Electronic Journal of Differential Equations, Vol. 2009(2009), No. 66, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLE NONNEGATIVE SOLUTIONS FOR SECOND-ORDER
BOUNDARY-VALUE PROBLEMS WITH SIGN-CHANGING
NONLINEARITIES
SHOULIANG XI, MEI JIA, HUIPENG JI
Abstract. In this paper, we study the existence of multiple nonnegative solutions for second-order boundary-value problems of differential equations with
sign-changing nonlinearities. Our main tools are the fixed-point theorem in
double cones and the Leggett-Williams fixed point theorem. We present also
the integral kernel associated with the boundary-value problem.
1. Introduction
Boundary-value problems with nonnegative solutions describe many phenomena
in the applied science, and they are widely used in fields, such as chemistry, biological, etc.; see for example [2, 4, 5, 6, 7, 8]. Problems with integral boundary
conditions have been applied in heat conduction, chemical engineering, underground
water flow-elasticity, etc. The existence of nonnegative solutions to these problems
have received a lot of attention; see [3, 8, 9, 10, 11, 12] and reference therein.
Recently, by constructing a special cone and using the fixed point index theory,
Liu and Yan [9] proved the existence of multiple solutions to the singular boundaryvalue problem
(p(t)x0 (t))0 + λf (t, x(t), y(t)) = 0
(p(t)y 0 (t))0 + λg(t, x(t), y(t)) = 0
αx(0) − βx0 (0) = γx(1) + δx0 (1) = 0
αy(0) − βy 0 (0) = γy(1) + δy 0 (1) = 0,
where the parameter λ in R+ , p ∈ C([0, 1], R+ ), α, β, γ, δ ≥ 0, βγ + αδ + αγ > 0,
f ∈ C((0, 1) × R+ × R, R+ ), g ∈ C((0, 1) × R+ × R, R), but g must be controlled
by f .
2000 Mathematics Subject Classification. 34B10, 34B18.
Key words and phrases. Nonnegative solutions; fixed-point theorem in double cones;
integral kernel; integral boundary conditions.
c
2009
Texas State University - San Marcos.
Submitted December 8, 2008. Published May 14, 2009.
Supported by grant 05EZ53 from the Foundation of Educational Commission of Shanghai.
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S. XI, M. JIA, H. JI
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By using fixed point index theory in a cone, Yang [10] studied the existence of
positive solutions to a system of second-order nonlocal boundary value problems
−u00 = f (t, u, v)
−v 00 = g(t, u, v)
u(0) = v(0) = 0
Z 1
u(1) = H1
u(τ )dα(τ )
v(1) = H2
Z
0
1
v(τ )dβ(τ ) ,
0
where α and β are increasing nonconstant functions defined on [0, 1] with α(0) =
0 = β(0) and f, g ∈ C((0, 1) × R+ × R+ , R+ ), Hi ∈ C(R+ , R+ ).
By using fixed point theory in a cone, Feng [12] studied positive solutions for the
boundary-value problem, with integral boundary conditions in Banach spaces,
x00 + f (t, x) = 0
with
Z
x(0) =
1
g(t)x(t)dt,
x(1) = 0
0
or
Z
x(0) = 0, x(1) =
1
g(t)x(t)dt,
0
where f ∈ C([0, 1] × P, P ), g ∈ L1 [0, 1], and P is a cone of E. All of these, we can
find the nonlinear term f is nonnegative.
In this paper, by using the fixed point theorem in double cones and the LeggettWilliams fixed point theorem, we study the existence of multiple nonnegative solutions to the boundary value problem
u001 (t) + f1 (t, u1 (t), u2 (t)) = 0
u002 (t) + f2 (t, u1 (t), u2 (t)) = 0
Z
u1 (1) =
u1 (0) = u2 (0) = 0
Z
1
g1 (s)u1 (s)ds, u2 (1) =
0
(1.1)
1
g2 (s)u2 (s)ds,
0
where f1 , f2 ∈ C((0, 1) × R+ × R+ , R), and g1 , g2 are nonnegative functions in
L1 [0, 1].
In this paper we assume that the following conditions:
(H1) fi ∈ C((0, 1) × R+ × R+ , R), gi ∈ L1 [0, 1] is nonnegative, i = 1, 2;
R1
(H2) 1 − 0 sgi (s)ds > 0;
(H3) f1 (t, 0, u2 (t)) ≥ 0(6≡ 0), f2 (t, u1 (t), 0) ≥ 0(6≡ 0), t ∈ [0, 1].
2. Preliminaries
Let X be a Banach space with norm k · k and K ⊂ X be a cone. For a constant
r > 0, denote Kr = {x ∈ K : kxk < r}, ∂Kr = {x ∈ K : kxk = r}. Suppose
α : K → R+ is a continuously increasing functional; i.e. α is continuous and
α(λx) ≤ α(x) for λ ∈ (0, 1). Let
K(b) = {x ∈ K : α(x) < b}, ∂K(b) = {x ∈ K : α(x) = b}.
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MULTIPLE NONNEGATIVE SOLUTIONS
3
and Ka (b) = {x ∈ K : kxk > a, α(x) < b}. φ : K → R+ is a continuously concave
functional. Denote
K(φ, a, b) = {x ∈ K : φ(x) ≥ a, kxk ≤ b}.
We will use the following two theorem.
Theorem 2.1 ([1]). 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+ is a continuously increasing functional
satisfying α(x) ≤ kxk ≤ M α(x) for all x ∈ 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.
Theorem 2.2 (Leggett-Williams fixed point theorem [13]). Let A : Kc → Kc be
completely continuous and φ be a nonnegative continuous concave functional on K
such that φ(x) ≤ kxk for all x ∈ Kc . Suppose that there exist 0 < d < a < b ≤ c
such that
(C4) {x ∈ K(φ, a, b) : φ(x) > a} =
6 ∅ and φ(Ax) > a for x ∈ K(φ, a, b);
(C5) kAxk < d for kxk ≤ d;
(C6) φ(Ax) > a for x ∈ K(φ, a, c) with kAxk > b.
Then A has at least three fixed points x1 , x2 , x3 in Kc satisfying
kx1 k < d,
a < φ(x2 ),
kx3 k > d,
φ(x3 ) < a.
Lemma 2.3. Assume that (H2) holds. Then for any yi ∈ C[0, 1], the boundary
value problem
u00i (t) + yi (t) = 0
Z 1
ui (0) = 0, ui (1) =
gi (s)ui (s)ds,
(2.1)
(2.2)
0
has a unique solution
Z
ui (t) =
1
Hi (t, s)yi (s)ds,
i = 1, 2,
(2.3)
0
where
Hi (t, s) = G(t, s) +
t
R1
gi (r)G(r, s)dr
,
R1
1 − 0 sgi (s)ds
0
(
t(1 − s),
G(t, s) =
s(1 − t),
i = 1, 2,
if 0 ≤ t ≤ s ≤ 1,
if 0 ≤ s ≤ t ≤ 1,
The proof is similar to [12, Lemma 2.1], and is omitted.
Lemma 2.4. Assume that (H2) holds. Let δ ∈ (0, 21 ), then for all t ∈ [δ, 1−δ], σ, s ∈
[0, 1], we have
Hi (σ, s) ≥ 0, Hi (t, s) ≥ δHi (σ, s).
(2.4)
4
S. XI, M. JIA, H. JI
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Proof. It is clear that Hi (σ, s) ≥ 0, From the properties of G(t, s), we obtain
G(t, s) ≥ δG(σ, s),
t ∈ [δ, 1 − δ], σ, s ∈ [0, 1],
then
R1
gi (r)G(r, s)dr
R1
1 − 0 sgi (s)ds
R1
δ 0 gi (r)G(r, s)dr
≥ δG(σ, s) +
R1
1 − 0 sgi (s)ds
R1
δσ 0 gi (r)G(r, s)dr
≥ δG(σ, s) +
= δHi (σ, s)
R1
1 − 0 sgi (s)ds
Hi (t, s) = G(t, s) +
t
0
The proof is complete.
Lemma 2.5. Assume that (H2) holds. If yi ∈ C[0, 1], yi ≥ 0, then the unique
solution ui (t) of the boundary-value problem (2.1)-(2.2) satisfies ui (t) ≥ 0 and
mint∈[δ,1−δ] ui (t) ≥ δkui k, i = 1, 2.
Proof. It is clear that ui (t) ≥ 0, for all t ∈ [0, 1], i = 1, 2. In fact, from (2.3) and
(2.4), for any t ∈ [δ, 1 − δ], s, σ ∈ [0, 1], i = 1, 2, we have
Z 1
Z 1
ui (t) =
Hi (t, s)yi (s)ds ≥
δHi (σ, s)yi (s)ds = δui (σ).
0
0
Hence,
ui (t) ≥ δ max |ui (σ)| = δkui k,
0≤σ≤1
and minδ≤t≤1−δ ui (t) ≥ δkui k. The proof is complete.
Let X = C[0, 1] × C[0, 1] with the norm k(u1 , u2 )k := ku1 k + ku2 k, K =
{(u1 , u2 ) ∈ X : ui ≥ 0, i = 1, 2} and K 0 = {(u1 , u2 ) ∈ K: ui (t) is concave in
[0, 1], mint∈[δ,1−δ] ui (t) ≥ δkui k, i = 1, 2}.
Clearly, K, K 0 ⊂ X are cones with K 0 ⊂ K. Let Ti : K → C[0, 1], i = 1, 2 be
defined by
Z 1
+
T1 (u1 , u2 )(t) =
H1 (t, s)f1 (s, u1 (s), u2 (s))ds , t ∈ [0, 1],
0
T2 (u1 , u2 )(t) =
1
Z
H2 (t, s)f2 (s, u1 (s), u2 (s))ds
+
t ∈ [0, 1],
,
0
where (B)+ = max{B, 0}. Let
T (u1 , u2 )(t) = (T1 (u1 , u2 )(t), T2 (u1 , u2 )(t)),
Z 1
A1 (u1 , u2 )(t) =
H1 (t, s)f1 (s, u1 (s), u2 (s))ds, t ∈ [0, 1],
0
Z
A2 (u1 , u2 )(t) =
1
H2 (t, s)f2 (s, u1 (s), u2 (s))ds,
t ∈ [0, 1],
0
A(u1 , u2 )(t) = (A1 (u1 , u2 )(t), A2 (u1 , u2 )(t)).
For (u1 , u2 ) ∈ X, define θ : X → K by
(θ(u1 , u2 ))(t) = (max{u1 (t), 0}, max{u2 (t), 0}),
then T = θ ◦ A.
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MULTIPLE NONNEGATIVE SOLUTIONS
Let Ti∗ : K 0 → C[0, 1], i = 1, 2 be defined by
Z 1
H1 (t, s)f1+ (s, u1 (s), u2 (s))ds,
T1∗ (u1 , u2 )(t) =
t ∈ [0, 1],
0
T2∗ (u1 , u2 )(t)
Z
=
5
(2.5)
1
H2 (t, s)f2+ (s, u1 (s), u2 (s))ds,
t ∈ [0, 1],
0
and
T ∗ (u1 , u2 )(t) = (T1∗ (u1 , u2 )(t), T2∗ (u1 , u2 )(t)).
Define α : K 0 → R+ by
α(u1 , u2 ) =
min
δ≤t≤1−δ
u1 (t) +
min
δ≤t≤1−δ
u2 (t).
It is clear that α is a continuous increasing functional and α(u1 , u2 ) ≤ k(u1 , u2 )k.
For u ∈ K 0 , we have
α(u1 , u2 ) =
min
δ≤t≤1−δ
u1 (t) +
min
δ≤t≤1−δ
u2 (t) ≥ δku1 k + δku2 k = δk(u1 , u2 )k.
Therefore,
1
α(u1 , u2 ).
δ
Lemma 2.6. Suppose A : K → X is completely continuous. Then θ ◦ A : K → K
is also a completely continuous operator.
α(u1 , u2 ) ≤ k(u1 , u2 )k ≤
Proof. The complete continuity of A implies that A is continuous and maps each
bounded subset of K to a relatively compact set of X. Let D ⊂ K be a bounded
set, for any > 0, there exist Pi (xi , yi ) ∈ X, i = 1, 2, . . . , m, such that
AD ⊂ ∪m
i=1 B(Pi , ),
where B(Pi , ) := {(u1 , u2 ) ∈ K : ku1 − xi k + ku2 − yi k < }. Then for any
∗
) ∈ (θ ◦ A)(D), there exists Q(xQ , yQ ) ∈ AD, such that
Q∗ (x∗Q , yQ
∗
(x∗Q , yQ
) = (max{xQ , 0}, max{yQ , 0}).
We choose a Pi ∈ {P1 , P2 , . . . , Pm }, such that
kxQ − xi k + kyQ − yi k < .
Since
∗
kx∗Q − x∗i k + kyQ
− yi∗ k ≤ kxQ − xi k + kyQ − yi k < ,
∗
we have Q∗ (x∗Q , yQ
) ∈ B(Pi∗ , ), and so (θ ◦ A)(D) is relatively compact.
For each > 0, there exists η > 0, such that kA(x1 , y1 ) − A(x2 , y2 )k < , for
kx1 − x2 k + ky1 − y2 k < η. Since
k(θ ◦ A)(x1 , y1 ) − (θ ◦ A)(x2 , y2 )k
= k max{A1 (x1 , y1 ), 0} − max{A1 (x2 , y2 ), 0},
max{A2 (x1 , y1 ), 0} − max{A2 (x2 , y2 ), 0} k
≤ kA(x1 , y1 ) − A(x2 , y2 )k < .
We have k(θ ◦ A)(x1 , y1 ) − (θ ◦ A)(x2 , y2 )k < , for kx1 − x2 k + ky1 − y2 k < η.
Hence, θ ◦ A is continuous in K and θ ◦ A is completely continuous. The proof
is complete.
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S. XI, M. JIA, H. JI
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Since fi is continuous, it is clear that A : K → X and T ∗ : K 0 → X are
completely continuous. From Lemmas 2.6 and 2.5, we have T : K → K and
T ∗ : K 0 → K 0 are completely continuous.
Lemma 2.7. If (u1 , u2 ) is a fixed point of T , then (u1 , u2 ) is a fixed point of A.
Proof. Suppose (u1 , u2 ) is a fixed point of T , obviously, we just need to prove that
Ai (u1 , u2 )(t) ≥ 0, i = 1, 2, for t ∈ [0, 1].
If there exist t0 ∈ (0, 1) and an i (i = 1, 2) such that ui (t0 ) = Ti (u1 , u2 )(t0 ) = 0
but Ai (u1 , u2 )(t0 ) < 0. Without loss of generalization, let i = 1 and (t1 , t2 ) be the
maximal interval and contains t0 such that A1 (u1 , u2 )(t) < 0 for all t ∈ (t1 , t2 ).
Obviously, (t1 , t2 ) 6= (0, 1). Or else, T1 (u1 , u2 )(t) = u1 (t) = 0, for all t ∈ [0, 1]. This
is in contradiction with (H3).
Case i: If t2 < 1, then A1 (u1 , u2 )(t2 ) = 0. Thus, A01 (u1 , u2 )(t2 ) ≥ 0, We obtain
A001 (u1 , u2 )(t) = −f1 (t, 0, u2 ) ≤ 0,
for t ∈ (t1 , t2 ).
So
A01 (u1 , u2 )(t) ≥ 0,
for t ∈ [t1 , t2 ]
A01 (u1 , u2 )(0)
We obtain t1 = 0, and
≥ 0, A1 (u1 , u2 )(0) < 0. This is in contradiction
with A1 (u1 , u2 )(0) = 0.
Case ii: If t1 > 0, we have A1 (u1 , u2 )(t1 ) = 0. Thus A01 (u1 , u2 )(t1 ) ≤ 0. We
obtain
A001 (u1 , u2 )(t) = −f1 (t, 0, u2 ) ≤ 0,
for t ∈ (t1 , t2 ).
So
A01 (u1 , u2 )(t) < 0,
for t ∈ [t1 , t2 ].
We obtain t2 = 1, A01 (u1 , u2 )(1) ≤ 0.
On the other hand, A1 (u1 , u2 )(t) < 0, for t ∈ (t1 , t2 ), A01 (u1 , u2 )(1) ≤ 0 imply
R1
A1 (u1 , u2 )(1) < 0. By (H1), A1 (u1 , u2 )(1) = 0 g1 (s)u1 (s)ds ≥ 0. This is a
contradiction. The proof is complete.
3. Main result
Denote
Z
Mi = max
t∈[0,1]
1
Z
Hi (t, s)ds,
0
mi =
min
t∈[δ,1−δ]
1−δ
Hi (t, s)ds, i = 1, 2
δ
Theorem 3.1. Suppose that condition (H1)–(H3) hold. Assume that there exist
positive numbers δ, a, b, λi , µi , i = 1, 2, such that δ ∈ (0, 21 ), 0 < a < δb < b,
λ1 + λ2 ≤ 1, µ1 + µ2 > 1, and satisfy
(H4) fi (t, u1 , u2 ) ≥ 0, for t ∈ [0, 1], u1 + u2 ∈ [0, b];
(H5) fi (t, u1 , u2 ) < λMi ai , for t ∈ [0, 1], u1 + u2 ∈ [0, a];
(H6) fi (t, u1 , u2 ) ≥ µmi δb
, for t ∈ [δ, 1 − δ], u1 + u2 ∈ [δb, b].
i
Then, (1.1) has at least two nonnegative solutions (u1 , u2 ) and (u∗1 , u∗2 ) such that
0 ≤ k(u1 , u2 )k < a < k(u∗1 , u∗2 )k, α(u∗1 , u∗2 ) < δb.
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MULTIPLE NONNEGATIVE SOLUTIONS
7
Proof. For all (u1 , u2 ) ∈ ∂Ka , from (H5) we have
Z 1
Hi (t, s)fi (s, u1 (s), u2 (s))ds)+
kTi (u1 , u2 )k = max (
t∈[0,1]
0
Z
1
= max max{
Hi (t, s)fi (s, u1 (s), u2 (s))ds, 0}
t∈[0,1]
λi a
max
Mi t∈[0,1]
<
Z
0
1
Hi (t, s)ds = λi a.
0
Therefore,
kT (u1 , u2 )k = kT1 (u1 , u2 )k + kT2 (u1 , u2 )k < λ1 a + λ2 a ≤ a.
So (C1) of Theorem 2.1 is satisfied.
For (u1 , u2 ) ∈ ∂Ka0 , from (H5), we have
Z 1
kTi∗ (u1 , u2 )k = max
Hi (t, s)fi+ (s, u1 (s), u2 (s))ds
t∈[0,1]
0
λi a
<
max
Mi t∈[0,1]
Z
1
Hi (t, s)ds = λi a.
0
We also obtain
kTi∗ (u1 , u2 )k = kT1∗ (u1 , u2 )k + kT2∗ (u1 , u2 )k < λ1 a + λ2 a ≤ a.
For (u1 , u2 ) ∈ ∂K 0 (δb), i.e., α(u1 , u2 ) = δb, For t ∈ [δ, 1 − δ], by Lemma 2.5, we
have δb ≤ u1 (t) + u2 (t) ≤ b. From (H6), we obtain
Z 1
α(T ∗ (u1 , u2 )) = min
H1 (t, s)f1+ (s, u1 (s), u2 (s))ds
t∈[δ,1−δ]
0
1
Z
+
t∈[δ,1−δ]
0
1−δ
Z
≥
H2 (t, s)f2+ (s, u1 (s), u2 (s))ds
min
H1 (t, s)f1+ (s, u1 (s), u2 (s))ds
min
t∈[δ,1−δ]
δ
1−δ
Z
+
H2 (t, s)f2+ (s, u1 (s), u2 (s))ds
min
t∈[δ,1−δ]
δ
Z 1−δ
Z 1−δ
µ1 δb
µ2 δb
≥
min
H1 (t, s)ds +
min
H2 (t, s)ds
m1 t∈[δ,1−δ] δ
m2 t∈[δ,1−δ] δ
= µ1 δb + µ2 δb > δb.
Therefore (C2) of Theorem 2.1 is satisfied.
Finally, we show that (C3) of Theorem 2.1 is satisfied. Let (u1 , u2 ) ∈ Ka0 (δb) ∩
{(u1 , u2 ) : T ∗ (u1 , u2 ) = (u1 , u2 )}, we have
α(u1 , u2 ) < δb, k(u1 , u2 )k > a.
From Lemma 2.5, we know that
1
α(u1 , u2 ) < b,
δ
0 ≤ u1 (t) + u2 (t) < b.
k(u1 , u2 )k ≤
8
S. XI, M. JIA, H. JI
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From (H4), we obtain
fi+ (s, u1 (s), u2 (s)) = fi (s, u1 (s), u2 (s)).
This implies that T (u1 , u2 ) = T ∗ (u1 , u2 ) for
(u1 , u2 ) ∈ Ka0 (δb) ∩ {(u1 , u2 ) : T ∗ (u1 , u2 ) = (u1 , u2 )}.
By Theorem 2.1 and Lemma 2.7, we know that (1.1) has at least two nonnegative
solutions (u1 , u2 ) and (u∗1 , u∗2 ) such that
0 ≤ k(u1 , u2 )k < a < k(u∗1 , u∗2 )k, α(u∗1 , u∗2 ) < b.
The proof is complete.
Define φ : K → R+ by
φ(u1 , u2 ) =
min
δ≤t≤1−δ
u1 (t) +
min
δ≤t≤1−δ
u2 (t)
Theorem 3.2. Suppose that condition (H1)–(H3) hold. There exist δ ∈ (0, 21 ),
a, b, λi , µi > 0, i = 1, 2, such that 0 < a < δb < b, λ1 + λ2 ≤ 1, µ1 + µ2 > 1, and
(H5), (H6) hold, and satisfy
(H7) fi (t, u1 , u2 ) ≥ 0, for t ∈ [0, 1], u1 + u2 ∈ [δb, b].
(H8) fi (t, u1 , u2 ) ≤ λMi bi , for t ∈ [0, 1], u1 + u2 ∈ [0, b].
∗∗
Then, (1.1) has at least three nonnegative solutions (u1 , u2 ), (u∗1 , u∗2 ), (u∗∗
1 , u2 ),
∗
∗
∗
∗
∗∗
∗∗
such that 0 ≤ k(u1 , u2 )k < a < k(u1 , u2 )k, φ(u1 , u2 ) < b, φ(u1 , u2 ) ≥ b.
Proof. Firstly, we prove T : Kb → Kb is a completely continuous operator. From
(H8), for i = 1, 2, we obtain
kTi (u1 , u2 )k = max
t∈[0,1]
Z
1
Hi (t, s)fi (s, u1 (s), u2 (s))ds
0
= max max
λi b
max
Mi t∈[0,1]
1
Z
t∈[0,1]
<
+
Hi (t, s)fi (s, u1 (s), u2 (s))ds, 0
0
Z
1
Hi (t, s)ds = λi b.
0
Therefore,
kT (u1 , u2 )k = kT1 (u1 , u2 )k + kT2 (u1 , u2 )k < λ1 b + λ2 b ≤ b.
From Lemma 2.6, we know T : Kb → Kb is a completely continuous operator. For
the operator T and any u1 + u2 ∈ [0, a], from (H5) and Theorem 3.1, we know (C5)
of Theorem 2.2 is satisfied.
Next, we show that (C4) of Theorem 2.2 is satisfied. Clearly,
{(u1 , u2 ) ∈ K(φ, δb, b) : φ(u1 , u2 ) > δb} =
6 ∅.
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MULTIPLE NONNEGATIVE SOLUTIONS
9
Assume (u1 , u2 ) ∈ K(φ, δb, b), for any t ∈ [δ, 1 − δ], we have δb ≤ u1 + u2 ≤ b. From
(H6) and (H7) we obtain
Z 1
+
H1 (t, s)f1 (s, u1 (s), u2 (s))ds
φ(T (u1 , u2 )) = min
t∈[δ,1−δ]
+
0
min
t∈[δ,1−δ]
H1 (t, s)f1 (s, u1 (s), u2 (s))ds
0
Z
+
H2 (t, s)f2 (s, u1 (s), u2 (s))ds
0
1−δ
Z
min
t∈[δ,1−δ]
H1 (t, s)f1 (s, u1 (s), u2 (s))ds
δ
Z
+
1
min
t∈[δ,1−δ]
≥
+
0
min
t∈[δ,1−δ]
H2 (t, s)f2 (s, u1 (s), u2 (s))ds
1
Z
≥
1
Z
min
t∈[δ,1−δ]
1−δ
H2 (t, s)f2 (s, u1 (s), u2 (s))ds
δ
Z 1−δ
Z 1−δ
µ1 δb
µ2 δb
min
min
H1 (t, s)ds +
H2 (t, s)ds
m1 t∈[δ,1−δ] δ
m2 t∈[δ,1−δ] δ
= µ1 δb + µ2 δb > δb.
≥
Finally, for (u1 , u2 ) ∈ K(φ, δb, b) and kT (u1 , u2 )k > b, it is easy to prove that
φ(T (u1 , u2 )) ≥ δkT (u1 , u2 )k > δb.
Then (C6) of Theorem 2.2 is satisfied. Therefore from Theorem 2.2 and Lemma
2.7 we know that (1.1) has at least three nonnegative solutions (u1 , u2 ), (u∗1 , u∗2 ),
∗∗
(u∗∗
1 , u2 ), such that
0 ≤ k(u1 , u2 )k < a < k(u∗1 , u∗2 )k,
α(u∗1 , u∗2 ) < b,
The proof is complete.
∗∗
α(u∗∗
1 , u2 ) ≥ b.
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College of Science, University of Shanghai for Science and Technology, Shanghai,
200093, China
E-mail address, S.Xi: xishouliang@163.com
E-mail address, M.Jia: jiamei-usst@163.com
E-mail address, H.Ji: jihuipeng1983@163.com