## Existence of Positive Solutions for Semilinear Elliptic Equation with Variable Exponent

Chu Changmu,, Meng Lu

School of Data Science and Information Engineering, Guizhou Minzu University, Guiyang 550025

 基金资助: 国家自然科学基金.  11861021国家自然科学基金.  11861052

 Fund supported: the NSFC.  11861021the NSFC.  11861052

Abstract

This paper is devoted to study a class of semilinear elliptic equation with variable exponent. By means of perturbation technique, variational methods and a priori estimation, the existence of positive solutions to this problem is obtain.

Keywords： Semilinear elliptic equation ; Variable exponent ; Variational methods ; Priori estimate ; Positive solution

Chu Changmu, Meng Lu. Existence of Positive Solutions for Semilinear Elliptic Equation with Variable Exponent. Acta Mathematica Scientia[J], 2021, 41(6): 1779-1790 doi:

## 1 问题及主要结果

$0\in\Omega\subset{{\Bbb R}} ^N(N\geq 3)$是具有光滑边界$\partial\Omega$的有界域. 该文考虑一类带有变指数增长的半线性椭圆型方程

$$$\left\{\begin{array}{ll} -\Delta u = u^{q(x)-1}, & x\in \Omega, \\ u>0, & x\in \Omega \\ u = 0, & x\in \partial \Omega, \end{array}\right.$$$

${\rm (Q_1)} $$q\in C(\overline{\Omega}) , q(0) = 2 , 当 x\neq 0 时, 2<q(x)\leq \mathop {\max }\limits_{x\in \overline{\Omega}}\{q(x)\} = q^+<2^* = \frac{2N}{N-2} ; {\rm (Q_2)} 存在 \alpha\in\left(0, \ \frac{N+2}{2}\right)$$ B_{\delta_0} = \{x||x|<\delta_0\}\subset\Omega$, 使得对任意$x\in B_{\delta_0}$, $q(x)\geq 2+|x|^{\alpha}$.

1973年, Ambrosetti和Rabinowitz在文献[1]中考虑了如下超线性椭圆问题非平凡解的存在性

$$$\left\{\begin{array}{ll} -\Delta u = f(x, u), &x\in\Omega, \\ u = 0, &x\in \partial \Omega. \end{array}\right.$$$

(f) 存在$\theta>2$, 使得

设$\{u_n\}\subset H^1_{0}(\Omega) $$I_{\mu} 的(PS)序列, 即存在 C>0 , 使得当 n\rightarrow \infty $$|I_{\mu}(u_{n})|\leq C, \ \ \ \ \ I_{\mu}' (u_{n})\rightarrow 0.$$ 由(2.5)式和引理2.1, 可以推出 这意味着 \frac{r}{2(2+r)}\|u_{n}\|^{2}\leq C+C_{\mu}+o(\|u_{n}\|) . 因此, \{u_{n}\}$$ H^1_{0}(\Omega)$中有界. 故存在$\{u_{n}\}$的一个子列(仍用$\{u_{n}\}$表示)和$u\in H^{1}(\Omega)$, 使得当$n\rightarrow \infty$

$$$\langle I' _{\mu}(u_i)-I' _{\mu}(u_j), u_i-u_j\rangle\rightarrow 0.$$$

$\begin{eqnarray} &&\left|\int_{\Omega}(k_{\mu}(x, u_i^+)-k_{\mu}(x, u_j^+))(u_i-u_j){\rm d}x\right|\\ &\leq& C\int_{\Omega}\left(|u_i|+|u_j|+|u_i|^{q^++r-1}+|u_j|^{q^++r-1}\right)|u_i-u_j|\rightarrow 0. \end{eqnarray}$

$\rm (1)$存在$m, \ \rho>0$, 使得对任意$u\in H^1_{0}(\Omega) $$\|u\| = \rho , I_{\mu}(u)> m ; \rm (2) 存在 w\in H^1_{0}(\Omega) 满足 \|w\|> \rho$$ I_{\mu}(w)<0$.

由$k_{\mu}$的定义知

$\begin{eqnarray} \left|\int_{\Omega}K_{\mu}(x, u^+)\, {\rm d}x\right|\leq C_{\mu} \int_{\Omega}(|u|^{q(x)}+|u|^{q(x)+r})\, {\rm d}x. \end{eqnarray}$

$\begin{eqnarray} \int_{\Omega}|u|^{q(x)+r}\, {\rm d}x\leq \int_{\Omega}(|u|^{2+r}+|u|^{2^*})\, {\rm d}x\leq C(\|u\|^{2+r}+\|u\|^{2^*}). \end{eqnarray}$

$\Omega_{\varepsilon} = \{x\in \Omega|2\leq q(x)< 2+\varepsilon\}$, 由Hölder不等式和Sobolev嵌入定理, 就有

$\begin{eqnarray} \int_{\Omega}|u|^{q(x)}\, {\rm d}x & = &\int_{\Omega_{\varepsilon}}|u|^{q(x)}\, {\rm d}x+\int_{\Omega\backslash\Omega_{\varepsilon}}|u|^{q(x)}\, {\rm d}x\\ &\leq&\int_{\Omega_{\varepsilon}}(|u|^{2}+|u|^{2+\varepsilon})\, {\rm d}x+\int_{\Omega\backslash\Omega_{\varepsilon}}(|u|^{2+\varepsilon}+|u|^{2^*})\, {\rm d}x\\ &\leq&\int_{\Omega_{\varepsilon}}|u|^{2}\, {\rm d}x+\int_{\Omega}(|u|^{2+\varepsilon}+|u|^{2^*})\, {\rm d}x\\ &\leq&C|\Omega_{\varepsilon}|^{\frac{2^*-2}{2^*}}\|u\|^2+C(\|u\|^{2+\varepsilon}+\|u\|^{2^*}). \end{eqnarray}$

$\Omega_0 = \{0\}$, 可得当$\varepsilon\rightarrow 0$时, $|\Omega_{\varepsilon}|\rightarrow 0$. 固定$\mu\in (0, 1]$, 存在$\varepsilon_0>0$使得对任意$\varepsilon\in (0, \varepsilon_0)$,

$\begin{eqnarray} |\Omega_{\varepsilon}|^{\frac{2^*-2}{2^*}}<\frac{1}{4CC_{\mu}}. \end{eqnarray}$

$u_{\mu}^- = 0$. 所以$u_{\mu}\geq 0$.$I_{\mu}(u_{\mu}) = c_{\mu}>0 = I_{\mu}(0)$, 可得$u_{\mu}\neq 0$. 由强极大值原理[18]$u_{\mu}$是问题(2.2)的正解, 又由(2.12)式可得

## 3 先验估计和定理1.1的证明

由引理2.1, $I_{\mu}(v)\leq L $$I_{\mu}'(v) = 0 \begin{eqnarray} L&\geq&I_{\mu}(v)- \frac{1}{2}\langle I_{\mu}' (v), v \rangle\\ &\geq&\int_{\Omega}\left(\frac{1}{2}-\frac{1}{q(x)}\right)k_{\mu}(x, v)v {\rm d}x\\ &\geq&\int_{\Omega_{\frac{\delta}{2}}}\left(\frac{1}{2}-\frac{1}{q(x)}\right)k_{\mu}(x, v)v{\rm d}x. \end{eqnarray} 由条件 {\rm (Q_1)} 知, 对任意 \delta\in (0, \delta_0) , 存在 m_{\delta}>0 , 使得对任意 x\in\Omega_{\frac{\delta}{2}} , \frac{1}{2}-\frac{1}{q(x)}\geq m_{\delta} . 因此 k_{\mu} 的定义, 当 t>1 时有 k_{\mu}(x, t)t\geq t^{q(x)}\geq t^{2} . 于是 \begin{eqnarray} \int_{\Omega_{\frac{\delta}{2}}}v^2{\rm d}x\leq \int_{\Omega_{\frac{\delta}{2}}}(1+k_{\mu}(x, v)v){\rm d}x\leq m_{\delta}^{-1}L+|\Omega|. \end{eqnarray} I_{\mu}'(v) = 0 知, v 是问题(2.2)的解. 对任意 \delta\in(0, \ \delta_0) , 令 \psi\in C^{\infty}_0 (\Omega, [0, 1]) 满足: 当 x\in\Omega_{\delta} 时, \psi(x) = 1 ; 当 |x|\leq \frac{\delta}{2} 时, \psi(x) = 0 . 此外, 对任意 x\in \Omega , |\nabla \psi|\leq C . v\psi^{2} 乘以问题(2.2)的两边并积分, 可得 \begin{eqnarray} \int_{\Omega}\nabla v \nabla (v\psi^2){\rm d}x = \int_{\Omega}k_{\mu}(x, v)v\psi^2 {\rm d}x\leq \int_{\Omega_{\frac{\delta}{2}}}k_{\mu}(x, v)v{\rm d}x\leq m_{\delta}^{-1}L+|\Omega|. \end{eqnarray} 由Young不等式知 \begin{eqnarray} \int_{\Omega}\nabla v\nabla (v\psi^2) {\rm d}x & = &\int_{\Omega}|\nabla v|^{2}\psi^2 {\rm d}x+2\int_{\Omega} v\psi^2\nabla v\nabla \psi^2 {\rm d}x \\ &\geq&\frac{1}{2}\int_{\Omega}|\nabla v|^{2}\psi^2 {\rm d}x-C\int_{\Omega}v^2|\nabla \psi|^{2} {\rm d}x \\ &\geq&\frac{1}{2}\int_{\Omega_{\delta}}|\nabla v|^{2}{\rm d}x-C\int_{\Omega_{\frac{\delta}{2}}}v^2{\rm d}x. \end{eqnarray} 由(3.2), (3.3)和(3.4)式知, 存在一个不依赖于 \mu$$ C_{\delta}>0$, 使得$\int_{\Omega_{\delta}}|\nabla v|^{2}{\rm d}x\leq C_{\delta}$.

$\begin{eqnarray} \int_{\Omega}\nabla v \nabla (v\phi^2){\rm d}x = \int_{\Omega}k_{\mu}(x, v)v\phi^2 {\rm d}x\leq \int_{B_{2\delta}}k_{\mu}(x, v)v{\rm d}x. \end{eqnarray}$

$\begin{eqnarray} \int_{\Omega}\nabla v\nabla (v\phi^2) {\rm d}x & = &\int_{\Omega}|\nabla v|^{2}\phi^2 {\rm d}x+2\int_{\Omega} v\phi\nabla v\nabla \phi {\rm d}x \\ &\geq&\frac{1}{2}\int_{\Omega}|\nabla v|^{2}\phi^2 {\rm d}x-C\int_{\Omega}v^2|\nabla \phi|^{2} {\rm d}x \\ &\geq&\frac{1}{2}\int_{B_{\delta}}|\nabla v|^{2}{\rm d}x-C\int_{B_{2\delta}}v^2{\rm d}x. \end{eqnarray}$

$k_{\mu}$的定义知

$\begin{eqnarray} |k_{\mu}(x, v)v|\leq C \left(|v|^{2}+|v|^{q^{+}+r}\right). \end{eqnarray}$

$\begin{eqnarray} \int_{B_{\delta}}|\nabla v|^{2}{\rm d}x\leq C\int_{B_{2\delta}}k_{\mu}(x, v)v{\rm d}x+C\int_{B_{2\delta}}v^2{\rm d}x\leq C\left(1+\int_{B_{2\delta}}|v|^{q^++r}{\rm d}x\right). \end{eqnarray}$

$\begin{eqnarray} \|v\|_{L^{\frac{Np}{N-2p}}(B_{\delta'})}&\leq& C\left(\|v\|_{L^{\frac{(N-2)p}{N-2p}}(B_{\delta}\backslash B_{\delta'})}+\|k_{\mu}(x, v)\|_{L^p(B_{\delta})}\right)\\ &\leq&C\left(\|v\|_{L^{\frac{(N-2)p}{N-2p}}(\Omega_{\delta'})}+\|k_{\mu}(x, v)\|_{L^p(B_{\delta})}\right)\\ &\leq&C\left(\int_{\Omega_{\delta'}}|\nabla v|^2{\rm d}x\right)^{\frac{1}{2^*}}+C\|k_{\mu}(x, v)\|_{L^p(B_{\delta})}\\ &\leq& C, \end{eqnarray}$

$\zeta_1 = \frac{Np}{N-2p}\geq q^++r$, 应用Holder不等式, 便能获得所需证明的不等式. 否则, 利用(2.1)式导出的不等式$r<\frac{1}{4N}<\frac{2}{N-2}$, 可选择$\sigma_1\in (0, \ \delta)\subset(0, \ \delta_3)$使得$\tau_1 = q_{\sigma_1}^++r-1<\frac{N}{N-2}$. 因此, 对任意$\mu\in (0, 1]$, $x\in B_{\sigma_1}$, 有

$\zeta_2\geq q^++r$, 应用Holder不等式, 便能获得所需证明的不等式. 若不然, 通过重复上述迭代过程, 便能在有限次迭代后, 得到$\zeta_k>0$, $\sigma_{k}\in (0, \sigma_{k-1})$和不依赖于$\mu$的常数$C>0$, 使得$\zeta_k\geq 2^*> q^++r $$\|v\|_{L^{\zeta_k}(B_{\sigma_k})}\leq C . 应用Holder不等式, 就有 \begin{eqnarray} \|v\|_{L^{q^++r}(B_{\sigma_k})}\leq C. \end{eqnarray} \delta_1 = \frac{\zeta_k}{2} . 由(3.16)和(3.23)式知: 对任意 \delta\in(0, \delta_1) , 引理3.3证毕. 命题3.1的证明 由引理3.1和引理3.3知, 存在不依赖于 \mu 的常数 C>0 , 使得 \begin{eqnarray} \int_{\Omega}|\nabla v|^{2}{\rm d}x\leq C. \end{eqnarray} 应用Sobolev嵌入定理, 有 \begin{eqnarray} \int_{\Omega}|v|^{2^*}{\rm d}x\leq C\left(\int_{\Omega}|\nabla v|^{2}{\rm d}x\right)^{\frac{2^*}{2}}\leq C. \end{eqnarray} s>0$$ t = q^++r$.$v^{2s+1}$乘上(2.2)式的两边并积分, 由$k_{\mu}$的定义, 可得

$\begin{eqnarray} \int_{\Omega}|\nabla v|^{2} v^{2s} {\rm d}x = \frac{1}{2s+1}\int_{\Omega}\nabla v\nabla v^{2s+1} {\rm d}x\leq C\left(1+\int_{\Omega}|v|^{2s+t}{\rm d}x\right). \end{eqnarray}$

$\begin{eqnarray} \int_{\Omega}|\nabla v|^{2} v^{2s} {\rm d}x = \frac{1}{(1+s)^{2}}\int_{\Omega}|\nabla v^{1+s}|^{2} {\rm d}x\geq\frac{C}{(1+s)^{2}}\left(\int_{\Omega}| v|^{(1+s)2^*} {\rm d}x\right)^{\frac{2}{2^*}}. \end{eqnarray}$

$\begin{eqnarray} \int_{\Omega}|v|^{2s+t}{\rm d}x&\leq& \left(\int_{\Omega}|v|^{2^*}{\rm d}x\right)^{\frac{t-2}{2^*}}\left(\int_{\Omega}|v|^{2(1+s)\frac{2^*}{2^*-t+2}}{\rm d}x\right)^{\frac{2^*-t+2}{2^*}}\\ &\leq&C\left(\int_{\Omega}|v|^{(1+s)\frac{2^*}{d}}{\rm d}x\right)^{\frac{2d}{2^*}}, \end{eqnarray}$

$\begin{eqnarray} \left(\int_{\Omega}| v|^{(1+s)2^*} {\rm d}x\right)^{\frac{1}{(1+s)2^*}}\leq (C(1+s))^{\frac{1}{1+s}} \left(\int_{\Omega}|v|^{(1+s)\frac{2^*}{d}}{\rm d}x\right)^{\frac{d}{(1+s)2^*}}. \end{eqnarray}$

$k = 1, 2, \cdots$, 令$s_k = d^{k}-1$. 由(3.29)式, 通过迭代可得

$\begin{eqnarray} \left(\int_{\Omega}| v|^{d^k2^*} {\rm d}x\right)^{\frac{1}{d^k2^*}}&\leq& (Cd^k)^{\frac{1}{d^k}} \left(\int_{\Omega}|v|^{d^{k-1}2^*}{\rm d}x\right)^{\frac{1}{d^{k-1}2^*}} \\&\leq&\prod\limits_{j = 1}^k(Cd^{j})^{\frac{1}{d^j}} \left(\int_{\Omega}|v|^{2^*}{\rm d}x\right)^{\frac{1}{2^*}} \\&\leq&C^{\sum\limits_{j = 1}^kd^{-j}}\cdot d ^{\sum\limits_{j = 1}^kjd^{-j}} \left(\int_{\Omega}|v|^{2^*}{\rm d}x\right)^{\frac{1}{2^*}}. \end{eqnarray}$

$d>1$知: $\sum\limits_{j = 1}^\infty d^{-j}$$\sum\limits_{j = 1}^\infty jd^{-j}$是收敛的. 令$k\to\infty$, 由(3.25)和(3.30)式可以推出$\|v\|_{L^{\infty}(\Omega)}\leq M$. 命题3.1证毕.

$u_{\mu}$是问题(1.1)的正解. 定理1.1得证.

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