Nodal Solutions for Gauged Schrödinger Equation with Nonautonomous Asymptotically Quintic Nonlinearity

Abstract

In this paper, we are dedicated to study the existence and asymptotic behavior of infinitely many nodal solutions of gauged Schrödinger equation with an asymptotically quintic nonlinear term. Based on variational methods, we show, for any integer $k\geqslant 1$, the existence of a radial nodal solution that changes sign exactly k times. Meanwhile, we prove that the energy of such solutions is an increasing function of k. Furthermore, we verify the asymptotic behavior of these solutions upon varying the parameter $\lambda$. In particular, some analytical techniques are applied, which allows us to overcome the difficulties resulting from the complicated competition between the nonlocal term and the asymptotically quintic nonlinearity. Our results enrich the previous ones in the literature involved asymptotically quintic case.

Appendix

Lemma A.1

Assume that $u_n\in {{\mathcal {M}}}_{r_n}^\lambda ,u\in \mathcal{M}_r^\lambda$ for all n and $\{u_n\},\{u_i^n\}$ weakly converges $u,u_i$, respectively, in $H_r^1(\mathbb {R}^2)$. Then for $i=1,2,\ldots ,k+1$, it holds that $\langle B'(u_n),u_i^n\rangle \rightarrow \langle B'(u),u_i\rangle$ as $n\rightarrow \infty$.

Proof

Since

$$\begin{aligned}{} & {} \langle B'(u_n),u_i^n\rangle -\langle B'(u),u_i\rangle \\= & {} \left( \langle B'(u_n),u_i^n\rangle -\langle B'(u_n),u_i\rangle \right) +\left( \langle B'(u_n),u_i\rangle -\langle B'(u),u_i\rangle \right) , \end{aligned}$$

using the Lemma 3.2 in [2], for $i=1,2,\ldots ,k+1$, we know that $\langle B'(u_n),u_i\rangle -\langle B'(u),u_i\rangle \rightarrow 0$ as $n\rightarrow \infty$.

Now let us prove that $\langle B'(u_n),u_i^n\rangle -\langle B'(u_n),u_i\rangle \rightarrow 0$ as $n\rightarrow \infty$. In fact, combining the polar transformation, the Hölder inequality, and Sobolev compact embedding theorem, for each $i=1,2,\ldots ,k+1$, we have

$$\begin{aligned}{} & {} |\langle B'(u_n),u_i^n\rangle -\langle B'(u_n),u_i\rangle |\\\leqslant & {} \sum _{j,l=1}^{k+1}\int _{\mathbb {R}^2}\frac{|(u_i^n)^2(|x|)-u_i^2(|x|)|}{|x|^2}\int _0^{|x|}\frac{s}{2}(u_j^n)^2(s)\textrm{d}s\int _0^{|x|}\frac{s}{2}(u_l^n)^2(s)\textrm{d}s\textrm{d}x\\{} & {} +2\sum _{j,l=1}^{k+1}\int _{\mathbb {R}^2}\frac{(u_j^n)^2(|x|)}{|x|^2}\int _0^{|x|}\frac{s}{2}(u_l^n)^2(s)\textrm{d}s\int _0^{|x|}\frac{s}{2}|(u_i^n)^2(s)-u_i^2(s)|\textrm{d}s\textrm{d}x\\\leqslant & {} \frac{1}{4\pi ^2}\left[ \sum _{j,l=1}^{k+1}|(u_i^n)^2-u_i^2|_2\left| \frac{1}{|x|^2}\int _{B_{|x|}}(u_j^n)^2\right| _2|u_l^n|_2^2\right. \\{} & {} \left. +2\sum _{j,l=1}^{k+1}|(u_j^n)^2|_2\left| \frac{1}{|x|^2}\int _{B_{|x|}}(u_l^n)^2\right| _2|(u_i^n)^2-u_i^2|_2^2\right] \\\leqslant & {} C_1|u_i^n-u_i|_4(|u_i^n|_4+|u_i|_4)\sum _{j,l=1}^{k+1}(|u_j^n|_6^2+|u_j^n|_2^2)|u_l^n|_2^2\rightarrow 0,\ \ n\rightarrow \infty . \end{aligned}$$

Thus, the proof is complete. $\square$

Now, we present a simple implicit function theorem. Let us denote $D=[c_1,d_1]\times [c_2,d_2]\times \cdots \times [c_k,d_k]$.

Lemma A.2

Let $F\in C([a,b]\times D)$. Assume that for each $y\in D$, $F(\cdot ,y)$ is decreasing on [a, b] and there exists $x_0\in [a,b]$ such that $F(x_0,y)=0$. Then there exists an implicit function $f\in C(D)$ such that $F(f(y),y)=0$ for $y\in D$.

Proof

According to the condition of Lemma A.2, there exists an implicit function $f:D\rightarrow [a,b]$ such that $F(f(y),y)=0$ for $y\in D$. Now, we prove the continuity of f. In fact, suppose that f is discontinuous at some $y_0\in D$. Then there exists $\{y_n\}\subset D$ satisfying $y_n\rightarrow y_0,x_n:=f(y_n)\rightarrow z_0$, but $z_0\ne f(y_0)$. However, according to the continuity of F, we have that

$$\begin{aligned} F(z_0,y_0)=\lim _{n\rightarrow \infty }F(f(y_n),y_n)=0. \end{aligned}$$

Thus $z_0=f(y_0)$. This leads to a contradiction, which completes the proof. $\square$