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Motivation: I'm trying to understand the proof of Theorem 3.1 in Antonelli, Saut, and Sparber - Well-Posedness and averaging of NLS with time-periodic dispersion management. Though in the following I'm raising kind of abstract question.

Consider non-linear Schrödinger equation (NLS): $$i\partial_tu +\Delta u = |u|^{2}u, \quad u(t_0, x)= \varphi(x).$$

Suppose $X$ (space of functions on $\mathbb R^d$) is Banach space and assume that $$\|u\|_{L_I^{\infty}X} \lesssim \|\varphi\|_{X} + |I|\|u\|^3_{L_{I}^{\infty}X}$$ here $I$ is small time interval.

Questions: (1) How to use the standard continuity argument to say that: there exists a solution $u$ to NLS in $I \times \mathbb R^d$ such that $$\|u\|_{L^{\infty}_{I} X} \leq C \|\varphi \|_{X}$$ for sufficiently small $I$? (2) What is standard continuity argument?

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I understand continuity argument a bit differently than the other post. Suppose you've shown the inequality $$\|u\|_{L^\infty(I,X)} \leq C(\|\varphi\|_{X} + |I|\|u\|_{L^\infty(I,X)}^3) \tag{1},$$ for some absolute constant $C>0$ and all intervals $I$ with $|I|\leq\delta$.

Since $u(t_0)=\varphi$ and your solution $t\mapsto u(t)$ is continuous with values in $X$, there exists a $t_*>t_0$ such that for $t\in [t_0,t_*]$, $\|u(t)\|_{X}\leq \frac{3C}{2}\|\varphi\|_{X}$. Let $T_*>t_0$ be the maximal such $t_*$, that is $$\forall t_0\leq t\leq T_*, \enspace \|u(t)\|_{X}\leq \frac{3C}{2}\|\varphi\|_{X}$$ and there exists a sequence $t_n\rightarrow T_*^{+}$ such that $\|u(t_n)\|_{X}>\frac{3C}{2}\|\varphi\|_{X}$. Note this implies $\|u(T_*)\|_{X} = \frac{3C}{2}\|\varphi\|_{X}$. If no such $T_*$ exists (i.e., $\|u(t)\|_{L^\infty([t_0,\infty),X)}\leq \frac{3C}{2}\|\varphi\|_{X}$), then there is nothing to prove, so we may assume otherwise.

We can use inequality (1) to get a lower bound for $T_*-t_0$. Indeed, setting $I=[t_0,T_*]$, if $|I|>\delta$, then there is nothing to prove. If $|I|\leq \delta$, then $$\|u\|_{L^\infty(I,X)} \leq C\|\varphi\|_{X} + \frac{27C^4(T_*-t_0)}{8}\|\varphi\|_{X}^3.$$ This implies that $\frac{27C^3(T_*-t_0)}{8}\|\varphi\|_{X}^2 \geq \frac{1}{2}$, otherwise we would have $$\|u\|_{L^\infty(I,X)} < \frac{3C}{2}\|\varphi\|_{X} \Longrightarrow \|u(T_*)\|_{X} < \frac{3C}{2}\|\varphi\|_{X},$$ which contradicts our choice of $T_*$.

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  • $\begingroup$ Certainly the correct interpretation, unlike mine. $\endgroup$
    – username
    Sep 21, 2022 at 15:13
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The way I understand it is that because of this bound you can derive a solution by a fixed point. Set $u_0=0$ and $$ u_{n+1} = A + \int_0^t B |u_n|^{p-1}u_n ds $$ (A,B represents the various quantities appearing in the paper) The bound found gives $\|u_{n+1}\| \leq \alpha + \beta |I| \|u_{n}\|^3$.

If $\|u_n\|\leq K$, then $\|u_{n+1}\|\leq K$ provided $$ K\leq \alpha + \beta |I| K^3. $$ For example take $K=10\alpha$. Then this is true provided $|I|\leq \frac{9}{\beta 10^3\alpha^2}$.

So now you have a bounded sequence, and by the same token a contraction, provided $$ 3 \beta |I| K^2 <1. $$ so the sequence you constructed converges and satisfies the $K$ bound, which is exactly what you wanted.

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  • $\begingroup$ thanks. Are you saying eventually that it just follows by contraction mapping principle? $\endgroup$
    –  Analyst
    Sep 20, 2022 at 12:43
  • $\begingroup$ @Analyst at least that's how i would do it. $\endgroup$
    – username
    Sep 21, 2022 at 6:20

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