Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 240, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF SOLUTIONS FOR FOUR-POINT RESONANCE
BOUNDARY-VALUE PROBLEMS ON TIME SCALES
NING WANG, HUI ZHOU, LIU YANG
Abstract. By using Mawhin’s continuation theorem, we prove the existence
of solutions for a class of multi-point boundary-value problem under different
resonance conditions on time scales.
1. Introduction
Consider multi-point boundary-value problem on time scales
x∆∇ (t) = f (t, x(t), x∆ (t)) + e(t), t ∈ (0, 1) ∩ T,
(1.1)
subject to boundary condition
x(0) = αx(η), x(1) = βx(ξ),
(1.2)
2
where T is a time scale such that 0, 1 ∈ T, η, ξ ∈ (0, 1) ∩ T, f : T × R → R is a
continuous function and e(t) ∈ L1 [0, 1].α, β ∈ R hold
α = 1, β = 1
(1.3)
αη(1 − β) + (1 − α)(1 − βξ) = 0.
(1.4)
or
By condition (1.3) or (1.4), the differential operator in (1.1) is not invertible, which
is called problem at resonance. The study on multi-point nonlocal boundary-value
problems for linear second-order ordinary differential equations was initiated by
Il’in and Moiseev [6, 7]. Since then some existence results were obtained for general
boundary-value problems by several authors. We refer the reader to some recent
results, such as [3, 5, 9, 10, 11] at non-resonance, and [4, 13, 14, 15] at resonance
and reference therein.
For problem (1.1), (1.2) in the continuous setting, by using upper and lower solution method Rachünková [13, 14, 15] obtained excellent results about the problem
x00 (t) = f (t, x(t), x0 (t)) + e(t),
t ∈ (0, 1),
(1.5)
subject to boundary condition
x(0) = x(η), x(1) = x(ξ)
2010 Mathematics Subject Classification. 34N05, 34B10.
Key words and phrases. Boundary-value problem; time scale; coincidence degree.
c
2015
Texas State University - San Marcos.
Submitted March 6, 2015. Published September 21, 2015.
1
(1.6)
2
N. WANG, H. ZHOU, L. YANG
EJDE-2015/240
Bai [2] developed the upper and lower solution method and obtained existence and
multiplicity results for the resonance problem
x00 (t) = f (t, x(t), x0 (t)) + e(t),
t ∈ (0, 1),
(1.7)
subject to boundary condition
x(0) = αx(η),
x(1) = βx(ξ)
(1.8)
where condition (1.4) is satisfied. But the existence results for resonance cases
on time scales is rare (see [8]). Motivated by works above, we apply coincidence
degree theory [12] to establish existence results for resonance problems (1.1)–(1.4).
Considering that the main tools we used are different from [2, 13, 14, 15], the results
we establish follows are new even in the continuous setting.
The article is organized as follows. Some basic definitions and conclusions on
time scales and the main tools we used are introduced in section 2. In section 3, we
discuss the existence of solutions under condition (1.3) and the existence results of
case (1.4) is considered in section 4.
2. Preliminaries
First we present some basic definitions on time scales (see [1]). A time scale T is
a closed nonempty subset of R. For t < sup T and r > inf T , we define the forward
jump operator σ and the backward jump operator ρ respectively by
σ(t) = inf{τ ∈ T |τ > t},
ρ(t) = sup{τ ∈ T |τ < t},
for all t ∈ T . If σ(t) > t, t is said to be right scattered, and if σ(t) = t, t is said to
be right dense. If ρ(t) < t, t is said to be left scattered, and if ρ(t) = t, t is said to
be left dense. A function f is left-dense continuous, if f is continuous at each left
dense point in T and its right-sided limits exists at each right dense points in T .
For u : T → R and t ∈ T , we define the delta-derivative of u(t), u∆ (t), to be
the number (when it exists), with the property that for each ε > 0, there is a
neighborhood, U , of t such that for all s ∈ U ,
|u(σ(t)) − u(s) − u∆ (t)(σ(t) − s)| ≤ ε|σ(t) − s|.
For u : T → R and t ∈ T , we define the nabla-derivative of u(t), u∇ (t), to be
the number (when it exists), with the property that for each ε > 0, there is a
neighborhood, U , of t such that for all s ∈ U ,
|u(ρ(t)) − u(s) − u∇ (t)(ρ(t) − s)| ≤ ε|σ(t) − s|.
Then we recall some notations and an abstract existence result briefly.
Let X, Y be real Banach spaces and let L : dom L ⊂ X → Y be a Fredholm
operator with index zero, here dom L denotes the domain of L. This means that
Im L is closed in Y and dim ker L = dim(Y / Im L) < +∞. Consider the supplementary subspaces X1 and Y1 such that X = ker L ⊕ X1 and Y = Im L ⊕ Y1
and let P : X → ker L and Q : Y → Y1 be the natural projections. Clearly,
ker L ∩ (dom L ∩ X1 ) = {0}, thus the restriction Lp := L|dom L∩X1 is invertible.
Denote by K the inverse of Lp .
Let Ω be an open bounded subset of X with dom L ∩ Ω 6= ∅. A map N : Ω → Y
is said to be L-compact in Ω if QN (Ω) is bounded and the operator K(I − Q)N :
Ω → X is compact. We first give the famous Mawhin continuation theorem.
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EXISTENCE OF SOLUTIONS
3
Lemma 2.1 (Mawhin [12]). Suppose that X and Y are Banach spaces, and L :
dom L ⊂ X → Y is a Fredholm operator with index zero.Furthermore, Ω ⊂ X is an
open bounded set and N : Ω → Y is L-compact on Ω. If
(1) Lx 6= λN x for all x ∈ ∂Ω ∩ dom L, λ ∈ (0, 1);
(2) N x 6∈ Im L for all x ∈ ∂Ω ∩ ker L;
(3) deg{JQN, Ω ∩ ker L, 0} =
6 0, where J : ker L → Im Q is an isomorphism,
then the equation Lx = N x has a solution in Ω ∩ dom L.
3. Existence results under condition (1.3)
In this section problem (1.1), (1.2) is considered under the assumption that α = 1
and β = 1. We introduce the space
X = {x : [0, 1] → R : x∆ ∈ AC[0, 1], x∆∇ ∈ L1 [0, 1]}
endowed with the norm kxk = sup{kxk0 , kx∆ k0 }, where kxk0 = supt∈[0,1] |x(t)|.
Let Y = L1 [0, 1] with the norm
Z 1
kxk1 =
|x(t)|∇t.
0
Let a linear mapping L : dom L ⊂ X → Y with
dom L = {x ∈ X : x(0) = x(η), x(1) = x(ξ)}
be defined by Lx = x∆∇ (t). Define the mapping N : X → Y by
N x(t) = f (t, x(t), x∆ (t)) + e(t)
Then problem (1.1), (1.2) can be written as Lx = N x, here L is a linear operator.
Lemma 3.1. If β = 1 and α = 1 then
(1)
Z η
Z ξ
Z
Im L = {y(t)|
(ξ−s)y(s)∇s+
η(ξ−1)y(s)∇s−
0
η
1
η(1−s)y(s)∇s = 0}; (3.1)
0
(2) L : dom L ⊂ X → Y is a Fredholm operator with index zero;
(3) Define projector operator P : X → ker L as P x = x(0), then the generalized
inverse of operator L, KP : Im L → dom L ∩ ker P can be written as
Z
Z t
t η
KP (y) = −
(η − s)y(s)∇s +
(t − s)y(s)∇s,
(3.2)
η 0
0
satisfying kKP (y(t))k1 ≤ 2kyk1 .
Proof. (1) First we show that
Z η
Z
Im L = {y(t) ∈ Y |
(ξ − s)y(s)∇s +
0
ξ
Z
η(ξ − 1)y(s)∇s −
η
1
η(1 − s)y(s)∇s = 0}.
0
First suppose y(t) ∈ Im L, then there exists x(t) such that Lx = y. That is
Z t
x(t) =
(t − s)y(s)∇s + x∆ (0)t + x(0).
0
Condition x(0) = x(η), x(1) = x(ξ) imply that
Z η
Z ξ
Z
(ξ − s)y(s)∇s +
η(ξ − 1)y(s)∇s −
0
η
0
1
η(1 − s)∇s = 0.
4
N. WANG, H. ZHOU, L. YANG
Next we suppose that
Z η
Z
y(t) ∈ {y(t)|
(ξ − s)y(s)∇s +
0
EJDE-2015/240
ξ
1
Z
η(ξ − 1)y(s)∇s −
η(1 − s)∇s = 0}.
η
0
Let x(t) ∈ X, where
Z
x(t) =
0
t
t
(t − s)y(s)∇s −
η
Then Lx = x∆∇ = y(t). because
Z η
Z
y(t) ∈ {y(t)|
(ξ − s)y(s)∇s +
0
Z
η
(η − s)y(s)∇s .
0
ξ
Z
η(ξ − 1)y(s)∇s −
η
1
η(1 − s)y(s)∇s = 0},
0
by a simple computation we have x(0) = x(η), x(1) = x(ξ). Thus y(t) ∈ Im L.
Summing up two steps above we obtain
Z η
Z ξ
Z 1
Im L = {y(t) ∈ Y |
(ξ − s)y(s)∇s +
η(ξ − 1)y(s)∇s −
η(1 − s)y(s)∇s = 0}.
0
η
0
(2) We claim that L is a Fredholm operator with index zero. It is easy to see
that ker L = R. Next we prove that Y = Im L ⊕ ker L. Suppose y(t) ∈ Y , define
the projector operator Q as
Rη
Rξ
R1
(ξ − s)y(s)∇s + η η(ξ − 1)y(s)∇s − 0 η(1 − s)y(s)∇s
0
Q(y) =
Rη
Rξ
R1
(ξ − s)∇s + η η(ξ − 1)∇s − 0 η(1 − s)∇s
0
Let y ∗ = y(t) − Q(y(t)), by a simple computation, y ∗ ∈ Im L. Hence Y = Im L +
ker L, furthermore considering Im L ∩ ker L = {0}, we have Y = Im L ⊕ ker L. Thus
dim ker L = codim Im L, which means L is a Fredholm operator with index zero.
(3) Define the projector operator P : X → ker L as P x = x(0), for y(t) ∈ Im L,
(LKP )(y(t)) = y(t),
and for x(t) ∈ dom L ∩ ker P , we know
(KP L)(x(t)) = KP (x∆∇ (t))
Z
Z tZ s
t η
(η − s)x∆∇ (s)∇s +
x∆∇ (τ )∇τ ∇s
=−
η 0
0
0
= x(t).
This shows that KP = (Ldom L∩ker P )−1 . Furthermore from the definition of the
norms in the X, Y ,
kKP (y(t))k1 ≤ 2kyk1 .
The above arguments complete the proof.
Theorem 3.2. Let f : [0, 1] × R2 → R be a continuous function. Assume that
following conditions are satisfied:
(C1) There exist functions a, b, c, r ∈ L1 [0, 1] and constant θ ∈ [0, 1) such that
for all (x, y) ∈ R2 , either
|f (t, x, y)| ≤ a(t)|x| + b(t)|y| + c(t)|y(t)|θ + r(t),
(3.3)
|f (t, x, y)| ≤ a(t)|x| + b(t)|y| + c(t)|x(t)|θ + r(t).
(3.4)
or
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(C2) There exists a constant M > 0 such that for x ∈ dom L, |x(ρ(t))| > M , for
all t ∈ [0, 1],
Z η
Z ξ
(ξ − s)(f (s, x, x∆ ) + e(s))∇s +
η(ξ − 1)(f (s, x, x∆ ) + e(s))∇s
0
η
(3.5)
Z 1
η(1 − s)(f (s, x, x∆ ) + e(s))∇s 6= 0
−
0
(C3) There exists M ∗ > 0 such that for d ∈ R, if |d| > M ∗ , then either
Z η
(f (t, d, 0) + e(t))∇t > 0
d×
(3.6)
0
or else
Z
d×
η
(f (t, d, 0) + e(t))∇t < 0
(3.7)
0
Then for each e ∈ L1 [0, 1], resonance problem (1.1), (1.2) with α = 1 and
β = 1 has at least one solution provided that
1
kak1 + kbk1 < .
3
Proof. We divide the proof into two steps.
Step 1. Let
Ω1 = {x ∈ dom L \ ker L : Lx = λN x}
for some λ ∈ [0, 1],
Then Ω1 is bounded. Suppose that x ∈ Ω1 , Lx = λN x, thus λ 6= 0, N x ∈ Im L =
ker Q, hence
Z η
Z ξ
∆
(ξ − s)(f (s, x, x ) + e(s))∇s +
η(ξ − 1)(f (s, x, x∆ ) + e(s))∇s
0
η
(3.8)
Z 1
η(1 − s)(f (s, x, x∆ ) + e(s))∇s = 0
−
0
Then (3.8) and condition (C2) imply that there exists t0 ∈ T such that |x(t0 )| < M .
Rt
In view of x(0) = x(ρ(t0 )) − 0 0 x∆ (s)∇s, we obtain that
|x(0)| < M + kx∆ k0 ,
t ∈ T.
(3.9)
∆
For x(1) = x(η), there exists t1 ∈ (η, 1) ∩ T such that x (t1 ) = 0. Then from
Z t
∆
∆
x (t) = x (t1 ) +
x∆∇ (s)∇s,
t1
we have
kx∆ (t)k0 ≤ kN xk1
Hence, from (3.9) and (3.10), we have
(3.10)
|x(0)| ≤ M + kN xk1 .
Again for x ∈ Ω1 , x ∈ dom L \ ker L, then (I − P )x ∈ dom L ∩ ker P, LP x = 0, thus
from Lemma 3.1, we have
k(I − P )xk = kKP L(I − P )xk ≤ 2kL(I − P )xk1 = 2kLxk1 ≤ 2kN xk1 .
Then
kxk ≤ kP xk + k(I − P )k ≤ M + 3kN xk1 .
(3.11)
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N. WANG, H. ZHOU, L. YANG
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If assumption (3.3) holds, we obtain
kxk ≤ M + 3kN xk1 = M + 3|f (t, x(t), x∆ (t)) + e(t)|1
≤ M + 3(kak1 |x| + kbk1 |x∆ | + kck1 |x∆ |θ + krk1 + kek1 )
≤ 3kak1 kxk0 + 3[kbk1 kx∆ k0 + kck1 kx∆ kθ0 + krk1 + kek1 +
M
].
3
Thus
3
M
[kbk1 kx∆ k0 + kck1 kx∆ kθ0 + krk1 + kek1 +
].
1 − 3kak1
3
kxk0 ≤
Then
3kak1
M
+ 1]3[kbk1 kx∆ k0 + kck1 kx∆ kθ0 + krk1 + kek1 +
]
1 − 3kak1
3
3
M
kbk1
]
kx∆ k0 +
[kck1 kx∆ kθ0 + krk1 + kek1 +
=3
1 − 3kak1
1 − 3kak1
3
kx∆ k0 ≤ kxk ≤ [
By a simple computation,
kx∆ k0 ≤
3kck1
3
M
].
kx∆ kθ0 +
[krk1 + kek1 +
1 − 3kak1 − 3kbk1
1 − 3kak1 − 3kbk1
3
Since θ ∈ [0, 1) and kak1 + kbk1 < 1/3, there exists positive constant M1 such
that kx∆ k0 ≤ M1 . Then from (3.9), there exists positive constant M2 such that
kxk0 ≤ M1 . Hence kxk = max{kxk0 , kx∆ k0 } ≤ max{M1 , M2 }, which means that
Ω1 is bounded. If (3.4) hold, similar to the above argument, we can prove Ω1 is
also bounded.
Step 2 The set Ω2 = {x ∈ ker L : N x ∈ Im L} is bounded. The fact that x ∈ Ω2
implies that x = d, N (x) = f (t, d, 0) + e(t) and QN x = 0, thus
Z ξ
Z η
Q(N x) =
(ξ − s)(f (s, d, 0) + e(s))∇s +
η(ξ − 1)(f (s, d, 0) + e(s))∇s
0
Z
η
1
−
η(1 − s)(f (s, d, 0) + e(s))∇s
0
÷
Z
0
η
Z
(ξ − s)∇s +
ξ
Z
η(ξ − 1)∇s −
η
1
η(1 − s)∇s = 0
0
(3.12)
So |d| < M ∗ , thus x = d is bounded. The proof of step 2 is complete.
Let Ω = {x ∈ X : kxk < N1 }, where N1 > max{M1 , M2 , M ∗ }. Then Ω1 ⊂ Ω
and Ω2 ⊂ Ω. From argument above, it is obviously that conditions (1), (2) of
Lemma 2.1 are satisfied. Furthermore, by using Ascolli-Arezela theorem, it is easy
to see that KP (I − Q)N : Ω → Y is compact, thus N is L-compact on Ω.
Next, we claim that condition (3) of Lemma 2.1 is also satisfied. If the first part
of condition (C3) is satisfied, define the isomorphism J : Im Q → ker L by J(a) = a
and let
H(λ, x) = λx + (1 − λ)JQN x, (λ, x) ∈ Ω × [0, 1].
By a simple calculation, we obtain, for (λ, x) ∈ ∂(Ω ∩ ker L) × [0, 1],
xH(λ, x) = λx2 + (1 − λ)xQN x > 0.
Thus H(λ, x) 6= 0.
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EXISTENCE OF SOLUTIONS
7
If the second part of condition (C3) is satisfied, define
H(λ, x) = −λx + (1 − λ)JQN x, (λ, x) ∈ Ω × [0, 1],
where J : Im Q → ker L defined by J(a) = −a, similar with above, we can obtain
that H(λ, x) 6= 0. Thus
deg(JQN, Ω ∩ ker L, 0) = deg(H(x, 0), Ω ∩ ker L, 0)
= deg(H(x, 1), Ω ∩ ker L, 0)
= deg(I, Ω ∩ ker L, 0) 6= 0.
Then by Lemma 2.1, Lx = N x has at least one solution in dom L ∩ Ω, which means
resonance problem (1.1), (1.2) has at least one solution. The proof is complete. 4. Existence result under condition (1.4)
This section we consider problem (1.1),(1.2) under the assumption that
αη(1 − β) + (1 − α)(1 − βξ) = 0.
The normed spaces X, Y and operators L, N are defined as in Section 3.
Lemma 4.1. If αη(1 − β) + (1 − α)(1 − βξ) = 0, then
(1)
Z η
Z ξ
Im L = {y(t)|α(1 − β)
(η − s)y(s)∇s + (1 − α)β
(ξ − s)y(s)∇s
0
0
1
Z
− (1 − α)
(1 − s)y(s)∇s = 0};
0
(2) L : dom L ⊂ X → Y is a Fredholm operator with index zero.
(3) Define projector operator P : X → ker L as P x = x(0), then the generalized
inverse of operator L, KP : Im L → dom L ∩ ker P can be written as
Z ξ
Z 1
Z t
t
[β
(ξ − s)y(s)∇s −
(1 − s)y(s)∇s],
KP (y) =
(t − s)y(s)∇s +
1 − βξ
0
0
0
satisfying
β+1
kKP (y(t))k1 ≤ (1 + |
|)kyk1 .
1 − βξ
Proof. (1) First we show that
Z η
Z ξ
n
Im L = y(t)|α(1 − β)
(η − s)y(s)∇s + (1 − α)β
(ξ − s)y(s)∇s
0
Z
− (1 − α)
1
0
o
(1 − s)y(s)∇s = 0
0
First suppose y(t) ∈ Im L, then there exists x(t) such that Lx = y. That is
Z t
x(t) =
(t − s)y(s)∇s + u∆ (0)t + u(0).
0
Then boundary condition x(0) = αx(η), x(1) = βx(ξ) together with αη(1 − β) +
(1 − α)(1 − βξ) = 0 imply that
Z η
Z ξ
Z 1
α(1−β)
(η −s)y(s)∇s+(1−α)β
(ξ −s)y(s)∇s−(1−α)
(1−s)y(s)∇s = 0.
0
0
0
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N. WANG, H. ZHOU, L. YANG
EJDE-2015/240
Next we assume that
Z ξ
Z η
n
(ξ − s)y(s)∇s
(η − s)y(s)∇s + (1 − α)β
y(t) ∈ y(t)|α(1 − β)
0
0
1
Z
− (1 − α)
o
(1 − s)y(s)∇s = 0 .
0
Let x(t) ∈ X, where
Z t
(t − s)y(s)∇s +
x(t) =
0
t
[β
1 − βξ
Z
ξ
Z
1
(1 − s)y(s)∇s] .
(ξ − s)y(s)∇s −
0
0
Then Lx = x∆∇ = y(t). Since
Z ξ
Z η
n
(ξ − s)y(s)∇s
(η − s)y(s)∇s + (1 − α)β
y(t) ∈ y(t)|α(1 − β)
0
0
1
Z
− (1 − α)
o
(1 − s)y(s)∇s = 0 ,
0
by a simple computation we have
x(0) = αx(η),
x(1) = βx(ξ)
Thus y(t) ∈ Im L. Summing up the two steps above we obtain that: (1)
Z η
Z ξ
Im L = {y(t)|α(1 − β)
(η − s)y(s)∇s + (1 − α)β
(ξ − s)y(s)∇s
0
Z
− (1 − α)
0
1
(1 − s)y(s)∇s = 0}.
0
(2) Next we claim that L is a Fredholm operator with index zero. It is easy to
see that ker L = R. Next we prove that Y = Im L ⊕ ker L. Suppose y(t) ∈ Y , define
the projector operator Q as
Z η
Z ξ
Q(y) = α(1 − β)
(η − s)y(s)∇s + (1 − α)β
(ξ − s)y(s)∇s
0
1
Z
− (1 − α)
0
(1 − s)y(s)∇s
0
Z
η
÷ α(1 − β)
Z
(η − s)∇s + (1 − α)β
0
ξ
Z
(ξ − s)∇s − (1 − α)
0
∗
1
(1 − s)∇s
0
∗
Let y = y(t) − Q(y(t)), by a simple computation, y ∈ Im L.Hence Y = Im L +
ker L, furthermore considering Im L ∩ ker L = {0}, we have Y = Im L ⊕ ker L. Thus
dim ker L = codim Im L,
which means L is a Fredholm operator with index zero.
(3) Define the projector operator P : X → ker L as P x = x(0), for y(t) ∈ Im L,
we have
(LKP )(y(t)) = y(t),
and for x(t) ∈ dom L ∩ ker P , we know
(KP L)(x(t))
= KP (x∆∇ (t))
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Z
=
EXISTENCE OF SOLUTIONS
t
t
[β
1 − βξ
(t − s)x∆∇ ∇s +
0
Z
ξ
(ξ − s)x∆∇ ∇s −
Z
0
9
1
(1 − s)x∆∇ ∇s]
0
= x(t)
These shows that KP = (Ldom L∩ker P )−1 . Furthermore from the definition of the
norms in the X, Y , we have
kKP (y(t))k1 ≤ (1 + |
β+1
|)kyk1 .
1 − βξ
Theorem 4.2. Let f : [0, 1] × R2 → R be a continuous function. Condition C1 , C2
are satisfied and:
(C5) There exists a constant M > 0 such that for x ∈ dom L, |x(ρ(t))| > M , for
all t ∈ [0, 1],
Z η
α(1 − β)
(η − s)(f (s, x, x∆ ) + e(s))∇s
0
ξ
Z
(ξ − s)(f (s, x, x∆ ) + e(s))∇s
+ (1 − α)β
Z
− (1 − α)
(4.1)
0
1
(1 − s)(f (s, x, x∆ ) + e(s))∇s 6= 0.
0
Then for each e ∈ L1 [0, 1], resonance problem (1.1), (1.2) with αη(1 − β) + (1 −
α)(1 − βξ) = 0 has at least one solution provided that
1
.
β+1
2 + | 1−βξ
|
kak1 + kbk1 <
The proof is similar to that of Theory 3.1 and it is omitted here.
5. An example
In this section we give an example to illustrate the main results of this article.
1
} ∪ [ 21 , 1], (n = 1, 2, . . . ). We consider the four-point boundaryLet T = {0} ∪ { 2n+1
value problem
x∆∇ (t) =
1
1
1
x(t) + x∆ (t) +
sin(x∆ (t))1/5 ,
18
18
12
t ∈ (0, 1) ∩ T,
(5.1)
subject to the boundary condition
1
x(0) = x( ),
3
1
x(1) = x( ),
2
(5.2)
It is easy to see that α = β = 1, η = 1/3, ξ = 1/2, a(t) = b(t) =
ka(t)k + kb(t)k < 1/3 and
f (t, x, y) ≤
1
1
1
|x| + |y| + |y|1/5 ,
18
18
12
f (t, x, y) ≥
1
18 ,
c(t) = 1/12,
1
1
1
|x| − |y| − .
18
18
12
Then all conditions of Theorem 3.2 hold. Hence, the problem has at least one
nontrival solution at resonance.
10
N. WANG, H. ZHOU, L. YANG
EJDE-2015/240
Acknowledgements. The authors would like to thank the anonymous referees
for their careful reading of this manuscript and for suggesting useful changes. This
work was sponsored by the NSFC (11201109), Anhui Provincial Natural Science
Foundation (1408085QA07), the Higher School Natural Science Project of Anhui
Province (KJ2014A200) and the outstanding talents plan of Anhui High school.
References
[1] F. M. Atici, G. Sh. Guseinov; On Green’s functions and positive solutions for boundary-value
problems on time scales, J. comput. Appl. Math., 141 (2002) 75-99.
[2] Z. B. Bai, W. G. Li, W. G. Ge; Existence and multiplicity of solutions for four point boundary
value problems at resonance, Nonlinear Anal., 60 (2005), 1151-1162.
[3] Z. B. Bai, W. G. Ge, Y. F. Wang; Multiplicity results for some second-order four-point
boundary-value problems, Nonlinear Anal., 60 (2004) 491-500.
[4] W. Feng, J. R. L. Webb; Solvability of three-point boundary value problems at resonance,
Nonlinear Anal., 30 (6) 1997, 3227-3238.
[5] Y. P. Guo, W. G. Ge; Positive solutions for three-point boundary-value problems with dependence on the first order derivative, J. Math. Anal. Appl., 290 (2004), 291-301.
[6] V. A. Il’in, E. I. Moiseev; Nonlocal boundary value problem of the second kind for a SturmLiouville operator, Differential Equation, 1987 23(8), 979-987.
[7] V. A. Il’in, E. I. Moiseev; Nonlocal boundary value problem of the first kind for a SturmLiouville operator in its differential and finite difference aspects, Differential Equation, 1987
23 (7), 803-810.
[8] N. Kosmatov; Multi-point boundary value problems on time scales at resonance, J. Math.
Anal. Appl., 323 (2006), 253-266.
[9] B. Liu; Positive solutions of a nonlinear four-point boundary value problems, Appl. Math.
Comput., 155 (2004), 179-203.
[10] R. Y. Ma; Positive solutions for a nonlinear three-point boundary value problem, Electronic
Journal of Differential Equations, 1999, No. 34 (1999), 1-8.
[11] R. Y. Ma, N. Cataneda; Existence of solution for nonlinear m-point boundary value problem,
J. Math. Anal. Appl., 256 (2001), 556-567.
[12] J. Mawhin; Topological degree methods in nonlinear boundary value problems, in NSFCBMS
regional Conference series in Math, American Mathematics Society, Providence, RI 1979.
[13] I. Rachunková; Multiplicity results for four-point boundary value problems, Nonlinear Anal.,
18 (1992), 495-505.
[14] I. Rachunková; On the existence of two solutions of the four-point problem, J. Math. Anal.
Appl., 193 (1995), 245-254.
[15] I. Rachunková; Upper and lower solutions and topological degree, J. Math. Anal. Appl., 234
(1999), 311-327.
Ning Wang
College of Mathematics and Statistics, Institute of Applied Mathematics, Hefei Normal University, Hefei 230061, China
E-mail address: Ningw@hftc.edu.cn
Hui Zhou
College of Mathematics and Statistics, Institute of Applied Mathematics, Hefei Normal University, Hefei 230061, China
E-mail address: zhouhui0309@126.com
Liu Yang
College of Mathematics and Statistics, Institute of Applied Mathematics, Hefei Normal University, Hefei 230061, China
E-mail address: yliu722@163.com