DOI: 10.1515/auom-2015-0042
An. Şt. Univ. Ovidius Constanţa
Vol. 23(2),2015, 279–304
Positive solutions for singular nonlocal
boundary value problems involving integral
conditions with derivative dependence
Baoqiang Yan, Donal O’Regan and Ravi P. Agarwal
Abstract
In this paper using a fixed point theory on a cone we present some
new results on the existence of multiple positive solutions for singular
nonlocal boundary value problems involving integral conditions with
derivative dependence.
1. Introduction.
In this paper we consider the existence of positive solutions of nonlinear
nonlocal boundary value problems (BVP) of the form
−x00 = q(t)f (t, x(t), x0 (t)), t ∈ (0, 1)
(1.1)
with integral boundary conditions
x0 (0) = 0, x(1) = α[x] =
Z
1
x(s)dA(s)
0
involving a Stieltjes integral, where A ∈ BV [0, 1].
Il’in and Moiseev first considered the existence of a solution to
x00 (t) = f (t, x(t), x0 (t)), t ∈ (0, 1),
Key Words: Cone, compact operator, nonlocal boundary value problems.
2010 Mathematics Subject Classification: 34B10, 34B16.
Received: June, 2014.
Revised: July, 2014.
Accepted: July, 2014.
279
(1.2)
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
x(0) = 0, x(1) =
m
X
280
αi x(ηi )
i=1
(see [10, 11]). Using degree-theoretic arguments, Gupta et al obtained conditions on the existence of solutions for the m-point boundary problem
x00 (t) = f (t, x(t), x0 (t)) + e(t), t ∈ (0, 1),
x(0) = 0, x(1) =
m
X
αi x(ηi )
i=1
(see [7, 8]). In [12] using a Leray-Schauder alternative Ma showed the existence
of at least one solution of
x00 (t) = f (t, x(t), x0 (t)) + e(t), t ∈ (0, 1),
0
x (0) = 0, x(1) =
m
X
αi x(ηi ).
i=1
In [22, 24], Webb and Infante considered
−x00 = q(t)f (t, x(t)), t ∈ (0, 1)
(1.3)
with boundary conditions
0
Z
x (0) = 0, x(1) = α[x] =
1
x(s)dA(s),
(1.4)
0
where dA(s) has a signed measure, and established the existence of positive
solutions and multiple positive solutions for BVP (1.3)-(1.4) when f is continuous and independent of x0 .
The boundary condition in BVP (1.1)-(1.2) generalizes the boundary conditions in [7-8, 10-13] and (1.1) generalizes the equations in [9,13-18,21-24]
(there f is independent of x0 ). One goal in this paper is to attempt to fill
a gap in the theory of singular nonlocal boundary value problems involving
integral conditions with derivative dependence.
Our paper is organized as follows. In Section 2, we present some lemmas
and preliminaries. In Section 3, two theorems are listed to show that x0 of
f (t, x, x0 ) can lead to BVP (1.1)-(1.2) having no positive solutions. In section
4, we discuss the existence of multiple positive solutions for BVP (1.1)-(1.2)
when f has no singularities. Section 5 presents the multiplicity of positive
solutions for BVP (1.1)-(1.2) when f is singular at x = 0 but not at x0 = 0. In
Section 6, we discuss the multiplicity of positive solutions for BVP (1.1)-(1.2)
when f is singular at x0 = 0 but not at x = 0. In Section 7, we consider the
case f is singular at x = 0 and x0 = 0.
281
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
2. Preliminaries
+
Let R = (−∞, +∞), R+ = (0, +∞), R = [0, +∞), R− = (−∞, 0) and we
list the following conditionsR for convenience.
R1
1
(C1 ) A ∈ BV [0, 1] with 0 G(t, s)dA(t) ≥ 0 for a.e. s ∈ [0, 1], 0 ≤ 0 dA(s)
R1
and 0 |dA(t)| < 1, where
1 − t, 0 ≤ s ≤ t ≤ 1
G(t, s) =
1 − s, 0 ≤ t ≤ s ≤ 1,
(C2 )
+
q ∈ C[0, 1], q(t) > 0 on (0, 1) and f ∈ C([0, 1] × R+ × R− , R ) with
0 < f (t, x, y) ≤ [h(x) + w(x)][r(|y|) + v(|y|)] on [0, 1] × R+ × R− ,
where w, v ∈ C(R+ , R+ ) are nonincreasing or w ≡ 0, v ≡ 0 and
Z 1
1
+
+
h, r ∈ C(R , R ) are nondecreasing with
)ds < +∞,
q(s)r(k0
1
−
s
0
for all k0 > 0,
(C3 )
there exists a constant a ∈ (0, 21 ) such that
f (t, x, y)
= +∞,
lim
x→+∞
x
uniformly for (t, y) ∈ [a, 1 − a] × (−∞, 0).
Let p(t) = 1 − t, t ∈ [0, 1] and Cp1 [0, 1] = {x : [0, 1] → R x is continuous on [0, 1] and continuously differentiable on (0, 1) with sup p(t)|x0 (t)| =
t∈(0,1)
sup (1 − t)|x0 (t)| < +∞}. For x ∈ Cp1 , define kxk = max{kxk1 , kxk2 } where
t∈(0,1)
kxk1 = max |x(t)|, kxk2 = sup (1 − t)|x0 (t)|.
t∈[0,1]
t∈(0,1)
Lemma 2.1 Cp1 is a Banach space. Also for any x ∈ Cp1 , |x0 (t)| ≤
t ∈ (0, 1).
Let
kxk
,
1−t
P = {x ∈ Cp1 |x(t) is concave and nonincreasing on
[0, 1] and x(0) ≥ kxk2 , α[x] ≥ 0}.
(2.1)
It is easy to see that P is a cone in Cp1 [0, 1].
We note the definition of the fixed point index i(A, Ω∩P, P ). Suppose that
Ω is a bounded open set in real Banach space E with θ ∈ Ω as its vectoreal zero,
P is a cone of E and A : Ω ∩ P → P is continuous and compact. Assume that
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
282
r : E → P is a retraction mapping, i.e., r is continuous and r(x) = x for all
x ∈ P . Choose R > 0 big enough such that TR = {x ∈ E : kxk < R} ⊇ Ω ∩ P .
Then
i(A, Ω ∩ P, P ) := deg(I − A · r, TR ∩ r−1 (Ω ∩ P ), θ),
where I : E → E is an identity operator and deg(I − A · r, TR ∩ r−1 (Ω ∩ P), θ)
is the Leray-Schauder degree (see [6]). The following lemmas are needed in
Section 4-7.
Lemma 2.2(see[6]) Let Ω be a bounded open set in real Banach space E,
P be a cone of E, θ ∈ Ω and A : Ω ∩ P → P be continuous and compact.
Suppose
λAx 6= x, ∀x ∈ ∂Ω ∩ P, λ ∈ (0, 1].
Then
i(A, Ω ∩ P, P ) = 1.
Lemma 2.3(see[6]) Let Ω be a bounded open set in real Banach space E,
P be a cone of E, θ ∈ Ω and A : Ω ∩ P → P be continuous and compact.
Suppose
Ax 6≤ x, ∀x ∈ ∂Ω ∩ P.
Then
i(A, Ω ∩ P, P ) = 0.
Remark: Ax 6≤ x ⇐⇒ x − Ax 6∈ P .
Lemma 2.4 If x ∈ P (defined above in (2.1)), then kxk = kxk1 .
Proof. If x ∈ P , one has
kxk1 = max{|x(t)|t ∈ [0, 1]} = x(0) ≥ kxk2 .
Then
kxk = max{kxk1 , kxk2 } = kxk1 .
The proof is complete.
Z 1
Lemma 2.5 Assume Φ ∈ C((0, 1), R+ ) with
Φ(t)dt < ∞ and
0
Z 1
F (t) =
G(t, s)Φ(s)ds+c, where c ≥ 0 is a constant. Then F ∈ P .
0
283
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Proof. From the definition of F , we have
00
F (t) ≤ 0, t ∈ (0, 1),
F 0 (0) = 0, F (1) = c ≥ 0,
which means that
F is nonincreasing and concave down on [0, 1]
(2.2)
with F (1) = c ≥ 0. Now (2.2) implies that for t ∈ [0, 1], F (t) = F ((1 − t)0 +
t · 1) ≥ (1 − t)F (0) + tF (1) ≥ (1 − t)F (0) = (1 − t)kF k1 , ∀t ∈ (0, 1) and
F (0) = max F (t) = kF k1 .
t∈[0,1]
Then
kF k2 = sup |(1 − t)F 0 (t)|
t∈(0,1)
Z
t∈(0,1)
t
Φ(s)ds| ≤ sup |F (t)| = kF k1 = F (0).
= sup |(1 − t)
(2.3)
t∈(0,1)
0
Moreover, from (C1 ), one has
1
Z
α[F ] = c
Z
dA(s) +
0
Z
=c
1
Z
0
0
F (s)dA(s)
0
1
Z
1
G(s, τ )dA(s)dτ ≥ 0.
Φ(τ )
dA(s) +
1
(2.4)
0
Hence, (2.2), (2.3) and (2.4) guarantee that F ∈ P . The proof is complete.
For x ∈ P , define an operator by
Z 1
(Bx)(t) = α[x] +
G(t, s)q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds,
∀t ∈ [0, 1],
0
(2.5)
where γ1 ∈ C[0, 1], γ2 ∈ C[0, 1] with mint∈[0,1] γ1 (t) > 0 and maxt∈[0,1] γ2 (t) <
0.
Lemma 2.6 Assume that (C1 ) and (C2 ) hold. Then B : P → P is continuous
and compact.
284
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Proof. First we show B : P → P is well defined. For x ∈ P , from (C1 ) and
(C2 ), we have
Z 1
|(Bx)(t)| = |α[x] +
G(t, s)q(s)f (s, x(s) + γ(s), x0 (s) + γ2 (s))ds|
0
Z 1
Z 1
≤ kxk
|dA(s)| +
(1 − s)q(s)|f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))|ds
Z0 1
Z0 1
≤ kxk
|dA(s)| +
(1 − s)q(s)[h(x(s) + γ1 (s))
0
0
0
0
+w(x(s) + γ1 (s))][r(|x0 (s) + γ2 (s)|) + v(|x0 (s) + γ2 (s)|)]ds
Z 1
Z 1
≤ kxk
|dA(s)| +
(1 − s)q(s)[h(kxk + kγ1 k) + w( min γ1 (s))]
s∈[0,1]
kxk + kγ2 k
) + v( min |γ2 (s)|)]ds < +∞, t ∈ [0, 1]
·[r(
1−s
s∈[0,1]
and
(1 − t)|(Bx)0 (t)| ≤ |(Bx)0 (t)|
Z t
=|
q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds|
0
Z 1
≤
q(s)[h(kxk + kγ1 k) + w( min γ1 (s))]
s∈[0,1]
0
kxk + kγ2 k
) + v( min |γ2 (s)|)]ds < +∞, t ∈ [0, 1].
·[r(
1−s
s∈[0,1]
Then B is well defined. For every x ∈ P , let Φ(t) = q(t)f (t, x(t)+γ1 (t), x0 (t)+
R1
γ2 (t)), c = α[x] and F (t) = 0 G(t, s)Φ(s)ds + c. It is easy to see that all
conditions of Lemma 2.5 hold, which implies that Bx ∈ P . As a result,
BP ⊆ P . Moreover, since
|(Bx)0 (t1 ) − (Bx)0 (t2 )|
Z t2
=|
q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds|
t
1
Z t2
≤
q(s)[h(kxk + kγ1 k) + w( min γ1 (s))]
s∈[0,1]
t1
2k
·[r( kxk+kγ
) + v(mins∈[0,1] |γ2 (s)|)]ds,
1−s
the Cauchy Principle guarantees that
lim (Bx)0 (t) exists and
t→0+
which means that Bx ∈ C 1 [0, 1].
lim (Bx)0 (t) exists,
t→1−
285
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Next we show that B : P → P is continuous. Assume that {xm }∞
m=1 ⊆ P
and x0 ∈ P with lim xm = x0 . Then, there exists an M > 0 such that
m→+∞
kxm k < M for all m ∈ {1, 2, · · · } (Lemma 2.1 guarantees that |x0m (t)| ≤
∀t ∈ (0, 1)). Thus,
(
lim (xm (t) + γ1 (t)) = x0 (t) + γ1 (t),
m→+∞
lim (x0m (t)
m→+∞
+ γ2 (t)) =
x00 (t)
+ γ2 (t),
M
,
1−t
t ∈ [0, 1],
t ∈ (0, 1)
(2.6)
and
|f (t, xm (t) + γ1 (t), x0m (t) + γ2 (t))|
≤ [h(xm (t) + γ1 (t)) + w(xm (t) + γ1 (t))][r(|x0m (t) + γ2 (t)|) + v(|x0m (t) + γ2 (t)|)]
kxk + kγ2 k
≤ [h(M + kγ1 k) + w( min γ1 (t))][r(
) + v( min |γ2 (t)|)].
t∈[0,1]
t∈[0,1]
1−t
(2.7)
From (2.6) and (2.7), the Lebesgue Dominated Convergence Theorem guarantees that
kBxm − Bx0 k1
Z
≤ |α[xm − x0 ]| + max |
t∈[0,1]
1
G(t, s)q(s)[f (s, xm (s) + γ1 (s), x0m (s) + γ2 (s))
0
−f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))]ds|
Z 1
Z 1
≤ kxn − x0 k
|dA(s)| +
(1 − s)q(s)|f (s, xm (s) + γ2 (s), x0m (s) + γ2 (s))
0
0
−f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))|ds
→ 0, as m → +∞
and
kBxm − Bx0 k2
Z
t
= sup (1 − t)| −
t∈(0,1)
q(s)[f (s, xm (s) + γ1 (s), x0m (s) + γ2 (s))
0
−f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))]ds|
1
Z
≤
q(s)|f (s, xm (s) + γ1 (s), x0m (s) + γ1 (s)) − f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))|ds
0
→ 0, as m → +∞,
which imply that
lim kBxm − Bx0 k = 0.
m→+∞
Hence, B : P → P is continuous.
Finally we show for any bounded D ⊆ P , B(D) is relatively compact.
Since D is bounded, there exists an M > 0 such that kxk ≤ M for all x ∈ D
286
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
(Lemma 2.1 guarantees that |x0 (t)| ≤
Z
M
, ∀t ∈ (0, 1) ). Thus, (C2 ) gives
1−t
1
= max |α[x] +
G(t, s)q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds|
t∈[0,1]
0
Z 1
Z 1
≤ kxk
|dA(s)| +
(1 − s)q(s)|f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))|ds
0
0
Z 1
Z 1
≤M
|dA(s)| +
(1 − s)q(s)[h(x(s) + γ1 (s)) + w(x(s) + γ1 (s))]
kBxk1
0
0
0
0
·[r(|x0 (s) + γ2 (s)|) + v(|x0 (s) + γ2 (s)|)]ds
Z 1
Z 1
≤M
|dA(s)| +
(1 − s)q(s)[h(M + kγ1 k) + w( min γ1 (s))]
s∈[0,1]
M + kγ2 k
) + v( min |γ2 (s)|)]ds
·[r(
1−s
s∈[0,1]
and
sup |(Bx)0 (t)|
t∈(0,1)
Z 1
= sup | −
q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds|
t∈[0,1]
0
Z 1
≤
q(s)|f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))|ds
0
Z 1
≤
q(s)[h(x(s) + γ1 (s)) + w(x(s) + γ1 (s))]
0
·[r(|x0 (s) + γ2 (s)|) + v(|x0 (s) + γ2 (s)|)]ds
Z 1
≤
q(s)[h(M + kγ1 k) + w( min γ1 (s))]
0
s∈[0,1]
M + kγ2 k
·[r(
) + v( min |γ2 (s)|)]ds.
1−s
s∈[0,1]
Consequently,
the functions belonging to {(BD)(t)} are uniformly bounded on [0, 1] (2.8)
and
the functions belonging to {(BD)0 (t)} are uniformly bounded on [0, 1] (2.9)
and so
functions belonging to {(Bx)(t), x ∈ D} are equicontinuous on [0, 1]. (2.10)
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
287
Since
|(Bx)0 (t1 ) − (Bx)0 (t2 )|
Z t2
q(s)f (s, x(s) + γ1 (s), x0 (s) + γ1 (s))ds|
=|
t
1
Z t2
M + kγ2 k
≤|
q(s)[h(M + kγ1 k) + w( min γ1 (s))][r(
) + v( min |γ2 (s)|)]ds,
1−s
s∈[0,1]
s∈[0,1]
t1
for any ε > 0, there exists a δ > 0 such that
|(Bx)0 (t1 ) − (Bx)0 (t2 )| < ε,
∀|t1 − t2 | < δ, x ∈ D,
which means that
the functions from {(Bx)0 (t), x ∈ D} are equicontinuous on (0, 1). (2.11)
Now {Bx, x ∈ D} ⊆ C 1 [0, 1], where C 1 [0, 1] = {y : [0, 1] → R : y(t)
is continuously differentible on [0, 1]} is a Banach space with norm kyk0 =
max{maxt∈[0,1] |y(t)|,
maxt∈[0,1] |y 0 (t)|}.
From (2.8)-(2.11), the Arzela-Ascoli theorem guarantees that B(D) is relatively compact in C 1 [0, 1]. Then, for any {xn } ⊆ D, there exists a y0 ∈ C 1 [0, 1]
and a subsequence {xni } of {xn } such that
lim kBxni − y0 k0 = 0.
ni →+∞
Now since
kBxn1 − y0 k
= max{max{|(Bxni )(t)−y0 (t)|t ∈ [0, 1]}, sup{p(t)|(Bxni )0 (t)−y00 (t)|t ∈ (0, 1)}}
≤ max{max{|(Bxni )(t) − y0 (t)|t ∈ [0, 1]}, max{|(Bxni )0 (t) − y00 (t)|t ∈ [0, 1])}}
= kBxni − y0 k0 ,
we have
lim kBxni − y0 k = 0,
ni →+∞
i.e., BD is relatively compact in Cp1 .
Hence, B : P → P is continuous and compact.
Remark 1: We use the two functions γ1 ∈ C[0, 1], γ2 ∈ C[0, 1] with
mint∈[0,1] γ1 (t) > 0 and maxt∈[0,1] γ2 (t) < 0 to help us remove the singularity
of f (t, x, y) at x = 0 and y = 0. If f (t, x, y) is continuous at x = 0 and y = 0,
we would take γ1 ≡ 0, γ2 ≡ 0, t ∈ [0, 1].
288
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Remark 2: Suppose that x ∈ P satisfies x = Bx, i.e.,
Z 1
x(t) = α[x] +
G(t, s)q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds,
∀t ∈ [0, 1].
0
Obviously, x(1) = α[x] and direct differentiating yields that
x0 (t) = −
Z
t
q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s))ds, t ∈ (0, 1),
0
which together with q(s)f (s, x(s) + γ1 (s), x0 (s) + γ2 (s)) ∈ C[0, 1] means that
x0 (0) = 0
and
x00 (t) = −q(t)f (t, x(t) + γ1 (t), x0 (t) + γ2 (t)), t ∈ (0, 1).
Hence, x(t) satisfies
x00 (t) = −q(t)f (t, x(t) + γ1 (t), x0 (t) + γ2 (t)), t ∈ (0, 1)
with
Z
0
x (0) = 0, x(1) =
1
x(s)dA(s).
0
Lemma 2.7 Assume that (C1 ), (C2 ) and (C3 ) hold. Then there exists R0 > 0
such that
i(B, ΩR ∩ P, P ) = 0, ∀R ≥ R0 ,
where ΩR = {x ∈ Cp |kxk < R}.
Proof. Let N ∗ =
a
R 1−a
a
2
.
(1−s)q(s)ds
From (C3 ), there exists R0 > 0 such that
f (t, x, y) ≥ N ∗ x, ∀x ≥ R0 , y ∈ (−∞, 0), t ∈ [a, 1 − a].
Let R0 =
R0
a .
(2.12)
For all R > R0 , set
ΩR = {x ∈ Cp |kxk < R}.
Now we show that
Bx 6≤ x, ∀x ∈ P ∩ ∂ΩR .
(2.13)
In fact, suppose that there is a x0 ∈ P ∩ ∂ΩR with Bx0 ≤ x0 . Lemma 2.4
implies that x0 (t) ≥ (1 − t)kx0 k, ∀t ∈ (0, 1), and so x0 (t) ≥ akx0 k ≥ aR ≥ R0
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
289
for all t ∈ [a, 1 − a]. Thus, x0 (t) + γ1 (t) ≥ akx0 k ≥ aR ≥ R0 for all t ∈ [a, 1 − a]
also. Then, from (2.12), we have
f (t, x0 (t) + γ1 (t), x00 (t) + γ2 (t)) ≥ N ∗ (x0 (t) + γ1 (t))
≥ N ∗ x0 (t) ≥ N ∗ aR, ∀t ∈ [a, 1 − a],
and so
x0 (0)
≥ (Bx0 )(0)
Z 1
(1 − s)q(s)f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))ds
= α[x0 ] +
0
Z 1−a
(1 − s)q(s)f (s, x0 (s) + γ1 (s), x00 (s) + γ2 (s))ds
≥
a
Z 1−a
≥
(1 − s)q(s)N ∗ (x0 (s) + γ1 (s))ds
Za1−a
≥
(1 − s)q(s)dsN ∗ aR
a
> R,
which implies that kx0 k ≥ kx0 k1 > R, a contradiction to x0 ∈ P ∩ ∂Ω. Then,
(2.13) is true. From Lemma 2.3, it is easy to see that
i(B, P ∩ Ω, P ) = 0.
The proof is complete.
3. Nonexistence of positive solutions to BVP(1.1)-(1.2)
In this section, we notice that the presence of z in f (t, x, z) can lead to the
nonexistence of positive solutions to (1.1)-(1.2).
Theorem 3.1 Suppose that there is a β ∈ C((0, 1), (0, +∞)) and δ > 0 such
that
f (t, y, z) ≤ −β(t), ∀(t, y, z) ∈ (0, 1) × (0, +∞) × [−δ, 0).
(3.1)
Then (1.1)-(1.2) has no positive solutions.
Proof. Suppose y0 (t) is a positive solution to (1.1)-(1.2). Then
y000 (t) + q(t)f (t, y0 (t), y00 (t)) = 0, t ∈ (0, 1)
y00 (0) = 0, y0 (1) = α[y],
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
290
which means that there is a t0 ∈ (0, 1) with y00 (t0 ) < 0, y0 (t0 ) > 0 (otherwise y 0 (t) ≥ 0 for all t ∈ (0, 1) which would contradict y(1) = α[y] ≤
R1
maxt∈[0,1] y(t) 0 |dA(s)| < maxt∈[0,1] y(t)). Let t∗ = inf{t < t0 |y00 (s) < 0 for
all s ∈ [t, t0 ]}. Clearly,
t∗ ≥ 0 and y00 (t∗ ) = 0, y00 (t) < 0 for all t ∈ (t∗ , t0 ].
(3.2)
The continuity of y00 (t) implies that there is a γ > 0 such that 0 > y00 (t) > −δ
for all t ∈ (t∗ , t∗ + γ]. Then (3.1) guarantees that f (t, y0 (t), y00 (t)) ≤ −β(t) for
all t ∈ (t∗ , t∗ + γ], which implies that
y000 (t) ≥ β(t) > 0, ∀t ∈ (t∗ , t∗ + γ],
and so
y00 (t) > 0, ∀t ∈ (t∗ , t∗ + γ],
which contradicts (3.2).
Consequently, (1.1)-(1.2) has no positive solutions.
Theorem 3.2 Suppose q ∈ C[0, 1] with q(t) > 0 for all t ∈ (0, 1) and here are
two functions h ∈ C((0, +∞), (0, +∞)), g ∈ C((−∞, 0), (0, +∞)) with
|f (t, y, z)| ≤ h(x)g(z), ∀(t, y, z) ∈ [0, 1] × (0, +∞) × (−∞, 0),
Z 0
1
dr = +∞ for all z < 0.
where
g(r)
z
(3.3)
Then (1.1)-(1.2) has no positive solutions.
Proof. Suppose y0 (t) is a positive solution to (1.1)-(1.2). Then
y000 (t) + q(t)f (t, y0 (t), y00 (t)) = 0, t ∈ (0, 1)
y00 (0) = 0, y0 (1) = α[y],
which means that there is a t0 ∈ (0, 1) with y00 (t0 ) < 0, y0 (t0 ) > 0 (otherwise y 0 (t) ≥ 0 for all t ∈ (0, 1) which would contradict y(1) = α[y] ≤
R1
maxt∈[0,1] y(t) 0 |dA(s) < maxt∈[0,1] y(t)). Let t∗ = inf{t < t0 |y00 (s) < 0 for
all s ∈ [t, t0 ]}. Clearly,
t∗ ≥ 0 and y00 (t∗ ) = 0, y00 (t) < 0 for all t ∈ (t∗ , t0 ].
Then, from (3.3),
−y000 (t)
= q(t)f (t, y0 (t), y00 (t))
≤ q(t)h(y0 (t))g(y00 (t)), t ∈ (t∗ , t0 ],
(3.4)
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
and so
−
291
y000 (t)
≤ q(t)h(y0 (t)), t ∈ (t∗ , t0 ].
g(y00 (t))
Integration from t∗ to t0 yields
Z 0
Z t0
1
+∞ =
dr ≤
q(s)h(y0 (s))ds < +∞.
y00 (t0 ) g(r)
t∗
This is a contradiction.
Consequently, (1.1)-(1.2) has no positive solutions.
Example 3.1. Consider the boundary value problems
x00 + (1 − t)a [1 − (|x0 |)−a ][xb + x−d ] = 0, t ∈ (0, 1),
x0 (0) = 0, x(1) =
(3.5)
1
Z
x(s)dA(s),
(3.6)
0
where dA(t) = 81 sin 2πtdt, a > 0, b > 0, d > 0.
It is easy to see that f (t, x, z) = (1−t)a (1−(|z|)a )[xb +x−d ] for all (t, x, z) ∈
[0, 1]×[0, +∞)×(−∞, +∞). Since limx→0+ (xb +x−d ) = limx→+∞ (xb +x−d ) =
+∞, there is a c0 > 0 such that xb + x−d ≥ c0 for all x ∈ (0, +∞). Then
f (t, x, z) ≤ (1 − t)a (1 − 2a )c0 for all (t, x, z) ∈ (0, 1) × (0, +∞) × [− 21 , 0). Then
Theorem 3.1 guarantees that (3.5)-(3.6) has no positive solutions.
Example 3.2. Consider the boundary value problems
x00 + (−x0 )a [1 + xb ] = 0, t ∈ (0, 1),
Z
0
x (0) = 0, x(1) =
(3.7)
1
x(s)dA(s),
(3.8)
0
where dA(s) = 81 sds, a > 1, b > 1, µ > 0, t ∈ [0, 1].
Let h(x) = 1 + xb , g(y) = (−y)a . It is easy to see that
|f (t, x, y)| = [1 + xb ](−y)a = h(x)g(y), ∀(t, x, y) ∈ [0, 1] × [0, +∞) × (−∞, 0]
and
Z
z
0
1
dy = +∞, ∀z < 0.
g(y)
Theorem 3.2 implies that (3.7)-(3.8) has no positive solutions.
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
292
4. Multiple positive solutions to BVP(1.1)-(1.2) without
singularities
In this section, f is continuous for (t, x, y) ∈ [0, 1] × [0, +∞) × (−∞, 0] and
Z 1
|dA(s)|.
c0 =
0
Theorem 4.1 Suppose (C1 ) − (C3 ) hold with w(t) ≡ 0 and v(t) ≡ 0 for all
t ∈ [0, +∞) and
(1 − c0 )c
> 1,
R1
Z z0 q(s)ds)
1
I(z) =
du, z ∈ R+ , I(+∞) = +∞.
r(u)
0
sup
c∈(0,+∞) I −1 (h(c)
where
(4.1)
Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1)
with x0,1 (t) ≥ 0 and x0,2 (t) > 0 on (0, 1).
Proof. From (4.1), choose an R1 > 0 with
R1 (1 − c0 )
R1
−1
I (h(R1 ) 0 q(s)ds)
> 1.
(4.2)
0
From (C3 ), choose an R2 > max{R1 , Ra } (R0 is defined as in (2.12)).
For x ∈ P , define
Z 1
(T x)(t) = α[x] +
G(t, s)q(s)f (s, x(s), x0 (s))ds, t ∈ [0, 1].
(4.3)
0
It is easy to see that Lemma 2.6 guarantees that the operator T in (4.3) is
continuous and compact from P to P (note here γ1 (t) ≡ 0 and γ2 (t) ≡ 0, for
t ∈ [0, 1]).
Let
Ω1 = {x ∈ Cp1 |kxk < R1 }
and
Ω2 = {x ∈ Cp1 |kxk < R2 }.
Then, we claim that
µT x 6= x,
∀µ ∈ (0, 1], x ∈ P ∩ ∂Ω1 .
(4.4)
Now we show that (4.4) is true. Suppose there exists an x0 ∈ P ∩ ∂Ω1 and
a µ0 ∈ (0, 1] such that x0 = µ0 T x0 . Then
00
x0 (t) + µ0 q(t)f (t, x0 (t), x00 (t)) = 0, t ∈ (0, 1)
(4.5)
x00 (0) = 0, x0 (1) = α[x0 ],
293
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
which means that x0 (t) ≥ 0 is on (0, 1) with x00 (0) = 0 and x00 (t) is nonincreasing on (0, 1). Without loss of generality, we assume that x00 (t) < 0 for all
t ∈ (0, 1) (obviously, (4.7) is true for x00 (t) = 0). From (4.5), we have
−x000 (t) ≤ q(t)f (t, x0 (t), x00 (t)) ≤ q(t)h(x0 (t))r(−x00 (t)), ∀t ∈ (0, 1),
which means that
−x000 (t)
≤ h(x0 (t))q(t) ≤ h(R1 )q(t), ∀t ∈ (0, 1).
r(−x00 (t))
(4.6)
Integration from 0 to t yields
I(−x00 (t)) − I(−x00 (0)) = I(−x00 (t)) ≤ h(R1 )
1
Z
q(s)ds,
0
and so
−x00 (t) ≤ I −1 (h(R1 )
Z
1
q(s)ds), ∀t ∈ (0, 1).
(4.7)
0
Now integrate from 0 to 1 to obtain
R1 (1 − c0 ) ≤ x0 (0) − x0 (1) ≤ I −1 (h(R1 )
Z
1
q(s)ds),
0
a contradiction to (4.2). Then, (4.4) is true. Lemma 2.1 implies that
i(T, Ω1 ∩ P, P ) = 1.
(4.8)
From Lemma 2.7, we have
i(T, P ∩ Ω2 , P ) = 0,
and so
i(T, P ∩ (Ω2 − Ω1 ), P ) = −1.
(4.9)
As a result, there exist x1 ∈ P ∩Ω1 and x2 ∈ P ∩(Ω2 −Ω1 ) such that x1 = T x1
and x2 = T x2 .
Consequently, BVP(1.1)-(1-2) has at least two different nonnegative solutions x1 (t) and x2 (t) with kx1 k < R1 < kx2 k.
Example 4.1. Consider the boundary value problems
x00 + µ[1 + |x0 |a ][1 + xb ] = 0, t ∈ (0, 1)
(4.10)
294
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Z
0
1
x(s)dA(s),
x (0) = 0, x(1) =
(4.11)
0
with dA(t) =
1
8
sin 2πtdt, 0 < a < 1, b > 1 and µ > 0. If
R (1−c0 )c
µ<
0
sup
1
c∈(0,+∞)
1
1+sa ds
,
+ cb
(4.12)
then BVP(4.10)-(4.11) has at least two different positive solutions x0,1 , x0,2 ∈
C 1 [0, 1] ∩ C 2 (0, 1).
It is easy to see that all conditions of Theorem 4.1 hold. Then, Theorem
4.1 guarantees that BVP(4.10)-(4.11) has at least two different positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
5. Multiple positive solutions to BVP(1.1)-(1.2) with
singularity at x = 0 but not at x0 = 0
0
In this section
Z 1 our nonlinearity f may be singular at x = 0 but not at x = 0
and c0 =
|dA(s)| .
0
Theorem 5.1 Suppose (C1 ) − (C3 ) hold with w ∈ C((0, +∞), (0, +∞)) ∩
Lloc [0, +∞) and v(t) ≡ 0 for all t ∈ [0, +∞) and
(1 − c0 )c
Rc
> 1, where
+ 0 w(s)ds])
sup
−1
c∈(0,+∞)
Z Iz (kqk0 [ch(c)
I(z) =
0
u
du, z ∈ (0, +∞), I(+∞) = +∞, kqk0 = max q(t).
r(u)
t∈[0,1]
(5.1)
Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1)
with x0,1 (t) > 0 and x0,2 (t) > 0 on (0, 1).
Proof. From (5.1) and the continuity of I −1 and h, choose an R1 > 0, and a
R1
with
ε > 0 with ε <
2
R1 (1 − c0 )
I −1 ((R1
+ ε)h(R1 + ε)kqk0 + kqk0
0
R R1 +ε
0
> 1.
w(s)ds)
From (C3 ), choose a R2 > max{R1 , Ra } (R0 is defined as in (2.12)).
(5.2)
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Let n0 ∈ {1, 2, · · · } be chosen so that
295
1
< ε, and let N0 = {n0 , n0 +1, · · · }.
n0
For each n ∈ N0 , for x ∈ P , define
Z 1
1
(Tn x)(t) = α[x] +
G(t, s)q(s)f (s, x(s) + , x0 (s))ds, t ∈ [0, 1].
n
0
(5.3)
Lemma 2.6 implies that Tn : P → P is continuous and compact (here
γ1 (t) ≡ n1 and γ2 (t) ≡ 0, for t ∈ [0, 1]).
Let
Ω1 = {x ∈ Cq1 |kxk < R1 }
and
Ω2 = {x ∈ Cq1 |kxk < R2 }.
Then, for each n ∈ N0 , we claim that
µTn x 6= x,
∀µ ∈ (0, 1], x ∈ P ∩ ∂Ω1 .
(5.4)
Now we show that (5.4) is true. Suppose there exists an x0 ∈ P ∩ ∂Ω1 and
a µ0 ∈ (0, 1] such that x0 = µ0 Tn x0 . Then
(
1
x000 (t) + µ0 q(t)f (t, x0 (t) + , x00 (t)) = 0, t ∈ (0, 1)
(5.5)
n
x0 (0) = 0, x0 (1) = α[x0 ],
which means that x0 (t) > 0 is on (0, 1) with x00 (0) = 0 and x00 (t) is nonincreasing on (0, 1). Without loss of generality, we assume that x00 (t) < 0 for all
t ∈ (0, 1) (obviously, (5.7) is true for x00 (t) = 0). From (5.5), we have
1
1
1
−x000 (t) ≤ q(t)f (t, x0 (t)+ , x00 (t)) ≤ q(t)[h(x0 (t)+ )+w(x0 (t)+ )]r(−x00 (t)),
n
n
n
∀t ∈ (0, 1),
which means that
−x000 (t)
1
1
≤ [h(x0 (t) + ) + w(x0 (t) + )]q(t), ∀t ∈ (0, 1)
0
r(−x0 (t))
n
n
and so
1
1
x00 (t)x000 (t)
≤ [h(x0 (t) + ) + w(x0 (t) + )](−x00 (t))q(t), ∀t ∈ (0, 1). (5.6)
0
r(−x0 (t))
n
n
Integration from 0 to t yields
I(−x00 (t)) − I(−x00 (0)) = I(−x00 (t))
296
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
≤ kqk0 h(x0 (0) +
1
1
)(x0 (0) + ) + kqk0
n
n
Z
1
x0 (0)+ n
w(s)ds,
1
α[x0 ]+ n
and so
−x00 (t) ≤ I −1 (kqk0 h(R1 + ε)(R1 + ε) + kqk0
Z
R1 +ε
w(s)ds), ∀t ∈ (0, 1], (5.7)
0
Now integrate from 0 to 1 to obtain
R1 (1−c0 ) ≤ x0 (0)−x0 (1) ≤ I −1 (kqk0 h(R1 +ε)(R1 +ε)+kqk0
Z
R1 +ε
w(s)ds),
0
a contradiction to (5.2). Then, (5.4) is true.
From Lemma 2.2, for each n ∈ N0 , we have
i(Tn , P ∩ Ω1 , P ) = 1.
(5.8)
Lemma 2.7 implies that
i(Tn , P ∩ Ω2 , P ) = 0,
and so
i(Tn , P ∩ (Ω2 − Ω1 ), P ) = −1,
n ∈ N0 .
(5.9)
As a result, for each n ∈ N0 , there exist xn,1 ∈ P ∩Ω1 and xn,2 ∈ P ∩(Ω2 −Ω1 )
such that xn,1 = Tn xn,1 and xn,2 = Tn xn,2 .
Now we consider {xn,1 }n∈N0 and {xn,2 }n∈N0 . Obviously, since {xn,1 }n∈N0
is bounded, it is easy to see that
{xn,1 (t)} is uniformly bounded on [0, 1]
(5.10)
with max |xn,1 (t)| ≤ R1 , n ∈ N0 .
t∈[0,1]
Using xn,1 instead of x0 in (5.2), from (5.7), one has
−x0n,1 (t) ≤ I −1 (kqk0 h(R1 + ε)(R1 + ε) + kqk0
Z
R1 +ε
w(s)ds), ∀t ∈ (0, 1],
0
which yields that
{x0n,1 (t)} is uniformly bounded on [0, 1]
(5.11)
{xn,1 (t)} is equicontinous on [0, 1].
(5.12)
and so
297
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
Using xn,1 instead of x0 in (5.6), we have
x0n,1 (t)x00n,1 (t)
1
1
≤ [h(xn,1 (t) + ) + w(xn,1 (t) + )]q(t), ∀t ∈ (0, 1).
r(−x0n,1 (t))
n
n
Integration from t1 to t2 yields
|I(−x0n,1 (t1 )))
−
I(−x0n,1 (t2 ))|
Z
−xn,1 (t2 )
=|
r
dr|
g(r)
−xn,1 (t1 )
Z −xn,1 (t2 )+ n1
≤ [h(R1 + ε)|xn,1 (t1 ) − xn,1 (t2 )| + |
w(r) dr|]kqk0 .
1
−xn,1 (t1 )+ n
Since w ∈ Lloc [0, +∞), from (5.12), we have
{I(−x0n,1 (t))} is equicontinous on [0, 1].
(5.13)
Since
|x0n,1 (t1 ) − x0n,1 (t2 )| = |I −1 (I(−x0n,1 (t1 ))) − I −1 (I(−x0n,1 (t2 )))|
and I −1 is uniformly continuous on [0, I(R1 )], we have
{x0n,1 (t)} is equicontinuous on [0, 1].
(5.14)
From (5.10), (5.11), (5.12) and (5.14), the Arzela-Ascoli Theorem guarantees the existence of a subsequence N = (nj ) of N0 and a function x0,1 ∈
C 1 [0, 1] with lim xnj ,1 → x0,1 with
j→+∞
x00,1 (0) = 0.
(5.15)
From
1 0
,x
(t)) = f (t, x0,1 (t), x00,1 (t)), ∀t ∈ (0, 1),
nj nj ,1
1
1
1
0
xnj ,1 (t) = xnj ,1
+ xnj ,1
t−
2
2
2
Z t
1
+
(s − t) q(s) f (s, xnj ,1 (s) + , x0nj ,1 (s)) ds
1
n
j
2
lim f (t, xnj ,1 (t) +
j→+∞
for t ∈ (0, 1) and
Z
xnj ,1 (1) =
1
xnj ,1 (s)dA(s),
0
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
298
the Lebesgue Dominated Convergence Theorem guarantees that
Z t
1
1
1
x0,1 (t) = x0,1
+x00,1
t−
+
(s−t) q(s) f (s, x0,1 (s), x00,1 (s)) ds
1
2
2
2
2
(5.16)
Z 1
x0,1 (s)dA(s) = α[x0,1 ].
(5.17)
x0,1 (1) =
0
Hence, from (5.15)-(5.17), x0,1 (t) is a positive solution to BVP(1.1)-(1.2) with
kx0,1 k ≤ R1 . Also (5.2) guarantees that kx0,1 k < R1 .
For the set {xn,2 }n∈N0 ⊆ (Ω2 − Ω1 ) ∩ P , from the same proof for the set
{xn,1 }n∈N0 , we can obtain a convergent subsequence {xni ,2 } of {xn,2 } with
lim xni ,2 = x0,2 ∈ C 1 . Moreover, x0,2 is a positive solution to equation (1.1)
i→+∞
with R1 < kx0,2 k < R2 .
Consequently, BVP(1.1)-(1.2) has at least two different positive solutions
x0,1 (t) and x0,2 (t) with kx0,1 k < R1 < kx0,2 k.
Example 5.1. Consider the boundary value problems
x00 + µ[1 + |x0 |a ][1 + xb + x−d ] = 0, t ∈ (0, 1)
Z 1
x0 (0) = 0, x(1) =
x(s)dA(s),
(5.18)
(5.19)
0
with dA(t) =
1
sin 2πtdt, 0 < a < 1, b > 1, 0 < d < 1 and µ > 0. If
8
R (1−c0 )c s
1+sa ds
0
µ < sup
,
1
b+1
+ 1−d
c1−d
c∈(0,+∞) c + c
(5.20)
then BVP(5.18)-(5.19) has at least two different positive solutions x0,1 , x0,2 ∈
C 1 [0, 1] ∩ C 2 (0, 1).
It is easy to see that all conditions of Theorem 5.1 hold. Then, Theorem
5.1 guarantees that BVP(5.18)-(5.19) has at least two different positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
6. Multiple positive solutions to BVP(1.1)-(1.2) with
singularity at x0 = 0 but not at x = 0
0
In this
Z section our nonlinearity f may be singular at x = 0 but not x = 0 and
1
|dA(s)|.
c0 =
0
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
299
Theorem 6.1 Suppose (C1 ) − (C3 ) hold with v ∈ C((0, +∞), (0, +∞)) and
w(t) ≡ 0 for all t ∈ [0, +∞) and
(1 − c0 )c
R1
−1
c∈(0,+∞) I
Z z (h(c) 0 q(s)ds)
sup
I(z) =
0
> 1, where
1
du, z ∈ (0, +∞), I(+∞) = +∞.
r(u) + v(u)
(6.1)
Then BVP(1.1)-(1.2) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩
C 2 (0, 1) with x0,1 (t) ≥ 0 and x0,2 (t) > 0 on (0, 1).
Proof. From (6.1) and the continuity of I −1 and h, choose an R1 > 0, and a
R1
with
ε > 0 with ε <
2
I −1 (I(ε)
R1 (1 − c0 )
> 1.
R1
+ h(R1 ) 0 q(s)ds)
0
From (C3 ), choose a R2 > max{R1 , Ra } (R0 is defined as in (2.12)).
1
Let n0 ∈ {1, 2, · · · } be chosen so that
< ε, and let N0 = {n0 , n0 +1, · · · }.
n0
For each n ∈ N0 , for x ∈ P , define
Z 1
1
(Tn x)(t) = α[x] +
G(t, s)q(s)f (s, x(s), x0 (s) − )ds, t ∈ [0, 1].
n
0
Lemma 2.6 guarantees that for each n ∈ N0 , Tn : P → P is continuous
and compact (here γ1 (t) ≡ 0 and γ2 (t) ≡ − n1 , for t ∈ [0, 1]).
Let
Ω1 = {x ∈ Cq1 |kxk < R1 }
and
Ω2 = {x ∈ Cq1 |kxk < R2 }.
An argument similar to that in the proof of (5.4) shows that for each
n ∈ N0 , we have that
µTn x 6= x,
∀µ ∈ (0, 1], x ∈ P ∩ ∂Ω1 ,
which together with Lemma 2.2 implies
i(Tn , P ∩ Ω1 , P ) = 1, n ∈ N0 .
Since Lemma 2.7 guarantees that
i(Tn , P ∩ Ω2 , P ) = 0, n ∈ N0
300
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
we have
i(Tn , P ∩ (Ω2 − Ω1 ), P ) = −1,
n ∈ N0 .
As a result, for each n ∈ N0 , there exist xn,1 ∈ P ∩Ω1 and xn,2 ∈ P ∩(Ω2 −Ω1 )
such that xn,1 = Tn xn,1 and xn,2 = Tn xn,2 .
An argument similar to that in the proof of (5.10)-(5.12) and (5.14) shows
that
{xn,1 (t)}, {xn,2 (t)} are uniformly bounded on [0, 1],
{x0n,1 (t)}, {x0n,2 (t)} are uniformly bounded on [0, 1],
{xn,1 (t)}, {xn,2 (t)} are equicontinous on [0, 1],
and
{−x0n,1 (t)}, {x0n,2 (t)} are equicontinuous on [0, 1],
which together with the Arzela-Ascoli Theorem guarantees the existence of a
subsequence {nj } of N0 and a function x0,1 ∈ C 1 [0, 1] with lim xnj ,1 → x0,1
j→+∞
with
x00,1 (0) = 0
and the existence of a subsequence {nk } of N0 and a function x0,2 ∈ C 1 [0, 1]
with lim xnk ,2 → x0,2 with
k→+∞
x00,2 (0) = 0
An argument similar to that in the proof of (5.16)-(5-17) shows that x0,1 (t) and
x0,2 (t) are two different positive of BVP(1.1)-(1.2) with kx0,1 k < R1 < kx0,2 k.
Example 6.1. Consider the boundary value problems
x00 + µ[1 + |x0 |a + |x0 |−d ][1 + xb ] = 0, t ∈ (0, 1)
Z 1
x0 (0) = 0, x(1) =
x(s)dA(s),
(6.2)
(6.3)
0
with dA(t) =
1
sin 2πtdt, 0 < a < 1, b > 1, 0 < d and µ > 0. If
8
R (1−c0 )c
sd
ds
0
sd +sa+d +1
,
µ < sup
b
1+c
c∈(0,+∞)
then BVP(6.2)-(6.3) has at least two different positive solutions x0,1 , x0,2 ∈
C 1 [0, 1] ∩ C 2 (0, 1).
301
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
It is easy to see that all conditions of Theorem 6.1 hold. Then, Theorem
6.1 guarantees that BVP(6.2)-(6.3) has at least two different positive solutions
x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
7. Multiple positive solutions to BVP(1.1)-(1.2) with
singularity at x = 0 and x0 = 0
0
In this
Z 1section our nonlinearity f may be singular at x = 0 and x = 0 and
c0 =
|dA(s)|.
0
Theorem 7.1 Suppose (C1 ) − (C3 ) hold and
(1 − c0 )c
R
> 1, where
−1 (kqk [ch(c) + c w(s)ds])
I
c∈(0,+∞)
0
0
Z z
u
du, z ∈ (0, +∞), I(+∞) = +∞, kqk0 = max q(t).
I(z) =
r(u)
+
v(u)
t∈[0,1]
0
(7.1)
Then (1.1) has at least two positive solutions x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1)
with x0,1 (t) ≥ 0 and x0,2 (t) > 0 on (0, 1).
sup
Proof. From (7.1) and the continuity of I −1 , choose an R1 > 0, and a ε > 0
R1
with
with ε <
2
R1 (1 − c0 )
I −1 (I(ε)
+ (R1 + ε)h(R1 + ε)kqk0 + kqk0
R R1 +ε
0
> 1.
w(s)ds)
0
From (C3 ), choose a R2 > max{R1 , Ra } (R0 is defined as in (2.12)).
2
Let n0 ∈ {1, 2, · · · } be chosen so that
< ε, and let N0 = {n0 , n0 +1, · · · }.
n0
For each n ∈ N0 , for x ∈ P , define
Z 1
1
1
1
(Tn x)(t) = α[x]+
G(t, s)q(s)f (s, x(s)+ (1−s)+ , x0 (s)− )ds, t ∈ [0, 1].
n
n
n
0
Lemma 2.6 guarantees that Tn : P → P is continuous and compact (here
γ1 (t) = n1 (1 − t) + n1 and γ2 (t) ≡ − n1 , for t ∈ [0, 1]).
Set
Ω1 = {x ∈ Cp |kxk < R1 }
and
Ω1 = {x ∈ Cp |kxk < R2 }.
302
SINGULAR NONLOCAL BOUNDARY VALUE PROBLEMS
An argument argument to that in the proof in Theorem 6.1 yields two different
positive solutions x0,1 (t) and x0,2 (t) with x0,1 ∈ Ω1 ∩P and x0,2 ∈ (Ω2 −Ω1 )∩P .
Example 7.1. Consider the boundary value problems
x00 + µ[1 + |x0 |a + |x0 |−e ][1 + xb + x−d ] = 0, t ∈ (0, 1)
Z 1
x0 (0) = 0, x(1) =
x(s)dA(s),
(7.2)
(7.3)
0
with dA(t) =
1
sin 2πtdt, 0 < a < 1, b > 1, 0 < d < 1, e > 0 and µ > 0. If
8
R (1−c0 )c se+1
se +sa+ +1 ds
0
µ < sup
,
1
b+1
+ 1−d
c1−d
c∈(0,+∞) c + c
then BVP(7.2)-(7.3) has at least two different positive solutions x0,1 , x0,2 ∈
C 1 [0, 1] ∩ C 2 (0, 1).
It is easy to see that all conditions of Theorem 7.1 hold. Now Theorem
7.1 guarantees that BVP(7.2)-(7.3) has at least two different positive solutions
x0,1 , x0,2 ∈ C 1 [0, 1] ∩ C 2 (0, 1).
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Baoqiang Yan,
Department of Mathematics,
Shandong Normal University,
Jinan, 250014, P.R.China.
Email: yanbqcn@aliyun.com
Donal O’Regan,
School of Mathematics, Statistics and Applied Mathematics,
National University of Ireland,
Galway, Ireland.
Email: donal.oregan@nuigalway.ie
Ravi P. Agarwal,
Department of Mathematics,
Texas A and M University-Kingsville,
Texas 78363, USA.
Email: Ravi.Agarwal@tamuk.edu