Sobolev spaces are function spaces generalising the Lebesgue spaces. Whereas elements of Lebesgue spaces have certain integrability condition imposed on them, functions in a Sobolev space have also differentiability conditions: that is, we require all partial derivatives of the function up to a ...
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What happens when you change space of test functions associated with weak derivatives?
Recall that $u \in L^2(0,T;H^1)$ has weak derivative $u' \in L^2(0,T;H^{-1})$ iff
$$\int_0^T uv' = -\int_0^T u'v$$
holds for all $v \in C_0^\infty(0,T).$
What happens if we only require that this ...
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20 views
Proof that function is in $H^1$
Let $\Omega = \{(x,y)\in\mathbb R^2: \sqrt{x^2+y^2}<\frac{1}{2}\}$ be a bounded domain and $v(x,y) = \ln\left|\ln\sqrt{x^2+y^2}\right|$. I'm trying to show that $v\in H^1(\Omega)$ where ...
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24 views
if the region $\Omega'\subset\Omega$ and $f\in H^k (\Omega)$ then$ f\in H^k (\Omega')$
prove that if the region $\Omega'\subset\Omega$ and $f\in H^k (\Omega)$ then $f\in H^k (\Omega')$
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1answer
49 views
There exist a function $u(x)\in C([0,1]),~ u(0)=u(1)=0$ such that $u \notin H_{0}^{1}((0,1))$?
There exist a function $u(x)\in C([0,1]),~ u(0)=u(1)=0$ such that $u \notin H_{0}^{1}((0,1))$?
Justify your answer!
Thanks.
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1answer
16 views
Mean value theorem in sobolov space under integral
Sorry this seems like a basic question, but I'm having trouble figuring out the answer.
Let g(x) be the step function over [-1,1] and f(x) a function with $f\in H^1[-1,1]$, that is it has a square ...
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32 views
$H^s(\mathbb{T})$ space is a Banach algebra with pointwise product
I have ran across the following theorem but the given proof does not convince me.
Theorem Let $u, v \in H^s(\mathbb{T})$ with $s>1/2$. Then the pointwise product $uv$ is in $H^s(\mathbb{T})$ and ...
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1answer
28 views
Sobolev spaces - embedding - exercise
I have to show that $H^{2}(\Omega) \subset \subset H^1(\Omega)$. I think the Arzela - Ascoli theorem can help.. .I dont know how to start this exercise . I am beginner in Sobolev spaces.. someone can ...
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1answer
67 views
Proving an alternative norm on Sobolev space is equivalent to usual norm
I have this exercice and my problel is only in item 4, and i will desespere.
Let $f \in L^2(\mathbb{R}^n).$
1- Why the equation $\Delta u - u = \dfrac{\partial f}{\partial x_i}$ admits a unique ...
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1answer
66 views
Is $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$ if $g$ is a Sobolev function on the real line?
If we are given a function $g\in W_2^k(\mathbb{R})$ (even consider $k=1$ for simplicity), then is it true or not that $\{g(n)\}_{n\in\mathbb{Z}}\in\ell_2$? That is, do we have ...
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1answer
59 views
Why are weak solutions to PDEs good enough?
Obviously solutions in $C^k$ are nicer but it seems people are happy with obtaining weak solutions in some Sobolev space that only satisfy the weak formulation. Why should this, in the real world, be ...
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1answer
36 views
Use difference quotient with not uniform bound to appoximate weak derivative
Suppose U is an open set,not necessarily bounded or has Lipschitz boundary, $f\in L^p(U)$ ,define the difference as usual: $$D^h_i f=\frac{f(x+he_i)-f(x)}{h},\ \ \forall x\in U'\subset\subset U$$
...
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31 views
about a proof in chap 5 in PDE evans book
in the proof of the theorem 5 page 281 of Evans ( PDE - Evans ) , he write the Morrey estimate. he writes in the estimate : $y \in B(x,r)$
After in the proof we have
$$ |v(y) - v(x)| = |u(y) - ...
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0answers
14 views
“Compensated weak compactness” in $W^{1,1}$
In a problem I'm working on I want to show that a bounded sequence
$\{u_n\} \subset W_0^{1,1}((0,1))$ converges weakly. Of course, since $W^{1,1}$ is not reflexive I don't get this for free.
The ...
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0answers
43 views
Typical problem in functional analysis #2
I need help with this problem from my homework
Let $\Omega$ be a bounded open subset of $\mathbb{R}^n$ and let $\kappa : \Omega\rightarrow\mathbb{R}$ be a continuous function, such that there's ...
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2answers
39 views
Sobolev space boundary value in PDE
I often read this:
Let $\Omega$ be a open bounded set. There is a unique $u \in H^1(\Omega)$ such that
$$-\Delta u = f \text{ on $\Omega$}$$
$$u|_{\partial \Omega} = g$$
But how can we write ...
