Abstract
In this paper, we consider the existence of a least energy nodal solution and a ground state solution, energy doubling property and asymptotic behavior of solutions of the following critical problem:{−(a+b∫R3|∇u|2dx)Δu+V(x)u+λϕu=|u|4u+kf(u),x∈R3,−Δϕ=u2,x∈R3.\\documentclass[12pt]{minimal}\t\t\t\t\\usepackage{amsmath}\t\t\t\t\\usepackage{wasysym}\t\t\t\t\\usepackage{amsfonts}\t\t\t\t\\usepackage{amssymb}\t\t\t\t\\usepackage{amsbsy}\t\t\t\t\\usepackage{mathrsfs}\t\t\t\t\\usepackage{upgreek}\t\t\t\t\\setlength{\\oddsidemargin}{-69pt}\t\t\t\t\\begin{document}$$ \\textstyle\\begin{cases} -(a+ b\\int _{\\mathbb{R}^{3}} \\vert \\nabla u \\vert ^{2}\\,dx)\\Delta u+V(x)u+\\lambda \\phi u= \\vert u \\vert ^{4}u+ k f(u),&x\\in \\mathbb{R}^{3}, \\\\ -\\Delta \\phi =u^{2},&x\\in \\mathbb{R}^{3}. \\end{cases} $$\\end{document} By nodal Nehari manifold method, for each b>0, we obtain a least energy nodal solution u_{b} and a ground-state solution v_{b} to this problem when kgg1, where the nonlinear function fin C(mathbb{R},mathbb{R}). We also give an analysis on the behavior of u_{b} as the parameter bto 0.
Highlights
R3 k k≤ s2k u+ 2 + tk u– 2 + Cs4 u+ 4 + Ct4 u– 4 k k for some constants C > 0
1 Introduction and main results Our goal of this paper is to consider the existence of nodal solution and ground state solution of the following Kirchhoff–Schrödinger–Poisson system:
Schrödinger–Poisson equation reduces to the undermentioned Schrödinger–Poisson system
Summary
≤ s2k u+ 2 + tk u– 2 + Cs4 u+ 4 + Ct4 u– 4 k k for some constants C > 0. Lemma 2.4 There exists k > 0 such that, for all k ≥ k , the infimum ckb is achieved. Proof In view of the definition of ckb, we deduce that there exists a sequence {un} ⊂ Mkb satisfying lim Jkb(un) = ckb. Suu+ + tuu– ∈ Mbk. By (2.18), (2.19), and (2.20), we deduce ckb ≥ Jkb suu+b + tuu–b + s22u A1 – s66u B1 + t2u2 A2 – t6u6 B2 > Jkb suu+b + tuu–b ≥ ckb. Suu+ + tuu– ∈ Mbk. by using (2.19), we obtain ckb ≥ Jkb suu+b + tuu–b + s22u A1 – s66u B1 + t2u2 A2 – t6u6 B2 > Jkb suu+b + tuu–b ≥ ckb, which is a contradiction. Since un ∈ Mkb, B1 = B2 = 0 and u is lower semicontinuous, it follows that ckb ≤ Jkb(u)) – 14 Jkb (u), u.
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