New answers tagged

0

I think the best answer is to look at the paper Di Fazio, G., Hakim, D.I. & Sawano, Y. Elliptic equations with discontinuous coefficients in generalized Morrey spaces, European Journal of Mathematics (2017) 3: 728. DOI: 10.1007/s40879-017-0168-y and the references therein. As a quick summary the paper says in the introduction $a_{ij} \in W^{1,n}(\Omega)...


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I finished by point 3. I promized. The paper is disponible here for more information. And I recal that $u^\varepsilon$ are regular engouh (continuous) so that measurability in any subsequent space is valid. The case $p=1$. The authors say the sequence $u^\varepsilon$ uniformely bounded in $L^\infty(0,T;L^1(\mathbb R^n))$ is weakly relatively compact in $L^\...


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You have already solved it with $(u,v)=\int \hat{u}(\xi)\tilde{v}(\xi)d\xi=\int\hat{u}(\xi)<\xi>^s\tilde{v}(\xi)<\xi>^{-s}d\xi$ is a isomorphism of $H^{-s}$ and the dual of $H^{s}$


1

For $\Rightarrow$, let $\alpha$ be such that $|\alpha| \leq k$, let $f=D^{\alpha}u$, let $\hat{u}$ be the $L^2$ (and distributional) Fourier transform of $u$. $f$ is both a $\mathcal{S}’$ distribution and a $L^2$ function, hence its (distributional) Fourier transform is (up to a scalar function) $\xi^{\alpha}\hat{u}$, and it must also be a $L^2$ function. ...


1

It is easy to see that $u^+$ or $u^-$ have as much weak derivatives as $u$, just set $$\Omega^+:=\{x\in\Omega: u(x)\ge 0\},\quad\Omega^-:=\{x\in\Omega: u(x)\le 0\}$$ Hence $$\int_{\Omega} u^+(x)\partial_j\varphi(x)\, dx=\int_{\Omega} u(x)\partial_j\varphi(x)\chi_{\Omega^+}(x)\, dx=-\int_{\Omega}\partial_j u(x)\chi_{\Omega^+}(x)\varphi(x)\, dx$$ where $\...


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The second definition is bogus. In fact, if you only assure that $u(t) \in W^{1,\infty}$ for (almost) all $t \in (0,T)$, the function $$ t \mapsto \|u(t)\|_{W^{1,\infty}}$$ might fail to be measurable, hence the left-hand side of (1) is not even well defined. Moreover, the separability condition is crucial. There are weakly measurable $u$, for which (1) ...


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For the poisson equation the natural space of the weak formulation is $H^{-1}$. However it is wrong or at least bad style to write it as an integral. It is advised to use the bracket notation $(f,\phi)_{H^{-1}}$. Notice that for all functions $f\in L^2$ there exists an associated element $f^*\in H^{-1}$ given by $$ (f^*,\phi)_{H^{-1}}=\int f \phi dx $$ To ...


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Here is an idea. If you would have equality, then you get $$ \| \frac{\mathrm d}{\mathrm d t} \Pi_C\gamma (t) \| = \|\dot\gamma(t)\|$$ for a.a. $t$. Moreover, the projection is firmly non-expansive. Hence, $$ \| \gamma(t_2) - \gamma(t_1) \|^2 \ge \| \Pi_C \gamma(t_2) - \Pi_C \gamma(t_1) \|^2 + \| \Pi_C \gamma(t_2) - \gamma(t_2) - \Pi_C \gamma(t_1) + \gamma(...


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Why not take a concrete example $L^2(0,1)$ is the inner product space $\langle f,g\rangle= \int_0^1 f(x)\overline{g(x)}dx$, $H^1(0,1)$ is the inner product space $( f,g) =\langle f,g\rangle+\langle f',g'\rangle$ $e_n = \frac{e^{2i \pi nx}}{\sqrt{1+4\pi^2 n^2}}$ is an orthonormal basis of $H^1(0,1)$. The map $f \mapsto (g \mapsto (g,\overline{f}))$ is an ...


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There exists a compact embedding $H^1_0(\Omega)\to H^{-1}(\Omega)$. There also exists an isometric isomorphism $H^1_0(\Omega)\to H^{-1}(\Omega)$. There is nothing contradictory about this, because these are two different maps. (If you view a map from a normed space to its dual as a bilinear form on the space, the first map corresponds to the $L^2$ inner ...


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Then answer is yes. Here is a proof: For each $j=1,\ldots,n$ put $f_j:=\partial_jg_1 \chi_{\Omega_1}+\partial_j g_2 \chi_{\Omega_2}$. Clearly, $f_j\in L^2(\Omega')$. If we can show that the weak (distributional) derivative of $f$ coincides with $f_j$, we obtain $f\in H^1(\Omega')$. To this end, we simply verify for any $\phi\in C^\infty_0(\Omega')$ that: \...


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Sure. If $M$ is a compact manifold, and $k$ is a non-negative integer, then $H^k(M)$ is the space of function $u\in L^2(M)$ with the property that for any $\ell$ smooth vector fields $X_1,\cdots, X_\ell$ on $M$, with $\ell\leq k$, we have $X_1\cdots X_\ell u\in L^2(M)$. For a reference, you can see Michael Taylor's PDE I text.


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This does not seem to be true. Consider the analogous problem in one dimension, where $B$ reduces to $(-1, 1)$ and $B_+=(0,1)$. Let $f=-2$. Then the unique solution to $$ \begin{cases} u''=-2, & (0, 1), \\ u(0)=u(1)=0, \end{cases} $$ is $u(x)=x(1-x)$, and its even extension is $$ v(x)=|x|(1-|x|), $$ which is NOT the solution to $$ \begin{cases} w''=-...


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It's not true that $\mathcal C_c^\infty (I)$ is dense in $W^{1,p}(I)$ when $I\neq \mathbb R$. What is true is $\mathcal C_c^\infty (\mathbb R)$ is dense in $W^{1,p}(\mathbb R)$. And indeed, if $I\neq \mathbb R$ is an interval, we define $W_0^{1,p}(I)$ as the closure of $\mathcal C_c^\infty (I)$ in $W^{1,p}(I)$.


2

By Poincaré, $$\int{|u|^2} \leq C\int{|\nabla u|^2} \leq C\int{|fu|}.$$ By Cauchy-Schwarz, $$\|u\|^2_{L^2} \leq C\|f\|_{L^2}\|u\|_{L^2}.$$ Thus $\|u\|_{L^2} \leq C\|f\|_{L^2}$. Now, with the same estimation, $$\|\nabla u\|_{L^2}^2 \leq \int{|fu|} \leq \|f\|_{L^2}\|u\|_{L^2} \leq C\|f\|^2_{L^2}.$$ Therefore, $$\|u\|^2_{W^{1,2}_0} = \|u\|^2_{L^2}+\|\...


2

You have to use the Poincaré inequality again. This gives $$\lVert u\rVert_{L^2}^2\le C_1\lVert \nabla u\rVert^2_{L^2}\le C_2\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}, $$ so $\lVert u\rVert_{L^2}^2 + \lVert \nabla u\rVert^2_{L^2}\le C_3\lVert f\rVert_{L^2}\lVert \nabla u\rVert_{L^2}$, for some big enough constant $C_3>0$. This is the inequality you ...


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To motivate Sobolev spaces, let me pose a motivating problem. Let $\Omega$ be a smooth, bounded domain in ${\Bbb R}^n$ and let $f$ be a $C^\infty$ function on $\Omega$. Prove that there exists a $C^2$ function $u$ satisfying $-\Delta u = f$ in $\Omega$ and $u = 0$ on the boundary of $\Omega$. As far as PDE's go, this is the tamest of the tame: it's a ...


1

No. Consider $f \in \cap_{s < 0}H^s(\mathbb{R})$ defined by $\hat f(t) = (1+t^2)^{-1/4}$. This function is not in $L^2$.


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I can't tell if you've miscopied or misread, but your statement about $w'$ is wrong. What Brezis actually writes (p. 213, section 8.2) is $$ w' = G'(v)v' = p|v|^{p-1}v' $$ Now, $w' = G'(v)v'$ is just the chain rule like you say, and for $G'(v)$ we calculate: $$ G'(v) = (p-1)|v|^{p-2}\cdot \mathop{sgn} v\cdot v + |v|^{p-1}$$ which simplifies to $(p-1)|v|^{p-...


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We know that $u\in W^{2,2}(\Omega')$ because $\|D^2 u\|_{L^2(\Omega')}<\infty$ and $u\in W^{1,2}(\Omega')$. Now, take a cut-off function $\eta\in C_c^{\infty}(\Omega)$ such that $0\leq \eta \leq 1$, $\eta=1$ in $\Omega'$, $\|\nabla \eta\|_{L^{\infty}(\Omega)}\leq C_1(\Omega',\Omega)$, $\|D^2 \eta\|_{L^{\infty}(\Omega)}\leq C_2(\Omega',\Omega)$. Then $...


1

Thank you so much for your answer! I actually managed to resolve my confusion. When calculating the given example in the book, one comes to a point where the decisive question is: Is $$ \int_0^a \frac{r}{(r \log(r))^p} \, \mathrm{d} r < \infty$$ for $a < 1$. But this integral is only finite if $p \leq 2$, so actually the counterexample in the ...


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