Volume 2010, Article ID 874959,12pages doi:10.1155/2010/874959
Research Article
Monotone Positive Solution of Nonlinear Third-Order BVP with Integral Boundary Conditions
Jian-Ping Sun and Hai-Bao Li
Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, China
Correspondence should be addressed to Jian-Ping Sun,[email protected] Received 7 September 2010; Accepted 31 October 2010
Academic Editor: Michel C. Chipot
Copyrightq2010 J.-P. Sun and H.-B. Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
This paper is concerned with the following third-order boundary value problem with integral boundary conditionsut ft, ut, ut 0, t∈0,1;u0 u0 0, u1 1
0gtutdt, wheref ∈ C0,1×0,∞×0,∞,0,∞ andg ∈ C0,1,0,∞. By using the Guo- Krasnoselskii fixed-point theorem, some sufficient conditions are obtained for the existence and nonexistence of monotone positive solution to the above problem.
1. Introduction
Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross section, a three-layer beam, electromagnetic waves or gravity driven flows and so on 1.
Recently, third-order two-point or multipoint boundary value problems BVPs for shorthave attracted a lot of attention2–17. It is known that BVPs with integral boundary conditions cover multipoint BVPs as special cases. Although there are many excellent works on third-order two-point or multipoint BVPs, a little work has been done for third-order BVPs with integral boundary conditions. It is worth mentioning that, in 2007, Anderson and Tisdell 18developed an interval ofλvalues whereby a positive solution exists for the following third-order BVP with integral boundary conditions
pu
t λft, ut, t∈t1, t3,
αut1−βut1
ξ2
ξ1
gtutdt,
ut2 0, pu
t3
η2
η1
ht pu
tdt
1.1
by using the Guo-Krasnoselskii fixed-point theorem. In 2008, Graef and Yang19studied the third-order BVP with integral boundary conditions
ut gtfut, t∈0,1, u0 u
p
1
q
wtutdt0. 1.2
For second-order or fourth-order BVPs with integral boundary conditions, one can refer to 20–24.
In this paper, we are concerned with the following third-order BVP with integral boundary conditions
ut f
t, ut, ut
0, t∈0,1, u0 u0 0, u1
1
0
gtutdt. 1.3
Throughout this paper, we always assume that f ∈ C0,1×0,∞×0,∞,0,∞ andg ∈ C0,1,0,∞. Some sufficient conditions are established for the existence and nonexistence of monotone positive solution to the BVP1.3. Here, a solutionuof the BVP 1.3is said to be monotone and positive ifut≥ 0,ut≥ 0 andut/≡0 fort ∈0,1. Our main tool is the following Guo-Krasnoselskii fixed-point theorem25.
Theorem 1.1. LetEbe a Banach space and letKbe a cone inE. Assume thatΩ1andΩ2are bounded open subsets ofEsuch thatθ ∈ Ω1, Ω1 ⊂ Ω2, and letT : K∩Ω2\Ω1 → Kbe a completely continuous operator such that either
1Tu ≤ u foru∈K∩∂Ω1andTu ≥ u foru∈K∩∂Ω2, or 2Tu ≥ u foru∈K∩∂Ω1andTu ≤ u foru∈K∩∂Ω2.
ThenT has a fixed point inK∩Ω2\Ω1.
2. Preliminaries
For convenience, we denoteμ1
0tgtdt.
Lemma 2.1. Letμ /1. Then for anyh∈C0,1, the BVP
−ut ht, t∈0,1, u0 u0 0, u1
1
0
gtutdt 2.1
has a unique solution
ut 1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
hsds, t∈0,1, 2.2
where
G1t, s 1 2
⎧⎨
⎩
2t−t2−s
s, 0≤s≤t≤1, 1−st2, 0≤t≤s≤1,
G2t, s
⎧⎨
⎩
1−ts, 0≤s≤t≤1, 1−st, 0≤t≤s≤1.
2.3
Proof. Letube a solution of the BVP2.1. Then, we may suppose that
ut 1
0
G1t, shsdsAt2BtC, t∈0,1. 2.4
By the boundary conditions in2.1, we have
A 1
2 1−μ
1
0
hs 1
0
G2τ, sgτdτds andBC0. 2.5
Therefore, the BVP2.1has a unique solution
ut 1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
hsds, t∈0,1. 2.6
Lemma 2.2see12. For anyt, s∈0,1×0,1,
t2
21−ss≤G1t, s≤ 1
21−ss. 2.7
Lemma 2.3see26. For anyt, s∈0,1×0,1,
0≤G2t, s≤1−ss. 2.8
In the remainder of this paper, we always assume thatμ <1,α∈0,1andβα2/2.
Lemma 2.4. Ifh∈C0,1andht≥0 fort ∈0,1, then the unique solutionuof the BVP2.1 satisfies
1ut≥0,t∈0,1,
2ut≥0,t∈0,1and mint∈α,1ut≥βu, whereumax{u∞,u∞}.
Proof. Since1is obvious, we only need to prove2. By2.2, we get
ut 1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
hsds, t∈0,1, 2.9
which indicates thatut≥0 fort∈0,1.
On the one hand, by2.9andLemma 2.3, we have
u
∞≤ 1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
hsds. 2.10
On the other hand, in view of2.2andLemma 2.2, we have
u∞≤ 1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
hsds. 2.11
It follows from2.10and2.11that
u ≤ 1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
hsds, 2.12
which together withLemma 2.2implies that
t∈α,1min ut min
t∈α,1
1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
hsds
≥ min
t∈α,1
t2 2
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
hsds
α2 2
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
hsds
≥βu.
2.13
Let E C10,1be equipped with the norm u max{u∞,u∞}. ThenE is a Banach space. If we denote
K
u∈E:ut≥0, ut≥0, t∈0,1,min
t∈α,1ut≥βu
, 2.14
then it is easy to see thatKis a cone inE. Now, we define an operatorTonKby
Tut 1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
f
s, us, us
ds, t∈0,1. 2.15
Obviously, ifuis a fixed point of T, thenuis a monotone nonnegative solution of the BVP 1.3.
Lemma 2.5. T:K → Kis completely continuous.
Proof. First, byLemma 2.4, we know thatTK⊂K.
Next, we assume thatD ⊂ K is a bounded set. Then there exists a constantM1 > 0 such thatu ≤M1for anyu∈D. Now, we will prove thatTDis relatively compact inK.
Suppose that{yk}∞k1⊂TD. Then there exist{xk}∞k1⊂Dsuch thatTxkyk. Let
M2sup f
t, x, y :
t, x, y
∈0,1×0, M1×0, M1 ,
M3 1 1−μ
1
0
G2τ, sgτdτds.
2.16
Then for anyk, byLemma 2.2, we have ykt|Txkt|
1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
f
s, xks, xks ds
≤ M2 2
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
ds
M2
2 1
6 M3
, t∈0,1,
2.17
which implies that {yk}∞k1 is uniformly bounded. At the same time, for any k, in view of Lemma 2.3, we have
yktTxkt
1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, xks, xks ds
≤M2
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
ds
M2 1
6M3
, t∈0,1,
2.18
which shows that {yk}∞k1 is also uniformly bounded. This indicates that {yk}∞k1 is equicontinuous. It follows from Arzela-Ascoli theorem that {yk}∞k1 has a convergent subsequence inC0,1. Without loss of generality, we may assume that{yk}∞k1 converges inC0,1. On the other hand, by the uniform continuity ofG2t, s, we know that for any ε >0, there existsδ1>0 such that for anyt1, t2∈0,1with|t1−t2|< δ1, we have
|G2t1, s−G2t2, s|< ε
2M21, s∈0,1. 2.19
Letδmin{δ1, ε/2M2M31}. Then for anyk,t1, t2 ∈0,1with|t1−t2|< δ, we have ykt1−ykt2Txkt1−Txkt2
≤ 1
0
|G2t1, s−G2t2, s||t1−t2| 1−μ
1
0
G2τ, sgτdτ
f
s, xks, xks ds
≤M2 1
0
|G2t1, s−G2t2, s|dsM2M3|t1−t2|
≤ M2ε
2M21M2M3|t1−t2|
< ε,
2.20
which implies that{yk}∞k1 is equicontinuous. Again, by Arzela-Ascoli theorem, we know that{yk}∞k1 has a convergent subsequence in C0,1. Therefore, {yk}∞k1 has a convergent subsequence inC10,1. Thus, we have shown thatT is a compact operator.
Finally, we prove thatTis continuous. Suppose thatum, u∈Kandum−u → 0m →
∞. Then there existsM4 >0 such that for anym,um ≤M4. Let
M5sup f
t, x, y :
t, x, y
∈0,1×0, M4×0, M4
. 2.21
Then for anymandt∈0,1, in view of Lemmas2.2and2.3, we have
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
f
s, ums, ums
≤ M5 2
1 1
1−μ 1
0
gτdτ
1−ss, s∈0,1,
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, ums, ums
≤M5
1 1 1−μ
1
0
gτdτ
1−ss, s∈0,1.
2.22
By applying Lebesgue Dominated Convergence theorem, we get
mlim→ ∞Tumt lim
m→ ∞
1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
f
s, ums, ums ds
1
0
G1t, s t2 2
1−μ 1
0
G2τ, sgτdτ
f
s, us, us ds Tut, t∈0,1,
m→ ∞limTumt lim
m→ ∞
1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, ums, ums ds
1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, us, us ds Tut, t∈0,1,
2.23
which indicates thatTis continuous. Therefore,T :K → Kis completely continuous.
3. Main Results
For convenience, we define
f0lim sup
xy→0 max
t∈0,1
f t, x, y
xy , f0lim inf
xy→0 min
t∈α,1
f t, x, y xy , f∞lim sup
xy→∞max
t∈0,1
f t, x, y
xy , f∞ lim inf
xy→∞min
t∈α,1
f t, x, y xy , H12
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
ds,
H2 β 2
1
α
1−ss 1 1−μ
1
0
G2τ, sgτdτ
ds.
3.1
Theorem 3.1. IfH1f0<1< H2f∞, then the BVP1.3has at least one monotone positive solution.
Proof. In view ofH1f0<1, there existsε1>0 such that
H1
f0ε1
≤1. 3.2
By the definition off0, we may chooseρ1 >0 so that
f t, x, y
≤
f0ε1 xy
, fort∈0,1, xy
∈ 0, ρ1
. 3.3
LetΩ1{u∈E:u< ρ1/2}. Then for anyu∈K∩∂Ω1, in view of3.2and3.3, we have
Tut 1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, us, us ds
≤ 1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
f0ε1
us us ds
≤H1
f0ε1 u
≤ u, t∈0,1.
3.4
By integrating the above inequality on0, t, we get
Tut≤ u, t∈0,1, 3.5
which together with3.4implies that
Tu ≤ u, u∈K∩∂Ω1. 3.6
On the other hand, since 1< H2f∞, there existsε2>0 such that
H2
f∞−ε2
≥1. 3.7
By the definition off∞, we may chooseρ2> ρ1, so that
f t, x, y
≥
f∞−ε2 xy
, fort∈α,1, xy
∈
ρ2,∞
. 3.8
LetΩ2{u∈E:u< ρ2/β}. Then for anyu∈K∩∂Ω2, in view of3.7and3.8, we have
Tu1 1
0
G11, s 1 2
1−μ 1
0
G2τ, sgτdτ
f
s, us, us ds
≥ 1 2
1
α
1−ss 1 1−μ
1
0
G2τ, sgτdτ
f∞−ε2
us us ds
≥H2
f∞−ε2 u
≥ u,
3.9
which implies that
Tu ≥ u, u∈K∩∂Ω2. 3.10
Therefore, it follows from3.6,3.10, andTheorem 1.1that the operator T has one fixed pointu∈K∩Ω2\Ω1, which is a monotone positive solution of the BVP1.3.
Theorem 3.2. IfH1f∞<1< H2f0, then the BVP1.3has at least one monotone positive solution.
Proof. The proof is similar to that ofTheorem 3.1and is therefore omitted.
Theorem 3.3. IfH1ft, x, y<xyfort∈0,1andxy∈0,∞, then the BVP1.3has no monotone positive solution.
Proof. Suppose on the contrary thatuis a monotone positive solution of the BVP1.3. Then ut≥0 andut≥0 fort∈0,1, and
ut 1
0
G2t, s t 1−μ
1
0
G2τ, sgτdτ
f
s, us, us ds
≤ 1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
f
s, us, us ds
< 1 H1
1
0
1−ss 1 1−μ
1
0
G2τ, sgτdτ
us us ds
≤ u, t∈0,1.
3.11
By integrating the above inequality on0, t, we get
ut<u, t∈0,1, 3.12
which together with3.11implies that
u<u. 3.13
This is a contradiction. Therefore, the BVP1.3has no monotone positive solution.
Similarly, we can prove the following theorem.
Theorem 3.4. IfH2ft, x, y>xyfort∈α,1andxy∈0,∞, then the BVP1.3has no monotone positive solution.
Example 3.5. Consider the following BVP:
ut 1 1t
ut ut
eutut 1000ut ut2 1ut ut
0, t∈0,1,
u0 u0 0, u1 1
0
tutdt.
3.14
Sinceft, x, y 1/1txy/exy 1000xy2/1xyandgt t, if we chooseα1/2, then it is easy to compute that
f01, f∞500, H1 11
24, H2 91
12288, 3.15
which shows that
H1f0<1< H2f∞. 3.16
So, it follows from Theorem 3.1 that the BVP 3.14 has at least one monotone positive solution.
Acknowledgment
This work was supported by the National Natural Science Foundation of China10801068.
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