Indian J. pure appl. Math., 38(3): 163-184, June 2007
c Printed in India.
°
A NECESSARY AND SUFFICIENT CONDITION FOR THE EXISTENCE OF C 4n−1 [0, 1]
POSITIVE SOLUTIONS OF HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY
VALUE PROBLEMS∗
C HENGLONG Z HAO
Department of Mathematics and Physics, Jinan Engineering Vocational Technical College,
Shandong, 250200, Peoples’ Republic of China
E-mail: [email protected]
AND
YANSHENG L IU
Institute of Science of Mathematics, Shandong Normal University,
Jinan, 250014, Peoples’ Republic of China
E-mail: [email protected]
(Received 10 April 2006; after final revision 18 October 2006; accepted 6 December 2006)
This paper investigates the existence of positive solutions to a class of higher order singular
sublinear boundary value problems. Some necessary and sufficient conditions for the existence
of C 4n−1 [0, 1] positive solutions are presented. In order to illustrate the applications of our
main results, some examples are presented.
Key Words : Singular Sublinear Boundary Value Problem; Positive Solution; Fixed Point
Theorem; Cone; Higher Order Differential Equation
1. I NTRODUCTION AND P RELIMINARY
Singular boundary value problems (SBVP) for ordinary differential equations arise in the field of
gas dynamics, fluid mechanics, the theory of boundary layer and so on. Meanwhile, SBVP is an
important branch in the field of differential equations (see [1–12, 14, 15]) as well. In recent years,
the positive solutions of SBVP to higher order nonlinear differential equations have been studied
extensively (see [4–11, 14, 15]).
∗
The Project Supported by NNSF of P. R. China (10571111), China Scholarship Council and Natural Science Foundation of Shandong Province (Y2006A22).
164
CHENGLONG ZHAO AND YANSHENG LIU
For instance, in superlinear case, Shi and Chen [4] obtained some necessary and sufficient conditions for the existence of C 2 [0, 1] or C 3 [0, 1] positive solutions of differential equations under some
given conditions. Wei [6] investigated the existence of positive solutions to the singular superlinear
boundary value problem

(4)

0 < t < 1;

 x (t) = f (t, x(t)),

 x(0) = x(1) = 0;

ax00 (0) − bx000 (0) = 0;



 cx00 (1) + dx000 (1) = 0,
where f, a, b, c, d satisfy
(A1 )
a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, a + b > 0, c + d > 0, 4 = ac + ad + bc > 0.
(A2 )
f ∈ C[(0, 1) × [0, +∞), [0, +∞)], f (t, t(1 − t)) > 0, t ∈ (0, 1),
and there exist constants λ, µ (1 < λ ≤ µ < ∞) such that for ∀ t ∈ (0, 1), x ∈ (0, ∞),
cµ f (t, x) ≤ f (t, cx) ≤ cλ f (t, x),
0 ≤ c ≤ 1;
cλ f (t, x) ≤ f (t, cx) ≤ cµ f (t, x),
c ≥ 1,
here λ > 1 means f (t, x) is superlinear on x. In the literature, some necessary and sufficient
conditions for the existence of C 2 positive solutions as well as C 3 positive solutions are given by
means of the fixed point theorem on cone.
In sublinear case, Wei [5] gave some necessary and sufficient conditions for the existence of
2
C positive solutions as well as C 3 positive solutions by means of the method of lower and upper
solutions with the maximum principle for SBVP:
(
x(4) (t) = f (t, x(t)),
0 < t < 1,
00
00
x(0) = x(1) = x (0) = x (1) = 0.
In [8], Zhang and Sun discussed the following boundary value problem:
 (4n)

(t) = f (t, x(t)),
 x


0 < t < 1,
∗
x(2k) (0)
=
x(2k) (1)
= 0, k = 0, 1, 2, . . . , 2n − 1,
by using the method of lower and upper solutions, where f ∈ C[(0, 1)×(0, +∞), [0, +∞)], f (t, x) 6≡
0, is quasi-homogeneous with respect to x, −∞ < λ ≤ 0 ≤ µ < 1, 2(µ−λ)
1+µ < 1. The main results
of [8] are the following Theorem A and Theorem B.
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
165
Theorem A — SBVP(∗) has a C 4n−2 positive solution if and only if
Z
1
0<
t(1 − t)f (t, t(1 − t))dt < +∞,
0
Z
1
lim t
(1 − s)f (s, s(1 − s))ds = 0,
t→0+
t
Z
lim (1 − t)
t→1−
t
sf (s, s(1 − s))ds = 0.
0
Theorem B — SBVP(∗) has a C 4n−1 positive solution if and only if
Z
0<
1
f (t, t(1 − t))dt < +∞.
0
When n = 1, Theorem A and Theorem B are the results of [4].
In paper [14], the singular fourth-order boundary value problem
(
y (4) (t) = p(t)y λ (t), 0 < t < 1, λ ∈ (0, 1),
y(0) = y(1) = y 0 (0) = y 0 (1) = 0
was studied via upper and lower solutions methods combined with the schauder fixed point theorem.
The authors established necessary and sufficient conditions and a sufficient condition.
Furthermore, in paper [15], the authors studied the existence of positive solutions of the differential equation (−1)m y (2m) (t) = f (t, y(t), y 00 (t), · · · , y (2(m−1)) (t)) with the boundary condition
y (2i) (0) = 0 = y (2i) (1), 0 ≤ i ≤ m − 1, and y (2i) (0) = 0 = y (2i+1) (1), 0 ≤ i ≤ m − 1 . They obtained sufficient conditions guaranteeing the existence of positive solutions if f is either superlinear
or sublinear by using fixed point theory.
Inspired by above papers, this paper investigates the following boundary value problem:


u(4n) (t) = f (t, u(t), u(4n−2) (t)),
0 < t < 1,



 u(0) = u(1) = 0,
(1.1)
(2k) (0) − bu(2k+1) (0) = 0,

R
(u)
=
au
1



 R (u) = cu(2k) (1) + du(2k+1) (1) = 0, k = 1, 2, . . . , 2n − 1.
2
where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, 4 = ac + ad + bc > 0, and f ∈ C[(0, 1) × (0, +∞) ×
(−∞, 0], [0, +∞)], f is quasi-homogeneous with respect to the last two variables, that is, there
are constants λ, µ, α, β; N1 , M1 , N2 , M2 with −∞ < λ ≤ 0 ≤ µ < +∞, 0 ≤ α ≤ β < 1,
µ + β < 1; 0 < N1 ≤ 1 ≤ M1 , 0 < N2 ≤ 1 ≤ M2 such that for all 0 < t < 1, u > 0, v ≤ 0
satisfying
(H1 )
c̄µ f (t, u, v) ≤ f (t, c̄u, v) ≤ c̄λ f (t, u, v),
0 < c̄ ≤ N1 ,
166
CHENGLONG ZHAO AND YANSHENG LIU
c̄λ f (t, u, v) ≤ f (t, c̄u, v) ≤ c̄µ f (t, u, v),
c̄ ≥ M1 ;
c̄β f (t, u, v) ≤ f (t, u, c̄v) ≤ c̄α f (t, u, v),
0 < c̄ ≤ N2 ,
c̄α f (t, u, v) ≤ f (t, u, c̄v) ≤ c̄β f (t, u, v),
c̄ ≥ M2 .
A typical function is f (t, u, v) =
n
P
i=1
pi (t)uαi (−v)βi , where pi (t) ∈ C[(0, 1), R+ ], λ = α1 ≤
α2 ≤ · · · ≤ αk < 0 < αk+1 ≤ · · · ≤ αn = µ, 0 ≤ βi < 1, k = 1, 2, · · · , n − 1, i = 1, 2, · · · , n.
Meanwhile, we also consider the following special case of (1.1):


u(4n) (t) = f (t, u(t)),
0 < t < 1,



 u(0) = u(1) = 0,
 R1 (u) = au(2k) (0) − bu(2k+1) (0) = 0,



 R (u) = cu(2k) (1) + du(2k+1) (1) = 0, k = 1, 2, . . . , 2n − 1.
2
(1.1’)
We obtain some necessary and sufficient conditions for the existence of C 4n−1 positive solutions
of SBVP(1.1) and (1.1’). In the remark of the paper, the results of (1.1) can be extended. It is
remarkable that this work is a proceeding one of reference [11]. So the similar techniques are
employed. Comparing with the previous works, the main features here are as follows. Firstly, this
paper is concerned with the existence of C 4n−1 positive solutions while [11] is on C 4n−2 positive
solutions. Secondly, the nonlinear term f is more extensive and may include u00 , u(4) , · · · , u(4n−2) .
Thirdly, the singularity of f on u is arbitrary and the boundary value conditions are more extensive
than previous works. Finally, the main results here are more general and more complicated than
other works such as in reference [4] and [8], which studied the existence of C 2 and C 3 positive
solutions. At the same time, the relations between the conditions (2.1)–(2.7) and the Green functions
subordinating to the homogeneous problem of SBVP(1.1) are revealed. 88
T
We say u ∈ C 4n−2 [0, 1] C 4n (0, 1) is a C 4n−2 [0, 1] positive solution of SBVP(1.1) if and only
if u(t) satisfies (1.1) and u(t) > 0 for t ∈ (0, 1). In addition, if u is a C 4n−2 [0, 1] positive solution
of SBVP(1.1), and both u4n−1 (0+ ) and u4n−1 (1− ) exist, then u is said to be a C 4n−1 [0, 1] positive
solution of SBVP(1.1).
Let E = {u ∈ C 4n−2 [0, 1] : u(0) = u(1) = 0}, and define the norm
kuk = max{kuk0 , kuk4n−2 }, ∀ u ∈ E,
where
kuk0 = sup |u(t)|,
0≤t≤1
kuk4n−2 = sup |u(4n−2) (t)|,
∀u ∈ E.
0≤t≤1
Then (E, k · k) is a Banach space.
For convenience, at the end of this section we list the following Lemmas which will be used in
section 2.
Lemma 1.1 ([13], P. 22) — Let K be a cone of real Banach space E, Ω1 , Ω2 be bounded open
T
sets of E, θ ∈ Ω1 ⊂ Ω2 . Suppose that A : K (Ω2 \ Ω1 ) → K is completely continuous such that
one of the following two conditions is satisfied:
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
(i) kAxk ≤ kxk, x ∈ K
or
T
∂Ω1 ; kAxk ≥ kxk, x ∈ K
T
167
∂Ω2 ,
T
T
(ii) kAxk ≤ kxk, x ∈ K ∂Ω2 ; kAxk ≥ kxk, x ∈ K ∂Ω1 .
T
Then A has a fixed point in K (Ω2 \ Ω1 ).
Lemma 1.2 ([11], P.3) Suppose (H1 ) holds.
(1) If b = d = 0, then SBVP(1.1) has a C 4n−2 [0, 1] positive solution if and only if
Z 1
0<
t(1 − t)f (t, t(1 − t), −1)dt < +∞;
(1.2’)
0
Z
1
lim t
(1 − s)f (s, s(1 − s), −1)ds = 0;
t→0+
(1.3’)
t
Z
t
lim (1 − t)
sf (s, s(1 − s), −1)ds = 0.
t→1−
(1.4’)
0
T
(2) If b = 0, d > 0, then SBVP(1.1) has a C 4n−2 [0, 1] C 4n−1 (0, 1] positive solution if and
only if
Z 1
0<
tf (t, t(1 − t), −1)dt < +∞,
(1.5’)
0
Z
1
lim t
f (s, s(1 − s), −1)ds = 0.
t→0+
(1.6’)
t
T
(3) If b > 0, d = 0, then SBVP(1.1) has a C 4n−2 [0, 1] C 4n−1 [0, 1) positive solution if and
only if
Z 1
0<
(1 − t)f (t, t(1 − t), −1)dt < +∞,
0
Z
t
lim (1 − t)
t→1−
f (s, s(1 − s), −1)ds = 0.
0
From the proof of Lemma 1.2 (see Theorem 2.1-Theorem 2.3 in [11, P.6–17], it is easy to see
the following Lemma holds.
Lemma 1.3 — Suppose (H1 ) holds. SBVP(1.10 ) has a C 4n−2 positive solution if and only if
Z 1
0<
t(1 − t)f (t, t(1 − t))dt < +∞,
0
Z
lim t
t→0+
1
(1 − s)f (s, s(1 − s))ds = 0,
t
Z
lim (1 − t)
t→1−
t
sf (s, s(1 − s))ds = 0.
0
168
CHENGLONG ZHAO AND YANSHENG LIU
2. M AIN R ESULTS
For convenience, we first give the the following condition.
(H2 ) f (t, u, v) is nondecreasing with respect to the last variable.
Our main results are as follows.
Theorem 2.1 — Suppose (H1 ) holds, and b = d = 0 in (1.1). Then SBVP(1.1) has a C 4n−1 [0, 1]
positive solution if and only if
Z
1
0<
f (t, t(1 − t), −t(1 − t))dt < +∞.
(2.1)
0
Theorem 2.2 — Suppose (H1 ) and (H2 ) hold, and b = c = 0 in (1.1). Then SBVP(1.1) has a
positive solution if and only if
C 4n−1 [0, 1]
Z
1
0<
f (t, t(1 − t), −t)dt < +∞.
(2.2)
0
Theorem 2.20 — Suppose (H1 ) holds, and b = 0, d > 0 in (1.1). Then
Z
1
0<
f (t, t(1 − t), −t)dt < +∞
(2.3)
0
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP(1.1).
Theorem 2.3 — Suppose (H1 ) and (H2 ) hold, and a = d = 0 in (1.1). Then SBVP(1.1) has a
positive solution if and only if
C 4n−1 [0, 1]
Z
1
f (t, t(1 − t), −(1 − t))dt < +∞.
0<
(2.4)
0
Theorem 2.30 — Suppose (H1 ) holds, and b > 0, d = 0 in (1.1). Then
Z
1
0<
f (t, t(1 − t), −(1 − t))dt < +∞
(2.5)
0
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP (1.1).
Theorem 2.4 — Suppose (H1 ) holds. If
Z
0<
1
f (t, t(1 − t), −1)dt < +∞,
(2.6)
0
then SBVP (1.1) has a C 4n−1 [0, 1] positive solution.
Theorem 2.5 — Suppose (H1 ) holds. Then SBVP (1.1’) has a C 4n−1 [0, 1] positive solution if
and only if
Z 1
0<
f (t, t(1 − t))dt < +∞.
(2.7)
0
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
It is well known that
1
G(t, s) =
4
(
(b + as)[d + c(1 − t)], s < t;
(b + at)[d + c(1 − s)], t ≤ s,
is the Green function of homogeneous boundary value problem

00

0 ≤ t ≤ 1,
 −u (t) = 0
0
au(0) − bu (0) = 0,


cu(1) + du0 (1) = 0.
169
(2.8)
(2.9)
Furthermore, it is easy to see that
Since
G(t, s) ≥
[c(1 − s)(b + as) + ads][c(1 − t)(b + at) + adt]
,
42
G(t, s) ≤
(b + as)[d + c(1 − s)]
.
4



















(2.10)
(b + as)[d + c(1 − t)]
, τ < s < t;
(b + aτ )[d + c(1 − s)]
d + c(1 − t)
, s ≤ t, τ ;
d + c(1 − τ )
G(t, s)
=
G(τ, s) 


b + at


, t, τ ≤ s;


b + aτ








(b + at)[d + c(1 − s)]


, t < s < τ,

(b + as)[d + c(1 − τ )]
we know
G(t, s) ≥ e(t)G(τ, s),
where
e(t) =
(b + at)[d + c(1 − t)]
.
(b + a)(c + d)
(2.11)
(2.12)
It follows from (2.8) that some special Green functions of different homogeneous boundary
value problems corresponding to (2.9) are
(
s(1 − t), s < t;
G1 (t, s) =
(b = 0, d = 0)
t(1 − s), t ≤ s,
(
s[d + c(1 − t)], s < t;
1
G2 (t, s) =
(b = 0, d > 0)
c+d
t[d + c(1 − s)], t ≤ s,
(
(b + as)(1 − t), s < t;
1
G3 (t, s) =
(b > 0, d = 0)
a+b
(b + at)(1 − s), t ≤ s,
170
CHENGLONG ZHAO AND YANSHENG LIU
(
s, s < t;
t, t ≤ s,
G4 (t, s) =
(
(b = 0, c = 0)
(1 − t), s < t;
(1 − s), t ≤ s,
G5 (t, s) =
(a = 0, d = 0)
(2.13)
Proof of Theorem 2.1 —
Sufficiency — Suppose (2.1) holds. Let c be a number such that c ≥ M2 and
Noticing that tβ (1 − t)β ≥ t(1 − t) for β < 1, then for t ∈ (0, 1), we have
f (t, t(1 − t), −t(1 − t)) = f (t, t(1 − t), −c ·
t(1−t)
c
≤ N2 .
t(1 − t)
)
c
t(1 − t) β
) f (t, t(1 − t), −1)
c
= cα−β tβ (1 − t)β f (t, t(1 − t), −1)
≥ cα (
≥ cα−β t(1 − t)f (t, t(1 − t), −1).
(2.14)
It is easy to see that (2.1) implies (1.2)–(1.4). Consequently, if (2.1) holds, then by Lemma 1.2,
we know SBVP (1.1) (b = d = 0) has a C 4n−2 positive solution u∗ (t).
T
Now we show u∗ (t) ∈ C 4n−1 [0, 1] C 4n (0, 1). By the proof of Theorem 2.1-Theorem 2.3 in
[11], we can obtain that there exist 0 < r1 < r < 1 < R < r2 such that
l12n−2
t(1 − t)R,
2
k1 t(1 − t)r ≤ u∗ (t) ≤
r1 t(1 − t) ≤ −u∗ (4n−2) (t) ≤ r2 t(1 − t),
∀ t ∈ [0, 1],
n
o
l2n−2 R
where the definition of k1 , l1 are the same as the definition in [11, p.5]. Choose c1 ≥ max M1 , 12N1
o
n
and c2 ≥ max M2 , Nr22 . Then we have
Z
0
Z
1
|u∗
(4n)
1
(t)|dt =
0
Z
f (t, u∗ (t), u∗ (4n−2) (t))dt
=
f
0
Z
1
≤
0
Z
≤
0
1
!
−u∗ (4n−2) (t)
u∗ (t)
t(1 − t), c2
(−t(1 − t)) dt
t, c1
c1 t(1 − t)
c2 t(1 − t)
Ã
1
µ
cµ1
µ
cµ1
u∗ (t)
c1 t(1 − t)
k1 r
c1
¶λ
¶λ
µ
cβ2
Z
= k1λ cµ−λ
cβ−α
rλ r2α
1
2
Ã
cβ2
r2
c2
−u∗ (4n−2) (t)
c2 t(1 − t)
!α
f (t, t(1 − t), −t(1 − t))dt
¶α
f (t, t(1 − t), −t(1 − t))dt
1
f (t, t(1 − t), −t(1 − t))dt < +∞.
0
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
171
(4n)
Namely, u∗ (t) is absolutely integrable on [0, 1]. This guarantees u∗ (t) ∈ C 4n−1 [0, 1]. Therefore, u∗ (t) is a C 4n−1 [0, 1] positive solution of SBVP (1.1).
Necessity — Let u(t) be a C 4n−1 [0, 1] positive solution of SBVP(1.1)(b = d = 0). Then
u(4n−1) (0+ ) ≤ 0 and u(4n−1) (1− ) ≥ 0. Also, the proof similar to the Lemma 1.2 (see Theorem
2.1-Theorem 2.3 in [11]) guarantees that there exist positive numbers m1 , m2 , q1 , q2 with m1 <
1 < m2 and q1 < 1 < q2 satisfying
m1 t(1 − t) ≤ u(t) ≤ m2 t(1 − t),
q1 t(1 − t) ≤ −u(4n−2) (t) ≤ q2 t(1 − t),
∀ t ∈ [0, 1].
1
1
} and c2 ≤ min{N2 ,
}. Then
M1 m2
M2 q2
¶µ µ
¶β
Z 1
Z 1 µ
t(1 − t)
λ t(1 − t)
α
f (t, t(1 − t), −t(1 − t))dt ≤
c1
c2
f (t, u(t), u(4n−2) (t))dt
c1 u(t)
−c2 u(4n−2) (t)
0
0
Choose positive numbers c1 ≤ min{N1 ,
µ
≤
cλ1
1
c1 m1
¶µ
µ
cα2
1
c2 q1
¶β Z
1
f (t, u(t), u(4n−2) (t))dt
0
−β (4n−1) −
= cλ−µ
cα−β
m−µ
(1 ) − u(4n−1) (0+ )] < +∞.
1
2
1 q1 [u
This guarantees (2.1) holds.
¤
Proof of Theorem 2.2 —
Sufficiency — Similar to (2.14), we get
f (t, t(1 − t), −t) ≥ cα−β tf (t, t(1 − t), −1).
R1
R1
Consequently, if 0 f (t, t(1 − t), −t)dt < ∞, we have 0 tf (t, t(1 − t), −1)dt < ∞. This means
that (2.2) implies (1.5)–(1.6).
So by Lemma 1.2 SBVP(1.1) (b = c = 0) has a C 4n−2 positive solution u∗ (t).
T
In the following, we show u∗ (t) ∈ C 4n−1 [0, 1] C 4n (0, 1). By the proof of Lemma 1.2 (see
Theorem 2.1 -Theorem 2.3 in [11]), we know
l22n−2
t(1 − t)ku∗ k,
2
where the definition of k2 and l2 is the same as the one in [11, p.5].
Moreover, since −u∗ (4n−2) (t) is concave and nondecreasing function, it is easy to prove
k2 t(1 − t)ku∗ k ≤ u∗ (t) ≤
−u∗ (4n−2) (1)t ≤ −u∗ (4n−2) (t) ≤ ku∗ k.
½
¾
½
¾
1
k2 ku∗ k
N2
,
Choose positive numbers c1 ≤ min N1 ,
and c2 ≤ min
.
M1
−u∗ (4n−2) (1) M2
Noticing (H2 ), we get
Z 1
Z 1
|u∗ (4n) (t)|dt =
f (t, u∗ (t), u∗ (4n−2) (t))dt
0
0
172
CHENGLONG ZHAO AND YANSHENG LIU
Z
1
=
f
0
Z
1
≤
0
Z
≤
µ
t, c1
µ
cµ1
¶
u∗ (t)
1
t(1 − t), (−c2 u∗ (4n−2) (t))(−t) dt
c1 t(1 − t)
c2
u∗ (t)
c1 t(1 − t)
1
k2 ku∗ k λ
cµ1 (
)
c1
0
¶λ µ
µ
1
c2
1
c2
¶β
¶β
(−c2 u∗ (4n−2) (1))α f (t, t(1 − t), −t)dt
(−c2 u∗ (4n−2) (1))α f (t, t(1 − t), −t)dt
= k2λ cµ−λ
cα−β
ku∗ kλ (−u∗ (4n−2) (1))α f (t, t(1 − t), −t)dt < +∞.
1
2
This implies u∗ (t) ∈ C 4n−1 [0, 1]. Therefore, u∗ (t) is a C 4n−1 [0, 1] positive solution of SBVP (1.1).
Necessity — Let u(t) be a C 4n−1 [0, 1] positive solution of SBVP(1.1)(b = c = 0). Then
u(4n−1) (0+ ) ≤ 0 and −u(4n−2) (t) is concave and nondecreasing function. It is easy to prove
−u(4n−2) (1)t ≤ −u(4n−2) (t) ≤ kuk.
And the proof of Lemma 1.2 (see Theorem 2.1-Theorem 2.3 in [11]) guarantees that there exist
positive numbers m1 , m2 with 0 < m1 < 1 < m2 such that
m1 t(1 − t) ≤ u(t) ≤ m2 t(1 − t),
Choose positive numbers c1 ≤ min{N1 ,
Z
Z
1
f (t, t(1 − t), −t)dt ≤
0
0
1
≤
1
1
} and c2 ≤ min{N2 ,
}. Then
(4n−2)
M1 m2
−u
(1)M2
¶
µ
t
t(1 − t)
(4n−2)
u(t), c2
u
(t) dt
f t, c1
c1 u(t)
−c2 u(4n−2) (1)
µ
cλ1
∀ t ∈ [0, 1].
1
c1 m1
¶µ
µ
cα2
1
c2 (−u(4n−2) (1))
¶β Z
1
f (t, u(t), u(4n−2) (t))dt
0
(4n−2)
= cλ−µ
cα−β
m−µ
(1))−β (−u(4n−1) (0)) < +∞.
1
2
1 (−u
This guarantees (2.2) holds.
¤
Proof of Theorem 2.20 :
Let u(t) be a C 4n−1 [0, 1] positive solution of SBVP(1.1)(b = 0, d > 0). Then u(4n−1) (0+ ) ≤ 0
and −u(4n−2) (t) is concave and nondecreasing function. It is easy to prove that there exists a
constant Ru such that
−u(4n−2) (1)t ≤ −u(4n−2) (t) ≤ Ru t.
By a similar process to the proof of necessity of Theorem 2.2, there exist positive numbers m1 , m2
with 0 < m1 < 1 < m2 such that
m1 t(1 − t) ≤ u(t) ≤ m2 t(1 − t), ∀ t ∈ [0, 1].
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
½
Choose positive numbers c1 ≤ min N1 ,
Z
Z
1
1
f (t, t(1 − t), −t)dt ≤
0
0
1
M1 m2
¾
1
}. Then
Ru M2
and c2 ≤ min{N2 ,
µ
¶
t(1 − t)
t
(4n−2)
f t, c1
u
(t) dt
u(t), c2
c1 u(t)
−c2 u(4n−2) (t)
µ
≤ cλ1
1
c1 m1
¶µ
µ
cα2
¶β Z
1
−c2 u(4n−2) (1)
1
f (t, u(t), u(4n−2) (t))dt
0
(4n−2)
= cλ−µ
cα−β
m−µ
(1))−β (−u(4n−1) (0+ )) < +∞.
1
2
1 (−u
This implies (2.3) holds.
¤
Proof of Theorem 2.3 —
The proof is very similar to Theorem 2.2. So it is omitted.
Proof of Theorem 2.30 —
Similar to the proof of Theorem 2.20 , it is easy to see that Theorem 2.30 holds.
Proof of Theorem 2.4 —
In the following, the proof of Theorem 2.4 will be divided into four cases: 1) b = d = 0;
2) b = 0, d > 0; 3) b > 0, d = 0; 4) b > 0, d > 0.
Case 1) b = d = 0.
Obviously, (2.6) implies (1.2)–(1.4). Consequently, by Lemma 1.2, we get that SBVP(1.1)
(b = d = 0) has a C 4n−2 positive solution u∗ (t).
Now we show u∗ (t) ∈ C 4n−1 [0, 1]. By the proof of Lemma 1.2 (see Theorem 2.1-Theorem 2.3
in [11]), we know
k1 t(1 − t)ku∗ k ≤ u∗ (t) ≤
l12n−2
t(1 − t)ku∗ k,
2
e1 (t)ku∗ k ≤ −u∗ (4n−2) (t) ≤ ku∗ k,
∀ t ∈ [0, 1],
1
where k1 = 302n−1
, l1 = 16 , e1 (t) = t(1 − t) are the same as the ones in [11, P.5]. Choose
n
o
o
n
ku∗ k
ku∗ k
c1 ≤ min N1 , 302n−1
. Then we get
and
c
≥
max
M
,
2
2 N2
M1
Z
0
Z
1
|u∗
(4n)
1
(t)|dt =
0
Z
f (t, u∗ (t), u∗ (4n−2) (t))dt
Ã
1
=
f
0
Z
≤
0
1
!
u∗ (t)
−u∗ (4n−2) (t)
t, c1
t(1 − t), c2
(−1) dt
c1 t(1 − t)
c2
µ
cλ1
u∗ (t)
c1 t(1 − t)
µ
≤
cλ−µ
cβ−α
1
2
l12n−2
2
¶µ
Ã
cβ2
−u∗ (4n−2) (t)
c2
¶µ
Z
ku∗ kµ+α
!α
f (t, t(1 − t), −1)dt
1
f (t, t(1 − t), −1)dt < +∞.
0
173
174
CHENGLONG ZHAO AND YANSHENG LIU
This guarantees u∗ (t) ∈ C 4n−1 [0, 1]. Therefore, u∗ (t) is a C 4n−1 [0, 1] positive solution of SBVP
(1.1).
Case 2) b = 0, d > 0 and Case 3) b > 0, d = 0.
The proofs are very similar to Case 1).
Case 4) b > 0, d > 0.
Let E = {u ∈ C 4n−2 [0, 1] : u(0) = u(1) = 0} with norm kuk4n−2 = sup |u(4n−2) (t)| for
0≤t≤1
u ∈ E. Then (E, k · k4n−2 ) is a Banach space. Evidently, if un , u ∈ E and kun − uk4n−2 → 0 as
n → ∞, it follows that kun − uk0 → 0, where |kun − uk0 = max |un (t) − u(t)|.
0≤t≤1
Define
P = {u ∈ E : R1 (u) = R2 (u) = 0,
u(t) ≥ 0, u(4n−2) (t) ≤ e(t)u(4n−2) (s) ≤ 0,
u(t) ≥ −kt(1 − t)u(4n−2) (s)}.
where e(t) is given by (2.12), and
k =
(2ac + 5bc + 5ad)(15abcd + 15b2 cd + 15abd2 + 10b2 c2 + 5abc2 + 5a2 cd + a2 c2 + 10a2 d2 )
1800(a + b)(c + d)
µ
×
5abc2 + 10b2 c2 + 10abcd + a2 c2 + 10a2 d2 + 5a2 cd
30
¶2n−3
·
1
.
44n−4
(2.15)
It is easy to see that P is a cone of E. From
Z
u(t)
=
···
0
Z
0
···
0
G1 (t, s2n−1 )G(s2n−1 , s2n−2 ) · · · G(s2 , s1 )(−u(4n−2) (s1 ))ds1 · · · ds2n−1
Z
G1 (t, s2n−1 )ds2n−1
0
Z
1
1
≤
=
Z
1
1
0
1
(b + as2n−2 )[d + c(1 − s2n−2 )]
ds2n−2
4
(b + as1 )[d + c(1 − s1 )]
ds1 · kuk4n−2
4
l2n−2
t(1 − t)kuk4n−2 ,
2
(2.16)
where
ac + 3ad + 3bc + 6bd
.
64
For fixed u ∈ P , by (2.16) together with (2.16), we have
l=
kt(1 − t)kuk4n−2 ≤ u(t) ≤
(2.17)
l2n−2
t(1 − t)kuk4n−2 ,
2
e(t)kuk4n−2 ≤ −u(4n−2) (t) ≤ kuk4n−2 ,
(2.18)
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
175
where k and l are defined by (2.15) and (2.17), respectively.
Suppose (2.6) holds. Then SBVP(1.1) has a C 4n−1 [0, 1] positive solution if and only if u is the
positive solution of the following integral equation
Z 1
u(t) = (Au)(t) =
h(t, s)f (s, u(s), u(4n−2) (s))ds, ∀ u ∈ P \ {θ},
0
where
Z
Z
1
h(t, s) =
···
0
0
1
G1 (t, s2n−1 )G(s2n−1 , s2n−2 ) · · · G(s2 , s1 )G(s1 , s)ds1 · · · ds2n−1 ;
and G1 (t, s) and G(t, s) are given by (2.13) and (2.8), respectively. For u ∈ P \ {θ}, it follows
from (2.11) and (2.12) that
Z 1
(Au)(4n−2) (t) = −
G(t, τ )f (τ, u(τ ), u(4n−2) (τ ))dτ
0
Z
≤ −e(t)
1
G(s, τ )f (τ, u(τ ), u(4n−2) (τ ))dτ
0
= e(t)(Au)(4n−2) (s) ≤ 0,
∀ t, s ∈ [0, 1].
This together with (2.10) implies
Z 1
Z 1
(Au)(t) =
···
G1 (t, s2n−1 )G(s2n−1 , s2n−2 )
0
0
· · · G(s2 , s1 )(−(Au)(4n−2) (s1 ))ds1 ds2 · · · ds2n−1
Z
Z
1
≥
0
t(1 − t)s2n−1 (1 − s2n−1 )ds2n−1
1
0
[c(1 − s2n−1 )(b + as2n−1 ) + ads2n−1 ][c(1 − s2n−2 )(a + bs2n−2 ) + ads2n−2 ]
ds2n−2
42
Z
1
[c(1 − s3 )(a + bs3 ) + ads3 ][c(1 − s2 )(a + bs2 ) + ads2 ]
···
ds2
42
0
Z 1
[c(1 − s2 )(a + bs2 ) + ads2 ][c(1 − s1 )(a + bs1 ) + ads1 ]
·
· (Au)(4n−2) (s1 )ds1
2
4
0
≥ −t(1 − t)
(2ac + 5bc + 5ad)(15abcd + 15b2 cd + 15abd2 + 10b2 c2 + 5abc2 + 5a2 cd + a2 c2 + 10a2 d2 )
1800(a + b)(c + d)
µ
·
5abc2 + 10b2 c2 + 10abcd + a2 c2 + 10a2 d2 + 5a2 cd
30
= −kt(1 − t)(Au)(4n−2) (s).
¶2n−3
·
1
44n−4
(Au)(4n−2) (s),
176
CHENGLONG ZHAO AND YANSHENG LIU
Thus, A(P \ {θ}) ⊂ P .
Now we show A : P \ {θ} → P is completely continuous.
First, we show A is bounded. In fact, let V ⊂ P \ {θ} be a bounded set. Then there½
exists a pos¾
kL1
itive constant L1 satisfying kuk ≤ L1 for u ∈ V . Choose positive constants c1 ≤ min N1 ,
M1
½
¾
kuk4n−2
and c2 ≥ max M2 ,
, where k is defined by (2.15). By (2.1) and (2.18), we get
N2
Z
(4n−2)
|(Au)
1
(t)| =
G(t, s)f (s, u(s), u(4n−2) (s))ds
0
Z
Ã
1
≤
G(s, s)f
0
≤
4 + bd
4
< +∞,
µ
u(s)
−u(4n−2) (s)
s, c1
s(1 − s), (−1)c2
c1 s(1 − s)
c2
l2n−2
2
¶µ
Z
cλ−µ
cβ−α
Lµ+α
1
2
1
∀ t ∈ [0, 1],
!
ds
1
f (s, s(1 − s), −1)ds
0
∀ u ∈ P \ {θ}.
(2.19)
Namely, AV is uniformly bounded.
Next, noticing (2.19) and Ascoli-Arzela theorem, we need to show only that AV is equicontinuous on [0, 1].
In fact, notice that
Z 1
−(Au)(4n−2) (t) =
G(t, s)f (s, u(s), u(4n−2) (s))ds
0
=
1
4
½
Z t
[d + c(1 − t)] (b + as)f (s, u(s), u(4n−2) (s))ds
0
Z
+
(b + at)
1
¾
[d + c(1 − s)]f (s, u(s), u(4n−2) (s))ds .
t
For ∀ t1 , t2 ∈ [0, 1] with t1 < t2 , ∀ u ∈ V , we get
(Au)(4n−2) (t2 ) − (Au)(4n−2) (t1 ) =
Z
t1
·
1
{[d + c(1 − t1 )]
4
(b + as)f (s, u(s), u(4n−2) (s))ds.
0
Z
+(b + at1 )
Z
·
0
t2
1
t1
[d + c(1 − s)]f (s, u(s), u(4n−2) (s))ds − [d + c(1 − t2 )]
(b + as)f (s, u(s), u(4n−2) (s))ds
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
Z
−(b + at2 )
1
= {
4
Z
t2
+
t1
Z
1
+
t2
Z
t1
Z
0
Z
t2
·
c(t2 − t1 )(b + as)f (s, u(s), u(4n−2) (s))ds
(b + at1 )[d + c(1 − s)]f (s, u(s), u(4n−2) (s))ds
[d + c(1 − t2 )](b + as)f (s, u(s), u(4n−2) (s))ds}
t1
≤
0
t1
[d + c(1 − s)]f (s, u(s), u(4n−2) (s))ds}
t2
(b + at2 )[d + c(1 − s)]f (s, u(s), u(4n−2) (s))ds
t2
−
Z
1
177
·
¸
bd
(t2 − t1 )f (s, u(s), u(4n−2) (s))ds + 2 1 +
4
f (s, u(s), u(4n−2) (s))ds
t1
Z
1
+
t2
(t1 − t2 )f (s, u(s), u(4n−2) (s))ds.
This together with (2.19) implies |(Au)(4n−2) (t2 ) − (Au)(4n−2) (t1 )| → 0(|t1 − t2 | → 0).
R1
And similar to the proof of (2.19), we get 0 f (s, u(s), u(4n−2) (s))ds < ∞.
The two above equalities imply that AV is relatively compact.
Finally, it remains to show A is continuous. Suppose un , u0 ∈ P , with
kun − u0 k4n−2 → 0 (n → ∞). Then {un } is a bounded set and kun − u0 k0 → 0(n → ∞). Let
M = sup{ku
½ n k4n−2 ,¾n = 0, 1, 2, · · ·}.
½ Then we¾may still choose positive constants
Mk
M
c1 ≤ min N1 ,
and c2 ≤ min N2 ,
, where k is defined by (2.15). Similar to the
M1
M2
proof of (2.19), we get
f (t, un (t), u(4n−2)
(t)) ≤ (
n
l2n−2 µ λ−µ β−α µ+α
) c1 c2 L1 f (s, s(1 − s), −1)),(2.20)
2
Z
(4n−2)
|(Aun )
(t) − (Au0 )
(4n−2)
1
(t)| ≤
0
4 + bd
|f (s, un (s), un(4n−2) (s))
4
−f (s, u0 (s), u0 (4n−2) (s))|ds,
Z
|(Aun )(t) − (Au0 )(t)| ≤
0
1
|f (s, un (s), un(4n−2) (s))
−f (s, u0 (s), u0 (4n−2) (s))|ds.
(2.21)
178
CHENGLONG ZHAO AND YANSHENG LIU
The above inequality, (2.1), (2.20), (2.21), the Lebesgue dominated convergence theorem and AscoliArzela theorem guarantee
kAun − Au0 k4n−2 → 0 (n → ∞),
that is, A is continuous.
To sum up, A : P \ θ → P is completely continuous.
Let
¯
n
o
¯
Pr =
u ∈ P ¯kuk4n−2 ≤ r ,
¯
n
o
¯
u ∈ P ¯kuk4n−2 ≤ R , 0 < r < 1 < R.
PR =
Choose r satisfying
0 < r ≤ min{[
Z
3
4
1
4
(ac + 4bc + 4ad)2 (a + 4b)(c + 4d) β
)
(
25642
16(b + a)(c + d)
1
k µ f (s, s(1 − s), −1)ds] 1−(µ+β) ,
2N1
, N2 },
l2n−2
where k and l are defined by (2.15) and (2.18), respectively. Then for u ∈ ∂Pr , we have
krt(1 − t) ≤ u(t) ≤
rl2n−2
t(1 − t) ≤ N1 t(1 − t), t ∈ [0, 1]
2
(a + 4b)(c + 4d)
r ≤ e(t)r ≤ −u(4n−2) (t) ≤ r ≤ N2 ,
16(b + a)(c + d)
1 3
∀ t ∈ [ , ].
4 4
Let c1 ≥ M1 . By virtue of the property of G(t, s), (2.1), and (H1 ), we get
Z
− (Au)(4n−2) (t) =
1
G(t, s)f (s, u(s), u(4n−2) (s))ds
0
Z
≥
3
4
1
4
[c(1 − s)(b + as) + ads][c(1 − t)(b + at) + adt]
42
f (s,
Z
≥
3
4
1
4
u(s)
s(1 − s), −1 · (−u(4n−2) (s)))ds
s(1 − s)
(ac + 4bc + 4ad)2 u(s) µ (a + 4b)(c + 4d) β
(
) (
r) f (s, s(1 − s), −1)ds
25642
s(1 − s)
16(a + b)(c + d)
≥
(ac + 4bc + 4ad)2 (a + 4b)(c + 4d) β
(
r)
25642
16(a + b)(c + d)
=
(ac + 4bc + 4ad)2 (a + 4b)(c + 4d) β
(
)
25642
16(a + b)(c + d)
Z
3
4
1
4
Z
3
4
1
4
(kr)µ f (s, s(1 − s), −1)ds
k µ rµ+β f (s, s(1 − s), −1)ds
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
≥ r = kuk4n−2 ,
179
∀ u ∈ ∂Pr .
This guarantees
kAuk ≥ kuk,
∀ u ∈ ∂Pr .
(2.22)
On the other hand, choose R satisfying
M1
4 + bd l2n−2 µ α−β
R ≥ max{
, N2 M2 , [
(
) N2
k
4
2
Let c =
N2
R .
Z
1
1−µ−α
1
f (s, s(1 − s), −1)ds]
}.
0
Then for u ∈ ∂PR , we have
M1 t(1 − t) ≤ kRt(1 − t) ≤ u(t) ≤
l2n−2
Rt(1 − t),
2
−cu(4n−2) (t) ≤ ckuk4n−2 = cR = N2 .
Therefore,
Z
− (Au)(4n−2) (t) =
1
G(t, s)f (s, u(s), u(4n−2) (s))ds
0
Z
1
u(s)
1
(b + as)[d + c(1 − s)]
f (s,
s(1 − s), −1 · · (−cu(4n−2) (s))ds
4
s(1 − s)
c
1
4 + bd u(s) µ 1 β
(
) ( ) (−cu(4n−2) (s))α f (s, s(1 − s), −1)ds
4
s(1 − s)
c
1
4 + bd l2n−2 R µ α−β α
(
) c
R f (s, s(1 − s), −1)ds
4
2
≤
0
Z
≤
0
Z
≤
0
=
4 + bd l2n−2 µ µ+α
(
) R
4
2
≤ R = kuk4n−2 ,
Z
1
f (s, s(1 − s), −1)ds
0
∀ u ∈ ∂PR .
This means
kAuk ≤ kuk,
∀ u ∈ ∂PR .
(2.23)
By the complete continuity of A, (2.22), (2.23), and Lemma 1.1, we obtain that A has a fixed
point u∗ (t) in PR \ Pr . Consequently, SBVP(1.1) has a C 4n−1 [0, 1] positive solution u∗ (t) in
PR \ Pr .
¤
Proof of Theorem 2.5 —
Sufficiency — By Lemma 1.3, the proof of sufficiency of Theorem 2.5 will be divided into four
cases: 1) b = d = 0; 2) b = 0, d > 0; 3) b > 0, d = 0; 4) b > 0, d > 0.
180
CHENGLONG ZHAO AND YANSHENG LIU
The discussion of sufficiency of Case 1) b = d = 0, Case 2) b = 0, d > 0 and Case 3)
b > 0, d = 0 are very similar to the proof of sufficiency of Theorem 2.1, Theorem 2.2 and Theorem
2.3.
As to case 4) b > 0, d > 0, the proof resembles that of case 4) in Theorem 2.4.
Necessity — Let u(t) be a C 4n−1 [0, 1] positive solution of SBVP(1.10 ). Then u(4n−1) (0+ ) ≤ 0
and u(4n−1) (1− ) ≥ 0. Also, from the proof of Lemma 1.2 (see Theorem 2.1-Theorem 2.3 in [11]),
it is easy to see that there exist q1 and q2 with 0 < q1 < 1 < q2 such that
q1 t(1 − t) ≤ u(t) ≤ q2 t(1 − t),
Choose positive number c1 ≤ min{N1 ,
Z
1
}. Then
M1 q2
Z
1
1
f (t, t(1 − t))dt ≤
0
∀ t ∈ [0, 1].
0
≤ cλ1 (
cλ1 (
t(1 − t) µ
) f (t, u(t))dt
c1 u(t)
1 µ
)
c1 q1
Z
1
f (t, u(t))dt
0
= cλ−µ
q1−µ [u(4n−1) (1− ) − u(4n−1) (0+ )] < +∞.
1
This guarantees that (2.7) holds.
¤
Remark : The results of this paper can be extended to the following boundary value problem:


u(4n) (t) = f (t, u(t), u00 (t), u(4) (t), · · · , u(4n−2) (t)),
0 < t < 1,



 u(0) = u(1) = 0,
(2.24)
(2k) (0) − bu(2k+1) (0) = 0,

R
(u)
=
au
1



 R (u) = cu(2k) (1) + du(2k+1) (1) = 0, k = 1, 2, . . . , 2n − 1.
2
where a ≥ 0, b ≥ 0, c ≥ 0, d ≥ 0, 4 = ac+ad+bc > 0, and f ∈ C[(0, 1)×(0, +∞)×R− ×R+ ×
R− · · ·×RS(2n−1) , [0, +∞)], S(2n−1) express the sign of (−1)(2n−1) , R+ = [0, ∞), R− = (−∞, 0].
f is quasi-homogeneous, that is, there are constants
Ni , Mi , N2n−1 , M2n−1 , λi , µi , α, β
−∞ < λi ≤ 0 ≤ µi < +∞, 0 < Ni ≤ 1 ≤ Mi , 0 < N2n−1 ≤ 1 ≤ M2n−1 , 0 ≤ α ≤ β < 1
2n−2
X
µi + β < 1, i = 0, 1, 2, . . . , 2n − 2
i=0
such that for all 0 < t < 1, (−1)i ui ≥ 0, u2n−1 < 0 satisfying
(H10 )
ci µi f (t, u0 , · · · , ui , · · · , u2n−1 ) ≤ f (t, u0 , · · · , ci ui , · · · , u2n−1 )
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
≤ ci λi f (t, u0 , · · · , ui , · · · , u2n−1 ),
0 < ci ≤ Ni ,
ci λi f (t, u0 , · · · , ui , · · · , u2n−1 ) ≤ f (t, u0 , · · · , ci ui , · · · , u2n−1 )
≤ ci µi f (t, u0 , · · · , ui , · · · , u2n−1 ),
ci ≥ Mi ;
c2n−1 β f (t, u0 , · · · , ui , · · · , u2n−2 , u2n−1 ) ≤ f (t, u0 , · · · , ui , · · · , u2n−2 , c2n−1 u2n−1 )
≤ c2n−1 α f (t, u0 , · · · , ui , · · · , u2n−2 , u2n−1 ),
0 < c2n−1 ≤ N2n−1 ,
c2n−1 α f (t, u0 , · · · , ui , · · · , u2n−2 , u2n−1 ) ≤ f (t, u0 , · · · , ui , · · · , u2n−2 , c2n−1 , u2n−1 )
≤ c2n−1 β f (t, u0 , · · · , ui , · · · , u2n−2 , u2n−1 ),
c2n−1 ≥ M2n−1 .
(H20 ) f (t, u0 , u1 , · · · , ui , · · · , u2n−1 ) is nondecreasing with respect to the last variable.
The main results on SBVP(2.24) are as follows.
Theorem 2.1 — Suppose (H10 ) holds, and b = d = 0 in (2.24). Then SBVP(2.24) has a
C 4n−1 [0, 1] positive solution if and only if
Z
1
0<
f (t, t(1 − t), −t(1 − t), t(1 − t), · · · , (−1)2n−1 t(1 − t))dt < +∞.
0
Theorem 2.2 — Suppose (H10 ) and (H20 ) hold, and b = c = 0 in (2.24). Then SBVP(2.24) has
a C 4n−1 [0, 1] positive solution if and only if
Z
1
0<
f (t, t(1 − t), −t(1 − t), t(1 − t), · · · (−1)(2n−2) t(1 − t), −t)dt < +∞.
0
Theorem 2.20 — Suppose (H10 ) holds, and b = 0, d > 0 in (2.24). Then
Z
1
0<
f (t, t(1 − t), −t(1 − t), t(1 − t) · · · , (−1)2n−2 t(1 − t), −t)dt < +∞
0
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP(2.24).
Theorem 2.3 — Suppose (H10 ) and (H20 ) hold, and a = d = 0 in (2.24). Then SBVP(2.24) has
a C 4n−1 [0, 1] positive solution if and only if
Z
1
0<
f (t, t(1 − t), −t(1 − t), t(1 − t), · · · , (−1)2n−2 t(1 − t), −(1 − t))dt < +∞.
0
Theorem 2.30 — Suppose (H10 ) holds, and b > 0, d = 0 in (2.24). Then
Z
0<
1
f (t, t(1 − t), −t(1 − t), t(1 − t), · · · , (−1)2n t(1 − t), −(1 − t))dt < +∞,
0
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP(2.24).
181
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CHENGLONG ZHAO AND YANSHENG LIU
Theorem 2.4 — Suppose (H10 ) holds. If
Z 1
0<
f (t, t(1 − t), −t(1 − t), t(1 − t), · · · , (−1)2n−2 t(1 − t), −1)dt < +∞,
0
then SBVP(2.24) has a C 4n−1 [0, 1] positive solution.
In the following we give some examples to illustrate the applications of our main results.
Example 1 — Consider SBVP(1.1)(b = d = 0) with
1
2
2
4
f (t, u, v) = a1 (t)u− 3 (−v 3 ) + a2 (t)u 5 (−v 7 ),
where ai ∈ C[(0, 1), R+ ](i = 1, 2).
By Theorem 2.1, one can see that SBVP(1.1)(b = d = 0) has a C 4n−1 positive solution if and
only if
Z 1
1
2
2
4
[a1 (t)(t(1 − t))− 3 (−t(1 − t)) 3 + a2 (t)(t(1 − t)) 5 (−t(1 − t)) 7 ]dt < +∞.
0
Example 2 — Consider SBVP(1.1) (b = c = 0) with
1
1
2
f (t, u, v) = b1 (t)u−4 (−v) 8 + b2 (t)u 4 (−v) 7 ,
where bi ∈ C[(0, 1), R+ ](i = 1, 2).
It is easy to see that f satisfies (H1 ) and( H2 ). Consequently, by Theorem 2.2, SBVP(1.1)
(b = c = 0) has a C 4n−1 positive solution if and only if
Z 1
1
1 2
[b1 (t)(t(1 − t))−4 t 8 + b2 (t)(t(1 − t)) 4 t 7 ]dt < +∞.
0
Example 3 — Consider SBVP(1.1)(a = d = 0) with
1
8
1
f (t, u, v) = c1 (t)u−8 (−v) 10 + c2 (t)u 9 (−v) 10 ,
where ci ∈ C[(0, 1), R+ ](i = 1, 2).
Obviously, f satisfies (H1 ) and( H2 ). So by Theorem 2.3, SBVP(1.1) (a = d = 0) has a C 4n−1
positive solution if and only if
Z 1
8
1
1
[c1 (t)(t(1 − t))−8 (1 − t) 10 + c2 (t)(t(1 − t)) 9 (1 − t) 10 ]dt < +∞.
0
Example 4 — Consider SBVP(1.1)(b = 0, d > 0) with
1
6
1
18
f (t, u, v) = d1 (t)u− 17 (−v) 7 + d2 (t)u 19 (−v) 21 ,
where di ∈ C[(0, 1), R+ ](i = 1, 2).
By Theorem 2.20 ,
Z 1
6
1
18
1
[d1 (t)(t(1 − t))− 17 (−t) 7 + d2 (t)(t(1 − t)) 19 (−t) 21 ]dt < +∞
0
HIGHER ORDER SINGULAR SUBLINEAR BOUNDARY VALUE PROBLEMS
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP(1.1)(b = 0, d > 0) .
Example 5 — Consider SBVP(1.1)(b > 0, d = 0) with
2
1
9
f (t, u, v) = h1 (t)u−5 (−v) 3 + h2 (t)u 20 (−v) 10 ,
where hi ∈ C[(0, 1), R+ ](i = 1, 2).
By Theorem 2.30 ,
Z 1
1
9
2
[h1 (t)(t(1 − t))−5 (−(1 − t)) 3 + h2 (t)(t(1 − t)) 20 (−(1 − t)) 10 ]dt < +∞
0
is a necessary condition of a C 4n−1 [0, 1] positive solution of SBVP(1.1)(b > 0, d = 0).
Example 6 — Consider SBVP(1.1) with
1
1
1
2
f (t, u, v) = f1 (t)u− 3 (−v) 16 + f2 (t)u 47 (−v) 23 ,
where fi ∈ C[(0, 1), R+ ](i = 1, 2), and
Z 1
1
1
[f1 (t)(t(1 − t))− 3 + f2 (t)(t(1 − t)) 47 ]dt < +∞.
0
Then by Theorem 2.4, one can see that SBVP(1.1) has a C 4n−1 positive solution.
Example 7 — Consider SBVP(1.10 ) with
3
f (t, u, v) = g1 (t)u−81 + g2 (t)u 101 ,
where gi ∈ C[(0, 1), R+ ](i = 1, 2).
By Theorem 2.5, SBVP(1.10 ) has a C 4n−1 positive solution if and only if
Z 1
3
[g1 (t)(t(1 − t))−81 + g2 (t)(t(1 − t)) 101 ]dt < +∞.
0
ACKNOWLEDGEMENT
The authors are grateful to the referees for their helpful comments.
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