Citation: Lennart Ante. Bitcoin transactions, information asymmetry and trading volume[J]. Quantitative Finance and Economics, 2020, 4(3): 365-381. doi: 10.3934/QFE.2020017
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We study the following Neumann problem of Kirchhoff type equation with critical growth
{−(a+b∫Ω|∇u|2dx)Δu+u=Q(x)|u|4u+λP(x)|u|q−2u,in Ω,∂u∂v=0,on ∂Ω, | (1.1) |
where Ω ⊂ R3 is a bounded domain with a smooth boundary, a,b>0, 1<q<2, λ>0 is a real parameter. We assume that Q(x) and P(x) satisfy the following conditions:
(Q1) Q(x)∈C(ˉΩ) is a sign-changing;
(Q2) there exists xM∈Ω such that QM=Q(xM)>0 and
|Q(x)−QM|=o(|x−xM|)asx→xM; |
(Q3) there exists 0∈∂Ω such that Qm=Q(0)>0 and
|Q(x)−Qm|=o(|x|)asx→0; |
(P1) P(x) is positive continuous on ˉΩ and P(x0)=maxx∈ˉΩP(x);
(P2) there exist σ>0, R>0 and 3−q<β<6−q2 such that P(x)≥σ|x−y|−β for |x−y|≤R, where y is xM∈Ω or 0∈∂Ω.
In recent years, the following Dirichlet problem of Kirchhoff type equation has been studied extensively by many researchers
{−(a+b∫Ω|∇u|2dx)Δu=f(x,u),in Ω,u=0,on ∂Ω, | (1.2) |
which is related to the stationary analogue of the equation
utt−(a+b∫Ω|∇u|2dx)Δu=f(x,u) | (1.3) |
proposed by Kirchhoff in [13] as an extension of the classical D'Alembert's wave equation for free vibrations of elastic strings. Kirchhoff's model takes into account the changes in length of the string produced by transverse vibrations. In (1.2) and (1.3), u denotes the displacement, b is the initial tension and f(x,u) stands for the external force, while a is related to the intrinsic properties of the string (such as Young's modulus). We have to point out that such nonlocal problems appear in other fields like biological systems, such as population density, where u describes a process which depends on the average of itself (see Alves et al. [2]). After the pioneer work of Lions [18], where a functional analysis approach was proposed. The Kirchhoff type Eq (1.2) with critical growth began to call attention of researchers, we can see [1,9,14,17,23,24,28,30] and so on.
Recently, the following Kirchhoff type equation has been well studied by various authors
{−(a+b∫R3|∇u|2dx)Δu+V(x)u=f(x,u),inR3,u>0,u∈H1(R3). | (1.4) |
There has been much research regarding the concentration behavior of the positive solutions of (1.4), we can see [10,11,12,25,33]. Many papers studied the existence of ground state solutions of (1.4), for example [5,8,15,16,21,22,24]. In addition, the authors established the existence of sign-changing solutions of (1.4) in [20,31]. In papers [27,32] proved the existence and multiplicity of nontrivial solutions of (1.4) by using mountain pass theorem.
In particular, Chabrowski in [6] studied the solvability of the Neumann problem
{−Δu=Q(x)|u|2∗−2u+λf(x,u),in Ω,∂u∂v=0,on ∂Ω, |
where Ω ⊂ RN is a smooth bounded domain, 2∗=2NN−2(N≥3) is the critical Sobolev exponent, λ>0 is a parameter. Assume that Q(x)∈C(¯Ω) is a sign-changing function and ∫ΩQ(x)dx<0, under the condition of f(x,u). Using the space decomposition H1(Ω)=span1⊕V, where V={v∈H1(Ω):∫Ωvdx=0}, the author obtained the existence of two distinct solutions by the variational method.
In [14], Lei et al. considered the following Kirchhoff type equation with critical exponent
{−(a+b∫Ω|∇u|2dx)Δu=u5+λuq−1|x|β,in Ω,u=0,on ∂Ω, |
where Ω ⊂ R3 is a smooth bounded domain, a,b>0, 1<q<2, λ>0 is a parameter. They obtained the existence of a positive ground state solution for 0≤β<2 and two positive solutions for 3−q≤β<2 by the Nehari manifold method.
In [34], Zhang obtained the existence and multiplicity of nontrivial solutions of the following equation
{−(a+b∫Ω|∇u|2dx)Δu+u=λ|u|q−2u+f(x,u)+Q(x)u5,in Ω,∂u∂v=0,on ∂Ω, | (1.5) |
where Ω is an open bounded domain in R3, a,b>0, 1<q<2, λ≥0 is a parameter, f(x,u) and Q(x) are positive continuous functions satisfying some additional assumptions. Moreover, f(x,u)∼|u|p−2u with 4<p<6.
Comparing with the above mentioned papers, our results are different and extend the above results to some extent. Specially, motivated by [34], we suppose Q(x) changes sign on Ω and f(x,u)≡0 for (1.5). Since (1.1) is critical growth, which leads to the cause of the lack of compactness of the embedding H1(Ω)↪L6(Ω), we overcome this difficulty by using P.Lions concentration compactness principle [19]. Moreover, note that Q(x) changes sign on Ω, how to estimate the level of the mountain pass is another difficulty.
We define the energy functional corresponding to problem (1.1) by
Iλ(u)=12‖u‖2+b4(∫Ω|∇u|2dx)2−16∫ΩQ(x)|u|6dx−λq∫ΩP(x)|u|qdx. |
A weak solution of problem (1.1) is a function u∈H1(Ω) and for all φ∈H1(Ω) such that
∫Ω(a∇u∇φ+uφ)dx+b∫Ω|∇u|2dx∫Ω∇u∇φdx=∫ΩQ(x)|u|4uφdx+λ∫ΩP(x)|u|q−2uφdx. |
Our main results are the following:
Theorem 1.1. Assume that 1<q<2 and Q(x) changes sign on Ω. Then there exists Λ0>0 such that for every λ∈(0,Λ0), problem (1.1) has at least one nontrivial solution.
Theorem 1.2. Assume that 1<q<2, 3−q<β<6−q2 and Q(x) changes sign on Ω, there exists Λ∗>0 such that for all λ∈(0,Λ∗). Then problem (1.1) has at least two nontrivial solutions.
Throughout this paper, we make use of the following notations:
● The space H1(Ω) is equipped with the norm ‖u‖2H1(Ω)=∫Ω(|∇u|2+u2)dx, the norm in Lp(Ω) is denoted by ‖⋅‖p.
● Define ‖u‖2=∫Ω(a|∇u|2+u2)dx for u∈H1(Ω). Note that ‖⋅‖ is an equivalent norm on H1(Ω) with the standard norm.
● Let D1,2(R3) is the completion of C∞0(R3) with respect to the norm ‖u‖2D1,2(R3)=∫R3|∇u|2dx.
● 0<QM=maxx∈ˉΩQ(x), 0<Qm=maxx∈∂ΩQ(x).
● Ω+={x∈Ω:Q(x)>0} and Ω−={x∈Ω:Q(x)<0}.
● C,C1,C2,… denote various positive constants, which may vary from line to line.
● We denote by Sρ (respectively, Bρ) the sphere (respectively, the closed ball) of center zero and radius ρ, i.e. Sρ={u∈H1(Ω):‖u‖=ρ}, Bρ={u∈H1(Ω):‖u‖≤ρ}.
● Let S be the best constant for Sobolev embedding H1(Ω)↪L6(Ω), namely
S=infu∈H1(Ω)∖{0}∫Ω(a|∇u|2+u2)dx(∫Ω|u|6dx)1/3. |
● Let S0 be the best constant for Sobolev embedding D1,2(R3)↪L6(R3), namely
S0=infu∈D1,2(R3)∖{0}∫R3|∇u|2dx(∫R3|u|6dx)1/3. |
In this section, we firstly show that the functional Iλ(u) has a mountain pass geometry.
Lemma 2.1. There exist constants r,ρ,Λ0>0 such that the functional Iλ satisfies the following conditions for each λ∈(0,Λ0):
(i) Iλ|u∈Sρ≥r>0; infu∈BρIλ(u)<0.
(ii) There exists e∈H1(Ω) with ‖e‖>ρ such that Iλ(e)<0.
Proof. (i) From (P1), by the H¨older inequality and the Sobolev inequality, for all u∈H1(Ω) one has
∫ΩP(x)|u|qdx≤P(x0)∫Ω|u|qdx≤P(x0)|Ω|6−q6S−q2‖u‖q, | (2.1) |
and there exists a constant C>0, we get
|∫ΩQ(x)|u|6dx|≤C∫Ω|u|6dx≤CS−3‖u‖6. | (2.2) |
Hence, combining (2.1) and (2.2), we have the following estimate
Iλ(u)=12‖u‖2+b4(∫Ω|∇u|2dx)2−16∫ΩQ(x)|u|6dx−λq∫ΩP(x)|u|qdx≥12‖u‖2−C6∫Ω|u|6dx−λqP(x0)|Ω|6−q6S−q2‖u‖q≥‖u‖q(12‖u‖2−q−C6S−3‖u‖6−q−λqP(x0)|Ω|6−q6S−q2). |
Set h(t)=12t2−q−C6S−3t6−q for t>0, then there exists a constant ρ=(3(2−q)S3C(6−q))14>0 such that maxt>0h(t)=h(ρ)>0. Letting Λ0=qSq2P(x0)|Ω|6−q6h(ρ), there exists a constant r>0 such that Iλ|u∈Sρ≥r for every λ∈(0,Λ0). Moreover, for all u∈H1(Ω)∖{0}, we have
limt→0+Iλ(tu)tq=−λq∫ΩP(x)|u|qdx<0. |
So we obtain Iλ(tu)<0 for every u≠0 and t small enough. Therefore, for ‖u‖ small enough, one has
m≜ |
(\mathrm{ii}) Let v\in H^1(\Omega) be such that supp v\subset\Omega^+ , v\not\equiv 0 and t > 0 , we have
\begin{equation*} I_{\lambda}(tv) = \frac{t^2}{2}\|v\|^2+\frac{bt^4}{4}\left(\int_{\Omega}|\nabla v|^2dx\right)^{2}-\frac{t^6}{6}\int_{\Omega}Q(x)|v|^{6}dx-\frac{\lambda t^q}{q}\int_{\Omega}P(x)|v|^qdx\\ \rightarrow -\infty \end{equation*} |
as t\rightarrow\infty , which implies that I_{\lambda}(tv) < 0 for t > 0 large enough. Therefore, we can find e\in H^1(\Omega) with \|e\| > \rho such that I_{\lambda}(e) < 0 . The proof is complete.
Denote
\begin{equation*} \begin{cases} \Theta_1 = \frac{abS_0^3}{4Q_M}+\frac{b^3S_0^6}{24Q_M^2}+\frac{aS_0\sqrt{b^2S_0^4+4aS_0Q_M}}{6Q_M}+\frac{b^2S_0^4\sqrt{b^2S_0^4+4aS_0Q_M}}{24Q_M^2}, \\ \Theta_2 = \frac{abS_0^3}{16Q_m}+\frac{b^3S_0^6}{384Q_m^2}+\frac{aS_0\sqrt{b^2S_0^4+16aS_0Q_m}}{24Q_m}+\frac{b^2S_0^4\sqrt{b^2S_0^4+16aS_0Q_m}}{384Q_m^2}. \end{cases} \end{equation*} |
Then we have the following compactness result.
Lemma 2.2. Suppose that 1 < q < 2 . Then the functional I_{\lambda} satisfies the (PS)_{c_{\lambda}} condition for every c_{\lambda} < c_* = min \{\Theta_1-D\lambda^{\frac{2}{2-q}}, \Theta_2-D\lambda^{\frac{2}{2-q}}\} , where D = \frac{2-q}{3q}(\frac{6-q}{4}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}})^{\frac{2}{2-q}} .
Proof. Let \{u_n\}\subset H^1(\Omega) be a (PS)_{c_\lambda} sequence for
\begin{equation} I_{\lambda}(u_n)\rightarrow c_{\lambda}\; \mathrm{and}\; I'_{\lambda}(u_n)\rightarrow 0\; \mathrm{as}\; n\rightarrow\infty. \end{equation} | (2.3) |
It follows from (2.1), (2.3) and the H \ddot{\mathrm{o}} lder inequality that
\begin{equation*} \begin{aligned} c_{\lambda}+1+o{(\|u_n\|)}&\geq I_{\lambda}(u_n)-\frac{1}{6}\langle I'_{\lambda}(u_n), u_n\rangle\\ &\geq \frac{1}{3}\|u_n\|^2+\frac{b}{12}\left(\int_{\Omega}|\nabla u_n|^2dx\right)^{2}\\ &\quad- \lambda\left(\frac{1}{q}-\frac{1}{6}\right)P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\|u_n\|^q\\ &\geq \frac{1}{3}\|u_n\|^2-\frac{\lambda(6-q)}{6q}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\|u_n\|^q. \end{aligned} \end{equation*} |
Therefore \{u_n\} is bounded in H^1(\Omega) for all 1 < q < 2 . Thus, we may assume up to a subsequence, still denoted by \{u_n\} , there exists u\in H^{1}(\Omega) such that
\begin{align} \begin{cases} u_n\rightharpoonup u, \; \; \; \mathrm{weakly\; in}\; H^{1}(\Omega), \\ u_n\rightarrow u, \; \; \; \mathrm{strongly\; in}\; L^{p}(\Omega)\; (1\leq p \lt 6), \\ u_n(x)\rightarrow u(x), \; \; \; \mathrm{a.e.\; in}\; \Omega, \end{cases} \end{align} | (2.4) |
as n\rightarrow\infty . Next, we prove that u_n\rightarrow u strongly in H^1(\Omega) . By using the concentration compactness principle (see [19]), there exist some at most countable index set J , \delta_{x_j} is the Dirac mass at x_j\subset \bar{\Omega} and positive numbers \{\nu_j\} , \{\mu_j\} , j\in J , such that
\begin{equation*} \left. \begin{array}{rcl} & |u_n|^6dx\rightharpoonup d\nu = |u|^6dx+\sum\limits_{j\in J}\nu_j\delta_{x_j}, \\ & |\nabla u_n|^2dx\rightharpoonup d\mu \geq |\nabla u|^2dx+\sum\limits_{j\in J}\mu_{j}\delta_{x_j}. \end{array} \right. \end{equation*} |
Moreover, numbers \nu_{j} and \mu_{j} satisfy the following inequalities
\begin{equation} \left. \begin{array}{rcl} & S_0\nu_{j}^\frac{1}{3}\leq \mu_{j}\; \; \mathrm{if}\; x_{j}\in \Omega, \\ & \frac{S_0}{2^\frac{2}{3}}\nu_{j}^\frac{1}{3}\leq \mu_{j}\; \; \; \mathrm{if}\; x_{j}\in \partial \Omega. \end{array} \right. \end{equation} | (2.5) |
For \varepsilon > 0 , let \phi_{\varepsilon, j}(x) be a smooth cut-off function centered at x_j such that 0\leq\phi_{\varepsilon, j}\leq 1, |\nabla \phi_{\varepsilon, j}|\leq\frac{2}{\varepsilon} , and
\begin{equation*} \phi_{\varepsilon, j}(x) = \begin{cases} 1, &\text{ in $B(x_j, \frac{\varepsilon}{2})\cap\bar{\Omega}$}, \\ 0, &\text{ in $\Omega \setminus B(x_j, \varepsilon)$.} \end{cases} \end{equation*} |
There exists a constant C > 0 such that
\begin{equation*} \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\int_{\Omega}P(x)|u_n|^q\phi_{\varepsilon, j}dx\leq P(x_0)\lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\int_{B(x_j, \varepsilon)}|u_n|^{q}dx = 0. \end{equation*} |
Since |\nabla \phi_{\varepsilon, j}|\leq\frac{2}{\varepsilon} , by using the H \ddot{\mathrm{o}} lder inequality and L^2(\Omega) -convergence of \{u_n\} , we have
\begin{equation*} \begin{aligned} &\quad\lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\left(a+b\int_{\Omega}|\nabla u_n|^2dx\right)\int_{\Omega}\langle\nabla u_n, \nabla \phi_{\varepsilon, j}\rangle u_n dx\\ &\leq C \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty} \left(\int_{\Omega}|\nabla u_n|^2dx\right)^{\frac{1}{2}}\left(\int_{\Omega}|u_n|^2|\nabla \phi_{\varepsilon, j}|^2dx\right)^{\frac{1}{2}}\\ &\leq C \lim\limits_{\varepsilon\rightarrow 0} \left(\int_{B(x_j, \varepsilon)}|u|^6dx\right)^{\frac{1}{6}}\left(\int_{B(x_j, \varepsilon)}|\nabla \phi_{\varepsilon, j}|^3dx\right)^{\frac{1}{3}}\\ &\leq C \lim\limits_{\varepsilon\rightarrow 0} \left(\int_{B(x_j, \varepsilon)}|u|^6dx\right)^{\frac{1}{6}}\left(\int_{B(x_j, \varepsilon)}\left(\frac{2}{\varepsilon}\right)^3dx\right)^{\frac{1}{3}}\\ &\leq C_1 \lim\limits_{\varepsilon\rightarrow 0} \left(\int_{B(x_j, \varepsilon)}|u|^6dx\right)^{\frac{1}{6}}\\ & = 0, \end{aligned} \end{equation*} |
where C_1 > 0 , and we also derive that
\begin{equation*} \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow\infty} \int_{\Omega}|\nabla u_n|^2 \phi_{\varepsilon, j}dx \geq\lim\limits_{\varepsilon\rightarrow 0}\int_{\Omega}|\nabla u|^2\phi_{\varepsilon, j}dx+\mu_j = \mu_j, \end{equation*} |
\begin{equation*} \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow\infty} \int_{\Omega}Q(x)|u_n|^6 \phi_{\varepsilon, j}dx = \lim\limits_{\varepsilon\rightarrow 0}\int_{\Omega}Q(x)|u|^6\phi_{\varepsilon, j}dx+Q(x_j)\nu_j = Q(x_j)\nu_j, \end{equation*} |
\begin{equation*} \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow\infty} \int_{\Omega}u_n^2 \phi_{\varepsilon, j}dx = \lim\limits_{\varepsilon\rightarrow 0}\int_{\Omega}u^2\phi_{\varepsilon, j}dx \leq\lim\limits_{\varepsilon\rightarrow 0}\int_{B(x_j, \varepsilon)}u^2dx = 0. \end{equation*} |
Noting that u_n\phi_{\varepsilon, j} is bounded in H^1(\Omega) uniformly for n , taking the test function \varphi = u_n\phi_{\varepsilon, j} in (2.3), from the above information, one has
\begin{equation*} \begin{aligned} 0& = \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\langle I'_{\lambda}(u_n), u_n\phi_{\varepsilon, j}\rangle\\ & = \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\bigg\{\left(a+b\int_{\Omega}|\nabla u_n|^2dx\right)\int_{\Omega}\langle\nabla u_n, \nabla(u_n\phi_{\varepsilon, j})\rangle dx+\int_{\Omega}u_n^2 \phi_{\varepsilon, j}dx\\ &\quad- \int_{\Omega}Q(x)|u_n|^6\phi_{\varepsilon, j}dx -\lambda\int_{\Omega}P(x)|u_n|^q\phi_{\varepsilon, j}dx \bigg\}\\ & = \lim\limits_{\varepsilon\rightarrow 0}\lim\limits_{n\rightarrow \infty}\bigg\{\left(a+b\int_{\Omega}|\nabla u_n|^2dx\right)\int_{\Omega}\left(|\nabla u_n|^2 \phi_{\varepsilon, j}+\langle\nabla u_n, \nabla \phi_{\varepsilon, j}\rangle u_n\right)dx\\ &\quad-\int_{\Omega}Q(x)|u_n|^6\phi_{\varepsilon, j}dx\bigg\}\\ &\geq \lim\limits_{\varepsilon\rightarrow 0}\bigg\{\left(a+b\int_{\Omega}|\nabla u|^2dx+b\mu_{j}\right)\left(\int_{\Omega}|\nabla u|^2 \phi_{\varepsilon, j}dx+\mu_{j}\right)\\ &\quad- \int_{\Omega}Q(x)|u|^6\phi_{\varepsilon, j}dx-Q(x_{j})\nu_{j}\bigg\}\\ &\geq \left(a+b\mu_{j}\right)\mu_{j}-Q(x_{j})\nu_{j}, \end{aligned} \end{equation*} |
so that
Q(x_{j})\nu_{j}\geq(a+b\mu_{j})\mu_{j}, |
which shows that \{u_n\} can only concentrate at points x_{j} where Q(x_{j}) > 0 . If \nu_{j} > 0 , by (2.5) we get
\begin{equation} \left. \begin{array}{rcl} & \nu_{j}^\frac{1}{3}\geq\frac{bS_0^2+\sqrt{b^2S_0^4+4aS_0Q_M}}{2Q_M}\; \; \mathrm{if}\; x_{j}\in \Omega, \\ & \nu_{j}^\frac{1}{3}\geq\frac{bS_0^2+\sqrt{b^2S_0^4+16aS_0Q_m}}{2^\frac{7}{3}Q_m}\; \; \mathrm{if}\; x_{j}\in \partial \Omega. \end{array} \right. \end{equation} | (2.6) |
From (2.5) and (2.6), we have
\begin{equation} \left. \begin{array}{rcl} & \mu_{j}\geq\frac{bS_0^3+\sqrt{b^2S_0^6+4aS_0^3Q_M}}{2Q_M}\; \; \mathrm{if}\; x_{j}\in \Omega, \\ & \mu_{j}\geq\frac{bS_0^3+\sqrt{b^2S_0^6+16aS_0^3Q_m}}{8Q_m}\; \; \mathrm{if}\; x_{j}\in \partial \Omega. \end{array} \right. \end{equation} | (2.7) |
To proceed further we show that (2.7) is impossible. To obtain a contradiction assume that there exists j_0\in J such that \mu_{j_0}\geq\frac{bS_0^3+\sqrt{b^2S_0^6+4aS_0^3Q_M}}{2Q_M} and x_{j_0}\in \Omega . By (2.1), (2.3) and (2.4), one has
\begin{equation*} \begin{aligned} c_\lambda& = \lim\limits_{n\rightarrow \infty}\bigg\{I_{\lambda}(u_n)-\frac{1}{6}\langle I'_{\lambda}(u_n), u_n\rangle\bigg\}\\ & = \lim\limits_{n\rightarrow \infty}\bigg\{\frac{a}{3}\int_{\Omega}|\nabla u_n|^2dx+\frac{b}{12}\left(\int_{\Omega}|\nabla u_n|^2dx\right)^2\\ &\quad+ \frac{1}{3}\int_{\Omega}u_n^2dx-\lambda\frac{6-q}{6q}\int_{\Omega}P(x)|u_n|^qdx\bigg\}\\ &\geq \frac{a}{3}\left(\int_{\Omega}|\nabla u|^2dx+\sum\limits_{j\in J}\mu_j\right)+\frac{b}{12}\left(\int_{\Omega}|\nabla u|^2dx+\sum\limits_{j\in J}\mu_j\right)^2\\ &\quad+ \frac{1}{3}\int_{\Omega}u^2dx-\lambda\frac{6-q}{6q}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\|u\|^q\\ &\geq \frac{a}{3}\mu_{j_0}+\frac{b}{12}\mu^2_{j_0}+\frac{1}{3}\|u\|^2-\lambda\frac{6-q}{6q}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\|u\|^q. \end{aligned} \end{equation*} |
Set
\begin{equation*} g(t) = \frac{1}{3}t^2-\lambda\frac{6-q}{6q}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}t^{q}, \; \; t \gt 0, \end{equation*} |
then
\begin{equation*} g'(t) = \frac{2}{3}t-\lambda \frac{6-q}{6}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}t^{q-1} = 0, \end{equation*} |
we can deduce that \min_{t\geq0}g(t) attains at t_0 > 0 and
\begin{equation*} t_0 = \left(\lambda\frac{6-q}{4}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\right)^{\frac{1}{2-q}}. \end{equation*} |
Consequently, we obtain
\begin{equation*} \begin{aligned} c_\lambda&\geq \frac{abS_0^3}{4Q_M}+\frac{b^3S_0^6}{24Q_M^2}+\frac{aS_0\sqrt{b^2S_0^4+4aS_0Q_M}}{6Q_M}\\ &\quad+\frac{b^2S_0^4\sqrt{b^2S_0^4+4aS_0Q_M}}{24Q_M^2}- D\lambda^{\frac{2}{2-q}}\\ & = \Theta_1-D\lambda^{\frac{2}{2-q}}, \end{aligned} \end{equation*} |
where D = \frac{2-q}{3q}\left(\frac{6-q}{4}P(x_0)S^{-\frac{q}{2}}|\Omega|^{\frac{6-q}{6}}\right)^{\frac{2}{2-q}} . If \mu_{j_0}\geq\frac{bS_0^3+\sqrt{b^2S_0^6+16aS_0^3Q_m}}{8Q_m} and x_{j_0}\in \partial\Omega , then, by the similar calculation, we also get
\begin{equation*} \begin{aligned} c_\lambda&\geq \frac{abS_0^3}{16Q_m}+\frac{b^3S_0^6}{384Q_m^2}+\frac{aS_0\sqrt{b^2S_0^4+16aS_0Q_m}}{24Q_m}\\ &\quad+\frac{b^2S_0^4\sqrt{b^2S_0^4+16aS_0Q_m}}{384Q_m^2}-D\lambda^{\frac{2}{2-q}}\\ & = \Theta_2-D\lambda^{\frac{2}{2-q}}. \end{aligned} \end{equation*} |
Let c_* = \min\{\Theta_1-D\lambda^{\frac{2}{2-q}}, \Theta_2-D\lambda^{\frac{2}{2-q}}\} , from the above information, we deduce that c_\lambda\geq c_* . It contradicts our assumption, so it indicates that \nu_j = \mu_j = 0 for every j\in J , which implies that
\begin{equation} \int_{\Omega}|u_n|^6dx\rightarrow\int_{\Omega}|u|^6dx \end{equation} | (2.8) |
as n\rightarrow\infty . Now, we may assume that \int_{\Omega}|\nabla u_n|^2dx\rightarrow A^2 and \int_{\Omega}|\nabla u|^2dx\leq A^2 , by (2.3), (2.4) and (2.8), one has
\begin{equation*} \begin{aligned} 0& = \lim\limits_{n\rightarrow \infty}\langle I_\lambda'(u_n), u_n-u\rangle\\ & = \lim\limits_{n\rightarrow\infty}\bigg[\left(a+b\int_{\Omega}|\nabla u_n|^2dx\right)\left(\int_{\Omega}|\nabla u_n|^2dx-\int_{\Omega}\nabla u_n \nabla udx\right)\\ &\quad+\int_{\Omega}u_n(u_n-u)dx-\int_{\Omega}Q(x)|u_n|^5(u_n-u)dx-\lambda\int_{\Omega}P(x)|u_n|^{q-1}(u_n-u)dx\bigg]\\ & = \left(a+bA^2\right)\left(A^2-\int_{\Omega}|\nabla u|^2dx\right). \end{aligned} \end{equation*} |
Then, we obtain that u_n\rightarrow u in H^1(\Omega) . The proof is complete.
As well known, the function
\begin{equation*} U_{\varepsilon, y}(x) = \frac{(3\varepsilon^2)^{\frac{1}{4}}}{(\varepsilon^2+|x-y|^2)^{\frac{1}{2}}}, \; \mathrm{for\; any}\; \varepsilon \gt 0, \end{equation*} |
satisfies
\begin{equation*} -\Delta U_{\varepsilon, y} = U_{\varepsilon, y}^{5}\; \; \mathrm{in}\; \mathbb{R}^3, \end{equation*} |
and
\begin{equation*} \int_{\mathbb{R}^3}|\nabla U_{\varepsilon, y}|^2dx = \int_{\mathbb{R}^3}|U_{\varepsilon, y}|^6dx = S_0^{\frac{3}{2}}. \end{equation*} |
Let \phi\in C^1(\mathbb{R}^3) such that \phi(x) = 1 on B(x_M, \frac{R}{2}) , \phi(x) = 0 on \mathbb{R}^3-B(x_M, R) and 0\leq\phi(x)\leq 1 on \mathbb{R}^3 , we set v_\varepsilon(x) = \phi(x)U_{\varepsilon, x_M}(x) . We may assume that Q(x) > 0 on B(x_M, R) for some R > 0 such that B(x_M, R)\subset \Omega . From [4], we have
\begin{equation} \begin{cases} \|\nabla v_\varepsilon\|_2^2 = S_0^\frac{3}{2}+O(\varepsilon), \\ \|v_\varepsilon\|_6^6 = S_0^\frac{3}{2}+O(\varepsilon^{3}), \\ \|v_\varepsilon\|_2^2 = O(\varepsilon), \\ \|v_\varepsilon\|^2 = aS_0^{\frac{3}{2}}+O(\varepsilon). \end{cases} \end{equation} | (2.9) |
Moreover, by [28], we get
\begin{equation} \begin{cases} \|\nabla v_\varepsilon\|_2^4\leq S_0^3+O(\varepsilon), \\ \|\nabla v_\varepsilon\|_2^8\leq S_0^6+O(\varepsilon), \\ \|\nabla v_\varepsilon\|_2^{12}\leq S_0^9+O(\varepsilon). \end{cases} \end{equation} | (2.10) |
Then we have the following Lemma.
Lemma 2.3. Suppose that 1 < q < 2 , 3-q < \beta < \frac{6-q}{2} , Q_M > 4Q_m , (Q_1) and (Q_2) , then \sup_{t\geq0}I_{\lambda}(tv_\varepsilon) < \Theta_1-D\lambda^{\frac{2}{2-q}}.
Proof. By Lemma 2.1, one has I_{\lambda}(tv_{\varepsilon})\rightarrow-\infty as t\rightarrow\infty and I_{\lambda}(tv_{\varepsilon}) < 0 as t\rightarrow 0 , then there exists t_{\varepsilon} > 0 such that I_{\lambda}(t_{\varepsilon} v_{\varepsilon}) = \sup_{t > 0}I_{\lambda}(tv_{\varepsilon})\geq r > 0 . We can assume that there exist positive constants t_1, t_2 > 0 and 0 < t_1 < t_{\varepsilon} < t_2 < +\infty . Let I_{\lambda}(t_{\varepsilon}v_{\varepsilon}) = \beta(t_{\varepsilon}v_{\varepsilon})-\lambda\psi(t_{\varepsilon}v_{\varepsilon}) , where
\begin{equation*} \beta(t_{\varepsilon}v_{\varepsilon}) = \frac{ t_{\varepsilon}^2}{2}\| v_{\varepsilon}\|^2+\frac{b t_{\varepsilon}^4}{4}\|\nabla v_{\varepsilon}\|_2^{4}-\frac{t_{\varepsilon}^6}{6}\int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx, \end{equation*} |
and
\begin{equation*} \psi(t_{\varepsilon}v_{\varepsilon}) = \frac{ t_{\varepsilon}^q}{q}\int_{\Omega}P(x)|v_{\varepsilon}|^qdx. \end{equation*} |
Now, we set
\begin{equation*} h(t) = \frac{t^2}{2}\| v_{\varepsilon}\|^2+\frac{b t^4}{4}\|\nabla v_{\varepsilon}\|_2^{4}-\frac{t^6 }{6}\int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx. \end{equation*} |
It is clear that \lim_{t\rightarrow0}h(t) = 0 and \lim_{t\rightarrow\infty}h(t) = -\infty . Therefore there exists T_{1} > 0 such that h(T_{1}) = \max_{t\geq0}h(t) , that is
\begin{equation*} h'(t)|_{T_{1}} = T_{1}\| v_{\varepsilon}\|^2+bT_{1}^3\|\nabla v_{\varepsilon}\|_2^{4}- T_{1}^5 \int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx = 0, \end{equation*} |
from which we have
\begin{equation} \| v_{\varepsilon}\|^2+bT_{1}^2\|\nabla v_{\varepsilon}\|_2^{4} = T_{1}^4 \int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx. \end{equation} | (2.11) |
By (2.11) we obtain
\begin{equation*} T_{1}^2 = \frac{b\|\nabla v_{\varepsilon}\|_2^4+\sqrt{b^2\|\nabla v_{\varepsilon}\|_2^8+4\| v_{\varepsilon}\|^2\int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx}}{2 \int_{\Omega}Q(x)|v_{\varepsilon}|^{6}dx}. \end{equation*} |
In addition, by (Q_2) , for all \eta > 0 , there exists \rho > 0 such that |Q(x)-Q_M| < \eta|x-x_M| for 0 < |x-x_M| < \rho , for \varepsilon > 0 small enough, we have
\begin{align*} \left|\int_{\Omega}Q(x)v_\varepsilon^6dx-\int_{\Omega}Q_Mv_\varepsilon^6dx\right| &\leq \int_{\Omega}|Q(x)-Q_M|v_\varepsilon^6dx\\ & \lt \int_{B(x_M, \rho)}\eta|x-x_M|\frac{(3\varepsilon^2)^\frac{3}{2}}{(\varepsilon^2+|x-x_M|^2)^3}dx\\ &\quad+ C\int_{\Omega\setminus{B(x_M, \rho)}}\frac{(3\varepsilon^2)^\frac{3}{2}}{(\varepsilon^2+|x-x_M|^2)^3}dx\\ &\leq C \eta\varepsilon^3\int_{0}^{\rho}\frac{r^3}{(\varepsilon^2+r^2)^{3}}dr+C\varepsilon^3\int_\rho^{R}\frac{r^2}{(\varepsilon^2+r^2)^{3}}dr\\ &\leq C \eta\varepsilon\int_{0}^{\rho/\varepsilon}\frac{t^3}{(1+t^2)^3}dt+C\int_{\rho/\varepsilon}^{R/\varepsilon}\frac{t^2}{(1+t^2)^3}dt\\ &\leq C_1 \eta\varepsilon+C_2\varepsilon^3, \end{align*} |
where C_1, C_2 > 0 (independent of \eta , \varepsilon ). From this we derive that
\begin{equation} \limsup\limits_{\varepsilon\rightarrow 0}\frac{|\int_{\Omega}Q(x)v_\varepsilon^6dx-\int_{\Omega}Q_Mv_\varepsilon^6dx|}{\varepsilon}\leq C_1\eta. \end{equation} | (2.12) |
Then from the arbitrariness of \eta > 0 , by (2.9) and (2.12), one has
\begin{equation} \int_{\Omega}Q(x)|v_\varepsilon|^6dx = Q_M\int_{\Omega}|v_\varepsilon|^6dx+o(\varepsilon) = Q_MS_0^{\frac{3}{2}}+o(\varepsilon). \end{equation} | (2.13) |
Hence, it follows from (2.9), (2.10) and (2.13) that
\begin{align*} \beta(t_{\varepsilon}v_{\varepsilon})&\leq h(T_{1})\\ & = T_{1}^2\left(\frac{1}{3}\| v_{\varepsilon}\|^2+\frac{b T_{1}^2}{12}\|\nabla v_{\varepsilon}\|_2^{4}\right)\\ & = \frac{b\|\nabla v_\varepsilon\|_2^4\|v_\varepsilon\|^2}{4\int_{\Omega}Q(x)|v_\varepsilon|^6dx}+\frac{b^3\|\nabla v_\varepsilon\|_2^{12}}{24(\int_{\Omega}Q(x)|v_\varepsilon|^6dx)^2}\\ &\quad+ \frac{\|v_\varepsilon\|^2\sqrt{b^2\|\nabla v_\varepsilon\|_2^8+4\|v_\varepsilon\|^2\int_{\Omega}Q(x)|v_\varepsilon|^6dx}}{6\int_{\Omega}Q(x)|v_\varepsilon|^6dx}\\ &\quad+ \frac{b^2\|\nabla v_\varepsilon\|_2^8\sqrt{b^2\|\nabla v_\varepsilon\|_2^8+4\|v_\varepsilon\|^2\int_{\Omega}Q(x)|v_\varepsilon|^6dx}}{24(\int_{\Omega}Q(x)|v_\varepsilon|^6dx)^2}\\ &\leq \frac{b(S_0^3+O(\varepsilon))(aS_0^\frac{3}{2}+O(\varepsilon))}{4(Q_MS_0^\frac{3}{2}+o(\varepsilon))} +\frac{b^3(S_0^9+O(\varepsilon))}{24(Q_MS_0^\frac{3}{2}+o(\varepsilon))^2}\\ &\quad+ \frac{(aS_0^\frac{3}{2}+O(\varepsilon))\sqrt{b^2(S_0^6+O(\varepsilon)) +4(aS_0^\frac{3}{2}+O(\varepsilon))(Q_MS_0^\frac{3}{2}+o(\varepsilon))}}{6(Q_MS_0^\frac{3}{2}+o(\varepsilon))}\\ &\quad+ \frac{b^2(S_0^6+O(\varepsilon))\sqrt{b^2(S_0^6+O(\varepsilon)) +4(aS_0^\frac{3}{2}+O(\varepsilon))(Q_MS_0^\frac{3}{2}+o(\varepsilon))}}{24(Q_MS_0^\frac{3}{2}+o(\varepsilon))^2}\\ &\leq \frac{abS_0^3}{4Q_M}+\frac{b^3S_0^6}{24Q_M^2}+\frac{aS_0\sqrt{b^2S_0^4+4aS_0Q_M}}{6Q_M}\\ &\quad+ \frac{b^2S_0^4\sqrt{b^2S_0^4+4aS_0Q_M}}{24Q_M^2}+C_3\varepsilon\\ & = \Theta_1+C_3\varepsilon, \end{align*} |
where the constant C_3 > 0 . According to the definition of v_\varepsilon , from [29], for \frac{R}{2} > \varepsilon > 0 , there holds
\begin{align} \psi(t_{\varepsilon}v_{\varepsilon}) &\geq \frac{1}{q}3^{\frac{q}{4}}t_1^q\int_{B(x_M, \frac{R}{2})} \frac{\sigma\varepsilon^\frac{q}{2}}{(\varepsilon^2+|x-x_M|^2)^{\frac{q}{2}}|x-x_M|^\beta}dx \\ &\geq C\varepsilon^\frac{q}{2}\int_0^{R/2}\frac{r^2}{(\varepsilon^2+r^2)^{\frac{q}{2}}r^\beta}dr \\ & = C\varepsilon^{\frac{6-q}{2}-\beta}\int_0^{R/2\varepsilon}\frac{t^2}{(1+t^2)^{\frac{q}{2}}t^\beta}dt \\ &\geq C\varepsilon^{\frac{6-q}{2}-\beta}\int_0^1t^{2-\beta}dt \\ & = C_4 \varepsilon^{\frac{6-q}{2}-\beta}, \end{align} | (2.14) |
where C_4 > 0 (independent of \varepsilon, \lambda ). Consequently, from the above information, we obtain
\begin{equation*} \begin{aligned} I_{\lambda}(t_{\varepsilon}v_{\varepsilon})& = \beta(t_{\varepsilon}v_{\varepsilon})-\lambda\psi(t_{\varepsilon}v_{\varepsilon})\\ &\leq \Theta_1+C_3\varepsilon-C_4\lambda \varepsilon^{\frac{6-q}{2}-\beta}\\ & \lt \Theta_1-D\lambda^{\frac{2}{2-q}}. \end{aligned} \end{equation*} |
Here we have used the fact that \beta > 3-q and let \varepsilon = \lambda^{\frac{2}{2-q}} , 0 < \lambda < \Lambda_1 = \min\{1, (\frac{C_3+D}{C_4})^{\frac{2-q}{6-2q-2\beta}}\} , then
\begin{equation} \begin{aligned} C_3\varepsilon-C_4\lambda \varepsilon^{{\frac{6-q}{2}-\beta}}& = C_3\lambda^{\frac{2}{2-q}}-C_4\lambda^{\frac{8-2q-2\beta}{2-q}}\\ & = \lambda^{\frac{2}{2-q}}(C_3-C_4\lambda^{\frac{6-2q-2\beta}{2-q}})\\ & \lt -D\lambda^{\frac{2}{2-q}}. \end{aligned} \end{equation} | (2.15) |
The proof is complete.
We assume that 0\in\partial\Omega and Q_m = Q(0) . Let \varphi\in C^1(\mathbb{R}^3) such that \varphi(x) = 1 on B(0, \frac{R}{2}) , \varphi(x) = 0 on \mathbb{R}^3-B(0, R) and 0\leq\varphi(x)\leq 1 on \mathbb{R}^3 , we set u_\varepsilon(x) = \varphi(x)U_\varepsilon(x) , the radius R is chosen so that Q(x) > 0 on B(0, R)\cap \Omega . If H(0) denotes the mean curvature of the boundary at 0 , then the following estimates hold (see [6] or [26])
\begin{equation} \begin{cases} \|u_\varepsilon\|_2^2 = O(\varepsilon), \\ \frac{\|\nabla u_{\varepsilon}\|_2^2}{\|u_{\varepsilon}\|_6^2}\leq\frac{S_0}{2^\frac{2}{3}}-A_3H(0)\varepsilon \log\frac{1}{\varepsilon}+O(\varepsilon), \end{cases} \end{equation} | (2.16) |
where A_3 > 0 is a constant. Then we have the following lemma.
Lemma 2.4. Suppose that 1 < q < 2 , 3-q < \beta < \frac{6-q}{2} , Q_M\leq4Q_m , H(0) > 0 , Q is positive somewhere on \partial\Omega , (Q_1) and (Q_3) , then \sup_{t\geq0}I_{\lambda}(tu_\varepsilon) < \Theta_2-D\lambda^{\frac{2}{2-q}}.
Proof. Similar to the proof of Lemma 2.3, we also have by Lemma 2.1, there exists t_{\varepsilon} > 0 such that I_{\lambda}(t_{\varepsilon} u_{\varepsilon}) = \sup_{t > 0}I_{\lambda}(tu_{\varepsilon})\geq r > 0 . We can assume that there exist positive constants t_1, t_2 > 0 such that 0 < t_1 < t_{\varepsilon} < t_2 < +\infty . Let I_{\lambda}(t_{\varepsilon}u_{\varepsilon}) = A(t_{\varepsilon}u_{\varepsilon})-\lambda B(t_{\varepsilon}u_{\varepsilon}) , where
\begin{equation*} A(t_{\varepsilon}u_{\varepsilon}) = \frac{ t_{\varepsilon}^2}{2}\| u_{\varepsilon}\|^2+\frac{b t_{\varepsilon}^4}{4}\|\nabla u_{\varepsilon}\|_2^{4}-\frac{t_{\varepsilon}^6}{6}\int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx, \end{equation*} |
and
\begin{equation*} B(t_{\varepsilon}u_{\varepsilon}) = \frac{ t_{\varepsilon}^q}{q}\int_{\Omega}P(x)|u_{\varepsilon}|^qdx. \end{equation*} |
Now, we set
\begin{equation*} f(t) = \frac{t^2}{2}\| u_{\varepsilon}\|^2+\frac{b t^4}{4}\|\nabla u_{\varepsilon}\|_2^{4}-\frac{t^6}{6}\int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx. \end{equation*} |
Therefore, it is easy to see that there exists T_{2} > 0 such that f(T_{2}) = \max_{f\geq0}f(t) , that is
\begin{equation} f'(t)|_{T_{2}} = T_{2}\| u_{\varepsilon}\|^2+bT_{2}^3\|\nabla u_{\varepsilon}\|_2^{4}- T_{2}^5 \int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx = 0. \end{equation} | (2.17) |
From (2.17) we obtain
\begin{equation*} T_{2}^2 = \frac{b\|\nabla u_{\varepsilon}\|_2^4+\sqrt{b^2\|\nabla u_{\varepsilon}\|_2^8+4\| u_{\varepsilon}\|^2 \int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx}}{2\int_{\Omega}Q(x)|u_{\varepsilon}|^{6}dx}. \end{equation*} |
By the assumption (Q_3) , we have the expansion formula
\begin{equation} \int_{\Omega}Q(x)|u_\varepsilon|^6dx = Q_m\int_{\Omega}|u_\varepsilon|^6dx+o(\varepsilon). \end{equation} | (2.18) |
Hence, combining (2.16) and (2.18), there exists C_5 > 0 , such that
\begin{align*} A(t_{\varepsilon}u_{\varepsilon})&\leq f(T_{2})\\ & = T_{2}^2\left(\frac{1}{3}\| u_{\varepsilon}\|^2+\frac{b T_{2}^2}{12}\|\nabla u_{\varepsilon}\|_2^4\right)\\ & = \frac{b\|\nabla u_{\varepsilon}\|_2^4\|u_{\varepsilon}\|^2}{4\int_{\Omega}Q(x)|u_\varepsilon|^6dx} +\frac{b^3\|\nabla u_\varepsilon\|_2^{12}}{24(\int_{\Omega}Q(x)|u_{\varepsilon}|^6dx)^2}\\ &\quad+ \frac{\|u_\varepsilon\|^2\sqrt{b^2\|\nabla u_\varepsilon\|_2^8 +4\|u_\varepsilon\|^2\int_{\Omega}Q(x)|u_\varepsilon|^6dx}}{6\int_{\Omega}Q(x)|u_\varepsilon|^6dx}\\ &\quad+ \frac{b^2\|\nabla u_\varepsilon\|_2^8\sqrt{b^2\|\nabla u_\varepsilon\|_2^8 +4\|u_\varepsilon\|^2\int_{\Omega}Q(x)| u_\varepsilon|^6dx}}{24(\int_{\Omega}Q(x)|u_\varepsilon|^6dx)^2}\\ &\leq \frac{ab}{4Q_m}\left(\frac{\|\nabla u_{\varepsilon}\|_2^6}{\int_{\Omega}|u_\varepsilon|^6dx}+O(\varepsilon)\right)+\frac{b^3}{24Q_m^2}\left(\frac{\|\nabla u_\varepsilon\|_2^{12}}{(\int_{\Omega}|u_\varepsilon|^6dx)^2}+O(\varepsilon)\right)\\ &\quad+ \frac{a}{6Q_m}\left(\frac{\|\nabla u_\varepsilon\|_2^2}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{1}{3}}\sqrt{\frac{b^2\|\nabla u_\varepsilon\|_2^8}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{4}{3}}+\frac{4aQ_m\|\nabla u_\varepsilon\|_2^2}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{1}{3}}}+O(\varepsilon)\right)\\ &\quad+ \frac{b^2}{24Q_m^2}\left(\frac{\|\nabla u_\varepsilon\|_2^8}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{4}{3}}\sqrt{\frac{b^2\|\nabla u_\varepsilon\|_2^8}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{4}{3}}+\frac{4aQ_m\|\nabla u_\varepsilon\|_2^2}{(\int_{\Omega}|u_\varepsilon|^6dx)^\frac{1}{3}}}+O(\varepsilon)\right)\\ &\leq \frac{abS_0^3}{16Q_m}+\frac{b^3S_0^6}{384Q_m^2}+\frac{aS_0\sqrt{b^2S_0^4+16aS_0Q_m}}{24Q_m}\\ &\quad+ \frac{b^2S_0^4\sqrt{b^2S_0^4+16aS_0Q_m}}{384Q_m^2}+C_5\varepsilon\\ & = \Theta_2+C_5\varepsilon. \end{align*} |
Consequently, by (2.14) and (2.15), similarly, there exists \Lambda_2 > 0 such that 0 < \lambda < \Lambda_2 , we get
\begin{equation*} \begin{aligned} I_{\lambda}(t_{\varepsilon}u_{\varepsilon})& = A(t_{\varepsilon}u_{\varepsilon})-\lambda B(t_{\varepsilon}u_{\varepsilon})\\ &\leq \Theta_2+C_5\varepsilon-C_6\lambda \varepsilon^{\frac{6-q}{2}-\beta}\\ & \lt \Theta_2-D\lambda^{\frac{2}{2-q}}. \end{aligned} \end{equation*} |
where C_6 > 0 (independent of \varepsilon, \lambda ). The proof is complete.
Theorem 2.5. Assume that 0 < \lambda < \Lambda_0 ( \Lambda_0 is as in Lemma 2.1) and 1 < q < 2 . Then problem (1.1) has a nontrivial solution u_\lambda with I_\lambda(u_\lambda) < 0 .
Proof. It follows from Lemma 2.1 that
\begin{equation*} m\triangleq\inf\limits_{u\in \overline{B_\rho(0)}}I_\lambda(u) \lt 0. \end{equation*} |
By the Ekeland variational principle [7], there exists a minimizing sequence \{u_n\}\subset {\overline{B_\rho(0)}} such that
\begin{equation*} I_\lambda(u_n)\leq \inf\limits_{u\in{\overline{B_\rho(0)}}}I_\lambda(u)+\frac{1}{n}, \; \; I_\lambda(v)\geq I_\lambda(u_n)-\frac{1}{n}\|v-u_n\|, \; \; v\in{\overline{B_\rho(0)}}. \end{equation*} |
Therefore, there holds I_\lambda(u_n)\rightarrow m and I_\lambda'(u_n)\rightarrow 0 . Since \{u_n\} is a bounded sequence and {\overline{B_\rho(0)}} is a closed convex set, we may assume up to a subsequence, still denoted by \{u_n\} , there exists u_{\lambda}\in{\overline{B_\rho(0)}}\subset H^{1}(\Omega) such that
\begin{align*} \begin{cases} u_n\rightharpoonup u_{\lambda}, \; \; \; \mathrm{weakly\; in}\; H^{1}(\Omega), \\ u_n\rightarrow u_{\lambda}, \; \; \; \mathrm{strongly\; in}\; L^{p}(\Omega), \; 1\leq p \lt 6, \\ u_n(x)\rightarrow u_{\lambda}(x), \; \; \; \mathrm{a.e.\; in}\; \Omega. \end{cases} \end{align*} |
By the lower semi-continuity of the norm with respect to weak convergence, one has
\begin{equation*} \begin{aligned} m&\geq\liminf\limits_{n\rightarrow \infty}\left[I_\lambda(u_n)-\frac{1}{6}\langle I_\lambda'(u_n), u_n\rangle\right]\\ & = \liminf\limits_{n\rightarrow \infty}\bigg[\frac{1}{3}\int_\Omega \left(a|\nabla u_n|^2+u_n^2\right)dx +\frac{b}{12}\left(\int_\Omega|\nabla u_n|^2dx\right)^2\\ &\quad+ \lambda\left(\frac{1}{6} -\frac{1}{q}\right)\int_\Omega P(x)|u_n|^{q}dx\bigg]\\ &\geq \frac{1}{3}\int_\Omega \left(a|\nabla u_\lambda|^2+u_\lambda^2\right)dx +\frac{b}{12}\left(\int_\Omega|\nabla u_\lambda|^2dx\right)^2\\ &\quad+ \lambda\left(\frac{1}{6} -\frac{1}{q}\right)\int_\Omega P(x)|u_\lambda|^{q}dx\\ & = I_\lambda(u_\lambda)-\frac{1}{6}\langle I_\lambda'(u_\lambda), u_\lambda\rangle = I_\lambda(u_\lambda) = m. \end{aligned} \end{equation*} |
Thus I_{\lambda}(u_\lambda) = m < 0 , by m < 0 < c_\lambda and Lemma 2.2, we can see that \nabla u_n\rightarrow \nabla u_\lambda in L^2(\Omega) and u_\lambda\not\equiv0 . Therefore, we obtain that u_\lambda is a weak solution of problem (1.1). Since I_\lambda(|u_\lambda|) = I_\lambda(u_\lambda) , which suggests that u_\lambda\geq0 , then u_\lambda is a nontrivial solution to problem (1.1). That is, the proof of Theorem 1.1 is complete.
Theorem 2.6. Assume that 0 < \lambda < \Lambda_{*} (\Lambda_{*} = \min\{\Lambda_0, \Lambda_1, \Lambda_2\}) , 1 < q < 2 and 3-q < \beta < \frac{6-q}{2} . Then the problem (1.1) has a nontrivial solution u_{1}\in H^1(\Omega) such that I_{\lambda}(u_{1}) > 0 .
Proof. Applying the mountain pass lemma [3] and Lemma 2.2, there exists a sequence \{u_n\}\subset H^1(\Omega) such that
\begin{equation*} I_{\lambda}(u_n)\rightarrow c_\lambda \gt 0\; \; \mathrm{and}\; \; I'_\lambda(u_n)\rightarrow0\; \mathrm{as}\; n\rightarrow \infty, \end{equation*} |
where
c_\lambda = \inf\limits_{\gamma\in\Gamma}\max\limits_{t\in[0, 1]}I_{\lambda}(\gamma(t)), |
and
\Gamma = \left\{\gamma\in C([0, 1], H^{1}(\Omega)): \gamma(0) = 0, \gamma(1) = e\right\}. |
According to Lemma 2.2, we know that \{u_n\}\subset H^1(\Omega) has a convergent subsequence, still denoted by \{u_n\} , such that u_n\rightarrow u_{1} in H^1(\Omega) as n\rightarrow\infty ,
\begin{equation*} I_{\lambda}(u_{1}) = \lim\limits_{n\rightarrow\infty}I_{\lambda}(u_n) = c_\lambda \gt r \gt 0, \end{equation*} |
which implies that u_{1}\not\equiv0 . Therefore, from the continuity of I'_\lambda , we obtain that u_{1} is a nontrivial solution of problem (1.1) with I_{\lambda}(u_{1}) > 0 . Combining the above facts with Theorem 2.5 the proof of Theorem 1.2 is complete.
In this paper, we consider a class of Kirchhoff type equations with Neumann conditions and critical growth. Under suitable assumptions on Q(x) and P(x) , using the variational method and the concentration compactness principle, we proved the existence and multiplicity of nontrivial solutions.
This research was supported by the National Natural Science Foundation of China (Grant Nos. 11661021 and 11861021). Authors are grateful to the referees for their very constructive comments and valuable suggestions.
The authors declare no conflict of interest in this paper.
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