggested, you should integrate using polar coordinates. – 75064 May 04 '13 at 03:10
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For (1) with polar coordinates I have $I_{1}=\int_{0}^{2\pi}\int_{0}^{\frac{1}{2}}r|log r^2|^{2\alpha}drd\theta$ – Jack Clorado May 04 '13 at 03:13
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@JackClorado Which is actually a proper integral, because $r (\log r)^{2\alpha}\to 0$ as $r\to 0$. So there is no issue of convergence here, no matter what $\alpha$ is. Logarithms of $r$ is slower than any power of $r$. – 75064 May 04 '13 at 03:15