# Nontrivial Solutions for the2nth Lidstone Boundary Value Problem

Authors: Yaohong Li; Jiafa Xu; Yongli Zan
Journal: Journal of Mathematics
 (2020)

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## Abstract

In this paper, we study the existence of nontrivial solutions for the2nth Lidstone boundary value problem with a sign-changing nonlinearity. Under some conditions involving the eigenvalues of a linear operator, we use the topological degree theory to obtain our main results.


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## Body

## 1. Introduction

In this paper, we investigate the existence of nontrivial solutions for the following2nth Lidstone boundary value problem with a sign-changing nonlinearity:(1)−1nu2nt=ft,ut,−u″t,…,−1n−1u2n−2t,0<t<1,u2i0=u2i1=0,i=0,1,…,n−1,where the nonlinearity f satisfies the following condition:(i)
(C0)f∈C0,1×ℝn,ℝ and there exist three nonnegative functions at,bt,andKx1,x2,…,xn≡0 such that(2)ft,x1,x2,…,xn≥−at−btKx1,x2,…,xn,fort∈0,1,xi∈ℝ,i=1,2,…,n.The Lidstone boundary value problem arises in many different areas of applied mathematics and physics. Whenn=2, problem (1) describes the deformation of an elastic beam in which both ends are simply supported. Recently, this problem has been extensively studied, and the authors refer the reader to [1–11] and references cited therein. For example, in [1], the authors used a cone-theoretic fixed point theorem to study the existence of nontrivial solutions for the nonlinear Lidstone boundary value problem:(3)y2mt=λatfyt,…,y2jt,…y2m−1t,0<t<1,y2i0=0=y2i1,i=0,…m−1,where −1mf>0 is continuous and a is nonnegative. In [2], the authors investigated the existence and uniqueness of positive solutions for the following generalized Lidstone boundary value problem:(4)−1nu2n=ft,u,−u″,…,−1n−1u2n−2,α0u2i0−β0u2i+10=0i=0,1,2,…,n−1,α1u2i1−β1u2i+11=0i=0,1,2,…,n−1,where αj≥0,βj≥0j=0,1 and α0α1+α0β1+α1β0>0. In view of symmetry, these results demonstrate that problem (4) is essentially identical with Dirichlet boundary condition (1).Meanwhile, we also note that there are a large number of papers in the literature devoted to sign-changing nonlinearities, and some results can be found in a series of papers [12–32] and the references cited therein. For example, in [12], the authors studied the following higher-order nonlinear fractional boundary value problem involving Riemann–Liouville fractional derivatives:(5)D0+αut=−ft,ut,D0+β1ut,D0+β2ut,…,D0+βn−1ut,0<t<1,u0=u′0=⋯=un−20=D0+βu1=0,where f is a sign-changing nonlinearity. Under some appropriate conditions involving the eigenvalues of the relevant linear operators, they utilized the topological degree to obtain a nontrivial solution for (5). In [13], the authors adopted the similar method in [12] to study the existence of nontrivial solutions for the following system of fractional q-difference equations with q-integral boundary conditions:(6)Dqαxt+f1t,yt=0,t∈0,1,Dqαyt+f2t,xt=0,t∈0,1,x0=0,Dqx0=0,Dqvx1=∫01htDqvxtdqt,y0=0,Dqy0=0,Dqvy1=∫01htDqvytdqt,where α∈2,3,v∈1,2,andDqα is the α-order Riemann–Liouville’s fractional q-derivative.Inspired by the aforementioned works, in this paper, we study the existence of nontrivial solutions for (1) where the nonlinearity f is sign-changing. Under some conditions involving the eigenvalues of the revelent linear operators, we use the topological degree to obtain our results.


## 2. Preliminaries

LetE:=C0,1,ℝ,v:=maxt∈0,1vt,P:=v∈E:vt≥0,∀t∈0,1,Br=v∈E:v<r for r>0. Clearly, E,⋅ is a real Banach space and P is a solid cone in E. In (1), let −1n−1u2n−2t=vt and from [2, P224], we can obtain that (1) is equivalent to the following integral equation:(7)vt=∫01G1t,sfs,∫01Gn−1s,τvτdτ,…,∫01G1s,τvτdτ,vsds,where(8)G1t,s=t1−s,0≤t≤s≤1,s1−t,0≤s≤t≤1,Git,s=∫01G1t,τGi−1τ,sdτ,i=2,3,….Next, we provide a lemma, which expresses some vital properties of the functionsGii=1,2,….Lemma 1.
(i)
Gi are nonnegative continuous functions on 0,12, and Git,s>0,t,s∈0,12(ii)
G1 has the inequalities t1−tG1s,s≤G1t,s≤G1s,s,∀t,s∈0,1(iii)
Git,s=Gis,t, ∀t,s∈0,1(iv)
∫01Git,ssinπsds=1/π2isinπt,t∈0,1, and ∫01Git,ssinπtdt=1/π2isinπs,s∈0,1Proof.
We only prove (iii) and (iv). For (iii),i=1 holds obviously. From the definition of Gi, we have(9)Gis,t=∫01G1s,τ1Gi−1τ1,tdτ1=∫01∫01G1s,τ1G1τ1,τ2Gi−2τ2,tdτ2dτ1=⋯=∫01⋯∫01︸i−1G1s,τ1G1τ1,τ2G1τ2,τ3⋯G1τi−1,tdτi−1⋯dτ2dτ1=∫01⋯∫01︸i−1G1t,τi−1⋯G1τ3,τ2G1τ2,τ1G1τ1,sdτi−1⋯dτ2dτ1=Git,s,fort,s∈0,1.
For (iv), wheni=1, we have(10)∫01G1t,ssinπsds=∫0ts1−tsinπsds+∫t1t1−ssinπsds=1π2sinπt.
Noting thatG1s,t=G1t,s, we have(11)∫01G1t,ssinπtdt=∫01G1s,tsinπtdt=1π2sinπs.
Wheni≥1, we have(12)∫01Git,ssinπsds=∫01⋯∫01︸iG1t,τi−1⋯G1τ3,τ2G1τ2,τ1G1τ1,ssinπsdsdτi−1⋯dτ2dτ1=1π2∫01⋯∫01︸i−1G1t,τi−1⋯G1τ3,τ2G1τ2,τ1sinπτ1dτi−1⋯dτ2dτ1=⋯=1π2isinπt.
Using the symmetry ofGi, we easily have(13)∫01Git,ssinπtdt=1π2isinπs.
This completes the proof.
Letαi≥0 with ∑i=1nαi2≠0. Then, we have the following equations:(14)∫01Gα1,…,αnt,ssinπsds=α11π2n+⋯+αn1π21sinπt,t∈0,1,(15)∫01Gα1,…,αnt,ssinπtdt=α11π2n+⋯+αn1π21sinπs,s∈0,1,where(16)Gα1,…,αnt,s=α1Gnt,s+⋯+αnG1t,s,fort,s∈0,1.Lemma 2.
LetLvt=∫01G1t,svsds, for t∈0,1. Then, if v∈P, we have Lv∈P0, where(17)P0=v∈P:vt≥t1−tv,∀t∈0,1.This is a direct result from Lemma1 (ii), so we omit its proof.Remark 1.
sinπt∈P0, for t∈0,1.Lemma 3 (see [33], Theorem 1. [3]).
LetΩ be a bounded open set in a Banach space E and T:Ω⟶E be a continuous compact operator. If there exists x0∈E\0 such that(18)x−Tx≠μx0,∀x∈∂Ω,μ≥0,then the topological degree degI−T,Ω,0=0.Lemma 4 (see [33], Lemma 4. [1]).
LetΩ be a bounded open set in a Banach space E with 0∈Ω and T:Ω⟶E be a continuous compact operator. If(19)Tx≠μx,∀x∈∂Ω,μ≥1,then the topological degree degI−T,Ω,0=1.


## 3. Main Results

Define the operatorA:E⟶E by(20)Avt=∫01G1t,sfs,∫01Gn−1s,τvτdτ,…,∫01G1s,τvτdτ,vsds.Moreover, the continuity off implies that A is completely continuous and the existence of solutions for (1) is equivalent to that of fixed points of A.Now, we list some assumptions for the functionsf and K:(C1)
There existβi≥0i=1,2,…,n with ∑i=1nβi2≠0 such that(21)limβ1x1+β2x2+⋯+βnxn⟶+∞Kx1,x2,…,xnβ1x1+β2x2+⋯+βnxn=0.(C2)
liminfβ1x1+β2|x2+⋯+βnxn⟶+∞ft,x1,x2,…,xn/β1x1+β2x2+⋯+βnxn>λβ1,β2,…,βn, uniformly for t∈0,1.(C3)
There existγi≥0i=1,2,…,n with ∑i=1nγi2≠0 such that(22)limsupγ1x1+γ2x2+⋯+γnxn⟶0ft,x1,x2,…,xnγ1x1+γ2x2+⋯+γnxn<λγ1,γ2,…,γn,where(23)λβ1,β2,…,βn−1=β11π2n+⋯+βn1π21,λγ1,γ2,…,γn−1=γ11π2n+⋯+γn1π21.Theorem 1.
Suppose that (C0)–(C3) hold. Then, (1) has at least one nontrivial solution.Proof.
We divide the following two steps:(i)
Step 1. By (C3), there exist ε1∈0,λγ1,γ2,…,γn and r>0 such that(24)ft,x1,x2,…,xn≤λγ1,γ2,…,γn−ε1γ1x1+γ2x2+⋯+γnxn,for allt∈0,1, xi∈ℝi=1,2,…,n with 0≤γ1x1+γ2x2+⋯+γnxn≤r. Substituting this inequality into (20), we have(25)Avt≤∫01G1t,sfs,∫01Gn−1s,τvτdτ,…,∫01G1s,τvτdτ,vsds≤λγ1,γ2,…,γn−ε1∫01G1t,sγ1∫01Gn−1s,τvτdτ+⋯+γn−1∫01G1s,τvτdτ+γnvsds≤λγ1,γ2,…,γn−ε1∫01γ1Gnt,s+⋯+γn−1G2t,s+γnG1t,svsds=λγ1,γ2,…,γn−ε1∫01Gγ1,…,γnt,svsds.Now for thisr, we claim(26)Av≠λv,∀v∈∂Br,λ≥1.Suppose the contrary. Then, there existv0∈∂Br and λ0≥1 such that Av0=λ0v0 Therefore, we obtain(27)v0t≤λ0v0t=Av0t≤λγ1,γ2,…,γn−ε1∫01Gγ1,…,γnt,sv0sds.Multiply bysinπt on both sides and integrate over 0,1 and use (15) to obtain(28)∫01v0tsinπtdt≤λγ1,γ2,…,γn−ε1∫01∫01Gγ1,…,γnt,sv0ssinπtdtds=λγ1,γ2,…,γn−ε1λγ1,γ2,…,γn−1∫01v0ssinπsds.This indicates that∫01v0tsinπtdt=0, and thus, v0t≡0, t∈0,1. This contradicts to v0∈∂Br. Consequently, (26) holds and Lemma 4 yields that(29)degI−A,Br,0=1.Step 2. By virtue of (C2), there exist ε2>0 and X0>0 such that(30)ft,x1,x2,…,xn≥λβ1,β2,…,βn+ε2β1x1+β2x2+⋯+βnxn,for all t∈0,1 and β1x1+β2x2+⋯+βnxn>X0. For any given ε with ε2−bε>0, and using (C1), there exists X1>X0 such that(31)Kx1,x2,…,xn≤εβ1x1+β2x2+⋯+βnxn,for allβ1x1+β2x2+⋯+βnxn>X1.
Consequently, whent∈0,1,β1x1+β2x2+⋯+βnxn>X1, we have(32)ft,x1,x2,…,xn≥λβ1,β2,…,βn+ε2β1x1+β2x2+⋯+βnxn−at−btKx1,x2,…,xn≥λβ1,β2,…,βn+ε2β1x1+β2x2+⋯+βnxn−at−btεβ1x1+β2x2+⋯+βnxn≥λβ1,β2,…,βn+ε2−bεβ1x1+β2x2+⋯+βnxn−at.
Noting that whent∈0,1, β1x1+β2x2+⋯+βnxn≤X1, ft,x1,…,xn,andKx1,…,xn are bounded, we can let(33)CX1=max0≤t≤1,β1x1+β2x2+⋯+βnxn≤X1ft,x1,…,xn+λβ1,β2,…,βn+ε2−bεX1,K∗=maxβ1x1+β2x2+⋯+βnxn≤X1Kx1,…,xn.
Then, we can obtain(34)ft,x1,x2,…,xn≥λβ1,β2,…,βn+ε2−bεβ1x1+β2x2+⋯+βnxn−at−CX1,for all t∈0,1,xi∈ℝ,i=1,2,…,n. Note that ε can be chosen arbitrarily small, and let(35)R>max∫01G1s,sas+K∗bs+CX1ds1−bεmaxt∈0,1∫01Gβ1,…,βnt,τdτ,S1,where(36)S1=∫01λβ1,β2,…,βn+ε2−bε+ε2−bεG1y,yay+K∗by+CX1dyε2−bε1−bεmaxt∈0,1∫01Gβ1,…,βnt,τdτ−ελβ1,β2,…,βn+ε2−bε∫01by∑i=1n∫01βiGn−iy,τdτdy,and G0y,τ≡1. Now, we claim(37)v−Av≠μφ,∀v∈∂BR,μ≥0,where φt=sinπt,t∈0,1. Suppose that (37) is not satisfied. Then, there exist v1∈∂BR and μ1>0 such that(38)v1−Av1=μ1φ.
Let(39)v˜t=∫01G1t,sas+bsK∫01Gn−1s,τv1τdτ,…,∫01G1s,τv1τdτ,v1s+CX1ds.
Then, from Lemma2, we have v˜∈P0. Now, we estimate the norm of v˜. Noting that v1∈∂BRv1=R, we obtain(40)v˜t≤∫01G1t,sas+bsεβ1∫01Gn−1s,τv1τdτ+⋯+βn−1∫01G1s,τv1τdτ+βn|v1s+K∗+CX1ds≤∫01G1s,sas+K∗bs+CX1ds+bε∫01Gβ1,…,βnt,τv1τdτ≤∫01G1s,sas+K∗bs+CX1ds+bv1εmaxt∈0,1∫01Gβ1,…,βnt,τdτ<R.
We calculatev1+v˜. By (38) we have(41)v1t+v˜t=Av1t+μ1φt+v˜t=∫01G1t,sfs,∫01Gn−1s,τv1τdτ,…,∫01G1s,τv1τdτ,v1s+as+bsK∫01Gn−1s,τv1τdτ,…,∫01G1s,τv1τdτ,v1s+CX1ds+μ1φt.
Using (C0), Lemma2 and Remark 1, we find(42)v1+v˜∈P0.
Therefore, (34) enables us to calculate(43)Av1t+v˜t≥λβ1,β2,…,βn+ε2−bε∫01G1t,sβ1∫01Gn−1s,τv1τdτ+⋯+βn−1∫01G1s,τv1τdτ+βnv1sds−∫01G1t,sas+CX1ds+∫01G1t,sas+bsK∫01Gn−1s,τv1τdτ,…,∫01G1s,τv1τdτ,v1s+CX1ds≥λβ1,β2,…,βn+ε2−bε∫01G1t,sβ1∫01Gn−1s,τv1τdτ+⋯+βn−1∫01G1s,τv1τdτ+βnv1sds=λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv1sds≥λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv1sds.
Consequently, we have(44)λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv1s+v˜sds=λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv1s+v˜sds−λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv˜sds=λβ1,β2,…,βn∫01Gβ1,…,βnt,sv1s+v˜sds+ε2−bε∫01Gβ1,…,βnt,sv1s+v˜sds−λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv˜sds.
Now, we estimateε2−bε∫01Gβ1,…,βnt,sv1s+v˜sds−λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv˜sds. From (42), we have v1t+v˜t≥t1−tv1+v˜≥t1−tv1−v˜,t∈0,1, and using G1t,s≤t1−t,t,s∈0,1, we have(45)ε2−bε∫01Gβ1,…,βnt,sv1s+v˜sds−λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,sv˜sds≥ε2−bε∫01Gβ1,…,βnt,ss1−sv1−v˜ds−λβ1,β2,…,βn+ε2−bε∫01Gβ1,…,βnt,s⋅∫01s1−say+byεβ1∫01Gn−1y,τv1τdτ+⋯+βn−1∫01G1y,τv1τdτ+βnv1y+K∗+CX1dyds≥∫01Gβ1,…,βnt,ss1−sds⋅ε2−bεv1−v˜−λβ1,β2,…,βn+ε2−bε∫01ay+byεv1∑i=1n∫01βiGn−iy,τdτ+K∗+CX1dy≥0.As a result, by means of (43) and (44), we obtain(46)Av1t+v˜t≥λβ1,β2,…,βn∫01Gβ1,…,βnt,sv1s+v˜sds.
Combining this with (38), we have(47)v1t+v˜t=Av1t+μ1φt+v˜t≥λβ1,β2,…,βn∫01Gβ1,…,βnt,sv1s+v˜sds+μ1φt≥μ1φt.
Defineμ∗=supS=supμ>0:v1+v˜≥μφ. Then, S≠∅μ1∈S and μ∗≥μ1. Hence, from (14), we have(48)v1t+v˜t≥λβ1,β2,…,βn∫01Gβ1,…,βnt,sv1s+v˜sds+μ1φt≥λβ1,β2,…,βn∫01Gβ1,…,βnt,sμ∗φsds+μ1φt=λβ1,β2,…,βnλβ1,β2,…,βn−1μ∗φt+μ1φt=μ∗+μ1φt,which contradicts the definition of μ∗. Therefore, (37) holds, and from Lemma 3, we obtain(49)degI−A,BR,0=0.
Now, (29) and (49) together imply that(50)degI−A,BRB¯r,0=degI−A,BR,0−degI−A,Br,0=−1.
Therefore, the operatorA has at least one fixed point in BR/B¯r. Equivalently, (1) has at least one nontrivial solution. This completes the proof.


## 4. Conclusion

In this paper, we use the topological degree to study the nontrivial solutions for the2nth Lidstone boundary value problem (1). To the best of our knowledge, there are few works that deal with the problem where the nonlinear terms may be unbounded and sign-changing. Moreover, it is remarked that the main result is discussed under some conditions concerning the first eigenvalues corresponding to the relevant linear operators. These mean that our main result is an improvement in some related works.



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