Electronic Journal of Differential Equations, Vol. 2004(2004), No. 143, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
TWIN POSITIVE SOLUTIONS FOR FOURTH-ORDER
TWO-POINT BOUNDARY-VALUE PROBLEMS WITH SIGN
CHANGING NONLINEARITIES
YU TIAN, WEIGAO GE
Abstract. A new fixed point theorem on double cones is applied to obtain
the existence of at least two positive solutions to
(Φp (y 00 (t))00 − a(t)f (t, y(t), y 00 (t)) = 0,
0 < t < 1,
y(0) = y(1) = 0 = y 00 (0) = y 00 (1),
where f : [0, 1] × [0, ∞) × (−∞, 0] → R, a ∈ L1 ([0, 1], (0, ∞)). We also give
some examples to illustrate our results.
1. Introduction
We study the existence of multiple positive solutions for the fourth-order twopoint boundary-value problem
(Φp (y 00 (t))00 − a(t)f (t, y(t), y 00 (t)) = 0,
00
0 < t < 1,
00
y(0) = y(1) = 0 = y (0) = y (1),
(1.1)
(1.2)
where the nonlinear term f is allowed to change sign, a ∈ L1 ([0, 1], (0, ∞)), Φp x =
|x|p−2 x, 1/p + 1/q = 1, p > 1. When p = 2 Problem (1.1)-(1.2) describes the
deformations of an elastic beam. The boundary conditions are given according to
the control at the ends of the beam.
A great deal of research has been devoted to the existence of solutions for the
fourth-order two point boundary value problem. We refer the reader to [1, 3, 4, 5,
6, 2] and their references. Aftabizadeh [1], [2], del Pino and Manasevich [3], Gupta
[4, 5], Ma and Wang [6], Liu [9] have studied the existence problem of positive
solutions of the following fourth-order two-point boundary-value problem
y (4) (t) − f (t, y(t), y 00 (t)) = 0,
0 < t < 1,
00
y(0) = y(1) = 0 = y (0) = y 00 (1).
All the above works were done under assumption that the nonlinear term f is
nonnegative.
2000 Mathematics Subject Classification. 34B10, 34B15.
Key words and phrases. Fourth-order two-point boundary-value problem; fixed point theorem;
double cones; positive solution.
c
2004
Texas State University - San Marcos.
Submitted August 28, 2004. Published December 3, 2004.
Supported by grant 10371006 from the National Natural Sciences Foundation of China.
1
2
Y. TIAN, W. GE
EJDE-2004/143
In this paper, we will impose growth conditions on f which ensure the existence
of at least two positive solutions for (1.1)-(1.2). The key tool in our approach is
the following fixed point theorem on double cones.
For a cone K in a Banach space X with norm k · k and a constant r > 0, let
Kr = {x ∈ K : kxk < r},
∂Kr = {x ∈ K : kxk = r}.
+
Suppose α : K → 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}
and Ka (b) = {x ∈ K : a < kxk, α(x) < b}. The origin in X is denoted by θ.
Theorem 1.1 ([10]). Let X be a real Banach space with norm k · k and K, K 0 ⊂ X
two solid cones with K 0 ⊂ K. Suppose T : K → K and T 0 : K 0 → K 0 are
two completely continuous operators and α : K 0 → R+ a continuously increasing
functional satisfying α(x) ≤ kxk ≤ M α(x) for all x in K 0 , where M ≥ 1 is a
constants b > a > 0 such that
(C1) kT xk < a, for x ∈ ∂Ka
0
(C2) kT 0 xk < a, for x ∈ ∂Ka0 and α(T 0 x) > b for x ∈ ∂K (b)
(C3) T x = T 0 x, for x ∈ Ka0 (b) ∩ {u : T 0 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.
2. Existence of positive solutions
Lemma 2.1. Suppose g(·) ∈ C[0, 1], then
(Φp (y 00 (t)))00 − g(t) = 0,
0 < t < 1,
00
(2.1)
00
(2.2)
G(s, τ )g(τ )dτ ds,
(2.3)
y(0) = y(1) = 0 = y (0) = y (1),
has a unique solution
Z
y(t) =
1
Z
G(t, s)Φq
0
where
1
0
(
t(1 − s),
G(t, s) =
s(1 − t),
0 ≤ t ≤ s ≤ 1;
0 ≤ s ≤ t ≤ 1.
Proof. Let Φp y 00 (t) = u(t), then (2.1)-(2.2) becomes
u00 − g(t) = 0,
0 < t < 1;
u(0) = u(1) = 0.
It is clear that the above boundary-value problem has a unique solution,
Z 1
u(t) =
G(t, s)g(s)ds,
0
(
t(1 − s), 0 ≤ t ≤ s ≤ 1;
. Then Φp y 00 (t) = u(t), i.e., y 00 (t) =
where G(t, s) =
s(1 − t), 0 ≤ s ≤ t ≤ 1.
(Φq u)(t). By the boundary condition we know that
Z 1
Z 1
Z 1
y(t) =
G(t, s)(Φq u)(s)ds =
G(t, s)Φq
G(s, r)g(r)dr ds.
0
0
0
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TWIN POSITIVE SOLUTIONS
3
The proof is completed.
In this paper, we assume the following conditions:
(H1) f : [0, 1] × [0, ∞) × (−∞, 0] → R is continuous, a ∈ L1 ([0, 1], (0, ∞))
(H2) a(t)f (t, 0, 0) 6≡ 0, f (t, 0, 0) ≥ 0 for t ∈ [0, 1].
Let X = {x ∈ C 2 [0, 1] : x(0) = x(1) = 0 = x00 (0) = x00 (1)}. Then X is a Banach
space with the norm kxk = supt∈[0,1] |x00 (t)|. Define
K = {x ∈ X : x is nonnegative and concave on [0, 1]},
0
and K = {x ∈ X : x is nonnegative and concave on [0, 1], α(x) ≥ δ q−1 kxk,
δ ∈ (0, 1/2)}, where α(x) = mint∈[δ,1−δ] {−x00 (t)}. Obviously K, K 0 ⊂ X are two
1
cones with K 0 ⊂ K. From the definition K 0 , we know that α(x) ≤ kxk ≤ δq−1
α(x),
δ ∈ (0, 1/2). Denote
Z 1
hZ 1
i+
(T x)(t) =
G(t, s)Φq
G(s, τ )a(τ )f (τ, x(τ ), x00 (τ ))dτ ds ,
0
0
where B + = max{B, 0}.
Z 1
Z
(Ax)(t) =
G(t, s)Φq
0
1
G(s, τ )a(τ )f (τ, x(τ ), x00 (τ ))dτ ds.
0
For x ∈ X, define θ : X → K by (θx)(t) = max{x(t), 0}, then T = θ ◦ A. For
x ∈ K 0 , let
Z 1
Z 1
(T 0 x)(t) =
G(t, s)Φq
G(s, τ )a(τ )f + (τ, x(τ ), x00 (τ ))dτ ds,
0
0
where f + (t, x(t), x00 (t)) = max{f (t, x(t), x00 (t)), 0}.
Lemma 2.2. For x ∈ X, we have kxk∞ ≤ kx00 k∞ and kx0 k∞ ≤ kx00 k∞ where
kxk∞ = supt∈[0,1] |x(t)|.
Proof. From
1
Z
x(t) =
1
Z
G(s, r)g(r)dr ds
G(t, s)Φq
0
0
and
x00 (t) = −Φq
Z
1
G(t, r)g(r)dr ,
0
we have
|x(t)| = 1
Z
Z
G(t, s)Φq
0
=
1
G(s, r)g(r)dr ds
0
1
Z
G(t, s)|x00 (s)|ds
0
≤ kx00 k∞
1
Z
G(t, s)ds
0
= kx00 k∞
Z
= kx k∞
Z
s(1 − t)ds +
0
00
t
1
t(1 − s)ds
t
t2 (1 − t) t(1 − t)2 +
.
2
2
4
Y. TIAN, W. GE
EJDE-2004/143
So kxk∞ = supt∈[0,1] |x(t)| ≤ 18 kx00 k∞ < kx00 k∞ . At the same time, from x0 (t) =
R t 00
x (s)ds, we have kx0 k∞ = supt∈[0,1] |x0 (t)| ≤ kx00 k∞ . The proof is complete. 0
Note that X is a Banach space with the norm kxk = supt∈[0,1] |x00 (t)|.
Lemma 2.3. T 0 (K 0 ) ⊂ K 0 .
Proof. For any x ∈ K 0 , it is clear that (T 0 x)(t) is nonnegative from the definition of
R1
T 0 . From (T 0 x)00 (t) = −Φq 0 G(t, s)f + (s, x(s), x00 (s))ds , we know (T 0 x)00 (t) ≤ 0.
So T 0 x is concave on [0, 1]. Then
Z 1
−(T 0 x)00 (t) = Φq
G(t, s)f + (s, x(s), x00 (s))ds
0
1
Z
G(s, s)f + (s, x(s), x00 (s))ds ,
≤ Φq
0
which implies
1
Z
k − (T 0 x)00 k∞ ≤ Φq
G(s, s)f + (s, x(s), x00 (s))ds ,
0
and
α(T 0 x) =
min [−(T 0 x)00 (t)]
t∈[δ,1−δ]
1
Z
=
G(t, s)f + (s, x(s), x00 (s))ds
min Φq
t∈[δ,1−δ]
=
0
min Φq
Z
t∈[δ,1−δ]
t
Z
+
1
t(1 − s)f + (s, x(s), x00 (s))ds
t
s(1 − t)f + (s, x(s), x00 (s))ds
0
≥
min Φq
Z
t∈[δ,1−δ]
1
δs(1 − s)f + (s, x(s), x00 (s))ds
t
t
Z
δs(1 − s)f + (s, x(s), x00 (s))ds
+
0
Z
= Φq δ
≥δ
q−1
1
G(s, s)f + (s, x(s), x00 (s))ds
0
0
kT xk.
The proof is complete.
For convenience, we denote
Q = max Φq
t∈[0,1]
m=
min
min
t∈[δi ,1−δi ]
It is clear that 0 < m < Q < ∞.
G(t, s)a(s)ds
,
0
t∈[δ,1−δ]
mi =
1
Z
1−δ
Z
Φq
G(t, s)a(s)ds
,
δ
Z
1−δi
Φq
G(t, s)a(s)ds
δi
.
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TWIN POSITIVE SOLUTIONS
5
From the continuity of f, a ∈ L1 ([0, 1], (0, ∞)). It is easy to see A : K → X and
T : K 0 → K 0 are completely continuous. So T : K → K is completely continuous.
0
Theorem 2.4. Suppose (H1) and (H2) are satisfied and there exist a, b, d such that
0 < d < δ q−1 a < a < δ q−1 b < b. Assume that f satisfies the following conditions:
(H3) For (t, u, v) ∈ [0, 1] × [0, b] × [−b, −d], f (t, u, v) ≥ 0
a
(H4) For (t, u, v) ∈ [0, 1] × [0, a] × [−a, 0], f (t, u, v) < Φp ( Q
).
b
q−1
(H5) For (t, u, v) ∈ [0, 1] × [0, b] × [−b, −δ
b], f (t, u, v) ≥ Φp ( m
).
Then (1.1)-(1.2) has at least two positive solutions y1 , y2 such that
0 < ky1 k < a < ky2 k,
α(y1 ) < δ q−1 b,
ky1 k∞ < a,
ky2 k∞ < b
(2.4)
Proof. First we show that T has a fixed point y1 ∈ K with ky1 k ≤ a. In fact, for
any y ∈ ∂Ka , we have kyk = a. So 0 ≤ y(t) ≤ a, −a ≤ y 00 (t) < 0, t ∈ [0, 1]. Let
I = {t ∈ [0, 1] : f (t, y(t), y 00 (t)) ≥ 0}.
kT yk = max |(T y)00 (t)|
t∈[0,1]
1
G(t, s)a(s)f (s, y(s), y 00 (s))ds , 0
= max max Φq
t∈[0,1]
0
Z
≤ max Φq
G(t, s)a(s)f (s, y(s), y 00 (s))ds
t∈[0,1]
I
Z
≤ Φq
max
f (t, u, v) max
G(t, s)a(s)ds
Z
t∈[0,1],0≤u≤a,−a≤v≤0
a
max Φq
Q t∈[0,1]
= a.
Z
G(t, s)a(s)ds
<
t∈[0,1]
I
1
0
The existence of y1 is proved by using condition (C1) of Theorem 1.1 and 0 ≤ y1 ≤ a,
−a ≤ y100 ≤ 0. Obviously, y1 is a solution of (1.1)-(1.2). Suppose this is not true,
then there is t0 ∈ (0, 1) such that y1 (t0 ) 6= (Ay1 )(t0 ). It must be (Ay1 )(t0 ) < 0 =
y1 (t0 ). Let (t1 , t2 ) be the maximum interval such that (Ay1 )(t0 ) < 0 for t ∈ (t1 , t2 ).
We claim [t1 , t2 ] 6= [0, 1] because of a(t)f (t, 0, 0) 6≡ 0 for t ∈ [0, 1].
If t2 < 1. y1 (t) = 0, t ∈ [t1 , t2 ]. (Ay1 )(t2 ) = 0, (Ay1 )(t) < 0, for t ∈ (t1 , t2 ). Then
(Ay1 )0 (t2 ) ≥ 0. For t ∈ (t1 , t2 ), we have
Z 1
00
(Ay1 ) (t) = −Φq (
G(t, s)a(s)f (s, y(s), y 00 (s))ds)
0
Z
= −Φq (
1
G(t, s)a(s)f (s, 0, 0)ds) < 0.
0
So (Ay1 )0 (t) is decreasing for t ∈ (t1 , t2 ). Noticing (Ay1 )0 (t2 ) ≥ 0, so t1 = 0 and
(Ay1 )0 (t) > 0, t ∈ [0, t2 ), (Ay1 )(0) < 0, which contradicts (1.2). If t1 > 0. So
y1 (t) = 0, (Ay1 )(t1 ) = 0, (Ay1 )(t) < 0 for t ∈ (t1 , t2 ).(Ay1 )0 (t1 ) ≤ 0.
Z 1
00
(Ay1 ) (t) = −Φq (
G(t, s)a(s)f (s, y(s), y 00 (s))ds)
0
Z
= −Φq (
0
1
G(t, s)a(s)f (s, 0, 0)ds) ≤ 0.
6
Y. TIAN, W. GE
EJDE-2004/143
So (Ay1 )0 (t) is decreasing for t ∈ (t1 , t2 ). So t2 = 1, (Ay1 )(1) < 0, which contradicts
boundary condition (1.2). So y1 is a solution of (1.1)-(1.2). We now show that
0
(C2) of Theorem 1.1 is satisfied. For x ∈ ∂Ka , i.e., kxk = a, then 0 < x(t) < a,
00
−a < x (t) < 0 for t ∈ [0, 1].
kT 0 yk = max |(T 0 y)00 (t)|
t∈[0,1]
Z 1
= max Φq
G(t, s)a(s)f + (s, y(s), y 00 (s))ds
t∈[0,1]
0
h
+
≤ Φq max f (t, u, v) : t ∈ [0, 1], 0 ≤ u ≤ a, −a ≤ v ≤ 0
Z 1
i
G(t, s)a(s)ds
× max Φq
t∈[0,1]
a
<
Q
= a.
0
1
Z
G(t, s)a(s)ds
0
For y ∈ ∂K 0 (δ q−1 b), we have α(y) = δ q−1 b. So δ q−1 b ≤ kyk ≤ b, i.e., −b ≤ y 00 (t) ≤
−δ q−1 b for t ∈ [δ, 1 − δ], at the same time kyk∞ ≤ b. Then
α(T 0 y) = − min (T 0 y)00 (t)
t∈[δ,1−δ]
Z
=
min Φq
t∈[δ,1−δ]
min
t∈[δ,1−δ]
G(t, s)a(s)f + (s, y(s), y 00 (s))ds
0
Z
≥ Φq
1
1−δ
G(t, s)a(s)f + (s, y(s), y 00 (s))ds
δ
> Φq min f (t, u, v) : t ∈ [0, 1], u ∈ [0, b], v ∈ [−b, −δ q−1 b]
Z 1−δ
× min
G(t, s)a(s)ds
t∈[δ,1−δ]
=b>δ
q−1
δ
b.
0
Finally we show that (C3 ) of Theorem 1.1 is also satisfied. Let x ∈ Ka (δ q−1 b) ∩ {u :
1
1
T 0 u = u}, then kxk < δq−1
α(x). From α(x) ≤ kxk ≤ δq−1
α(x), we have
min {−x00 (t)} = α(x) ≥ δ q−1 kxk > δ q−1 a > d .
t∈[δ,1−δ]
So −x00 ∈ [d, b]. From (H3), we have f (t, u, v) = f + (t, u, v), which implies T y = T 0 y.
Therefore, there exist two positive solutions y1 , y2 satisfying (2.2).
Remark. When p = 2, a(t) ≡ 1, f (t, u, v) > 0, δ = 1/4, Theorem 2.4 reduces to [9,
Theorem 3.1]. But our result shows at least two positive solutions, whereas there
is at least one positive solution in B. Liu [9, Theorem 3.1].
Theorem 2.5. Suppose (H1),(H2) hold. Also assume that
R 1−δ
(H6) δi ∈ (0, 1/2), i = 1, 2, . . . , n, 0 < δi i a(s)ds < ∞
EJDE-2004/143
TWIN POSITIVE SOLUTIONS
7
(H7) There exists constants ai , bi , di > 0, i = 1, 2, . . . , n, where 0 < di <
δ q−1 ai < ai < δiq−1 bi < bi < di+1 such that for i = 1, 2, . . . , n, we have
f (t, u, v) ≥ 0
for (t, u, v) ∈ [0, 1] × [0, bi ] × [−bi , −di ],
ai
f (t, u, v) < Φq ( ) for (t, u, v) ∈ [0, 1] × [0, ai ] × [−ai , 0],
Q
bi
f (t, u, v) ≥ Φq ( ) for (t, u, v) ∈ [0, 1] × [0, bi ] × [−bi , −δiq−1 bi ].
mi
Then (1.1)-(1.2) has at least n + 1 positive solutions y1 , y2 , . . . , yn+1 satisfying
α(y2 ) < δ1q−1 b1 ,
0 ≤ ky1 k < a1 < ky2 k ≤ b1 ,
0 < ky2 k∞ < b1 ,
an < kyn+1 k,
a2 < ky3 k,
α(y3 ) <
...
α(yn+1 ) < δnq−1 bn ,
0 < ky1 k∞ < a1 ,
δ2q−1 b2 ,
0 < ky3 k∞ < b2 ,
0 < kyn+1 k∞ < bn .
Theorem 2.6. Suppose (H1), (H2), (H6) hold. Also assume
(H8) There exists constants ai , bi > 0, d, i = 1, 2, . . . , n, where 0 < d < δ q−1 ai <
ai < δiq−1 bi < bi , such that for i = 1, 2, . . . , n, we have:
f (t, u, v) ≥ 0 for (t, u, v) ∈ [0, 1] × [0, bn ] × [−bn , −d],
ai
for (t, u, v) ∈ [0, 1] × [0, ai ] × [−ai , 0],
f (t, u, v) <
Q
bi
for (t, u, v) ∈ [0, 1] × [0, bi ] × [−bi , −δiq−1 bi ],
f (t, u, v) ≥
mi
Then (1.1)-(1.2) has at least 2n positive solutions y1 , y2 , . . . , y2n satisfying
0 ≤ ky1 k < a1 < ky2 k,
0 < ky1 k∞ < a1 ,
ky3 k < a2 < ky4 k,
α(y4 ) <
ky2n−1 k < an < ky2n k,
δ2q−1 b2
α(y2n ) <
α(y2 ) < δ1q−1 b1 < α(y3 ),
0 < ky2 k∞ < b1 ,
< α(y5 ),
δnq−1 bn ,
ky3 k∞ < a2 ,
ky2n−1 k∞ < an ,
ky4 k∞ < b2 , . . .
ky2n k∞ < bn .
Example. Consider the boundary-value problem
y + π/6 √
5 19
t 3 cos(y 00 + π)
+
= 0, 0 < t < 1,
(Φ3 y 00 (t))00 −
(2.5)
12
10
6
y(0) = y(1) = 0 = y 00 (0) = y 00 (1),
√
5
t
where a(t) = t, f (t, u, v) = u+π/6
π))19 + 10
, p = 3, q = 3/2. Clearly
( 3 cos(v + 12
6
f is allowed to change sign on [0, 1] × [0, ∞) × (−∞, 0).
√
Z 1
1/2
5
t
3 1/2
2
Q = max Φ3/2
G(t, s)a(s)ds = max
)
= 3− 4 .
(−t + 1)
=(
27
t∈[0,1]
t∈[0,1] 6
0
Note that
Z
1−δ
Z
G(t, s)a(s)ds = (1 − t)
δ
2
Z
1−δ
(1 − s)sds
s ds + t
δ
=
t
t
1
[−t3 + t(4δ 3 − 3δ 2 + 1) − 2δ 3 ].
6
8
Y. TIAN, W. GE
EJDE-2004/143
Let δ = 1/4, d = π/36, a = π/12, b = π/2. It is clear that d < δ 1/2 a < a < δ 1/2 b.
Then
r
Z 1−δ
1
m = min Φ3/2
G(t, s)a(s)ds >
.
24
t∈[δ,1−δ]
δ
√
For (t, u, v) ∈ [0, 1]×[0, π/2]×[−π/2, −π/36], we have f (t, u, v) = u+π/6
( 3 cos(v+
6
5
t
19
+ 10
> 0. So (H3) holds. For (t, u, v) ∈ [0, 1] × [0, π/12] × [−π/12, 0],
12 π))
√
19 t
u+π/6 √
5
π
1
π
f (t, u, v) = 6
3 cos(v+ 12
π) + 10
< 24
×( 23 )19 + 10
< 0.6 < ( 12
×35/4 )2 =
Φ3 (a/Q). So (H4) holds. For (t, u, v) ∈ [0, 1] × [0, π/2] × [−π/2, −π/4], f (t, u, v) =
√
19
u+π/6 √
5
t
3 cos(v + 12
π)
+ 10
> (π 6)2 > Φ3 (b/m). So (H5) holds. Thus by
6
Theorem 2.4, this boundary-value problem has at least two positive solutions y1 , y2
such that
π
π
π
π
0 < ky1 k <
< ky2 k, α(y1 ) <
, ky1 k∞ <
, ky2 k∞ < .
12
16
12
2
References
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[3] M. A. Del Pino, R. F. Manasevich; Existence for fourth-order boundary value problem under
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Yu Tian
Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081,
China
E-mail address: tianyu2992@163.com
Weigao Ge
Department of Applied Mathematics, Beijing Institute of Technology, Beijing 100081,
China
E-mail address: gew@bit.edu.cn