Existence of Multiple Solutions for a Class of Quasilinear Schrödinger Equations

Xue Yanfang,*, Zhu Xincai,

College of Mathematics and Statistics, Xinyang Normal University, Henan Xinyang 464000

 基金资助: 国家自然科学基金(11901499)国家自然科学基金(11901500)信阳师范学院南湖学者青年项目(201912)

 Fund supported: The NSFC(11901499)The NSFC(11901500)Nanhu Scholar Program for Young Scholars of XYNU(201912)

Abstract

The multiple solutions are studied for a class of Quasilinear Schrödinger equation under the coercive potential. By using the method of variable substitution, the quasilinear problem is transformed into a semilinear one, then infinitely many high energy solutions of the equation are obtained with the help of the fountain theorem.

Keywords： Quasilinear Schrödinger equations; Fountain theorem; Coercive potential

Xue Yanfang, Zhu Xincai. Existence of Multiple Solutions for a Class of Quasilinear Schrödinger Equations[J]. Acta Mathematica Scientia, 2023, 43(1): 93-100

1 引言

$$$\label{eq1z} i\frac{\partial \psi}{\partial t}=-\triangle\psi+W(x)\psi-l(|\psi|^2)\psi-\gamma[\triangle\rho(|\psi|^2)] \rho^\prime(|\psi|^2)\psi,$$$

$$$\label{eq2z} -\Delta u+V(x)u-\gamma\Delta \rho(u^2) \rho^\prime(|u|^2)u=h(x,u), \ \ x\in \Bbb R^N,$$$

$$$\label{eq3h} -\Delta u+V(x)u-\gamma\Delta (u^2)u=h(x,u), \ \ x\in \Bbb R^N.$$$

$$$\label{eq4h} -\Delta u+V(x)u-\frac{\gamma u}{2\sqrt{1+u^2}}\Delta (\sqrt{1+u^2})=h(x,u), \ \ x\in \Bbb R^N,$$$

$(V^\prime)$$V(x)\in C(\Bbb R^N,\Bbb R), \inf\limits_{x\in\Bbb R^N}V(x)\geq a_0>0, \lim\limits_{|x|\rightarrow \infty}V(x)=+\infty. 文献[16]中的强制位势如下 (V^{\prime\prime})$$V(x)\in C(\Bbb R^N,\Bbb R)$, $\inf\limits_{x\in\Bbb R^N}V(x)\geq a_0>0$. 而且, 对任意 $M>0$, ${\rm meas}\{x\in\Bbb R^N:V(x)\leq M\}<\infty$.

$(V)$$V(x)\in C(\Bbb R^N,\Bbb R), \inf\limits_{x\in\Bbb R^N}V(x)\geq a_0>0. 而且, 存在常数 d_0>0, 使得对任意的 M>0, 有 \lim\limits_{|y|\rightarrow \infty} {\rm meas}\{x\in\Bbb R^N: |x-y|\leq d_0, V(x)\leq M\}=0. 强制位势可以保证嵌入的紧性, 这三种强制位势中, (V^\prime) 是最强的, 其次是 (V^{\prime\prime}), 最弱的是 (V). 存在满足 (V) 但不满足 (V^\prime), (V^{\prime\prime}) 的函数, 例如 \begin{eqnarray*} V(x)= \left\{\begin{array}{ll} 2n|x|-2n(n-1)+b_0, & |x|\in (n-1,\frac{2n-1}{2}),\\ -2n|x|+2n^2+b_0, & |x|\in (\frac{2n-1}{2},n), \end{array}\right. \end{eqnarray*} 其中n\in N, b_0\in \Bbb R.d_0=\sup\limits_{x\in\Bbb R^N}|V(x)|, 则 V(x) 满足 (V) 但不满足 (V^\prime), (V^{\prime\prime}). 到目前为止, 针对方程(1.4)的结果不多(见文献[2,4-5,8,12,14]及其参考文献). 其中文献[2]在径向位势下, 借助 Jeanjean 的单调技巧得到一个正解; 文献[4]在势阱位势下, 通过扰动方法得到一个正的基态解. 文献[5]讨论多解问题, 在周期位势下, 通过定义 Nehari 流形和 H^1(\Bbb R^N) 中单位球面之间的同胚, 得到无穷多个解; 文献[8] 在周期位势下, 通过山路引理, 得到一个正解; 文献[12] 在势阱位势下, 借助 Pohozaev 恒等式得到一个非平凡的基态解. 然而, 对于强制位势下, 解的多重性结果, 目前还没有. 受文献[9,16,18]的启发, 我们考虑在最弱的强制位势 (V) 下, 方程(1.4)解的多重性问题, 其中非线性项 h(x,s) 满足如下条件 (h_1)$$\lim\limits_{s\rightarrow 0}\frac{h(x,s)}{s}=0,$ 关于 $x\in \Bbb R^N$ 一致成立.

$(h_2)$$\lim\limits_{s\rightarrow +\infty}\frac{h(x,s)}{s^{p-1}}=0, 关于 x\in \Bbb R^N 一致成立, 其中 2<p<2^*=\frac{2N}{N-2}. (h_3) 存在 \mu>2, 使得 sh(x,s)-\mu H(x,s)\geq0. (h_4)$$h(x,-s)=-h(x,s)$.

$\begin{eqnarray*}\label{for21b} \gamma^*= \left\{\begin{array}{ll} \frac{16(\mu-2)}{(\mu-4)^2}, \ \ &\mu<4,\\ +\infty,\ \ &\mu\geq 4. \end{array}\right. \end{eqnarray*}$

$H:=\bigg\{u\in H^{1}(\Bbb R^N):\int_{\Bbb R^N}(|\nabla u|^{2}+V(x)u^{2})<\infty\bigg\},$

$\|u\|=\bigg(\int_{\Bbb R^N}(|\nabla u|^{2}+V(x)u^2)\bigg)^\frac{1}{2}.$

2 预备知识

$\begin{eqnarray*} J(u)&=&\frac{1}{2}\int_{\Bbb R^N}\left[(1+\frac{\gamma u^2}{2(1+u^2)})|\nabla u|^{2}\right]+\frac{1}{2} \int_{\Bbb R^N}V(x)u^{2}-\int_{\Bbb R^N}H(x,u). \end{eqnarray*}$

$\begin{matrix}\label{ch2} f(t)=\sqrt{1+\frac{\gamma t^2}{2(1+t^2)}}. \end{matrix}$

$I(v)=\frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2} \int_{\Bbb R^N}V(x)|F^{-1}(v)|^2-\int_{\Bbb R^N}H(x,F^{-1}(v)),$

$I$$H 上有定义, 且 I(v)=J(u)=J(F^{-1}(v)). 在假设条件 (h_1)-(h_3) 下, I\in C^1(H,\Bbb R) 并且如果 v$$I$ 的临界点, 那么 $u=F^{-1}(v)$ 是方程(1.4)的解[13,14].

$$$\label{eq33d} -{\rm div}\bigg[(1+\frac{\gamma u^2}{2(1+u^2)})\nabla u\bigg]+V(x)u+\frac{\gamma u}{2(1+u^2)^2}|\nabla u|^2=h(x,u).$$$

$\begin{matrix}\label{for131d} \int_{\Bbb R^N}\bigg[(1+\frac{\gamma u^2}{2(1+u^2)})\nabla u\cdot\nabla \varphi+\frac{\gamma u}{2(1+u^2)^2}|\nabla u|^2\varphi +V(x) u\varphi-h(x,u)\varphi\bigg]=0, \end{matrix}$

(2.3)式等价于下面的等式

$\begin{matrix}\label{for231d} \langle I^\prime(v),\psi\rangle&=&\int_{\Bbb R^N}\bigg(\nabla v\cdot\nabla \psi +V(x) \frac{F^{-1}(v)}{f(F^{-1}(v))}\psi-\frac{h(x,F^{-1}(v))}{f(F^{-1}(v))}\psi\bigg)=0. \end{matrix}$

$\begin{matrix}\label{eq5h} -\triangle v+V(x)\frac{F^{-1}(v)}{f(F^{-1}(v))}=\frac{h(x,F^{-1}(v))}{f(F^{-1}(v))}, \ \ x\in \Bbb R^N. \end{matrix}$

(1) 对任意 $t\in \Bbb R$, 当 $\gamma>0$ 时, 有 $1\leq f(t)\leq \sqrt{\frac{2+\gamma}{2}}$;

$-2<\gamma<0$ 时, 有 $\sqrt{\frac{2+\gamma}{2}} \leq f(t)\leq 1$;

(2) 对任意 $t\in \Bbb R$, 当 $\gamma>0$ 时, 有 $\sqrt{\frac{2}{2+\gamma}}|t|\leq |F^{-1}(t)|\leq |t|$;

$-2<\gamma<0$ 时, 有 $|t|\leq |F^{-1}(t)|\leq \sqrt{\frac{2}{2+\gamma}}|t|$;

(3) 当 $t\rightarrow0$ 时, 有 $\frac{F^{-1}(t)}{t}\rightarrow 1$;

(4) 当 $t\rightarrow\infty$ 时, 有 $\frac{F^{-1}(t)}{t}\rightarrow \sqrt{\frac{2}{2+\gamma}}$;

(5) 对任意 $t\in \Bbb R$, 当 $\gamma>0$ 时, 有 $0\leq \sup\limits_{t\in R}\frac{f^\prime(t)t}{f(t)}\leq 1+\frac{4-2\sqrt{4+2\gamma}}{\gamma}$;

$-2<\gamma<0$ 时, 有 $1+\frac{4-2\sqrt{4+2\gamma}}{\gamma}\leq \inf\limits_{t\in R}\frac{f^\prime(t)t}{f(t)}\leq 0$.

$(A_2)$$a_k:=\max\limits_{u\in Y_k,\|u\|=\rho_k}I(u)\leq0, (A_3)$$b_k:=\inf\limits_{u\in Z_k,\|u\|=r_k}I(u)\rightarrow+\infty, k\rightarrow+\infty$,

$(A_4)$ 对任意的 $c>0$, $I$ 满足 $(PS)_c$ 条件.

$I$ 有一列无界的临界值序列.

3 主要结果的证明

$$$\label{1h} H(x,s)\geq C |s|^\mu.$$$

$\gamma>0$ 时, 根据(3.1)式以及引理 2.1(2) 得

$\begin{eqnarray*}\label{Q6} I(v)&=&\frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2} \int_{\Bbb R^N}V(x)|F^{-1}(v)|^2- \int_{\Bbb R^N}H(x,F^{-1}(v))\\ &\leq& \frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2} \int_{\Bbb R^N}V(x)|v|^2-C \int_{\Bbb R^N}|F^{-1}(v)|^\mu\\ &\leq& \frac{1}{2}\|v\|^2-C(\sqrt{\frac{2}{2+\gamma}})^{\mu} \int_{\Bbb R^N}|v|^\mu. \end{eqnarray*}$

$-2<\gamma<0$ 时, 可得到类似结论, 从而喷泉定理的条件 $(A_2)$ 成立.

$$$\label{2h} H(x,s)\leq \frac{\epsilon}{2} |s|^2+C_\epsilon |s|^{p}.$$$

$\gamma>0$ 时, 根据条件 $(V)$,(3.2)式, 引理 2.1(2) 可得

$\begin{eqnarray*} I(v) & = &\frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2} \int_{\Bbb R^N}V(x)|F^{-1}(v)|^2-\int_{\Bbb R^N}H(x,F^{-1}(v))\\ & \geq & \frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2+\gamma} \int_{\Bbb R^N}V(x)| v|^2-\frac{\epsilon}{2} \int_{\Bbb R^N}|F^{-1}(v)|^2-C_\epsilon \int_{\Bbb R^N}|F^{-1}(v)|^p\\ & \geq &\frac{1}{2}\int_{\Bbb R^N}|\nabla v|^{2}+\frac{1}{2+\gamma} \int_{\Bbb R^N}V(x)|v|^2-\frac{\epsilon}{2\alpha_0} \int_{\Bbb R^N}V(x)|v|^2-C_\epsilon |v|^p_p\\ & \geq & C_1\|v\|^2-C_\epsilon |v|^p_p, \end{eqnarray*}$

$I(v)\geq C_1\|v\|^2-C_\epsilon\beta_k^p\|v\|^p.$

$r_k=(\frac{2C_1}{C_\epsilon\beta_k^p p})^{\frac{1}{p-2}}$, 若 $v\in Z_k$$\|v\|=r_k, 则有 b_k:=\inf\limits_{u\in Z_k,\|u\|=r_k}I(u)\geq C_1r_k^2-C_\epsilon\beta_k^pr_k^p =C_1 (1-\frac{2}{p})(\frac{2C_1}{C_\epsilon\beta_k^p p})^{\frac{p}{p-2}}. 因为 p>2, 而且当 k\rightarrow+\infty 时, \beta_k\rightarrow0, 所以当 k\rightarrow+\infty 时, 有 b_k\rightarrow+\infty. -2<\gamma<0 时, 可类似证明, 从而喷泉定理的条件 (A_3) 成立. 第四步, 证明泛函 I 满足喷泉定理的条件 (A_4), 即对任意的 c>0, I 满足 (PS)_c 条件. 设 \{v_n\}\subset H$$I$ 在临界水平 $c>0$ 处的 $(PS)_c$ 序列, 即 $I(v_n)\rightarrow c,\ \ I^\prime(v_n)\rightarrow 0,$ 也就是

$\begin{matrix}\label{for31d} \frac{1}{2}\int_{\Bbb R^N}|\nabla v_n|^{2}+\frac{1}{2} \int_{\Bbb R^N}V(x)|F^{-1}(v_n)|^2-\int_{\Bbb R^N}H(x,F^{-1}(v_n))=c+o_n(1), \end{matrix}$

$\begin{eqnarray*} \langle I^\prime(v_n),\varphi\rangle=\int_{\Bbb R^N}\bigg(\nabla v_n\cdot\nabla \varphi + V(x)\frac{F^{-1}(v_n)}{f(F^{-1}(v_n))}\varphi -\frac{h(x,F^{-1}(v_n))}{f(F^{-1}(v_n))}\varphi\bigg) =o_n(1). \end{eqnarray*}$

$|\nabla \varphi_n|=\bigg| \bigg(1+\frac{F^{-1}(v_n)f^\prime(F^{-1}(v_n))}{f(F^{-1}(v_n))}\bigg)\nabla v_n\bigg| \leq C|\nabla v_n|.$

$k(x,v)=V(x)v-V(x)\frac{F^{-1}(v)}{f(F^{-1}(v))}+\frac{h(x,F^{-1}(v))}{f(F^{-1}(v))}$, 则方程(2.5)转化成半线性薛定谔方程

$\begin{matrix}\label{eq6h} -\triangle v+V(x)v=k(x,v), \ \ x\in \Bbb R^N. \end{matrix}$

$\begin{eqnarray*} \frac{k(x,s)}{s}=V(x)\bigg(1-\frac{F^{-1}(s)}{s}\cdot \frac{1}{f(F^{-1}(s))}\bigg)+\frac{h(x,F^{-1}(s))}{F^{-1}(s)}\cdot\frac{F^{-1}(s)}{s}\cdot\frac{1}{f(F^{-1}(s))} \rightarrow0, \end{eqnarray*}$

$s\rightarrow +\infty$ 时, 有

$\begin{eqnarray*} \frac{k(x,s)}{s^{p-1}}=V(x)\bigg(\frac{1}{s^{p-2}}-\frac{F^{-1}(s)}{s}\cdot\frac{1}{s^{p-2}} \frac{1}{f(F^{-1}(s))}\bigg)+\frac{h(x,F^{-1}(s))}{s^{p-1}}\frac{1}{f(F^{-1}(s))} \rightarrow0. \end{eqnarray*}$

$\lim\limits_{n\rightarrow \infty}\int_{\Bbb R^N}(k(x,v_n)-k(x,v))(v_n-v)=0.$

$\begin{eqnarray*} o_n(1)&=&\langle I^\prime(v_n)-I^\prime(v),v_n-v\rangle \\ &=&\int_{\Bbb R^N}\bigg( |\nabla (v_n-v)|^{2}+V(x)(v_n-v)^2\bigg)-\int_{\Bbb R^N}(k(x,v_n)-k(x,v))(v_n-v)\\ &=&\|v_n-v\|^2+o_n(1). \end{eqnarray*}$

(II) $-2<\gamma<0$ 的情形. 由(3.4)式以及引理 2.1(2)(5) 知

$\begin{eqnarray*} \mu c+o_n(1)\geq\int_{\Bbb R^N}\frac{\mu-2}{2}|\nabla v_n|^{2}+\frac{\mu-2}{2}\int_{\Bbb R^N}V(x)|v_n|^2=\frac{\mu-2}{2}\|v_n\|^2. \end{eqnarray*}$

$\begin{eqnarray*} \frac{F^{-1}(v_n)}{f(F^{-1}(v_n))}-\frac{F^{-1}(v)}{f(F^{-1}(v))} =\frac{1-\frac{F^{-1}(\xi)f^\prime(F^{-1}(\xi))}{f(F^{-1}(\xi))}}{f^2(F^{-1}(\xi))}(v_n-v). \end{eqnarray*}$

$-2<\gamma<0$ 时, $f(s)\leq 1$, 再结合引理 2.1(5) 知

$\begin{matrix}\label{ch7h} \bigg[ \frac{F^{-1}(v_n)}{f(F^{-1}(v_n))}-\frac{F^{-1}(v)}{f(F^{-1}(v))}\bigg](v_n-v) \geq(v_n-v)^2. \end{matrix}$

$\begin{matrix}\label{ch8h} \int_{\Bbb R^N}\bigg[ \frac{h(x,F^{-1}(v_n))}{f(F^{-1}(v_n))}-\frac{h(x,F^{-1}(v))}{f(F^{-1}(v))}\bigg](v_n-v) \rightarrow 0. \end{matrix}$

$\begin{eqnarray*} o_n(1)&=&\langle I^\prime(v_n)-I^\prime(v),v_n-v\rangle \\ &=&\int_{\Bbb R^N}\bigg[ |\nabla (v_n-v)|^{2}+V(x)\bigg(\frac{F^{-1}(v_n)}{f(F^{-1}(v_n))}-\frac{F^{-1}(v)}{f(F^{-1}(v))}\bigg) (v_n-v) \bigg] \nonumber \\ &&-\int_{\Bbb R^N}\bigg[ \frac{h(x,F^{-1}(v_n))}{f(F^{-1}(v_n))}-\frac{h(x,F^{-1}(v))}{f(F^{-1}(v))}\bigg](v_n-v) \\ &\geq&\int_{\Bbb R^N}\big( |\nabla (v_n-v)|^{2}+V(x)(v_n-v)^2\big)+o_n(1)\\ &=&\|v_n-v\|^2+o_n(1). \end{eqnarray*}$

$n\rightarrow\infty$ 时, 有 $\|v_n-v\|\rightarrow 0$, 此时 $I$ 满足 $(PS)_c$ 条件.

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