Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 158, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS
WITH INTEGRAL BOUNDARY CONDITIONS ON INFINITE
INTERVALS
CHANGLONG YU, JUFANG WANG, YANPING GUO, HUIXIAN WEI
Abstract. In this article, we consider the existence of positive solutions for a
class of boundary value problems with integral boundary conditions on infinite
intervals
(ϕp (x0 (t)))0 + φ(t)f (t, x(t), x0 (t)) = 0, 0 < t < +∞,
Z +∞
x(0) =
g(s)x0 (s)ds,
lim x0 (t) = 0,
t→+∞
0
where ϕp (s) = |s|p−2 s, p > 1. By applying the Avery-Peterson fixed point
theorem in a cone, we obtain the existence of three positive solutions to the
above boundary value problem and give an example at last.
1. Introduction
Boundary value problems (BVPs) on infinite intervals often appear in applied
mathematics and physics and so on. The existence and multiplicity of positive
solutions for such problems have become an important area of investigation in
recent years. There are many papers concerning the existence of solutions on the
half-line for the boundary value problems, see [1, 2, 4, 5, 8, 9, 11, 12, 15, 16, 17]
and the references therein.
At the same time, we notice that a class of BVPs with integral boundary conditions appeared in heat conduction, chemical engineering, underground water flow,
thermo-elasticity, and plasma physics. Such problems include two-, three-, multipoint and nonlocal BVPs as special cases and attract more attention see [7, 10, 14]
and the references therein. For more information about the general theory of integral equations and their relation with BVPs, we refer to the book of Corduneanu
[6] and Agarwal and O’Regan [2].
Recently, Lian et al [12], studied the existence of positive solutions for the
boundary-value problem with a p-Laplacian operator
(ϕp (x0 (t)))0 + φ(t)f (t, x(t), x0 (t)) = 0,
0
0 < t < +∞,
0
αx(0) − βx (0) = 0, x (∞) = 0.
2000 Mathematics Subject Classification. 34B18, 34B15.
Key words and phrases. Cone; Avery-Peterson fixed point theorem; positive solution;
integral boundary conditions; infinite interval.
c
2012
Texas State University - San Marcos.
Submitted July 5, 2012. Published September 18, 2012.
1
2
C. YU, J. WANG, Y. GUO, H. WEI
EJDE-2012/158
Guo and Yu [8] establish the existence of three positive solutions for m-point
BVPs on infinite intervals
(ϕp (x0 (t)))0 + φ(t)f (t, x(t), x0 (t)) = 0,
x(0) =
m−2
X
i=1
αi x0 (ηi ),
0 < t < +∞,
lim x0 (t) = 0,
t→+∞
using the Avery-Peterson fixed point theorem in a cone.
Due to the fact that an infinite interval is noncompact, the discussion about
BVPs on the half-line is more complicated, in particular, for the BVPs with integral
boundary conditions on infinite intervals, few works were done.
Motivated by the results [8, 12], we will study the following BVPs with integral
conditions:
(ϕp (x0 (t)))0 + φ(t)f (t, x(t), x0 (t)) = 0, 0 < t < +∞,
Z +∞
(1.1)
x(0) =
g(s)x0 (s)ds,
lim x0 (t) = 0,
0
t→+∞
p−2
where ϕp (s) = |s| s, p > 1, φ : R+ → R+ , f (t, u, v) : R3+ → R+ is a continuous
function, R+ = [0, +∞), g ∈ L1 [0, +∞) is nonnegative.
In this article, we use the following conditions:
(H1) φ ∈ C(R+ , R+ ), φ 6≡ 0 on any interval of the form (t0 , +∞) and
Z +∞
Z +∞
Z +∞
φ(s)ds < +∞,
ϕ−1
φ(s)ds dτ < +∞;
0
0
τ
3
C(R+ , R+ ), f (t, 0, 0) 6=
(H2) f (t, (1 + t)u, v) ∈
0 on any subinterval of (0, +∞)
and when u, v are bounded f (t, (1 + t)u, v) is bounded on [0, +∞).
2. Preliminary results
In this section, we present some definitions, theorems and lemmas, which will
be needed in the proof of the main results. We first give the Avery-Peterson fixed
point theorem.
Definition 2.1. Let E be a real Banach space. A nonempty closed convex set
P ⊂ E is called a cone if it satisfies the following two conditions:
(i) x ∈ P and λ ≥ 0 imply λx ∈ P ;
(ii) x ∈ P and −x ∈ P imply x = 0.
Definition 2.2. Given a cone P in a real Banach space E. A continuous map ψ
is called a concave (resp. convex) functional on P if for all x, y ∈ P and 0 ≤ λ ≤ 1,
it holds ψ(λx + (1 − λ)y) ≥ λψ(x) + (1 − λ)ψ(y), (resp. ψ(λx + (1 − λ)y) ≤
λψ(x) + (1 − λ)ψ(y)).
Let α, γ, θ, ψ be nonnegative continuous maps on P with α concave, γ, θ convex.
Then for positive numbers a, b, c, d, we define the following subsets of P :
P (γ d ) = {x ∈ P : γ(x) < d};
P (αb , γ d ) = {x ∈ P (γ d ) : b ≤ α(x)};
P (αb , θc , γ d ) = {x ∈ P (γ d ) : b ≤ α(x), θ(x) ≤ c};
R(ψa , γ d ) = {x ∈ P (γ d ) : a ≤ ψ(x)}.
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POSITIVE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS
3
It is obvious that P (γ d ), P (αb , γ d ), P (αb , θc , γ d ) are convex and R(ψa , γ d ) is closed.
Theorem 2.3 ([3]). Let P be a cone of a real Banach space E. Let γ, θ be nonnegative convex functional on P satisfying
ψ(λx) ≤ λψ(x),
α(x) ≤ ψ(x),
∀0 ≤ λ ≤ 1,
kxk ≤ M γ(x)
∀x ∈ P (γ d )
with M, d some positive numbers. Suppose that T : P (γ d ) → P (γ d ) is completely
continuous and there exist positive numbers a, b, c with a < b such that
(1) {x ∈ P (αb , θc , γ d ) : α(x) > b} =
6 ∅ and α(T x) > b for x ∈ P (αb , θc , γ d );
d
(2) α(T x) > b for x ∈ P (αb , γ ) with θ(T x) > c;
(3) 0 6∈ R(ψa , γ d ) and ψ(T x) < a for x ∈ R(ψa , γ d ) with ψ(x) = a.
Then T has at least three fixed points x1 , x2 , x3 ∈ P (γ d ) such that γ(xi ) ≤ d,
i = 1, 2, 3; ψ(x1 ) < a; ψ(x2 ) > a with α(x2 ) < b; α(x3 ) > b.
Consider the space
X = x ∈ C 1 [0, +∞),
|x(t)|
< +∞, lim x0 (t) = 0
t→+∞
0≤t<+∞ 1 + t
sup
(2.1)
with the norm kxk = max{kxk1 , kx0 k∞ }, where kxk1 = sup0≤t<+∞ |x(t)|/(1 + t),
kx0 k∞ = sup0≤t<+∞ |x0 (t)|. By using the standard arguments, we can obtain that
(X, k · k) is a Banach space. Define the P ⊂ X by
n
P = x ∈ X : x(t) ≥ 0, t ∈ [0, +∞),
Z +∞
(2.2)
o
x(0) =
g(s)x0 (s)ds, x is concave on [0, +∞) .
0
Remark 2.4. If x satisfies (1.1), then (ϕp (x0 (t)))0 = −φ(t)f (t, x(t), x0 (t)) ≤ 0 on
[0, +∞), which implies that x is concave on [0, +∞). Moreover, if limt→+∞ x0 (t) =
0, then x0 (t) ≥ 0, t ∈ [0, +∞) and so x is monotone increasing on [0, +∞).
Let k > 1 be a constant. For x ∈ P , define the nonnegative continuous functionals:
k
α(x) =
min x(t), γ(x) = sup x0 (t),
k + 1 k1 ≤t<+∞
0≤t<+∞
Z +∞
x(t)
ψ(x) = θ(x) = sup
, A=
g(s)ds,
(2.3)
0≤t<+∞ 1 + t
0
Z t
Z +∞
Z +∞
−1
−1
C = ϕp
φ(s)ds , C1 (t) =
ϕp
φ(s)ds dτ.
0
0
τ
Since the Arzela-Ascoli theorem does not apply in the space X, we need a modified compactness criterion to prove T is compact. In the following, we present an
explicit one. For more general cases, we refer the readers to [2, 17] and the reference
therein.
Definition 2.5. For l > 0, let V = {x ∈ X : kxk < l}, and
V1 := {
x(t)
, x ∈ V } ∪ {x0 (t), x ∈ V }
1+t
4
C. YU, J. WANG, Y. GUO, H. WEI
EJDE-2012/158
which is called equiconvergent at infinity if for all ε > 0 there exists T = T (ε) > 0
such that for all x ∈ V1 ,
x(t1 )
x(t2 ) −
< ε, |x0 (t1 ) − x0 (t2 )| < ε, ∀t1 , t2 ≥ T.
1 + t1
1 + t2
0
Lemma 2.6 ([13]). If { x(t)
1+t , x ∈ V } and {x (t), x ∈ V } are both equicontinuous
on any compact interval of [0, +∞) and equiconvergent at infinity . Then V is
relatively compact on X.
Lemma 2.7. Let g ∈ L1 [0, +∞) and g is nonnegative, if v(t) is nonnegative and
continuous on [0, +∞) and limt→+∞ v(t) exists. Then there exists at least one η,
0 ≤ η ≤ +∞ such that
Z +∞
Z +∞
g(s)v(s)ds = v(η)
g(s)ds.
(2.4)
0
0
Proof. It is obvious that the function v(t) exists and has maxima and minima which
are nonnegative and noted by M ∗ , m∗ on [0, +∞), then for all t ∈ [0, +∞), we have
m∗ ≤ v(t) ≤ M ∗ , so
Z +∞
Z +∞
Z +∞
m∗
g(s)ds ≤
g(s)v(s)ds ≤ M ∗
g(s)ds.
0
If
R +∞
0
0
0
R +∞
g(s)ds = 0, the result is clear; If 0 g(s)ds > 0, there is
R +∞
g(s)v(s)ds
∗
m ≤ 0R +∞
≤ M ∗.
g(s)ds
0
Therefore, there exists at least one η, 0 ≤ η ≤ +∞ such that
Z +∞
Z +∞
g(s)v(s)ds = v(η)
g(s)ds.
0
(2.5)
0
R +∞
Lemma 2.8. Let y ∈ C[R+ , R+ ], and 0 y(t)dt < ∞, then the boundary-value
problem
(ϕp (x0 (t)))0 + y(t) = 0, 0 < t < +∞,
Z +∞
(2.6)
x(0) =
g(s)x0 (s)ds,
lim x0 (t) = 0,
t→+∞
0
has a unique solution
Z +∞
Z
−1
x(t) =
g(s)ϕp
0
+∞
Z
y(τ )dτ ds +
s
0
t
ϕ−1
p
Z
+∞
y(τ )dτ ds.
s
1
Define the operator T : P → C [0, +∞) by
Z +∞
Z +∞
(T x)(t) =
g(s)ϕ−1
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds
p
0
s
Z t
Z +∞
−1
+
ϕp
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds, t ∈ [0, +∞).
0
(2.7)
s
Lemma 2.9. For x ∈ P , kxk1 ≤ M kx0 k∞ , where M = max
R +∞
0
g(s)ds, 1 .
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POSITIVE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS
Proof. Since x ∈ P , by Lemma 2.7,
Z t
Z +∞
x(t)
1
=
(
x0 (s)ds +
g(s)x0 (s)ds)
1+t
1+t 0
0
R +∞
t + 0 g(s)ds 0
kx k∞ ≤ M kx0 k∞ .
≤
1+t
The result follows.
Lemma 2.10 ([8]). For x ∈ P , α(x) ≥
5
1
k+1 θ(x).
Lemma 2.11. Let (H1)–(H2) hold. Then T : P → P is completely continuous.
Proof. It is easy to see that T : P → P is well defined. Now we prove that T is
continuous and compact respectively. Let xn → x as n → +∞ in P , then there
exists r0 such that supn∈N \{0} kxn k < r0 . Set Br0 = sup{f (t, (1 + t)u, v), (t, u, v) ∈
[0, +∞) × [0, r0 ]2 }. Then we have
Z +∞
Z +∞
0
0
φ(s)|f (s, xn , xn ) − f (s, x, x )|ds ≤ 2Br0
φ(s)ds.
0
0
Therefore, by the Lebesgue dominated convergence theorem,
Z
+∞
0
|ϕp ((T xn )0 (t)) − ϕp ((T x)0 (t))| = φ(s)(f (s, xn , xn ) − f (s, x, x0 ))ds
t
Z +∞
0
≤
φ(s)|f (s, xn , xn ) − f (s, x, x0 )|ds
0
→ 0,
as n → +∞.
0
Furthermore, kT xn − T xk ≤ M k(T xn ) − (T x)0 k∞ → 0, as n → +∞. Hence, T is
continuous.
T is compact provided that it maps bounded sets into relatively compact sets.
Let Ω be any bounded subset of P . Then there exists r > 0 such that kxk ≤ r for
all x ∈ Ω. Obviously,
Z +∞
k(T x)0 k∞ = ϕ−1
(
φ(s)f (s, x(s), x0 (s))ds) ≤ Cϕ−1
p
p (Br )
0
for all x ∈ Ω. Hence, kT Ωk ≤ M Cϕp−1 (Br ). So T Ω is bounded.
Moreover, for any L ∈ (0, +∞) and t1 , t2 ∈ [0, L],
(T x)(t ) (T x)(t ) 1
2 −
1 + t1
1 + t2
Z +∞
Z +∞
1
1 0
g(s)ϕ−1
φ(τ
)f
(τ,
x(τ
),
x
(τ
))dτ
ds
−
≤
p
1 + t1
1 + t2
0
s
Z t2
Z +∞
1
1 ϕ−1
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds
+
−
p
1 + t1
1 + t2
0
s
Z t2
Z +∞
1
+
|
ϕ−1
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds|
1 + t1 t1 p
s
≤ ϕ−1
p (Br )(AC + C1 (L))|t1 − t2 | + |C1 (t1 ) − C1 (t2 )|)
→ 0,
uniformly as t1 → t2 ,
6
C. YU, J. WANG, Y. GUO, H. WEI
EJDE-2012/158
and
0
0
|ϕp ((T x) (t1 )) − ϕp ((T x) (t2 ))| = Z
t2
φ(s)(f (s, x, x0 )ds
t1
Z
t2
≤ Br |
φ(s)ds| → 0,
uniformly as t1 → t2 ,
t1
for all x ∈ Ω. So T Ω is equicontinuous on any compact interval of [0, +∞).
Finally, for any x ∈ Ω,
Z t
Z +∞
1
(T x)(t)
| = lim
ϕ−1
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds
lim |
p
t→+∞ 1 + t 0
t→+∞
1+t
s
Z +∞
−1
−1
≤ M ϕp (Br ) lim ϕp
φ(s)ds = 0,
t→+∞
t
Z +∞
0
−1
lim |(T x) (t)| = lim ϕp (
φ(s)f (s, x(s), x0 (s))ds)
t→+∞
t→+∞
t
Z +∞
−1
−1
≤ ϕp (Br ) lim ϕp
φ(s)ds = 0.
t→+∞
t
So T Ω is equiconvergent at infinity. By using Lemma 2.6, we obtain that T Ω
is relatively compact, that is, T is a compact operator. Hence, T : P → P is
completely continuous. The proof is complete.
3. Main results
For the main result of this article we sue the hypothesis
(H3) f (t, (1 + t)u, v) ≤ ϕp (d/C), for (t, u, v) ∈ [0, +∞) × [0, M d] × [0, d];
2
b
(H4) f (t, (1 + t)u, v) > ϕp (b/N ), for (t, u, v) ∈ [ k1 , k] × [ kb , (k+1)
km ] × [0, d];
(H5) f (t, (1 + t)u, v) < ϕp (a/M C), for (t, u, v) ∈ [0, +∞) × [0, a] × [0, d]; where
Z k
Z k
1
−1
m = min{A, 1}, N =
(g(s)
+
1)ϕ
φ(τ )dτ ds.
p
2
(k + 1) k1
s
Theorem 3.1. Let A > 0. Suppose (H1)–(H5) hold. Suppose further that there
exist numbers a, b, d such that 0 < ka < b ≤ M mkd/(k + 1)2 . Then (1.1) has at
least three positive solutions x1 , x2 , x3 such that
sup
0≤t<+∞
x1 (t)
< a,
0≤t<+∞ 1 + t
sup
a<
x0i (t) ≤ d,
x2 (t)
(k + 1)2 b
<
,
km
0≤t<+∞ 1 + t
sup
x3 (t)
< M d,
0≤t<+∞ 1 + t
sup
i = 1, 2, 3;
min x3 (t) >
1
k ≤t<k
min x2 (t) <
1
k <t<k
(k + 1)
b;
k
(k + 1)
b.
k
(3.1)
Proof. Let X, P, α, γ, θ, ψ and T be defined as (2.1)-(2.3) and (2.7) respectively. It
is easy to prove that the fixed points of T coincide with the solution of BVP (1.1).
So it is enough to show that T has three positive fixed points.
In fact, for any x ∈ P (γ d ), sup0≤t<+∞ x0 (t) ≤ d and so sup0≤t<+∞ x(t)
1+t ≤
M d. Condition (H3) implies that f (t, x(t), x0 (t)) ≤ ϕp (d/C) for all t ∈ [0, +∞).
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POSITIVE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS
7
Therefore,
γ(T x) =
sup (T x)0 (t) = (T x)0 (0)
0≤t<+∞
=
ϕ−1
p
Z
+∞
φ(s)f (s, x(s), x0 (s))ds
0
d
≤ ϕ−1
p
C
P (γ d )
Z
+∞
φ(s)ds = d.
0
P (γ d )
→
is completely continuous.
Hence T :
Obviously, α, γ, θ, ψ satisfy the assumptions in Theorem 2.3. Next we show that
conditions (1)-(3) in Theorem 2.3 hold.
2
k
1
Firstly, choose the function x(t) = (1 − k+1
e− A t ) (k+1)
b, 0 ≤ t < +∞. It
k
2
can be checked that x ∈ P (αb , θc , γ d ) with α(x) > b, where c = (k+1)
km b. So
c
d
c
d
{x ∈ P (αb , θ , γ )|α(x) > b} =
6 ∅. For any x ∈ P (αb , θ , γ ), we obtain
h1 i
x(t)
1
x(t)
(k + 1)2
b
≤
min x(t) ≤
≤
≤
b, t ∈
,k ,
1
k
1 + k k ≤t≤k
1+k
1+t
km
k
and 0 ≤ x0 (t) ≤ d, t ∈ [0, +∞). In view of assumption (H4) together with Lemma
2.10, we obtain
1
(T x)(t)
1
θ(T x) =
sup
α(T x) ≥
k+1
k + 1 0≤t<+∞ 1 + t
Z
Z +∞
1
1 h +∞
0
=
sup
g(s)ϕ−1
φ(τ
)f
(τ,
x(τ
),
x
(τ
))dτ
ds
p
(k + 1) 0≤t<+∞ 1 + t 0
s
Z t
Z +∞
i
−1
+
ϕp
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds
0
s
hZ k
Z k
1
−1
0
≥
g(s)ϕ
φ(τ
)f
(τ,
x(τ
),
x
(τ
))dτ
ds
p
(k + 1)2 k1
s
Z k
i
Z k
−1
ϕp
φ(τ )f (τ, x(τ ), x0 (τ ))dτ ds
+
1
k
s
b
1
>
N (k + 1)2
Z
k
(g(s) +
1
k
1)ϕ−1
p
Z
k
φ(τ )dτ ds = b.
s
c
d
Hence, α(T x) > b for x ∈ P (αb , θ , γ ).
Next we will verify that the condition (2) of Theorem 2.3 is satisfied. Let x ∈
P (αb , γ d ) with θ(T x) > c, it follows from Lemma 2.10 that
1
1
1 (k + 1)2
(k + 1)
θ(T x) >
c=
b=
b > b,
k+1
k+1
k + 1 km
km
Thus α(T x) > b for all x ∈ P (αb , γ d ) with θ(T x) > c.
Finally, we show that condition (3) of theorem 2.3 is satisfied. It is clear that
0 ∈ R(ψa , γ d ). Suppose that x ∈ R(ψa , γ d ) with ψ(x) = a, then by condition (H5)
and Lemma 2.9, we obtain
α(T x) ≥
ψ(T x) ≤ M γ(T x) = M (T x)0 (0)
Z +∞
−1
= M ϕp (
φ(s)f (s, x(s), x0 (s))ds)
0
8
C. YU, J. WANG, Y. GUO, H. WEI
≤M·
EJDE-2012/158
Z
a −1 +∞
ϕp (
φ(s)ds = a.
MC
0
Therefore, T has at least three fixed points x1 , x2 , x3 ∈ P (γ d ) such that
ψ(x1 ) < a,
ψ(x2 ) > a with α(x2 ) < b, α(x3 ) > b.
In addition, condition (H2) guarantees that those fixed points are positive. So
(1.1) has at least three positive solutions x1 , x2 , x3 satisfying (3.1) and the proof is
complete.
4. Example
Consider the boundary-value problem with integral boundary value conditions
(|x0 |x0 )0 + e−t f (t, x(t), x0 (t)) = 0,
Z +∞
x(0) =
e−2s x0 (s)ds,
lim x0 (t) = 0,
(4.1)
t→+∞
0
where
(
f (t, u, v) =
| sin t|
100
| sin t|
100
+ 104
+ 104
u 10
1+t 1 10
1+t
+
+
1
v
100 200 ,
1
v
100 200 ,
u ≤ 1,
u ≥ 1.
Set φ(t) = e−t and it is easy to verify that (H1) and (H2) hold. Choose k = 4,
a = 41 , b = 2, d = 200. Then by simple calculations, we can obtain M = 1, m = 12 ,
Z 4
Z 4p
p
1
1
1
−2s
−s
−4
e−s − e−4 ds >
C = 1, N =
(e
+ 1) e − e ds ≥
.
1
1
25 4
25 4
48
So the nonlinear term f satisfies
(1) f (t, (1 + t)u, v) ≤ 0.01 + 104 + 0.01 < 4 × 104 = ϕ3 (d/C), for (t, u, v) ∈
[0, +∞) × [0, 200]2 ;
(2) f (t, (1 + t)u, v) ≥ 104 > 962 = ϕ3 (b/N ), for (t, u, v) ∈ [ 41 , 4] × [ 12 , 25] ×
[0, 200];
10
1
(3) f (t, (1+t)u, v) ≤ 0.01+ 104 × 14
+0.01 < 16
= ϕ3 (a/M C), for (t, u, v) ∈
1
[0, +∞) × [0, 4 ] × [0, 200].
Therefore, the conditions in Theorem 3.1 are all satisfied. So (4.1) has at least
three positive solutions x1 , x2 , x3 such that
sup
0≤t<+∞
x1 (t)
1
≤ ,
4
0≤t<+∞ 1 + t
sup
sup
0≤t<+∞
x0i (t) ≤ 200,
i = 1, 2, 3;
1
x2 (t)
< sup
< 25,
2 0≤t<+∞ 1 + t
x3 (t)
≤ 200
1+t
min x3 (t) >
1
k ≤t≤k
min x2 (t) ≤
1
k ≤t≤k
5
;
2
5
.
2
Acknowledgements. This research was supported by grants: 10901045 from the
Natural Science Foundation of China, A2009000664 and A2011208012 from the
Natural Science Foundation of Hebei Province, and XL201047 from the Foundation
of Hebei University of Science and Technology.
EJDE-2012/158
POSITIVE SOLUTIONS FOR BOUNDARY-VALUE PROBLEMS
9
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Changlong Yu
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
China
E-mail address: changlongyu@126.com
Jufang Wang
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
China
E-mail address: wangjufang1981@126.com
Yanping Guo
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, 050018,
China
Huixian Wei
Department of Baise, Shijiazhuang Institute of Railway Technology, Shijiazhuang, 050041,
China