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.