Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 106, pp. 1–14.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
MULTIPLE SOLUTIONS FOR PERTURBED p-LAPLACIAN
BOUNDARY-VALUE PROBLEMS WITH IMPULSIVE EFFECTS
MASSIMILIANO FERRARA, SHAPOUR HEIDARKHANI
Abstract. We establish the existence of three distinct solutions for a perturbed p-Laplacian boundary value problem with impulsive effects. Our approach is based on variational methods.
1. Introduction
In this work, we show the existence of at least three solutions for the nonlinear
perturbed problem
−(ρ(x)Φp (u0 (x)))0 + s(x)Φp (u0 (x)) = λf (x, u(x)) + µg(x, u(x))
0
+
α1 u (a ) − α2 u(a) = 0,
0
a.e. x ∈ (a, b),
−
β1 u (b ) + β2 u(b) = 0
(1.1)
with the impulsive conditions
∆(ρ(xj )Φp (u0 (xj ))) = Ij (u(xj )),
j = 1, 2, . . . , l
(1.2)
where a, b ∈ R with a < b, p > 1, Φp (t) = |t|p−2 t, ρ, s ∈ L∞ ([a, b]) with ρ0 :=
ess inf x∈[a,b] ρ(x) > 0, s0 := ess inf x∈[a,b] s(x) > 0, ρ(a+ ) = ρ(a) > 0, ρ(b− ) =
ρ(b) > 0, α1 , α2 , β1 , β2 are positive constants, f, g : [a, b] × R → R are two
L1 -Carathéodory functions, x0 = a < x1 < x2 < · · · < xl < xl+1 = b,
−
0 +
0 −
∆(ρ(xj )Φp (u0 (xj ))) = ρ(x+
j )Φp (u (xj )) − ρ(xj )Φp (u (xj ))
where z(y + ) and z(y − ) denote the right and left limits of z(y) at y,
Prespectively,
p
Ij : R → R for j = 1, . . . , l are continuous satisfying the condition j=1 (Ij (t1 ) −
Ij (t2 ))(t1 − t2 ) ≥ 0 for every t1 , t2 ∈ R, λ is a positive parameter and µ is a
non-negative parameter.
The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Processes with such a
character arise naturally and often, especially in phenomena studied in mechanical
systems with impact, biological systems such as heart beats, population dynamics, theoretical physics, radiophysics, pharmacokinetics, mathematical economy,
chemical technology, electric technology, metallurgy, ecology, industrial robotics,
biotechnology processes, chemistry, engineering, control theory and so on. For the
2000 Mathematics Subject Classification. 34B15, 34B18, 34B37, 58E30.
Key words and phrases. Multiple solutions; perturbed p-Laplacian; critical point theory;
boundary-value problem with impulsive effects; variational methods.
c
2014
Texas State University - San Marcos.
Submitted August 8, 2013. Published April 15, 2014.
1
2
M. FERRARA, S. HEIDARKHANI
EJDE-2014/106
background, theory and applications of impulsive differential equations, we refer
the interest readers to [3, 4, 10, 13, 14, 16, 19, 21].
Existence and multiplicity of solutions for impulsive differential equations have
been studied by several authors and, for an overview on this subject, we refer the
reader to the papers [1, 2, 15, 18, 23, 24, 26, 27]. For instance, Tian and Ge in
[23], using variational methods, have studied the existence of at least two positive
solutions for the nonlinear impulsive boundary-value problem
−(ρ(t)Φp (u0 (t)))0 + s(t)Φp (u0 (t)) = f (t, u(t))
0
∆(ρ(ti )Φp (u (tj ))) = Ii (u(ti )),
0
αu (a) − βu(a) = A,
a.e. t 6= ti , t ∈ (a, b),
i = 1, 2, . . . , l
0
γu (b) + σu(b) = B,
where a, b ∈ R with a < b, p > 1, Φp (t) = |t|p−2 t, ρ, s ∈ L∞ ([a, b]) with
ess inf t∈[a,b] ρ(t) > 0, ess inf t∈[a,b] s(t) > 0, 0 < ρ(a), ρ(b) < +∞, A ≤ 0, B ≥ 0,
α, β, γ, σ are positive constants, Ii ∈ C([0, +∞), [0, +∞)) for i = 1, . . . , l,
f ∈ C([a, b] × [0, +∞), [0, +∞)), f (t, 0) 6= 0 for t ∈ [a, b], t0 = a < t1 < t2 · · · <
−
+
0 +
0 −
tl < tl+1 = b, ∆(ρ(ti )Φp (u0 (ti ))) = ρ(t+
i )Φp (u (ti )) − ρ(ti )Φp (u (ti )) where x(ti )
−
(respectively x(ti )) denotes the right limit (respectively left limit) of x(t) at t = ti
for i = 1, . . . , l. Also, Tain and Ge in [24] have studied the existence of positive
solutions to the linear and nonlinear Sturm-Liouville impulsive problem by using
variational methods. In fact they have generalized the results of [18, 23]. In [1],
Bai and Dai by using critical point theory, some criteria have obtained to guarantee
that the impulsive problem
−(ρ(t)Φp (u0 (t)))0 + s(t)Φp (u0 (t)) = λf (t, u(t))
0
∆(ρ(ti )Φp (u (tj ))) = Ii (u(ti )),
0
αu (a) − βu(a) = A,
a.e. t 6= ti , t ∈ (a, b),
i = 1, 2, . . . , l
0
γu (b) + σu(b) = B,
where a, b ∈ R with a < b, p > 1, Φp (t) = |t|p−2 t, ρ, s ∈ L∞ ([a, b]) with
ess inf t∈[a,b] ρ(t) > 0, ess inf t∈[a,b] s(t) > 0, 0 < ρ(a), ρ(b) < +∞, λ is a positive
parameter, A, B are constant, α, β, γ, σ are positive constants, f : [a, b] × R → R is
a continuous function, Ii : R → R for i = 1, . . . , l are continuous functions, t0 = a <
−
0 +
0 −
t1 < t2 · · · < tl < tl+1 = b, ∆(ρ(ti )Φp (u0 (ti ))) = ρ(t+
i )Φp (u (ti )) − ρ(ti )Φp (u (ti ))
+
−
where x(ti ) (respectively x(ti )) denotes the right limit (respectively left limit) of
x(t) at t = ti for i = 1, . . . , l, has at least one solution, two solutions and infinitely
many solutions when the parameter lies in different intervals. In particular, in [2],
Bai and Dai, employing a three critical points theorem due to Ricceri [20] have
ensured the existence of at least three solutions for (1.1)-(1.2) in the case µ = 0.
In this article, motivated by [2], employing a three critical points theorem obtained in [5] which we recall in the next section (Theorem 2.1), we ensure the
existence of at least three weak solutions for the problem (1.1)-(1.2). We explicitly observe that in [2], µ = 0 and no exact estimate of λ for which the problem
(1.1)-(1.2) admits multiple solutions is ensured. The aim of this work is to establish
precise values of λ and µ for which the problem (1.1)-(1.2) admits at least three
weak solutions.
Theorem 2.1 has been used for establishing the existence of at least three solutions for eigenvalue problems in the papers [6, 7, 8, 12]. Fora review on the subject,
we refer the reader to [11].
EJDE-2014/106
MULTIPLE SOLUTIONS
3
2. Preliminaries
Our main tool is the following three critical points theorem.
Theorem 2.1 ([5, Theorem 2.6]). Let X be a reflexive real Banach space, Φ : X →
R be a coercive continuously Gâteaux differentiable and sequentially weakly lower
semicontinuous functional whose Gâteaux derivative admits a continuous inverse
on X ∗ , and Ψ : X → R be a continuously Gâteaux differentiable functional whose
Gâteaux derivative is compact, such that Φ(0) = Ψ(0) = 0. Assume that there exist
r > 0 and x ∈ X, with r < Φ(x) such that
(a1)
1
r
Ψ(x)
Φ(x) ,
Φ(x)
Λr :=] Ψ(x) , sup r Ψ(x) [
Φ(x)≤r
supΦ(x)≤r Ψ(x) <
(a2) for each λ ∈
the functional Φ − λΨ is coercive.
Then, for each λ ∈ Λr the functional Φ − λΨ has at least three distinct critical
points in X.
Let X := W 1,p ([a, b]) equipped with the norm
Z b
Z b
1/p
kuk :=
ρ(x)|u0 (x)|p dx +
s(x)|u(x)|p dx
a
a
which is equivalent to the usual one. The following lemma is useful for proving our
main result.
Lemma 2.2 ([23, Lemma 2.6]). Let u ∈ X. Then
kuk∞ = max |u(x)| ≤ M kuk
(2.1)
x∈[a,b]
where
M = 21/q max
1
n
(b −
1/p
a)1/p s0
,
(b − a)1/p o
,
1/p
ρ0
1 1
+ = 1.
p q
By a classical solution of the problem (1.1)-(1.2), we mean a function u ∈ {u(x) ∈
X : ρ(x)Φp (u0 )(.) ∈ W 1,∞ (xj , xj+1 ), j = 0, 1, . . . , l} such that u satisfies (1.1)-(1.2).
We say that a function u ∈ X is a weak solution of the problem (1.1)-(1.2) if
Z b
Z b
ρ(x)Φp (u0 (x))v 0 (x)dx +
s(x)Φp (u(x))v(x)dx
a
a
l
α u(a) β u(b) X
2
2
v(a) + ρ(b)Φp
v(b) +
Ij (u(xj ))v(xj )
+ ρ(a)Φp
α1
β1
j=1
Z b
Z b
−λ
f (x, u(x))v(x)dx − µ
g(x, u(x))v(x)dx = 0
a
a
for every v ∈ X.
For the sake of convenience, in the sequel, we define
Z t
F (x, t) =
f (x, ξ)dξ for all (x, t) ∈ [a, b] × R,
0
Z t
G(x, t) =
g(x, ξ)dξ for all (x, t) ∈ [a, b] × R,
0
C1 =
M p ρ(a)α2p−1
ρ(b)β2p−1 +
p
α1p−1
β1p−1
4
M. FERRARA, S. HEIDARKHANI
EJDE-2014/106
l
C2 =
1 X bj
−
M γj +1 ,
p j=1 γj + 1
C3 =
1 X bj
+
M γj +1 ,
p j=1 γj + 1
l
C4 =
l X
aj M +
j=1
bj
M γj +1 .
γj + 1
For given constants δ1 , δ2 , η1 and η2 put
1
δ
δ2 α1
β1 1
1
+
+
δ1 + δ2 / (b − a)
+
−1 ,
K1 := (b − a)
η1
η2
α2
β2
η1
η2
b−a
Z a+ b−a
Z
Z
b−
b
η1
η2
K2 := |δ1 |p
ρ(x)dx + |K1 |p
ρ(x)dx + |δ2 |p
ρ(x)dx,
a+ b−a
η
a
b− b−a
η
1
1
b − a β o
α1 β1
1
|δ1 |,
+
|δ1 |,
+
|δ2 |,
|δ2 | ,
K3 = max
α2
η1
α2
η2
β2
β2
Z b
Z b
1/p
K4 := (C1 + C3 ) K2 + K3p
s(x)dx + C4 K2 + K3p
s(x)dx
,
nα
1
b − a
a
h1 (x) = δ1
a
b − a α b − a
α1
1
− a , h2 (x) = K1 x − a −
+ δ1
+
,
x+
α2
η1
η1
α2
β1
h3 (x) = δ2 x −
−b ,
β2
and
K F :=
Z
a
a+ b−a
η
1
Z
F (x, h1 (x))dx +
b− b−a
η
2
a+ b−a
η
1
Z
F (x, h2 (x))dx +
b
b− b−a
η
F (x, h3 (x))dx.
2
In this article, we assume throughout, and without further mention, that the
following condition holds:
(A1) The impulsive functions Ij have sublinear growth, i.e., there exist constants
aj > 0, bj > 0, and γj ∈ [0, p − 1) for j = 1, 2, . . . , l such that
|Ij (t)| ≤ aj + bj |t|γj for very t ∈ R, j = 1, 2, . . . , l.
R
Moreover, set Gθ := Ω max|t|≤θ G(x, t)dt for all θ > 0, and Gη := inf Ω×[0,η] G for
all η > 0. If g is sign-changing, then clearly, Gθ ≥ 0 and Gη ≤ 0.
A special case of our main results is the following theorem, whose proof we delay
until the end of the paper.
γj +1
Pl
bj
2 q > 0. Let f : R → R be a
Theorem 2.3. Assume that C20 := p1 − j=1 γj +1
Rt
non-negative continuous function. Put F (t) = 0 f (ξ)dξ for each t ∈ R. Suppose
that
F (ξ)
F (ξ)
lim inf C 0
= lim sup C 0
=0
C40
C40
2
2
p
ξ→0
ξ→+∞ p/q ξ p − 1/q
ξ
−
ξ
ξ
p/q
1/q
2
2
2
2
where
l X
γj +1 bj
C40 :=
aj 21/q +
2 q .
γj + 1
j=1
EJDE-2014/106
MULTIPLE SOLUTIONS
5
Then, there is λ∗ > 0 such that for each λ > λ∗ and for every L1 -Carathéodory
function g : [0, 1] × R → R satisfying the condition
Rt
supx∈[0,1] 0 g(x, s)ds
lim sup
< +∞,
C20 p
C40
|t|→∞
t − 21/q
t
2p/q
∗
∗
there exists δλ,g
> 0 such that, for each µ ∈ [0, δλ,g
[, the problem
−(Φp (u0 (x)))0 + Φp (u0 (x)) = λf (u(x)) + µg(x, u(x))
0
+
u (0 ) − u(0) = 0,
0
a.e. x ∈ (0, 1),
−
u (1 ) + u(1) = 0
with the impulsive conditions
∆(ρ(xj )Φp (u0 (xj ))) = Ij (u(xj )),
j = 1, 2, . . . , l
admits at least three weak solutions.
We need the following proposition in the proof our main result.
Proposition 2.4. Let T : X → X ∗ be the operator defined by
Z b
Z b
0
0
T (u)v =
ρ(x)Φp (u (x))h (x)dx +
s(x)Φp (u(x))h(x)dx
a
a
β u(b) α u(a) 2
2
h(a) + ρ(b)Φp
h(b)
+ ρ(a)Φp
α1
β1
l
X
+
Ij (u(xj ))v(xj )
j=1
for every u, h ∈ X. Then T admits a continuous inverse on X ∗ .
Proof. For any u ∈ X \ {0},
hT (u), ui
kuk
kuk→∞
R b ρ(x)Φ (u0 (x))u0 (x)dx + R b s(x)Φ (u(x))u(x)dx
p
p
a
a
= lim
kuk
kuk→∞
Pl
α2 u(a)
β2 u(b)
ρ(a)Φp
u(a)
+
ρ(b)Φ
u(b) + j=1 Ij (u(xj ))u(xj ) p
α1
β1
+
kuk
R b ρ(x)|u0 (x)|p dx + R b s(x)|u(x)|p dx
a
a
= lim
kuk
kuk→∞
Pl
β2 u(b)
ρ(a)Φp α2αu(a)
u(a)
+
ρ(b)Φ
u(b) + j=1 Ij (u(xj ))u(xj ) p
β1
1
+
kuk
α
u(a)
u(b)
kukp + ρ(a)Φp 2α1
u(a) + ρ(b)Φp β2β1
u(b)
= lim
kuk
kuk→∞
Pl
j=1 Ij (u(xj ))u(xj )
+
= ∞.
kuk
lim
Thus, the map T is coercive.
6
M. FERRARA, S. HEIDARKHANI
EJDE-2014/106
For any u ∈ X and v ∈ X, we have
hT (u) − T (v), u − vi
Z b
=
ρ(x)(Φp (u0 (x)) − Φp (v 0 (x)))(u0 (x) − v 0 (x))
a
+ s(x)(Φp (u(x)) − Φp (u(x)))(u(x) − v(x)) dx
α u(a) α v(a) β u(b) 2
2
2
+ ρ(a)(Φp
− Φp
)(u(a) − v(a)) + ρ(b)(Φp
α1
α1
β1
l
β v(b) X
2
)(u(b) − v(b)) +
− Φp
(Ij (u(xj )) − Ij (v(xj )))(u(xj ) − v(xj )).
β1
j=1
Hence, from our assumptions on the data, we have
Z b
hT (u) − T (v), u − vi ≥
ρ(x)(Φp (u0 (x)) − Φp (v 0 (x)))(u0 (x) − v 0 (x))
a
+ s(x)(Φp (u(x)) − Φp (u(x)))(u(x) − v(x)) dx.
Now, taking into account [22, (2.)], there exist cp , dp > 0 such that
hT (u) − T (v), u − vi
 R cp b ρ(x)|u0 (x) − v 0 (x)|p + s(x)|u(x) − v(x)|p dx
a
≥
0
s(x)|u(x)−v(x))|2
(x)−v 0 (x))|2
dp R b ρ(x)|u
+
dx
0
0
2−p
2−p
(|u(x)|+|v(x)|)
a (|u (x)|+|v (x)|)
if p ≥ 2,
(2.2)
if 1 < p < 2.
At this point, if p ≥ 2, then it follows that
hT (u) − T (v), u − vi ≥ cp ku − vkp ,
so T is uniformly monotone. By [25, Theorem 26.A (d)], T −1 exists and is continuous on X ∗ . On the other hand, if 1 < p < 2, by Hölder’s inequality, we obtain
Z b
s(x)|u(x) − v(x)|p dx
a
b
p/2 Z b
(2−p)/2
s(x)|u(x) − v(x)|2
p
≤
dx
s(x)(|u(x)|
+
|v(x)|)
dx
2−p
a (|u(x)| + |v(x)|)
a
p/2 (p−1)(2−p) Z b
Z b s(x)|u(x) − v(x)|2
2−p
2
p
p
2
2
≤
dx
s(x)(|u(x)|
+
|v(x)|
)dx
2−p
a (|u(x)| + |v(x)|)
a
Z b s(x)|u(x) − v(x)|2
p/2
(2−p)p/2
(p−1)(2−p)
2
≤2
dx
kuk + kvk
.
2−p
a (|u(x)| + |v(x)|)
(2.3)
Similarly, one has
Z b
ρ(x)|u0 (x) − v 0 (x)|p dx
a
(2.4)
Z b ρ(x)|u0 (x) − v 0 (x)|2
p/2
(2−p)p/2
(p−1)(2−p)
2
≤2
dx
kuk + kvk
.
0
0
2−p
a (|u (x)| + |v (x)|)
Z
Then, relation (2.2) together with (2.3) and (2.4), yields
hT (u) − T (v), u − vi
EJDE-2014/106
MULTIPLE SOLUTIONS
7
(p−1)(2−p)
Z
2/p
2
2
dp b
0
0
p
≥
ρ(x)|u
(x)
−
v
(x)|
dx
(kuk + kvk)2−p
a
Z b
2/p +
s(x)|u(x) − v(x)|p dx
a
p−2
Z b
Z b
2/p
2 dp
0
0
p
ρ(x)|u
(x)
−
v
(x)|
dx
+
s(x)|u(x) − v(x)|p dx
2−p
(kuk + kvk)
a
a
ku − vk2
p−2
= 2 dp
.
(kuk + kvk)2−p
≥
Thus, T is strictly monotone. By [25, Theorem 26.A (d)], T −1 exists and is
bounded. Moreover, given g1 , g2 ∈ X ∗ , by the inequality
hT (u) − T (v), u − vi ≥ 2p−2 dp
ku − vk2
2−p ,
(kuk + kvk)
choosing u = T −1 (g1 ) and v = T −1 (g2 ) we have
kT −1 (g1 ) − T −1 (g2 )k ≤
1
2p−2 dp
(kT −1 (g1 )k + kT −1 (g2 )k)2−p kg1 − g2 kX ∗ .
So T −1 is locally Lipschitz continuous and hence continuous. This completes the
proof.
3. Main results
To introduce our result, we fix three constants θ > 0, δ1 and δ2 such that
C2 p
C4
K4
Mp θ − M θ
<
R
b
KF
sup|t|≤θ F (x, t)dx
a
and taking
λ ∈ Λ :=
C4
C2 p
K4
Mp θ − M θ
,
,
R
b
KF
sup|t|≤θ F (x, t)dx
a
we set
δλ,g := min
n
C2 p
Mp θ
−
C4
Mθ
−λ
Rb
sup|t|≤θ F (x, t)dx
a
Gθ
,
K4 − λK F o
(b − a)Gη
(3.1)
o
,
(3.2)
and
1
n
δ λ,g := min δλ,g ,
max{0, (b − a) lim sup|t|→∞
where we define
r
0
supx∈[a,b] G(x,t)
C2 p
C4
Mp t − M
t
= +∞, so that, for instance, δ λ,g = +∞ when
lim sup
|t|→∞
supx∈[a,b] G(x, t)
C2 p
Mp t
and Gη = Gθ = 0.
Now, we formulate our main result.
−
C4
Mt
≤ 0,
}
8
M. FERRARA, S. HEIDARKHANI
EJDE-2014/106
Theorem 3.1. Assume that C2 > 0 and there exist constants δ1 and δ2 , and
positive constants θ, η1 and η2 with δ12 + δ22 6= 0, η1 + η2 < η1 η2 and
1/p
K2
such thatR
(A2)
b
a
sup|t|≤θ F (x,t)dx
C2
Mp
θp −
C4
M
θ
(A3) lim sup|t|→+∞
<
KF
K4
>
θ
C4
> ( )1/(p−1)
M
C1
;
supx∈[a,b] F (x,t)
C2 p
C4
Mp t − M
t
≤ 0.
Then, for each
λ ∈ Λ :=
C2 p
C4
K4
Mp θ − M θ
,
R
b
F
K
sup|t|≤θ F (x, t)dx
a
and for every L1 -Caratéodory function g : [a, b] × R → R satisfying the condition
lim sup
|t|→∞
supx∈[a,b] G(x, t)
C2 p
Mp t
−
C4
Mt
< +∞,
there exists δ λ,g > 0 given by (3.2) such that, for each µ ∈ [0, δ λ,g [, the problem
(1.1)-(1.2) admits at least three distinct weak solutions in X.
Proof. To apply Theorem 2.1 to our problem, we introduce the functionals Φ, Ψ :
X → R for each u ∈ X, as follows
l Z u(xj )
X
1
ρ(a)α2p−1
ρ(b)β2p−1
p
p
p
Φ(u) = kuk +
Ij (t)dt +
|u(a)|
+
p−1
p−1 |u(b)| , ,
p
pα
pβ
0
1
1
j=1
Z b
µ
Ψ(u) =
[F (x, u(x)) + G(x, u(x))]dx.
λ
a
Now we show that the functionals Φ and Ψ satisfy the required conditions. It is well
known that Ψ is a differentiable functional whose differential at the point u ∈ X is
Z b
µ
Ψ0 (u)(v) =
[f (x, u(x)) + g(x, u(x))]v(x)dx,
λ
a
for every v ∈ X, as well as, is sequentially weakly upper semicontinuous. Furthermore, Ψ0 : X → X ∗ is a compact operator. Indeed, it is enough to show that Ψ0
is strongly continuous on X. For this, for fixed u ∈ X, let un → u weakly in X
as n → +∞. Then we have un converges uniformly to u on [a, b] as n → +∞ (see
[25]). Since f and g are L1 -Carathéodory functions, f and g are continuous in R for
every x ∈ [a, b]. So f (x, un ) + µλ g(x, un ) → f (x, u) + µλ g(x, u) strongly as n → +∞,
from which follows Ψ0 (un ) → Ψ0 (u) strongly as n → +∞. Thus we have established
that Ψ0 is strongly continuous on X, which implies that Ψ0 is a compact operator
by Proposition 26.2 of [25]. Moreover, Φ is continuously differentiable and whose
differential at the point u ∈ X is
Z b
Z b
0
0
0
Φ (u)v =
ρ(x)Φp (u (x))v (x)dx +
s(x)Φp (u(x))v(x)dx
a
a
l
β u(b) α u(a) X
2
2
v(a) + ρ(b)Φp
v(b) +
Ij (u(xj ))v(xj )
+ ρ(a)Φp
α1
β1
j=1
EJDE-2014/106
MULTIPLE SOLUTIONS
9
for every v ∈ X, while Proposition 2.4 gives that Φ0 admits a continuous inverse
on X ∗ . Furthermore, Φ is sequentially weakly lower semicontinuous. Indeed, let
for fixed u ∈ X, assume un → u weakly in X as n → +∞. The continuity and
convexity of kukp imply kukp is sequentially weakly lower semicontinuous, which
combining the continuity of Ij for j = 1, . . . , l yields that
l Z un (xj )
1
X
ρ(b)β2p−1
ρ(a)α2p−1
p
p
p
|u
(a)|
+
|u
(b)|
lim
kun k +
Ij (t)dt +
n
n
n→+∞ p
pα1p−1
pβ1p−1
j=1 0
l Z u(xj )
X
1
ρ(b)β2p−1
ρ(a)α2p−1
p
p
p
≥ kuk +
|u(a)|
+
Ij (t)dt +
p−1
p−1 |u(b)| ,
p
pα
pβ
0
1
1
j=1
namely
lim inf Φ(un ) ≥ Φ(u)
n→+∞
which means Φ is sequentially weakly lower semicontinuous. Clearly, the weak
solutions of the problem (1.1) are exactly the solutions of the equation Φ0 (u) −
C2 p
C4
λΨ0 (u) = 0. Put r = M
p θ − M θ and

b−a

h1 (x), x ∈ [a, a + η1 ),
b−a
(3.3)
w(x) = h2 (x), x ∈ [a + b−a
η1 , b − η1 ],


b−a
h3 (x), x ∈ (a + η1 , b].
It is easy to see that w ∈ X and, in particular, in view of
Z b
Z b
Z
ρ(x)|w0 (x)|p dx = K2 and 0 ≤
s(x)|w(x)|p dx ≤ K3p
a
a
b
s(x)dx,
a
we have
b
Z
kwk ≤ K2 + K3p
1/p
s(x)dx
,
a
which in conjunction with the inequality
Φ(u) ≤ (C1 + C3 )kukp + C4 kuk
(3.4)
for all u ∈ X (see[2]), yields
Φ(w) ≤ K4 .
(3.5)
Moreover, by the same reasoning as given given in the proof [2, Lemma 5], using
(3.5), from the condition
C4 1/(p−1)
θ
1/p
>
K2 >
M
C1
one has 0 < r < Φ(w). Taking (2.1) into account, by the same arguing as given in
the proof [2, Lemma 5] we have
Φ−1 (] − ∞, r]) ⊆ {u ∈ X; kuk∞ ≤ θ} ,
and it follows that
Z
sup
Ψ(u) =
u∈Φ−1 (]−∞,r])
sup
u∈Φ−1 (]−∞,r])
Z
≤
b
[F (x, u(x)) +
a
b
sup F (x, t)dx +
a |t|≤θ
µ θ
G .
λ
µ
G(x, u(x))]dx
λ
10
M. FERRARA, S. HEIDARKHANI
EJDE-2014/106
On the other hand, from the definition of Ψ, we infer
Z b
Z
µ b
Ψ(w) =
F (x, w(x))dx +
G(x, w(x))dx
λ a
a
Z
µ b
G(x, w(x))dx
= KF +
λ a
µ
≥ K F + (b − a)
inf
G
λ [a,b]]×[0,η]
µ
= K F + (b − a) Gη .
λ
Therefore, owing to Assumption (A2) and (3.5), we have
Rb
supu∈Φ−1 (]−∞,r]) a [F (x, u(x)) + µλ G(x, u(x))]dx
supu∈Φ−1 (]−∞,r]) Ψ(u)
=
r
r
Rb
(3.6)
µ θ
sup|t|≤θ F (x, t)dx + λ G
a
≤
C2 p
C4
Mp θ − M θ
and
Rb
K F + µλ a G(x, w(x))dx
Ψ(w)
≥
Φ(w)
K4
Rb
F (x, w(x))dx + (b − a) µλ Gη
≥ a
.
K4
Since µ < δλ,g , one has
Rb
C2 p
C4
M p θ − M θ − λ a sup|t|≤θ F (x, t)dx
,
µ<
Gθ
which means
Rb
sup|t|≤θ F (x, t)dx + µλ Gθ
1
a
< .
C2 p
C4
λ
θ
−
θ
Mp
M
(3.7)
Furthermore,
µ<
and this means
K4 − λK F
,
(b − a)Gη
K F + (b − a) µλ Gη
1
> .
K4
λ
Then
Rb
a
sup|t|≤θ F (x, t)dx + µλ Gθ
C2 p
Mp θ
−
C4
Mθ
<
K F + (b − a) µλ Gη
1
<
.
λ
K4
Hence from (3.6)-(3.8), the condition (a1) of Theorem 2.1 is verified.
Finally, since µ < δ λ,g , we can fix l > 0 such that
lim sup
|t|→∞
supx∈[a,b] G(x, t)
C2 p
Mp t
−
C4
Mt
<l
and µl < M p . Therefore, there exists a function h ∈ L1 ([a, b]) such that
G(x, t) ≤ l(
C2 p C4
t −
t) + h(x)
Mp
M
for all x ∈ [a, b] and for all t ∈ R.
(3.8)
EJDE-2014/106
MULTIPLE SOLUTIONS
Now, fix 0 < <
Mp
λ
F (x, t) ≤ (
−
µl
λ.
11
From (A3) there is a function h ∈ L1 ([a, b]) such that
C2 p C4
t −
t) + h (x)
Mp
M
for all x ∈ [a, b]and for all t ∈ R.
Using (3.4), it follows that, for each u ∈ X,
Φ(u) − λΨ(u)
l Z u(xj )
X
1
ρ(a)α2p−1
ρ(b)β2p−1
p
p
|u(a)|
kukp +
Ij (t)dt +
+
p−1
p−1 |u(b)|
p
pα
pβ
0
1
1
j=1
Z
µ
− λ [F (x, u(x)) + G(x, u(x))]dx
λ
Ω
C2
C2
C4
C4
≥ (C2 − λ p − µl p )kukp − (C4 + λ
+ µl )kuk − λkh k1 − µkhk1 ,
M
M
M
M
and thus
lim (Φ(u) − λΨ(u)) = +∞,
=
kuk→+∞
which means the functional Φ − λΨ is coercive, and the condition (a2) of Theorem
2.1 is satisfied. Since, from (3.6) and (3.8),
λ∈
Φ(w)
r
,
,
Ψ(w) supΦ(x)≤r Ψ(x)
Theorem 2.1, with x = w, assures the existence of three critical points for the
functional Φ − λΨ, and the proof is complete.
Here, we exhibit an example whose construction is motivated by [2, Example 1],
in which the hypotheses of Theorem 3.1 are satisfied.
Example 3.2. Consider the problem
−((x + 3)|u0 (x)|u0 (x))0 + (2x + 2)|u(x)|u(x) = λf (x, u(x)) + µg(x, u(x))
a.e. x ∈ (1, 2),
0
u0 (2− ) + u(2) = 0,
1
5
∆((x1 + 3)|u0 (x1 )|u0 (x1 ) = −( + |u(x1 )|3/2 ), x1 ∈ (1, 2)
12 24
+
u (1 ) − u(1) = 0,
(3.9)
where
(
f (x, t) =
x(3t2 − 2t) if (x, t) ∈ [1, 2] × (−∞, 1],
xt
if (x, t) ∈ [1, 2] × [1, +∞).
5
1
+ 24
|u(x1 )|3/2 )
g(x, t) = ex−t t3 for all x ∈ [1, 2] and t ∈ R, and I1 (u(x1 )) = −( 12
3/2
3/2
satisfying the condition (|v(x1 )|
− |u(x1 )| )(u(x1 ) − v(x1 )) ≥ 0 for all u, v ∈
W 1,3 ([1, 2]). A direct calculation shows
(
x(t3 − t2 ) if (x, t) ∈ [1, 2] × (−∞, 1],
F (x, t) = x 2
if (x, t) ∈ [1, 2] × [1, +∞).
2 (t − 1)
In view of Lemma 2.2, M = 1. Choose η1 = η2 = 4, δ1 = 1, δ2 = −1 and θ = 1. We
observe that C1 = 3, C2 = 41 , C3 = 5/12, C4 = 1/6, K1 = 0, K2 = 9/4, K3 = 5/4,
12
M. FERRARA, S. HEIDARKHANI
K4 ≈
1
12×2.011×10−3 ,
R2
K F ≈ 3.125 × 10−1 and
lim sup
EJDE-2014/106
sup|t|≤θ F (x, t)dx ≤ 0. So, since
1
supx∈[1,2] F (x, t)
t3
4
|t|→+∞
−
t
6
= 0,
we see that all assumptions of Theorem 3.1 are satisfied. Hence, for each λ >
1
12×2.011×10−3
3.125×10−1
and every µ ≥ 0 (since g∞ = 0), the problem (3.9) has at least three
solutions in W 1,3 ([1, 2]).
The following example illustrates the result in Theorem 2.3.
Example 3.3. Consider the problem
+
−(|u0 (x)|u0 (x))0 + |u(x)|u(x) = λe−u(x) u2 (x)(3 − u(x)) + µex−u(x) (u(x)+ )γ ,
a.e. x ∈ (0, 1)
0
u0 (1− ) + u(1) = 0,
1
5
∆((x1 + 3)|u0 (x1 )|u0 (x1 ) = −( + |u(x1 )|3/2 ), x1 ∈ (0, 1)
12 24
+
u (0 ) − u(0) = 0,
(3.10)
where u = max{u, 0}, I1 (u(x1 )) =
+
satisfying the condition
(|v(x1 )|3/2 − |u(x1 )|3/2 )(u(x1 ) − v(x1 )) ≥ 0 for all u, v ∈ W 1,3 ([1, 2]) and γ is a
positive real number. It is obvious that C20 = 1/4 and C40 = 1/6. Also a direct
calculation shows F (t) = e−t t3 for all t ∈ R. So, one has
1
−( 12
+
lim inf
ξ→0
5
3/2
)
24 |u(x1 )|
F (ξ)
= lim sup
1
− 6√
ξ→+∞
3 ξ
4
1 3
16 ξ
F (ξ)
= 0.
1
− 6√
3 ξ
4
1 3
16 ξ
Hence, using Theorem 2.3, there is λ∗ > 0 such that, since g∞ = 0, for each λ > λ∗
and µ ≥ 0, the problem (3.10) admits at least three solutions.
K0
Proof of Theorem 2.3. Fix λ > λ∗ := K 0F4 for some constants δ1 and δ2 , and positive constants η1 and η2 with δ12 + δ22 6= 0, η1 + η2 < η1 η2 where
|δ |p
5p
|δ2 |p
5
1
K40 := (C10 + C30 )
+ p+1 (|δ1 | + |δ2 |)p +
+ ( max{|δ1 |, |δ2 |})p
4
2
4
4
1/p
|δ |p
p
p
5
|δ
|
5
2
1
+ p+1 (|δ1 | + |δ2 |)p +
+ ( max{|δ1 |, |δ2 |})p
+ C40
4
2
4
4
γj +1
Pl
bj
2p
1
0
0
where C1 := p and C3 = p + j=1 γj +1 2 q , and
Z 1/4
Z 3/4 5
1
5|δ1 | K 0F :=
F (|δ1 |(x + 1))dx +
F − (|δ1 | + |δ2 |)(x − ) +
dx
2
4
4
0
1/4
Z 1
+
F (|δ2 |(x − 2))dx.
3/4
Recalling that
lim inf
F (ξ)
= 0,
C40
− 21/q
ξ
there is a sequence {θn } ⊂]0, +∞[ such that limn→∞ θn = 0 and
ξ→0
C20 p
ξ
2p/q
sup|ξ|≤θn F (ξ)
lim
0
0
n→∞ C2 θ p − C4 θ
2p/q n
21/q n
= 0.
EJDE-2014/106
MULTIPLE SOLUTIONS
13
Indeed, one has
lim
sup|ξ|≤θn F (ξ)
0
n→∞ C2 θ p
2p/q n
−
C40
θ
21/q n
= lim
F (ξθn )
n→∞ C2 ξ p
2p/q θn
−
C40
ξ
21/q θn
C20 p
ξ
2p/q θn
C20 p
θ
2p/q n
−
−
C40
ξ
21/q θn
C40
θ
21/q n
= 0,
where F (ξθn ) = sup|ξ|≤θn F (ξ). Hence, there exists θ > 0 such that
sup|ξ|≤θ F (ξ)
C20
2p/q
p
θ −
C40
21/q
θ
< min
K 0F
1
;
0
(b − a)K4 (b − a)λ
and
|δ |p
|δ2 |p 1/p
C0
5p
θ
1
+ p+1 (|δ1 | + |δ2 |)p +
> 1/q > ( 40 )1/(p−1) .
4
2
4
C1
2
The conclusion follows by using Theorem 3.1 with η1 = η2 = 4.
Remark 3.4. The methods used here can be applied studying discrete boundary
value problems as in [9], and also non-smooth variational problems as in [17].
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Massimiliano Ferrara
Department of Law and Economics, University Mediterranea of Reggio Calabria, Via
dei Bianchi, 2 - 89131 Reggio Calabria, Italy
E-mail address: massimiliano.ferrara@unirc.it
Shapour Heidarkhani
Department of Mathematics, Faculty of Sciences, Razi University, 67149 Kermanshah,
Iran
E-mail address: s.heidarkhani@razi.ac.ir