# What is standard continuity argument for well-posedness?

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?

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 $$T_*$$.

• Certainly the correct interpretation, unlike mine. Sep 21, 2022 at 15:13

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.

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