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I'm doing the calculations about the following assertion

Let $\Omega$ be $\{(x,y):0. The function $u(x,y)=\log (x^2+y^2)$ belongs to $W^{1,2}(\Omega)$, which you can check by integrating $|\nabla u|^2\approx 1/x^2$ within $\Omega$. We have $\Delta u=0$, which is the nicest equation one could ask for. However, $u$ does not belong to $W^{1,3}(\Omega)$, which you also can check by integration.

My efforts:

$| \nabla u |^2 = 4/(x^2+y^2)$ and \begin{equation} \int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{x^2+y^2}dy dx = \int_{0}^{1} \dfrac{\arctan}{x}dx < \infty \end{equation} because $\lim_{x\rightarrow 0}\dfrac{\arctan}{x} = 1$ and $\dfrac{\arctan}{x}$ is bounded in(0,1). Hence $\nabla u \in L^{2}(\Omega)$.

  1. Am I right here?

  2. I don't know how $\int_{0}^{1}\int_{0}^{x^2} \dfrac{1}{(x^2+y^2)^{^3/2}}dy dx$ diverges. Thank you.

3. \begin{eqnarray} \int_{0}^{1}\int_{0}^{x^2} \ln(x^2+y^2)dydx &=& \int_{0}^{1}\int_{0}^{x^4} u\ln udu dx\\ & = &\int_{0}^{1} \left \{\dfrac{1}{2} u^2 \ln + \dfrac{1}{4}u^2 \right \}_{0}^{x^4}dx\\ &=& \int_{0}^{1} (x^8 \ln(x^4) -\dfrac{1}{4}x^8) dx. \end{eqnarray} In the second equality used that $\lim_{x\rightarrow 0}x^2 \ln x = 0$. Moreover, we obatain that $x^2 \ln(x^2)$ is finite in $(0,\infty)$ and the integral above is finite. Am I right here too?

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    The assertio$n$ was made here: http://math.stackexchange.com/questions/176991/understanding-a-corollary-to-a-theorem-concerning-sobolev-spaces/178110#1781102012-08-04

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  1. It's correct. Maybe you could justify why is $u$ is $L^2$.
  2. Let $I(x):=\int_0^{x^2}\frac{dy}{(x^2+y^2)^{3/2}}$. Put $tx=y$, ($xdt=dy$) to get $I(x)=\int_0^x\frac{x dt}{(x^2+x^2t^2)^{3/2}}=\frac x{x^3}\int_0^x\frac{dt}{(1+t^2)^{3/2}}\geq \frac 1{x^2}\frac 1{2^{3/2}}x=\frac 1{2^{3/2}}\frac 1x,$ and the integral over $(0,1)$ of the last term is infinite.
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    Ok. Is correct my argument that $u \in L^2$?2012-08-03