4
$\begingroup$

I was referred to the paper C. Kenig and F. Merle, "Global well-posedness, scattering, and blow-up for the energy critical, focusing non-linear Schrodinger equation in the radial case", in which one of the propositions (Proposition 4.2) concerns compactness modulo symmetries of the flow of a certain critical element to the energy-critical NLS

$$\begin{cases}iu_{t}+\Delta u \pm |u|^{\frac{4}{N-2}}u=0 & {(x,t)\in\mathbb{R}^{N}\times\mathbb{R}}\\ u|_{t=0}=u_{0}\in\dot{H}^{1}(\mathbb{R}^{N}) & {}\end{cases} \tag{CP}$$

Proposition 4.1, which I have omitted because I do not think it essential for my question, gives the existence of a critical initial datum $u_{0,C}\in\dot{H}^{1}$ with corresponding solution $u_{C}$ to the Cauchy problem, such that $u_{C}$ has maximal time interval of existence $I$ containing the origin, and $\|u_{C}\|_{S(I)}=+\infty$, where $S(I)$ is a certain space-time norm.

Proposition 4.2 Assume $u_{C}$ is as in Proposition 4.1 and that $\|u_{C}\|_{S(I_{+})}=+\infty$, where $I_{+}=(0,+\infty)\cap I$. Then there exists $x(t)\in \mathbb{R}^{N}$ and $\lambda(t)\in\mathbb{R}^{+}$, for $t\in I_{+}$, such that $$K = \{v(x,t) : v(x,t) = \frac{1}{\lambda(t)^{(N-2)/2}}u_{C}(\frac{x-x(t)}{\lambda(t)},t)\}$$ has the property that $\overline{K}$ is compact in $\dot{H}^{1}$.

I took a look at the proof on pg. 13, and I am having trouble arriving at the first claim the authors present, assuming that Proposition 4.2 is false. The aforementioned claim is reproduced below.

Using the notation $u(x,t)=u_{C}(x,t)$, the authors claim that if Proposition 4.2 is false, then there exists an $\eta_{0}>0$ and a sequence $\{t_{n}\}_{n=1}^{\infty}$ of times $t_{n}\geq 0$, such that, for all $\lambda_{0}\in\mathbb{R}^{+}$, $x_{0}\in\mathbb{R}^{N}$, we have $$\|\frac{1}{\lambda_{0}^{(N-2)/2}}u(\frac{x-x_{0}}{\lambda_{0}}, t_{n})-u(x,t_{n}')\|_{\dot{H}_{x}^{1}}\geq \eta_{0}, \quad n\neq n' \tag{*}$$

If Proposition 4.2 is false, then we know that for all functions $x:I_{+}\rightarrow\mathbb{R}^{N}$ and $\lambda:I_{+}\rightarrow\mathbb{R}^{+}$, there exist a sequence of times $\{t_{n}\}$ such that $$\{\frac{1}{\lambda(t_{n})^{(N-2)/2}}u(\frac{x-x(t_{n})}{\lambda(t_{n})},t_{n})\}_{n=1}^{\infty}$$ has no convergent subsequence. However, it's mot clear to me how to obtain what Kenig-Merle write. Rather, it seems to me that (*) follows from assuming that that flow $\{u(t) : t\in I\}$ is not relatively compact in $G \backslash\dot{H}^{1}$, where $G$ is the group of translations and dilations associated to the equation and we equip this space with the quotient metric.

Perhaps something related to what I am asking is the following from pg. 6 of the paper T. Tao, M. Visan, and X. Zhang, "Minimal-mass blowup solutions of the mass-critical NLS".

A function $u\in C_{t,loc}^{0}L_{x}^{2}(I\times\mathbb{R}^{N})$ is almost periodic modulo $G$ if the quotiented orbit $\{Gu(t) : t\in I\}$ is a precompact subset of $G\backslash L_{x}^{2}(\mathbb{R}^{N})$. Equivalently, $u$ is almost periodic modulo $G$ if there exists a compact subset $K\subset L_{x}^{2}(\mathbb{R}^{N})$ such that $u(t)\in GK$, $t\in I$.

Above, $G$ is the group of symmetries now associated to the mass-critical NLS. Proving the equivalence assertion seems relevant to what I am asking in regards to the Kenig-Merle paper. I feel like I'm missing some standard functional analysis/topology argument implicit in the cited works, but it's not clear to me what such an argument is. Any help would be greatly appreciated.

$\endgroup$
5
  • $\begingroup$ If $\{u(t): t \in I\}$ were totally bounded in $G \backslash \dot H^1$, then there would exist $g(t) \in G$ such that $\{ g(t) u(t): t \in I \}$ was totally bounded in $\dot H^1$. Now take contrapositives (and use Heine-Borel). $\endgroup$
    – Terry Tao
    Apr 28, 2017 at 5:59
  • $\begingroup$ @TerryTao Dear Professor Tao, thanks for your comment. My difficulty with your suggestion, which I think is just a reformulation of my difficulty as stated in my original question, is that I do not know how to show the proposition contained in your first sentence. It's clear to me that for each, say $2^{-n}$, I can find a function $g_{n}(t)\in G$ such that $\{g_{n}(t)u(t):t\in I\}$ is contained in finitely many balls of radius $2^{-n}$ in $\dot{H}^{1}$. But I do not see why I can extract a limiting $g$ such that that $\{g(t)u(t) :t \in I\}$ is totally bounded in $\dot{H}^{1}$. $\endgroup$ Apr 28, 2017 at 6:43
  • 1
    $\begingroup$ Given any two vectors $f_1, f_2$ in $\dot H^1$, $\|g(t) f_1 - f_2\|_{\dot H^1}$ will converge to $(\|f_1\|_{\dot H^1}^2 + \|f_2\|_{\dot H^1}^2)^{1/2}$ as $t \to \infty$. As a consequence, if $g_n(t) u(t)$ is within $2^{-n}$ of some fixed $f_n$, and $g_{n+1}(t) u(t)$ is within $2^{-n-1}$ of some fixed $f_{n+1}$, and $u(t)$ has norm larger than, say, $10 \times 2^{-n}$, then $g_{n+1}(t)^{-1} g_n(t)$ is constrained to a compact set that depends only on $f_n$ and $f_{n+1}$. One can use this to adjust the $g_{n+1}$ and $f_{n+1}$ so that $g_{n+1}(t)$ is within, say, $2^{-n}$ of $g_n(t)$ ... $\endgroup$
    – Terry Tao
    Apr 28, 2017 at 16:06
  • $\begingroup$ (increasing the number of balls needed to cover at the $2^{-n-1}$ stage if necessary). Iterating this, the $g_n(t)$ now converge to a usable limit $g(t)$. $\endgroup$
    – Terry Tao
    Apr 28, 2017 at 16:07
  • $\begingroup$ [Correction: it is $g_{n+1}(t) g_n(t)^{-1}$ that is constrained, not $g_{n+1}(t)^{-1} g_n(t)$.] $\endgroup$
    – Terry Tao
    Apr 28, 2017 at 16:12

1 Answer 1

2
$\begingroup$

The following is based on Professor Tao's very helpful comments. I have included them in a separate answer, rather than editing the original question, due to length, not for claim of originality.

Abbreviate the critical element $u_{C}(x,t)=u(x,t)$. By the small data theory and the property $\|u\|_{S([0,T_{+}(u_{0})))}=+\infty$, we know that there exists an $A_{0}>0$ such that $\inf_{t\in[0,T_{+}(u_{0}))}\|u(t)\|_{\dot{H}^{1}}\geq A_{0}$. Set $I:=[0,T_{+}(u_{0}))$.

Suppose that we have constructed the functions $g_{1},\ldots,g_{n}: I \rightarrow G$ such that for $1\leq j\leq n$, there exists $f_{j,1},\ldots,f_{j,N_{j}}\in\dot{H}^{1}$ such that for every $t\in I:=[0,T_{+}(u_{0}))$, there exists $k$ such that $$|g_{j}(t)u(t)-f_{j,k}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-j};$$ $$\sup_{t\in I}\|g_{j+1}u(t)-g_{j}(t)u(t)\|_{\dot{H}^{1}} \leq C\epsilon(A_{0})2^{-j}, \quad \forall j=1,\ldots,n-1;$$ and $$|g_{j+1}(t)-g_{j}(t)|\leq \epsilon(A_{0})2^{-j}, \quad \forall t\in I, \enspace \forall j=1,\ldots,n-1$$ where $\epsilon(A_{0})$ is some small constant depending only on the lower bound $A_{0}$.

We proceed to construct $g_{n+1}: I\rightarrow G$. Let $f_{n,1},\ldots,f_{n,N_{n}}$ be the elements of $\dot{H}^{1}$ such that for all $t\in I$, there exists $j\in\{1,\ldots,N_{n}\}$ such that $$\|g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}}\leq 2^{-n}$$ By compactness of the orbits, we know that there exists a function $\tilde{g}_{n+1}:I\rightarrow G$ and elements $\tilde{f}_{n+1,1},\ldots,\tilde{f}_{n+1,\tilde{N}_{n+1}}$ such that for all $t\in I$, there exists $j\{1,\ldots,\tilde{N}_{n+1}\}$ such that $$\|\tilde{g}_{n+1}(t)u(t)-\tilde{f}_{n+1,j}\|_{\dot{H}^{1}}\leq 2^{-n-3}$$ Divide $I$ into finitely many intervals $I(f_{n,j},\tilde{f}_{n+1,k})$ such that $$\forall t\in I(f_{n,j},\tilde{f}_{n+1,k}), \quad \|g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}}\leq 2^{-n}, \enspace \|\tilde{g}_{n+1}(t)u(t)-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\leq 2^{-n-3}$$

We claim that for all $t\in I(f_{n,j},\tilde{f}_{n+1,k})$, $\tilde{g}_{n+1}(t)g_{n}(t)^{-1}$ lies in a compact subset depending only on $f_{n,j}$ and $\tilde{f}_{n+1,k}$. Indeed, otherwise there is a subsequence of times $t_{m}$ such that $\tilde{g}_{n+1}(t_{m})g_{n}(t_{m})\rightharpoonup 0$ and therefore \begin{align*} 2^{-n+1}&\geq \|\tilde{g}_{n+1}(t_{m})u(t_{m})-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}+\|g_{n}(t_{m})u(t_{m})-f_{n,j}\|_{\dot{H}^{1}}\\ &\geq \|g_{n}(t_{m})^{-1}f_{n,j}-\tilde{g}_{n+1}(t_{m})^{-1}\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\\ &=\|\tilde{g}_{n+1}(t_{m})g_{n}(t_{m})^{-1}f_{n,j}-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\\ &\stackrel{m\rightarrow\infty}\longrightarrow (\|f_{n,j}\|_{\dot{H}^{1}}^{2}+\|\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}^{2})^{1/2}\\ &\geq 2^{-n+2}, \end{align*} provided $n$ is sufficiently big depending on $A_{0}$ (this isn't an issue provided we choose $\epsilon(A_{0})$). Since the set $\{\tilde{g}_{n+1}(t)g_{n}(t)^{-1} : t\in I(f_{n,j},\tilde{f}_{n+1,k})\}$ lies in a compact set. We can cover it by finitely many balls $B_{1},\ldots,B_{M}$ of diameter at most $\min_{j,k}\delta(f_{n,j},\tilde{f}_{n+1},k)>0$, where $\delta(f_{n}-f_{n+1})$ is chosen so that for any $g\in G$ with $|g-id|\leq \delta(f_{n},f_{n+1})$, $$\|gf_{n,j}-f_{n,j}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-n}, \quad \delta(f_{n,j},\tilde{f}_{n+1,k})\leq \epsilon(A_{0})2^{-n}$$ For each $l=1,\ldots,M$, let $g_{l}^{*}$ denote the center of $B_{l}$. Define the intervals $$I(f_{n,j},\tilde{f}_{n+1,k},l) := \{t\in I(f_{n,j},\tilde{f}_{n+1,k}) : \tilde{g}_{n+1}(t)g_{n}(t) \in B_{l}\},$$ where we make a choice in the event of overlap so that the $I(f_{n,j},\tilde{f}_{n+1,k},l)$ form a partition of $I(f_{n,j},\tilde{f}_{n+1,k})$. Now define the new function $g_{n+1}$ on $I(f_{n,j}, \tilde{f}_{n+1,k},l)$ by $$g_{n+1}(t) := ({g}_{l}^{*})^{-1}\tilde{g}_{n+1}(t), \quad t\in I(f_{n,j}, \tilde{f}_{n+1,k},l)$$ Next, define new elements $f_{n+1,k,l}$ of $\dot{H}^{1}$ by $$f_{n+1,k,l} := (g_{l}^{*})^{-1}f_{n+1}$$ so that for every $t\in I(f_{n,j},\tilde{f}_{n+1,k},l)$, $$\|g_{n+1}(t)u(t)-{f}_{n+1,k,l}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-n-1}$$ Additionally, we have that for all $t\in I(f_{n,j},\tilde{f}_{n+1,k},l)$, \begin{align*} \|g_{n+1}(t)u(t)-g_{n}(t)u(t)\|_{\dot{H}^{1}} &= \|(g_{n+1}(t)g_{n}(t)^{-1})g_{n}(t)u(t)-g_{n}(t)u(t)\|_{\dot{H}^{1}}\\ &\leq \|(g_{n+1}(t)g_{n}(t)^{-1})g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}} + \|f_{n,j}-g_{n}(t)u(t)\|_{\dot{H}^{1}}\\ &=\|g_{n}(t)u(t)-g_{n}(t)g_{n+1}(t)^{-1}f_{n,j}\|_{\dot{H}^{1}} + \epsilon(A_{0})2^{-n}\\ &\leq \epsilon(A_{0})2^{-n+1}+\|((g_{l}^{*})^{-1}-g_{n}(t)g_{n+1}(t)^{-1})f_{n,j}\|_{\dot{H}^{1}}\\ &\leq 3\epsilon(A_{0})2^{-n} \end{align*}

We now repeat this procedure on each of the finitely remaining intervals $I(f_{n,j'},\tilde{f}_{n+1,k'})$, each time adjoining finitely many new elements $f_{n+1,k',1},\ldots,f_{n+1,k',M_{j',k'}}$ to the net. In the end, we have constructed the function $g_{n+1}$ on the entire interval $I$ and adjoined finitely many elements to the $(n+1)$-th stage net.

Now define $$g(t) := \lim_{n\rightarrow\infty}g_{n}(t), \quad t\in I$$ To see that the set $\{g(t)u(t) : t\in I\}$ is totally bounded, we note that the sequence $g_{n}(t)u(t)\rightarrow g(t)u(t)$ in $\dot{H}^{1}$, uniformly in $t\in I$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.