# Positive Solutions for Two-Point Boundary Value Problems for Fourth-Order Differential Equations with Fully Nonlinear Terms

Authors: Yixin Zhang; Yujun Cui
Journal: Mathematical Problems in Engineering
 (2020)

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## Abstract

In this paper, we consider the existence of positive solutions for the fully fourth-order boundary value problemu4t=ft,ut,u′t,u″t,u‴t,0≤t≤1,u0=u1=u″0=u″1=0, where f:0,1×0,+∞×−∞,+∞×−∞,0×−∞,+∞⟶0,+∞ is continuous. This equation can simulate the deformation of an elastic beam simply supported at both ends in a balanced state. By using the fixed-point index theory and the cone theory, we discuss the existence of positive solutions of the fully fourth-order boundary value problem. We transform the fourth-order differential equation into a second-order differential equation by order reduction method. And then, we examine the spectral radius of linear operators and the equivalent norm on continuous space. After that, we obtain the existence of positive solutions of such BVP.


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## Body

## 1. Introduction

In this paper, we study the existence of positive solutions for the fully fourth-order boundary value problem:(1)u4t=ft,ut,u′t,u″t,u‴t,0≤t≤1,u0=u1=u″0=u″1=0,where f:0,1×0,+∞×−∞,+∞×−∞,0×−∞,+∞⟶0,+∞ is continuous. This boundary value problem can simulate the deformation of an elastic beam, whose one end is fixed and the other end is free in a balanced state. In mechanics, BVP (1) is called a cantilever beam equation. In this equation, each derivative of ut has its physical meaning: u′t is the slope, u″t is the bending moment stiffness, u‴t is the shear force stiffness, and u4t is the load density stiffness. The nonlinear fourth-order differential equation boundary value problem can simulate the deformation of an elastic beam under external force, and different boundary value conditions can show its force under different conditions. Because of its importance in mechanics, many scholars have done a lot of research on the existence of solutions for fourth-order ordinary differential equations using various nonlinear methods [1–11].As the nonlinear term does not contain the derivative term of the unknown function, equation (1) becomes(2)u4t=ft,ut,0≤t≤1,u0=u1=u″0=u″1=0.Ifft,u is superlinear or sublinear growth on u, the authors in [1] used the fixed-point theorem on the cone to obtain the existence of the positive solution of equation (2). In [3], the author used the fixed-point theorem and topological degree theory to study the existence of one or two positive solutions for the fourth-order differential equation boundary value problem:(3)u4t−λft,ut=0,0≤t≤1,u0=u1=u″0=u″1=0.Under the fixed-point index method on the cone, the authors of [11] discussed the existence of positive solutions for fourth-order boundary value problems with two parameters. Among them, the assumption condition of the nonlinear term f is related to the first eigenvalue of the corresponding linear operator. It is noteworthy that the nonlinear term in the abovementioned boundary value problems does not include the higher-order derivative u‴. When the nonlinear term contains the higher-order derivative of the unknown function, the authors of [8, 9] used the upper and lower solution method to study the existence of solutions for fully fourth-order nonlinear boundary value problems with nonlinear boundary conditions. In [10], the author discussed a fourth-order boundary value problem with fully form:(4)u4t=ft,ut,u′t,u″t,u‴t,0≤t≤1,u0=u′0=u″1=u‴1=0.When the nonlinear term satisfies superlinear growth and sublinear growth, the author used the fixed-point index method, combined with the positivity of linear operators and spectral radius, to get the positive solutions for the boundary value problem. But the linear operator in [10] does not involve the first and second derivatives of unknown functions.In this paper, by using cone theory and the fixed-point index, combined with the spectral radius of linear integral operators, and the application of equivalent norms, we discuss the existence of positive solutions for boundary value problems (1).


## 2. Preliminaries

In this section, we give some assumptions that are important to our main results:(i)
H1: f:0,1×0,+∞×−∞,+∞×−∞,0×−∞,+∞⟶0,+∞ is continuous(ii)
H2: there exist nonnegative constants a1,c1≥0 and r>0 such that a1x1+c1x3≤ft,x1,x2−x3,x4, x1,x3∈0,r,x2∈−r,r,andx4∈−∞,+∞.(iii)
H3: there exist nonnegative constants a,b,c,d,M≥0 and 0<q<1 such that(5)ft,x1,x2,−x3,x4≤ax1+bx2+cx3+dx4q+M,x1,x3∈0,+∞,x2,x4∈−∞,+∞.LetC0,1 denote the Banach space of continuous functions from 0,1 into ℝ with norm u=max0≤t≤1ut. Let P=u∈C0,1utis concave on0,1,u0=u1=0. Then, P is a positive cone on C0,1.The functions on coneP have the following properties:Lemma 1 (see [12]).
Every functionu∈P on the cone P is differentiable almost everywhere on 0,1 and satisfies(6)ut≥t1−tu,t∈0,1,u′t≤utt1−t,a.e. on0,1.Letvt=−u″t, then the differential equation BVP (1) can be transformed into the following two second-order differential equations:(7)u″t=−vt,u0=u1=0,(8)−v″t=ft,ut,u′t,u″t,u″t,v0=v1=0.LetGt,s=s1−t0≤s≤t≤1t1−s0≤t≤s≤1, which is the corresponding Green’s function of BVP (7). Thus, BVP (7) can be transformed into an equivalent integral equation:(9)ut=∫01Gt,svsds.By using (9), BVP (8) can be reduced to(10)−v″t=ft,Avt,Av′t,−vt,−v′t,v0=v1=0,where operator A is defined as Avt=∫01Gt,svsds. Thus, BVP (10) can be reduced to the equivalent integral equation:(11)vt=∫01Gt,sfs,Avs,Av′s,−vs,−v′sds.From the expression of Green’s functionGt,s, we know that Gt,s is continuous on 0,1×0,1, and we have(12)t1−ts1−s≤Gt,s≤s1−sort1−t,0≤t,s≤1.From the standard proof, we can easily obtain the following statement.Lemma 2.
A:L0,1⟶C0,1 is a completely continuous operator.For∀v∈P, we define an operator Fvt=ft,Avt,Av′t,−vt,−v′t. It follows from Lemma 1 and condition H3 that(13)Fvt=ft,Avt,Av′t,−vt,−v′t≤aAvt+bAv′t+cvt+dv′tq+M≤a∫01Gt,svsds+b∫01Gtt,svsds+cvt+dvqtq1−tq+M.It shows thatF is an operator which is defined on P to L0,1. It is easy to conclude that F is the continuous bounded operator mapping from P to L0,1. Therefore, T=A°F:P⟶P is a completely continuous operator. So, BVP (1) is equivalent to the operator equation u=Tu, where the operator T=P⟶P is given by(14)Tvt=∫01Gt,sfs,Avs,Av′s,−vs,−v′sds.Leta˜=a,b,c with a,b,c≥0. We define an operator Ta˜ on cone P:(15)Ta˜ut=a∫01Gt,s∫01Gs,τuτdτds+b∫01Gt,s∫01Gss,τuτdτds+c∫01Gt,susds,u∈P.It is easy to see thatTa˜:P⟶P is a linear operator.Letu0t=t1−t, by (15), we obtain(16)Ta˜u0t≥at1−t∫01s1−s∫01Gs,ττ1−τdτds+bt1−t∫01s1−s∫01Gss,ττ1−τdτds+ct1−t∫01s1−ss1−sds=t1−ta∫01s1−s∫01Gs,ττ1−τdτds+b∫01s1−s∫01Gss,ττ1−τdτds+c∫01s21−s2ds.Then, if scalarsa,b,c≥0, not all equal to zero, there exists a constant α>0 satisfies Ta˜u0≥αu0. Thus, from Krein–Rutman theorem, we know rTa˜>0, and there is va˜∈P/θ such that λa˜Ta˜va˜=va˜, where λa˜=rTa˜−1 is the first eigenvalue of the operator Ta˜.Now, we estimate the range ofλa˜=rTa˜−1. For any u∈P, from (15), we get(17)Ta˜ut≤a∫01s1−s∫01Gs,τuτdτds+b∫01s1−s∫01Gss,τuτdτds+c∫01s1−susds,a,b,c≥0,a˜=a,b,c,thereby(18)Ta˜u≤a∫01s1−s∫01Gs,τuτdτds+b∫01s1−s∫01Gss,τuτdτds+c∫01s1−susds.By the expression of Green’s functionGt,s and (15), we infer that(19)Ta˜ut≥t1−ta∫01s1−s∫01Gs,τuτdτds+b∫01s1−s∫01Gss,τuτdτds+c∫01s1−susds,a,b,c≥0,a˜=a,b,c.According to the above two inequalities, we have(20)Ta˜ut≥t1−tTa˜u.Recalled thatu0t=t1−t. By (20), we have(21)Ta˜u0≥Ta˜u0u0.Note the positivity of operatorTa˜, we have Ta˜2u0≥Ta˜u02u0. Hence, for any n∈N, using the recursion method to the above inequality, we get Ta˜nu0≥Ta˜u0nu0. Therefore, we have(22)Ta˜nu0≥Ta˜u0nu0.Thus,(23)Ta˜n≥Ta˜nu0u0≥Ta˜u0n.From this inequality and Gelfand formula on spectral radius, we obtain(24)rTa˜=limn⟶∞Ta˜nn≥Ta˜u0>0.From (24), we get(25)Ta˜≥rTa˜≥Ta˜u0.Byλa˜=rTa˜−1, we can get(26)1Ta˜u0≤λa˜≤1Ta¯.Now, we calculateTa˜ and Ta˜u0 as follows. We have known that(27)Gs,τ=s1−τ,0≤s≤τ≤1,τ1−s,0≤τ≤s≤1,Gss,τ=1−τ,0≤s≤τ≤1,−τ,0≤τ<s≤1.We first calculateTa˜. From (15), we get(28)Ta˜=maxt∈0,1a∫01Gt,s∫01Gs,τuτdτds+b∫01Gt,s∫01Gss,τuτdτds+c∫01Gt,susds.Through integration, we have(29)Ta˜=maxt∈0,1t1−t112a+13b+12c=148a+112b+18c.Next, we calculateTa˜u0. Because(30)Ta˜u0=maxt∈0,1a∫01Gt,s∫01Gs,τu0τdτds+b∫01Gt,s∫01Gss,τu0τdτds+c∫01Gt,su0sds.Through integration, we have(31)Ta˜u0=maxt∈0,1t1−t160a+115b+112t−16t3+112t4c=1240a+160b+5192c.Combining this with (29), we can obtain(32)48a+4b+6c≤λa˜≤9604a+16b+25c,a,b,c>0,a˜=a,b,c.Now, we consider the special case withb˜=a1,0,c1. We make a definition that(33)Tb˜ut=a1∫01Gt,s∫01Gs,τuτdτds+c1∫01Gt,susds,b˜=a1,0,c1.Because ofλb˜Tb˜vb˜t=vb˜t, it is equivalent to λb˜ and vb˜t satisfying the differential equation:(34)u4t=λb˜a1ut−c1u″t,t∈0,1,u0=u1=u″0=u″1=0.We notice that the functionvb˜t is nonnegative. After solving the above differential equations, we may take vb˜t=sinπt. Hence, we conclude the value of λb˜ is λb˜=π4/a1+π2c1.To prove our main results, we also need the two following lemmas.Lemma 3 (see [13, 14]).
LetE is a Banach space, P is a cone in E, and ΩP is a bounded open set in P. Assume that A:ΩP¯⟶P is a completely continuous operator. If there exists x0∈P/θ such that(35)x−Ax≠μx0,∀x∈∂ΩP,μ≥0.
Then,iA,ΩP,P=0.Lemma 4 (see [13, 14]).
LetE is a Banach space, P is a cone in E, ΩP is a bounded open set in P, and θ∈ΩP. Assume thatA:ΩP¯⟶P is a completely continuous operator. If(36)Ax≠μx,∀x∈∂ΩP,μ≥1,then iA,ΩP,P=1.


## 3. Main Results

Theorem 1.
Suppose that conditionsH1,H2, and H3 are satisfied, and(37)λa˜>1,(38)λb˜<1.λa˜ and λb˜ are the first eigenvalues of operator Ta˜ and operator Tb˜, respectively. Then, BVP (1) has at least one positive solution.Proof. It follows fromH1 and H2 that, for some r>0, the function f satisfies(39)ft,x1,x2,−x3,x4≥a1x1+c1x3,x1,x3∈0,r,x2∈−r,r,x4∈−∞,+∞.It is known thatvb˜ is the positive eigenvector of Tb˜ corresponding to λb˜, and we have vb˜=λb˜Tb˜vb˜. According to (14), (38), and (39), for any v∈∂Br∩P, we have(40)Avt=∫01Gt,svsds≤r∫01Gt,sds<r,Av′t=∫01Gtt,svsds≤r∫01Gtt,sds<r.Thus, we get(41)Tvt=∫01Gt,sfs,Avs,Av′s,−vs,−v′sds≥∫01Gt,sa1Avs+c1vsds=a1∫01Gt,s∫01Gs,τvτdτds+c1∫01Gt,svsds=Tb˜vt,t∈0,1.Let us suppose thatT has no fixed point on ∂Br∩P; otherwise, Theorem 1 is proved.Now, we prove(42)v−Tv≠ζvb˜,∀v∈∂Br∩P,ζ≥0,where Br=x∈C0,1|x≤r. If otherwise, there exists v0∈∂Br∩P and ζ0≥0 such that v0−Tv0=ζ0vb˜. Then, we have ζ0>0 and v0=Tv0+ζ0vb˜≥ζ0vb˜. Let ζ∗=supζ|v0≥ζvb˜, we can know ζ∗≥ζ0>0 and v0≥ζ∗vb˜. By Tb˜P⊂P, we get λb˜Tb˜v0≥ζ∗λb˜Tb˜vb˜=ζ∗vb˜. Based on (41), we have(43)v0=Tv0+ζ0vb˜≥Tb˜v0+ζ0vb˜≥ζ∗λb˜+ζ0vb˜.Then, by the definition ofζ∗, we conclude that ζ∗≥ζ∗/λb˜+ζ0. By (15), we notice that λb˜<1 and ζ0>0. So, the above inequality contradicts the definition of ζ∗. Hence, (42) holds. By Lemma 3, we have(44)iT,Br∩P,P=0.FromH3, we have(45)Tvt≤∫01Gt,sa∫01Gs,τvτdτ+b∫01Gss,τvτdτ+cvs+dv′sq+Mds≤a∫01Gt,s∫01Gs,τvτdτds+b∫01Gt,s∫01Gss,τvτdτds+c∫01Gt,svsds+d∫01Gt,svsqsq1−sqds+M=Ta˜vt+d∫01Gt,svsqsq1−sqds≤Ta˜vt+d∫01Gt,svsqsq1−sqds+M≤Ta˜vt+d∫01s1−svsqsq1−sqds+M≤Ta˜vt+d∫01vsqds+M.Note thatrTa˜−1>1. Take(46)ε=1−rTa˜2.According to Gelfand formula,rTa˜=limn⟶∞Ta˜nn, and there is a natural number N, so that, for n≥N, Ta˜n≤rTa˜+εn.For anyv∈C0,1, we define(47)v∗=∑i=1NrTa˜+εN−iTa˜i−1v,where T10=I is an identity operator. It is not difficult to verify that ·∗ is a norm in C0,1, and we have(48)∫01vqsds∗≤Kv∗q,v∈C0,1,where K is a constant. Take(49)r1>max1,r,4M∗ε−1,4dKε−11/1−q.Because ofv∗≥rTa˜+εN−1v, we can take r2>r1 so that when v≥r2, we have v∗>r1.Next, we prove(50)Tv≠μv,∀v∈∂Br2∩P,μ≥1.If there existsv1∈∂Br2∩P and μ0≥1, such that Tv1=μ0v1, by Lemma 1 and (45), we can obtain(51)Tv1t=∫01Gt,sfs,Av1s,Av1′s,v1s,v1′sds≤Ta˜v1t+d∫01v1qsds+M.Hence, we have(52)0≤μ0v1t=Tv1t≤Ta˜v1t+d∫01v1qsds+M,∀t∈0,1.Furthermore,Ta˜P⊂P, and we have(53)0≦Ta˜jTv1t≤Ta˜jTa˜v1+d∫01v1qsds+Mt,∀t∈0,1,j=1,2,…,N−1,and we also get(54)Ta˜jTv1≤Ta˜jTa˜v1+d∫01v1qsds+M,j=1,2,…,N−1.Therefore, from (47), we have(55)Tv1∗=∑i=1NrTa˜+εN−iTa˜i−1Tv1≤∑i=1NrTa˜+εN−iTa˜i−1Ta˜v1+d∫01v1qsds+M=Ta˜v1+d∫01v1qsds+M∗.Whenv1=r2, we have v1∗>r1. Hence, from (48) and (55), it follows that(56)μ0v1∗=Tv1∗≤Ta˜v1+d∫01v1qsds+M∗≤Ta˜v1∗+d∫01v1qsds∗≤Ta˜v1∗+dKv1∗q+ε4v1∗=∑i=1N−1rTa˜+εN−iTa˜iv1+dKv1∗q+ε4v1∗≤rTa˜+ε∑i=1N−1rTa˜+εN−i−1Ta˜iv1+Ta˜Nv1+ε4v1∗qr11−q+ε4v1∗≤rTa˜+ε∑i=1N−1rTa˜+εN−i−1Ta˜iv1rTa˜+εNv1+ε4v1∗+ε4v1∗≤rTa˜+ε∑i=1N−1rTa˜+εN−iTa˜i−1v1+ε2v1∗=rTa˜+εv1∗+ε2v1∗≤rTa˜+3ε2v1∗.Byμ0≥1 and (56), we can get 1≤rTa˜+3ε/2. But it is contradictory to the definition of ε=1−rTa˜/2. Hence, (50) holds. According to Lemma 4, we obtain that(57)iT,Br2∩P,P=1.Now, from (44) and (57), it follows that(58)iT,Br2∩PBr¯∩P,P=iT,Br2∩P,P−iT,Br∩P,P=1.Hence,T has at least one fixed point in Br2∩P/B¯r∩P. That is to say, BVP (1) has at least one positive solution. The proof of Theorem 1 is completed.



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*Source: 8813287-2020-11-06.xml*