R E S E A R C H
Open Access
Existence of positive solutions for boundary
value problems of
p
-Laplacian difference
equations
Fatma T Fen
1,2*and Ilkay Y Karaca
2*Correspondence:
1Department of Mathematics, Gazi
University, Teknikokullar, Ankara, 06500, Turkey
2Department of Mathematics, Ege
University, Bornova, Izmir, 35100, Turkey
Abstract
In this paper, by using the Avery-Peterson fixed point theorem, we investigate the existence of at least three positive solutions for a third orderp-Laplacian difference equation. An example is given to illustrate our main results.
MSC: 34B10; 34B15; 34J10
Keywords: positive solutions; fixed point theorem;p-Laplacian difference equation; boundary value problem
1 Introduction
The aim of this paper is to study the existence of positive solutions for a third orderp -Laplacian difference equation
[φp(y(n))] +q(n)f(n,y(n),y(n)) = , n∈[,N],
ay() –by() = , cy(N+ ) +dy(N+ ) = , y() = , (.)
where
• N> an integer;
• a,c> , andb,d≥withad+ac(N+ ) +bc> ; • f andqare continuous and positive;
• φpis calledp-Laplacian,φp(x) =|x|p–xwithp> , its inverse function is denoted by
φq(x)withφq(x) =|x|q–xwith/p+ /q= ;
• si=rx(i) = ifr,s∈Zands<r, whereZis the integer set, denote
[r,s] ={r,r+ , . . . ,s}forr,s∈Zwithr≤s.
Difference equations, the discrete analog of differential equations, have been widely used in many fields such as computer science, economics, neural network, ecology, cybernet-ics,etc.[]. In the past decade, the existence of positive solutions for the boundary value problems (BVPs) of the difference equations has been extensively studied; to mention a few references, see [–] and the references therein. Also there has been much interest shown in obtaining the existence of positive solutions for the third orderp-Laplacian dy-namic equations on time scales. To mention a few papers along these lines, see [–].
We now discuss briefly several of the appropriate papers on the topic.
Liu [] studied the following second orderp-Laplacian difference equation with multi-point boundary conditions:
⎧ ⎪ ⎨ ⎪ ⎩
[φ(x(n))] +f(n,x(n+ ),x(n),x(n+ )) = , n∈[,N],
x() –mi=αix(ηi) =A,
x(N+ ) –βix(ηi) =B.
The sufficient conditions to guarantee the existence of at least three positive solutions of the above multi-point boundary value problem were established by using a new fixed point theorem obtained in [].
Liu [] studied the following boundary value problem:
⎧ ⎪ ⎨ ⎪ ⎩
[φ(x(n))] +f(n,x(n+ ),x(n),x(n+ )) = , n∈[,N],
x() –mi=αix(ηi) = ,
x(N+ ) –βix(ηi) =B.
By using the five functionals fixed point theorem [], Liu obtained the existence criteria of at least three positive solutions.
Therefore, in this paper, we will consider the existence of at least three positive solutions for the third orderp-Laplacian difference equation (.) by using the Avery-Peterson fixed point theorem [].
Throughout this paper we assume that the following condition holds:
(C) f: [,N]×[, +∞)×R→(, +∞)andq: [,N]→(, +∞)are continuous.
This paper is organized as follows. In Section , we give some preliminary lemmas which are key tools for our proof. The main result is given in Section . Finally, in Section , we give an example to demonstrate our result.
2 Preliminaries
In this section we present some lemmas, which will be needed in the proof of the main result.
Letγ andθ be nonnegative continuous convex functionals onP,αbe a nonnegative continuous concave functional onPandψbe a nonnegative continuous functional onP. Then for positive real numberst,v,w, andz, we define the following convex sets ofP:
P(γ,z) =y∈P:γ(y) <z ,
P(α,v;γ,z) =y∈P:v≤α(y),γ(y)≤z ,
P(α,v;θ,w;γ,z) =y∈P:v≤α(y),θ(y)≤w,γ(y)≤z ,
and a closed set
R(ψ,t;γ,z) =y∈P:t≤ψ(y),γ(y)≤z .
Lemma .([]) LetBbe a real Banach space andP⊂Bbe a cone inB.Letγ,θ be nonnegative continuous convex functionals onP,letαbe a nonnegative continuous concave functional onP,and letψbe a nonnegative continuous functional onPsatisfyingψ(λy)≤
λψ(y)for all≤λ≤,such that for some positive numbers z and M,
α(y)≤ψ(y), y ≤Mγ(y), for all y∈P(γ,z).
Suppose that T:P(γ,z)→P(γ,z)is completely continuous and there exist positive num-bers t,v,and w with <t<v<w such that
(i) {y∈P(α,v;θ,w;γ,z)|α(y) >v} =∅andα(Ty) >vfory∈P(α,v;θ,w;γ,z); (ii) α(Ty) >vfory∈P(α,v;γ,z)withθ(Ty) >w;
(iii) /∈R(ψ,t;γ,z)andψ(Ty) <tfory∈R(ψ,t;γ,z)withψ(y) =t. Then T has at least three fixed points y,y,y∈P(γ,z)such that
γ(yi)≤z, i= , , , v<α(y), t<ψ(y) withα(y) <v andψ(y) <t.
Leth(n) (n∈[,N]) be a positive sequence. Consider the following BVP:
[φp(y(n))] +h(n) = , n∈[,N],
ay() –by() = , cy(N+ ) +dy(N+ ) = , y() = . (.) Lemma . If y is a solution of BVP(.),then there exists unique n∈[,N+ ]such that
y(n) > andy(n+ )≤.
Proof Supposeysatisfies (.). It follows that
y(n) =y() +ny() – n–
r=
r–
i=
φq
i–
s=
h(s)
, n∈[,N+ ]. (.)
The BCs in (.) imply that
ay() =by()
and
cy() +c(N+ )y() –c
N+
r=
r–
i=
φq
i–
s=
h(s)
+dy() –d
N+
i=
φq
i–
s=
h(s)
= .
It follows that
y() =b
ay()
and
y() = a
ad+ac(N+ ) +bc
c
N+
r=
r–
i=
φq
i–
s=
h(s)
+d
N+
i=
φq
i–
s=
h(s)
Similarly, we get
y(n) =y(N+ ) – (N+ –n)y(N+ ) – N+
r=n N+
i=r
φq
i–
s=
h(s)
,
n∈[,N+ ]. (.)
The BCs in (.) imply that
y(N+ ) = –d
cy(N+ )
and
y(N+ )
= – c
ad+ac(N+ ) +bc
a
N+
r=
N+
i=r
φq
i–
s=
h(s)
+b
N+
i=
φq
i–
s=
h(s)
.
Sincea,b≥, andc,d> withad+ac(N+ ) +bc> andh(n) a positive sequence, one can easily see thaty() > andy(N+ )≤. It follows fromy() > ,y(N+ )≤, and the fact thaty(n) is decreasing on [,N+ ] that there exists uniquen∈[,N+ ]
such thaty(n) > andy(n+ )≤. The proof is complete.
Lemma . If y is a solution of BVP(.),then y()≥,y(N+ )≥,and y(n) > for all n∈[,N+ ].
Proof We get from Lemma . the result that there exists uniquen∈[,N] such that
y(n) > andy(n+ )≤. It follows from (.) that
y(n) =
y(n+ ) – n–
i=n+φq( i–
s=h(s)), n∈[n+ ,N+ ],
y(n) + n–
i=n φq(
i–
s=h(s)), n∈[,n].
Then
y(n) =
⎧ ⎪ ⎨ ⎪ ⎩
y(N+ ) – (N+ –n)y(n+ ) + N+
r=n
r–
i=n+φq( i–
s=h(s)),
n∈[n+ ,N+ ],
y() +ny(n) +nr=– n–
i=r φq(
i–
s=h(s)), n∈[,n+ ],
with
y(n+ ) =y() + (n+ )y(n) +
n
r=
n–
i=r
φq
i–
s=
h(s)
=y(N+ ) – (N+ –n)y(n+ )
N+
r=n+
r–
i=n+
φq
i–
s=
h(s)
.
It follows fromh(n) being positive,y(n) > , andy(n+ )≤ that
y(n) >
y(N+ ), n∈[n+ ,N+ ],
Soy(n)≥min{y(),y(N+ )}for alln∈[,N+ ]. From BCs in BVP (.), we get
ay() =by()≥,
cy(N+ ) = –dy(N+ )≥.
Then
miny(),y(N+ ) ≥.
Hencey(n) > for alln∈[,N+ ]. The proof is complete.
Lemma . If y is a solution of BVP(.),then
y(n)≥σn max
n∈[,N+]y(n) for all n∈[,N+ ], (.)
whereσn=min{Nn+,NN+–+n}.
Proof It follows from Lemma . and Lemma . thaty(n)≥ forn∈[,N+ ]. Suppose thaty(n) =max{y(n) :n∈[,N+ ]}. Sincey() > andy(N+ )≤, we getn∈
[,N+ ]. Forn∈[,n], it is easy to see that
y(n) –y()
n
n+y() –y(n) = n
n–
s= y(s) –n n–
s=y(s)
n
=–(n–n)
n–
s=y(s) +n n–
s=n y(s)
n
.
Sincey(n) = –φq(
n–
s=h(s)) < for alln∈[,N+ ], we gety(s)≤y(j) for alls≥j.
Then –(n–n) n–
s=y(s) +n n–
s=n y(s)≤. It follows that y(n)–y()
n n+y() –y(n)≤.
Then
y(n)≥ n
n
y(n) +
– n
n
y()≥ n
N+ n∈max[,N+]y(n) for alln∈[,n].
Similarly, ifn∈[n,N+ ], we get
y(n)≥N+ –n
N+ n∈max[,N+]y(n) for alln∈[n,N+ ].
Then
y(n)≥min
n N+ ,
N+ –n N+
max
n∈[,N+]y(n) for alln∈[,N+ ].
Lemma . If y is a solution of BVP(.),then
y(n) =b+na
a Ah–
n–
r=
r–
i=
φq
i–
s=
h(s)
where Ahsatisfies the equation
Proof The proof follows from Lemma . and is omitted.
We get
On the other hand, by a discussion similar to Lemma . and Lemma ., we haveAh=
y(N+ ),Bh=y(N+ ) with
It follows that
One has from Lemma .Bh=y(N+ )≥. Therefore Define the norm
y=max
max
n∈[,N+]y(n),n∈max[,N+]
y(n).
fory∈P,n∈[,N+ ]. Then
(Ty)(n) = b+na
ad+ac(N+ ) +bc
c
N+
r=
r–
i=
φq
i–
s=
q(s)fs,y(s),y(s)
+d
N+
i=
φq
i–
s=
q(s)fs,y(s),y(s)
– n–
r=
r–
i=
φq
i–
s=
q(s)fs,y(s),y(s)
.
Lemma . Suppose that(C)holds.Then (i) Tysatisfies the following:
⎧ ⎪ ⎨ ⎪ ⎩
[φp((Ty)(n))] +q(n)f(n,y(n),y(n)) = , <n<N,
a(Ty)() –b(Ty)() = , c(Ty)(N+ ) +d(Ty)(N+ ) = ,
(Ty)() = .
(.)
(ii) Ty∈Pfor eachy∈P.
(iii) yis a solution of BVP(.)if and only ifyis a solution of the operator equation
y=Ty.
(iv) T:P→P is completely continuous.
Proof
(i) By the definition ofTy, we get(.).
(ii) Note the definition ofP. Since (C) holds, fory∈P,(.), Lemma ., Lemma . and Lemma . imply that(Ty)(n)is decreasing on[,N+ ]and
(Ty)(n)≥σnmaxn∈[,N+](Ty)(n)for alln∈[,N+ ]. Together with(.), it follows thatTy∈P.
(iii) It is easy to see from(.)thatyis a solution of BVP(.)if and only ifyis a solution of the operator equationy=Ty.
(iv) It suffices to prove thatTis continuous onPandTis relative compact.
We divide the proof into three steps:
Step . For each bounded subsetD⊂P, prove thatAyis bounded inRfory∈D Denote
L=max
max n∈[,N+]y(n),
max n∈[,N+]
y(n):x∈D
and
L= max
j∈[,N]fL(j) =j∈max[,N]umax,|x|≤L
q(j)f(j,u,x).
It follows from (.) in the proof of Lemma . that
≤Ay≤ N
i=
φq
i–
s=
q(s)fs,y(s),y(s)
≤(N+ )φq(NL).
Step . For each bounded subsetD⊂P, and eachy∈D, it is easy to prove thatT is
continuous aty.
Step . For each bounded subsetD⊂P, prove thatTis relative compact onD. In fact, for each bounded subset⊆Dandy∈. Suppose
y=max max
n∈[,N+]y(n),n∈max[,N+]
y(n)<M,
and Step implies that there exists a constantM> such thatAy<M. Then
(Ty)(n) = b+na
a Ay–
n–
r=
r–
i=
φq
i–
s=
q(s)fs,y(s),y(s)
≤ b+na
a M+
N+
r=
N+
i=
φq
N
s=
q(s)fs,y(s),y(s)
≤ b+ (N+ )a
a M+ (N+ )(N+ )φq
N
s=
fM(s)
:=M,
wherefM(s) =maxu≤M,|x|≤Mq(j)f(s,u,x). Similarly, one has
(Ty)(n)=
Ay–
n–
i=
φq
i–
s=
q(s)fs,y(s),y(s)
≤M+
N+
i=
φq
N
s=
fM(s)
=M+ (N+ )φq
N
s=
fM(s)
:=M.
It follows thatTis bounded. SinceB=RN+, one knows thatTis relative compact.
Steps , , and imply thatTis completely continuous.
3 Main result
In this section, our objective is to establish the existence of at least three positive solutions for BVP (.) by using the Avery-Peterson fixed point theorem [].
Choose [N+] >k> , where [x] denotes the largest integer not greater thanx, and denote
σk=min{Nk+,NN+–+k}.
Define the functionals onP:P→[, +∞) by
γ(y) = max n∈[,N+]
y(n),
θ(y) =ψ(y) = max n∈[,N+]y(n),
Fory∈P andn∈[,N+ ] we have
y(n) =y(n) –y() +y()
≤ n–
i=
y(i)
+y()
=
n–
i=
y(i)
+
b ay()
≤ N+
i=
y(i)
+
b ay()
≤
N+ + b
a
max n∈[,N+]
y(n),
i.e.,
max
n∈[,N+]y(n)≤
N+ +b
a
max n∈[,N+]
y(n). (.)
So,
y=max
max
n∈[,N+]y(n),n∈max[,N+]
y(n)
≤
N+ + b
a
max n∈[,N+]
y(n).
Hence, we obtain
y ≤
N+ + b
a
γ(y), y∈P. (.)
Let
= N+
i=
φq
i–
s=
q(s)
,
=σkmin
[N+]
r=k
[N+]–
i=r
φq
i–
s=k
q(s)
, N+–k
r=[N+]
r–
i=[N+]
φq
i–
s=k
q(s)
.
Theorem . Suppose that(C)holds.If there are positive numbers
t<v< v
σk
<z withv<z,
such that the following conditions are satisfied:
(C) f(n,y(n),y(n))≤φp(z)for all(n,y(n),y(n))∈[,N+ ]× [, (N+ + ba)z]×[–z,z];
(C) f(n,y(n),y(n)) >φp(v)for all(n,y(n),y(n))∈[k,N+ –k]×[v,σv
Proof We choose positive numberst,v,w=σv
Now the proof is divided into four steps. Step . We will show that (C) implies that
T:P(γ,z)→P(γ,z).
This implies that (C) holds. Then one has from (C) and (.) in the proof of Lemma .
γ(Ty) = max
Step . We show that condition (i) in Lemma . holds. Choosey(n) = σv
It follows from (C) that
Similarly to Lemma . there existsn∈[k,N+–k] such thaty(n) > andy(n+)≤
We conclude that condition (i) of Lemma . holds.
Step . We prove that condition (ii) of Lemma . holds. Ify∈P(α,v;γ,z) andθ(Ty) >
w=σv
k, then we have
α(Ty) = min
n∈[k,N+–k](Ty)(n)≥σkn∈max[,N+](Ty)(n) =σkθ(Ty) >v.
Then condition (ii) of Lemma . is satisfied.
Sup-≤
N+ + b
a
N+
i=
φq i–
s=
q(s)fs,y(s),y(s)
<
N+ + b
a
a
(N+ )a+b
t
N+
i=
φq
i–
s=
q(s)
=t.
Thus, condition (iii) of Lemma . is satisfied.
From Steps - together with Lemma . we find that the operatorT has at least three fixed points which are positive solutionsy,y, andybelonging toP(γ,z) of (.) such
that
γ(yi)≤z, i= , , , v<α(y), t<ψ(y) withα(y) <v,ψ(y) <t.
4 An example
Example . Consider the following BVP:
y(n) +f(n,y(n),y(n)) = , n∈[, ],
y() –y() = , y() +y() = , y() = , (.)
where f(n,y(n),y(n)) is continuous and positive for all (n,y(n),y(n))∈[,N]×[, +∞)×R. Corresponding to BVP (.), we haveN= ,p= ,q(n) = ,n∈[,N],a=c= ,
b=d= ,φ(y) =y.
It is easy to see that (C) holds.
Choose the constantk= , thenσ=min{,}=,= ,,=. Taking
t= ,v= , andz= ,, it is easy to check that
=t<v= < v
σ
=,
< ,, v= , <z= ,.. If
f(n,y(n),y(n))≤,, , for all(n,y(n),y(n))∈[, ]×[, ,,]× [–,, ,];
f(n,y(n),y(n)) >, for all(n,y(n),y(n))∈[, ]×[,, ]× [–,, ,];
f(n,y(n),y(n)) <, , for all(n,y(n),y(n))∈[, ]×[, ]× [–,, ,],
then Theorem . implies that BVP (.) has at least three positive solutions such that
max n∈[,]
yi(n)≤,, i= , , , < min n∈[,]y(n),
< max
n∈[,]y(n) with n∈min[,]y(n) < andn∈max[,]y(n) < .
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Acknowledgements
The authors would like to thank the referees for their valuable suggestions and comments.
Received: 17 July 2014 Accepted: 25 September 2014 Published:13 Oct 2014
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