# On Linear and Nonlinear Fourth-Order Eigenvalue Problems with Nonlocal Boundary Condition

Authors: Dongming Yan
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We determine the principal eigenvalue of the linear problemu(4)(t)+βu′′(t)=μ[u(t)−u′′(t)],t∈(0,1),  u(0)=u(1)=∫01p(s)u(s)ds, u′′(0)=u′′(1)=∫01q(s)u′′(s)ds, where 0<β<π2 and p,q∈L[0,1]. Moreover, we investigate the existence of positive solutions for the corresponding nonlinear problem. The proofs of our main results are based upon the Krein-Rutman theorem and fixed point index theory.


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## Body

## 1. Introduction

The deformations of an elastic beam can be described by the boundary value problems of the fourth-order ordinary differential equations. For example, an elastic beam in an equilibrium state whose both ends are simply supported can be described by the fourth-order boundary value problem of the form(1)u(4)(t)=f(t,u(t),u′′(t)),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0,
see Gupta [1, 2]. Owing to its significance in physics, it has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, and the method of lower and upper solutions; see, for example, [3–16].Recently, Bai [3] investigated the existence of positive solutions for more general fourth-order nonlocal boundary value problem
(2)u(4)(t)+βu′′(t)=λf(t,u(t),u′′(t)),t∈(0,1),u(0)=u(1)=∫01‍p(s)u(s)ds,u′′(0)=u′′(1)=∫01‍q(s)u′′(s)ds.
By using the Krasnoselskii fixed point theorem, the sufficient conditions for the existence of positive solutions of (2) are obtained. We can find that, however, those conditions do not involve the eigenvalues with respect to the relevant linear operator, and those sufficient conditions are not optimal.The likely reason is that the spectrum structure of the linear eigenvalue problem(3)u(4)(t)+βu′′(t)=μ[u(t)-u′′(t)],t∈(0,1),u(0)=u(1)=∫01‍p(s)u(s)ds,u′′(0)=u′′(1)=∫01‍q(s)u′′(s)ds
is not clear.It is the purpose of this paper to investigate the first eigenvalue of (3) by using the Krein-Rutman theorem [17, 18], and then we use this spectrum result to establish the existence of positive solutions of nonlinear nonlocal problem (2). The existence of positive solution is obtained by means of fixed point index theory under some conditions concerning the first eigenvalue with respect to the relevant linear operator. The obtained sufficient conditions in this paper are optimal. For the concepts and properties of fixed point index theory, we refer the reader to [19].The rest of the paper is arranged as follows: in Section2, we study the spectrum of (3). Finally, in Section 3, we give an application of our main result.


## 2. Spectrum of (3)

Let us assume that(A1)
0<β<π2, p,q∈L[0,1], p(t)≥0, q(t)≥0, t∈[0,1],  ∫01‍p(s)ds<1,  ∫01‍q(s)sinβsds+∫01‍q(s)sinβ(1-s)ds<sinβ.(A2)
f∈C([0,1]×[0,∞)×(-∞,0],[0,∞)).Definition 1.
We sayμ is an eigenvalue of linear problem (see (3)), if (3) has nontrivial solutions.To study the spectrum of (3), we need several preliminary results.Let(4)G1(t,s)={t(1-s),0≤t≤s≤1;s(1-t),0≤s≤t≤1;G2(t,s)={sinβtsinβ(1-s)βsinβ,0≤t≤s≤1;sinβssinβ(1-t)βsinβ,0≤s≤t≤1.Lemma 2 (see [7]).
Assume (A1) holds. Then one has(i)
Gi(t,s)≥0,  for t,s∈[0,1]; Gi(t,s)>0,  for t,s∈(0,1);(ii)
biGi(t,t)Gi(s,s)≤Gi(t,s)≤CiGi(t,t),  for t,s∈[0,1],where C1=1, b1=1; C2=1/sinβ, b2=βsinβ.Lemma 3 (see [3]).
Assume (A1) holds. Then for anyg∈C[0,1],  u solves the problem
(5)u(4)(t)+βu′′(t)=g(t),t∈(0,1),u(0)=u(1)=∫01‍p(s)u(s)ds,u′′(0)=u′′(1)=∫01‍q(s)u′′(s)ds,
if and only if
(6)u(t)=∫01‍∫01‍K1(t,s)K2(s,τ)g(τ)dτds,
where
(7)K1(t,s)=G1(t,s)+11-∫01‍p(x)dx∫01‍G1(s,x)p(x)dx,K2(s,τ)=G2(s,τ)+ρ(s)∫01‍G2(τ,x)q(x)dx,ρ(s)=sinβs+sinβ(1-s)sinβ-∫01‍q(x)sinβxdx-∫01‍q(x)sinβ(1-x)dx.Let(8)e(t)≔-t2+t+∫01‍p(s)(s-s2)ds1-∫01‍p(s)ds,t∈[0,1].
Then, there exist c,c~>0 such that
(9)∫01‍K1(t,s)e(s)ds≤ce(t),t∈[0,1],(10)∫01‍K2(t,s)e(s)ds≤c~e(t),t∈[0,1],
respectively.In fact, from (ii) of Lemma2, we can obtain
(11)∫01‍K1(t,s)e(s)ds=∫01[G1(t,s)+11-∫01‍p(x)dx‍×∫01‍G1(s,x)p(x)dx11-∫01‍p(x)dx11-∫01‍p(x)dx]e(s)ds≤∫01‍[G1(t,t)+11-∫01‍p(x)dx×∫01‍G1(s,s)p(x)dx11-∫01‍p(x)dx11-∫01‍p(x)dx]e(s)ds=G1(t,t)∫01‍e(s)ds+∫01‍p(x)dx1-∫01‍p(x)dx×∫01‍G1(s,s)e(s)ds=(-t2+t)∫01‍e(s)ds+∫01‍p(x)dx1-∫01‍p(x)dx×∫01‍(s-s2)e(s)ds≤ce(t),t∈[0,1],
where we can take
(12)c=max{∫01‍e(s)ds,∫01‍p(s)ds∫01‍(s-s2)p(s)ds×∫01‍(s-s2)e(s)ds∫01‍p(s)ds∫01‍(s-s2)p(s)ds∫01‍p(s)ds∫01‍(s-s2)p(s)ds},
and also we obtain
(13)∫01‍K2(t,s)e(s)ds=∫01‍[G2(t,s)+ρ(t)∫01‍G2(s,x)q(x)dx]e(s)ds≤1sinβ∫01‍[G2(t,t)+ρ(t)∫01‍G2(x,x)q(x)dx]×e(s)ds≤c1sinβtsinβ(1-t)+c2≤c~e(t),
where
(14)c1=1βsin2β∫01‍e(s)ds,c2=1sinβ∫01‍e(s)ds∫01‍G2(s,s)q(s)dsmax0≤t≤1ρ(t),c~=1-∫01‍p(s)ds∫01‍(s-s2)p(s)ds(c1+c2).Let(15)X={u∈C2[0,1]:u(0)=u(1)=∫01‍p(s)u(s)dsu′′(0)=u′′(1)=∫01‍q(s)u′′(s)ds-εe(t)≤-u′′(t)≤εe(t)},
for a certain ε>0, and let Y be the Banach space C[0,1] with the norm ∥u∥∞=max0≤t≤1|u(t)|. For u∈X, we have
(16)u(t)=∫01‍K1(t,s)(-u′′(s))ds.
Combining (16) and (9) with the fact that -εe(s)≤-u′′(s)≤εe(s), we conclude that
(17)-cεe(t)≤u(t)≤cεe(t),t∈[0,1].
So, we may define the norm of u∈X by
(18)∥u∥X≔inf{ε∣ε>0,-εe(t)≤-u′′(t)≤εe(t),t∈[0,1]}.
This norm is so called u0-norm.It is easy to get the following lemmas.Lemma 4.
(
X
,
|
|
·
|
|
X
) is a Banach space.Proof.
Let{un} be Cauchy sequence of (X,||·||X). Then, we have
(19)∥un-um∥X→0,(n,m→∞).
From (18) and (19), for all ε>0, there exists N>0 such that
(20)-εe(t)≤-(un′′(t)-um′′(t))≤εe(t),t∈[0,1],∀n,m>N.
Hence,
(21)∥un′′-um′′∥∞≤ε∥e∥∞→0,(n,m→∞),
which implies that {un′′} is a Cauchy sequence of (C[0,1],||·||∞). According to the completeness of (C[0,1],||·||∞), there exists u0∈C2[0,1] such that
(22)∥un′′-u0′′∥∞→0,(n→∞).
Let m→∞ in (20). Then, we have
(23)-εe(t)≤-(un′′(t)-u0′′(t))≤εe(t),t∈[0,1],(n>N),
which implies that
(24)∥un-u0∥X≤ε,(n>N).
Hence, for Cauchy sequence {un}⊂(X,||·||X), there exists u0∈X satisfying
(25)∥un-u0∥X→0,(n→∞).
Therefore, (X,||·||X) is a Banach space.Lemma 5.
X
∩
C
4
[
0,1
] is compactly embedded in X.Proof.
Let{un}⊂X∩C4[0,1] be bounded. Then, according to the fact that C4[0,1] is compactly embedded in C2[0,1], there exist {unj}⊂{un} and u0∈C2[0,1] such that
(26)∥unj′′-u0′′∥∞→0,j→∞.
From (26), for all ε>0, there exists N>0 such that
(27)-ε≤-(unj′′(t)-u0′′(t))≤ε,t∈[0,1],nj>N.
Since
(28)e(t)≥∫01‍(s-s2)p(s)ds1-∫01‍p(s)ds,t∈[0,1],
combining this fact with (27) we have
(29)-1-∫01‍p(s)ds∫01‍(s-s2)p(s)dsεe(t)≤-(unj′′(t)-u0′′(t))≤1-∫01‍p(s)ds∫01‍(s-s2)p(s)dsεe(t),t∈[0,1],nj>N,
which implies
(30)∥unj-u0∥X≤1-∫01‍p(s)ds∫01‍(s-s2)p(s)dsε→0,nj→∞.
Therefore, X∩C4[0,1] is compactly embedded in X.Let(31)P≔{u∈X∣u(t)≥0,u′′(t)≤0,t∈[0,1]}.
Then, the cone P is normal and has nonempty interior intP.In the rest of this section, we will prove the existence of the first eigenvalue of (3). To wit, we get the following.Theorem 6.
Assume (A1) holds. Then (3) has an algebraically simple eigenvalue μ1, with an eigenfunction φ1∈intP, and there is no other eigenvalue with a positive eigenfunction.Remark 7.
Ifp=q≡0, then μ1 can be explicitly given by
(32)μ1=π4-βπ21+π2
and the corresponding eigenfunction φ1(t)=sinπt, t∈[0,1].Proof of Theorem6.
Foru∈X, define a linear operator T:X→Y by
(33)(Tu)(t)≔∬01K1(t,s)K2(s,τ)[u(τ)-u′′(τ)]dτds.
Then, by the definition of K1, we have
(34)(Tu)(t)=∫01‍{[G1(t,s)+11-∫01‍p(x)dx∫01‍G1(s,x)p(x)dx]×∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ11-∫01‍p(x)dx11-∫01‍p(x)dx}ds=∫0t‍{[s(1-t)+11-∫01‍p(x)dx∫01‍G1(s,x)p(x)dx]×∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ11-∫01‍p(x)dx11-∫01‍p(x)dx}ds+∫t1‍{[t(1-s)+11-∫01‍p(x)dx∫01‍G1(s,x)p(x)dx]×∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ11-∫01‍p(x)dx11-∫01‍p(x)dx}ds.
Therefore, we can obtain
(35)(Tu)′(t)=∫0t‍{-s∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ}ds+∫t1‍{(1-s)∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ}ds,(36)(Tu)′′(t)=-t∫01‍K2(t,τ)[u(τ)-u′′(τ)]dτ+(t-1)∫01‍K2(t,τ)[u(τ)-u′′(τ)]dτ=-∫01‍K2(t,τ)[u(τ)-u′′(τ)]dτ.
We claim thatT:P→P. In fact, for u∈X, let ||u||X=α. Then, from (18) and (17) we have, for t∈[0,1],
(37)-αe(t)≤-u′′(t)≤αe(t),-cαe(t)≤u(t)≤cαe(t).
On the one hand, from (36)-(37) and (10) we have
(38)-(Tu)′′(t)=∫01‍K2(t,s)[u(s)-u′′(s)]ds≤∫01‍K2(t,s)[cα+α]e(s)ds≤c~(cα+α)e(t).
On the other hand, from (36)-(37) and (10) we have
(39)-(Tu)′′(t)=∫01‍K2(t,s)[u(s)-u′′(s)]ds≥-∫01‍K2(t,s)[cα+α]e(s)ds≥-c~(cα+α)e(t).
Therefore, T(X)⊆X. If u∈P, then u(s)≥0, u′′(s)≤0 on [0,1]. According to (36) and the fact that
(40)K2(t,s)≥0,0≤t,s≤1,
we have
(41)(Tu)′′(t)=-∫01‍K2(t,s)[u(s)-u′′(s)]ds≤0.
Furthermore, according to (41), the definition of T, and the fact that
(42)K1(t,s)≥0,0≤t,s≤1,
we have
(43)(Tu)(t)=∫01‍{K1(t,s)∫01‍K2(s,τ)[u(τ)-u′′(τ)]dτ}ds=∫01‍K1(t,s)[-(Tu)′′(s)]ds≥0.
Then u∈P, and accordingly, T(P)⊆P.
Now, sinceT(X)⊂X∩C4[0,1] and X∩C4[0,1] is compactly embedded in X, we have that T:X→X is compact.
Next, we show thatT:P→P is strongly positive.
Foru∈P∖{θ}, it is easy to check that there exist ρ1u,ρ2u>0, such that for t∈[0,1](44)(Tu)(t)≥ρ1ue(t),-(Tu)′′(t)≥ρ2ue(t).
In fact, for every u∈P∖{θ}, it follows from Lemma 2 that
(45)(Tu)(t)=∬01[G1(t,s)+11-∫01‍p(x)dx∫01‍G1(s,x)p(x)dx]×[G2(s,τ)+ρ(s)∫01‍G2(τ,x)q(x)dx]×[u(τ)-u′′(τ)]dτds≥∬01[G1(s,s)1-∫01‍p(x)dx∫01‍G1(s,s)1-∫01‍p(x)dxG1(t,t)G1(s,s)+G1(s,s)1-∫01‍p(x)dx∫01‍G1(x,x)p(x)dx]×[G2(s,τ)+ρ(s)∫01‍G2(τ,x)q(x)dx]×[u(τ)-u′′(τ)]dτds=ρ1ue(t),
where
(46)ρ1u=∬01G1(s,s)×[G2(s,τ)+ρ(s)∫01‍G2(τ,x)q(x)dx]×[u(τ)-u′′(τ)]dτds>0,(47)-(Tu)′′(t)=∫01‍[G2(t,s)+ρ(t)∫01‍G2(s,x)q(x)dx]×[u(s)-u′′(s)]ds≥βsinβ∫01‍[∫01‍G2(t,t)G2(s,s)+ρ(t)G2(s,s)∫01‍G2(x,x)q(x)dx]×[u(s)-u′′(s)]ds=βsinβ[G2(t,t)+ρ(t)∫01‍G2(x,x)q(x)dx]×[∫01‍G2(s,s)[u(s)-u′′(s)]ds]≥ρ(t)[∫01‍sinβxsinβ(1-x)q(x)dx]×[∫01‍G2(s,s)[u(s)-u′′(s)]ds]≥ρ2ue(t),
where
(48)ρ2u=mint∈[0,1]ρ(t)[∫01‍∫01‍sinβxsinβ(1-x)q(x)dx×∫01‍G2(s,s)[u(s)-u′′(s)]ds]×(14+∫01‍p(s)(s-s2)ds1-∫01‍p(s)ds)-1>0.
Therefore, it follows from (44) that Tu∈intP.
Now, on the one hand, by the Krein-Rutman theorem ([17, Theorem 7.C] and [18, Theorem 19.3]), T has an algebraically simple eigenvalue r(T)>0 with an eigenfunction φ1∈intP. Moreover, there is no other eigenvalue with a positive eigenfunction. On the other hand, we have from the definition of T and Lemma 3 that (3) is equivalent to the integral equation
(49)u(t)=μ(Tu)(t).
Therefore, μ1=(r(T))-1 with a positive eigenfunction φ1 is a simple eigenvalue of (3). Moreover, for (3), there is no other eigenvalue with a positive eigenfunction.


## 3. An Application of Theorem6

For convenience, we introduce the following notations:(50)f¯∞=limsup|u|+|v|→+∞maxt∈[0,1]f(t,u,v)|u|+|v|,f_0=liminf|u|+|v|→0+mint∈[0,1]f(t,u,v)|u|+|v|,f¯0=limsup|u|+|v|→0+maxt∈[0,1]f(t,u,v)|u|+|v|,f_∞=liminf|u|+|v|→+∞mint∈[0,1]f(t,u,v)|u|+|v|,Br={u∈X∣∥u∥X<r},∂Br={u∈X∣∥u∥X=r}(r>0).Theorem 8.
Assume that (A1) and (A2) hold, and0≤f¯∞<f_0≤+∞. Then, for each λ satisfying
(51)μ1f_0<λ<μ1f¯∞,
there exists at least one positive solution of (2).Theorem 9.
Assume that (A1) and (A2) hold, and0≤f¯0<f_∞≤+∞. Then, for each λ satisfying
(52)μ1f_∞<λ<μ1f¯0,
there exists at least one positive solution of (2).Remark 10.
Bai [3] proved existence of positive solutions via Guo-Krasnoselskii fixed point theorem under some conditions which do not involve the eigenvalue of (3). While our Theorems 8 and 9 are established under (51) or (52) which are related to the eigenvalue of (3). Our Theorems 8 and 9 cover an undefined case in [3]. Consider the following boundary value problem:
(53)u(4)(t)+π24u′′(t)=3π4+14(1+π2)[u(t)-u′′(t)]-24(1+π2)sin[u(t)-u′′(t)],t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0.
In this case, μ1=3π4/4(1+π2), f(t,u,v)=((3π4+1)/4(1+π2))(u-v)-(2/4(1+π2))sin(u-v), λ=1, and
(54)μ1f_∞=3π43π4+1<1=λ<3π43π4-1=μ1f¯0.
According to Theorem 9, the above boundary value problem has at least one positive solution. For the above boundary value problem, however, we cannot obtain the above conclusion by [3, Theorem 3.1] since
(55)f_∞=3π4+14(1+π2)<20<36<π3(4+22+2)22=η1λ.
Moreover, (51) and (52) are optimal. In order to illustrate this point, consider the problem
(56)u(4)(t)+βu′′(t)=π4-βπ21+π2[u(t)-u′′(t)]+ρ(u(t),u′′(t)),t∈(0,1),u(0)=u(1)=u′′(0)=u′′(1)=0,
where
(57)ρ(x,y)={(x-y)2,0≤x-y≤1,x-y,1≤x-y<∞.
In this case, μ1=(π4-βπ2)/(1+π2), f(t,u,v)=((π4-βπ2)/(1+π2))[u-v]+ρ(u,v), λ=1, and
(58)μ1f_0=λ=μ1f¯∞,μ1f_∞=λ=μ1f¯0.
However, (56) has no positive solution. In fact, suppose on the contrary that (56) has a positive solution u. Multiplying the first equation of (56) with sinπt and integrating from 0 to 1, we get
(59)0=∫01‍ρ(u(t),u′′(t))sinπtdt>0,
which is a contradiction.To prove Theorems8 and 9, we need the following preliminary results.Lemma 11.
For everyu∈X, there exist L0,L1>0 such that
(60)L0∥u∥∞≤∥u′∥∞≤∥u′′∥∞≤L1∥u∥X.Proof.
(
1
) By u(0)=∫01‍p(s)u(s)ds, there is u(t)=∫0t‍u′(s)ds+∫01‍p(s)u(s)ds, so
(61)|u(t)|≤∫01‍|u′(s)|ds+∫01‍p(s)|u(s)|ds≤∥u′∥∞+∥u∥∞∫01‍p(s)ds.
Hence
(62)L0∥u∥∞≤∥u′∥∞,
where L0=1-∫01‍p(s)ds.
(
2
) By u(0)=u(1), there is a ξ∈(0,1) such that u′(ξ)=0, and so, for t∈[0,1](63)-u′(t)=∫tξ‍u′′(s)ds.
Hence |u′(t)|≤|∫tξ‍|u′′(s)|ds|≤∫01‍|u′′(s)|ds≤||u′′||∞, and accordingly,
(64)∥u′∥∞≤∥u′′∥∞.
(
3
) We have from (18) that |u′′(t)|≤||u||Xe(t), which implies that
(65)|u′′(t)|≤(14+∫01‍p(s)(s-s2)ds1-∫01‍p(s)ds)∥u∥X,
and consequently
(66)∥u′′∥∞≤L1∥u∥X,
where L1=(1/4)+(∫01‍p(s)(s-s2)ds/(1-∫01‍p(s)ds)).Let(67)(Au)(t)≔λ∬01K1(t,s)K2(s,τ)f(τ,u(τ),u′′(τ))dτds,t∈[0,1],λ>0.
It is easy to show that A:P→P is a completely continuous operator. In addition, we can verify that the nonzero fixed points of the operator A are positive solutions of the problem (2).Lemma 12 (see [19]).
LetE be Banach space, P a cone in E, and Ω(P) a bounded open set in P. Suppose that A:Ω(P)¯→P is a completely continuous operator. If there exists u0∈P∖{θ} such that
(68)u-Au≠μu0,∀u∈∂Ω(P),μ≥0,
then the fixed point index i(A,Ω(P),P)=0.Lemma 13 (see [19]).
LetE be Banach space, P a cone in E, and Ω(P) a bounded open set in P with θ∈Ω(P). Suppose that A:Ω(P)¯→P is a completely continuous operator. If
(69)Au≠μu,∀u∈∂Ω(P),μ≥1,
then the fixed point index i(A,Ω(P),P)=1.Proof of Theorem8.
It follows from the first inequality of (51) that there exists r0>0, such that
(70)λf(t,u,v)≥μ1(|u|+|v|),∀t∈[0,1],0≤|u|+|v|≤r0.
Let φ1 be the positive eigenfunction of T (see (33)) corresponding to μ1. Thus φ1=μ1Tφ1.
Letr1=L0r0/(1+L0)L1. Then for every u∈∂Br1∩P, we have from Lemma 11 that
(71)0≤u(t)-u′′(t)≤∥u∥∞+∥u′′∥∞≤(1+L0)L0∥u∥X≤r0.
It follows from (70) that
(72)(Au)(t)≥μ1∬01K1(t,s)K2(s,τ)[u(τ)-u′′(τ)]dτds=μ1(Tu)(t),t∈[0,1].
We may suppose thatA has no fixed points on ∂Br1∩P (otherwise, the proof is finished). Now we show that
(73)u-Au≠μφ1,∀u∈∂Br1∩P,μ≥0.
Suppose the contrary, that there existsu1∈∂Br1∩P and τ1≥0 such that u1-Au1=τ1φ1. Hence τ1>0 and
(74)u1=Au1+τ1φ1≥τ1φ1.
Put
(75)τ*≔sup{τ∣u1≥τφ1}.
It is easy to see that τ*≥τ1>0 and u1≥τ*φ1. We find from T(P)⊂P that
(76)μ1Tu1≥τ*μ1Tφ1=τ*φ1.
Therefore, by (72), we have
(77)u1=Au1+τ1φ1≥μ1Tu1+τ1φ1≥τ*φ1+τ1φ1,
which contradicts the definition of τ*. Hence (73) is true, and we have from Lemma 12 that
(78)i(A,Br1∩P,P)=0.
It follows from the second inequality of (51) that there exists 0<σ<1 and r2>r1 such that
(79)λf(t,u,v)≤σμ1(|u|+|v|),∀t∈[0,1],|u|+|v|≥r2.
LetT1u=σμ1Tu, u∈X. Then T1:X→X is a bounded linear operator and T1(P)⊂P.
Let(80)M=[max0≤t,s,τ≤1K1(t,s)K2(s,τ)]×[supu∈B¯r3∩Pλ∫01‍f(τ,u(τ),u′′(τ))dτ],
where r3=L0r2/(1+L0)L1. It is clear that M<+∞.
Let(81)W={u∈P∣u=μAu,0≤μ≤1}.
In the following, we prove that W is bounded.
For anyu∈W, set E(τ)={τ∈[0,1]∣u(τ)-u′′(τ)≥r2}. Then,
(82)u(t)=μ(Au)(t)≤λ∬01K1(t,s)K2(s,τ)f(τ,u(τ),u′′(τ))dτds=λ∫01‍∫E(τ)‍K1(t,s)K2(s,τ)f(τ,u(τ),u′′(τ))dτds+λ∫01‍∫[0,1]∖E(τ)‍K1(t,s)K2(s,τ)f×(τ,u(τ),u′′(τ))dτds≤σμ1∬01K1(t,s)K2(s,τ)[u(τ)-u′′(τ)]dτds+λ∫01‍∫[0,1]∖E(τ)‍K1(t,s)K2(s,τ)f×(τ,u(τ),u′′(τ))dτds≤(T1u)(t)+M,t∈[0,1].
Thus, ((I-T1)u)(t)≤M, for every t∈[0,1]. Since μ1 is the first eigenvalue of T and 0<σ<1, the first eigenvalue of T1, (r(T1))-1>1. Therefore, the inverse operator (I-T1)-1 exists and
(83)(I-T1)-1=I+T1+T12+⋯+T1n+⋯.
It follows from T1(P)⊂P that (I-T1)-1(P)⊂P. So, we have u(t)≤(I-T1)-1M, t∈[0,1], and we conclude that W is bounded.
Selectr4>max{r1,supW}. Then, from the invariance property of the fixed point index, we have
(84)i(A,Br4∩P,P)=i(θ,Br4∩P,P)=1.
By (78) and (84), we have that
(85)i(A,(Br4∩P)∖(B¯r1∩P),P)=i(A,Br4∩P,P)-i(A,Br1∩P,P)=1.
Then A has at least one fixed point on (Br4∩P)∖(B¯r1∩P). This means that the boundary value problem (2) has at least one positive solution.Proof of Theorem9.
It follows from the second inequality of (52) that there exists r0>0, such that
(86)λf(t,u,v)≤μ1(|u|+|v|),∀t∈[0,1],0≤|u|+|v|≤r0.
Let r1=L0r0/(1+L0)L1. Then, for every u∈∂Br1∩P, we have from Lemma 11 that
(87)0≤u(t)-u′′(t)≤∥u∥∞+∥u′′∥∞≤(1+L0)L1L0∥u∥X≤r0.
It follows from (86) that
(88)(Au)(t)≤μ1∬01K1(t,s)K2(s,τ)[u(τ)-u′′(τ)]dτds=μ1(Tu)(t),t∈[0,1].
Suppose there exists u2∈∂Br1∩P and τ2≥1 satisfying Au2=τ2u2. We may suppose that A has no fixed points on ∂Br1∩P (otherwise, the proof is finished). So τ2>1. By (88), we have τ2u2=Au2≤μ1Tu2. By induction, we have τ2nu2≤μ1nTnu2, for all n∈N. Thus
(89)∥Tn∥≥∥Tnu2∥X∥u2∥X≥τ2n∥u2∥Xμ1n∥u2∥X=τ2nμ1n.
By Gelfand’s formula, we have
(90)r(T)=limn→∞∥Tn∥n≥τ2μ1>1μ1,
which is a contradiction with r(T)=1/μ1. Hence
(91)Au≠τu,∀u∈∂Br1∩P,τ≥1,
and we have from Lemma 13 that
(92)i(A,Br1∩P,P)=1.
It follows from the first inequality of (52) that there exist ε>0 and R0>0 such that
(93)λf(t,u,v)≥(μ1+ε)(|u|+|v|),|u|+|v|≥R0.
From (A2), we know that there exists d≥0 such that
(94)λf(t,u,v)≥(μ1+ε)(|u|+|v|)-d,t∈[0,1],0≤|u|+|v|<+∞.
We can takeR>r1 large enough, such that
(95)ρ1u≥d∬01K1(s,s)K2(s,τ)dτdsε(∫01‍(s-s2)p(s)ds/(1-∫01‍p(s)ds)),
where ρ1u is defined as in (46). Then by (94) and (33), for all u∈P,  ∥u∥X≥R, t∈[0,1], we have
(96)(Au)(t)=λ∬01K1(t,s)K2(s,τ)f(τ,u(τ),u′′(τ))dτds≥(μ1+ε)∬01K1(t,s)K2(s,τ)×[u(τ)-u′′(τ)]dτds-d∬01K1(t,s)K2(s,τ)dτds≥μ1(Tu)(t)+ερ1ue(t)-d∬01K1(s,s)K2(s,τ)dτds≥μ1(Tu)(t)+ερ1u∫01‍(s-s2)p(s)ds1-∫01‍p(s)ds-d∬01K1(s,s)K2(s,τ)dτds≥μ1(Tu)(t).
Hence, for everyu∈P,  ∥u∥X≥R, we have
(97)(Au)(t)≥μ1(Tu)(t),t∈[0,1].
Let φ1 be the positive eigenfunction of T corresponding to μ1. Thus φ1=μ1Tφ1.
We may suppose thatA has no fixed points on ∂BR∩P (otherwise, the proof is finished). Now, we show that
(98)u-Au≠τφ1,∀u∈∂BR∩P,τ≥0.
Suppose the contrary, that there existsu3∈∂BR∩P and τ3≥0 such that
(99)u3-Au3=τ3φ1.
Hence τ3>0 and
(100)u3=Au3+τ3φ1≥τ3φ1.
Put
(101)τ*:=sup{τ∣u3≥τφ1}.
It is easy to see that τ*≥τ3>0 and u3≥τ*φ1. We find from T(P)⊂P that
(102)μ1Tu3≥τ*μ1Tφ1=τ*φ1.
Therefore by (97), we have
(103)u3=Au3+τ3φ1≥μ1Tu3+τ3φ1≥τ*φ1+τ3φ1,
which contradicts the definition of τ*. Hence (98) is true, and we have from Lemma 12 that
(104)i(A,BR∩P,P)=0.
By (92) and (104), we have that
(105)i(A,(BR∩P)∖(B¯r1∩P),P)=i(A,BR∩P,P)-i(A,Br1∩P,P)=-1.
Then, A has at least one fixed point on (BR∩P)∖(B¯r1∩P). This means that the boundary value problem (2) has at least one positive solution.



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*Source: 932145-2013-03-24.xml*