On the global well-posedness of the quadratic NLS on H1(T)+L2(R)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$H^1({{\mathbb {T}}}) + L^2({{\mathbb {R}}})$$\end{document}

We study the one dimensional nonlinear Schrödinger equation with power nonlinearity uα-1u\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\left| u \right| ^{\alpha - 1} u$$\end{document} for α∈[1,5]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha \in [1,5]$$\end{document} and initial data u0∈H1(T)+L2(R)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$u_0 \in H^1({{\mathbb {T}}}) + L^2({{\mathbb {R}}})$$\end{document}. We show via Strichartz estimates that the Cauchy problem is locally well-posed. In the case of the quadratic nonlinearity (α=2\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\alpha = 2$$\end{document}) we obtain global well-posedness in the space C(R,H1(T)+L2(R))\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C({{\mathbb {R}}}, H^1({{\mathbb {T}}}) + L^2({{\mathbb {R}}}))$$\end{document} via Gronwall’s inequality.