# Tag Info

3

We can (for example) consider two cases: Case 1 - $q> p$ We have that $\|u\|_p^p=\int_I |u|^p$. Because $q>p$, we have that $\frac{q}{p}>1$, hence we can apply Holder inequality, to conclude that $\|u\|_p^p\leq C \|u\|_q^{p}$, where $C>0$ is a positive constant (that can change in every line). From it we have that ...

3

As you guessed, the first step is an integration by parts: $$\int_U |Du|^p dx = \int_U \nabla u \cdot \nabla u |Du|^{p-2} dx = - \int_U u \nabla \cdot ( \nabla u |Du|^{p-2}) dx = - \int_U u \left( \Delta u |Du|^{p-2} + (p-2) (\nabla u^T D^2 u \nabla u) |Du|^{p-4})\right) \leq C \int_U u |Du|^{p-2} |D^2 u| dx$$ Now the next step is an invocation of the ...

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Consider $\Omega$ the unit ball of $\Bbb R^d$ and the function constant equal to $1$: it's clearly an element of $H^1(\Omega)$. However, it's not in $H^1_0(\Omega)$, otherwise we would contradict Poincaré's inequality ($\lVert u\rVert_{L^2}\leqslant C\lVert \nabla u\rVert_{L^2}$). In particular, this proves that these space are not equal when such an ...

3

There is a characterization of $W_0^{1,2}$ in terms of pointwise traces: a function $u \in W^{1,2}(\Omega)$ belongs to $W_0^{1,2}(\Omega)$ if and only if $$\lim_{r \to 0^+} \int_{B(x,r) \cap \Omega} |u(y)| \, dy = 0$$ for $(1,2)$-quasievery point $x \in \partial \Omega$. Since you assume $u \in W^{1,2}_0(\Omega)$ and point out that $|(u-v)^+| \le |u|$, ...

2

Just to continue Chris Wong's argument. Let $\Omega = \{x\in \mathbb{R}^n: \|x\|_{\infty}= \max|x_i|<1 \}$ be an $n$-cube with side length $2$, then it can be verified that: $$u = \prod_{i=1}^n \sin(2\pi x_i)$$ is in $H^1_0(\Omega)\cap H^2(\Omega)$, since on boundary of the $n$-cube, there is some $x_i = \pm 1$. However for $\partial u/\partial x_i$ ...

2

It's possible that $u \in H^1(\Omega)$ isn't even continuous on $\Omega$, so it's not even clear what we'd mean by $u(\text{some interior point})$. But if $\partial\Omega$ is $C^1$, then the trace of $u$ on $\partial\Omega$ is a well-defined function in $L^2(\partial\Omega)$, and this is what we mean when we write $u|_{\partial\Omega}$. No claim is being ...

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I don't think so. You can imagine an infinite sequence of pyramids of height 1 with ever tinier hypersquare bases of area $A_m$, which will clearly be in $L^2$, so long as $\sum |A_m| < \infty$. The question is whether there's enough room for the gradient to remain square integrable. The gradient is like the height over the distance, which is like ...

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This is true for $n = 1$, but false for $n \ge 2$. Recall, that for $n \ge 2$, there is a positive function $h \in H^1(\mathbb R^n) \setminus L^\infty(\mathbb R^n)$. Now, take a sequence $\{x_i\}$ with $|x_i| \to \infty$ and define $$f(x) = \sum_{i=1}^\infty 2^{-i} \, h(x - x_i).$$ Then you have $f(x_i) = +\infty$ for all $i$ (if you want finite function ...

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Nothing changes at all. Define $W(0,T) = L^2(0,T; H^1) \cap H^1(0,T; H^{-1})$. Note that for $u \in W(0,T)$ the mapping $$v \mapsto \int_0^T u \, v' + u' \, v \, \mathrm{d}t$$ is continuous from $H_0^1(0,T)$ to $H^{-1}$. Hence, if the mapping is zero on the dense subspace $C_0^\infty(0,T)$, it is zero on the whole space $H_0^1(0,T)$, in particular on the ...

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Usually there are sufficiently nice regularity results. Sobolev spaces are often a useful way to show that a weak solution exists, where previously existence theory was harder to arrive at (although there are other techniques). In other cases, the Sobolev space is sufficient. For instance, in the case of the heat equation, often it is desired to know how ...

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I agree with gerw's comment: this does not look possible. For example, the constant function $w\equiv 1$ on the unit ball $D\subset \mathbb R^n$ is in $H^1(D)$ (and of course in all $H^m(D)$). So, for the trace operator to be defined, we would need to make sense out of $\int_{\partial D} v$ for $v\in L^2(D)$. But for functions such as $v(x)=\sin ... 1 You may want to look at Navier-Stokes Equations: Theory and Numerical Analysis by Temam chapter 3. He defines the problem (3D NSE) to be to find$u$such that $$u \in L^2(0,T;V), \, u^{\prime} \in L^1(0,T; V^*)$$ and then shows, in Thm. 3.3, that $$u \in L^{8/3}(0,T; L^4(\Omega)) \\ u^{\prime} \in L^{4/3}(0,T;V^*)$$ and in Thm. 3.4 that there is at ... 1 No. Suppose there is such a$\theta$. Then there is a$\varphi \in L^\infty$so that $$-\frac{\Delta \theta}{\theta} = \varphi \text{ a.e. in } \Omega,$$ which is equivalent to the PDE $$\begin{cases} -\Delta \theta = \varphi \theta & \text{in }\Omega \\ \theta =0 &\text{on } \partial \Omega. \end{cases}$$ Multiply by$\theta$, integrate over ... 1 Aha! This is actually a bit of a tricksy problem in Fourier analysis. What you want to do is, consider the Fourier transform of$g$, call it$\hat{g}$. Since$g$is in$L^2(\mathbb{R})$, so is$\hat{g}$, by the properties of the Fourier transform. Now let's periodize$\hat{g}$into a function we will suggestively call$\hat{G}$with $$\hat{G} (k) = ... 1 Notice that the second expression with m = 1 says that u and its derivative are both square-integrable. When m is a positive integer, u \in H^m means that u and all its derivatives up to order m are square-integrable. However, this can be extended to m real, by using the fact that the Fourier transform of D^m u is the same as (ik)^m \hat{u} ... 1 Do you know anything about the trace operator? There is a theorem that says that if U is a bounded domain and \partial U is C^1 and u \in W^{1,p}, then$$ u \in W_0^{1,p}(U) \iff Tu=0 \text{ on } \partial U, $$where T is the trace operator. You want p = 2 here. 1 I think that there is a typo in the equation for w. Here's how I would approach this issue. First note that we can integrate by parts with sufficiently smooth functions to deduce the equality $$\int_\Omega \text{curl}X \cdot Y = \int_\Omega X \cdot \text{curl} Y + \int_{\partial \Omega} n \times X \cdot Y.$$ Now, if we view ... 1 Got it, thanks to Shuhao Cao. Let \Omega be the domain under consideration. From |v| = 0, we conclude that v=0 a.e. . If this was not the case, i.e., if there was a \Omega_0\subset \Omega of nonzero measure where v\neq 0, then, |v| > 0 and we would get a contradiction. Thus, v=0 almost everywhere on \Omega. Then, for every coordinate ... 1 The reason the definition of a distribution involves infinitely differentiable functions is to make it possible to define derivatives of all orders of any distribution. However, if you stick to distributions of orders at most 1 (which is usual in the context of 2nd order elliptic problems), then the space of test functions can be taken to be C^1, as ... 1 First$$ \nabla f = -\alpha (-\ln|x|)^{\alpha - 1} \frac{x}{|x|^2} $$and$$ |\nabla f | = \alpha (-\ln|x|)^{\alpha - 1} \frac{1}{|x|} $$Let r = |x|, then 0<r<1:$$ \int_{B(0,1)} |f|^n \,dx \sim c(n)\int^1_0 |\ln r|^{\alpha n} r^{n-1} dr $$This integral might behave singular when r\to 0. For \alpha n< n-1, and$$ r^{n-1}<|\ln ... 1 The integral you have to prove convergence is actually$\int_0^{1/2}\frac 1{r\log^2(r)}dr$(don't forget the determinant of Jacobian). First look at$\int_\varepsilon^{1/2}\frac 1{r\log^2(r)}dr$for$\varepsilon>0$, then make the substitution$s:=\log r$. It's a good example of an element of$H^1(\Omega)$which is not essentially bounded. 1 I'm assuming that your function$u$belongs to$W^{3,2}(\Omega)$. We have the following result that can be found in any good book of Sobolev spaces (for example in Leoni's book or even in Brezi's book, but in the later you have to iterate the estimates that he find only to$W^{1,p}(\Omega)$). If$\Omega$is a bounded regular domain,$p\geq 1$, ... 1 I think people use this rather vague statement just to avoid presenting extra technicalities in Sobolev spaces theory. The full picture should be: Restriction on boundary is valid for any$C^{\infty}(\overline \Omega)$. Hence we define the trace:$T:C^{\infty}(\overline \Omega)\rightarrow L^2(\partial \Omega) $.$C^{\infty}(\overline \Omega)$is dense in ... 1 The duality is misplaced. If you have a function from$(0,T)$to a vector space, its derivative, strong or weak, is also a (generalized) function from$(0,T)$to the same vector space. This is in fact clear from the definition you give of a weak derivative. Here, both$u$and$u'$are in$L^2((0,T), H^1(\Omega))$. On the other hand, if$u \in H^1((0,T), ...

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I found a nice elementary proof for this: All the equalities are meant to hold a.e.. Lemma: Let $u\in W^{1,p}(\Omega)$ and $N=\left\{x\in\Omega:u(x)=0\right\}$. Then $D_iu=0$ a.e. in $N$. Proof: Let $u^+=\max(u,0)$ and $u^-=\min(u,0)$. By the chain rule, $u^+,u^-\in W^{1,p}(\Omega)$ and $D_iu^+(x)=D_iu(x)$ if $u(x)>0$ and $D_i u^+(x)=0$ otherwise, ...

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Using the definitions, you should be able to prove that $H^2_0(\Omega) \subset H_0^1(\Omega) \cap H^2(\Omega)$. The reverse inclusion does not hold; as a counterexample, try picking a really nice $\Omega$ (such as a disc) and writing down a well-behaved function that vanishes at the boundary, but whose first derivatives do not. Then such a function will be ...

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