PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 63 (77), 1998, 66–74

OSCILLATORY AND ASYMPTOTIC BEHAVIOUR OF SOME DIFFERENCE EQUATIONS

A. Sternal, Z. Szafrański and B. Szmanda

Communicated by Gradimir Milovanović

Abstract. We consider the oscillation and asymptotic behaviour of nonoscillatory solutions of a class of nonlinear difference equations.

1. Introduction

We consider a nonlinear difference equation

Δ(r_{n}Δ(u_{n} + p_{n}u_{n−k})) = q_{n}f(u_{n−l}),  n = 0, 1, 2,…  (1)

where Δ denotes the forward difference operator, i.e., Δv_{n} = v_{n+1} − v_{n} for any sequence (v_{n}) of real number, k and l are nonnegative integers, (p_{n}) and (q_{n}) are sequences of real numbers with q_{n} ≥ 0 eventually, (r_{n}) is a sequence of positive numbers and

∑_{n=0}^{∞} \frac{1}{r_{n}} = ∞.  (2)

The function f is real valued function satisfying uf(u) > 0 for u ≠ 0.

By a solution of (1) we mean a sequence (u_{n}) which is defined for n ≥ − max{k, l} and satisfies (1) for all large n. A nontrivial solution (u_{n}) of (1) is said to be oscillatory if for every positive integer n_{0} there exists n ≥ n_{0} such that u_{n}u_{n+1} ≤ 0. Otherwise it is called nonoscillatory.

Recently, there has been considerable interest in the study of oscillation and asymptotic behaviour of solutions of difference equations; see for example [2], [3], [5–15] and the references cited therein. For the general theory of difference equations one can refer to [1] and [4].

AMS Subject Classification (1991): Primary 39A10 Key Words: nonoscillatory solution, difference equation, asymptotic properties.

Our purpose is to study the oscillatory and asymptotic behaviour of nonoscillatory solutions of equation (1). The obtained results extend those contained in [14].

2. Main results

Here we give some oscillatory and asymptotic properties of solution of (1). We will need the following assumptions:

f(u) is bounded away from zero if u is bounded away from zero,  (3)

∑_{n=0}^{∞} q_{n} = ∞.  (4)

The following lemma describes some asymptotic properties of the sequence (z_{n}) defined as follows:

z_{n} = u_{n} + p_{n}u_{n−k},  (5)

where (u_{n}) is a nonoscillatory solution of (1).

Lemma. Assume that (3) and (4) hold and there exists a constant P_{1} such that P_{1} ≤ p_{n} ≤ 0.

(a) If (u_{n}) is an eventually positive solution of (1), then the sequences (z_{n}) and (r_{n}Δz_{n}) are eventually monotonic and either

lim_{n→∞} z_{n} = lim_{n→∞} r_{n}Δz_{n} = ∞  (6)

or

lim_{n→∞} z_{n} = lim_{n→∞} r_{n}Δz_{n} = 0,  Δz_{n} < 0 and z_{n} > 0.  (7)

(b) If (u_{n}) is an eventually negative solution of (1), then the sequences (z_{n}) and (r_{n}Δz_{n}) are eventually monotonic and either

lim_{n→∞} z_{n} = lim_{n→∞} r_{n}Δz_{n} = −∞  (8)

or

lim_{n→∞} z_{n} = lim_{n→∞} r_{n}Δz_{n} = 0,  Δz_{n} > 0 and z_{n} < 0.  (9)

Proof. Let (u_{n}) be an eventually positive solution of (1), say u_{n−k} > 0 and u_{n−l} > 0 for n ≥ n_{0}. From (1) we have

Δ(r_{n}Δz_{n}) = q_{n}f(u_{n−l}) ≥ 0  for n ≥ n_{0}  (10)

that is (r_{n}Δz_{n}) is nondecreasing, which implies that (Δz_{n}) is eventually of constant sign and in consequence (z_{n}) is eventually monotonic.

First suppose there exists n_{1} ≥ n_{0} such that Δz_{n_{1}} ≥ 0, then since q_{n} ≡ 0 eventually, there exists n_{2} ≥ n_{1} such that r_{n}Δz_{n} ≥ r_{n_{2}}Δz_{n_{2}} = c > 0 for n ≥ n_{2}. Summing the above inequality, by (2) we have

z_{n} ≥ z_{n_{2}} + c ∑_{i=n_{2}}^{n−1} \frac{1}{r_{i}} → ∞ n → ∞,

hence z_{n} → ∞ as n → ∞.

Since u_{n} ≥ z_{n}, so u_{n} → ∞ as n → ∞. Then summing (10) we get

r_{n}Δz_{n} = r_{n_{2}}Δz_{n_{2}} + ∑_{i=n_{2}}^{n−1} q_{i}f(u_{i−l})

which in view of (3) and (4), implies that r_{n}Δz_{n} → ∞ as n → ∞, and thus (6) holds.

Now, if Δz_{n} < 0 for n ≥ n_{0}, then r_{n}Δz_{n} → L ≤ 0 as n → ∞. Summing (10) from n to m and letting m → ∞ gives

∑_{i=n}^{∞} q_{i}f(u_{i−l}) = L − r_{n}Δz_{n} < ∞.

The last inequality together with (3) and (4) implies

lim_{n→∞} inf u_{n} = 0.  (11)

Suppose that L < 0. Then we have r_{n}Δz_{n} ≤ L for n ≥ n_{0}. Also, we can choose n_{3} ≥ n_{0} such that z_{n_{3}} < 0. Summing the above inequality we get

z_{n} ≤ z_{n_{3}} + L ∑_{i=n_{3}}^{n−1} \frac{1}{r_{i}} < L ∑_{i=n_{3}}^{n−1} \frac{1}{r_{i}} for n > n_{3}

and, by assumption, we obtain

P_{1}u_{n−k} ≤ p_{n}u_{n−k} < z_{n} < L ∑_{i=n_{3}}^{n−1} \frac{1}{r_{i}} , n > n_{3}

so

u_{n−k} > \frac{L}{P_{1}} ∑_{i=n_{3}}^{n−1} \frac{1}{r_{i}} → ∞ n → ∞,

which contradicts (11). Thus lim_{n→∞} r_{n}Δz_{n} = 0. Next we show that z_{n} > 0 for n ≥ n_{0}. If not, then there exists n_{4} ≥ n_{0} such that z_{n_{4}} ≤ 0, then since Δz_{n} < 0 for n ≥ n_{0} z_{n} < z_{n_{5}} < 0 for n ≥ n_{5} ≥ n_{4} that is

u_{n} < z_{n_{5}} − p_{n}u_{n−k} for n ≥ n_{5}  (12)

By (11), there is an increasing sequence of positive integers (n_{i}) such that u_{n_{i}−k} → 0 as i → ∞. This together with the assumption about (p_{n}) and (12) implies that there exists i_{0} such that u_{n_{i_{0}}} < z_{n_{5}}/2 < 0, contradicting u_{n} > 0 eventually.

Since (z_{n}) is decreasing, z_{n} → L_{1} ≥ 0. If L_{1} > 0, then u_{n} ≥ z_{n} ≥ L_{1}, contradicting (11) Thus (7) holds and (a) is proved.

The proof of (b) is similar to that of (a) and hence will be omitted.

Theorem 1. Suppose that (3) and (4) holds. If there exists a constant P_{2} such that P_{2} ≤ p_{n} ≤ −1, then every nonoscillatory solution (u_{n}) of (1) satisfies |u_{n}| → ∞ as n → ∞.

Proof. If (u_{n}) is an eventually positive solution of (1) such that (u_{n}) does not tend to ∞ as n → ∞, then (6) cannot hold since z_{n} ≤ u_{n} eventually. Thus, by Lemma (a) (7) holds. Moreover, from the proof of (7) we have (11) holding. But

0 < z_{n} = u_{n} + p_{n}u_{n−k} ≤ u_{n} − u_{n−k},

so u_{n} > u_{n−k} which contradicts (11). This completes the proof for u_{n} > 0. The proof is similar when (u_{n}) is eventually negative.

From Theorem 1 we immediately obtain

Corollary 1. Under the assumptions of Theorem 1 all bounded solutions of (1) are oscillatory.

Theorem 2. Suppose that there exists a constant P_{3} such that −1 < P_{3} ≤ p_{n} ≤ 0 and that f is a nondecreasing continuous function such that

∫_{0}^{±a} \frac{du}{f(u)} < ∞,  a > 0.  (13)

If

∑_{n=n_{0}}^{∞} \frac{1}{r_{n−l}} ∑_{i=n−l}^{n} q_{i} = ∞,  (14)

then every nonoscillatory solution (u_{n}) of (1) satisfies either |u_{n}| → ∞ or u_{n} → 0 as n → ∞.

Proof. Assume that (u_{n}) is an eventually positive solution of (1) which does not satisfy our assertion. Then for (z_{n}) defined in (5) we see from (1), that Δ(r_{n}Δz_{n}) ≥ 0 eventually that is (r_{n}Δz_{n}) is nondecreasing and (z_{n})is eventually monotonic. Now if (z_{n}) is eventually nonpositive, then the assumption concerning (p_{n}) implies u_{n} ≤ −p_{n}u_{n−k} ≤ −P_{3}u_{n−k} so u_{n+k} ≤ −P_{3}u_{n} for all n sufficiently large, say for n ≥ n_{0}. It then follows by induction that for all n ≥ n_{0} we have u_{n+ik} ≤ (−P_{3})^{i}u_{n} for every positive integer i. Since 0 < −P_{3} < 1, the last inequality implies that u_{n} → 0 as n → ∞ which contradicts our assumption. Also, if there exists n_{1} ≥ n_{0} such that Δz_{n_{1}} ≥ 0, then there is n_{2} ≥ n_{1} such that r_{n}Δz_{n} ≥ r_{n_{2}}Δz_{n_{2}} > 0 for n ≥ n_{2} which, by (2), implies that z_{n} → ∞ as n → ∞. Since u_{n} ≥ z_{n} we have u_{n} → ∞ as n → ∞, again a contradiction to our assumptions on (u_{n}).

Therefore we have z_{n} > 0 and Δz_{n} < 0 for n ≥ n_{0}. Since 0 < z_{n} ≤ u_{n} and f is nondecreasing from (1) we get

Δ(r_{n}Δz_{n}) ≥ q_{n}f(z_{n−l}) for n ≥ n_{1} = n_{0} + l

Summing the above inequality we obtain

r_{n+1}Δz_{n+1} − r_{n−l}Δz_{n−l} ≥ ∑_{i=n−l}^{n} q_{i}f(z_{i−l})

and so

∑_{i=n−l}^{n} q_{i}f(z_{i−l}) ≤ −r_{n−l}Δz_{n−l} n ≥ n_{1.}

In view of monotonicity of (z_{n}) and f we see that

\frac{f(z_{n−l})}{r_{n−l}} ∑_{i=n−l}^{n} q_{i} ≤ −Δz_{n−l},

and further

\frac{1}{r_{n−l}} ∑_{i=n−l}^{n} q_{i} ≤ \frac{−Δz_{n−l}}{f(z_{n−l})} ≤ ∫_{z_{n+1−l}}^{z_{n−l}} \frac{du}{f(u)}, n ≥ n_{1}.

Summing the last inequality from n_{1} to n by (13) we get

∑_{j=n_{1}}^{n} \frac{1}{r_{j−l}} ∑_{i=n−l}^{n} q_{i} ≤ ∫_{z_{n+1−l}}^{z_{n_{1}−l}} \frac{du}{f(u)} < ∫_{0}^{z_{n_{1}−l}} \frac{du}{f(u)} < ∞,

which contradicts (14). The proof is similar when (u_{n}) is eventually negative.

Corollary 2. Under the assumptions of Theorem 2 any bounded solution of (1) is either oscillatory or tends to zero as n → ∞.

Theorem 3. Assume that there exist constants P_{3} and P_{4} such that either

−1 < P_{3} ≤ p_{n} ≤ 0  (15)

or

0 ≤ p_{n} ≤ P_{4} < 1.  (16)

Then every unbounded solution (u_{n}) of (1) is either oscillatory or satisfies |u_{n}| → ∞ as n → ∞.

Proof. Let (u_{n}) be an unbounded solution of (1) which is eventually positive, say u_{n−k} > 0 and u_{n−l} > 0 for n ≥ n_{0}. Then as before we have Δ(r_{n}Δz_{n}) ≥ 0 for n ≥ n_{0}, so (r_{n}Δz_{n}) is nondecreasing and hence (z_{n}) is monotonic.

First assume that (15) holds. Then it follows that z_{n} > 0 for n ≥ n_{1} ≥ n_{0}. Otherwise, there exists n_{2} ≥ n_{1} such that u_{n} + p_{n}u_{n−k} = z_{n} ≤ 0 for n ≥ n_{2} and (15) implies that u_{n} ≤ −P_{3}u_{n−k} ≤ u_{n−k}. This implies that (u_{n}) is bounded, a contradiction.

Further we claim that (Δz_{n}) is eventually positive. Otherwise, (z_{n}) is decreasing and hence is bounded from above, say 0 < z_{n} ≤ M for some constant M. Therefore u_{n} = z_{n} − p_{n}u_{n−k} ≤ M − P_{3}u_{n−k}. Since (u_{n}) is unbounded there is an increasing sequence of positive integers (n_{i}) such that u_{n_{i}} → ∞ as i → ∞ and u_{n_{i}} = max_{n_{1}≤n≤n_{i}} u_{n}. Then we have

u_{n_{i}} ≤ M − P_{3}u_{n_{i}−k} ≤ M − P_{3}u_{n_{i}},

so (1 + P_{3})u_{n_{i}} ≤ M for all i which is impossible in view of (15)

Finally, observe, as in the proof of Lemma, that (r_{n}Δz_{n}) nondecreasing and (Δz_{n}) eventually positive implies that z_{n} → ∞ as n → ∞ and hence u_{n} → ∞ as n → ∞ since u_{n} ≥ z_{n}.

Now assume that (16) holds. Then it is clear that z_{n} > 0 for n ≥ n_{0}. Also we see that (Δz_{n}) is eventually positive. In fact, if not, then (z_{n}) is decreasing and so is bounded from above and since z_{n} ≥ u_{n} (u_{n}) is bounded, a contradiction.

As previously we conclude that z_{n} → ∞ as n → ∞. Since z_{n} ≤ u_{n}+P_{4}z_{n−k} ≤ u_{n}+P_{4}z_{n} we have (1−P_{4})z_{n} ≤ u_{n} which in view of (16), implies u_{n} → ∞ as n → ∞.

A similar argument treats the case of eventually negative solution.

Theorem 4. Suppose that there exist constants P_{5} and P_{6} such that P_{5} ≤ p_{n} ≤ P_{6} < −1 and f is a nondecreasing continuous function such that

∫_{ε}^{∞} \frac{du}{f(u)} < ∞,  ∫_{−ε}^{−∞} \frac{du}{f(u)} < ∞,  ε > 0.  (17)

If

∑_{n=n_{0}}^{∞} \frac{1}{r_{n−l}} ∑_{i=n−k+1}^{∞} q_{i} = ∞  when l ≥ k,  (18)

or

∑_{n=n_{0}}^{∞} \frac{1}{r_{n}} ∑_{i=n}^{∞} q_{i} = ∞  when l < k  (19)

then all bounded solutions of (1) are oscillatory.

Proof. Assume that there exists a bounded nonoscillatory solutions (u_{n}) of (1) and let u_{n} > 0 eventually, say u_{n−k−l} > 0 for n ≥ n_{0}. Then as before for the sequence (z_{n}) defined in (5) it follows that (r_{n}Δz_{n}) is a nondecreasing sequence and in consequence (z_{n}) is eventually monotonic. We show first that (z_{n}) is eventually negative. If there exists n_{1} ≥ n_{0} such that z_{n_{1}} > 0, then by the assumptions we get u_{n_{1}} = z_{n_{1}} − p_{n_{1}}u_{n_{1}−k} > −P_{6}u_{n_{1}−k}. Then it follows by induction that u_{n_{1}+ik} > (−P_{6})^{i}u_{n_{1}}, which implies u_{n_{i}+ik} → ∞ as i → ∞ contradicting the boundedness of (u_{n}). Therefore (z_{n}) is eventually negative, say for n ≥ n_{0}. Now we observe that Δz_{n} < 0 for n ≥ n_{0}. If not, then a similar argument as in the proof of Lemma leads to the fact that z_{n} → ∞ contradicting z_{n} < 0 for n ≥ n_{0}. By assumption, we have P_{5}u_{n−k} ≤ p_{n}u_{n−k} < z_{n} < 0, which implies that 0 < z_{n+k}/P_{5} < u_{n} for n ≥ n_{0}.

In view of monotonicity of f from (1) we see that

Δ(r_{n}Δz_{n}) ≥ q_{n}f(\frac{z_{n+k−l}}{P_{5}} )  for n ≥ n_{1} = n_{0} + l  (20)

Summing (20) from n − k to m > n − k we obtain

r_{m+1}Δz_{m+1} − r_{n−k}Δz_{n−k} ≥ ∑_{i=n−k}^{m} q_{i}f(\frac{z_{i+k−l}}{P_{5}} ).

After letting m → ∞, we have

−r_{n−k}Δz_{n−k} ≥ ∑_{i=n−k}^{∞} q_{i}f(\frac{z_{i+k−l}}{P_{5}} ) ≥ ∑_{i=n−k+1}^{∞} q_{i}f(\frac{z_{i+k−l}}{P_{5}} ),

from which we get

−r_{n−k}Δz_{n−k} ≥ f(\frac{z_{n+1−l}}{P_{5}} ) ∑_{i=n−k+1}^{∞} q_{i}.  (21)

Since (r_{n}Δz_{n}) is nondecreasing, for l ≥ k we have r_{n−l}Δz_{n−l} ≤ r_{n−k}Δz_{n−k} and further from (21) we obtain

\frac{1}{r_{n−l}} ∑_{i=n−k+1}^{∞} q_{i} ≤ −\frac{Δz_{n−l}}{f(\frac{z_{n+1−l}}{P_{5}} )} for n ≥ n_{1}.  (22)

In view of monotonicity of (z_{n}) and f for z_{n−l}/P_{5} ≤ u ≤ z_{n+1−l}/P_{5} we have

\frac{1}{f(u)} ≥ \frac{1}{f(\frac{z_{n+1−l}}{P_{5}} )}

and so

∫_{z_{n−l}/P_{5}}^{z_{n+1−l}/P_{5}} \frac{du}{f(u)} ≥ \frac{1}{P_{5}} \frac{Δz_{n−l}}{f(\frac{z_{n+1−l}}{P_{5}} )}  for n ≥ n_{1}.  (23)

Now using (23) in (22) and summing both sides from n_{1} to n we get

∑_{j=n_{1}}^{n} \frac{1}{r_{j−l}} ∑_{i=j−k+1}^{∞} q_{i} ≤ −P_{5} ∫_{z_{n_{1}−l}/P_{5}}^{z_{n+1−l}/P_{5}} \frac{du}{f(u)}, n ≥ n_{1}

which in view of (17) contradicts the condition (18).

If l < k, then summing (20) from n to m > n and letting m → ∞ we obtain

−r_{n}Δz_{n} ≥ ∑_{i=n}^{∞} q_{i}f(\frac{z_{i+k−l}}{P_{5}} ) ≥ f(\frac{z_{n+k−l}}{P_{5}} ) ∑_{i=n}^{∞} q_{i}.  (24)

Since n + k − l ≥ n + 1, it follows that

f(\frac{z_{n+1}}{P_{5}} ) ≤ f(\frac{z_{n+k−l}}{P_{5}} ).

Therefore from (24) we get

\frac{1}{r_{n}} ∑_{i=n}^{∞} q_{i} ≤ −\frac{Δz_{n}}{f(\frac{z_{n+1}}{P_{5}} )} for n ≥ n_{1}

and the rest of the proof follows analogously to that as above.

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Institute of Mathematics (Received 10 12 1997) Poznań University of Technology 60-965 Poznań Poland