Low Perturbations and Combined Effects of Critical and Singular Nonlinearities in Kirchhoff Problems

Abstract

In this paper, we study three-dimensional Kirchhoff equations with critical growth and singular nonlinearity. We are concerned with the qualitative analysis of solutions to the following nonlocal problem

$$\begin{aligned} {\left\{ \begin{array}{ll} -\left( a+b\displaystyle \int _{\Omega }|\nabla u|^2dx\right) \Delta u=\lambda u^{-\gamma }+u^5, &{}\mathrm {in}\ \ \Omega , \\ u>0, &{}\mathrm {in}\ \ \Omega , \\ u=0, &{}\mathrm {on}\ \ \partial \Omega , \end{array}\right. } \end{aligned}$$

where $\Omega \subset {\mathbb {R}}^3$ is a bounded domain with smooth boundary, $0<\gamma <1$, and $a,\,b,\,\lambda$ are positive constants. By combining variational methods with some delicate decomposition techniques, we obtain the existence of two positive solutions in the case of low perturbations of the singular nonlinearity, namely for small values of the parameter $\lambda$.

Appendix

This appendix contains two lemmas. In Lemma $A_1$, we give the proof of Lemma 3.6, which is an application of the Ekeland’s variational principle, and adapted from [31]. In Lemma $A_2$, we show that the functional $I_\lambda$ fails to satisfy the concrete Palais-Smale condition at the level $\mu _1$, by finding a sequence $\{u_n\}$ of $I_\lambda$ such that $I_\lambda (u_n)\rightarrow \mu _1$, $|dI_\lambda |(u_n)\rightarrow 0$, but $\{u_n\}$ possesses no convergent subsequence in $H_0^1(\Omega )$. Consequently, $\mu _1$ is exactly the threshold value for $I_\lambda$, since we have proved in Proposition 3.4 that below $\mu _1$, the functional $I_\lambda$ satisfies the concrete Palais-Smale condition.

Lemma

$A_1$. Let $c_1^*$ be the Mountain Pass value as defined in (3.29). Then there exists a concrete Palais-Smale sequence of $I_\lambda$ at the level $c_1^*$, that is a sequence $\{u_n\}$ such that $I_\lambda (u_n)\rightarrow c_1^*$ and $|dI_\lambda |(u_n)\rightarrow 0$ as $n\rightarrow \infty$.

Proof

We first recall (3.30) as follows:

$$\begin{aligned} \Gamma =\left\{ \sigma |\sigma \in C([0, 1], P): \sigma (0)=u, I_\lambda (\sigma (1))\le 0, \Vert \sigma (1)\Vert \ge 100\rho \right\} . \end{aligned}$$

As a closed subset of C([0, 1], P), $\Gamma$ is a complete metric space. For $g\in \Gamma$, define

$$\begin{aligned} F(g)=\sup _{t\in [0,1]}I_\lambda (g(t)). \end{aligned}$$

Then F is continuous in $\Gamma$. By relation (3.26), we have

$$\begin{aligned} F(g)\ge \inf _{u\in \partial B_\rho }I_\lambda (u)\ge \frac{1}{6}a\rho ^2. \end{aligned}$$

Therefore, F(g) is bounded from below.

Given $\varepsilon >0$, by Ekeland’s variational principle, there exists $g\in \Gamma$ such that

$$\begin{aligned} {\left\{ \begin{array}{ll} F(g)\le \inf _{h\in \Gamma }F(h)+\varepsilon =c_1^*+\varepsilon , \\ F(g)\le F(h)+\varepsilon \Vert g-h\Vert , ~~~ h\in \Gamma . \end{array}\right. } \end{aligned}$$

Denote

$$\begin{aligned} {\widetilde{M}}(g)=\left\{ t\in [0,1]|I_\lambda (g(t))=F(g)=\sup _{s\in [0,1]}I_\lambda (g(s))\right\} . \end{aligned}$$

Then

$$\begin{aligned} c_1^*\le I_\lambda (g(t))\le c_1^*+\varepsilon ,~~\mathrm {for}~t\in {\widetilde{M}}(g). \end{aligned}$$

We claim that there exists $t_\varepsilon \in {\widetilde{M}}(g)$ such that $|dI_\lambda |(g(t_\varepsilon ))\le \varepsilon$, which completes the proof. Otherwise, for all $t\in {\widetilde{M}}(g)$, $|dI_\lambda |(g(t_\varepsilon ))>\varepsilon$. By the definition, for $t\in {\widetilde{M}}(g)$, there exists $v(t)\in P$ such that

$$\begin{aligned} \displaystyle \lambda \int _{\Omega }g^{-\gamma }(t)(v(t)-g(t))dx> & {} \displaystyle M\left( \int _{\Omega }|\nabla g(t)|^2dx\right) \int _{\Omega }\nabla g(t)\nabla (v(t)-g(t))dx\nonumber \\&\displaystyle -\int _{\Omega }g^5(t)(v(t)-g(t))dx+\varepsilon \Vert v(t)-g(t)\Vert .\qquad \,\, \end{aligned}$$

(4.1)

By the Fatou lemma, in a neighborhood $B_{\delta (t)}(t)$ of t in [0, 1], it holds that

$$\begin{aligned} \displaystyle \lambda \int _{\Omega }g^{-\gamma }(s)(v(t)-g(s))dx> & {} \displaystyle M\left( \int _{\Omega }|\nabla g(s)|^2dx\right) \int _{\Omega }\nabla g(s)\nabla (v(t)-g(s))dx\\&\displaystyle -\int _{\Omega }g^5(s)(v(t)-g(s))dx+\varepsilon \Vert v(t)-g(s)\Vert \end{aligned}$$

for $s\in B_{\delta (t)}(t)$.

We may assume $B_{\delta (t)}(t)\cap \{0,1\}=\emptyset$, since $I_\lambda (g(0))<0$, $I_\lambda (g(1))<0$, $\{B_{\delta (t)}(t)|t\in {\widetilde{M}}(g)\}$ is an open covering of ${\widetilde{M}}(g)$. There exists a finite covering $B_i=B_{\delta (t_i)}(t_i)$, $i=1,2,...,n$. Let

$$\begin{aligned} \varphi _0(t)=\mathrm {dist}\left( t,\bigcup _{i=1}^nB_i\right) ,\ \varphi _i(t)=\mathrm {dist}\left( t,[0,1]\backslash B_i\right) , i=1,2,...,n. \end{aligned}$$

$\varphi _0(t)=0$ for $t\in {\widetilde{M}}(g)$ and $\varphi _i(0)=\varphi _i(1)=0$ for $i=1,2,...,n$. Also define

$$\begin{aligned} \psi _i(t)=\frac{\varphi _i(t)}{\sum _{i=0}^n\varphi _i(t)},~t\in [0,1];\ \omega (t)=\sum _{i=1}^n\psi _i(t)(v(t_i)-g(t)),~t\in [0,1]. \end{aligned}$$

For $t\in {\widetilde{M}}(g)$, by (4.1) we have

$$\begin{aligned}&\displaystyle M\left( \int _{\Omega }|\nabla g(t)|^2dx\right) \int _{\Omega }\nabla g(t)\nabla \omega (t)dx-\int _{\Omega }g^5(t)\omega (t)dx-\lambda \int _{\Omega }g^{-\gamma }(t)\omega (t)dx\\&\quad <\displaystyle -\varepsilon \sum _{i=1}^n\psi _i(t)\Vert v(t_i)-g(t)\Vert \\&\quad \le \displaystyle -\varepsilon \Vert \sum _{i=1}^n\psi _i(t)(v(t_i)-g(t))\Vert =\displaystyle -\varepsilon \Vert \omega (t)\Vert . \end{aligned}$$

Hence $\omega (t)\ne 0$ for $t\in {\widetilde{M}}(g)$. There exists $\delta >0$ such that $\Vert \omega (t)\Vert \ge \delta$ for $t\in {\widetilde{M}}(g)$. Let $\varphi (t)=\min \left\{ 1, \frac{\delta }{\Vert \omega (t)\Vert }\right\}$, $t\in [0,1]$, then $\varphi \in C([0,1], {\mathbb {R}}^+)$. Define

$$\begin{aligned} {\left\{ \begin{array}{ll} h(t)=\varphi (t)\omega (t), ~~ t\in [0,1],\\ \Vert h\Vert =\delta ,~\Vert h(t)\Vert =\delta , ~~~ t\in {\widetilde{M}}(g). \end{array}\right. } \end{aligned}$$

Since $h(0)=h(1)=0$, for $\tau$ small enough, $g+\tau h\in \Gamma$, we have

$$\begin{aligned} \displaystyle F(g) \le \displaystyle F(g+\tau h)+\varepsilon \Vert \tau h\Vert =\displaystyle F(g+\tau h)+\varepsilon \tau \delta . \end{aligned}$$

(4.2)

Choose $t=t(\tau )\in {\widetilde{M}}(g+\tau h)$, one has

$$\begin{aligned} I_\lambda (g(t(\tau ))+\tau h(t(\tau )))\ge I_\lambda (g(s)+\tau h(s)),~~\mathrm {for}~s\in [0,1]. \end{aligned}$$

Let $\tau _n\rightarrow 0^+$, $t_n=t(\tau _n)\rightarrow t_\varepsilon$, we have

$$\begin{aligned} I_\lambda (g(t_\varepsilon ))\ge I_\lambda (g(s)),~~\mathrm {for}~s\in [0,1]. \end{aligned}$$

Thus, $t_\varepsilon \in {\widetilde{M}}(g)$. It follows from (4.2) that

$$\begin{aligned} \displaystyle -\varepsilon \delta \le \displaystyle \frac{1}{\tau _n}[F(g+\tau _n h)-F(g)] \le \displaystyle \frac{1}{\tau _n}[I_\lambda (g(t_n)+\tau _nh(t_n))-I_\lambda (g(t_n))]. \end{aligned}$$

(4.3)

Taking the limit $n\rightarrow \infty$ in (4.3), by the Fatou’s lemma we obtain

$$\begin{aligned} \displaystyle -\varepsilon \delta\le & {} \displaystyle M\left( \int _{\Omega }|\nabla g(t_\varepsilon )|^2dx\right) \int _{\Omega }\nabla g(t_\varepsilon )\nabla h(t_\varepsilon )dx\\&\displaystyle -\int _{\Omega }g^5(t_\varepsilon )h(t_\varepsilon )dx-\lambda \int _{\Omega }g^{1-\gamma }(t_\varepsilon )h(t_\varepsilon )dx\\\le & {} \displaystyle \varphi (t_\varepsilon )\bigg \{M\left( \int _{\Omega }|\nabla g(t_\varepsilon )|^2dx\right) \int _{\Omega }\nabla g(t_\varepsilon )\nabla \omega (t_\varepsilon )dx\\&\displaystyle -\int _{\Omega }g^5(t_\varepsilon )\omega (t_\varepsilon )dx-\lambda \int _{\Omega }g^{1-\gamma }(t_\varepsilon )\omega (t_\varepsilon )dx\bigg \}\\< & {} \displaystyle -\varphi (t_\varepsilon )\cdot \varepsilon \Vert \omega (t_\varepsilon )\Vert =-\varepsilon \delta , \end{aligned}$$

which is a contradiction. Hence, the proof is completed. $\square$

Here we use the notations $U, U_\varepsilon , \eta , \varphi _\varepsilon =\eta U_\varepsilon$ and $t_0$ as in Lemma 3.7. In particular, $U, t_0$ satisfy the equation (3.33), and let u be the local minimizer of $I_1$ obtained in Lemma 3.3.

Lemma

$A_2$. Let $u_\varepsilon =u+t_0\varphi _\varepsilon$. Then $I_\lambda (u_\varepsilon )\rightarrow \mu _1$, $|dI_\lambda |(u_\varepsilon )\rightarrow 0$, $u_\varepsilon \rightharpoonup u$, but $\int _{\Omega }|\nabla u_\varepsilon |^2dx=\int _{\Omega }|\nabla u|^2dx+t_0^2\int _{{\mathbb {R}}^3}|\nabla U|^2dx+o(1)$. Hence $u_\varepsilon$ ($\varepsilon \rightarrow 0$) is a concrete Palais-Smale sequence of $I_\lambda$ at the level $\mu _1$, but possesses no convergent subsequence in $H_0^1(\Omega )$.

Proof

Using the estimate for the integrals involving $\varphi _\varepsilon$, we have

$$\begin{aligned} \int _{\Omega }|\nabla u_\varepsilon |^2dx=\int _{\Omega }|\nabla u|^2dx+t_0^2\int _{{\mathbb {R}}^3}|\nabla U|^2dx+o(1). \end{aligned}$$

Hence we deduce as $\varepsilon \rightarrow 0$

$$\begin{aligned} I_\lambda (u_\varepsilon )\rightarrow & {} \displaystyle \frac{1}{2}{\mathcal {M}}\left( \int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\right) -\frac{1}{6}\int _{\Omega }u^6dx\\&\displaystyle -\frac{\lambda }{1-\gamma }\int _{\Omega }u^{1-\gamma }dx-\frac{1}{6}S^{-3}{\mathcal {F}}_1^3\left( \int _{\Omega }|\nabla u|^2dx\right) =\displaystyle \mu _1. \end{aligned}$$

For $v\in P$, denote

$$\begin{aligned} \omega _\varepsilon =v-u_\varepsilon . \end{aligned}$$

By estimating, we show

$$\begin{aligned} M\left( \int _{\Omega }|\nabla u_\varepsilon |^2dx\right) \int _{\Omega }\nabla u_\varepsilon \nabla \omega _\varepsilon dx=\displaystyle \int _{\Omega }u_\varepsilon ^5\omega _\varepsilon dx+\lambda \int _{\Omega }u_\varepsilon ^{-\gamma }\omega _\varepsilon dx+o(1)\Vert \omega _\varepsilon \Vert , \end{aligned}$$

which means

$$\begin{aligned} |dI_\lambda |(u_\varepsilon )=o(1)~~\mathrm {as}~\varepsilon \rightarrow 0. \end{aligned}$$

Note that $u_\varepsilon =u+t\varphi _\varepsilon$ and $\varphi _\varepsilon =\eta U_\varepsilon$. Then we have

$$\begin{aligned}&\displaystyle \bigg |M\Big (\int _{\Omega }|\nabla u_\varepsilon |^2dx\Big )\int _{\Omega }\nabla u_\varepsilon \nabla \omega _\varepsilon dx\nonumber \\&\qquad \displaystyle -M\Big (\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\Big )\Big (\int _{\Omega }\nabla u\nabla \omega _\varepsilon dx+t_0\int _{\Omega }\nabla U_\varepsilon \nabla \omega _\varepsilon dx\Big )\bigg |\nonumber \\&\quad \le \displaystyle \bigg |M\Big (\int _{\Omega }|\nabla u_\varepsilon |^2dx\Big )-M\Big (\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\Big )\bigg |\left| \int _{\Omega }\nabla u_\varepsilon \nabla \omega _\varepsilon dx\right| \nonumber \\&\qquad \displaystyle +\bigg |M\Big (\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\Big )\int _{\Omega }\nabla (u_\varepsilon -u)\nabla \omega _\varepsilon dx\nonumber \\&\qquad \displaystyle -M\Big (\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\Big )t_0\int _{\Omega }\nabla U_\varepsilon \nabla \omega _\varepsilon dx\bigg |\nonumber \\&\quad \le \displaystyle o(1)\Vert \omega _\varepsilon \Vert +M\Big (\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1\Big (\int _{\Omega }|\nabla u|^2dx\Big )\Big )t_0\nonumber \\&\qquad \times \left| \int _{\Omega }[\nabla (\eta U_\varepsilon )-\nabla U_\varepsilon ]\nabla \omega _\varepsilon dx\right| \nonumber \\&\quad \le \displaystyle o(1)\Vert \omega _\varepsilon \Vert +C\left( \int _{\{|x|\ge \delta \}}(|\nabla U_\varepsilon |^2+U^2_\varepsilon )dx\right) ^{\frac{1}{2}}\Vert \omega _\varepsilon \Vert =\displaystyle o(1)\Vert \omega _\varepsilon \Vert . \end{aligned}$$

(4.4)

In the above we assume $\eta (x)=1$ for $|x|\le \delta$. Moreover, we also have

$$\begin{aligned}&\displaystyle \left| \int _{\Omega }u_\varepsilon ^5\omega _\varepsilon dx-\int _{\Omega }u^5\omega _\varepsilon dx-t^5_0\int _{\Omega }U_\varepsilon ^5\omega _\varepsilon dx\right| \nonumber \\&\quad \le \displaystyle C\int _{\Omega }u^4\varphi _\varepsilon |\omega _\varepsilon |dx+C\int _{\Omega }u^3\varphi _\varepsilon ^2|\omega _\varepsilon | dx+C\int _{\Omega }u^2\varphi _\varepsilon ^3|\omega _\varepsilon | dx\nonumber \\&\qquad \displaystyle +C\int _{\Omega }u\varphi _\varepsilon ^4|\omega _\varepsilon |dx+C\int _{\Omega }|(\eta U_\varepsilon )^5-U_\varepsilon ^5||\omega _\varepsilon |dx\nonumber \\&\quad \le \displaystyle o(1)\Vert \omega _\varepsilon \Vert +C\int _{\{|x|\ge \delta \}}U_\varepsilon ^5|\omega _\varepsilon |dx =\displaystyle o(1)\Vert \omega _\varepsilon \Vert , \end{aligned}$$

(4.5)

and

$$\begin{aligned} \bigg |\int _{\Omega }u_\varepsilon ^{-\gamma }\omega _\varepsilon dx-\int _{\Omega }u^{-\gamma }\omega _\varepsilon dx\bigg |\le & {} \displaystyle C\int _{\Omega }u^{-\gamma -1}\varphi ^2_\varepsilon |\omega _\varepsilon |dx\nonumber \\\le & {} \displaystyle C\int _{\Omega }\varphi ^2_\varepsilon |\omega _\varepsilon |dx =\displaystyle o(1)\Vert \omega _\varepsilon \Vert . \end{aligned}$$

(4.6)

Since $u, U_\varepsilon$ solve the system:

$$\begin{aligned} {\left\{ \begin{array}{ll} \displaystyle M(A)\int _{\Omega }\nabla u\nabla \omega _\varepsilon dx-\int _{\Omega }u^5\omega _\varepsilon dx-\lambda \int _{\Omega }u^{-\gamma }\omega _\varepsilon dx=0, \\ \displaystyle M(A)t_0\int _{{\mathbb {R}}^3}\nabla U_\varepsilon \nabla \omega _\varepsilon dx=t_0^5\int _{{\mathbb {R}}^3}U_\varepsilon ^5\omega _\varepsilon dx, \end{array}\right. } \end{aligned}$$

where $A=\int _{\Omega }|\nabla u|^2dx+{\mathcal {F}}_1(\int _{\Omega }|\nabla u|^2dx)$, the estimates (4.4) and (4.5) follow from (4.6). The proof is thus complete. $\square$