# Unilateral Global Interval Bifurcation for the Hessian Equation and Its Applications

Authors: Wenguo Shen
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

In this paper, we establish a unilateral global bifurcation result from the interval for thek-Hessian equations with nondifferentiable nonlinearity. By applying the above result, we shall prove the existence of the principal half-eigenvalues for the half quasilinear problems. Furthermore, we shall determine the interval of γ, in which there exist one-sign solutions for the following k-Hessian equations: SkD2u=αx−u+k+βx−u−k+γaxfu,inB,ux=0,on∂B.


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## Body

## 1. Introduction

We first consider the followingk-Hessian equation:(1)SkD2u=λax−uk+gx,−u,λ,inB,ux=0,on∂B,where D2u=∂2u/∂xi∂xj is the Hessian matrix of u, B is the unit ball of ℝN, ax is a weighted function, λ is a positive parameter, and g∈CB¯×ℝ2. Here, SkD2u=σkλD2u=∑1≤i1<⋯<ik≤Nλi1…λik is the k-Hessian operator of u, where λD2u≔λ=λ1,…,λk is the eigenvalues of D2u.Thek-Hessian equations are a type of important fully nonlinear equations. Historically, there are a large amount of papers in the literature on the existence, regularity, and the qualitative properties of solutions for the k-Hessian equation; see [1–13] and the references therein. Wang [14] has proved the existence of the first eigenvalue λ1 of problem (1) with λax=λkandg=0. When k=1, the k-Hessian equation is just the semilinear elliptic equation. The N-Hessian equation is just the Monge–Ampère equation.Recently, in [15], the radial solutions of problem (1) is equivalent to the solutions of the following boundary value problem:(2)rN−ku′k′=NCNkrN−1λar−uk+gr,−u,λ,r∈0,1,u′0=u1=0,where r=x with x∈B. By a solution of problem (2), we understand that it is a function which belongs to C20,1 and satisfies (2). Dai [15] established a global bifurcation result for the k-Hessian equation (2) with λar−uk+gr,−u,λ equal λk−uk+g−u.However, in [15], the nonlinearities are differentiable at the origin. In [16], Berestycki established an important global bifurcation theorem from intervals for a class of second-order problems involving nondifferentiable nonlinearity. Recently, Dai and Ma [17, 18] considered one-dimensional and high-dimensional p-Laplacian problems involving nondifferentiable nonlinearity, respectively. In 2018, Shen [19, 20] established interval bifurcation theorems for the Kirchhoff-type problems and the Monge–Ampère equations involving nondifferentiable nonlinearity, respectively.Motivated by the above papers, in this paper, we shall establish a unilateral global interval bifurcation result for the followingk-Hessian equations with nondifferentiable nonlinearity:(3)SkD2u=λax−uk+Fx,−u,λ,inB,ux=0,on∂B,where B is the unit open ball of ℝN and the nonlinear term F has the form F=f+g, where f,g∈CB¯×ℝ2 are radially symmetric with respect to x.It is clear that the radial solutions of (3) is equivalent to the solutions of the following problem:(4)rN−ku′k′=NCNkrN−1λar−uk+Fr,−u,λ,r∈0,1,u′0=u1=0,where r=x with x∈B.On the other hand, half-linear or half-quasilinear problems have attracted the attention of some specialists (see [16–20]). On the above basis of the unilateral global interval bifurcation result, we shall study the following half-quasilinear eigenvalue problem:(5)SkD2u=λax−uk+αx−u+k+βx−u−k,inB,ux=0,on∂B,where u+=maxu,0,u−=−minu,0, and αx,βx∈CB¯ are radially symmetric.It is clear that the radial solutions of (5) is equivalent to the solutions of the following problem:(6)rN−ku′k′=NCNkrN−1λar−uk+αr−u+k+βr−u−k,r∈0,1,u′0=u1=0.Following the above eigenvalue theory (see Theorem2), we shall investigate the existence of radial one-sign solutions for the following problem:(7)SkD2u=αx−u+k+βx−u−k+λaxfu,inB,ux=0,on∂B.It is clear that the radial solutions of (7) is equivalent to the solutions of the following problem:(8)rN−ku′k′=NCNkrN−1αr−u+k+βr−u−k+γarfu,r∈0,1,u′0=u1=0.The rest of this paper is arranged as follows. In Section2, we give some preliminaries. In Section 3, we establish the unilateral global interval bifurcation result for k-Hessian equation (4) (see Theorem 1). In Section 4, on the basis of the unilateral global interval bifurcation result, we shall prove the existence of the principal eigenvalues for k-Hessian equation (6) (see Theorem 2). In Section 5, we shall investigate the existence of one-sign solutions for k-Hessian equation (8).


## 2. Preliminaries

LetY=C0,1 with the norm u∞=maxr∈0,1ur. Let E≔ur∈C10,1u′0=u1=0 with the norm u=maxu∞,u′∞. Let P+=u∈E:ur>0,r∈0,1 and P−=−P+, P=P+∪P−. Let K+=ℝ×P+ and K=ℝ×P under the product topology.By Sections 5 and 6 in [15], we firstly consider the following problem:(9)rN−k−ν′k′=hr,r∈0,1,ν′0=ν1=0.Let us define the operatorGkh:E⟶E by(10)Gkh=vr=∫1rsk−N∫s0hτdτ1/kds,0≤r≤1.For a givenh∈Y, Gkh:Y⟶E is a completely continuous and (10) is equivalent to (9).With a simple transformationv=−u, problem (2) can be equivalently written as(11)rN−k−ν′k′=NCNkrN−1λarvk+gr,v,λ,r∈0,1,ν′0=ν1=0,where g∈C0,1×ℝ+2 and a satisfies the following conditions:(A0)ax∈CB¯ is radially symmetric and ar≥0, ar≢0 on any subinterval of 0,1, where r=x with x∈B¯(A1)gr,s,λ=osk near s=0 uniformly in r∈0,1 and λ on bounded setsAndg also satisfies the following subcritical growth restriction (see [15], p.4-(1.5)):(12)gr,s,λ≤C1+sq−1,uniformly for r∈0,1, λ on bounded sets, and some q∈0,k∗. Here,(13)k∗=Nk+1N−2k,ifk<N2,+∞,ifk≥N2,is the critical exponent for the k-Hessian operator, which was first determined in [21].Define the Nemytskii operatorH:ℝ×E⟶Y by(14)Hμ,vr≔NCNkrN−1μarvk+gr,v,μ.Then, it is clear thatH is a continuous (compact) operator and problem (11) can be equivalently written as(15)v=GkHμ,v≔Fμ,v,where F:ℝ×E⟶E is a completely continuous and Fμ,0=0,∀μ∈ℝ.Now, we consider the following eigenvalue problem:(16)rN−k−v′k′=λNCNkrN−1arvk,r∈0,1,v′0=v1=0.By the same proof as in Theorem 4.1 of [14], we can show that problem (16) possesses the first eigenvalue λ1 which is positive, simple, unique and the corresponding eigenfunctions are positive in (0, 1).By Rabinowitz [22], using the same method to prove ([15], Theorem 1.1) with obvious changes, we may get the following global bifurcation result.Lemma 1 (see [15], Theorem 1.1).
Assume that (A0), (A1), and (12) hold. Then, λ1,0 is the unique bifurcation point of problem (11) and there exists an unbounded continuum C⊆K∪λ1,0 of solutions to problem (11) emanating from λ1,0.By Dancer ([23], Theorem 2), under the conditions (A0), (A1), and (12), one can obtain that problem (11) has two distinct unbounded subcontinua C+ and C−, consisting of the bifurcation branch C emanating from λ1,0, which satisfy:Lemma 2 (see ([23], Theorem 2)).
BothC+ and C− are unbounded and(17)Cν⊂ℝ×Pν∪λ1,0,where ν∈+,−.Next, we give the following results.Lemma 3.
LetI be an interval and if y and z are functions such that y, z, φky′ and φkz′ are differentiable on I and yt>0,zt>0,y′t<0,z′t<0 for t∈I. Then, we have the following identity:(18)ddtyφkzφkyφk−z′−φkzφk−y′=yφkzφkyLkz−φkzLky+rN−k−y′k+1+k−yz′zk+1+k+1yky′−z′zk,where φks=rN−ksk,Lky=φk−y′′.Proof.
Similar the proof of Lemma 2.3 of [15] with obvious change, the left-hand side of (18) equals(19)ddtyk+1rN−k−z′kzk−yrN−k−y′k=k+1yky′rN−k−z′k+yk+1rN−k−z′k′zk−yk+1rN−k−z′kkzk−1z′z2k−y′rN−k−y′k−yrN−k−y′k′=yφkzφkyLkz−φkzLky+rN−k−y′k+1+k−yz′zk+1+k+1yky′−z′zk.Lemma 4.
In (18) of Lemma 3, by Young’s inequality, we get(20)−y′k+1+k−yz′zk+1+k+1yky′−z′zk≥0,and the equality holds if and only if sgny=sgnz and y′/yk+1=z′/zk+1.Proof.
Now recall Young’s inequality:(21)AB≤Aαα+Bββ,where A,B∈ℝ+,α,β>1,1/α+1/β=1.
Letα=k+1,β=k+1/k, A=−k+11/k+1y′,andB=k+1k/k+1yk−z′/zk in (21). We obtain inequality (20) holds.
By Lemma3 and Lemma 4, we have the following result:Lemma 5.
In (18) of Lemma 3, we have(22)∫01yφkzφkyLkz−φkzLkydr≤0.Proof.
By (18) of Lemma 3, it follows that(23)∫01yφkzφkyφk−z′−φkzφk−y′′dr=∫01rN−k−y′k+1+k−yz′zk+1+k+1yky′−z′zkdr+∫01yφkzφkyLkz−φkzLkydr.
Similar to the proof as (2.3) in Lemma 2.3 of [15], we can show that the left-hand side of (23) equals 0. By Lemma 4, we have the result.
In order to treat the problems with nonasymptotic nonlinearity at 0 and∞, we use Whyburn-type superior limit theorems. However, if the collection of the infinite sequence of sets is unbounded, Whyburn’s limit theorem ([24], Theorem 9.1) cannot be used directly because the collection may not be relatively compact. In order to overcome this difficulty, Dai [25] established the following results.Lemma 6 (see [25], Lemma 2.5).
LetX be a normal space and let Cnn=1,2,… be a sequence of unbounded connected subsets of X. Assume that(i)
There existsz∗∈liminfn⟶+∞Cn with z∗<+∞.(ii)
For everyR>0, ∪n=1+∞Cn∩B¯R is a relatively compact set of X, where(24)BR=x∈Xx≤R.Then,D≔limsupn⟶∞Cn is unbounded, closed, and connected.Lemma 7 (see [25], Theorem 1.2).
LetX be a normal vector space and let Cnn=1,2,… be a sequence of unbounded connected subsets of X. Assume that(i)
There existsz∗∈liminfn⟶+∞Cn with z∗=+∞.(ii)
There exists a homeomorphismT:X⟶X such that Tz∗<+∞ and TCn be a sequence of unbounded connected subsets in X.(iii)
For everyR>0, ∪n=1+∞Cn∩B¯R is a relatively compact set of X, where(25)BR=x∈Xx≤R.Then,D≔limsupn⟶∞Cn is unbounded, closed, and connected.Lemma 8 (see [25], Corollary 2.3).
LetX,ρ be a metric space. If Cii∈ℕ is a sequence of sets whose limit superior is L and there exists a homeomorphism T:X⟶X such that for every R>0, ∪i=1+∞TCi∩B¯R is a relatively compact set; then, for each ϵ>0, there exists an m such that for every n>m,Cn⊂VϵL, where VϵL denotes the set of all points p with ρp,x<ϵ for any x∈L.


## 3. Global Bifurcation from an Interval

With a simple transformationv=−u, problem (4) can be equivalently written as(26)rN−k−v′k′=NCNkrN−1λarvk+Fr,v,λ,r∈0,1,v′0=v1=0,where a satisfying (A0), F=f+g, f,g∈C0,1×ℝ+2, g satisfying (A1) and (12), and f satisfying(A2)fr,s,λ/sk≤M, for all r∈0,1, 0<s≤1, and all λ∈ℝ, where M is a positive constant.LetSν denote the closure in ℝ×E of the set of nontrivial solutions λ,v of (26) with v∈Pν. By an argument similar to that of ([15], Lemma 5.1) with obvious changes, we can show that the following existence and uniqueness theorem is valid for problem (26).Lemma 9 (see [15], Lemma 5.1).
Ifλ,v is a solution of (26) under the assumptions of (A0), (A1), (12), and (A2), and v has a double zero, then ν≡0.The first main result for (26) is the following theorem.Theorem 1.
Let (A0), (A1), (12), and (A2) hold. Let d=M/a0, where a0=minr∈0,1ar, and let I=λ1−d,λ1+d. The component Cν of Sν∪I×0, containing I×0, is unbounded and lies in Kν∪I×0 for ν∈+,−.To prove Theorem1, we introduce the following auxiliary approximate problem:(27)rN−k−v′k′=NCNkrN−1λarvk+fr,vvϵ,λ+gr,v,λ,r∈0,1,v′0=v1=0.To prove Theorem1, the following lemma will play a key role.Lemma 10.
Letϵn, 0<ϵn<1, be a sequence converging to 0. If there exists a sequence λn,vn∈Kν such that it is a nontrivial solution of problem (27) corresponding to ϵ=ϵn and λn,vn converges to λ,0 in ℝ×E, then λ∈I.Proof.
Letwn=vn/vn, then wn satisfies the following problem:(28)rN−k−wn′k′=NCNkrN−1λarwnk+fr,νnνnϵ,λnνnk+gr,νn,λnνnk,r∈0,1,wn′0=wn1=0.
Let(29)g¯r,v,λ=max0≤s≤vgr,s,λ,for allr∈0,1andλon  bounded  sets,then, g¯ is nondecreasing with respect to v and(30)limv⟶0+g¯r,v,λvk=0,uniformly for r∈0,1 and λ on bounded sets. Furthermore, it follows from (30) that(31)gr,v,λvk≤g¯r,v,λvk≤g¯r,v∞,λvk≤g¯r,v,λvk⟶0,asv⟶0,uniformly for r∈0,1 and λ on bounded sets.
Clearly,A2 implies that(32)fr,vnvnεn,λnvnk=ft,vnvnεn,λnvnkvnkεn⋅vnkvnkεnvnk≤M1⋅vnkεn⟶M,for all r∈0,1.
Note thatwn=1 implies wn∞≤1. Using this fact with (31) and (32), we have that λnNrN−1arwnk+NrN−1fr,vnvnϵn,λn/vnk+NrN−1gr,vn,λn/vnk is bounded in E for n large enough. The compactness of Gk implies that wn is converges in E. Without loss of generality, we may assume that wn⟶w in E with w=1. Clearly, we have w∈Pν¯.
We claim thatw∈Pν. On the contrary, suppose that w∈∂Pν, by Lemma 9, then, w≡0, which is a contradiction with w=1.
Now, we deduce the boundedness ofλ. Let ψ be an eigenfunction of problem (16) corresponding to λ1.
We know thatwn satisfies rN−k−wn′k′=N/CNkrN−1λarwnk+fr,vnvnεn,λn/vnk+gr,vn,λn/vnk, r∈0,1, wn′0=wn1=0, and ψ satisfies rN−k−ψ′k′=N/CNkrN−1λ1arψk,r∈0,1,ψ′0=ψ1=0.
We can assume without loss of generality thatν=+. By Lemma 5, it follows that(33)∫01wnφkψφkwnLψ−φkψLwndr=∫01λ1−λnar−fr,vnvnεn,λnvnkwnk−gr,vn,λnvnNwnkNCNkrN−1wnk+1dr≤0.
Similarly, we can also show that(34)∫01λn−λ1ar+fr,vnvnεn,λnvnkwnk+gr,vn,λnvnkwnkNCNkrN−1ψk+1dr≤0.
Ifλ≤λ1, considering (33), (A1), and (A2), we have that(35)∫01λ1−λarNCNkrN−1wk+1dr≤limn⟶∞∫01fr,vnvnϵn,λnvnkwnkNCNkrN−1wnk+1dr≤limn⟶∞∫01fr,vnvnϵn,λnvnkvnkϵnvnkϵnNCNkrN−1wnk+1dr≤∫01MNCNkrN−1wk+1dr.
Hence, we get that(36)∫01λ1−λa0NCNkrN−1wk+1dr≤∫01MNCNkrN−1wk+1dr,which implies λ≥λ1−d.
Ifλ≥λ1, considering (34), (A1), and (A2), we have that(37)∫01λ−λ1arNCNkrN−1ψk+1dr≤limn⟶∞∫01−fr,vnvnεn,λnvnkwnkNCNkrN−1ψk+1dr≤limn⟶∞∫01−fr,vnvnεn,λnvnkvnkεnvnkεnNCNkrN−1ψk+1dr≤∫01MNCNkrN−1ψk+1dr.
Hence, we get that(38)∫01λ−λ1a0NCNkrN−1ψk+1dr≤∫01MNCNkrN−1ψk+1dr,which implies λ≤λ1−d.
Therefore, we have thatλ∈I.Proof of Theorem 1.
We only prove the case ofν=+ since the proof of the case ν=− can be given similarly. Let C+ be the component of S+∪I×0.
First, using the similar method to prove ([17], Theorem 1.1) with obvious changes, we may prove that C+⊂K+∪I×0.
In the following, we prove thatC+ is unbounded. Suppose on the contrary that C+ is bounded. Using the similar method to prove Theorem 1 in [16] with obvious changes, we can find a neighborhood O of C+ such that ∂O∩P+=∅.
In order to complete the proof of this theorem, we consider problem (27). For ϵ>0, it is easy to show that nonlinear term fr,vvϵ,λ+gr,v,λ satisfies the condition (H2). Let(39)Pϵ=λ,v:λ,vsatisfies27andv≢0¯ℝ×E.
By Lemma1, there exists an unbounded continuum Cϵ of Sϵ bifurcating from λ1,0 such that Cϵ⊂K+∪λ1,0. Furthermore, by Lemma 2, there are two unbounded continua Cϵ+ and Cϵ− of Sϵ bifurcating from λ1,0, which are both unbounded and Cϵν⊂ℝ×Pν∪λ1,0.
So there existsλϵ,vϵ∈Cϵ+∩∂O for all ϵ>0. Since o is bounded in K+, equation (27) shows that λϵ,vϵ is bounded in ℝ×C2 independently of ϵ. By the compactness of Gk, one can find a sequence ϵn⟶0 such that λϵn,vϵn converges to a solution λ,v of (27). So v∈P+¯. If v∈∂P+, then from Lemma 9, it follows that v≡0. By Lemma 10, λ∈I, which contradicts the definition of O. On the other hand, if v∈P+, then λ,v∈P+∩∂O which contradicts P+∩∂O=∅.
From Theorem1 and its proof, we can easily get the following corollaries.Corollary 1.
There exist two unbounded subcontinuaD+ and D− of solutions of (26) in ℝ×Y, bifurcating from I×0 and Dν⊂ℝ×Pν∪I×0 for ν=+ and ν=−.We relax the assumption ofax as the following.B
0
a
x
∈
C
B
¯ is a radially symmetric sign-changing weight.Corollary 2.
LetA3 hold and f≡0. Then, there exist two unbounded subcontinua C+ and C− of solutions of (26) in ℝ×Y, bifurcating from I×0, and Cν⊂ℝ×Pν∪I×0 for ν=+ and ν=−.


## 4. Spectrum of Half-Quasilinear Eigenvalue Problems

With a simple transformationv=−u, problem (7) can be equivalently written as(40)rN−k−v′k′=NCNkrN−1λarvk+αrv+k+βrv−k,r∈0,1,v′0=v1=0,where v+=maxv,0,v−=−minv,0, and αrandβr satisfy the following:A
3
α
r
,
β
r
∈
C
0,1 are radially symmetric.Problem (40) is called half quasi-linear because it is positive k homogeneous in the cones u>0 and u<0. Similar to that of [16], we say that λ is a half-eigenvalue of problem (40), if there exists a nontrivial solution λ,uλ. λ is also said to be simple if v=cuλ,c>0 for all solutions λ,v of problem (40). A half-eigenvalue is called the principal half-eigenvalue if the corresponding eigenfunction is positive or negative.By an argument similar to that of [17, 18], we can obtain the following theorem.Theorem 2.
There exist two simple half-eigenvaluesλ+ and λ− for problem (40). The corresponding half-linear solutions are in λ+×P+ and λ−×P−. Furthermore, aside from λ+ and λ−, there is no other half-eigenvalue with positive or negative eigenfunction.Proof.
Applying the similar method to prove Theorem 1.2 in [17] or Theorem 3.1 in [18] with obvious changes, by Theorem 1 and Lemma 3–5, we can prove the conclusion.
Naturally, we can consider the bifurcation structure of the perturbation of problem (40) of the following form:(41)rN−k−v′k′=NCNkrN−1λarvk+αrv+k+βrv−k+gr,v,λ,r∈0,1,v′0=v1=0,where g satisfies (A1) and (12).Remark 1.
By some simple computation, we can show that ifβ≡0, then λ−=λ1, α≡0 implies λ+=λ1, and α=β≡0 implies λ+=λ−=λ1.Remark 2.
Note thatminλ+,λ− is the least half-eigenvalue for (40).
Using a similar method to prove Theorem 3.1 in [17] or Theorem 3.2 in [18] with obvious changes, by Corollary 1, we may obtain the following result.Theorem 3.
Forν=+,−, λν,0 is a bifurcation point for problem (41). Moreover, there exists an unbounded continuum Dν of solutions of problem (41), such that Dν⊂ℝ×Pν∪λν,0.Remark 3.
Theorem3 indicates that the bifurcation interval I=λ+,λ−, i.e., for problem (41), the bifurcation interval I is a finite point set.


## 5. One-Sign Solutions for the Hessian Equation

With a simple transformationv=−u, problem (8) can be equivalently written as(42)rN−k−v′k′=NCNkrN−1αrv+k+βrv−k+γarfv,r∈0,1,v′0=v1=0,where f satisfies the following subcritical growth restriction (see [15], (6.1)):(43)fs≤C1+sp,for some p∈k,k∗. Here,(44)k∗=Nk+1N−2k,ifk<N2,+∞,ifk≥N2.These are the lower critical exponent for the k-Hessian operator (see [26, 27]).We assume thatf satisfies the following assumptions:(H1)sfs>0 for s≠0(H2)f0,f∞∈0,+∞(H3)f0=∞ and f∞∈0,∞(H4)f0∈0,∞ and f∞=∞(H5)f0=0 and f∞∈0,∞.(H6)f0∈0,∞ and f∞=0(H7)f0=∞ and f∞=0(H8)f0=0 and f∞=∞(H9)f0=∞ and f∞=∞(H10)f0=0 and f∞=0Here,(45)f0=lims⟶0fssk,f∞=lims⟶+∞fssk.Obviously, any one-sign solution of problem (42) is a one-sign solution of problem (8). Conversely, we claim that any one-sign solution of problem (8) is a one-sign solution of problem (42).Define the Nemytskii operatorH1:ℝ×E⟶Y by(46)H1μ,vr≔NCNkrN−1αrv+k+βrv−k+μarfv.Then, it is clear thatH1 is a continuous (compact) operator and problem (42) can be equivalently written as(47)v=GkH1μ,v≔F1μ,v,where F1:ℝ×E⟶E is completely continuous and F1μ,0=0,∀μ∈ℝ.The main results of this section are the following theorems.Theorem 4.
LetA0, A3, H1, and H2 hold. Then, for any γ∈λν/f∞,λν/f0∪λν/f0,λν/f∞ with ν=+,−, problem (8) possesses two solutions u+ and u− such that νuν>0.Remark 4.
FromH1 and H2, we can see that there exists a positive constant Q such that fs/s≥Q for all s≠0.Remark 5.
Note that ifαt=βt≡0, the result of Theorem 4 is equivalent to those of Theorem 6.3 in [15]; Theorem 4 extends the corresponding results of [15].
In order to prove Theorem4, similar to the proof of Lemma 2.3 of [15] and Lemma 3.1 of [17] with obvious change, we need the following Sturm-type comparison result.Lemma 11.
Letb2r≥maxb1r,b1r+α+β,b1r−α−β for r∈0,1 and bir∈C0,1,i=1,2. Also, let v1andv2 be solutions of the following differential equations:(48)rN−k−v′k′=NCNkrN−1birvk+αrv+k+βrv−k,r∈0,1,v′0=v1=0,respectively. If c,d⊆0,1, v1c=v1d=0, and v1t≠0 in c,d, then either there exists τ∈c,d such that v2τ=0 or b2=b1 or b2=b1+α+β or b2=b1−α−β and v2r=μv1r for some constant μ≠0.Proof.
Ifu2r=0 on c,d, we discuss two cases.Case 1.
u
1
r
u
2
r
>
0 in c,d, similar to the proof of Lemma 10. By Lemma 3 and 4, we have that u2=μu1 and b2=b1.Case 2.
u
1
r
u
2
r
<
0 in c,d, similar to the proof of Case 1. We have that u2=μu1 and b2=b1+α+β or b2=b1−α−β.
By Lemma11, we obtain the following result that will be used later.Lemma 12.
Letbir∈C0,1,i=1,2 and b2r≥maxb1r,b1r+α+β,b1r−α−β and b2r≢maxb1r,b1r+α+β,b1r−α−β on 0,1. Also let v1andv2 be solutions of problem (48). If v1r≠0 in 0,1, then v2 has at least one zero in (0, 1).Proof of Theorem 4.
It suffices to prove that problem (42) possesses two solutions v+andv− such that νvν>0 for ν=+,−. Note that f0∈0,+∞ implies f0=0. Hence, v=0 is always the solution of problem (42). Let ζ∈Cℝ,ℝ such that fu=f0vk+ζv with lims⟶0ζs/sk=0. For ν=+,−, applying Theorem 3 to problem (42), we know that there is an unbounded continua Dν of solutions to problem (42), bifurcating from λν/f0,0, such that(49)Dν⊂Pν∪λνf0,0.
To complete this theorem, it will be enough to show thatDν joins λν/f0,0 to λν/f∞,∞. Let μn,vn∈Dν\λν/f0,0 satisfy μn+vn⟶∞. We note that μn>0 for all n∈ℕ since 0,0 is the only solution of problem (42) for γ=0 and Dν∩0×Y=∅.
We divide the rest of proofs into two steps.Step 1. We show that there exists a constant numberM>0 such that μn∈0,M, for all n.On the contrary, we suppose thatlimn⟶+∞μn=+∞. Since μn,vn∈Dν, it follows from the compactness of Gk that(50)rN−k−vn′k′=NCNkrN−1μnarfn˜rvnk+αrvn+k+βrvn−k,r∈0,1,v′0=v1=0,wherefn˜:0,1×0,∞ by(51)fn˜r=fvnvnk,vn≠0,r∈0,1,f0,vn=0,r∈0,1.By Remark4, we have(52)limn⟶+∞μnfn˜r=±∞.Letψν be an eigenfunction corresponding to λν. But if limn⟶+∞μnfn˜r=−∞, applying Lemma 12 to vn and ψν, we have that ψν must change sign for n large enough, by ψν∈Pν, which is impossible. So limn⟶+∞μnfn˜r=+∞, applying Lemma 12 to ψν and vn, we get that vn must change sign for n large enough, and this contradicts the fact that vn∈Pν.Step 2. We show thatDν joins λν/f0,0 to λν/f∞,∞.It follows from Step 1 thatvn⟶+∞. Let ξ∈Cℝ,ℝ be such that fv=f∞vk+ξv. Then, limv⟶+∞ξv/vk=0. Let ξ¯v=maxξs:0≤s≤v, then  ξ¯ is nondecreasing. Set ξ˜v=maxξs:v/2≤s≤v. Then, we have that(53)limv⟶+∞ξ˜vvk=0,ξ¯v≤ξ¯v2+ξ˜v.
It follows that(54)limv⟶+∞ξ¯vvk=0.
We divide the equation(55)rN−k−vn′k′=NCNkrN−1μnarf∞νn+αrvn+k+βrvn−k+μnarξvn,r∈0,1,νn0=νn1=0,by vnk and set wn=vn/vn. Since wn is bounded in C20,1, choosing a subsequence and relabeling if necessary, we have that wn⟶w for some w∈E. Moreover, from (54) and the fact that ξ¯ is nondecreasing, we have that(56)limn⟶∞ξvntvnk=0,∀t∈0,1,since(57)ξvnrvnk≤ξ¯vnrvnk≤ξ¯vnr∞vnk≤ξ¯vnrvnk⟶0,n⟶∞,∀r∈0,1.
By the compactness ofGk, we obtain that(58)rN−k−wn′k′=NCNkrN−1μnarf∞wn+αrwn+k+βrwn−k,r∈0,1,wn′0=wn1=0,where μ≔limn⟶∞μn again choosing a subsequence and relabeling if necessary. Thus, it is clear that w=1 and w∈Dν¯⊆Dν since Dν is closed in ℝ×E. Moreover, by Theorem 2, μf∞=λν, so that(59)μ=λνf∞.
Thus,Dν joins λν/f0,0 to λν/f∞,∞.Remark 6.
From theProof of Theorem 4, we can see that if f0,f∞∈0,+∞, then there exist λ2>0 and λ3>0 such that problem (8) possesses two one-sign solutions u+ and u−νuν>0,ν=+,− for all λ∈−λ3,−λ2∪λ2,λ3 and has no one-sign solution for all λ∈−λ2,0∪0,λ2∪−∞,−λ3∪λ3,∞.Proof.
We shall only prove the case ofλ∈0,λ2 since the proofs of the other cases are completely analogous.
Clearly,f0,f∞∈0,+∞ implies that there exists a positive constant M such that(60)fssk≤M,for anys≠0.
It is sufficient to show that there existsλ2>0 such that problem (42) has no solution for all λ∈0,λ2. Suppose on the contrary that there exists one pair μ,v∈D+ such that μ∈0,CNk−α0−β0/γa0M. Let w=v/v. Obviously, one has that(61)1=w=∫r1NCNksk−N∫0sτN−1ατwn+k+βτwn−k+μγaτfvvnkdτ1/kds≤α0+β0+μγa0MCNk1/k<1.
This is a contradiction.
From theProof of Theorem 4 and Remark 6, we can deduce the following two corollaries.Corollary 3.
Assume that there exists a positive constantρ>0 such that(62)fssk≥ρ,for anys≠0.
Then, there existsς∗>0 such that problem (8) has no one-sign solution for any λ∈−∞,−ς∗∪ς∗,+∞.Corollary 4.
Assume that there exists a positive constantϱ>0 such that(63)fssk≤ϱ,for anys≠0.
Then, there existsη∗>0 such that problem (8) has no one-sign solution for any λ∈−η∗,0∪0,η∗.Theorem 5.
LetA0, A3, H1, and H3 hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈0,λν/f∞,forλν>0(ii)
γ∈0,λ+/f∞∪λ−/f∞,0,forνλν>0(iii)
γ∈0,λ−/f∞∪λ+/f∞,0,forνλν<0(iv)
γ∈λν/f∞,0,forλν<0
Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
We shall only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
Inspired by the idea of [28], we define the cutoff function of f as follows:(64)fns≔nsk,s∈−1n,1n,f2n−1nk−1ns+2nk−1−f2n,s∈1n,2n,−f−2n−−1knk−1ns+2−1knk−1−f−2n,s∈−2n,−1n,fs,s∈−∞,−2n∪2n,+∞.
We consider the following problem:(65)rN−k−v′k′=NCNkrN−1αrv+k+βrv−k+γarfnv,r∈0,1,v′0=v1=0.
Clearly, we can see thatlimn⟶+∞fns=fs, fn0=n, and fn∞=f∞.
Similar to theProof of Theorem 4, there exists an unbounded continuum Dνn of solutions of problem (65) emanating from λν/n,0 and joining to λν/f∞,∞, such that Dνn⊂ℝ×Pν∪λν/n,0 and λν/f∞,∞∈Dνn.
Takingz∗=0,0, we easily obtain that z∗∈liminfn⟶+∞Dνn. So condition (i) in Lemma 6 is satisfied with z∗=0,0.
SinceF1 is completely continuous from ℝ×E⟶E, we have that ∪n=1+∞Dνn∩B¯R is precompact, and accordingly (ii) in Lemma 6 holds.
Therefore, by Lemma6, Dν=limsupn⟶∞Dνn is unbounded, closed, and connected such that z∗∈Dν.
Clearly,ν is the solution of problem (42) for any λ,u∈Dν. From the definition of superior limit (see [24, p. 7]), we can easily see that Dν⊆∪n=1∞Dνn. So one has that Dν⊆ℝ×Pν∪0,0.
We may claim thatz∗=0,0 is the unique bifurcation point of Dν.
Suppose on the contrary that there exists a sequenceλn,νn∈Dν/0,0=limsupn⟶∞Dνn/0,0 such that limn⟶∞λn=μ≠0 and limn⟶∞ν=0. Hence, for any N0∈ℕ, there exists n0≥N0 such that λn,νn∈Dνn0. By (54), it follows that λn0=λν/n0 for n0≥N0. From the arbitrary of N0, it implies that n0⟶∞, i.e., μ=0, which contradicts the assumption of μ≠0.
Similar to the method of the proof of Step 2 in Theorem4, we can obtain that λν/f∞,∞∈Dν.
Thus,Dν is an unbounded component of solutions of problem (42) emanating from 0,0, such that Dν⊂ℝ×Pν∪0,0, and Dν joins 0,0 to λν/f∞,∞.Theorem 6.
LetA0, A3, H1, H4, and (43) hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈0,λν/f0,forλν>0(ii)
γ∈0,λ+/f0∪λ−/f0,0,forνλν>0(iii)
γ∈0,λ−/f0∪λ+/f0,0,forνλν<0(iv)
γ∈λν/f0,0,forλν<0
Then problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
We shall only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
Ifλ,v is any nontrivial solution of problem (42), dividing problem (42) by v2k and setting w=v/v2 yields(66)rN−k−w′k′=NCNkrN−1αrw+k+βrw−k+γarfvv2k,r∈0,1,w′0=w1=0.
Define(67)f˜w≔w2kfww2,ifw≠0,0,ifw=0.
Evidently, problem (66) is equivalent to(68)rN−k−w′k′=NCNkrN−1αrw+k+βrw−k+γarf˜w,r∈0,1,w′0=w1=0.
It is obvious thatγ,0 is always the solution of problem (68). By simple computation, we can show that f˜0=f∞=∞ and f˜∞=f0∈0,∞. Now, applying Theorem 5, there exists an unbounded continuum Cν of solutions of problem (68) emanating from 0,0, such that Cν⊂ℝ×Pν∪0,0, and Dν joins 0,0 to λν/f˜∞,∞.
Under the inversionw⟶w/w2=v, we obtain Cν⟶Dν being an unbounded component of solutions of problem (42) emanating from λν/f0,0. Applying a similar method as the proof of Theorem 6.5 in [15], by (43), we can get that 0,∞∈Dν.Theorem 7.
LetA0, A3, H1, and H5 hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈λν/f∞,+∞,forλν>0(ii)
γ∈λ+/f∞,+∞∪−∞,λ−/f∞,forνλν>0(iii)
γ∈λ−/f∞,+∞∪−∞,λ+/f∞,forνλν<0(iv)
γ∈−∞,λν/f∞,forλν<0
Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
We shall only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
Define(69)fns≔1nsk,s∈−1n,1n,f2n−1nk+1ns+2nk+1−f2n,s∈1n,2n,−f−2n−−1knk+1ns+2−1knk+1−f−2n,s∈−2n,−1n,fs,s∈−∞,−2n∪2n,+∞.
We consider the following problem:(70)rN−k−v′k′=NCNkrN−1αrv+k+βrv−k+γarfnv,r∈0,1,v′0=v1=0.
Clearly, we can see thatlimn⟶+∞fns=fs, fn0=1/n, and fn∞=f∞.
Similar to theProof of Theorem 4, there exists an unbounded continuum Dνn of solutions of problem (70) emanating from λνn,0 and joining to λν/f∞,∞, such that Dνn⊂ℝ×Pν∪λνn,0 and λν/f∞,∞∈Dνn.
Takingz∗=∞,0, we easily obtain that z∗∈liminfn⟶+∞Dνn with z∗ℝ×E=+∞ So condition (i) in Lemma 7 is satisfied with z∗=+∞,0.
Define a mappingT:R×E⟶R×E such that(71)Tγ,v=1γ,v,γ∈−∞,0∪0,+∞,0,ν,γ=∞,∞,ν,γ=0.
It is easy to verify thatT is a homeomorphism and Tz∗ℝ×E=0. Obviously, TDνn be a sequence of unbounded connected subsets in E. So (ii) in Lemma 7 holds.
SinceF1 is completely continuous from ℝ×E⟶E, we have that ∪n=1+∞TDνn∩B¯R is s precompact, and accordingly (iii) in Lemma 7 holds.
Therefore, by Lemma7, Dν=limsupn⟶∞Dνn is unbounded, closed, and connected such that z∗∈Dν. Obviously, v is the solution of problem (42) for any γ,v∈Dν. From the definition of superior limit (see [24], P. 7), we can easily see that Dν⊆∪n=1∞Dνn. So one has that Dν⊂ℝ×Pν∪0.
Next, we show that the projection ofDν on ℝ is nonempty. By Theorem 4, we have known that Dνn has unbounded projection on ℝ for any fixed n∈ℕ. By Lemma 8, for each fixed ϵ>0, there exists an m such that for every n>m,Dνn⊂VϵDν. This implies that(72)λνn,∞⊆ProjℝDνn⊆ProjℝVϵDν,where ProjℝDν denotes the projection of Dν on ℝ. It follows that the projection of Dν is nonempty on ℝ.Theorem 8.
LetA0, A3, H1, and H6 hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈λν/f0,+∞,forλν>0(ii)
γ∈−∞,λ+/f0∪λ−/f0,+∞,forνλν>0(iii)
γ∈−∞,λ−/f0∪λ+/f0,+∞,forνλν<0(iv)
γ∈−∞,λν/f0,forλν<0
Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
Applying a similar method as the proof of Theorem6 and the conclusion of Theorem 7, we can easily get the desired conclusion.Theorem 9.
LetA0, A3, H1, and H7 hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈0,+∞,forλν>0(ii)
orγ∈0,+∞∪−∞,0, forνλν>0,orνλν<0(iii)
γ∈−∞,0,forλν<0
Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.
Define(73)fns≔1nsk,s∈−∞,−2n∪2n,+∞.−−2knk−1+f−nns+n+f−n,s∈−2n,−n,2knk−1−fnns−n+fn,s∈n,2n,fs,s∈−n,−2n∪2n,n,−f−2n−−1knk−1ns+2−1knk−1−f−2n,s∈−2n,−1n,f2n−1nk−1ns+2nk−1−f2n,s∈1n,2n,nsk,s∈−1n,1n.
We consider the following problem(74)rN−k−v′k′=NCNkrN−1αrv+k+βrv−k+γarfnv,r∈0,1,v′0=v1=0.Clearly, we can see thatlimn⟶+∞fns=fs, fn0=n, and fn∞=1/n.Similar to theProof of Theorem 4, there exists an unbounded continuum Dνn of solutions of problem (74) emanating from λν/n,0, such that Dνn⊂ℝ×Pν∪λν/n,0 and λνn,∞∈Dνn.Takingzn=λν/n,0 and z∗=0,0, we have that zn⟶z∗. Similar to the proof of Theorem 5, by Lemma 6, Dν=limsupn⟶∞Dνn is unbounded, closed, and connected such that z∗=0,0∈Dν, and z∗=0,0 is the unique bifurcation point of Dν. By Theorem 7, we have z∗=∞,∞∈Dν.Thus, we can obtain the results.Theorem 10.
LetA0, A3, H1, H8, and (43) hold. For ν∈+,−, assume that one of the following conditions holds:(i)
γ∈0,+∞,forλν>0(ii)
orγ∈0,+∞∪−∞,0, forνλν>0,orνλν<0(iii)
γ∈−∞,0, forλν<0Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
Applying a similar method as the proof of Theorem6 and the conclusion of Theorem 9, we can easily get the desired conclusion.Theorem 11.
LetA0, A3, H1, H9, and (43) hold. For ν∈+,−, assume that one of the following conditions holds:(i)
There exists aλνν>0forλν>0, such that γ∈0,λνν(ii)
There exists aνλνν>0forνλν>0, such that γ∈λ−−,0∪0,λ++(iii)
There exists aνλνν<0forνλν<0, such that,  γ∈λ++,0∪0,λ−−(iv)
There exists aλνν<0forλν<0, such that γ∈λνν,0Then, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
Define(75)fns≔nsk,s∈−∞,−2n∪2n,+∞.−−2knk+1+f−nns+n+f−n,s∈−2n,−n,2knk+1−fnns−n+fn,s∈n,2n,fs,s∈−n,−2n∪2n,n,−f−2n−−1knk−1ns+2−1knk−1−f−2n,s∈−2n,−1n,f2n−1nk−1ns+2nk−1−f2n,s∈1n,2n,nsk,s∈−1n,1n.
We consider the following problem:(76)rN−k−v′k′=NCNkrN−1αrv+k+βrv−k+γarfnv,r∈0,1,v′0=v1=0.
Clearly, we can see thatlimn⟶+∞fns=fs, fn0=n and fn∞=n.
Similar to theProof of Theorem 4, there exists an unbounded continuum Dνn of solutions of problem (48) emanating from λν/n,0, such that Dνn⊂ℝ×Pν∪λν/n,0 and λν/n,+∞∈Dνn.
Takingzn=λν/n,0 and z∗=0,0, we have that zn⟶z∗. Similar to the proof of Theorem 5, by Lemma 6, Dν=limsupn⟶∞Dνn is unbounded, closed, and connected such that 0,0∈Dν. By Theorem 6, we have that 0,∞∈Dν.
Thus, we can easily get the results of this theorem.Theorem 12.
LetA0, A3, H1, and H10 hold. For ν∈+,−, assume that one of the following conditions holds:(i)
There exists aλνν>0forλν>0, such that γ∈λνν,+∞(ii)
There exists aνλνν>0forνλν>0, such that γ∈−∞,λ−−∪λ++,+∞(iii)
There exists aνλνν<0forνλν<0, such that γ∈−∞,λ++∪λ−−,+∞(iv)
There exists aλνν<0forλν<0, such that γ∈−∞,λννThen, problem (8) possesses two solutions u+ and u−, such that νuν>0 for ν=+,−.Proof.
Using an argument similar to that of Theorem6, in view of Theorem 11, we can easily get the results of this theorem.



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*Source: 2625683-2020-04-13.xml*