Table Of ContentPOSITIVE SOLUTIONS OF A NONLINEAR FOURTH-ORDER
INTEGRAL BOUNDARY VALUE PROBLEM
6
1
0
SLIMANEBENAICHA,FAOUZIHADDOUCHI
2
y
a Abstract. In this paper, the existence of positive solutions for a nonlinear
M fourth-order two-point boundary value problem with integral condition is in-
vestigated. By using Krasnoselskii’s fixed point theorem on cones, sufficient
conditions fortheexistence ofatleastonepositivesolutionsareobtained.
4
2
]
A
1. Introduction
C
. Inthiswork,westudytheexistenceofpositivesolutionsofanonlineartwo-point
h
boundaryvalueproblem(BVP)forthe followingfourth-orderdifferentialequation:
t
a
m
u′′′′(t)+f(u(t))=0, t (0,1), (1.1)
[ ∈
1
4 u′(0)=u′(1)=u′′(0)=0, u(0)= a(s)u(s)ds, (1.2)
v Z0
3 where
7
(H1) f C([0, ),[0, ));
7 ∈ ∞ ∞
1
3 (H2) a C([0,1],[0, )) and 0< a(s)ds<1.
∈ ∞ 0
0
Fourth-order ordinary differential equations are models for bending or deforma-
. R
1 tionofelasticbeams,andthereforehaveimportantapplicationsinengineeringand
0 physical sciences. Recently, the two-point and multi-point boundary value prob-
6
lems for fourth-order nonlinear differential equations have receivedmuch attention
1
from many authors. Many authors have studied the beam equation under various
:
v boundary conditions and by different approaches.
i
X In 2009, Graef et al.[5] considered the fourth order three-point boundary value
problem
r
a u′′′′(t)=g(t)f(u(t)), t (0,1), (1.3)
∈
together with the boundary conditions
u(0)=u′(0)=u′′(β)=u′′(1)=0. (1.4)
In 2006, Anderson and Avery [2], studied the following fourth order right focal
four-point boundary value problem
u′′′′(t)+f(u(t))=0, 0<t<1, (1.5)
u(0)=u′(q)=u′′(r)=u′′′(1)=0. (1.6)
2000 Mathematics Subject Classification. 34B15,34B18.
Key words and phrases. Positive solutions; Krasnoselskii’sfixed point theorem; Fourth-order
integralboundaryvalueproblems;Existence; Cone.
1
2 S.BENAICHA,F.HADDOUCHI
In 2011, Xiading Han et al.[6], considered the nonlocal fourth-order boundary
value problem with variable parameter
u′′′′(t)+B(t)u′′(t)=λf(t,u(t),u′′(t)), 0<t<1, (1.7)
1 1
u(0)=u(1)= p(s)u(s)ds, u′′(0)=u′′(1)= q(s)u′′(s)ds. (1.8)
Z0 Z0
In 2013,Yan Sun and Cun Zhu [14], considered the singular fourth-order three-
point boundary value problem
u′′′′(t)+f(t,u(t))=0, 0<t<1, (1.9)
u(0)=u′(0)=u′′(0)=0, u′′(1) αu′′(η)=λ. (1.10)
−
In 2014,Xiaorui Liu and Dexiang Ma [10], considered the third-order two-point
boundary value problem
u′′′(t)+f(u(t))=0, 0<t<1, (1.11)
1
u′(0)=u′(1)=0, u(0)= k(s)u(s)ds, (1.12)
Z0
and in 2015, Wenguo shen [13], studied the fourth-order second-point nonhomoge-
neous singular boundary value problem
u′′′′(t)+a(t)f(u(t))=0, 0<t<1, (1.13)
u(0)=α, u(1)=β, u′(0)=λ, u′(1)= µ. (1.14)
−
For some other results on boundary value problem, we refer the reader to the
papers [1, 3, 4, 7, 8, 11, 12, 18, 15, 17, 16, 19].
To the authors knowledge, there are few papers that have considered the exis-
tence of solutions for fourth-order two-pointboundary value problem with integral
condition. Motivatedby the worksmentionedabove,the aimofthis paper isto es-
tablishsomesufficientconditions for the existence ofatleastone positivesolutions
of the BVP (1.1) and (1.2).
We shall first construct the Green’s function for the associated linear boundary
value problem and then determine the properties of the Green’s function for as-
sociated linear boundary value problem. Finally, existence results for at least one
positive solution for the above problem are established when f is superlinear or
sublinear. As applications, some interesting examples are presented to illustrate
the main results.
2. Preliminaries
We shall consider the Banach space C([0,1]) equipped with the sup norm
u =supt∈[0,1] u(t)
k k | |
Defnition 2.1. Let E be a real Banach space. A nonempty, closed, convex set
K E is a cone if it satisfies the following two conditions:
⊂
(i) x K, λ 0 imply λx K;
∈ ≥ ∈
(ii) x K, x K imply x=0.
∈ − ∈
Defnition2.2. AnoperatorT :E E iscompletelycontinuousifitiscontinuous
→
and maps bounded sets into relatively compact sets.
Toprovesomeofourresults,wewillusethefollowingfixedpointtheorem,which
is due to Krasnoselskii’s [9].
POSITIVE SOLUTIONS 3
Theorem 2.3. [9]. Let E be a Banach space, and let K E, be a cone. Assume
⊂
that Ω and Ω are open subsets of E with 0 Ω , Ω Ω and let
1 2 1 1 2
∈ ⊂
A:K (Ω Ω ) K
2 1
∩ \ →
be a completely continuous operator such that
(i) Au u , u K ∂Ω , and Au u , u K ∂Ω ; or
1 2
k k≤k k ∈ ∩ k k≥k k ∈ ∩
(ii) Au u , u K ∂Ω , and Au u , u K ∂Ω .
1 2
k k≥k k ∈ ∩ k k≤k k ∈ ∩
Then A has a fixed point in K (Ω Ω ).
2 1
∩ \
Consider the two-point boundary value problem
u′′′′(t)+y(t)=0, t (0,1), (2.1)
∈
1
u′(0)=u′(1)=u′′(0)=0, u(0)= a(s)u(s)ds. (2.2)
Z0
Lemma 2.4. The problem (2.1)-(2.2) has a unique solution
1 1 1
u(t)= G(t,s)+ a(τ)G(τ,s)dτ y(s)ds,
1 α
Z0 (cid:16) − Z0 (cid:17)
where G(t,s):[0,1] [0,1] R is the Green’s function defined by
× →
1 t3(1 s)2 (t s)3, 0 s t 1;
G(t,s)= − − − ≤ ≤ ≤ (2.3)
6(t3(1 s)2, 0 t s 1,
− ≤ ≤ ≤
and
1
α= a(t)dt.
Z0
Proof. Integrating (2.1) over the interval [0,t] for t [0,1], we obtain
∈
t
u′′′(t)= y(s)ds+C ,
1
−
Z0
t
u′′(t)= (t s)y(s)ds+C t+C ,
1 2
− −
Z0
1 t 1
u′(t)= (t s)2y(s)ds+ C t2+C t+C ,
1 2 3
−2 − 2
Z0
1 t 1 1
u(t)= (t s)3y(s)ds+ C t3+ C t2+C t+C . (2.4)
1 2 3 4
−6 − 6 2
Z0
From the boundary conditions (2.2) we get
1
C =C =0, C = (1 s)2y(s)ds,
2 3 1
−
Z0
4 S.BENAICHA,F.HADDOUCHI
and
C = u(0)
4
1 1 τ 1 1
= a(τ) (τ s)3y(s)ds+ τ3 (1 s)2y(s)ds+C dτ
4
− 6 − 6 −
Z0 (cid:16) Z0 Z0 (cid:17)
1 1 τ 1 1 1
= a(τ) (τ s)3y(s)ds dτ + a(τ)τ3 (1 s)2y(s)ds dτ
−6 − 6 −
Z0 (cid:16)Z0 (cid:17) Z0 (cid:16)Z0 (cid:17)
1
+C a(τ)dτ,
4
Z0
which implies
1 1 1 1 τ
C = a(τ)τ3 (1 s)2y(s)ds dτ a(τ) (τ s)3y(s)ds dτ .
4 6(1 α) − − −
− (cid:16)Z0 (cid:16)Z0 (cid:17) Z0 (cid:16)Z0 (cid:17) (cid:17)
Replacing these expressions in (2.4), we get
1 t 1 1
u(t) = (t s)3y(s)ds+ t3 (1 s)2y(s)ds
−6 − 6 −
Z0 Z0
1 1 1
+ a(τ)τ3 (1 s)2y(s)ds dτ
6(1 α) −
− hZ0 (cid:16)Z0 (cid:17)
1 τ
a(τ) (τ s)3y(s)ds dτ
− −
Z0 (cid:16)Z0 (cid:17) i
1 t 1 1
= t3(1 s)2 (t s)3 y(s)ds+ t3(1 s)2y(s)ds
6 − − − 6 −
Z0 h i Zt
1 1 τ
+ a(τ) τ3(1 s)2 (τ s)3 y(s)ds
6(1 α) − − −
− hZ0 (cid:16)Z0 h i
1
+ τ3(1 s)2y(s)ds dτ
−
Zτ (cid:17) i
1 t 1 1
= t3(1 s)2 (t s)3 y(s)ds+ t3(1 s)2y(s)ds
6 − − − 6 −
Z0 h i Zt
1 1 1 τ
+ a(τ) τ3(1 s)2 (τ s)3 y(s)ds
1 α 6 − − −
− Z0 (cid:16) Z0 h i
1 1
+ a(τ)τ3(1 s)2y(s)ds dτ
6 −
Zτ (cid:17)
1 1 1 1
= G(t,s)y(s)ds+ a(τ)G(τ,s)y(s)ds dτ
1 α
Z0 − Z0 (cid:16)Z0 (cid:17)
1 1 1
= G(t,s)+ a(τ)G(τ,s)dτ y(s)ds.
1 α
Z0 (cid:16) − Z0 (cid:17)
(cid:3)
Lemma 2.5. Let θ ]0,1[ be fixed. Then
∈ 2
(i) G(t,s) 0, for all t,s [0,1];
≥ ∈
(ii) 1θ3s(1 s)2 G(t,s) 1s(1 s)2, for all (t,s) [θ,1 θ] [0,1].
6 − ≤ ≤ 6 − ∈ − ×
POSITIVE SOLUTIONS 5
Proof. (i) We will show that G(t,s) 0, for all (t,s) [0,1] [0,1]. Since it is
≥ ∈ ×
obviousfort s,weonlyneedtoprovethecases t. Nowwesupposethats t.
≤ ≤ ≤
Then
1 1
G(t,s)= t3(1 s)2 (t s)3 = t(t ts)2 (t s)3
6 − − − 6 − − −
1h i h i
t(t s)2 (t s)3
≥ 6 − − −
(2.5)
1h i
(t s)2 t (t s)
≥ 6 − − −
1 h i
= s(t s)2 0.
6 − ≥
(ii) If s t, from (2.3) we have
≤
1 1
G(t,s)= t3(1 s)2 (t s)3 t3(1 s)3 (t s)3
6 − − − ≥ 6 − − −
1h i h i
= (t ts)3 (t s)3
6 − − −
(2.6)
1h i
= s(1 t) t2(1 s)2+t(1 s)(t s)+(t s)2
6 − − − − −
1 h i
t2(1 t)s(1 s)2.
≥ 6 − −
On the other hand
1 1 1 1
G(t,s) s(1 s)2 = t3(1 s)2 (t s)3 s(1 s)2
− 6 − 6 − − 6 − − 6 −
1
= s 2t3+t3s+3t2 3ts 1+2s
6 − − −
1 (cid:16) (cid:17)
= s(t 1)2 (s 2)t+2s 1 (2.7)
6 − − −
1 h i
s(t 1)2 (s 2)t+2t 1
≤ 6 − − −
1 h i
= s(t 1)2(st 1) 0.
6 − − ≤
If t s, from (2.3), we have
≤
1 1
G(t,s)= t3(1 s)2 t3s(1 s)2, (2.8)
6 − ≥ 6 −
and,
1 1 1
G(t,s)= t3(1 s)2 s3(1 s)2 s(1 s)2. (2.9)
6 − ≤ 6 − ≤ 6 −
Let
1 1 t3, t 1;
ρ(t)= min t3,t2(1 t) = ≤ 2
6 { − } 6(t2(1 t), t 1.
− ≥ 2
From (2.6), (2.7), (2.8) and (2.9) we have
1
ρ(t)s(1 s)2 G(t,s) s(1 s)2, for all (t,s) [0,1] [0,1].
− ≤ ≤ 6 − ∈ ×
For θ ]0,1[, we have
∈ 2
θ3 1
s(1 s)2 G(t,s) s(1 s)2, for all (t,s) [θ,1 θ] [0,1].
6 − ≤ ≤ 6 − ∈ − ×
6 S.BENAICHA,F.HADDOUCHI
(cid:3)
Lemma 2.6. Let y(t) C([0,1],[0, )) and θ ]0,1[. The unique solution of
∈ ∞ ∈ 2
(2.1)-(2.2) is nonnegative and satisfies
min u(t) θ3(1 α+β) u ,
t∈[θ,1−θ] ≥ − k k
where β = 1−θa(t)dt, α= 1a(t)dt.
θ 0
Proof. FromR Lemma 2.4 andRLemma 2.5, u(t) is nonnegative. For t [0,1], from
∈
Lemma 2.4 and Lemma 2.5, we have that
1 1 1
u(t) = G(t,s)+ a(τ)G(τ,s)dτ y(s)ds
1 α
Z0 (cid:16) − Z0 (cid:17)
1 1 α
s(1 s)2 1+ y(s)ds
≤ 6 − 1 α
Z0 (cid:16) − (cid:17)
1 1
= s(1 s)2y(s)ds.
6(1 α) −
− Z0
Then
1 1
u s(1 s)2y(s)ds, (2.10)
k k≤ 6(1 α) −
− Z0
and for t [θ,1 θ], we have
∈ −
1 1 1
u(t)= G(t,s)+ a(τ)G(τ,s)dτ y(s)ds
1 α
Z0 (cid:16) − Z0 (cid:17)
θ3 1 1 1−θ
s(1 s)2+ s(1 s)2a(τ)dτ y(s)ds
≥ 6 − 1 α −
Z0 h − Zθ i (2.11)
θ3 1 β
= s(1 s)2 1+ y(s)ds
6 − 1 α
Z0 (cid:16) − (cid:17)
θ3 1 α+β 1
= − s(1 s)2y(s)ds.
6 · 1 α −
− Z0
From (2.10), (2.11), we obtain
min u(t) θ3(1 α+β) u .
t∈[θ,1−θ] ≥ − k k
(cid:3)
Let θ ]0,1[. We define the cone
∈ 2
K = u C([0,1],R),u 0: min u(t) θ3(1 α+β) u ,
∈ ≥ t∈[θ,1−θ] ≥ − k k
(cid:26) (cid:27)
and the operator A:K C[0,1] by
→
1 1 1
Au(t)= G(t,s)+ a(τ)G(τ,s)dτ f(u(s))ds. (2.12)
1 α
Z0 (cid:16) − Z0 (cid:17)
Remark 2.7. By Lemma 2.4, problem (1.1), (1.2) has a positive solution u(t) if
and only if u is a fixed point of A.
Lemma 2.8. The operator A defined in (2.12) is completely continuous and satis-
fies AK K.
⊂
POSITIVE SOLUTIONS 7
Proof. From Lemma 2.6, we obtain AK K. A is completely continuous by an
application of Arzela-Ascoli theorem. ⊂ (cid:3)
In what follows, we will use the following notations
f(u) f(u)
f0 =ul→im0+ u , f∞ =ul→im∞ u .
We note that the case f0 =0 and f∞ = correspondsto the superlinear case and
∞
f0 = and f∞ =0 corresponds to the sublinear case.
∞
3. Existence of positive solutions
In this section, we will state and prove our main results.
Theorem 3.1. Assume that f0 =0 and f∞ = . Then BVP (1.1) and (1.2) has
∞
at least one positive solution.
Proof. Since f =0,there exists ρ >0suchthat f(u) ǫu,for 0<u ρ , where
0 1 1
≤ ≤
ǫ>0 satisfies
ǫ
1.
6(1 α) ≤
−
Thus, if we let
Ω = u C[0,1]: u <ρ ,
1 1
{ ∈ k k }
then, for u K ∂Ω , we have
1
∈ ∩
1 1 1
Au(t)= G(t,s)+ a(τ)G(τ,s)dτ f(u(s))ds
1 α
Z0 (cid:16) − Z0 (cid:17)
1 1 1 1
s(1 s)2+ s(1 s)2a(τ)dτ ǫu(s)ds
≤ 6 − 1 α −
Z0 (cid:16) − Z0 (cid:17)
1 ǫ 1 (3.1)
u s(1 s)2ds
≤ 6 · 1 αk k −
− Z0
1 ǫ
u
≤ 6 · 1 αk k
−
u .
≤k k
Therefore
Au u , u K ∂Ω .
1
k k≤k k ∈ ∩
Further, since f∞ = , there exists ρ2 > 0 such that f(u) δu, for u > ρ2,
∞ ≥
where δ >0 is chosen so that
b b
θ6 (1 α+β)2 1
δ − (1 2θ)( +θ θ2) 1.
36 · 1 α − 2 − ≥
−
Let ρ = max 2ρ , ρb2 and Ω = u C[0,1]: u <ρ . Then u
2 1 θ3(1−α+β) 2 { ∈ k k 2} ∈
K ∂Ω2 implies tnhat o
∩
min u(t) θ3(1 α+β) u =θ3(1 α+β)ρ ρ ,
2 2
t∈[θ,1−θ] ≥ − k k − ≥
so, by (2.12) and for t [θ,1 θ], we obtain
∈ − b
8 S.BENAICHA,F.HADDOUCHI
1 1 1
Au(t)= G(t,s)+ a(τ)G(τ,s)dτ f(u(s))ds
1 α
Z0 (cid:16) − Z0 (cid:17)
1−θ θ3 1 1−θ θ3
s(1 s)2+ s(1 s)2a(τ)dτ δu(s)ds
≥ 6 − 1 α 6 −
Zθθ3 h1−θ − βZθ i
= δ s(1 s)2 1+ u(s)ds
6 − 1 α
θ3δ Zθ(1 α+β) (cid:16)1−θ − (cid:17) (3.2)
= − s(1 s)2u(s)ds
6 · 1 α −
− Zθ
θ3δ (1 α+β) 1−θ
− min u(t) s(1 s)2ds
≥ 6 · 1−α t∈[θ,1−θ] Zθ −
θ6 (1 α+β)2 1
δ − (1 2θ)( +θ θ2) u
≥ 36 · 1 α − 2 − k k
−
u .
≥k k
Hence, Au u , u K ∂Ω . By Theorem 2.3, the operator A has a fixed
2
point in Kk (Ωk≥kΩk) su∈ch th∩at ρ u ρ . (cid:3)
2 1 1 2
∩ \ ≤k k≤
Theorem 3.2. Assume that f0 = and f∞ =0. Then BVP (1.1) and (1.2) has
∞
at least one positive solution.
Proof. Since f = , there exists ρ > 0 such that f(u) λu, for 0 < u ρ ,
0 1 1
∞ ≥ ≤
where λ>0 is chosen so that
θ6 (1 α+β)2 1
λ − (1 2θ)( +θ θ2) 1.
36 · 1 α − 2 − ≥
−
Thus, for u K ∂Ω with
1
∈ ∩
Ω = u C[0,1]: u <ρ ,
1 1
{ ∈ k k }
we have from (3.2)
1 1 1
Au(t) = G(t,s)+ a(τ)G(τ,s)dτ f(u(s))ds
1 α
Z01−(cid:16)θ θ3 − Z0 1 1−θ θ3 (cid:17)
s(1 s)2+ s(1 s)2a(τ)dτ λu(s)ds
≥ 6 − 1 α 6 −
Zθ h − Zθ i
θ6 (1 α+β)2 1
λ − (1 2θ)( +θ θ2) u
≥ 36 · 1 α − 2 − k k
−
u .
≥ k k
Then, Au(t) u for t [θ,1 θ] , which implies that
≥k k ∈ −
Au u , u K ∂Ω .
1
k k≥k k ∈ ∩
Next we construct the set Ω . We discuss two possible cases:
2
Case 1. If f is bounded. Then, there exists L > 0 such that f(u) L. Set
≤
Ω = u C[0,1]: u <ρ , where ρ =max 2ρ , L .
2 { ∈ k k 2} 2 1 6(1−α)
n o
POSITIVE SOLUTIONS 9
If u K ∂Ω , similar to the estimates of (3.1), we obtain
2
∈ ∩
1 L 1
Au(t) s(1 s)2ds (3.3)
≤ 6 · 1 α −
− Z0
1 L
ρ = u , (3.4)
≤ 6 · 1 α ≤ 2 k k
−
and consequently, Au u , u K ∂Ω .
2
k k≤k k ∈ ∩
Case2. Supposethatf isunbounded,sincef∞ =0,thereexistsρ2 >0(ρ2 >ρ1)
such that f(u) ηu for u>ρ , where η >0 satisfies
2
≤
η b b
1.
b 6(1 α) ≤
−
Onthe other hand, from condition (H1), there is σ >0 such that f(u) ησ, with
≤
0 u ρ .
2
≤ ≤
Now, set Ω = u C[0,1]: u <ρ , where ρ =max σ,ρ .
2 2 2 2
{ ∈ k k } { }
If u K ∂Ω , then we have f(u) ηρ . Similar to (3.1), we obtain
b 2 2
∈ ∩ ≤
1 ηρ 1 b
Au(t) 2 s(1 s)2ds (3.5)
≤ 6 · 1 α −
− Z0
1 ηρ
2 ρ = u , (3.6)
≤ 6 · 1 α ≤ 2 k k
−
so, Au u , for u K ∂Ω . Therefore by Theorem 2.3, A has at least one
2
fixedk pokin≤t, wkhkich is a p∈ositi∩ve solution of (1.1) and (1.2). (cid:3)
4. Examples
Example 4.1. Consider the boundary value problem
u′′′′(t)+u2(e−u+1)=0, 0<t<1, (4.1)
1
u′(0)=u′(1)=u′′(0)=0, u(0)= s2u(s)ds, (4.2)
Z0
where f(u) = u2(e−u + 1) C([0, ),[0, )) and a(t) = t2 0, 1a(s)ds =
∈ ∞ ∞ ≥ 0
1s2ds= 1.
0 3 R
We have
R
f(u) u2(e−u+1)
lim = lim =0,
u→0+ u u→0+ u
f(u) u2(e−u+1)
lim = lim =+ .
u→+∞ u u→+∞ u ∞
From Theorem 3.1, theproblem (4.1)and (4.2)has at leastonepositive solution.
Example 4.2. Consider the boundary value problem
u′′′′(t)+√1+u+sinu=0, 0<t<1, (4.3)
1
u′(0)=u′(1)=u′′(0)=0, u(0)= su(s)ds, (4.4)
Z0
where f(u) = √1+u+sinu C([0, ),[0, )) and a(t) = t 0, 1a(s)ds =
∈ ∞ ∞ ≥ 0
1sds= 1.
0 2 R
R
10 S.BENAICHA,F.HADDOUCHI
We have
f(u) √1+u+sinu
lim = lim =+ ,
u→0+ u u→0+ u ∞
f(u) √1+u+sinu
lim = lim =0.
u→+∞ u u→+∞ u
Therefore, by Theorem 3.2, the problem (4.3) and (4.4) has at least one positive
solution.
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Slimane Benaicha, Laboratory of Fundamental and Applied Mathematics of Oran
(LMFAO), Department of Mathematics, University of Oran 1 Ahmed Benbella, 31000
Oran,Algeria
E-mail address: [email protected]
