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Research article Special Issues

Tracking control for a class of fractional order uncertain systems with time-delay based on composite nonlinear feedback control

  • Received: 27 January 2024 Revised: 07 March 2024 Accepted: 20 March 2024 Published: 08 April 2024
  • MSC : 34H05, 93C15

  • In this paper, we dealt with the tracking control problem of a class of fractional-order uncertain systems with time delays. In order to handle the effects brought by the uncertainties, external disturbances, time-delay terms, and to overcome the obstacles caused by inputs saturation, the tracking controller, which consisted of linear control law, nonlinear law, and robust control law proposed in this paper, was designed by combining the composite nonlinear feedback control method and the properties of fractional order operators. Furthermore, the validation of this tracking controller was proved.

    Citation: Guijun Xing, Huatao Chen, Zahra S. Aghayan, Jingfei Jiang, Juan L. G. Guirao. Tracking control for a class of fractional order uncertain systems with time-delay based on composite nonlinear feedback control[J]. AIMS Mathematics, 2024, 9(5): 13058-13076. doi: 10.3934/math.2024637

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  • In this paper, we dealt with the tracking control problem of a class of fractional-order uncertain systems with time delays. In order to handle the effects brought by the uncertainties, external disturbances, time-delay terms, and to overcome the obstacles caused by inputs saturation, the tracking controller, which consisted of linear control law, nonlinear law, and robust control law proposed in this paper, was designed by combining the composite nonlinear feedback control method and the properties of fractional order operators. Furthermore, the validation of this tracking controller was proved.



    An expedient feature of the p-Laplacian eigenvalue problem is that the eigenfunctions may be multiplied by constant factors (in other words, the fact that if u is an eigenfunction, so is ku). Unfortunately, the p(x)-Laplacian eigenvalue problem does not possess this expedient property. It is important to stress that the loss of the property under consideration is not only a consequence of the dependence on x, but it can also occur in presence of unbalanced growth. For example, the double phase operator (that does not depend on x)

    div(|u|p2u+μ(x)|u|q2u), (1.1)

    loses this property. In this paper we are interested in considering that the operator has both peculiarities: It depends on x and it is unbalanced.

    Let ΩRN(N2) be a bounded domain with Lipschitz boundary Ω. This article studies an eigenvalue problem coming from the minimization of the Rayleigh quotient:

    ||u||Hn||u||Hn, (1.2)

    among all uW1,Hn0(Ω),u0. These functions belong to an appropriate Musielak-Orlicz Sobolev space with variable exponents; see its definition in section two. The function a:ˉΩ[0,+) is a C1 differentiable function.

    Put

    Kn(u):=||u||Hn,kn(u):=||u||Hn,Sn(u):=Ω[pn(x)|uKn(u)|pn(x)2+qn(x)a(x)|uKn(u)|qn(x)2]dxΩ[pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2]dx,Hn:=tpn(x)+a(x)tqn(x) (1.3)

    and define the first eigenvalue as

    λ1(pn(),qn())=infuW1,Hn0(Ω){0}||u||Hn||u||Hn. (1.4)

    By a similar proof of Proposition 3.1 in [1], we can show that the following equation

    div[(pn(x)|uKn(u)|pn(x)2+qn(x)a(x)|uKn(u)|qn(x)2)uKn(u)]=λ(pn(),qn())Sn(u)ukn(u)(pn(x)|ukn(u)|pn(x)2+qn(x)a(x)|ukn(u)|qn(x)2),uW1,Hn0(Ω) (1.5)

    is the Euler-Lagrange equation corresponding to the minimization of the Rayleigh quotient (1.2), where λ(pn(),qn())=λ1(pn(),qn()).

    Here, we impose the following hypotheses on the variable exponents pn(x) and qn(x).

    (H1): Assume that pn(x) and qn(x) are two sequences of C1 functions in ¯Ω, qn()>pn() for every n1 and

    pn(x),qn(x)+,uniformly for allxΩ, (1.6)
    pn(x)pn(x)ξ1(x),uniformly for allxΩ, (1.7)
    qn(x)qn(x)ξ2(x),uniformly for allxΩ. (1.8)

    (H2): The following two quotients are bounded, namely,

    lim supn+p+npnk1,lim supn+q+nqnk2, (1.9)

    where for a function g we denote

    g=minx¯Ωg(x),g+=maxx¯Ωg(x).

    (H3): We also assume that

    pn>1,qn>1,q+npn<1+1N, (1.10)

    then we can find a positive and continuous function θ(0<θ<+), such that

    limnqn(x)pn(x)=θ(x) (1.11)

    uniformly for all xΩ.

    The differential operator in (1.5) is the double-phase operator with variable exponents, which can be given by

    div(|u|pn(x)2u+μ(x)|u|qn(x)2u). (1.12)

    This operator is the classical double phase operator (1.1) when pn(x) and qn(x) are constants. Moreover, special cases of (1.12), studied extensively in the literature, occur when inf¯Ωμ>0 (the weighted (q(x),p(x))-Laplacian) or when μ0 (the p(x)-Laplacian).

    The energy functional related to the double-phase operator (1.12) is given by

    Ω|u|pn(x)+μ(x)|u|qn(x)dx, (1.13)

    whose integrand switches two different elliptic behaviors. The integral functional (1.13) was first introduced by Zhikov [2,3,4,5], who obtained that the energy density changed its ellipticity and growth properties according to the point in order to provide models for strongly anisotropic materials. Moreover, double phase differential operators (1.12) and corresponding energy functionals (1.13) have several physical applications. We refer to the works of [6] on transonic flows, [7] on quantum physics and [8] on reaction diffusion systems. Finally, we mention a recent paper that is very close to our topic. For related works dealing with the double phase eigenvalue problems, we refer to the works of Colasuonno-Squassina [9], who proved the existence and properties of related variational eigenvalues. By using the Rayleigh quotient of two norms of Musielak-Orlicz space, the author of this paper has defined the eigenvalue, which has the same properties as the p-Laplace operator. Recently, Liu-Papageorgiou has considered an eigenvalue problem for the Dirichlet (p,q())Laplacian by using the Nehari method (see [10]), a nonlinear eigenvalue problem for the Dirichlet (p,q)Laplacian with a sign-changing Carathéodory reaction (see [11]) and a nonlinear eigenvalue problem driven by the anisotropic (p(),q())Laplacian (see [12]). Motivated by [9], Yu[13] discuss the asymptotic behavior of an eigenvalue for the double phase operator. However, to the author's knowledge, the eigenvalue problem for variable exponents double phase operator has remained open. Our article fits into this general field of investigation.

    Assume that δ:Ω[0,) is the distance function, which is given by

    δ(x):=dist(x,Ω)=infyΩ|xy|.

    This function is a Lipschitz continuous function. For all xΩ, we get |δ|=1. Define

    Λ:=infφW1,0(Ω){0} (1.14)

    It is known from the paper [1] that

    \begin{equation} \Lambda_{\infty} = \frac{\|\nabla\delta\|_{L^{\infty}(\Omega)}}{\|\delta\|_{L^{\infty}(\Omega)}} = \frac{1}{\max\nolimits_{x\in\Omega}\{{\rm dist}(x, \partial\Omega)\}}. \end{equation} (1.15)

    Define

    \begin{align*} \triangle_{\infty}u_{\infty}: = \sum\limits_{i, j = 1}^{N}(u_{\infty})_{x_{i}}(u_{\infty})_{x_{j}}(u_{\infty})_{x_{i}x_{j}}, \\ k_{\infty}(u): = ||u||_{L^{\infty}(\Omega)} = {\rm{ess}}\sup\limits_{x\in\Omega}|u|, \;\;\;\; \end{align*} (1.16)
    \begin{equation} K_{\infty}(u): = ||\nabla u||_{L^{\infty}(\Omega)} = {\rm{ess}}\sup\limits_{x\in\Omega}|\nabla u|, \end{equation} (1.17)

    and

    \begin{equation*} k_{\infty}(u_{\infty}): = ||u_{\infty}||_{L^{\infty}(\Omega)} = {\rm{ess}}\sup\limits_{x\in\Omega}|u_{\infty}|. \end{equation*}

    The following are the main results of this paper.

    Theorem 1.1. Let u\in C(\Omega) be a weak solution of problem (1.5), then it is also a viscosity solution of the problem (3.2).

    Theorem 1.2. Let hypotheses (H1)–(H3) be satisfied, \lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))} and \Lambda_{\infty} be defined by (1.4) and (1.14), respectively. In addition, assume that u_{n} normalized by \|u_{n}\|_{\mathcal{H}_{n}} = 1 is the positive first eigenfunction, then,

    (1)

    \begin{equation} \lim\limits_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))} = \Lambda_{\infty}; \end{equation} (1.18)

    (2) there exists a nonnegative function u_{\infty} such that u_{\infty}\in C^{\alpha}(\Omega)\setminus\{0\} and ||u_{\infty}||_{L^{\infty}(\Omega)} = 1;

    (3) we can extract a subsequence, which is still denoted by u_{n} , such that

    \begin{equation} u_{n}\rightarrow u_{\infty}, \end{equation} (1.19)

    in the space C^{\alpha}(\Omega) , where \alpha \, (0 < \alpha < 1) is a constant;

    (4) we can obtain that the function u_{\infty}(x) is a nontrivial viscosity solution of the problem

    \begin{equation} \left\{ \begin{array}{ll} \min\bigg\{-\Lambda_{\infty}u_{\infty}+|\nabla u_{\infty}|, -\Lambda_{\infty}(u_{\infty})^{\theta(x_{0})}+|\nabla u_{\infty}|, \\ -\triangle_{\infty}u_{\infty} -[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{2}(x_{0})\bigg\} = 0, & { x\in\Omega }, \\ \qquad\qquad\qquad\qquad u_{\infty} = 0, & { x\in\partial \Omega }. \end{array} \right. \end{equation} (1.20)

    To the best of our knowledge, this is the first work dealing with the double phase eigenvalue problem (1.5). The rest of this paper is organized as follows. In section two, we collect some notations and facts about the Musielak-Orlicz space L^{\mathcal{H}}(\Omega) and W_{0}^{1, \mathcal{H}}(\Omega) , which will be used in this paper. Section three and section four are devoted to prove Theorems 1.1 and 1.2, respectively.

    In this section, we recall some known results about the Musielak-Orlicz spaces L^{\mathcal{H}}(\Omega) and W_{0}^{1, \mathcal{H}}(\Omega) . For more detail, please see references [9,14,15,16,17].

    We follow the notation of [9]. Let N(\Omega) denote the set of all generalized N -functions. Let us introduce the nonlinear function \mathcal{H}:\Omega\times[0, +\infty)\rightarrow [0, +\infty) defined as

    \begin{equation*} \mathcal{H}(x, t): = t^{p(x)}+a(x)t^{q(x)}, \quad {\rm for\; all\;} (x, t)\in \Omega\times[0, +\infty), \end{equation*}

    with 1 < p(x) < q(x) and 0\leq a(\cdot)\in L^{1}(\Omega) . It is clear that \mathcal{H}\in N(\Omega) is a locally integrable and generalized N -function. In addition, it fulfills the \Delta_{2} condition, namely,

    \begin{equation*} \mathcal{H}(x, 2t)\leq2^{q^{+}}\mathcal{H}(x, t). \end{equation*}

    Therefore, in correspondence to \mathcal{H} , we define the Musielak-Orlicz space L^{\mathcal{H}}(\Omega) as

    \begin{equation*} L^{\mathcal{H}}(\Omega): = \{u:\Omega\rightarrow \mathbb{R}\; {\rm measurable}:\rho_{\mathcal{H}}(u) < +\infty\}, \end{equation*}

    which can be equipped with the norm

    \begin{equation*} \|u\|_{\mathcal{H}}: = \inf\{\gamma > 0:\rho_{\mathcal{H}}(u/\gamma)\leq1\}, \end{equation*}

    where

    \begin{equation*} \rho_{\mathcal{H}}(u): = \int_{\Omega}\mathcal{H}(x, |u|)dx, \end{equation*}

    which is called \mathcal{H} -modular.

    Similarly, we can define the Musielak-Orlicz Sobolev spaces. The space W^{1, \mathcal{H}}(\Omega) is given by

    W^{1, \mathcal{H}}(\Omega) = \left\{u \in L^{\mathcal{H}}(\Omega)\, {\rm such \, that}\, |\nabla u| \in L^{\mathcal{H}}(\Omega)\right\},

    with the norm

    \begin{equation*} \|u\|_{1, \mathcal{H}}: = \|u\|_{\mathcal{H}}+\|\nabla u\|_{\mathcal{H}}. \end{equation*}

    We denote by W_{0}^{1, \mathcal{H}}(\Omega) the completion of C_{0}^{\infty}(\Omega) in W^{1, \mathcal{H}}(\Omega) . With these norms, the spaces L^{\mathcal{H}}(\Omega) , W^{1, \mathcal{H}}(\Omega) and W_{0}^{1, \mathcal{H}}(\Omega) are separable, reflexive and uniformly convex Banach spaces.

    From Proposition 2.16 (ⅱ) in [18], if

    \frac{q^{+}}{p^{-}} < 1+\frac{1}{N},

    then the following Poincar \acute{e} -type inequality

    \begin{equation*} \|u\|_{\mathcal{H}}\leq C\|\nabla u\|_{\mathcal{H}} \end{equation*}

    holds for all u\in W_{0}^{1, \mathcal{H}}(\Omega) , where C is a positive constant independent of u . Therefore, in this paper, we equip W_{0}^{1, \mathcal{H}}(\Omega) with the equivalent norm \|\nabla u\|_{\mathcal{H}} for all u\in W_{0}^{1, \mathcal{H}}(\Omega) .

    Proposition 2.1. [18] If u\in L^{\mathcal{H}}(\Omega) and \rho_{\mathcal{H}}(u) is the \mathcal{H} -modular, then the following properties hold.

    (1) If\: u\neq0 , then \|u\|_{\mathcal{H}} = \lambda if, and only if, \varrho_{\mathcal{H}}(\frac{u}{\lambda}) = 1;

    (2) \|u\|_{\mathcal{H}} < 1 \ ( = 1; > 1) if, and only if, \varrho_{\mathcal{H}}(u) < 1 \ ( = 1; > 1);

    (3) If \|u\|_{\mathcal{H}} \leq 1 , then \|u\|_{\mathcal{H}}^{q^{+}} \leq \rho_{\mathcal{H}}(u)\leq \|u\|_{\mathcal{H}}^{p^{-}};

    (4) If \|u\|_{\mathcal{H}} \geq 1 , then \|u\|_{\mathcal{H}}^{p^{-}} \leq \rho_{\mathcal{H}}(u)\leq \|u\|_{\mathcal{H}}^{q^{+}};

    (5) \|u\|_{\mathcal{H}}\rightarrow 0 if, and only if, \rho_{\mathcal{H}}(u)\rightarrow 0;

    (6) \|u\|_{\mathcal{H}}\rightarrow 0 if, and only if, \rho_{\mathcal{H}}(u)\rightarrow 0.

    Given u\in C(\Omega)\bigcap W^{1, \mathcal{H}_{n}}_{0}(\Omega) and \phi\in C^{2}(\Omega) . Define

    \begin{align*} \triangle_{p_{n}(x)}\phi &: = {\rm div}(|\nabla \phi|^{p_{n}(x)-2}\nabla\phi)\\ & = |\nabla \phi|^{p_{n}(x)-4}\{|\nabla \phi|^{2}\triangle \phi+(p_{n}(x)-2)\triangle_{\infty}\phi+|\nabla \phi|^{2}{\rm ln}(|\nabla \phi|)\nabla\phi\cdot\nabla p_{n}\}, \end{align*}
    \begin{align*} \triangle_{q_{n}(x)}\phi &: = {\rm div}(|\nabla \phi|^{q_{n}(x)-2}\nabla\phi)\\ & = |\nabla \phi|^{q_{n}(x)-4}\{|\nabla \phi|^{2}\triangle \phi+(q_{n}(x)-2)\triangle_{\infty}\phi+|\nabla \phi|^{2}{\rm ln}(|\nabla \phi|)\nabla\phi\cdot\nabla q_{n}\}, \end{align*}

    and

    \begin{equation*} \triangle_{\infty}\phi: = \sum\limits_{i, j = 1}^{N}\frac{\partial\phi}{\partial x_{i}}\frac{\partial\phi}{\partial x_{j}}\frac{\partial^{2}\phi}{\partial x_{i}\partial x_{j}}, \end{equation*}

    where \triangle_{\infty}\phi is the \infty -Laplacian.

    Here, we are now in a position to give the following definition of weak solutions to problem (1.5).

    Definition 3.1. We call u\in W^{1, \mathcal{H}_{n}}_{0}(\Omega)\backslash\{0\} a weak solution of problem (1.5) if

    \begin{align} &\int_{\Omega}\left(p_{n}(x)\left|\frac{\nabla u}{K(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{\nabla u}{K(u)}\right|^{q_{n}(x)-2}\right)\frac{\nabla u\cdot\nabla v}{K(u)} dx \\ = &\lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} S(u) \int_{\Omega}\left(p_{n}(x)\left|\frac{u}{k(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{u}{k(u)}\right|^{q_{n}(x)-2}\right)\frac{uv}{k(u)}dx \end{align} (3.1)

    is satisfied for all test functions v \in W^{1, \mathcal{H}_{n}}_{0}(\Omega) . If u\neq 0 , we say that \lambda_{(p_{n}(\cdot), q_{n}(\cdot))} is an eigenvalue of (1.5) and that u is an eigenfunction corresponding to \lambda_{(p_{n}(\cdot), q_{n}(\cdot))} .

    In (1.5), we replace u by \phi and keep S_{n} , K_{n} and k_{n} unchanged, then

    \begin{align*} \left\{ \begin{array}{ll} -p_{n}(x)(K(u))^{1-p_{n}(x)}\triangle_{p_{n}(x)}\phi-q_{n}(x)a(x)(K(u))^{1-q_{n}(x)}\triangle_{q_{n}(x)}\phi\\ -q_{n}(x)(K(u))^{1-q_{n}(x)}|\nabla \phi(x)|^{q_{n}(x)-2}\nabla\phi(x)\cdot\nabla a(x) \\ -(K(u))^{1-p_{n}(x)}|\nabla \phi|^{p_{n}(x)-2}\nabla\phi(x)\cdot\nabla p_{n}(x)\\ -a(x)(K(u))^{1-q_{n}(x)}|\nabla \phi(x)|^{q_{n}(x)-2}\nabla\phi(x)\cdot\nabla q_{n}(x)\\ +p_{n}(x)(K(u))^{1-p_{n}(x)}{\rm ln}(K(u))|\nabla \phi(x)|^{p_{n}(x)-2}\nabla\phi(x)\cdot\nabla p_{n}(x)\\ +q_{n}(x)a(x)(K(u))^{1-q_{n}(x)}{\rm ln}(K(u))|\nabla \phi(x)|^{q_{n}(x)-2}\nabla\phi(x)\cdot\nabla q_{n}(x)\\ -\lambda_{(p_{n}(\cdot), q_{n}(\cdot))}S(u)(p_{n}(x)(k(u))^{1-p_{n}(x)}|\phi|^{p_{n}(x)-2}\phi\\ +q_{n}(x)a(x)(k(u))^{1-q_{n}(x)}|\phi(x)|^{q_{n}(x)-2}\phi(x)) = 0, & { x\in\Omega }, \\ \qquad\qquad\qquad\qquad\qquad\qquad\phi = 0, & { x\in\partial \Omega }. \end{array} \right. \end{align*}

    We first recall the definition of viscosity solutions. Assume we are given a continuous function

    \begin{equation*} F: \mathbb{R}^N \times \mathbb{R} \times \mathbb{R}^N \times \mathcal{S}(N) \rightarrow \mathbb{R}, \end{equation*}

    where \mathcal{S}(N) denotes the set of N \times N symmetric matrices.

    Consider the problem

    \begin{equation} F(x, u, \nabla u, D^{2}u) = 0, \end{equation} (3.2)

    where

    \begin{align} F(x, u, \nabla u, D^{2}u) = & -p_{n}(x)(K(u))^{1-p_{n}(x)}\{|\nabla u|^{p_{n}(x)-4}[|\nabla u|^{2}\triangle u\\ &+(p_{n}(x)-2)\triangle_{\infty}u +|\nabla u|^{2}{\rm ln}(|\nabla u|)\nabla u\cdot\nabla p_{n}(x)]\}\\ &-q_{n}(x)a(x)(K(u))^{1-q_{n}(x)}\{|\nabla u|^{q_{n}(x)-4}[|\nabla u|^{2}\triangle u\\ &+(q_{n}(x)-2)\triangle_{\infty}u +|\nabla u|^{2}{\rm ln}(|\nabla u|)\nabla u\cdot\nabla q_{n}(x)]\}\\ &-q_{n}(x)(K(u))^{1-q_{n}(x)}|\nabla u|^{q_{n}(x)-2}\nabla u\cdot\nabla a(x) -(K(u))^{1-p_{n}(x)}|\nabla u|^{p_{n}(x)-2}\nabla u\cdot\nabla p_{n}(x)\\ &-a(x)(K(u))^{1-q_{n}(x)}|\nabla u|^{q_{n}(x)-2}\nabla u\cdot\nabla q_{n}(x)\\ &+p_{n}(x)(K(u))^{1-p_{n}(x)}{\rm ln}(K(u))|\nabla u|^{p_{n}(x)-2}\nabla u\cdot\nabla p_{n}(x)\\ &+q_{n}(x)a(x)(K(u))^{1-q_{n}(x)}{\rm ln}(K(u))|\nabla u|^{q_{n}(x)-2}\nabla u\cdot\nabla q_{n}(x)\\ &-\lambda_{(p_{n}(\cdot), q_{n}(\cdot))}S(u)(p_{n}(x)(k(u))^{1-p_{n}(x)}|u|^{p_{n}(x)-2}u\\ &+q_{n}(x)a(x)(k(u))^{1-q_{n}(x)}|u|^{q_{n}(x)-2}u). \end{align} (3.3)

    Definition 3.2. Assume that x_{0}\in \Omega , u\in C(\Omega) , \psi\in C^{2}(\Omega) and \varphi\in C^{2}(\Omega) .

    (1) Let u(x_{0}) = \psi(x_{0}) and suppose that u-\psi attains its strict maximum value at x_{0} . If

    \begin{equation*} F(x_{0}, \psi(x_{0}), \nabla \psi(x_{0}), D^{2}\psi(x_{0}))\leq0 \end{equation*}

    for all of such x_{0} , then the function u is said to be a viscosity subsolution of Eq (3.2).

    (2) Let u(x_{0}) = \varphi(x_{0}) and suppose that u-\varphi attains its strict minimum value at x_{0} . If

    \begin{equation*} F(x_{0}, \varphi(x_{0}), \nabla \varphi(x_{0}), D^{2}\varphi(x_{0}))\geq0 \end{equation*}

    for all of such x_{0} , then the function u is said to be a viscosity supersolution of Eq (3.2).

    (3) If u is both a subsolution and a supersolution of the problem (3.2), then u is a viscosity solution of the problem (3.2).

    Proof of Theorem 1.1. Claim: u is a viscosity supersolution of (3.2).

    Let x_{0}\in \Omega and \varphi\in C^{2}(\Omega) . Assume that u(x_{0}) = \varphi(x_{0}) and the function u-\varphi obtains its strict minimum value at the point x_{0} . Our goal is to show that

    \begin{equation} F(x_{0}, u(x_{0}), \nabla \varphi(x_{0}), D^{2}\varphi(x_{0}))\geq0. \end{equation} (3.4)

    If

    \begin{equation*} F(x_{0}, u(x_{0}), \nabla \varphi(x_{0}), D^{2}\varphi(x_{0})) < 0, \end{equation*}

    then by continuity there exists a positive constant r such that B(x_{0}, 2r)\subset\Omega , u > \varphi in this ball, except for the point x_{0} and

    \begin{equation*} F(x, u(x), \nabla \varphi(x), D^{2}\varphi(x)) < 0, \end{equation*}

    for all x\in B(x_{0}, 2r) . Thus, if x\in B(x_{0}, r) , we have

    \begin{equation} \begin{array}{ll} - \text{div} \left[\left(p_{n}(x)\left|\frac{\nabla \varphi(x)}{K_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{\nabla \varphi(x)}{K_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{\nabla \varphi(x)}{K_{n}(u)}\right]\nonumber\\ -\lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} S_{n}(u) \left(p_{n}(x)\left|\frac{u(x)}{k_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{u(x)}{k_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{u(x)}{k_{n}(u)} < 0. \end{array} \end{equation}

    If x\in \partial B(x_{0}, r) , the minimum value of the function u-\varphi is defined as m . Let \Phi(x): = \varphi(x)+\frac{m}{2} . Note that m > 0 and the above inequality still holds if the function \varphi(x) is replaced by \Phi(x) , namely,

    \begin{equation} \begin{array}{ll} - \text{div} \left[\left(p_{n}(x)\left|\frac{\nabla \Phi(x)}{K_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{\nabla \Phi(x)}{K_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{\nabla \Phi(x)}{K_{n}(u)}\right]\\ -\lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} S_{n}(u) \left(p_{n}(x)\left|\frac{u(x)}{k_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{u(x)}{k_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{u(x)}{k_{n}(u)} < 0. \end{array} \end{equation} (3.5)

    Define \eta(x): = (\Phi-u)^{+}\geq0 , then if x\in \partial B(x_{0}, r) , we have \eta(x)\equiv0 .

    Let

    \Omega_{1} = \{x|x\in B(x_{0}, r)\, \text{and}\: \Phi(x) > u(x)\}.

    We multiply (3.5) by the function \eta(x) and integrate over B(x_{0}, r) , then the inequality

    \begin{align} &\int_{\Omega_{1}}\left(p_{n}(x)\left|\frac{\nabla \Phi}{K_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{\nabla \Phi}{K_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{\nabla \Phi}{K_{n}(u)}\cdot\nabla(\Phi-u)dx\\ &-\int_{\Omega_{1}}\lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} S_{n}(u) \left(p_{n}(x)\left|\frac{u}{k_{n}(u)}\right|^{p_{n}(x)-2} +q_{n}(x)a(x)\left|\frac{u}{k_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{u}{k_{n}(u)}(\Phi-u)dx < 0 \end{align} (3.6)

    is true.

    If we define

    \begin{equation*} \eta_{1}(x) = \left\{ \begin{array}{ll} (\Phi-u)^{+}, & { x\in B(x_{0}, r) }, \\ 0, & { x\in \Omega \setminus B(x_{0}, r) }, \end{array} \right. \end{equation*}

    and use \eta_{1}(x) as a test function in (3.1), then we get

    \begin{align} &\int_{\Omega_{1}}\left(p_{n}(x)\left|\frac{\nabla u}{K_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{\nabla u}{K_{n}(u)}\right|^{q_{n}(x)-2}\right)\frac{\nabla u}{K_{n}(u)} \cdot\nabla(\Phi-u)dx\\ &-\int_{\Omega_{1}}\lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} S_{n}(u) \left(p_{n}(x)\left|\frac{u}{k_{n}(u)}\right|^{p_{n}(x)-2}+q_{n}(x)a(x)\left|\frac{u}{k_{n}(u)}\right|^{q_{n}(x)-2}\right) \frac{u}{k_{n}(u)}(\Phi-u)dx = 0. \end{align} (3.7)

    Subtracting (3.7) from (3.6), we arrive at

    \begin{align} &\int_{\Omega_{1}}p_{n}(x)\left(\left|\frac{\nabla \Phi}{K_{n}(u)}\right|^{p_{n}(x)-2}\frac{\nabla \Phi}{K_{n}(u)}-\left|\frac{\nabla u}{K_{n}(u)}\right|^{p_{n}(x)-2}\frac{\nabla u}{K_{n}(u)}\right)\cdot\nabla(\Phi-u)dx\\ &+\int_{\Omega_{1}}q_{n}(x)a(x)\left(\left|\frac{\nabla \Phi}{K_{n}(u)}\right|^{q_{n}(x)-2}\frac{\nabla \Phi}{K_{n}(u)}-\left|\frac{\nabla u}{K_{n}(u)}\right|^{q_{n}(x)-2}\frac{\nabla u}{K_{n}(u)}\right) \nabla(\Phi-u)dx < 0. \end{align} (3.8)

    The first integral is nonnegative due to the elementary inequality

    \begin{equation} \langle|a|^{p-2}a-|b|^{p-2}b, a-b\rangle\geq0, \end{equation} (3.9)

    which holds for all p > 1 . Here, we take p = p_{n}(x) . We get a contradiction. Hence, (3.4) holds. Similarly, we conclude that u is a viscosity subsolution of (3.2) and we omit the details.

    Let n\in N be large enough such that p_{n}\geq r > N , which results in W_{0}^{1, \mathcal{H}_{n}}(\Omega)\hookrightarrow W_{0}^{1, r}(\Omega) (see Proposition 2.16 (1) of Blanco, Gasiński, Harjulehto and Winkert [18]). It follows that u_{n} are continuous functions. The reason is that the space W_{0}^{1, r}(\Omega)\hookrightarrow\hookrightarrow C^{\alpha}(\Omega) , 0 < \alpha < 1 . Moreover, it is known (see [9]) that for each n\in \mathbb{N} fixed, we have u_{n} > 0 .

    In order to prove Theorem 1.2, we only need to prove the following conclusions.

    Lemma 4.1. Let h:\overline{\Omega}\rightarrow (1, \infty) be a given continuous function, then

    \begin{equation} ||\nabla v|^{s}|^{\frac{1}{s}}_{\frac{p(x)}{s}}\leq ||\nabla v||_{\mathcal{H}}, \end{equation} (4.1)

    for all v \in W_{0}^{1, \mathcal{H}}(\Omega) and s\in (1, p^{-}) .

    Proof. Since \left\|\frac{\nabla v}{||\nabla v||_{\mathcal{H}}}\right\|_{\mathcal{H}} = 1 , it follows from Proposition 2.1 that

    \begin{equation} \int_{\Omega}\left[\left(\frac{|\nabla v|}{||\nabla v||_{\mathcal{H}}}\right)^{p(x)}+a(x)\left(\frac{|\nabla v|}{||\nabla v||_{\mathcal{H}}}\right)^{q(x)}\right]dx = 1. \end{equation} (4.2)

    Thus,

    \begin{equation} \int_{\Omega}\left[\left(\frac{|\nabla v|}{||\nabla v||_{\mathcal{H}}}\right)^{s}\right]^{\frac{p(x)}{s}}\frac{dx}{\frac{p(x)}{s}}\leq1. \end{equation} (4.3)

    Invoking Proposition 2.1 again, we conclude that

    \begin{equation*} \left|\left(\frac{|\nabla v|}{||\nabla v||_{\mathcal{H}}}\right)^{s}\right|_{\frac{p(x)}{s}}\leq1, \end{equation*}

    which implies (4.1).

    Lemma 4.2. If u\in L^{\infty}(\Omega) , then we have

    \begin{equation} \lim\limits_{n\rightarrow \infty}k_{n}(u) = k_{\infty}(u). \end{equation} (4.4)

    Proof. Step1: To show that the following inequality holds,

    \begin{equation} \limsup\limits_{n\rightarrow \infty}k_{n}(u)\leq k_{\infty}(u). \end{equation} (4.5)

    If k_{n}(u)\leq k_{\infty}(u) , the above inequality is true. Thus, we can assume that k_{n}(u) > k_{\infty}(u) , and since q_{n}(x) > p_{n}(x) > 1 , we have

    \begin{align*} 1& = \left(\int_{\Omega}\left|\frac{u}{k_{n}(u)}\right|^{p_{n}(x)}+a(x)\left|\frac{u}{k_{n}(u)}\right|^{q_{n}(x)}dx\right)^{\frac{1}{p^{-}_{n}}}\nonumber\\ &\leq\left[\int_{\Omega}\left(\frac{k_{\infty}(u)}{k_{n}(u)}\right)^{p_{n}(x)}+a(x)\left(\frac{k_{\infty}(u)}{k_{n}(u)}\right)^{q_{n}(x)}dx\right]^{\frac{1}{p^{-}_{n}}}\nonumber\\ &\leq\left[\int_{\Omega}\left(\frac{k_{\infty}(u)}{k_{n}(u)}\right)^{p^{-}_{n}}+a(x)\left(\frac{k_{\infty}(u)}{k_{n}(u)}\right)^{p^{-}_{n}}dx\right]^{\frac{1}{p^{-}_{n}}}\nonumber\\ & = \frac{k_{\infty}(u)}{k_{n}(u)}\left(|\Omega|+\int_{\Omega}a(x)dx\right)^{\frac{1}{p^{-}_{n}}}, \end{align*}

    which implies (4.5) holds.

    Step2: To show that the following inequality holds,

    \begin{equation} \liminf\limits_{n\rightarrow \infty}k_{n}(u)\geq k_{\infty}(u). \end{equation} (4.6)

    Case1: k_{\infty}(u) = 0 . It is easy to find that (4.6) holds.

    Case2: k_{\infty}(u) > 0 . Given \varepsilon > 0 , there exists a nonempty set \Omega_{\varepsilon}\subset\Omega such that, for all x\in \Omega_{\varepsilon} , |u| > k_{\infty}(u)-\varepsilon . Ignoring those indices n that k_{n}(u)\geq k_{\infty}(u)-\varepsilon , we have

    \begin{align*} 1& = \left(\int_{\Omega}\left|\frac{u}{k_{n}(u)}\right|^{p_{n}(x)}+a(x)\left|\frac{u}{k_{n}(u)}\right|^{q_{n}(x)}dx\right)^{\frac{1}{p^{-}_{n}}}\nonumber\\ &\geq\left(\int_{\Omega_{\varepsilon}}\left|\frac{u}{k_{n}(u)}\right|^{p_{n}(x)} +a(x)\left|\frac{u}{k_{n}(u)}\right|^{q_{n}(x)}dx\right)^{\frac{1}{p^{-}_{n}}}\nonumber\\ &\geq\left(\int_{\Omega_{\varepsilon}}\left|\frac{k_{\infty}(u)-\varepsilon}{k_{n}(u)}\right|^{p_{n}(x)} +a(x)\left|\frac{k_{\infty}(u)-\varepsilon}{k_{n}(u)}\right|^{q_{n}(x)}dx\right)^{\frac{1}{p^{-}_{n}}}\nonumber\\ &\geq\left(\int_{\Omega_{\varepsilon}}\left|\frac{k_{\infty}(u)-\varepsilon}{k_{n}(u)}\right|^{p^{-}_{n}} +a(x)\left|\frac{k_{\infty}(u)-\varepsilon}{k_{n}(u)}\right|^{p^{-}_{n}}dx\right)^{\frac{1}{p^{-}_{n}}}\nonumber\\ & = \frac{k_{\infty}(u)-\varepsilon}{k_{n}(u)}\left(|\Omega_{\varepsilon}|+\int_{\Omega_{\varepsilon}}a(x)dx\right)^{\frac{1}{p^{-}_{n}}}, \end{align*}

    which gives

    \begin{equation*} \liminf\limits_{n\rightarrow \infty}k_{n}(u)\geq k_{\infty}(u)-\varepsilon. \end{equation*}

    The arbitrariness of \varepsilon implies that (4.6) is true. Consequently, (4.4) holds.

    Remark 4.1. If |\nabla u|\in L^{\infty}(\Omega) , we can argue as Lemma 4.2 to obtain that

    \begin{equation} \lim\limits_{n\rightarrow \infty}K_{n}(u) = K_{\infty}(u). \end{equation} (4.7)

    Lemma 4.3. If the assumptions of Theorem 1.2 hold, then

    (1) (1.18) holds;

    (2) there exists a nonnegative function u_{\infty} such that u_{\infty}\in C^{\alpha}(\Omega)\setminus\{0\} and ||u_{\infty}||_{L^{\infty}(\Omega)} = 1;

    (3) we can extract a subsequence, which is still denoted by u_{n} , such that

    \begin{equation*} u_{n}\rightarrow u_{\infty} \end{equation*}

    in the space C^{\alpha}(\Omega) , where \alpha \, (0 < \alpha < 1) is a constant.

    Proof. Assume for simplicity that the following inequality holds

    \begin{equation*} \int_{\Omega}dx = 1. \end{equation*}

    Step 1: To show that,

    \begin{equation} \limsup\limits_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}\leq\Lambda_{\infty}. \end{equation} (4.8)

    Inserting u(x) = \delta(x) into (1.4) gives

    \begin{equation*} \lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}\leq \frac{||\nabla \delta||_{\mathcal{H}_{n}}}{||\delta||_{\mathcal{H}_{n}}}. \end{equation*}

    Note that by Lemma 4.2 and Remark 4.1, we have

    \begin{equation*} \limsup\limits_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}\leq\frac{\|\nabla\delta\|_{L^{\infty}(\Omega)}}{\|\delta\|_{L^{\infty}(\Omega)}} = \Lambda_{\infty}. \end{equation*}

    Step 2: We now claim that u_{\infty}\in W_{0}^{1, \infty}(\Omega) .

    Since (4.8) holds, for all n\in N sufficiently large, we can assume that \lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}\leq\Lambda_{\infty}+1 . Thus, we have

    \begin{equation*} \Lambda_{\infty}+1\geq\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))} = \frac{||\nabla u_{n}||_{\mathcal{H}_{n}}}{|| u_{n}||_{\mathcal{H}_{n}}} = ||\nabla u_{n}||_{\mathcal{H}_{n}}. \end{equation*}

    Note that the sequence \{||\nabla u_{n}||_{\mathcal{H}_{n}}\} is bounded.

    Let r\in[1, \infty) be arbitrary. We can find an integer n_{r} , for all n\geq n_{r} , such that p_{n}(\cdot)\geq r and

    \begin{equation*} W_{0}^{1, \mathcal{H}_{n}}(\Omega)\hookrightarrow W_{0}^{1, r}(\Omega)\hookrightarrow\hookrightarrow L^{r}(\Omega). \end{equation*}

    Hence, the sequence \{u_{n}\} is bounded in the reflexive Banach space W_{0}^{1, r}(\Omega) . We can find a subsequence, still defined by \{u_{n}\} , and a function u_{\infty}\in W_{0}^{1, r}(\Omega) , such that \nabla u_{n}\rightharpoonup\nabla u_{\infty} in W_{0}^{1, r}(\Omega) and u_{n}\rightarrow u_{\infty} in L^{r}(\Omega) .

    Define

    s_{n}(x): = \frac{p_{n}(x)}{p_{n}(x)-r}, x\in \Omega,

    and it follows that

    s_{n}^{+} = \frac{p_{n}^{-}}{p_{n}^{-}-r}, s_{n}^{-} = \frac{p_{n}^{+}}{p_{n}^{+}-r}

    and

    \begin{equation} |1|_{s_{n}(x)}\leq\max\{|\Omega|^{\frac{1}{s_{n}^{+}}}, |\Omega|^{\frac{1}{s_{n}^{-}}}\}. \end{equation} (4.9)

    Using Hölder's inequality and the above inequality, we have

    \begin{align} \left(\int_{\Omega}|\nabla u_{n}|^{r}dx\right)^{\frac{1}{r}} &\leq \left(\frac{1}{s_{n}^{-}}+\frac{r}{p_{n}^{-}}\right)|1|^{\frac{1}{r}}_{s_{n}(x)}||\nabla u_{n}|^{r}|^{\frac{1}{r}}_{\frac{p_{n}(x)}{r}}\\ &\leq \left(\frac{1}{s_{n}^{-}}+\frac{r}{p_{n}^{-}}\right)\max\{|\Omega|^{\frac{1}{s_{n}^{+}}}, |\Omega|^{\frac{1}{s_{n}^{-}}}\}^{\frac{1}{r}}||\nabla u_{n}|^{r}|^{\frac{1}{r}}_{\frac{p_{n}(x)}{r}}. \end{align} (4.10)

    Thus, (4.1) and (4.10) ensure that

    \begin{equation} \|\nabla u_{n}\|_{L^{r}(\Omega)}\leq 2(1+|\Omega|)||\nabla u_{n}||_{\mathcal{H}_{n}}\leq 2(1+|\Omega|)\Lambda_{\infty}+1. \end{equation} (4.11)

    We choose an arbitrary positive real number r_{1} such that B(x, r_{1})\subset\Omega , where the point x\in\Omega is a Lebesgue point such that |\nabla u_{\infty}|\in L^{1}(\Omega) , then we find that

    \begin{align} \frac{1}{|B(x, r_{1})|}\int_{B(x, r_{1})}|\nabla u_{\infty}(y)|dy &\leq\liminf\limits_{n\rightarrow \infty}\frac{1}{|B(x, r_{1})|}\int_{B(x, r_{1})}|\nabla u_{n}(y)|dy\\ &\leq\liminf\limits_{n\rightarrow \infty}|B(x, r_{1})|^{-\frac{1}{r}}||\nabla u_{n}||_{L^{r}(\Omega)}\\ &\leq|B(x, r)|^{-\frac{1}{r}}2(1+|\Omega|)(\Lambda_{\infty}+1). \end{align} (4.12)

    Passing to the limit as r\rightarrow +\infty in the above inequality, gives

    \begin{equation*} \frac{1}{|B(x, r_{1})|}\int_{B(x, r_{1})}|\nabla u_{\infty}(y)|dy\leq2(1+|\Omega|)(\Lambda_{\infty}+1). \end{equation*}

    Letting r_{1}\rightarrow0^{+} in the above inequality, gives

    \begin{equation*} |\nabla u_{\infty}(x)|\leq 2(1+|\Omega|)(\Lambda_{\infty}+1), \end{equation*}

    for a.e. x\in\Omega , which implies that \nabla u_{\infty}\in L^{\infty}(\Omega) , as claimed.

    Step 3: We want to prove that u_{n}\rightarrow u_{\infty} in C^{\alpha}(\Omega) ( 0 < \alpha < 1 ) and \|u_{\infty}\|_{L^{\infty}(\Omega)} = 1 .

    Keeping in mind that r\in[1, \infty) is an arbitrary constant, we can assume that r > N . Therefore, this combined with the fact that W^{1, r}_{0}(\Omega)\hookrightarrow\hookrightarrow C^{\alpha}(\Omega)\, (0 < \alpha < 1) implies that there exists a nonnegative function u_{\infty}\in C^{\alpha}(\Omega)\setminus\{0\} , such that u_{n}\rightarrow u_{\infty} in C^{\alpha}(\Omega) and u_{n} converges uniformly to u_{\infty} in \Omega . Given \varepsilon\in(0, 1) , we can find a constant N_{\varepsilon}\in \mathbb{N} such that

    \begin{equation} |u_{n}(x)-u_{\infty}(x)| < \varepsilon, \end{equation} (4.13)

    for all x\in\Omega, n\geq N_{\varepsilon} . It follows that

    \begin{align} [\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{p^{-}_{n}}} & = \left[\int_{\Omega}|u_{n}-u_{\infty}|^{p_{n}(x)}+a(x)|u_{n}-u_{\infty}|^{q_{n}(x)}dx\right]^{\frac{1}{p^{-}_{n}}}\\ &\leq\left[\int_{\Omega}\varepsilon^{p_{n}(x)}+a(x)\varepsilon^{q_{n}(x)}dx\right]^{\frac{1}{p^{-}_{n}}}\\ &\leq\varepsilon\left[\int_{\Omega}(1+a(x))dx\right]^{\frac{1}{p^{-}_{n}}}\\ &\leq\left[\int_{\Omega}(1+a(x))dx\right]^{\frac{1}{p^{-}_{n}}} \end{align} (4.14)

    and

    \begin{align} [\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{q^{+}_{n}}} &\leq\varepsilon^{\frac{p^{-}_{n}}{q^{+}_{n}}}\left[\int_{\Omega}(1+a(x))dx\right]^{\frac{1}{q^{+}_{n}}}\\ &\leq\left[\int_{\Omega}(1+a(x))dx\right]^{\frac{1}{q^{+}_{n}}}, \end{align} (4.15)

    for all n\geq N_{\varepsilon} . Letting n\rightarrow \infty in (4.14) and (4.15) yields

    \begin{equation} \lim\limits_{n\rightarrow \infty}[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{p^{-}_{n}}} = \lim\limits_{n\rightarrow \infty}[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{q^{+}_{n}}} = 0. \end{equation} (4.16)

    Thus, the inequality

    \begin{align*} |||u_{n}||_{\mathcal{H}_{n}}-\|u_{\infty}\|_{L^{\infty}(\Omega)}| &\leq|||u_{n}||_{\mathcal{H}_{n}}-||u_{\infty}||_{\mathcal{H}_{n}}|+|||u_{\infty}||_{\mathcal{H}_{n}} -\|u_{\infty}\|_{L^{\infty}(\Omega)}|\nonumber\\ &\leq||u_{n}-u_{\infty}||_{\mathcal{H}_{n}}+|||u_{\infty}||_{\mathcal{H}_{n}} -\|u_{\infty}\|_{L^{\infty}(\Omega)}|\nonumber\\ &\leq\left\{[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{p^{-}_{n}}} +[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{q^{+}_{n}}}\right\} +|||u_{\infty}||_{\mathcal{H}_{n}}-\|u_{\infty}\|_{L^{\infty}(\Omega)}|.\nonumber\\ &\leq\left\{[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{p^{-}_{n}}} +[\rho_{\mathcal{H}_{n}}(u_{n}-u_{\infty})]^{\frac{1}{q^{+}_{n}}}\right\} +|k_{n}(u_{\infty})-\|u_{\infty}\|_{L^{\infty}(\Omega)}| \end{align*}

    holds. In view of Lemma 4.2 and (4.16), we can get

    \begin{equation} \|u_{\infty}\|_{L^{\infty}(\Omega)} = \lim\limits_{n\rightarrow \infty}||u_{n}||_{\mathcal{H}_{n}} = 1. \end{equation} (4.17)

    Step 4: To show that \liminf_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}\geq\Lambda_{\infty} .

    Since \nabla u_{n}\rightharpoonup\nabla u_{\infty} in W_{0}^{1, r}(\Omega) , ||u_{n}||_{\mathcal{H}_{n}} = 1 and the inequality (4.11) holds, we have

    \begin{align*} \|\nabla u_{\infty}\|_{L^{r}(\Omega)} \leq\liminf\limits_{n\rightarrow \infty}\|\nabla u_{n}\|_{L^{r}(\Omega)} \leq\liminf\limits_{n\rightarrow \infty}\|\nabla u_{n}\|_{\mathcal{H}_{n}} = \liminf\limits_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}. \end{align*}

    Letting r\rightarrow \infty and using Proposition 7 in [19] and equality (4.17), we get

    \begin{equation} \Lambda_{\infty}\leq\frac{\|\nabla u_{\infty}\|_{L^{\infty}(\Omega)}}{\|u_{\infty}\|_{L^{\infty}(\Omega)}}\leq\liminf\limits_{n\rightarrow \infty}\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}. \end{equation} (4.18)

    Thus, (4.8) and (4.18) imply that (1.18) holds. The proof is complete.

    Remark 4.2. We can again argue with Step 3 to obtain

    \begin{equation} \|\nabla u_{\infty}\|_{L^{\infty}(\Omega)} = \lim\limits_{n\rightarrow \infty}||\nabla u_{n}||_{\mathcal{H}_{n}}. \end{equation} (4.19)

    The function u_{\infty}(x) also has the following property.

    Lemma 4.4. If the assumptions of Theorem 1.2 hold, we can deduce that u_{\infty}(x) is a nontrivial viscosity solution of the problem (1.20).

    Proof. For the first part we only need to show that u_{\infty} is a viscosity subsolution of (1.20). Let x_{0}\in \Omega and \psi\in C^{2}(\Omega) . Assume that u_{\infty}-\psi attains its strict maximum value of zero at x_{0} , namely, u_{\infty}(x_{0})-\psi(x_{0}) = 0 .

    Claim: We want to show that

    \begin{align} &\max\bigg\{\Lambda_{\infty}\psi(x_{0})-|\nabla \psi(x_{0})|, (\psi(x_{0}))^{\theta(x_{0})}K_{\infty}(u_{\infty})-|\nabla \psi(x_{0})|, \\ &\triangle_{\infty}\psi(x_{0})+[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{2}(x_{0})\bigg\}\leq0. \end{align} (4.20)

    By Lemma 4.3, we know that the convergence of u_{n} to u_{\infty} in \Omega is uniform. Therefore, there exists a sequence \{x_{n}\}\subset \Omega such that x_{n}\rightarrow x_{0} (as n\rightarrow \infty ), u_{n}(x_{n}) = \psi(x_{n}) and u_{n}-\psi attains its strict maximum value at x_{n} .

    Employing Theorem 1.1, it turns out that for any n\in \mathbb{N} large enough, u_{n} are continuous viscosity solutions of (1.5) with \lambda_{(p_{n}(\cdot), \, q_{n}(\cdot))} = \lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))} . Thus, we have

    \begin{align} &-p_{n}(x_{n})(K_{n}(u_{n}))^{1-p_{n}(x_{n})}|\nabla \psi(x_{n})|^{p_{n}(x_{n})-4}\{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})+(p_{n}(x_{n})-2)\triangle_{\infty}\psi(x_{n})\\ &+[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\nabla p_{n}(x_{n})\}\\ &-q_{n}(x_{n})a(x_{n})(K_{n}(u_{n}))^{1-q_{n}(x_{n})}|\nabla \psi(x_{n})|^{q_{n}(x_{n})-4}\{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})+(q_{n}(x_{n})-2)\triangle_{\infty}\psi(x_{n})\\ &+[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\nabla q_{n}(x_{n})\}\\ &-q_{n}(x_{n})(K_{n}(u_{n}))^{1-q_{n}(x_{n})}|\nabla \psi(x_{n})|^{q_{n}(x_{n})-2}\nabla \psi(x_{n})\cdot\nabla a(x_{n}) \\ &-(K_{n}(u_{n}))^{1-p_{n}(x_{n})}|\nabla \psi(x_{n})|^{p_{n}(x_{n})-2}\nabla\psi(x_{n})\cdot\nabla p_{n}(x_{n})\\ &-a(x_{n})(K_{n}(u_{n}))^{1-q_{n}(x_{n})}|\nabla \psi(x_{n})|^{q_{n}(x_{n})-2}\nabla\psi(x_{n})\cdot\nabla q_{n}(x_{n})\\ &-\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}S_{n}(u_{n})p_{n}(x_{n})(k_{n}(u_{n}))^{1-p_{n}(x_{n})}|\psi(x_{n})|^{p_{n}(x_{n})-2}\psi(x_{n})\\ &-\lambda^{1}_{(p_{n}(\cdot), \, q_{n}(\cdot))}S_{n}(u_{n})q_{n}(x_{n})a(x_{n})(k_{n}(u_{n}))^{1-q_{n}(x_{n})}|\psi(x_{n})|^{q_{n}(x_{n})-2}\psi(x_{n})\geq0. \end{align} (4.21)

    Case 1: \psi(x_{0}) = u_{\infty}(x_{0}) > 0 .

    Continuing (4.21), for n\in \mathbb{N} sufficiently large, we have |\nabla \psi(x_{n})| > 0 . Let us assume the assertion is not true, then by (4.21) and continuity, we have \psi(x_{0})\leq0 . This leads to a contradiction.

    Dividing both sides of (4.21) by

    p_{n}(x_{n})(p_{n}(x_{n})-2)(K_{n}(u_{n}))^{1-p_{n}(x_{n})}|\nabla \psi(x_{n})|^{p_{n}(x_{n})-4},

    we see that the following inequality holds

    \begin{align} &\quad-\frac{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})}{p_{n}(x_{n})-2}-\triangle_{\infty}\psi(x_{n})-[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\frac{\nabla p_{n}(x_{n})}{p_{n}(x_{n})-2}\\ &\quad-\frac{q_{n}(x_{n})}{p_{n}(x_{n})}\bigg|\frac{\nabla \psi(x_{n})}{K_{n}(u_{n})}\bigg|^{q_{n}(x_{n})-p_{n}(x_{n})} \bigg\{a(x_{n})\frac{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})}{p_{n}(x_{n})-2}+a(x_{n})\bigg(\frac{q_{n}(x_{n})-2}{p_{n}(x_{n})-2}\bigg)\triangle_{\infty}\psi(x_{n})\\ &\quad+\frac{|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\nabla a(x_{n})}{p_{n}(x_{n})-2}+a(x_{n})\frac{|\nabla \psi(x_{n})|^{2}}{p_{n}(x_{n})-2}\nabla \psi(x_{n})\cdot\frac{\nabla q_{n}(x_{n})}{q_{n}(x_{n})}\\ &\quad+a(x_{n})[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\frac{\nabla q_{n}(x_{n})}{p_{n}(x_{n})-2}\bigg\}-\frac{|\nabla \psi(x_{n})|^{2}}{p_{n}(x_{n})}\frac{\nabla \psi(x_{n})\cdot\nabla p_{n}(x_{n})}{p_{n}(x_{n})-2}\\ &\geq\bigg(\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\bigg)^{3}S_{n}(u_{n})\bigg|\frac{\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\psi(x_{n})}{\nabla \psi(x_{n})}\bigg|^{p_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2}\\ &\quad+\bigg(\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\bigg)^{3}S_{n}(u_{n})\frac{q_{n}(x_{n})}{p_{n}(x_{n})}a(x_{n})\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2} \bigg[\bigg(\frac{|\psi(x_{n})|}{k_{n}(u_{n})}\bigg)^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{p_{n}(x_{n})-4}\\ &\geq0. \end{align} (4.22)

    Now, letting n\rightarrow \infty , we deduce that

    \begin{align*} &-\frac{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})}{p_{n}(x_{n})-2}-\triangle_{\infty}\psi(x_{n})\rightarrow -\triangle_{\infty}\psi(x_{0}), \nonumber\\ &-[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\frac{\nabla p_{n}(x_{n})}{p_{n}(x_{n})-2}\nonumber\\ \rightarrow &-[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{1}(x_{0}), \nonumber\\ &-\frac{q_{n}(x_{n})}{p_{n}(x_{n})}\bigg\{a(x_{n})\frac{|\nabla \psi(x_{n})|^{2}\triangle \psi(x_{n})}{p_{n}(x_{n})-2}+a(x_{n})\bigg(\frac{q_{n}(x_{n})-2}{p_{n}(x_{n})-2}\bigg)\triangle_{\infty}\psi(x_{n}) +\frac{|\psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\nabla a(x_{n})}{p_{n}(x_{n})-2}\nonumber\\ &+a(x_{n})[{\rm ln}(|\nabla \psi(x_{n})|)-{\rm ln}(K_{n}(u_{n}))]|\nabla \psi(x_{n})|^{2}\nabla \psi(x_{n})\cdot\frac{\nabla q_{n}(x_{n})}{p_{n}(x_{n})-2}\nonumber\\ &+a(x_{n})\frac{|\nabla \psi(x_{n})|^{2}}{p_{n}(x_{n})-2}\nabla \psi(x_{n})\cdot\frac{\nabla q_{n}(x_{n})}{q_{n}(x_{n})}\bigg\}\nonumber\\ \rightarrow &-\theta^{2}(x_{0})a(x_{0})\left\{\triangle_{\infty}\psi(x_{0}) +[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{2}(x_{0})\right\}, \nonumber\\ &-\frac{|\nabla \psi(x_{n})|^{2}}{p_{n}(x_{n})}\frac{\nabla \psi(x_{n})\cdot\nabla p_{n}(x_{n})}{p_{n}(x_{n})-2}\rightarrow 0.\nonumber \end{align*}

    Taking the lower limit in inequality (4.22) and employing the limits above, we have

    \begin{align} &-\bigg|\frac{\nabla \psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\limits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\theta^{2}(x_{0})a(x_{0})\\ &\cdot\left\{\triangle_{\infty}\psi(x_{0})+[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{2}(x_{0})\right\}\\ &-\left\{\triangle_{\infty}\psi(x_{0})+[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{1}(x_{0})\right\}\\ = &-\bigg(\bigg|\frac{\nabla \psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\limits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\theta^{2}(x_{0})a(x_{0})+1\bigg)\triangle_{\infty}\psi(x_{0})\\ &-[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0}) \left(\xi_{1}(x_{0})+\theta^{2}(x_{0})a(x_{0})\bigg|\frac{\nabla \psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\limits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\xi_{2}(x_{0})\right)\\ \geq&(\Lambda_{\infty})^{3}\liminf\limits_{n\rightarrow \infty}S_{n}(u_{n})\bigg|\frac{\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\psi(x_{n})}{\nabla \psi(x_{n})}\bigg|^{p_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2}\\ &+(\Lambda_{\infty})^{3}\theta (x_{0})a(x_{0}) \liminf\limits_{n\rightarrow \infty}S_{n}(u_{n})\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2} \bigg[\bigg(\frac{|\psi(x_{n})|}{k_{n}(u_{n})}\bigg)^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{p_{n}(x_{n})-4}\\ \geq&0. \end{align} (4.23)

    Note that by (4.17), (4.19) and u_{\infty}(x_{0}) = \psi(x_{0}) > 0 , we have

    \begin{align} \bigg|\frac{\nabla\psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} & = \bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}k_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\\ &\leq\bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}u_{\infty}(x_{0})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\\ & = \bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}\psi(x_{0})}\bigg|^{\lim\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} \end{align} (4.24)

    and

    \begin{align} \bigg|\frac{\nabla\psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} & = \bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}(k_{\infty}(u_{\infty}))^{\theta(x_{0})}}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\\ &\leq\bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}(u_{\infty}(x_{0}))^{\theta(x_{0})}}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}\\ & = \bigg|\frac{\nabla \psi(x_{0})}{\Lambda_{\infty}(\psi(x_{0}))^{\theta(x_{0})}}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))}. \end{align} (4.25)

    Claim:

    \begin{equation} \Lambda_{\infty}\psi(x_{0})-|\nabla \psi(x_{0})|\leq0. \end{equation} (4.26)

    Assume that \Lambda_{\infty}\psi(x_{0}) > |\nabla \psi(x_{0})| , then (4.24) and (1.11) imply

    \begin{equation} \bigg|\frac{\nabla\psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} = 0 \end{equation} (4.27)

    and

    \begin{equation} \lim\limits_{n\rightarrow \infty}\left|\frac{\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\psi(x_{n})}{\nabla \psi(x_{n})}\right|^{(p_{n}(x_{n})-4)\setminus(q_{n}(x_{n})-4)} = \bigg(\frac{ \Lambda_{\infty}\psi(x_{0})}{|\nabla \psi(x_{0})|}\bigg)^{\frac{1}{\theta(x_{0})}} > 1. \end{equation} (4.28)

    Thus, choosing \varepsilon > 0 small enough, we have

    \begin{equation} \left|\frac{\lambda^{1}_{(p_{n}(\cdot), q_{n}(\cdot))}\psi(x_{n})}{\nabla \psi(x_{n})}\right|^{(p_{n}(x_{n})-4)\setminus(q_{n}(x_{n})-4)}\geq1+\varepsilon, \end{equation} (4.29)

    for all n\in \mathbb{N} sufficiently large. By (4.29), we get

    \begin{align} &\liminf\limits_{n\rightarrow \infty}\left|\frac{\lambda^{1}_{(p_{n}, q_{n})}\psi(x_{n})}{\nabla \psi(x_{n})}\right|^{p_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2}\\ = &\liminf\limits_{n\rightarrow \infty}\frac{\left(\left|\frac{\lambda^{1}_{(p_{n}, q_{n})}\psi(x_{n})}{\nabla \psi(x_{n})}\right|^{(p_{n}(x_{n})-4)\setminus(q_{n}(x_{n})-4)}\right)^{q_{n}(x_{n})-4}}{q_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{\frac{p_{n}(x_{n})-2}{q_{n}(x_{n})-4}}\\ \geq& R\psi(x_{0})^{3}\lim\limits_{n\rightarrow \infty}\frac{(1+\varepsilon)^{q_{n}(x_{n})-4}}{q_{n}(x_{n})-4}\\ = &+\infty. \end{align} (4.30)

    From (4.23), (4.27) and (4.30), we see that

    \begin{equation} -\left\{\triangle_{\infty}\psi(x_{0})+[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{1}(x_{0})\right\}\geq+\infty, \end{equation} (4.31)

    which is a contradiction. Hence, (4.26) holds.

    Claim:

    \begin{equation} (\psi(x_{0}))^{\theta(x_{0})}K_{\infty}(u_{\infty})-|\nabla \psi(x_{0})|\leq0. \end{equation} (4.32)

    Suppose that the above inequality is not true, then we have

    \begin{align*} &\quad\lim\limits_{n\rightarrow \infty}\bigg[\bigg(\frac{\psi(x_{n})}{k_{n}(u_{n})}\bigg)^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{(p_{n}(x_{n})-4)/(q_{n}(x_{n})-4)}\nonumber\\ & = \lim\limits_{n\rightarrow \infty}\bigg[(\psi(x_{n}))^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{(p_{n}(x_{n})-4)/(q_{n}(x_{n})-4)}\nonumber\\ & = \bigg[(\psi(x_{0}))^{\theta(x_{0})}\frac{K_{\infty}(u_{\infty})}{|\nabla \psi(x_{0})|}\bigg]^{\frac{1}{\theta(x_{0})}} > 1. \end{align*}

    Thus, choosing \varepsilon_{1} > 0 small enough, we have

    \begin{equation} \bigg[\bigg(\frac{\psi(x_{n})}{k_{n}(u_{n})}\bigg)^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{(p_{n}(x_{n})-4)/(q_{n}(x_{n})-4)}\geq1+\varepsilon_{1}, \end{equation} (4.33)

    for all n\in \mathbb{N} sufficiently large. We are led to

    \begin{align} &\quad\liminf\limits_{n\rightarrow \infty}\bigg[\bigg(\frac{|\psi(x_{n})|}{k_{n}(u_{n})}\bigg)^{(q_{n}(x_{n})-4)/(p_{n}(x_{n})-4)}\frac{K_{n}(u_{n})}{|\nabla \psi(x_{n})|}\bigg]^{p_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{p_{n}(x_{n})-2}\\ &\geq \liminf\limits_{n\rightarrow \infty}\frac{(1+\varepsilon_{1})^{q_{n}(x_{n})-4}}{q_{n}(x_{n})-4}\frac{|\psi(x_{n})|^{2}\psi(x_{n})}{\frac{p_{n}(x_{n})-2}{q_{n}(x_{n})-4}}\\ & = \theta(x_{0})\psi(x_{0})^{3}\lim\limits_{n\rightarrow \infty}\frac{(1+\varepsilon_{1})^{q_{n}(x_{n})-4}}{q_{n}(x_{n})-4}\\ & = +\infty. \end{align} (4.34)

    In view of (\psi(x_{0}))^{\theta(x_{0})}K_{\infty}(u_{\infty})-|\nabla \psi(x_{0})| > 0 and (4.25),

    \begin{equation*} \bigg|\frac{\nabla\psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} = 0. \end{equation*}

    Therefore, this fact along with (4.23) shows that (4.31) holds. This is a contradiction. Thus we deduce that (4.32) holds.

    Claim:

    \begin{equation} \triangle_{\infty}\psi(x_{0})+[{\rm ln}(|\nabla \psi(x_{0})|)-{\rm ln}(K_{\infty}(u_{\infty}))]|\nabla \psi(x_{0})|^{2}\nabla \psi(x_{0})\cdot\xi_{2}(x_{0})\leq0. \end{equation} (4.35)

    Taking (4.24) and (4.26) into account, we have

    \begin{equation} \bigg|\frac{\nabla\psi(x_{0})}{K_{\infty}(u_{\infty})}\bigg|^{\liminf\nolimits_{n\rightarrow \infty}(q_{n}(x_{n})-p_{n}(x_{n}))} = +\infty. \end{equation} (4.36)

    At the same time, by (4.25) and (4.32), we also deduce that (4.36) holds. If we assume that inequality (4.35) does not hold, then by (4.23) and (4.36), there is a contradiction. Thus, we deduce that (4.35) holds.

    Case 2: \psi(x_{0}) = u_{\infty}(x_{0}) = 0 .

    Note that if |\nabla \psi(x_{0})| = 0 (in this case, we have \triangle_{\infty}\psi(x_{0}) = 0), the inequality (4.20) trivially holds. Hence, let us assume that |\nabla \psi(x_{0})| > 0 , then |\nabla \psi(x_{n})| > 0 for n\in \mathbb{N} large enough. We can use very similar arguments as Case 1 to conclude that (4.20) holds. The same argument can be used in order to show that u_{\infty} is a viscosity supersolution of (1.20).

    By Lemmas 4.3 and 4.4, it follows that Theorem 1.2 holds.

    Remark 4.3. In the particular case where p_{n}(x) = np(x) and q_{n}(x) = nq(x) , Theorems 1.1 and 1.2 are also true.

    In this paper, we studied a double-phase eigenvalue problem with large variable exponents. As we know, for p -Laplace operator eigenvalue problems, there is an important feature that if u is an eigenfunction, so is ku , where k is an arbitrary constant. However, the double-phase operator with variable exponents looses this property. To overcome the above mentioned shortcoming, we defined the eigenvalue by using the Rayleigh quotient of two norms of Musielak-Orlicz space. Moreover, in the particular case where p_{n}(\cdot) = p_{n} and q_{n}(\cdot) = q_{n} , Theorems 1.1 and 1.2 are also true (see [13]).

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    This work was supported by the National Natural Science Foundation of China (No.12001196) and the Natural Science Foundation of Henan (No. 232300421143).

    The authors declare that they have no competing interests.



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