# Green’s Function and Positive Solutions of a Third-Order Equation with Periodic Boundary Conditions

Authors: Nickolai Kosmatov
Journal: Discrete Dynamics in Nature and Society
 (2020)

---

## Abstract

We apply the fixed point index to obtain positive solutions of a nonresonant periodic boundary value problem for a third-order differential equationu‴+ρ3u=λfu.


---

## Body

## 1. Introduction

Third-order differential equations arise in many areas of physics and engineering [1] and describe, for example, deflection of a curved beam having a constant or varying cross section, a three-layered beam, and electromagnetic waves. Boundary value problems for third-order differential equations have been studied by many authors, for example, [2–9] just to name a few. In this paper, we consider a well-known [10, 11] boundary value problem:(1)u‴t+ρ3ut=λfut,tϵ0,2π,(2)ui0=ui2π,i=0,1,2.We improve the results of [10, 11] and obtain positive solutions using the fixed point index. The solutions to (1) and (2) will be sought in the Banach space ℬ=C0,2π endowed with the max-norm. In order to obtain positive solutions, we apply the fixed point theorem of Guo and Lakshmikantham [12] stated in Section 2.Green’s function ofu′t+ρut=0 with u0=u2π is(3)G1t,s=1e2πρ−1eρ2π+s−t,0≤s≤t≤2π,eρs−t,0≤t≤s≤2π.Green’s function ofu″t−ρu′t+ρ2ut=0 with ui0=ui2π, i=0,1, is(4)G2t,s=23ρeπρ+e−πρ−2cos3πρg1t−s,0≤s≤t≤2π,g2s−t,0≤t≤s≤2π,where(5)g1x=e1/2ρxe−πρ−cos3πρsin32ρx+sin3πρcos32ρx,(6)g2x=e−1/2ρxeπρ−cos3πρsin32ρx+sin3πρcos32ρx,where x∈0,2π. To ensure that G2t,s≥0, we need ρ∈0,1/3.We maximizeg1 and g2. Introduce, for convenience,(7)Aρ=e−πρ−cos3πρ,Bρ=sin3πρ,Cρ=eπρ−cos3πρ.We formally find thatg1′x1=0 if(8)x1ρ=23ρcot−13Bρ−Aρ3Aρ+Bρ,provided x1∈0,2π.Note that(9)limρ⟶0+AρBρ=−13,limρ⟶1/3−AρBρ=∞.Also(10)AB′ρ=π3−e−πρsin3πρ+3cos3πρsin23πρ.Denoting(11)αρ=3−e−πρsin3πρ+3cos3πρ,we have α′ρ=4πe−πρsin3πρ>0, ρ∈0,1/3, and α0=0 and we have A/B′ρ>0. Hence,(12)ϕρ=3Bρ−Aρ3Aρ+Bρ,is a decreasing function on 0,1/3, limρ⟶0+ϕρ=∞, and limρ⟶1/3−ϕρ=−1/3. We have(13)limρ⟶1/3−x1ρ=limρ⟶1/3−23ρcot−13Bρ−Aρ3Aρ+Bρ=4π3,limρ⟶0+x1ρ=limρ⟶0+23ρcot−1ϕρ=−23limρ⟶0+ϕ′ρ1+ϕ2ρ=23limρ⟶0+A′ρBρ−B′ρAρA2ρ+B2ρ=23limρ⟶0+3−e−πρsin3πρ+3cos3πρe−2πρ+1−2e−πρcos3πρ=2π3limρ⟶0+2sin3πρcos3πρ+3sin3πρ−e−πρ=2π3limρ⟶0+23cos3πρ3cos3πρ−3sin3πρ+e−πρ=π.It is a lengthy exercise to verify thatx1ρ is increasing on 0,1/3. We infer(14)π<x1ρ<4π3.It follows from (8) that(15)cos32ρx1=3B−A2A2+B2,sin32ρx1=3A+B2A2+B2.In fact,g1″x1<0; thus,(16)g1x≤g1x1=e1/2ρx1Asin32ρx1+Bcos32ρx1=32e1/2ρx1A2+B2.By a similar argument, one can show thatg2x≤g2x2, where(17)x2ρ=23ρtan−13Cρ−Bρ3Bρ+Cρ∈2π3,π,so that(18)g2x≤g2x2=e−1/2ρx2Csin32ρx2+Bcos32ρx2=32e−1/2ρx2C2+B2.We are in position to state and prove our first lemma.Lemma 1.
Green’s functionsG1t,s and G2t,s satisfy(19)M1=maxt,s∈0,2πG1t,s=e2πρe2πρ−1,(20)m1=maxt,s∈0,2πG1t,s=1e2πρ−1,(21)M2=maxt,s∈0,2πG2t,s=1eπρ+e−πρ−2cos3πρmaxe1/2ρx1−π,e−1/2ρπ−x2,(22)m2=maxt,s∈0,2πG2t,s=2sin3πρ3ρeπρ+e−πρ−2cos3πρ.Proof.
Identities (19) and (20) are easy to check.
The following identities are useful:(23)A2+B2=e−2πρ−2e−πρcos3πρ+1,C2+B2=e2πρ−2eπρcos3πρ+1,A2+B2=e−2πρC2+B2,eπρ+e−πρ−2cos3πρ=e−πρC2+B2.
Then, using (16) and (18), we have(24)maxt,s∈0,2πG2t,s=23ρeπρ+e−πρ−2cos3πρmaxg1x1,g2x2=2eρπρ3C2+B2maxg1x1,g2x2=eπρρC2+B2maxe1/2ρx1A2+B2,e−1/2ρx2C2+B2=eπρρC2+B2maxe1/2ρx1−ρπ,e−1/2ρx2=e1/2πρρeπρ+e−πρ−2cos3πρmaxe1/2ρx1−ρπ,e−1/2ρx2=1ρeπρ+e−πρ−2cos3πρmaxe1/2ρx1−π,e1/2ρπ−x2,where x1 and x2 are given by (8) and (17), respectively. In addition,(25)minx∈0,2πg1x=ming10,g12π=sin3πρ=ming20,g22π=minx∈0,2πg2x.
Hence,(26)mint,s∈0,2πG2t,s=2sin3πρ3ρeπρ+e−πρ−2cos3πρ.
Observe that (21) and (22) are exact and improve the corresponding estimates stated in Lemma 3 of [10], namely,(27)2sin3πρ3ρeπρ+12≤G2t,s≤23ρsin3πρ.
Indeed,eπρ+12>eπρ+e−πρ−2cos3πρ, so (22) improves the first inequality. From (14) and (17), it is clear that(28)maxe1/2ρx1−π,e1/2ρπ−x2<e1/2πρ,maxe1/2ρx1−π,e1/2ρπ−x2eπρ+e−πρ−2cos3πρ<11+e−2πρ−2e−πρcos3πρ=1e−πρ−cos3πρ2+sin23πρ<23sin3πρ,that is, (21) is preferred to the constant in the second inequality.
Now, Green’s function ofu‴t+ρ3ut=0, ui0=ui2π, i=0,1,2, is determined from (3)–(6) by(29)Gt,s=∫02πG1t,τG2τ,sdτ=L1H1t,s+L2H2t,s,where(30)H1t,s=e1/2ρt−ssin32ρt−s−π6−eπρsin32ρt−s−2π−π6,s≤t,e1/2ρt−s+2πsin32ρt−s+2π−π6−eπρsin32ρt−s−π6,t≤s,H2t,s=eρs−t,s≤t,eρs−t−2π,t≤s,L1=23ρ21+e2πρ−2eπρcos3πρ,L2=13ρ21−e−2πρ.
One can find (29) in [11]. The structure of (29) is too complex to be analyzed the same way as G2t,s. Instead, by Lemma 1,(31)Gt,s=∫02πG1t,τG2τ,sdτ≤M2∫02πG1t,τdτ=M2ρ,and, similarly,(32)Gt,s≥m2ρ.
Also,(33)Gt,s=∫02πG1t,τG2τ,sdτ≤M1∫02πG2t,τdτ=M1ρ2,Gt,s≥m1ρ2.
Thus,(34)1ρ2maxm1,ρm2=m≤Gt,s≤M=1ρ2minM1,ρM2.Comparing m1 to ρm2 and M1 to ρM2, respectively, we find that the constants “compete” and we cannot simplify (34). Finally, note that Theorem 2.6 in [11] contains similar estimates only for 0<ρ<2/33 while (34) is fulfilled for 0<ρ<1/3.
In the next section, as an application of (34), we obtain a positive solution of (1) and (2).


## 2. Main Results

Assume that (H1) f:0,∞⟶0,∞ is continuous and λ>0.It is clear that the mapT:ℬ⟶ℬ defined by(35)Tut=λ∫02πGt,sfusds,is completely continuous. Define(36)C=u∈ℬ:ut≥mMu.Obviously, C is a cone in ℬ. By (34), T:C⟶C and u∈ℬ is a solution of (1) and (2) if u∈ℬ is a fixed point of T.Let(37)CR=u∈C:u<R,∂CR=u∈C:u=R.We apply the following theorem [12].Theorem 1.
Letℬ be a Banach space. Assume that T:C¯r⟶C is completely continuous such that Tu≠u for u∈∂Cr.(A1) If Tu≥u for u∈∂Cr, then iT,Cr,C=0.(A2) If Tu≤u for u∈∂Cr, then iT,Cr,C=1.We will restrict our attention to the case.(H2) f0,f∞≠0, where f0=limx⟶0fx/xandf∞=limx⟶∞fx/x.Recall that(38)∫02πGt,sds=1ρ3.If there existsδ>0 such that fx≥δx, then for u∈C,(39)Tut=λ∫02πGt,sfusds≥λδ∫02πGt,susds≥λδmM∫02πGt,sdsu=λδmMρ3u.Similarly, if there existsδ>0 such that fx≤γx, then, for u∈C,(40)Tut=λ∫02πGt,sfusds≤λγ∫02πGt,susds≤λγ∫02πGt,sdsu=λγρ3u.From these inequalities, by Theorem 1, we have the following result.Theorem 2.
IfH1 and H2 are satisfied, then (1) and (2) have at least one positive solution provided:(41)λ∈Mρ3mmaxf∞,f0,ρ3minf∞,f0.Proof.
Considerf∞>f0. Let f∞>ϵ>0 be such that(42)λ∈Mρ3mf∞−ϵ,ρ3f0+ϵ.
There existsR1<0 such that fx≤f0+ϵx, x∈0,R1, that is, fut≤f0+ϵut, u∈∂CR1. Hence,(43)Tu≤λf0+ϵ1ρ3u<u,u∈∂CR1.
There existsR>R1 such that fx≥f∞−ϵx, x≥R. Let R2=max2R1,MR/m. Then, ut≥m/Mu≥R, u∈∂CR2. As above, for u∈∂CR2,(44)Tu≥λf∞−ϵmMρ3u>u.It follows that iT,CR1,C=1 and iT,CR2,C=0 so that iT,CR2/C¯R1,C=−1, that is, T has a fixed point in CR2/C¯R1.
The casef∞<f0 is handled similarly.



---

*Source: 7917434-2020-04-01.xml*