Research article

Infinitely many sign-changing solutions for a semilinear elliptic equation with variable exponent

  • Received: 03 January 2021 Accepted: 18 March 2021 Published: 26 March 2021
  • MSC : 35J20, 35J62, 35Q55

  • This paper is devoted to study a class of semilinear elliptic equations with variable exponent. By means of perturbation technique, variational methods and a priori estimation, the existence of infinitely many sign-changing solutions to this class of problem is obtained.

    Citation: Changmu Chu, Yuxia Xiao, Yanling Xie. Infinitely many sign-changing solutions for a semilinear elliptic equation with variable exponent[J]. AIMS Mathematics, 2021, 6(6): 5720-5736. doi: 10.3934/math.2021337

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  • This paper is devoted to study a class of semilinear elliptic equations with variable exponent. By means of perturbation technique, variational methods and a priori estimation, the existence of infinitely many sign-changing solutions to this class of problem is obtained.



    Let 0ΩRN(N3) be a bounded domain with smooth boundary Ω. In this paper, we are interested in establishing the multiplicity of sign-changing solutions to the following semilinear elliptic equations with variable exponent

    {Δu=|u|q(x)2u,  in  Ω,u=0,  on  Ω, (1.1)

    where q(x) satisfies the following assumptions.

    (Q1)qC(¯Ω), q(0)=2 and 2<q(x)maxx¯Ω{q(x)}=q+<2=2NN2 for x0;

    (Q2) there exist α(0, N+22) and Bδ0={x||x|<δ0}Ω such that q(x)2+|x|α for any xBδ0.

    In 1973, Ambrosetti and Rabinowitz in [2] obtained a positive and a negative solution to the following superlinear elliptic problem

    {Δu=f(x,u),in Ω,u=0,on Ω. (1.2)

    The existence of the third solution to problem (1.2) was established by Wang in [17]. Castro, Cossio and Neuberger in [6] proved that the third solution to problem (1.2) obtained in [17] changes sign only once. Bartsch and Wang in [3] obtained the existence of sign-changing solution. In addition, Bartsch, Weth and Willem in [4] showed that problem (1.2) possesses a least energy sign-changing solution. In order to study the sign-changing critical points of even functionals, Li and Wang in [11] established a Ljusternik-Schnirelmann theory and showed that problem (1.2) possesses infinitely many sign-changing solutions. Subsequently, the existence of infinitely many sign-changing solutions to problem (1.2) was also obtained by some versions of the symmetric mountain pass lemma(see [15] and [19]).

    In fact, these papers required f(x,t) to satisfy the following condition ((AR)-condition, for short)

    f(x,t)tθF(x,t)>0,  for all xΩ and |t| sufficiently large,

    where θ>2 and F(x,t)=t0f(x,s)ds. It is well known that (AR)-condition is important to guarantee the boundedness of Palais-Smale sequence of the Euler-Lagrange functional associated to problem (1.2) which plays a crucial role in applying the critical point theory. For more than 40 years, several researchers studied problem (1.2) trying to drop the above (AR)-condition. For example, a weaker super-quadratic condition ((SQ)-condition, for short) is that

    lim|t|F(x,t)|t|2=  uniformly in  xΩ.

    Under (SQ)-condition or some extra assumptions, the existence and multiplicity of nontrivial solution for problem (1.2) were obtained, see [7,8,12,14,16] and the references therein.

    Recently, the special case of problem (1.1) as problem (1.2) is also concerned by some scholars(see [1,5,9,10,13]). They obtained the existence or multiplicity of the nontrivial solution of problem (1.1) from the discussion the compact embedding from H10(Ω) to Lq(x)(Ω) with a variable critical or supercritical exponent. In particular, Cao, Li and Liu in [5] obtained that problem (1.1) has infinitely many nodal solutions when q(x)=2+|x|α2(0<α<min{N2,N2}) and B1 is the unit ball in RN. In addition, Hashizume and Sano in [9] proposed that essinfxΩ{q(x)}=2 is another critical case. Indeed, if there exists x0Ω such that q(x0)=infxΩ{q(x)}=2, then the conditions (AR) and (SQ) do not hold. Therefore, the problem we intend to study is a new phenomenon. To the best of our knowledge, for either p-Laplacian equation(including semilinear elliptic equation) or p(x)-Laplacian equation, there are no results in this case. The main difficulty with problem (1.1) is that the corresponding functional may possess unbounded Palais-Smale sequences. To overcome this difficulty, we will use the perturbation technique and the Moser's iteration.

    The main result of this paper reads as follows.

    Theorem 1.1. Suppose that (Q1) and (Q2) hold. Then, for every integer k1, problem (1.1) has k sign-changing solutions.

    Remark 1.2. In [5] and [9], it is crucial to require the space is radially symmetric. However, we do not need the domain to be radial.

    To end this section, we describe the basic ideas in the proof of Theorem 1.1. Noticing that q(0)=2, inspired by [18], we first modify the nonlinear term to guarantee the boundedness of Palais-Smale sequence of the corresponding functional and obtain infinitely many sign-changing solutions of auxiliary problem by a version of the symmetric mountain pass lemma. Subsequently, we use the Moser iteration to obtain the existence of infinitely many sign-changing solutions for problem (1.1).

    Throughout this paper, let Bδ={x||x|<δ}Ω and Ωδ=Ω Bδ. We use to denote the usual norms of H10(Ω). The letter C stands for positive constant which may take different values at different places.

    According to q(0)=2, it seems to be difficult to confirm whether the energy function I corresponding to (1.1) satisfies the Palais-Smale condition or not.To apply variational methods, the first step in proving Theorem 1.1 is modifying the nonlinear term to obtain the perturbation equation. Since q(x) is a continuous function and q+<2, we can choose r>0 such that

    r<min{2q+, 14N}. (2.1)

    Let ψ(t)C0(R,[0,1]) be a smooth even function with the following properties: ψ(t)=1 for |t|1, ψ(t)=0 for |t|2 and ψ(t) is monotonically decreasing on the interval (0,+). Define

    bμ(t)=ψ(μt),         mμ(t)=t0bμ(τ)dτ,

    for μ(0,1]. We will deal with the modified problem

    {Δu=(umμ(u))r|u|q(x)2u,  in  Ω,u=0,  on  Ω. (2.2)

    Theorem 2.1. Suppose that (Q1) and (Q2) hold. Then, for any μ(0,1], problem (2.2) has infinitely many sign-changing solutions.

    Let E:=H10(Ω) be the usual Sobolev space endowed with the inner product u,v=Ωuvdx for u, vE and the norm u:=u,u12. Let P be the positive cone of E, and Y,M be two subspaces of E with dimY<, dimYcodimM1. For any δ>0, define ±D(δ):={uE:dist(u,±P)<δ}. Set D:=D(δ)(D(δ)) and S=ED. Let GC1(E,R) and the gradient G be of the form G(u)=uKG(u), where KG:EE is a continuous operator. Let K={uE:G(u)=0} and K[a,b]={uK:G(u)[a,b]}. We assume that there is another norm of E such that uCu for all uE, where C is a positive constant. Moreover, we assume that unu0 whenever unu weakly in (E,). Write E=M1M. Let

    Q(ρ)={uM:upu2+uuu+Du=ρ},

    where ρ>0, D>0 and p>2 are fixed constants. Let us assume that Q=Q(ρ)GβS and γ=infQG, where β=supYG and Gβ={uE:G(u)β}. It is easy to see βγ. In addition, we assume that

    (A)KG(±D(δ))±D(δ);

    (A1) Assume that for any a, b>0, there is a c1=c1(a,b)>0 such that G(u)a and ubuc1;

    (A2)limuY, uG(u)=, supYG:=β.

    Now we recall the following Palais-Smale condition and abstract critical point theorem (see Definition 3.3 and Theorem 5.6 in [19]).

    Definition 2.2. The functional G is said to satisfy the (wPS) condition if for any sequence {un} such that {G(un)} is bounded and G(un)0, we have either {un} is bounded and has a convergent subsequence or G(un)un. In particular, if {G(un)}c, we say that (wPS)c is satisfied.

    Theorem 2.3. Assume that (A), (A1) and (A2) hold. If the even functional G satisfies the (wPS)c condition at level c for each c[γ,β], then K[γε,β+ε](E(P(P)) for all ε>0 small.

    Let 0<λ1<<λk< denote the distinct Dirichlet eigenvalues of the eigenvalue problem

    {Δu=λu,  in  Ω,u=0,  on  Ω.

    Then each λk has finite multiplicity. In addition, the principal eigenvalue λ1 is simple with a positive eigenfunction φ1, and the eigenfunctions φk corresponding to λk(k2) are sign-changing. Let Nk denote the eigenspace of λk. Then dimNk<. We fix k and let Ek:=N1N2Nk.

    The formal energy functional Iμ:H10(Ω)R associated with (2.2) is defined by

    Iμ(u)=12Ω|u|2dxΩFμ(x,u)dx,

    where fμ(x,t)=(tmμ(t))r|t|q(x)2t, Fμ(x,t)=t0fμ(x,τ)dτ. Then IμC1(E,R) and Iμ=id(Δ)1fμ=idKIμ. Obviously, the critical points of Iμ are just the weak solutions of problem (2.2).

    Lemma 2.4. The function Fμ(x,t) defined above satisfies the following inequalities:

    Fμ(x,t)1q(x)tfμ(x,t),       Fμ(x,t)1q(x)+rtfμ(x,t)+Cμ,

    for t>0, where Cμ>0 is a positive constant.

    Proof. Since bμ(t) to is monotonically decreasing on the interval (0,+), we have

    ddt(tmμ(t))=mμ(t)tbμ(t)m2μ(t)=t(bμ(ξ)bμ(t))m2μ(t)0,

    for t>0, where ξ(0,t). Therefore, tmμ(t) is monotonically increasing on the interval (0,+). Hence, fμ(x,t)tq(x)1=(tmμ(t))r is also monotonically increasing on the interval (0,+). It follows that

    Fμ(x,t)=t0fμ(x,τ)dτt0fμ(x,t)tq(x)1τq(x)1dτ=1q(x)tfμ(x,t), (2.3)

    for t>0.

    By definition of the function mμ, we have mμ(t)=Aμ for t2μ, where A=1+21ψ(τ)dτ. For t>2μ, one has

    Fμ(x,t)=2μ0fμ(x,τ)dτ+t2μ(μA)rτq(x)+r1dτ=2μ0(fμ(x,τ)(μA)rτq(x)+r1)dτ+t0(μA)rτq(x)+r1dτCμ+tfμ(x,t)q(x)+r. (2.4)

    It implies from (2.3) and (2.4) that

    Fμ(x,t)1q(x)+rtfμ(x,t)+Cμ

    for t>0.

    Lemma 2.5. Suppose that (Q1) and (Q2) hold. Then, for any μ(0,1], Iμ satisfies the (PS) condition.

    Proof. Let {un} be a (PS) sequence of Iμ in E. This means that there exists C>0 such that

    |Iμ(un)|C,     Iμ(un)0 as  n. (2.5)

    From (2.1) and Lemma 2.4, we derive that

    Iμ(un)12+rIμ(un),un=r2(2+r)un2+Ω(12+r1q(x)+r)fμ(x,un)undxCμr2(2+r)un2Cμ,

    which implies that r2(2+r)un2C+Cμ+o(un). We obtain {un} is bounded in E. Up to a subsequence, we may assume that

    {unu,  in  E,unu,  in  Ls(Ω),  1s<2.

    For any integer pair (i,j), one has

    uiuj2=Iμ(ui)Iμ(uj),uiuj+Ω(fμ(x,ui)fμ(x,uj))(uiuj)dx.

    It follows from (2.5) that

    Iμ(ui)Iμ(uj),uiuj0,     as   i, j+. (2.6)

    It is easy to see that

    |fμ(x,t)||t|q(x)1+(μA)r|t|q(x)+r1.

    Note that 2q(x)<q(x)+rq++r<2. It implies that

    |Ω(fμ(x,ui)fμ(x,uj))(uiuj)dx|CΩ(|ui|+|uj|+|ui|q++r1+|uj|q++r1)|uiuj|0 (2.7)

    as i and j tend to +. From (2.6) and (2.7), we have uiuj0 as i, j+, which implies that {un} contains a strongly convergent subsequence in E. Hence Iμ satisfies the (PS) condition.

    G, Y and M are taken to be Iμ, Ek and Ek1 in Theorem 2.3, respectively. Next we will complete the proof of Theorem 2.1 by verifying the conditions of Theorem 2.3 one by one.

    Lemma 2.6. Suppose that (Q1) holds. If we replace G, Y and M with Iμ, Ek and Ek1, respectively, then conditions (A1) and (A2) are satisfied.

    Proof. Consider another norm u=us of E, s(2,2). Then usCu for all uE, where C>0 is a constant and unu0 whenever unuweakly in (E,). Define βk=supEkIμ. Let

    Qk(ρ)={uEk1:ussu2+uusu+λβkkus=ρ},

    it is easy to obtain that there exists a constant c2>0 such that usc2 for any uQk(ρ). By assumption (Q1) and definition of the function mμ, we have

    |Fμ(x,t)||t|q(x)q(x)+|t|q(x)+rq(x)+r|t|q(x)+|t|q(x)+r.

    It implies that

    |ΩFμ(x,u)dx|Ω(|u|q(x)+|u|q(x)+r)dx. (2.8)

    By the Sobolev imbedding theorem, it implies from 2q(x)<q(x)+rq++r<2 that

    Ω|u|q(x)+rdxΩ(|u|2+r+|u|q++r)dx. (2.9)

    Set Ωε={xΩ|2q(x)<2+ε}. By the Hölder inequality and the Sobolev imbedding theorem, we have

    Ω|u|q(x)dx=Ωε|u|q(x)dx+ΩΩε|u|q(x)dxΩε(|u|2+|u|2+ε)dx+ΩΩε(|u|2+ε+|u|q+)dxΩε|u|2dx+Ω(|u|2+ε+|u|q+)dxC|Ωε|222u2+Ω(|u|2+ε+|u|q+)dx. (2.10)

    Since Ω0={0}, we obtain |Ωε|0 as ε0. Therefore, there exists ε0>0 such that

    |Ωε|222<14C (2.11)

    for any ε(0,ε0). From (2.8)-(2.11), for any a, b>0, there is a c1=c1(a,b)>0 such that Iμ(u)a and uq++rbuc1. That is, condition (A1) is satisfied.

    For t>max{1,2μ}, one has

    Fμ(x,t)=2μ0fμ(x,τ)dτ+t2μ(μA)rτq(x)+r1dτ=2μ0(fμ(x,τ)(μA)rτq(x)+r1)dτ+t0(μA)rτq(x)+r1dτ1q(x)+r(μA)rtq(x)+r1q++r(μA)rt2+r.

    Set Y=Ek. Noticing that dimEk< and all norms of finite dimensional space are equivalent, it implies that

    Iμ(u)u212ΩF(x,u)u2dx

    as u, uEk. Therefore, limuEk, uIμ(u)=. So condition (A2) is satisfied.

    Let Qk=Qk(ρ)IβkμS and γk=infQIμ. Set P:={uE:u(x)0 for a.e xΩ. Then, P(P) is the positive(negative) cone of E and weakly closed. By Lemma 5.4 in [19], there is a η=η(βk)>0 such that dist(Q,P)=η>0. We define ±D0(δ0):={uE:dist(u,±P)<δ0}, where δ0 is determined by the following lemma.

    Lemma 2.7. Under the assumption (Q1), there is a δ0(0,η) such that KIμ(±D0(δ0))±D0(δ0). Therefore, condition (A) is satisfied.

    Proof. Write u±=max{±u,0}. For any uE and each s(2.2], there exists a Cs>0 such that

    u±sCsdist(u,P). (2.12)

    Let v=KIμ(u). Similar to the derivation of (2.8), (2.9) and (2.10), we have

    Ωfμ(x,u+)v+dxΩε|u+||v+|dx+Ω(|u+|1+ε+|u+|q++r1)|v+|dx. (2.13)

    From (2.12) and (2.13), by the Hölder inequality and the Sobolev imbedding theorem, we obtain

    dist(v,P)v+v+2=v+,v+=Ωfμ(x,u+)v+dxC((Ωε|u+|2|dx)12+u+1+ε2+ε+u+q++r1q++r)v+C(|Ωε|2222u+2+u+1+ε2+ε+u+q++r1q++r)v+C(|Ωε|2222dist(u,P)+(dist(u,P))1+ε+(dist(u,P))q++r1)v+.

    That is,

    dist(KIμ(u),P)C(|Ωε|2222dist(u,P)+(dist(u,P))1+ε+(dist(u,P))q++r1).

    It follows from (2.11) that there exists a δ0(0,η) such that dist(KIμ(u),P)<δ0 for every uD0(δ0). Similarly, dist(KIμ(u),P)<δ0 for every uD0(δ0). The conclusion follows.

    Now we are in a position to prove the main result of this section.

    Proof of Theorem 2.1. By Theorem 2.3, Lemmas 2.5, 2.6 and 2.7, we obtain

    K[γkε,βk+ε](E(P(P))

    for all ε>0 small. That is, there exists a uk,μE(P(P) (sign-changing critical point) such that

    Iμ(uk,μ)=0,   Iμ(uk,μ)[γk1,βk+1],

    where γk=infQkIμ. Next we show the γk as k. Recall the Gagliardo-Nirenberg inequality,

    uscsuαu1α2,     uE. (2.14)

    where s(2,2) and α(0,1) is defined by

    1s=(121N)α+12(1α). (2.15)

    In addition, for uEk, we see that u21λ1/2ku. Combine (2.14) with (2.15), we have

    us2scs2sus2λ(1α)(s2)/2k,     uEk. (2.16)

    For uQk(ρ), by the Sobolev imbedding theorem, we deduce from (2.16) that

    ρ=ussu2+uusu+λβkkusuus2(uλβkkus)1/2+u2su2us2s(uus)1/22(λβkk)1/2+C2us2sC1/2u2(λβkk)1/2+C2us2sC1/2u2(λβkk)1/2+C2cs2sus2λ(1α)(s2)/2kmax{C1/2λβk/2ku,2C2cs2sλ(1α)(s2)/2kus2}. (2.17)

    It implies that

    uΛsTkT, (2.18)

    where Λs=min{C1/2,21/(s2)C2/(s2)c1s}, Tk=min{λβk/2k,λ(1α)/2k} and T=min{ρ,ρ1/(s2)}. From (2.8)-(2.11), we know that

    |ΩFμ(x,u)dx|14u2+CΩ|u|q++rdx,    uE.

    We can choose that ρ>0 such that ρ<18C. For any uQk(ρ), we see that uss/u2ρ. Therefore, for any uQk(ρ), it implies from (2.18) that

    Iμ(u)=12u2ΩFμ(x,u)dx14u2CΩ|u|q++rdxu2(14Cuq++rq++ru2)(14Cρ)u218u218(Λp++rTkT)2.

    Since λk as k, we obtain Tk=min{λβk/2k,λ(1α)/2k} as k. Therefore, γk as k. Hence, for any μ(0,1], problem (2.2) has infinitely many sign-changing solutions. The proof is complete.

    In this section, we will show that solutions of auxiliary problem (2.2) are indeed solutions of original problem (1.1). For this purpose, we need the following uniform L-estimate for critical points of the functional Iμ.

    Proposition 3.1. Suppose that (Q1) and (Q2) hold. If v is a critical point of Iμ with Iμ(v)L, then there exists a positive constant M=M(L) independent of μ such that vL(Ω)M.

    In order to prove Proposition 3.1, we need some preliminaries.

    Lemma 3.2. Suppose that (Q1) and (Q2) hold. If Iμ(v)L and Iμ(v)=0, then, for any δ(0,δ0), there exists Cδ>0 independent of μ such that Ωδ|v|2dxCδ.

    Proof. By Lemma 2.4 and (Q1), we have

    LIμ(v)Iμ(v),vq(x)=Ω(121q(x))|v|2dx+Ω(fμ(x,v)vq(x)Fμ(x,v))dxΩ(121q(x))|v|2dxΩδ(121q(x))|v|2dx. (3.1)

    According to (Q1), for any δ(0,δ0), we know that there exists mδ>0 such that 121q(x)mδ for any xΩδ. Therefore, we have

    Ωδ|v|2dxm1δΩδ(121q(x))|v|2dxm1δL=Cδ.

    The proof is complete.

    Lemma 3.3. Let 1<p<N2 and 0<r<R. Suppose that the nonnegative functions w(x) and g(x) satisfy gLp(BR) and

    Δwg,   in  BR. (3.2)

    Then, we have

    wLNpN2p(Br)C(wL(N2)pN2p(BRBr)+gLp(BR)), (3.3)

    where C=C(N,p,R,r)>0.

    Proof. Set ξ=N(p1)N2p. Then, we have the following identity

    N(1+ξ)N2=NpN2p=pξp1. (3.4)

    Let φC0(RN,[0,1]) satisfies φ(x)=1 for |x|r and φ(x)=0 for |x|R. For any θ>0, multiply inequality (3.2) by the test function ((w+θ)ξθξ)φ2 and integrate to obtain

    BRw(((w+θ)ξθξ)φ2)dxBRg((w+θ)ξθξ)φ2dx. (3.5)

    By the Young inequality, we hace

    BRw(((w+θ)ξθξ)φ2)dx=ξBR|w|2(w+θ)ξ1φ2dx+2BR((w+θ)ξθξ)φwφdx4ξ(ξ+1)2BR|(w+θ)ξ+12|2φ2dxCBR(w+θ)ξ+12(w+θ)ξ+12φφdxCBR|(((w+θ)ξ+12θξ+12)φ)|2dxCBR(w+θ)ξ+1|φ|2dxC(BR(((w+θ)ξ+12θξ+12)φ)2NN2dx)N2NCBR(w+θ)ξ+1|φ|2dx. (3.6)

    According to 1<p<N2, we have 2pp1>2NN2. It implies that

    BRg((w+θ)ξθξ)φ2dxgLp(BR)(BR((w+θ)ξθξ)pp1φ2pp1dx)p1pgLp(BR)(BR((w+θ)ξ)pp1φ2NN2dx)p1p. (3.7)

    Letting θ0, we conclude from (3.5), (3.6) and (3.7) that

    (BRwN(ξ+1)N2φ2NN2dx)N2NCBRwξ+1|φ|2dx+CgLp(BR)(BRwpξp1φ2NN2dx)p1p,

    which implies that

    (BRwN(ξ+1)N2φ2NN2dx)N2NCBRwξ+1|φ|2dx, (3.8)

    or

    (BRwN(ξ+1)N2φ2NN2dx)N2NCgLp(BR)(BRwpξp1φ2NN2dx)p1p. (3.9)

    From (3.4), (3.8) and (3.9), we have

    (BRwNpN2pφ2NN2dx)N2pNpC(BRw(N2)pN2p|φ|2dx)N2p(N2)p=C(BRBrw(N2)pN2pdx)N2p(N2)p,

    or

    (BRwNpN2pφ2NN2dx)N2pNpCgLp(BR).

    Therefore, we obtain

    wLNpN2p(Br)(BRwNpN2pφ2NN2dx)N2pNpC(wL(N2)pN2p(BRBr)+gLp(BR)).

    The proof is complete.

    Lemma 3.4. Suppose that (Q1) and (Q2) hold. If Iμ(v)L and Iμ(v)=0, then there exist δ1(0, δ0) and C>0 independent of μ such that Bδ|v|2dxC for any δ(0, δ1).

    Proof. It follows from Iμ(v)=0 that v is a solution of problem (2.2). For any δ(0, δ0) and B2δΩ, let ϕC0(Ω,[0,1]) satisfies ϕ(x)=1 for |x|δ, ϕ(x)=0 for |x|2δ and |ϕ|C for xΩ. Multiply equation (2.2) by vϕ2 and integrate to obtain

    Ωv(vϕ2)dx=Ωkμ(x,v)vϕ2dxB2δkμ(x,v)vdx. (3.10)

    By the Young inequality, we have

    Ωv(vφ2)dx=Ω|v|2φ2dx+2Ωvφvφdx12Ω|v|2φ2dxCΩv2|ϕ|2dx12Bδ|v|2dxCB2δv2dx. (3.11)

    By definition of the function mμ, we know that mμ(t)=t for t1μ and mμ(t)1μ for t>1μ. Therefore, we have

    |kμ(x,v)|Cμr|v|q(x)+r1C|v|q(x)+r1C|v|q++r1, (3.12)

    for any μ(0, 1] and xΩ. From (3.10), (3.11) and (3.12), we obtain

    Bδ|v|2dxCB2δkμ(x,v)vdx+CB2δv2dxCB2δ|v|q++rdx. (3.13)

    In order to complete our proof, we just need to prove that there exists δ2>0 such that Bδ2|v|q++rdxC.

    By definition of the function kμ, we obtain kμ(x,t)tq(x)1. According to Lemma 2.4, (Q1) and (Q2), for any δ(0, δ0), we have

    LIμ(v)12Iμ(v),v=Ω(kμ(x,v)v2Kμ(x,v))dxΩ(121q(x))kμ(x,v)vdx12q+Bδ|x|α|v|q(x)dx. (3.14)

    Noticing that N3, from (Q2) and (2.1), we have

    0<α(1+r)2(1+r)=α(1+r)1r<α(4N+1)4N1<(N+2)(4N+1)2(4N1)<N.

    Therefore, we can choose p(1,2NN+1) satisfying

    p(1r)p1>2    and     0<pα(1+r)2+rp(1+r)<7N+28<N. (3.15)

    Let q+δ=sup{q(x)|xBδ}. It follows from (Q1) and (3.15) that there exists δ3<min{1, δ0} such that

    q+δp(1r)p1    and     0<pα(q+δ1+r)q+δp(q+δ1+r)<N (3.16)

    for any δ(0, δ3). Using the Young inequality, we deduce from (3.14) and (3.16) that

    Bδ|v|p(q(x)+r1)dx=Bδ(|x|α|v|q(x))p(q(x)+r1)q(x)|x|pα(q(x)+r1)q(x)dxCBδ|x|α|v|q(x)dx+CBδ|x|pα(q(x)+r1)q(x)p(q(x)+r1)dxCBδ|x|α|v|q(x)dx+CBδ|x|pα(q+δ+r1)q+δp(q+δ+r1)dx2q+CL+CδNpα(q+δ+r1)q+δp(q+δ+r1), (3.17)

    for any δ(0, δ3). According to (3.12) and (3.17), for any δ(0, δ3), we obtain

    kμ(x,v)Lp(Bδ)Cδ. (3.18)

    Since v=kμ(x,v) in Bδ. By Lemma 3.2 and Lemma 3.3, for any δ(0, δ), it implies from (3.18) that

    vLNpN2p(Bδ)C(vL(N2)pN2p(BδBδ)+kμ(x,v)Lp(Bδ))C(vL(N2)pN2p(Ωδ)+kμ(x,v)Lp(Bδ))C(Ωδ|v|2dx)12+Ckμ(x,v)Lp(Bδ)C, (3.19)

    where δ(0, δ3) and C=C(δ,δ,N,p)>0 is independent of μ.

    If ζ1=NpN2pq++r, using the Hölder inequality, we are done. Otherwise, using the fact r<14N<2N2 provided by (2.1), we can choose σ1(0, δ)(0, δ3) such that τ1=q+σ1+r1<NN2. It follows from (3.12) that

    |kμ(x,v)|C|v|τ1,

    for any μ(0, 1] and xBσ1. Noticing that ζ1>NN2, we have p1=ζ1τ1>1. According to (3.19), we obtain kμ(x,v)Lp1(Bσ1). Similar to (3.19), we can choose σ2(0,σ1) to obtain

    vLζ2(Bσ2)C,

    where C=C(σ1,σ2,N,p1)>0 is independent of μ and

    ζ2=Np1N2p1=Nζ1Nτ12ζ1N(N2)τ1ζ1=d1ζ1,

    here d1=N(N2)τ1>1. If ζ2q++r, using the Hölder inequality, we are done. Otherwise, repeating the above process and using a finite number of iterations, we obtain that there exist ζk>0 and σk(0,σk1) such that ζk2>q++r and vLζk(Bσk)C, where C>0 is independent of μ. Using the Hölder inequality, we have

    vLq++r(Bσk)C. (3.20)

    Let δ1=ζk2. It implies from (3.13) and (3.20) that

    Bδ|v|2dxCB2δ|v|q++rdxCBσk|v|q++rdxC,

    for any δ(0,δ1).

    Proof of Proposition 3.1. By Lemma 3.2 and Lemma 3.4, we obtain that there exists C>0 independent of μ such that

    Ω|v|2dxC. (3.21)

    Using the Sobolev embedding theorem, we have

    Ω|v|2dxC(Ω|v|2dx)22C. (3.22)

    Let s>0 and t=q++r. According to (3.12), multiply equation (2.2) by v2s+1 and integrate to obtain

    Ωvv2s+1dx=Ωkμ(x,v)v2s+1dxCΩ|v|2s+tdx.

    It implies that

    Ω|v|2v2sdx=12s+1Ωvv2s+1dxCΩ|v|2s+tdx. (3.23)

    On the one hand, by the Sobolev embedding theorem, we have

    Ω|v|2v2sdx=1(1+s)2Ω|v1+s|2dxC(1+s)2(Ω|v|(1+s)2dx)22. (3.24)

    On the other hand, by the Hölder inequality and (3.22), we have

    Ω|v|2s+tdx(Ω|v|2dx)t22(Ω|v|2(1+s)22t+2dx)2t+22C(Ω|v|(1+s)2ddx)2d2, (3.25)

    where d=2t+22>1. According to (3.23), (3.24) and (3.25), we obtain

    (Ω|v|(1+s)2dx)22(C(1+s))2(Ω|v|(1+s)2ddx)2d2,

    which implies that

    (Ω|v|(1+s)2dx)1(1+s)2(C(1+s))11+s(Ω|v|(1+s)2ddx)d(1+s)2. (3.26)

    Now we carry out an iteration process. Set sk=dk1 for k=1,2,. By (3.26), we have

    (Ω|v|dk2dx)1dk2(Cdk)1dk(Ω|v|dk12dx)1dk12Πkj=1(Cdj)1dj(Ω|v|2dx)12Ckj=1djdkj=1jdj(Ω|v|2dx)12. (3.27)

    Since d>1, the series j=1dj and j=1jdj are convergent. Letting k, we conclude from (3.22) and (3.27) that vL(Ω)M. The proof is complete.

    Proof of Theorem 1.1. By the proof of Theorem 2.1, for every integer k1, we know that problem (2.2) has k sign-changing solutions uk,μ satisfying γk1<Iμ(uk,μ)<βk+1. Consider the functional

    J(u)=12Ω|u|2dxΩ|u|p(x)p(x)dx.

    By definition of the function fμ, we obtain |fμ(x,t)||t|q(x)1. It is easy to see that Iμ(u)J(u). Therefore, there exists a sequence of positive numbers {Υk} independent of μ such that βk+1Υk. Let Lk=max{Υ1,Υ2,,Υk}. By Proposition 3.1, there exists a positive constant Mk=Mk(Lk) independent of μ such that uk,μL(Ω)Mk. By definition of the function mμ, we have mμ(t)=t for t1μ. Hence, problem (2.2) reduces to problem (1.1) for |t|1μ. Let μ<12Mk. It is easy to see that uk,μ is indeed a sign-changing solution of problem (1.1).

    Thanks to Professor Jiaquan Liu of Peking University for his great help and valuable advice in this paper. Supported by National Natural Science Foundation of China (No.11861021).

    The authors declare no conflict of interest.



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