Electronic Journal of Differential Equations, Vol. 2011 (2011), No. 120, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLVABILITY OF A SECOND-ORDER MULTI-POINT
BOUNDARY-VALUE PROBLEMS AT RESONANCE ON A
HALF-LINE WITH DIM KER L=2
WEIHUA JIANG, BIN WANG, ZHENJI WANG
Abstract. We show the existence of solutions for a second-order multi-point
boundary-value problem at resonance on a half-line, where the dimension of
the kernel of the differential operator is 2. Our main tools are the coincidence
degree theory due to Mawhin, suitable operators, and algebraic methods. Our
results are illustrated with an example.
1. Introduction
In this article, we show the existence of solutions for the boundary-value problem
x00 (t) = f (t, x(t), x0 (t)) + e(t),
x(0) =
m
X
i=1
αi x(ξi ),
t ∈ (0, +∞),
lim x0 (t) =
t→+∞
n
X
βj x0 (ηj ),
(1.1)
(1.2)
j=1
where f : [0, +∞) × R2 → R, e ∈ L1 [0, +∞), 0 < ξ1 < ξ2 < · · · < ξm < +∞,
0 < η1 < η2 < · · · < ηn < +∞, m ≥ 2, n ≥ 1.
Multi-point boundary value problems of ordinary differential equations arise in
a variety of different areas of Applied Mathematics and Physics. For example, the
vibrations of a guy wire of a uniform cross-section being composed of N parts of
different densities can be set up as a multi-point boundary-value problem (see [19]);
many problems in the theory of elastic stability can be handled by the method
of multi-point problems(see [26]). Bridges of small size are often designed with
two supported points, which leads to a standard two-point boundary condition
and bridges of large size are sometimes contrived with multi-point supports, which
corresponds to a multi-point boundary condition (see [30]).
Boundary-value problem (1.1)-(1.2) is called a problem at resonance if Lx :=
x00 (t) = 0 has non-trivial solutions under the boundary condition (1.2); i.e., when
dim ker L ≥ 1.
On the finite interval [0,1], the first-order, second-order and high-order multipoint boundary-value problems at resonance have been studied by many authors
(see [2, 4, 5, 6, 7, 8, 9, 10, 14, 15, 16, 18, 20, 21, 22, 24, 25]), where dim ker L = 1.
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Resonance; Fredholm operator; multi-point boundary-value problem;
coincidence degree theory.
c 2011 Texas State University - San Marcos.
Submitted June 30, 2011. Published September 19, 2011.
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In [11, 28, 29], the second-order multi-point boundary-value problems at resonance
have been discussed when dim ker L = 2 on the finite interval [0,1]. Recently, the
boundary-value problems at resonance on the infinite interval with dim ker L = 1
has been investigated by many authors, see [12, 13, 27, 17]and references cited
therein. Although the existing literature on solutions of multi-point boundaryvalue problems is quite wide, to the best of our knowledge, there is few paper to
investigate the resonance case with dim ker L = 2 on the infinite interval.
Motivated by the above results, by constructing the suitable operators and getting help from the algebraic methods, we will show the existence of solutions for the
second-order multi-point boundary-value problem at resonance on a half-line with
dim ker L = 2, which brings many difficulties. And we give an example to illustrate
our results. Some methods used in this paper are new and they can be used to
solve the nth-order boundary-value problems at resonance with 1 < dim ker L ≤ n.
This paper is organized as follows. In Section 2, some necessary backgrounds
will be stated and some lemmas be proved. In Section 3, the main results will be
given and proved. In Section 4, an example is given to illustrate our results.
In this article, we will assume the following conditions:
(C1) f : [0, +∞) × R2 → R is a S-Carathéodory function; i.e.,
(i) f (t, ·) is continuous on R2 for a.e. t ∈ [0, +∞).
(ii) f (·, x) is Lebesgue measurable on [0, +∞) for each x ∈ R2 .
(iii) For each r > 0, there exists a function ϕr ∈ L1 [0, +∞), ϕr (t) ≥ 0,
R +∞
t ∈ [0, +∞) satisfying 0 sϕr (s)ds < +∞ such that
|f (t, x)| ≤ ϕr (t), a. e. t ∈ [0, +∞), ||x|| < r.
Pn
Pm
(C2)
i=1 αi =
j=1 βj = 1,
i=1 αi ξi = 0.
Q1 e−t Q2 e−t
a11 a12
:=
6= 0, where
(C3) ∆ =
Q1 te−t Q2 te−t
a21 a22
Z +∞
Z ξi
m
n
X
X
(ξi − s)y(s)ds, Q2 y =
βj
y(s)ds.
αi
Q1 y =
Pm
i=1
0
j=1
ηj
2. Preliminary
To obtain our results, we introduce some notation and two theorems.
Let X and Y be real Banach spaces and let L : dom(L) ⊂ X → Y be a Fredholm
operator with index zero, P : X → X, Q : Y → Y be projectors such that
Im P = ker L,
ker Q = Im L,
X = ker L ⊕ ker P,
Y = Im L ⊕ Im Q.
It follows that
L|dom L∩ker P : dom L ∩ ker P → Im L
is invertible. We denote the inverse by KP .
If Ω is an open bounded subset of X, dom L ∩ Ω 6= ∅, the map N : X → Y
will be called L-compact on Ω if QN (Ω) is bounded and KP (I − Q)N : Ω → X is
compact.
Theorem 2.1 (citem4). Let L : dom L ⊂ X → Y be a Fredholm operator of index
zero and N : X → Y L-compact on Ω. Assume that the following conditions are
satisfied:
(1) Lx 6= λN x for every (x, λ) ∈ [(dom L \ ker L) ∩ ∂Ω] × (0, 1);
(2) N x ∈
/ Im L for every x ∈ ker L ∩ ∂Ω;
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(3) deg(QN |ker L , Ω ∩ ker L, 0) 6= 0, where Q : Y → Y is a projection such that
Im L = ker Q.
Then the equation Lx = N x has at least one solution in dom L ∩ Ω.
Let
|x(t)|
and
t→+∞ 1 + t
with norm kxk = max{kxk0 , kx0 k∞ }, where
X = {x ∈ C 1 [0, +∞) : lim
|x(t)|
,
t∈[0,∞) 1 + t
kxk0 = sup
kxk∞ =
lim x0 (t) exist}
t→+∞
sup
|x(t)|.
t∈[0,+∞)
It is easy to prove that (X, k · k) is a Banach space.
Theorem 2.2 ([1]). Let M ⊂ X. Then M is relatively compact if the following
conditions hold:
(a) M is bounded in X;
(b) the functions belonging to M are equi-continuous on any compact interval
of R+ ;
(c) the functions from M are equi-convergent at +∞.
R +∞
Let Y = L1 [0, +∞) with the norm kyk1 = 0 |y(s)|ds. Define Lx = x00 , with
domain
m
n
X
X
dom L = {x ∈ X : x00 ∈ L1 [0, +∞), x(0) =
αi x(ξi ), lim x0 (t) =
βj x0 (ηj )}.
t→+∞
i=1
j=1
Obviously, ker L = {a + bt : a, b ∈ R}. Now, we will prove that
Im L = {y ∈ Y : Q1 y = Q2 y = 0}.
In fact, if Lx = y, then y ∈ Y and
x(t) = x(0) + x0 (0)t +
Z
t
(t − s)y(s)ds.
0
It follows from (1.2) that Q1 y = Q2 y = 0.
On the other hand, assume y ∈ Y satisfying Q1 y = Q2 y = 0. Take
Z t
x(t) =
(t − s)y(s)ds.
0
Then x ∈ X, x00 (t) = y(t) and x satisfies (1.2). So, x ∈ dom L; i.e., y ∈ Im L.
Define operators T1 , T2 : Y → Y as follows:
1
1
T1 y = (∆11 Q1 y + ∆12 Q2 y)e−t , T2 y = (∆21 Q1 y + ∆22 Q2 y)e−t ,
∆
∆
where ∆ij is the algebraic cofactor of aij . Define the operator Q : Y → Y by
Qy = T1 y + (T2 y) · t.
By a simple calculation, we obtain T1 (T1 y) = T1 y, T1 (T2 yt) = 0, T2 (T1 y) = 0,
T2 (T2 yt) = T2 y. So, Q2 y = Qy; i.e., Q : Y → Y is a linear projector. Obviously, Q
is continuous.
For y ∈ Y , y = (y − Qy) + Qy, we have Qy ∈ Im Q and Q(y − Qy) = 0. It
follows from Q(y − Qy) = 0, the definitions of Q, T1 , T2 and condition (C3), that
Q1 (y − Qy) = Q2 (y − Qy) = 0; i.e., y − Qy ∈ Im L. So, Y = Im L + Im Q.
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W. JIANG, B. WANG, Z. WANG
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Take y ∈ Im L ∩ Im Q, then y = Qy = 0; i.e., Y = Im L ⊕ Im Q, So, we have
dim ker L = codim Im L = 2, thus L is a Fredholm operator with index zero.
Define the continuous projection P : X → ker L by
(P x)(t) = x(0) + x0 (0)t,
t ∈ [0, +∞).
Then X = ker L ⊕ ker P .
Define the operator KP : Im L → dom L ∩ ker P by
Z t
KP y =
(t − s)y(s)ds.
0
Then KP is the inverse operator of L|dom L∩ker P and
kKP yk ≤ kyk1 .
(2.1)
Rt
In fact, for x ∈ dom L ∩ ker P , KP L(x) = 0 (t − s)x00 (s)ds = x(t). On the other
Rt
hand, for y ∈ Im L, LKP (y) = ( 0 (t − s)y(s)ds)00 = y(t). By
Z +∞
|KP y|
≤
|y(s)|ds = kyk1 ,
1+t
0
Z t
|(KP y)0 (t)| = |
y(s)ds| ≤ kyk1 ,
0
we obtain (2.1).
Let the nonlinear operator N : X → Y be defined by
N x = f (t, x(t), x0 (t)) + e(t),
t ∈ [0, +∞).
Then problem (1.1)–(1.2) is equivalent to
x ∈ dom L.
Lx = N x,
Lemma 2.3. Suppose that Ω is an open bounded subset of X such that dom L∩Ω 6=
Φ. Then N is L-compact on Ω.
Proof. Since Ω is bounded, there exists a constant r > 0 such that kxk ≤ r for any
x ∈ Ω. For x ∈ Ω, by (C1), we obtain
Z ξi
m
X
|Q1 N x| = |
αi
(ξi − s)[f (s, x(s), x0 (s)) + e(s)]ds|
≤
0
i=1
m
X
+∞
Z
|αi ξi |
ϕr (s) + |e(s)|ds := l1
0
i=1
and
|Q2 N x| = |
n
X
Z
j=1
≤
n
X
j=1
+∞
f (s, x(s), x0 (s)) + e(s)ds|
βj
ηj
Z
|βj | ·
+∞
ϕr (s) + |e(s)|ds := l2 .
0
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SOLVABILITY OF A SECOND-ORDER DE
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Thus,
Z
+∞
kQN xk1 =
|QN x(s)|ds
0
Z
≤
+∞
Z
|T1 N x(s)|ds +
0
+∞
|T2 N x(s)|sds
0
1
[|∆11 | · |Q1 N x| + |∆12 | · |Q2 N x|]
|∆|
1
+
[|∆21 | · |Q1 N x| + |∆22 | · |Q2 N x|]
|∆|
1
[(|∆11 | + |∆21 |)l1 + (|∆12 | + |∆22 |)l2 ].
≤
|∆|
≤
(2.2)
So, QN (Ω) is bounded. Now, we will prove that KP (I − Q)N (Ω) is compact.
(a). Obviously, KP (I − Q)N : Ω → Y is continuous. For x ∈ Ω, since
Z +∞
Z +∞
kN xk1 =
|f (s, x(s), x0 (s)) + e(s)|ds ≤
ϕr (s) + |e(s)|ds := l3 , (2.3)
0
0
t
|KP (I − Q)N x(t)|
1
=
|
(t − s)(I − Q)N x(s)ds|
1+t
1+t 0
Z +∞
≤
|N x(s)| + |QN x(s)|ds
Z
0
= kN xk1 + kQN xk1 ,
and
Z t
|[KP (I − Q)N x]0 (t)| = |
(I − Q)N x(s)ds|
0
Z +∞
≤
|N x(s)| + |QN x(s)|ds
0
= kN xk1 + kQN xk1 ,
by (2.2) and (2.3), we obtain that KP (I − Q)N (Ω) is bounded.
(b). For any T ∈ [0, +∞), we will prove that functions belonging to KP (I −
Q)N (Ω) are equi-continuous on [0, T ]. In fact, for x ∈ Ω, we have
|N x(s)| ≤ ϕr (s) + |e(s)|. s ∈ [0, ∞),
1
|QN x(s)| ≤
[(|∆11 |l1 + |∆12 |l2 ) + (|∆21 |l1 + |∆22 |l2 )s]e−s .
|∆|
For any t1 , t2 ∈ [0, T ], t1 < t2 , we have
KP (I − Q)N x(t1 ) KP (I − Q)N x(t2 )
−
1 + t1
1 + t2
R t1
R t2
(t1 − s)(I − Q)N x(s)ds
(t2 − s)(I − Q)N x(s)ds
0
− 0
|
=|
1 + t1
1 + t2
Z t1
Z t2
t1
t2
(I − Q)N x(s)ds −
(I − Q)N x(s)ds|
≤|
1 + t1 0
1 + t2 0
Z t1
Z t2
1
1
+|
s(I − Q)N x(s)ds −
s(I − Q)N x(s)ds|
1 + t1 0
1 + t2 0
(2.4)
(2.5)
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W. JIANG, B. WANG, Z. WANG
≤|
t2
t1
−
|·
1 + t1
1 + t2
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+∞
Z
t2
Z
|N x(s)| + |QN x(s)|ds +
|N x(s)| + |QN x(s)|ds
0
t1
Z +∞
1
1
−
|·T
|N x(s)| + |QN x(s)|ds
+|
1 + t1
1 + t2
0
Z t2
+T ·
|N x(s)| + |QN x(s)|ds
t1
t2
1
1
t1
−
|+|
−
| · T )(kN xk1 + kQN xk1 )
= (|
1 + t1
1 + t2
1 + t1
1 + t2
Z t2
+ (1 + T )
|N x(s)| + |QN x(s)|ds.
t1
and
Z
|[KP (I − Q)N x]0 (t1 ) − [KP (I − Q)N x]0 (t2 )| = |
t2
(I − Q)N x(s)ds|
Z
t1
t2
≤
|N x(s)| + |QN x(s)|ds.
t1
t
1
By (2.2)–(2.5), the continuity of 1+t
and 1+t
and the absolute continuity of integral,
we obtain that functions from KP (I − Q)N (Ω) are equi-continuous on [0, T ].
(c). Now, we will show that functions in KP (I − Q)N (Ω) are equi-convergent at
+∞. For x ∈ Ω, we have
Z +∞
KP (I − Q)N x(t)
=
(I − Q)N x(s)ds.
lim
t→+∞
1+t
0
Z +∞
lim [KP (I − Q)N x]0 (t) =
(I − Q)N x(s)ds.
t→+∞
0
By
Z +∞
KP (I − Q)N x(t)
−
(I − Q)N x(s)ds
1+t
0
Z t
Z +∞
t
≤|
(I − Q)N x(s)ds −
(I − Q)N x(s)ds
1+t 0
0
Z t
1
+
|s(I − Q)N x(s)|ds
1+t 0
Z +∞
≤
|(I − Q)N x(s)|ds
t
Z
Z +∞
i
1 h +∞
+
|s(I − Q)N x(s)|ds
|(I − Q)N x(s)|ds +
1+t 0
0
Z +∞
Z +∞
1
≤
|N x(s)| + |QN x(s)|ds +
(1 + s)[|N x(s)| + |QN x(s)|]ds,
1+t 0
t
and
0
Z
|[KP (I − Q)N x] (t) −
+∞
Z
0
+∞
|N x(s)| + |QN x(s)|ds,
(I − Q)N x(s)ds| ≤
t
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SOLVABILITY OF A SECOND-ORDER DE
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From (2.4) and (2.5), we can get that functions from KP (I − Q)N (Ω) are equicontinuous at +∞. By Theorem 2.2, we obtain that KP (I − Q)N (Ω) is compact.
Therefore, N is L−compact on Ω.
3. Main results
The following theorem is our main result.
Theorem 3.1. Assume that (C1)–(C3) and the following conditions hold:
(H1) There exist functions α(t), β(t), γ(t), δ(t) ∈ L1 [0, +∞), and θ ∈ [0, 1) such
that either
|u|
|u| θ
|f (t, u, v)| ≤ α(t) + β(t)
+ γ(t)|v| + δ(t)(
)
1+t
1+t
or
|u|
+ γ(t)|v| + δ(t)|v|θ ;
|f (t, u, v)| ≤ α(t) + β(t)
1+t
(H2) There exist constants A > 0, B > 0 such that, if |x(t)| > A for every
t ∈ [0, B] or |x0 (t)| > A for every t ∈ [0, +∞), then either Q1 N x 6= 0 or
1
;
Q2 N x 6= 0, where kβk1 + kγk1 < 2+B
(H3) There exists a constant C > 0 such that, if |a| > C or |b| > C, then either
(1) aQ1 N (a + bt) + bQ2 N (a + bt) < 0, or
(2) aQ1 N (a + bt) + bQ2 N (a + bt) > 0.
Then the boundary-value problem (1.1)–(1.2) has at least one solution in X.
Proof. We divide the proof into four steps.
Step1. Let
Ω1 = {x ∈ dom L \ ker L : Lx = λN x, for some λ ∈ [0, 1]}.
We will prove that Ω1 is bounded. In fact, x ∈ Ω1 means λ 6= 0 and N x ∈ Im L.
Thus
Q1 N x = Q2 N x = 0.
By (H2), there exist t0 ∈ [0, B], t1 ∈ [0, +∞) such that
|x(t0 )| ≤ A,
|x0 (t1 )| ≤ A.
So,
|x0 (t)| = |x0 (t1 ) −
Z
t1
x00 (s)ds| ≤ A +
t
Z
t1
|N x(s)|ds ≤ A + kN xk1 ;
t
i.e., kx0 k∞ ≤ A + kN xk1 . Considering
Z t0
Z
|x(0)| = |x(t0 ) −
x0 (s)ds| ≤ A + |
0
t0
x0 (s)ds|
0
≤ A + kx0 k∞ · B ≤ A(1 + B) + B · kN xk1 ,
we have
kP xk ≤ |x(0)| + |x0 (0)| ≤ A(2 + B) + (1 + B)kN xk1 .
By LP x = 0, (2.1) and (H1), we obtain
kxk = kP x + (I − P )xk ≤ kP xk + kKP L(I − P )xk
≤ kP xk + kLxk1 ≤ kP xk + kN xk1
≤ (2 + B)(A + kN xk1 )
≤ (2 + B)(A + kαk1 + kβk1 · kxk + kγk1 · kxk + kδk1 · kxkθ + kek1 ).
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W. JIANG, B. WANG, Z. WANG
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So,
2+B
(A + kαk1 + kek1 + kδk1 · kxkθ ).
1 − (2 + B)(kβk1 + kγk1 )
It follows from θ ∈ [0, 1) that Ω1 is bounded.
Step2. Set Ω2 = {x ∈ ker L : N x ∈ Im L}. Then Ω2 is bounded. In fact,
x ∈ Ω2 implies x = a + bt and Q1 N (a + bt) = Q2 N (a + bt) = 0. By (H3 ), we obtain
|a| ≤ C, |b| ≤ C. So, Ω2 is bounded.
Step3. Define the isomorphism J : ker L → Im Q by
kxk ≤
J(a + bt) =
1
[∆11 a + ∆12 b + (∆21 a + ∆22 b)t]e−t .
∆
Assume (H3)(1) holds. Let
Ω3 = {x ∈ ker L : −λJx + (1 − λ)QN x = 0, for some λ ∈ [0, 1]}.
Then Ω3 is bounded.
In fact, x ∈ Ω3 means that there exist constants a, b ∈ R, λ ∈ [0, 1] such that
x = a + bt and λJx = (1 − λ)QN x. If λ = 0, then QN x = 0. So,
∆11 Q1 N x + ∆12 Q2 N x = 0,
∆21 Q1 N x + ∆22 Q2 N x = 0.
It follows from ∆ 6= 0 that Q1 N x = Q2 N x = 0. By (H3), we obtain |a| ≤ C,
|b| ≤ C.
If λ = 1, we can similarly get a = b = 0. For λ ∈ (0, 1), by λJx = (1 − λ)QN x,
we obtain
λ∆11 a + λ∆12 b = (1 − λ)∆11 Q1 N (a + bt) + (1 − λ)∆12 Q2 N (a + bt),
λ∆21 a + λ∆22 b = (1 − λ)∆21 Q1 N (a + bt) + (1 − λ)∆22 Q2 N (a + bt).
It follows from ∆ 6= 0 that
λa = (1 − λ)Q1 N (a + bt),
λb = (1 − λ)Q2 N (a + bt).
If |a| > C, |b| > C, by (H3)(1), we obtain
λ(a2 + b2 ) = (1 − λ)[aQ1 N (a + bt) + bQ2 N (a + bt)] < 0,
a contradiction. So, Ω3 is bounded.
Remark 3.2. If (H3)(2) holds, take
Ω3 = {x ∈ ker L : λJx + (1 − λ)QN x = 0, for some λ ∈ [0, 1]}.
We can similarly prove that Ω3 is bounded.
S
S3
Step4. Take an open bounded set Ω ⊃ i=1 Ωi {0}. We will prove that
(1.1)–(1.2) has at least one solution in dom L ∩ Ω.
By Step1 and Step2, we obtain
(1) Lx 6= λN x, for every (x, λ) ∈ [(dom L \ ker L) ∩ ∂Ω] × (0, 1);
(2) N x 6∈ Im L, for every x ∈ ker L ∩ ∂Ω.
Now we will show that
(3) deg(QN |ker L , Ω ∩ ker L, 0) 6= 0.
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Let H(x, λ) = ±λJx + (1 − λ)QN x. By step 3, we know that H(x, λ) 6= 0, for every
(x, λ) ∈ (ker L ∩ ∂Ω) × [0, 1]. Thus, by the homotopy property of degree, we obtain
deg(QN |ker L , Ω ∩ ker L, 0) = deg(H(·, 0), Ω ∩ ker L, 0)
= deg(H(·, 1), Ω ∩ ker L, 0)
= deg(±J, Ω ∩ ker L, 0) = ±1 6= 0.
By Theorem 2.1, we can get that Lx = N x has at least one solution in dom L∩Ω;
i.e. , (1.1)–(1.2) has at least one solution in X. The prove is completed.
4. Example
Let’s consider the boundary-value problem
x00 (t) = f (t, x(t), x0 (t)) + e(t),
x(0) = 2x(1) − x(2),
0
t ∈ [0, ∞),
0
x (∞) = x (2),
(4.1)
(4.2)
where
(
−e−10t x(0),
0 ≤ t ≤ 2,
p
f (t, x(t), x (t)) =
−10t
0
−t 3 0
x (t), t > 2.
e
sinx (t) + e
(
0,
0 ≤ t ≤ 2,
e(t) =
te−t , t > 2.
0
Corresponding to problem (1.1)-(1.2), we have that m = 2, n = 1, α1 = 2, α2 = −1,
ξ1 = 1, ξ2 = 2, β1 = 1, η1 = 2. Obviously, (C1) and (C2) are satisfied. By simple
calculation, we obtain a11 = −(1 − e−1 )2 , a21 = 6e−1 − 2 − 4e−2 , a12 = e−2 ,
a22 = 3e−2 .
a12
a
∆ = 11
= e−4 − e−2 6= 0 .
a21 a22
So, (C3) is satisfied. Take α(t) = 0, θ = 31 ,
(
(
(1 + t)e−10t , 0 ≤ t ≤ 2,
0,
0 ≤ t ≤ 2,
β(t) =
γ(t) =
−10t
0,
t > 2,
e
, t > 2,
(
0,
0 ≤ t ≤ 2,
δ(t) =
e−t , t > 2.
1 11
31 −20
Then f satisfies (H1). We can easily get that kβk1 = 10
[ 10 − 10
e ], kγk1 =
1 −20
e
.
So,
we
have
kβk
+
kγk
<
1/5.
1
1
10
Let B = 2, A = e−54 /1000. We get that Q1 N x 6= 0 if |x(t)| > A, for any
t ∈ [0, 2] and Q2 N x 6= 0 if |x0 (t)| > A, for any t ∈ [0, ∞). This means that (H2) is
satisfied.
Set C = 100. We can easily get that
aQ1 N (a + bt) + bQ2 N (a + bt) > 0
if |a| > C or |b| > C. So, (H3) is satisfied.
By theorem 3.1, we obtain that problem (4.1)–(4.2) has at least one solution.
10
W. JIANG, B. WANG, Z. WANG
EJDE-2011/??
Acknowledgments. This work is supported by grants 10875094 and 60874003
from the Natural Science Foundation of China; 08M007, 11171088 and A2009000664
from the Natural Science Foundation of Hebei Province; (2008153) from the Foundation of Hebei Education Department; XL200814 from the Foundation of Hebei
University of Science and Technology.
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Weihua Jiang
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
Hebei, China
E-mail address: weihuajiang@hebust.edu.cn
Bin Wang
Department of Basic Courses, Hebei Professional and Technological College of Chemical and Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei, China
E-mail address: wb@hebcpc.cn
Zhenji Wang
Department of Basic Courses, Hebei Professional and Technological College of Chemical and Pharmaceutical Engineering, Shijiazhuang, 050026, Hebei, China
E-mail address: wzj@hebcpc.cn