superquadraticとは 意味・読み方・使い方

語源

From super- +‎ quadratic.

発音

形容詞

superquadratic (not comparable)

1. (mathematics) Describing an extension of a quadratic function related to the superellipsoid
   • 2015, Wenmin Gong, Guangcun Lu, “Existence results for coupled Dirac systems via Rabinowitz-Floer theory”, in arXiv‎:

     In this paper, we construct the Rabinowitz-Floer homology for the coupled Dirac system \begin{equation*} \left\{ \begin{aligned} Du=\frac{\partial H}{\partial v}(x,u,v)\hspace{4mm} {\rm on} \hspace{2mm}M,\\ Dv=\frac{\partial H}{\partial u}(x,u,v)\hspace{4mm} {\rm on} \hspace{2mm}M, \end{aligned} \right. \end{equation*} where ${\displaystyle M}$ is an ${\displaystyle n}$-dimensional compact Riemannian spin manifold, ${\displaystyle D}$ is the Dirac operator on ${\displaystyle M}$, and ${\displaystyle H:\Sigma M\oplus \Sigma M\to \mathbb {R} }$ is a real valued superquadratic function of class ${\displaystyle C^{1}}$ with subcritical growth rates.