Volume 10 (2009), Issue 1, Article 9, 10 pp.
POSITIVE SOLUTIONS FOR SECOND-ORDER BOUNDARY VALUE PROBLEM
WITH INTEGRAL BOUNDARY CONDITIONS AT RESONANCE ON A
HALF-LINE
AIJUN YANG AND WEIGAO GE
D EPARTMENT OF A PPLIED M ATHEMATICS
B EIJING I NSTITUTE OF T ECHNOLOGY,
B EIJING , 100081, P. R. C HINA .
yangaij2004@163.com
gew@bit.edu.cn
Received 03 February, 2009; accepted 25 February, 2009
Communicated by R.P. Agarwal
A BSTRACT. This paper deals with the second order boundary value problem with integral boundary conditions on a half-line:
(p(t)x0 (t))0 + g(t)f (t, x(t)) = 0, a.e. in (0, ∞),
Z ∞
x(0) =
x(s)g(s)ds,
lim p(t)x0 (t) = p(0)x0 (0).
t→∞
0
A new result on the existence of positive solutions is obtained. The interesting points are: firstly,
the boundary value problem involved in the integral boundary condition on unbounded domains;
secondly, we employ a new tool – the recent Leggett-Williams norm-type theorem for coincidences and obtain positive solutions. Also, an example is constructed to illustrate that our result
here is valid.
Key words and phrases: Boundary value problem; Resonance; Cone; Positive solution; Coincidence.
2000 Mathematics Subject Classification. 34B10; 34B15; 34B45.
1. I NTRODUCTION
In this paper, we study the existence of positive solutions to the following boundary value
problem at resonance:
(1.1)
(p(t)x0 (t))0 + g(t)f (t, x(t)) = 0,
Z
a.e.in (0, ∞),
∞
lim p(t)x0 (t) = p(0)x0 (0),
0
R∞
1
where g ∈ L [0,
∞)
with
g(t)
>
0
on
[0,
∞)
and
g(s)ds = 1, p ∈ C[0, ∞) ∩ C 1 (0, ∞),
0
R
∞ 1
1
∈ L1 [0, ∞), 0 p(t)
dt ≤ 1 and p(t) > 0 on [0, ∞).
p
(1.2)
031-09
x(0) =
x(s)g(s)ds,
t→∞
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A IJUN YANG AND W EIGAO G E
Second-order boundary value problems (in short: BVPs) on infinite intervals, arising from the
study of radially symmetric solutions of nonlinear elliptic equations and models of gas pressure
in a semi-infinite porous medium [10], have received much attention, to identify a few, we refer
the readers to [9] – [11] and references therein. For example, in [9], Lian and Ge studied the
following second-order BVPs on a half-line
(1.3)
x00 (t) = f (t, x(t), x0 (t)),
(1.4)
x(0) = x(η),
0 < t < ∞,
lim x0 (t) = 0
t→∞
and
(1.5)
(1.6)
x00 (t) = f (t, x(t), x0 (t)) + e(t),
x(0) = x(η),
0 < t < ∞,
0
lim x (t) = 0,
t→∞
By using Mawhin’s continuity theorem, they obtained the existence results.
N. Kosmanov in [11] considered the second-order nonlinear differential equation at resonance
(1.7)
(p(t)u0 (t))0 = f (t, u(t), u0 (t)),
a.e. in (0, ∞)
with two sets of boundary conditions:
(1.8)
u0 (0) = 0,
n
X
i=1
κi ui (Ti ) = lim u(t)
t→∞
and
(1.9)
u(0) = 0,
n
X
i=1
κi ui (Ti ) = lim u(t).
t→∞
The author established
Pn existence theorems by the coincidence degree theorem of Mawhin under
the condition that i=1 κi = 1.
Although the existing literature on solutions of BVPs is quite wide, to the best of our knowledge, only a few papers deal with the existence of positive solutions to BVPs at resonance. In
particular, there has been no work done for the boundary value problems with integral boundary conditions on a half-line, such as the BVP (1.1) – (1.2). Moreover, our main approach is
different from the existing ones and our main ingredient is the Leggett-Williams norm-type theorem for coincidences obtained by O’Regan and Zima [4], which is a new tool used to study
the existence of positive solutions for nonlocal BVPs at resonance. An example is constructed
to illustrate that our result here is valid and almost sharp.
2. R ELATED L EMMAS
For the convenience of the reader, we review some standard facts on Fredholm operators
and cones in Banach spaces. Let X, Y be real Banach spaces. Consider a linear mapping
L : dom L ⊂ X → Y and a nonlinear operator N : X → Y . Assume that
1◦ L is a Fredholm operator of index zero, i.e., Im L is closed and dim Ker L = codim Im L <
∞.
The assumption 1◦ implies that there exist continuous projections P : X → X and Q : Y →
Y such that Im P = Ker L and Ker Q = Im L. Moreover, since dim Im Q = codim Im L, there
exists an isomorphism J : Im Q → Ker L. Denote by Lp the restriction of L to Ker P ∩ dom L.
Clearly, Lp is an isomorphism from Ker P ∩ dom L to Im L, we denote its inverse by Kp :
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 9, 10 pp.
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P OSITIVE S OLUTIONS FOR BVP AT R ESONANCE ON A H ALF -L INE
3
Im L → Ker P ∩ dom L. It is known (see [3]) that the coincidence equation Lx = N x is
equivalent to
x = (P + JQN )x + KP (I − Q)N x.
Let C be a cone in X such that
(i) µx ∈ C for all x ∈ C and µ ≥ 0,
(ii) x, −x ∈ C implies x = θ.
It is well known that C induces a partial order in X by
xy
if and only if
y − x ∈ C.
The following property is valid for every cone in a Banach space X.
Lemma 2.1 ([7]). Let C be a cone in X. Then for every u ∈ C \ {0} there exists a positive
number σ(u) such that
||x + u|| ≥ σ(u)||x|| for all x ∈ C.
Let γ : X → C be a retraction, that is, a continuous mapping such that γ(x) = x for all
x ∈ C. Set
Ψ := P + JQN + Kp (I − Q)N and Ψγ := Ψ ◦ γ.
In order to prove the existence result, we present here a definition.
Definition 2.1. f : [0, ∞) × R → R is called a g-Carathéodory function if
(A1)
(A2)
(A3)
for each u ∈ R, the mapping t 7→ f (t, u) is Lebesgue measurable on [0, ∞),
for a.e. t ∈ [0, ∞), the mapping u 7→ f (t, u) is continuous on R,
1
for
R ∞ each l > 0 and g ∈ L [0, ∞), there exists αl : [0, ∞) → [0, ∞) satisfying
g(s)αl (s)ds < ∞ such that
0
|u| ≤ l
implies
|f (t, u)| ≤ αl (t) for a.e.
t ∈ [0, ∞).
We make use of the following result due to O’Regan and Zima.
Theorem 2.2 ([4]). Let C be a cone in X and let Ω1 , Ω2 be open bounded subsets of X with
Ω1 ⊂ Ω2 and C ∩ (Ω2 \ Ω1 ) 6= ∅. Assume that 1◦ and the following conditions hold.
2◦ N is L-compact, that is, QN : X → Y is continuous and bounded and Kp (I − Q)N :
X → X is compact on every bounded subset of X,
3◦ Lx 6= λN x for all x ∈ C ∩ ∂Ω2 ∩ Im L and λ ∈ (0, 1),
4◦ γ maps subsets of Ω2 into bounded subsets of C,
5◦ degB {[I − (P + JQN )γ]|Ker L , Ker L ∩ Ω2 , 0} =
6 0, where degB denotes the Brouwer
degree,
6◦ there exists u0 ∈ C \ {0} such that ||x|| ≤ σ(u0 )||Ψx|| for x ∈ C(u0 ) ∩ ∂Ω1 , where
C(u0 ) = {x ∈ C : µu0 x for some µ > 0} and σ(u0 ) such that ||x+u0 || ≥ σ(u0 )||x||
for every x ∈ C,
7◦ (P + JQN )γ(∂Ω2 ) ⊂ C,
8◦ Ψγ (Ω2 \ Ω1 ) ⊂ C.
Then the equation Lx = N x has a solution in the set C ∩ (Ω2 \ Ω1 ).
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 9, 10 pp.
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4
A IJUN YANG AND W EIGAO G E
For simplicity of notation, we set
Z
∞
s
Z
ω :=
(2.1)
0
0
1
dτ
p(τ )
g(s)ds
and
G(t, s) =



























1
ω
1
ω
hR
∞ 1 R∞
g(r)drdτ
s p(τ ) τ
i
R∞ 1 R∞
Rτ
− 0 p(τ ) τ g(r)dr 0 g(r)drdτ
Rt
Rτ
Rt
+1 + 0 p(τ1 ) 0 g(r)drdτ − s p(τ1 ) dτ,
Rt
1
dτ
0 p(τ )
0 ≤ s < t < ∞,
hR
∞ 1 R∞
g(r)drdτ
s p(τ ) τ
i
R∞ 1 R∞
Rτ
− 0 p(τ ) τ g(r)dr 0 g(r)drdτ
Rt
Rτ
+1 + 0 p(τ1 ) 0 g(r)drdτ,
Rt
1
dτ
0 p(τ )
Note that G(t, s) ≥ 0 for t, s ∈ [0, 1], and set



(2.2)
0 < κ ≤ min 1,


0 ≤ t ≤ s < ∞.



1
.
sup G(t, s) 

t,s∈[0,∞)
3. M AIN R ESULT
We work in the Banach spaces
(3.1)
n
o
X = x ∈ C[0, ∞) : lim x(t) exists
t→∞
and
(3.2)
Z
Y = y : [0, ∞) → R :
∞
g(t)|y(t)|dt < ∞
0
with the norms ||x||X = sup |x(t)| and ||y||Y =
t∈[0,∞)
R∞
0
g(t)|y(t)|dt, respectively.
Define the linear operator L : dom L ⊂ X → Y and the nonlinear operator N : X → Y
with
n
(3.3) dom L = x ∈ X : lim p(t)x0 (t) exists, x, px0 ∈ AC[0, ∞)
t→∞
Z ∞
0 0
1
and gx, (px ) ∈ L [0, ∞), x(0) =
x(s)g(s)ds
0
o
and lim p(t)x0 (t) = p(0)x0 (0)
t→∞
by Lx(t) =
1
− g(t)
(p(t)x0 (t))0
and N x(t) = f (t, x(t)), t ∈ [0, ∞), respectively. Then
Ker L = {x ∈ dom L : x(t) ≡ c
and
on [0, ∞)}
∞
Im L = y ∈ Y :
g(s)y(s)ds = 0 .
0
R∞
Next, define the projections P : X → X by (P x)(t) = 0 g(s)x(s)ds and Q : Y → Y by
Z ∞
(Qy)(t) =
g(s)y(s)ds.
Z
0
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P OSITIVE S OLUTIONS FOR BVP AT R ESONANCE ON A H ALF -L INE
5
Clearly, Im P = Ker L and Ker Q = Im L. So dim Ker L = 1 = dim Im Q = codim Im L.
Notice that Im L is closed, L is a Fredholm operator of index zero.
Note that the inverse Kp : Im L → dom L ∩ Ker P of Lp is given by
Z ∞
(Kp y)(t) =
k(t, s)g(s)y(s)ds,
0
where
(3.4)
k(t, s) :=


1
ω
Rt

1
ω
Rt
1
dτ
0 p(τ )
1
dτ
0 p(τ )
R∞Rτ
s
1
drg(τ )dτ
s p(r)
−
Rt
1
dτ,
s p(τ )
R∞Rτ
s
1
drg(τ )dτ,
s p(r)
0 ≤ s < t < ∞,
0 ≤ t ≤ s < ∞.
R∞ 1
It is easy to see that |k(t, s)| ≤ 2 0 p(s)
ds.
In order to apply Theorem 2.2, we have to prove that N is L-compact, that is, QN is continuous and bounded and Kp (I − Q)N is compact on every bounded subset of X. Since the
Arzelà-Ascoli theorem fails in the noncompact interval case, we will use the following criterion.
n
o
Theorem 3.1 ([10]). Let M ⊂ x ∈ C[0, ∞) : lim x(t) exists . Then M is relatively compact
t→∞
if the following conditions hold:
(B1) all functions from M are uniformly bounded,
(B2) all functions from M are equicontinuous on any compact interval of [0, ∞),
(B3) all functions from M are equiconvergent at infinity, that is, for any given ε > 0, there
exists a T = T (ε) > 0 such that |f (t) − f (∞)| < ε for all t > T and f ∈ M .
Lemma 3.2. If f : [0, ∞) × R → R is a g-Carathéodory function, then N is L-compact.
Proof. Suppose that Ω ⊂ X is a bounded set. Then there exists l > 0 such that ||x||X ≤ l for
a g-Carathéodory function, there exists αl ∈ L1 [0, ∞) satisfying αl (t) > 0,
x ∈ Ω. Since f is
R∞
t ∈ (0, ∞) and 0 g(s)αl (s)ds < ∞ such that for a.e. t ∈ [0, ∞), |f (t, x(t))| ≤ αl (t) for
x ∈ Ω. Then for x ∈ Ω,
Z ∞
Z ∞
Z ∞
||QN x||Y =
g(t) g(s)f (s, x(s))ds dt ≤
g(s)αl (s)ds < ∞,
0
0
0
which implies that QN is bounded on Ω.
Next, we show that Kp (I − Q)N is compact, i.e., Kp (I − Q)N maps bounded sets into
relatively compact ones. Furthermore, denote KP,Q = KP (I − Q)N (see [9], [11]). For x ∈ Ω,
one gets
Z ∞
Z ∞
|(KP,Q x)(t)| ≤
g(τ )f (τ, x(τ ))dτ ds
k(t, s)g(s) f (s, x(s)) −
0
0
Z ∞
Z ∞
1
≤2
dτ
g(s)|f (s, x(s))|ds
p(τ )
0
0
Z ∞
Z ∞
+
g(s)
g(τ )|f (τ, x(τ ))|dτ ds
0
0
Z ∞
Z ∞
1
≤4
dτ
g(s)αl (s)ds < ∞,
p(τ )
0
0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 9, 10 pp.
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6
A IJUN YANG AND W EIGAO G E
that is, KP,Q (Ω) is uniformly bounded. Meanwhile, for any t1 , t2 ∈ [0, T ] with T a positive
constant,
|(KP,Q x)(t1 ) − (KP,Q x)(t2 )|
Z ∞Z ∞ Z τ
1
1
= drg(τ )dτ g(s) f (s, x(s))
ω 0
s
s p(r)
Z t1
Z ∞
1
−
g(τ )f (τ, x(τ ))dτ ds
dτ
0
t2 p(τ )
Z t1 Z t1
Z ∞
1
−
dτ g(s) f (s, x(s)) −
g(τ )f (τ, x(τ ))dτ ds
p(τ )
0
s
0
Z ∞
Z t2 Z t2
1
dτ g(s) f (s, x(s)) −
g(τ )f (τ, x(τ ))dτ ds −
p(τ )
0
0
s
Z ∞
Z s
Z τ
Z ∞
1
1
g(s)
g(r)|f (r, x(r))|drdτ ds +
g(τ )|f (τ, x(τ ))|dτ
≤
ω 0
0 p(τ ) 0
0
Z t1
Z τ
Z ∞
Z s
1
1
·
g(s)
g(r)drdτ ds] · dτ 0
0 p(τ ) 0
t2 p(τ )
Z t1
Z s
Z s
Z ∞
1
+ g(τ )|f (τ, x(τ ))|dτ +
g(τ )
g(r)|f (r, x(r))|drdτ ds
p(s) 0
t
0
0
Z ∞ 2
Z t1
Z ∞
Z ∞
1
≤
g(r)|f (r, x(r))|dr +
g(τ )|f (τ, x(τ ))|dτ ·
g(r)dr dτ 0
0
0
t2 p(τ )
Z t1
Z ∞
1
+2
g(r)|f (r, x(r))|dr dτ 0
t2 p(τ )
Z t1
Z ∞
1
≤4
g(s)αl (s)ds · dτ → 0,
uniformly as |t1 − t2 | → 0,
0
t2 p(τ )
which means that KP,Q (Ω) is equicontinuous. In addition, we claim that KP,Q (Ω) is equiconvergent at infinity. In fact,
|(KP,Q x)(∞) − (KP,Q x)(t)|
Z Z Z
Z ∞
Z ∞
1 ∞ ∞ τ 1
1
≤
drg(τ )dτ g(s) |f (s, x(s))| +
g(τ )|f (τ, x(τ ))|dτ ds ·
dτ
ω 0
p(τ )
s
s p(r)
0
t
Z s
Z ∞
Z s
Z ∞
1
+
ds
g(τ )|f (τ, x(τ )|dτ +
g(τ )
g(r)|f (r, x(r))|drdτ ds
p(s)
t
0
0
0
Z ∞
Z ∞
1
≤4
g(s)αl (s)ds ·
dτ → 0, uniformly as t → ∞.
p(τ )
0
t
Hence, Theorem 3.1 implies that Kp (I − Q)N (Ω) is relatively compact. Furthermore, since f
satisfies g-Carathéodory conditions, the continuity of QN and Kp (I − Q)N on Ω follows from
the Lebesgue dominated convergence theorem. This completes the proof.
Now, we state our main result on the existence of positive solutions for the BVP (1.1) – (1.2).
Theorem 3.3. Assume that
(H1) f : [0, ∞) × R → R is a g-Carathéodory function,
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P OSITIVE S OLUTIONS FOR BVP AT R ESONANCE ON A H ALF -L INE
7
(H2) there exist positive constants b1 , b2 , b3 , c1 , c2 , B with
Z ∞
c2
b2 c 2 b3
1
(3.5)
B>
+2
+
ds
c1
b1 c 1 b1
p(s)
0
such that
−κx ≤ f (t, x),
f (t, x) ≤ −c1 x + c2 ,
f (t, x) ≤ −b1 |f (t, x)| + b2 x + b3
for t ∈ [0, ∞), x ∈ [0, B],
(H3) there exist b ∈ (0, B), t0 ∈ [0, ∞), ρ ∈ (0, 1] and δ ∈ (0, 1). For each t ∈ [0, ∞),
is non-increasing on x ∈ (0, b] with
Z ∞
1−δ
f (s, b)
(3.6)
G(t0 , s)g(s)
ds ≥
.
b
δρ
0
Then the BVP (1.1) – (1.2) has at least one positive solution on [0, ∞).
f (t,x)
xρ
Proof. Consider the cone
C = {x ∈ X : x(t) ≥ 0 on
[0, ∞)}.
Let
Ω1 = {x ∈ X : δ||x||X < |x(t)| < b
on
[0, ∞)}
and
Ω2 = {x ∈ X : ||x||X < B}.
Clearly, Ω1 and Ω2 are bounded and open sets and
Ω1 = {x ∈ X : δ||x||X ≤ |x(t)| ≤ b
on
[0, ∞)} ⊂ Ω2
(see [4]). Moreover, C ∩ (Ω2 \ Ω1 ) 6= ∅. Let J = I and (γx)(t) = |x(t)| for x ∈ X. Then γ is
a retraction and maps subsets of Ω2 into bounded subsets of C, which means that 4◦ holds.
In order to prove 3◦ , suppose that there exist x0 ∈ ∂Ω2 ∩ C ∩ dom L and λ0 ∈ (0, 1) such that
Lx0 = λ0 N x0 , then (p(t)x00 (t))0 + λ0 g(t)f (t, x0 (t)) = 0 for all t ∈ [0, ∞). In view of (H2), we
have
1
1
−
(p(t)x00 (t))0 = f (t, x0 (t)) ≤ −b1
|(p(t)x00 (t))0 | + b2 x0 (t) + b3 .
λ0 g(t)
λ0 g(t)
Hence,
(3.7)
−(p(t)x00 (t))0 ≤ −b1 |(p(t)x00 (t))0 | + λ0 b2 g(t)x0 (t) + λ0 b3 g(t).
Integrating both sides of (3.7) from 0 to ∞, one gets
Z ∞
0=−
(p(t)x00 (t))0 dt
0
Z ∞
Z ∞
Z
0
0
≤ −b1
g(t)x0 (t)dt + λ0 b3
|(p(t)x0 (t)) |dt + λ0 b2
0
0
∞
g(t)dt,
0
which gives
Z
(3.8)
∞
|(p(t)x00 (t))0 |dt
0
Similarly, from (H2), we also obtain
Z
(3.9)
b2
<
b1
Z
∞
g(t)x0 (t)dt +
0
∞
g(t)x0 (t)dt ≤
0
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 9, 10 pp.
b3
.
b1
c2
.
c1
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8
A IJUN YANG AND W EIGAO G E
On the other hand,
∞
Z
(3.10)
x0 (t) =
k(t, s)(p(s)x00 (s))0 ds
g(t)x0 (t)dt +
0
Z
≤
(3.11)
∞
Z
0
∞
∞
Z
|k(t, s)| · |(p(s)x00 (s))0 |ds.
g(t)x0 (t)dt +
0
0
Then, (3.8)-(3.9) yield
c2
B = ||x0 ||X ≤
+2
c1
b2 c 2 b3
+
b1 c 1 b1
Z
0
∞
1
ds,
p(s)
which contradicts (3.5).
To prove 5◦ , consider x ∈ Ker L ∩ Ω2 . Then x(t) ≡ c on [0, ∞). Let
Z ∞
H(c, λ) = c − λ|c| − λ
g(s)f (s, |c|)ds
0
for c ∈ [−B, B] and λ ∈ [0, 1]. It is easy to show that 0 = H(c, λ) implies c ≥ 0. Suppose
0 = H(B, λ) for some λ ∈ (0, 1]. Then, (3.5) leads to
Z ∞
0 ≤ B(1 − λ) = λ
g(s)f (s, B)ds ≤ λ(−c1 B + c2 ) < 0,
0
which is a contradiction. In addition, if λ = 0, then B = 0, which is impossible. Thus,
H(x, λ) 6= 0 for x ∈ Ker L ∩ ∂Ω2 and λ ∈ [0, 1]. As a result,
degB {H(·, 1), Ker L ∩ Ω2 , 0} = degB {H(·, 0), Ker L ∩ Ω2 , 0}.
However,
degB {H(·, 0), Ker L ∩ Ω2 , 0} = degB {I, Ker L ∩ Ω2 , 0} = 1.
Then,
degB {[I − (P + JQN )γ]Ker L , Ker L ∩ Ω2 , 0} = degB {H(·, 1), Ker L ∩ Ω2 , 0} =
6 0.
Next, we prove 8◦ . Let x ∈ Ω2 \ Ω1 and t ∈ [0, ∞),
Z ∞
Z ∞
(Ψγ x)(t) =
g(s)|x(s)|ds +
g(s)f (s, |x(s)|)ds
0
0
Z ∞
Z ∞
+
k(t, s)g(s) f (s, |x(s)|) −
g(τ )f (τ, |x(τ )|)dτ ds
0
0
Z ∞
Z ∞
=
g(s)|x(s)|ds +
G(t, s)g(s)f (s, |x(s)|)ds
0
0
Z ∞
≥
(1 − κG(t, s))g(s)|x(s)|ds ≥ 0.
0
Hence, Ψγ (Ω2 \ Ω1 ) ⊂ C, i.e. 8◦ holds.
Since for x ∈ ∂Ω2 ,
Z ∞
Z ∞
(P + JQN )γx =
g(s)|x(s)|ds +
g(s)f (s, |x(s)|)ds
0
0
Z ∞
≥
(1 − κ)g(s)|x(s)|ds ≥ 0,
0
then, (P + JQN )γx ⊂ C for x ∈ ∂Ω2 , and 7◦ holds.
J. Inequal. Pure and Appl. Math., 10(1) (2009), Art. 9, 10 pp.
http://jipam.vu.edu.au/
P OSITIVE S OLUTIONS FOR BVP AT R ESONANCE ON A H ALF -L INE
9
It remains to verify 6◦ . Let u0 (t) ≡ 1 on [0, ∞). Then u0 ∈ C \ {0}, C(u0 ) = {x ∈ C :
x(t) > 0 on [0, ∞)} and we can take σ(u0 ) = 1. Let x ∈ C(u0 ) ∩ ∂Ω1 . Then x(t) > 0 on
[0, ∞), 0 < ||x||X ≤ b and x(t) ≥ δ||x||X on [0, ∞). For every x ∈ C(u0 ) ∩ ∂Ω1 , by (H3)
Z ∞
Z ∞
(Ψx)(t0 ) =
g(s)x(s)ds +
G(t0 , s)g(s)f (s, x(s))ds
0
0
Z ∞
f (s, x(s)) ρ
x (s)ds
≥ δ||x||X +
G(t0 , s)g(s)
xρ (s)
0
Z ∞
f (s, b)
ρ
ρ
≥ δ||x||X + δ ||x||X
G(t0 , s)g(s) ρ ds
b
0
1−ρ Z ∞
b
f (s, b)
= δ||x||X + δ ρ ||x||X
G(t0 , s)g(s)
ds
1−ρ
b
||x||X
0
Z ∞
f (s, b)
ρ
≥ δ||x||X + δ ||x||X
G(t0 , s)g(s)
ds
b
0
≥ ||x||X .
Thus, ||x||X ≤ σ(u0 )||Ψx||X for all x ∈ C(u0 ) ∩ ∂Ω1 .
In addition, 1◦ holds and Lemma 3.2 yields 2◦ . Then, by Theorem 2.2, the BVP (1.1) – (1.2)
has at least one positive solution x∗ on [0, ∞) with b ≤ ||x∗ ||X ≤ B. This completes the proof
of Theorem 3.3.
Remark 1. Note that with the projection P (x) = x(0), Conditions 7◦ and 8◦ of Theorem 2.2
are no longer satisfied.
To illustrate how our main result can be used in practice, we present here an example.
Example 3.1. Consider the following BVP

 2(et x0 (t))0 + e−t f (t, x(t)) = 0, a.e. in (0, ∞),
(3.12)
R
 x0 (0) = lim et x0 (t), x(0) = ∞ e−s x(s)ds.
0
t→∞
Corresponding to the BVP (1.1) – (1.2), p(t) = 2et , g(t) = e−t and f (t, x) = (t − 12 )e−2t x +
e−t x2 . We can get ω = 14 and
( 13 1 −t
+ 6 (e − 3e−s ) + 14 (e−2t + 2e−2s ) − 12 e−(t+2s) , 0 ≤ s ≤ t < ∞,
12
(3.13) G(t, s) =
13
− 13 e−t + 14 (e−2t + 2e−2s ) − 21 e−(t+2s) ,
0 ≤ t ≤ s < ∞.
12
3
Obviously, G(t, s) ≥ 0 for t, s ∈ [0, +∞). Choose κ = 12 , B = 5, c1 = 25 , c2 = 12 e− 2 , b1 = 12 ,
3
b2 = 32 and b3 = 32 e− 2 such that (H2) holds, and take b = 54 , t0 = 0, ρ = 1 and δ = 49 such that
(H3) is satisfied. Then thanks to Theorem 3.3, the BVP (3.12) has a positive solution on [0, ∞).
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