# Sobolev spaces, weak derivative

• Dec 3rd 2011, 06:52 AM
AlexanderW
Sobolev spaces, weak derivative
Hi,

in which sobolev spaces $H^{1,p}(\Omega)$, $\ 1\leq p\leq \infty$, are the functions

a) $u(x)=x^\frac{3}{2}\sin\left ( \tfrac{1}{x} \right )$ with $\Omega=(0,1)$

b) $u(x)= \left |\ln x \right |^{-1}$ with $\Omega=\left ( 0, \frac{1}{2} \right )$ ?

So, do I have to find out in which spaces $L^p(\Omega)$ the functions above has a first weak derivative?

And if yes: How can I find it out?

Alexander
• Dec 3rd 2011, 08:31 AM
girdav
Re: Sobolev spaces, weak derivative
If the function is differentiable in the usual sense, then it's also weakly differentiable and these two derivatives are the same.
• Dec 3rd 2011, 11:27 AM
Jose27
Re: Sobolev spaces, weak derivative
Quote:

Originally Posted by AlexanderW
So, do I have to find out in which spaces $L^p(\Omega)$ the functions above has a first weak derivative?
Alexander

Yes, so as girdav noted, you have to check for which $p$ do the derivatives lie in $L^p$.

For the first one, notice that $u'(x)=\frac{3}{2}x^{1/2} \sin(1/x) - x^{-1/2}\cos(1/x)$ so that $u'\in L^p(0,1)$ iff $f(x)=x^{-1/2}\cos(1/x) \in L^p(0,1)$. Now $|f(x)|\leq x^{-1/2}$ so $f\in L^q$ for all $1\leq q<2$, and $\int_0^1 f^2(x)dx=\infty$ (this is easy to see: change $x=1/y$ and bound below by a harmonic series). So that $u\in W^{1,p}(0,1)$ iff $1\leq p<2$.

The second one is easier: $u'(x)=\frac{\ln^2(x)}{x}\geq 0$ and $\int _0^{1/2} u'(x)=\infty$ so that $u$ doesn't belong to any Sobolev space $W^{1,p}$ for $1\leq p \leq \infty$.
• Dec 3rd 2011, 02:07 PM
AlexanderW
Re: Sobolev spaces, weak derivative