Hot answers tagged

4

If we assume that $v$ is twice-differentiable, we can integrate the $u'v'$ term by parts, so the equation becomes $$ u(0) = u(1)v'(1)-u(0)v'(0) + \int_0^1 (-v''(x)+v(x))u(x) \, dx. $$ Therefore a natural thing to do now is find a function that satisfies $v'(0)=-1$, $v'(1)=0$, and $-v''+v=0$ in between. Some elementary differential equation solving later, we ...


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 ...


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 ...


2

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


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

(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 ...


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( ...


2

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 ...


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

The question is misleading in that if you expand the scalar product, then the integral over each term is zero, so it is really a statement about scalar functions I would argue. To prove the scalar valued version, try for $f$ (now a scalar valued function) the identity and then generalize. Ask if you have problems in executing this plan, but please add more ...


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

Typically you may study the heat equation $$u_t- \Delta u = 0.$$ We can't always expect the solution $u$ to be twice weakly differentiable, so we ask that $\Delta u$ in fact means $$(\Delta u)(v) = -\int \nabla u \nabla v$$ that is, it is a linear functional $\Delta: H^1_0 \to H^{-1}$ which is bounded. Hence from $u_t = \Delta u$ (the equation) we see that ...


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 ...


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 ...


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 ...


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 := ...


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

If $u \in L^2(0,T;H^1_0)$ with $u' \in L^2(0,T;H^{-1})$ then $u \in C^0([0,T];L^2)$, so $u(t)$ makes sense for all $t$. And yes that identity does hold, since $(u(t), u(t)_{L^2}$ is absolutely continuous wrt. $t$. I think the $c_n$ are actually classically differentiable since if I am right they come from the Galerkin approximations. Applying Gronwall's ...


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.


1

The answer is no, basically because the discrete symmetry group that fixes even functions is finite, so given any badly behaved sequence $(u_n(x)) \subset H^s$, its image after symmetrisation $(u_n(x)+u_n(-x))$ is also in $H^s$, and can have the same problems. For example, take $u_n(x) = \exp{(-(x-n)^2)}$: then $(u_n)$ has no convergent subsequence, and ...


1

You don't need to introduce Sobolev spaces, but the point here is that $W^{1,2}$ is the completion of $C^\infty$ with the norm $\|f\|_{W^{1,2}}:=\|f\|_{L^2}+\|f'\|_{L^2}$. So if you prove the above inequality for all smooth $f$ with appropriate conditions, then by continuity of the integral you automatically have the result for all $f\in W^{1,2}$.


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 ...


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

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

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)



Only top voted, non community-wiki answers of a minimum length are eligible