1
c
Applied Mathematics E-Notes, 14(2014), 57-68 Available free at mirror sites of http://www.math.nthu.edu.tw/∼amen/
ISSN 1607-2510
2
Existence of Positive Solutions For A Fourth-Order
p-Laplacian Boundary Value Problem∗
3
4
Shoucheng Yu†, Zhilin Yang‡, Lianlong Sun§
5
6
Received 10 April 2013
7
Abstract
This article is concerned with the existence of positive solutions of a fourthorder p-Laplacian boundary value problem. Based on a priori estimates achieved
by utilizing Jensen’s integral inequalities for convex and concave functions, we
use fixed point index theory to establish the existence of positive solutions for the
above problem.
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11
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14
15
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1
Introduction
This article is concerned with the existence of positive solutions for the p-Laplacian
boundary value problem

00 p−1 00 00

u ) = f(t, u, −u00 ),
(|u |
(1)
a1 u(0) − b1 u0 (0) = c1 u(1) + d1 u0 (1) = 0,


00 p
00 p 0
00 p
00 p 0
a2 (−u ) (0) − b2 ((−u ) ) (0) = c2 (−u ) (1) + d2 ((−u ) ) (1) = 0,
where p > 0, f ∈ C([0, 1] × R2 , R+ ), ai , bi , ci , di > 0, and δi = ai di + bi ci + ai ci > 0 for
i = 1, 2.
Fourth order boundary value problems, including those with the p-Laplacian operator, have their origin in beam theory, ice formation, fluids on lungs, brain warping,
designing special curves on surfaces, etc. In our problem (1), the nonlinearity f involves the second-order derivative u00 . Such nonlinearity may be encountered in some
physical models. For example, the equation
∂4u
∂2u
∂u
=
−
p
− a(x)u + b(x)u3
∂t
∂x4
∂x2
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25
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is known in the studies of phase transitions near a Lifschitz point [16].
The p-Laplacian boundary value problems arise in non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow
∗ Mathematics
Subject Classifications: 34B18, 45M20, 47N20.
of Mathematics, Qingdao Technological University, Qingdao, P. R. China
‡ Department of Mathematics, Qingdao Technological University, Qingdao, P. R. China
§ Department of Mathematics, Qingdao Technological University, Qingdao, P. R. China
† Department
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Existence of Positive Solutions for p-Laplacian Equation
36
laws; see [5, 6]. That explains why many authors have extensively studied the existence of positive solutions for p-Laplacian boundary value problems, by using topological degree theory, monotone iterative techniques, coincidence degree theory, and the
Leggett-Williams fixed point theorem or its variants; see [1, 2, 3, 4, 8,10, 11, 12, 13, 14, 15]
and the references therein.
In [14], by using the method of upper and lower solutions, Zhang and Liu obtained
the existence of positive solutions for the fourth-order singular p-Laplacian boundary
value problem
p−1 00 00
(|u00 |
u ) = f(t, u(t)) for 0 < t < 1,
(2)
37
subject to the boundary conditions
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u(0) = u(1) − au(ξ) = u00 (0) = u00 (1) − bu00 (η) = 0,
where p > 1, 0 < ξ, η < 1, and f ∈ C((0, 1) × (0, ∞), (0, ∞)) may be singular at t = 0
and/or at t = 1 and u = 0.
In [15], Zhang and Liu obtained the existence of positive solutions for (2) with the
boundary conditions
u(0) −
m−2
X
ai u(ξi ) = u(1) = u00 (0) −
i=1
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64
m−2
X
bi u(ηi ) = u00 (1) = 0,
(4)
i=1
where m > 3, ai , bi , ξi , ηi ∈ (0, 1)(i = 1, 2, . . . , m − 2) are nonnegative constants and
Pm−2
Pm−2
i=1 ai < 1,
i=1 bi < 1, and f ∈ C((0, 1)×R+ , R+ ) may be singular at t = 0 and/or
at t = 1. By using the monotone iterative method, they established the existence of
positive solutions of pseudo-C 3 [0, 1] for the above problem.
In [8], Guo et al. investigated the existence and multiplicity of positive solutions
for the fourth-order p-Laplacian boundary value problem
p−2
50
(3)
(|u00 |
u00 )00 = λg(t)f(u) for 0 < t < 1,
(5)
where λ is a positive parameter. By using fixed point index theory and the method
of upper and lower solutions, they obtained the following result: there exists λ∗ < ∞
such that (5) has at least two positive solutions for λ ∈ (0, λ∗ ), (5) has at least one
positive solution for λ = λ∗ , and (5) have no positive solution at all for λ > λ∗ .
The presence of the second-order derivative u00 contributes to the difficulty to obtain
a priori estimates of positive solutions for some problems associated with (1). To
facilitate the establishment of such estimates, by using the reduction of order, we
transform (1) into a boundary value problem for an equivalent second-order integrodifferential equation (see the next section for more details). More importantly, we
observe that if p = 1, then (1) reduces to the semilinear fourth-order boundary value
problem

(4)
00

u = f(t, u, −u ),
0
(6)
a1 u(0) − b1 u (0) = c1 u(1) + d1 u0 (1) = 0,


00
000
00
000
a2 u (0) − b2 u (0) = c2 u (1) + d2 u (1) = 0.
Motivated by [11, 12, 13], we regard (6) as a perturbation of (1). In fact, we make
repeated use of the Jensen integral inequalities for convex and concave functions in
59
Yu et al.
72
order to derive a priori estimates of positive solutions for some operator equations
associated with (1), these estimates based on which we use fixed point index theory to
establish the existence of positive solutions for the above problem. Our main results
extend the corresponding ones in [11, 12, 13]. Also, some relations between (1) and (6)
may be seen from the Jensen inequalities for convex and concave functions.
This article is organized as follows. In Section 2, we provide some preliminary results. Our main results, namely Theorem 3.1 and 3.2, followed by two simple examples,
are stated and proved in Section 3.
73
2
74
Let
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Preliminaries
E := C[0, 1], kuk := max |u(t)|, P := {u ∈ E : u(t) > 0 for t ∈ [0, 1]}.
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06t61
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Clearly (E, k · k) is a real Banach space and P is a cone in E. Define Bρ := {u ∈ E :
kuk < ρ} for all ρ > 0. Substituting v := −u00 into (1), we have

R1
p−1 00

v) (t) = f(t, 0 k1 (t, s)v(s)ds, v(t)),
−(|v|
(8)
a2 vp (0) − b2 (vp )0 (0) = 0,

 p
c2 v (1) + d2 (vp )0 (1) = 0,
where
(
1
k1 (t, s) :=
δ1
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(b1 + a1 s)(c1 (1 − t) + d1 ), 0 6 s 6 t 6 1,
(b1 + a1 t)(c1 (1 − s) + d1 ), 0 6 t 6 s 6 1.
Moreover, (8) is equivalent to the nonlinear integral equation
v(t) =
82
Z
1
k2 (t, s)f(s,
0
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where
1
k2 (t, s) :=
δ2
(
Z
88
89
90
1
0
1p
k1 (s, τ )v(τ )dτ, v(s))ds
,
(9)
(b2 + a2 s)(c2 (1 − t) + d2 ), 0 6 s 6 t 6 1,
(b2 + a2 t)(c2 (1 − s) + d2 ), 0 6 t 6 s 6 1.
Define the operator A : P −→ P by
(Av)(t) :=
Z
1
k2 (t, s)f(s,
0
87
(7)
Z
1
0
p1
k1 (s, τ )v(τ )dτ, v(s))ds
.
(10)
Now the condition f ∈ C([0, 1] × R2+ , R+ ) implies that A : P → P is a completely
continuous operator, and the existence of positive solutions for (1) is equivalent to that
of positive fixed points of A. Let
Z 1
k3 (t, τ ) :=
k2 (t, s)k1 (s, τ )ds.
0
60
91
Existence of Positive Solutions for p-Laplacian Equation
For any given nonnegative constants α, β, let
Gα,β (t, s) := αk3 (t, s) + βk2 (t, s)
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and
(Lα,β v)(t) :=
94
(11)
1
Z
Gα,β (t, s)v(s)ds.
(12)
0
95
96
97
Clearly Lα,β : E → E is a completely continuous positive linear operator. If α + β > 0,
then the spectral radius r(Lα,β ) is positive. The Krein-Rutmann theorem then implies
that there exists ϕα,β ∈ P \ {0} such that r(Lα,β )ϕα,β = L∗α,β ϕα,β , i.e.
r(Lα,β )ϕα,β (s) =
98
Z
1
Gα,β (t, s)ϕα,β (t)dt,
(13)
0
99
100
101
where L∗α,β : E → E is the dual operator of A. Note that we may normalize ϕα,β so
that
Z 1
ϕα,β (t)dt = 1.
(14)
0
102
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LEMMA 2.1. For any given nonnegative constants α,β with α + β > 0, let
Z 1
Z 1
2
κα,β :=
tϕα,β (t)dt +
(1 − t)ϕα,β (t)dt,
1
2
0
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105
where ϕα,β is given in (13) and (14). Then for every concave function φ ∈ P , we have
Z 1
φ(t)ϕα,β (t)dt > κα,β kφk.
0
106
The proof can be carried out as that of Lemma 2.4 in [11]. Thus we omit it.
107
LEMMA 2.2 (see [9]). Let a ∈ R+ , b ∈ R+ . If σ ∈ (0, 1], then
(a + b)σ > 2σ−1 (aσ + bσ ).
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If σ ∈ [1, +∞), then
(a + b)σ 6 2σ−1 (aσ + bσ ).
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LEMMA 2.3 (see [9]). Suppose g ∈ C[a, b] with I := g([a, b]) and h ∈ C(I). If h is
convex on I, then
!
Z b
Z b
1
1
h
g(t)dt 6
h(g(t))dt.
b−a a
b−a a
If h is concave on I, then
h
1
b−a
Z
b
a
g(t)dt
!
>
1
b−a
Z
b
h(g(t))dt.
a
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Yu et al.
116
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LEMMA 2.4. Let E and P be defined in (7). Suppose that Ω ⊂ E is a bounded
open set and that T : Ω ∩ K −→ K is a completely continuous operator. If there exist
u0 ∈ K\{0} and µ > 0 such that
uµ − (T u)µ 6= λu0 for all λ > 0 and u ∈ ∂Ω ∩ K,
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then i(T, Ω ∩ K, K) = 0 where i indicates the fixed point index on K.
PROOF. Note the operator Sλ u := ((T u)µ + λu0 )1/µ : P → P is a completely
continuous operator for all λ > 0. If i(T, Ω ∩ K, K) = i(S0 , Ω ∩ K, K) 6= 0, then the
homotopy invariance implies
i(Sλ , Ω ∩ K, K) = i(S0 , Ω ∩ K, K) 6= 0
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for all λ > 0, and, in turn, the fixed point equation u = Sλ u have at least one solution
on K ∩ P for all λ > 0, contradicting the complete continuity of T and the boundedness
of K. Thus we have i(T, Ω ∩ K, K) = 0, as desired. This completes the proof.
130
LEMMA 2.5 (see [7]). Let E be a real Banach space and K be a cone in E. Suppose
that Ω ⊂ E is a bounded open set, 0 ∈ Ω, and T : Ω ∩ K −→ K is a completely
continuous operator. If
131
u − λT u 6= 0 for all λ ∈ [0, 1] and u ∈ ∂Ω ∩ K,
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129
132
then i(T, Ω ∩ K, K) = 1.
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3
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135
Main Results
Let p∗ := min{1, p}, p∗ := min{1, p}, and mi := maxt,s∈[0,1] ki (t, s) for i = 1, 2, 3. Now
we list our hypotheses on f and ai , bi , ci , di for i = 1, 2:
136
(H1) f ∈ C([0, 1] × R2 , R+ ).
137
(H2) ai , bi , ci , di > 0 and δi := ai di + bi ci + ai ci > 0 for i = 1, 2.
138
(H3) There are α1 , β1 > 0 and c > 0, such that r(Ln1 ,n2 ) > 1 and
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f(t, x, y) > α1 xp + β1 yp − c for all t ∈ [0, 1] and x, y > 0,
where Ln1 ,n2 is defined as in (11) and (12),
n1 := 2
p∗
p
−1
p∗
p∗
α1p mp1∗ −1 m2p
−1
and n2 := 2
p∗
p
−1
p∗
p∗
β1p m2p
−1
.
(H4) There are α2 , β2 > 0 and r1 > 0 such that r(Ln3 ,n4 ) < 1 and
f(t, x, y) 6 α2 xp + β2 yp for all t ∈ [0, 1] and x, y ∈ [0, r1],
where Ln3 ,n4 is defined as in (11) and (12),
n3 := 2
p∗
p
p∗
−1
∗
α2p mp1
p∗
−1
m2p
−1
and n4 := 2
p∗
p
p∗
−1
p∗
β2p m2p
−1
.
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146
Existence of Positive Solutions for p-Laplacian Equation
(H5) There are α3 , β3 > 0 and r2 > 0 such that r(Ln5 ,n6 ) > 1 and
f(t, x, y) > α3 xp + β3 yp for all t ∈ [0, 1] and x, y ∈ [0, r2],
147
where Ln5 ,n6 is defined as in (11) and (12),
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n5 := 2
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157
158
159
160
161
162
163
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165
166
167
168
169
170
171
172
p∗
−1
and n6 := 2
p∗
p
−1
p∗
p∗
β3p m2p
−1
.
where Ln7 ,n8 is defined as in (11) and (12),
n7 := 4
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156
p∗
α3p mp1∗ −1 m2p
f(t, x, y) 6 α4 xp + β4 yp + c for all t ∈ [0, 1] and x, y > 0,
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155
−1
(H6) There are α4 , β4 > 0 and c > 0 such that r(Ln7 ,n8 ) < 1 and
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154
p∗
p
p∗
p
p∗
−1
α4p mp1
∗
p∗
−1
m2p
−1
and n8 := 4
p∗
p
p∗
−1
p∗
β4p m2p
−1
.
REMARK 3.1. Notice that the expression (10) implies that if v ∈ P \ {0} is a fixed
point of the operator, then v(t) > 0 holds for all t ∈ (0, 1) with vp ∈ P ∩ C 2 [0, 1]. This,
together with the substitution v := −u00 , in turn, implies that if u is a positive solution
p
of (1), then (−u00 ) ∈ (P \ {0}) ∩ C 2 [0, 1] and hence u ∈ (P \ {0}) ∩ C 4 (0, 1).
THEOREM 3.1. If (H1)-(H4) hold, then (1) has at least one positive solution u ∈
(P \{0}) ∩ C 4 (0, 1).
PROOF. It suffices to prove that A has at least one fixed point v ∈ P \ {0}. To
this end, let
M1 := {v ∈ P : vp∗ = (Av)p∗ + λ, λ > 0}.
We show that M1 is bounded. Indeed, if v ∈ M1 , then vp∗ is concave on [0, 1] and
there exists λ > 0 such that vp∗ = (Av)p∗ + λ. Thus vp∗ (t) > (Av)p∗ (t). Note
p∗ , p∗/p ∈ (0, 1]. By (H3) and the Jensen integral inequality for concave functions
(Lemma 2.3 ), we have that, for all v ∈ M1 ,
p∗
v (t)
>
>
Z
Z
1
k2 (t, s)f(s,
0
1
k2 (t, s)f
0
1
Z
p∗
p
p∗
p
p∗
Z
1
0
(s,
Z
pp∗
k1 (s, τ )v(τ )dτ, v(s))ds
1
k1 (s, τ )v(τ )dτ ), v(s))ds
0
Z
n
[α1
1
o
p∗
p∗
k1p (s, τ )vp (τ )dτ + β1 vp (s)] p − c p ds
0
0
Z 1
Z 1
n p∗
p∗
p∗
p∗
−1
p −1
>
k2 (t, s)m2
2 p [α1p
k1p∗ (s, τ )vp∗ (τ )dτ + β1p vp∗ (s)]
0
0
o
p∗
p
ds
−c
Z 1
Z 1
n p∗
p∗
p∗
p∗
−1
>
k2 (t, s)m2p
2 p −1 [α1p mp1∗ −1
k1 (s, τ )vp∗ (τ )dτ + β1p vp∗ (s)]
>
k2 (t, s)m2p
0
−1
0
63
Yu et al.
−c
173
= 2
174
= 2
176
180
181
Z
=
178
−1
p∗
p
p∗
p
+2
177
179
p∗
p
+2
175
o
p∗
p
1
p∗
p∗
−1
Z
1
p∗
−1
α1p mp1∗ −1 m2p
p∗
p
−1
−1
p∗
p
ds
p∗
p
β1 m2
−1
p∗
p∗
187
188
189
k2 (t, s)v (s)ds − c
Z
193
195
197
1
k2 (t, s)vp∗ (s)ds − c
p∗
p
0
so that
Z
Recall that v
Z
1
Z
1
k2 (t, s)ds
0
p∗
m2p
−1
p∗
p
p∗
m2p
−1
k2 (t, s)ds
0
m3 .
0
p∗
1
p∗
p∗
−1
c p m2p m3
:= N1 for all v ∈ M1 .
v (t)ϕn1 ,n2 (t)dt 6
r(Ln1 ,n2 ) − 1
p∗
is concave on [0, 1]. By Lemma 2.1, we have
R1 p
v ∗ (t)ϕn1 ,n2 (t)dt
N1
p∗
kv k 6 0
6
κn1,n2
κn1 ,n2
for all v ∈ M1 . This proves the boundedness of M1 . Taking R > sup{kvk : v ∈ M1 },
we have
vp∗ 6= (Av)p∗ + λ for v ∈ ∂BR ∩ P and λ > 0.
Now Lemma 2.4 yields
i(A, BR ∩ P, P ) = 0.
(15)
Let
M2 := {v ∈ B r1 ∩ P : v = λAv, 0 6 λ 6 1}.
We claim that M2 = {0}. Indeed, if v ∈ M2 , then there exists λ ∈ [0, 1] such that
v(t) = λAv(t). Thus we have
v(t) 6 (Av)(t) =
Z
1
k2 (t, s)f(s,
Z
1
0
p1
k1 (s, τ )v(τ )dτ, v(s))ds
for all v ∈ B r1 ∩ P.
Note p∗ , p∗/p > 1. By (H4) and the Jensen integral inequality for convex functions
(Lemma 2.3), we have that, for all v ∈ M2 ,
∗
198
m2
−1
k3 (t, s)vp∗ (s)ds
0
0
196
p∗
p
1
Gn1 ,n2 (t, s)vp∗ (s)ds − c
192
194
p∗
p
0
Z
−1
190
191
0
0
185
186
k2 (t, s)k1 (s, τ )vp∗ (τ )dτ ds
Multiply the above inequality by ϕn1 ,n2 (t) and integrate over [0, 1] and use (13) and
(14) to obtain
Z 1
Z 1
p∗
p∗
−1
vp∗ (t)ϕn1 ,n2 (t)dt − c p m2p m3 ,
vp∗ (t)ϕn1 ,n2 (t)dt > r(Ln1 ,n2 )
183
184
1
p∗
0
182
Z
0
p∗
β1p m2p
1
0
α1p mp1∗ −1 m2p
−1
Z
vp (t)
6
Z
1
k2 (t, s)f(s,
0
Z
1
0
pp∗
k1 (s, τ )v(τ )dτ, v(s))ds
64
Existence of Positive Solutions for p-Laplacian Equation
6
199
Z
1
Z
1
0
6
200
6
2
p∗
p
−1
0
1
Z
6
202
2
p∗
p
k1 (s, τ )v(τ )dτ, v(s))ds
−1
α2
Z
1
0
p∗
p
k2 (t, s)m2
k2 (t, s)m2p
−1
−1
i
vp (s) ds
∗
h
k1p (s, τ )vp (τ )dτ
p∗
p
α2
1
Z
0
p∗
[α2p mp1
∗
6
pp∗
ds
+ β2 v (s)
p
∗
∗
k1p (s, τ )vp (τ )dτ
−1
1
Z
p∗
p
+ β2
v (s) ds
p∗
∗
k1 (s, τ )vp (τ )dτ
0
+β2
Z 1 ∗
h p∗ ∗ Z
p∗
p
−1
2 p −1 k2 (t, s)m2p
α2p mp1 −1
203
204
1
Z
p∗
−1
0
p∗
p
(s,
0
p∗
p
1
Z
p∗
p
k2 (t, s)m2
0
201
p∗
k2p (t, s)f
0
1
∗
k1 (s, τ )vp (τ )dτ
0
p∗
p
i
p∗
+β2 v (s)] ds
205
=
206
p∗
p
2
207
+2
208
p∗
p
=
2
+2
209
Z
=
210
1
−1
p∗
p
∗
p
∗
−1
α2 mp1 −1 m2p
−1
−1
p∗
p
p∗
p
p∗
p
p∗
p
β2 m2
p∗
p
∗
α2 mp1
−1
p∗
p
−1
β2 m2
1
0
Z
1
∗
k2 (t, s)k1 (s, τ )vp (τ )dτ ds
0
∗
k2 (t, s)vp (s)ds
0
−1
p∗
p
1
Z
Z
p∗
p
m2
−1
Z
−1
1
Z
1
k3 (t, s)vp (s)ds
∗
0
∗
k2 (t, s)vp (s)ds
0
Gn3,n4 (t, s)vp (s)ds.
∗
0
211
212
213
Multiply the above inequality by ϕn3 ,n4 (t) and integrate over [0, 1] and use (13) and
(14) to obtain
Z 1
Z 1
∗
∗
vp (t)ϕn3 ,n4 (t)dt 6 r(Ln3 ,n4 )
vp (t)ϕn3 ,n4 (t)dt,
0
214
215
216
217
218
219
220
221
222
R1
0
so that 0 vp (t)ϕn3 ,n4 (t)dt = 0, whence vp (t) ≡ 0 and M2 = {0}, as claimed. A
consequence of that is
∗
∗
v 6= λAv for all v ∈ B r1 ∩ P and λ ∈ [0, 1].
Now Lemma 2.5 yields
i(A, B r1 ∩ P, P ) = 1.
(16)
Note that we may assume R > r1 . Combining (15) and (16) gives
i(A, (BR \ B r1 ) ∩ P, P ) = 0 − 1 = −1.
Therefore A has at least one fixed point on (BR \ B r1 ) ∩ P , and thus (1) has at least
one positive solution. This completes the proof.
65
Yu et al.
223
224
225
226
227
228
229
230
THEOREM 3.2. If (H1), (H2), (H5) and (H6) hold, then (1) has at least one positive
solution u ∈ (P \{0}) ∩ C 4 (0, 1).
PROOF. It suffices to prove that A has at least one fixed point v ∈ P \ {0}. To
this end, let
M3 := {v ∈ B r2 ∩ P : vp∗ = (Av)p∗ + λ, λ > 0}.
We shall now prove that M3 ⊂ {0}. Indeed, if v ∈ M3 , then there exists λ > 0 such
that vp∗ = (Av)p∗ + λ. Thus we have
p∗
p∗
v (t) > (Av) (t) =
Z
1
k2 (t, s)f(s,
0
231
232
233
234
1
0
pp∗
k1 (s, τ )v(τ )dτ, v(s))ds
for all v ∈ B r2 ∩P.
Note p∗ , p∗/p ∈ (0, 1]. By (H5) and the Jensen integral inequality for concave functions
(Lemma 2.3), we obtain that, for all v ∈ B r2 ∩ P,
p∗
v (t)
>
>
Z
>
Z
>
237
>
k2 (t, s)f(s,
p∗
p
k2 (t, s)f
1
Z
p∗
p
p∗
p
k2 (t, s)m2
0
236
1
0
1
0
235
Z
1
Z
1
0
(s,
Z
pp∗
k1 (s, τ )v(τ )dτ, v(s))ds
1
k1 (s, τ )v(τ )dτ, v(s))ds
0
−1
Z
α3
1
0
k1p (s, τ )vp (τ )dτ
p Z 1
p∗
∗
p∗
p
p
p∗
p∗
2
α3
k1 (s, τ )v (τ )dτ + β3 v (s) ds
k2 (t, s)m2
0
0
Z 1
Z 1
p∗
p∗
p∗
p∗
p∗ −1
−1
p −1
p
p
p∗
p∗
p
α3 m1
k1 (s, τ )v (τ )dτ + β3 v (s) ds
2
k2 (t, s)m2
Z
p∗
p
−1
p∗
p
−1
p∗
p
p∗
p
−1
0
0
238
=
240
p∗
p
2
=
−1
α3 mp1∗ −1 m2
Z
0
1Z 1
k2 (t, s)k1 (s, τ )vp∗ (τ )dτ ds
0
Z 1
p∗
p∗
−1
β3p m2p
k2 (t, s)vp∗ (s)ds
0
p
Z 1
Z
p∗
p∗
∗
p∗
−1
2 p −1 m2p
α3p mp1∗ −1
k3 (t, s)vp∗ (s)ds + β3p
+2
239
p∗
p
−1
0
241
=
pp∗
+ β3 v (s)
ds
p
Z
0
1
1
k2 (t, s)vp∗ (s)ds
Gn5,n6 (t, s)vp∗ (s)ds.
0
242
243
244
Multiply the above inequality by ϕn5 ,n6 (t) and integrate over [0, 1] and use (13) and
(14) to obtain
Z 1
Z 1
vp∗ (t)ϕn5 ,n6 (t)dt > r(Ln5 ,n6 )
vp∗ (t)ϕn5 ,n6 (t)dt,
0
245
246
247
0
R1
so that 0 vp∗ (t)ϕn5 ,n6 (t)dt = 0, whence vp∗ (t) ≡ 0 and M3 ⊂ {0}, as required. As a
result of that, we have
vp∗ 6= (Av)p∗ + λ for all v ∈ ∂Br2 ∩ P and λ > 0.
66
248
Existence of Positive Solutions for p-Laplacian Equation
Now Lemma 2.4 yields
i(A, Br2 ∩ P, P ) = 0.
249
250
(17)
Let
251
M4 := {v ∈ P : v = λAv, 0 6 λ 6 1}.
252
We are going to prove that M4 is bounded. Indeed, if v ∈ M4 ,then vp is concave and
v(t) 6 (Av)(t) =
253
Z
1
Z
k2 (t, s)f(s,
0
1
0
p1
k1 (s, τ )v(τ )dτ, v(s))ds
for all v ∈ M4 .
255
Note p∗ , pp > 1. By (H6) and the Jensen integral inequality for convex functions
(Lemma 2.3), we have
256
vp (t)
∗
254
∗
257
6
6
Z
6
259
6
260
6
261
262
6
263
264
=
Z
266
=
p∗
p
k2 (t, s)f
Z
pp∗
k1 (s, τ )v(τ )dτ, v(s))ds
1
0
Z
s,
p∗
p
1
0
k1 (s, τ )v(τ )dτ, v(s) ds
Z 1
pp∗
1
−1
p
p
p
k2 (t, s)m2
α4
k1 (s, τ )v (τ )dτ + β4 v (s) + c
ds
0
0
∗
Z 1
Z 1
p∗
p
p∗
p∗
p
p −1
−1
p
p
p
p
p
k2 (t, s)m2
2
[(α4
k1 (s, τ )v (τ )dτ + β4 v (s)) + c ] ds
0
0
Z 1
Z 1
p∗
p∗
p∗
p∗
∗
−1
2 p −1 [2 p −1 α4p
k1 (s, τ )vp (τ )dτ
k2 (t, s)m2p
0
0
p∗
p∗
p∗
∗
+2 p −1 β4p vp (s) + c p ] ds
∗
Z 1
Z 1
p∗
p∗
p∗
∗
p
p∗
∗
∗
−1
k2 (t, s)m2p
4 p −1 α4p mp1 −1
k1 (s, τ )vp (τ )dτ + 4 p −1 β4p vp (s)
0
0
p∗
p∗
+2 p −1 c p ds
p∗
p
Z
p∗
p
4
+4
265
k2 (t, s)f(s,
0
1
0
258
1
Z
1
p∗
−1
p∗
p
α4p mp1
−1
∗
p∗
p
p∗
−1
p∗
p
β4 m2
m2p
−1
−1
Z
0
Z
1
1Z 1
0
p∗
k2 (t, s)v (s)ds + 2
268
269
p∗
p
−1
c
p∗
p
0
Gn7,n8 (t, s)vp∗ (s)ds + 2
p∗
p
0
267
∗
k3 (t, s)vp (τ )dτ ds
−1
c
p∗
p
p∗
m2p
−1
p∗
p
m2
−1
Z
1
k2 (t, s)ds
0
m3 .
Multiply the above inequality by ϕn7 ,n8 (t) and integrate over [0, 1] and use (13) and
(14) to obtain
Z
0
1
vp (t)ϕn7 ,n8 (t) 6 r(Ln7 ,n8 )
∗
Z
0
1
vp∗ (t)ϕn7 ,n8 (t) + 2
p∗
p
−1
c
p∗
p
p∗
m2p
−1
m3 ,
67
Yu et al.
270
so that
Z
271
1
p∗
v (t)ϕn7 ,n8 (t) 6
2
p∗
p
273
Z
1
p
v (t)ϕn7 ,n8 (t)dt
0
pp∗
so that
Z
278
1
0
289
290
291
292
293
294
295
kϕn7 ,n8 k
p∗
p
−1
N2 kϕn7 ,n8 k
p∗
p
Z
1
∗
vp (t)ϕn7 ,n8 (t)dt
0
−1
,
(18)
N2p∗ kϕn7,n8 k1−p∗
.
κn7 ,n8
v 6= λAv for all v ∈ ∂BR ∩ P and λ ∈ [0, 1].
Now Lemma 2.5 implies
i(A, BR ∩ P, P ) = 1.
(19)
Note that we may assume R > r2 . Combining (17) and (19) gives
i(A, (BR \ B r2 ) ∩ P, P ) = 1 − 0 = 1.
286
288
p∗
vp (t)ϕnp7 ,n8 (t)dt
This proves the boundedness of M4 . Taking R >sup{kvk : v ∈ M4 }, we have
284
287
1
vp (t)ϕn7 ,n8 (t)dt 6 N2p∗ kϕn7 ,n8 k1−p∗ .
kvp k 6
282
285
Z
Note vp is concave. By Lemma 2.1, we have
280
283
−1
0
6
276
281
6
6
275
279
p∗
Now p∗ /p > 1 and the Jensen integral inequality for convex functions (Lemma 2.3)
imply
274
277
p∗
c p m2p m3
:= N2 .
1 − r(Ln7 ,n8 )
0
272
−1
Therefore the operator A has at least one fixed point on (BR \ B r2 ) ∩ P . Thus (1) has
at least one positive solution. This completes the proof.
REMARK 3.2. (H3) and (H4) describe the p-superlinear growth of f, as exemplified
by f(t, x, y) := xq1 + yq2 with q1 > p and q2 > p.
REMARK 3.3. (H5) and (H6) describe the p-sublinear growth of f, as exemplified
by f(t, x, y) := xq3 + yq4 with 0 < q3 < p, 0 < q4 < p.
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