ON A CLASS OF FOURTH-ORDER NONLINEAR DIFFERENCE EQUATIONS MAŁGORZATA MIGDA, ANNA MUSIELAK, AND EWA pptx

Necessary and sufficient conditions are obtained for the difference equation to admit the existence of nonoscillatory solutions with special asymptotic properties.. A solution of 1.1 is cal

DIFFERENCE EQUATIONS

MAŁGORZATA MIGDA, ANNA MUSIELAK, AND EWA SCHMEIDEL

Received 18 August 2003 and in revised form 22 October 2003

We consider a class of fourth-order nonlinear difference equations The classification of nonoscillatory solutions is given Next, we divide the set of solutions of these equations into two types:F+- andF−-solutions Relations between these types of solutions and their nonoscillatory behavior are obtained Necessary and sufficient conditions are obtained for the difference equation to admit the existence of nonoscillatory solutions with special asymptotic properties

1 Introduction

Consider the difference equation

∆
a n∆
b n∆
c n∆yn+ f

n, y n

whereN= {0, 1, 2, },∆ is the forward difference operator defined by ∆yn = y n+1 − y n, and (a n), (b n), and (c n) are sequences of positive real numbers Function f :N×R→R

By a solution of (1.1) we mean a sequence (y n) which satisfies (1.1) forn sufficiently large.

We consider only such solutions which are nontrivial for all largen A solution of (1.1)

is called nonoscillatory if it is eventually positive or eventually negative Otherwise it is called oscillatory

In the last few years there has been an increasing interest in the study of oscillatory and asymptotic behavior of solutions of difference equations Compared to second-order difference equations, the study of higher-order equations, and in particular fourth-order equations (see, e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14]), has received consider-ably less attention An important special case of fourth-order difference equations is the discrete version of the Schr¨odinger equation

The purpose of this paper is to establish some necessary and sufficient conditions for the existence of solutions of (1.1) with special asymptotic properties

Throughout the rest of our investigations, one or several of the following assumptions will be imposed:

Copyright©2004 Hindawi Publishing Corporation

Advances in Di fference Equations 2004:1 (2004) 23–36

2000 Mathematics Subject Classification: 39A10

URL: http://dx.doi.org/10.1155/S1687183904308083

i =1(1/a i)=∞ i =1(1/b i)=∞ i =1(1/c i)= ∞;

(H2) y f (n, y) > 0 for all y =0 andn ∈N;

(H3) the function f (n, y) is continuous onRfor each fixedn ∈N

In [14] we can find the following existence theorem (some modification of Schauder’s theorem) which will be used in this paper

Lemma 1.1 Suppose Ω is a Banach space and K is a closed, bounded, and convex subset Suppose T is a continuous mapping such that T(K) is contained in K, and suppose that T(K) is uniformly Cauchy Then T has a fixed point in K.

2 Main results: existence of nonoscillatory solutions

In this section, we obtain necessary and sufficient conditions for the existence of nonoscil-latory solutions of (1.1) with certain asymptotic properties We start with the following Lemma

Lemma 2.1 Assume that (H1) and (H2) hold Let ( y n ) be an eventually positive solution of ( 1.1 ) Then exactly one of the following statements holds for all sufficiently large n:

(i) y n > 0, ∆y n > 0, ∆(c n∆yn)> 0, and ∆(b n∆(cn∆yn))> 0;

(ii) y n > 0, ∆y n > 0, ∆(c n∆yn)< 0, and ∆(b n∆(cn∆yn))> 0.

Proof Let (y n) be an eventually positive solution of (1.1) Then, by assumption (H2), (∆(an∆(bn∆(cn∆yn)))) is eventually negative Therefore, it is easy to see that the sequences (a n∆(bn∆(cn∆yn))), (b n∆(cn∆yn)), and (c n∆yn) are all monotone and of one sign, say for

n ≥ n1

Suppose thata n2∆(bn2∆(cn2∆yn2))= − c1< 0 for some n2≥ n1 Hence,

a n∆
b n∆
c n∆yn≤ − c1 forn ≥ n2, (2.1) then

∆
b n∆
c n∆yn≤ − c1

Summing both sides of the last inequality fromn2ton −1, we have

b n∆
c n∆yn− b n2∆
c n2∆yn2



≤ −

n−1

i = n2

c1

Thenb n∆(cn∆yn)≤ −n −1

i = n2(c1/a i), which tends to−∞asn → ∞ Then there existsc2> 0 and n3≥ n2such that

b n∆
c n∆yn≤ − c2, forn ≥ n3. (2.4) So,

∆
c n∆yn≤ − c2

Summing both sides of the last inequality formn3ton −1, we obtain

c n∆yn − c n3∆yn3≤ −

n−1

i = n3

c2

which tends to−∞, asn → ∞

Then there exists c3> 0 and n4≥ n3 such that (c n∆yn)≤ − c3, for n ≥ n4 Hence,

∆yn ≤ − c3/c n A final summation yieldsy n − y n3≤ −n −1

i = n4(1/c i)→ −∞, which implies limn →∞ y n = −∞ This contradiction impliesa n∆(bn∆(cn∆yn))> 0 eventually.

Next, assume that there exists n5∈N such that b n∆(cn∆yn)< 0, for n ≥ n5, then (c n∆yn) must be eventually positive for otherwise we are again led to conclude thatc n∆yn

− c n3∆yn3≤ −n −1

i = n3(c2/b i), limn →∞ y n = −∞ Thus, case (ii) is verified

Next, suppose thatb n∆(cn∆yn)> 0 for all n ≥ n1 Thenb n∆(cn∆yn)> b n1∆(cn1∆yn1)=

c4> 0.

Divide the above inequality byb nand sum fromn1ton −1 to get

c n∆yn − c n1∆yn1> c4

n−1

i = n1

1

Now we introduce an operator which divides the set of solutions of a special case of (1.1) into two disjoint subsets We will prove that, for nonoscillatory solution, the first of them equals type (ii) solution and the second equals type (i) solution We assume that

c n = a n+1 Hence (1.1) takes the form

∆
a n∆
b n∆
a n+1∆yn= − f

n, y n

We introduce an operator as follows:

F n = x n −1

a n∆
b n∆
a n+1∆xn−
a n∆xn −1



b n∆
a n+1∆xn. (2.9) Hence

∆Fn = x n∆
a n∆
b n∆an+1∆xn− b n+1∆
a n∆xn −1



∆(an+2∆xn+1. (2.10)

It is clear, by (H2), that the operatorF nis nonincreasing for every nonoscillatory solution (y n) of (2.8)

IfF n ≥0 for alln ∈ N, then a solution (y n) of (2.8) is called anF+-solution IfF n < 0

for somen, then (y n) is called anF−-solution

The operatorF divides the set of solutions of (2.8) into two disjoint subsets:F+- and

F−-solutions

Theorem 2.2 Assume that ( b n ) is a bounded sequence Let y be an F+-solution of ( 2.8 ), then

∞



n =1

b n+1∆
a n∆xn −1



∆
a n+2 ∆xn+1< ∞, (2.11) lim

n →∞ b n∆
a n+1∆yn=0. (2.12)

Proof Let (y n) be anF+-solution of (2.8) Then, from (2.8), we obtain

∆Fk = − y k f

k, y k

− b k+1∆
a k∆yk −1



∆
a k+2∆yk+1. (2.13)

By summation, we obtain

F n = F1− n −

1



k =1

y k f

k, y k

− n −

1



k =1

b k+1∆
a k∆yk −1



∆
a k+2∆yk+1. (2.14)

SinceF n ≥0, we have

n−1

k =1

b k+1∆
a k∆yk −1



∆
a k+2∆yk+1≤ F1. (2.15)

Thereforen −1

k =1b k+1∆(ak∆yk −1)∆(ak+2∆yk+1)< ∞

Because (b n) is a bounded sequence, then (1/b n) is bounded away from zero Hence, from (2.11) and the equality

b n+1∆
a n∆yn −1



∆
a n+2∆yn+1= 1

b n −1

b n −1∆
a n∆yn −1



b n+1∆
a n+2∆yn+1, (2.16)

we obtain

lim

n →∞ b n −1∆
a n∆yn −1



then

lim

n →∞ b n∆
a n+1∆yn=0. (2.18)

Theorem 2.3 Assume that ( b n ) is a bounded sequence Every nonoscillatory solution ( y n)

of ( 2.8 ) is an F+-solution if and only if (y n ) is type (ii) solution.

Proof We prove this theorem for an eventually positive solution.

Let (y n) be an eventually positiveF+-solution Suppose for the sake of contradiction that it is type (i) solution

Then from∆(bn∆(an+1∆yn))> 0, we get b n∆(an+1∆yn)> b M ∆(aM+1∆yM)> 0 for

suf-ficiently largeM and n > M.

This inequality contradicts condition (2.12) of Theorem 2.2 So, (y n) is type (ii) solution

Let (y n) be type (ii) solution We will show the positivity of the operatorF on the whole

sequence Choose m sufficiently large Then, from the definition of type (ii) solution,

we haveF n > 0 for n ≥ m Because the operator F is nonincreasing, hence F j ≥ F m > 0

for all j < m Since m was taken arbitrary, then F n > 0 for all n ∈N So, (y n) is anF+

Remark 2.4 Assume that (b n) is a bounded sequence Then every nonoscillatory solution

of (2.8) is anF−-solution if and only if (y n) is type (i) solution

Now we turn our attention to (1.1) We introduce the notation

P n,N =

n



k = N+2

1

a k

k



j = N+2

1

b j

j



i = N+2

1

c i, Q n,N =

n−1

k = N

1

c k

k−1

j = N

1

b j

j −1



i = N

1

a i . (2.19)

Note thatQ n,Ncan be written in the form

Q n,N = n

−1



i = N

1

a i

n−1

j = i+1

1

b j

n−1

k = j+1

1

Lemma 2.5 Assume conditions (H1) and (H2) hold If ( y n ) is an eventually positive solution

of ( 1.1 ), then there exist positive constants C1and C2and integer N such that

for n ≥ N + 3.

Proof Let ( y n) be an eventually positive solution of (1.1) Theny n > 0 for large n From

Lemma 1.1,∆yn > 0 eventually, and so y n ≥ C1> 0.

Now we prove the right-hand side of (2.21) From (1.1) and (H2), there existsN such

that

∆
a n∆
b n∆
c n∆yn< 0, forn ≥ N. (2.22) Summing the above inequality formN to n −1, we get

∆
b n∆
c n∆yn< A0

whereA0is a constant

Summing again, we have

b n∆
c n∆y< A0

n−1

i = N

1

a i+b N∆
c N∆yN, (2.24) and therefore,

∆
c n∆yn< A0

b n

n−1

i = N

1

a i+A1

b n, forn ≥ N + 1. (2.25) Summing the last inequality, we obtain

c n∆yn < A0

n−1

j = N

1

b j

j −1



l = N

1

a i+A1

n−1

j =1

1

b j +A2, n ≥ N + 2, (2.26) whereA andA are constants

∆yn < A0

c n

n−1

j = N

1

b j

j −1



i = N

1

a i+

A1

c n

n−1

j =1

1

b j+

A2

A final summation yields

y n < A0

n−1

k = N

1

c k

k



j = N

1

b j

j −1



i = N

1

a i+A1

n−1

k = N

1

c k

k−1

j = N

1

b k +A2

n−1

k = N

1

c k+A3, n ≥ N + 3. (2.28)

It is easy to see that every term on the right-hand side of the last inequality is less than

Q n,N Therefore, we obtainy n ≤ C2Q n,Nforn ≥ N + 3, where C2is a positive constant

We say that a nonoscillatory solution (y n) of (1.1) is asymptotically constant if there exist some positive constantα such that y n → α and asymptotically Q n,N if there is some positive constantβ such that y n /Q n,N → β.

According to Lemma 2.5, we may regard an asymptotically constant solution as a

“minimal” solution, and an asymptoticallyQ n,Nsolution as a “maximal” solution Now, we present a necessary and sufficient condition for the existence of an asymptot-icallyQ n,Nsolution

Theorem 2.6 Assume that (H1), (H2), and (H3) hold and f is a nondecreasing function

in another argument, that is, “ f (n,t1) ≤ f (n,t2) for t1 < t2 and each fixed n.” Then a nec-essary and su fficient condition for ( 1.1 ) to have a solution ( y n ) satisfying

lim

n →∞

y n

is that

∞



n =1

f

for some integer N ≥ 1 and some nonzero constant C.

Proof

Necessity Let (y n) be a nonoscillatory solution of (1.1) which satisfies (2.29) Without loss of generality, we may assume thatβ > 0 Then there exist positive numbers d1andd2

such that

d1Q n,N ≤ y n ≤ d2Q n,N, n ≥ N + 3, (2.31) whereN is a sufficiently large integer Then

f

n, y n



≥ f

n,d1Q n



On the other hand, summing (1.1) fromN to n −1, and fromLemma 1.1, we get

0< a n∆
b n∆
c n∆yn= a N∆
b N∆
c N∆yN− n

−1



i = N

f

i, y i

which implies that

∞



i = N

f

i, y i

≤ a N∆
b N∆
c N ∆yN< ∞ (2.34)

So, by (2.32), we have

∞



i = N

f

i,d1Q i,N

Sufficiency Assume that (2.30) holds withC > 0 since a similar argument holds if C < 0.

LetN be large enough that

∞



i = N −3

f

i,CQ n,N

<1

Consider the Banach spaceB N of all real sequencesy =(y n) defined forn ≥ N + 3 such

that

 y  = sup

n ≥ N+3

y n

Q2

n,N

LetS be the subset of B Ndefined by

S =

y n

∈ B N:C

2Q n,N ≤ y n ≤ CQ n,N,n ≥ N + 3



It is not difficult to see that S is a bounded, convex, and closed subset of BN

We define a mappingT : S → B Nas follows:

(T y) n = C

2Q n,N+Q n,N

∞



i = n −1

F(i) +

n−1

j = N F( j −1)Q j,N

+

n−1

i = N

1

c i

i −1



j = N F( j)

j −1



k = N

1

b k

k−1

s = N

1

a s

+

n−1

i = N

1

c i

i −1



j = N

1

b j

j −1



k = N F(k + 1)

k−1

s = N

1

a s, forn ≥ N + 3,

(2.39)

where we have used the notationF(k) for denoting f (k −2,y(k −2))

We first show that T(S) ⊂ S Indeed, if y ∈ S, it is clear from (2.39) that (T y) n ≥

(C/2)Q n,Nforn ≥ N + 3 Furthermore, for n ≥ N + 3, we have

(T y) n ≤ C

2Q n,N+Q n,N

∞



i = n −1

F(i) + Q n,N

n−1

j = N F( j −1)

+

n−1

i = N

1

c i

i −1



k = N

1

b k

k−1

s = N

1

a s

i −1



j = k+1 F( j)

+

n−1

i = N

1

c i

i −1



j = N

1

b j

j −1



s = N

1

a s

j −1



k = s+1 F(k + 1)

2Q n,N+Q n,N

∞



i = N −3

F(i + 2) + Q n,N

∞



j = N −3

F( j + 2)

+Q n,N

∞



j = N −1

F( j + 2) + Q n,N

∞



k = N F(k + 2)

2Q n,N+ 4Q n,N

∞



i = N −3

F(i + 2).

(2.40)

So, we have

(T y) n ≤ C

2Q n,N+ 4Q n,N

∞



i = N −3

f

i, y i



Therefore, by (2.36), we get

(T y) n ≤ C

2Q n,N+ 4Q n,N

∞



i = N −3

f

i,CQ i,N

ThusT maps S into itself.

Next we prove thatT is continuous Let (y(m)) be a sequence inS such that y(m) → y

asm → ∞ BecauseS is closed, y ∈ S Now, by (2.41), we get


T y(m)

n −(T y) n  ≤4Q n,N

∞



i = N −3

f

i, y(i m)

− f

i, y i, n ≥ N + 3, (2.43)

and therefore,


T y(m)

n −(T y) n  ≤ 4

Q n,N

∞



i = N −3

f

i, y i(m)

− f

i, y i. (2.44)

lim

m →∞f

i, y(i m)

− f

i, y i  =0,

f

i, y i(m)

− f

i, y i  ≤2f

i,CQ i,N

, fori ≥ N + 3, (2.45)

we see from Lebesgue’s dominated convergence theorem that

lim

m →∞T y(m) − T y  =0. (2.46) This means thatT is continuous.

Finally, we need to show thatT(S) is uniformly Cauchy To see this, we have to show

that, given any
> 0, there exists an integer N1such that, form > n > N1,





(Q T y)2 m

m,N

−(T y) n

Q2

n,N





for anyy ∈ S Indeed, by (2.41) and (2.36), we have





(Q T y)2 m

m,N

−(T y) n

Q2

n,N





 ≤ Q C n,N+Q8n,N

∞



i = N −3

f

i, y i

≤ 2C

Q n,N −→0. (2.48) Therefore, byLemma 1.1, there existsy ∈ S such that y n =(T y) n, forn ≥ N + 3 It is easy

to see that (y n) is a solution of (1.1) Furthermore, by Stolz’s theorem, we have

lim

n →∞

y n

Q n,N =lim

n →∞

∆yn

∆Qn,N = nlim→∞

c n∆yn

c n∆Qn,N = nlim→∞

∆
c n∆yn

∆
c n∆Qn,N

=lim

n →∞

b n∆
c n∆yn

b n∆
c n∆Qn,N  = nlim→∞

∆
b n∆
c n∆yn

∆ n −1

i =1

1/a i

=lim

n →∞ a n∆
b n∆
c n∆yn,

(2.49)

so,

lim

n →∞

y n

Q n,N =lim

n →∞

C +

∞



s = n+2 G(s)

=lim

n →∞

C +

∞



s = n+2

f

i, y i



Theorem 2.6extends [13, Theorem 1] and [14, Theorem 3]

Example 2.7 Consider the difference equation

∆ 1

n∆
(n −1)∆
(n −1)∆yn+ 1

n5/3(n + 1) y n

1/3 =0, forn ≥2. (2.51)

It is easy to calculate thatQ n,N =(1/8)n(n + 1), n ≥4 Hence the above equation has a solution (y n) such that limn →∞(y n /Q n,N)= C =0 In fact, y n = n2 is a solution of this equation with limn →∞(y n /Q n,N)=8

Next we derive a necessary and sufficient condition for the existence of an asymptoti-cally constant solution of (1.1)

Theorem 2.8 Assume that (H1), (H2), and (H3) hold and the function f is a monotonic function in the second argument Then a necessary and su fficient condition for ( 1.1 ) to have

a solution ( y n ) which satisfies

lim

is that

∞



i =1

P i,Nf (i,c)< ∞, (2.53)

for some integer N ≥ 1 and some nonzero constant c.

Proof

Necessity Without loss of generality, we assume that (y n) is an eventually positive solu-tion of (1.1) such that

lim

Then there exist positive constantsd3andd4such that

Letz n = b n∆(cn∆yn) It is clear that if condition (H1) is satisfied, then solution (y n) of (1.1) of type (i) tends to infinity Since (y n) satisfies condition (ii) ofLemma 2.1, hence

y n > 0, z n < 0, ∆y n > 0, and ∆z n > 0 eventually Let N be so large that (2.55) and (ii) hold forn ≥ N We will use (1.1) in the following form:

∆
a n −2∆zn −2



= − f

n −2,y(n −2)



Multiplying the above equation byP i −2,N −2, and summing both sides of it fromi = N to

n −2, we obtain

n−2

i = N

P i −2,N −2f

i −2,y(i −2) 

= − n

−2



i = N

P i −2,N −2∆
a i −2∆zi −2 

It is easy to calculate thatQ n,N =(1/8)n(n + 1), n ≥4... hold and the function f is a monotonic function in the second argument Then a necessary and su fficient condition for ( 1.1 ) to have

a solution ( y n ) which satisfies... n,N)=8

Next we derive a necessary and sufficient condition for the existence of an asymptoti-cally constant solution of (1.1)

Theorem 2.8 Assume that (H1), (H2), and (H3)