# The Existence and Application of Unbounded Connected Components

Authors: Hua Luo; Ruyun Ma
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

LetX be a Banach space and Cn a family of connected subsets of R×X. We prove the existence of unbounded components in superior limit of {Cn}, denoted by lim¯ Cn, which have prescribed shapes. As applications, we investigate the global behavior of the set of positive periodic solutions to nonlinear first-order differential equations with delay, which can be used for modeling physiological processes.


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## Body

## 1. Introduction and the Main Results

The connectivity result on the fixed set of a1-parameter family of maps, which goes back to Leray and Schauder [1] and was proved in its full generality by Browder [2], is a useful tool in the study of global continua of solutions on nonlinear differential equations. Costa and Gonçalves [3] stated and proved a suitable version for the study of nonlinear boundary value problems at resonance. Massabò and Pejsachowicz [4] generalized the main results of [1, 2] to the n-parameter family of compact vector fields. The above results were established when the parameter(s) changes in a bounded set. Sun and Song [5] proved the existence of unbounded connected component of 1-parameter family of compact vector fields, where the parameter varies on whole real line. All of these results play important roles in the study of nonlinear functional analysis and nonlinear differential equations.For clearly reading, we firstly recall Kuratowski's definitions and notations in [6].Let𝕄 be a metric space. Let {Cn∣n=1,2,…} be a family of subsets of 𝕄. Then thesuperior limit𝒟 of {Cn} is defined by
(1)𝒟:=lim⁡¯Cn={x∈M∣∃{ni}⊂ℕ,xni∈Cni,mmmmmmmmmmmsuchthatxni⟶x}.Acomponent of a set 𝕄 means a maximal connected subset of 𝕄.Definition 1.
LetX be a Banach space with the norm ∥·∥. Let ζ be a component of solutions in ℝ×X. ζ meets (a,0) and infinity means that there existed a sequence {(λk,uk)}⊂[ζ∖{(a,0)}] such that (λk,uk)→(a,0) as k→∞.
Forρ,β∈(0,∞), let us denote
(2)Bρ:={u∈X∣∥u∥≤ρ},Ωβ,ρ:=([0,∞)×X)∖{(η,u)∈[β,∞)×X∣∥u∥≤ρ}.
Let{𝒞n} be a family of connected subsets of ℝ×X. The purpose of this paper is to study the existence of unbounded components in lim⁡¯𝒞n which have prescribed shapes.
More precisely, we will prove the following theorems.Theorem 2.
LetX be a Banach space and let {𝒞n} be a family of connected subsets of [0,∞)×X. Assume that(A1)
there exist0<σ<r<∞ and λ*∈(0,∞), such that
(3)𝒞n∩{(μ,y)∣0<μ≤λ*+σ,r-σ≤∥y∥≤r+σ}=∅;(A2)
ηn→0+ and 𝒞n meets (ηn,0) and infinity;(A3)
for everyR>0, (⋃n=1∞‍𝒞n)⋂‍𝔹R is a relatively compact set of ℝ×X, where
(4)𝔹R={(μ,x)∈ℝ×X∣|μ|<R,∥x∥≤R}.Then there exists a component 𝒞 in lim⁡¯𝒞n satisfying(a)
𝒞 meets (0,0) and infinity;(b)
𝒞∩{(μ,y)∣0<μ≤λ*,∥y∥=r}=∅.Theorem 3.
Leta∈ℝ be a constant. Let X be a Banach space, and let {𝒞n} be a family of connected subsets of ℝ×X. Assume that(H1)
𝒞n∩((-∞,a]×X)=∅;(H2)
there exist0<σ<r<∞ and b∈(a,∞), such that
(5)𝒞n∩{(μ,y)∣μ≥b-σ,r-σ≤∥y∥≤r+σ}=∅;(H3)
ηk>a for all k∈ℕ, ηk→+∞ and 𝒞n meets (ηn,0) and infinity in ([a,∞)×X)∖Ωb,r;(H4)
for everyR>0, (⋃n=1∞‍𝒞n)∩𝔹R is a relatively compact set of ℝ×X. Then there exists a component 𝒞 in lim⁡¯𝒞n such that(a)
both𝒞∩Ωb,r and 𝒞∩(([a,∞)×X)∖Ωb,r) are unbounded;(b)
𝒞∩{(μ,y)∣μ≥b,∥y∥=r}=∅.


## 2. Proofs of the Main Results

To prove Theorems2 and 3, we need the following preliminary result, which is proved by Ma and An [7] by using Whyburn lemma, and the method of Sun and Song to prove [6, Lemma 2.2].Lemma 4 (see [7, Lemma 2.2]).
LetE be a Banach space with the norm ∥·∥E. Let {Dn} be a family of connected subsets of E. Assume that(i)
there existzn∈Dn, n=1,2,…, and z*∈E, such that zn→z*;(ii)
lim⁡n→∞rn=∞, where rn=sup⁡{∥x∥E∣x∈Dn};(iii)
for everyR>0, (⋃n=1∞‍Dn)∩BR is a relatively compact set of E, where
(6)BR={x∈E∣∥x∥E≤R}.Then there exists an unbounded component 𝒞 in lim⁡¯Dn and z*∈𝒞.Proof of Theorem2.
(a) It is a direct consequence of Lemma4.
(b) Assume, on the contrary, that the conclusion is not true. Then there exists(μ*,u*)∈𝒞 with μ*≤λ* and ∥u*∥=r. Hence, there exists {(ηnk,unk)}⊂𝒞n, such that
(7)lim⁡k→∞ηnk=μ*,lim⁡k→∞unk=u*.
Thus, there exists k0∈ℕ, such that, for k≥k0,
(8)ηnk<λ*+σ2,r-σ2≤∥unk∥≤r+σ2.
However, this contradicts (3).Proof of Theorem3.
(a) Sinceηk→∞, we may assume that
(9)ηk>b,k∈ℕ.
So, it follows from conditions (ii) and (iii) that 𝒞n meets {b}×Br and infinity in ([a,∞)×X)∖Ωb,r.
For each(b,v)∈({b}×Br)∩𝒟, let ℰ(b,v)(⊂𝒟) be a component containing (b,v). Let
(10)ς(b,v):=sup⁡{λ∣(λ,u)∈ℰ(b,v),u∈Br}.
Set(11)Π:={(b,v)∣(b,v)∈({b}×Br)∩𝒟,ℰ(b,v)mmmmmmisunboundedin([a,∞)×X)∖Ωb,r}.
Then Π≠∅ since
(12)(𝒞j∩({b}×Br))⊆Π,j∈ℕ.
From Lemma 4, it follows that Π is closed in [0,∞)×X, and, furthermore, Π is compact in [0,∞)×X.
Let(13)Σ⋄:=⋃‍(b,v)∈Πℰ(b,v).
By Lemma 4, lim⁡¯Cn∩(([a,∞)×X)∖Ωb,r) contains a component ζ which meets {b}×Br and infinity in (([a,∞)×X)∖Ωb,r). Obviously
(14)ζ∈Σ⋄.
Ifς(b,v)=+∞ for some (b,v)∈Π, then Theorem 3 holds.
Assume, on the contrary, thatς(b,v)<+∞ for all (b,v)∈Π.
For every(b,v)∈Π, let ℰ⊳(b,v) be the component in ℰ(b,v)∩([b,∞)×Br) which contains (b,v). Using the standard method, we can find a bounded open set U(b,v) in [b,∞)×Br, such that
(15)ℰ⊳(b,v)⊂U(b,v),∂U(v)∩Σ⋄=∅
and for every (b,v)∈Π,
(16)sup⁡{λ∣(λ,u)∈U¯(b,v)}<∞,
where ∂U(v) and U¯(v) are the boundary and closure of U(v) in [b,∞)×Br, respectively.
Evidently, the following family of the open sets of{b}×Br(17){U(b,v)∩({b}×Br)∣(b,v)∈Π}
is an open covering of Π. Since Π is compact set in {b}×Br, there exist v1,…,vm such that (b,vi)∈Π(i=1,…,m), and the family of open sets in {b}×Br:(18){U(b,vi)∩({b}×Br)∣i=1,…,m}
is a finite open covering of Π. This implies that
(19)Π⊆{U(b,vi)∩({b}×Br)∣i=1,…,m}.
Let(20)U1=⋃‍i=1mU(b,vi).
Then U1 is a bounded open set in [b,∞)×Br,
(21)∂U1∩(⋃(b,v)∈Π‍ℰ⊳(b,v))=∅,∂U1∩Σ⋄=∅
and by (16), we have
(22)sup⁡{λ∣(λ,u)∈U¯1}<∞,
where ∂U1 and U¯1 are the boundary and closure of U1 in [b,∞)×Br, respectively.
Now, (22) together with (19) and (21) implied that
(23)sup⁡{λ∣(λ,u)∈Σ⋄,u∈Br}<∞.
However, this contradicts λn→∞.
Therefore, there exists(b,v*)∈Π such that ζ:=ℰ(b,v*) which is unbounded in both [b,∞)×Br and ([a,∞)×X)∖([b,∞)×Br).
(b) By a fully analogous argument as in the proof of Theorem2(b) (with minor modifications), one can immediately obtain the desired results.


## 3. Application to Functional Differential Equations

In recent years, there has been considerable interest in the existence ofω-periodic solutions of the equation
(24)u′(t)=a(t)g(u(t))u(t)-λh(t)f(u(t-τ(t))),t∈ℝ,
where a,h∈C(ℝ,[0,∞)) are ω-periodic functions and τ is a continuous ω-periodic function. Equation (24) has been proposed as a model for a variety of physiological processes and conditions including production of blood cells, respiration, and cardiac arrhythmias. See, for example, [8–20] and the references therein.Recently, Wang [18] used the fixed point index [20, 21] to study the existence, multiplicity, and nonexistence of positive solutions of (24) under the following assumptions.(C1)
a,h∈C(ℝ,[0,∞)) are ω-periodic functions, ∫0ω‍a(t)dt>0, h(t)>0 on [0,ω]; τ∈C(ℝ,ℝ) is ω-periodic functions.(C2)
f,g:[0,∞)→[0,∞) are continuous. 0<l≤g(s)≤L for s>0, l,L are given positive constants. f(s)>0 for s>0.Letσ:=e-∫0ω‍a(t)dt and denote
(25)M(r):=max⁡0≤t≤r{f(t)},m(r):=min⁡{f(t):σL(1-σl)1-σLσL(1-σl)1-σLr≤t≤r};f0:=lim⁡u→0+f(u)u,f∞:=lim⁡u→+∞f(u)u.
His results provide no any information about the global behavior of the set of positive solutions of (24).In this section, we will use Theorems2 and 3 to establish several results on the global behavior of the set of positive solutions of (24), and, accordingly, we get some existence and multiplicity results of positive solutions of (24).We will work essentially in the Banach spaceX={u∈C[0,ω]∣uisω-periodic.} with sup norm ∥·∥.By a positive solution of (24), we mean a pair (λ,u), where λ>0 and u is a solution of (24) with u>0 on [0,ω].LetΣ⊂ℝ+×E be the closure of the set of positive solutions of (24).We extend the functionf to a continuous function f~ defined on ℝ in such a way that f~>0 for all s<0. For λ>0, we then look at arbitrary solutions u of the eigenvalue problem
(26)-u′(t)+a(t)g(u(t))u(t)=λh(t)f~(u(t-τ(t))),hmmmmmmmmmmmmmmt∈ℝ,uisω-periodic.
It was shown in [18] that (26) is equivalent to
(27)u(t)=λ∫tt+ω‍Gu(t,s)h(s)f~(u(s-τ(s)))ds,
where
(28)Gu(t,s)=e-∫ts‍a(θ)g(u(θ))dθ1-e-∫0ω‍a(θ)g(u(θ))dθ,t≤s≤t+ω.
By the positivity of Green's function Gu(·,·), h(·), and f(·), such solutions are positive. Therefore, the closure of the set of nontrivial solutions (λ,u) of (24) in ℝ+×X is exactly Σ.Next, we consider the spectrum of the linear eigenvalue problem(29)-u′(t)+a(t)cu(t)=λh(t)u(t-τ(t)),mmmmmmmmt∈ℝ,uisω-periodic.Lemma 5.
Letc be a positive constant. Then the linear problem (29) has a unique eigenvalue λ⋄(c), which is positive and simple, and the corresponding eigenfunction ψ(·) is of one sign.Proof.
Define an operatorT:X→X by
(30)Tu(t)=λ∫tt+ω‍Gu(t,s)h(s)f~(u(s-τ(s)))ds.
Let K be the cone
(31)K={u∈X∣u≥0}.
Then it follows from [18, Lemma 2.2] that T:K→K is strongly positive and completely continuous. Thus, the desired result is a direct consequence of Krein-Rutman Theorem, cf. [22, Theorem 19.3].Theorem 6.
Let (C1)-(C2) hold. Assume that(C3)
f0=∞=f∞.Then Σ contains a component 𝒞 satisfying(1)
𝒞 meets (0,0) and (0,∞);(2)
for everyr>0, there exists μ(r)>0, such that
(32)𝒞∩{(μ,u)∈(0,μ(r))×X∣∥u∥=r}=∅.Corollary 7.
Let (C1)–(C3) hold. Then there is a constantη*>0 such that (24) has at least two positive solutions as 0<λ<η*.Proof.
Let(33)B:=sup⁡{λ∣(λ,u)∈𝒞},η*:=sup⁡{μ(r)∣r∈(0,∞)}.
Then 0<η*≤B. It is easy to see from Theorem 6 that (24) has at least two positive solutions as 0<λ<η*.Denote the coneK in X by
(34)K={u∈X∣u(t)≥0on⁡[0,ω],u(t)≥δ∥u∥},mmmmmmmδ:=σL(1-σl)1-σL,
and for r>0, let
(35)Kr={u∈K∣∥u∥<r}.Define an operatorTλ:K→X by
(36)Tλu(t)=λ∫tt+ω‍Gu(t,s)h(s)f~(u(s-τ(s)))ds,mmmmmmmmmmmmmmmmt∈[0,ω].Lemma 8 (see [18]).
Assume that (C1)-(C2) hold. ThenTλ:K→K is completely continuous.Lemma 9.
Let (C1)-(C2) hold. Ifu∈∂Kr,r>0, then
(37)∥Tλu∥≤λMr1-σl∫0ω‍h(s)ds,
where Mr=1+max⁡δr≤s≤r{f(s)}.Proof.
It is well known from Wang [18] that
(38)σL1-σL≤Gu(t,s)≤11-σl,t≤s≤t+ω.
This together with the fact that h is ω-periodic and u(t)≥δr on [0,ω] implies that
(39)∥Tλu∥≤λMr1-σl∫tt+ω‍h(s)ds=λMr1-σl∫0ω‍h(s)ds.To prove above Theorem6, we define f[n]:[0,∞)→[0,∞) by
(40)f[n](s)={f(s),s∈(1n,∞),nf(1n)s,s∈[0,1n].
Then f[n]∈C([0,∞),[0,∞)) with
(41)f[n](s)>0,∀s∈(0,∞),(f[n])0=nf(1n)>0.
By (C3), it follows that
(42)lim⁡n→∞(f[n])0=∞.Now let us consider the auxiliary family of the equations(43)u′(t)=a(t)g(u(t))u(t)-λh(t)f[n](u(t-τ(t))),mmmmmmmmmmmmmmt∈ℝ,uisω-periodic.
Let ξ,χ∈C[0,∞) be such that
(44)f[n](s)=(f[n])0s+ξ(s)=nf(1n)s+ξ(s),g(s)=g(0)+χ(s).
Note that
(45)lim⁡s→0+ξ(s)s=0,lim⁡s→0+χ(s)=0.
Define a linear operator A:D(A)⊂X→X(46)(Au)(t)=-u′(t)+a(t)g(0)u(t),u∈D(A),
with
(47)D(A)={u∈C1[0,ω]∣uisω-periodic.}.
From [18], it follows that A-1:X→X is compact and continuous.Now (43) can be rewritten to the form
(48)u(t)=λA-1[h(·)(f[n])0u(·-τ(·))](t)+A-1N(λ,u(·))(t),
where
(49)N(λ,u(·))(t):=[a(·)χ(u(·))u(·)hh-λh(·)ξ(u(·-τ(·)))](t).
It is easy to check that
(50)lim⁡∥u∥→0N(λ,u)∥u∥=0,uniformlyon⁡boundedλintervals.
It is easy to check that (48) is equivalent to
(51)Au=λh(·)(f[n])0u(·-τ(·))(t)+N(λ,u(·))(t).
Let us consider (48) as a bifurcation problem from the trivial solution u≡0.Since(f[n])0>0, the results of Nonlinear Krein-Rutman Theorem (see Dancer [22] and Zeidler [23, Corollary 15.12]) for (48) can be stated as follows: there exists a continuum 𝒞+[n] of positive solutions of (48) joining (λ⋄(g(0))/(f[n])0,0) to infinity in K. Moreover, 𝒞+[n]∖{(λ⋄(g(0))/(f[n])0,0)}⊂int⁡K and (λ⋄(g(0))/(f[n])0,0) is the only bifurcation point of (48) lying on trivial solutions line u≡0.Proof of Theorem6.
Let us verify that{𝒞+[n]} satisfies all of the conditions of Theorem 2.
It follows from (42) that λ⋄(g(0))/(f[n])0=:ηn→0 as n→∞. Therefore, (A2) holds.
Letr>0 be fixed. Then there exists n0∈ℕ, such that
(52)1n≤δr.
Thus
(53)Mr(f)=Mr(f[n]),n≥n0.
From this and Lemma 9, it follows that there exists λ*(r) with 0<λ*(r)<r(1-σl)/Mr∫0ω‍h(s)ds, such that (48) has no solution (λ,u) with
(54)∥u∥=r,0<λ<λ*(r).
Since r is arbitrary, we see that (A1) is satisfied.(A3)
can be deduced directly from the Arzela-Ascoli Theorem and the definition off[n].
Therefore, the superior limit of{𝒞+[n]} contains an unbounded component 𝒞 with (0,0)∈𝒞.
Moreover,𝒞∩{(μ,y)∣0<μ≤λ*(r),∥y∥=r}=∅.
Finally, we show that𝒞 meets (0,0) and (0,∞).
Let{μn,yn}⊂𝒞 with
(55)|μn|+∥yn∥⟶∞.
We claim that μn→0+.
Assume on the contrary thatμn→+∞. Let
(56)vn(t)=yn(t)∥yn∥.
Then
(57)vn′(t)=a(t)g(yn(t))vn(t)-μnh(t)×f(yn(t-τ(t)))yn(t-τ(t))vn(t-τ(t)),t∈ℝ,(58)vn∈X.
So vn′(t)<0 on [0,ω] as n→∞. This contradicts (58). Thus (55) implies that
(59)∥yn∥⟶∞.
Assume on the contrary thatμn→α>0 ( after taking a subsequence and relabeling if necessary).
Sinceyn(t)≥δ∥yn∥ on [0,ω], it follows from (59) that yn(t)→∞ uniformly on [0,ω]. This together with f∞=∞ implies vn′(t)<0 on [0,ω] as n→∞. This contradicts (58) again.Theorem 10.
Let (C1)-(C2) hold. Assume that(C4)
f0=0;f∞=0.Then Σ contains a component 𝒞 satisfying the following.(1)
For givenb^>0,
(60)sup⁡{λ∣(λ,u)∈(𝒞∩Ωb^,1)}=+∞.(2)
For everyr>0, there exists η(r)>0, such that
(61)𝒞∩{(μ,u)∈[μ(r),∞)×X∣∥u∥=r}=∅.Corollary 11.
Let (C1)-(C2) and (C4) hold. Then there is a constantη*>0 such that (24) has at least two positive solutions as λ>η*.To prove Theorem10, we define f[n] as in (40). Notice that (C4) implies that
(62)lim⁡n→∞(f[n])0=0.
Let ξ,χ be the function satisfying (44)-(45).Now (43) can be rewritten to the form
(63)u(t)=λA-1[h(·)(f[n])0u(·-τ(·))](t)+A-1N(λ,u(·))(t),
where
(64)N(λ,u(·))(t):=[a(·)χ(u(·))u(·)v-λh(·)ξ(u(·-τ(·)))](t).
It is easy to check that
(65)lim⁡∥u∥→0N(λ,u)∥u∥=0,uniformlyon⁡boundedλintervals.
Let us consider (63) as a bifurcation problem from the trivial solution u≡0.By similar method to deal with (48) in which f0=∞, we have from (f[n])0>0 and [23, Corollary 15.12]) that there exists a continuum 𝒞+[n] of positive solutions of (63) joining (λ⋄(g(0))/(f[n])0,0) to infinity in K. Moreover, 𝒞+[n]∖{(λ⋄(g(0))/(f[n])0,0)}⊂intK and (λ⋄(g(0))/(f[n])0,0) is the only bifurcation point of (48) lying on trivial solutions line u≡0.Lemma 12.
Assume that (C1)-(C2) hold. Ifu∈∂Kr,r>0, then
(66)∥Tλu∥≥λσLm^r1-σL∫0ω‍h(s)ds,
where
(67)m^r=min⁡δr≤x≤r{f(x)}.Proof.
SinceσL/(1-σL)≤Gu(t,s) and f(u(t))≥m^r for t∈[0,ω], it follows that
(68)∥Tλu∥≥λ∫tt+ω‍Gu(t,s)h(s)f(u(s))ds≥λσLm^r1-σL∫tt+ω‍h(s)ds≥λσLm^r1-σL∫0ω‍h(s)ds.Lemma 13.
Assume that (C1)-(C2) and (C4) hold, and letI be a compact subinterval of (0,∞). Then
(69)∥u∥≤MI,(λ,u)∈{(λ,u)∈C+[n]∣λ∈I}
for some positive constant MI, independent of n,λ, and u.Proof.
Assume on the contrary that there exists a sequence{(μk,yk)}⊂𝒞+[n]∩(I×K) such that
(70)∥yk∥⟶∞.
Set vk(t)=yk(t)/∥yk∥. Then
(71)∥vk∥=1,(72)vn′(t)=a(t)g(yn(t))vn(t)-μnh(t)×f(yn(t-τ(t)))yn(t-τ(t))vn(t-τ(t)),t∈ℝ,(73)vn∈X.
By (72), {vn′} is bounded in X. This together with the fact that {μn}⊂I implies that there exists (μ*,v*)∈I×X with
(74)∥v*∥=1,
such that
(75)lim⁡k→∞(μk,vk)=(μ*,v*),inℝ×X,
(after choosing a subsequence and relabeling if necessary). Since {g(un)} is bounded in X, {g(un)} is bounded in L2(0,ω), and subsequently, g(un)⇀g^ for some g^∈L2(0,ω). By the standard method, we can prove that
(76)l≤g^(t)≤L,a.e.t∈[0,ω].
Moreover, combining (75) and (76) with the assumption f∞=0 and the corresponding integral equations of (72) and (73) and using Lebesgue dominated convergence theorem, we conclude that
(77)v*′(t)-a(t)g^(t)v*=0,a.e.t∈ℝ,(78)v*∈X.
Note that a(t)g^(t)v*≥0 and a(t)g^(t)v*≢0 on [0,ω]. This means that v*(0)<v*(ω). However, this contradicts (78).Now, we are in the position to prove Theorem10.Proof of Theorem10 (sketched).
fixedr>0. From Lemma 12, it follows that (63) has no solution if
(79)λ>λ*(r)=:1+r(σLm^r/(1-σL))∫0ω‍h(s)ds.
Applying Lemma13, it is easy to verify that {𝒞+[n]} satisfies all of the conditions of Theorem 3. So, there exists a component 𝒞 in lim⁡¯𝒞+[n] such that(a)
both𝒞∩Ωλ*(1),1 and (𝒞∩(([a,∞)×X)∖Ωλ*(1),1)) are unbounded;(b)
𝒞∩{(μ,y)|μ≥λ*(1),∥y∥=1}=∅. Lemma 13 guarantees that 𝒞 satisfies
(80)sup⁡⁡{λ∣(λ,u)∈(𝒞∩Ωλ*(1),1)}=+∞.



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