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Let

  • $T\ge 0$
  • $d\in\mathbb N$
  • $\Omega\subseteq\mathbb R^d$ be open and $\Lambda\subseteq\mathbb R^d$ be bounded and open with $\overline\Lambda\subseteq\Omega$
  • $u\in C^{1,\:2}([0,T]\times\Omega,\mathbb R^d)$ and $$\tilde u(t):=u(t,\;\cdot\;)\;\;\;\text{for }t\in[0,T]$$

Suppose that $$\left.\tilde u(t)\right|_{\partial\Lambda}=0\;\;\;\text{for all }t\in[0,T]\;.\tag 1$$

How can we show that $$\tilde u(t)\in H_0^1(\Lambda,\mathbb R^d)\;\;\;\text{for all }t\in[0,T]\tag 2$$ and that $$[0,T]\to H_0^1(\Lambda,\mathbb R^d)\;,\;\;\;t\mapsto\tilde u(t)\tag 3$$ is continuous?

I've tried the following for the continuity of $(3)$: Let $(t_n)_{n\in\mathbb N}\subseteq[0,T]$ with $$|t_n-t|\:\xrightarrow{n\to\infty}\:0\tag 4$$ for some $t\in[0,T]$, $$v_n:=\left|\tilde u(t_n)\right|^2\;\;\;\text{for }n\in\mathbb N$$ and $v:=\left|\tilde u(t)\right|^2$. By $(2)$, $$v_n\in L^1(\Lambda,\mathbb R^d)\;\;\;\text{for all }n\in\mathbb N\;.\tag 5$$ Moreover, $$|v_n(x)-v(x)|\:\xrightarrow{n\to\infty}\:0\;\;\;\text{for all }x\in\Lambda\;,\tag 6$$ since $u$ is continuous in the first variable. Since $u$ is continuous in the second variable and $\overline\Lambda$ is compact, $$|v(x)|\le M\;\;\;\text{for all }x\in\Lambda\tag 7$$ for some $M\ge 0$. Let $\varepsilon>0$. By $(4)$ and $(6)$, $$|v_n(x)-v(x)|<\varepsilon\;\;\;\text{for all }n\ge N\tag 8$$ for some $N\in\mathbb N$. Thus, $$|v_n(x)|<M+\varepsilon\;\;\;\text{for all }n\ge N\tag 9$$ by combination of $(7)$ and $(8)$. Thus, we can apply Lebesgue's dominated convergence theorem and conclude that $$\left\|v_n-v\right\|_{L^1(\Lambda,\mathbb R^d)}\:\xrightarrow{n\to\infty}\:0\;.\tag{10}$$ If I'm not terribly wrong, we can apply the same argumentation to $$w_n:=\left|\nabla\tilde u(t_n)\right|^2\;\;\;\text{for }n\in\mathbb N$$ and $w:=\left|\nabla\tilde u(t)\right|^2$ in order to finally obtain the continuity of $(3)$.

Any doubts? And in either case: How can we show $(2)$. It seems to be obvious by $(1)$, but how do we need to argue in detail?

0xbadf00d
  • 14,208

1 Answers1

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You have now edited the question multiple times, each time adding assumptions. At this point the statement is obvious, since $C^{1,2}([0,T]\times \Lambda) \subset C([0,T],H^1(\Lambda))$ with compact embedding.

Hans Engler
  • 16,092
  • (a) When you say, the statement is not correct, you only mean statement $(2)$, right? Do you agree that my proof of the continuity of $(3)$ is correct (under the assumption that $(2)$ holds)? – 0xbadf00d Sep 02 '16 at 20:17
  • (b) Consider $t=1/2$. Then, I guess, the crucial point in your counterexample is that $x\mapsto|u_x(x,t)|^2$ is not continuous at $x\in\partial\Lambda=\left{-1,1\right}$. So, what if we add the assumption that for each $t\in[0,T]$ each partial derivative $\frac{\partial u(t,;\cdot;)}{\partial x_i}$ is continuous from $\overline\Lambda$ to $\mathbb R^d$? Do you think that in that case $(2)$ is true too? – 0xbadf00d Sep 02 '16 at 21:05
  • My point is that $u$ does not have values in $H^1$ for $t \le \frac{1}{2}$. Your assumptions do not imply (2) and therefore certainly not (3). I doubt that (2) implies (3). I suggest you post this as a separate question. – Hans Engler Sep 03 '16 at 01:45
  • (b)' But you say that $u(;\cdot;,t)\not\in H^1([-1,1])$, since $x\mapsto|u_x(x,t)|^2$ is not integrable for $t\le\frac12$. But if we add the assumption that for each $t\in[0,T]$ each partial derivative $\frac{\partial u(t,;\cdot;)}{\partial x_i}$ is continuous from $\overline\Lambda$ to $\mathbb R^d$, then $(2)$ should follow, if we assume boundedness of $\Lambda$. Do you agree? (a)' In either case, my question (a) was if you think that $(3)$ is correct if $(2)$ holds. – 0xbadf00d Sep 03 '16 at 09:29
  • You are assuming here that $\nabla u(\cdot,t) \in L^\infty(\Lambda)$ for all $t$. This implies of course that $u(\cdot,t) \in H^1$. As to concluding (3) from (2), I think your arguments has gaps. For example, (5) does not follow from the assumptions. – Hans Engler Sep 07 '16 at 12:40
  • By $(2)$, $$\tilde u(t)\in L^2(\Lambda,\mathbb R^d);;;\text{for all }t\in (0,T)$$ and that's equivalent to $$\left|\tilde u(t)\right|^2\in L^1(\Lambda,\mathbb R^d);;;\text{for all }t\in (0,T);.$$ So, by definition of the $v_n$, it's clear that $(5)$ holds. Where do you see a problem? – 0xbadf00d Sep 08 '16 at 10:50
  • The main problem is that (2) does not imply (3). For a counterexample, consider the function $u(t,x) = \left(x^2 + (t-1)^2 \right)^{1/4}$ for $x \in \Lambda = [0,1]$ and $0 \le t \le T = 2$. Then $u \in H^1(\Lambda_T)$ but $|u(t,\cdot)|_{H^1(\Lambda)}$ becomes infinite as $t \to 1$. Thus $t \mapsto \tilde u(t)$ is not continuous as a $H^1$-valued function. The fact that this $u$ does not vanish on the lateral boundary of $\Lambda_T$ is easy to fix. – Hans Engler Sep 08 '16 at 18:19
  • As I said in my third comment: I want to assume that $u\in C^{1,:2}(\Omega,\mathbb R^d)$ with $\Omega$ being open, $\overline\Lambda\subseteq\Omega$ and $\Lambda$ being bounded. This assumption will guarantee that ${\rm D}^\alpha u(t,;\cdot;)$ is bounded in $\Lambda$, for all $|\alpha|\le 2$ and $t\in[0,T]$. I've added this to the question. Your $u$ doesn't satisfy these assumptions, cause $\frac{\partial u}{\partial x}(1,;\cdot;)$ is not continuous at $x=0$. – 0xbadf00d Sep 08 '16 at 19:10
  • If $(2)$ doesn't imply $(3)$, there must be something wrong with my proof. Please tell me what my mistake is and I believe you. – 0xbadf00d Sep 08 '16 at 19:10
  • You have proved that $t \mapsto \tilde u(t)$ is continuous from $[0,T]$ to $L¹(\Lambda)$.The mistake is in your assertion that the same proof works for $w_n$. Several things break down in this case. For example, $\nabla \tilde u(t)$ need not be bounded on $\Lambda$ and $\nabla u(\cdot,t_n)$ need not converge uniformly, since you are not assuming that $u \in C^{1,2}$ on the closure of $\Lambda_T$. – Hans Engler Sep 09 '16 at 02:02
  • Actually, that was a leftover of the original formulation of the question (without the additional assumption). I've edited that. What do you think now? – 0xbadf00d Sep 09 '16 at 11:53
  • Still wrong. (2) does not imply (3) as my example shows. You haven't addressed that at all. Und jetzt machen wir mal Schluss mit dieser Unterhaltung. Ich bin nicht Dein Professor. – Hans Engler Sep 09 '16 at 11:58
  • Your examples don't satisfy the assumptions, cause they have points of discontinuity. You agree to me that $(v_n)_{n\in\mathbb N}$ converges to $v$ in $L^1$ and I've fixed the problem you've mentioned which prevented us from applying the same argumentation to $w_n,w$. So, we can conclude that $t\mapsto\tilde u(t)$ is continuous from $[0,T]$ to $H^1(\Lambda,\mathbb R^d)$. Where is the problem? I'm continuing the discussion, cause I think the conclusion is obvious, but you're saying the statement is wrong. And that's why I ask you where my mistake is. – 0xbadf00d Sep 09 '16 at 12:23
  • See my edited answer.. – Hans Engler Sep 09 '16 at 14:39
  • The only assumption I've added is the one I've introduced in my third comment. With the other edits I've corrected typos. – 0xbadf00d Sep 09 '16 at 14:47