# On the Existence of Positive Solutions for a Fourth-Order Boundary Value Problem

Authors: Yumei Zou
Journal: Journal of Function Spaces
 (2017)

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## Abstract

By using the method of order reduction and the fixed point index, the existence of positive solutions for a fourth-order boundary value problem is studied. We provide conditions under which the existence results hold. Such conditions are related to the first eigenvalue corresponding to the relevant linear differential equation with dependence on the derivatives of unknown function.


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## Body

## 1. Introduction

In this paper, we consider the following fourth-order ordinary differential equation boundary value problem (BVP):(1)u4t=ft,ut,u′t,u′′t,u′′′t,t∈0,1,u0=u′0=u′′1=u′′′1=0,where f:[0,1]×R+3×R-→R+ is continuous.BVP (1) is used to model such phenomena as the deformations of an elastic beam in equilibrium state, whose one end-point is fixed and the other is freed. In mechanics, BVP (1) is called cantilever beam equation. Owing to its importance in mechanics, the existence of solutions to this problem has been studied by many authors; see [1–7] and references therein.Very recently, Li [3] used the theory of the fixed point index to discuss the existence of solutions of BVP (1) when f(t,x0,x1,x2,x3) is superlinear or sublinear growth on x0,x1,x2,x3. It should be remarked that two constants 1 and λ1, where λ1 is the first eigenvalue of the linear eigenvalue problem corresponding to BVP (1), play an important role in the discussion. The first eigenvalue principles were also used in [8–24].However, none of these known results use the first eigenvalues of the corresponding linear differential equation that contains the derivative of the unknown function. This is because the presence of derivatives in linear eigenvalue problem will make the study extremely difficult. To overcome this difficulty, we employ the method of order reduction to develop spectral properties of associated linear differential equation that contains the derivative of the unknown function. Then, we use the fixed point index theory to investigate the existence results of positive solutions for BVP (1) under some conditions concerning the first eigenvalue corresponding to the relevant linear eigenvalue problem that contains the derivative of the unknown function. It should be noted that the method of order reduction was also used in [7, 25] to transform higher order boundary value problem to a lower order integrodifferential equation.


## 2. Main Results

LetR+=[0,+∞) and R-=(-∞,0]. We use the classical Banach space C[0,1],C1[0,1] with the norm uC=maxt∈[0,1]|u(t)|,uC1=max⁡uC,u′C, respectively. Let P denote the cone of all nonnegative functions in C[0,1].We assume the following hypothesis holds throughout this paper.H1
f∈C0,1×R+3×R-,R+.Firstly, we consider the existence of solutions to BVP (1) when the nonlinearity f satisfies the sublinear growth property. To this end, we use the method of order reduction to transform BVP (1). For v∈C[0,1], we set(2)T3vt=∫t1vsds,T2vt=∫01G2t,svsds,T1vt=∫01G1t,svsds,where(3)G2t,s=t,0≤t≤s≤1,s,0≤s≤t≤1,G1t,s=t22,0≤t≤s≤1,2ts-s22,0≤s≤t≤1.Then we have T1v′′′=T2v′′=T3v′=-v. From the above expressions, we easily see that Ti:C[0,1]→C[0,1](i=1,2,3) are three completely continuous linear operators, and G1 and G2 have the following properties: (4)t22s≤G1t,s≤s,ts≤G2t,s≤s,∀t,s∈0,1. Hence we conclude that(5)T1P1⊂P1=u∈C0,1:ut≥t22uC,t∈0,1.Using the above transformationsTi(i=1,2,3), BVP (1) can be converted into the equivalent nonlinear integral equation:(6)vt=∫t1fs,T1vs,T2vs,T3vs,-vsds,v∈C0,1.Define an operatorA by (7)Avt=∫t1fs,T1vs,T2vs,T3vs,-vsds,v∈C0,1.Then the existence of a positive solution of BVP (1) is equivalent to the existence of a nontrivial fixed point of A on P. Now condition (H1) implies that A:P→P is a completely continuous operator.Fora=(a1,b1,c1,d1)∈R+4 with a12+b12≠0, we define linear operator Ka by(8)Kavt=∫t1a1T1vs+b1T2v+c1T3vs+d1vsds,v∈C0,1.Clearly, Ka:P→P is a completely continuous linear operator.Lemma 1.
For the operatorKa defined by (8), the spectral radius r(Ka)≠0 and Ka had unique positive unit eigenfunction φa corresponding to its first eigenvalue λ1=(r(Ka))-1.Proof.
Takev(t)=1-t; clearly v∈P; by (3) we have (9)T1vt≥∫0t2ts-s221-sds≥∫0tts21-sds=t3123-2t≥t312,T2vt≥∫0ts1-sds=t263-2t≥t26.So we have (10)Kavt≥a1∫t1T1v0sds+b1∫t1T2v0sds≥a148+b118vt.Thus there exists a constant c>0 such that cKavt≥v(t), t∈[0,1]. From Lemma 2 in [22], we know that the spectral radius r(Ka)≠0 and Ka had a positive eigenfunction corresponding to its first eigenvalue λ1=(r(Ka))-1. Hence Lemma 1 holds.
It follows from the definition of operatorKa that the function φa, corresponding to the first eigenvalue λ1=(r(Ka))-1 of the operator Ka, belongs to C4[0,1] and satisfies the equation (11)T1φa4t=λ1a1T1φa+b1T1φa′+c1T1φa′′-d1T1φa′′′,t∈0,1,T1φa0=T1φa′0=T1φa′′1=T1φa′′′1=0.
In particular, fora=(a1,b1,0,0)∈R+4, the positive eigenfunction φa of the operator Ka corresponding to the first eigenvalue λ1=(rKa)-1 belongs to C4[0,1] and satisfies the equation(12)T1φa4t=λ1a1T1φa+b1T1φa′,t∈0,1,T1φa0=T1φa′0=T1φa′′1=T1φa′′′1=0.Theorem 2.
Iff satisfies (H1) and the following conditions, then BVP (1) has at least one positive solution.(H2)
There area=(a1,b1,c1,d1)∈R+4 with a12+b12≠0 and c>0 such that r(Ka)<1 and (13)ft,x1,x2,x3,x4≤a1x1+b1x2+c1x3+d1x4+c,x1,x2,x3,x4∈R+3×R-.(H3)
There areb=(a2,b2,c2,d2)∈R+4 with a22+b22≠0 and r>0 such that r(Kb)>1 and (14)ft,x1,x2,x3,x4≥a1x1+b1x2+c1x3+d1x4,x1,x2,x3,x4∈0,r3×-r,0.Proof.
LetW1={v∈P:v=λAv,for  someλ∈[0,1]}. We will prove that W1 is a bounded set. In fact, if v∈W1, by the definition of W1,v=λAv for some λ∈[0,1]. Hence we have (15)vt≤Avt=∫t1fs,T1vs,T2vs,T3vs,-vsds,v∈P,t∈0,1.By (H2), we have v(t)≤(Kav)(t)+c(1-t), which can be rewritten in the form (I-Ka)v≤v~0, where v~0(t)=c(1-t). The condition r(Ka)<1 ensures that I-Ka has a bounded inverse operator (I-Ka)-1 which is given by (I-Ka)-1=I+Ka+Ka2+⋯+Ka+⋯. It follows from Ka(P)⊂P that (I-Ka)-1(P)⊂P. So we have v≤(I-Ka)-1v~0 and W1 is bounded. Select R>max⁡{r,sup⁡{vC:v∈W1}}; we have v≠λAv,v∈∂ΩR,λ∈[0,1], where ΩR={u∈C[0,1]:uC<R}. Now Lemma 2.3.1 in [26] implies(16)iA,ΩR∩P,P=1.
Letφb be the positive eigenvalue function of Kb (with replacement of a by b in Lemma 1). We may suppose that A has no fixed point on ∂Ωr∩P (otherwise, the proof is finished). Now we shall show that(17)v-Av≠λφb,∀v∈∂Ωr∩P,λ≥0,where Ωr={u∈C[0,1]:uC<r}. In fact, if (17) is not valid, there exist v0∈∂Ωr∩P and λ0≥0 such that v0-Av0=λ0φb. Hence we have λ0>0. Since v0∈∂Ωr∩P, by the definitions of P and Ti(i=1,2,3), we have (18)0≤v1t,T1vt,T2vt,-T3vt≤r,t∈0,1.Hence from (H3), we obtain that(19)ft,T1v0t,T2v0t,T3v0t,-v0t≥a2T1v0t+b2T2v0t+c2T3v0t+d2v0tfrom which it follows that(20)v0=Av0+λ0φb≥Kbv0+λ0φb.Set λ∗=sup⁡{λ:v0≥λφb}. It is easy to see that λ∗>0 and v0≥λ∗φb. It follows from Kb(P)⊂P that Kbv0≥λ∗Kbφb=λ∗r(Kb)φb. Therefore, by (20), (21)v0=Av0+λ0φb≥rKbλ∗φb+λ0φbwhich contradicts the definition of λ∗. Hence (17) is true and we have from Corollary 2.3.1 in [26] that(22)iA,Ωr∩P,P=0.By (16) and (22), we have that(23)iA,ΩR∩P∖Ωr¯∩P,P=iA,ΩR∩P,P-iA,Ωr∩P,P=1.Then A has at least one fixed point in (ΩR∩P)∖(Ωr¯∩P) which means that BVP (1) has a positive solution. The proof of Theorem 2 is completed.Next, we consider the existence of solutions to BVP (1) when the nonlinearity f satisfies the superlinear growth condition. In this case, we assume that f(t,x0,x1,x2,x3) is independent of x2 and x3; that is, we consider the following simplified BVP:(24)u4t=ft,ut,u′tu0=u′0=u′′1=u′′′1=0.As in [3], a positive solution of BVP (24) is equivalent to a nontrivial fixed point of B defined by(25)But=∫01Gt,sfs,us,u′sds,u∈C10,1,where G(t,s) is Green’s function of the homogeneous linear problem u(4)(t)=0,u(0)=u′0=u′′1=u′′′1=0, which is explicitly given by(26)Gt,s=16t23s-t,0≤t≤s≤1,16s23t-s,0≤s≤t≤1.From (26) we can easily verify that G(t,s) has the following properties:(27)13t2s2≤Gt,s≤12t2or12s2,t,s∈0,1.For c=(a1,b1)∈R+2 with a12+b12≠0 (we also use c denoting (a1,b1,0,0)). Define a linear integral operator L by(28)Lcut=∫01Gt,sa1us+b1u′sds,u∈C10,1.In what follows we work on the Banach spaceC1[0,1] and utilize the cone (29)P3=u∈C10,1:u0=0,ut≥12t2uC,u′t≥tu′C,0≤t≤1.Lemma 3 (see [3, 6]).
Ifu∈P3, then u′C1=uC and (1/2)uC1≤uC≤uC1.Lemma 4 (see [3, 6]).
B
,
L
:
P
3
→
P
3 are two completely continuous operators.Based on the definition of operatorLc, (12) can be rewritten in the following form:(30)T1φc=rKc-1LcT1φcwhich means that the operator Lc has an eigenvalue (r(Kc))-1, and its related eigenfunction can be taken by T1φc. In fact, we have r(Kc)=r(Lc) from equivalent differential equation (12).Theorem 5.
Iff satisfies (H1) and the following conditions, then BVP (24) has at least one positive solution.(H4)
There arec=(a1,b1)∈R+2 with a12+b12≠0 and c>0 such that r(Lc)>1 and (31)ft,x1,x2≥a1x1+b1x2-c,x1,x2∈R2,t∈0,1.(H5)
There ared=(a2,b2)∈R+2 with a22+b22≠0 and δ>0 such that r(Ld)<1 and (32)ft,x1,x2≤a2x1+b2x2,x1,x2∈0,δ2,t∈0,1.Proof.
Setε>0 such that (1-ε)r(Lc)>1. Take R>max⁡{δ,30c/ε(2a1+5b1)}. For every u∈∂ΩR∩P3, it follows from (27) and Lemma 3 that(33)Bu≥Lcu-c∫01Gt,sds≥1-εLcu+εLcu-c2t2≥1-εLcu+ε∫0113t2s2a12s2uC1+b1suC1ds-c2t2≥1-εLcu+t2εuC1a130+b112-c2≥1-εLcu.We may suppose that B has no fixed point on ∂ΩR∩P3 (otherwise, the proof is finished). Now we show that(34)u-Bu≠μT1φc,∀u∈∂ΩR∩P3,μ≥0,where φc is the positive eigenfunction of operator Sc related to its first eigenvalue (r(Sc))-1, and T1φc∈P3∖{θ} follows from (5).
In fact, if (34) is not valid, there exist u0∈∂ΩR∩P3 and μ0≥0 such that(35)u0-Bu0=μ0T1φc.Hence we have μ0>0. Set μ∗=sup⁡{μ:u0≥μT1φc}. It is easy to see that μ∗>0 and u0≥μ∗T1φc. Since u0∈∂ΩR∩P3, by (33), we obtain that(36)u0=Bu0+μ0T1φc≥1-εLcu0+μ0T1φc.Therefore by (30), u0≥μ∗(1-ε)r(Lc)T1φc>μ∗T1φc, which contradicts the definition of μ∗. Hence (34) is true and we have from Corollary 2.3.1 in [26] that(37)iA,ΩR∩P3,P3=0.
Next, we prove that(38)u≠μBu,u∈∂Ωδ∩P3,μ∈0,1.If this is false, then there exist u1∈∂Ωδ∩P3 and μ1∈[0,1] such that u1=μ1Bu1. It follows from Lemma 3 that 0≤u1(t),u1′(t)≤δ,t∈[0,1]. Then by assumption (H5), we have u1=μ1Bu1≤μ1Ldu1≤Ldu1. Notice that r(Ld)<1; we obtain u1≤(I-Ld)-1θ=θ, which is a contradiction with u1∈∂Ωδ∩P3. Hence, (38) holds. Then by Lemma 2.3.1 in [26], we have(39)iA,Ωδ∩P3,P3=1.
By (37) and (39), we have that(40)iB,ΩR∩P3∖Ωδ¯∩P3,P3=iB,ΩR∩P3,P3-iB,Ωδ∩P3,P3=-1.Then B has at least one fixed point in (ΩR∩P3)∖(Ωδ¯∩P3) which means that BVP (24) has a positive solution. The proof of Theorem 5 is completed.



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*Source: 4946198-2017-08-14.xml*