Periodic solutions for second order differential equations with indefinite singularities

Theorem 1.1

Suppose that P_v > P_∞ and

Then there exists at least one positive T–periodic solution to the Rayleigh-Plesset equation (1.4) in the case of k ∈ ( 13 , +∞).

Lemma 2.1

([20, 26]). Assume that there exist positive constants M₀, M₁ and M₂ with 0 < M₀ < M₁, such that the following conditions hold.

1. For any λ ∈ (0, 1], each possible positive T–periodic solution x to the equation

   u″(t)+λf(u(t))u′(t)+λφ(t)um(t)−λα(t)xμ(t)+λβ(t)xy(t)=0

   satisfies the inequalities M₀ < x(t) < M₁ and |x′(t)| < M₂, for all t ∈ [0, T].

2. Each possible solution c to the equation

   α¯cμ−β¯cy−φ¯cm=0

   satisfies the inequality M₀ < c < M₁.

3. The inequality

   (α¯M0μ−β¯M0y−φ¯M0m)(α¯M1μ−β¯M1y−φ¯M1m)<0

   holds.

Then, equation (1.3) has at least one T–periodic solution u such that M₀ < u(t) < M₁ for all t ∈ [0, T].

Remark 2.1

If φ̄ > 0 and ᾱ > 0, then we have from the condition of m ≥ 0 and μ > y > 0 that there are two constants D₁ and D₂ with 0 < D₁ < D₂ such that

and

Lemma 2.2

([5]). Let u ∈ [0, ω] → ℝ be an arbitrary absolutely continuous function with u(0) = u(ω). Then the inequality

holds.

Lemma 2.3

Assume ᾱ > 0 and φ̄ > 0, then for each u ∈ D, there are constants τ₁, τ₂ ∈ [0, T] such that

and

where

m, μ and y are determined in equation (1.3)

Proof

Let u ∈ D, then

which together with the fact of u(t) > 0 for all t ∈ [0, T] gives

Integrating the above equality over the interval [0, T], we obtain

i.e.,

Since

it follows from (2.9) that

From (2.11), we can conclude that there is a point ξ ∈ [0, T] such that

If (2.12) does not hold, then

which implies that

and

From (2.11), (2.13) and μ > y, and by using mean value theorem of integrals, we have that there exists a point η ∈ [0, T] such that

i.e.,

which contradicts to (2.14). This contradiction verifies (2.12), and so (2.5) holds. Meanwhile, we can assert that (2.6) is true. In fact, multiplying two sides of (2.8) with u^μ(t) and integrating it over the interval [0, T], we obtain that

which together with

yields

Thus, there is a point η ∈ [0, T] such that

where A₁ is defined by (2.7). This implies that (2.6) holds.□

Lemma 2.4

Assume ᾱ > 0 and φ̄ > 0, and suppose that the following assumptions

and

hold, where F(x) is defined by (2.3).

Then there exists a constant y₀ > 0, such that

Proof

For each u ∈ D, we have

Since u(t) is a T–periodic function, there are two points t₁, t₂ ∈ ℝ such that 0 < t₂ – t₁ ≤ T with u(t₁) = max_t∈[0,T] u(t) and u(t₂) = min_t∈[0,T] u(t). Assumptions of ᾱ > 0 and φ̄ > 0 implies that (2.6) in Lemma 2.3 holds. This gives

to which by applying assumption (2.16) yields

By integrating (2.18) over the interval [t₁, t₂], we get

It follows from (2.19), (2.20) and the condition φ̄ > 0 that

On the other hand, by integrating (2.18) on [0, T], we get

i.e.,

Substituting it into (2.21), we have from (2.19) that

and so

Assumption (2.17) ensures that there exists a constant y₀ > 0 such that

Combining (2.23) with (2.22), we can get

□

Lemma 2.5

Assume ᾱ > 0 and φ̄ > 0. Suppose further that the following assumptions

and

hold. Then there exists a constant y₁ > 0, such that

Proof

Suppose that u ∈ D, then u satisfies (2.18). Let t₁ and t₂ be defined as same as the ones in the proof of Lemma 2.4. From (2.6) in Lemma 2.3, we see that

which together with assumption (2.25), yields that

By integrating (2.18) over the interval [t₁, t₂], we get

i.e.,

From (2.27), we obtain

Assumption (2.26) implies that there exists a constant y₁ > 0 such that

which together with (2.28) gives

□

Theorem 3.1

Assume φ̄ > 0 and ᾱ > 0. Suppose further

and

Then equation(1.3) has at least one positive T–periodic solution.

Proof

Firstly, we will prove that there exist two constants y₂ > 0 and y₃ > 0 with y₃ > y₂, such that

and

For each u ∈ D, u satisfies (2.18). Let t₁ and t₂ be defined as same as the ones in the proof of Lemma 2.4, that is 0 < t₂ – t₁ ≤ T, u(t₁) = max_t∈[0,T] u(t) and u(t₂) = min_t∈[0,T] u(t). Take t₃ = t₁ + T, then u(t₃) = max_t∈[0,T] u(t) and 0 ≤ t₃ – t₂ < T. It is easy to see that (2.16) can be deduced from (3.2), and (3.1) verifies (2.17). By using Lemma 2.4, as well as (2.5) in Lemma 2.3, we know that

Integrating (2.18) over the interval [t₂, t₃], we have

i.e.,

which together with (3.5) gives

It follows from (3.2) that there is a constant y₃ > 0 such that

From (3.5) and (3.6), we see that the conclusions of (3.3) and (3.4) hold. Next, we will show that there exists a positive constant ρ₁ such that

In fact, if u ∈ D, then

Let u attain its maximum over [0, T] at t₁ ∈ [0, T], then u′(t₁) = 0 and we deduce from (3.8) that

for all t ∈ [t₁, t₁ + T]. Thus,

It follows from (3.4) and (3.3) that

and so

which implies that (3.7) holds. Let m₀ = min{D₁, y₂} and m₁ = max{y₃, D₂} be two constants, where D₁ and D₂ are constants determined in Remark 2.1, then from (3.4), (3.3) and (3.7), we see that each possible positive T–periodic solution u to equation(2.1) satisfies

This implies that condition 1 of Lemma 2.1 is satisfied. Also, we can deduce from Remark 2.1 that

and

which gives that condition 2 of Lemma 2.1 holds. Furthermore, we have

So condition 3 of Lemma 2.1 holds, too. By using Lemma 2.1, we see that equation(1.3) has at least one positive T–periodic solution.□

Theorem 3.2

Assume φ̄ > 0 and ᾱ > 0. Suppose further

and

Then equation(1.3) has at leat one positive T–periodic solution.

Proof

From the proof of Theorem 3.1, we see that it suffices for us to verify (3.4) and (3.3). In order to do it, let u ∈ D, u satisfies (2.18). Let t₂ and t₃ be defined as same as the ones in the proof of Theorem 3.1, that is 0 ≤ t₃ – t₂ < T, u(t₃) = max_t∈[0,T] u(t) and u(t₂) = min_t∈[0,T] u(t). It is easy to see that (2.25) can be deduced from (3.10), and (3.9) verifies (2.26). By using Lemma 2.5 and (2.5) in Lemma 2.4, we know that

Integrating (2.18) over the interval [t₂, t₃], we have

i.e.,

which together with (3.11) and gives

It follows from (3.10) and the condition φ̄ > 0 that there is a constant ρ₂ > 0 such that

From (3.11) and (3.13), we see that the conclusions of (3.3) and (3.4) hold.□

Theorem 3.3

Assume φ̄ > 0, ᾱ > 0 and

If

then equation(1.3) has at least one positive T–periodic solution, where G(x) is defined by (2.4) and

.

Remark 3.1

Assumption (3.14) implies that C₀ ∈ (–∞, +∞) is a constant.

Proof

Similar to the proof of Theorem 3.2, we need only prove the estimates of (3.4) and (3.3). Let u ∈ D, then u satisfies (2.18). Let t₁ and t₂ be as same as the ones in the proof of Lemma 2.4. Multiplying (2.18) with u^μ(t) and integrating it on [t₄, t₂], we get

Since ∫t1t2u″(t)uμ(t)dt≤0, it follows from (3.17) and (3.16) that

By using (2.6) and (3.16), we get

which together with (3.15) verifies (3.4). Thus, according to (2.5), we arrive at

On the other hand, by (3.18) again, we obtain from (3.19) that