# Bifurcation of Sign-Changing Solutions form-Point Boundary Value Problems

Authors: Yulian An; Maoan Han
Journal: ISRN Mathematical Analysis
 (2012)

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## Abstract

With the help of bifurcation techniques, some multiplicity results and global structure for sign-changing solutions of somem-point boundary value problems are obtained when the nonlinear term is sublinear at 0.


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## Body

## 1. Introduction and Main Results

In this paper, we consider them-point boundary value problems:(1.1)u′′+λf(u)=0,t∈(0,1),(1.2)u(0)=0,u(1)=∑i=1m-2αiu(ηi),
where integer m≥3,ηi∈(0,1), λ is a positive parameter, and f∈C1(ℝ,ℝ). We study the multiplicity and global structure of sign-changing solutions of (1.1) and (1.2) under the assumptions:(A1)
αi>0 for i=1,…,m-2 with 0<∑i=1m-2αi<1;(A2)
f∈C1(ℝ,ℝ) satisfies sf(s)>0 for s≠0;(A3)
f0:=lim⁡|s|→0f(s)/s=0;(A4)
f∞:=lim⁡|s|→∞f(s)/s=0.Multipoint boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics. The existence of solutions of second-order multipoint boundary value problems has been extensively studied in the literature, see [1–4] and the references therein. Particularly, many authors have studied the existence of sign-changing solutions for various nonlinear boundary value problems, see for example [5–10].Recently, the global structure of solutions of nonlinear multipoint boundary value problems has also been investigated by several authors using bifurcation methods, see [7–10]. These papers dealt with the case f0∈(0,∞), and relatively little is known about the global structure of solutions when f satisfying f0=0. The main reason is that the global bifurcation techniques cannot be used directly in this case. Very recently, [11] investigated the global structure of positive solutions for a class of boundary value problems with f0=0. However, to our knowledge there is no paper studying the global structure of sign-changing solutions for nonlinear multipoint boundary value problems under the assumption f0=0. The purpose of present paper is to fill this gap.In this paper, we consider the global structure of nodal solutions of (1.1) and (1.2), a kind of sign-changing having a given number of zeros, when f0=f∞=0. We find that the discussion is more complicated, when sign-changing solutions are concerned. Eigenvalue theory and Sturm’s comparison theorem play important roles in our discussion.Now, we introduce some notations as follows.LetY=C[0,1] with the norm ∥u∥∞=max⁡t∈[0,1]|u(t)|. Let X={u∈C1[0,1]∣u(0)=0,u(1)=∑i=1m-2αiu(ηi)}, and E={u∈C2[0,1]∣u(0)=0,u(1)=∑i=1m-2αiu(ηi)} equipped with the norm:(1.3)‖u‖X=max⁡{‖u‖∞,‖u′‖∞},‖u‖=max⁡{‖u‖∞,‖u′‖∞,‖u′′‖∞}.For anyC1 function u, if u(x0)=0 and u′(x0)≠0, then x0 is called a simple zero of u. For any integer k≥1 and any ν∈{+,-}, let Skν,Tkν⊂C2[0,1] be sets consisting of functions u∈C2[0,1] satisfying the following conditions:Skν:
(i)u(0)=0,νu′(0)>0;(ii)u has only simple zeros in [0,1] and has exactly k-1 zeros in (0,1).Tkν:
(i)u(0)=0,νu′(0)>0 and u′(1)≠0;(ii)u′ has only simple zeros in (0,1) and has exactly k zeros in (0,1);(iii)u has a zero strictly between each two consecutive zeros of u′.Remark 1.1.
Ifu∈Tkν, then u∈Skν or u∈Sk+1ν. The sets Tkν(k=1,2,…) are open in E and disjoint [8].Lemma 1.2 (See [8]).
Let (A1) and (A2) hold. If(μ,u) is a nontrivial solution of (1.1) and (1.2). Then, u∈Tkν for some k,ν.
Let𝕏=ℝ×X with the product topology. As in [12], we add the point {(λ,∞)∣λ∈ℝ} to the space 𝕏. Denote θ∈X,θ(t)≡0,t∈[0,1].The main results of this paper are as follows.Theorem 1.3.
Let (A1)–(A4) hold. Then, there exists a component𝒞kν⊂(0,∞)×Tkν of solutions of (1.1) and (1.2), which joins (∞,θ) to (∞,∞)(see Figure 1(a)) such that Projℝ𝒞kν=[ρkν,∞) for some ρkν>0. Here, 𝒞kν joins (∞,θ) to (∞,∞) meaning that:
(1.4)lim⁡(λ,u)∈Ckν,‖u‖≤1,λ→+∞‖u‖=0,lim⁡(λ,u)∈Ckν,‖u‖>1,λ→+∞‖u‖=+∞.(a)(b)Corollary 1.4.
Let (A1)–(A4) hold. Then, there existsλkν≥ρkν>0 such that (1.1) and (1.2) have at least two solutions in Tkν for  λ∈(λkν,∞).Remark 1.5.
Theorem1.3 extends the result stated in [11]. Meanwhile, Theorem 1.3 and Corollary 1.4 do not only obtain the multiplicity of nodal solutions of (1.1) and (1.2), but also describe the global structure of these solutions.


## 2. Preliminary Lemmas

The following definition and lemmas about superior limit and component are important to prove Theorem1.3.Definition 2.1 (See [13]).
LetW be a Banach space, and {Cn∣n=1,2,…} be a family of subsets of W.Then, the superior limit 𝒟 of {Cn} is defined by:
(2.1)D≔limsup⁡n→∞Cn={x∈W∣∃{ni}⊂N,xni∈Cni,suchthatxni⟶x}.Lemma 2.2 (See [13]).
Each connected subset of metric spaceW is contained in a component, and each component of W is closed.Lemma 2.3 (See [11]).
LetW be a Banach space and Cn a family of closed connected subsets of W. Assume that:
(i) there existzn∈Cn, n=1,2,…, and z*∈W such that zn→z*;
(ii)rn=∞, where  rn=sup⁡{∥x∥∣x∈Cn};
(iii) for allR>0,(∪n=1∞Cn)∩BR is a relative compact set of W, where
(2.2)BR={x∈W∣‖x‖≤R}.
Then, there exists an unbounded connected component 𝒞 in 𝒟 such that z*∈𝒞.Define a linear operatorL:E→Y by:(2.3)Lu:=-u′′,u∈E.
We consider the linear eigenvalues problem:(2.4)Lu=λu,u∈E.
Let λk be the kth eigenvalue of (2.4), and φk an eigenfunction corresponding to λk. The following lemma or similar result can be found in [7–9].Lemma 2.4.
Let (A1) hold. Then,(2.5)0<λ1<λ2<⋯<λk<λk+1<⋯,lim⁡k→∞λk=∞.
For each k∈N, algebraic multiplicity of λk is equal to 1, and the corresponding eigenfunction φk∈Tk+ and is strictly positive on (0,1).Define a mapTλ:Y→E by:(2.6)Tλu(t)=λ∫01H(t,s)f(u(s))ds,
where(2.7)H(t,s)=G(t,s)+∑i=1m-2αiG(ηi,s)1-∑i=1m-2αiηit,G(t,s)={(1-t)s,0≤s≤t≤1,t(1-s),0≤t≤s≤1.It is clear thatTλ:Y→X is completely continuous provided that (A1) and (A2) hold.Lemma 2.5.
Let (A1) and (A2)  hold, and{(μl,yl)}⊂(0,∞)×Tkν be a sequence of solutions of (1.1) and (1.2). Assume that μl≤C0 for some constant C0>0,  and  lim⁡l→∞⁡∥yl∥=∞. Then,
(2.8)lim⁡l→∞⁡‖yl‖∞=∞.Proof.
From the relationyl(t)=μl∫01H(t,s)f(yl(s))ds,we conclude that yl′(t)=μl∫01Ht(t,s)f(yl(s))ds. Then,
(2.9)‖yl′‖∞≤C0(1+∑i=1m-2αi1-∑i=1m-2αiηi)∫01|f(yl(s))|ds.
Equations (2.9) and (1.1) imply that {∥yl′∥∞},{∥yl′′∥∞} are bounded, whenever {∥yl∥∞} is bounded.


## 3. Proof of the Main Results

We will construct a sequence of functions{f[n]} which is asymptotic linear at 0 and satisfies(3.1)lim⁡n→∞sup⁡s∈R|f[n](s)-f(s)|=0,lim⁡n→∞(f[n])0:=lim⁡n→∞(lim⁡|s|→0f[n](s)s)=0.
By means of some corresponding auxiliary equations, we can obtain a sequence of unbounded components {Ckν[n]} via Rabinowitz’s global bifurcation theorem [14]. Based on the sequence, we can find an unbounded component Ckν satisfying:(3.2)Ckν⊂limsup⁡n→∞Ckν[n],
and joining (∞,θ) with (∞,∞). We do it as follows.For eachn∈ℕ, define f[n](s):ℝ→ℝ by:(3.3)f[n](s)={f(s),s∈(1n,∞)∪(-∞,-1n),nf(1n)s,s∈[-1n,1n].
Then, f[n]∈C(ℝ,ℝ)⋂C1(ℝ∖{±1/n},ℝ) with(3.4)sf[n](s)>0,∀s≠0,(f[n])0=nf(1n).
By (A3), it follows that(3.5)lim⁡n→∞(f[n])0=0.
Now let us consider the auxiliary family of problems:(3.6)u′′+λf[n](u)=0,t∈(0,1),u(0)=0,u(1)=∑i=1m-2αiu(ηi).
From Proposition  4.1 in [8], we obtain the following.Lemma 3.1.
Let (A1) and (A2) hold. If(μ,u) is a nontrivial solution of (3.6). Then, u∈Tkν for some k,ν.Letg[n]∈C(ℝ,ℝ) such that:(3.7)f[n](u)=(f[n])0u+g[n](u)=nf(1n)u+g[n](u).
Note that(3.8)lim⁡|s|→0g[n](s)s=0.
Let us consider(3.9)Lu-λ(f[n])0u=λg[n](u),
as a bifurcation problem from the trivial solution u≡θ.Equation (3.9) can be converted to the equivalent form:(3.10)u(t)=∫01H(t,s)[λ(f[n])0u(s)+λg[n](u(s))]ds:=λL-1[(f[n])0u(⋅)](t)+λL-1[g[n](u(⋅))](t).
Note that ∥L-1[g[n](u)]∥=o(∥u∥) for u near θ in X. Applying Lemma 2.4, the global bifurcation result of Rabinowitz [14] for (3.9) can be stated as follows: for each integer k≥1, ν∈{+,-}, there exists a continuum Ckν[n] of solutions of (3.9) joining (λk/(f[n])0,θ) to infinity in 𝕏. Moreover, Ckν[n]∖(λk/(f[n])0,θ)⊂(0,∞)×Tkν.For properties ofCkν[n], we give the following lemmas.Lemma 3.2.
Let (A1)–(A4) hold. Then for each fixedn,  Ckν[n] joins (λk/(f[n])0,θ) to  (∞,∞) in 𝕏 (see Figure 1(b)).Proof.
We divide the proof into two steps.
Step 1.
We show thatsup⁡{λ∣(λ,u)∈Ckν[n]}=∞. Assume on the contrary that sup⁡{λ∣(λ,u)∈Ckν[n]}=:C0<∞. Let {(μl,yl)}⊂Ckν[n] be such that:
(3.11)|μl|+‖yl‖→∞.
Similar to the argument of Lemma 2.5, we conclude that ∥yl∥∞→∞.
Since(μl,yl)∈𝒞kν[n], we have
(3.12)yl′′(t)+μlf[n](yl(t))=0,t∈(0,1),yl(0)=0,yl(1)=∑i=1m-2αiyl(ηi).
Set vl(t)=yl(t)/∥yl∥∞. Then, ∥vl∥∞=1, and
(3.13)vl′′(t)+μlf[n](yl(t))‖yl‖∞=0,t∈(0,1).
Using lim⁡|u|→0f(u)/u=0, we can show that
(3.14)lim⁡l→∞|f[n](yl(t))|‖yl‖∞=0.
The proof is similar to that of Theorem  1 in [12], and therefore we omit it. Equations (3.13) and (3.14) imply that ∥vl′′∥∞≤M for some constant M>0, independent of l. Hence, {vl} has a convergent subsequence in X. Without loss of generality, we assume that there exists (μ*,v*)∈[0,C0]×X with:
(3.15)‖v*‖∞=1,
such that
(3.16)lim⁡l→∞(μl,vl)=(μ*,v*),inR×Y.
Note that (3.12) is equivalent to
(3.17)vl(t)=μl∫01H(t,s)f[n](yl(s))‖yl‖∞ds,t∈(0,1).
Combining this with (3.16) and using (3.14) and the Lebesgue dominated convergence theorem, we have
(3.18)v*(t)=μ*∫01H(t,s)0ds=0,t∈(0,1).
This contradicts (3.15). Therefore,
(3.19)sup⁡{λ∣(λ,y)∈Ckν[n]}=∞.
Step 2.
We show thatsup⁡{∥u∥∞∣(λ,u)∈Ckν[n]}=∞. On the contrary, assume that sup⁡{∥u∥∞∣(λ,u)∈Ckν[n]}=M0<∞. Then, there exists a sequence {(μl,yl)}⊂Ckν[n] such that
(3.20)μl⟶∞,‖yl‖∞≤M0.
From Remark 1.1, we can take a subsequences of {(μl,yl)}, still denoted by {(μl,yl)}, such that {yl}⊂Tkν⋂Skν or {yl}⊂Tkν⋂Sk+1ν. Without loss of generality, we suppose that {yl}⊂Tkν⋂Skν. When {yl}⊂Tkν⋂Sk+1ν is considered, the proof is similar. We omit it.
Note that(μl,yl) satisfies the autonomous equation:
(3.21)yl′′+μlf[n](yl)=0,t∈(0,1).
Therefore, the graph of yl consists of a sequence of positive and negative bumps, together with a truncated bump at the right end of the interval [0,1], with the following properties (ignoring the truncated bump) (see [8]): all the positive (respectively, negative) bumps (i) have the same shape (the shapes of the positive and negative bumps may be different); (ii) attain the same maximum (minimum) value.
Let(3.22)0=τl0<τl1<⋯<τlk-1
denote the zeros of yl in [0,1]. Then, after taking a subsequence if necessary, lim⁡l→∞τlj:=τ∞j,j∈{0,1,…,k-1}. Clearly, τ∞0=0. Set τ∞k=1. We can choose at least one subinterval (τ∞j,τ∞j+1)≜I∞j which is of length at least 1/k for some j∈{0,1,…,k-1}. Then, for this j, τlj+1-τlj>3/4k if l is large enough. Put (τlj,τlj+1)≜Ilj.
Obviously, for the above givenk,ν, and j, yl(t) have the same sign on Ilj for all l. Without loss of generality, we assume
(3.23)yl(t)>0,t∈Ilj.
Armed with the information on the shape of yl, it is easy to show that for the above given Ilj, ∥yl∥Ilj,∞:=max⁡Iljyl(t)≤M0,l=1,2,….
Letσ be a constant with 0<σ<3/8k. Since yl is concave on Ilj, we have
(3.24)yl(t)≥σ‖yl‖Ilj,∞,∀t∈[τlj+σ,τlj+1-σ].
Then, there must exist constants α,β with [α,β]⊂I∞j and l0 such that
(3.25)yl(t)≥σ‖yl‖Ilj,∞>0,uniformlyfort∈[α,β]andl>l0.
On the other hand, note that
(3.26)f[n](yl(t))yl(t)≥inf⁡{f[n](s)s∣0<s≤M0}>0,t∈(τlj,τlj+1).
Using the relation:
(3.27)yl′′(t)+μlf[n](yl(t))yl(t)yl(t)=0,t∈(τlj,τlj+1),
and Sturm’s comparison theorem, we deduce that yl must change its sign on (α,β) if l is sufficiently large, contradicting (3.25). Therefore,
(3.28)lim⁡l→∞‖yl‖∞=∞.
Hence, Ckν[n] joins (λk/(f[n])0,θ) to (∞,∞) in 𝕏.Lemma 3.3.
Let (A1)–(A4)  hold. Then, there existsρkν>0 such that
(3.29)(⋃n=1∞Ckν[n])⋂((0,ρkν)×X)=∅.Proof.
The proof is similar to that of Lemma  4.3 in [11]. We omit it.Lemma 3.4.
Let (A1)–(A4) hold, and letρkν be as in Lemma 3.3. Then, there exist n0∈ℕ and λkν̂≥ρkν>0 such that for any λ>λkν̂ and u∈Ckν[n]:
(3.30)Ckν[n]∩{(λ,u)∣λ≥λkν̂;‖u‖∞=1}=∅,∀n>n0.Proof.
Suppose on the contrary that there exists{(μl,yl)}⊂(⋃n=1∞Ckν[n])⋂((0,∞)×X) such that
(3.31)lim⁡l→∞μl=∞,‖yl‖∞=1.
Now, the method used in the proof of Lemma3.2, Step 2, is still valid. Let σ be a constant with 0<σ<3/8k. Taking subsequences again if necessary, still denoted by {(μl,yl)}, such that {yl}⊂Tkν⋂Skν. Without loss of generality, we can also derive an interval [α,β]⊂I∞j and l0 such that
(3.32)1≥yl(t)≥σ‖yl‖Ilj,∞=σ,uniformlyfort∈[α,β]andl>l0.
It is easy to find an integer n0∈ℕ such that 1/n0<σ. This implies that
(3.33)inf⁡{f[n](s)s∣σ<s≤1}=inf⁡{f(s)s∣σ<s≤1},∀n>n0.
Note that for n>n0,
(3.34)f[n](yl(t))yl(t)≥inf⁡{f(s)s∣σ<s≤1}>0,uniformlyfort∈(α,β)andl>l0.
Combining these facts and the relation:
(3.35)yl′′(t)+μlf[n](yl(t))yl(t)yl(t)=0,t∈(α,β),
and Sturm’s comparison theorem, we conclude that yl must change its sign on (α,β) if l is large enough. This contradicts (3.32), and the proof is done.Lemma 3.5.
Let (A1)–(A4) hold, and letn0 be as in Lemma 3.4. Then, there exist λkν≥ρkν>0 and ϵ∈(0,1/2) such that for any λ>λkν and u∈Ckν[n]:
(3.36)Ckν[n]∩{(λ,u)∣λ≥λkν;1-2ϵ≤‖u‖∞≤1+2ϵ}=∅,∀n>n0.Proof.
Similar to the proof of Lemma3.4, we can find a constant λkν̃>0 such that ∥u∥∞≠1 provided that (λ,u)∈(λkν̃,∞)×Tkν being a solution of (1.1) and (1.2).
Letλkν=max⁡{λkν̂,λkν̃}+1. We claim that there exists ϵ∈(0,1/2) such that (3.36) holds. Suppose on the contrary that there exists
(3.37){(μl,yl)}⊂(⋃n=1∞Ckν[n])⋂((λkν,∞)×X),
satisfying
(3.38)lim⁡l→∞μl=μ*≥λkνlim⁡l→∞‖yl‖∞=1.
We can discuss two cases.
Case 1.
Ifμ*<∞. {yl} is compact in X implies that there exists a subsequence, still denoted by {yl}, such that
(3.39)lim⁡l→∞yl=y*∈Tkν,‖y*‖∞=1.
Obviously, (μ*,y*) is a solution of (1.1) and (1.2). It is impossible.Case 2.
Ifμ*=∞. Taking subsequences again if necessary, still denoted by {(μl,yl)}, such that 1/2≤∥yl∥∞≤3/2. Using the same argument as Lemma 3.4, we can find a contradiction.Proof of Theorem1.3.
We will prove that the superior limit ofCkν[n] contains an unbounded component 𝒞kν⊂(0,∞)×Tkν of solutions of (1.1) and (1.2), which joins (∞,θ) to (∞,∞). For r>0, introduce
(3.40)Ωr={u∈Y∣‖u‖∞<r}.
Set
(3.41)Γkν≔([0,∞)×Tkν)∖{(η,u)∣η≥λkν;u∈Tkν,‖u‖∞≤1+ϵ},Σkν:={(η,u)∣η≥λkν;u∈Tkν,‖u‖∞≤1-ϵ}.
Let n0 and ϵ be as in Lemma 3.5. Firstly, for each given nonnegative integer p=0,1,2,…, and n≥n0 with (λk/(f[n])0)≥λkν+p, we define the connected subset, (ζkν[n])p, in Ckν[n] satisfying (see Figure 2(a)):
(i)(ζkν[n])p⊂(Ckν[n]∖(λkν+p,∞)×Ω1-ϵ);
(ii)(ζkν[n])p joins {λkν+p}×Ω1-ϵ with infinity in Γkν.By Lemmas 2.2 and 2.3, limsup⁡n→∞(ζkν[n])p contains a component (ζkν)p joining {λkν+p}×Ω1-ϵ with infinity in Γkν (see Figure 2(b)).
It is easy to verify that if(λ,u)∈(ζkν)p(p=0,1,2,…), then (λ,u) is a solution of (1.1) and (1.2), and u∈Tkν.
Next, by using Lemma2.3 and the method in [11] (see (4.22)–(4.30) in [11]), we can find a component 𝒞kν in limsup⁡p→∞(ζkν)p, which is unbounded both in Γkν and Σkν.
Finally, we show that𝒞kν joins (∞,θ) with (∞,∞). This will be done by the following three steps.
Step 1.
We show thatlim⁡λ→+∞∥u∥∞=0 for (λ,u)∈(𝒞kν∩Σkν).
Suppose on the contrary that there exists{(μl,yl)}⊂𝒞kν with ∥yl∥∞≤1-ϵ, and
(3.42)μl→+∞,‖yl‖∞≥a,
for some constant a>0. Applying the method of proving Lemma 3.2, we can deduce a contradiction.Step 2.
We show thatsup⁡{λ∣(λ,u)∈(𝒞kν∩Γkν)}=∞. By a similar argument as Lemma 3.2, we can get the conclusion.Step 3.
We show thatlim⁡λ→+∞∥u∥∞=+∞ for (λ,u)∈(𝒞kν∩Γkν).
On the contrary, suppose that there exists{(μl,yl)}⊂(𝒞kν∩Γkν) with
(3.43)μl→+∞,1<‖yl‖∞≤M,
for some constant M>0. The proof can be done by the same argument as Lemma 3.2.
This completes the proof of Theorem1.3.(a)(b)Proof of Corollary1.4.
The result can be directly obtained by Theorem1.3.



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