Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 401923, 14 pages
doi:10.1155/2012/401923
Research Article
Positive Solutions for Neumann Boundary Value
Problems of Second-Order Impulsive Differential
Equations in Banach Spaces
Xiaoya Liu and Yongxiang Li
Department of Mathematics, Northwest Normal University, Lanzhou 730070, China
Correspondence should be addressed to Xiaoya Liu, liuxy135@163.com
Received 8 September 2011; Revised 28 December 2011; Accepted 6 January 2012
Academic Editor: Yuriy Rogovchenko
Copyright q 2012 X. Liu and Y. 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.
The existence of positive solutions for Neumann boundary value problem of second-order
tk , −Δu |ttk Ik utk ,
impulsive differential equations −u t Mut ft, ut, t ∈ J, t /
k 1, 2, . . . , m, u 0 u 1 θ, in an ordered Banach space E was discussed by employing
the fixed point index theory of condensing mapping, where M > 0 is a constant, J 0, 1,
f ∈ CJ ×K, K, Ik ∈ CK, K, k 1, 2, . . . , m, and K is the cone of positive elements in E. Moreover,
an application is given to illustrate the main result.
1. Introduction
The theory of impulsive differential equations is a new and important branch of differential
equation theory, which has an extensive physical, chemical, biological, engineering background and realistic mathematical model, and hence has been emerging as an important area
of investigation in the last few decades; see 1. Correspondingly, boundary value problems
of second-order impulsive differential equations have been considered by many authors, and
some basic results have been obtained; see 2–7. But many of them obtained extremal solutions by monotone iterative technique coupled with the method of upper and lower solutions;
see 2–4. The research on positive solutions is seldom and most in real space R; see 5–7.
In this paper, we consider the existence of positive solutions to the second-order impulsive differential equation Neumann boundary value problem in an ordered Banach space E:
−u t Mut ft, ut,
−Δu |ttk Ik utk ,
t ∈ J, t / tk ,
k 1, 2, . . . , m,
u 0 u 1 θ,
1.1
2
Abstract and Applied Analysis
where M > 0 is a constant, f ∈ CJ × E, E, J 0, 1; 0 < t1 < t2 < · · · < tm < 1; Ik ∈ CE, E
is an impulsive function, k 1, 2, . . . , m. Δu |ttk denotes the jump of u t at t tk , that is,
Δu |ttk u tk − u t−k , where u tk and u t−k represent the right and left limits of u t at
t tk , respectively.
In the special case where E R 0, ∞, Ik 0, k 1, 2, . . . , m, NBVP 1.1 has
been proved to have positive solutions; see 8, 9. Motivated by the aforementioned facts, our
aim is to study the positive solutions for NBVP 1.1 in a Banach space by fixed point index
theory of condensing mapping. Moreover, an application is given to illustrate the main result.
As far as we know, no work has been done for the existence of positive solutions for NBVP
1.1 in Banach spaces.
2. Preliminaries
Let E be an ordered Banach space with the norm · and partial order ≤, whose positive
cone K {x ∈ E | x ≥ θ} is normal with normal constant N. Let J 0, 1; 0 t0 < t1 <
t2 < · · · < tm < tm1 1; Jk tk−1 , tk , k 1, 2, . . . , m 1, J J \ {t1 , t2 , . . . , tm }. Let
P C1 J, E {u ∈ CJ, E | u t is continuous at t / tk , and left continuous at t tk , and
u tk exists, k 1, 2, . . . , m}. Evidently, PC1 J, E is a Banach space with the norm uPC1 max{utC , u PC }, where uC supt∈J ut, u PC supt∈J u t; see 2. An abstract
function u ∈ PC1 J, E ∩ C2 J , E is called a solution of NBVP 1.1 if ut satisfies all the
equalities of 1.1.
Let CJ, E denote the Banach space of all continuous E-value functions on interval
J with the norm uC supt∈J ut. Let α· denote the Kuratowski measure of noncompactness of the bounded set. For the details of the definition and properties of the
measure of noncompactness, see 10, 11. For any B ⊂ CJ, E and t ∈ J, set Bt {ut | u ∈ B} ⊂ E. If B is bounded in CJ, E, then Bt is bounded in E, and αBt ≤
αB.
Now, we first give the following lemmas in order to prove our main results.
Lemma 2.1 see 12. Let B ⊂ CJ, E be equicontinuous. Then αBt is continuous on J, and
αB max αBt αBJ.
t∈J
2.1
Lemma 2.2 see 13. Let B {un } ⊂ CJ, E be a bounded and countable set. Then αBt is
Lebesgue integral on J, and
α
un tdt | n ∈ N
≤ 2 αBtdt.
J
2.2
J
Lemma 2.3 see 14. Let D ⊂ E be bounded. Then there exists a countable set D0 ⊂ D, such that
αD ≤ 2αD0 .
Abstract and Applied Analysis
3
To prove our main results, for any h ∈ CJ, E, we consider the Neumann boundary
value problem NBVP of linear impulsive differential equation in E:
−u t Mut ht,
−Δu |ttk yk ,
t ∈ J ,
2.3
k 1, 2, . . . , m,
u 0 u 1 θ,
where M > 0, yk ∈ E, k 1, 2, . . . , m.
Lemma 2.4. For any h ∈ CJ, E, M > 0, and yk ∈ E, k 1, 2, . . . , m, the linear NBVP 2.3 has
a unique solution u ∈ P C1 J, E ∩ C2 J , E given by
ut 1
Gt, shsds 0
m
Gt, tk yk ,
2.4
k1
where
√
√
⎧
cosh M1 − t cosh Ms
⎪
⎪
⎪
, 0 ≤ s ≤ t ≤ 1,
√
√
⎪
⎨
M sinh M
Gt, s √
√
⎪
⎪
cosh Mt cosh M1 − s
⎪
⎪
⎩
, 0 ≤ t < s ≤ 1.
√
√
M sinh M
2.5
Proof. Suppose that ut is a solution of 2.3; then
u t − Mut −ht,
√
√
Mt
Mt
e−2
e
√
Mt
Let yt e−2
e
√
√
√
ut
−Me− Mt ut e− Mt u t −e− Mt ht.
√
Mt
2.6
ut ; then
y t −e−
√
Mt
ht,
Δy|ttk −e−
√
Mtk
2.7
yk .
Integrating 2.7 from 0 to t1 , we have
yt1 − y0 −
t1
e−
√
Ms
2.8
hsds.
0
Again, integrating 2.7 from t1 to t, where t ∈ t1 , t2 , then
yt y t1 −
t
t1
e−
√
Ms
hsds y0 −
t
0
e−
√
Ms
hsds − e−
√
Mt1
y1 .
2.9
4
Abstract and Applied Analysis
Repeating the aforementioned procession, for t ∈ J, we have
yt y0 −
t
e−
√
Ms
hsds −
0
e−
√
Mtk
2.10
yk .
0<tk <t
Hence,
e
√
Mt
t √
√
√
2 Mt
− Ms
− Mtk
y0 − e
ut e
hsds −
e
yk .
0
2.11
0<tk <t
For t ∈ J, integrating 2.11 from 0 to t, we have
ut e
√
− Mt
t
u0 √
Ms
e2
y0ds −
0
−
t
e
e
√
− Mt
√
Ms
e2
0
√
2 Ms
0
t
e
√
− Mtk
s
e−
√
Mτ
hτdτ ds
0
yk ds
0<tk <s
t √
√
√
1
y0 e2 Mt − 1 − e2 Mt e− Ms hsds
u0 √
2 M
0
t
e
√
Ms
hsds −
0
Notice that y0 √
e
√
2 Mt
−e
√
√
2 Mtk
− Mtk
e
.
yk
0<tk <t
2.12
Mu0 u 0; thus, for t ∈ J, we have
√
√
√
√
1
e− Mt 2 Mu0 ut √
Mu0 u 0 e Mt
2 M
−
t √
√
√
√
Mu0 u 0 e− Mt e− Mt e Ms hsds
0
−e
√
Mt
t
e
0
√
− Ms
hsds −
e
0<tk <t
√
2 Mt
−e
√
√
2 Mtk
− Mttk e
√
√
√
√
1
√
Mu0 − u 0 e− Mt Mu0 u 0 e Mt
2 M
yk
Abstract and Applied Analysis
e−
5
√
Mt
t
e
√
Ms
hsds − e
√
Mt
t
0
−
e
e−
0
√
Mt−tk yk − e
√
Mtk −t
yk
√
Ms
hsds
,
0<tk <t
2.13
u t √
√
√
√
1
−
Mu0 − u 0 e− Mt Mu0 u 0 e Mt
2
− e−
√
Mt
t
e
√
Ms
hsds − e
√
Mt
t
0
−
e
e−
0
√
Mt−tk yk e
√
− Mt−tk yk
√
Ms
2.14
hsds
.
0<tk <t
In view of that u 0 u 1 θ, we have
u0 1
0
1
0
√
√
√
√
m
e M1−s e− M1−s
e M1−tk e− M1−tk √ hsds √ √ √
√ √
M e M − e− M
M e M − e− M
k1
√
√
m
cosh M1 − s
cosh M1 − tk hsds yk .
√
√
√
√
M sinh M
M sinh M
k1
2.15
Substituting 2.15 into 2.13, for t ∈ J, we obtain
√
√
√
√
t e M1−t e− M1−t e Ms e− Ms
ut hsds
√ √ √
0
2 M e M − e− M
0<tk <t
e
√
M1−t
√
√
√
e− M1−t e Mtk e− Mtk
yk
√ √ √
2 M e M − e− M
√
√
cosh Mt cosh M1 − s
hsds
√
√
M sinh M
t
√
√
cosh Mt cosh M1 − tk yk
√
√
M sinh M
t≤tk <1
1
1
0
Gt, shsds 2.16
m
Gt, tk yk .
k1
Inversely, we can verify directly that the function u ∈ P C1 J, E ∩ C2 J , E defined by
2.4 is a solution of the linear NBVP 2.3. Therefore, the conclusion of Lemma 2.4 holds.
6
Abstract and Applied Analysis
By 2.5, it is easy to verify that Gt, s has the following property:
√
cosh2 M
≤ Gt, s ≤ √
√
√
√ .
M sinh M
M sinh M
1
2.17
Evidently, CJ, E is also an ordered Banach space with the partial order ≤ reduced by
the positive cone CJ, K {u ∈ CJ, E | ut ≥ θ, t ∈ J}. CJ, K is also normal with the
same normal constant N.
Define an operator A : CJ, K → CJ, K as follows:
Aut 1
Gt, sfs, usds 0
m
Gt, tk Ik utk .
2.18
k1
Clearly, A : CJ, K → CJ, K is continuous, and the positive solution of NBVP 1.1 is the
nontrivial fixed point of operator A. However, the integral operator A is noncompactness in
general Banach space. In order to employ the topological degree theory and the fixed point
theory of condensing mapping, there demands that the nonlinear f and impulsive function
Ik satisfy some noncompactness measure condition. Thus, we suppose the following.
P 0 For any R > 0, fJ × KR and Ik KR are bounded and
α ft, D ≤ LαD,
αIk D ≤ Mk αD,
k 1, 2, . . . , m,
2.19
where KR K ∩ Bθ, R, D ⊂ K
countable
set,
√ is arbitrarily
√
√ L > 0 and Mk ≥ 0 are counsants
M
/
M
sinh
M < 1.
and satisfy 4L/M 2cosh2 M m
k
k1
Lemma 2.5. Suppose that condition P 0 is satisfied; then A : CJ, K → CJ, K is condensing.
Proof. Since AB is bounded and equicontinuous for any bounded and nonrelative compact
set B ⊂ CJ, K, by Lemma 2.3, there exists a countable set B1 {un } ⊂ B, such that
αAB ≤ 2αAB1 .
2.20
By assumption P 0 and Lemma 2.1,
1
m
αAB1 t α
Gt, sfs, un sds Gt, tk Ik un tk 0
k1
1
m
≤α
Gt, sfs, un sds
Gt, tk Ik un tk α
0
≤2
1
0
k1
m
Gt, sα fs, B1 s ds Gt, tk αIk B1 tk k1
Abstract and Applied Analysis
≤2
1
7
Gt, sLαB1 sds 0
≤ 2L
m
Gt, tk Mk αB1 tk k1
1
Gt, sdsαB1 m
Mk Gt, tk αB1 0
k1
√
2
cosh M m
2L
k1 Mk
αB1 √
αB1 √
M
M sinh M
√ m
2
2L cosh M k1 Mk
αB1 .
√
√
M
M sinh M
≤
2.21
Since AB1 is equicontinuous, by Lemma 2.1, we have
αAB1 max αAB1 t ≤
t∈J
√ m
2
2L cosh M k1 Mk
αB1 .
√
√
M
M sinh M
2.22
Combining 2.20 and P 0, we have
αAB ≤ 2αAB1 ≤
√ m
2
4L 2cosh M k1 Mk
αB.
√
√
M
M sinh M
2.23
Hence, A : CJ, K → CJ, K is condensing.
Let P be a cone in CJ, K defined by
P {u ∈ CJ, K | ut ≥ σuτ, ∀t, τ ∈ J},
2.24
√
where σ 1/cosh2 M.
Lemma 2.6. For any fJ, K ⊂ K, ACJ, K ⊂ P .
Proof. For any u ∈ CJ, K, t, τ ∈ J, by 2.18 and the second inequality of 2.17, we have
Auτ 1
Gτ, sfs, usds 0
√
2
M
≤√
√
M sinh M
cosh
1
0
m
Gτ, tk Ik utk k1
m
fs, usds Ik utk .
k1
2.25
8
Abstract and Applied Analysis
By this, 2.18, and the first inequality of 2.17, we have
Aut 1
Gt, sfs, usds 0
≥√
1
√
M sinh M
1
m
Gt, tk Ik utk k1
fs, usds 0
m
Ik utk 2.26
k1
≥ σAuτ.
Hence, ACJ, K ⊂ P .
Thus, for any fJ, K ⊂ K, A : P → P is condensing mapping; the positive solution
of NBVP 1.1 is equivalent to the nontrivial fixed point of A in P . For 0 < r < R < ∞, let
Pr {u ∈ P | uC < r}, and ∂Pr {u ∈ P | uC r}, which is the relative boundary bound
of Pr in P . Denote that Pr,R PR \ Pr ; then the fixed point of A in Pr,R is the positive solution
of NBVP 1.1. We will use the fixed point theory of condensing mapping to find the fixed
point of A in Pr,R .
Let X be a Banach space and let P ⊂ X be a cone in X. Assume that Ω is a bounded open
subset of X and let ∂Ω be its bound. Let Q : P ∩ Ω → P be a condensing mapping. If Qu /u
for every u ∈ P ∩ ∂Ω, then the fixed point index iQ, P ∩ Ω, P is defined. If iQ, P ∩ Ω, P /
0,
then Q has a fixed point in P ∩ Ω. As the fixed point index theory of completely continuous
mapping, see 10, 11, we have the following lemmas that are needed in our argument for
condensing mapping.
Lemma 2.7. Let Q : P → P be condensing mapping; if
u/
λAu,
∀u ∈ ∂Pr , 0 < λ ≤ 1,
2.27
then iQ, Pr , P 1.
Lemma 2.8. Let Q : P → P be condensing mapping; if there exists v0 ∈ P, v0 /
θ, such that
u − Au /
τv0 ,
∀u ∈ ∂Pr , τ ≥ 0,
2.28
then iQ, Pr , P 0.
3. Main Results
P 1 i There exist δ > 0, a, ak > 0, such that for all x ∈ Pδ and t ∈ J, ft, x ≤
2
ax, Ik x ≤ ak x, and a m
k1 ak /σ < M.
ii There exist b, bk > 0, h0 ∈ CJ, K, and yk ∈ K, such that for all x ∈ P and
t ∈ J, ft, x ≥ bx − h0 t, Ik x ≥ bk x − yk , and b σ 2 m
k1 bk > M.
P 2 i There exist δ > 0, b, bk > 0, such that for all x ∈ Pδ and t ∈ J, ft, x ≥ bx, Ik x ≥
bk x, and b σ 2 m
k1 bk > M.
ii There exist a, ak > 0, h0 ∈ CJ, K, and yk ∈ K, such that for all x ∈ P and
2
t ∈ J, ft, x ≤ ax h0 t, Ik x ≤ ak x yk , and a m
k1 ak /σ < M.
Abstract and Applied Analysis
9
Theorem 3.1. Let E be an ordered Banach space, whose positive cone K is normal, f ∈ CJ × K, K,
and Ik ∈ CK, K, k 1, 2, . . . , m. Suppose that conditions P 0 and P 1 or P 2 are satisfied; then
the NBVP 1.1 has at least one positive solution.
Proof. We show, respectively, that the operator A defined by 2.18 has a nontrivial fixed point
in two cases that P 1 is satisfied and P 2 is satisfied.
Case 1. Assume that P 1 is satisfied; let 0 < r < δ, where δ is the constant in condition P 1,
to prove that A satisfies
∀u ∈ ∂Pr , 0 < λ ≤ 1.
u/
λAu,
3.1
If 3.1 is not true, then there exist u0 ∈ ∂Pr and 0 < λ0 ≤ 1, such that u0 λ0 Au0 ; by the
definition of A, u0 t satisfies
−u0 t Mu0 t λ0 ft, u0 t,
−Δu0 |ttk λ0 Ik u0 tk ,
t ∈ J, t /
tk ,
k 1, 2, . . . , m,
3.2
u0 0 u0 1 θ.
Integrating 3.2 from 0 to 1, using i of assumption P 1, we have
M − a
1
u0 tdt ≤
0
m
ak u0 tk .
3.3
k1
Since u0 ∈ P , for any t, s ∈ J, by the definition of P , we have u0 t ≥ σu0 s, u0 tk ≤
1/σu0 s, and thus
m
σM − au0 s ≤
k1
σ
ak
u0 s,
3.4
2
that is; M − a m
k1 ak /σ u0 s ≤ θ. So we obtain that u0 s ≤ θ in J, which contracts
with u0 ∈ ∂Pr . Hence 3.1 is satisfied; by Lemma 2.7, we have
iA, Pr , P 1.
3.5
Let e ∈ CJ, K, e 1, v0 t ≡ e, and obviously v0 ∈ P . We show that if R is large enough,
then
u − Au /
τv0 ,
∀u ∈ ∂PR , τ ≥ 0.
3.6
10
Abstract and Applied Analysis
In fact, if there exist u0 ∈ ∂PR , τ0 ≥ 0 such that u0 − Au0 τ0 v0 , then Au0 u0 − τ0 v0 ; by the
definition of A, u0 t satisfies
−u0 t Mu0 t − Mτ0 v0 ft, u0 t,
−Δu0 |ttk Ik u0 tk ,
t ∈ J, t /
tk ,
3.7
k 1, 2, . . . , m,
u0 0 u0 1 θ.
By ii of assumption P 1, we have
−u0 t Mu0 t ft, u0 t Mτ0 v0 ≥ bu0 t − h0 t,
t ∈ J .
3.8
Integrating on J and using ii of assumption P 1, we have
b − M
1
u0 tdt 0
m
bk u0 tk ≤
1
h0 tdt m
0
k1
yk .
3.9
k1
If b > M, for any t, s ∈ J, by the definition of P , we have u0 t ≥ σu0 s; u0 tk ≥ σu0 s; thus
σb − M σ
m
bk u0 s ≤
1
k1
By σb − M σ
m
k1
h0 tdt 0
m
yk .
3.10
k1
bk > 0 and the normality of cone K, we have
N h0 C m
k1 yk R1 .
u0 C ≤
σb − M σ m
k1 bk
3.11
If b ≤ M, then for any t, s ∈ J, by the definition of P , we have u0 t ≤ 1/σu0 s, u0 tk ≥
σu0 s; thus
By b σ 2
m
k1
1
m
m
b − M
σ bk u0 s ≤
h0 tdt yk .
σ
0
k1
k1
3.12
bk > M, and the normality of K, we have
N h0 C m
k1 yk R2 .
u0 C ≤
b − M/σ σ m
k1 bk
3.13
Let R > max{R1 , R2 , r}; then 3.6 is satisfied; by Lemma 2.8, we have
iA, PR , P 0.
3.14
Abstract and Applied Analysis
11
Combining 3.5, 3.14, and the additivity of fixed point index, we have
iA, Pr,R , P iA, PR , P − iA, Pr , P −1 /
0.
3.15
Therefore A has a fixed point in Pr,R , which is the positive solution of NBVP 1.1.
Case 2. Assume that P 2 is satisfied; let 0 < r < δ, where δ is the constant in condition P 2,
to proof that A satisfies
u − Au /
τv0 ,
∀u ∈ ∂Pr , τ ≥ 0,
3.16
where v0 t e ∈ P, e / θ. In fact, if there exists u0 ∈ ∂Pr and τ0 ≥ 0, such that u0 − Au0 τ0 v0 ,
then u0 satisfies 3.7 and i of condition P 2, and we have
−u0 t Mu0 t ft, u0 t Mτ0 v0 ≥ bu0 t,
t ∈ J .
3.17
Integrating on J and using i of assumption P 2, we have
b − M
1
u0 tdt 0
m
bk u0 tk ≤ θ.
3.18
k1
If b > M, for any t, s ∈ J, by the definition of P , we have u0 t ≥ σu0 s, u0 tk ≥
σu0 s, for all t, s ∈ J, and thus
σb − M σ
m
bk u0 s ≤ θ.
3.19
k1
By σb − M σ m
k1 bk > 0, we obtain that u0 s ≤ θ, which contracts with u0 ∈ ∂Pr . Hence
3.16 is satisfied.
If b ≤ M, for any t, s ∈ J, by the definition of P , we have u0 t ≤ 1/σu0 s, u0 tk ≥
σu0 s, for all t, s ∈ J, and thus
m
1
b − M σ bk u0 s ≤ θ.
σ
k1
3.20
By b σ 2 m
k1 bk > M, we obtain that u0 s ≤ θ, which contracts with u0 ∈ ∂Pr . Hence 3.16
is satisfied.
Hence, by Lemma 2.8, we have
iA, Pr , P 0.
3.21
Next, we show that if R is large enough, then
u/
λAu,
∀u ∈ ∂PR , 0 < λ ≤ 1.
3.22
12
Abstract and Applied Analysis
In fact, if there exists u0 ∈ ∂PR and 0 < λ0 ≤ 1 such that u0 λ0 Au0 , then u0 satisfies 3.2.
Integrating 3.2 on J, and using ii of P 2, we have
M − a
1
u0 tdt −
0
1
m
m
ak u0 tk ≤
h0 tdt yk .
0
k1
3.23
k1
For any t, s ∈ J, by the definition of P , we have u0 t ≥ σu0 s, u0 tk ≤ 1/σu0 s, for
all t, s ∈ J, and thus
By a m
k1
1
m
m
1
σM − a −
ak u0 s ≤
h0 tdt yk .
σ k1
0
k1
3.24
ak /σ 2 < M, we have
1
u0 s ≤
0
h0 tdt m
yk
m
k1
σM − a − 1/σ
k1
ak
.
3.25
By the normality of K, we have
1
y
N 0 h0 tdt m
k
k1
.
m
u0 s ≤
σM − a − 1/σ k1 ak
3.26
N h0 C m
k1 yk
R3 .
u0 C ≤
σM − a − 1/σ m
k1 ak
3.27
Thus
Let R > max{R3 , r}; then 3.22 is satisfied; by Lemma 2.7, we have
iA, PR , P 1.
3.28
Combining 3.21, 3.28, and the additivity of fixed index, we have
iA, Pr,R , P iA, PR , P − iA, Pr , P 1.
3.29
Therefore A has a fixed point in Pr,R , which is the positive solution of NBVP 1.1.
Remark 3.2. The conditions P 1 and P 2 are a natural extension of suplinear condition and
sublinear condition in Banach space E. Hence if Ik θ, then Theorem 3.1 improves and
extends the main results in 8, 9.
Abstract and Applied Analysis
13
4. Example
We provide an example to illustrate our main result.
Example 4.1. Consider the following problem:
∂2
− 2 wt, x wt, x ∂t
−Δ
1
1
wt, x, t ∈ J , x ∈ I,
100
1
1
w
, x , x ∈ I,
100
2
et−s w2 s, xds 0
∂
wt, x|t1/2
∂t
∂
∂
w0, x w1, x 0,
∂t
∂t
4.1
x ∈ I,
where J 0, 1, J J \ {1/2}, I 0, T , and T > 0 is a constant.
Conclusion
Problem 4.1 has at least one positive solution.
Proof. Let E CI, and K {w ∈ CI | wx ≥ 0, x ∈ I}; then E is a Banach space with
norm w maxt∈I |wx|, and K is a positive cone of E. Let ut wt, ·; then the problem
1
4.1 can be transformed into the form of NBVP 1.1, where M 1, ft, u 0 et−s u2 sds
1/100ut, and I1 u1/2 1/100u1/2. Evidently CJ, E is a Banach space with norm
uC maxt∈J ut, and CJ, K is positive cone of CJ, E. Let P {u ∈ CJ, K | ut ≥
σus, t, s ∈ J}, where σ 1/cosh2 1; then P is a cone in CJ, K, and for any u ∈ P, t ∈ J,
we have σuC ≤ ut ≤ uC .
Next, we will verify that the conditions P 0 and P 1 in Theorem 3.1 are satisfied.
It is easy to verify that for any R > 0, fJ × KR and I1 KR are bounded. Let gt, u 1 t−s 2
e
u sds; then g is completely continuous. So for any countable bounded set D ⊂ K,
0
we have
1
1
αD αD,
α ft, D ≤ α gt, D 100
100
αI1 D 1
αD,
100
and for L √
1/100, and M1 1/100, by simple calculations, 4L/M 2cosh2
√
M sinh M < 1. So condition P 0 is satisfied.
Let δ 1/100; then for u ∈ Pδ , we have
ft, u ≤
δ
1
1
δe − 1
1
ut ut ≤
ut,
σ
100
σ
100
1
1
1
1
u
ut.
I1 u
≤
2
100
2
100σ
0
4.2
√
M M1 /
et−s ds
4.3
Let a δe − 1/σ 1/100, a1 1/100σ; by simple calculations, we have a a1 /σ 2 < 1.
So i of condition P 1 is satisfied.
14
Abstract and Applied Analysis
Let R 100, for u ∈ P, uC ≥ R; we have ft, u ≥ 1/2 σ 2 R 1/100ut. For
u ∈ P , 0 ≤ uC ≤ R, we have ft, u ≤ R2 et 1. Hence, let h0 t 10000et 1, y1 0; then
for any u ∈ P , we have
1 2
1
σ R
ut − h0 t
2
100
1
1
σut.
I1 u
≥
2
100
ft, u ≥
4.4
Let b 1/2σ 2 R 1/100, and b1 1/100σ; by simple calculations, we have b σ 2 b1 > 1,
so ii of condition P 1 is satisfied.
By Theorem 3.1, Problem 4.1 has at least one positive solution.
Acknowledgment
This work was supported by NNSF of China no. 10871160, 11061031.
References
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