Research article Special Issues

Wind energy development and perceived real estate values in Ontario, Canada

  • Received: 03 October 2014 Accepted: 19 November 2014 Published: 27 November 2014
  • This paper focuses on public concerns about real estate value loss in communities in the vicinity of wind turbines. There are some conflicting results in recent academic and non-academic literatures on the issue of property values in general—yet little has been studied about how residents near turbines view the value of their own properties. Using both face-to-face interviews (n = 26) and community survey results (n = 152) from two adjacent communities, this exploratory mixed-method study contextualizes perceived property value loss. Interview results suggest a potential connection between perceived property value loss and actual property value loss, whereby assumed property degradation from turbines seem to lower both asking and selling prices. This idea is reinforced by regression results which suggest that felt property value loss is predicted by health concerns, visual annoyances and community-based variables. Overall, the findings point to the need for greater attention to micro-level local, and interconnected impacts of wind energy development.

    Citation: Chad Walker, Jamie Baxter, Sarah Mason, Isaac Luginaah, Danielle Ouellette. Wind energy development and perceived real estate values in Ontario, Canada[J]. AIMS Energy, 2014, 2(4): 424-442. doi: 10.3934/energy.2014.4.424

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  • This paper focuses on public concerns about real estate value loss in communities in the vicinity of wind turbines. There are some conflicting results in recent academic and non-academic literatures on the issue of property values in general—yet little has been studied about how residents near turbines view the value of their own properties. Using both face-to-face interviews (n = 26) and community survey results (n = 152) from two adjacent communities, this exploratory mixed-method study contextualizes perceived property value loss. Interview results suggest a potential connection between perceived property value loss and actual property value loss, whereby assumed property degradation from turbines seem to lower both asking and selling prices. This idea is reinforced by regression results which suggest that felt property value loss is predicted by health concerns, visual annoyances and community-based variables. Overall, the findings point to the need for greater attention to micro-level local, and interconnected impacts of wind energy development.


    Our goal of this paper is to consider the existence of nodal solution and ground state solution for the following fractional Kirchhoff equation:

    $ {(a+bu2K)LKu+V(x)u=|u|2α2u+kf(x,u),xΩ,u=0,xR3Ω,
    $
    (1)

    where $ a,b, k $ are positive real numbers. Let $ \alpha\in(\frac34,1) $ such that $ 2^{\ast}_{\alpha} = \frac{6}{3-2\alpha}\in(4,6) $ and $ V(x)\in C(\mathbb{R}^3,\mathbb{R}^+) $, $ \Omega\subset\mathbb{R}^{3} $ is a bounded domain with a smooth boundary $ \partial\Omega $. $ \|u\|_{K}^{2}: = \int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|u(x)-u(y)|^{2}K(x-y)dxdy $. The non-local integrodifferential operator $ \mathcal{L}_K $ is defined as follows:

    $ \mathcal{L}_Ku(x) = \frac12\int_{\mathbb{R}^{3}}(u(x+y)+u(x-y)-2u(x))K(y)dy,\;x\in\mathbb{R}^{3}, $

    the kernel $ K:\mathbb{R}^{3}\rightarrow (0,\infty) $ is a function with the properties that

    (ⅰ) $ m K\in L^1(\mathbb{R}^{3}) $, where $ m(x) = \min\{|x|^2,1\} $;

    (ⅱ) there exists $ \lambda>0 $ such that $ K(x)\geq\lambda|x|^{-(3+2\alpha)} $ for any $ x\in\mathbb{R}^{3}\backslash\{0\} $;

    (ⅲ) $ K(x) = K(-x) $ for any $ x\in\mathbb{R}^{3}\backslash\{0\} $.

    We note that when $ K(x) = \frac 1{|x|^{3+2\alpha}} $, the integrodifferential operator $ \mathcal{L}_K $ is the fractional Laplacian operator $ (-\Delta)^\alpha $:

    $ (-\Delta)^\alpha u(x) = -\frac{C(\alpha)}{2}\int_{\mathbb{R}^{3}}\frac{(u(x+y)+u(x-y)-2u(x))}{|y|^{3+2\alpha}}dy $

    and in this case

    $ \int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|u(x)-u(y)|^{2}K(x-y)dxdy = \frac{2}{C(\alpha)}\int_{\mathbb{R}^{3}}|(-\Delta)^{\alpha/2} u(x)|^2dx, $

    where $ C(\alpha) = \left(\int_{\mathbb{R}^{3}}\frac{1-\cos \xi_1}{|\xi|^{3+2\alpha}}d \xi\right)^{-1} $ is the normalized constant.

    When $ a = 1 $, $ b = 0 $ and $ K(x) = \frac 1{|x|^{3+2\alpha}} $, the fractional Kirchhoff equation is reduced to the undermentioned fractional nonlocal problem

    $ {(Δ)αu+V(x)u=|u|2α2u+kf(x,u),xΩ,u=0,xR3Ω.
    $
    (2)

    Equation (2) is derived from the fractional Schrödinger equation and the nonlinearity $ f(x, u) $ represents the particles interacting with each other. On the other hand, recently a great attention has been given to the so called fractional Kirchhoff equation (see [3,10] etc., ):

    $ (a+bR3R3|u(x)u(y)|2|xy|3+2αdxdy)(Δ)αu=f(x,u),
    $
    (3)

    where $ \Omega\subset \mathbb{R}^{N} $ is a bounded domain or $ \Omega = \mathbb{R}^{N} $, $ a>0, b>0 $ and $ u $ satisfies some boundary conditions. Problem (3) is called to the stationary state of the fractional Kirchhoff equation

    $ utt+(a+bR3R3|u(x)u(y)|2|xy|3+2αdxdy)(Δ)αu=f(x,u).
    $
    (4)

    As a special significant case, the nonlocal aspect of the tension arises from nonlocal measurements of the fractional length of the string. For more mathematical and physical background on Schrödinger-Kirchhoff type problems, we refer the readers to [6] and the references therein.

    In the remarkable work of Caffarelli and Silvestre [2], the authors express this nonlocal operator $ (-\Delta)^{\alpha} $ as a Dirichlet-Neumann map for a certain elliptic boundary value problem with local differential operators defined on the upper half space. This technique is a valid tool to deal with the equations involving fractional operator in the respects of regularity and variational methods. However, we do not know whether the Caffarelli-Silvestre extension method can be applied to the general integro-differential operator $ \mathcal{L}_K $. So the nonlocal feature of the integro-differential operator brings some difficulties to applications of variational method to problem (1). Some of these additional difficulties were overcome in [12,13]. More studies on this integro-differential operator equation, for examples, positive, negative, sign-changing or ground state solutions, we refer the papers [4,5,9,16] and their references. In [1], the authors considered the obstacle problems for $ \mathcal{L}_K $ about the higher regularity of free boundaries. The obstacle problems for nonlinear integro-differential operators and applications are also available in papers [7,8]. The appearance of nonlocal term not only makes it playing an important role in many physical applications, but also brings some difficulties and challenges in mathematical analysis. This fact makes the study of fractional Kirchhoff equation or similar problems particularly interesting. A lot of interesting results on the existence of nonlocal problems were obtained recently, specially on the existence of positive solutions, multiple solutions, ground states and semiclassical states, for examples, we refer [3,15,19] and the cited references.

    In past few years, some researchers began to search for nodal solutions of Schrödinger type equation with critical growth nonlinearity and have got some interesting results. For example, Zhang [20] considered the following Schrödinger-Poisson system:

    $ {Δu+u+k(x)ϕu=a(x)|u|p2u+u5,xR3,Δϕ=k(x)u2,xR3,
    $
    (5)

    where $ p\in(4, 6) $, $ k(x) $ and $ a(x) $ are nonnegative functions. By using variational method, a ground state solution and a nodal solution for problem (5) were obtained.

    Wang [17] studies the following Kirchhoff-type equation:

    $ {(a+bΩ|u|2dx)Δu=|u|4u+λf(x,u), xΩ,u=0, xΩ,
    $
    (6)

    where $ \Omega\subset \mathbb{R}^{3} $ is a bounded domain, $ \lambda, a, b>0 $ are fixed parameters, and $ f(x,\cdot) $ is continuously differentiable for a.e. $ x\in\Omega $. By combining constrained variational method with the degree theory, the existence of ground state and nodal solutions for above problem were obtained.

    However, as for fractional Kirchhoff types equation, to the best of our knowledge, few results involved the existence and asymptotic behavior of ground state and nodal solutions in case of critical growth. If $ k(x)\equiv1 $, the method used in [20] seems not valid for problem (1), because their result depends on the case $ k\in L^{p}(\mathbb{R}^{3})\cap L^{\infty}(\mathbb{R}^{3}) $ for some $ p\in (1,2^{\ast}_{\alpha}) $. We employ minimization arguments on suitable nodal Nehari manifold $ \mathcal{M}_{k} $ to construct a nodal solution by using various constrained method and qualitative deformation lemma given in [14].

    It's worth noting that, the Brouwer degree method used in [18] strictly depends on the nonlinearity $ f\in C^{1}(\mathbb{R}^{3}\times\mathbb{R},\mathbb{R}) $, so we have to find new tricks to solve our modeling where we only allow $ f:\mathbb{R}^{3}\times\mathbb{R}\rightarrow \mathbb{R} $ is a Carathéodory function. On the other hand, in our modeling, both of the nonlocal fractional operator $ \mathcal{L}_K $ and nonlocal terms $ \int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|u(x)-u(y)|^{2}K(x-y)dxdy $ appear, we need to overcome the difficulties caused by the nonlocal terms under a suitable variational framework. In brief, our discussion is based on the standard nodal Nehari manifold method used in [14].

    Throughout this paper, we let

    $ E = \{u\in X: \int_{\mathbb{R}^{3}}V(x)u^{2}{\text{d}}x < \infty, u = 0\; \rm{a.e. in}\; \mathbb{R}^{3}\backslash\Omega\}, $

    where the space X introduced by Servadei and Valdinoci ([12,13]) denotes the linear space of Lebesgue measurable functions $ u: \mathbb{R}^{3} \to \mathbb{R} $ such that its restriction to $ \Omega $ $ u|_{\Omega}\in L^2(\Omega) $ and

    $ ((x, y)\rightarrow (u(x)-u(y))\sqrt{K(x-y)}\in L^2((\mathbb{R}^{3}\times\mathbb{R}^{3}) \setminus(\Omega^c\times\Omega^c),dxdy) $

    with the following norm

    $ ||u||^2_X = ||u||^2_{L^2}+\int_{Q}|u(x)-u(y)|^{2}K(x-y)dxdy, $

    where $ Q = (\mathbb{R}^{3}\times\mathbb{R}^{3}) \setminus(\Omega^c\times\Omega^c) $. Then, $ E $ is a Hilbert space with inner product

    $ \langle u,v\rangle = \frac a2\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}(u(x)-u(y))(v(x)-v(y))K(x-y)dxdy +\int_{\Omega}V(x)uvdx, \forall \;u,\;v\in E $

    and the norm $ \|\cdot\| $ defined by

    $ \|u\|^{2} = \frac a2\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|u(x)-u(y)|^{2}K(x-y)dxdy +\int_{\Omega}V(x)u^{2}dx. $

    $ X_0 = \{u\in X: u = 0\; \rm{a.e. in}\; \mathbb{R}^{3}\backslash\Omega\} $ is a Hilbert space with inner product $ (u,v)_K = \frac 12\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}(u(x)-u(y))(v(x) -v(y))K(x-y)dxdy $ and the Gagliardo norm $ \|u\|_{K}^{2}: = (u,u)_K = \frac 12\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}|u(x)-u(y)|^{2}K(x-y)dxdy $. In $ X_0 $ the norms $ \|\cdot\|_K $ and $ \|\cdot\|_X $ are equivalent. Thus due to $ V(x) \ge 0 $ in $ \Omega $, the inner product $ \langle\cdot,\cdot\rangle $ and the norm $ \|\cdot\| $ defined above still make sense for the Hilbert space $ E $. Moreover, $ \|u\|^2_K\le a\|u\|^2 $. For more details, we refer to [12].

    The following result for the space $ X_0 $ will be used repeatedly.

    Lemma 1.1. ([12]) Let $ s\in(0, 1) $, $ N>2s $, $ \Omega $ be an open bounded set of $ \mathbb{R}^{N} $, $ X_0\hookrightarrow\hookrightarrow L^{p}(\Omega) $ for $ 2<p<2^{\ast}_{\alpha} $, and $ X_0\hookrightarrow L^{2^{\ast}_{\alpha}}(\Omega) $ is continuous. Then, $ E\subset X_0 $ has the same embedding properties.

    A weak solution $ u\in E $ of (1) is defined to satisfy the following equation

    $ 12(a+bu2K)R3R3(u(x)u(y))(v(x)v(y))K(xy)dxdy+ΩV(x)u(x)v(x)dxΩ|u(x)|2α2u(x)v(x)dxkΩf(x,u(x))v(x)dx=0,
    $

    for any $ v\in E $, i.e.

    $ \langle u,v\rangle+b\|u\|_K^{2}(u,v)_K - k\int_{\Omega}f(x,u)vdx-\int_{\Omega}|u|^{2^{\ast}_{\alpha}-2}uvdx = 0,\;\forall v\in E. $

    As for the function $ f $, we assume $ f:\Omega\times\mathbb{R}\rightarrow \mathbb{R} $ is a Carathéodory function and satisfing the following hypotheses:

    (f1): $ f(x,t)\cdot t>0 $ for $ t\ne 0,\forall x\in \Omega $ and $ \lim_{t\rightarrow0}\frac{f(x,t)}{t} = 0\;\;\rm{for}\; x\in \Omega\; \rm{uniformly} $;

    (f2): There exists $ q\in(4,2^{\ast}_{\alpha}) $ such that $ \lim_{t\rightarrow \infty}\frac{f(x,t)}{t^{q-1}} = 0 $ for all $ t\in \mathbb{R}\setminus\{0\} $ and for $ x\in \Omega $ uniformly;

    (f3): $ \frac{f(x,t)}{|t|^3} $ is an nondecreasing function with respect to $ t $ in $ (-\infty,0) $ and $ (0,+\infty) $ for a.e. $ x\in \Omega $.

    Remark 1. We note that under the conditions ($ f_1 $)-($ f_3 $), it is easy to see the function $ f(x,t) = t^3 $ is an example satisfying all conditions ($ f_1 $)-($ f_3 $).

    The main results can be stated as follows.

    Theorem 1.2. Suppose that $ (f_{1}) $-$ (f_{3}) $ are satisfied. Then, there exists $ k^{\star}>0 $ such that for all $ k\geq k^{\star} $, the problem (1) has a ground state nodal solution $ u_k $.

    Remark 2. The ground state nodal solution $ u_{k} $ with $ u_{k}^{\pm}\ne 0 $ is a solution of (1) satisfying

    $ J_{ k}(u_{k}) = c_k: = \inf\limits_{u\in\mathcal{M}_{k}}J_{ k}(u), $

    where $ \mathcal{M}_{k} $ is defined by (7) in next section. For $ u\in E $, $ u^{\pm} $ is defined by

    $ u^{+} = \max\{u(x),0\},\quad u^{-} = \min\{u(x),0\}. $

    We recall that the nodal set of a continuous function $ u:\mathbb{R}^3\to \mathbb{R} $ is the set $ u^{-1}(0) $. Every connected component of $ \mathbb{R}^3\setminus u^{-1}(0) $ is called a nodal domain. For this equation talking about the nodal domain of $ u_{k} $ in somewhat is difficult for us, we leave it as future topics.

    Theorem 1.3. Suppose that $ (f_{1}) $-$ (f_{3}) $ are satisfied. Then, there exists $ k^{\star\star}>0 $ such that for all $ k\geq k^{\star\star} $, the $ c^{\ast} = \inf_{u\in \mathcal{N}_{ k} }J _{ k}(u)>0 $ is achieved by $ v_k $, which is a ground state solution of (1) and

    $ J _{ k}({ u_k}) > 2c^{\ast}, $

    where $ \mathcal{N}_{ k} = \{u\in E\backslash\ \{0\}|\langle J'_k(u),u \rangle = 0 \} $, $ u_k $ is the ground state nodal solution obtained in Theorem 1.1.

    Comparing with the literature, the above two results can be regarded as a supplementary of those in [3,4,14,17].

    The remainder of this paper is organized as follows. In Section 2, we give some useful preliminaries. In Section 3, we study the existence of ground state and nodal solutions of (1) and we prove Theorems 1.1-1.2.

    We define the energy functional associated with equation (1) as follows:

    $ Jk(u)=12u2+b4u4KkΩF(x,u)dx12αΩ|u|2αdx,uE.
    $

    According to our assumptions on $ f(x,t) $, $ J_{ k}(u) $ belongs to $ C^{1}(E,\mathbb{R}) $ (see [10,12]), then by direct computations, we have

    $ Jk(u),v=u,v+bu2K(u,v)KkΩf(x,u)vdxΩ|u|2α2uvdx,u,vE.
    $

    Note that, since (1) involves pure critical nonlinearity $ |u|^{2^{\ast}_{\alpha}-2}u $, it will prevent us using the standard arguments as in [14]. Hence, we need some tricks to overcome the lack of compactness of $ E\hookrightarrow L^{2^{\ast}_{\alpha}}(\mathbb{R}^{3}) $.

    For fixed $ u\in E $ with $ u^{\pm}\neq0 $, the function $ \varphi_{u}:[0,\infty)\times[0,\infty)\rightarrow \mathbb{R} $ is well defined by $ \varphi_{u}(s,t) = J _{ k}(su^{+}+tu^{-}). $ We set

    $ H(s,t) = \langle J'_k(su^{+}+tu^{-}),su^{+}\rangle,\;\; G(s,t) = \langle J'_k(su^{+}+tu^{-}),tu^{-}\rangle. $

    The argument generally used is to modify the method developed in [11]. The nodal Nehari manifold is defined by

    $ Mk={uE,u±0andJk(u),u+=Jk(u),u=0},
    $
    (7)

    which is a subset of the Nehari manifold $ \mathcal{N}_{ k} $ and contains all nodal solutions of (1). Then, one needs to show the minimizer $ u\in \mathcal{M}_{k} $ is a critical point of $ J_k $. Because the problem (1) contains a nonlocal term $ \|u\|_{K}^{2} $, the corresponding functional $ J_k $ does not have the decomposition

    $ J _{ k}(u) = J _{ k}(u^{+})+J _{ k}(u^{-}), $

    it brings difficulties to construct a nodal solution.

    The following result describes the shape of the nodal Nehari manifold $ \mathcal{M}_{k} $.

    Lemma 2.1. Assume that $ (f_{1}) $-$ (f_{3}) $ are satisfied, if $ u\in E $ with $ u^{\pm}\neq0 $, then there is a unique pair $ (s_{u}, t_{u})\in(0,\infty)\times(0,\infty) $ such that $ s_{u}u^{+}+t_{u}u^{-}\in\mathcal{M}_{k} $, which is also the unique maximum point of $ \varphi_{u} $ on $ [0,\infty)\times[0,\infty) $. Furthermore, if $ \langle J_k'(u),u^\pm\rangle\le 0 $, then $ 0<s_u,\; t_u\le 1 $.

    Proof. From $ (f_{1}) $ and $ (f_{2}) $, for any $ \varepsilon>0 $, there is $ C_{\varepsilon}>0 $ satisfying

    $ |f(x,t)|ε|t|+Cε|t|q1,tR.
    $
    (8)

    By above equality and Sobolev's embedding theorems, we have

    $ H(s,t):=s2u+2+bsu++tu2K(su++tu,su+)KΩ|su+|2αdxkΩf(x,su+)su+dxast2R3R3[u+(x)u(y)+u(x)u+(y)]K(xy)dxdys2u+2C1s2αu+2αkεC2s2u+2kCεC3squ+q.
    $
    (9)

    By choosing $ \varepsilon>0 $ small enough, we can deduce $ H(s,t)>0 $ for $ 0<s\ll 1 $ and all $ t\geq0 $. Similarly,

    $ G(s,t):=t2u2+bsu++tu2K(su++tu,tu)KΩ|tu|2αdxkΩf(x,tu)tudxast2R3R3[u+(x)u(y)+u(x)u+(y)]K(xy)dxdy>0,
    $
    (10)

    for $ 0<t\ll 1 $ and all $ s\geq0 $. We denote $ D(u) = \langle u^+,u^-\rangle = a( u^+,u^-)_K = -\frac a2\int_{\mathbb{R}^{3}}\int_{\mathbb{R}^{3}}[u^+(x)u^-(y)+u^-(x)u^+(y)]K(x-y)dxdy\ge 0 $. Hence, by choosing $ \delta_{1}>0 $ small, we have

    $ H(δ1,t)>0,G(s,δ1)>0.
    $
    (11)

    For any $ \delta_{2}>\delta_{1} $, by condition $ (f_{1}) $, it is easy to see

    $ H(δ2,t)δ22u+2+bδ2u++tu2K(δ2u++tu,δ2u+)Kδ22αR3|u+|2αdx+δ2tD(u).
    $

    Similarly, we have

    $ G(s,δ2)δ22u2+bsu++δ2u2K(su++δ2u,δ2u)Kδ22αR3|u|2αdx+sδ2D(u).
    $

    By choosing $ \delta_{2}\gg1 $, we deduce

    $ H(δ2,t)<0,G(s,δ2)<0
    $
    (12)

    for all $ s,t\in[\delta_{1},\delta_{2}] $.

    Following (11) and (12), we can use Miranda's Theorem (see Lemma 2.4 in [17]) to get a positive pair $ (s_{u}, t_{u})\in(0,\infty)\times(0,\infty) $ such that $ s_{u}u^{+}+t_{u}u^{-}\in\mathcal{M}_{k} $. Similar to the standard argument in [17], we can prove the pair $ (s_{u}, t_{u}) $ is unique maximum point of $ \varphi_{u} $ on $ [0,+\infty)\times [0,+\infty) $.

    Lastly, we will prove that $ 0<s_{u},t_{u}\leq1 $ when $ \langle J'_k(u),u^{\pm}\rangle\leq0 $. Following from $ s_{u}\geq t_{u}>0 $, by a direct computation, we have

    $ s2uu+2+s2uD(u)+s4ub(u+2K+u2K+2D(u))(u+2K+D(u))s2uu+2+sutuD(u)+b(s2uu+2K+t2uu2K+2sutuD(u))(s2uu+2K+sutuD(u))=s2αuR3|u+|2αdx+kR3f(x,suu+)suu+dx.
    $
    (13)

    On the other hand, $ \langle J'_k(u),u^{+}\rangle\leq0 $ implies that

    $ u+2+D(u)+b(u+2K+u2K+2D(u))(u+2K+D(u))R3|u+|2αdx+kR3f(x,u+)u+dx.
    $
    (14)

    From (13) - (14), we can see that

    $ (1s2u1)(u+2+D(u))(s2α4u1)R3|u+|2αdx+kR3[f(x,suu+)(suu+)3f(x,u+)(u+)3](u+)4dx.
    $

    So we have $ s_{u}\leq1 $, which implies that $ 0<s_{u},t_{u}\leq1 $.

    Lemma 2.2. There exists $ k^{\star}>0 $ such that for all $ k\geq k^{\star} $, the infimum $ c_{k} = \inf_{u\in \mathcal{M}_k}J_k(u) $ is achieved.

    Proof. For any $ u\in\mathcal{M}_{k} $, in view of the definitions of $ \mathcal{M}_k $, it follows that

    $ u±2+bu2K(u,u±)K=kR3f(x,u±)u±dx+R3|u±|2αdx.
    $

    Hence, in view of (8), we have

    $ u±2kεC1u±2+kC2u±q+C3u±2α.
    $

    By choosing $ \varepsilon>0 $ small enough, we can deduce

    $ u±ρ
    $
    (15)

    for some $ \rho>0 $. From assumption $ (f_{3}) $, we have

    $ f(x,t)t4F(x,t)0,
    $
    (16)

    and $ f(x,t)t-4F(x,t) $ is nondecreasing in $ (0,+\infty) $ and non-increasing in $ (-\infty,0) $ with respect to the variable $ t $. Hence, combining with $ \langle J'_k(u),u\rangle = 0 $, we have

    $ Jk(u)=14u2+(1412α)R3|u|2αdx+k4R3[f(x,u)u4F(x,u)]dx14u2.
    $

    So $ c_{k} = \inf_{u\in \mathcal{M}_k}J_k(u)\ge 0 $ is well defined.

    Let $ u\in E $ with $ u^{\pm}\neq0 $ be fixed. According to Lemma 2.1, for each $ k>0 $, there exists $ s_{ k}, t_{ k}>0 $ such that $ s_{ k}u^{+}+t_{ k}u^{-}\in\mathcal{M}_{k} $. Hence, by $ (f_{1}) $ and the Sobolev's embedding theorem, we have

    $ s2αkR3|u+|2αdx+t2αkR3|u|2αdxsku++tku2+bsku++tku4K2s2ku+2+2t2ku2+4bs4ku+4K+4bt4ku4K,
    $

    which implies $ (s_{ k},t_{ k}) $ is bounded and furthermore by taking any sequence $ (s_{ k_{n}}, t_{ k_{n}})\rightarrow (s_{0}, t_{0}) $ as $ k_n\rightarrow \infty $, we claim that $ s_0 = t_0 = 0 $. In fact, if $ s_{0} > 0 $ or $ t_{0}> 0 $. Thanks to $ s_{ k_{n}}u^{+}+t_{ k_{n}}u^{-}\in \mathcal {M}_{k_n} $, we get

    $ sknu++tknu2+bsknu++tknu4K=R3|sknu++tknu|2αdx+knR3f(sknu++tknu)(sknu++tknu)dx.
    $
    (17)

    Because $ k_{n}\rightarrow \infty $ and $ \{s_{ k_{n}}u^{+}+t_{ k_{n}}u^{-}\} $ is bounded in $ E $, we have a contradiction with the equality (17). On the other hand, we have

    $ 0ckJk(sku++tku)s2ku+2+t2ku2+2bs4ku+4K+2bt4ku4K,
    $

    so $ \lim_{ k\rightarrow \infty} c_k = 0 $. By the definition of $ c_k $, we can find a sequence $ \{u_{n}\}\subset\mathcal{M}_{k} $ satisfing $ \lim_{n\rightarrow \infty}J _{ k}(u_{n}) = c_k $, which converges to $ u_k = u^++u^-\in E. $ By standard arguments, we have

    $ u±nu±inLp(R3)p(2,2α),u±n(x)u±(x)a.e.xR3.
    $

    Denote $ \beta: = \frac{s}{3}(S)^{\frac{3}{2s}} $, where

    $ S: = \inf\limits_{u\in E\backslash\{0\}}\frac{\|u\|^{2}}{(\int_{\mathbb{R}^{3}}|u|^{2^{\ast}_{\alpha}}dx)^{\frac{2}{2^{\ast}_{\alpha}}}}. $

    Sobolev embedding theorem insures that $ \beta>0 $. So that there exists $ k^{\star}>0 $ such that $ c_k<\beta $ for all $ k\geq k^{\star} $. Fix $ k\geq k^{\star} $, in view of Lemma 2.1, we have $ J _{ k}(su^{+}_{n}+tu ^{-}_{n})\leq J _{ k}(u_{n}). $

    By $ u^{\pm}_{n}\rightharpoonup u^{\pm} \; \; \rm{in}\; E $, we have

    $ u±n2u±nu±2=2u±n,u±u±2.
    $

    By taking $ n\rightarrow \infty $ in both sides of above equality, there holds

    $ \lim\limits_{n\to \infty}\|u^{\pm}_{n}\|^2 = \lim\limits_{n\to \infty}\|u^{\pm}_{n}-u^{\pm}\|^{2}+\|u^{\pm}\|^{2}. $

    On the other hand, by (8) we have

    $ R3F(x,su±n)dxR3F(x,su±)dx.
    $

    Then,

    $ lim infnJk(su+n+tun)s22limn(u+nu+2+u+2)+t22limn(unu2+u2)+bs44[limnu+nu+2K+u+2K]2+bt44[limnunu2K+u2K]2+stlim infnD(un)+bs2t24a2lim infnD2(un)+bs2t22lim infn(u+n2Kun2K)+bs3t2alim infn(D(un)u+n2K)+bst32alim infn(D(un)un2K)s2α2αlimn(|u+nu+|2α2α+|u+|2α2α)t2α2αlimn(|unu|2α2α+|u|2α2α)kR3F(x,su+)dxkR3F(x,tu)dx,
    $

    where $ |\ast|_{2} $ and $ |\ast|_{2^{\ast}_{\alpha}} $ are the norm in $ L^{2}(\mathbb{R}^{3}) $ and $ L^{2^{\ast}_{\alpha}}(\mathbb{R}^{3}) $ repeatedly. So, Fatou's Lemma follows that

    $ lim infnJk(su+n+tun)Jk(su++tu)+s22limnu+nu+2+t22limnunu2s2α2αlimn|u+nu+|2α2αt2α2αlimn|unu|2α2α+bs42limnu+nu+2Ku+2K+bs44(limnu+nu+2K)2+bt42limnunu2Ku2K+bt44(limnunu2K)2=Jk(su++tu)+s22A1s2α2αB1+t22A2t2α2αB2+bs42A3u+2K+bs44A23+bt42A4u2K+bt44A24,
    $

    where

    $ A1=limnu+nu+2,A2=limnunu2,B1=limn|u+nu+|2α2α,B2=limn|unu|2α2α,A3=limnu+nu+2K,A4=limnunu2K.
    $

    From the inequality above, we deduce that

    $ Jk(su++tu)+s22A1s2α2αB1+t22A2t2α2αB2+bs42A3u+2K+bs44A23+bt42A4u2K+bt44A24ck.
    $
    (18)

    We next prove $ u^{\pm}\neq0 $. Because the claim $ u^{-}\neq0 $ is similar, so we only prove $ u^+\neq0 $. Indirectly, we suppose that $ u^{+} = 0 $ and so $ A_1\ge \rho $ from (15). By letting $ t = 0 $ in (18), the case $ B_{1} = 0 $ is done. So we only study the case $ B_{1}>0 $ at length. In this case, by the definition of $ S $, we deduce

    $ β=α3S32αα3(A1(B1)22α)32α.
    $

    It happens that,

    $ α3(A1(B1)22α)32α=maxs0{s22A1s2α2αB1}maxs0{s22A1s2α2αB1+bs42A3u+2K+bs44A23}.
    $

    The inequality (18) and $ c_k<\beta $ follows that

    $ βmaxs0{s22A1s2α2αB1+bs42A3u+2K+bs44A23}ck<β,
    $

    which is a contradiction. Thus $ u^{+}\neq0 $ are claimed.

    Then, we consider the key point to the proof of Theorem 1.1, that is $ B_{1} = B_{2} = 0 $ and then $ c_{ k} $ is achieved by $ u_k = u^{+}+u^{-}\in \mathcal{M}_k $.

    Similarly, we only prove $ B_{1} = 0 $. Indirectly, we suppose that $ B_{1}>0 $. We have two cases.

    Case 1: $ B_{2}>0 $. $ \quad $ Let $ \bar{s} $ and $ \bar{t} $ be the numbers such that

    $ ˉs22A1ˉs2α2αB1+bˉs42A3u+2K+bˉs44A23=maxs0{s22A1s2α2αB1+bs42A3u+2K+bs44A23},
    $
    $ ˉt22A2ˉt2α2αB2+bˉt42A4u2K+bˉt44A24=maxt0{t22A2t2α2αB2+bt42A4u2K+bt44A24}.
    $

    Since $ \varphi_{u} $ is continuous, we have $ (s_{u},t_{u})\in[0, \bar{s}]\times[0, \bar{t}] $ satisfying

    $ φu(su,tu)=max(s,t)[0,ˉs]×[0,ˉt]φu(s,t).
    $

    If $ t>0 $ small enough, then, $ \varphi_{u}(s,0)<J _{ k}(su^{+})+J _{ k}(tu^{-})\leq \varphi_{u}(s,t) $ for all $ s\in[0, \bar{s}]. $ Thus there is $ t_{0}\in[0, \bar{t}] $ such that $ \varphi_{u}(s,0)\leq \varphi_{u}(s,t_{0}) $ for all $ s\in[0,\bar{s}]. $ Thus, $ (s_{u},t_{u})\not\in[0, \bar{s}]\times\{0\}. $ Similarly, $ (s_{u},t_{u})\not\in\{0\}\times[0,\bar{t}] $.

    By direct computation, we get

    $ s22A1s2α2αB1+bs42A3u+2K+bs44A23>0,
    $
    (19)
    $ t22A2t2α2αB2+bt42A4u2K+bt44A24>0,
    $
    (20)

    for all $ (s,t)\in(0,\bar{s}]\times(0,\bar{t}] $. Hence, there holds

    $ βˉs22A1ˉs2α2αB1+bˉs42A3u+2K+bˉs44A23+t22A2t2α2αB2+bt42A4u2K+bt44A24,
    $
    $ βˉt22A2ˉt2α2αB2+bs42A3u+2K+bs44A23+s22A1s2α2αB1+bˉt42A4u2K+bˉt44A24.
    $

    In view of (18), it follows that $ \varphi_{u}(s,\bar{t})\leq 0\;\;\forall s\in[0,\bar{s}]\;\; $ and $ \varphi_{u}(\bar{s},t)\leq 0\;\;\forall t\in[0,\bar{t}]. $ That is, $ (s_{u},t_{u})\not\in \{\bar{s}\}\times[0,\bar{t}] $ and $ (s_{u},t_{u})\not\in [0,\bar{s}]\times\{\bar{t}\} $. Hence, we can deduce that $ (s_{u},t_{u})\in(0, \bar{s})\times(0, \bar{t}) $. By (18), (19) and (20), we deduce

    $ ckJk(suu++tuu)+s2u2A1s2αu2αB1+t2u2A2t2αu2αB2+bs4u2A3u+2K+bs4u4A23+bt4u2A4u2K+bt4u4A24>Jk(suu++tuu)ck.
    $

    It is impossible. The proof of Case 1 is completed.

    Case 2: $ B_{2} = 0 $. $ \quad $ From the definition of $ J _{ k} $, it is easy to show that there exists $ t_{0}\in[0,\infty) $ such that $ \varphi_{u}(s,t)\leq0, $ for all $ (s,t)\in[0,\bar{s}]\times[t_{0},\infty) $. Thus, there is $ (s_{u},t_{u})\in[0,\bar{s}]\times[0,\infty) $ satisfying

    $ φu(su,tu)=max(s,t)[0,ˉs]×[0,)φu(s,t).
    $

    We need to prove that $ (s_{u},t_{u})\in(0,\bar{s})\times(0,\infty) $. Similarly, it is noticed that $ \varphi_{u}(s,0)<\varphi_{u}(s,t) $ for $ s\in [0,\bar{s}] $ and $ 0<t\ll 1 $, that is $ (s_{u},t_{u})\not\in[0,\bar{s}]\times\{0\} $. Also, for $ s $ small enough, we get $ \varphi_{u}(0,t)<\varphi_{u}(s,t) $ for $ t\in [0,\infty) $, that is $ (s_{u},t_{u})\not\in\{0\}\times[0,\infty). $ We note that

    $ βˉs22A1ˉs2α2αB1+bˉs42A3u+2K+bˉs44A23+t22A2t2α2αB2+bt42A4u2K+bt44A24.
    $

    Thus also from (20) and $ B_2 = 0 $, we have $ \varphi_{u}(\bar{s},t)\leq 0 $ for all $ t\in[0,\infty) $. Hence, $ (s_{u},t_{u})\not\in\{\bar{s}\}\times[0,\infty) $. That is, $ (s_{u},t_{u}) $ is an inner maximizer of $ \varphi_{u} $ in $ [0,\bar{s})\times[0,\infty) $. Hence by using (19), we obtain

    $ ckJk(suu++tuu)+s2u2A1s2αu2αB1+t2u2A2t2αu2αB2+bs4u2A3u+2K+bs4u4A23+bt4u2A4u2K+bt4u4A24>Jk(suu++tuu)ck,
    $

    which is a contradiction.

    Since $ u^{\pm}\neq0 $, by Lemma 2.1, there are $ s_{u}, t_{u}>0 $ such that $ \widetilde{u}: = s_{u}u^{+}+t_{u}u^{-}\in \mathcal{M}_{ k} $. On the other hand, Fatou's Lemma follows that

    $ Jk(u),u±lim infnu±n2+blim infnun2K(un,u±n)KlimnR3f(x,u±n)u±ndxlimnR3|u±n|2α+lim infnL(un)limnJk(un),u±n=0.
    $

    By Lemma 2.1, we know $ 0<s_{u},t_{u}\leq1 $. Since $ u_{n}\in \mathcal{M}_k $ and $ B_{1} = B_{2} = 0 $, we have

    $ ckJk(˜u)14Jk(˜u),˜u=14(suu+2+tuu2)+(1412α)(|suu+|2α2α+|tuu|2α2α)+k4R3[f(x,suu+)(suu+)4F(x,suu+)]dx+k4R3[f(x,tuu)(tuu)4F(x,tuu)]dxlim infn[Jk(un)14Jk(un),un]=ck.
    $

    So, we have completed proof of Lemma 2.2.

    In this section, we will prove main results.

    Proof. Since $ u_k\in \mathcal{M}_{ k} $ and $ J _{ k}(u_k^{+}+u_k^{-}) = c_k $, by Lemma 2.1, for $ (s,t)\in (\mathbb{R}_{+}\times\mathbb{R}_{+})\backslash(1,1) $, we have

    $ Jk(su+k+tuk)<ck.
    $
    (21)

    If $ J'_k(u_k)\neq0 $, then there exist $ \delta>0 $ and $ \theta>0 $ such that

    $ \|J'_k(v)\|\geq\theta,\; {\text{for}}\; {\text{all}}\; \|v-u_k\|\leq3\delta. $

    We know by the result (15), if $ u\in\mathcal{M}_{k} $, there exists $ L>0 $ such that $ \|u^{\pm}\|>L $ and we can assume $ 6\delta<L $. Let $ Q: = (\frac 12,\frac 32)\times (\frac 12,\frac 32) $, and $ g(s,t) = su_k^{+}+tu_k^{-} $, $ (s,t)\in Q $. In view of (21), it is easy to see that

    $ ¯ck:=maxQIg<ck.
    $
    (22)

    Let $ \varepsilon: = min\{(c_k-\overline{c}_k)/4,\theta\delta/8\} $ and $ S_{\delta}: = B(u_k,\delta) $, there exists a deformation $ \eta\in C([0,1]\times E,E) $ satisfying

    $ (a) $] $ \eta(t,v) = v $ if $ t = 0 $, or $ v\notin (J _{ k})^{-1}([c_k-2\varepsilon,c_k+2\varepsilon])\cap S_{2\delta} $;

    $ (b) $] $ \eta(1,(J _{ k})^{c_k+\varepsilon}\cap S_{\delta})\subset (J _{ k})^{c_k-\varepsilon} $;

    $ (c) $]$ J _{ k}(\eta(1,v))\leq J _{ k}(v) $ for all $ v\in E $;

    $ (d) $]$ J _{ k}(\eta(\cdot,v)) $ is non increasing for every $ v\in E $.

    To finish the proof of Theorem 1.1, one of the key points is to prove that

    $ max(s,t)ˉQJk(η(1,g(s,t)))<ck.
    $
    (23)

    The other is to prove that $ \eta(1,g(Q))\cap\mathcal{M}_k\neq\emptyset $. Let us begin this work. In fact, it follows from Lemma 2.1 that $ g(s,t)\in (J _{ k})^{c_k+\varepsilon}. $ On the other hand, from (a) and (d), we get

    $ Jk(η(1,v))Jk(η(0,v))=Jk(v),vE.
    $
    (24)

    For $ (s,t)\in Q $, when $ s\neq 1 $ or $ t\neq 1 $, according to (21) and (24),

    $ Jk(η(1,g(s,t)))Jk(g(s,t))<ck.
    $

    If $ s = 1 $ and $ t = 1 $, that is, $ g(1,1) = u_k $, so that it holds $ g(1,1)\in J _{ k}^{c_k+\varepsilon}\cap S_{\delta} $, then by (b)

    $ Jk(η(1,g(1,1)))ckε<ck.
    $

    Thus (23) holds. Then, let $ \varphi(s,t): = \eta(1,g(s,t)) $ and

    $ Υ(s,t):=(1sJk(φ(s,t)),(φ(s,t))+,1tJk(φ(s,t)),(φ(s,t))).
    $

    The claim holds if there exists $ (s_{0},t_{0})\in Q $ such that $ \Upsilon(s_{0},t_{0}) = (0,0) $. Since

    $ g(s,t)uk2=(s1)u+k+(t1)uk2|s1|2u+k2>|s1|2(6δ)2,
    $

    and $ |s-1|^2(6\delta)^{2}>4\delta^{2}\Leftrightarrow s<2/3\;\;\rm{or}\;\;s>4/3 $, using (ⅰ) and the range of $ s $, for $ s = \frac{1}{2} $ and for every $ t\in[\frac{1}{2},\frac{3}{2}] $ we have $ g(\frac{1}{2},t)\notin S_{2\delta} $, so from (a), we have $ \varphi(\frac{1}{2},t) = g(\frac{1}{2},t) $. Thus

    $ Υ(12,t)=(2Jk(12u+k+tuk),12u+k,1tJk(12u+k+tuk),tu).
    $

    On the other hand, from (9) and $ u\in \mathcal{M} _{k} $, we have that

    $ H(t,t)=(t2t4)(u+2+D(u))+(t4t2α)R3|u+|2αdx+kt4R3(f(x,u+)f(x,tu+)t3)u+dx.
    $

    According to $ (f_{3}) $, when $ 0<t<1 $, $ H(t,t)>0, $ while in the case $ t>1 $, $ H(t,t)<0. $ Similarly, it is easy to get $ G(t,t)>0\;\rm{for}\;t\in(0,1),\;\; G(t,t)<0\;\rm{for}\;t>1. $ By above discussions, due to $ (f_{3}) $ and $ 2_\alpha^*>4 $, we have

    $ H(12,t)=12u+k2+t2D(uk)+b(14u+k2K+t2uk2K+taD(uk))(14u+k2K+t2aD(uk))(12)2αR3|u+k|2αdxkR3f(x,12u+k)12u+kdxH(12,12)>0,
    $

    which implies that

    $ H(12,t)>0,t[12,32].
    $
    (25)

    Analogously, $ \varphi(\frac{3}{2},t) = g(\frac{3}{2},t) $ implies that

    $ H(32,t)=32u+k2+3t2D(uk)+b(94u+k2K+t2uk2K+3taD(uk))(94u+k2K+3t2aD(uk))(32)2αR3|u+k|2αdxkR3f(x,32u+k)32u+kdxH(32,32)<0,
    $

    that is,

    $ H(32,t)<0,t[12,32].
    $
    (26)

    By the same way,

    $ G(s,12)>0,s[12,32],andG(s,32)>0,s[12,32].
    $
    (27)

    From (25)-(27), the assumptions of Miranda's Theorem (see Lemma 2.4 in [17]) are satisfied. Thus, there exists $ (s_{0},t_{0})\in Q $ such taht $ \Upsilon(s_{0},t_{0}) = 0 $, i.e. $ \eta(1,g(s_{0},t_{0}))\in \mathcal{M}_{ k} $. Comparing to (23), $ u_k $ is a ground state nodal solution of problem (1).

    Proof. Recall that $ \beta = \frac{\alpha}{3}S^{\frac{3}{2\alpha}} $, where $ S: = \inf_{u\in E\backslash\{0\}}\frac{\|u\|^{2}}{(\int_{\mathbb{R}^{3}}|u|^{2^{\ast}_{\alpha}}dx)^{\frac{2}{2^{\ast}_{\alpha}}}}. $ Similar to the proof of Lemma 2.2, we claim that there exists $ k_{1}^{\star}>0 $ such that for all $ k\geq k_{1}^{\star} $, there exists $ v_k\in \mathcal{N}_k $ such that $ J _{ k}(v_k) = c^{\ast}>0 $ and there is $ k^{\star}>0 $ such that $ c^{\ast}<\beta $ for all $ k\geq k^{\star} $. By standard processes, we can assume $ v_{n}\rightharpoonup v_k\in E. $ Therefore, $ \liminf_{n\rightarrow \infty}J _{ k}(tv_{n})\geq c^*\geq J _{ k}(tv_k)+\frac{t^{2}}{2}A-\frac{t^{2^{\ast}_{\alpha}}}{2^{\ast}_{\alpha}}B +\frac{bt^{4}}{4}(C^2+2C\|v_k\|_{K}^2) $, where $ A = \lim_{n\rightarrow \infty}\|v_{n}-v_k\|^{2}, B = \lim_{n\rightarrow \infty}|v_{n}-v_k|_{2^{\ast}_{\alpha}}^{2^{\ast}_{\alpha}} $ and $ C = \|v_{n}-v_k\|_{K}^2 $.

    Firstly, we prove that $ v_k\neq0 $. By contradiction, we suppose $ v_k = 0 $. The Case $ B = 0 $ is contained in above equality. So we only consider the case $ B>0 $. The fact $ \beta\leq\frac{\alpha}{3}\left(\frac{A_{1}}{(B_{1})^{\frac{2}{2^{\ast}_{\alpha}}}}\right)^{\frac{3}{2\alpha}} $ follows that

    $ β˜t22A˜t2α2αB:=maxt0{t22At2α2αB}maxt0{t22At2α2αB+bt44(C2+2Cvk2K)}c<β.
    $

    Which is a contradiction.

    Then, we prove that $ B = 0 $ and $ c^{\ast}\; is\; achieved \; by \; v_k $. Indirectly, we suppose that $ B>0 $. Firstly, we can maximize $ \varphi_{v_k}(t) = J _{ k}(tv_k) $ in $ [0,\infty) $ at $ t_{v} $, which be an inner maximizer. So we get $ \frac{t_v^{2}}{2}A-\frac{t_v^{2^{\ast}_{\alpha}}}{2^{\ast}_{\alpha}}B +\frac{bt_v^{4}}{4}(C^2+2C\|v_k\|_{K}^2)>0 $. Thus from $ t_{v}v_k\in \mathcal{N}_{b}^{ k} $ we get a contradiction by

    $ c^*\le J _{ k}(t_vv_k) < J _{ k}(t_vv_k)+\frac{t_v^{2}}{2}A-\frac{t_v^{2^{\ast}_{\alpha}}}{2^{\ast}_{\alpha}}B +\frac{bt_v^{4}}{4}(C^2+2C\|v_k\|_{K}^2)\le c^*. $

    From the above arguments we know $ \widetilde{v}: = t_{v}v_k\in \mathcal{N}_k $. Furthermore, we have

    $ cJk(˜v)14Jk(˜v),˜v14vk2+(1412α)|vk|2α2α+k4R3[f(x,vk)vk4F(x,vk)]dx=lim infn[Jk(vn)14Jk(vn),vn]=c.
    $

    Therefore, $ t_{v} = 1 $, and $ c^{\ast} $ is achieved by $ v_k\in \mathcal{N}_k $.

    By standard arguments, we can find $ v_k\in E $, a ground state solution of problem (1). For all $ k\geq k^{\star} $, the problem (1) also has a ground state nodal solution $ u_k $. Let $ k^{\star\star} = \max\{ k^{\star}, k^{\star}_{1}\}. $ Suppose that $ u_k = u^{+}+u^{-} $, we let $ s_{u^{+}},t_{u^{-}}\in(0,1) $ such that

    $ s_{u^{+}}u^{+}\in \mathcal{N}_k,\;\;\; t_{u^{-}}u^{-}\in \mathcal{N}_k. $

    Thus, the above fact follows that

    $ 2cJk(su+u+)+Jk(tuu)Jk(su+u++tuu)<Jk(u++u)=ck.
    $
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