New answers tagged

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 + ...


1

Take a positive, nonzero $u \in C^{\infty}_c(\mathbb{R}^d)$ and consider $\{u_{\epsilon}\}$, where $u_{\epsilon}(x) = u(x\epsilon^{-1})$. For simplicity, let $s = 1$ and notice that (by a change of variables) $$\|u_{\epsilon}\|_{L^{2^*}} = \Big(\int |u_{\epsilon}|^{2^*}\Big)^{\frac{1}{2^*}} = \epsilon^{\frac{d}{2^*}}\|u\|_{L^{2^*}}$$ $$\|\nabla u\|_{L^2} = ...


1

If you have $y \in H^1(\Omega) \cap C(\bar\Omega)$ you can use a truncation argument to obtain $x \in H^1(\Omega) \cap C_c(\bar\Omega)$ such that $\|x-y\|_{H^1}$ is arbitrarily small.


0

As suggested by daw, take $u(x) = 1$. If you have some regularity of $\Omega$, then $$\int_{\mathbb R^n} u \nabla v \, \mathrm{d}x = \int_{\partial\Omega} \frac\partial{\partial n} v \, \mathrm{d}s$$ for $v \in C_c^\infty(\mathbb R^n)$. The right hand side cannot be written as $\int_{\mathbb{R}^n} w \, v \, \mathrm{d}x$ for some $w \in ...


1

If $\Omega$ is bounded, it suffices to take $u(x)\equiv 1$. Then $\hat u$ cannot be in $W^{1,p}(\mathbb R^n)$ for $p>n$ since it is discontinuous. (Sobolev embedding)


0

Continuous but not in any Sobolev space: let $f(x)=g(x_1)$ where $g$ is a continuous but not absolutely continuous: Cantor staircase, or a Weierstrass-type nowhere differentiable function. Since Sobolev functions are absolutely continuous on almost every line segment parallel to a coordinate axis, $f$ is not in $W^{1,p}$. In $W^{1,p}(U)$ but without a ...


0

It does indeed hold. Well, if $\Omega$ is bounded, $u=0$ on $\Omega^c$ not just $\partial \Omega$, and with a slightly bigger integral on the right. On the plus side you only need one $k$. Your idea is already quite good. Now, you write $$u(x) = -h \sum_{l=0}^{N_h} D^h_k u(x+l h e_k),$$ where $N_h$ is such that $N_h h \geq diam(\Omega)$ and $N_h h \leq C$ ...


0

In your original problem you need two derivatives of $u$, whereas for the weak formulation one suffices. So you have reduced the regularity required for $u$ by integrating by parts. Sobolev spaces have been created specifically to be the appropriate spaces where solutions of PDEs in work form live.


1

The functions $$ \varphi_{\lambda}(x)= \cos(\sqrt{\lambda}x),\;\;\;\psi_{\lambda}(x)=\cos(\sqrt{\lambda}(x-l)) $$ satisfy the differential equation $-f''+\lambda f = 0$ subject to the conditions $$ \varphi_{\lambda}(0)=1,\;\varphi_{\lambda}'(0)=0,\;\;\;\;\;\;\psi_{\lambda}(l)=0,\;\psi_{\lambda}'(l)=0. $$ These solutions are linearly independent ...


2

Yes, it does. Since $W^{1,p}(U)$ is reflexive, there exists a further subsequence $\{u_{k_{j_i}}\}$ which converges weakly to some $v\in W^{1,p}(U)$. It must hence also converge weakly to $v$ in $L^p$, and since $u_{k_{j_i}}\to u$ in $L^p$ it follows that $u=v$ and hence has a weak derivative.


1

Notice that we can write $$c_1^2 + c_2^2 = |\nabla v|^2 = (\nabla v \cdot \tau)^2 + (\nabla v \cdot \nu)^2,$$ where $\tau$ is the direction of the tangent line. By assumption, we know that the second term on the RHS is $0$ on $\Gamma_0$. What is left to show is that also the tangential part of the gradient is $0$. Do you see why this is true? EDIT: take ...


2

The first result you need is the following: (i) If $u \in \mathcal{D}'(\mathbb{R}^n)$ and $\varphi \in \mathcal{D}(\mathbb{R}^n)$, then $u \ast \varphi \in \mathcal{E}(\mathbb{R}^n)$ and $(u \ast \varphi) \ast \psi = u \ast (\varphi \ast \psi)$ $\forall \psi \in C^{\infty}_c(\mathbb{R}^n)$ We show that $\mathcal{D}(\mathbb{R}^n)$ is dense in ...


0

The usual trick when you have boundary data is to get rid of it by shifting it to the inside of your domain. For instance in your case, you would want to find a function $$G : \Omega \times (0,T) \to \mathbb{R}$$ such that $$G(x,t) = g(x,t)$$ for $$x \in \partial \Omega$$. Then instead of your problem you consider $$ \begin{cases} v_t - \Delta v = f - ...


0

Using the reverse triangle inequality the way you did, too much stuff goes into the subtracted term. You need a more careful estimate, namely $$ \left|\sum_{|\alpha|\le m} a_\alpha(x)\xi^\alpha\right|\geq \frac{\gamma}{2}|\xi|^m - B \tag{1} $$ where $\gamma$ is from the definition of ellipticity and $B$ is a constant. When $(1)$ is applied to estimate ...


1

Due to your boundary conditions, you have $$a(u;v) = \int_\Omega -\Delta u \, v - 2 \, u \, v \, \mathrm{d}S.$$ We define $B\,u := -\Delta u - 2 \, u$. Hence, $B : H^2 \to L^2$ is continuous. Since the embedding of $L^2$ into $(H^2)^*$ is compact, $B : H^2 \to (H^2)^*$ is compact. Now your $A$ is just the composition of $B$ with the inverse of the Riesz map ...


1

As remarked by Guiseppe Negro, this is, in general not possible. For instance, the function $\varphi := (t \mapsto |t|^\alpha)$ on $\Omega=(-1,1)$ belong to $C(\bar \Omega)$, but not to $H^1(\Omega)$ if $\alpha > 0$ is sufficiently small. Now, for $u \in X$ with $u(t) = 1$ for all $t \in (-1/2, 1/2)$ you will not have $u \, \varphi \in H^1(\Omega)$.


1

The following result tells that $u \in H^1$ is not less restrictive than absolutely continuous. Theorem: Let $\Omega \subset \mathbb{R}$ be an open set, let $1\le p < \infty$. Then $u \in W^{1,p}(\Omega)$ if and only if it admits an absolutely continuous representative whose classical derivative belongs to $L^p(\Omega)$. A proof of this theorem can ...


2

Okay, so let's write out the condition $f(x)=\langle x,y\rangle$... $$\int_{-1}^1 x(t)y(t)+x'(t)y'(t)\,\mathrm{d}t=\int_{-1}^1 e^{2t}x(t)\,\mathrm{d}t.$$ Observe we may write $$\begin{array}{ll} \displaystyle \int_{-1}^1 x'(t)y'(t)\,\mathrm{d}t & \displaystyle =\int_{-1}^1 \left[\frac{\mathrm{d}}{\mathrm{d}t}\big( ...


0

If you define $u^-=\max \{-u,0\}$ then by [Section 4.2.2, Th 4][L. C. Evans and R. F. Gariepy, Measure theory and fine properties of functions] you have $\nabla u^-=\chi _{\{u<0\}}\nabla (-u)=(\chi _{\{u<0\}})^2\nabla (-u)$ and the result follows...


0

Here is a sketch for a proof. Depending on what you already know, you might skip the first step. Show that $u$ is absolutely continuous with $u(x) = \int_0^x u'(s) \,\mathrm{d}s$. There is $\hat x$ with $u(\hat x) = \bar u$. Use $u(x) = u(\hat x) + \int_{\hat x}^x u'(s) \, \mathrm{d}s$ to get an estimate for $\|u\|_{L^2(0,1)}$.


1

In general, if $u \in C^2(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, we have $\widehat{\Delta u}(\xi)=\sum_{i=1}^n \widehat{\partial_{xi} \partial_{xi} u}(\xi)=-4\pi^2 |\xi|^2 \widehat{u}(\xi)$ Now, $H^s(\mathbb{R}^n):=\lbrace u \in L^2(\mathbb{R}^n) : \Lambda^s u \in L^2(\mathbb{R}^n) \rbrace$ with $\omega_s(\xi):=(1+|\xi|^2)^{s/2}$, and $\Lambda^s u := ...


0

The link is that if $f\in C^2 \cap L^2$, $$\mathcal{F}[\Delta f](\xi) = -|\xi|^2 \mathcal{F}[f](\xi)$$ So taking the laplacian is like multiplying by $-|\xi|^2$ in Fourier So, by extension, you can define an operator $\Delta$ by $\Delta f (x) = \mathcal{F^{-1}} \left[ -|\xi|^2 \mathcal{F}[f](\xi) \right](x)$$ Now, about your operator $\Lambda^s$, the ...


0

The points do not have measure zero. On the (topological) boundary you typically have to consider the (so called Hausdorff-) measure of one dimension less (for, e.g., defining equivalence classes). And the zero dimensional measure is the "counting measure". Every point has measure one.


0

Yes, the boundary is the usual topological boundary. To answer your second question, as stated, yes, it means that there is a representative that attains the value $0$ on the boundary of the interval. As you probably recall, every Sobolev function in one variable has a representative that is absolutely continuous, hence the statement you quoted is correct, ...


0

Pick $f\in V$, then \begin{align*} \|f\|_{H^1}^2 &= \int_{0}^1 f(x)^2\, \mathrm{d} x + \int_{0}^1 f'(x)^2\, \mathrm{d} x\\ &= \int_{0}^1 f'(x)\left(\int_{x}^1 f(\xi)\, \mathrm{d} \xi \right)\, \mathrm{d} x + \int_0^1 f'(x)^2\, \mathrm{d} x\\ &\le \left(\int_0^1 f'(\xi)^2\, \mathrm{d}x \right)^{1/2} \left(\int_0^1 \left( \int_x^1 f(\xi)\, ...


0

Let me call attention to a technical (but important) detail. Since functions in Sobolev classes are technically only defined "almost everywhere", how can we impose a condition $v(0)=0$ in a meaningful way? You have tried to get the required estimate (using the fact that $(0,1)$ has length one), assuming "naively" that we can compute $v(x) = v(0) + \int_0^x ...


1

It is true almost in the way you stated it with $a_\epsilon = \epsilon$. You only have to restrict the norm on the left hand side to $\Omega_\epsilon =\{x \in \Omega \colon dist(x, \Omega^c)>\epsilon \}$, since this is where $u_\epsilon$ is defined. On the plus side, boundary conditions and boundary regularity are irrelevant. The proof is rather ...


0

One approach is to use the Fourier transform. Since $u$ is in $H^1_0$, its zero extension to $\mathbb{R}^n$ is in $H^1(\mathbb{R}^n)$. Consider the Fourier transform $\hat u$. The squared $H^1$ norm of $u$ is $$\int (1+|\xi|^2) |\hat u|^2\tag1$$ The squared $L^2$ norm of $u-u_\epsilon$ is $$\int |1-\hat \eta_\epsilon|^2 |\hat u|^2\tag2$$ where ...


2

(This was supposed to be a comment but it is a bit too long) If you take the $\|\cdot\|_{W^{1,p}}$ closure of compactly supported functions you only get the functions that vanish at the boundary. On $\mathbb{R}^n$ there is no boundary, so no problem. Similarly with higher order Sobolev spaces: on $W^{k,p}$ the closure of $C^\infty_c$ consists of those ...


3

A seemingly natural way would be to define a new function that is the value of $v$ on $e_i$ and the value of zero everywhere else. Not really. If a function has value $v$ on $e_i$ that is non-zero on any compactly supported subset of $e_i$, and $0$ elsewhere, it does not have $H^{1/2}$ regularity on the whole boundary: $$ \int_{\partial ...


0

Yes. This can be obtained by the typical approach via mollification: Extend $f$ to $\mathbb{R}^n \setminus \Omega$ by $1$. Let $f_k$ be the convolution of $f$ with a smooth convolution kernel. This directly yields $f_k > 0$ a.e. and $f_k \to f$ in $L^2(\Omega)$ if the convolution kernels are appropriately chosen.


2

No, piecewise constants are not in $H^{1/2}$. Begin with $d=2$ case: then $u$ is a piecewise constant function on the circle. Its Fourier coefficients $\hat u(n)$ decay like $1/n$ and no faster (either by direct calculation, or considering that $u'$ is a combination of Dirac deltas). Hence, $$ \sum_n |n| |\hat u(n)|^2 = \infty $$ A narrow miss; the ...



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