Filomat 27:3 (2013), 499–513
DOI 10.2298/FIL1303499D
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
Exact number of positive solutions for quasilinear boundary value
problems with p−convex nonlinearity
Mohammed Derhaba , Hafidha Sebbaghb
a Department
b Department
of Mathematics, Faculty of Sciences, University Abou-Bekr Belkaid Tlemcen, B.P.119, Tlemcen, 13000, Algeria
of Mathematics, Faculty of Sciences, University Abou-Bekr Belkaid Tlemcen, B.P.119, Tlemcen, 13000, Algeria,
Abstract. By using the quadrature method, we study the exact number of positive solutions of the
following quasilinear boundary value problem :
)′
 (

− φp (u′ )




u


 u (0) = u (1)
=
>
=
λ f (u) in (0, 1) ,
0 in (0, 1) ,
0,
( ) p−2
where φp y = y y, y ∈ R, p > 1, λ > 0 and f : R+ → R+ is of class C2 and p-convex function.
1. Introduction
The purpose of this work is to study the exact number of positive solutions of the following quasilinear
boundary value problem















(
)′
− φp (u′ )
= λ f (u) in (0, 1) ,
u
> 0 in (0, 1) ,
u (0) = u (1) = 0,
(1)
( ) p−2
where φp y = y y, y ∈ R, p > 1, λ > 0 and f : R+ → R+ is of class C2 and p-convex function.
Boundary value problems with convex nonlinearity or p-convex nonlinearity have been studied by
several authors. Let us recall some of them.
In [15], the author consider the following problem


−u′′
=



u
>



 u (0) = u (1) =
λ f (u) in (0, 1) ,
0 in (0, 1) ,
0.
2010 Mathematics Subject Classification. Primary 34B15-34C10.
Keywords. p−Laplacian; positive solutions; quadrature method; p−convex nonlinearity.
Received: 30 November 2012; Revised: 13 February, 2013; Accepted: 03 March 2013
Communicated by Eberhard Malkowsky
Research partially supported by a MESRS-DRS grant B02020100075 and PNR project N0 43/28/2011
Email addresses: derhab@yahoo.fr (Mohammed Derhab), hfsb87@hotmail.com (Hafidha Sebbagh)
(2)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
500
By using the quadrature method, he proved when f is convex, with f (0) > 0, then (2) has for each
λ > 0, either zero, one, or two positive solutions on the interval (0, 1) . For such f, the authors determined a
number µ1 ≥ 0 and proved the existence of a number λ∗ ≥ µ1 such that
(i) If µ1 = 0, then λ∗ > 0 and (2) has two solutions for 0 < λ < λ∗ , one for λ = λ∗ , and none for λ > λ∗ ;
(ii) If 0 < µ1 < λ∗ , then (2) has one solution for 0 < λ ≤ µ1 , and λ = λ∗ , two for µ1 < λ < λ∗ , and none for
λ > λ∗ ;
(iii) If 0 < µ1 = λ∗ < +∞, then (2) has one solution for 0 < λ < λ∗ and none for λ > λ∗ ;
(iv) If µ1 = +∞, then λ∗ = +∞ and (2) has solutions for all λ > 0.
The author in [15] gave also a necessary and sufficient condition for λ∗ > µ1 .
In [11], the authors studied the following boundary value problem


−u′′
=



u
>



 u (−R) = u (R) =
f (u) in (−R, R) ,
0 in (−R, R) ,
0,
(3)
where f is a strictly convex C2 function on [0, +∞) and R > 0.
By using the time-map technique ( see e.g. [17] ), the authors proved that if lim
u→+∞
have the following results
f (u)
= +∞, then we
u
(i) If f (u) > 0 (u ∈ [0, +∞)) , then there exists Rsup > 0 such that (3) has two solutions for R < Rsup , one
solution for R = Rsup , and none for R > Rsup ;
(ii) If f (0) = 0 and f ′ (0) > 0, then there exists Rsup > 0 such that (3) has one solution for R < Rsup and
none for R ≥ Rsup ;
(iii) If f (0) < 0, then there exists Rsup > 0 such that (3) has one solution for R ≤ Rsup and none for R ≥ Rsup .
f (u)
= L, with L > 0.
u
In [12], the authors investigated the exact number of positive solutions for the following quasilinear
boundary value problem
The authors studied also the case when lim
u→+∞
)′
 (

− φp (u′ )
=




u
>


 u (−R) = u (R) =
f (u) in (−R, R) ,
0 in (−R, R) ,
0,
(4)
p−2
( )
where φp y = y y, y ∈ R, p > 2 and f is a C1 function on [0, +∞) whose roots are isolated and
p−convex.
f (u)
= +∞,
By using the time-map technique, the authors proved that if f is strictly p−convex and lim
u→+∞ φp (u)
then we have the following results
f (u)
= 0, then for all R > 0 problem (4) has a unique solution,
u→0 φp (u)
(i) If f (0) = 0 and lim
f (u)
= m ∈ (0, +∞) , then there exists R0 > 0 such that (4) has a unique solution
(u)
for R < R0 and none for R ≥ R0 ;
(ii) If f (0) = 0 and lim
u→0 φp
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
501
(iii) If f (0) = 0 and f ′ (0) < 0 or f (0) < 0, then there exists R1 > 0 such that (4) has a unique solution for
R < R1 and none for R ≥ R1 .
The authors in [12] gave also a classification of positive solutions when the nonlinearity f is strictly
f (u)
p−convex, f (0) > 0 and has one root or two roots and lim
= +∞.
u→+∞ φp (u)
One of the methods used to study problems of type (1) is the time- map approach. This method is used
to study uniqueness and multiplicity of solutions for boundary value problems of type (1).
We note that if the time map is monotone, problems of type (1) has at most one solution and if the time
map is convex, or concave, then the problem (1) has at most two solutions ( see [1], [11], [14], [17] and [19] ).
In this work we prove that if the nonlinearity f is super p-linear at 0 and at +∞ and p-convex, then the
time map is concave. Our results improve and generalize those obtained in the literature.
The paper is organized as follows: In section 2, we state our main result. In section 3, we state the
method used to prove our main result. Some preliminary lemmas are the aim of section 4. Section 5 is
devoted to the proof of our main result. Finally in section 6, we give an example
2. Main result
We consider the following boundary value problem















(
)′
− φp (u′ )
= λ f (u) in (0, 1) ,
u
>
0 in (0, 1) ,
u (0) = u (1) =
0,
( ) p−2
where φp y = y y, y ∈ R, p > 1, λ > 0 and f : R+ → R+ satisfying the following hypothesis
f (u)
(H1) lim+ φp (u) = lim
u→0
′
f (u)
u→+∞ φp (u)
(5)
= +∞,
(H2) f (u) > 0, for all u > 0,
(H3) f ′′ (u) > 0, for all u > 0,
(H4) lim ((p − 1) f ′ (u) − u f ′ (u)) < 0,
u→+∞
(H5) (p − 2) f ′ (u) − u f ′′ (u) < 0, for all u > 0 and p > 2.
∫u
(
)
(H6) lim pF(u) − u f (u) < 0, where F (u) = f (s) ds, for all u > 0.
u→+∞
0
Remark 2.1. We note that if 1 < p ≤ 2, then by (H2) and (H3), we have (p − 2) f ′ (u) − u f ′′ (u) < 0, for all u > 0.
To state our result, define
{
}
S+ = u ∈ C1 ([0, 1]) ; u > 0 in (0, 1) , u (0) = u (1) = 0 and u′ (0) > 0 .
Let A+ be the subset of S+ composed by the functions u satisfying:
• u is symmetrical about 12 .
• The derivative of u vanishes once and only once in (0, 1) .
The main result of this work is
Theorem 2.2. Assume that p = 2 or p ≥ 4, λ > 0 and f satisfies the hypothesis (H1)- (H6), then there exists λ∗ > 0
such that
i) If λ > λ∗ , then the problem (5)admits no positive solution,
ii) If λ = λ∗ , then the problem (5) admits a unique positive solution and it belongs to A+ ,
iii) If λ < λ∗ , then the problem (5) admits exactly two positive solutions and they belong to A+ .
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
502
3. Time-mapping approach
In this section we introduce the well-know time mapping approach (see for instance [2]-[10], [15]-[16]
and the monograph [14] for further developments of this topic).
Consider the following boundary value problem
{
(
)′
− φp (u′ ) = 1 (u) in (0, 1) ,
u (0) = u (1) = 0,
(6)
where 1 ∈ C (R+ , R) .
∫s
Define G (s) := 1 (t) dt.
0
For any E ≥ 0 and p > 1, let
}
{
p
(
)
p
X+ p, E = s > 0; E −
G (ζ) > 0, ∀ ζ, 0 < ζ < s ,
p−1
(
)
S+ p, E =
{
(
)
0 if X+ (p, E ) = ∅,
SupX+ p, E otherwise,
{
(
)
( (
))
}
D = E ≥ 0; 0 < S+ p, E < +∞ and 1 S+ p, E > 0 ,
and we define the following time-map
(
)
]− 1
p
p
Ep −
G (u)
du.
p−1
S+ (p,E)[
∫
T+ p, E =
0
We now state the following well know theorem without proof ( See for instance [10] ).
Theorem 3.1. Assume that 1 ∈ C (R+ , R) , E ≥ 0 and p >( 1. Then
the problem (6) admits a solution u ∈ A+
)
1
′ (0)
satisfying u
= E if and
if E ∈ D ∩ (0, +∞) and T+ p, E = 2 . In this case the solution is unique and its
( only
)
sup-norm is equal to S+ p, E .
4. Preliminary lemmas
Lemma 4.1. Consider the equation in s ∈ R+
Ep −
p
λF(s) = 0,
p−1
where E > 0, p > 1 and λ > 0.
(
)
Then for any E > 0, equation (7) admits a unique positive zero r+ p, λ, E . Moreover
(
)
(i) The function E 7→ r+ p, λ, E is C1 in (0, +∞) and
(
)
p − 1 Ep−1
)
∂r+ (
(
) > 0, ∀p > 1, ∀λ > 0 and ∀E > 0.
p, λ, E =
∂E
λ f (r+ p, λ, E )
(7)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
503
(
)
(ii) lim+ r+ p, λ, E = 0.
E→0
(
)
(iii) lim r+ p, λ, E = +∞.
E→+∞
Proof. The proof of this lemma is similar to that of Lemma 4.1 in [4] or Lemma 8 in [2]. So, it is omitted.
(
)
Now, we are ready, for any p > 1, λ > 0 and E > 0, to compute X+ p, λ, E as defined in section 3.
In fact
(
) ]
)[
(
X+ p, λ, E = 0, r+ p, λ, E .
Then
(
)
(
)
S+ p, λ, E = r+ p, λ, E .
On other hand, we have
{
(
)
( (
))
}
D = E > 0, 0 < S+ p, λ, E < +∞ and f S+ p, λ, E > 0
= ]0, +∞[ .
By lemma 4.1, we have
(
)
lim+ S+ p, λ, E = 0,
(8)
E→0
(
)
lim S+ p, λ, E = +∞,
(9)
E→+∞
and
(
)
p − 1 Ep−1
)
∂S+ (
( (
)) > 0, ∀p > 1, ∀λ > 0 and ∀E > 0.
p, λ, E =
∂E
λ f S+ p, λ, E
At present, we define, for any p > 1, λ > 0 and E > 0, the time-map T+ by
T+
(
)
p, λ, E =
]− 1
p
p
Ep −
λF(s)
du.
p−1
S+ (p,λ,E)[
∫
0
(10)
(
)
Now if we put the change of variables u = S+ p, λ, E t in (10), we obtain that
T+
(
)
p, λ, E =
(
p
λ
p−1
) −1p ∫1
[
F(s+ (p, λ, E)) − F(s+ (p, λ, E)t)
1
]− p
dt.
0
We observe that
(
)
(
(
))
T+ p, λ, E = G p, λ, S+ p, λ, E , for all p > 1, λ > 0 and E > 0,
where
1
) −1p ∫1
p
[
]− p
λ
G p, λ, ρ =
F(ρ) − F(ρt)
dt,
p−1
(
)
(
0
(11)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
504
for all p > 1, λ > 0 and ρ >( 0.
)
Since the function E 7→ S+ p, λ, E is an increasing C1 −diffeomorphism from (0, +∞) onto itself it follows
that if we put, for all p > 1 and λ > 0,
{
}
(
)
(
) 1
J1 p, λ := E > 0 : T+ p, λ, E =
,
2
and
}
{
)
(
) 1
(
,
J2 p, λ := ρ > 0 : G p, λ, ρ =
2
then
( (
))
( (
))
Card J1 p, λ = Card J2 p, λ , for all p > 1 and λ > 0.
( Hence,
) from now, we will focus our attention in counting
( the )number of solutions of the equation
G p, λ, ρ = 21 in the variable ρ > 0, instead of the equation T+ p, λ, E = 12 in the variable E > 0.
Proposition 4.2. If u is a positive solution of problem (5) , then u ∈ A+ .
Proof. The proof is similar to that of Proposition 4.2 in [7]. So, it is omitted.
Lemma 4.3. For any p > 1 and λ > 0 , we have
(
)
(i) lim+ G p, λ, ρ = 0.
ρ→0
(
)
(ii) lim G p, λ, ρ = 0.
ρ→+∞
Proof. Let p > 1 and λ > 0 be fixed.
(i) By assumption (H1), we have for all A1 > 0, there exists A2 > 0 such that
f (u) > A1 φp (u) , for all 0 < u < A2 .
Let t ∈ (0, 1) and ρ > 0, we have
∫ρ
F(ρ)−F(ρt)
ρp
=
f (u)du
ρt
ρp
>
∫1 f (ρτ)
dτ
φ p (ρ )
t
∫1 f (ρτ)
τp−1 dτ
φp (ρτ)
t
∫1
A1 τp−1 dτ, for all 0 < ρ < A2
=
A1
p
=
=
t
(1 − tp ) .
Therefore for all 0 < ρ < A2 , we have
( )
( )
F ρ − F ρt
A1
(1 − tp ) .
>
p
ρ
p
(12)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
505
By (11) and (12), one has
(
G p, λ, ρ
)
(
≤
=
p
λ
p−1
−1
A 1 p λ− p
1
) −1p
∫1 [ A
0
(p−1)
p
1
p
1
p
(1 − tp )
]− 1p
dt, for all 0 < ρ < A2
(
)
β 1 − 1p , 1p , for all 0 < ρ < A2 ,
where β (., .) is the Euler Beta function defined by
( )
β x, y :=
∫1
tx−1 (1 − t) y−1 dt, for all x > 0 and y > 0.
0
Then for all A1 > 0, there exists A2 > 0 such that
(
)
− p1
G p, λ, ρ ≤ A1 λ
− 1p
)
)1 (
(
p−1 p
1 1
β 1 − , , for all 0 < ρ < A2 .
p
p p
Which means that
(
)
lim+ G p, λ, ρ = 0.
ρ→0
(ii) The proof is similar to that of (i). So, it is omitted.
Lemma 4.4. For any p > 1 and λ > 0, we have
(
)
(
)
(i) There exists ρ1 > 0 such that the function ρ 7→ G p, λ, ρ is strictly increasing on 0, ρ1 .
(
)
(
)
(ii) There exists ρ2 > ρ1 such that he function ρ 7→ G p, λ, ρ is strictly decreasing on ρ2 , +∞ .
Proof. Let p > 1 and λ > 0 be fixed.
Differentiating (11) with respect to ρ, we obtain
(
)− 1p ∫ρ
λp
)
∂G (
1
p, λ, ρ =
pρ p − 1
∂ρ
0
( )
H ρ − H (u)
du,
p+1
[
]
F(ρ) − F(u) p
(13)
where
H (u) = pF (u) − u f (u) .
We have
(
)
H′ (u) = p − 1 f (u) − u f ′ (u) ,
and
(
)
H′′ (u) = p − 2 f ′ (u) − u f ′′ (u) .
Since f satisfies (H3), (H4), (H5) and (H6) and by Remark 2.1, it follows that there exists two real numbers
ρ1 and ρ2 with 0 < ρ1 < ρ2 such that
(
)
(
)
( )
H′ (u) > 0 on 0, ρ1 , H′ ρ1 = 0 and H′ (u) < 0 on ρ1 , +∞ ,
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
506
and
(
)
( )
(
)
H (u) > 0 on 0, ρ2 , H ρ2 = 0 and H (u) < 0 on ρ2 , +∞ .
So, it follows that
)
(
)
∂G (
p, λ, ρ > 0 for all p > 1, λ > 0 and ρ ∈ ρ1 , +∞ ,
∂ρ
and
)
(
)
∂G (
p, λ, ρ < 0 for all p > 1, λ > 0 and ρ ∈ ρ2 , +∞ .
∂ρ
(
)
(
)
Which means that (for all p)> 1 and λ > 0, the function ρ 7→ G p, λ, ρ is strictly increasing on 0, ρ1 and
strictly decreasing on ρ2 , +∞ .
(
)
Now we are going to prove that the time map ρ 7→ G p, λ, ρ admits a unique critical point on (ρ1 , ρ2 ).
For this, we consider the following Cauchy problem

(
)′

′

 −( φ) p (u ) =( λ)f (u)


 u 1 = ρ, u′ 1 = 0,
2
2
(14)
where ρ > 0.
We note that (since each
positive solution of the problem (5) belongs to A+ , then we can define the
)
time-map ρ 7→ G p, λ, ρ by the following implicit equation
[
)
( ( )
)
(
)
1 ( )
u G ρ , λ, ρ = 0 and u x, λ, ρ > 0, for all x ∈ , G ρ ,
2
(
)
( )
(
)
where x 7→ u x, λ, ρ is a positive solution of the Cauchy problem (14) and G ρ := G p, λ, ρ .
(
)′
)
)
∂u (
∂2 u (
Now if we put by definition v (x) :=
x, λ, ρ , ω (x) :=
x, λ, ρ and since φp (u′ ) = (p−1) |u′ |p−2 u′′
2
∂ρ
∂ρ
for all p ≥ 2, then if p = 2 or p ≥ 3, we have
{
′ p−2 ′ ′
−(p
( )− 1)(|u | v )
v 21 = 1, v′ ( 21 )
=
=
λ f ′ (u)v,
0,
and if p = 2 or p ≥ 4, we have

′ p−2 ′ ′
′ p−4 ′ ′2 ′
′′
2
′

 (p (− )1)(|u | ω ) + (p − 1)(p − 2)(|u | u v ) + λ f (u)v + λ f (u)ω =

 ω 1 = 0, ω′ ( 1 ) = 0
2
2
0,
We have the following results
Proposition
[
( )] 4.5. Assume that p = 2 or p ≥ 3 and f satisfies the hypothesis (H1)-(H6), then v has at most one root
on 0, G ρ .
Proof. The proof is similar to that of Proposition 11 in [12]. So, it is omitted.
( ( ))
Proposition
( ( ))4.6. Assume that p = 2 or p ≥ 4, f satisfies the hypothesis (H1)-(H6) and suppose that v G ρ = 0,
then ω G ρ < 0.
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
Proof. We consider the following problem


(p − 1)(|u′ |p−2 ω′ )′



 +(p − 1)(p − 2)(|u′ |p−4 u′ v′2 )′ + λ f ′′ (u)v2 + λ f ′ (u)ω = 0,


 (1)

 ω 2 = 0, ω′ ( 12 ) = 0.
(15)
Multiplying the equation in (15) by u′ and integrating the resulting equation over ( 21 , x), we obtain
∫x
J
= (p − 1)
:
∫x
′ p−2
(|u |
′ ′ ′
1
2
2
1
2
∫x
∫x
′′
+λ
(|u′ |p−4 u′ v′ )′ u′ (t)dt
ω ) u (t)dt + (p − 1)(p − 2)
′
f ′ (u)u′ (t)ω (t) dt
f (u)v (t) u (t)dt + λ
2
1
2
1
2
= (p − 1)I1 + (p − 1)(p − 2)I2 + I3 + I4 ,
where
∫x
I1 = (|u′ |p−2 ω′ )′ u′ (t)dt,
1
2
∫x
I2 = (|u′ |p−4 u′ v′ )′ u′ (t)dt,
2
1
2
I3 =
∫x
f ′′ (u)v2 (t) u′ (t)dt,
1
2
and
I4 =
∫x
f ′ (u)u′ (t)ω (t) dt.
1
2
Easy computations shows that
∫x
I1
= |u′ (x)|p−2 u′ (x)ω′ (x) −
I2
λ ∫x
= |u′ (x)|p−2 u′ (x)ω′ (x) +
f (u (t)) ω′ (t)dt,
p−1 1
2
∫x
p−2 ′2
p−4 ′
′
′
= |u (x)| v (x) − |u (t)| u (t)u′′ (t)v′2 (t)dt
1
p−1
(|u′ (t)|p−2 u′ (t))′ ω′ (t)dt
1
2
1
2
1 ∫x ′ p−2 ′ ′2
(|u (t)| ) v (t)dt,
p−2 1
2
∫x
2
′
′
= λv (x) f (u(x)) − λ f (ρ) − 2λ f ′ (u(t))v(t)v′ (t)dt
= |u′ (x)|p−2 v′2 (x) −
I3
1
2
(
) ∫x
= λv2 (x) f ′ (u(x)) − λ f ′ (ρ) + 2 p − 1 (|u′ (t)|p−2 v′ (t))′ v′ (t)dt
1
2
]
(
) ∫x [ ′ p−2 ′ ′2
= λv2 (x) f ′ (u(x)) − λ f ′ (ρ) + 2 p − 1
(|u (t)| ) v (t) + |u′ (t)|p−2 v′′ (t) v′ (t) dt,
1
2
507
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
508
and
I4
= λ f (u(x))ω(x) − λ
∫x
f (u(t))ω′ (t)dt.
1
2
Then, we have
= (p − 1) |u′ (x)|p−2 u′ (x)ω′ (x) + (p − 1)(p − 2) |u′ (x)|p−2 v′2 (x) + λv2 (x) f ′ (u(x)) − λ f ′ (ρ)
+λ f (u(x))ω(x) + (p − 1) |u′ (x)|p−2 v′2 (x) .
J
Since J = 0 and using the previous equality, we obtain
Ẽ (x) = E (x) ,
(16)
[
( )]
where Ẽ and E are defined on 12 , G ρ by
(
)
2
Ẽ (x) = (p − 1)(p − 2) |u′ (x)|p−2 v′ (x) + λ f ′ (u(x))v2 + p − 1 |u′ (x)|p−2 v′ (x),
(17)
and
E (x) = (p − 1) |u′ (x)|p−2 u′ (x)ω′ (x) − λ f ′ (ρ) + λ f (u(x))ω(x),
[
( )]
Let x ∈ 12 , G ρ be fixed, we have
(18)
(p − 1)(p − 2)2 |u′ (x)|p−4 u′ (x)u′′ (x)v′ (x) + (p − 1)(p − 2) |u′ (x)|p−2 v′′ (x)
Ẽ (x) =
+λ f ′′ (u(x))u′ (x)v2 (x) + 2λ f ′ (u(x))v(x)v′ (x) + (p − 1)(p − 2) |u′ (x)|p−4 u′ (x)u′′ (x)v′2 (x)
+2(p − 1) |u′ (x)|p−2 v′ (x)v′′ (x)
(p − 1)(p − 2)2 |u′ (x)|p−4 u′ (x)u′′ (x)v′ (x) + (p − 1)(p − 2) |u′ (x)|p−2 v′′ (x)
=
+λ f ′′ (u(x))u′ (x)v2 (x) − (p − 1)(p − 2) |u′ (x)|p−4 u′ (x)u′′ (x)v′2 (x)
= −(p − 2)λ f ′ (u(x))v(x) + λ f ′′ (u(x))u′ (x)v2 (x) − (p − 1)(p − 2) |u′ (x)|p−4 u′ (x)u′′ (x)v′2 (x).
( )
Since p ≥ 2, u′ (x) ≤ 0 for all x in [ 12 , G ρ ] and using the hypothesis (H2) and (H3) and by the proposition
4.5, we obtain that
′
1 ( )
Ẽ′ (x) ≤ 0, for all x ∈ [ , G ρ ],
2
which implies that
1 ( )
1
Ẽ′ (x) ≤ Ẽ′ ( ), for all x ∈ [ , G ρ ].
2
2
( )
Which means that for all x ∈ [ 21 , G ρ ], we have
(
)
2
(p − 1)(p − 2) |u′ (x)|p−2 v′ (x) + λ f ′ (u(x))v2 + p − 1 |u′ (x)|p−2 v′ (x) ≤ λ f ′ (ρ).
By using this inequality and (16), (17) and (18), we obtain
1 ( )
(p − 1) |u′ (x)|p−2 u′ (x)w′ (x) + λ f (u(x))w(x) ≥ 0, for all x ∈ [ , G ρ ].
2
(
( )]
p−2
Dividing this last inequality by |u′ (x))| u′ (x) and since u′ < 0 on 12 , G ρ , we have
(p − 1)w′ (x) +
λ f (u(x))
|u′ (x))|p−2 u′ (x)
w(x) ≤ 0, for all x ∈
(
]
1 ( )
,G ρ .
2
(19)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
509
Now by using the following energy equality
[
]
p − 1 ′ p ′ 1 p
1 ( )
|u (x)| − u ( ) = F(ρ) − F(u(x)), for all x ∈ [ , G ρ ],
p
2
2
we obtain that
(
p
u (x) = −
p−1
′
) 1p
[
]1
1 ( )
F(ρ) − F(u(x)) p , for all x ∈ [ , G ρ ].
2
(20)
By (19) and (20), one has
(p − 1)w′ (x) − (
p
p−1
) p−1
p
(
]
λ f (u(x))
1 ( )
w(x) ≤ 0, for all x ∈ , G ρ .
2
[
] p−1
F(ρ) − F(u(x)) p
That is
(
]
)
1 ( )
d ( −A(x)
e
.w(x) ≤ 0, for all x ∈ , G ρ ,
dx
2
(21)
where
1
A(x) =
p−1
∫x
(
ε
p
p−1
) p−1
p [
f (u)
F(ρ) − F(u)
] p−1
p
du, for all (ϵ, x) ⊂
(
]
1 ( )
,G ρ .
2
( )
Now, we are going to prove that w(G ρ ) < 0. By (15), one has
(p−1)(|u′ |p−2 ω′ )′ (x)+(p−1)(p−2)(|u′ |p−4 u′ v′ )′ (x)+λ f ′′ (u (x))v2 (x)+λ f ′ (u (x))ω (x) = 0, for all x ∈
2
That is
(p − 1)(|u′ |p−2 ω′ )′ (x) =
−(p
[ − 1)(p − 2)×
]
× (p − 3) |u′ (x)|p−4 u′′ (x) v′2 (x) + 2 |u′ (x)|p−4 u′ (x) v′ (x) v′′ ((x))
(
)
( )
−λ f ′′ (u (x))v2 (x) − λ f ′ (u (x))ω (x) , for all x ∈ 12 , G ρ .
( )
( )
( )
1
= ω 12 = 0 and v 12 = 1, we obtain that
2
( )
( )
1
1
− λ f ′ (ρ)ω
lim+ (p − 1)(|u′ |p−2 ω′ )′ (x) = −λ f ′′ (ρ)v2
1
2
2
x→ 2
Since p = 2 or p ≥ 4, u′
=
−λ f ′′ (ρ) < 0,
which implies that there exists α > 0 such that
′ p−2
(p − 1)(|u |
(
)
1 1
ω ) (x) < 0, for all x ∈ , + α .
2 2
′ ′
Then, we have
( )p−2 ( )
1 1
1
1 ω′
= 0, for all x ∈ ( , + α),
|u′ (x)|p−2 ω′ (x) < u′
2 2
2 2
(
)
1 ( )
,G ρ .
2
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
510
which implies that
1 1
w′ (x) < 0, for all x ∈ ( , + α).
2 2
( )
Since ω 21 = 0, we obtain
1 1
w(x) < 0, for all x ∈ ( , + α).
2 2
Now, if we choose ε ∈ ( 12 , 12 + α), we have
w(ε) < 0.
By (21) and (22), it follows that
e−A(x) .w(x) ≤ w(ε) < 0, for all x ∈
( )
If we put x = G ρ , we obtain that
( )
w(G ρ ) < 0.
(22)
(
]
1 ( )
,G ρ .
2
Remark 4.7. The proof of this proposition is a generalization to that of Lemma 2.4 in [8].
Theorem 4.8. If there exists ρ∗ ∈ (0, +∞) such that G′ (ρ∗ ) = 0, then G′′ (ρ∗ ) < 0.
Proof. Since the time map G is defined by the following implicit equation
u(G(ρ), λ, ρ) = 0 and u(x, λ, ρ) > 0, for all x ∈
[
)
1 ( )
,G ρ ,
2
then differentiating the equation u(G(ρ), λ, ρ) = 0 with respect to ρ, we obtain that
∂u(G(ρ), λ, ρ) ′
∂u(G(ρ), λ, ρ)
G (ρ) +
= 0.
∂x
∂ρ
Differentiating the resulting equation, one has
(
)
∂2 u(G(ρ), λ, ρ) ′
∂2 u(G(ρ), λ, ρ) ′
∂u(G(ρ), λ, ρ) ′′
G
(ρ)
+
G (ρ) +
G (ρ)+
2
∂ρ∂x
∂x
∂x
∂2 u(G(ρ), λ, ρ) ′
∂2 u(G(ρ), λ, ρ)
G (ρ) +
= 0.
∂x∂ρ
∂ρ2
Now if there exists ρ∗ ∈ (0, +∞) such that G′ (ρ∗ ) = 0, we have
∂2 u(G(ρ∗ ), λ, ρ∗ )
∂u(G(ρ∗ ), λ, ρ∗ ) ′′
G (ρ∗ ) +
= 0.
∂x
∂ρ2∗
∂2 u(G(ρ∗ ), λ, ρ∗ )
∂u(G(ρ∗ ), λ, ρ∗ )
( )
< 0 and by proposition 4.6, we have w(G ρ∗ ) =
< 0, then it
∂x
∂ρ2∗
follows that
Since
G′′ (ρ∗ ) < 0.
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
511
5. Proof of theorem 2.2
Assume that p = 2 or p ≥ 4 and λ > 0.
From the preceding lemmas one has the following picture about the function ρ 7→ G(p, λ, ρ) which is
defined on (0, +∞)
(
)
(
)
• lim+ G p, λ, ρ = lim G p, λ, ρ = 0,
ρ→+∞
ρ→0
and
• G(p, λ, .) admits a unique critical point ρ∗ at which admits its maximum value G(p, λ, ρ∗ ).
So, by theorem 3.1, one has
i) If G(p, λ, ρ∗ ) < 12 , problem (5) admits no positive solution,
ii) If G(p, λ, ρ∗ ) = 12 , problem (5 ) admits a unique positive solution and it’s belongs to A+ ,
iii) If G(p, λ, ρ∗ ) > 12 , problem (5) admits exactly two positive solutions and they belong to A+ .


1 p
 1

−

∫
[
]

p−1 
If we put λ∗ = ( p ) 2 F(ρ∗ ) − F(ρ∗ t) p dt , then by theorem 3.1, it follows that
 0

(i) If λ < λ∗ , problem (5) admits no positive solution,
(ii) If λ = λ∗ , problem (5) admits a unique positive solution and it belongs to A+ ,
(iii) If λ > λ∗ , problem (5) admits exactly two positive solutions and they belong to A+ .
Remark: We conjecture that the result obtained in this work remain valid if p ∈ ]1, 2[ ∪ ]2, 4[ .
6. Example
In this section, we apply the previous result to the following problem
)′
 (
′)

(u
−
φ
=

p



u
>


 u (0) = u (1) =
uα + c in (0, 1) ,
0 in (0, 1) ,
0,
( ) p−2
where φp y = y y, y ∈ R, p = 2 or p ≥ 4, α > p − 1 and c > 0.
It is easy to check that
lim+
u→0
f (u)
f (u)
= lim
= +∞,
φp (u) u→+∞ φp (u)
f ′ (u) = αuα−1 > 0, for all u > 0,
f ′′ (u) = α(α − 1)uα−2 > 0, for all u > 0,
(23)
Mohammed Derhab, Hafidha Sebbagh / Filomat 27:3 (2013), 499–513
512
(p − 1) f (u) − u f ′ (u) = (p − 1 − α)uα + (p − 1)c, for all u > 0.
Then
lim (p − 1) f (u) − u f ′ (u) = −∞,
u→+∞
(p − 2) f ′ (u) − u f ′′ (u) = α(p − α − 1)uα+1 , for all u > 0.
Then
(p − 2) f ′ (u) − u f ′′ (u) < 0, for all u > 0,
and
pF(u) − u f (u) =
(
)
p − α − 1 α+1
u
+ (p − 1)cu, for all u > 0.
α+1
Then
lim pF(u) − u f (u) = −∞.
u→+∞
Then by theorem 2.2, it follows that there exists λ∗ > 0 such that
i) If λ > λ∗ , then the problem (23) admits no positive solution,
ii) If λ = λ∗ , then the problem (23) admits a unique positive solution and it’s belongs to A+ ,
iii) If λ < λ∗ , then the problem (23) admits exactly two positive solutions and they belong to A+ .
Acknowledgments
The authors thanks the anonymous referee for several comments and suggestions which contributed to
improve this paper.
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