# Positive Solutions for a Fourth-Order Boundary Value Problem

Authors: Kun Wang; Zhilin Yang
Journal: Journal of Mathematics
 (2013)

---

## Abstract

This paper deals with the existence and multiplicity of positive solutions for the
fourth-order boundary value problemu(4)=f(t,u,u′,−u′′,u′′′),u(0)=u′(1)=u′′′(0)=u′′(1)=0. Here f∈C([0,1]×ℝ+4,ℝ+)(ℝ+:=[0,+∞)). We use fixed point index theory to establish our main results based on a priori estimates achieved by utilizing some integral identities and integral inequalities.


---

## Body

## 1. Introduction

The purpose of this paper is to study the existence and multiplicity of positive solutions for the fourth-order boundary value problem(1)u(4)=f(t,u,u′,-u′′,u′′′),u(0)=u′(1)=u′′′(0)=u′′(1)=0,
where f∈C([0,1]×ℝ+4,ℝ+). By a positive solution of (1), we mean a function u∈C4([0,1],ℝ+) that solves (1) and satisfies u(t)>0 for all t∈(0,1].In recent years, fourth-order boundary value problems for nonlinear ordinary differential equations have been extensively studied by using diverse methods, including fixed point theorems on cones [1–6], the method of lower and upper solutions [7–10], the iterative method [11–14], critical point theory [15], and the shooting method [16]. It should be remarked that none of the results cited above involves derivatives of all orders in their nonlinearities. In [9], Ma and Yang studied the fourth-order four-point boundary value problem
(2)u(4)=f(t,u,-u′′),u(0)=0,u(1)=au(η),u′′(0)=0,u′′(1)=bu(ξ),
where η,ξ∈(0,1), a,b⩾0. The authors used the method of upper and lower solutions and a new maximum principle to establish their existence results. In [14], Pei and Chang studied a class of fourth-order boundary value problem
(3)y(4)=f(t,y),y(0)=y(1)=y′(0)=y′(1)=0.
By using the monotone iterative technique, the authors proved that problem (3) has at least one symmetric positive solution under certain conditions. In [17], Pang et al. studied the existence and multiplicity of nontrivial solutions for the fourth-order boundary value problem(4)x(4)=f(x,-x′′),x(0)=x(1)=x′′(0)=x′′(1)=0,
where f∈C(ℝ2,ℝ). Making use of the theory of Leray-Schauder degree, under appropriate conditions on the nonlinearity f, the authors proved that problem (4) has at least six different nontrivial solutions, including two positive ones and two negative ones.In [18], Yang and Sun studied the fourth-order boundary value problem
(5)u(4)=f(t,u,u′,-u′′,-u′′′),u(0)=u′(1)=u′′(0)=u′′′(1)=0,
where f∈C([0,1]×ℝ+4,ℝ+). By using fixed point index theory based on a priori estimates, the authors established the main results on the existence, multiplicity and uniqueness of positive solutions for (5).Motivated by [18], in this paper, we will discuss the existence and multiplicity of positive solutions for problem (1). As in [18], we first use the method of order reduction to transform (1) into a boundary value problem for a second-order integro-differential equation and then seek the existence and multiplicity of positive solutions for the resultant problem. To overcome the difficulties stemming from the presence of derivatives of all orders and the difference between (1) and (5) in their boundary value conditions, we have to prove that the maximum of every nonnegative concave function v can be dominated by the integrals(6)∫01v(t)(1-t)e1-tdt,∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt
(see Lemmas 5 and 6 below for more details). Based on a priori estimates achieved by utilizing some integral identities and integral inequalities, we use fixed point index theory to prove the existence and multiplicity of positive solutions for (1).This paper is organized as follows. In Section2, we transform (1) into a boundary value problem for a second-order integro-differential equation and then establish some basic integral identities and integral inequalities that are useful in deriving the priori estimates in the next section. Our main results, namely Theorems 10–12, are stated and proved in Section 3.


## 2. Preliminaries

LetE:=C1([0,1],ℝ) and ∥u∥:=max{∥u∥0,∥u′∥0}, where ∥u∥0:=max{|u(t)|:t∈[0,1]}. Furthermore, let
(7)P:={u∈E:u(t)⩾0,u′(t)0,∀t∈[0,1]}.
Clearly (E,∥·∥) is a real Banach space and P is a cone in E.Let(8)k1(t,s):=min{t,s},k2(t,s):=min{1-s,1-t}.Define the linear integral operatorsB1 and B2 by
(9)(Biv)(t):=∫01‍ki(t,s)v(s)ds,v∈P(i=1,2).
Substituting v(t):=-u′′(t) into (1), we transform (1) into the following second-order boundary value problem for the integro-differential equation
(10)-v′′(t)=f(t,(B1v)(t),(B1v)′(t),v(t),-v′(t)),v′(0)=v(1)=0,
which is equivalent to the nonlinear integral equation
(11)v(t)=∫01k2(t,s)f(s,(B1v)(s),(B1v)′(s),v(s),-v′(s))ds.Define the operatorA:P→P by
(12)(Av)(t):=∫01k2(t,s)f(s,(B1v)(s),(B1v)′(s),v(s),-v′(s))ds,v∈P.Nowf∈C([0,1]×ℝ+4,ℝ+) implies that A is a completely continuous operator. In our setting, the existence of positive solutions for (1) is equivalent to that of positive fixed points of A:P→P.To establish the priori estimates of positive solutions for some problems associated with (11), we need several integral identities and integral inequalities below.Lemma 1.
Ifv∈P∩C2[0,1],v′(0)=v(1)=0, then
(13)∫01-v′′(t)(1-t)e1-tdt=∫01(v(t)-2v′(t))(1-t)e1-tdt.Proof.
Integrating by parts and usingv′(0)=v(1)=0, we have
(14)∫01(-v′′(t))(1-t)e1-tdt=∫01v′(t)(t-2)e1-tdt=∫01(-v′(t))te1-tdt-2∫01v′(t)(1-t)e1-tdt,(15)∫01-v′(t)te1-tdt=∫01‍v(t)(1-t)e1-tdt,
from which (13) follows. This completes the proof.Lemma 2.
Ifv∈C([0,1],ℝ+), then
(16)∫01((B1v)(t)+2(B1v)′(t)tetdt)=∫01‍v(t)tetdt.Proof.
Notice(B1v)(0)=0. Integrating by parts, we obtain
(17)∫01(B1v)(t)tetdt=∫01(1-t)et(B1v)′(t)dt,
so that
(18)∫01((B1v)(t)+2(B1v)′(t))tetdt=∫01(1+t)et(B1v)′(t)dt.
Notice (B1v)′(1)=0 and (B1v)′′(t)=-v(t). Integrating the right hand of the above by parts again gives (16). This completes the proof.If bothh1 and h2 are continuous and decreasing on [0,1], then Chebyshev's inequality for h1 and h2 is given by
(19)∫01h1(t)h2(t)dt⩾∫01h1(t)dt∫01h2(t)dt.
In addition, if both h1 and h2 are continuous, with one of them decreasing on [0,1] and the other increasing on [0,1], then Chebyshev's inequality for h1 and h2 becomes
(20)∫01h1(t)h2(t)dt⩽∫01h1(t)dt∫01h2(t)dt.Lemma 3.
Ifv∈P,v(1)=0, and v′ is decreasing on [0,1], then
(21)∫01v(t)(1-t)e1-tdt⩾(e-2)∫01-v′(t)(1-t)e1-tdt.Proof.
Noticev(1)=0 and v′⩽0. Since -v′ is increasing on [0,1] and (1-t)e1-t is decreasing on [0,1], it follows from Chebyshev's inequality that
(22)∫01-v′(t)(1-t)e1-tdt⩽∫01-v′(t)dt∫01(1-t)e1-tdt=v(0).
Notice thatte1-t is increasing on [0,1]. Chebyshev's inequality again implies
(23)∫01-v′(t)te1-tdt⩾∫01-v′(t)dt∫01te1-tdt=(e-2)v(0).
Note that (15) is guaranteed by v∈P,v(1)=0. By virtue of (15), we have
(24)∫01v(t)(1-t)e1-tdt⩾(e-2)v(0).
Combining this with (22) yields (21). This complete the proof.Lemma 4.
Ifv∈P,v(1)=0, then
(25)v(0)⩽∫01‍(v(t)-2v′(t))(1-t)e1-tdt.Proof.
By (15), we have
(26)∫01(v(t)-2v′(t))(1-t)e1-tds=∫01-v′(t)(2-t)e1-tdt⩾∫01-v′(t)dt=v(0).
This completes the proof.Lemma 5.
Ifv∈P∩C2[0,1] and v′′⩽0, then
(27)(e-2)∥v∥0⩽∫01v(t)(1-t)e1-tdt⩽e∥v∥0.Proof.
It is easy to see that∫01‍v(t)tetdt⩽e∥v∥0. The condition v′⩽0 implies ∥v∥0=v⩽(0). By the concavity of v, we have
(28)∫01v(t)(1-t)e1-tdt⩾∫01(1-t)e1-tv(t·1+(1-t)·0)dt⩾v(0)∫01(1-t)2e1-tdt=(e-2)∥v∥0.
This completes the proof.Lemma 6.
Ifv∈P∩C2[0,1] and v′′⩽0, then we have
(29)3-e6∥v∥0⩽∫01‍((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt⩽3e∥v∥0.Proof.
By the concavity ofB1v and v, we have
(30)∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt⩾∫01(B1v)(t)(1-t)e1-tdt=∫01(B1v)(t·1+(1-t)·0)(1-t)e1-tdt⩾(B1v)(1)∫01‍t(1-t)e1-tdt=(3-e)∫01sv(s)ds⩾(3-e)∫01s(1-s)v(0)ds=3-e6∥v∥0,∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt⩽e∫01(B1v)(t)+2(B1v)′(t)dt=e∫01(∫01k1(t,s)v(s)ds+2∫01v(s)ds)dt⩽3e∥v∥0.
This completes the proof.Lemma 7 (see [19]).
LetE be a real Banach space and P a cone in E. Suppose that Ω⊂E is a bounded open set and that T:Ω¯∩P→P is a completely continuous operator. If there exists ω0∈P∖{0}  such that
(31)ω-Tω≠λω0,∀λ⩾0,ω∈∂Ω∩P,
then i(T,Ω∩P,P)=0, where i indicates the fixed point index on P.Lemma 8 (see [19]).
LetE be a real Banach space and P a cone in E. Suppose that Ω⊂E is a bounded open set with 0∈Ω and that T:Ω¯∩P→P is a completely continuous operator. If
(32)ω-λTω≠0,∀λ∈[0,1],ω∈∂Ω∩P,
then i(T,Ω∩P,P)=1.


## 3. Existence and Multiplicity of Positive Solutions for (1)

For simplicity, we denote byx:=(x1,x2,x3,x4)∈ℝ+4 and Iρ:=[0,ρ] for ρ>0 in this section. Now we list our hypotheses.(H1)
f∈C([0,1]×ℝ+4,ℝ+).(H2)
There exista1⩾0, b1⩾0, c>0, such that d1:=a1(3-e(e-2))/6e2+b1>1, and
(33)f(t,x)⩾a1(x1+2x2)+b1(x3+2x4)-c
holds for all (t,x)∈[0,1]×ℝ+4.(H3)
There exists a functionΦ∈C(ℝ+,ℝ+) such that
(34)f(t,x)⩽Φ(x4)
for all x∈[0,1]×Im3×ℝ+ and
(35)∫0∞ξdξΦ(ξ)+2me>m.
where m:=c/(d1-1)>0 and c,d1 are given as in (H2).(H4)
There exista2⩾0, b2⩾0, r>0 such that d2:=3a2e/(e-2)+b2<1, and
(36)f(t,x)⩽a2(x1+2x2)+b2(x3+2x4)
holds for all (t,x)∈[0,1]×Ir3.(H5)
There exista3⩾0, b3⩾0, c>0 such that d3:=3a3e∕(e-2)+b2<1, and
(37)f(t,x)⩽a3(x1+2x2)+b3(x3+2x4)+c
holds for all (t,x)∈[0,1]×ℝ+4.(H6)
There are nonnegative constantsa4⩾0, b4⩾0, r>0, such that d4:=a4(3-e)(e-2)/6e2+b4>1, and
(38)f(t,x)⩾a4(x1+2x2)+b4(x3+2x4)
holds for all (t,x)∈[0,1]×Ir4.(H7)
There is a constantω>0 such that f(t,x) is increasing on Iω4 in x, and
(39)∫01f(s,ω,ω,ω,ω)ds<ω.Remark 9.
f
(
t
,
x
) is said to be increasing in x if f(t,x)⩽f(t,y) holds for every pair x,y∈ℝ+4 with x⩽y and for all t∈[0,1], where the partial ordering in ⩽ℝ4 is understood componentwise.We denoteΩρ:={v∈E:∥v∥<ρ} for ρ>0 in the sequel.Theorem 10.
If (H1)–(H4) hold, then (1) has at least one positive solution.Proof.
Let(40)ℳ1:={v∈P:v=Av+μφ,μ⩾0},
where φ(t):=(1-t)et-1∈P. We are now going to prove that ℳ1 is bounded. Indeed, if v∈ℳ1, then there exist μ⩾0 such that v=Av+μφ, which can be written in the form
(41)v(t)=∫01k2(t,s)f({(B1v)′(s)}s,(B1v)(s),(B1v)′(s),v(s),-v′(s))ds+μφ(t),
which is equivalent to
(42)-v′′(t)=f((t),(B1v)(t),(B1v)′(t),v(t),-v′(t))+μ(1+t)et-1.
By (H2), we have
(43)-v′′(t)⩾a1((B1v)(t)+2(B1v)′(t))+b1(v(t)-2v′(t))-c.
Note (27) and (29). Multiply the above by (1-t)e1-t, integrate over [0,1], and use Lemmas 3, 5, and 6 to obtain
(44)∫01(v(t)-2v′(t))(1-t)e1-tdt⩾a1∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt+b1∫01(v(t)2v′(t))(1-t)e1-tdt-c⩾a1(3-e)6∥v∥0+b1∫01(v(t)-2v′(t))(1-t)e1-tdt-c⩾a1(3-e)6e∫01v(t)(1-t)e1-tdt+b1∫01(v(t)-2v′(t))(1-t)e1-tdt-c⩾d1∫01(v(t)-2v′(t))(1-t)e1-tdt-c,
so that
(45)∫01(v(t)-2v′(t))(1-t)e1-tdt⩽m,∀v∈ℳ1.
Now Lemma 4 implies
(46)∥v∥0=v(0)⩽m,∀v∈ℳ1,
where m is given in (H3). Furthermore, this estimate leads to
(47)∥(B1v)(t)∥0=(B1v)(1)=∫01sv(s)ds⩽m,∥(B1v)′(t)∥0=(B1v)′(0)=∫01v(s)ds⩽m
for all v∈ℳ1. Let
(48)Λ:={μ⩾0:thereexistsomev∈Psuchthatv=Av+μφ}.
Now (46) implies that μ⩽me for all μ∈Λ. By (H3), there is a function Φ∈C(ℝ+,ℝ+) such that
(49)f(t,(B1v)(t),(B1v)′(t),v(t),-v′(t))⩽Φ(-v′(t)),∀v∈ℳ1.
Thus
(50)-v′′(t)=f(t,(B1v)(t),(B1v)′(t),v(t),-v′(t))+μ(1+t)et-1⩽Φ(-v′(t))+μ(1+t)et-1⩽Φ(-v′(t))+2me,∀v∈ℳ1,μ∈Λ,∫0-v′(1)ξdξΦ(ξ)+2me⩽∫01-v′(t)dt=v(0)⩽m,∀v∈ℳ1.
Combining the last inequality with (35), we see that there exists M>0 such that -v(1)⩽M for all v∈ℳ1. This means that ℳ1 is bounded. Taking R>sup{∥v∥:v∈ℳ1}, we have
(51)v≠Av+μφ,∀v∈ΩR∩P,μ≥0.
Now Lemma7 yields
(52)i(A,ΩR∩P,P)=0.
Let(53)ℳ2:={v∈Ω¯r∩P:v=λAv,λ∈[0,1]}.
Now we want to prove thatℳ2={0}. In fact, if v∈ℳ2, then v∈Ω¯r∩P and there is λ∈[0,1] such that v=λAv, that is,
(54)v(t)=λ∫01k2(t,s)×f(s,(B1v)(s),(B1v)′(s),v(s),-v′(s))ds,
which can be written in the form(55)-v′′(t)=λf(t,(B1v)(t),(B1v)′(t),v(t),-v′(t)).
By (H4), we have(56)-v′′(t)⩽a2((B1v)(t)+2(B1v)′(t))+b2(v(t)-2v′(t)).
Note (27) and (29). Multiply (56) by (1-t)e1-t, integrate over [0,1], and use Lemmas 5 and 6 to obtain
(57)∫01(v(t)-2v′(t))(1-t)e1-tdt⩽a2∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt+b2∫01(v(t)-2v′(t))(1-t)e1-tdt⩽3ea2∥v∥0+b2∫01(v(t)-2v′(t))(1-t)e1-tdt⩽3ea2e-2∫01v(t)(1-t)e1-tdt+b2∫01(v(t)-2v′(t))(1-t)e1-tdt⩽d2∫01(v(t)-2v′(t))(1-t)e1-tdt,
so that ∫01‍(v(t)-2v′(t))(1-t)e1-t=0, whence v(t)≡0 and ℳ2={0}, as required. A consequence of this is
(58)v≠λAv,∀v∈∂Ωr∩P,λ∈[0,1].
Now Lemma 8 yields
(59)i(A,Ωr∩P,P)=1.
This together with (52) implies
(60)i(A,(ΩR∖Ω¯r)∩P,P)=0-1=-1.
Therefore A has at least one fixed point on (ΩR∖Ω¯r)∩P and thus (1) has at least one positive solution. This completes the proof.Theorem 11.
If (H1), (H5), and (H6) hold, then (1) has at least one positive solution.Proof.
Let(61)ℳ3:={v∈P:v=λAv,λ∈[0,1]}.
We now assert thatℳ3 is bounded. Indeed, if v∈ℳ3, then there is λ∈[0,1] such that v=λAv, which is equivalent to
(62)-v′′(t)=λf(t,(B1v)(t),(B1)′(t),v(t),-v′(t)).
Now (H5) implies(63)-v′′(t)⩽a3((B1v)(t)+2(B1v)′(t))+b3(v(t)-2v′(t))+c.
Note (27) and (29). Multiply the above by (1-t)e1-t, integrate over [0,1], and use Lemmas 5 and 6 to obtain
(64)∫01(v(t)-2v′(t))(1-t)e1-tdt⩽a3∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt+b3∫01(v(t)-2v′(t))(1-t)e1-tdt+c⩽3ea3∥v∥0+b3∫01(v(t)-2v′(t))(1-t)e1-tdt+c⩽d3∫01(v(t)-2v′(t))(1-t)e1-tdt+c,
so that ∫01‍(v(t)-2v′(t))(1-t)e1-tdt⩽c1 for all v∈ℳ3, where c1:=c/(1-d3)>0. By Lemma 4, we obtain
(65)∥v∥0=v(0)⩽∫01(v(t)-2v′(t))(1-t)e1-tdt⩽c1,∀v∈ℳ3.
Furthermore, this estimate leads to(66)∥B1v∥0=(B1v)(1)=∫01sv(s)ds⩽c1,∥(B1v)′∥0=(B1v)′(0)=∫01v(s)ds⩽c1
for all v∈ℳ3. By (63), we have
(67)-v′′(t)⩽(3a3+b3)c1-2b3v′(t)+c,∀v∈ℳ3.
Notev(1)=v′(0)=0 and v′′⩽0. Integrating the above over [0,1], we obtain
(68)∥v′∥0=-v′(1)⩽(3a3+b3)c1+2b3v(0)+c⩽3(a3+b3)c1+c.
This proves the boundedness ofℳ3. Taking R>sup{∥v∥:v∈ℳ3}, we have
(69)v≠λAv,∀v∈∂ΩR∩P,λ∈[0,1].
Now Lemma8 yields
(70)i(A,ΩR∩P,P)=1.
On the other hand, let
(71)ℳ4:={v∈Ω¯r∩P:v=Av+λφ,λ⩾0},
where φ∈P∖{0} is given by φ(t):=(1-t)et-1. Now we want to prove that ℳ4⊂{0}. Indeed, if v∈ℳ4, then there exists λ⩾0 such that v=Av+λφ, which can be written as
(72)-v′′(t)=f(t,(B1v)(t),(B1v)′(t),v(t),-v′(t))+λ(t+1)et-1.
By (H6), we have
(73)-v′′(t)a4⩾((B1v)(t)+2(B1v)′(t))+b4(v(t)-2v′(t)).
Note (27) and (29). Multiply the above by (1-t)e1-t, integrate over [0,1], and use Lemmas 3, 5, and 6 to obtain
(74)∫01(v(t)-2v′(t))(1-t)e1-tdt⩾a4∫01((B1v)(t)+2(B1v)′(t))(1-t)e1-tdt+b4∫01(v(t)-2v′(t))(1-t)e1-tdt,⩾a4(3-e)6e∥v∥0+b4∫01(v(t)-2v′(t))(1-t)e1-tdt,⩾a4(3-e)6e∫01v(t)(1-t)e1-tdt+b4∫01(v(t)-2v′(t))(1-t)e1-tdt,⩾d4∫01(v(t)-2v′(t))(1-t)e1-tdt,
so that ∫01‍(v(t)-2v′(t))(1-t)e1-tdt=0, whence v(t)≡0. This proves ℳ4⊂{0}, as required. As a result of this, we obtain
(75)v≠Av+λφ,∀v∈∂Ωr∩P,λ⩾0.
Lemma7 implies
(76)i(A,Ωr∩P,P)=0.
Now (70) and (76) combined lead to
(77)i(A,(ΩR∖Ω¯r)∩P,P)=1-0=1.
Consequently,A has at least one fixed point on (ΩR∖Ω¯r)∩P. Thus (1) has at least one positive solution. This completes the proof.Theorem 12.
If (H1), (H2), (H3), (H6), and (H7) hold, then (1) has at least two positive solutions.Proof.
Ifv∈P and ∥v∥⩽ω, then we have
(78)∥B1v∥0=(B1v)(1)=∫01sv(s)ds⩽ω,∥(B1v)′∥0=(B1v)(0)=∫01v(s)ds⩽ω.
By (H7), we have(79)f(t,x)⩽f(t,ω,ω,ω,ω),∀t∈[0,1],x∈Iω4.
Consequently, for allv∈∂Ωω∩P, we have
(80)∥Av∥0=(Av)(0)=∫01(1-s)f(s,(B1v)(s),(B1v)′(s)v(s),-v′(s))ds⩽∫01f(s,ω,ω,ω,ω)ds<ω=∥v∥,(81)∥(Av)′∥0=-(Av)′(1)=∫01f(s,(B1v)(s),(B1v)′(s),v(s),-v′(s))ds⩽∫01f(s,ω,ω,ω,ω)ds<ω=∥v∥.
The preceding inequalities imply
(82)∥Av∥<ω=∥v∥,∀v∈∂Ωω∩P
and thus
(83)v≠λAv,∀v∈∂Ωω∩P,0⩽λ⩽1.
Now Lemma 5 yields
(84)i(A,Ωω∩P,P)=1.By (H2), (H3), and (H6), we know that (52) and (76) hold (see the proofs of Theorems 10 and 11). Note that we can choose R>ω>r in (52) and (76). Combining (52), (76), and (84), we obtain
(85)i(A,(ΩR∖Ω¯ω)∩P,P)=0-1=-1,i(A,(Ωω∖Ω¯r)∩P,P)=1-0=1.
Therefore, A has at least two fixed points, with one on (ΩR∖Ω¯ω)∩P and the other on (Ωω∖Ω¯r)∩P. Hence (1) has at least two positive solutions. This completes the proof.



---

*Source: 316576-2013-01-13.xml*