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What is the proof of the local well-posedness of the quadratic nonlinear Schrödinger equation $\mathrm{i} \,\partial_t u + \Delta u \pm \left|u\right| u = 0$ on the 1D torus in $H^s$ for $s > 1$ (a good reference would suffice)?

$H^s(\mathbb{T})$ is an algebra, but $\left|u\right| u$ is not of the form $u^2$, $\overline{u}u$ or $\overline{u}^2$ and so the LWP doest not immediately follow from the Banach contraction mapping principle.

The LWP should hold according to Tao's webpage (even for $s > 0$). However, the above problem is not covered by Theorem I in [Bo1993] (reference as on Tao's webpage) and in fact Remark (ii) after Proposition 5.73 states that uniqueness of the solutions is unclear.

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On a 1D torus, $H^s$, $s>1$, is embedded into $L^\infty$. A simple proof can be based on the semigroup $e^{i t\Delta}$, i.e., find a solution $u$ satisfying $$ u(t)=e^{i t\Delta}u_0 + i \int_0^t e^{i (t-s)\Delta} |u(s)|u(s) \, d s ,\quad t\in(0,T] , $$ where $u_0\in H^s$ is a given initial value. Given $v\in C([0,T];H^s)$ one can define $u=Mv$ as the solution of $$ u(t)=e^{i t\Delta}u_0 + i \int_0^t e^{i (t-s)\Delta} |v(s)|v(s) \, d s ,\quad t\in(0,T] . $$ Then one can show that the map $M$ is a contraction on the metric space $X$ if $T$ is small, where $$ X=\{v\in C([0,T];H^s): \|v-u_0\|_{H^s}\le 1 \} . $$

In fact, if $u_1=Mv_1$ and $u_2=Mv_2$, with $v_1,v_2\in X$, then $$ u_1(t)-u_2(t) =i \int_0^t e^{i (t-s)\Delta} (|v_1(s)|v_1(s)-|v_2(s)|v_2(s)) \, d s ,\quad t\in(0,T] . $$ Therefore \begin{align} \|u_1-u_2\|_{C([0,T];H^s)} &\le CT \sup_{0\le s\le t\le T} \|e^{i (t-s)\Delta} (|v_1(s)|v_1(s)-|v_2(s)|v_2(s))\|_{H^s} \\ &\le CT \sup_{0\le s\le t\le T} \||v_1(s)|v_1(s)-|v_2(s)|v_2(s) \|_{H^s} \\ &\le CT \|v_1-v_2\|_{C([0,T];H^s)} . \end{align} Therefore, if $T$ is sufficiently small then the map $M$ is a contraction. The key step is that for $s\in[0,2]$ and $\|v_1\|_{H^s}+\|v_2\|_{H^s}\le \|u_0\|_{H^s}+1$ there holds (since $|v_1(s)|v_1(s)$ is a quadratic term) $$ \||v_1(s)|v_1(s)-|v_2(s)|v_2(s) \|_{H^s} \le C \|v_1(s)-v_2(s)\|_{H^s} . $$ However, it may not be true for $s>2$.

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  • $\begingroup$ Thanks! Could you elaborate more on how to obtain the last estimate above? $\endgroup$ – Leonid Chaichenets Jan 11 at 14:56

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