Electronic Journal of Differential Equations, Vol. 2010(2010), No. 163, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE OF POSITIVE SOLUTIONS FOR
BOUNDARY-VALUE PROBLEMS WITH INTEGRAL BOUNDARY
CONDITIONS AND SIGN CHANGING NONLINEARITIES
SHU LIU, MEI JIA, YINGQIANG TIAN
Abstract. In this article, we show the existence of positive solutions for
boundary value problems with integral boundary conditions and sign changing nonlinearities. By using a fixed point theorem in double cones, we obtain
sufficient conditions for the existence of two positive solutions.
1. Introduction
The theory of boundary value problems with integral boundary conditions for
ordinary differential equations arises in different areas of applied mathematics and
physics. For example, heat conduction, chemical engineering, underground water
flow, thermo-elasticity, and plasma physics can all be reduced to nonlocal problems
with integral boundary conditions. For boundary value problems with integral
boundary conditions and comments on their importance, we refer the reader to
[7, 10, 11] and the references therein. For more information about the general
theory of integral equations and their relation with boundary value problems we
refer to [1, 4].
Moreover, boundary value problems with integral boundary conditions constitute
a very interesting and important class of problems. They include two, three, multipoint and nonlocal boundary value problems as special cases. The existence and
multiplicity of positive solutions for such problems have received a great deal of
attention. To identify a few, we refer the reader to [2, 9, 12, 13, 14, 15] and
references therein. But the corresponding theory for boundary-value problem with
integral boundary conditions and sign changing nonlinearities of one-dimensional
p-Laplacian is not investigated till now. By using the fixed point theorem in double
cones, Guo in [8] discussed the existence of positive solutions for second-order threepoint boundary-value problem
x00 + f (t, x) = 0,
0
x(0) − βx (0) = 0,
0 ≤ t ≤ 1,
x(1) = αx(η),
2000 Mathematics Subject Classification. 34B15, 34B18.
Key words and phrases. Boundary value problems; p-Laplacian; integral boundary conditions;
fixed point theorem in double cones; positive solutions.
c
2010
Texas State University - San Marcos.
Submitted June 15, 2010. Published November 12, 2010.
Supported by grant 10ZZ93 from the Innovation Program of Shanghai Municipal
Education Commission.
1
2
S. LIU, M. JIA, Y. TIAN
EJDE-2010/163
where f is allowed to change sign. Sufficient conditions are obtained by imposing
growth conditions on f which ensure the existence of at least two positive solutions
for the above boundary value problems. Meanwhile, they proved a fixed point
theorem in double cones which generalize the fixed point theorem in a cones to
some degree. By using a theorem similar to the one in [8], Cheung [3] proved the
existence of two positive solutions for the problem
(Φp (u0 ))0 + h(t)f (t, u) = 0,
0 < t < 1,
with each of the following two sets of boundary conditions
u0 (0) = 0,
m−2
X
u(1) =
αi u(ξi );
i=1
u(0) =
m−2
X
u0 (1) = 0,
αi u(ξi ),
i=1
where h : [0, 1] → R+ and f : [0, 1] × [0, ∞) → R are continuous functions.
Recently, by using the classical fixed-point index theorem for compact maps,
Feng [5] established sufficient conditions for the existence of multiple positive solutions for the second-order impulsive differential equations, with p-Laplacian and
integral boundary conditions,
−(φp (u0 (t)))0 = f (t, u(t)),
t 6= tk , t ∈ (0, 1)
−∆u|t=tk = Ik (u(tk )),
u0 (0) = 0,
t = 1, 2, . . . , n,
Z
1
u(1) =
g(t)u(t)dt .
0
Motivated by the above, we consider the existence of positive solutions for the
boundary-value problem, with integral boundary conditions and sign changing nonlinearities of one-dimensional p-Laplacian,
(ϕp (u0 ))0 + f (t, u) = 0,
au(0) − bu0 (0) =
m−2
X
0 ≤ t ≤ 1,
ai u(ξi ),
i=1
Z
u(1) =
(1.1)
1
g(s)u(s)ds,
0
where a, b ∈ [0, +∞), ai ∈ (0, +∞), i = 1, 2, . . . , m − 2, 0 < ξ1 < · · · < ξm−2 < 1,
m ≥ 3, ϕp (u) = |u|p−2 u, p > 1, ϕq = (ϕp )−1 , and 1q + p1 = 1.
We will assume the following conditions:
(H1) f : [0, 1] × [0, +∞) → R is continuous;
(H2) f (t, 0) ≥ 0 (f 6≡ 0) for t ∈ [0, 1];
R1
Pm−2
(H3) a > i=1 ai ; g is non-negative, integrable, and σ := 0 g(t)dt ∈ [0, 1).
2. Preliminaries
For a cone K in a Banach space X with norm k · k and a constant r > 0, let
Kr = {x ∈ K : kxk < r}, ∂Kr = {x ∈ K : kxk = r}. Suppose α : K → [0, +∞)
is a continuously increasing functional, i.e., α is continuous and α(λx) ≤ λα(x) for
λ ∈ (0, 1). Denote K(b) = {x ∈ K : α(x) < b}, ∂K(b) = {x ∈ K : α(x) = b} and
Ka (b) = {x ∈ K : a < kxk, α(x) < b}.
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EXISTENCE OF POSITIVE SOLUTIONS
3
Lemma 2.1 ([8]). Let X be a real Banach space with norm k · k and K, K 0 ⊂ X
two cones with K 0 ⊂ K. Suppose T : K → K and T ∗ : K 0 → K 0 are two completely
continuous operators and α : K 0 → R+ a continuously increasing functional satisfying α(x) ≤ kxk ≤ M α(x) for all x in K 0 , where M ≥ 1 is a constant. If there
are constants b > a > 0 such that
(C1) kT xk < a, for x ∈ ∂Ka ;
(C2) kT ∗ xk < a, for x ∈ ∂Ka0 and α(T ∗ x) > b for x ∈ ∂K 0 (b);
(C3) T x = T ∗ x, for x ∈ ∂Ka0 (b) ∩ {u : T ∗ u = u},
then T has at least two fixed points y1 and y2 in K such that
0 ≤ ky1 k < a < ky2 k,
α(y2 ) < b.
We also denote C + [0, 1] = {u ∈ C[0, 1] : u(t) ≥ 0, t ∈ [0, 1]}.
Lemma 2.2. For any h ∈ C + [0, 1], the boundary-value problem
(ϕp (u0 ))0 + h(t) = 0,
au(0) − bu0 (0) =
0 ≤ t ≤ 1,
m−2
X
ai u(ξi ),
(2.1)
i=1
Z
u(1) =
1
g(s)u(s)ds,
0
has a unique solution of the form
Rs
Pm−2 R ξ
Z t
Z s
bϕq (Ah ) + i=1 ai 0 i ϕq (Ah − 0 h(τ )dτ )ds
u(t) =
+
ϕq (Ah −
h(τ )dτ )ds,
Pm−2
a − i=1 ai
0
0
(2.2)
or
R1
R1
Rr
Z 1
Z s
g(s)( s ϕq ( 0 h(τ )dτ − Ah )dr)ds
0
u(t) =
+
ϕ
(
h(τ )dτ − Ah )ds, (2.3)
R1
q
1 − 0 g(s)ds
t
0
where Ah satisfies
R1
Rr
g(s)( s ϕq ( 0 h(τ )dτ − Ah )dr)ds
bϕq (Ah ) =
R1
1 − 0 g(s)ds
Z s
Z s
Z 1
m−2
X Z 1
+a
ϕq (
h(τ )dτ − Ah )ds −
ai
ϕq (
h(τ )dτ − Ah )ds.
(a −
Pm−2
i=1
0
ai )
R1
0
0
ξi
i=1
0
(2.4)
Proof. If u is a solution of (2.1), it is easy to see that
Z t
ϕp (u0 (t)) − ϕp (u0 (0)) = −
h(s)ds.
0
0
Let Ah = ϕp (u (0)). Integrating, we can have
Z t
0
u (t) = ϕq (Ah −
h(s)ds).
0
Integrating from t to 1, we have
Z
u(t) = u(1) −
1
Z
ϕq (Ah −
t
s
h(τ )dτ )ds.
0
(2.5)
4
S. LIU, M. JIA, Y. TIAN
Substituting u(1) =
EJDE-2010/163
R1
g(s)u(s)ds, we obtain
Z
Z 1
Z 1
u(t) =
g(s)u(s)ds −
ϕq (Ah −
0
0
t
s
h(τ )dτ )ds.
0
R1
Using the above equation and u(1) = 0 g(s)u(s)ds, we have
Z r
Z
Z 1
Z 1
Z 1
Z 1
g(s)u(s)ds
g(s)ds −
g(s)(
ϕq (Ah −
h(τ )dτ )dr)ds =
0
0
0
s
0
1
g(s)u(s)ds,
0
i.e.,
Z
R1
1
0
g(s)u(s)ds =
0
R1
Rr
g(s)( s ϕq ( 0 h(τ )dτ − Ah )dr)ds
.
R1
1 − 0 g(s)ds
So
R1
u(t) =
0
R1
Rr
Z 1
Z s
g(s)( s ϕq ( 0 h(τ )dτ − Ah )dr)ds
+
ϕ
(
h(τ )dτ − Ah )ds. (2.6)
R1
q
1 − 0 g(s)ds
t
0
Integrating (2.5) from 0 to t, we obtain
Z t
Z
u(t) = u(0) +
ϕq (Ah −
0
s
h(τ )dτ )ds.
0
Pm−2
Using the boundary condition au(0) − bu0 (0) = i=1 ai u(ξi ), we have
Rs
Pm−2 R ξ
Z t
Z s
bϕq (Ah ) + i=1 ai 0 i ϕq (Ah − 0 h(τ )dτ )ds
u(t) =
+
ϕq (Ah −
h(τ )dτ )ds,
Pm−2
a − i=1 ai
0
0
where A can be easily obtained from the boundary condition au(0) − bu0 (0) =
Pm−2 h
i=1 ai u(ξi ) and (2.6).
Lemma 2.3. Assume that (H3) holds. Then for each h ∈ C + [0, 1], there exists a
R1
unique Ah ∈ (0, 0 h(τ )dτ ) satisfying (2.2) and (2.3). Moreover, there is a unique
Rσ
σh ∈ (0, 1) such that Ah = 0 h h(τ )dτ .
Proof. For h ∈ C + [0, 1], define
R1
R1
Rr
Pm−2
(a − i=1 ai ) 0 g(s)( s ϕq ( 0 h(τ )dτ − η)dr)ds
Hh (η) = bϕq (η) −
R1
1 − 0 g(s)ds
Z 1
Z s
Z s
m−2
X Z 1
−a
ϕq (
h(τ )dτ − η)ds +
ai
ϕq (
h(τ )dτ − η)ds,
0
0
i=1
where η ∈ (−∞, +∞). Hence
Z 1
Z
m−2
X
ϕq (
Hh (η) = (
ai − a)
i=1
+ bϕq (η) −
0
(a −
0
s
h(τ )dτ − η)ds −
ξi
m−2
X
0
Z
ξi
ai
0
i=1
R1
R1
Rr
Pm−2
i=1 ai ) 0 g(s)( s ϕq ( 0 h(τ )dτ
R1
1 − 0 g(s)ds
Z
ϕq (
s
h(τ )dτ − η)ds
0
− η)dr)ds
.
By the properties of the function ϕq (·), we know that Hh (η) is strictly increasing
R1
function on (−∞, +∞). We also easily know that Hh (0) < 0, Hh ( 0 h(τ )dτ ) > 0,
so we can
R 1prove the lemma by the continuity of Hh . Namely, there is a unique
Ah ∈ (0, 0 h(τ )dτ ) satisfying (2.2) and (2.3). Meanwhile, we could have that there
EJDE-2010/163
EXISTENCE OF POSITIVE SOLUTIONS
is unique σh ∈ (0, 1) such that Ah =
[0, 1].
R σh
0
h(τ )dτ by the continuity of
5
Rt
0
h(τ )dτ on
Lemma 2.4. Assume that (H3) holds. If h ∈ C + [0, 1], then the solution of boundary value problem (2.1) has the following properties:
(i) u(t) is a concave function;
(ii) u(t) ≥ 0, t ∈ [0, 1];
(iii) u(σh ) = max0≤t≤1 u(t), where σh is defined in lemma 2.3.
Proof. Suppose that u(t) is the solution of (2.1).
(i) From the fact that (ϕp (u0 ))0 (t) = −h(t) ≤ 0 we know that ϕp (u0 ) is nonincreasing. It follows that u0 (t) is also non-increasing. Thus, u(t) is concave down
on [0, 1].
(ii) From (i), we know that the minimum of u(t) is obtained at 0 or 1. So we
only have to prove that u(0) ≥ 0 and also u(1) ≥ 0.
Firstly, we prove that u(0) ≥ 0. If u(0) < 0 and u0 (0) ≥ 0, by the condition
Pm−2
au(0) − bu0 (0) = i=1 ai u(ξi ), we have
au(0) −
m−2
X
ai u(ξi ) = bu0 (0) ≥ 0,
i=1
so
au(0) ≥
m−2
X
ai u(ξi ).
(2.7)
ai u(0).
(2.8)
i=1
From a >
Pm−2
i=1
ai > 0, we have
au(0) <
m−2
X
i=1
From (2.7) and (2.8), we obtain
m−2
X
ai u(0) >
i=1
m−2
X
ai u(ξi ).
i=1
Hence, there is a j, 1 ≤ j ≤ m − 2 such that 0 > u(0) > u(ξj ). Then we obtain
that u(1) = mint∈[0,1] u(t) < 0 and that there is η1 ≤ η2 ∈ [0, 1] such that u(t) ≤ 0
when t ∈ [0, η1 ] ∪ [η2 , 1] and u(t) ≥ 0 when t ∈ [η1 , η2 ].
R1
From the boundary condition u(1) = 0 g(s)u(s)ds, we have
Z η1
Z η2
Z 1
g(s)u(s)ds −
g(s)u(s)ds −
g(s)u(s)ds
|u(1)| = −u(1) = −
0
Z
≤
η1
η1
Z
η2
g(s)|u(s)|ds +
0
= |u(1)|
1
g(s)|u(s)|ds +
η1
Z
η2
Z
g(s)|u(s)|ds
(2.9)
η2
1
g(s)ds.
0
R1
We have 0 g(s)ds ≥ 1 which is contradiction to with (H3), so u(0) ≥ 0.
If u(0) < 0 and u0 (0) < 0, we can prove that u(0) ≥ 0 by using the same way as
in (2.9). Similarly, we can prove u(1) ≥ 0. Therefore, we obtain that u(t) ≥ 0 for
t ∈ [0, 1].
6
S. LIU, M. JIA, Y. TIAN
EJDE-2010/163
(iii) From the boundary conditions, we can easily show that the maximum of u(t)
can not be at 0 or 1. If u(0) is the maximum, then u0 (0) ≤ 0 which is contradicted
Pm−2
with boundary condition au(0) − bu0 (0) = i=1 ai u(ξi ). If u(1) is the maximum,
R1
it is contradicts the boundary condition u(1) = 0 g(s)u(s)ds.
Rσ
From Lemmas 2.2 and 2.3, we have u0 (t) = ϕq ( t h h(τ )dτ ). By the concavity of
u(t), we know u(σh ) = max0≤t≤1 u(t).
Lemma 2.5. If u ∈ C[0, 1] and u(t) ≥ 0 is a concave function, then for any
δ ∈ (0, 12 ), we have
min
δ≤t≤1−δ
u(t) ≥ δkuk.
Proof. Let u(σ) = kuk. We discuss three cases:
(i) If σ < δ, then minδ≤t≤1−δ u(t) = u(1 − δ). By the concavity of u(t), we have
u(σ) − u(1)
u(1 − δ) − u(1)
≤
;
1−σ
δ
i.e.,
u(1 − δ)
u(σ)
1
1
≥
+( −
)u(1).
δ
1−σ
δ
1−σ
For σ < δ < 1 − δ, we have 1 − σ > δ. So
u(1 − δ) ≥
δ
u(σ) ≥ δu(σ).
1−σ
(ii) If σ > 1 − δ, then minδ≤t≤1−δ u(t) = u(δ) and σ > δ. By the concavity of
u(t), we have
u(σ) − u(0)
u(δ) − u(0)
≤
;
σ
δ
i.e.,
u(δ)
u(σ)
1
1
≥
+ ( − )u(0).
δ
σ
δ
σ
So
u(δ) ≥
δ
u(σ) ≥ δu(σ).
σ
(iii) For δ ≤ σ ≤ 1 − δ, if minδ≤t≤1−δ u(t) = u(1 − δ), the proof is the same as
(i); if minδ≤t≤1−δ u(t) = u(δ), the proof is the same as (ii).
From the three cases above, the proof is complete.
Let X = C[0, 1] and kuk = max0≤t≤1 |u(t)|, denote K = {u ∈ X : u(t) ≥
0, t ∈ [0, 1]} and K 0 = {u ∈ K : u(t) is a concave function on [0, 1]}. Obviously,
K, K 0 ⊂ X are two cones of X with K 0 ⊂ K. For u ∈ K, from Lemma 2.5 we know
mint∈[ k1 ,1− k1 ] u(t) ≥ k1 kuk with k > max{ ξ11 , 1−ξ2m−2 }, and we can show 1 − k1 > ξ1 .
Denote
α(u) =
min
1
1
t∈[ k
,1− k
]
u(t),
for u ∈ K 0 .
EJDE-2010/163
EXISTENCE OF POSITIVE SOLUTIONS
We have α(u) ≤ kuk ≤ kα(u). Define T : K → K by
R
R ξi
R σu
h
P
bϕq ( 0σu f (τ,u(τ ))dτ )+ m−2
f (τ,u(τ ))dτ )ds
i=1 ai 0 ϕq ( s

Pm−2


a− i=1 ai

i


+ R t ϕ (R σu f (τ, u(τ ))dτ )ds + ,

q
0
s
T u(t) = h R 1 g(s)(R 1 ϕ (R r f (τ,u(τ ))dτ )dr)ds
q

0
s

Rσu


1− 01 g(s)ds

i


+ R 1 ϕ (R s f (τ, u(τ ))dτ )ds + ,
q σu
t
where B + = max{B, 0} and σu is defined in Lemma 2.3.
Define A : K → X by

R
R ξi
R σu
P
bϕq ( 0σu f (τ,u(τ ))dτ )+ m−2
f (τ,u(τ ))dτ )ds

i=1 ai 0 ϕq ( s
P

m−2

a− i=1 ai


R
R

+ t ϕq ( σu f (τ, u(τ ))dτ )ds,
0
s
Au(t) = R 1
R1
Rr
g(s)(
ϕ

q ( σ f (τ,u(τ ))dτ )dr)ds
s

Ru
 0

1− 01 g(s)ds


R
R
1
s

+ t ϕq ( σu f (τ, u(τ ))dτ )ds,
7
0 ≤ t ≤ σu ,
σu ≤ t ≤ 1,
0 ≤ t ≤ σu ,
σu ≤ t ≤ 1.
For u ∈ X, denote θ : X → K while (θu)(t) = max{u(t), 0}, then T = θ ◦ A. For
u ∈ K 0 , define T ∗ : K 0 → K 0 by

R
R ξi
R σu +
P
f (τ,u(τ ))dτ )ds
bϕq ( 0σu f + (τ,u(τ ))dτ )+ m−2

i=1 ai 0 ϕq ( s
P


a− m−2
ai

i=1

R
R

+ t ϕq ( σu f + (τ, u(τ ))dτ )ds,
0 ≤ t ≤ σu ,
0
s
∗
T u(t) = R 1
R1
Rr +
g(s)(
ϕ
(
f
(τ,u(τ
))dτ
)dr)ds

q
0
s
σu

R


1− 01 g(s)ds


R
R
1
s

+ t ϕq ( σu f + (τ, u(τ ))dτ )ds,
σu ≤ t ≤ 1.
Lemma 2.6. T ∗ : K 0 → K 0 is completely continuous.
Proof. Obviously, we know T ∗ (u) ≥ 0. By [ϕp ((T ∗ u)0 (t))]0 = −f + (t, u(t)) ≤ 0, it
is easy to show ϕp ((T ∗ u)0 (t)) is non-increasing. Then (T ∗ u)0 (t) is also monotone
non-increasing, i.e., (T ∗ u)(t) is concave function. Thus T ∗ : K 0 → K 0 . We can have
T ∗ is completely continuous from Ascoli-Arzela theorem and concavity of f + . From Lemma 2.2, we have the following lemma.
Lemma 2.7. A function u(t) is a solution of boundary value problem (1.1) if and
only if u(t) is a fixed point of the operator A.
Lemma 2.8 ([8]). If A : K → X is completely continuous, then T = θ ◦A : K → K
is also completely continuous.
Lemma 2.9. If u is a fixed point of operator T , then u is also a fixed point of
operator A.
Proof. Let u be a fixed point of operator of T . Obviously, if Au(t) ≥ 0, t ∈ [0, 1],
then u(t) is a fixed point of operator A.
We prove that if T u(t) = u(t), then Au(t) ≥ 0 for t ∈ [0, 1].
Suppose this is not true, then there is a t0 ∈ (0, 1) such that Au(t0 ) < 0 = u(t0 ).
Let (t1 , t2 ) be the maximal interval which contain t0 and such that Au(t) < 0,
t ∈ (t1 , t2 ). It follows [t1 , t2 ] 6= [0, 1] from (H2).
8
S. LIU, M. JIA, Y. TIAN
EJDE-2010/163
If t2 < 1, we have u(t) = 0 for t ∈ [t1 , t2 ], Au(t) < 0 for t ∈ (t1 , t2 ) and Au(t2 ) =
0. Thus (Au)0 (t2 ) ≥ 0. From (H2), we know [ϕp ((Au)0 (t))]0 = −f (t, 0) ≤ 0, so
t1 = 0.
On the other hand, from the fact that Au(t) < 0 for t ∈ [0, t2 ) and Au(t2 ) =
0, (Au)0 (t2 ) ≥ 0 and (Au)00 (t) ≤ 0, we can obtain (Au)0 (0) > 0. Without loss of
generality, we suppose that there exists i = k such that u(ξi ) ≤ 0 for 1 ≤ i ≤
Pm−2
k and u(ξi ) > 0 for k < i ≤ m − 2. Since (Au)0 (0) > 0, a >
i=1 ai and
Pm−2
0
aAu(0) − bAu (0) = i=1 ai Au(ξi ) < 0, we have
0
|aAu(0) − bAu (0)| > a|Au(0)| >
k
X
ai |Au(0)| +
i=1
>
k
X
ai |Au(0)|
i=k+1
m−2
X
ai |Au(ξi )| −
i=1
m−2
X
ai Au(ξi )
i=k+1
m−2
X
= −(
ai Au(ξi ))
i=1
=|
m−2
X
ai Au(ξi )|,
i=1
a contradiction. So t2 = 1.
If t1 > 0, then we have Au(t) = 0 for t ∈ [t1 , t2 ], Au(t) < 0 for t ∈ (t1 , 1) and
Au(t1 ) = 0. Thus (Au)0 (t1 ) ≤ 0. We have [ϕp ((Au)0 (t))]0 = −f (t, 0) ≤ 0 by (H2).
This implies Au(t) < 0 for t ∈ (t1 , 1] and Au(1) = mint∈[t1 ,1] Au(t).
We can prove that Au(t) ≥ 0 for t ∈ [0, t1 ]. If there exists a t3 6∈ [t1 , 1] such
that Au(t3 ) < 0 and there is a maximal interval [t4 , t5 ] which contains t3 such that
Au(t) < 0 for t ∈ (t4 , t5 ). Obviously [t4 , t5 ) ∩ [t1 , 1] = ∅, so 1 6∈ (t4 , t5 ); i.e., t5 < 1,
this is a contradiction with the above discussion. Thus we can show Au(t) ≥ 0 for
t ∈ [0, t1 ].
For Au(1) < 0, we have
1
Z
Au(1) =
g(s)Au(s)ds.
0
Then
Z
|Au(1)| = −Au(1) = −
t1
Z
0
Z
≤
1
g(s)Au(s)ds −
t1
Z
1
g(s)|Au(1)|ds +
0
g(s)Au(s)ds
t1
g(s)|Au(1)|ds
t1
Z
= |Au(1)|
1
g(s)ds.
0
R1
So |Au(1)| ≤ |Au(1)| 0 g(s)ds which is a contradiction with (H3 ). Thus t1 = 0.
The above also contradicts [t1 , t2 ] 6= [0, 1]. Thus the proof is complete.
EJDE-2010/163
EXISTENCE OF POSITIVE SOLUTIONS
9
3. Main result
Denote
ξm−2 + 1
t∗ =
,
2
N0 = min
M=
1 1 − ξm−2
1
(
− )q ,
q
2
k
N=
1
max
a+b
P
, R 11
a− m−2
i=1 ai 1− 0 g(s)ds
,
Z
n b+a ξ
1
1
1 ξ1
1 1
(ξ1 − )q ,
g(s)((ξ1 − )q − (ξ1 − s)q )ds,
(b + aξ1 )q
k
q k1
k
Z 1− k1
o
1
1
b + a1 ξ1
q 1
(1 − − ξ1 ) ,
g(s)((1 − ξ1 − )q − (s − ξ1 )q )ds .
(b + aξ1 )q
k
q ξ1
k
It is easy to see that N0 > 0 and
1
k
< t∗ < 1 − k1 .
Theorem 3.1. Suppose (H1)–(H3) hold, that there exist nonnegative constants
c1 , c2 , c3 such that
N0
1
0 < c1 < max{ , 1}c2 < c3 < c3
N
k
and that f satisfies the following growth conditions:
(H4) f (t, u) ≥ 0 for (t, u) ∈ [0, 1] × [c1 , c3 ];
(H5) f (t, u) < ϕp ( cN2 ) for (t, u) ∈ [0, 1] × [0, c2 ];
c3
(H6) f (t, u) ≥ ϕp ( M
) for (t, u) ∈ [ k1 , 1 − k1 ] × [ k1 c3 , c3 ];
c1
(H7) f (t, u) ≥ ϕp ( N0 ) for (t, u) ∈ [ k1 , 1 − k1 ] × [ k1 c2 , c3 ].
Then (1.1) has at least two positive solutions u1 and u2 such that
0 < ku1 k < c2 ≤ ku2 k,
α(u2 ) <
1
c3 .
k
Proof. For any u ∈ ∂Kc2 , from (H5 ) we have
kT uk = max |T u(t)| = T u(t).
0≤t≤1
If t < σu , we have
h bϕ (R σu f (τ, u(τ ))dτ ) + Pm−2 a R ξi ϕ (R σu f (τ, u(τ ))dτ )ds
q 0
i
q s
i=1
T u(t) =
Pm−2 0
a − i=1 ai
Z t
Z σu
i+
f (τ, u(τ ))dτ )ds
+
ϕq (
0
s
R1
R1
Pm−2 R 1
bϕq ( 0 ϕp ( cN2 )dτ ) + i=1 ai 0 ϕq ( 0 ϕp ( cN2 )dτ )ds
<
Pm−2
a − i=1 ai
Z 1
Z 1
c2
ϕq (
ϕp ( )dτ )ds
+
N
0
0
c2
a+b
=
≤ c2 ,
P
N a − m−2
i=1 ai
and if t > σu , we have
R
R
R
Z s
h 1 g(s)( 1 ϕq ( r f (τ, u(τ ))dτ )dr)ds Z 1
i+
0
s
σu
T u(t) =
+
ϕq (
f (τ, u(τ ))dτ )ds
R1
1 − 0 g(s)ds
t
σu
10
S. LIU, M. JIA, Y. TIAN
EJDE-2010/163
R1
R1
R1
Z 1
Z 1
g(s)( 0 ϕq ( 0 ϕp ( cN2 )dτ )dr)ds
c2
<
+
ϕq (
ϕp ( )dτ )ds
R1
N
1 − 0 g(s)ds
0
0
1
c2
=
≤ c2 .
R
N 1 − 1 g(s)ds
0
0
From the above, we have kT uk < c2 , So (C1) of Lemma 2.1 is satisfied.
For u ∈ ∂Kc0 2 ; i.e., u ∈ K 0 and kuk = c2 . From Lemma 2.4 and (H5), we can
prove that kT ∗ uk < c2 using the same way as the T u above.
For u ∈ ∂K 0 ( k1 c3 ); i.e., u ∈ K 0 and α(u) = k1 c3 . So k1 c3 ≤ u(t) ≤ c3 for
1
1
k ≤ t ≤ 1 − k . From Lemmas 2.6, 2.7 and (H6), we have
1 ∗
1
kT uk = T ∗ u(σu )
k
k
R σu +
R σu
Pm−2 R ξi
1 bϕq ( 0 f (τ, u(τ ))dτ ) + i=1 ai 0 ϕq ( s f + (τ, u(τ ))dτ )ds
=
Pm−2
k
a − i=1 ai
Z σu
Z σu
+
ϕq (
f + (τ, u(τ ))dτ )ds
α(T ∗ u) =
min
1
1
]
t∈[ k ,1− k
0
T ∗ u(t) ≥
s
R1
Rr
g(s)( s ϕq ( σu f + (τ, u(τ ))dτ )dr)ds
0
=
R1
k
1 − 0 g(s)ds
Z 1
Z s
+
ϕq (
f + (τ, u(τ ))dτ )ds
1
R1
σu
σu
Z t∗
Z 1
Z s
n Z t∗
o
1
≥ min
ϕq (
f + (τ, u(τ ))dτ )ds,
ϕq (
f + (τ, u(τ ))dτ )ds
k
0
s
t∗
t∗
Z t∗
Z 1− k1
Z s
Z t∗
n
o
c3
c3
1
ϕq (
ϕp ( )dτ )ds
ϕp ( )dτ )ds,
≥ min
ϕq (
1
k
M
M
t∗
t∗
s
k
c3
1 1 + ξm−2
1
1 − ξm−2
1
min {(
− )q , (
− )q }
kM
q
2
k
2
k
1
= c3 .
k
≥
Finally, we show that (C3) of Lemma 2.1 is also satisfied.
Let u ∈ ∂Kc0 2 ( k1 c3 ) ∩ {u : u = T ∗ u}, then u = T ∗ u, u ∈ K 0 , kuk > c2 and
α(u) ≤ k1 c3 .
We know kuk ≤ kα(u) ≤ c3 and α(u) ≥ k1 kuk > k1 c2 by Lemma 2.5. By
∗
T : K 0 → K 0 and the definition of T ∗ , we can get
min u(t) = min{u(0), u(1)} ≥ 0.
0≤t≤1
From the boundary condition au(0) − bu0 (0) =
au(0) = bu0 (0) +
m−2
X
Pm−2
i=1
ai u(ξi )
i=1
≥ bu0 (0) + a1 u(ξ1 )
≥ bu0 (ξ) + a1 u(ξ1 )
ai u(ξi ), we have
EJDE-2010/163
EXISTENCE OF POSITIVE SOLUTIONS
11
b
(u(ξ1 ) − u(0)) + a1 u(ξ1 ),
ξ1
=
where ξ ∈ (0, ξ1 ). So we can obtain
u(0) ≥
b + a1 ξ1
u(ξ1 ).
b + aξ1
From the definition T ∗ and the use of condition (H7), we have
Z ξ1
Z ξ1
Z σu
Z ξ1
+
u(ξ1 ) >
ϕq (
ϕq (
f + (τ, u(τ ))dτ )ds
f (τ, u(τ ))dτ )ds >
0
1
k
s
c1
1
>
(ξ1 − )q ,
qN0
k
s
if σu > ξ1 ,
and
1
1− k
Z
u(ξ1 ) >
Z
ϕq (
ξ1
>
s
Z
+
1
1− k
f (τ, u(τ ))dτ )ds >
σu
Z
ϕq (
ξ1
1
c1
(1 − − ξ1 )q ,
qN0
k
s
f + (τ, u(τ ))dτ )ds
ξ1
if σu ≤ ξ1 .
Hence,
b + a1 ξ1 c1
1
1
min{(ξ1 − )q , (1 − − ξ1 )q } ≥ c1 .
·
b + a1 ξ qN0
k
k
R1
From the boundary condition u(1) = 0 g(s)u(s)ds, we have
u(0) ≥
Z
ξ1
u(1) >
ξ1
Z
g(s)u(s)ds >
1
k
0
ξ1
Z
g(s)(
Z
ϕq (
1
k
ξ1
f + (τ, u(τ ))dτ )dr)ds
r
Z ξ1
c1
1
g(s)((ξ1 − )q − (ξ1 − s)q )ds),
(
qN0 k1
k
>
if σu > ξ1 ,
and
Z
1
1− k
Z
1
1− k
g(s)u(s)ds >
u(1) >
ξ1
Z
g(s)(
ξ1
s
1
1− k
Z
ϕq (
r
f + (τ, u(τ ))dτ )dr)ds
ξ1
Z 1− k1
c1
1
>
(
g(s)((1 − ξ1 − )q − (s − ξ1 )q )ds),
qN0 ξ1
k
if σu ≤ ξ1 .
So, u(1) ≥ c1 . Therefore, for u ∈ ∂Kc0 2 ( k1 c3 ) ∩ {u : u = T ∗ u}, we have
c1 ≤ u(t) ≤ kuk ≤ kα(u) = c3 .
It follows f (t, u(t)) ≥ 0, t ∈ [0, 1] from (H4). Thus, T ∗ u = T u. So the condition
of Lemma 2.1 is satisfied. Then by Lemma 2.7, we know that operator T has two
fixed points u1 and u2 satisfying
0 < ku1 k < c2 ≤ ku2 k,
The proof is complete.
α(u2 ) <
1
c3 .
k
12
S. LIU, M. JIA, Y. TIAN
EJDE-2010/163
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College of Science, Shanghai University for Science and Technology, Shanghai
200093, China
E-mail address, Shu Liu: liu2008shu@126.com
E-mail address, Mei Jia: jiamei-usst@163.com
E-mail address, Yingqiang Tian: tian0013@163.com