# Local well-posedness of the quadratic NLS on the 1D torus

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.

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 $$s>2$$.