New answers tagged

0

The following is an answer by Terry Tao, which is similar to Winther's comment. If $f$ is dimensionless, then $D^\alpha f$ has the units of $L^{-|\alpha|}$, and so $\int_\Omega |D^\alpha f|^2\ dx$ has the units of $L^n \times (L^{-\alpha})^2$. As $\Omega$ has units of $L^n$, each summand in $\|f\|_{m,\Omega}^2$ has the units of ...


0

The related question Scalar property of $ C(\Omega)=\sum_{|\alpha|\leq m}\color{blue}{\big|\Omega\big|^{\dfrac{2|\alpha|-n}{n}}} \int_{\Omega}|D^\alpha f|^2\ dx $ and its answer gives a direct answer for the second question. For the first question, one can use the calculation of $\|T_\delta f\|_{m,\delta\Omega}$ to find the correct exponents.


0

For the calculation of $C(\delta\Omega)$, when the domain changes from $\Omega$ to $\delta\Omega$, $f$ should changes to be $T_\delta f$ where $$ T_\delta f(y):=f(\frac{y}{\delta}),\quad y\in\delta\Omega. $$ It does not make sense at all to compare $\|f\|_{m,\Omega}$ with $\|f\|_{m,\delta\Omega}$ since if $f$ is defined on $\Omega$, it is not necessarily ...


1

Let us assume that $f_k$ converges to $F \in D'(\Omega)$ in the sense that $$\int_\Omega f_k \, v \, \mathrm{d}x \to F(v) \qquad\forall v \in D(\Omega).$$ Now, since $W^{1,q}(\Omega)$ is reflexive, you get $f \in W^{1,q}(\Omega)$ and a subsequence such that $f_{n_k} \rightharpoonup f$ in $W^{1,q}(\Omega)$. In particular, $f_{n_k} \rightharpoonup f$ in ...


1

The footnote 31 on page 305 explains the logic of the choice of $h$: the support of $w$ is at some distance from the top and side surface of the cylinder $ Q_+$, but it need not be separated from the bottom surface, which is $x_N=0$. This is how the chosen partition of unity works: in order to cover the domain by finitely many sets, we need these sets to ...


0

I have good news and bad news. The good news is that you were correct. The bad news is that it was when you said: I'm really confused with norms and semi norms in $H^1$ and $L^2$. I'll try to clarify it a bit. First of all... The basic difference between a norm $\Vert\cdot\Vert$ and a seminorm $\vert \cdot\vert$ is that a norm can only be zero if ...


0

One way is by contradiction: suppose that for all $n\in\mathbb N$, there exists $u_n\in H^1(\Omega)$ such that $$\|u_n\|_{L^2(\Omega)}> n\|\nabla u_n\|_{L^2(\Omega)}+n\|u_n\|_{L^2(\partial\Omega)}.$$ Then $\|u_n\|_{L^2(\Omega)}>0$ for all $n$, and if we set $v_n=u_n/\|u_n\|_{L^2(\Omega)}$, then $\|v_n\|_{L^2(\Omega)}=1$ for all $n$, and also ...


0

Here the author probably meant that $T\mu$ is the restriction map $$ T\mu(\xi_1,\cdots, \xi_{n})=\mu(\xi_{1},\cdots, \xi_{n-1},0) $$ Now writing out explicitly we have $$ \widehat{T\mu}(\xi')=\int\mu(\xi,0)e^{-i \xi\cdot \xi'}d\xi $$ But recall that formally we have (for 1D functions) $$ f(0)=\frac{1}{2\pi}\int ...


0

$V=B_{r/2}(x^0) ∩U$; Note that $\bigcup_{v∈ V}[ B_{ε}(v)+2ε e_1 ]$ is pushed up from $\partial U$ by about $ε$. The choice of $r/2$ stops this set from 'spilling out' of $B_r$.


1

Actually, you have $H_\perp^1 = \{0\}$, the set containing only the zero function. Indeed, suppose that $u \in H^1(\Omega)$ satisfies $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in H_0^1(\Omega)$. Then, using the density of $H_0^1(\Omega)$ in $L^2(\Omega)$ you get $\int_\Omega u \, v \, \mathrm{d}x = 0$ for all $v \in L^2(\Omega)$ and this shows $u = ...


0

For the identification (or related) see Theorem 4.9 in Rudin1991 ("Functional Analysis") However, this identification (and also the closedness of $V$) need not be invoked for the particular result. I assume you are talking about Corollary 2.5 in Girault and Raviart. Namely, the Hahn-Banach theorem provides that any functional $\mathbf{f}\in V'$ that ...


2

Following Daniel Fischer comments I'm trying to post an answer: Let $G = (a,b)$ We have that $$\lVert u\rVert_{L^\infty(G)}^2 = \lVert u^2\rVert_{L^\infty(G)} \le \int_a^b 2 |u(t)u'(t)|dt \le 2 \lVert u\rVert_{L^2(G)}\lVert u'\rVert_{L^2(G)}$$ where the first inequality is justified by the fact that $u^2$ is absolutely continuos and $u(a) = 0$ so that we ...


0

I'm essentially copying my answer to another closely related question. Note that the definition of $H^{-1}(\Omega)$ is different here. Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich: $$ H^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*} $$ where ...


0

Let me restate your question as follows. Let $\Omega$ be a nonempty open subset of $\mathbb{R}^d$. On the one hand, we have the sandwich: $$ H_0^1(\Omega)\subset L^2(\Omega)\subsetneq H^{-1} (\Omega),\tag{*} $$ where $H^{-1}(\Omega)$ is defined as the dual of $H_0^1(\Omega)$. On the other hand, by Riesz representation theorem, $H^{-1}(\Omega)$ can be ...


3

For $u$ to be a classical solution, both $u$ and the coefficients in the partial differential operator $P$ in the LHS of your first formula need to be regular enough so that $P$ does not map u away from $C^0(\Omega)$, for in this case you can integrate by parts the LHS of your second formula. The path from weak solutions to classical solutions relies on two ...


-1

If $f\in W^{1,p}(\mathbb{R}^n)$, then $f\in L^r$ for all $r\in [p,\infty)$ such that $$\frac{1}{r}> \frac{1}{p} -\frac1n \tag{1}$$ This is a consequence of the Sobolev-Morrey embedding theorem, which I stated in a way that does not require separate consideration of $p<n,p=n,p>n$. A consequence can be stated as follows: If $f\in ...


4

Using Hölder, we deduce that $$\int\nabla f\cdot v\ dx:= \int f\mathrm{div }v\ dx\leq\|f\|_{L^2}\|\mathrm{div }v\|_{L^2} \leq\|f\|_{L^2}\|v\|_{H^1} $$ for $v\in H_0^1(\Omega)$, so $\nabla f\in H^{-1}$.


1

Let $\Omega \subset \mathbb{R}^n$ open. If $u \in \mathscr{C}(\Omega)$, and the distributional derivatives of $u$ are given by integration against continuous functions, that is, there exist $v_1,\dotsc, v_n \in \mathscr{C}(\Omega)$ such that for all $\varphi \in \mathscr{D}(\Omega)$ we have $$- \int_{\Omega} u(x)D_k\varphi(x)\,dx = \int_{\Omega} ...


-1

Pseudo proof: Suppose $f,g$ are continuous functions on $\mathbb R,$ with $g$ the weak derivative of $f.$ Let $0<x.$ Define piecewise linear functions $\varphi_n$ such that $\varphi_n$ connects $(0,0), (1/n,1), (x-1/n,1), (x,0)$ with line segments, $\varphi_n =0$ elsewhere. Now $\varphi_n$ is not a smooth test function, but pretend it is. Then $$\tag 1 ...


2

First, complement (P2) with the "boundary condition" $|u(x)|\rightarrow0$ as $|x|\rightarrow\infty$. Next, remind that the set $C_0^\infty{(\mathbb{R}^N)}$ of all compactly supported smooth functions defined in $\mathbb{R}^N$ is dense in $H^1(\mathbb{R}^N)$. Now, $$\displaystyle\int_{\Omega}\nabla u \nabla v = - \displaystyle\int_{\Omega} u\Delta v + ...


1

An identity that is originally proved for smooth functions can be extended to a Sobolev space provided that both sides are continuous with respect to the Sobolev norm. This is a general topological fact: if two continuous functions agree on a dense subset, then they agree everywhere. In the case of linear expressions like $u\mapsto \int \varphi \nabla u$, ...


1

Means that $\partial_{x_j} u \in L^p(\Omega)$ $\forall j=1,...,n$, i.e. $\nabla u \in L^p(\Omega) \times \cdot \cdot \cdot \times L^p(\Omega)$.


2

Let $B\subset \mathbb R^3$ denote the open unit ball and consider the function $$ w(x)=|x|^{-\lambda}-1,\quad \lambda <\frac12.$$ This function is clearly not continuous but $w\in H^1_0(B)$. Define $f=-\Delta w$ and consider the problem $$ \begin{cases} -\Delta u = f, & \text{on }B\\ u\in H^1_0(B) \end{cases}$$ This equation has the unique ...


1

This is a consequence of the Sobolev Embedding Theorems, which is explained, e.g., here (and it's not trivial, but well known. If this is a textbook what you are reading it should be mentioned somewhere).


0

My thought is that it is just Holder + Jensen. Write $$ \int_{\Omega} v^{\alpha_1}u \,d x \leq \|v^{\alpha_1}\|_{L^p(\Omega)}\|u\|_{L^q(\Omega)}. $$ Now note that \begin{align} \|v^{\alpha_1}\|_{L^p(\Omega)}^p = \int_{\Omega}\left(v^{p}\right)^{\alpha_1}\,dx &= |\Omega|\int_{\Omega}\left(v^{p}\right)^{\alpha_1}\,\frac{dx}{|\Omega|}\\ &\leq ...


0

In general you can not obtain $f\in D(A)$ for each $b:\Omega\rightarrow\mathbb{R}$. Consider $f_k(x)=\sum_{l\neq i}h_k(x_l)$, where $h_k$ is a smooth sequence which converges to a non smooth function in $L^2$ in one dimension subspace. This is possible, for instance using mollifier. Then $f_k$ converges to a function in $L^2\setminus H^1$, and you still ...


1

You can use the Fourier transform for a nice class of functions to solve for the resolvent. \begin{align} (\lambda I-\Delta)^{-1}f & = \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}\frac{\hat{f}(\xi)}{\lambda+|\xi|^2}e^{ix\cdot\xi}d\xi \\ & = ...


3

Let $u \in H^1_0(\Omega)$ be any function, then $u + c \in H^1(\Omega) \setminus H^1_0(\Omega)$ (provided $\Omega$ is bounded) And yes, you need the function in $H^1_0$ for this to work. The intuition is easy (picture one dimension): if you know how big the derivative is (how fast the function is growing), you can know more or less which level will the ...


1

Take $u \equiv 1$. Then, $u \in H^1(\Omega) \setminus H_0^1(\Omega)$ and your inequality fails (as long as $\Omega \ne \emptyset$).


1

Let $u \in H^2 \cap H^1_0$. According to ellipticity we know that $$ C_0 \|\nabla u \|_{L^2}^2 = \int_\Omega C_0 \partial_i u \partial_i u \le \int_\Omega a^{ij} \partial_i u \partial_j u. $$ Next we integrate by parts, using the fact that $u=0$ on $\partial \Omega$: $$ \int_\Omega a^{ij} \partial_i u \partial_j u = \int_\Omega -\partial_j(a^{ij} ...


0

Yes, you can have problems between two elements. However, it is easy to show that $v_h$ is continuous. Using the continuity of $v_h$ and the fact that $v_h$ is piecewise smooth, you can calculate its weak derivative. Contrary to the hint, I would start with $\int_\Omega v_h \, \partial_i w \, \mathrm{d}x$, to obtain the $i$th weak derivative of $v_h$.


0

In order to show that $f \in W^{1,2}(\Omega)$, we have to show that it possesses weak derivatives. Recall that $v_i$ is $\partial_i f$ (in a weak sense), if $$\int_\Omega f \, \partial_i \varphi \, \mathrm{d}x = - \int_\Omega v_i \, \varphi \, \mathrm{d}x$$ for all $\varphi \in C_0^\infty(\Omega)$. Now, let $\varphi \in C_0^\infty(\Omega)$ be given. We have ...


2

D1X, this fact has been said here What is precisely the definition of Elliptic Partial Differential Equation? We remember that if $\Omega \subset \mathbb{R}^n$ is a limited open with boundary $\partial \Omega$ of class $C^1$, the following are true Green's identity (with $u,v \in C^2(\Omega)\cap C^1(\overline{\Omega})$) \begin{align*} \displaystyle ...


1

Do you know the weak derivative of $\gamma(nu)$? Using this, you can show for $V = \gamma(n u)$ $$\int_I u' v' \ge 0$$ and, then, the claim follows by simply $n \to \infty$ (and the dominated convergence theorem). What about $v = \gamma(n \, u)^q$ for suitably chosen $q$?


1

Keep in mind that there is a constant $C$ with the property that $\|v\|_\infty \le C \|v\|_{H^1}$ for all $v \in H^1(I)$. If $f \in L^1(I)$ then $$\left| \int_I fv \right| \le \|f\|_1\|v\|_\infty \le C \|f\|_1 \|v\|_{H^1}.$$


2

You know that if $$ u(0) = \int_{0}^{1}u'v_0'+uv_0 dx,\;\;\; u\in H^1(0,1), $$ then $$ 0=\int_{0}^{1}\varphi' v_0'+\varphi v_0 dx, \;\; \varphi\in\mathcal{C}_{c}^{\infty}(0,1), $$ which implies that $v_0'$ has a weak derivative and $v_0''=v_0 \in L^2$. That's enough to imply to $v_0$ is twice absolutely continuous, and \begin{align} ...


0

Your differential equation is wrong. Using integration by parts, we get $$u(0) = \int_0^1 -u \, v_0'' + u \, v_0 \, \mathrm{d}x + u(0) \, v_0'(0) - u(1) \, v_0'(1).$$ This equation has to be satisfied for all $u \in H^1$. Hence...


1

Short answers. Q1: yes. Q2: no, but yes for almost every $r$. Q3: sharp forms of trace theorem also exists, and the target space in it (a Besov space) has lower regularity. Explanation: Q1. You can apply trace theorem on a smaller disk. Q2-3. The reason we may lose regularity when restricting to a surface is that the surface may pass through a singularity ...


1

For part (1), to not stack subscripts too deeply, let us assume that the full sequence $(u_n)$ converges uniformly to the continuous function $u\in C([0,1])$. If $1 < p < \infty$, by the reflexivity of $W^{1,p}(I)$ we know that $(u_n)$ has a subsequence $(u_{n_m})$ converging weakly to some $v \in W^{1,p}(I)$. Then $(u_{n_m}')$ converges weakly to ...


1

Concerning 3., do not estimate the integrals $\int f_j(x) \phi(x+n)dx$ in your way but approximate $f_j$ by functions with compact support (just multiply with an indicator function). If $f_j$ has compact support the integral is $0$ for $n$ large enough. Concerning 2. the limit of a convergent subsequence would be necessarily $0$ because of 3. But the norms ...


0

It all seems to be right except for some fine details. $\dot{\text{res}}$ can be interpreted in two ways: First, as a function $$ \dot{\text{res}} : [0,T] \rightarrow H^1(\Omega),$$ which maps time-points to (state)functions in $ H^1(\Omega)$. Under this interpretation $$\dot{\text{res}} \in L^2\left(0,T,H^1(\Omega)\right).$$ Second, as a linear form $$ ...


0

You seem to be looking for domains $\Omega$ for which $W^{1,p}_0(\Omega) = W^{1,p}(\Omega)$. This will be the case if $\partial \Omega$ has $p$-capacity equal to $0$. Look e.g. in Measure Theory and Fine Properties of Functions by Evans and Gariepy, or Function Spaces and Potential Theory by Adams and Hedberg.


2

It depends on your definition of $\|f\|_{H^{-1}}$. Let us equip $H_0^1$ with the scalar product $$(u,v)_{H_0^1} = \int \nabla u \nabla v \, \mathrm dx.$$ Then, the weak formulation of Poisson's equation is $$(\phi,v)_{H_0^1} = f(v) \quad\forall v \in H_0^1.$$ Hence, the solution $\phi \in H_0^1$ is just the Riesz representative of $f \in H^{-1} = (H_0^1)'$. ...


1

You can show your regulary if $g \in W^{1,p}(\Omega)$, $p > d \ge 2$, by using a trick by Stampacchia. First, we note $$a(u - g,v) + \int_\Omega f(u) v \, \mathrm{d} x = a(-g,v).$$ Now, let $K > 0$ and set $$(u - g)_K = \begin{cases} u - g & \text{if } |u - g| \le K, \\ \operatorname{sign}(u-g) \, K & \text{else}.\end{cases}$$ Then, $(u-g)_K ...


1

Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...


2

Actually $u_n\to u$ in $W^{1,2}$ is $$ \lVert{u_n-u}\rVert_{L^2}^2 + \lVert{u'_n-u'}\rVert_{L^2}^2 \to 0\ , $$ which is quite different from what you thought.


2

Convergence in $W^{1,2}$ is much stronger than that. It means $$\|u_n - u\|_{L^2}^2 + \|u_n' - u'\|_{L^2}^2 \to 0.$$


1

This seems to work: The hyperbolicity implies $$ \theta \int_\Omega |Du|^2 \, dx \leq \int \sum_{i,j} a^{ij} u_i u_j \, dx = B[u,u] - \int_\Omega c u^2 \, dx \leq B[u,u]\,, $$ since $c \geq 0$. For $u \in H_0^1(\Omega)$ the Poincaré inequality implies that there exists $C \geq 0$ such that $$ \|u\|_{H^1(\Omega)}^2 = \|u\|_{L^2(\Omega)}^2 + ...



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