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

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:


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: where  is a compact linear operator and  : R ×  →  is compact with  = (‖‖) at  = 0 uniformly on bounded  intervals, where  is a real Banach space with the norm ‖ ⋅ ‖.
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][24][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  = 0 and  = ∞ 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][24][25] studied the bifurcation from intervals for Sturm-Liouville problems and its applications and established the unilateral global interval bifurcation for -Laplacian with non- − 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: where  is a parameter, the nonlinear term  has the form  =  + , where  and  are continuous functions on [0, 1] × R 3 , and , ,  satisfy the following conditions: where  1 is a positive constant.
Let S denote the closure of the set of nontrivial solutions of (6) in R × , and S ]  denote the subset of S with  ∈  ]  for ] ∈ {+, −} and S  = S +  ∪ S −  .Under assumptions (1)-(3), we will show that (6) and there are two distinct unbounded subcontinua, C +  and C −  , consisting of the bifurcation branch where   is given in Lemma 1.
On the basis of the above unilateral global interval bifurcation result, we study the following half-linear eigenvalue problem: where  + = max{, 0},  − = min{, 0} and () and () satisfy the following: We will show that there exist two sequences of simple half-eigenvalues for problem (7): The corresponding half-linear solutions are in { +  } ×  +  and { −  } ×  −  .Furthermore, aside from these solutions and the trivial ones, there is no other solutions of problem (7).
Following the above eigenvalue theory (see Theorem 19), we will investigate the existence of nodal solutions for the following fourth-order problem: For  =  = 0, Dai and Han [16] have established the existence of nodal solutions for problem (8) with crossing nonlinearity.In this paper, we assume that  satisfies the following assumptions: (5) () > 0 for  ̸ = 0.
(11)  0 = 0 and  ∞ = 0, where lim The rest of this paper is arranged as follows.In Section 2, 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).

Hypotheses and Lemmas
We define the linear operator  : () ⊂  →  : with From [26, p. 439-440], we consider the following auxiliary problem: for a given ℎ ∈ [0, 1].We can get that problem (11) can be equivalently written as where Then  is a closed operator and  −1 :  →  is completely continuous.Define the Nemytskii operator  : R ×  →  by Then it is clear that  is continuous operator and problem (3) can be equivalently written as Clearly,  : R ×  →  is completely continuous and (, 0) = 0, ∀ ∈ R.
In the following, we summarize some preliminary results from [10,16].
Lemma 6 (see [20,Lemma 2]).Let  and  be two integers such that  ≥  ≥ 2. Suppose that there exist two families of real numbers: Then, if  1 ≤  1 , there exist two integers  and  having the same parity: 1 ≤  ≤  − 1 and 1 ≤  ≤  − 1, such that Definition 7 (see [27]).Let  be a Banach space and let {  |  = 1, 2, . ..} be a family of subsets of .Then the superior limit D of {  } is defined by Lemma 8 (see [27]).Each connected subset of metric space  is contained in a component, and each connected component of  is closed.

Unilateral Global Bifurcation
The main result for problem (6) is the following theorem.In order to prove Theorem 10, we need the following results.

Lemma 11. If C ]
is bounded, we can find a neighborhood O of C ]  such that O ∩ S ]  = 0, where C ]  (] = +, −) is given by Theorem 10.
Proof.We only prove the case of C +  since the case of C −  is similar.
Let U be a uniform neighborhood of C +  in R × .We discuss two cases.
Case 1 (if U ∩ S +  ̸ = 0).Since the solutions of problem (6) and then Case 2 (if U ∩ S +  = 0).In this case, we take O = U.It is obvious that the result holds.
Consider the following auxiliary approximate problem: For  > 0, it is easy to show that nonlinear term () + (, ||  , ) + (, , ) is differ-entiable at the origin.Let By Lemma 1, there are two unbounded continua C + , of S +  and C − , of S −  bifurcating from (  , 0), consisting of the bifurcation branch C , , which satisfy the following result.
Proof.By Lemma 1.24 of [4], there exists a bounded open neighborhood O  of (  , 0) such that It follows that By Lemma 4, we show that C ] , \ {(  , 0)} cannot leave Φ ]  outside of a neighborhood of (  , 0).Thus, we have Next, we will prove that both C + , and C − , are unbounded.Without loss of generality, we may suppose that C − , is bounded.Therefore, in view of (32) and Lemma 1, there exists This contradicts the definitions of  +  and  −  .To prove Theorem 10, the next lemma will play a key role.Lemma 13.Let   , 0 <   < 1, be a sequence converging to 0. If there exists a sequence (  ,   ) ∈ R ×  ]  such that (  ,   ) is a nontrivial solution of problem (27) corresponding to  =   , and (  ,   ) converges to (, 0) in R × , then  ∈   .

Clearly, we have 𝑦 ∈ 𝑆 ]
.We claim that  ∈  ]  .On the contrary, supposing that  ∈  ]  , by Lemma 4, then  ≡ 0, which is a contradiction with ‖‖ = 1.Now, we deduce the boundedness of .Let  ]  ∈  ]  be an eigenfunction of problem (19) corresponding to   . Let be, respectively, the sequences of generalized simple zeros of   and  ]  .Suppose  1 ≤  1 ; then we deduce from Lemma 6 existence of integers  and  having the same parity such that Therefore, without loss of generality, we choose  1 =  0 ,  1 =  1 ,  2 =   , and  2 =  +1 and since  and  have the same parity   and  ]  do not vanish and have the same sign in both the intervals We can assume without loss of generality that   > 0 and  ]  > 0 in ( 1 ,  1 ).By the Picone identity in [29, Theorem 4], we have that The left-hand side of (39) equals We prove that ( 1 ) = 0, noting the conclusion of Lemma 3, then ( ]  )  ( 1 ) > 0. By L'Hospital rule, we have that In the following, we will prove that Let and V  =   V.Then we have and  ≥ 0 be such that |  = V,  ≥ 0 in (, ) and () = () =   () =   () = 0 and   =   .Set  fl   ; then  should be a solution of the problem The Strong Maximum Principle implies that  < 0 in (, ).This follows that Next, we will prove that ( 1 ) = 0, noting the conclusion of Lemma 3, then ( ]  )  ( 1 ) < 0. By L'Hospital rule, we have that By (42), we can show that   1 ≤ 0. Therefore, the left-hand side of (39) ≤ 0. By (42), we have  1 ≥ 0.
It follows that Similarly, we can also show that By ( 35), (46), and (47), taking the limit as  → +∞, we can obtain that So, by (48), we have that Furthermore, it follows that if  ≤   , if  ≥   , Therefore, we have that  ∈   .
) is given by Theorem 10.
Proof.We only prove the case of C +  since the case of C −  is similar.For any (, ) ∈ C +  , there are two possibilities: (i)  ∈  +  or (ii)  ∈  +  .It is obvious that (, ) ∈ Φ +  in the case of (i).While case (ii) implies that  has at least one double zero in [0, 1], Lemma 4 follows that  ≡ 0. Hence, there exists a sequence (  ,   ) ∈ Φ +  such that (  ,   ) is a solution of problem (27) corresponding to  = 0, and (  ,   ) converges to (, 0) in R×.By Lemma 13, we have  ∈   : that is, (, ) ∈   × {0} in the case of (ii).Hence, So there exists (  ,   ) ∈ C + , ∩ O for all  > 0. Since O is bounded in R×, (27) shows that (  ,   ) is bounded in R× 4 independently of .By the compactness of  −1 , one can find a sequence   → 0 such that (  ,   ) converges to a solution (, ) of ( 6).So  ∈  From Theorem 10 and its proof, we can easily get the following two corollaries.

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  > 0 and  < 0. Similar to that of [20], we say that  is a half-eigenvalue of problem (7) if there exists a nontrivial solution (,   ). is said to be simple if V =   ,  > 0 for all solutions (, V) of problem (7).
Proof.We discuss four cases.
By Lemma 17, we obtain the following result that will be used later.
After taking a subsequence if necessary, we may assume that as  → +∞, where   is the th eigenvalue of the following problem: Let   be the corresponding eigenfunction of   .It is easy to check that the distance between any two consecutive zeros of   is ( − )/ (also see [30]).Hence, the number of zeros of   |  goes to infinity as  → +∞.Note that the conclusion of Lemma 17 also is valid if  =  = 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 halfeigenvalues for problem (7), The corresponding half-linear solutions are in { +  }× +  and { −  }× −  .Furthermore, aside from these solutions and the trivial ones, there is no other solutions of problem (7).
We divide the proof into three steps.
Step 1 (we show that  =  ]  ).We may assume without loss of generality that the first generalized simple zero of  ]  to occur in (0, 1) is a generalized simple zero of .That is, there exists  ∈ (0, 1] such that () =   () = 0 and   () ̸ = 0 or   () ̸ = 0, and  and  ]  do not vanish and have the same sign in (0, ).It follows that Similar to the proof of Lemma 3.1 of [16], we can obtain the left-hand side of (64) ≥ 0. So, one has that  ]  ≤ .On the other hand, similar to the proof of Lemma 13, by Lemma 6, there must exist an interval (, ) ⊂ (0, 1) such that  and  ]  do not vanish and have the same sign in (, ), and Similar to the proof of Lemma 3.1 of [16], we can obtain the left-hand side of (65) ≥ 0. So, one has that  ≤  ]  .Hence  =  ]  .
Step 3 (we prove that  ]  (] = + or ] = −) are increasing).In fact, if ( ]  , ) and ( ]  , V) are the solutions of problem (7) with  ∈  ]  , V ∈  ]  , and  < , the first generalized simple zero of V 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  ]  <  ]  , which is impossible, since the half-eigenvalues were shown to be simple.Therefore, by Lemma 17, we can get  ]  <  ]  .
Naturally, we can consider the bifurcation structure of the perturbation of problem (7) of the form where (, , ) satisfies (4).
Step 1.We show that if there exists a constant number  > 0 such that for  large enough, then D ]  joins ( ]  / 0 , 0) to ( ]  / ∞ , ∞).In this case Let (⋅) ∈ (R, R) be such that By the compactness of  −1 , we obtain that where  fl lim →∞   , again choosing a subsequence and relabeling if necessary.
Proof.We will only prove the case of (i) since the proofs of the cases for (ii), (iii), and (iv) are completely analogous.
Proof.Define   (114) Remark 35.The nonlinear term of ( 8) is not necessarily homogeneous linearizable at the origin and infinity because of the influence of the term  + +  − .Clearly, so the bifurcation results of [7,8,[10][11][12][13]16] cannot be applied directly to obtain our results.

Lemma 12 .
C + , and C − , are both unbounded and
). Proof of Theorem 10.We only prove the case of C +  since the case of C −  is similar.Let C +  be the component of S +  ∪ (  × {0}), containing   ×{0}.By Lemma 14, we can show that C + Remark 5 it follows that  ≡ 0. Note that (, ) ∈ C + , ∩ O since C + , ∩O is closed subset of R×.By Lemma 13,  ∈   , which contradicts the definition of O. On the other hand, if  ∈  +  , then (, ) ∈ S +  ∩ O which contradicts S +  ∩ O = 0.
be an eigenfunction corresponding to  ]  .But if lim →+∞    f () = −∞, applying Lemma 17 to   and  ]  , we have that  ]  must change sign for  large enough, which is impossible.So lim →+∞    f () = +∞.By Lemma 18, we get that   must change sign for  large enough, and this contradicts the fact that   ∈  ]  .

)
Let   =   /‖  ‖;   should be the solutions of problem Since   is bounded in , choosing a subsequence and relabeling if necessary, we have that   →  for some  ∈ .