Electronic Journal of Differential Equations, Vol. 2013 (2013), No. 36, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
POSITIVE SOLUTIONS FOR A 2nTH-ORDER p-LAPLACIAN
BOUNDARY VALUE PROBLEM INVOLVING ALL
DERIVATIVES
YOUZHENG DING, JIAFA XU, XIAOYAN ZHANG
Abstract. In this work, we are mainly concerned with the positive solutions
for the 2nth-order p-Laplacian boundary-value problem
−(((−1)n−1 x(2n−1) )p−1 )0 = f (t, x, x0 , . . . , (−1)n−1 x(2n−2) , (−1)n−1 x(2n−1) ),
x(2i) (0) = x(2i+1) (1) = 0,
(i = 0, 1, . . . , n − 1),
R2n
+ , R+ )(R+
where n ≥ 1 and f ∈ C([0, 1] ×
:= [0, ∞)). To overcome the
difficulty resulting from all derivatives, we first convert the above problem
into a boundary value problem for an associated second order integro-ordinary
differential equation with p-Laplacian operator. Then, by virtue of the classic
fixed point index theory, combined with a priori estimates of positive solutions,
we establish some results on the existence and multiplicity of positive solutions
for the above problem. Furthermore, our nonlinear term f is allowed to grow
superlinearly and sublinearly.
1. Introduction
In this paper, we investigate the existence and multiplicity of positive solutions for the following 2nth-order p-Laplacian boundary value problem involving
all derivatives
−(((−1)n−1 x(2n−1) )p−1 )0 = f (t, x, x0 , . . . , (−1)n−1 x(2n−2) , (−1)n−1 x(2n−1) ),
x(2i) (0) = x(2i+1) (1) = 0,
(i = 0, 1, . . . , n − 1),
(1.1)
where f ∈ C([0, 1]×R2n
+ , R+ ). Here, by a positive solution (1.1) we mean a function
u ∈ C 2n [0, 1] that solves (1.1) and satisfies u(t) > 0 for all t ∈ (0, 1].
We are here interested in the case where f depends explicitly on all derivatives.
When f involves all even derivatives explicitly, many researchers [1, 2, 4, 12, 15]
study the Lidstone boundary value problem
(−1)n u(2n) = f (t, u, −u00 , . . . , (−1)n−1 u(2n−2) ),
u(2i) (0) = u(2i) (1) = 0,
n ≥ 2,
i = 0, 1, 2, . . . , n − 1.
2000 Mathematics Subject Classification. 34B18, 45J05, 47H11.
Key words and phrases. Integro-ordinary differential equation; a priori estimate; index;
fixed point; positive solution.
c
2013
Texas State University - San Marcos.
Submitted September 10, 2012. Published January 30, 2013.
1
(1.2)
2
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
Yang [19] considered the existence and uniqueness of positive solutions for the
following generalized Lidstone boundary value problem
(−1)n u(2n) = f (t, u, −u00 , . . . , (−1)n−1 u(2n−2) ),
α0 u(2i) (0) − β0 u(2i+1) (0) = 0, α1 u(2i) (1) − β1 u(2i+1) (1) = 0, i = 0, 1, 2, . . . , n − 1,
(1.3)
where αj ≥ 0, βj ≥ 0 (j = 0, 1) and α0 α1 + α0 β1 + α1 β0 > 0. In view of the symmetry, the results in [1, 2, 4, 12, 15, 19] demonstrate that problems (1.2) and (1.3)
are essentially identical with second-order Dirichlet problem and Sturm-Liouville
problem (the case n = 1), respectively.
Yang, O’Regan and Agarwal [20] studied the existence and multiplicity of positive solutions for the second-order boundary value problem depending on the firstorder derivative u0
u00 + f (t, u, u0 ) = 0,
(1.4)
u(0) = u0 (1) = 0.
In order to overcome the difficulty resulting from the first-order derivative, they imposed the Bernstein-Nagumo condition [3, 13] on the nonlinear term f to establish
several existence theorems for (1.4).
Yang and O’Regan [21] studied the existence, multiplicity and uniqueness of positive solutions for the 2nth-order boundary value problem involving all derivatives
of odd orders
(−1)n u(2n) = f (t, u, u0 , −u000 , . . . , (−1)n−1 u(2n−1) ),
u(2i) (0) = u(2i+1) (1) = 0,
i = 0, 1, 2, . . . , n − 1,
(1.5)
where n ≥ 2 and f ∈ C([0, 1] × Rn+1
+ , R+ ) depends on u and all derivatives of odd
orders. As application, they utilized their results to discuss the positive symmetric
solutions for a Lidostone problem involving an open question posed by Eloe [5].
Yang [22] discussed a 2nth-order ordinary differential equation involving all derivatives, and the results improved and extended the corresponding ones in [19, 20, 21].
Equations of the p-Laplacian form occur in the study of non-Newtonian fluid
theory and the turbulent flow of a gas in a porous medium. Since 1980s, there exist
a very large number of papers devoted to the existence of solutions for differential
equations with p-Laplacian, for instance, see [6, 7, 10, 11, 16, 17, 18, 23, 24, 25] and
the references therein.
Yang and his coauthors [17, 23, 24] studied some boundary value problems with
the p-Laplacian operator. Yang and O’Regan [23] studied the existence and multiplicity of positive solutions for the focal problem involving both the p-Laplacian
and the first order derivative
((u0 )p−1 )0 + f (t, u, u0 ) = 0,
t ∈ (0, 1),
0
u(0) = u (1) = 0,
(1.6)
where p > 1 and f ∈ C([0, 1] × R2+ , R+ ). Moreover, they applied their main
results obtained here to establish the existence of positive symmetric solutions to
the Dirichlet problem
(|u0 |p−2 u0 )0 + f (u, u0 ) = 0,
t ∈ (−1, 0) ∪ (0, 1),
u(−1) = u(1) = 0.
(1.7)
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POSITIVE SOLUTIONS
3
However, the existence of positive solutions for p-Laplacian equation with the
nonlinear term involving the derivatives, such as Lidstone problem, has not been
extensively studied yet. To the best of our knowledge, only [8, 14, 26] is devoted to
this direction. Guo and Ge [8] considered the following boundary-value problem
(Φ(y (2n−1) ))0 = f (t, y, y 00 , . . . , y (2n−2) ),
y (2i) (0) = y (2i) (1) = 0,
0 ≤ t ≤ 1,
(1.8)
i = 0, 1, 2, . . . , n − 1,
where f ∈ C([0, 1]×Rn , R)(R := (−∞, +∞)). Some growth conditions are imposed
on f which yield the existence of at least two symmetric positive solutions by using
a fixed point theorem in cones. An interesting feature in [8] is that the nonlinearity
f may be sign-changing.
Motivated by the works mentioned above, in particular [17, 19, 20, 21, 22, 23, 24],
in this work, we discuss the existence and multiplicity of positive solutions for
(1.1). To overcome the difficulty resulting from all derivatives, we first transform
(1.1) into a boundary value problem for an associated second order integro-ordinary
differential equation. Then, we will use fixed point index theory to establish our
main results based on a priori estimates achieved by utilizing some properties of
concave functions, properties including Jensen’s inequalities and our inequality (2.4)
below. The results obtained here improve some existing results in the literature.
2. Preliminaries
Let E := C [0, 1], kuk := max{kuk0 , ku0 k0 }, where kuk0 := maxt∈[0,1] |u(t)|.
Furthermore, let P := {u ∈ E : u(t) ≥ 0, u0 (t) ≥ 0, ∀t ∈ [0, 1]}. Then E is a real
Banach space and P a cone on E. For any positive integer i ≥ 2, we denote
Z 1
k1 (t, s) := min{t, s}, ki (t, s) :=
ki−1 (t, τ )k1 (τ, s) dτ, ∀t, s ∈ [0, 1].
1
0
Define
Z
1
(Bi u)(t) :=
ki (t, s)u(s) ds,
hi (t, s) := ∂ki (t, s)/∂t,
i = 1, 2, . . . ,
0
Then
((Bi u)(t))0 :=
Z
1
hi (t, s)u(s) ds,
i = 1, 2, . . . ,
0
and Bi , Bi0 : E → E are completely continuous linear operators and Bi , Bi0 are also
positive operators.
Let (−1)n−1 x(2n−2) := u, it is easy to see that (1.1) is equivalent to the following
system of integro-ordinary differential equations
−((u0 )p−1 )0 = f (t, (Bn−1 u)(t), ((Bn−1 u)(t))0 , . . . , u, u0 ),
(2.1)
u(0) = u0 (1) = 0.
Furthermore, the above system can be written in the form
Z tZ 1
1
p−1
ds. (2.2)
u(t) =
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
0
s
Denote by
(Au)(t) :=
Z tZ
0
s
1
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
1
p−1
ds.
4
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
Hence, f ∈ C([0, 1] × R2n
+ , R+ ) implies that A : P → P is a completely continuous
operator, and the existence of positive solutions for (2.1) is equivalent to that of
positive fixed points of A.
Lemma 2.1. Let κ := 1 − 2/e and ψ(t) := tet , t ∈ [0, 1]. Then ψ(t) is nonnegative
on [0, 1] and
Z 1
κψ(s) ≤
k1 (t, s)ψ(t) dt ≤ ψ(s).
(2.3)
0
Lemma 2.2. Let u is concave, increasing and nonnegative on [0, 1]. Then
Z 1
u(t)ψ(t) dt ≥ κekuk.
(2.4)
0
Proof. The concavity of u and maxt∈[0,1] u(t) = u(1) = kuk imply
Z 1
Z 1
Z 1
u(t)ψ(t) dt =
u(t · 1 + (1 − t) · 0)ψ(t) dt ≥ u(1)
tψ(t) dt = κekuk.
0
0
0
This completes the proof.
Lemma 2.3 ([21]). Let u ∈ P and q > 0. Then
Z 1h
Z
n−2
i
X
(Bn−1 uq )(t) + 2
((Bn−1−i uq )(t))0 ψ(t) dt =
0
1
uq (t)ψ(t) dt.
(2.5)
0
i=0
Lemma 2.4 ([9]). Let Ω ⊂ E be a bounded open set and A : Ω ∩ P → P is a
completely continuous operator. If there exists v0 ∈ P \ {0} such that v − Av 6= λv0
for all v ∈ ∂Ω ∩ P and λ ≥ 0, then i(A, Ω ∩ P, P ) = 0, where i is the fixed point
index on P .
Lemma 2.5 ([9]). Let Ω ⊂ E be a bounded open set with 0 ∈ Ω. Suppose A :
Ω ∩ P → P is a completely continuous operator. If v 6= λAv for all v ∈ ∂Ω ∩ P and
0 ≤ λ ≤ 1, then i(A, Ω ∩ P, P ) = 1.
Lemma 2.6 (Jensen’s inequalities). Let θ > 0 and ϕ ∈ C([0, 1], R+ ). Then
Z 1
θ Z 1
ϕ(t) dt ≤
(ϕ(t))θ dt, if θ ≥ 1,
0
Z
1
0
θ Z 1
ϕ(t) dt ≥
(ϕ(t))θ dt,
0
if 0 < θ ≤ 1.
0
3. Main results
p−2
For brevity, we define y = (y1 , y2 , . . . , y2n−1 , y2n ) ∈ R2n
},
+ , γp := max{1, 2
∗
p∗ = min{1, p − 1}, p = max{1, p − 1}, Ki := maxt,s∈[0,1] ki (t, s) > 0, Hi :=
maxt,s∈[0,1] hi (t, s) > 0,
n
h
n−1
p∗ −1
io1−p/p∗
X
βp := 2p∗ −1 κ (n − 1)
(Ki + Hi )
+1
,
i=1
n ∗ h
n−1
p∗ −1
io1−p/p∗
X
αp := 2p −1 (n − 1)
(Ki + Hi )
+1
.
i=1
(H1) f ∈ C([0, 1] ×
R2n
+ , R+ ).
EJDE-2013/36
POSITIVE SOLUTIONS
5
(H2) There exist a1 > βp and c > 0 such that
f (t, y) ≥ a1
n−1
p−1
X
(y2i−1 +2(n−i)y2i )+y2n−1
−c,
for all y ∈ R2n
+ and t ∈ [0, 1].
i=1
(H3) For any M0 > 0 there is a function ΦM0 ∈ C(R+ , R+ ) such that
p−1
), ∀(t, y) ∈ [0, 1] × [0, M0 ]2n−1 × R+ ,
f (t, y) ≤ ΦM0 (y2n
Z ∞
dξ
= ∞ for any δ > 0.
Φ
M0 (ξ)
δ
(H4) There exist b1 ∈ (0, αp ) and r > 0 such that
f (t, y) ≤ b1
n−1
p−1
X
(y2i−1 + 2(n − i)y2i ) + y2n−1
for all y ∈ [0, r]2n and t ∈ [0, 1].
i=1
(H5) There exist a2 > βp and r > 0 such that
f (t, y) ≥ a2
n−1
X
(y2i−1 + 2(n − i)y2i ) + y2n−1
p−1
for all y ∈ [0, r]2n and t ∈ [0, 1].
i=1
(H6) There exist b2 ∈ (0, αp ) and c > 0 such that
f (t, y) ≤ b2
n−1
p−1
X
(y2i−1 + 2(n − i)y2i ) + y2n−1
+ c for all y ∈ R2n
+ and t ∈ [0, 1].
i=1
(H7) f is increasing in y and there is a constant ω > 0 such that
Z 1
∗
f p /(p−1) (s, ω, . . . , ω) ds < ω.
0
Remark 3.1. A function f is said to be increasing in y if f (t, x) ≤ f (t, y) holds for
2n
every pair x, y ∈ R2n
+ with x ≤ y, where the partial ordering ≤ in R+ is understood
componentwise.
Theorem 3.2. If (H1)–(H4) hold, then (1.1) has at least one positive solution.
Proof. Let
M1 := {u ∈ P : u = Au + λϕ, for some λ ≥ 0},
where ϕ(t) := te−t . Clearly, ϕ(t) is nonnegative and concave on [0, 1]. We claim
M1 is bounded. We first establish the a priori bound of kuk0 for M1 . Indeed,
u ∈ M1 implies u is concave (by the concavity of A and ϕ) and u(t) ≥ (Au)(t). By
definition we obtain
Z tZ 1
1
p−1
ds (3.1)
u(t) ≥
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
0
s
for all u ∈ M1 . Note that p∗ , p∗ /p − 1 ∈ [0, 1]. Now, by (H2), we find
p∗
h n−1
p−1 i p−1
X
p∗
p∗
p∗
a1
(y2i−1 +2(n−i)y2i )+y2n−1
≤ (f (t, y)+c) p−1 ≤ f p−1 (t, y)+c p−1 .
i=1
(3.2)
6
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
Combining this and Jensen’s inequality, we obtain
up∗ (t)
hZ tZ
≥
0
1
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
i p∗
ds
s
1
Z tZ
p∗
f p−1 (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ ds
≥
s
0
1
Z
1
p−1
p∗
k1 (t, s)f p−1 (s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds
=
0
p∗
i p∗
c p−1
ds −
k1 (t, s)
((Bi u)(s) + 2(n − i)((Bi u)(s)) ) + u(s)
≥ a1
2
0
i=1
Z
n−1
1
hX
ip∗
p∗
≥ 2p∗ −1 a1p−1
k1 (t, s)
(Bi u)(s) + 2(n − i)((Bi u)(s))0
ds
p∗
p−1
Z
h n−1
X
1
0
0
i=1
Z
p∗
1
c p−1
k1 (t, s)u (s) ds −
+2
a1
2
0
Z 1
h n−1
i p∗
p∗
XZ 1
k1 (t, s)
(ki (s, τ ) + 2(n − i)hi (s, τ ))u(τ ) dτ
ds
= 2p∗ −1 a1p−1
p∗ −1
p∗
p−1
p∗
0
i=1
Z
0
p∗
1
c p−1
+2
a1
k1 (t, s)u (s) ds −
2
0
Z 1
h Z 1 Pn−1 (k (s, τ ) + 2(n − i)h (s, τ )) n−1
p∗
X
i
i
i=1
= 2p∗ −1 a1p−1
(Ki
k1 (t, s)
Pn−1
0
0
i=1 (Ki + Hi )
i=1
p∗
Z 1
ip∗
p∗
c p−1
p∗ −1 p−1
p∗
+ Hi )u(τ ) dτ
ds + 2
a1
k1 (t, s)u (s) ds −
2
0
p∗ −1 p∗ Z 1
n−1
X
≥ 2
(Ki + Hi )
a1p−1
k1 (t, s)
p∗ −1
p∗
p−1
p∗
0
i=1
×
h n−1
XZ 1
i
(ki (s, τ ) + 2(n − i)hi (s, τ ))up∗ (τ ) dτ ds
0
i=1
Z
p∗
1
c p−1
+2
a1
k1 (t, s)u (s) ds −
2
0
n−1
p∗ −1 p∗ Z 1
X
= 2
(Ki + Hi )
a1p−1
k1 (t, s)
p∗ −1
p∗
p−1
p∗
0
i=1
×
h n−1
X
i
((Bi up∗ )(s) + 2(n − i)((Bi up∗ )(s))0 ) ds
i=1
+2
p∗ −1
p∗
p−1
Z
a1
0
1
p∗
c p−1
k1 (t, s)u (s) ds −
.
2
p∗
(3.3)
EJDE-2013/36
POSITIVE SOLUTIONS
7
Multiply both sides of the above expression by ψ(t) and integrate over [0,1] and use
(2.3) and (2.5) to obtain
Z 1
ψ(t)up∗ (t) dt
0
n−1
p∗ −1 p∗ Z
X
≥ 2
(Ki + Hi )
a1p−1 κ
1
h n−1
X
ψ(t)
(Bi up∗ )(t)
0
i=1
i=1
p∗
1
Z
c p−1
ψ(t)u (t) dt −
2
0
p∗
Z
n−1
1
h
i
p∗
X
p∗ −1
c p−1
p∗ −1 p−1
p∗
=2
a1 κ (n − 1)
.
(Ki + Hi )
+1
ψ(t)u (t) dt −
2
0
i=1
p∗
i
+ 2(n − i)((Bi up∗ )(t))0 dt + 2p∗ −1 a1p−1 κ
Therefore,
Z 1
ψ(t)up∗ (t) dt ≤
0
(3.4)
p∗
p∗
c p−1
:= N1 .
i
h
P
p∗ −1
p∗
n−1
2p∗ a1p−1 κ (n − 1)
+
1
−
2
(K
+
H
)
i
i
i=1
Recall that every u ∈ M1 is concave and increasing on [0, 1]. So is up∗ with
p∗ ∈ (0, 1]. Now Lemma 2.2 yields
1/p∗
kuk0 ≤ (κe)−1/p∗ N1
(3.5)
for all u ∈ M1 , which implies the a priori bound of kuk0 for M1 , as claimed. It
follows, from the boundedness of kuk0 for M1 , that there is λ0 > 0 such that λ ≤ λ0
for all λ ∈ Λ, where
Λ := {λ ≥ 0 : there exists u ∈ M1 such that u = Au + λϕ}.
If u ∈ M1 , then
1
Z 1
p−1
u0 (t) =
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
+ λ(1 − t)e−t
t
for some λ ≥ 0, and by (H3),
Z 1
(u0 )p−1 (t) ≤ γp
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ + γp λp−1
0
t
1
Z
ΦM0 ((u0 )p−1 (τ )) dτ + γp λp−1
.
0
≤ γp
t
0 p−1
Let v(t) := (u )
(t). Then v(t) ∈ C([0, 1], R+ ) and v(1) = 0. Moreover,
Z 1
v(t) ≤ γp
ΦM0 (v(τ )) dτ + γp λp−1
.
0
t
Let F (t) :=
R1
t
ΦM0 (v(τ )) dτ . Then
−F 0 (t) = ΦM0 (v(t)) ≤ ΦM0 (γp F (t) + γp λ0p−1 ).
Therefore,
Z
v(t)
γp λp−1
0
dξ
≤
ΦM0 (ξ)
Z
γp F (t)+γp λp−1
0
γp λp−1
0
dξ
≤ γp (1 − t).
ΦM0 (ξ)
8
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
Hence there is N2 > 0 such that
k(u0 )p−1 k0 = kvk0 = v(0) ≤ N2 .
1/p∗
Let N3 := max{(κe)−1/p∗ N1
1/p−1
, N2
}. Then
kuk ≤ N3 ,
∀u ∈ M1 .
This proves the boundedness of M1 . As a result of this, for every R > N3 , we have
u − Au 6= λψ,
∀u ∈ ∂BR ∩ P, λ ≥ 0.
Now by Lemma 2.4, we obtain
i(A, BR ∩ P, P ) = 0.
(3.6)
Let
M2 := {u ∈ B r ∩ P : u = λAu for some λ ∈ [0, 1]}.
We shall prove M2 = {0}. Indeed, if u ∈ M2 , we have for any u ∈ B r ∩ P
Z tZ 1
1
p−1
u(t) ≤
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
ds. (3.7)
0
s
∗
Notice that p , p∗ /p−1 ≥ 1. Now, similar to (3.3), by Jensen’s inequality and (H4),
we obtain
∗
up (t)
hZ tZ
≤
0
1
Z
1
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
i p∗
ds
s
k1 (t, s)f p
≤
1
p−1
∗
/(p−1)
(s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds
0
p∗ /(p−1)
Z
≤ b1
1
h n−1
i p∗
X
ds
k1 (t, s)
(Bi u)(s) + 2(n − i)((Bi u)(s))0 + u(s)
0
≤ 2p
∗
∗
−1 p /(p−1)
b1
i=1
h n−1
ip∗
XZ 1
k1 (t, s)
(ki (s, τ ) + 2(n − i)hi (s, τ ))u(τ ) dτ
ds
1
Z
0
+ 2p
∗
∗
−1 p /(p−1)
b1
0
i=1
1
Z
∗
k1 (t, s)up (s) ds
0
1
1
Pn−1
(ki (s, τ ) + 2(n − i)hi (s, τ ))
Pn−1
0
0
i=1 (Ki + Hi )
Z 1
n−1
ip∗
X
∗
∗
p∗ −1 p /(p−1)
k1 (t, s)up (s) ds
×
(Ki + Hi )u(τ ) dτ
ds + 2
b1
= 2p
∗
∗
−1 p /(p−1)
b1
Z
hZ
k1 (t, s)
i=1
0
i=1
≤ 2
n−1
X
i=1
p
(Ki + Hi )
∗
−1
p∗ /(p−1)
Z
1
h n−1
X
0
∗
((Bi up )(s)
k1 (t, s)
b1
i=1
Z
i
∗
p∗
0
p∗ −1 p /(p−1)
+ 2(n − i)((Bi u )(s)) ) ds + 2
b1
1
∗
k1 (t, s)up (s) ds.
0
(3.8)
EJDE-2013/36
POSITIVE SOLUTIONS
9
Multiply both sides of the above expression by ψ(t) and integrate over [0, 1] and
use (2.3) and (2.5) to obtain
Z 1
∗
ψ(t)up (t) dt
0
Z
n−1
p∗ −1 ∗
X
p /(p−1)
≤ 2
(Ki + Hi )
b1
1
h n−1
X
∗
ψ(t)
(Bi up )(t)
0
i=1
p
∗
+ 2(n − i)((Bi u )(t))
0
i
dt +
∗
i=1
∗
p /(p−1)
2p −1 b1
Z
(3.9)
1
p∗
ψ(t)u (t) dt
0
= 2p
∗
∗
−1 p /(p−1)
b1
h
(n − 1)
n−1
p∗ −1
iZ
X
(Ki + Hi )
+1
∗
ψ(t)up (t) dt.
0
i=1
R1
1
∗
Therefore, 0 ψ(t)up (t) dt = 0, whence u(t) ≡ 0, ∀u ∈ M2 . As a result, M2 = {0},
as claimed. Consequently,
u 6= λAu,
∀u ∈ ∂Br ∩ P, λ ∈ [0, 1].
Now Lemma 2.5 yields
i(A, Br ∩ P, P ) = 1.
(3.10)
Combining this with (3.6) gives
i(A, (BR \B r ) ∩ P, P ) = 0 − 1 = −1.
Hence the operator A has at least one fixed point on (BR \ B r ) ∩ P and therefore
(1.1) has at least one positive solution. This completes the proof.
Theorem 3.3. If (H1), (H5), (H6) are satisfied, then (1.1) has at least one positive
solution.
Proof. Let
M3 := {u ∈ B r ∩ P : u = Au + λψ for some λ ≥ 0}.
We claim M3 ⊂ {0}. Indeed, if u ∈ M3 , then we have u ≥ Au by definition. That
is,
Z tZ 1
1
p−1
u(t) ≥
ds. (3.11)
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
0
s
Similar to M2 = {0}, we can also obtain M3 ⊂ {0}. As a result of this, we have
u − Au 6= λψ, ∀u ∈ ∂Br ∩ P, λ ≥ 0.
Now Lemma 2.4 gives
i(A, Br ∩ P, P ) = 0.
(3.12)
Let
M4 := {u ∈ P : u = λAu for some λ ∈ [0, 1]}.
We assert M4 is bounded. We first establish the a priori bound of kuk0 for M4 .
Indeed, if u ∈ M4 , then u is concave and u ≤ Au, which can be written in the form
Z tZ 1
1
p−1
u(t) ≤
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
ds, (3.13)
0
s
10
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
for all u ∈ M4 . Note that p∗ , p∗ /p − 1 ≥ 1. Now by (H6) and Jensen’s inequality,
we obtain
∗
up (t) ≤
hZ tZ
0
1
Z
1
f (τ, (Bn−1 u)(τ ), ((Bn−1 u)(τ ))0 , . . . , u(τ ), u0 (τ )) dτ
i p∗
ds
s
k1 (t, s)f p
≤
1
p−1
∗
(s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds
/(p−1)
0
1
Z
≤
n h n−1
ip−1
X
k1 (t, s) b2
(Bi u)(s) + 2(n − i)((Bi u)(s))0 + u(s)
0
i=1
+c
op∗ /(p−1)
p∗ /(p−1)
Z
ds
1
≤ b3
k1 (t, s)
h n−1
X
0
i p∗
ds
((Bi u)(s) + 2(n − i)((Bi u)(s))0 ) + u(s)
i=1
∗
+
≤
p /(p−1)
c1
2
∗
p∗ /(p−1)
2p −1 b3
1
Z
k1 (t, s)
0
+ 2p
∗
∗
−1 p /(p−1)
b3
h n−1
X
((Bi u)(s) + 2(n − i)((Bi u)(s))0 )
ip∗
ds
i=1
1
Z
p∗ /(p−1)
∗
k1 (t, s)up (s) ds +
c1
2
0
1
1
Pn−1
(ki (s, τ ) + 2(n − i)hi (s, τ ))
Pn−1
0
0
i=1 (Ki + Hi )
Z 1
n−1
∗
i
X
p
∗
∗
p∗ /(p−1)
×
(Ki + Hi )u(τ ) dτ
ds + 2p −1 b3
k1 (t, s)up (s) ds
= 2p
+
∗
∗
−1 p /(p−1)
b3
Z
k1 (t, s)
hZ
i=1
0
i=1
p∗ /(p−1)
c1
2
Z
n−1
p∗ −1 ∗
X
p /(p−1)
≤ 2
(Ki + Hi )
b3
1
k1 (t, s)
0
i=1
h n−1
X
∗
((Bi up )(s)
i=1
Z
i
∗
p∗
0
p∗ −1 p /(p−1)
+ 2(n − i)((Bi u )(s)) ) ds + 2
b3
1
k1 (t, s)up∗ (s) ds
0
p∗ /(p−1)
+
c1
2
,
(3.14)
for all u ∈ M4 , b3 ∈ (b2 , αp ) and c1 > 0 being chosen so that
(b2 z + c)p
∗
/(p−1)
p∗ /(p−1) p∗ /(p−1)
≤ b3
z
p∗ /(p−1)
+ c1
, ∀z ≥ 0.
EJDE-2013/36
POSITIVE SOLUTIONS
11
Multiply both sides of (3.14) by ψ(t) and integrate over [0, 1] and use (2.3) and
(2.5) to obtain
Z 1
∗
ψ(t)up (t) dt
0
Z
n−1
p∗ −1 ∗
X
p /(p−1)
≤ 2
(Ki + Hi )
b3
1
ψ(t)
0
i=1
p
∗
+ 2(n − i)((Bi u )(t))
0
i
dt +
h n−1
X
∗
(Bi up )(t)
i=1
∗
∗
p /(p−1)
2p −1 b3
Z
1
p∗ /(p−1)
p∗
ψ(t)u (t) dt +
c1
0
= 2p
∗
∗
−1 p /(p−1)
b3
h
n−1
p∗ −1
iZ
X
(n − 1)
(Ki + Hi )
+1
+
2
∗
ψ(t)up (t) dt
.
(3.15)
Therefore,
Z 1
∗
ψ(t)up (t) dt ≤
0
2
0
i=1
p∗ /(p−1)
c1
1
p∗ /(p−1)
c1
p∗ /(p−1) 2 − 2p∗ b3
(n − 1)
p∗ −1
:= N4 .
(K
+
H
)
+
1
i
i
i=1
Pn−1
This, together with Jensen’s inequality and ψ(t)/e ∈ [0, 1] (Note that p∗ ≥ 1), leads
to
Z 1
∗
Z 1 ∗
p∗ −1
ψ(t) p∗ 1/p
ψ(t)
1/p∗
≤ e p∗ N4
dt ≤ e
up (t)
dt
(3.16)
e
u(t)
e
e
0
0
for all u ∈ M4 . From Lemma 2.2, we find
1/p∗
∗
kuk0 ≤ κ−1 e−1/p N4
:= N5 , ∀u ∈ M4 ,
which implies the a priori bound of kuk0 for M4 , as claimed. Furthermore, for any
positive integer i ≥ 1, this estimate leads to
k(Bi u)k0 = (Bi u)(1) ≤ N5 ,
∀u ∈ M4
and for each positive integer i ≥ 2, we see
Z 1
0
0
k(Bi u) k0 = (Bi u) (0) =
(Bi−1 u)(t) dt ≤ N5 , ∀u ∈ M4 .
0
Moreover, for i = 1, we have
0
0
Z
k(B1 u) k0 = (B1 u) (0) =
1
u(t) dt ≤ N5 , ∀u ∈ M4 .
0
Combining these and (H6), we have
−((u0 )p−1 )0 ≤ f (t, (Bn−1 u)(t), ((Bn−1 u)(t))0 , . . . , u, u0 ),
u(0) = u0 (1) = 0.
Let (u0 )p−1 (t) := w0 (t). Then w ∈ C([0, 1], R+ ) and u0 (1) = 0 implies w0 (1) = 0.
Therefore,
−w00 (t) ≤ n2 N5 , ∀u ∈ M4 ,
so that
kw0 k0 = w0 (0) ≤ n2 N5 .
12
Y. DING, J. XU, X. ZHANG
EJDE-2013/36
Consequently,
1/p−1
ku0 k0 = kw0 k0
Let N6 := max{N5 , (n N5 )
2
1/p−1
≤ (n2 N5 )1/p−1 ,
∀u ∈ M4 .
}. Then
kuk ≤ N6 ,
∀u ∈ M4 .
This proves the boundedness of M4 . As a result of this, for every R > N6 , we have
u 6= λAu,
∀u ∈ ∂BR ∩ P, λ ∈ [0, 1].
Now Lemma 2.5 yields
i(A, BR ∩ P, P ) = 1.
Combining this with (3.12) gives
(3.17)
i(A, (BR \B r ) ∩ P, P ) = 1 − 0 = 1.
Hence the operator A has at least one fixed point on (BR \ B r ) ∩ P and therefore
(1.1) has at least one positive solution. This completes the proof.
Theorem 3.4. If (H1)–(H3), (H6), (H7) are satisfied. Then (1.1) has at least two
positive solutions.
Proof. By (H2), (H3), and (H6), we know that (3.6) and (3.12) hold. Note we may
choose R > ω > r in (3.6) and (3.12) (see the proofs of Theorems 3.2 and 3.3). By
(H7) and Jensen’s inequality, we have that for all u ∈ ∂Bω ∩ P ,
∗
[(Au)(t)]p
Z 1
∗
≤
k1 (t, s)f p /(p−1) (s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds
0
1
Z
fp
≤
∗
/(p−1)
(3.18)
(s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds < ω
0
and
∗
[((Au)(t))0 ]p
Z 1
p∗ /(p−1)
0
0
=
f (s, (Bn−1 u)(s), ((Bn−1 u)(s)) , . . . , u(s), u (s)) ds
Z
≤
t
1
fp
∗
/(p−1)
(3.19)
(s, (Bn−1 u)(s), ((Bn−1 u)(s))0 , . . . , u(s), u0 (s)) ds < ω.
0
Thus we obtain
kAuk < ω = kuk,
∀u ∈ ∂Bω ∩ P,
This implies
u 6= λAu,
∀u ∈ ∂Bω ∩ P, λ ∈ [0, 1].
Now Lemma 2.5 yields
i(A, Bω ∩ P, P ) = 1.
Combining this with (3.6) and (3.12) gives
i(A, (BR \B ω ) ∩ P, P ) = 0 − 1 = −1,
(3.20)
i(A, (Bω \B r ) ∩ P, P ) = 1 − 0 = 1.
Hence the operator A has at least two fixed points, with one on (BR \ B ω ) ∩ P and
the other on (Bω \ B r ) ∩ P . Therefore, (1.1) has at least two positive solutions.
This completes the proof.
EJDE-2013/36
POSITIVE SOLUTIONS
13
Acknowledgments. The authors would like to thank the anonymous referees for
their careful and thorough reading of the original manuscript.
This research supported by grants: 10971046 from the NNSF-China, ZR2012AQ007 from Shandong Provincial Natural Science Foundation, yzc12063 from GIIFSDU, and 2012TS020 from IIFSDU.
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Youzheng Ding
School of Mathematics, Shandong University, Jinan 250100, Shandong, China
E-mail address: dingyouzheng@139.com
Jiafa Xu
School of Mathematics, Shandong University, Jinan 250100, Shandong, China
E-mail address: xujiafa292@sina.com
Xiaoyan Zhang
School of Mathematics, Shandong University, Jinan 250100, Shandong, China
E-mail address: zxysd@sdu.edu.cn