# Positive Solutions to Periodic Boundary Value Problems for Four-Order Differential Equations

Authors: Huantao Zhu; Zhiguo Luo
Journal: ISRN Mathematical Analysis
 (2012)

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## Abstract

We apply fixed point theorem in a cone to
obtain sufficient conditions for the existence of single and multiple positive solutions of periodic boundary value problems for a class of four-order
differential equations.


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## Body

## 1. Introduction

In this paper, we investigate the existence of positive solutions of the following periodic boundary value problem:(1.1)y(4)(t)-Ay′′(t)+By(t)=f(t,y(t),y′′(t)),t∈I,y(i)(0)=y(i)(2π),i=0,1,2,3,
where I=[0,2π], A and B are positive constants with A2>4B, f∈C(I×(0,+∞)×R,[0,+∞)).In recent years, the nonlinear periodic boundary value problems have been widely studied by many authors, for example, see [1–7] and the references therein. Many theorems and methods of nonlinear functional analysis, for instance, upper and lower solutions method, fixed point theorems, variational method, and critical point theory, and so on, have been applied to their problems. When positive solutions are discussed, it seems that fixed point theorem in cones is quite effective in dealing with the problems with singularity. In [8], Zhang and Wang proved periodic boundary value problems with singularity(1.2)y′′(t)+k2y(t)=g(t,y(t)),t∈I,y(i)(0)=y(i)(2π),i=0,1,
have multiple positive solutions under some conditions, where g(t,y)  is singular at y=0, that is,(1.3)lim⁡y→0+g(t,y)=+∞.
Relying on a nonlinear alternative of Leray-Schauder type and fixed point theorem, Chu and Zhou [9] discussed the existence of positive solutions for the third-order periodic boundary value problem(1.4)y′′′(t)+k3y(t)=g(t,y(t)),t∈I,y(i)(0)=y(i)(2π),i=0,1,2,
where k∈(0,1/3).  However, relatively few papers have been published on the same problem for four-order differential equations. Recently, by using a maximum principle for operator Lu=u(4)-βu′′+αu  in periodic boundary condition and fixed point index theory in cones, Li [10] considered the existence of positive solution for the fourth-order periodic boundary value problem(1.5)y(4)(t)-βy′′(t)+αy(t)=g(t,y(t)),t∈[0,1],y(i)(0)=y(i)(1),i=0,1,2,3,
where f:[0,1]×R+→R+  is continuous, α, β∈R and satisfy 0<α<(β/2+2π2)2, β>-2π2, α/π4+β/π2>1. However, since there appears y′′  in nonlinear term g, the method in [9] cannot be directly applied to (1.1). The main aim of this paper is to establish sufficient conditions for the existence of positive solutions to the problem (1.1).To prove our main results, we present an existence theorem.Theorem 1.1 (see [11]).
LetE be a Banach space and P  a cone in E. Suppose Λ1 and Λ2 are open subsets of E such that 0∈Λ1⊂Λ̅1⊂Λ2  and suppose that
(1.6)T:P∩(Λ̅2Λ1)⟶P
is a completely continuous operator. If one of the following conditions is satisfied:(i)
∥Tx∥≤∥x∥for x∈P∩∂Λ1,∥Tx∥≥∥x∥  for x∈P∩∂Λ2,(ii)
∥Tx∥≥∥x∥  for x∈P∩∂Λ1,∥Tx∥≤∥x∥  for x∈P∩∂Λ2. Then T has a fixed point in P∩(Λ̅2/Λ1).


## 2. Preliminaries

In this section, we present some preliminary results which will be needed in Section3.Letα>0, and for any function u∈C(I),  we defined the operator(2.1)(Jαu)(t)=∫02πGα(t,s)u(s)ds,
where(2.2)Gα(t,s)={eα(t-s)+eα(2π-t+s)2α(e2πα-1),0≤s≤t≤2π,eα(s-t)+eα(2π-s+t)2α(e2πα-1),0≤t≤s≤2π.
By a direct calculation, we easily obtain(2.3)∫02πGα(t,s)ds=α-2,mα:=min⁡s,t∈IGα(t,s)=2eαπ2α(e2πα-1),Mα:=max⁡s,t∈IGα(t,s)=1+e2απ2α(e2πα-1).
Set(2.4)ρ=A+A2-4B2,λ=A-A2-4B2,orρ=A-A2-4B2,λ=A+A2-4B2,
then ρ, λ∈R. Now, we consider the problem(2.5)-u′′(t)+λ2u(t)=f(t,Jρu,ρ2Jρu-u),t∈I,u(i)(0)=u(i)(2π),i=0,1.Lemma 2.1.
Ifu∈C2(I)  is a (positive) solution of problem (2.5), then y=Jρu∈C4(I)  is a (positive) solution of problem (1.1). Moreover, the problem (2.5) is equivalent to integral equation
(2.6)u(t)=∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds.Proof.
Ifu∈C2(I),  then Jρu∈C4(I)  and
(2.7)(Jρu)′′(t)=-u(t)+ρ2(Jρu)(t),t∈I,(Jρu)(i)(0)=(Jρu)(i)(2π),i=0,1.
Thus,
(2.8)y′′(t)=-u(t)+ρ2(Jρu)(t),y′′′(t)=-u′(t)+ρ2(Jρu)′(t),y(4)(t)=(-u(t)+ρ2(Jρu)(t))′′=-u′′(t)-ρ2u(t)+ρ4(Jρu)(t).
Then,
(2.9)y(4)(t)-Ay′′(t)+By(t)=-u′′(t)+(A-ρ2)u(t)+(ρ4-Aρ2+B)(Jρu)(t)=-u′′(t)+λ2u(t)=f(t,Jρu,ρ2Jρu-u)=f(t,y(t),y′′(t)).
On the other hand, y(0)=y(2π),y′(0)=y′(2π),
(2.10)y′′(0)=-u(0)+ρ2(Jρu)(0)=-u(2π)+ρ2(Jρu)(2π)=y′′(2π),y′′′(0)=-u′(0)+ρ2(Jρu)′(0)=-u′(2π)+ρ2(Jρu)′(2π)=y′′′(2π).
Hence, if u∈C2(I) is a solution of problem (2.5), then y=Jρu∈C4(I) is a solution of problem (1.1). And, if u∈C2(I) is a positive function, noting that Gρ(t,s)>0 for any s, t∈I, we have
(2.11)y=∫02πGρ(t,s)u(s)ds>0.
Noting that, for any function h∈C(I), linear problem
(2.12)-u′′(t)+λ2u(t)=h(t),t∈I,u(i)(0)=u(i)(2π),i=0,1,
has a unique solution
(2.13)u(t)=∫02πGλ(t,s)h(s)ds,
one can easily obtain that (2.6) holds. The proof is complete.In the following application, we takeE=C2(I) with the supremum norm ∥·∥ and define(2.14)Pλ={u∈E:u(t)≥0forallt∈I,min⁡t∈Ju(t)≥δλ‖u‖},
where δλ=mλ/Mλ∈(0,1).One easily checks and verifies thatPλ  is a cone in E. For any r>0, let Prλ={u∈Pλ:∥u∥<r},  then ∂Prλ={u∈Pλ:∥u∥=r}.  For any u∈E, define mapping Tλ:E→E by(2.15)(Tλu)(t)=∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds,
then the fixed point of Tλ in E is a positive solution of (2.5).Lemma 2.2.
For anyς>τ>0, Tλ:P̅ςλ∖Pτλ→Pλ  is completely continuous.Proof.
For anyu∈P̅ςλ∖Pτλ, τ≤∥u∥≤ς   and δλτ≤u(t)≤ς for all t∈I. Thus, if u∈P̅ςλ∖Pτλ,
(2.16)(Jρu)(t)=∫02πGρ(t,s)u(s)ds≥δλτ∫02πGρ(t,s)ds≥δλτρ2.
It is easy to see that Tλ is continuous and completely continuous since f:I×(0,+∞)×R→[0,+∞)  is continuous. Next, we show that Tλ:P̅ςλ∖Pτλ→Pλ.  Since f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))≥0  for u∈P̅ςλ∖Pτλ, Tλu≥0. On the other hand,
(2.17)‖Tλu‖=max⁡t∈I∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≤Mλ∫02πf(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds,(Tλu)(t)≥mλ∫02πf(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≥δλ‖Tλu‖.
The proof is complete.


## 3. Positive Solutions of (1.1)

In this section, we make the following hypotheses.(H1λ)
There exist nonnegative functionsh(u), g(u)∈C((0,+∞)(0,+∞))  and p(t,v), q(t,v)∈C(J×R) such that
(3.1)f(t,u,v)≤p(t,v)h(u)+q(t,v)g(u)
for all (t,u,v)∈I×(0,∞)×R  and
(3.2)sup⁡u>0{ug(ρ-2u)max⁡(δλ-1)u≤v≤(1-δλ)u{∫02πp(t,v)dth(ρ-2u)+∫02πq(t,v)dtg(ρ-2u)}g(ρ-2δλu)}>Mλ,
where g is nonincreasing and h/g is nondecreasing on (0,∞).(H2λ)
One has(3.3)liminf⁡u→0+min⁡{∫02πf(t,w,v)dt,ρ-2δλu≤w≤ρ-2u,(δλ-1)u≤v≤(1-δλ)u}u>mλ-1.(H3λ)
One has(3.4)liminf⁡u→+∞min⁡{∫02πf(t,w,v)dt,ρ-2δλu≤w≤ρ-2u,(δλ-1)u≤v≤(1-δλ)u}u>mλ-1.Under the above hypotheses, we can obtain the following result.Theorem 3.1.
Assume that(H1λ) and (H2λ)  are satisfied, then there exist two positive constants α, β   such that (1.1) has at least positive solution y with
(3.5)ρ-2α<‖y‖<ρ-2β.
Assume(H1λ) and (H3λ) are satisfied, then there exist two positive constants β, γ such that (1.1) has at least positive solution y with
(3.6)ρ-2β<‖y‖<ρ-2γ.
Assume(H1λ), (H2λ), and (H3λ) are satisfied, then there exist positive constants α, β, γ   such that (1.1) has at least two positive solutions y1, y2 with
(3.7)ρ-2α<‖y1‖<ρ-2β<‖y2‖<ρ-2γ.Proof.
First, we assume that(H1λ) and (H2λ) are satisfied. From the condition (H1λ),  one can obtain that there exist a β>0 such that
(3.8)βg(ρ-2β)max⁡(δλ-1)β≤v≤(1-δλ)β{∫02πp(t,v)dth(ρ-2β)+∫02πq(t,v)dtg(ρ-2β)}g(ρ-2δλβ)>Mλ.
For any u∈∂Pβλ, δλβ≤u(t)≤β  for all t∈I and
(3.9)βδλρ2≤(Jρu)(t)=∫02πGρ(t,s)u(s)ds≤βρ2,δλβ-β≤ρ2(Jρu)(t)-u(t)≤β-δλβ.
Thus, for u∈∂Pβλ, from (H1λ),  we have
(3.10)(Tλu)(t)=∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≤Mλ∫02π[p(s,ρ2(Jρu)(s)-u(s))h((Jρu)(s))+q(s,ρ2(Jρu)(s)-u(s))g((Jρu)(s))]ds≤Mλ∫02π[p(s,ρ2(Jρu)(s)-u(s))h((Jρu)(s))g((Jρu)(s))+q(s,ρ2(Jρu)(s)-u(s))h((Jρu)(s))g((Jρu)(s))]g((Jρu)(s))ds≤Mλ∫02π[p(s,ρ2(Jρu)(s)-u(s))h(ρ-2β)g(ρ-2β)+q(s,ρ2(Jρu)(s)-u(s))]g(ρ-2δλβ)ds≤Mλg(ρ-2δλβ)g(ρ-2β)max⁡(δλ-1)β≤v≤(1-δλ)β{∫02πp(t,v)dth(ρ-2β)+∫02πq(t,v)dtg(ρ-2β)}<β=‖u‖,
which implies that
(3.11)‖(Tλu)‖<‖u‖,∀u∈∂Pβλ.
From(H2λ)  is satisfied, there exists a positive constant α<β  such that
(3.12)min⁡{∫02πf(t,w,v)dt,ρ-2δλα≤w≤ρ-2α,(δλ-1)α≤v≤(1-δλ)α}>mλ-1α.
For any u∈∂Pαλ, δλα≤u(t)≤α for all t∈I and
(3.13)δλαρ2≤(Jρu)(t)=∫02πGρ(t,s)u(s)ds≤αρ2,δλα-α≤ρ2(Jρu)(t)-u(t)≤α-δλα,(Tu)(t)=∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≥mλ∫02πf(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≥mλmin⁡{∫02πf(t,w,v)dt,ρ-2δλα≤w≤ρ-2α,(δλ-1)α≤v≤(1-δλ)α}>α=‖u‖,
which implies that
(3.14)‖(Tλu)‖>‖u‖,∀u∈∂Pαλ.
From (3.11) and (3.14) and Theorem 1.1, one can obtain that Tλ has a fixed point u in  P̅βλ∖Pαλ with α<∥u∥<β.  Hence, y=Jρu  is a positive solution of (1.1) with ρ-2α<∥y∥<ρ-2β.
Next, we assume that(H1λ) and (H3λ) are satisfied. In this case, we have (3.11).
Suppose that(H3λ) is satisfied, there exists a positive constant γ>β such that
(3.15)min⁡{∫02πf(t,w,v)dt,ρ-2δλγ≤w≤ρ-2γ,(δλ-1)γ≤v≤(1-δλ)γ}>mλ-1γ.
For any u∈∂Pγλ, δλγ≤u(t)≤γ  for all t∈I and
(3.16)δλγρ2≤(Jρu)(t)=∫02πGρ(t,s)u(s)ds≤γρ2,δλγ-γ≤ρ2(Jρu)(t)-u(t)≤γ-δλγ,(Tu)(t)=∫02πGλ(t,s)f(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≥mλ∫02πf(s,(Jρu)(s),ρ2(Jρu)(s)-u(s))ds≥mλmin⁡{∫02πf(t,w,v)dt,ρ-2δλγ≤w≤ρ-2γ,(δλ-1)γ≤v≤(1-δλ)γ}>γ=‖u‖,
which implies that
(3.17)‖(Tλu)‖>‖u‖,∀u∈∂Pγλ.
From (3.11) and (3.17) and Theorem 1.1, one can obtain that Tλ has a fixed point u in  P̅γλ∖Pβλ with β<∥u∥<γ.  Thus, y=Jρu   is a positive solution of (1.1) with ρ-2β<∥y∥<ρ-2γ.
Assume that(H1λ), (H2λ), and (H3λ) are satisfied. Repeating the above argument, one can obtain that Tλ has a fixed point u1 in  P̅βλ∖Pαλ  and a fixed point u2 in  P̅γλ∖Pβλ with
(3.18)α<‖u1‖<β<‖u2‖<γ.
Hence, y1=Jρu1, y2=Jρu2  are two positive solutions of (1.1) with
(3.19)ρ-2α<‖y1‖<ρ-2β<‖y2‖<ρ-2γ.
The proof is complete.


## 4. A Similar Problem

In this section, we use the idea in Sections2 and 3 to consider the following problem:(4.1)-y(4)(t)+k4y(t)=f(t,y(t),y′′(t)),t∈I,0<k<12,y(i)(0)=y(i)(2π),i=0,1,2,3,
where I=[0,2π], f∈C(I×(0,+∞)×R,[0,+∞)).Letu∈C2(I)  and y=J̃u=∫02πK(t,s)u(s)ds,  where(4.2)K(t,s)={cos⁡k(π-t+s)2ksin⁡kπ,0≤s≤t≤2π,cos⁡k(π+t-s)2ksin⁡kπ,0≤t≤s≤2π.
Then,(4.3)∫02πK(t,s)ds=k-2,(J̃u)′′(t)=u(t)-k2(J̃u)(t),t∈I,(J̃u)(i)(0)=(J̃u)(i)(2π),i=0,1,
and one easily check that (4.1) is equivalent to the problem(4.4)-u′′(t)+k2u(t)=f(t,J̃u,u-k2J̃u),t∈I,u(i)(0)=u(i)(2π),i=0,1.
If u∈C2(I) is a (positive) solution of problem (4.4), then y=J̃u∈C4(I)  is a (positive) solution of problem (4.1). Moreover, the problem (4.4) is equivalent to integral equation(4.5)u(t)=∫02πGk(t,s)f(s,J̃u(s),u(s)-k2J̃u(s))ds.
For any u∈E, define mapping T:E→E by(4.6)(Tu)(t)=∫02πGk(t,s)f(s,J̃u(s),u(s)-k2J̃u(s))ds.
For any ς>τ>0, one can obtain that T:P̅ςk∖Pτk→Pk  is completely continuous.Similar to the proof of Theorem3.1, we can obtain the following result.Theorem 4.1.
Assume that(H1k) and (H2k) are satisfied, then there exist two positive constants α, β  such that (4.1) has at least positive solution y with
(4.7)k-2α<‖y‖<k-2β.
Assume(H1k) and (H3k) are satisfied, then there exist two positive constants β, γ such that (4.1) has at least positive solution y with
(4.8)k-2β<‖y‖<k-2γ.
Assume(H1k), (H2k), and (H3k) are satisfied, then there exist positive constants α, β, γ such that (4.1) has at least two positive solutions y1, y2 with
(4.9)k-2α<‖y1‖<k-2β<‖y2‖<k-2γ,
where (Hjk)(j=1,2,3)  is condition obtained by replacing λ and ρ by k in the condition (Hjλ) defined in Section 3.Example 4.2.
Consider the differential equation(4.10)y(4)(t)-μy′′(t)+y(t)=120πμ(ey′′(t)+1+tμy(t)),t∈I,y(i)(0)=y(i)(2π),i=0,1,2,3,
where μ>4 is a constant.
Let(4.11)λ=μ-μ2-42,ρ=μ+μ2-42,f(t,u,v)=ev+(1+t)(μu)-120πμ,h(u)≡1,g(u)=1u,p(t,v)=ev20πμ,q(t,v)=1+t20πμ2.
Then, for all (t,u,v)∈I×(0,∞)×R,
(4.12)0<f(t,u,v)≤p(t,v)h(u)+q(t,v)g(u).
Noting that
(4.13)sup⁡u>0{umax⁡(δλ-1)u≤v≤(1-δλ)u{∫02πp(t,v)dth(ρ-2u)/g(ρ-2u)+∫02πq(t,v)dt}g(ρ-2δλu)}>10μδλe+3πμ-1ρ2,lim⁡μ→+∞Mλμδλ=14π,lim⁡μ→+∞ρ2μ=1,
we obtain that (H1λ) holds when μ is sufficiently large. On the other hand, it is easy to check that (H2λ) is satisfied since f(t,u,v)→+∞  as u→0+ for any t∈I and v∈R. Hence, (4.10) has at least a positive solution when μ is sufficiently large.



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