Electronic Journal of Differential Equations, Vol. 2009(2009), No. 13, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
SOLVABILITY FOR SECOND-ORDER M-POINT
BOUNDARY VALUE PROBLEMS AT RESONANCE
ON THE HALF-LINE
YANG LIU, DONG LI, MING FANG
Abstract. In this article, we investigate the existence of positive solutions for
second-order m-point boundary-value problems at resonance on the half-line
(q(t)x0 (t))0 = f (t, x(t), x0 (t)),
x(0) =
m−2
X
a.e. in (0, ∞),
lim q(t)x0 (t) = 0.
αi x(ξi ),
t→∞
i=1
Some existence results are obtained by using the Mawhin’s coincidence theory.
1. Introduction
In this article, we study the existence of positive solutions for the second-order
m-point boundary-value problems at resonance on the half-line
(q(t)x0 (t))0 = f (t, x(t), x0 (t)),
x(0) =
m−2
X
a.e. in (0, ∞),
lim q(t)x0 (t) = 0,
αi x(ξi ),
t→∞
i=1
(1.1)
(1.2)
where f : [0, ∞) × R2 → R is a Carathéodory function, αi ∈ R (1 ≤ i ≤ m − 2),
0 < ξ1 < ξ2 < · · · < ξm−2 < 1, q ∈ C[0, ∞) ∩ C 1 (0, ∞) with q > 0 on [0, ∞) and
1
q ∈ L1 [0, ∞).
In recent years, many authors have studied the existence of positive solutions
for some boundary value problems on the half-line (see [6, 7, 12, 13, 14, 15]) or at
resonance (see [2, 3, 4, 5, 9, 10]). However, to the best of our knowledge, only one
paper [8] studied the existence and uniqueness positive solutions for second-order
three-point boundary value problems at resonance on the half-line. There is little
research concerning (1.1)-(1.2), so it is worthwhile to investigate the problem.
Inspired by [2, 4, 5], the purpose of our paper is to discuss the existence of positive
solutions for the second-order m-point boundary value problem at resonance on the
half-line. Our method is based on the coincidence degree theory of Mawhin.
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. m-point boundary value problem; resonance; half-line;
coincidence degree theory.
c
2009
Texas State University - San Marcos.
Submitted July 8, 2008. Published January 12, 2009.
1
2
Y. LIU, D. LI, M. FANG
EJDE-2009/13
The remaining part of this paper is organized as follows. In section 2, we present
some preliminaries and lemmas. Section 3 is devoted to proving the existence of
positive solutions for (1.1)-(1.2).
2. Preliminaries and lemmas
Now, we briefly recall some notation and an abstract existence result.
Let X, Z be normed spaces, L : dom L ⊂ X → Z be a Fredholm operator
of index zero, and P : X → X, Q : Z → Z be continuous projectors such that
Im P = ker L, ker Q = Im L and X = ker L ⊕ ker P , Z = Im L ⊕ Im Q, It follows
that L|dom L∩ker P : domL ∩ ker P → Im L is invertible. We denote the inverse of
the mapping by KP : Im L → domL ∩ ker P . The generalized inverse of L denoted
by KP,Q : Z → domL ∩ ker P is defined by KP,Q = Kp (I − Q).
Definition 2.1. Let L : dom L ⊂ X → Z be a Fredholm mapping, E be a metric
space, and N : E → Z be a mapping. We say that N is L-compact on E if
QN : E → Z and KP,Q N : E → X are compact on E. In addition, we say that N
is L-completely continuous if it is L-compact on every bounded E ⊂ X.
Definition 2.2. We say that the map f : [0, ∞) × Rn → R, (t, x) → f (t, z) is
L1 [0, ∞)-Carathéodory, if the following conditions are satisfied
(i) for each z ∈ Rn , the mapping t → f (t, z) is Lebesgue measurable;
(ii) for a.e. t ∈ [0, ∞), the mapping z → f (t, z) is continuous on Rn ;
(iii) for each r > 0, there exists ϕr ∈ L1 [0, ∞) such that, for a.e. t ∈ [0, ∞) and
every z such that |z| ≤ r, we have |f (t, z)| ≤ ϕr (t).
Lemma 2.3 ([1]). Let X be the space of all bounded continuous vector-value functions on [0, ∞) and M ⊂ X. Then M is relatively compact in X if the following
conditions hold:
(i) M is bounded in X:
(ii) the functions from M are equicontinuous on any compact interval of [0, ∞);
(iii) the functions from M are equiconvergent, that is, given > 0, there exists
a T = T () > 0 such that |φ(t) − φ(∞)| < , for all t > T and all φ ∈ S.
Lemma 2.4 ([11]). Let Ω ⊂ X be open and bounded, L be a Fredholm mapping
of index zero and N be 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 6∈ Im L for every x ∈ ker L ∩ ∂Ω;
(3) deg(JQN |∂Ω∩ker L , Ω ∩ ker L, 0) 6= 0, with Q : Z → Z is a continuous
projection such that Im L = ker Q and J : Im Q → ker L is an isomorphism.
Then the equation Lx = N x has at least one solution in dom L ∩ Ω.
Let AC[0, ∞) denote the space of absolutely continuous functions on the interval
[0, ∞). In this paper, the following space X will be basic space to study (1.1)-(1.2),
which is denoted by
X = {x ∈ C 1 [0, ∞), x, qx0 ∈ AC[0, ∞) lim x(t)
t→∞
and lim x0 (t)exist, (qx0 )0 ∈ L1 [0, ∞)}
t→∞
endowed with the norm kxk = max{kxk∞ , kx0 k∞ }, where kxk∞ = supt∈[0,∞) |x(t)|.
Let Z = L1 [0, ∞), and denote the norm in L1 [0, ∞) by k · k1 .
EJDE-2009/13
SECOND-ORDER M-POINT BVPS AT RESONANCE
3
Define L to be the linear operator from L ⊂ X to Z with
dom L = {x ∈ X : x(0) =
m−2
X
αi x(ξi ), lim q(t)x0 (t) = 0}
t→∞
i=1
and Lx(t) = (q(t)x0 (t))0 , x ∈ dom L, t ∈ [0, ∞). We define N : X → Z by setting
N x(t) = f (t, x(t), x0 (t)),
t ∈ [0, ∞),
then (1.1)-(1.2) can be written
Lemma 2.5. If
Pm−2
i=1
Lx = N x
Pm−2 R ξi
αi = 1 and i=1 αi 0
e−s
q(s) ds
6= 0, then
(i) ker L = {x ∈ dom L : x(t) = c, c ∈ R, t ∈ [0, ∞)};
Pm−2 R ξ 1 R ∞
y(τ )dτ ds = 0};
(ii) Im L = {y ∈ Z : i=1 αi 0 i q(s)
s
(iii) L : dom L ⊂ X → X is a Fredholm operator of index zero. Furthermore,
the linear continuous projector operator Q : Z → Z can be defined
m−2
X Z ξi 1 Z ∞
αi
(Qy)(t) = h(t)
y(τ )dτ ds, t ∈ [0, ∞),
q(s) s
0
i=1
where
h(t) = Pm−2
i=1
e−t
Rξ
αi 0 i
e−s
q(s) ds
,
t ∈ [0, ∞).
(iv) The generalized inverse KP : Im L → dom L ∩ ker P of L can be written by
Z t
Z ∞
1
KP y(t) = −
y(τ )dτ ds.
0 q(s) s
(v) kKP yk ≤ max{kq −1 k∞ , kq −1 k1 }kyk1 , for all y ∈ ImL.
Proof. By direct calculations, we easily know that (i) and (ii) hold. (iii) For any
y ∈ Z, take the prosector
m−2
X Z ξi 1 Z ∞
αi
(Qy)(t) = h(t)
y(τ )dτ ds, t ∈ [0, ∞).
q(s) s
0
i=1
Let y1 = y − Qy, by direct calculations, we have
m−2
X Z ξi 1 Z ∞
y1 (τ )dτ ds
q(s) s
i=1 0
m−2
m−2
X Z ξi 1 Z ∞
X Z ξi 1 Z ∞
=
y(τ )dτ ds 1 −
αi
h(τ )dτ ds = 0.
q(s) s
q(s) s
0
i=1 0
i=1
So y1 ∈ Im L. Hence, Z = Im L + Im Q, since Im L ∩ Im Q = {0}, we obtain
Z = Im L ⊕ Im Q.
Thus, dim ker L = dim Im Q = 1.
Hence, L is a Fredholm operator of index zero.
(iv) Let P : Z → Z be defined by
P x(t) = x(0),
t ∈ [0, ∞).
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Y. LIU, D. LI, M. FANG
EJDE-2009/13
Then the generalized inverse KP : Im L → dom L ∩ ker P of L can be written as
Z t
Z ∞
1
KP y(t) = −
y(τ )dτ ds.
0 q(s) s
In fact, for any y ∈ ImL, we have
LKP y(t) = (q(t)KP y 0 (t))0 = y(t).
and for x ∈ dom L ∩ ker P , one has
0
t
Z
0
KP Lx(t) = KP (q(t)x (t)) = −
0
t
Z
=−
0
Z
1
q(s)
∞
Z
(q(τ )x0 (τ ))0 dτ ds
s
1 lim q(σ)x0 (σ) − q(s)x0 (s) ds
q(s) σ→∞
t
x0 (s)ds = x(t) − x(0),
=
0
in view of x(0) = 0 (since x ∈ ker P ), thus,
(KP L)x(t) = x(t),
t ∈ [0, ∞).
−1
Hence, KP = (L|dom L∩ker P ) .
(v) From the definition of KP , we have
Z ∞
Z t
1
|y(τ )|dτ ds ≤ kq −1 k1 kyk1 ,
kKP yk∞ = sup |KP y| ≤ sup
t∈[0,∞)
t∈[0,∞) 0 q(s) s
and
0
k(KP y) k∞
1
= sup |(KP y) | ≤ sup
t∈[0,∞)
t∈[0,∞) q(t)
0
∞
Z
|y(s)|ds ≤ kq −1 k∞ kyk1 .
t
Hence,
kKP yk ≤ max{kq −1 k1 , kq −1 k∞ }kyk1 .
Lemma 2.6. If f is a Carathéodory function and
N is L-compact.
Pm−2
i=1
|αi |
R ξi
0
1
q(s) ds
< ∞, then
Proof. Let M ⊂ X be bounded with r = sup{kxk : x ∈ M } and consider
KP,Q N (M ). By f : [0, ∞) × R2 → R satisfies the Carathéodory conditions with
respect to L1 [0, ∞), there exists a Lebesgue integrable function ϕr such that
|N x(t)| = |f (t, x(t), x0 (t))| ≤ ϕr (t)
a.e. in (0, ∞).
Then for all x ∈ M , we have
Z ∞
kQN xk1 ≤
|QN x(s)|ds
0
Z
∞
=
0
Z
≤
m−2
X Z
αi
h(s)
i=1
∞
|h(s)|
0
m−2
X
i=1
ξi
0
Z
|αi |
0
1
q(ς)
ξ1
∞
Z
ς
1
q(ς)
Z
f (τ, x(τ ), x0 (τ ))dτ dς ds
∞
ϕr (τ )dτ dςds
0
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SECOND-ORDER M-POINT BVPS AT RESONANCE
≤ khk1 kϕr k
m−2
X
i=1
Z
|αi |
0
ξi
1
dς < ∞.
q(ς)
Thus,
kKP,Q N xk∞
Z t
Z ∞
1
= sup
f (τ, x(τ ), x0 (τ ))
t∈[0,∞) 0 q(s) s
m−2
X Z ξi 1 Z ∞
− h(τ )
αi
f (ζ, x(ζ), x0 (ζ))dζdς dτ ds
q(ς)
0
ς
i=1
Z t
Z ∞
1
≤ sup
f (τ, x(τ ), x0 (τ ))
q(s)
t∈[0,∞) 0
s
m−2
X Z ξi 1 Z ∞
− h(τ )
αi
f (ζ, x(ζ), x0 (ζ))dζdς dτ ds
q(ς)
0
ς
i=1
Z ∞
Z ξi
Z ∞
Z ∞
m−2
X
1
1
|αi |
≤
ϕr (τ ) + |h(τ )|
ϕr (ζ)dζdς dτ ds
q(s) 0
q(ς) 0
0
0
i=1
Z ξi
m−2
X
1
dς < ∞,
≤ kϕr k1 kq −1 k1 1 + khk1
|αi |
q(ς)
0
i=1
and
k(KP,Q N x)0 k∞
1 Z ∞
= sup f (s, x(s), x0 (s))
t∈[0,∞) q(t) t
m−2
X Z ξi 1 Z ∞
αi
− h(s)
f (τ, x(τ ), x0 (τ ))dτ dς ds
q(ς)
0
ς
i=1
Z ∞
Z ξi
Z ∞
m−2
X
1
1
ϕr (s) + |h(s)|
|αi |
ϕr (τ )dτ dς ds
≤ sup
q(ς) 0
t∈[0,∞) q(t) 0
0
i=1
Z
m−2
ξi
X
1
≤ kq −1 k∞ kϕr k1 1 + khk1
|αi |
dς < ∞.
q(ς)
0
i=1
It follows that KP,Q N (M ) is uniformly bounded in X.
Let x ∈ M and t1 , t2 ∈ [0, T ] with T ∈ (0, ∞), we have
|KP,Q N x(t2 ) − KP,Q N x(t1 )|
Z t2 1 Z ∞ =
f (τ, x(τ ), x0 (τ ))
t1 q(s) s
m−2
X Z ξi 1 Z ∞
f (ζ, x(ζ), x0 (ζ))dζdς dτ ds
− h(τ )
αi
q(ς)
0
ς
i=1
Z ξi
Z t2
Z ∞
Z ∞
m−2
X
1
1
ϕr (τ ) + |h(τ )|
|αi |
ϕr (ζ)dζdς dτ ds
≤
q(ς) 0
0
t1 q(s) 0
i=1
5
6
Y. LIU, D. LI, M. FANG
Z
t2
≤
t1
Z ξi
m−2
X
1
1
kϕr k1 1 + khk1
dς ds → 0,
|αi |
q(s)
q(ς)
0
i=1
EJDE-2009/13
as t1 → t2 ,
and
|(KP,Q N x)0 (t2 ) − (KP,Q N x)0 (t1 )|
m−2
1 Z ∞
X Z ξi 1 Z ∞
f (s, x(s), x0 (s)) − h(s)
f (τ, x(τ ), x0 (τ ))dτ dς ds
=
αi
q(t2 ) t2
q(ς) ς
0
i=1
Z ∞
1
−
f (s, x(s), x0 (s))
q(t1 ) t1
m−2
X Z ξi 1 Z ∞
− h(s)
αi
f (τ, x(τ ), x0 (τ ))dτ dς ds
q(ς)
0
ς
i=1
Z
1
1 ∞ −
|f (s, x(s), x0 (s))|
≤
q(t2 ) q(t1 ) t2
Z ∞
Z ξi
m−2
X
1
|f (τ, x(τ ), x0 (τ ))|dτ dς ds
+ |h(s)|
|αi |
q(ς) ς
0
i=1
Z t2 1
+
|f (s, x(s), x0 (s))|
q(t1 ) t1
Z ∞
Z ξi
m−2
X
1
|f (τ, x(τ ), x0 (τ ))|dτ dς ds
+ |h(s)|
|αi |
q(ς) ς
0
i=1
Z ξi
m−2
X
1
dς
≤ kq −1 k2∞ |q(t1 ) − q(t2 )|kϕr k1 1 + khk1
|αi |
q(ς)
0
i=1
Z t2 Z ξi
m−2
X
1
dςkϕr k1 ds → 0, as t1 → t2 .
+ kq −1 k∞
ϕr (s)) + |h(s)|
|αi |
q(ς)
t1
0
i=1
So KP,Q N (E) is equicontinuous on every compact subset of [0, ∞).
We introduce the following notation:
KP,Q N x(∞) = lim KP,Q N x(t)
t→∞
Z ∞
Z ∞
1
=
f (τ, x(τ ), x0 (τ ))
q(s) s
0
m−2
X Z ξi 1 Z ∞
− h(τ )
αi
f (ζ, x(ζ), x0 (ζ))dζdς dτ ds,
q(ς) ς
0
i=1
and
(KP,Q N x)0 (∞) = lim (KP,Q N x)0 (t)
t→∞
Z ∞
1
= lim
f (s, x(s), x0 (s))
t→∞ q(t) t
m−2
X Z ξI 1 Z ∞
− h(s)
αi
f (τ, x(τ ), x0 (τ ))dτ dς ds = 0.
q(ς) ς
0
i=1
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SECOND-ORDER M-POINT BVPS AT RESONANCE
7
Thus,
|KP,Q N x(t) − KP,Q N x(∞)|
Z ∞ 1 Z ∞
=
f (τ, x(τ ), x0 (τ ))
q(s) s
t
m−2
X Z ξi 1 Z ∞
− h(τ )
αi
f (ζ, x(ζ), x0 (ζ))dζdς dτ ds
q(ς)
0
ς
i=1
Z ∞
Z ∞
Z ∞
Z ξi
m−2
X
1
1
≤
ϕr (τ ) + |h(τ )|
ϕr (ζ)dζdς dτ ds
|αi |
q(s) s
q(ς) ς
t
0
i=1
Z ∞
Z
m−2
ξi
X
1
1
≤
kϕk1 1 + khk1
dτ ds → 0, uniformly as t → ∞,
|αi |
q(s)
q(τ )
t
0
i=1
and
|(KP,Q N x)0 (t) − (KP,Q N x)0 (∞)|
m−2
1 Z ∞
X Z ξI 1 Z ∞
0
=
f (s, x(s), x (s)) − h(s)
αi
f (τ, x(τ ), x0 (τ ))dτ dς ds
q(t) t
q(ς)
0
ς
i=1
Z ∞
Z
m−2
ξI
X
1
1
ϕr (s) + |h(s)|
|αi |
dςkϕr k1 ds → 0,
≤
q(t) t
q(ς)
0
i=1
uniformly as t → ∞. Therefore, KP,Q N (M ) is equiconvergent. It follows from
Lemma 2.3 that KP,Q N (M ) is relatively compact for each bounded M ∈ X. The
continuity of KP,Q N (M ) follows from the Lebesgue Dominated Theorem. We can
easily see that QN is continuous and QN (M ) is relatively compact. Thus, by
Definition 2.1, we have that the mapping N : X → Z is L-completely continuous.
3. Main results
Theorem 3.1. Let f : [0, ∞) × R2 → R be a Carathéodory function, in addition,
assume
that
Rξ 1
Pm−2
Pm−2
Pm−2 R ξ e−s
(H0 ) i=1 αi = 1, i=1 |αi | 0 i q(s)
ds < ∞ and i=1 αi 0 i q(s)
ds 6= 0;
(H1) There exists a constant M > 0, such that for all x ∈ dom L \ ker L if
|x(t)| > M , t ∈ [0, ∞), then
h(t)
m−2
X
i=1
Z
αi
0
ξi
1
q(s)
Z
∞
f (τ, x(τ ), x0 (τ ))dτ ds 6= 0
(3.1)
s
(H2) There exist β, γ, δ, ρ : [0, ∞) → [0, ∞), β, γ, δ, ρ ∈ L1 [0, ∞), and constant
θ ∈ [0, 1), such that for all (x1 , x2 ) ∈ R2 , t ∈ [0, ∞) satisfying one of the
following inequalities
|f (t, x1 , x2 )| ≤ β(t)|x1 | + γ(t)|x0 | + δ(t)|x2 |θ + ρ(t),
0
θ
|f (t, x1 , x2 )| ≤ β(t)|x1 | + γ(t)|x | + δ(t)|x1 | + ρ(t),
(3.2)
(3.3)
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Y. LIU, D. LI, M. FANG
EJDE-2009/13
(H3) There exists a constant N ∗ > 0, such that for all c ∈ R, if |c| > N ∗ , then,
either
m−2
X Z ξi 1 Z ∞
c
f (τ, c, 0)dτ ds < 0,
(3.4)
αi
q(s) s
0
i=1
or
m−2
X Z ξi 1 Z ∞
c
f (τ, c, 0)dτ ds > 0.
(3.5)
αi
q(s) s
0
i=1
Then (1.1)-(1.2) has at least one solution if
max{2kq −1 k1 , kq −1 k1 + kq −1 k∞ }(kβk1 + kγk1 ) < 1.
Proof. Set
Ω1 = {x ∈ dom L \ ker L : Lx = λN x, λ ∈ [0, 1]}.
For x ∈ Ω1 , since Lx = λN x, thus, λ 6= 0, N x ∈ Im L = ker Q, hence,
m−2
X Z ξi
h(t)
αi
f (τ, x(τ ), x0 (τ ))dτ ds = 0.
0
i=1
Thus, by (H1), there exists t0 ∈ [0, ∞), such that |x(t0 )| ≤ M . In view of
Z t0
|x(0)| = |x(t0 ) −
x0 (s)ds| ≤ M + kx0 k1 .
0
In addition,
1
x (t) = −
q(t)
0
Z
∞
0
∞
Z
0
(q(s)x (s)) ds = −
t
Lx(s)ds,
t
which implies
1
kx0 k∞ = sup −
q(t)
t∈[0,∞)
Z
∞
Lx(s)ds ≤ kq −1 k∞ kLxk1 ≤ kq −1 k∞ kN xk1 ,
t
and
kx0 k1 =
Z
0
∞
Z ∞
1
Lx(s)dsdτ ≤ kq −1 k1 kLxk1 ≤ kq −1 k1 kN xk1 .
−
q(τ ) τ
Thus,
|x(0)| ≤ M + kq −1 k1 kN xk1 .
(3.6)
Again for all x ∈ Ω1 , (I − P )x ∈ dom L ∩ ker P , LP x = 0, thus, from Lemma 2.4,
we get
k(I − P )xk = kKP (I − P )xk ≤ max{kq −1 k1 , kq −1 k∞ }kL(I − P )xk1
= max{kq −1 k1 , kq −1 k∞ }kLxk1
≤ max{kq
−1
k1 , kq
−1
(3.7)
k∞ }kN xk1 .
Hence, we have from (3.1) that
kxk ≤ kP xk + k(I − P )xk
≤ M + kq −1 k1 kN xk1 + max{kq −1 k1 , kq −1 k∞ }kN xk1
≤ M + max{2kq
−1
k1 , kq
−1
k1 + kq
−1
(3.8)
k∞ }kN xk1 .
Let Λ = max{2kq −1 k1 , kq −1 k1 + kq −1 k∞ }. If (3.2) holds, then from (3.8), we get
kxk ≤ M + ΛkN xk1 ≤ M + Λ(kβk1 kxk∞ + kγk1 kx0 k∞ + kδk1 kx0 kθ∞ + kρk1 ). (3.9)
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SECOND-ORDER M-POINT BVPS AT RESONANCE
9
Thus, from kxk∞ ≤ kxk and (3.9), we have
kxk∞ ≤
M + Λ(kβk1 kxk∞ + kγk1 kx0 k∞ + kδk1 kx0 kθ∞ + kρk1 )
.
1 − Λkβk1
It follows from kx0 k∞ ≤ kxk, (3.9) and (3.10) that
M
kx0 k∞ ≤ Λkβk1 kxk∞ + Λ kγk1 kx0 k∞ + kδk1 kx0 kθ∞ + kρk1 +
Λ
Λkγk1
Λkδk
Λkρk
+
M
1
1
≤
kx0 k∞ +
kx0 kθ∞ +
.
1 − Λkβk1
1 − Λkβk1
1 − Λkβk1
So
Λkδk1
Λkρk1 + M
kx0 k∞ ≤
kx0 kθ∞ +
.
1 − Λ(kβk1 + kγk1 )
1 − Λ(kβk1 + kγk1 )
Since θ ∈ [0, 1), by (3.11), there exists M1 > 0, such that
kx0 k∞ ≤ M1 .
(3.10)
(3.11)
(3.12)
Similar, by (3.10) and (3.12), there exists M2 > 0, such that
kxk∞ ≤ M2 .
(3.13)
Hence,
kxk = max{kxk∞ , kx0 k∞ } ≤ max{M1 , M2 }.
Then Ω1 is bounded.
If (3.3) holds, similar to the above argument, we can prove that Ω1 is bounded
too. Let
Ω2 = {x ∈ ker L : N x ∈ Im L}.
For x ∈ Ω2 , then we have x = c ∈ R, thus,
m−2
X Z ξi 1 Z ∞
f (τ, c, 0)dτ ds = 0.
(3.14)
αi
q(s) s
0
i=1
Then, we have by (H3) and (3.14) that
kxk = |c| ≤ N ∗ ,
which implies that Ω2 is bounded. We define the isomorphism J : Im Q → ker L by
c ∈ R, t ∈ [0, ∞).
J(ch(t)) = c,
If (3.4) holds, set
Ω3 = {x ∈ ker L : −λx + (1 − λ)JQN x = 0, λ ∈ [0, 1]}.
For every c0 ∈ Ω3 , we obtain
λc0 = (1 − λ)
m−2
X
i=1
Z
αi
0
ξi
1
q(s)
Z
∞
f (τ, c0 , 0)dτ ds.
s
If λ = 1, then c0 = 0 and if |c0 | > N ∗ , in view of (3.4), one has
m−2
X Z ξi 1 Z ∞
λc20 = (1 − λ)c0
αi
f (τ, c0 , 0)dτ ds < 0,
q(s) s
0
i=1
which contradicts λc20 ≥ 0. Thus, Ω3 is bounded.
If (3.5) holds, then let
Ω3 = {x ∈ ker L : λx + (1 − λ)JQN x = 0, λ ∈ [0, 1]},
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Y. LIU, D. LI, M. FANG
EJDE-2009/13
similar to the above argument, we can show that Ω3 is bounded.
In the following, we shall prove that all conditions of Lemma 2.4 are satisfied.
Let Ω to be a bounded open subset of X such that ∪3i=1 Ωi ⊂ Ω. Then by the above
argument, we have
(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.
Lastly, we will prove that (3) of Lemma 2.4 is satisfied. Define
H(x, λ) = ±λx + (1 − λ)QN x.
It is obvious that H(x, λ) 6= 0 for every x ∈ ∂Ω ∩ ker L. Thus,
deg(JQN |ker L∩∂Ω , Ω ∩ ker L, 0) = deg(H(·, 0), Ω ∩ ker L, 0)
= deg(H(·, 1), Ω ∩ ker L, 0)
= deg(±I, Ω ∩ ker L, 0) 6= 0.
Then by Lemma 2.4, Lx = N x has at least one solution in dom L ∩ Ω. In other
words, (1.1)-(1.2) has at least one solution in C 1 [0, ∞).
Acknowledgements. The author would like to express his sincere appreciation to
the anonymous referee for his/her helpful comments in improving the presentation
and quality of this article.
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EJDE-2009/13
SECOND-ORDER M-POINT BVPS AT RESONANCE
Yang Liu
Department of Mathematics, Hefei Teachers College, Hefei, Anhui 230061, China
E-mail address: liuyang19830206@yahoo.com.cn
Dong Li
Department of Mathematics, Jiamusi University, Jiamusi, Heilongjiang 154007, China
E-mail address: ld09281117@sohu.com
Ming Fang
Department of Mathematics, Yanbian University, Yanji, Jilin 133000, China
E-mail address: fangming@ybu.edu.cn
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