GEORGIAN MATHEMATICAL JOURNAL: Vol. 2, No. 6, 1995, 593-602

MORE ON OSCILLATION OF nTH-ORDER EQUATIONS

WITOLD A. J. KOSMALA

Abstract. In this paper we prove that a higher-order differential equation with one middle term has every bounded solution oscillatory. Moreover, the behavior of unbounded solutions is given. Two other results dealing with positive solutions are also given.

1. Introduction

Little is known about the behavior of solutions of differential equations which involve a nonlinear term H(t, x), where H : ℝ^{+}×ℝ → ℝ is continuous (ℝ is the real line and ℝ^{+} is the interval (0, ∞)), decreasing in its second variable and is such that uH(t, u) < 0 for all u ≠ 0. Some properties of solutions of such equations are given by the author in [1]–[4]. In [3] the author gave two oscillation results for odd-order equations of the form

x^{(n)} + p(t)x^{(n−1)} + q(t)x^{(n−2)} + H(t, x) = 0.  (1)

In [4] bounded and eventually positive solutions of (1) were studied for different choices of p and q. The main difficulty in handling equations involving the indicated H is the lack of tools for equations involving H continuous, increasing in its second variable, and keeping the sign of the second variable. Some successful tools for equations with either H proved to be nonlinear functions which were used by Erbe [5], Heidel [6], Kartsatos [7], Kartsatos and Kosmala [8], Kosmala [3], and others. In most cases the method used on an even-ordered equation does not apply to an odd-ordered equation, and vice versa.

2. Preliminaries

A function x(t), t ∈ [t_{x}, ∞) ⊂ ℝ^{+}, is a solution of (1) if it is n times continuously differentiable and satisfies (1) on [t_{x}, ∞). The number t_{x} > 0 depends on the particular solution x(t) under consideration. We say that

593 a function is “oscillatory” if it has an unbounded set of zeros. Moreover, a property P holds “eventually” or “for all large t” if there exists T > 0 such that P holds for all t ≥ T. C^{n}(I) denotes the space of all n times continuously differentiable functions f : I → ℝ. And we write C(I) instead of C^{0}(I). From [3] we quote the following lemma.

1991 Mathematics Subject Classification. 34C10.

Key words and phrases. Higher-order equation; oscillatory, bounded, or positive solutions.

1072-947X/95/1100-0593$07.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1995 Plenum Publishing Corporation

Lemma 2.1. If x is an eventually positive solution of (1), p ∈ C^{1}[t_{0}, ∞), q ∈ C[t_{0}, ∞) with t_{0} > 0 where 2q(t) ≤ p′(t) for all t ≥ t_{0}, then either x^{(n−2)}(t) ≤ 0 or x^{(n−2)}(t) > 0 for all large t.

The main result from [2] we state as our next lemma.

Lemma 2.2. Suppose that x(t) is a nonoscillatory solution of (1) with n even. Suppose further that p ∈ C^{1}[t_{0}, ∞), q ∈ C[t_{0}, ∞) with t_{0} > 0 where p(t) ≤ 0 and 2q(t) ≤ p′(t) ≤ 0 for all t ≥ t_{0} and lim_{t→∞} p(t) is finite. Moreover, suppose that

− ∫^{∞}tH(t, ±k)dt = ±∞

for any constant k > 0. Suppose further that for v > 0 both

I(v) ≡ ∫_{v}^{∞}tq(t)dt and ∫^{∞}I(t)dt

are finite. Then x(t)x^{(n−2)} > 0 eventually.

3. Main Results

Theorem 3.1. Suppose that n is odd, p ∈ C[t_{0}, ∞) with t_{0} > 0 where p(t) for all t ≥ t_{0}, and that

∫_{t^{∗}}^{∞}t^{i}( exp ∫_{t^{∗}}^{t} p(s)ds)H(t, k)dt = −∞  (2)

for any t^{∗} ≥ 0, every positive real constant k, and some integer i where 1 ≤ i ≤ n − 1.

(a) Then every solution of

x^{(n)} + p(t)x^{(n−1)} + H(t, x) = 0  (3)

with bounded (n − i − 1)^{st} derivative is oscillatory. In particular, every bounded solution of (3) is oscillatory.

(b) If i = 1 and x(t) is an unbounded solution of (3), then x(t)x^{(j)}(t) > 0 for all j = 0, 1, 2, ⋯, n eventually.

OSCILLATION OF nTH ORDER EQUATIONS 595

Proof of part (a). First we assume that x(t) is a positive solution of (3) such that x^{(n−i−1)}(t) is bounded and p(t) ≤ 0 for t ≥ t_{1} ≥ t_{0}. Now we distinguish three cases.

Case 1. Suppose that x^{(n−1)}(t_{2}) for some t_{2} ≥ t_{1}. Then, from (3) we have

x^{(n)}(t_{2}) = −H(t_{2}, x(t_{2})) > 0.

Thus, x^{(n−1)}(t) is increasing at any t_{2}, for which it is zero. Therefore, x^{(n−1)}(t) cannot have any zeros larger than t_{2}.

Case 2. Suppose that x^{(n−1)}(t) > 0 for t ≥ t_{3} ≥ t_{2}. Then, from (3) we have that

x^{(n)}(t) = −p(t)x^{(n−1)}(t) − H(t, x(t)) > 0.

But, x^{(n)}(t) > 0 and x^{(n−1)}(t) > 0 imply that x^{(n−2)}(t) tends to +∞ as t tends to +∞. But, x^{(n−i−1)}(t) is bounded for some integer i, where 1 ≤ i ≤ n − 1, which contradicts lim_{t→∞} x^{(n−2)}(t) = +∞ no matter what i is. This takes us to the final case.

Case 3. Suppose that x^{(n−1)}(t) < 0 for t ≥ t_{3} ≥ t_{1}. Since n is odd there exists t_{4} ≥ t_{3} such that x′(t) > 0 for all t ≥ t_{4}. Thus, we have k ≡ x(t_{4}) ≤ x(t) and hence

−t^{i}( exp ∫_{t_{4}}^{t} p(s)ds)H(t, x(t)) ≥ −t^{i}( exp ∫_{t_{4}}^{t} p(s)ds)H(t, k)  (4)

for all t ≥ t_{4}. Now, define

F (t) = t^{i}( exp ∫_{t_{4}}^{t} p(s)ds)x^{(n−1)}(t)

for t ≥ t_{4}. Then, for t ≥ t_{4} we have

F′(t) = t^{i}( exp∫_{t_{4}}^{t}p(s)ds)x^{(n)}(t) +

+t^{i}( exp∫_{t_{4}}^{t}p(s)ds)p(t)x^{(n−1)}(t) + it^{i−1}( exp∫_{t_{4}}^{t}p(s)ds)x^{(n−1)}(t) =

= t^{i}( exp∫_{t_{4}}^{t}p(s)ds)( − p(t)x^{(n−1)}(t) − H(t, x(t))) +

+t^{i}( exp∫_{t_{4}}^{t}p(s)ds)p(t)x^{(n−1)}(t) + it^{i−1}( exp∫_{t_{4}}^{t}p(s)ds)x^{(n−1)}(t) =

= it^{i−1}( exp∫_{t_{4}}^{t}p(s)ds)x^{(n−1)}(t) − t^{i}( exp∫_{t_{4}}^{t}p(s)ds)H(t, x(t)).

Now we integrate above from t_{4} to t, t ≥ t_{4},and then use (4) to obtain

F (t) − F (t_{4}) = ∫_{t_{4}}^{t} is^{i−1}( exp ∫_{t_{4}}^{s} p(u)du)x^{(n−1)}(s)ds −

− ∫_{t_{4}}^{t} s^{i}( exp ∫_{t_{4}}^{s} p(u)du)H(s, x(s))ds ≥

≥ ∫_{t_{4}}^{t} is^{i−1}( exp ∫_{t_{4}}^{s} p(u)du)x^{(n−1)}(s)ds −

− ∫_{t_{4}}^{t} s^{i}( exp ∫_{t_{4}}^{s} p(u)du)H(s, k)ds.

Due to integral hypothesis (2) and the fact that F (t) < 0, we have that

lim_{t→∞} ∫_{t_{4}}^{t} s^{i−1}( exp ∫_{t_{4}}^{s} p(u)du)x^{(n−1)}(s)ds = −∞.

But, since p(t) ≤ 0 and x^{(n−1)}(t) < 0, which implies that x^{(n−2)}(t) > 0, we have that

t^{i−1}( exp ∫_{t_{4}}^{t} p(s)ds) ≥ t^{i−1}x^{(n−1)}(t).

Thus,

lim_{t→∞} ∫_{t_{4}}^{t} s^{i−1}x^{(n−1)}(s)ds = −∞.  (5)

Integration by parts gives

lim_{t→∞} [s^{i−1}x^{(n−2)}(s)|_{t_{4}}^{t} − ∫_{t_{4}}^{t}(i − 1)s^{i−2}x^{(n−2)}(s)ds] = −∞.

Since x^{(n−2)}(t) is positive, we have

lim_{t→∞} ∫_{t_{4}}^{t} s^{i−2}x^{(n−2)}(s)ds = +∞.

Integration by parts gives

lim_{t→∞} ∫_{t_{4}}^{t} s^{i−3}x^{(n−3)}(s)ds = −∞.

Continuing this process we get

(−1)^{m} ∫_{t_{4}}^{∞}s^{i−m}x^{(n−m)}(s)ds = +∞,

where m is an integer such that 1 ≤ m ≤ i. (This can be proved by induction.) Setting m = i, we obtain

(−1)^{i} ∫_{t_{4}}^{∞}x^{(n−i)}(s)ds = +∞,

which gives

lim_{t→∞}(−1)^{i}[x^{(n−i−1)}(t) − x^{(n−i−1)}(t_{4})] = +∞.

But, x^{(n−i−1)}(t) is bounded. This is a contradiction. Therefore assuming that x(t) is eventually positive prevents x^{(n−1)}(t) from existing. Similar steps to the above ones also bring a contradiction for an eventually negative solution of (3). Hence, the proof of part (a) is complete.

Proof of part (b). Suppose that x(t) is an unbounded positive solution of (3) and p(t) ≤ 0 for t ≥ 0. In view of case 1 above, x^{(n−1)}(t) cannot be oscillatory. From (5) we have that

∫^{∞}x^{(n−1)}(t)dt = −∞,

which gives lim_{t→∞} x^{(n−2)}(t) = −∞. This is, however, not possible due to the positiveness of x(t). Therefore, the only possibility is case 2 above, which yields the desired result.

Theorem 3.2. Suppose n is odd, p ∈ C^{1}[t_{0}, ∞), q ∈ C[t_{0}, ∞) with t_{0} >0 where p(t) ≤ 0 and q(t) ≤ p′(t) ≤ 0 for all t ≥ t_{0}, and lim_{t→∞} p(t) is finite. Suppose further that for v > 0 both

I(v) ≡ ∫_{v}^{∞}tq(t)dt and ∫^{∞}I(t)dt  (6)

are finite. Also, H^{∗}(t, u) ≡ \frac{d}{du}H(t, u) is continuous negative, and decreasing for u > 0, and

∫^{∞}H^{∗}(t, 0)dt = −∞.  (7)

If u(t) and v(t) are two bounded eventually positive solutions of (1), then the difference w(t) = u(t) − v(t) must be oscillatory.

Remark 3.3. The conditions of Theorem 3.2 imply that 2q(t) ≤ p′(t) for t ≥ t_{0} and that

∫^{∞}tH^{∗}(t, 0) = −∞.

This is needed so that the proof of Lemma 2.2 can be used in Case 2 below. Furthermore, functions in the differential equation

x^{(5)} + (\frac{1}{3t^{3}} − 1)x^{(4)} − \frac{2}{t^{4}} x′′′ + (1 − e^{tx}) = 0

satisfy all the conditions of Theorem 3.2.

Proof of Theorem 3.2. Suppose that u(t) and v(t) are bounded, positive solutions of (1) and that w(t) = u(t) − v(t) for all t ≥ t_{1} ≥ t_{0}. The case where w(t) is negative follows similar steps and is therefore omitted. Now, from (1) we have

w^{(n)}(t) + p(t)w^{(n−1)}(t) + q(t)w^{(n−2)}(t) +
+H(t, u(t)) − H(t, v(t)) = 0.  (8)

Note that since w(t) > 0, we have H(t, u(t)) < H(t, v(t)). Hence, we can follow the proof of Lemma 2.1 to reach the conclusion that w^{(n−2)}(t) ≤ 0 or w^{(n−2)}(t) > 0 for all t ≥ t_{2} ≥ t_{1}. In order to prove this theorem, we need to consider both cases and find a contradiction in each.

Case 1. Suppose that w^{(n−2)}(t) > 0 for t ≥ t_{2}. Since n is odd we have w′(t) > 0 for all t ≥ t_{3} ≥ t_{2}.Now, if w^{(n−2)}(t_{4}) > 0 for some t_{4} ≥ t_{3}, then from (8) we have

w^{(n)}(t_{4}) = −q(t_{4})w^{(n−2)}(t_{4}) − (H(t_{4})) − H(t_{4}, v(t_{4}))) > 0.

Therefore, w^{(n−1)}(t) is increasing at each of its zeros. Thus, w^{(n−1)}(t) > 0 or w^{(n−1)}(t) < 0 for all t ≥ t_{5} > t_{4}.If w^{(n−1)}(t) > 0, then together with w^{(n−2)}(t) > 0 we contradict the fact that w(t) is bounded. Thus, we must have w^{(n−1)}(t) < 0 for all t ≥ t_{5}. Since w′(t) > 0, we have 0 < k ≡ w(t_{5}) ≤ w(t) for all t ≥ t_{5}. So, by the mean value theorem there exists a function λ between u and v, hence positive, such that

\frac{H(t, u(t)) − H(t, v(t))}{u(t) − v(t)} = H^{∗}(t, λ(t)).

Therefore,

H(t, u(t)) − H(t, v(t)) = (u(t) − v(t))H^{∗}(t, λ(t)) =
= w(t)H^{∗}(t, λ(t)) ≤ kH^{∗}(t, 0)  (9)

for all t ≥ t_{5}. Now, integrate (8) from t_{5} to t ≥ t_{5} to obtain

w^{(n−1)}(t) + p(t)w^{(n−2)} = w^{(n−1)}(t_{5}) + p(t_{5})w^{(n−2)}(t_{5}) +

+ ∫_{t_{5}}^{t} (p′(s) − q(s))w^{(n−2)}(s)ds − ∫_{t_{5}}^{t} (H(s, u(s)) − H(s, v(s)))ds =

= M + f(t) − ∫_{t_{5}}^{t} (H(s, u)) − H(s, v(s)))ds,

where M is a constant and f(t) the first integral in the above expression. If z(t) = w^{(n−2)}(t), then z satisfies a first-order linear differential equation and thus can be written as

z(t) = exp [ − ∫_{t_{5}}^{t} p(s)ds]{z(t_{5}) + ∫_{t_{5}}^{t} [ exp ∫_{t_{5}}^{s} p(r)dr][M + f(s) −

− ∫_{t_{5}}^{s} (H(r, u(r)) − H(r, v(r)))dr]ds}.

But, f(t) ≥ 0 and exp [−∫^{t} ; t_{5}p(s)ds] ≥ 1; thus in view of (9) we have

z(t) ≥ ∫_{t_{5}}^{t} [ exp ( − ∫_{s}^{t} p(r)dr)][M − k ∫_{t_{5}}^{s} H^{∗}(r, 0)dr]ds.  (10)

Due to condition (7), there exists s_{0} ∈ ℝ^{+} such that

M − k ∫_{t_{5}}^{s_{0}}H^{∗}(t, 0)dt > 0.

Therefore, the right hand side of (10) tends to +∞. This implies that the left hand side of (10) also tends to +∞. But this is a contradiction to the positiveness of w(t). Hence, w^{(n−2)}(t) cannot be eventually positive.

Case 2. Suppose that w^{(n−2)}(t) ≤ 0 for t ≥ t_{2}. With a few minor changes, the proof of Lemma 2.2 goes through for equation (8), contradicting this situation as well. In this proof we need assumption (6). We will not give the detailed proof since it is very long and tedious. Also, see Remark 3.3.

Theorem 3.4. Suppose that n is odd, p ∈ C^{1}[t_{0}, ∞) for t_{0} > 0, p(t) > 0 for all t ≥ t_{0}, with lim_{t→∞} p(t) finite and

∫^{∞} \frac{1}{p(s)}ds = +∞.

Also,

∫^{∞}H(t, k)dt = −∞

for any positive real constant k. Then, the differential inequality

(p(t)x^{(n−1)})′ + H(t, x) ≥ 0  (11)

has no bounded eventually positive solutions.

Proof. Suppose that x(t) > 0 is a bounded solution of (11) and p(t) > 0 for t ≥ t_{0} > 0. Then

(p(t)x^{(n−1)})′ ≥ −H(t, x) > 0.  (12)

Therefore, if v(t) ≡ p(t)x^{(n−1)}(t), then v(t) is increasing for all t ≥ t_{0}. We distinguish two cases.

Case 1. Suppose v(t) > 0 for all t ≥ t_{1} ≥ t_{0}. Then, from (12) we have v′(t) > 0, and so v(t) ≥ v(t_{1}) > 0 for all t ≥ t_{1}. Therefore,

x^{(n−1)}(t) = \frac{v(t)}{p(t)} ≥ \frac{v(t_{1})}{p(t)}

for all t ≥ t_{1}. Integrating the above from t to t_{1}, t ≥ t_{1} we obtain

x^{(n−2)}(t) − x^{(n−2)}(t_{1}) ≥ v(t_{1}) ∫_{t_{1}}^{∞} \frac{1}{p(s)}ds → +∞ as t → +∞.

Therefore, lim_{t→∞} x^{(n−2)}(t) = +∞, which is a contradiction to the boundedness of x(t).

Case 2. Suppose v(t) < 0 for all t ≥ t_{2} ≥ t_{0}. Then, from (12) since p(t) > 0, we have x^{(n−1)}(t) < 0. Now, since n is odd we have x′(t) > 0 which gives H(t, x(t)) ≤ H(t, k) for all t ≥ t_{2} ≡ k. Therefore, integrating (11) from t_{2} to t, t ≥ t_{2}, we get

p(t)x^{(n−1)}(t) − p(t_{2})x^{(n−1)}(t_{2}) ≥

≥ − ∫_{t_{2}}^{t} H(s, x(s))ds ≥ − ∫_{t_{2}}^{t} H(s, k)ds → +∞

as t → +∞. Thus, lim_{t→∞} p(t)x^{(n−1)}(t) = +∞. Since lim_{t→∞} p(t) is finite, we have lim_{t→∞} x^{(n−1)}(t) = +∞, which is a contradiction to the positiveness of x(t). Hence, the proof is complete.

Aknowledgment

This paper is written during the author’s sabbatical at the Louisiana State University, Baton Rouge, Louisiana, USA.

References

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3. A. G. Kartsatos, The oscillation of a forced equation implies the oscillation of the unforced equation–small forcing. J. Math. Anal. Appl. 76(1980), 98-106.

4. A. G. Kartsatos and W. A. Kosmala, The behaviour of an nth-order equation with two middle terms. J. Math. Anal. Appl. 88(1982), 642-664.

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6. W. A. Kosmala, Properties of solutions of nth-order equations. Ordinary and delay differential equations, 101-105, Pitman, 1992.

7. W. A. Kosmala, Oscillation of a forced higher-order equation. Ann. Polonici Math. (to appear).

8. W. A. Kosmala, Behavior of bounded positive solutions of higher-order differential equations. Hiroshima Math. J. (to appear).

(Received 10.02.94)

Author’s address: Department of Mathematical Sciences Appalachian State University Boone, North Carolina 28608 USA