Forced oscillation of hyperbolic equations with mixed nonlinearities

In this paper, we consider the mixed nonlinear hyperbolic equations with forcing term via Riccati inequality. Some sufficient conditions for the oscillation are derived byusing Young's inequality and integral averaging method.

This paper is organized as follows.In the next section we introduce the main ideas by using Young's inequality (see, [4]), which plays an important role in establishing oscillation theorems for superlinear and sublinear hyperbolic equations.In Section 3 we use the well-known Riccati type transformation.The latest section contains the main results and an example which illustrates our results.In this section we reduce the multi-dimensional oscillation problems for (E) to one-dimensional oscillation problems.It is known that the first eigenvalue λ 1 of the eigenvalue problem −∆w = λw in G, w = 0 on ∂G is positive, and the corresponding eigenfunction Φ(x) can be chosen so that Φ(x) > 0 in G.The notations in this paper are as follows: where The following lemma is of basic importance for our later considerations.Lemma 1(Young's Inequality).If p > 1 and q > 1 are conjugate numbers, i.e. 1  p + 1 q = 1, then for any u, v ∈ R and equality holds iff u = |v| q−2 v.
have no eventually positive solutions, then every solution u(x, t) of the problem (E), (B1) is oscillatory in Ω, where EJQTDE, 2012 No. 33, p. 3 Proof.Suppose to the contrary that there is a nonoscillatory solution u of the problem (E), (B1).Without loss of generality we may assume that u(x, t) > 0 in G × [t 0 , ∞) for some t 0 > 0 because the case u(x, t) < 0 is handled similarily.It follows from Lemma 1 that and hence, Eq. (E) can be written in the form Multiplying ( 2) by K Φ Φ(x) and integrating over G, we obtain Using Green's formula, we see that Therefore U(t) is an eventually positive solution of (1) with +F Ψ (t).This contradicts the hypothesis and completes the proof.
EJQTDE, 2012 No. 33, p. 4 have no eventually positive solutions, then every solution u(x, t) of the problem (E), (B2) is oscillatory in Ω, where Proof.Suppose to the contrary that there is a nonoscillatory solution u of problem (E), (B2).Without loss of generality we may assume that u(x, t) > 0 in G × [t 0 , ∞) for some t 0 > 0. Dividing (2) by |G| and integrating over G, we have It follows from Green's formula that Combining ( 5) with (6) yields Hence Ũ (t) is an eventually positive solution of (4) with + FΨ (t).This contradicts the hypothesis and completes the proof.

Second order functional differential inequalities
We consider sufficient conditions for every solution y(t) of the functional differential inequality (r(t)y ′ (t)) ′ + p(t)y ′ (t) + q(t)y(t) ≤ f (t) (7) to have no eventually positive solution, where f (t) ∈ C([0, ∞); R).Theorem 3. Assume that (H2) for any T > 0 there exists an interval EJQTDE, 2012 No. 33, p. 5 If the Riccati inequality has no solution on [t k , t k+1 ], then (7) has no eventually positive solution, where φ(t) ∈ C 1 ((0, ∞); (0, ∞)) and Proof.Suppose that y(t) is a positive solution of ( 7) on [t 0 , ∞) for some t 0 > 0. From the hypothesis (H2) there exists an interval If we set then we obtain Multiplying (10) by φ(t), we obtain and hence Combining ( 12) with ( 13), we have We define z(t) = φ(t)w(t), then the above inequality reduces to Therefore z(t) is a solution of (8) on I.This contradicts the hypothesis and completes the proof.
Theorem 4. Assume that (H2) holds, and let EJQTDE, 2012 No. 33, p. 7 then (7) has no eventually positive solution, where Proof.Suppose that y(t) is a positive solution of ( 7) on [t 0 , ∞) for some t 0 ≥ T 0 > 0. Then there exists an interval [a, b] such that t 0 ≤ a < b, and hence y(t) > 0 in (a, b).Proceeding as in the proof of Theorem 3, there exists a positive solution w(s) of ( 11 Letting t → b − 0 in the above, we obtain On the other hand, multiplying ( 11) by H 1 (s, t) and integrating over [t, c] for t ∈ (a, c], we obtain and so Letting t → a + 0 in the above, we obtain Adding ( 16) and ( 17), we easily obtain the following which contradicts the condition (15).This contradiction proves that Theorem 4 holds.
Proof.For any T ≥ t 0 , let a = T .In (18) we choose T = a.Then there exists c > a such that Combining ( 20) and ( 21), we obtain (15).The conclusion follows from Theorem 4, and the proof is completed.

Oscillation results
Using the Riccati inequality, we derive sufficient conditions for every solution of the hyperbolic equation (E) to be oscillatory.We are going to use the following lemma which is due to Usami [5].