# Unilateral Global Bifurcation from Intervals for Fourth-Order Problems and Its Applications

Authors: Wenguo Shen; Tao He
Journal: Discrete Dynamics in Nature and Society
 (2016)

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## Abstract

We establish a unilateral global bifurcation result from interval for a class of fourth-order problems with nondifferentiable nonlinearity. By applying the above result, we firstly establish the spectrum for a class of half-linear fourth-order eigenvalue problems. Moreover, we also investigate the existence of nodal solutions for the following half-linear fourth-order problems:x″″=αx++βx-+ratfx, 0<t<1, x(0)=x(1)=x″(0)=x″(1)=0, where r≠0 is a parameter, a∈C([0,1],(0,∞)), x+=max⁡{x,0}, x-=-min⁡{x,0}, α,β∈C[0,1], and f∈C(R,R), sf(s)>0, for s≠0. We give the intervals for the parameter r which ensure the existence of nodal solutions for the above fourth-order half-linear problems if f0∈[0,∞) or f∞∈[0,∞], where f0=lims→0f(s)/s and f∞=lims→+∞f(s)/s. We use the unilateral global bifurcation techniques and the approximation of connected components to prove our main results.


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## Body

## 1. Introduction

In the past twenty years, fourth-order BVP have attracted the attention of many specialists in differential equations because of their interesting applications. For example, Bai and Wang [1], Ma and Wang [2], and Chu and O’Regan [3] have investigated the fourth-order BVP by the fixed point theory in cones. Meanwhile, by applying the bifurcation techniques of Rabinowitz [4, 5], Gupta and Mawhin [6], Lazer and McKenna [7], Rynne [8], Liu and O’Regan [9], Ma et. al [10, 11], Shen [12, 13], and Ma [14] studied the existence of nodal solutions for the fourth-order BVP.Now, consider the following operator equation:(1)u=λBu+Hλ,u,where B is a compact linear operator and H:R×E→E is compact with H=o(u) at u=0 uniformly on bounded λ intervals, where E is a real Banach space with the norm ·. If the characteristic value μ of B has multiplicity 1 and (2)S=λ,u:λ,usatisfies1,u≢0¯R×E,Dancer [15] has shown that there are two distinct unbounded continua Cμ+ and Cμ-, consisting of the bifurcation branch Cμ of S emanating from (μ,0), where either Cμ+ and Cμ- are both unbounded or Cμ+∩Cμ-≠{(μ,0)}. This result has been extended to the fourth-order problems by Dai and Han [16]. More specifically, Dai and Han [16] considered the following fourth-order problem:(3)x′′′′=λatx+gt,x,λ,0<t<1,x0=x1=x′′0=x′′1=0,where g:[0,1]×R2→R is continuous satisfying g(t,x,0)≡0, and(4)limx→0⁡gt,x,λx=0uniformly on [0,1] and μ on bounded sets.LetY=C[0,1] with the norm x∞=maxt∈[0,1]xt. Let E={x∈C3[0,1]∣x(0)=x(1)=x′′(0)=x′′(1)=0} with the norm x=x∞+x′∞+x′′∞+x′′′∞. Let E=R×E under the product topology. Let Sk+ denote the set of functions in E which have exactly k-1 generalized simple zeros in (0,1) and are positive near t=0, set Sk-=-Sk+, and Sk=Sk+∪Sk-. They are disjoint and open in E. Finally, let Φk±=R×Sk± and Φk=R×Sk. Let C denote the closure of the set of nontrivial solutions of (3) in R×E, and Ckν denote the subset of C with u∈Skν for ν∈+,- and Ck=Ck+∪Ck-.From Dancer [15], Dai and Han [16] obtained that problem (3) has two distinct unbounded subcontinua Ck+ and Ck-, consisting of the bifurcation branch Ck emanating from (λk,0), which satisfy the following.Lemma 1.
EitherCk+ and Ck- are both unbounded or Ck+∩Ck-≠{λk,0}, and (5)Ckν⊂R×Skν∪λk,0,where ν∈{+,-} and λk were defined as in [10] or [16].For the abstract unilateral global bifurcation theory, we refer the reader to [4, 5, 15, 17–19] and the references therein.However, among the above papers, the nonlinearities are linear in the zeros and infinity. The problems involving nondifferentiable nonlinearities have also been investigated by applying bifurcation techniques; see Berestycki [20], Schmitt and Smith [19], Rynne [21], Ma and Dai [22], and Dai et al. [23–25] and references therein. Among them, in 1977, Berestycki [20] studied the differential equations involving nondifferentiable nonlinearity. The above Berestycki’s ([20]) result has been improved partially by Schmitt and Smith [19] by applying a set-valued version of Rabinowitz global bifurcation theorem. In 1998, Rynne [21] established the interval bifurcation from u=0 and u=∞ and obtained sets of positive or negative solutions with the approximation technique from Berestycki [20]. Recently, Ma and Dai [22] established the global interval bifurcation for a Sturm-Liouville problem with a nonsmooth nonlinearity by [15]. Later, Dai et al. [23–25] studied the bifurcation from intervals for Sturm-Liouville problems and its applications and established the unilateral global interval bifurcation for p-Laplacian with non-p-1-linearization nonlinearity, respectively.On the other hand, half-linear or half-quasilinear problems have attracted the attention of some specialists; see [20, 22, 25]. Among them, Berestycki [20] studied the bifurcation structure for the half-linear equations. Recently, Ma and Dai [22] and Dai and Ma [25] studied the existence of nodal solutions for a class of half-linear or half-quasilinear eigenvalue problems and improved Berestycki’s result, respectively.Motivated by the above papers, in this paper, we will firstly establish some Dancer-type unilateral global bifurcation results about the continuum of solutions for the fourth-order problems:(6)x′′′′=λatx+Ft,x,λ,0<t<1,x0=x1=x′′0=x′′1=0,where λ is a parameter, the nonlinear term F has the form F=f+g, where f and g are continuous functions on [0,1]×R3, and a,f,g satisfy the following conditions:(H1)
a∈C([0,1],(0,∞)).(H2)
ft,x,λ/x≤M1, for all t∈[0,1], 0<x≤1, ∀λ∈R, where M1 is a positive constant.(H3)
g(t,x,λ)=o(x) near x=0, uniformly in t∈[0,1] and λ on bounded sets.LetS denote the closure of the set of nontrivial solutions of (6) in R×E, and Skν denote the subset of S with u∈Skν for ν∈+,- and Sk=Sk+∪Sk-.Under assumptions(H1)–(H3), we will show that Ik=[λk-d1,λk+d1] is a bifurcation interval of problem (6) and there are two distinct unbounded subcontinua, Ck+ and Ck-, consisting of the bifurcation branch Ck from Ik=[λk-d1,λk+d1], where λk is given in Lemma 1.On the basis of the above unilateral global interval bifurcation result, we study the following half-linear eigenvalue problem:(7)x′′′′=λatx+αx++βx-,0<t<1,x0=x1=x′′0=x′′1=0,where x+=max⁡{x,0}, x-=min⁡{x,0} and α(t) and β(t) satisfy the following:(H4)
α(t),β(t)∈C[0,1].We will show that there exist two sequences of simple half-eigenvalues for problem (7): λ1+<λ2+⋯<λk+<⋯ and λ1-<λ2-⋯<λk-<⋯. The corresponding half-linear solutions are in {λk+}×Sk+ and {λk-}×Sk-. Furthermore, aside from these solutions and the trivial ones, there is no other solutions of problem (7).Following the above eigenvalue theory (see Theorem19), we will investigate the existence of nodal solutions for the following fourth-order problem:(8)x′′′′=αx++βx-+λatfx,0<t<1,x0=x1=x′′0=x′′1=0.Let Σ denote the closure of the set of nontrivial solutions of (8) in R×E, with Σkν denoting the subset of Σ with u∈Skν for ν∈+,- and Σk=Σk+∪Σk-. For α=β=0, Dai and Han [16] have established the existence of nodal solutions for problem (8) with crossing nonlinearity. In this paper, we assume that f satisfies the following assumptions:(H5)
sf(s)>0 for s≠0.(H6)
f0∈(0,∞) and f∞∈(0,∞).(H7)
f0∈(0,∞) and f∞=∞.(H8)
f0∈(0,∞) and f∞=0.(H9)
f0=0 and f∞∈(0,∞).(H10)
f0=0 and f∞=∞.(H11)
f0=0 and f∞=0,where(9)limx→0⁡fxx=f0,∀t∈0,1,limx→∞⁡fxx=f∞,∀t∈0,1.The rest of this paper is arranged as follows. In Section2, we have given some preliminaries. In Section 3, we establish the unilateral global bifurcation result from the interval for problem (6). In Section 4, on the basis of the unilateral global interval bifurcation result, we will establish the spectrum for a class of the half-linear fourth-order eigenvalue problem (7) (see Theorem 19). In Section 5, following the above eigenvalue theory (see Theorem 19), we will investigate the existence of nodal solutions for a class of the half-linear fourth-order problem (8).


## 2. Hypotheses and Lemmas

We define the linear operatorL:D(L)⊂E→Y:(10)Lx=x′′′′,x∈DL,with D(L)={x∈C4[0,1]∣x(0)=x(1)=x′′(0)=x′′(1)=0}.From [26, p. 439-440], we consider the following auxiliary problem:(11)x′′′′=ht,t∈0,1,x0=x1=x′′0=x′′1=0,for a given h∈C[0,1]. We can get that problem (11) can be equivalently written as (12)xt=L-1ht=∫01kt,shsds,where(13)kt,s≔16s1-t2t-s2-t2,s≤t,16t1-s2s-t2-s2,s>t.Then L is a closed operator and L-1:Y→E is completely continuous.Define the Nemytskii operatorH:R×E→E by (14)Hλ,xt≔λL-1ax+L-1gt,x,λ.Then it is clear thatH is continuous operator and problem (3) can be equivalently written as (15)x=Hλ,x.Clearly, H:R×E→E is completely continuous and H(μ,0)=0, ∀μ∈R.Let(16)g¯t,x,λ=max0≤s≤x⁡gt,s,λ∀t∈0,1,λon  bounded  sets,and then g¯ is nondecreasing with respect to x, and(17)limx→0+⁡g¯t,x,λx=0uniformly for t∈[0,1] and λ on bounded sets. Further it follows from (17) that(18)gt,x,λx≤g¯t,x,λx≤g¯t,x∞,λx≤g¯t,x,λx⟶0asx⟶0uniformly for t∈[0,1] and λ on bounded sets.In the following, we summarize some preliminary results from [10, 16].Definition 2 (see [16]).
Letu∈E and t∗∈I such that u(t∗)=u′′(t∗)=0. We call that t∗ a generalized simple zero if u′(t∗)≠0 or u′′′(t∗)≠0. Otherwise, we call that t∗ a generalized double zero. If there is no generalized double zero of u, we call that u a nodal solution.Lemma 3 (see [10] or [16]).
Let(H1) hold. The linear eigenvalue problem(19)x′′′′t=λatx,0<t<1,x0=x1=x′′0=x′′1=0has an infinite sequence of positive eigenvalues: (20)0<λ1a<λ2a<⋯<λka⟶+∞k⟶+∞.Moreover, each eigenvalue is simple. To each eigenvalue λk(a) there corresponds an essential unique eigenfunction ψk which has exactly k-1 generalized simple zeros in (0,1) and is positive near 0.Lemma 4 (see [16]).
If(λ,x) is a nontrivial solution of (6) under assumptions (H2) and (H3) and x has a generalized double zero, then x≡0.Remark 5.
By Lemma4, we can see that if (λ,x) is a nontrivial solution of (6) under assumptions (H2) and (H3), then x∈⋃k=1∞Skν.Lemma 6 (see [20, Lemma 2]).
Letj and k be two integers such that j≥k≥2. Suppose that there exist two families of real numbers:(21)ξ0=0<ξ1<ξ2<⋯<ξk-1<ξk=1,η0=0<η1<η2<⋯<ηj-1<ηj=1.Then, if ξ1≤η1, there exist two integers p and q having the same parity: 1≤p≤k-1 and 1≤q≤j-1, such that (22)ξp≤ηq≤ηq+1≤ξp+1.Definition 7 (see [27]).
LetX be a Banach space and let {Cn∣n=1,2,…} be a family of subsets of X. Then the superior limit D of {Cn} is defined by(23)D≔limsupn→∞⁡Cn=x∈X∣∃ni⊂N,xni∈Cni,such  thatxni⟶x.Lemma 8 (see [27]).
Each connected subset of metric spaceX is contained in a component, and each connected component of X is closed.Lemma 9 (see [28]).
LetX be a Banach space and let {Cn∣n=1,2,…} be a family of closed connected subsets of X. Assume that(i)
there existzn∈Cn, n=1,2,…, and z∗∈X, such that zn→z∗;(ii)
rn=sup⁡{x∣x∈Cn}=∞;(iii)
for allR>0, ⋃n=1∞Cn∩BR is a relative compact set of X, where (24)BR=x∈X∣x≤R.Then there exists an unbounded component C in D and z∗∈C.


## 3. Unilateral Global Bifurcation

The main result for problem (6) is the following theorem.Theorem 10.
Let(H1), (H2), and (H3) hold. Let d1=M1/a0, where a0=mint∈[0,1]a(t), and let Ik=[λk-d1,λk+d1] for every k∈N. The component Ck+ of Sk+∪(Ik×{0}), containing Ik×{0}, is unbounded and lies in Φk+∪(Ik×{0}) and the component Ck- of Sk-∪(Ik×{0}), containing Ik×{0}, is unbounded and lies in Φk-∪(Ik×{0}).In order to prove Theorem10, we need the following results.Lemma 11.
IfCkν is bounded, we can find a neighborhood O of Ckν such that ∂O∩Skν=∅, where Ckν (ν=+,-) is given by Theorem 10.Proof.
We only prove the case ofCk+ since the case of Ck- is similar.
LetU be a uniform neighborhood of Ck+ in R×E.
We discuss two cases.
Case  1 (if
∂
U
∩
S
k
+
≠
∅
). Since the solutions of problem (6) are bounded in R×E, then Ck+ is compact in R×E. It follows that U¯∩Sk+ is compact metric space. Obviously, Ck+ and ∂U∩Sk+ are two disjoint closed subsets of U¯∩Sk+. Because of the maximal connectedness of Ck+, there does not exist a component C∗ of U¯∩Sk+ such that C∗∩Ck+≠∅, C∗∩(∂U∩Sk+)≠∅. By [27] or [4, Lemma 1.1], there exist two disjoint compact subsets K1,K2 of U¯∩Sk+, such that U¯∩Sk+=K1∪K2, Ck+⊂K1, ∂U∩Sk+⊂K2. Evidently, d(K1,K2)>0. Let r=min⁡{d(K1,K2),d(K1,∂U)}, and let V be the r/2-neighborhood of K1. Set (25)O=U∩V,and then(26)Ck+⊂O,∂O∩Sk+=∅.
Case  2 (if
∂
U
∩
S
k
+
=
∅
). In this case, we take O=U. It is obvious that the result holds.
Consider the following auxiliary approximate problem:(27)x′′′′=λatx+ft,xxϵ,λ+gt,x,λ,0<t<1,x0=x1=x′′0=x′′1=0.For ϵ>0, it is easy to show that nonlinear term λa(t)x+f(t,xxϵ,λ)+g(t,x,λ) is differ-entiable at the origin. Let (28)Sϵ=λ,x:λ,xsatisfies27,x≢0¯R×E.By Lemma1, there are two unbounded continua Ck,ϵ+ of Sϵ+ and Ck,ϵ- of Sϵ- bifurcating from (λk,0), consisting of the bifurcation branch Ck,ϵ, which satisfy the following result.Lemma 12.
C
k
,
ϵ
+ and Ck,ϵ- are both unbounded and (29)Ck,ϵν⊂Φkν∪λk,0ν=+,-.Proof.
By Lemma1.24 of [4], there exists a bounded open neighborhood Ok of (λk,0) such that (30)Ck,ϵν-λk,0∩Ok⊂Φkνν=+,-.It follows that (31)Ck,ϵν∩Ok⊂Φkν∪λk,0ν=+,-.By Lemma 4, we show that Ck,ϵν∖{(λk,0)} cannot leave Φkν outside of a neighborhood of (λk,0). Thus, we have(32)Ck,ϵν⊂Φkν∪λk,0ν=+,-.Next, we will prove that both Ck,ϵ+ and Ck,ϵ- are unbounded.
Without loss of generality, we may suppose thatCk,ϵ- is bounded. Therefore, in view of (32) and Lemma 1, there exists (λ∗,u∗)∈Ck,ϵ+∩Ck,ϵ- such that (λ∗,u∗)≠(λk,0) and u∗∈Sk+∩Sk-. This contradicts the definitions of Sk+ and Sk-.To prove Theorem10, the next lemma will play a key role.Lemma 13.
Letϵn, 0<ϵn<1, be a sequence converging to 0. If there exists a sequence (λn,xn)∈R×Skν such that (λn,xn) is a nontrivial solution of problem (27) corresponding to ϵ=ϵn, and (λn,xn) converges to (λ,0) in R×E, then λ∈Ik.Proof.
Without loss of generality, we may assume thatxn≤1. Let yn=xn/xn; then yn satisfies the problem(33)yn′′′′=λnatyn+fnt+gnt,0<t<1,yn0=yn1=yn′′0=yn′′1=0,where(34)fnt=ft,xnxnϵ,λnxn,gnt=gt,xn,λnxn.By (18), it follows that(35)limn→∞⁡gnt=0,uniformly for t∈[0,1] and λ on bounded sets. Furthermore, (H2) implies that(36)fnt=ft,xnxnϵ,λnxn≤xnϵM1xnxn=xnϵM1yn≤M1yn,for all t∈[0,1].
Note thatyn=1 implies yn∞≤1. Using this fact with (35) and (36), we have that λna(t)yn+fn(t)+gn(t) is bounded in C4[0,1] for n large enough. The compactness of L-1 implies that yn is convergence in C4[0,1]. Without loss of generality, we may assume that yn→y in E with y=1. Clearly, we have y∈Skν¯.
We claim thaty∈Skν. On the contrary, supposing that y∈∂Skν, by Lemma 4, then y≡0, which is a contradiction with y=1.
Now, we deduce the boundedness ofλ. Let φkν∈Skν be an eigenfunction of problem (19) corresponding to λk.
Let(37)t0=0<t1<t2<⋯<tk-1<tk=1,s0=0<s1<s2<⋯<sk-1<sk=1be, respectively, the sequences of generalized simple zeros of yn and φkν.
Supposet1≤s1; then we deduce from Lemma 6 existence of integers p and q having the same parity such that (38)tp≤sq≤sq+1≤tp+1.
Therefore, without loss of generality, we chooseξ1=t0, η1=t1, ξ2=sq, and η2=sq+1 and since p and q have the same parity yn and φkν do not vanish and have the same sign in both the intervals [ξ1,η1] and [ξ2,η2]. It follows that yn(ξ1)=yn(η1)=yn′′(ξ1)=yn′′(η1)=0, and φkν(ξ2)=φkν(η2)=φkν′′(ξ2)=φkν′′(η2)=0.
We can assume without loss of generality thatyn>0 and φkν>0 in (ξ1,η1). By the Picone identity in [29, Theorem 4], we have that(39)∫ξ1η1ynφkνynφkν′′′-φkνyn′′′+yn′φkνφkνyn′′-ynφkν′′-ynφkν′′ynφkν′′dt=A1+B1,A1=∫ξ1η1ynφkνynLφkν-φkνLyndt,B1=∫ξ1η1yn′′-ynφkνφkν′′2-2φkνφkν′′ynφkν′2dt.
The left-hand side of (39) equals(40)
-
lim
t
→
ξ
1
+
⁡
y
n
φ
k
ν
y
n
φ
k
ν
′
′
′
-
φ
k
ν
y
n
′
′
′
+
y
n
′
φ
k
ν
φ
k
ν
y
n
′
′
-
y
n
φ
k
ν
′
′
-
y
n
φ
k
ν
′
′
y
n
φ
k
ν
′
+
lim
t
→
η
1
-
⁡
y
n
φ
k
ν
y
n
φ
k
ν
′
′
′
-
φ
k
ν
y
n
′
′
′
+
y
n
′
φ
k
ν
φ
k
ν
y
n
′
′
-
y
n
φ
k
ν
′
′
-
y
n
φ
k
ν
′
′
y
n
φ
k
ν
′
≕
H
ξ
1
+
H
η
1
.
We prove thatHξ1≤0. If φkν(ξ1)≠0, then Hξ1=0. If φkν(ξ1)=0, noting the conclusion of Lemma 3, then φkν′(ξ1)>0. By L’Hospital rule, we have that(41)Hξ1=-limt→ξ1+⁡yn′ynφkν′′′-φkνyn′′′+ynynφkν′′′-φkνyn′′′′φkν′-limt→ξ1+⁡yn′′φkνyn′′-ynφkν′′+yn′φkν′yn′′+φkνyn′′′-yn′φkν′′-ynφkν′′′φkν′+φkν′′ξ1limt→ξ1+⁡yn′yn′φkν-ynφkν′+ynyn′′φkν-ynφkν′′2φkνφkν′=yn′ξ12φkν′′ξ1φkν′ξ1+φkν′′ξ12φkν′ξ1limt→ξ1+⁡yn′′yn′φkν-ynφkν′+2yn′yn′′φkν-ynφkν′′+ynyn′′φkν-ynφkν′′′φkν′=yn′ξ12φkν′′ξ1φkν′ξ1.
In the following, we will prove that(42)φkν′′t≤0,∀t∈ξ1,η1.Let u≔φkν(t)+1. Then u′′′′=λkaφkν and u≥1 in (ξ1,η1). For some ɛ>0 small enough, let v∈C4([-ɛ,1+ɛ]) and b≥0 be such that v(-ɛ)=v(1+ɛ)=v′′(-ɛ)=v′′(1+ɛ)=0, vI=u, and v′′′′=λkbv. Then we have(43)v′′′′=λkbv,t∈-ɛ,1+ɛ,v-ɛ=v1+ɛ=v′′-ɛ=v′′1+ɛ=0.Set m≔2ξ1-ɛ/2 and n≔2η1+ɛ/2. Let w∈C4([m,n]) and c≥0 be such that wI=v, w≥0 in (m,n) and w(m)=w(n)=w′′(m)=w′′(n)=0 and w′′′′=λkcw. Set z≔w′′; then z should be a solution of the problem(44)z′′=λkcw,t∈m,n,zm=zn=0.The Strong Maximum Principle implies that z<0 in (m,n). This follows that φkν′′(t)≤0 in [ξ1,η1]. It follows that Hξ1≤0.
Next, we will prove thatHη1≤0. If φkν(η1)≠0, then Hη1=0. If φkν(η1)=0, noting the conclusion of Lemma 3, then φkν′(η1)<0. By L’Hospital rule, we have that(45)Hη1=-yn′η12φkν′′η1φkν′η1.By (42), we can show that Hη1≤0. Therefore, the left-hand side of (39) ≤0.
By (42), we have B1≥0.
It follows that(46)0≥A1=∫ξ1η1ynφkνyn·Lφkν-φkν·Lyndt=∫ξ1η1λk-λnatyn2dt-∫ξ1η1ynfnt+gntdt.Similarly, we can also show that(47)0≥A2=∫ξ2η2φkνynφkν·Lyn-yn·Lφkνdt=∫ξ2η2λn-λkatφkν2dt+∫ξ2η2φkν2ynfnt+gntdt.
By (35), (46), and (47), taking the limit as n→+∞, we can obtain that(48)∫ξ1η1λk-λaty2dt-limn→∞⁡∫ξ1η1fntydt≤0,∫ξ2η2λ-λkatφkν2dt+limn→∞⁡∫ξ2η2φkν2ynfntdt≤0.
By (36), y and ϕkν have the same sign in (ξi,ηi)(i=1,2); one has (49)limn→∞⁡∫ξ1η1fntydt≤M1∫ξ1η1y2dt,limn→∞⁡∫ξ2η2φkν2ynfntdt≤M1∫ξ2η2φkν2dt.
So, by (48), we have that(50)∫ξ1η1λk-λaty2dt-M1∫ξ1η1y2dt≤0,∫ξ2η2λ-λkatφkν2dt-M1∫ξ2η2φkν2dt≤0.Furthermore, it follows thatifλ≤λk,(51)∫ξ1η1λk-λa0t-M1y2dt≤0,henceλ≥λk-d1,ifλ≥λk,(52)∫ξ2η2λ-λka0dt-M1φkν2dt≤0,henceλ≤λk+d1.
Therefore, we have thatλ∈Ik.Lemma 14.
Forν∈{+,-}, the component Ckν of Skν∪(Ik×{0}) satisfies Ckν⊂Φkν∪(Ik×{0}), where Ckν(ν=+,-) is given by Theorem 10.Proof.
We only prove the case ofCk+ since the case of Ck- is similar. For any (λ,x)∈Ck+, there are two possibilities: (i) x∈Sk+ or (ii) x∈∂Sk+. It is obvious that (λ,x)∈Φk+ in the case of (i). While case (ii) implies that x has at least one double zero in [0,1], Lemma 4 follows that x≡0. Hence, there exists a sequence (λn,xn)∈Φk+ such that (λn,xn) is a solution of problem (27) corresponding to ϵ=0, and (λn,xn) converges to (λ,0) in R×E. By Lemma 13, we have λ∈Ik: that is, (λ,x)∈Ik×{0} in the case of (ii). Hence, Ck+⊂Φk+∪(Ik×{0}).Proof of Theorem10.
We only prove the case ofCk+ since the case of Ck- is similar. Let Ck+ be the component of Sk+∪(Ik×{0}), containing Ik×{0}. By Lemma 14, we can show that Ck+⊂Φk+∪(Ik×{0}).
Suppose on the contrary thatCk+ is bounded. By Lemma 11, we can find a neighborhood O of Ck+ such that ∂O∩Sk+=∅.
In order to complete the proof of this theorem, we consider problem (27). By Lemma 1, there are two continua Ck,ϵ+ and Ck,ϵ-, consisting of the bifurcation branch Ck,ϵ. By Lemma 12, we have that Ck,ɛ+ and Ck,ɛ- are both unbounded and (53)Ck,ϵν∩Ok⊂Φkν∪λk,0.
So there exists(λϵ,xϵ)∈Ck,ϵ+∩∂O for all ϵ>0. Since O is bounded in R×E, (27) shows that (λϵ,xϵ) is bounded in R×C4 independently of ϵ. By the compactness of L-1, one can find a sequence ϵn→0 such that (λϵ,xϵ) converges to a solution (λ,x) of (6). So x∈Sk+¯. If x∈∂Sk+, then from Lemma 4 or Remark 5 it follows that x≡0. Note that (λ,x)∈Ck,ϵ+∩∂O since Ck,ϵ+∩∂O is closed subset of R×E. By Lemma 13, λ∈Ik, which contradicts the definition of O. On the other hand, if x∈Sk+, then (λ,x)∈Sk+∩∂O which contradicts Sk+∩∂O=∅.From Theorem10 and its proof, we can easily get the following two corollaries.Corollary 15.
There exist two unbounded subcontinuaDk,0+ and Dk,0- of solutions of (6) in R×E, bifurcating from Ik×{0}, and Dk,0ν⊂(R×Skν)∪(Ik×{0}) for ν=+ and ν=-.
We relax the assumption ofa as the following:(A1)
a(t)∈C[0,1] is a sign-changing weight.Corollary 16.
Let(A1) hold and f≡0. Then there exist two unbounded subcontinua £k+ and £k- of solutions of (6) in R×E, bifurcating from (λk,0), and £kν⊂(R×Skν)∪(I×{0}) for ν=+ and ν=-.


## 4. Spectrum of Half-Quasi-Linear Problems

In this section, we consider the half-linear problem (7). Problem (7) is called half-linear because it is positive homogeneous in the cones u>0 and u<0. Similar to that of [20], we say that λ is a half-eigenvalue of problem (7) if there exists a nontrivial solution (λ,uλ). λ is said to be simple if v=cuλ, c>0 for all solutions (λ,v) of problem (7).In order to prove our main results, we need the following Sturm type comparison result.Lemma 17.
Letb2(t)≥max⁡{b1(t),b1(t)+α(t)+β(t),b1(t)-α(t)-β(t)} for t∈(0,1) and bi(t)∈C(0,1), i=1,2. Also let u1,u2 be solutions of the following differential equations: (54)u′′′′=bitu+αu++βu-,t∈0,1,i=1,2,respectively. If (c,d)⊆(0,1), u1(c)=u1′′(c)=0, and u1′(c)≠0 or u1′′′(c)≠0 and u1(d)=u1′′(d)=0, and u1′(d)≠0 or u1′′′(d)≠0, and u1(t)≠0 in (c,d), then either there exists τ∈(c,d) such that u2(τ)=0 or b2=b1 or b2=b1+α+β or b2=b1-α-β and u2(t)=μu1(t) for some constant μ≠0.Proof.
We discuss four cases.
Case  1 (
u
1
(
t
)
>
0
,
u
2
(
t
)
>
0
in
(
c
,
d
)
). Then an easy calculation shows that(55)∫cdu1′′′′u2-u2′′′′u1dt=∫cdb1-b2u1u2dt<0.
Similar to the step of the proof of Lemma3.1 of [16], we can obtain the left-hand side of (55) ≥0. This is a contradiction.
Case  2 (
u
1
(
t
)
>
0
,
u
2
(
t
)
<
0
in
(
c
,
d
)
). Similar to (55), we can get(56)∫cdu1′′′′u2-u2′′′′u1dt=∫cdb1-b2+α+βu1u2dt>0.Similar to the step of the proof of Lemma 3.1 of [16], we can obtain that the left-hand side of (56) ≤0. This is a contradiction.
Case  3 (
u
1
(
t
)
<
0
,
u
2
(
t
)
<
0
in
(
c
,
d
)
). Similar to Case 1, we can get the result.
Case  4 (
u
1
(
t
)
<
0
,
u
2
(
t
)
>
0
in
(
c
,
d
)
). We can get(57)∫cdu1′′′′u2-u2′′′′u1dt=∫cdb1-b2-α-βu1u2dt>0.
Similar to the proof of Lemma3.1 of [16], we can obtain the left-hand side of (57) ≤0. This is a contradiction.By Lemma17, we obtain the following result that will be used later.Lemma 18.
LetI=[a,b] be such that I⊂[0,1] and (58)meas⁡I>0.
Letgn∈C([0,1],R) be such that(59)limn→+∞⁡gnt=+∞uniformlyonI.
Letyn∈E be a solution of the equation(60)yn′′′′t=gntyn+αyn++βyn-,t∈I.Then the number of zeros of yn in I goes to infinity as n→+∞.Proof.
Setα0≔maxt∈0,1αt and β0≔maxt∈0,1βt. By simple computation, we can show that (61)gnt+αyn+yn+βyn-yn≥gnt-α0-β0∀t∈0,1.After taking a subsequence if necessary, we may assume that (62)gnjt-α0-β0≥λj∀t∈Ias j→+∞, where λj is the jth eigenvalue of the following problem:(63)u′′′′=λu,0<t<1,ua=ub=u′′a=u′′b=0.Let φj be the corresponding eigenfunction of λj. It is easy to check that the distance between any two consecutive zeros of φj is (a-b)/j (also see [30]).
Hence, the number of zeros ofφjI goes to infinity as j→+∞. Note that the conclusion of Lemma 17 also is valid if α=β=0. Using these facts and Lemma 17, we can obtain the desired results.On the basis of the unilateral global interval bifurcation result, we establish the spectrum of the half-linear problem (8). More precisely, we will use Theorem 10 to prove the following result.Theorem 19.
There exist two sequences of simple half-eigenvalues for problem (7), λ1+<λ2+⋯<λk+<⋯ and λ1-<λ2-⋯<λk-<⋯. The corresponding half-linear solutions are in {λk+}×Sk+ and {λk-}×Sk-. Furthermore, aside from these solutions and the trivial ones, there is no other solutions of problem (7).Proof of Theorem19.
By Theorem10, we know that there exists at least one solution of problem (7), (λkν,xkν)∈R×Skν, for every k=1,2,…,ν=-, and ν=-. The positive of problem (7) implies that {(λkν,cxkν),c>0} are half-linear solutions in {λkν}×Skν. Lemma 4 implies that any nontrivial solution u of problem (7) lies in some Skν. We claim that, for any solution (λ,x) of problem (7) with x∈Skν, we have that λ=λkν and x=hxkν  k for some positive constant h.
Next, we only prove the case ofν=+ since the case of ν=- is similar.
We divide the proof into three steps.
Step  1 (we show that
λ
=
λ
k
ν
). We may assume without loss of generality that the first generalized simple zero of xxkν to occur in (0, 1) is a generalized simple zero of x. That is, there exists c∈(0,1] such that x(c)=x′′(c)=0 and x′(c)≠0 or x′′′(c)≠0, and x and xkν do not vanish and have the same sign in (0,c). It follows that(64)∫0cx′′′′xkν-xkν′′′′xdt=∫0cλ-λkνxxkνdt.Similar to the proof of Lemma 3.1 of [16], we can obtain the left-hand side of (64) ≥0. So, one has that λkν≤λ.
On the other hand, similar to the proof of Lemma13, by Lemma 6, there must exist an interval (d,e)⊂(0,1) such that x and xkν do not vanish and have the same sign in (d,e), and xkν(d)=xkν(e)=xkν′′(d)=xkν′′(e)=0. We have(65)∫dexkν′′′′x-x′′′′xkνdt=∫deλkν-λatxkνxdt.Similar to the proof of Lemma 3.1 of [16], we can obtain the left-hand side of (65) ≥0. So, one has that λ≤λkν. Hence λ=λkν.
Step  2 (we will prove that
x
=
h
x
k
ν
for some positive constant
h
). Without loss of generality, we may assume that x and xkν are positive in (0,c). By the Picone identity in [29, Theorem 4], noting λ=λkν, we have that(66)∫0cxxkνxxkν′′′-xkνx′′′+x′xkνxkνx′′-xxkν′′-xxkν′′xxkν′′dt=C1,where(67)C1=∫0cx′′-xxkνxkν′′2-2xkνxkν′′xxkν′2dt.Using similar methods of the proof of Lemma 13, we can show that the left-hand side of (66) ≤0. Hence, C1≤0. It follows that(68)∫0cx′′-xxkνxkν′′2dt≤2∫0cxkνxkν′′xxkν′2dt≤0.Thus, [(x/xkν)′]2=0. Furthermore, we obtain that x=h1xkν on (0,c) for some positive constant h1. We may assume without loss of generality that the first zero of xxk to occur in (c,1] is a generalized simple zero of x. That is, there exists c1∈(c,1] such that x(c1)=x′′(c1)=0 and x′(c1)≠0 or x′′′(c1)≠0, and x and xkν do not vanish and have the same sign in (c,c1]. Using method similar to the above, we can show that x=h2xkν on (c,c1] for some positive constant h2. Clearly, x′(c)=h1xkν′(c)=h2xkν′(c) and Lemma 3 imply h1=h2. Repeating the above process k times, we can show that x=hxkν for some positive constant h.
Step  3 (we prove thatλkν (ν=+ or ν=-) are increasing). In fact, if (λkν,u) and (λjν,v) are the solutions of problem (7) with u∈Skν, v∈Sjν, and k<j, the first generalized simple zero of uv to occur in (0, 1) is a generalized simple zero of v. Indeed, if this were not, by Lemma 6, using method similar to the proof in Step 1, we could obtain λkν<λjν, which is impossible, since the half-eigenvalues were shown to be simple. Therefore, by Lemma 17, we can get λkν<λjν.
Naturally, we can consider the bifurcation structure of the perturbation of problem (7) of the form(69)x′′′′=λatx+αx++βx-+gt,x,λ,0<t<1,x0=x1=x′′0=x′′1=0,where g(t,x,λ) satisfies (4).Theorem 20.
Forν=+,-, (λkν,0) is a bifurcation point for problem (69). Moreover, there exists an unbounded continuum Dk,0ν of solutions of problem (69) emanating from (λkν,0), such that Dk,0ν⊂((R×Skν)∪{(λkν,0)}).Proof.
Letα0≔maxt∈0,1αt and β0≔maxt∈0,1βt: (70)Ik=λk-α0+β0a0,λk+α0+β0a0.Corollary 15 shows that there exist two unbounded subcontinua Dk,0+ and Dk,0- of solutions of (69) in R×E, bifurcating from Ik×{0}, and Dk,0ν⊂(R×Skν)∪(Ik×{0}) for ν=+ and ν=-. Let us show that Dk,0ν∩(R×{0})=(λkν,0); that is, (λkν,0) is a bifurcation point for problem (69). Indeed, if there exists (λn,xn) being a sequence of solutions of problem (69) converging to (λ,0), let yn=xn/xn, and then yn should be a solution of problem(71)yn=L-1λnayn+αyn++βyn-+gt,xn,λxn.By (18) and the compactness of L-1, we obtain that for some convenient subsequence yn→y0 as n→+∞. Now y0 verifies the equation (72)Ly0=λaty0+αy0++βy0-and y0=1. This implies that λ=λkν for some k∈N and ν∈{+,-}.


## 5. Nodal Solutions for Half-Linear Eigenvalue Problems

We start this section by studying the following eigenvalue problem:(73)x′′′′=αx++βx-+λratfx,0<t<1,x0=x1=x′′0=x′′1=0,where λ>0 is a parameter.Letζ(x),ξ(x)∈C(R,R) be such that(74)fx=f0x+ζx,fx=f∞x+ξxwith(75)limx→0⁡ζxx=0,limx→∞⁡ξxx=0.Let us consider(76)x′′′′=λratf0x+αx++βx-+λratζx,0<t<1,x0=x1=x′′0=x′′1=0as a bifurcation problem from the trivial solution x≡0, and(77)x′′′′=λratf∞x+αx++βx-+λratξx,0<t<1,x0=x1=x′′0=x′′1=0as a bifurcation problem from infinity.Applying Theorem20 to problem (76), we have the following result.Lemma 21.
Let(H1), (H4), (H5), and (H6) hold. For ν=+,-, (λkν/rf0,0) is a bifurcation point for problem (76). Moreover, there exists an unbounded continuum Dk,0ν of solutions of problem (76) that joins (λkν/rf0,0) to infinity, such that Dk,0ν⊂((R×Skν)∪{(λkν/rf0,0)}).
We add the points{(λ,∞)∣λ∈R} to space R×E; by the results of Rabinowitz [31], we have the following lemma.Lemma 22.
Let(H1), (H4), (H5), and (H6) hold. For ν=+,-, (λkν/rf∞,∞) is a bifurcation point for problem (77). Moreover, there exists an unbounded continuum Dk,∞ν of solutions of problem (77) meeting (λkν/rf∞,∞), such that Dk,∞ν⊂((R×Skν)∪{(λkν/rf∞,∞)}).
We note that problem (76) and problem (77) are the same, and each of them is equivalent to problem (73). By Lemmas 21 and 22, we obtain the following lemma.Lemma 23.
Let(H1), (H4), (H5), and (H6) hold. There exists an unbounded continuum Dkν=Dk,0ν=Dk,∞ν of solutions of problem (73) emanating from (λkν/rf0,0) or (λkν/rf∞,∞), such that Dkν⊂((R×Skν)∪{(λkν/rf0,0)}) and Dkν⊂((R×Skν)∪{(λkν/rf∞,∞)}), and Dkν joins (λkν/rf0,0) to (λkν/rf∞,∞).Remark 24.
Any solution of (73) of the form (1,x) yields a solution x of (8). In order to prove our main results, one will only show that Dkν crosses the hyperplane {1}×E in R×E.Remark 25.
From(H5) and (H6), we can see that there exists a positive constant Q such that f(s)/s≥Q for all s≠0.Theorem 26.
Let(H1), (H4), (H5), and (H6) hold. For ν=+,-, either λkν/f∞<r<λkν/f0 or λkν/f0<r<λkν/f∞. Then problem (8) possesses a solution xkν such that xkν has exactly k-1 zeros in (0,1) and νxkν is positive near 0.Proof.
By Remark24, in order to prove Theorem 26, we only show that Dkν joins (λkν/f0,0) to (λkν/f∞,∞). Let (μn,xn)∈Dkν satisfy(78)μn+xn⟶∞.We note that μn>0 for all n∈N since (0,0) is the only solution of (73) for λ=0 and Dkν∩{0}×E=∅.
Case  1 (
λ
k
ν
/
f
∞
<
r
<
λ
k
ν
/
f
0
). In this case, we show that (79)λkνrf∞,λkνrf0⊆λ∈R∣λ,x∈Dkν.We divide the proof into two steps.
Step  1. We show that if there exists a constant number M>0 such that(80)μn∈0,M,for n large enough, then Dkν joins (λkν/rf0,0) to (λkν/rf∞,∞).
In this case(81)xn⟶∞.
Letξ(·)∈C(R,R) be such that(82)fx=f∞x+ξxwith(83)limx→∞⁡ξxx=0.Let ξ¯(x)=maxξs:0≤s≤x, then ξ¯ is nondecreasing, and(84)limx→+∞⁡ξ¯xx=0.Moreover, from (84) and the fact that ξ¯ is nondecreasing, we have that(85)limn→∞⁡ξxntxn=0,∀t∈0,1,since(86)ξxntxn≤ξ¯xntxn≤ξ¯xnt∞xn≤ξ¯xntxn⟶0,n⟶∞∀t∈0,1.We divide the equation(87)xn′′′′=μnratf∞xn+αxn++βxn-+μnratξxn,0<t<1,xn0=xn1=xn′′0=xn′′1=0by xn and set yn=xn/xn. Since yn is bounded in E, choosing a subsequence and relabeling if necessary, we have that yn→y for some y∈E with y=1.
By the compactness ofL-1, we obtain that(88)y′′′′=μratf∞y+αy++βy-,0<t<1,y0=y1=y′′0=y′′1=0,where μ≔limn→∞μn, again choosing a subsequence and relabeling if necessary.
It is clear thaty=1 and y∈Dkν¯⊆Dkν since Dkν is closed in R×E. Moreover, by Theorem 19, μrf∞=λkν, so that(89)μ=λkνrf∞.Thus, Dkν joins (λkν/rf0,0) to (λkν/rf∞,∞).
Step  2. We show that there exists a constant number M>0 such that μn∈(0,M], for n large enough.
On the contrary, we suppose thatlimn→+∞μn=+∞. Since (μn,xn)∈Dkν, it follows from the compactness of L-1 that(90)xn′′′′=μnratfn~txn+αxn++βxn-,0<t<1,xn0=xn1=xn′′0=xn′′1=0,where fn~:[0,1]×[0,∞)→R by(91)fn~t=fxnxn,xn≠0,t∈0,1,f0,xn=0,t∈0,1.By Remark 25, we have(92)limn→+∞⁡μnrfn~t=±∞.
Letφkν be an eigenfunction corresponding to λkν. But if limn→+∞μnrfn~(t)=-∞, applying Lemma 17 to xn and φkν, we have that φkν must change sign for n large enough, which is impossible. So limn→+∞μnrfn~(t)=+∞. By Lemma 18, we get that xn must change sign for n large enough, and this contradicts the fact that xn∈Skν.
Case  2 (
λ
k
ν
/
f
0
<
r
<
λ
k
ν
/
f
∞
). In this case, if (μn,xn)∈Dkν is such that(93)limn→∞⁡μn+xn=∞,limn→∞⁡μn=∞,then(94)λkνrf0,λkνrf∞⊆λ∈0,∞∣λ,x∈Dkνand, moreover, ({1}×E)∩Dkν≠∅.
Assume that there existsM>0 such that, for all n∈N,(95)μn∈0,M.Applying a similar argument to that used in Step  1 of Case  1, after taking a subsequence and relabeling if necessary, it follows that(96)μn,xn⟶λkνrf∞,∞,n⟶∞.Again Dkν joins (λkν/rf0,0) to (λkν/rf∞,∞) and the result follows.Theorem 27.
Let(H1), (H4), (H5), and (H7) hold. For ν∈{+,-}, assume that one of the following conditions holds.(i)
r∈(0,λkν/f0) for λkν>0.(ii)
r∈(0,λk+/f0)∪(λk-/f0,0) for νλkν>0.(iii)
r∈(0,λk-/f0)∪(λk+/f0,0) for νλkν<0.(iv)
r∈(λkν/f0,0) for λkν<0.
Then problem (8) possesses two solutions xk+ and xk- such that xk+ has exactly k-1 zeros in (0,1) and is positive near 0 and xk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
We will 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 [32], we define the cut-off function of f as the following:(97)fns≔fs,s∈-n,n,2n2-fnns-n+fn,s∈n,2n,2n2+f-nns+n+f-n,s∈-2n,-n,ns,s∈-∞,-2n∪2n,+∞.We consider the following problem:(98)x′′′′=αx++βx-+λratfnx,0<t<1,x0=x1=x′′0=x′′1=0.
Clearly, we can see thatlimn→+∞f[n](s)=f(s), (f[n])0=f0, and (f[n])∞=n.
Similar to the proof of Theorem26, by Lemma 23 and Remark 24, there exists an unbounded continuum Dkν[n] of solutions of problem (98) emanating from (λkν/rf0,0), such that Dkν[n]⊂(Φkν∪{(λkν/rf0,0)}), and Dkν[n] joins (λkν/rf0,0) to (λkν/rn,∞).
Takingzn=(λkν/rn,∞) and z∗=(0,∞), we have that zn→z∗.
So condition (i) in Lemma9 is satisfied with z∗=(0,∞).
Obviously,(99)rn=sup⁡λ+x∣λ,x∈Dkνn=∞,and accordingly, (ii) in Lemma 9 holds and (iii) in Lemma 9 can be deduced directly from the Arezela-Ascoli Theorem and the definition of f[n].
Therefore, by Lemma9, limsupn→∞Dkν[n] contains an unbounded connected component Dkν with (0,∞)∈Dkν.
Fromlimn→+∞f[n](s)=f(s), (98) can be converted to the equivalent equation (73). Since Dkν[n]⊂Φkν, we conclude that Dkν⊂Φkν. Moreover, Dkν⊂Σkν by (8).
In the following, we show that(λkν/rf0,0)∈Dkν.
Let(μn,xn)∈Dkν∖{(λkν/rf0,0)} satisfy limn→∞xn=0.
Letζ∈C(R,R) be such that(100)fs=f0s+ζswith lims→0ζ(s)/s=0, let ζ¯(x)=max0≤s≤xζs, then ζ¯ is nondecreasing, and(101)limx→0⁡ζ¯xx=0.
We divide the equation(102)xn′′′′=μnratf0xn+αxn++βxn-+λratζxn,0<t<1,xn0=xn1=xn′′0=xn′′1=0.
Letyn=xn/xn;yn should be the solutions of problem(103)yn′′′′=μnratf0yn+αyn++βyn-+μnratζxnxn,t∈0,1,yn0=yn1=yn′′0=yn′′1=0.
Sinceyn is bounded in E, choosing a subsequence and relabeling if necessary, we have that yn→y for some y∈E.
Furthermore, from (101) and the fact that ζ¯ is nondecreasing, we have that(104)limn→∞⁡ζxnxn=0,since(105)ζxnxn≤ζ¯xnxn≤ζ¯xn∞xn≤ζ¯xnxn⟶0,n⟶+∞.
By (104) and the compactness of L-1, we obtain that (106)Ly=μratf0y+αy++βy-.
It is clear thaty=1 and y∈Dkν¯⊆Dkν since Dkν is closed in R×E. Moreover, by Theorem 19, μrf0=λkν, so that(107)μ=λkνrf0.
Thus,Dkν joins (λkν/rf0,0) to (0,∞).Theorem 28.
Let(H1), (H4), (H5), and (H8) hold. For ν∈{+,-}, assume that one of the following conditions holds.(i)
r∈(λkν/f0,+∞) for λkν>0.(ii)
r∈(-∞,λk-/f0)∪(λk+/f0,+∞) for νλkν>0.(iii)
r∈(-∞,λk+/f0)∪(λk-/f0,+∞) for νλkν<0.(iv)
r∈(-∞,λkν/f0) for λkν<0.
Then problem (8) possesses two solutions xk+ and xk- such that xk+ has exactly k-1 zeros in (0,1) and is positive near 0 and xk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
Define(108)fns≔1ns,s∈-∞,-2n∪2n,+∞.2+f-nns+n+f-n,s∈-2n,-n,2-fnns-n+fn,s∈n,2n,fs,s∈-n,n.We consider the following problem:(109)x′′′′=αx++βx-+λratfnx,0<t<1,x0=x1=x′′0=x′′1=0.
Clearly, we can see thatlimn→+∞f[n](s)=f(s), (f[n])0=f0, and (f[n])∞=1/n.
Similar to the proof of Theorem26, by Lemma 23 and Remark 24, there exists an unbounded continuum Dkν[n] of solutions of problem (109) emanating from (λkν/rf0,0), such that Dkν[n]⊂(Φkν∪{(λkν/rf0,0)}), and Dkν[n] joins (λkν/rf0,0) to (nλkν/r,∞).
Takingzn=(nλkν/r,∞) and z∗=(∞,∞), we have that zn→z∗.
So condition (i) in Lemma9 is satisfied with z∗=(∞,∞).
Obviously,(110)rn=sup⁡λ+x∣λ,x∈Dνn=∞,and accordingly, (ii) in Lemma 9 holds and (iii) in Lemma 9 can be deduced directly from the Arezela-Ascoli Theorem and the definition of f[n].
Therefore, by Lemma9, limsupn→∞Dkν[n] contains an unbounded connected component Dkν with (∞,∞)∈Dkν.
Fromlimn→+∞f[n](s)=f(s), (109) can be converted to the equivalent equation (73). Since Dkν[n]⊂Φkν, we conclude Dkν⊂Φkν. Moreover, Dkν⊂Σkν by (8).
Similar to the proof of Theorem27, we can obtain that (λkν/rf0,0)∈Dkν.Theorem 29.
Let(H1), (H4), (H5), and (H9) hold. For ν∈{+,-}, assume that one of the following conditions holds.(i)
r∈(λkν/f∞,+∞) for λkν>0.(ii)
r∈(λk+/f∞,+∞)∪(-∞,λk-/f∞) for νλkν>0.(iii)
r∈(λk-/f∞,+∞)∪(-∞,λk+/f∞) for νλkν<0.(iv)
r∈(-∞,λkν/f∞) for λkν<0.
Then problem (8) possesses two solutions xk+ and xk- such that xk+ has exactly k-1 zeros in (0,1) and is positive near 0 and xk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
We will only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
If(λ,x) is any nontrivial solution of problem (73), dividing problem (73) by x2 and setting y=x/x2 yields(111)y′′′′=αy++βy-+λratfxx2,0<t<1,y0=y1=y′′0=y′′1=0.
Define(112)f~y≔y2fyy2,ify≠0,0,ify=0.
Evidently, problem (111) is equivalent to(113)y′′′′=αy++βy-+λratf~y,0<t<1,y0=y1=y′′0=y′′1=0.
It is obvious that(λ,0) is always the solution of problem (113). By simple computation, we can show that f0~=f∞∈(0,∞) and f∞~=f0=0. Similar to the proof of Theorem 28, there exists an unbounded continuum Ckν of solutions of problem (113) emanating from (λkν/rf0~,0), such that Ckν⊂((R×Skν)∪{(λkν/rf0~,0)}), and Ckν joins (λkν/rf0~,0) to (∞,∞).
Under the inversiony→y/y2=x, we obtain Ckν→Dkν being an unbounded component of solutions of problem (73) emanating from (∞,0), such that Dkν⊂((R×Skν)∪{(∞,0)}), and Dkν joins (∞,0) to (λkν/rf∞,∞).
Moreover, by (8), we can obtain that Dkν⊂Σkν.
Thus,Dkν is an unbounded component of solutions of problem (8) such that Dkν joins (∞,0) to (λkν/rf∞,∞).Theorem 30.
Let(H1), (H4), H5, and (H10) hold. For ν∈{+,-}, assume that one of the following conditions holds.(i)
r∈(0,+∞) for λkν>0.(ii)
r∈(0,+∞)∪(-∞,0) for νλkν>0, or νλkν<0.(iii)
r∈(-∞,0) for λkν<0.
Then problem (8) possesses two solutions xk+ and xk- such that xk+ has exactly k-1 zeros in (0,1) and is positive near 0 and xk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
Define(114)fns≔ns,s∈-∞,-2n∪2n,+∞.2n2+f-nns+n+f-n,s∈-2n,-n,2n2-fnns-n+fn,s∈n,2n,fs,s∈-n,-2n∪2n,n,-f-2n+1n2ns+2+f-2n,s∈-2n,-1n,f2n-1n2ns-2+f2n,s∈1n,2n,1ns,s∈-1n,1n.We consider the following problem:(115)x′′′′=αx++βx-+λratfnx,0<t<1,x0=x1=x′′0=x′′1=0.
Clearly, we can see thatlimn→+∞f[n](s)=f(s), (f[n])0=1/n, and (f[n])∞=n.
Applying the similar method used in the proof of Theorem26, by Lemma 23 and Remark 24, there exists an unbounded continuum Dkν[n] of solutions of problem (115) emanating from (nλν/r,0) or (λν/nr,∞), such that Dkν[n]⊂((R×Skν)∪{(nλν/r,0)}) and Dkν[n]⊂((R×Skν)∪{(λν/rn,∞)}), and Dkν[n] joins (nλν/r,0) to (λν/rn,∞).
Takingzn=(nλν/r,0) and z∗=(∞,0) or zn=(λν/rn,∞) and z∗=(0,∞), we have that zn→z∗. By Lemma 9, we obtain an unbounded component Dkν with (∞,0),(0,∞)∈Dkν.
Fromlimn→+∞f[n](s)=f(s), (115) can be converted to the equivalent equation (73). Thus, Dkν is an unbounded component of solutions of problem (73) emanating from (∞,0) or (0,∞), such that Dkν⊂((R×Skν)∪{(∞,0)}) and Dkν⊂(R×Skν)∪{(0,∞)}) and Dkν joins (∞,0) to (0,∞).
SinceDkν[n]⊂(R×Skν), we conclude Dkν⊂(R×Skν). Moreover, by Remark 24 and (8), we can obtain that Dkν⊂Σkν.
Thus,Dkν is an unbounded component of solutions of problem (8) emanating from (∞,0) or (0,∞), such that Dkν⊂((R×Skν)∪{(0,0)}) and Dkν⊂((R×Skν)∪{(∞,∞)}), and Dkν joins (∞,0) to (0,∞).
Applying the similar method used in the proof of Theorem27, we obtain an unbounded connected component Dkν⊂Ckν with (∞,0)∈Dkν.Theorem 31.
Let(H1), (H4), (H5), and (H11) hold. For ν∈{+,-}, assume that one of the following conditions holds.(i)
There exists aλνkν>0 for λkν>0, such that r∈(0,λνkν).(ii)
There exists aνλνkν>0 for νλkν>0, such that r∈(-∞,λ-k-)∪(λ+k+,+∞). r∈(λ-k-,0)∪(0,λ+k+).(iii)
There exists aνλνkν<0 for νλkν<0, such that r∈(λ+k+,0)∪(0,λ-k-).(iv)
There exists aλνkν<0 for λkν<0, such that r∈(λνkν,0).
Then problem (8) possesses two solutions xk+ and xk- such that xk+ has exactly k-1 zeros in (0,1) and is positive near 0 and xk- has exactly k-1 zeros in (0,1) and is negative near 0.Proof.
Define(116)fns≔1ns,s∈-∞,-2n∪2n,+∞.2+f-nns+n+f-n,s∈-2n,-n,2-fnns-n+fn,s∈n,2n,fs,s∈-n,-2n∪2n,n,-f-2n+1n2ns+2+f-2n,s∈-2n,-1n,f2n-1n2ns-2+f2n,s∈1n,2n,1ns,s∈-1n,1n.We consider the following problem:(117)x′′′′=αx++βx-+λratfnx,0<t<1,x0=x1=x′′0=x′′1=0.
Clearly, we can see thatlimn→+∞f[n](s)=f(s), (f[n])0=1/n, and (f[n])∞=1/n.
Applying the similar method used in the proof of Theorem30, there exists an unbounded continuum Dkν[n] of solutions of problem (117) emanating from (nλν/r,0) or (nλν/r,∞), such that Dkν[n]⊂((R×Skν)∪{(nλν/r,0)}) and Dkν[n]⊂((R×Skν)∪{(nλν/r,∞}), and Dkν[n] joins (nλν/r,0) to (nλν/r,∞).
Moreover, we can obtain the desired results.Remark 32.
Ifα=β=0 and f0,f∞∈(0,∞), Ma et al. [10, 11] studied problem (8) under the conditions (H5) and (H6); in the situation, the results of Theorem 26 improve on Theorem 3.1 of [10] and Theorem 1.1 of [11], respectively.Remark 33.
Whenα=β=0 and f0,f∞∈(0,∞), Dai and Han [16] investigated the existence of nodal solutions for problem (8). Thus, our results partially extend and improve the corresponding Theorem 4.1 of [16].Remark 34.
Whenα=β=0 and f0,f∞∈(0,∞), Lazer and McKenna [7] investigated the existence of nodal solutions for problem (8). Thus, our results partially extend and improve the corresponding Theorem 1 of [7].Remark 35.
The nonlinear term of (8) is not necessarily homogeneous linearizable at the origin and infinity because of the influence of the term αx++βx-. Clearly, so the bifurcation results of [7, 8, 10–13, 16] cannot be applied directly to obtain our results.



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*Source: 5956713-2016-03-09.xml*