# Global Structure of Positive Solutions for a Singular Fourth-Order Integral Boundary Value Problem

Authors: Wenguo Shen; Tao He
Journal: Discrete Dynamics in Nature and Society
 (2014)

---

## Abstract

We consider fourth-order boundary value problemsu′′′′(t)=λh(t)f(u(t)),0<t<1,u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0, where ∫01‍u(s)dα(s) is a Stieltjes integral with α(t) being nondecreasing and α(t) being not a constant on [0,1];h(t) may be singular at t=0 and t=1, h∈C((0,1),[0,∞)) with h(t)≢0 on any subinterval of (0,1); f∈C([0,∞),[0,∞)) and f(s)>0 for all s>0, and f0=∞,f∞=0,f0=lims→0+f(s)/s,f∞=lims→+∞f(s)/s. We investigate the global structure of positive solutions by using global bifurcation techniques.


---

## Body

## 1. Introduction

Recently, fourth-order boundary value problem(1)x′′′′+kx′′+lx=λh(t)f(x),0<t<1,x(0)=x(1)=x′(0)=x′(1)=0
has been investigated by the fixed point theory in cones, see [1–4] (k=l=0). By applying bifurcation techniques, see Rynne [5] (k=l=0), Korman [6] (k=l=0), Xu and Han [7] (k=0,l≠0), Shen [8, 9] (k≠0,l≠0), and references therein. However, these papers only studied the nonsingular boundary value problems.In 2008, Webb et al. [10] studied the existence of multiple positive solutions of nonlinear nonlocal boundary value problems (BVPs) for equations of the form
(2)u′′′′(t)=g(t)f^(t,u(t)),foralmosteveryt∈(0,1),u(0)=∫01‍u(s)dA(s),u′(0)=u(1)=u′(1)=0,
where g,f^ are continuous and nonnegative functions and A is a function of bounded variation. They treat many boundary conditions appearing in the literature in a unified way. The main tool they used is the fixed point index theory in cones. In 2009, Ma and An [11] studied the global structure for second-order nonlocal boundary value problem involving Stieltjes integral conditions by applying bifurcation techniques.Motivated by above papers, in this paper, we will use global bifurcation techniques to study the global structure of positive solutions of the singular problem(3)u′′′′(t)=λh(t)f(u(t)),0<t<1,u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0,
where h(t) may be singular at t=0 and t=1, and λ∈(0,∞) is a parameter.In order to prove our main result, let us make the assumptions as follows:(A1)
α:[0,1]→ℝ is nondecreasing and α(t) is not a constant on [0,1], ∫01‍k(t,s)dα(t)≥0 for s∈[0,1], and 0≤a<1 with a:=∫01‍γ(t)dα(t),γ(t)=(t-1)2(2t+1);(A2)
h∈C((0,1),[0,∞)) with h(t)≢0 on any subinterval of (0,1), and 0<∫01‍h(s)ds<∞;(A3)
f∈C([0,∞),[0,∞)) satisfies f(s)>0 for all s>0;(A4)
f0=lims→0+(f(s)/s)=∞;(A5)
f∞=lims→+∞(f(s)/s)=0.Remark 1.
For other results on the existence and multiplicity of positive solutions and nodal solutions for the boundary value problems of fourth-order ordinary differential equations based on bifurcation techniques, see Ma et al. [12–15] and Bai and Wang [16] and their references.
The rest of the paper is arranged as follows: In Section2, we state some properties of superior limit of certain infinity collection of connected sets. In Section 3, we will give some preliminary results. In Section 4, we state and prove our main results.


## 2. Superior Limit and Component

In order to treat the casef0=∞,f∞=0, we will need the following definition and lemmas.Definition 2 (see [17]).
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
(4)𝒟:=limsupn→∞Cn={x∈X∣∃{ni}⊂ℕ,xni∈Cni,suchthatxni⟶x}.Lemma 3 (see [17]).
Each connected subset of metric spaceX is contained in a component, and each connected component of X is closed.Lemma 4 (see [11]).
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
(5)BR={x∈X∣∥x∥≤R}.Then there exists an unbounded connected component C in 𝔻 and z*∈C.


## 3. Preliminaries

We consider the problem as follows:(6)u′′′′(t)=y(t),0<t<1,u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0.Lemma 5.
For anyy∈C[0,1], the problem (6) has a unique solution
(7)u(t)=∫01‍K(t,s)y(s)ds,
where
(8)K(t,s)=γ(t)1-a∫01‍k(t,s)dα(t)+k(t,s),(9)k(t,s)=16{t2(1-s)2[(s-t)+2(1-t)s],0≤t≤s≤1,s2(1-t)2[(t-s)+2(1-s)t],0≤s≤t≤1,a=∫01‍γ(t)dα(t),γ(t)=(t-1)2(2t+1).Proof.
By [10], the problem (6) can be equivalently written as
(10)u(t)=γ(t)∫01‍u(s)dα(s)+∫01‍k(t,s)y(s)ds.
Applying α to both sides of (10), we obtain
(11)∫01‍u(t)dα(t)=∫01‍[γ(t)∫01‍u(s)dα(s)]dα(t)+∫01‍[∫01‍k(t,s)y(s)ds]dα(t).
Thus, we have
(12)∫01‍u(t)dα(t)=∫01‍γ(t)dα(t)·∫01‍u(s)dα(s)+∫01‍[∫01‍k(t,s)dα(t)]y(s)ds.
Furthermore, it follows that
(13)∫01‍u(s)dα(s)=11-∫01‍γ(t)dα(t)∫01‍[∫01‍k(t,s)dα(t)]y(s)ds.
So, we obtain
(14)u(t)=∫01‍[γ(t)1-a∫01‍k(t,s)dα(t)+k(t,s)]y(s)ds.Lemma 6 (see [2–4]).
Green's functionk(t,s) defined by (9) satisfies the following:(i)
k(t,s)≥0 is continuous for all t,s∈[0,1];(ii)
c(t)k(τ(s),s)≤k(t,s)≤k(τ(s),s), forallt,s∈[0,1], and for any δ∈(0,1/2) and t∈[δ,1-δ], such that
(15)k(t,s)≥23δ2k(τ(s),s),∀s∈[0,1],
where
(16)τ(s)={13-2s,0≤s≤12,2s1+2s,12≤s≤1,k(τ(s),s)=maxt∈[0,1]k(t,s)={2s2(1-s)33(3-2s)2,0≤s≤12,2s3(1-s)23(1+2s)2,12≤s≤1,c(t)=23min{t2,(1-t)2},t∈[0,1],mint∈[δ,1-δ]c(t)=23δ2.Lemma 7.
Green's functionK(t,s) defined by (8) satisfies the following:(i)
K(t,s)≥0 is continuous for all t,s∈[0,1];(ii)
K(t,s)≤K(s), forallt,s∈[0,1], and for any δ∈(0,1/2), there exists a constant γδ>0, for any t∈[δ,1-δ], such that
(17)K(t,s)≥γδK(s),∀s∈[0,1],K(s)=1-a+α(1)-α(0)1-a·k(τ(s),s),γδ=23δ2·1-a+δ2(3-2δ)(α(1)-α(0))1-a+α(1)-α(0),
where k(t,s) is defined by (9), maxt∈[0,1]γ(t)=1, mint∈[δ,1-δ]γ(t)=δ2(3-2δ).Proof.
(i) From Lemma6 (i), we get the proof of Lemma 7 (i) immediately.
(ii) By Lemma6 (ii), we get
(18)K(t,s)≤11-a∫01‍k(τ(s),s)dα(t)+k(τ(s),s)≤1-a+α(1)-α(0)1-a·k(τ(s),s)=K(s),1-a+α(1)-α(0)1-a((s)s)∀t,s∈[0,1].
By Lemma 6 (ii), for any δ∈(0,1/2) and t∈[δ,1-δ],s∈[0,1], we obtain
(19)K(t,s)≥δ2(3-2δ)1-a∫01‍23δ2k(τ(s),s)dα(t)+23δ2k(τ(s),s)≥23δ2·1-a+δ2(3-2δ)(α(1)-α(0))1-a·k(τ(s),s)=γδK(s),∀s∈[0,1].Lemma 8.
Fory∈C[0,1] and y≥0, the unique solution of the problem (6) satisfies the following:(i)
u(t)≥0,  forallt∈[0,1];(ii)
mint∈[δ,1-δ]u(t)≥γδ∥u∥∞,where γδ is defined by Lemma 7 (ii), ∥u∥∞=maxt∈[0,1]|u|.Proof.
(i) From Lemma7 (i), we get the proof of Lemma 8 (i) immediately.
(ii) From (7) and Lemma 7, we have
(20)mint∈[δ,1-δ]u(t)=mint∈[δ,1-δ]∫01‍K(t,s)y(s)ds≥∫01‍mint∈[δ,1-δ]K(t,s)y(s)ds≥γδ∫01‍K(s)y(s)ds≥γδ∫01‍maxt∈[0,1]K(t,s)y(s)ds≥γδmaxt∈[0,1]∫01‍K(t,s)y(s)ds=γδ∥u∥∞.
Therefore, the proof of Lemma 8 is complete.LetY=C[0,1] be the Banach space with the norm ∥u∥∞=maxt∈[0,1]|u|.LetE={u∈C2[0,1]∣u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0} with the norm
(21)∥u∥E=max{∥u∥∞,∥u′∥∞,∥u′′∥∞}.Let(22)P={mint∈[δ,1-δ]u∈C[0,1]∣u(t)≥0,t∈[0,1],mint∈[δ,1-δ]u(t)≥γδ∥u∥∞},
and for r>0, let Ωr={u∈P∣∥u∥E<r}.In order to use bifurcation technique to study the problem (3), we consider the linear eigenvalue problem
(23)u′′′′(t)=λh(t)u(t),0<t<1,u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0.Let(24)Lλu(t)=λ∫01‍K(t,s)h(s)u(s)ds,t∈[0,1],(25)Tλu(t)=λ∫01‍K(t,s)h(s)f(u(s))ds,t∈[0,1].By [18], it is easy to show the following lemma.Lemma 9.
Assume that (A1)–(A3) hold the following.
L
λ
:
P
→
P is a completely continuous linear operator and Lλ(P)⊂P, and the fixed points of the operator Lλ in P are the positive solutions of the BVP (23).
T
λ
:
P
→
P is a completely continuous operator and Tλ(P)⊂P, and the fixed points of the operator Tλ in P are the positive solutions of the BVP (3).
By virtue of Krein-Rutman theorem (Theorem 2.5 in [19]), one has (see [18] or [20]) the following lemma.Lemma 10.
Suppose thatLλ:C[0,1]→C[0,1] is a completely continuous linear operator and Lλ(P)⊂P. If there exist ψ∈C[0,1]∖(-P) and a constant c>0 such that cLλψ≥ψ, then the spectral radius r(Lλ)≠0 and Lλ has a positive eigenfunction ϕ1 corresponding to its first eigenvalue λ1=1/r(Lλ), that is, ϕ1=λ1Lλϕ1.Lemma 11.
Suppose (A1) and (A2) are satisfied, then for the operatorLλ defined by (24), the spectral radius r(Lλ)≠0 and Lλ has a positive eigenfunction ϕ1∈intP corresponding to its first eigenvalue λ1=1/r(Lλ).Proof.
It is easy to see that there ist1∈(0,1) such that K(t1,t1)h(t1)>0. Thus there exists [α,β]⊂(0,1) such that t1∈(α,β) and K(t,s)h(s)>0,forallt,s∈[α,β]. Take ψ∈C[0,1] such that ψ(t)≥0,forallt∈[0,1],ψ(t1)>0, and ψ(t)=0,forallt∉[α,β]. Then for t∈[α,β],
(26)(Lλψ)(t)=λ∫01‍K(t,s)h(s)ψ(s)ds≥λ∫αβ‍K(t,s)h(s)ψ(s)ds>0.
So there exists a constant c>0 such that cLλψ≥ψ,forallt∈[0,1]. From Lemma 10, we know that the spectral radius r(Lλ)≠0 and Lλ has a positive eigenfunction corresponding to its first eigenvalue λ1=1/r(Lλ).Lemma 12.
Let (A1)–(A3) hold. The solutionu(t) of the problem (3) satisfies
(27)∥u∥∞≤∥u′∥∞≤∥u′′∥∞≤∥u′′′∥∞.Proof.
Fromu′(0)=u′(1)=0, there exists ξ∈(0,1), such that u′′(ξ)=0. Using a similar proof of (10) in [21, page 212], it is easy to show that
(28)|u(t)|=|∫t1‍u′(s)ds-u(1)|=|∫t1‍u′(s)ds|≤∫t1‍|u′(s)|ds≤∫01‍|u′(s)|ds,|u′(t)|=|u′(0)+∫0t‍u′′(s)ds|=|∫0t‍u′′(s)ds|≤∫01‍|u′′(s)|ds,|u′′(t)|=|u′′(ξ)+∫ξt‍u′′′(s)ds|=|∫ξt‍u′′′(s)ds|≤∫01‍|u′′′(s)|ds.
Furthermore, we obtain
(29)∥u∥∞≤∥u′∥∞≤∥u′′∥∞≤∥u′′′∥∞.Lemma 13.
Let (A1)–(A3) hold. Assume that{(μk,uk)}⊂(0,∞)×P is a sequence of positive solutions of (3). Assume that ∥μk∥≤c0 for some constant c0>0, and
(30)limk→∞∥uk∥E=∞.
Then
(31)limk→∞∥uk∥∞=∞.Proof.
Assume on the contrary that(32)∥uk∥∞≤M0
for some constant M0>0.
Since(μk,uk) is a solution of the problem (3), we have
(33)uk(t)=μk∫01‍K(t,s)h(s)f(uk(s))ds,t∈[0,1].
Thus,
(34)uk′′′(t)=μk∫01‍∂3∂t3K(t,s)·h(s)f(uk(s))ds,|∂3∂t3K(t,s)|=|121-a∫01‍k(t,s)dα(t)+∂3∂t3k(t,s)||∂3∂t3K(t,s)|≤|12(α(1)-α(0))1-ak(τ(s),s)+∂3∂t3k(t,s)||∂3∂t3K(t,s)|≤α(1)-α(0)16(1-a)+5:=M1,
where 0≤|k(τ(s),s)|≤1/192, sup0≤t,s≤1,t≠s|(∂3/∂t3)k(t,s)|≤5 (see [3]).
Furthermore, it follows that(35)∥uk′′′∥∞≤c0M1B0∫01‍h(s)ds,
where B0=maxs∈[0,M0]{f(s)}, together with (A2), which implies that ∥uk′′′∥∞ is bounded whenever ∥uk∥∞ is bounded.
Together with Lemma12, we obtain
(36)∥uk∥E≤M2
for some constant M2>0. This is a contradiction.


## 4. Main Results

LetΣ be the closure of the set of positive solutions of (3) in [0,∞)×E. The main results of this paper are the following.Theorem 14.
Let (A1)–(A5) hold, then (3) has at least one solution for any λ∈(0,∞).
LetL:D(L)⊂E→E be an operator defined by
(37)Lu=u′′′′,u∈D(L),
with
(38)D(L)={u∈C4[0,1]∣u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0∫01‍}.
Then L is a closed operator and L-1:Y→E is completely continuous.
For eachn∈ℕ, define f[n](s):[0,∞)→[0,∞) by
(39)f[n](s)={f(s),s∈(1n,∞),nf(1n)s,s∈[0,1n].
Then f[n]∈C([0,∞),[0,∞)) with
(40)f[n](s)>0,∀s∈(0,∞),(f[n])0=nf(1n).
By (A4), it follows that
(41)limn→∞(f[n])0=∞.
To apply the global bifurcation theorem, one extends f to an odd function g:ℝ→ℝ by
(42)g(s)={f(s),s≥0,-f(-s),s<0.
Similarly one may extend f[n] to an odd function g[n]:ℝ→ℝ for each n∈ℕ.
Now let one consider the auxiliary family of the equations(43)u′′′′(t)=λh(t)g[n](u(t)),0<t<1,u(0)=∫01‍u(s)dα(s),u′(0)=u(1)=u′(1)=0.
Let ζ[n]∈C(ℝ) be such that
(44)g[n](u)=(g[n])0u+ζ[n](u)=nf(1n)u+ζ[n](u).
Then
(45)lim|s|→0ζ[n](s)s=0.
Let one consider
(46)Lu=λh(t)(g[n])0u+λh(t)ζ[n](u)
as a bifurcation problem from the trivial solution u≡0.
From Lemma5, (46) can be converted to the equivalent equation
(47)u(t)=∫01‍K(t,s)[λh(s)(g[n])0u(s)+λh(s)ζ[n](u(s))]dsu(t):=λL-1[h(·)(g[n])0u(·)](t)+λL-1[h(·)ζ[n](u(·))](t).
Further one has that
(48)∥L-1[h(·)ζ[n](u(·))]∥E=o(∥u∥E),as∥u∥E⟶0.
Indeed, (8) implies that, for all (t,s)∈[0,1]×[0,1],
(49)|∂∂tK(t,s)|=|6t(t-1)1-a∫01‍k(t,s)dα(t)+∂∂tk(t,s)|≤α(1)-α(0)128(1-a)+3,|∂2∂t2K(t,s)|=|6(2t-1)1-a∫01‍k(t,s)dα(t)+∂2∂t2k(t,s)|≤α(1)-α(0)32(1-a)+8,
where 0≤|k(τ(s),s)|≤1/192, max0≤t,s≤1|(∂/∂t)k(t,s)|≤3, max0≤t,s≤1|(∂2/∂t2)k(t,s)|≤8 (see [3]).
So, the compactness ofL-1 together with (45) and (A2) imply that
(50)∥(L-1[h(·)ζ[n](u(·))])′∥∞=o(∥u∥∞),∥(L-1[h(·)ζ[n](u(·))])′′∥∞=o(∥u∥∞),
and consequently
(51)∥(L-1[h(·)ζ[n](u(·))])′∥∞=o(∥u∥E),∥(L-1[h(·)ζ[n](u(·))])′′∥∞=o(∥u∥E).
LetS0+={u∈E∣u(t)>0,forallt∈[0,1]}, S0-={u∈E∣u(t)<0,forallt∈[0,1]}, then S0+∩S0-=∅. Let S0=S0+∪S0-,Φ0±=R×S0±.
By Lemma11 and the fact (g[n])0>0, the global bifurcation result (see Rabinowitz [22]) for (46) can be stated as follows: there exists a continuum C+[n](⊂[0,∞)×P) of positive solutions of (46) joining (λ1/(g[n])0,0) to infinity in [0,∞)×P. Moreover, (λ1/(g[n])0,θ) is the only positive bifurcation point of (46) lying on trivial solutions line u≡θ. Moreover, C+[n]∖{(λ1/(g[n])0,0)}⊂Φ0+.
Since(52)limn→∞λ1(g[n])0=limn→∞λ1nf(1/n)=0,
condition (i) in Lemma 4 is satisfied with z*=(0,0). Obviously
(53)rn=sup{λ+∥u∥∣(λ,u)∈C+[n]}=∞,
and accordingly (ii) in Lemma 4 holds. (iii) in Lemma 4 can be deduced directly from the Arzela-Ascoli theorem and the definition of g[n]. Therefore, the superior limit of {C+[n]}, 𝒟, contained an unbounded connected component C with (0,0)∈C. Since C+[n]⊂Φ+, one concludes C⊂Φ+. Moreover, C⊂Σ by (3).Proof of Theorem14.
We firstly prove(54)ProjRC=(0,∞).
Assume on the contrary that(55)sup{λ∣(λ,u)∈C}<∞.
Then there exists a sequence (μk,uk)∈C such that
(56)limk→∞∥uk∥E=∞,μk≤c0,
for some positive constant c0 with doing not depend on k. From Lemma 13, we have
(57)limk→∞∥uk∥∞=∞.
This together with the fact(58)mint∈[δ,1-δ]uk(t)≥γδ∥uk∥∞,∀δ∈(0,12)
imply that for arbitrary δ∈(0,1/2)(59)limk→∞uk(t)=∞,uniformlyfort∈[δ,1-δ].
Since (μk,uk)∈C, we have that
(60)uk′′′′=μkh(t)f(uk),0<t<1,uk(0)=∫01‍uk(s)dα(s),uk′(0)=uk(1)=uk′(1)=0.
Setvk(t)=uk(t)/∥uk∥∞. Then
(61)∥vk(t)∥∞=1,(62)vk′′′′(t)=μkh(t)f(uk(t))uk(t)vk(t),0<t<1,vk(0)=∫01‍y(s)dα(s),vk′(0)=vk(1)=vk′(1)=0.
Now, choosing a subsequence and relabeling if necessary, it follows that there exists(μ*,v*)∈[0,c0]×E with
(63)∥v*∥∞=1
such that
(64)limk→∞(μk,vk)=(μ*,v*),inℝ×Y.
By (A3), let(65)f¯(u)=max0≤s≤uf(s).
Then f¯ is nondecreasing and
(66)limu→+∞f¯(u)u=0.
Further it follows from (66) that
(67)f(u)∥u∥∞≤f¯(u)∥u∥∞≤f¯(∥u∥∞)∥u∥∞⟶0,∥u∥∞⟶+∞.
Thus,(68)limk→∞f(uk)uk=0.
Notice that (62) is equivalent to
(69)vk(t)=μk∫01‍K(t,s)h(s)f(uk(s))uk(s)vk(s)ds,t∈[0,1].
Furthermore, by (59), (68), and (69), together with the Lebesgue dominated convergence theorem, it follows that
(70)v*(t)=μk∫01‍K(t,s)h(s)·0·v*(s)ds,t∈[0,1].
It follows that
(71)v*(t)≡0.
This contradicts (63). Therefore
(72)sup{λ∣(λ,u)∈C+[n]}=∞.
Noticing thatλ=0 is the only solution of the problem (3), thus
(73)ProjRC=(0,∞).
Furthermore, it follows the proof of Theorem14.



---

*Source: 614376-2014-01-08.xml*