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Global bifurcation of sign-changing radial solutions of elliptic equations of order 2m in annular domains

  • In this paper we study the global bifurcation of sign-changing radial solutions for some semilinear elliptic problems of order 2m in an annulus with Dirichlet boundary conditions.

    Citation: Ruyun Ma, Dongliang Yan, Liping Wei. Global bifurcation of sign-changing radial solutions of elliptic equations of order 2m in annular domains[J]. AIMS Mathematics, 2020, 5(5): 4909-4916. doi: 10.3934/math.2020313

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  • In this paper we study the global bifurcation of sign-changing radial solutions for some semilinear elliptic problems of order 2m in an annulus with Dirichlet boundary conditions.



    Let Ω(a,b) denote the annulus {xRn:a<|x|<b}, 0<a<b<, n2, and consider the semilinear elliptic problems

    (1)mΔmu=λg(|x|)f(u)      in Ω(a,b), (1.1)
    u=uν==m1uνm1=0    on Ω(a,b), (1.2)

    where λ>0 is a parameter, ν is the outward normal derivative, m is a positive integer and f,g satisfy at least the following assumptions:

    (H1) f:RR is a continuous function;

    (H2) g:[a,b][0,) is a continuous function such that g0 in [a,b].

    When m=1 and n=1, the existence of positive radial solutions of problem (1.1), (1.2) has been intensively studied in the case where f is superlinear/sublinear at 0 and , see Coffman and Marcus [1], Erbe and Wang [2], Erbe, Hu and Wang [3], Lan and Webb [4] and references therein. When m=1 and n1, the existence of radial positive solutions of (1.1), (1.2) was studied by Lin and Pai [5], Wang [6]. When n2 and m>1, Dalmasso [7] investigated the existence of positive radial solutions for (1.1), (1.2) in an annulus with Dirichlet boundary conditions. He considered a nonlinearity which is either sublinear or the sum of a sublinear and a superlinear term. All of these work are based upon the fixed point theorem in cones. Ma and Thompson [8] used bifurcation theorem [9] to study the multiplicity of nodal solutions of (1.1), (1.2) when m=1 and n=1 under the conditions: either

    λkf<r<λkf0

    or

    λkf0<r<λkf.

    Of course, the natural question is whether or not similar result can be established for (1.1), (1.2) in the case m>1 and n>1?

    The purpose of this paper is to generalize the main results in Ma and Thompson [8] to the case m>1 and n>1.

    Since we are interested in radial nodal solutions, problems (1.1), (1.2) reduce to the one-dimensional boundary value problems

    (1)mtn1Lmu=λtn1g(t)f(u)     t(a,b), (1.3)
    u(j)(a)=u(j)(b)=0      j=0,1,,m1, (1.4)

    where L denotes the polar form of the Laplacian, i.e.:

    Lv(t):=t1nddt(tn1dv(t)dt).

    Thus

    L0v(t)=v(t);    Ljv(t)=L(Lj1v(t)),   j=1,,m. (1.5)

    In particular,

    L2v(t)=L(Lv(t))=t1nddt(tn1ddt[Lv(t)])=t1nddt(tn1ddt[t1nddt(tn1dv(t)dt)]). (1.6)

    The rest of the paper is organized as follows. In Section 2 we state some properties of Green function Gm(t,s) due to Dalmasso [7] and the radial eigenvalues theory of the corresponding linear eigenvalue problem of (1.3), (1.4) due to Elias [10] and Rynne [11]. Section 3 we introduce Dancer's unilateral global bifurcation theorem of Dancer [9]. Finally in Section 4, we state and prove our main result via global bifurcation technique.

    The homogeneous Dirichlet problem

    (1)mtn1Lmv(t)=tn1e(t)     t(a,b),v(j)(a)=v(j)(b)=0     j=0,1,,m1 (2.1)

    has only the trivial solution. Then it is well-known(see e.g. M. A. Naimark [12,P.29]) that the operator (1)mLm with Dirichlet boundary conditions has one and only one Green's function Gm(t,s).

    Lemma 2.1.(Dalmasso [7,Theorem 2.1]) Gm(t,s)>0 for a<t,s<b.

    Proof. From (1.5) and (1.6), it follows that (1)mLm has a Pólya's factorization in [a,b]. So, it is a disconjugate operator on [a,b]. Therefore, Gm(t,s)>0 for a<t,s<b is an immediate consequence of Coppel [13,Theorem 11 on p.108], or Elias [14,Theorem 0.13].

    Lemma 2.2.(Dalmasso [7,Theorem 2.2]) (ⅰ) There exists a positive constant Cm, such that

    0Gm(t,s)Cm(sa)m(bs)m,    at,sb. (2.2)

    (ⅱ) For any δ(0,(ba)/2) there exists η(0,1) such that

    Gm(t,s)ηCm(sa)m(bs)m,    asb, a+δtbδ. (2.3)

    We consider the Banach spaces

    X={uC2m[a,b]|u satisfies (1.4)}

    and

    Y=C0[a,b].

    Denote

    E={uC2m1[a,b]|u satisfies (1.4)}

    with the norm ||||2m1 which, for convenience, we will write as ||||. Define an operator A:XY

    Au(t):=tn1(1)mLmu(t),      uX. (2.4)

    Then Au=0 is disconjugate on [a,b], and the boundary conditions (1.4) are such that A is formally self-adjoint, that is

    Au,v=u,Av,       u,vX, (2.5)

    where , denotes the standard L2(a,b) inner product. Moreover, Then A1:YE is compact.

    For each integer k1 and ν{+,}, let Sk,ν denote the set of functions uE such that:

    (1) u has only simple zeros in (a,b) and no quasi-derivative of u is zero at a or b, other than those specified in (1.4);

    (2) u has exactly k1 zeros in (a,b);

    (3) νu>0 in a deleted neighborhood of t=0.

    The following result is an immediate consequence of Rynne[11,Theorem 2.4].

    Lemma 2.3. Assume that

    (H0) pC0[a,b], and p0 on [a,b], while p0 on any interval of [a,b].

    Then for each k1 and each ν{+,}, problem

    Au(t)=λtn1p(t)u,      t(a,b),   (2.6)
    u(j)(a)=u(j)(b)=0,     j=0,1,,m1 (2.7)

    has a unique solution (λk,ψk)R+×Sk,ν with ||ψk||=1. In addition:

    (1) σ(L,p)={λk:k1};

    (2) If k>k1 then λk>λk>0;

    (3) limkλk=.

    The main point to prove our main result (see Theorem 4.1 below) consists in using the unilateral global bifurcation theorem of Dancer [9].

    Let E1 be a real Banach space with norm ||||E1. E will denote R×E1. Let the mapping G:EE1 satisfy

    Assumption A: if G(λ,0)=0 for λR, G is completely continuous and

    G(λ,0)=λLx+H(λ,0),

    where L is a completely continuous linear operator on E1 and ||H(λ,x)||E1/||x||E10 uniformly on bounded subsets of R as ||x||E10.

    Define Φ(λ):E1E1 by Φ(λ)(x)=xG(λ,x) and define L to be the closure of {(λ,x)E:x=G(λ,x),x0} in E. Then (cp. Rabinowitz [15]) L(R×{0})r(L)×{0}, where r(L) denotes the set of real characteristic value of L. If μr(L), define Cμ to be the component of L containing (μ,0).

    Assume now that μr(L) such that μ has multiplicity 1. Suppose that vE1{0} and lE1 such that

    v=μLv,     l=μLl,

    (where L is the adjoint of L) and l(v)=1. If y(0,1), define

    Ky={(λ,u)E:|l(u)|>y||u||E1},
    K+y={(λ,u)E:l(u)>y||u||E1},  Ky={(λ,u)E:l(u)<y||u||E1}.

    By [15,Lemma 1.24], there exists an S>0 such that

    (L{(μ,0)})ˉES(μ)Ky,

    where ES(μ)={(λ,u)E||λμ|+uE1<S} and ˉES(μ) denotes closure of ES(μ). For 0<ϵS and ν=±, define Dνμ,ϵ to be the component of {(μ,0)}(LˉEϵ(μ)Kνy) containing (μ,0), Cνμ,ϵ to be the component of ¯CμDνμ,ϵ containing (μ,0) (where ν is interpreted in the natural way), and Cμ,ν to be the closure of Sϵ>0Cνμ,ϵ. Then Cμ,ν is connected and, by [9], Cμ=Cμ,+Cμ,. By [15], Lemma 1.24], the definition of Cμ,ν is independent of y.

    Theorem A. [9,Theorem 2] Either Cμ,+ and Cμ, are both unbounded or

    Cμ,+Cμ,{(μ,0)}.

    We shall make use of the following assumptions

    (A1) fC(R,R) with sf(s)>0 for s0;

    (A2) there exist f0, f(0,) such that

    f0=lim|s|0f(s)s,f=lim|s|f(s)s.

    Theorem 4.1. Let (A1), (A2) and (H2) hold. Assume that for some kN, either

    λkf<1<λkf0 (4.1)

    or

    λkf0<1<λkf. (4.2)

    Then (1.3), (1.4) has two solutions u+k and uk such that u+k has exactly k1 zeros in (a,b) and is positive near 0, and uk has exactly k1 zeros in (a,b) and is negative near 0.

    Remark 4.1. Theorem 4.1 generalizes Dalmasso [7,Theorem 1.1] where only the existence of radial positive solution was studied under the conditions k=1,ν=+ and

    f0=,     f=0. (4.3)

    Obviously, (4.3) is a special case of (4.2).

    Remark 4.2. Theorem 4.1 generalizes Ma and Thompson [8,Theorem 1.1] where n=1 and m=1, i.e., the second order ODE version of Theorem 4.1 was considered.

    Remark 4.3. Conditions (4.1) and (4.2) in Theorem 4.1 are optimal. Let us consider the following counterexample.

    u=λ˜f(u),    u(0)=u(1)=0. (4.4)

    where

    ˜f(s):=π2s+s21+s4    sR.

    It is easy to check that

    f0=π2=f,    λ1=π2

    i.e.

    λ1f0=1=λ1f.

    We shall see that (4.4) has no positive solution when λ=1. In fact, suppose on the contrary that (1,u) is a positive solution of (4.4). Then, multiplying both sides of (4.4) by sinπt and integrating from 0 to 1, we get

    10u2(t)1+u4(t)sinπtdt=0.

    This is a contradiction. Therefore, the conditions (4.1) and (4.2) are optimal.

    Remark 4.4. For other related results on the fourth order problems, see Drábek and Holubová [16], Cabada and Enguiça [17] and Ma and Lu [18].

    Let ζ, ξC(R) be such that

    f(u)=f0u+ζ(u),f(u)=fu+ξ(u).

    Obviously

    lim|u|0ζ(u)u=0,lim|u|ξ(u)u=0.

    Let

    ˜ξ(u)=max0|s|u |ξ(s)|

    then ˜ξ is nondecreasing and

    limu˜ξ(u)u=0. (4.5)

    Let us consider

    Auλtn1g(t)f0u=λtn1g(t)ζ(u) (4.6)

    as a bifurcation problem from the trivial solution u0.

    Equation (4.6) can be converted to the equivalent equation

    u(t)=baGm(t,s)sn1[λg(s)f0u(s)+λg(s)ζ(u(s))]ds  :=λA1[σ()g()f0u()](t)+λA1[σ()g()ζ(u())](t) (4.7)

    where σ(t):=tn1. Further we note that ||A1[σ()g()ζ(u()]||=o(||u||) for u near 0 in E, since

    ||A1[σ()g()ζ(u())]||=maxt[a,b]|baGm(t,s)sn1g(s)ζ(u(s))ds|   +   +maxt[a,b]|bam1Gm(t,s)tm1sn1g(s)ζ(u(s))ds|Cmaxs[a,b]|g(s)|||ζ(u())||.

    The results of Dancer [9] for (4.6) can be stated as follows: For each integer k1, ν{+, }, Either C+k and Ck are both unbounded or

    C+kCk{(λk/f0,0)}.

    We shall show that

    (C+kCk){(λk/f0,0)}=. (4.8)

    Assume on the contrary that (ˉη,y)((C+kCk){(λk/f0,0)}) for some ˉηλk/f0.

    If y0 in [a,b], then y(Sk,+Sk,) implies that y(t)0 in a left neighborhood of t=a. This contradicts the fact y has exactly k1 simple zeros in (a,b).

    If y0 in [a,b], then there exists a sequence {(ηn,yn)}C+k or {(ηn,yn)}Ck, such that

    ηnˉη,    yn0   in E.

    we only deal with the first case.

    Set

    vn:=yn/||yn||.

    Then

    Avn=ηng(t)tn1f(yn(t))yn(t)vn(t).

    Using the fact that any zeros on vn in (a,b) are simple and the standard argument, we may assume that vnv for some vSk,+ with ||v||=1, and

    Av=ˉηtn1g(t)f0v.

    This implies ˉη=λk/f0. However, it is a contradiction. Therefore, (4.8) is valid, and accordingly, both C+k and Ck are unbounded.

    Proof of Theorem 4.1. It is clear that any solution of (4.6) of the form (1,u) yields a solution u of (1.3)-(1.4). We will show Cνk crosses the hyperplane {1}×E in R×E. To do this, it is enough to show that Cνk joins (λkf0,0) to (λkf,). Let (μn,yn)Cνk satisfy

    μn+||yn||.

    We note that μn>0 for all nN since (0,0) is the only solution of (4.6) for λ=0 and Cνk({0}×E)=.

    Using the same method to prove Ma and Thompson [8,Theorem 1.1] with obvious change, and replacing 'from Lemma 2.1' in [8,Page 714] with 'from the proof of Elias [10,Lemma 4]' we may deduce the desired result.

    We are concerned with determining values of λ, for which there exist nodal solutions of elliptic equations of order 2m in annular domains with Dirichlet boundary conditions. Furthmore, we have given an example to show that the interval of λ is optimal.

    The authors are very grateful to the anonymous referees for their very valuable suggestions. This work was supported by National Natural Science Foundation of China (No.11671322).

    All of the authors of this article claims that together they have no any competing interests each other.



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