# On a Fourth-Order Boundary Value Problem at Resonance

Authors: Man Xu; Ruyun Ma
Journal: Journal of Function Spaces
 (2017)

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## Abstract

We investigate the spectrum structure of the eigenvalue problemu4x=λux,x∈0,1;u0=u1=u′0=u′1=0. As for the application of the spectrum structure, we show the existence of solutions of the fourth-order boundary value problem at resonance -u4x+λ1ux+gx,ux=hx,x∈0,1;u0=u1=u′0=u′1=0, which models a statically elastic beam with both end-points being cantilevered or fixed, where λ1 is the first eigenvalue of the corresponding eigenvalue problem and nonlinearity g may be unbounded.


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## Body

## 1. Introduction

Starting from the seminal paper of Landesman and Lazer [1], the existence and multiplicity of solutions of nonlinear second-order boundary value problem at resonance, (1)u′′x+π2ux+gx,ux=ex,x∈0,1,u0=u1=0,and its general case have been extensively studied; see Gupta [2, 3], Iannacci and Nkashama [4, 5], Costa and Goncalves [6], Ambrosetti and Mancini [7], Fonda and Habets [8], Các [9], and Ahmad [10] and the references therein. Because of the linear operator L:D(L)→L2(0,1), (2)Lu≔u′′+π2u,u∈DL≔y∈L20,1:u0=u1=0is not reversible; this kind of problems as (1) is of problemsat resonance.In the past twenty years, the existence and multiplicity of solutions (or positive solutions) of nonlinear fourth-order boundary value problems atnonresonance case have been investigated by many authors. Especially, many works address the nonlinear fourth-order differential equation of the following form: (3)u4x=gx,ux,u′x,u′′x,u′′′x,x∈0,1,with one of the following sets of boundary conditions:(i)
Both end-points simply supported conditions:(4)u0=u1=u′′0=u′′1=0Navierboundarycondition.(ii)
Both end-points cantilevered or fixed conditions:(5)u0=u1=u′0=u′1=0Dirichletboundarycondition.(iii)
One end simply supported and the other end sliding clamped conditions:(6)u0=u′′0=u′1=u′′′1=0.See Rynne [11], Korman [12], Ma et al. [13–15], Cabada et al. [16, 17], Vrabel [18], Schröder [19], Drábek and Holubová [20], Webb et al. [21], Chu and O’Regan [22], and Cid et al. [23] for references along this line.However, relatively little is known about the fourth-order problem at resonance; see Gupta et al. [24, 25], Jurkiewicz [26], and Iannacci and Nkashama [27]. The likely reason is that the spectrum theory of fourth-order operators is not available.The purpose of this paper is to show the existence of solutions of the fourth-order boundary value problem at resonance:(7)-u4x+λ1ux+gx,ux=hx,x∈0,1,u0=u1=u′0=u′1=0,which models a statically elastic beam with both end-points being cantilevered or fixed, where g:(0,1)×R→R is a L2-Carathéodory function, h∈L2(0,1), and λ1 is the first eigenvalue of the corresponding linear eigenvalue problem. More precisely, we provide a sufficient condition for the solvability of problem (7) in which nonlinearity g is not necessarily needed to satisfy the Landesman-Lazer type condition or the monotonicity assumption.To do this, we investigate the spectrum structure of the linear eigenvalue problem:(8)u4x=λux,x∈0,1,u0=u1=u′0=u′1=0.We will show that the eigenvalues of (8) form a sequence: (9)0<λ1<λ2<λ3<⋯⟶+∞.Moreover, for each j∈N, λj (λj=mj4,  mj is the simple root of the equation cos⁡mcosh⁡m-1=0) is simple, and the corresponding eigenfunction is φj, which forms (with a suitable normalization) an orthogonal system of L2(0,1). Since(10)u4x=m14ux,x∈0,1,u0=u1=u′0=u′1=0has nontrivial solution sin⁡m1x-sinh⁡m1x+sin⁡m1-sinh⁡m1/cos⁡m1-cosh⁡m1(cosh⁡m1x-cos⁡m1x), problem (7) is a problem in resonance.We refer, for motivations and results, to the classical papers of [4] for the second-order boundary value problems at resonance and [24] for the fourth-order boundary value problems which are simply supported at both ends and are at resonance. In this paper, we will use the classical spaces Ck[0,1], Lk(0,1), and L∞(0,1); we shall make use, in what follows, of the Sobolev spaces Hk(0,1) and H0k(0,1); we refer the reader to see [28] for their definitions and properties.The rest of the paper is arranged as follows. In Section2, we investigate the spectrum structure of eigenvalue problem (8). In Section 3, we give some preliminary results that are needed to apply Leray-Schauder continuation method to obtain the existence of solutions for problem (7). Finally, Section 4 is devoted to stating and proving our main result.


## 2. The Eigenvalue Problem

In this section, we consider the linear eigenvalue problem:(11)u4x=λux,x∈0,1,u0=u1=u′0=u′1=0.Lemma 1.
The equation(12)cos⁡mcosh⁡m-1=0,m∈R+has infinitely many simple roots (13)0<m1<m2<m3⋯⟶+∞.Moreover, (14)m2k-1∈2k-12π,2kπ,m2k∈2kπ,2k+12πfor k∈N.Proof.
Let(15)γm=cos⁡mcosh⁡m-1,m∈R+.
It is easy to check that, fork∈N,(16)γ2k-1π<0,γ2kπ>0.We claim that γ(m) has exactly one root mj∈[jπ,(j+1)π]; moreover, for any j∈N, mj is simple. Assume that the claim is not true. Then, the following two cases must occur.
Case
1. There are three zeros in (j0π,(j0+1)π) for some j0∈N. In this case, we may find τ∈(j0π,(j0+1)π) such that (17)γ′′τ=0.However, this contradicts the fact that (18)γ′′m=-2sin⁡msinh⁡m.
Case
2. There is a double zero t^∈(j0π,(j0+1)π) for some j0∈N. In this case, we only deal with case j0 being odd. Case j0 is even and can be treated similarly.
Since(19)γj0π<0,γj0+1π>0,we may assume that there exists t^∈(j0π,(j0+1)π) such that (20)γm<0,m∈jπ,t^,γt^=γ′t^=0.Combining this with the fact γ′′t^>0, it concludes that γ(m)>0 in some left neighborhood of t^. However, this is a contradiction.Lemma 2.
The linear eigenvalue problem (11) has infinitely many eigenvalues: (21)λj=mj4j∈N,and the eigenfunction corresponding to λj is given by (22)φjx=sin⁡mjx-sinh⁡mjx+sin⁡mj-sinh⁡mjcos⁡mj-cosh⁡mjcosh⁡mjx-cos⁡mjx.Moreover, φk∈Sk,+, where Sk,+ denote the set of u∈C3[0,1] such that(i)
u has only simple zeros in (0,1) and has exactly k-1 such zeros;(ii)
u′′0>0 and u′′1≠0.Proof.
By [11, P. 308], we know problem (11) has a sequence of eigenvalues 0<λ1<λ2<⋯ with limk→∞⁡λk=+∞. For any given k∈N, each eigenvalue λk is simple and has a corresponding eigenfunction φk satisfying (i) and (ii). By a direct calculation, we have (21) and (22).


## 3. Preliminaries

Throughout the paper, we assume that(H0)
p∈L∞(0,1) such that, for a.e. x∈(0,1), 0≤p(x)≤m24-m14; moreover p(x)<m24-m14 on a subset of (0,1) of positive measure.Define a linear operatorL:D(L)⊂L2(0,1)→L2(0,1) by(23)Lu≔u4+λ1u,where D(L)=H4(0,1)∩H02(0,1)≔H. Then L is the linear self-adjoint operator, and thus L2(0,1) admits the orthogonal direct sum decomposition L2(0,1)=N⊕R, where N is the one-dimensional null space of L and R is the range space of L, namely, (24)N=y∈L20,1∣y=sφ1for  somes∈R,R=y∈L20,1∣∫01yxφ1xdx=0.Therefore each u∈H02(0,1)⊂L2(0,1) has a unique decomposition: (25)u=sφ1+w≔u¯x+u~x,where s∈R, w∈R, so that, with obvious notations, H02(0,1)=H¯02(0,1)+H~02(0,1).Sinceu∈H02(0,1)⊂L2(0,1), it follows that u has the Fourier series expansion:(26)ux=∑j=1∞sjφjx,sj=∫01uxφjxdx.By (25), we observe that(27)u¯x=s1φ1x,u~x=∑j=2∞sjφjx.Lemma 3.
Assume thatp satisfies (H0). Let σ>0 and q∈L∞(0,1) satisfy (28)0≤qx≤px+σfor  a.e.x∈0,1.Then there exists a constant δ=δ(p)>0 such that, for all u∈H, we have(29)∫01-u4x+λ1ux+qxuxu¯x-u~xdx≥δ-σu~H22.Proof.
We will divide the proof into three steps.
Step
1. It follows from Lemma 2 that, for all u∈H, (30)u¯4x=λ1u¯x,x∈0,1,u0=u1=u′0=u′1=0.Multiplying both sides of the equation in (30) by u¯ and integrating from 0 to 1, we get that (31)∫01u¯′′x2dx-λ1∫01u¯x2dx=0.This together with the orthogonality of u¯ and u~ in L2(0,1) implies that(32)∫01-u4x+λ1ux+qxuxu¯x-u~xdx=∫01qxu¯x2dx+∫01u~′′x2dx-∫01qx+λ1u~x2dx≥∫01u~′′x2dx-∫01qx+λ1u~x2dx≡Dqu~.Subsequently, by (28), we have(33)Dqu~≥Dpu~-σ∫01u~x2dx.
Step
2. We show that if u∈H, then there exists a constant δ=δ(p)>0 satisfying(34)Dpu~≥δu~H22.
Firstly, by (27), we observe that(35)u~′′x=∑j=2∞sjφj′′x.
By Lemma2, (mj4,φj) is a solution of (11). So that, substituting into (11) and multiplying both sides of the equation by φj and integrating from 0 to 1, we get that, for x∈(0,1),(36)∫01φj′′x2dx=mj4∫01φjx2dx.This fact together with (35) and using Parseval identity yields that(37)∫01u~x2dx=∑j=2∞sj2,∫01u~′′x2dx=∑j=2∞mj4sj2.Therefore, by (H0), we find that (38)Dpu~≥∫01u~′′x2-m24u~x2dx≥∑j=2∞sj2mj4-m24≥0,with equality if and only if sj2(mj4-m24)=0 for all j∈N and j≥2. Therefore, for j>2, one has sj=0, and, by using the series expansion, u~(x) reduces to u~(x)=s2φ2(x). But then, we have(39)Dpu~=s22∫01m24-m14-pxφ2x2dx=0.It follows from (H0) that s2=0, and hence u~=0.
Next we will prove that (34) is true. Suppose, on the contrary, that there exists a sequence {u~n}⊂H~02(0,1) and u~⊂H~02(0,1) such that(40)limn→∞Dpu~=0withu~nH2=1.It follows from the compact embedding of H~02(0,1) into C1[0,1] that (41)u~n⟶u~inC10,1,u~n⇀u~inH~020,1.Now (41) implies that(42)u~H22≤liminf⁡n→∞u~nH22.At the same time by (40) and (41), we obtain (43)limn→∞u~nH22=∫01px+λ1u~x2dx.This together with (42) implies that(44)u~H22≤∫01px+λ1u~x2dx,that is, Dp(u~)≤0. By the fact that Dp(u~)≥0 with equality if and only if u~=0, we know u~=0; this contradicts the fact that u~nH2=1.
Step
3. By a direct observation of (33) and (34), we obtain the desired results.Lemma 4.
Letξ∈(0,m24-m14) be fixed constant. Define a linear operator E:H→L2(0,1) by (45)Eu=u4+λ1u+ξu.Then E-1:L2(0,1)→H is completely continuous.Proof.
By the theory of linear fourth-order differential equations, the operatorE:H→L2(0,1) defined by (46)Eu=u4+λ1u+ξuis one-to-one and continuous obviously. It follows that E-1:L2(0,1)→H is completely continuous.


## 4. The Main Result and the Proof

The main result of the paper addresses the existence of solutions of fourth-order problem (7), when the nonlinearity is unbounded. For the sake of simplicity, we assume the following:(H1)
g:(0,1)×R→R is a L2-Carathéodory function; namely, g(·,u) is measurable on (0,1) for every u∈R, g(x,·) is continuous on R for a.e. x∈(0,1), for any constant r>0, and there exists a function Γr∈L2(0,1) such that (47)gx,u≤Γrxfor a.e. x∈(0,1) and all u∈R with |u|≤r.(H2)
ug(x,u)≥0 for a.e. x∈(0,1) and all u∈R.(H3)
For all constantσ>0, there exist a constant R=R(σ)>0 and a function b=b(σ)∈L∞(0,1) such that (48)gx,u≤px+σu+bxfor a.e. x∈(0,1) and all u∈R with |u|≥R, where p∈L∞(0,1) has given by (H0).Theorem 5.
Assume that (H0)–(H3) hold. Then problem (7) has at least one solution for any h∈L2(0,1) provided: (49)∫01hxφ1xdx=0.Proof.
Letδ>0 be associated with function p and ξ∈(0,m24-m14) be a fixed constant with ξ<δ/2. To study problem (7) using Leray-Schauder continuation method, we prove that each of the possible solutions of the homotopy (50)-u4+λ1u+1-λξu+λgx,u=λhx∈0,1,u0=u1=u′0=u′1=0has a priori bound. Therefore, we claim that if u∈H is a solution of (50), then there exists a constant ρ>0 independently of λ∈[0,1) such that (51)uH<ρ.
If we assume on the contrary that there exists a sequence{λn}⊂(0,1) and a sequence {un}⊂H with unH≥n for all n∈N such that (52)-un4+λ1un+1-λnξun+λngx,un=λnh,x∈0,1,un0=un1=un′0=un′1=0.Let vn=un/unH. Then(53)-vn4+λ1vn+ξvn=λnξvn+λnhunH-λngx,ununH,x∈0,1,vn0=vn1=vn′0=vn′1=0.Obviously, by Lemma 4, (53) is equivalent to (54)vn=E-1λnξvn+hunH-gx,ununH.
(47) together with (48) yields that there exists a function c∈L∞(0,1) depending only on R=R(δ) such that (55)gx,u≤px+δ2u+bx+cxfor  a.e.x∈0,1andallu∈R.Subsequently, the right-hand member of (54) is bounded in L2(0,1) independently of n. By Lemma 4, there exists v∈H such that limn→∞⁡vn=v in H. Moreover, vH=1.
On the other hand, (H3) yields that there existR=R(δ)>0 and b=b(δ)∈L∞(0,1) such that (56)gx,u≤px+δ4u+bxfor a.e. x∈(0,1) and all u∈R with |u|≥R, where R is chosen such that b(x)/u<δ/4. Let us define a function p~:(0,1)×R→R by (57)p~x,u=gx,uu,foru≥R,gx,RRuR+1-uRpx,for0≤u≤R,gx,-RRuR+1+uRpx,for-R≤u≤0.Then, this together with (H2) and (56) yields that(58)0≤p~x,u≤px+δ2fora.e.x∈0,1andallu∈R.Moreover, p~(x,u)u is a L2-Carathéodory function. Define f:(0,1)×R→R by (59)fx,u=gx,u-p~x,uu.By (H1), it yields that, for a.e. x∈(0,1) and all u∈R, there exists v∈L2(0,1), such that (60)fx,u≤vx.Observe that v depend only on Γ and γR.
Thus, problem (50) is equivalent to (61)-u4x+λ1ux+1-λξux+λp~x,uxux+λfx,ux=λhx,x∈0,1,u0=u1=u′0=u′1=0.The fact ξ∈(0,m24-m14) with ξ<δ/2 together with (58) yields that (62)0≤1-λξ+λp~x,u≤px+δ2fora.e.x∈0,1andallu∈R.Therefore, by Lemma 3, (60), and the compact embedding of H2(0,1) into L2(0,1), we have (63)0=∫01-u4x+λ1ux+1-λξux+λp~x,uxuxu¯x-u~xdx+∫01λfx,ux-λhxu¯x-u~xdx≥δ2u~H22-u~L2+u¯L2fL2+hL2≥δ2u~H22-Cu~H2+u¯H2for some constant C>0.
By (63), we deduce immediately that limn→∞⁡v~n=0 in H2(0,1). Therefore we can write vn=v¯n. Since vH=1, we shall suppose that(64)vx=cφ1xfor  somec>0.Now, using Lemma 2, we can get that there exists N such that, for n≥N, vn(x)>0 on (0,1). So that, for n≥N,(65)un0=un1=un′0=un′1=0,unx>0.
Multiplying both sides of the equation in (53) by v¯n and integrating from 0 to 1, by (31), (49), and the fact λn∈(0,1), we have(66)1-λnξ∫01v¯n2dx=-λnunH∫01gx,unv¯ndx.So that ∫01g(x,un)v¯ndx<0. By (H2) and (65), we conclude a contradiction.



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