# Sobolev spaces, weak derivative

• Dec 3rd 2011, 05:52 AM
AlexanderW
Sobolev spaces, weak derivative
Hi,

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

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

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

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

And if yes: How can I find it out?

Alexander
• Dec 3rd 2011, 07:31 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, 10:27 AM
Jose27
Re: Sobolev spaces, weak derivative
Quote:

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

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

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

The second one is easier: $\displaystyle u'(x)=\frac{\ln^2(x)}{x}\geq 0$ and $\displaystyle \int _0^{1/2} u'(x)=\infty$ so that $\displaystyle u$ doesn't belong to any Sobolev space $\displaystyle W^{1,p}$ for $\displaystyle 1\leq p \leq \infty$.
• Dec 3rd 2011, 01:07 PM
AlexanderW
Re: Sobolev spaces, weak derivative