# Existence of Positive Solutions for Nonlinear Third-Order Boundary Value Problem

Authors: Tiaoxia Dun; Pengyu Chen
Journal: Journal of Mathematics
 (2014)

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## Abstract

We are concerned with the existence of positive solutions for the nonlinear third-order three-point boundary value problemu′′′(t)+λg(t)f(u(t))=0, 0<t<1, u(0)=αu(η), u′(0)=u′′(1)=0, where 0<η<1, 0<α<1, λ is a positive parameter, g:(0,1)→[0,∞),andf:[0,∞)→[0,∞) is continuous. We construct Green’s function for the associated linear boundary value problem and obtain some useful properties of Green’s function. Finally, by using fixed-point index theorem in cones, we establish the existence results of positive solutions for the boundary value problem an example illustrates the application of the results obtained.


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## Body

## 1. Introduction

The existence of positive solutions for third-order three-point boundary value problems has attracted considerable attention by a number of authors. For example, Anderson [1], by using the well-known Guo-Krasnoselskii fixed point theorems [2, 3], obtained some existence results for positive solutions for the following boundary value problem (BVP for short):
(1)x‴(t)=f(t,x(t)),t1<t<t3,x(t1)=x′(t2),γx(t3)+δx′′(t3)=0.
In [4], Sun considered the existence of multiple positive solutions (at least three) to the BVP
(2)u‴(t)=a(t)f(t,u(t),u′(t),u′′(t)),0<t<1,u(0)=δu(η),u′(η)=0,u′′(1)=0,
where δ∈(0,1), η∈[1/2,1) are constants and a:(0,1)→[0,∞) and f:[0,1]×[0,∞)×R×R→[0,∞) are continuous. The methods used in the paper were based on an application of a fixed point theorem due to Graef and Yang [5]. In [6], the authors considered the existence of positive solution to the third-order three-point BVP
(3)u‴(t)+a(t)f(u(t))=0,t∈(0,1),u(0)=u′(0)=0,u′(1)=αu′(η),
where 0<η<1 and 1<α<1/η. The proof relies on Guo-Krasnoselskii fixed point theorem.Motivated by the abovementioned works, in this paper, the fixed-point index theorem in cones was used to discuss the third-order three-point BVP(4)u‴(t)+λg(t)f(u(t))=0,0<t<1,u(0)=αu(η),u′(0)=u′′(1)=0,
where 0<η<1, 0<α<1, and λ is a positive parameter. Throughout this paper, we assume that the following conditions are satisfied:(C1)
f:[0,∞)→[0,∞) is continuous;(C2)
g∈C([0,1],[0,∞)) and is not identically zero on any subinterval of [0,1];(C3)
0<∫01‍s(2-s)g(s)ds<∞, ∫η1‍g(s)ds>0.


## 2. Preliminaries

In this section, we will state some important preliminary lemmas.LetE=C[0,1], C+[0,1]={u∈C[0,1]:u(t)≥0,t∈[0,1]}; then E is a Banach space with norm ∥u∥=max⁡0≤t≤1⁡|u(t)|.Lemma 1.
Assume thath:(0,1)→[0,∞) is continuous, 0<η, α<1, 0<∫01‍s(2-s)h(s)ds<∞; then the BVP
(5)u‴(t)+h(t)=0,0<t<1,u(0)=αu(η),u′(0)=u′′(1)=0
has a unique solution
(6)u(t)=∫01G(t,s)h(s)ds,
where
(7)G(t,s)=k(t,s)+α1-αk(η,s)
is corresponding Green’s function, and
(8)k(t,s)={12t2-12(t-s)2,0≤s≤t≤1,12t2,0≤t≤s≤1.Proof.
We can reduce equationu‴(t)+h(t)=0 to an equivalent integral equation
(9)u(t)=-12∫0t‍(t-s)2h(s)ds+a2t2+a1t+a0.
By u′(0)=0, it follows that a1=0; from u′′(1)=0, we get that a2=(1/2)∫01‍h(s)ds, thanks to u(0)=αu(η); we have a0=-(α/2(1-α))∫0η‍(η-s)2h(s)ds+(αη2/2(1-α))∫01‍h(s)ds.
Thus, substitutinga0,a2 into (9), we have
(10)u(t)=12∫0t‍[t2-(t-s)2]h(s)ds+12∫t1‍t2h(s)ds+α2(1-α)∫η1‍η2h(s)ds+α2(1-α)∫0η‍[η2-(η-s)2]h(s)ds=∫01‍k(t,s)h(s)ds+α1-α∫01‍k(η,s)h(s)ds,
where
(11)k(t,s)={12t2-12(t-s)2,0≤s≤t≤1,12t2,0≤t≤s≤1.Lemma 2.
Let0<η, α<1, for G(t,s) be given as (7); then we obtain the following results:(i)
0≤G(t,s)≤((1+αη-α)/2(1-α))s(2-s), 0≤t, s≤1,(ii)
G(t,s)≥(η2/2(1-α))s(2-s), η≤t≤1, 0≤s≤1.Proof.
Ifs≤t, for all 0≤t, s≤1, it follows from (8) that
(12)k(t,s)=12t2-12(t-s)2=12s(2t-s)≤12s(2-s),(13)k(t,s)=12t2-12(t-s)2=12(t2s2-t2s2+2t2s-2t2s+2ts-s2)=12t2(2s-s2)+12[2ts(1-t)-s2(1-t2)]=12t2(2s-s2)+12s(1-t)(2t-s-st)≥12t2(2s-s2)+12s(1-t)(s-st)=12t2(2s-s2)+12s2(1-t)2≥12t2(2s-s2),
and therefore
(14)k(t,s)≥12t2s(2-s).
If t≤s, it follows from (8) that
(15)k(t,s)=12t2≤12s2≤12s(2-s),(16)k(t,s)=12t2≥12t2s(2-s).
By (8), we have
(17)k(η,s)={12η2-12(η-s)2,0≤s≤η≤1,12η2,0≤η≤s≤1,
and thus
(18)2k(η,s)s(2-s)={2η-s2-s≤η,0≤s≤η≤1,η2s(2-s)≤η2s≤η,0≤η≤s≤1.
So we obtain that
(19)k(η,s)≤12ηs(2-s).
From (14) and (16), it follows that
(20)k(η,s)≥12η2s(2-s).
Hence, by (7), (9), and (16) we get
(21)0≤G(t,s)≤1+αη-α2(1-α)s(2-s),∀0≤t,s≤1,
and by (7), (14), (16), and (20) we have
(22)G(t,s)≥η22(1-α)s(2-s),η≤t≤1,0≤s≤1.Remark 3.
Letφ(s)=(2η-s)/(2-s), 0≤s≤η; then φ′(s)=(2η-2)/(2-s)2<0, 0≤s≤η, and it follows that
(23)2k(η,s)s(2-s)=φ(s)≤φ(0)=η,that is,k(η,s)≤12ηs(2-s).Lemma 4.
Assume that0<η, α<1, h:(0,1)→[0,∞) is continuous, 0<∫01‍s(2-s)h(s)ds<∞; then the unique solution u of the BVP (4) is nonnegative and satisfies
(24)min⁡η≤t≤1u(t)≥ρ∥u∥,
where ρ=η2/(1+αη-α). By 0<α<1<(1-η2)/(1-η), it is easy to know 0<ρ<1.Proof.
From (7), (14), (16), and (20), for t∈[0,1], we have
(25)u(t)≥12(t2+αη21-α)∫01‍s(2-s)h(s)ds≥0,
and thus, u(t) is nonnegative.
Lett∈[0,1]; by (12) and (19), we get
(26)u(t)=∫01‍k(t,s)h(s)ds+α1-α∫01‍k(η,s)h(s)ds≤12∫01‍s(2-s)h(s)ds+αη2(1-α)∫01‍s(2-s)h(s)ds=1+αη-α2(1-α)∫01‍s(2-s)h(s)ds,
and therefore
(27)∥u∥≤1+αη-α2(1-α)∫01‍s(2-s)h(s)ds.
On the other hand, for t∈[η,1], from Lemma 2 (ii), we have
(28)u(t)≥η22(1-α)∫01‍s(2-s)h(s)ds≥η21+αη-α∥u∥,
and thus
(29)min⁡η≤t≤1u(t)≥ρ∥u∥,
where ρ=η2/(1+αη-α).Choose a coneK in E as follows:
(30)K={u∈C+[0,1]:min⁡η≤t≤1u(t)≥ρ∥u∥}.
Define an operator A:C+[0,1]→C+[0,1] by
(31)Au(t)=λ∫01‍G(t,s)g(s)f(u(s))ds.
By the definition of operator A, a positive solution of BVP (4) is equivalent to a nonzero fixed point of A.Lemma 5.
Assume that (C1)–(C3) hold; then the operator A:K→K is completely continuous.Proof.
By the definition of operatorA and Lemma 5, for u∈K, it is easy to prove that A(K)⊂K. According to the Ascoli-Arzela theorem, we can easily obtain that A:K→K is a completely continuous mapping.Lemma 6 (see [7]).
LetΩ be a bounded open subset of E with θ∈Ω, and let A:K∩Ω¯→K be a completely continuous mapping. If μAu≠u for every u∈K∩∂Ω and 0<μ≤1, then i(A,K∩Ω,K)=1.Lemma 7 (see [7]).
LetΩ be a bounded open subset of E, and let A:K∩Ω¯→K be a completely continuous mapping. If there exists an v∈K∖{θ}, such that u-Au≠τv for every u∈K∩∂Ω and τ≥0, then i(A,K∩Ω,K)=0.


## 3. Main Results

In this section, we state and prove our main results.To be convenient, we introduce the following notations:(32)f0=limsup⁡x→0+f(x)x,f0=liminf⁡x→0+f(x)x,f∞=limsup⁡x→∞f(x)x,f∞=liminf⁡x→∞f(x)x,B=(max⁡0≤t≤1∫01‍G(t,s)g(s)ds)-1,C=(min⁡η≤t≤1∫η1‍G(t,s)g(s)ds)-1.
Our main results are as follows.Theorem 8.
Assume that (C1)–(C3) hold and f∞>0, f0<∞, C/f∞<B/f0; then for any λ∈(C/f∞,B/f0), BVP (4) has at least one positive solution.Proof.
For everyλ∈(C/f∞,B/f0), according to the condition of f0<B/λ and the definition of f0, there exist constants R, ε1>0, when 0<u≤R; we have f(u)/u≤(1/λ)(B-ε1); namely,
(33)f(u)≤1λ(B-ε1)u≤1λ(B-ε1)R,0<u≤R.
Let Ω1={u∈C+[0,1]:∣∥u∥<R}, for u∈K∩∂Ω1; from (31), we get
(34)∥Au∥=max⁡0≤t≤1λ∫01‍G(t,s)g(s)f(u(s))ds≤1λ(B-ε1)Rmax⁡0≤t≤1λ∫01‍G(t,s)g(s)ds=R-ε1Rmax⁡0≤t≤1∫01‍G(t,s)g(s)ds<R.
Therefore, for every u∈K∩∂Ω1, 0<μ≤1, when ∥u∥=R, we have μAu≠u. In fact, if there exist u0∈K∩∂Ω1 and 0<μ0≤1, such that μ0Au0=u0, then ∥Au0∥=(1/μ0)∥u0∥≥∥u0∥=R; this is a contradiction. Hence A satisfies the condition of Lemma 6 in K∩∂Ω1. By Lemma 6, we have
(35)i(A,K∩Ω1,K)=1.
On the other hand, by the condition of C/f∞<λ and the definition of f∞, there exist H>ρR, ε2>0, and for u≥H, we get f(u(s))≥(1/λ)(C+ε2)u. Let R1=H/ρ>R, Ω2={u∈C+[0,1]:∥u∥<R1}; then, min⁡{u(t):t∈[η,1]}≥ρ∥u∥=H, for all u∈K∩∂Ω2.
Choosee(t)≡1, e∈K∖{θ}. We prove that A satisfies the condition of Lemma 7 in K∩∂Ω2; namely, u-Au≠τe for every u∈K∩∂Ω2 and τ≥0. If it is not true, there exist u0∈K∩∂Ω2 and τ0≥0, such that u0-Au0=τ0e. Let σ=min⁡{u0(t):t∈[η,1]}, σ≥ρ∥u0∥=ρR1, for t∈[η,1]; we have
(36)u0(t)=λ∫01‍G(t,s)g(s)f(u0(s))ds+τ0≥1λ(C+ε2)σλ∫η1‍G(t,s)g(s)ds+τ0=σ+ε2σ∫η1‍G(t,s)g(s)ds+τ0>σ,
and this is a contradiction. Thus A satisfies the condition of Lemma 7 in K∩∂Ω2. By Lemma 7 we get
(37)i(A,K∩Ω1,K)=0.
Now, from (35) and (37) it follows that
(38)i(A,K∩(Ω2∖Ω1¯),K)=i(A,K∩Ω2,K)-i(A,K∩Ω1,K)=-1.
Therefore, A has a fixed point in K∩(Ω2∖Ω1¯), which is a positive solution of BVP (4).Remark 9.
Iff∞=∞, let C/f∞=0; if f0=0, let B/f0=∞.Theorem 10.
Assume that (C1)–(C3) hold and f0>0, f∞<∞, C/f0<B/f∞; then for any λ∈(C/f0,B/f∞), BVP (4) has at least one positive solution.Proof.
Let the operatorA be defined as (31).
For everyλ∈(C/f0,B/f∞), according to the condition of f∞<B/λ and the definition of f∞, there exist constants R2, ε3>0, and when u>R2, we have f(u)≤(1/λ)(B-ε3)u. Let a=max⁡{f(u):0≤u≤R2}; then
(39)f(u)≤a+1λ(B-ε3)u,u∈[0,∞).
Choose R3>aB/ε3f∞; let Ω3={u∈C+[0,1]:∣∥u∥<R3}. Then for every u∈K∩∂Ω3, we have
(40)∥Au∥=max⁡0≤t≤1λ∫01‍G(t,s)g(s)f(u0(s))ds≤(a+1λ(B-ε3)∥u∥)max⁡0≤t≤1λ∫01‍G(t,s)g(s)ds≤B∥u∥max⁡0≤t≤1∫01‍G(t,s)g(s)ds+(a-1λε3∥u∥)λ∫01‍G(t,s)g(s)ds=R3+(a-1λε3R3)λ(1+αη-α)2(1-α)∫01‍s(2-s)g(s)ds<R3.
Thus, for u∈K∩∂Ω3 and 0<μ≤1, when ∥u∥=R3, we have μAu≠u. In fact, if there exist u0∈K∩∂Ω3 and 0<μ0≤1, such that μ0Au0=u0, then ∥Au0∥=(1/μ0)∥u0∥≥∥u0∥=R3; this is a contradiction. Hence A satisfies the condition of Lemma 6 in K∩∂Ω3. By Lemma 6, we have
(41)i(A,K∩Ω3,K)=1.
On the other hand, by the condition of C/f0<λ and the definition of f0, there exist R4, ε4>0, such that C/λ+ε4≤f0, 0<R4<R3, and when 0<u<R4, we have f(u)≥(C/λ+ε4)u. Let Ω4={u∈C+[0,1]:∥u∥<R4}. Choose e(t)≡1. The process of the proof is similar to Theorem 8; we can get u-Au≠τe for every u∈K∩∂Ω4 and τ≥0. By Lemma 7 we get
(42)i(A,K∩Ω1,K)=0.
Combining (41), (42), and Ω¯4⊂Ω3, it follows that
(43)i(A,K∩(Ω3∖Ω4¯),K)=i(A,K∩Ω3,K)-i(A,K∩Ω4,K)=1.
Therefore, A has a fixed point in K∩(Ω3∖Ω4¯), which is a positive solution of BVP (4).Remark 11.
Iff0=∞, let C/f0=0; if f∞=0, let B/f0=0.Example 12.
We consider the following third-order three-point BVP:(44)u‴(t)+λt(u+16u3)32(1+u2)=0,t∈(0,1),u(0)=13u(12),u′(0)=u′′(1)=0,
where λ is a positive parameter; clearly
(45)f(u)=u+16u332(1+u2),g(t)=t,α=13,η=12.
It is easy to obtain that
(46)∫01‍s(2-s)g(s)ds=512,∫1/21‍g(s)ds=38,f∞=12,f0=132,
and thus, (C1)–(C3) hold.
By calculating, we get(47)B=28877,C=38427,ρ=310.
By Theorem 8, for any C/f∞=256/9<λ<B/f0=9216/77, the BVP (44) has at least one positive solution.


## 4. Conclusions

In this work, Green’s function for the associated linear boundary value problem was constructed, and some useful properties of the Green’s function were obtained. On these bases, the existence results of positive solutions for the boundary value problem were established by using fixed-point index theorem in cones. To demonstrate the applications of our results, an example is given.



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