Electronic Journal of Differential Equations, Vol. 2006(2006), No. 74, pp. 1–25.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
MULTIPLE POSITIVE SOLUTIONS FOR SINGULAR
BOUNDARY-VALUE PROBLEMS WITH DERIVATIVE
DEPENDENCE ON FINITE AND INFINITE INTERVALS
BAOQIANG YAN
Abstract. In this paper, Krasnoselskii’s theorem and the fixed point theorem of cone expansion and compression are improved. Using the results
obtained, we establish the existence of multiple positive solutions for the singular second-order boundary-value problems with derivative dependance on
finite and infinite intervals.
1. Introduction
In [1], by an alternative method to Leray-Schauder and sequential technique,
Agarwal and O’Regan considered the singular boundary-value problem
1
(py 0 )0 + Φ(t)f (t, y, py 0 ) = 0,
p
αy(0) − β lim+ p(t)y 0 (t) = 0,
t ∈ (0, 1)
(1.1)
y(1) = 0
t→0
and obtained the existence of one solution to equation (1.1) when α = 0 or β = 0.
In [23], by a generalization of the Kneser’s property (continuum) of the crosssections of the solutions funnel, Palamides and Galanis considered the following
problems
1
(py 0 )0 + Φ(t)f (t, y, py 0 ) = 0, t ∈ (0, +∞)
p
(1.2)
y(0) = 0,
lim p(t)y 0 (t) = 0
t→+∞
and also obtained the existence of one positive and monotone unbounded solution.
There are some other results on the existence of at least one solution for equation
(1.1), (1.2), and we refer the reader also to [2, 3, 4, 5, 6, 7, 8, 11, 19, 20, 21, 22].
Moreover, under the condition that p ≡ 1, β = 0 and f has no singularity at
x = 0 and px0 = 0, in [15], using pairs of lower and upper solutions, Henderson
and Thompson considered the existence of three solutions for equation (1.1) and
2000 Mathematics Subject Classification. 34B16, 34B18, 34B40.
Key words and phrases. Krasnoselskii’s theorem; singular boundary value problems;
fixed point theorem of cone expansion and compression.
c
2006
Texas State University - San Marcos.
Submitted March 16, 2006. Published July 10, 2006.
Supported by grants 10571111 from the National Natural Science and Y2005A07 from the
natural science of Shandong Province.
1
2
B. YAN
EJDE-2006/74
in [24], by reducing the equation (1.1) to a quasi-linear one, I.Yermachenko and
F.Sadyrbaev obtained the existence of multiple solutions for equation (1.1) also.
Up to now, there are fewer results on the existence of multiple positive solutions
to equation (1.1), (1.2) if f (t, x, px0 ) is singular at x and is related to px0 . Motivated
by this, in this paper, we discuss the existence of multiple positive solutions to
equation (1.1), (1.2) when f (t, x, px0 ) is singular at x = 0.
There are three sections in our paper. In section 2, in order to overcome the
difficulty from px0 , we improve the Krasnoselskii’s theorem and fixed point theorem
of cone expansion and compression on unbounded set in a Banach space with a
special norm. In section 3, we establish special cones, and using obtained theorems,
present the existence of multiple positive solutions to equation (1.1). In section 4,
we consider the existence of multiple positive solutions to equation (1.2).
2. The improvement of the Krasnoselskii’s theorem and fixed point
theorem of cone expansion and compression
In this section, we improve the Krasnoselskii’s theorem and fixed point theorem
of cone expansion and compression in a Banach space with a special norm.
In [12], Granas and Dugudji presented the theory of fixed point index on unbounded open sets which has same basic properties as those in the theory of fixed
point index on bounded open sets [14]. The degree theory on bounded open sets
and unbounded open sets can be found in [9, 12, 13, 16, 17, 25].
According to the the theory of fixed point index on unbounded open sets in
Chapter 4 of [12], it is easy to obtain following result. Let E be a real Banach
space containing a cone P .
Lemma 2.1. Assume Ω ⊆ E, θ ∈ Ω, Ω ∩ P is a relatively open set in P . Let
A : P ∩ Ω → P be continuous with relatively compact A(P ∩ Ω). Suppose that
Ax 6= µx,
∀x ∈ P ∩ ∂Ω, µ ≥ 1.
(2.1)
Then i(A, P ∩ Ω, P ) = 1.
Proof. Let H(t, x) = tAx, t ∈ [0, 1] and x ∈ P ∩ ∂Ω. Then H : [0, 1] × (P ∩ Ω) → P
is continuous, and the continuity of H(t, x) in t is uniform with respect to x ∈
P ∩ Ω. Moreover, H(t, P ∩ Ω) is relatively compact for every t ∈ [0, 1]. Evidently,
H(t, x) 6= x for x ∈ P ∩ ∂Ω and 0 ≤ t ≤ 1. Hence, by the homotopy invariance and
normality of fixed point index, we have
i(A, P ∩ Ω, P ) = i(θ, P ∩ Ω, P ) = 1.
The proof is complete.
Now we consider a real Banach space in a special case. Assume that E is a linear
space and it satisfies three conditions:
(1) There is a norm x → kxk1 on x ∈ E and under k · k1 , E is a normed linear
space (not complete)
(2) There is another semi-norm k · k2
(3) Under kxk = max{kxk1 , kxk2 }, E is a Banach space.
For example, for x ∈ C 1 ([0, 1], R), under kxk1 = maxt∈[0,1] |x(t)|, C 1 ([0, 1], R) is an
incomplete normed linear space. Let kxk2 = maxt∈[0,1] |x0 (t)|. Obviously, k · k2 is
a semi-norm of C 1 ([0, 1], R). If we define kxk = max{kxk1 , kxk2 }, C 1 ([0, 1], R) is a
Banach space.
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MULTIPLE POSITIVE SOLUTIONS
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Assume that P is a cone of E and Ω ⊂ E is a open set with supx∈Ω kxk1 < +∞.
Since supx∈Ω kxk1 ≤ supx∈Ω kxk, it is possible that Ω is unbounded in E. We have
the following lemma(the ideas coming from [14]).
Lemma 2.2. With E and P as above, assume that Ω ⊆ E is an open set with
supx∈Ω kxk1 < +∞. Let A : P ∩ Ω → P be continuous with relatively compact
A(P ∩ Ω) and B : P ∩ ∂Ω → P be continuous with relatively compact B(P ∩ ∂Ω).
Suppose that
(a) inf x∈P ∩∂Ω kBxk1 > 0;
(b) x − Ax 6= tBx, for all x ∈ P ∩ ∂Ω, t ≥ 0.
Then, we have
i(A, P ∩ Ω, P ) = 0.
(2.2)
Proof. Suppose that the E1 is a Banach space completion of E under norm kxk1 . By
the extension theorem of Dugundji [10], we can extend B to a continuous operator
from P ∩ Ω into P such that
B(P ∩ Ω) ⊆ coB(P ∩ ∂(Ω)) ⊆ (coB(P ∩ ∂Ω))1 ,
(2.3)
where (coB(P ∩ ∂Ω))1 is the closure of B(P ∩ ∂Ω) under the norm k · k1 and the
followings are similar. Let F = B(P ∩ ∂Ω), then (coB(P ∩ ∂Ω))1 = (coF )1 = (M )1 ,
where
n
n
X
X
M = {y =
λi yi : yi ∈ F, λi ≥ 0,
λi = 1; n = 1, 2, . . . }.
i=1
i=1
We first prove
kyk1 > 0.
inf
(2.4)
y∈(M )1
Denote by E0 the subspace of E spanned by F under norm k · k1 . Since B(P ∩ ∂Ω)
is relatively compact in E under norm k · k, we know that B(P ∩ ∂Ω) is relatively
compact in E1 under norm k·k1 . Therefore, E0 is separable. Evidently, P0 = P ∩E0
is a cone of E0 and F ⊆ P0 . By property of the cone [14, Theorem 1.4.1], there
exists f0 ∈ E0∗ such that f0 (y) > 0 for any y ∈ P0 with y 6= θ. We claim that
inf f0 (y) = σ > 0.
y∈F
(2.5)
In fact, if σ = 0, then there exists {yk } ⊆ F such that f0 (yk ) → 0. By the
relative compactness of F in E, there is a subsequence {yki } of {yk } such that
yki → y0 ∈ P and y0 ∈ E0 . Then y0 ∈ P0 , and so f0 (yki ) → f0 (y0 ) = 0. Hence,
y0 = θ and kykiP
k1 → 0, which contradicts hypothesis (a). P
Thus, (2.5) holds.
n
n
For any y = i=1 λi yi ∈ M , where yi ∈ F , λi ≥ 0 and i=1 λi = 1, we have
f0 (y) =
n
X
λi f0 (yi ) ≥
i=1
n
X
λi σ = σ,
i=1
and therefore
f0 (y) ≥ σ, ∀y ∈ (M )1 .
(2.6)
Since (M )1 = (coF )1 is compact, there exists a z0 ∈ (M )1 such that
inf
y∈(M )1
kyk1 = kz0 k1 .
(2.7)
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B. YAN
EJDE-2006/74
By (2.6), f0 (z0 ) ≥ σ, and this implies that z0 6= θ. It follows therefore from (2.7)
that (2.4) holds. By (2.3) and (2.4), we get
inf kBxk1 = σ > 0.
(2.8)
x∈P ∩Ω
Now, it is easy to show that (2.2) holds. In fact, if i(A, P ∩ Ω, P ) 6= 0, then by the
hypothesis (b) and the homotopy invariance property of fixed point index, we have
i(A + tB, P ∩ Ω, P ) = i(A, P ∩ Ω, P ) 6= 0, ∀t > 0.
In particular, choosing t0 >
we have
a+c
σ ,
where a = supx∈Ω kxk1 and c = supx∈P ∩Ω kAxk1 ,
i(A + t0 B, P ∩ Ω, P ) 6= 0,
and so, by the solution property of fixed point index, there exists an x0 ∈ P ∩ Ω
such that Ax0 + t0 Bx0 = x0 . Hence
kx0 − Ax0 k1
a+c
t0 =
,
≤
kBx0 k1
σ
which is a contradiction. The proof is complete.
Corollary 2.3. Assume that Ω is an open set with supx∈Ω kxk1 < +∞. Let A :
P ∩ Ω → P be continuous with relatively compact A(P ∩ Ω). If there exists u0 > θ
such that
x − Ax 6= tu0 , ∀x ∈ P ∩ ∂Ω, t ≥ 0,
(2.9)
then (2.2) holds.
Proof. Since k · k1 is a norm, u0 > θ implies that ku0 k1 > 0. Hence, the corollary
follows directly from Lemma 2.2 by putting Bx = u0 for any x ∈ P ∩ ∂Ω.
Corollary 2.4. Assume that Ω is an open set with supx∈Ω kxk1 < +∞. Let A :
P ∩ Ω → P be continuous with relatively compact A(P ∩ Ω). If
Ax 6≤ x,
∀x ∈ P ∩ ∂Ω,
(2.10)
then (2.2) holds.
Proof. Choose an u0 > θ. Then,
x − Ax 6= tu0 , ∀x ∈ P ∩ ∂Ω, t ≥ 0.
By Corollary 2.3, (2.2) holds.
Lemma 2.5. Assume that Ω is an open set with supx∈Ω kxk1 < +∞. Let A :
P ∩ Ω → P be continuous with relatively compact A(P ∩ Ω). Suppose that
(i) inf x∈P ∩∂Ω kAxk1 > 0; and
(ii) Ax 6= µx, ∀x ∈ P ∩ ∂Ω, 0 < µ < 1.
Then (2.2) holds.
Proof. Taking B = A in Lemma 2.2, we see that condition (a) of Lemma 2.2 is
the same as condition (i) of Lemma 2.5. Also, condition (b) of Lemma 2.2 is true.
In fact, if there exist x0 ∈ P ∩ ∂Ω and t0 ≥ 0 such that x0 − Ax0 = t0 Ax0 , then
Ax0 = µx0 , where µ0 = (1 + t0 )−1 . Evidently 0 < µ0 ≤ 1, which contradicts the
condition (ii). Thus, (2.2) follows from Lemma 2.2.
Lemma 2.6. Assume that Ω is an open set with supx∈Ω kxk1 < +∞. Let A :
P ∩ Ω → P be continuous with relatively compact A(P ∩ Ω). Suppose that
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MULTIPLE POSITIVE SOLUTIONS
5
(i’) Ax 6= µx, ∀x ∈ P ∩ ∂Ω, 0 ≤ µ ≤ 1, and
(ii’) the set {kAxk−1
1 Ax|x ∈ P ∩ ∂Ω} is relatively compact.
Then (2.2) holds.
Proof. Let A1 x = α(kAxk1 )−1 Ax for x ∈ P ∩∂Ω, where α = supx∈P ∩∂Ω kAxk1 > 0.
Then, by hypotheses, A1 : P ∩ ∂Ω → P is continuous with relatively compact
A1 (P ∩∂Ω). By the extension theorem, A1 can be extended to a continuous operator
from P ∩Ω into P with relatively compact A1 (P ∩Ω). We now prove that A1 satisfies
the condition (i) and (ii) of Lemma 2.5. In fact, first we have
inf
x∈P ∩∂Ω
kA1 xk1 = σ > 0.
Secondly, if there exists x0 ∈ P ∩ ∂Ω and 0 < µ0 ≤ 1 such that A1 x0 = µ0 x0 ,
then Ax0 = λ0 x0 , where λ0 = µ0 α−1 kAx0 k1 . Evidently, 0 < λ0 ≤ µ0 ≤ 1, which
contradicts hypothesis (i’). Hence, by Lemma 2.5, we have
i(A1 , P ∩ Ω, P ) = 0.
(2.11)
(1 − t)Ax + tA1 x 6= x, ∀x ∈ P ∩ ∂Ω, 0 ≤ t ≤ 1.
(2.12)
Now, we prove
If there is an x1 ∈ P ∩∂Ω and a 0 ≤ t1 ≤ 1 such that (1−t1 )Ax1 +t1 A1 x1 = x1 , then
Ax1 = µ1 x1 , where µ1 = [1 + t1 (α/kAx1 k1 − 1)]−1 , 0 ≤ µ1 ≤ 1, in contradiction
with hypothesis (i’). Hence, by (2.11), (2.12) and the homotopy invariance of fixed
point index, we get
i(A, P ∩ Ω, P ) = i(A1 , P ∩ Ω, P ) = 0.
The proof is complete.
Theorem 2.7. Let Ω1 and Ω2 be two open in E such that θ ∈ Ω1 and Ω1 ⊆ Ω2
with supt∈Ω2 kxk1 < +∞. Let A : P ∩ (Ω2 − Ω1 ) → P be continuous with relatively
compact A(P ∩ (Ω2 − Ω1 )). Suppose that one of the two conditions
(H1) Ax 6≥ x, ∀x ∈ P ∩ ∂Ω1 and Ax 6≤ x, ∀x ∈ P ∩ ∂Ω2 ,
(H2) Ax 6≤ x, ∀x ∈ P ∩ ∂Ω1 and Ax 6≥ x, ∀x ∈ P ∩ ∂Ω2
is satisfied. Then A has at least one fixed point in P ∩ (Ω2 − Ω1 ).
Proof. By the extension theorem (Dugundji [10]), A has a completely continuous
extension (also noted by A) from P ∩ Ω2 from P ∩ Ω2 to P . First we assume that
(H1) is satisfied, i.e., it is the case of cone expansion. It is easy to see that
Ax 6= µx,
∀x ∈ P ∩ ∂Ω1 , µ ≥ 1,
(2.13)
since, otherwise, there exists x0 ∈ P ∩ ∂Ω1 and µ0 ≥ 1 such that Ax0 = µ0 x0 ≥ x0 ,
in contradiction with (H1). Now, from (2.1) and Lemma 2.1, we obtain
i(A, P ∩ Ω1 , P ) = 1.
(2.14)
On the other hand, by Corollary 2.4, we have
i(A, P ∩ Ω2 , P ) = 0.
(2.15)
It follows therefore from (2.14) and (2.15) and additivity property of fixed point
index that
i(A, P ∩ (Ω2 − Ω1 ), P ) = i(A, P ∩ Ω2 , P ) − i(A, P ∩ Ω1 , P ) = −1 6= 0.
(2.16)
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B. YAN
EJDE-2006/74
Hence, by the solution property of fixed point index, A has at least one fixed point
in P ∩ (Ω2 − Ω1 ).
Similarly, when (H2) is satisfied, instead of (2.14), (2.15), we have i(A, P ∩
Ω2 , P ) = 1, and i(A, P ∩ (Ω2 − Ω1 ), P ) = 1. As a result we also can assert that A
has at least one fixed point in P ∩ (Ω2 − Ω1 ). The proof is complete.
We remark that this theorem improves [14, theorem 2.3.3] because the condition
that Ω1 and Ω2 are bounded is not necessary.
Theorem 2.8. Let Ω1 = {x ∈ E|kxk1 < r} and Ω2 = {x ∈ E|kxk1 < R} be two
open in E with r < R. Let A : P ∩ (Ω2 − Ω1 ) → P be continuous with relatively
compact A(P ∩ (Ω2 − Ω1 )). Suppose that one of the two conditions
(H3) kAxk ≤ kxk, for all x ∈ P ∩ ∂Ω1 and kAxk1 ≥ kxk1 , for all x ∈ P ∩ ∂Ω2 ,
(H4) kAxk1 ≥ kxk1 , for all x ∈ P ∩ ∂Ω1 and kAxk ≤ kxk, for all x ∈ P ∩ ∂Ω2
is satisfied. Then A has at least one fixed point in P ∩ (Ω2 − Ω1 ).
Proof. We only need to prove this theorem under condition (H3), since the proof
is similar when (H4) is satisfied. By the extension theorem, A can be extended to
a continuous operator from P ∩ Ω2 into P with relatively compact A(P ∩ Ω2 ). We
may assume that A has no fixed points on P ∩ ∂Ω1 and P ∩ ∂Ω2 . It is easy to see
that (2.13) holds, since otherwise, there exist x0 ∈ P ∩ ∂Ω1 and µ0 > 1 such that
Ax0 = µ0 x0 and hence kAx0 k = µ0 kx0 k > kx0 k, in contradiction with (H3). Thus,
by (2.13), Lemma 2.1, (2.14) holds.
On the other hand, it is also easy to verify
Ax 6= µx, ∀x ∈ P ∩ ∂Ω2 , 0 < µ < 1.
(2.17)
In fact, if there are x1 ∈ P ∩ ∂Ω2 and 0 < µ1 < 1 such that Ax1 = µ1 x1 , then
kAx1 k1 = µ1 kx1 k1 < kx1 k1 ,
in contradiction with (H3). In addition, by (H3) we have
inf
x∈P ∩∂Ω2
kAxk1 ≥ inf kxk1 > 0.
x∈∂Ω2
(2.18)
It follows from (2.17), (2.18) and Lemma 2.5 that (2.15) holds. As before, (2.14) and
(2.15) imply (2.16), and therefore A has at least one fixed point in P ∩(Ω2 −Ω1 ). We remark that this theorem improves the the Krasnoselskii’s theorem in [14]
because the condition that Ω1 and Ω2 are bounded is not necessary.
3. The existence of multiple positive solutions to equation (1.1)
In this section, we consider (1.1) and suppose that f ∈ C([0, 1]×R0+ ×R, R+ ), p ∈
R1 1
C([0, 1], R) ∩ C((0, 1), R0+ ) ∩ C 1 ((0, 1), R) with 0 p(r)
dr < +∞, Φ ∈ C((0, 1), R0+ )
and α ≥ 0, β ≥ 0 (not equal to 0 at the same time); here R+ = [0, +∞), R0+ =
(0, +∞), R = (−∞, +∞). Let
Z 1
1
ρ2 = β + α
dr (ρ > 0),
p(r)
0
Z
Z t
1 1 1
1
1
u1 (t) =
dr, v1 (t) = (β + α
dr),
ρ t p(r)
ρ
p(r)
0
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MULTIPLE POSITIVE SOLUTIONS
7
(
u1 (t)v1 (s)p(s), 0 ≤ s ≤ t ≤ 1
G1 (t, s) =
v1 (t)u1 (s)p(s), 0 ≤ t ≤ s ≤ 1.
Assume that Cp1 [0, 1] = {x : [0, 1] → R| x(t) is continuous on [0, 1] and p(t)x0 (t) is
continuous on [0, 1] also with maxt∈[0,1] p(t)|x0 (t)| < +∞} (see [21]). For x ∈ Cp1 , let
kxk1 = maxt∈[0,1] |x(t)|, kxk2 = maxt∈[0,1] p(t)|x0 (t)| and kxk = max{kxk1 , kxk2 }.
It is easy to see that Cp1 satisfies the conditions (1), (2) and (3) of the Banach space
E in section 2.
Obviously, x(t) ∈ Cp1 is a solution to equation (1.1) if and only if x(t) is a solution
of the following integral equation
Z 1
G1 (t, s)Φ(s)f (s, x(s), p(s)x0 (s))ds, t ∈ [0, 1].
x(t) =
0
Let P = {x ∈
t ∈ [0, 1].
Cp1 |x(t)
1
≥ γ1 (t)kxk1 }, where γ1 (t) = u1 (t)v1 (t) u1 (0)v
for all
1 (1)
Lemma 3.1. Assume that l ∈ L1 [0, 1] with l(t) > 0 for all t ∈ (0, 1) and q(t) =
R1
G1 (t, s)l(s)ds, t ∈ [0, 1]. Then
0
q(t) ≥ γ1 (t) max q(s).
s∈[0,1]
Proof. Suppose q(t0 ) = maxs∈[0,1] q(s). Then
 v (t)u (s)p(s)
1
1


u1 (t0 )v1 (s)p(s) , 0 ≤ t ≤ s ≤ t0 ≤ 1

 u1 (t)v1 (s)p(s)

, 0 ≤ t0 ≤ s ≤ t ≤ 1
G1 (t, s)
0 )u1 (s)p(s)
= vv11(t(t)u
1 (s)p(s)
G1 (t0 , s) 
 v1 (t0 )u1 (s)p(s) , 0 ≤ t, t0 ≤ s ≤ 1


 u1 (t)v1 (s)p(s) , 0 ≤ s ≤ t, t ≤ 1
0
u1 (t0 )v1 (s)p(s)

u1 (s)
1

u1 (t)v1 (t) u1 (t0 ) v1 (s)u
, 0 ≤ t ≤ s ≤ t0 ≤ 1

1 (t)


v
(s)
1
u (t)v (t) 1
,
0 ≤ t0 ≤ s ≤ t ≤ 1
1
1
v1 (t0 ) u1 (s)v1 (t)
=
1

0 ≤ t, t0 ≤ s ≤ 1
u1 (t)v1 (t) u1 (t)v1 (t0 ) ,


u (t)v (t)
1
0 ≤ s ≤ t, t0 ≤ 1
1
1
u1 (t0 )v1 (t) ,

1

u1 (t)v1 (t) v1 (1)u
, 0 ≤ t ≤ s ≤ t0 ≤ 1

1 (0)


1
u (t)v (t)
1
1
u1 (0)v1 (1) , 0 ≤ t0 ≤ s ≤ t ≤ 1
≥
1

u
(t)v
(t)
1
1

u1 (0)v1 (1) , 0 ≤ t, t0 ≤ s ≤ 1


u (t)v (t)
1
, 0 ≤ s ≤ t, t ≤ 1
1
1
0
u1 (0)v1 (1)
= γ1 (t).
As a consequence,
Z
q(t) =
1
Z
G1 (t, s)l(s)ds =
0
0
Z
≥ γ1 (t)
1
G1 (t, s)
G1 (t0 , s)l(s)ds
G1 (t0 , s)
1
G1 (t0 , s)l(s)ds = γ1 (t) max q(s).
0
The proof is complete.
Now we will list some conditions for convenience:
s∈[0,1]
8
B. YAN
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(H1) There exists a k ∈ C([0, 1], R0+ ), a g ∈ C(R0+ , R0+ ) and a decreasing continuous function h ∈ C(R0+ , R0+ ) such that
∀x ∈ R0+ , z ∈ R, t ∈ [0, 1],
R1
g(x)
where h(x)
is an increasing function and 0 p(s)Φ(s)k(s)h(cγ1 (s))ds < +∞
for each c > 0;
f (t, x, z) ≤ k(t)g(x),
(H2)
ch(c)
sup
c∈R0+
u1 (0)v1 (1)
R1
0
> 1;
p(s)Φ(s)k(s)h(cγ1 (s))dsg(c)
(H3) There exists a k1 ∈ C([0, 1], R0+ ) and a g1 ∈ C(R0+ , R0+ ) with f (t, x, z) ≥
k1 (t)g1 (x), for all (t, x, z) ∈ [0, 1] × R0+ × (−∞, +∞) such that
lim
x→+∞
g1 (x)
= +∞,
x
R1
where 0 p(s)Φ(s)k1 (s)ds < +∞;
(H4) For any c > 0, there exists a ψc ∈ C([0, 1], R0+ ) such that f (t, x, z) ≥ ψc (t)
R1
for all (t, x, z) ∈ [0, 1] × (0, c] × (−∞, +∞) with 0 p(s)Φ(s)ψc (s)ds < +∞.
For given n ∈ {1, 2, . . . }, let fn (t, x, z) = f (t, max{ n1 , x}, z) and for x ∈ P , define
Z 1
(An x)(t) =
G1 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds, n ∈ {1, 2, . . . }, t ∈ [0, 1].
0
(3.1)
Lemma 3.2. Assume the condition (H1) holds. Then, for every n ∈ {1, 2, . . . },
An : P → P is continuous and for any r > 0 and Br = {x ∈ Cp1 |kxk1 ≤ r},
An (P ∩ Br ) is relatively compact.
Proof. First, for a given n ∈ {1, 2, . . . }, we show that An P ⊆ P . For any x ∈ P ,
we have
Z
1
|(An x)(t)| = G1 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
Z
0
1
≤
0
Z
1
G1 (t, s)Φ(s)f (s, max{ , x(s)}, p(s)x0 (s))ds
n
1
1
G1 (t, s)Φ(s)k(s)g(max{ , x(s)})ds
n
0
Z 1
g(max{ n1 , x(s)})
1
≤
G1 (t, s)Φ(s)k(s)h(max{ , x(s)})
ds
n
h(max{ n1 , x(s)})
0
Z 1
1 g(max{ n1 , kxk1 })
≤
G1 (t, s)Φ(s)k(s)dsh( )
n h(max{ n1 , kxk1 })
0
< +∞
≤
and
Z
1 t
|p(t)(An x) (t)| = | −
v1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
ρ 0
Z
α 1
+
u1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
ρ t
0
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
≤
K
ρ
1
Z
0
Z
9
1
p(s)Φ(s)f (s, max{ , x(s)}, p(s)x0 (s))ds
n
1
1
p(s)Φ(s)k(s)g(max{ , x(s)})ds
n
0
Z
g(max{ n1 , x(s)})
1
K 1
p(s)Φ(s)k(s)h(max{ , x(s)})
ds
≤
ρ 0
n
h(max{ n1 , x(s)})
Z
K 1
1 g(max{ n1 , kxk1 })
≤
p(s)Φ(s)k(s)dsh( )
ρ 0
n h(max{ n1 , kxk1 })
< +∞,
K
≤
ρ
where K = max{αu1 (0), v1 (1)} and the following is same as before. Then, An is
well defined. Moreover, from Lemma 3.1, for any x ∈ P , we have
(An x)(t) ≥ γ1 (t) max |(An x)(s)| = γ1 (t)kAn xk1 ,
∀t ∈ [0, 1].
s∈[0,1]
Consequently, An P ⊆ P .
Second, we show An : P → P is continuous. Assume that limm→+∞ xm =
x0 , which means there exists an M > 1/n such that kxm k ≤ M , for all m ∈
{0, 1, 2, . . . }. Then
1
fn (t, xm (t), p(t)x0 m (t)) ≤ k(t)g(max{ , xm (t)})
n
g(max{ n1 , xm (t)})
1
= k(t)h(max{ , xm (t)})
n
h(max{ n1 , xm (t)})
1 g(M )
, m ∈ {1, 2, . . . }.
≤ k(t)h( )
n h(M )
Since
fn (t, xm (t), p(t)x0m (t)) → fn (t, x0 (t), p(t)x00 (t)),
as m → +∞,
from (3.2), the Lesbegue Dominated Convergence Theorem guarantees
max |(An xm )(t) − (An x0 )(t)|
lim
m→+∞ t∈[0,1]
Z
=
lim
max |
m→+∞ t∈[0,1]
Z
1
G1 (t, s)Φ(s)fn (s, xm (s), p(s)x0m (s))ds
0
1
−
G1 (t, s)Φ(s)fn (s, x0 (s), p(s)x00 (s))ds|
0
Z
1
G1 (t, s)Φ(s)fn (s, xm (s), p(s)x0m (s))
m→+∞ t∈[0,1] 0
− fn (s, x0 (s), p(s)x00 (s))ds
Z 1
≤ lim u1 (0)v1 (1)
p(s)Φ(s)fn (s, xm (s), p(s)x0m (s))
m→+∞
0
− fn (s, x0 (s), p(s)x00 (s))ds = 0
≤
lim
max
and
lim
max |p(t)(An xm )0 (t) − p(t)(An x0 )0 (t)|
m→+∞ t∈[0,1]
(3.2)
10
B. YAN
EJDE-2006/74
1Z t
v1 (s)p(s)Φ(s)[fn (s, xm (s), p(s)x0m (s))
max −
m→+∞ t∈[0,1]
ρ 0
Z
α 1
0
− fn (s, x0 (s), p(s)x0 (s))]ds +
u1 (s)p(s)Φ(s)[fn (s, xm (s), p(s)x0m (s))
ρ t
− fn (s, x0 (s), p(s)x00 (s))]ds
Z
K 1
≤ lim
p(s)Φ(s)fn (s, xm (s), p(s)x0m (s)) − fn (s, x0 (s), p(s)x00 (s))ds
m→+∞ ρ 0
= 0,
=
lim
which mean that
lim kAn xm − An x0 k = 0.
m→+∞
Finally, we show An (Br ∩ P ) is relatively compact. Obviously, Br is an unbounded
set in Cp1 . Without loss of generality, we suppose r > 1/n. Then, for any x ∈ Br ∩P ,
we have
Z 1
max |(An x)(t)| = max |
G1 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
t∈[0,1]
t∈[0,1]
Z
0
1
G1 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
= max
t∈[0,1]
0
Z
1
≤ max
t∈[0,1]
0
Z
g(max{ n1 , x(s)})
1
G1 (t, s)Φ(s)k(s)h(max{ , x(s)})
ds
n
h(max{ n1 , x(s)})
1
1 g(r)
G1 (t, s)Φ(s)k(s)dsh( )
n
h(r)
0
Z 1
1 g(r)
≤ u1 (0)v1 (1)
p(s)Φ(s)k(s)dsh( )
n
h(r)
0
≤ max
t∈[0,1]
and
Z
1 t
max |p(t)(An x) (t)| = max | −
v1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
ρ 0
t∈[0,1]
t∈[0,1]
Z
α 1
u1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
+
ρ t
Z
K 1
≤
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
ρ 0
Z
g(max{ n1 , x(s)})
K 1
1
≤
p(s)Φ(s)k(s)h(max{ , x(s)})
ds
ρ 0
n
h(max{ n1 , x(s)})
Z
K 1
1 g(r)
≤
p(s)Φ(s)k(s)dsh( )
,
ρ 0
n h(r)
0
which means that An (Br ∩ P ) is bounded. Assume that t, t0 ∈ [0, 1]. Then, for
x ∈ Br ∩ P , we have
Z 1
0
|(An x)(t) − (An x)(t )| = |
G1 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
0
Z
−
0
1
G1 (t0 , s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
Z
11
1
|G1 (t, s) − G1 (t0 , s)|Φ(s)fn (s, x(s), p(s)x0 (s))ds
≤
0
Z
≤
0
1
1 g(r)
|G1 (t, s) − G1 (t0 , s)|Φ(s)k(s)dsh( )
n h(r)
and
|p(t)(An x)0 (t) − p(t0 )(An x)0 (t0 )|
Z
1 t
v1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
=|−
ρ 0
Z 0
1 t
+
v1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
ρ 0
Z
α 1
+
u1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
ρ t
Z
α 1
−
u1 (s)p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
ρ t0
Z t0
K
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))|ds
≤2 |
ρ t
Z t0
1 g(r)
K
p(s)Φ(s)k(s)ds|h( )
.
≤2 |
ρ t
n h(r)
Then, for any ε > 0, we can choose δ > 0 small enough such that
|(An x)(t) − (An x)(t0 )| < ε,
|p(t)(An x)0 (t) − p(t0 )(An x)0 (t0 )| < ε,
for all x ∈ Br ∩ P , |t − t0 | < δ, t, t0 ∈ [0, 1]. Consequently, {(An (Br ∩ P ))(t)} and
{p(t)(An (Br ∩ P ))0 (t)} is equicontinuous on [0, 1].
Consequently, from Arzela-Ascoli theorem, An (Br ∩ P ) is relatively compact.
The proof is complete.
Theorem 3.3. Assume that (H1)–(H4) hold. Then (1.1) has at least two positive
solutions.
Proof. From Lemma 3.2, for each n ∈ {1, 2, . . . }, An : P → P is continuous operator
and for any r > 0, An (Br ∩ P ) is relatively compact. From (H2), choose R0 > 0
such that
R0 h(R0 )
> 1.
(3.3)
R1
u1 (0)v1 (1) 0 p(s)Φ(s)k(s)h(R0 γ1 (s))dsg(R0 )
Without loss of the generality, suppose R0 ≥ 1/n, n ∈ {1, 2, . . . }. Set
Ω1 = {x ∈ Cp1 |kxk1 < R0 }.
Then for any x ∈ ∂Ω1 ∩ P , one has
x 6≤ An x, n ∈ {1, 2, . . . }.
If there exists an x0 ∈ ∂Ω1 ∩ P such that x0 ≤ An x0 , obviously,
x0 (t) ≥ γ1 (t)R0 . Then
R0 = max |x0 (t)|
t∈[0,1]
Z
≤ max
t∈[0,1]
0
1
1
G1 (t, s)Φ(s)f (s, max{ , x0 (s)}, p(s)x0 0 (s))ds
n
(3.4)
max{ n1 , x0 (t)}
≥
12
B. YAN
Z
EJDE-2006/74
1
1
p(s)Φ(s)k(s)g(max{ , x0 (s)})ds
n
0
Z 1
g(max{ n1 , x0 (s)})
1
≤ u1 (0)v1 (1)
p(s)Φ(s)k(s)h(max{ , x0 (s)})
)ds
n
h(max{ n1 , x0 (s)})
0
Z 1
g(R0 )
p(s)Φ(s)k(s)h(R0 γ1 (s))ds
≤ u1 (0)v1 (1)
h(R0 )
0
≤ u1 (0)v1 (1)
which implies
R0 h(R0 )
u1 (0)v1 (1)
R1
0
≤ 1.
Φ(s)k(s)h(R0 γ1 (s))dsg(R0 )
This contradicts (3.3). Then, (3.4) is true. From the proof of Theorem 2.7, the
(2.13) is true, which means
i(An , P ∩ Ω1 , P ) = 1,
n ∈ {1, 2, . . . }.
(3.5)
(1)
(1)
(1)
As a result, for n ∈ {1, 2, . . . }, there exists an xn ∈ P ∩ Ω1 with An xn = xn .
(1)
(1)
(1)
Since kxn k1 ≤ R0 , n ∈ {1, 2, . . . }, it is easy to see that {xn (t)} and {p(t)x0 n (t)}
are uniformly bounded. Moreover, from Lemma 3.1, the condition (H4), there exists
a ψR0 (t) such that
Z 1
1
0
x(1)
(t)
=
G1 (t, s)Φ(s)f (s, max{ , x(1)
(s)}, p(s)(x(1)
n
n ) (s))ds
n n
0
Z 1
(3.6)
≥
G1 (t, s)Φ(s)ψR0 (s)ds
0
≥ γ1 (t)k ∗ ,
R1
where k ∗ = maxt∈[0,1] 0 G1 (t, s)Φ(s)ψR0 (s)ds. Thus, for any t0 , t00 ∈ (0, 1) and
n ∈ {1, 2, . . . },
0
(1) 00
|x(1)
n (t ) − xn (t )|
Z 1
(1) 0
≤
|G1 (t0 , s) − G1 (t00 , s)|fn (s, x(1)
n (s), p(s)(xn ) (s))ds
(3.7)
0
1
Z
|G1 (t0 , s) − G1 (t00 , s)|Φ(s)k(s)h(γ1 (s)k ∗ )ds
≤
0
g(R0 )
,
h(R0 )
and
0 0
00
(1) 0 00
|p(t0 )(x(1)
n ) (t ) − p(t )(xn ) (t )|
00
Z
t
(1) 0
≤ 2
p(s)Φ(s)fn (s, x(1)
n (s), p(s)(xn ) (s))ds
(3.8)
t0
t00
Z
g(R0 )
, n ∈ {1, 2, . . . }.
p(s)Φ(s)k(s)h(γ1 (s)k ∗ )ds
h(R0 )
t0
Consequently, for any ε > 0, we can choose a δ > 0 such that
≤ 2
0
(1) 00
|x(1)
n (t ) − xn (t )| < ε,
0 0
00
(1) 0 00
|p(t0 )(x(1)
n ) (t ) − p(t )(xn ) (t )| < ε,
(1)
for all n ∈ {1, 2, . . . }, |t0 − t00 | < δ, t0 , t00 ∈ [0, 1], which implies that {xn (t)} and
(1)
{p(t)(xn )0 (t)} are equi-continuous on [0, 1].
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
13
(1)
The Arzela-Ascoli theorem guarantees that there is a subsequence {xnj } of
(1)
(1)
with limj→+∞ xnj = x0 . From (3.6), we have
(1)
{xn }
x0 (t) ≥ k ∗ γ1 (t),
∀t ∈ [0, 1].
Then, for t ∈ (0, 1), if j is big enough, we have
(1)
(1)
(1) 0
0
|fnj (t, x(1)
nj (t), p(t)(xnj ) (t)) − f (t, x0 (t), p(t)(x0 ) (t))|
(1)
(1)
(1) 0
0
= |f (t, x(1)
nj (t), p(t)(xnj ) (t)) − f (t, x0 (t), p(t)(x0 ) (t))| → 0,
as j → +∞
and
1 (1)
0
, x (t)}, p(t)(x(1)
nj ) (t))
n j nj
g(R0 )
.
≤ k(t)h(k ∗ γ1 (t))
h(R0 )
(1) 0
fnj (t, x(1)
nj (t), p(t)(xnj ) (t)) = f (t, max{
(3.9)
Then the Lesbegue Dominated Convergence Theorem guarantees that
(1)
x0 (t) = lim x(1)
nj (t)
j→+∞
Z 1
1
0
= lim
G1 (t, s)Φ(s)f (s, max{ , x(1)
(s)}, p(s)(x(1)
nj ) (s))ds
j→+∞ 0
n j nj
Z 1
(1)
(1)
=
G1 (t, s)Φ(s)f (s, x0 (s), p(s)(x0 )0 (s))ds.
(3.10)
0
(1)
kx0 k1
∗
(1)
Obviously
≤ R0 . Thus (3.3) can guarantee kx0 k1 < R0 . Let 0 < a∗ <
b∗ < 1, and 0 < c < mint∈[a∗ ,b∗ ] γ1 (t). Suppose
Z b∗
−1
∗
N∗ =
min
G
(t,
s)Φ(s)k
(s)dsc
+ 1.
1
1
∗ ∗
t∈[a ,b ]
a∗
From the condition (H3), there exists an R0 > R such that
g1 (x) > N ∗ x, ∀x ≥ R0 .
(3.11)
Now we define
R0
Ω2 = {x ∈ kxk1 < ∗ }.
c
R0
0
We might as well suppose that c∗ > 1, R > 1. Then we have
An x 6≤ x for all x ∈ ∂Ω2 ∩ P, n ∈ {1, 2, . . . }.
(3.12)
(3.13)
Otherwise, suppose there exists x0 ∈ ∂Ω2 ∩P with An x0 ≤ x0 . Since x0 ∈ ∂(Ω2 )∩P ,
min
x0 (t) ≥
∗ ∗
t∈[a ,b ]
min
γ1 (t)kxk1 > c∗
∗ ∗
t∈[a ,b ]
R0
= R0 > 1.
c∗
Then, for t ∈ [a∗ , b∗ ], from (3.11) and (3.14), we have
x0 (t) ≥ (An x0 )(t)
Z 1
=
G1 (t, s)Φ(s)fn (s, x0 (s), p(s)x00 (s))ds
0
Z
b∗
≥
a∗
1
G1 (t, s)Φ(s)f (s, max{ , x0 (s)}, p(s)x00 (s))ds
n
(3.14)
14
B. YAN
Z
b∗
≥
a∗
Z
EJDE-2006/74
1
G1 (t, s)Φ(s)k1 (s)g1 (max{ , x0 (s)})ds
n
b∗
>
a∗
Z b∗
>
G1 (t, s)Φ(s)k1 (s)N ∗ x0 (s)ds
G1 (t, s)Φ(s)k1 (s)dsN ∗ c∗
a∗
R0
> ∗,
c
R0
c∗
∀n ∈ {1, 2, . . . }
which implies kx0 k1 > R0 /c∗ . This contradicts to x0 ∈ P ∩ ∂Ω2 .
From (3.4) and (3.13), the Theorem 2.7 guarantees that An has a fixed point
(2)
xn ∈ (Ω2 − Ω1 ) ∩ P , n ∈ {1, 2, . . . }. It is easy to see that
(2)
x(2)
n (t) ≥ γ1 (t)kxn k1 ≥ γ1 (t)R0 ,
n ∈ {1, 2, . . . }.
(3.15)
(2)
By proof similar to (3.6), (3.7), (3.8), we know that {xn } is relatively compact
(2)
(2)
(2)
(2)
in Cp1 . Then there exists a subsequence {xni } of {xn } with limi→+∞ xni = x0 .
(2)
And moreover, by similar proof as (3.10), we get x0 (t) is a positive solution to
0
(2)
equation (1.1) with R
c∗ > kx0 k1 > R0 . Consequently, equation (1.1) has at least
(1)
(2)
two different positive solution x0 (t) and x0 (t). The proof is complete.
Example 3.4. Now we consider
x00 +
1 −1/2
t
(1 − t)−1/4 (x−1/4 + x2 )(1 + cos2 (x0 )) = 0,
16
lim x0 (t) = 0 = x(1).
t ∈ (0, 1)
(3.16)
t→0+
Then (3.16) has at least two positive solutions.
To prove that (3.16) has at least two positive solutions, we apply Theorem 3.3
1 −1/2
with Φ(t) = 16
t
(1 − t)−1/4 , p(t) ≡ 1, f (t, x, z) = (x−1/4 + x2 )(1 + cos2 (z)),
k(t) ≡ 1, g(x) = 2(x−1/4 + x2 ), h(x) = x−1/4 , γ1 (t) = 1 − t, k1 (t) ≡ 1, g1 (x) =
x−1/4 + x2 , Ψc (t) = c−1/4 . It is easy to verify that (H1)–(H4) hold. Hence, (3.16)
has at leat two positive solutions.
Example 3.5. Now we consider
x00 +
1 −1/4
t
(1 − t)−1/4 (x−1/4 + x2 )(π + arctan x0 ) = 0,
12π
x(0) = 0 = x(1).
t ∈ (0, 1)
(3.17)
Then (3.17) has at least two positive solutions.
To prove that (3.17) has at least two positive solutions, we apply Theorem 3.3
1 −1/4
with Φ(t) = 12π
t
(1 − t)−1/4 , p(t) ≡ 1, f (t, x, z) = (x−1/4 + x2 )(π + arctan x0 ),
−1/4
k(t) ≡ 1, g(x) = 3π
+ x2 ), h(x) = x−1/4 , γ1 (t) = t(1 − t), k1 (t) ≡ 1,
2 (x
−1/4
2
g1 (x) = x
+ x , Ψc (t) = c−1/4 . It is easy to verify that (H1)–(H4) hold. Hence,
(3.17) has at leat two positive solutions.
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
15
4. The existence of multiple positive solutions to equation (1.2)
In this section, we consider (1.2) and suppose that f ∈ C(R+ × R0+ × R+ , R+ ),
R +∞ 1
+
+
p ∈ C(R+ , R) ∩ C(R0+ , R0+ ) ∩ C 1 (R0+ , R) with 0
p(r) dr = +∞, Φ ∈ C(R0 , R );
here R+ = [0, +∞), R0+ = (0, +∞), R = (−∞, +∞). Let
(
u2 (t)v2 (s)p(s), a ≤ s ≤ t < +∞
G2 (t, s) =
v2 (t)u2 (s)p(s), 0 ≤ t ≤ s < +∞,
Rt 1
where u2 (t) = 1 and v2 (t) = 0 p(r) dr for all t ∈ R+ .
1
Let C∞
= x : [0, +∞) → R| x(t) is continuous on R+ and p(t)x0 (t) is continuous
x(t)
on R+ also with limt→+∞ 1+v
exists and supt∈[0,+∞) p(t)|x0 (t)| < +∞ . For
2 (t)
1
x ∈ C∞
, let
kxk1 =
sup
t∈[0,+∞)
|x(t)|
1 + v2 (t)
and kxk2 =
sup
p(t)|x0 (t)|.
t∈[0,+∞)
1
1
. Now
and k · k2 is a semi-norm of C∞
It is easy to see that k · k1 is a norm of C∞
1
satisfies (1), (2) and (3) of the Banach
Let kxk = max{kxk1 , kxk2 }. Obviously, C∞
space E in section 2.
1
It is easy to prove that if x(t) ∈ C∞
is a solution to integral equation
Z ∞
x(t) =
G2 (t, s)Φ(s)f (s, x(s), p(s)x0 (s))ds, t ∈ R+ ,
0
then x(t) is a solution to (1.2).
1
|x(t) ≥ γ2 (t)kxk1 , ∀t ∈ R+ }, where
Let P = {x ∈ C∞
(R t
1
dr, t ∈ [0, τ ]
0 p(r)
γ2 (t) =
1,
t ∈ (τ, +∞)
γ
e2 (t) =
γ2 (t)
,
1 + v2 (t)
t ∈ R+ ;
Rτ 1
dr = 1. Suppose that x = (1 + v2 (t))y, t ∈ R+ and F (t, y, z) =
here 0 p(r)
f (t, (1 + v2 (t))y, z) = f (t, x, z).
Now we will list some conditions for convenience:
(H1) There exists a k ∈ C(R+ , R0+ ), a g ∈ C(R0+ , R0+ ) and a decreasing continuous function h ∈ C(R0+ , R0+ ) such that
∀(y, z) ∈ R0+ × R0+ , t ∈ R+ ,
R∞
g(y)
where h(y)
is an increasing function and 0 p(s)Φ(s)k(s)h(cγ̃2 (s))ds < +∞
for each c > 0
F (t, y, z) ≤ k(t)g(y),
(H2)
sup R ∞
c∈R0+
0
ch(c)
>1
p(s)Φ(s)k(s)h(ce
γ2 (s))dsg(c)
(H3) There exists a k1 ∈ C(R+ , R0+ ) and a g1 ∈ C(R0+ , R0+ ) with F (t, y, z) ≥
k1 (t)g1 (y), for all (t, y, z) ∈ [0, +∞) × R0+ × R+ such that
lim
y→+∞
where
R∞
0
g1 (y)
= +∞,
y
p(s)Φ(s)k1 (s)ds < +∞
16
B. YAN
EJDE-2006/74
(H4) for any c > 0, there exists a ψc ∈ C(RR+ , R0+ ) such that F (t, y, z) ≥ ψc (t)
∞
for all (t, y, z) ∈ R+ × (0, c] × R+ with 0 p(s)Φ(s)ψc (s)ds < +∞.
Let Cl = x : R+ → R| x(t) is continuous on R+ and limt→+∞ x(t) exists with
norm kxkl = supt∈[0,+∞) |x(t)|. From [18], we know that Cl is a Banach space and
following theorem is true.
Theorem 4.1 ([18]). Let M ⊆ Cl (R+ , R). Then M is relatively compact in the
space Cl (R+ , R) if the following conditions hold:
(a) M is bounded in Cl
(b) the functions belonging to M are locally equi-continuous on R+ ;
(c) the functions from M are equiconvergent, that is, given ε > 0, there corresponds T (ε) > 0 such that |x(t) − x(+∞)| < ε for any t ≥ T (ε) and
x ∈ M.
1
Theorem 4.2 ([22]). Let M ⊆ C∞
(R+ , R). Then M is relatively compact in
1
+
C∞ (R , R) if the following conditions hold:
1
(a) M is bounded in C∞
;
x(t)
, x ∈ M } and the functions
(b) the functions belonging to {y y(t) = 1+v
2 (t)
0
belonging to {y y(t) = p(t)x (t), x ∈ M } are locally equi-continuous on R+ ;
x(t)
(c) the functions from {y y(t) = 1+v
, x ∈ M } and the functions from
2 (t)
{y y(t) = p(t)x0 (t), x ∈ M } are equi-convergent at +∞.
R +∞
Lemma 4.3 ([22]). Assume that Φ(t) ∈ C(R0+ , R+ ) with 0 p(s)Φ(t)dt < +∞
R∞
and F (t) = 0 G2 (t, s)Φ(s)ds. Then
F (t) ≥ γ2 (t)
F (τ )
,
1 + v2 (τ )
∀t ∈ R+ , τ ∈ R+ ,
and
F (t) ≥ γ2 (t)kF k1 ,
∀t ∈ R+ ,
lim
t→+∞
F (t)
= 0.
1 + v2 (t)
Let fn (t, x, z) = f (t, max{ n1 (1 + v2 (t)), x}, z), n ∈ {1, 2, . . . } and for x ∈ P ,
n ∈ {1, 2, . . . }, t ∈ R+ , define
Z ∞
(An x)(t) =
G2 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds .
(4.1)
0
Lemma 4.4. Assume that the condition (H1) holds. Then for each n ∈ {1, 2, . . . },
1
An : P → P is continuous and for any r > 0 and Br = {x ∈ C∞
|kxk1 ≤ r},
An (P ∩ Br ) is relatively compact for each n ≥ 1.
Proof. First, we show that An P ⊆ P . For any x ∈ P , we have
Z
∞
|(An x)(t)| = G2 (t, s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
Z
0
∞
1
G2 (t, s)Φ(s)f (s, max{ (1 + v2 (s)), x(s)}, p(s)x0 (s))ds
n
Z0 ∞
1
x(s)
=
G2 (t, s)Φ(s)F (s, max{ ,
}, p(s)x0 (s))ds
n
1
+
v2 (s)
0
Z ∞
1
x(s)
≤
G2 (t, s)Φ(s)k(s)g(max{ ,
})ds
n
1
+
v2 (s)
0
≤
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
Z
∞
x(s)
g(max{ n1 , 1+v
})
1
x(s)
2 (s)
G2 (t, s)Φ(s)k(s)h(max{ ,
})
ds
x(s)
n 1 + v2 (s) h(max{ 1 ,
})
∞
1 g(max{ n1 , kxk1 })
G2 (t, s)Φ(s)k(s)dsh( )
n h(max{ n1 , kxk1 })
≤
0
Z
≤
0
17
n 1+v2 (s)
< +∞
and
Z
∞
|p(t)(An x) (t)| =
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
t
Z ∞
1
p(s)Φ(s)f (s, max{ (1 + v2 (s)), x(s)}, p(s)x0 (s))ds
≤
n
Zt ∞
1
x(s)
p(s)Φ(s)F (s, max{ ,
=
}, p(s)x0 (s))ds
n
1
+
v
(s)
2
Zt ∞
x(s)
1
})ds
≤
p(s)Φ(s)k(s)g(max{ ,
n
1
+
v2 (s)
t
x(s)
Z ∞
g(max{ n1 , 1+v
})
1
x(s)
2 (s)
≤
p(s)Φ(s)k(s)h(max{ ,
})
ds
x(s)
n 1 + v2 (s) h(max{ 1 ,
t
n 1+v2 (s) })
Z ∞
1 g(max{ n1 , kxk1 })
≤
p(s)Φ(s)k(s)dsh( )
n h(max{ n1 , kxk1 })
0
< +∞.
0
Since (An x)(t) ≥ 0, it is easy to see that An is well defined. Moreover, from Lemma
4.3, for any x ∈ P , we have
(An x)(t) ≥ γ2 (t)kAn xk1 ,
∀t ∈ R+ .
Consequently, An P ⊆ P .
Second, we show An : P → P is continuous. Assume that limm→+∞ xm =
x0 , which means there exists an M > 1/n such that kxm k ≤ M , for all m ∈
{0, 1, 2, . . . }. Then
1 xm (t)
fn (t, xm (t), p(t)x0 m (t)) = F (t, max{ ,
}, p(t)x0m (t))
n 1 + v2 (t)
1 xm (t)
})
≤ k(t)g(max{ ,
n 1 + v2 (t)
xm (t)
g(max{ n1 , 1+v
})
1 xm (t)
2 (t)
≤ k(t)h(max{ ,
})
n 1 + v2 (t) h(max{ 1 , xm (t) })
n 1+v2 (t)
1 g(M )
≤ k(t)h( )
,
n h(M )
m ∈ {1, 2, . . . }.
Since
fn (t, xm (t), p(t)x0m (t)) → fn (t, x0 (t), p(t)x00 (t)),
as m → +∞,
from (4.2), the Lesbegue Dominated Convergence Theorem guarantees that
|(An xm )(t) − (An x0 )(t)|
m→+∞ t∈[0,+∞)
1 + v2 (t)
lim
sup
(4.2)
18
B. YAN
Z
EJDE-2006/74
∞
G2 (t, s)
Φ(s)(fn (s, xm (s), p(s)x0m (s))
m→+∞ t∈[0,+∞)
1 + v2 (t)
0
− fn (s, x0 (s), p(s)x00 (s)))ds
Z ∞
G2 (t, s)
≤ lim
sup
Φ(s)|fn (s, xm (s), p(s)x0m (s))
m→+∞ t∈[0,+∞) 0
1 + v2 (t)
=
lim
sup
− fn (s, x0 (s), p(s)x00 (s))|ds
Z ∞
p(s)Φ(s)|fn (s, xm (s), p(s)x0m (s))
= lim
m→+∞
0
− fn (s, x0 (s), p(s)x00 (s))|ds = 0
and
lim
sup
m→+∞ t∈[0,+∞)
=
lim
sup
m→+∞ t∈[0,+∞)
Z
−
≤
|p(t)(An xm )0 (t) − p(t)(An x0 )0 (t)|
Z
∞
p(s)Φ(s)fn (s, xm (s), p(s)x0m (s))ds
t
∞
p(s)Φ(s)fn (s, x0 (s), p(s)x00 (s))ds
t
Z ∞
lim
sup
p(s)Φ(s)|fn (s, xm (s), p(s)x0m (s))
m→+∞ t∈[0,+∞)
t
− fn (s, x0 (s), p(s)x00 (s))|ds
Z ∞
= lim
p(s)Φ(s)fn (s, xm (s), p(s)x0m (s)) − fn (s, x0 (s), p(s)x00 (s))ds = 0,
m→+∞
0
which implies
lim kAn xm − An x0 k = 0.
m→+∞
Finally, we show An (Br ∩P ) is relatively compact. Obviously, Br is a unbounded set
1
in C∞
. Without loss of generality, we suppose r > 1/n. Then, for any x ∈ Br ∩ P ,
we have
|(An x)(t)|
t∈[0,+∞) 1 + v2 (t)
Z ∞
G2 (t, s)
= sup |
Φ(s)fn (s, x(s), p(s)x0 (s))ds|
1 + v2 (t)
t∈[0,+∞)
0
Z ∞
G2 (t, s)
= sup
Φ(s)fn (s, x(s), p(s)x0 (s))ds
1
+
v
(t)
2
t∈[0,+∞) 0
x(s)
Z ∞
})
g(max{ n1 , 1+v
G2 (t, s)
1
x(s)
2 (s)
≤ sup
Φ(s)k(s)h(max{ ,
})
ds
x(s)
1
1 + v2 (t)
n 1 + v2 (s) h(max{ ,
t∈[0,+∞) 0
n 1+v2 (s) })
Z ∞
G2 (t, s)
1 g(r)
≤ sup
Φ(s)k(s)dsh( )
1 + v2 (t)
n h(r)
t∈[0,+∞) 0
Z ∞
1 g(r)
≤
p(s)Φ(s)k(s)dsh( )
n
h(r)
0
sup
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
19
and
|p(t)(An x)0 (t)|
sup
t∈[0,+∞)
Z
=
sup
t∈[0,+∞)
Z
=
∞
|
t
∞
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
sup
t∈[0,+∞)
t
Z
≤
∞
x(s)
})
g(max{ n1 , 1+v
1
x(s)
2 (s)
p(s)Φ(s)k(s)h(max{ ,
ds
})
n 1 + v2 (s) h(max{ 1 , x(s) })
∞
1 g(r)
p(s)Φ(s)k(s)dsh( )
n h(r)
sup
t∈[0,+∞)
t
Z
≤
sup
t∈[0,+∞)
Z ∞
=
0
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds|
t
n 1+v2 (s)
1 g(r)
p(s)Φ(s)k(s)dsh( )
,
n h(r)
which implies An (P ∩Br ) is bounded. Assume that t, t0 ∈ R+ . Then, for x ∈ Br ∩P ,
we have
(An x)(t) (An x)(t0 )
−
|
1 + v2 (t) 1 + v2 (t0 )
Z ∞
G2 (t, s)
=|
Φ(s)fn (s, x(s), p(s)x0 (s))ds
1
+ v2 (t)
0
Z ∞
G2 (t0 , s)
−
Φ(s)fn (s, x(s), p(s)x0 (s))ds|
0)
1
+
v
(t
2
Z ∞0
G2 (t0 , s)
G2 (t, s)
≤
−
|Φ(s)fn (s, x(s), p(s)x0 (s))ds
|
0)
1
+
v
(t)
1
+
v
(t
2
2
Z0 ∞
G2 (t0 , s)
1 g(r)
G2 (t, s)
≤
−
|Φ(s)k(s)dsh( )
|
1 + v2 (t) 1 + v2 (t0 )
n h(r)
0
|
and
|p(t)(An x)0 (t) − p(t0 )(An x)0 (t0 )|
Z
Z
1
0
=
p(s)Φ(s)fn (s, x(s), p(s)x (s))ds −
t0
t
=
1
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
t0
Z
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
t
≤
t0
Z
t
1 g(r)
.
p(s)Φ(s)k(s)dsh( )
n h(r)
For any ε > 0, T > 0, we can choose δ > 0 small enough such that
|
(An x)(t) (An x)(t0 )
−
| < ε,
1 + v2 (t) 1 + v2 (t0 )
|p(t)(An x)0 (t) − p(t0 )(An x)0 (t0 )| < ε,
(Br ∩P ))(t)
for all x ∈ Br ∩ P , |t − t0 | < δ, t, t0 ∈ [0, T ]. Consequently, { (An1+v
} and
2 (t)
{p(t)(An (Br ∩ P ))0 (t)} is locally equi-continuous on [0, +∞).
20
B. YAN
EJDE-2006/74
Moreover, Lemma 4.3 guarantees that
(Ax)(t) lim
sup t→+∞ x∈P ∩Br 1 + v2 (t)
Z ∞
G2 (t, s)
≤ lim
Φ(s)fn (s, x(s), p(s)x0 (s))ds
t→+∞ 0
1 + v2 (t)
Z ∞
G2 (t, s)
1
x(s)
≤ lim
sup
Φ(s)k(s)g(max{ ,
})ds
t→+∞ x∈P ∩Br 0
1 + v2 (t)
n 1 + v2 (s)
R∞
G2 (t, s)Φ(s)k(s)ds 1 g(r)
h( )
=0
≤ lim 0
t→+∞
1 + v2 (t)
n h(r)
and
lim
sup |p(t)(Ax)0 (t)| ≤ lim
Z
∞
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
t
Z ∞
1
≤ lim
sup
p(s)Φ(s)k(s)g(max{ , x(s)})ds
t→+∞ x∈P ∩Br t
n
Z ∞
1 g(r)
≤ lim
= 0,
p(s)Φ(s)k(s)dsh( )
t→+∞ t
n h(r)
t→+∞ x∈P ∩Br
t→+∞
∩Br ))(t)
which imply that the functions belonging to { (A(P
} and the functions be1+v2 (t)
0
longing to {p(t)(A(P ∩ Br )) (t)} are equi-convergent.
As a result, from Lemma 4.3, An (Br ∩ P ) is relatively compact. The proof is
complete.
Theorem 4.5. Assume that (H1)–(H4) hold. Then (1.2) has at least two positive
solutions.
Proof. From Lemma 4.4, for each n ∈ {1, 2, . . . }, An : P → P is continuous operator
and for any r > 0, An (Br ∩ P ) is relatively compact. From (H2), choose R0 > 0
such that
R0 h(R0 )
R∞
> 1.
(4.3)
p(s)Φ(s)k(s)h(R
0 γ̃2 (s))dsg(R0 )
0
Without loss of the generality, suppose R0 ≥ 1/n. Set
1
Ω1 = {x ∈ C∞
|kxk1 < R0 }.
(4.4)
Then for any x ∈ ∂Ω1 ∩ P , one has
x 6≤ An x,
n ∈ {1, 2, . . . }.
If there exists x0 ∈ ∂Ω1 ∩ P such that x0 ≤ An x0 , obviously,
x0 (t) ≥ γ2 (t)R0 and
1
max{ n
,x0 (t)}
1+v2 (t)
(4.5)
max{ n1 , x0 (t)}
≥ γ̃2 (t)R0 . Then
|x0 (t)|
1
t∈[0,+∞) + v2 (t)
Z ∞
G2 (t, s)
1
≤ sup
Φ(s)f (s, max{ , x0 (s)}, p(s)x0 0 (s))ds
1 + v2 (t)
n
t∈[0,+∞) 0
x0 (s)
Z ∞
1
max{ n , 1+v2 (s) }
≤
p(s)Φ(s)k(s)g(
)ds
1 + v2 (s)
0
R0 =
sup
≥
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
∞
Z
≤
0
x0 (s)
g(max{ n1 , 1+v
})
1
x0 (s)
2 (s)
p(s)Φ(s)k(s)h(max{ ,
})
ds
x
(s)
n 1 + v2 (s) h(max{ 1 , 0
})
n 1+v2 (s)
∞
Z
≤
21
p(s)Φ(s)k(s)h(R0 γ̃2 (s))ds
0
g(R0 )
,
h(R0 )
which implies
R0 h(R0 )
R +∞
0
p(s)Φ(s)k(s)h(R0 γ̃2 (s))dsg(R0 )
≤ 1.
This contradicts (4.3). Then (4.5) is true. From the proof of Theorem 2.7, the
(2.13) is true, which means
i(An , P ∩ Ω1 , P ) = 1,
n ∈ {1, 2, . . . }.
(1)
(1)
(4.6)
(1)
So for any n ≥ 1, there exists an xn ∈ P ∩ Ω1 with An xn = xn .
(1)
(1)
From {xn } ⊆ Ω1 ∩ P , the {xn } is bounded. From Lemma 4.3, the condition
(H4), there exists a ψR0 (t) such that
Z ∞
1
0
x(1)
(t)
=
G2 (t, s)Φ(s)f (s, max{ , x(1)
(s)}, p(s)(x(1)
n
n ) (s))ds
n n
0
Z ∞
(4.7)
≥
G2 (t, s)Φ(s)ψR0 (s)ds
0
≥ γ2 (t)k ∗ ,
R∞
where k ∗ = supt∈[0,+∞) 0
G2 (t,s)
1+v2 (t) Φ(s)ψR0 (s)ds.
Then, from Lemma 4.3, one has
x(1)
n (t) lim sup t→+∞ n≥1 1 + v2 (t)
Z ∞
G2 (t, s)
≤ lim
Φ(s)fn (s, x(s), p(s)x0 (s))ds
t→+∞ 0
1 + v2 (t)
Z ∞
(1)
G2 (t, s)
1 xn (s)
≤ lim sup
Φ(s)k(s)g(max{ ,
})ds
t→+∞ n≥1 0
1 + v2 (t)
n 1 + v2 (s)
R∞
G2 (t, s)Φ(s)k(s)h(k ∗ γ̃2 (s))ds g(R0 )
≤ lim 0
=0
t→+∞
1 + v2 (t)
h(R0 )
(4.8)
and
0
lim sup |p(t)(x(1)
n ) (t)|
t→+∞ n≥1
Z
∞
p(s)Φ(s)fn (s, x(s), p(s)x0 (s))ds
≤ lim sup
t→+∞ n≥1
t
Z
≤ lim sup
t→+∞ n≥1
Z
≤ lim
t→+∞
t
∞
(1)
1 xn (s)
})ds
p(s)Φ(s)k(s)g(max{ ,
n 1 + v2 (s)
∞
p(s)Φ(s)k(s)h(k ∗ γ̃2 (s))ds
t
(4.9)
g(R0 )
h(R0 )
= 0,
x(1) (t)
n
which imply that the functions belonging to { 1+v
} and the functions belonging
2 (t)
(1)
to {p(t)(xn )0 (t)} are equi-convergent. Moreover, for any t0 , t00 ∈ [0, +∞), n ∈
22
B. YAN
EJDE-2006/74
{1, 2, . . . },
(1)
(1)
xn (t0 )
xn (t00 )
−
|
1 + v2 (t0 ) 1 + v2 (t00 )
Z ∞
G2 (t0 , s)
G2 (t00 , s)
(1) 0
≤
|
−
|f (s, x(1)
n (s), p(s)(xn ) (s))ds
0
00 ) n
1
+
v
(t
)
1
+
v
(t
2
2
Z0 ∞
G2 (t0 , s)
G2 (t00 , s)
g(R0 )
≤
|
−
|Φ(s)k(s)h(γ̃2 (s)k ∗ )ds
,
0)
00 )
1
+
v
(t
1
+
v
(t
h(R
2
2
0)
0
|
(4.10)
and
0 0
00
(1) 0 00
|p(t0 )(x(1)
n ) (t ) − p(t )(xn ) (t )|
Z 00
t
(1) 0
≤
p(s)Φ(s)fn (s, x(1)
n (s), p(s)(xn ) (s))ds
t0
t00
≤
Z
t0
g(R0 )
,
p(s)Φ(s)k(s)h(γ̃2 (s)k ∗ )ds
h(R0 )
(4.11)
n ∈ {1, 2, . . . }.
Consequently, for any ε > 0, T > 0, we can choose a δ > 0 such that if |t0 − t00 | < δ
and t0 , t00 ∈ [0, T ], then
(1)
|
(1)
xn (t0 )
xn (t00 )
−
| < ε,
0
1 + v2 (t ) 1 + v2 (t00 )
0 0
00
(1) 0 00
|p(t0 )(x(1)
n ) (t ) − p(t )(xn ) (t )| < ε,
x(1) (t)
(1)
n
for all n ∈ {1, 2, . . . }, which implies that { 1+v
} and {p(t)(xn )0 (t)} are locally
2 (t)
equi-continuous on [0, +∞). Thus, Theorem 4.2 guarantees that there is a con(1)
(1)
(1)
(1)
vergent subsequence {xnj } of {xn } with limj→+∞ xnj = x0 . From (4.7), we
have
(1)
x0 (t) ≥ k ∗ γ2 (t), ∀t ∈ [0, +∞).
Then, for t ∈ (0, +∞), if j is big enough, we have
(1)
(1)
(1) 0
0
|fnj (t, x(1)
nj (t), p(t)(xnj ) (t)) − f (t, x0 (t), p(t)(x0 ) (t))|
(1)
(1)
(1) 0
0
= |f (t, x(1)
nj (t), p(t)(xnj ) (t)) − f (t, x0 (t), p(t)(x0 ) (t))| → 0,
as j → +∞, and
1
(1) 0
(1 + v2 (t)), x(1)
nj (t)}, p(t)(xnj ) (t))
nj
g(R0 )
.
≤ k(t)h(k ∗ γ̃2 (t))
h(R0 )
(1) 0
fnj (t, x(1)
nj (t), p(t)(xnj ) (t)) = f (t, max{
Then the Lesbegue Dominated Convergence Theorem guarantees that
(1)
x0 (t)
= lim x(1)
nj (t)
j→+∞
Z ∞
1
(1) 0
= lim
G2 (t, s)Φ(s)f (s, max{ (1 + v2 (s)), x(1)
nj (s)}, p(s)(xnj ) (s))ds
j→+∞ 0
nj
Z ∞
(1)
(1)
=
G2 (t, s)Φ(s)f (s, x0 (s), p(s)(x0 )0 (s))ds,
0
(4.12)
EJDE-2006/74
MULTIPLE POSITIVE SOLUTIONS
(1)
23
(1)
which implies x0 (t) is a positive solution to (1.2). Obviously, kx0 k1 ≤ R0 . Then
(1)
(4.3) can guarantee kx0 k1 < R0 .
Let τ < a∗ < b∗ < +∞, and 0 < c∗ < mint∈[a∗ ,b∗ ] γ̃2 (t). Suppose
Z b∗
−1
G2 (t, s)
∗
N∗ =
+ 1.
min
Φ(s)k
(s)dsc
1
t∈[a∗ ,b∗ ] a∗ 1 + v2 (t)
From condition (H2), there exists an R0 > R such that
g1 (y) > N ∗ y,
∀y ≥ R0 .
(4.13)
Now we define
R0
1 kxk1 < ∗ }.
Ω2 = {x ∈ C∞
c
R0
0
We might as well suppose that c∗ > 1, R > 1. Then we have
An x 6≤ x,
(4.14)
n ∈ {1, 2, . . . }
(4.15)
for all x ∈ ∂Ω2 ∩ P . Otherwise, suppose there exists x0 ∈ ∂Ω2 ∩ P with An x0 ≤ x0 .
Since x0 ∈ ∂(Ω2 ) ∩ P ,
min
∗ ∗
t∈[a ,b
0
|x(t)|
x0 (t)
∗R
≥ min
γ̃
(t)
sup
>
c
= R0 > 1.
2
c∗
] 1 + v2 (t)
t∈[a∗ ,b∗ ]
t∈[0,+∞) 1 + v2 (t)
Then for t ∈ [a∗ , b∗ ], from (4.13), we have
x0 (t)
(An x0 )(t)
≥
1 + v2 (t)
1 + v2 (t)
Z +∞
G2 (t, s)
=
Φ(s)fn (s, x0 (s), p(s)x00 (s))ds
1 + v2 (t)
0
Z b∗
G2 (t, s)
1
≥
Φ(s)f (s, max{ (1 + v2 (s)), x0 (s)}, p(s)x00 (s))ds
1
+
v
(t)
n
∗
2
a
Z b∗
1
x0 (s)
G2 (t, s)
Φ(s)k1 (s)g1 (max{ ,
})ds
≥
1
+
v
(t)
n
1
+
v2 (s)
∗
2
a
Z b∗
G2 (t, s)
x0 (s)
>
Φ(s)k1 (s)N ∗
ds
1 + v2 (s)
a∗ 1 + v2 (t)
Z b∗
R0
>
G2 (t, s)Φ(s)k1 (s)dsN ∗ c∗ ∗
c
a∗
0
R
> ∗,
c
which implies kx0 k1 > R0 /c∗ . This contradicts to x0 ∈ P ∩ ∂Ω2 . Then (4.15) is
true.
(2)
From (4.5) and (4.15), Theorem 2.7 guarantees that An has a fixed point xn ∈
(2)
(2)
(Ω2 − Ω1 ) ∩ P . For the set {xn }, since kxn k1 = supt∈[0,+∞)
guarantees that there is a ψR0 /c∗ (t) such that
f (t, x, z) = F (t, y, z) ≥ ψ R∗0 (t),
c
∀(t, y, z) ∈ [0, +∞) × (0,
|x(2)
n (t)|
1+v2 (t)
≤
R0
c∗ ,
(H4)
R0
] × [0, +∞). (4.16)
c∗
(2)
By proof as in (4.7), (4.8), (4.9), (4.10) and (4.11), we can prove that {xn } is
(2)
1
relatively compact in C∞
, which means that there exists a subsequence {xni } of
24
B. YAN
(2)
(2)
EJDE-2006/74
(2)
(2)
{xn } with limi→+∞ xni = x0 . By proof as in (4.12) x0 (t) is a positive solution
0
(2)
to equation (1.2) with R
c∗ > kx0 k1 > R0 .
(1)
Consequently, equation (1.2) has at least two different positive solutions x0 (t)
(2)
and x0 (t). The proof is complete.
Example 4.6. Now we consider
2
x00 +
1
x0
1 −t −1/4
3
x
) = 0, t ∈ (0, +∞)
e t
((1 + t)1/2 x−1/2 +
)(1
+
16
(1 + t)3
1 + x0 2
x(0) = 0,
lim x0 (t) = 0.
t→+∞
(4.17)
Then, equation (4.17) has at least two positive solutions.
To prove that (4.17) has at least two positive solutions, we apply Theorem 4.5
1 −t −1/4
1
x02
3
with Φ(t) = 16
e t
, p(t) ≡ 1, f (t, x, z) = ((1+t)1/2 x−1/2 + (1+t)
3 x )(1+ 1+x0 2 ),
(
t, t ∈ [0, 1]
k(t) ≡ 1, g(y) = 2(y −1/2 + y 3 ), h(x) = y −1/2 , γ1 (t) =
, γ̃2 (t) =
1, t ∈ (0, +∞)
γ2 (t)
1+t ,
k1 (t) ≡ 1, g1 (y) = y −1/4 + y 3 , Ψc (t) = c−1/4 . It is easy to verify that
(H1)–(H4) hold. Hence, (4.17) has at leat two positive solutions.
Acknowledgements. The author wants to thanks Professor Temoney for his guidance, and the anonymous referee for his/her suggestions.
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Baoqiang Yan
Department of Mathematics, Shandong Normal University, Ji-nan, 250014 China
and School of Mathematics, Trinity College, Dublin University, Dublin 2, Ireland
E-mail address: bqyan819@beelink.com