Mama Abdelli
email: abdelli-mama@yahoo.fr

Oscillatory Behavior near Blow-up of the Solutions to Some Nonlinear Singular Second Order ODE's

Keywords: AMS classification numbers: 34C10, 34C15, 34D05, 34G20, 35B44 Second order scalar ODE, nonlinear ODE, Oscillatory and non-Oscillatory Blow-up

In this paper, we study the oscillation properties of solutions for the scalar second order nonlinear ODE: (|u | l u ) + d|u| β u = c|u | α u , where α, β, l, c, d are positive constants.

Introduction

We consider the scalar second order nonlinear ODE

(|u | l u ) + d|u| β u = c|u | α u , (1.1)
where l, d, c, β, α are positive constants.

For dissipative ordinary differential equation of the type

(|u | l u ) + d|u| β u + c|u | α u = 0, (1.2)
In [START_REF] Abdelli | Global behavior of the solutions to a class of nonlinear second order ODE's[END_REF] Abdelli and Haraux proved the existence and uniqueness of a global solution u(t) of (1.2) with initial data (u 0 , u 1 ) ∈ R 2 . They established the decay rate and used a method introduced by Haraux [4] to study the oscillatory or non-oscillatory properties of nontrivial solutions. This method is based on a polar coordinate system and the oscillation properties appear to depend on the relation between α and β(l+1)+l β+2 . The results of [START_REF] Abdelli | Global behavior of the solutions to a class of nonlinear second order ODE's[END_REF] can be summarized as follows: Let (A 1 ), (A 2 ), (A 3 ) and (A 4 ) be the assumptions defined as follows: i) If (A 1 ) or (A 2 ) is satisfied, then any non-null solution u(t) of (1.2) and its derivative u (t) have non-constant sign on each interval (T, ∞).

(A 1 ) α > β(l+1)+l β+2 (A 2 ) α = β(l+1)+l
ii) If (A 3 ) is satisfied, any non-null solution u(t) of (1.2) has a finite number of zeroes on (0, ∞). Moreover, for t large enough, u(t) and u (t) have opposite sign and u(t) and u (t) have the same sign.

iii) If (A 4 ) is satisfied, then any non-null solution u(t) of (1.2) has at most one zero on (0, ∞).

We can also consider the equation

u + |u| β u = g(u ), (1.3)
where g is a locally Lipschitz continuous function satisfying the following hypotheses

∃c > 0, ∀v, |g(v)| ≤ c|v| α+1 (1.4) ∃η > 0, ∀v, g(v)v ≥ η|v| α+1 , (1.5)
The equation (1.3) has been studied by Aloui [START_REF] Aloui | Oscillatory behavior near blow-up of the solutions to some second-order nonlinear ODE[END_REF]. By using a method different from the ones from Souplet [START_REF] Souplet | Critical exponents, special large-time behavior and oscillatory blow-up in nonlinear ODE's[END_REF] and Balabane, Jazar and Souplet [START_REF] Balabane | Oscillatory blow-up in nonlinear second order ODE's: the critical case[END_REF], the author recovers the oscillation (or non-oscillation) properties of the solution of (1.3) near the blow-up time T by the same method as [START_REF] Haraux | Sharp decay estimates of the solutions to a class of nonlinear second order ODE's[END_REF] when 1 < α < β. Moreover, the author generalized the results to (1.3) with g a general function satisfying (1.4)- (1.5). The results of [START_REF] Aloui | Oscillatory behavior near blow-up of the solutions to some second-order nonlinear ODE[END_REF] can be summarized as follows:

i) The energy defined by E(t) = u 2 2 + |u| β+2 β+2 blows-up as soon as u ≡ 0 and we have, denoting by T the blow-up time

a) If 0 < α ≤ β β+2 , C 0 (T -t) -2 α ≤ E(t) ≤ C 1 (T -t) -2 α , b) If β β+2 < α < β, E(t) ≤ C (T -t) -(β+2)(α+1) β-α . as t → T , for some C 0 , C 1 , C > 0. ii) If 0 < α < β β+2 or α = β β+2 , c < (β + 2)( β+2 2β+2 ) β+1 β+2
, then all nontrivial solutions have an oscillatory finite-time blow-up T and

lim inf t→T u(t) = lim inf t→T u (t) = -∞, lim sup t→T u(t) = lim sup t→T u (t) = +∞ iii) If β β+2 < α < β, g ∈ C 1
and g > 0, then all nontrivial solutions have a non-oscillatory finite-time blow-up T and u, u have the same sign as t → T .

iv) If α = β β+2 , c ≥ c 0 = (β + 2)( β+2 2β+2 ) β+1 β+2
. Then any solution u(t) of (1.3) blows-up in finite time T and has a finite number of zeroes in [0, T ].

Note that (1.3) with g(v) = c|v| α v is a special case of (1.1) when l = 0. The objective of this paper is to recover the oscillatory (or non-oscillatory) properties of solutions of (1.1) when t ∈ [0, T ] by the same method as in [START_REF] Abdelli | Global behavior of the solutions to a class of nonlinear second order ODE's[END_REF] when l < α < β. Moreover, we use the techniques from [START_REF] Aloui | Oscillatory behavior near blow-up of the solutions to some second-order nonlinear ODE[END_REF].

The plan of the paper is as follows. In section 2 we prove the local existence of the solution of (1.1). In section 3, we show that any solution has an unbounded energy for any nontrivial initial data. In section 4 we show that, under natural conditions, all nontrivial solutions are blowing up and we obtain precise energy estimates of solutions when t → T , with T the blow-up time. Finally, oscillatory and non-oscillatory behavior's are delimited in section 5 and 6.

Local existence

In this section, we shall discuss the local existence for the initial value problem associated to equation (1.1) Proposition 2.1. assume that l ≤ inf{α, β}. Then for any (u 0 , u 1 ) ∈ R 2 , there exists T > 0 for which problem (1.1) has a solution on [0, T ] in the following sense:

u ∈ C 1 [0, T ], |u | l u ∈ C 1 [0, T ] and u 0 = u(0), u 1 = u (0). (2.6)
Proof. To show the existence of the solution for (2.6), we consider for ε ∈ (0, 1)

(ε + (l + 1)|u ε | l )u ε + d|u ε | β u ε = c|u ε | α u ε u ε (0) = u 0 , u ε (0) = u 1 .
(2.7)

The existence and uniqueness of u ε in the class C 2 [0, T ] for some T > 0 is classical. Multiplying (2.7) by u ε , we have the following energy identity

d dt ε 2 |u ε (t)| 2 + l + 1 l + 2 |u ε (t)| l+2 + d β + 2 |u ε (t)| β+2 = c|u ε | α+2 . (2.8) Introducing E ε (t) = ε 2 |u ε (t)| 2 + l + 1 l + 2 |u ε (t)| l+2 + d β + 2 |u ε (t)| β+2 ,
we have as a consequence of (2.8)

d dt E ε (t) ≤ c l + 2 l + 1 E ε (t) α+2 l+2 ≤ c l + 2 l + 1 E ε (t) 1+ α-l l+2 .
Then

- l + 2 α -l d dt E ε (t) -α-l l+2 ≤ c l + 2 l + 1 d dt E ε (t) -α-l l+2 ≥ -c α -l l + 1 By integrating over (0, t), we have E ε (t) -α-l l+2 ≥ -c α -l l + 1 t + 1 E ε (0) α-l l+2
.

Hence, we can estimate an existence time for u ε as a consequence of the inequality

E ε (t) ≤ -c α -l l + 1 t + 1 E ε (0) α-l l+2 -l+2 α-l , ∀0 ≤ t ≤ T ε = l + 1 c(α -l)E ε (0) α-l l+2
.

Introducing T 0 = l + 1 c(α -l)E ε (0) α-l l+2
, it is clear that T 0 < T ε and for ε small enough, we have

∀t ∈ [0, T 0 ], |u ε (t)| ≤ M 1 , |u ε (t)| ≤ M 2 .
(2.9)

where M 1 , M 2 are positive constants independent of ε. Then u ε , u ε are uniformly bounded. From (2.7), we obtain ∀t ∈ [0, T 0 ],

|u ε (t)| l u ε (t) = (l + 1)|u ε (t)| l |u ε (t)| ≤ (ε + (l + 1)|u ε (t)| l )u ε (t) ,
by using (2.9), we deduce

∀t ∈ [0, T 0 ], |u ε (t)| l u ε (t) ≤ M 3 . (2.10) Therefore the function w ε (t) := |u ε (t)| l u ε (t) is uniformly Lipshitz on [0, T 0 ] independently of ε. Then the family of functions u ε (t) = |w ε (t)| 1 l+1 sgnw ε (t) is uniformly equicontinous (actually Hölder continuous ) on [0, T 0 ].
We can now pass to the limit as ε → 0. As a consequence of Ascoli's theorem and a priori estimate (2.9), we may extract a subsequence which is still denoted for simplicity by (u ε ) for which

u ε → u in C 1 [0, T 0 ]
as ε tends to 0. Integrating (2.7) over (0, t), we get

|u ε (t)| l u ε (t) -|u ε (0)| l u ε (0) = c t 0 |u ε (s)| α u ε (s) ds -d t 0 |u ε (s)| β u ε (s) ds -ε t 0 u ε (s) ds = c t 0 |u ε (s)| α u ε (s) ds -d t 0 |u ε (s)| β u ε (s) ds -ε(u ε (t) -u 1 ).
(2.11)

From (2.11), we then have, as ε tends to 0

|u ε | l u ε → c t 0 |u (s)| α u (s) ds -d t 0 |u(s)| β u(s) ds + |u (0)| l u (0) in C 0 [0, T 0 ]. Hence |u | l u = c t 0 |u (s)| α u (s) ds -d t 0 |u(s)| β u(s) ds + |u (0)| l u (0), (2.12) and |u | l u ∈ C 1 [0, T 0 ].
Finally by differentiating (2.12) we conclude that u is a solution of (1.1). Hence, the result with T = T 0 .

The maximal solution

In this section, we still assume 0 ≤ l ≤ inf{α, β}. Then as a consequence of [START_REF] Abdelli | Global behavior of the solutions to a class of nonlinear second order ODE's[END_REF] the solution u of (1.1) with u(0) = u 0 and u (0

) = u 1 is unique on [0, T 0 ]. Moreover, if v is another solution of the same problem on [0, T 1 ] with T 1 > T 0 , then u = v on [0, T 0 ]
. This allows us to obtain a maximal solution on [0, T * ) with 0 < T * ≤ +∞.

Remark 3.1. Integrating (2.8) over (0, t), we then have, by passing to the limit as ε tends to 0

E(t) -E(0) = c t 0 |u (s)| α+2 ds,
where

E(t) = l + 1 l + 2 |u (t)| l+2 + d β + 2 |u(t)| β+2 .
It follows that E is differentiable at any point t ∈ [0, T * ) and

d dt E(t) = c|u (t)| α+2 .
Proposition 3.2. Let (u 0 , u 1 ) = (0, 0) be such that the unique solution of (2.6) is global. Then, u is unbounded and E(t) → ∞ as t tends to ∞.

Proof. Assuming u to be global and bounded, we can introduce the compact metric space

Z = t≥0 {u(t), u (t)} R 2
endowed with the distance associated to the euclidian norm in R 2 . Let {S(t)} t≥0 be the dynamical system such that

S(t) : Z → Z (v 0 , v 1 ) → (v(t), v (t))
.

where v is the solution of problem (1.1) with v 0 = v(0) and v (0) = v 1 . For (ϕ, ψ) ∈ Z, we set

Φ(ϕ, ψ) = -E(ϕ, ψ) = - l + 1 l + 2 |ψ| l+2 + d β + 2 |ϕ| β+2 .
Then if (ϕ(t), ψ(t)) = S(t)(ϕ 0 , ψ 0 ), we have as previously shown in remark 3.1:

d dt Φ(ϕ(t), ψ(t)) = -c|ψ(t)| α+2 ≤ 0. In particular Φ(S(t)(ϕ, ψ)) ≤ Φ(ϕ, ψ), ∀(ϕ, ψ) ∈ Z, ∀t ≥ 0.
Let ω(u 0 , u 1 ) be the ω-limit set of the (u(t), u (t)) as t → +∞. It is clear that

ω(u 0 , u 1 ) ⊂ {(v 0 , v 1 ) ∈ Z, (v(t), v (t)) is global and bounded where (v(t), v (t)) = S(t)(v 0 , v 1 )}.
Since Φ(u(t), u (t)) is non-increasing and bounded, it has a limit L as t → ∞. Hence,

∀(v 0 , v 1 ) ∈ ω(u 0 , u 1 ), Φ(S(t)(v 0 , v 1 )) = L, ∀t ≥ 0. Because d dt Φ(v(t), v (t)) = -c|v (t)| α+2 = 0, ∀t ≥ 0,
this implies v ≡ 0 on R + and by the equation (1.1) we derive v ≡ 0. We now know that ω(u 0 , u 1 ) = {0, 0}. In particular, as t → ∞ Φ(u(t), u (t)) → Φ(0, 0) = 0. But by hypothesis Φ(u(t), u (t)) is non-increasing and Φ(u(0), u (0)) < 0. This is contradictory hence E(t) cannot be bounded.

4 Blow-up of nontrivial solutions and energy estimates near blow-up Theorem 4.1. Let β > α > l and let u = 0 be a solution of (1.1), then u blows-up in a finite time. Moreover, if T > 0 denotes the blow-up time,

i) If l < α ≤ β(l+1)+l β+2 , then there exist C 0 , C 1 > 0 such that C 0 (T -t) -l+2 α-l ≤ E(t) ≤ C 1 (T -t) -l+2 α-l , as t → T, (4.13)
ii) If β(l+1)+l β+2 < α < β, then there exists C > 0 such that

E(t) ≤ C (T -t) -(β+2)(α+1) β-α , as t → T. (4.14)
Proof. We consider the functional:

F (t) = E(t) -|u| γ u|u | l u ,
where l > 0, γ > 0 and > 0. By using Young's inequality with exponents l + 2 and l+2 l+1 , we obtain

|u| γ u|u | l u ≤ c 1 |u| (γ+1)(l+2) + c 2 |u | l+2 ,
we assume that (γ + 1)(l + 2) ≤ β + 2, which reduces to the condition

γ ≤ β -l l + 2 , (4.15) therefore ∀u ∈ R, |u| (γ+1)(l+2) ≤ max{|u| β+2 , 1} ≤ |u| β+2 + 1.
Then, we obtain the existence of K > 0 such that

-C 1 + E(t)(1 -K ) ≤ F (t) ≤ E(t)(1 + K ) + C 2 ,
for small enough, we have

1 2 E(t) -C 1 ≤ F (t) ≤ 2E(t) + C 2 , ∀t ∈ [0, T ]. (4.16)
On the other hand

F (t) = d dt E(t) -(|u| γ u) |u | l u -|u| γ u(|u | l u ) = c|u | α+2 + d |u| γ+β+2 -(γ + 1)|u| γ |u | l+2 -c |u| γ u|u | α u .
(4.17)

By using Young's inequality in the third term with exponents α+2 α-l and α+2 l+2 , we obtain

|u| γ |u | l+2 ≤ δ|u| γ( α+2 α-l ) + c(δ)|u | α+2 , (4.18)
we assume that

γ α + 2 α -l ≤ γ + β + 2,
this is equivalent to the condition

γ ≤ (β + 2)( α -l l + 2 ), (4.19) in order that ∀u ∈ R, |u| γ( α+2 α-l ) ≤ |u| β+γ+2 + 1.
Taking δ small enough, we have for some P > 0 and ρ 1 > 0

-(γ + 1)|u| γ |u | l+2 ≥ - d 4 |u| β+γ+2 -P |u | α+2 -ρ 1 . (4.20)
By using Young's inequality in the last term with exponents α + 2 and α+2 α+1 , we obtain

|u| γ u|u | α u ≤ δ|u| (γ+1)(α+2) + c (δ)|u | α+2 ,
we assume that (γ + 1)(α + 2) ≤ β + γ + 2, which reduces to the condition

γ ≤ β -α α + 1 . (4.21) Then, we have ∀u ∈ R, |u| (γ+1)(α+2) ≤ |u| β+γ+2 + 1.
Taking δ small enough, we have for some P > 0 and ρ 2 > 0

-|u| γ u|u | α u ≥ - d 4 |u| β+γ+2 -P |u | α+2 -ρ 2 . (4.22)
Using (4.20) and (4.22), we have from (4.17)

F (t) ≥ (c -P -P )|u | α+2 + d 2 |u| β+γ+2 -M ≥ (c -Q )|u | α+2 + 2 |u| β+γ+2 -M,
where Q = P + P . we have for small enough,

F (t) ≥ 2 (|u | α+2 + |u| β+γ+2 ) -M, set γ = min (β + 2) α -l l + 2 , β -α α + 1 , β -l l + 2 , and σ = min α + 2 l + 2 , 1 + β -α (β + 2)(α + 1)
.

Then , by using (4.16) and the inequality (x + y) σ ≤ c(σ)(x σ + y σ ) for x, y ≥ 0 , we have

F (t) ≥ 2 c -1 (σ)c 1 E(t) σ -M ≥ 4 c 2 F (t) σ -M ,
where c 2 = c -1 (σ)c 1 and M > 0. First T max < ∞. Assuming T max = ∞, since E is unbounded and nondecreasing, E tends to infinity as t → T max and by (4.16) so is F , thus there exists T * for which 4 c 2 F (t) σ > 2M for t ≥ T * . Therefore,

F (t) ≥ 4 c 3 F (t) σ , (4.23)
a contradiction. Then T max = T < ∞.

Then, we distinguich two cases: i) l < α ≤ β(l+1)+l β+2 , so that (β + 2) α-l l+2 ≤ β-l l+2 and

β -α α + 1 - β -l l + 2 = (β -α)(l + 2) -(β -l)(α + 1) (α + 1)(l + 2) = β(l + 1) + l -α(β + 2) (α + 1)(l + 2) ≥ 0.
We choose

γ = (β + 2)(α -l) l + 2 and σ = α + 2 l + 2 .
By using (4.23), we obtain

d dt (F (t)) -α-l l+2 = - α -l l + 2 F (t)F (t) -α+2 l+2 ≤ - α -l 4(l + 2) c 3 ,
by integrating the above inequality from t to τ , we obtain

F (τ ) -α-l l+2 -F (t) -α-l l+2 ≤ -c 4 (τ -t),
where c 4 = α-l 4(l+2) c 3 . Since F (τ ) → +∞ if τ → T, then F (τ ) -α-l l+2 → 0. Therefore by letting τ → T, we obtain

F (t) ≤ -l+2 α-l c 4 (T -t) -l+2 α-l ,
assuming c 5 = -l+2 α-l c 4 , we have

F (t) ≤ c 5 (T -t) -l+2 α-l , using (4.16), we have E(t) ≤ C 1 (T -t) -l+2 α-l , (4.24)
with C 1 > 2c 5 .

For the converse inequality, we have

E (t) = c|u | α+2 ≤ cKE(t) α+2 l+2 . Then d dt E(t) -α-l l+2 = - α -l l + 2 d dt E(t)E(t) -α+2 l+2 ≥ - α -l l + 2 cK,
by integrating the above inequality from t to τ , we obtain

E(τ ) -α-l l+2 -E(t) -α-l l+2 ≥ -K α -l l + 2 c(τ -t).
Since E(τ ) → +∞ if τ → T, we have

E(t) ≥ C 0 (T -t) -l+2 α-l . (4.25)
Therefore by (4.24) and (4.25), we obtain

C 0 (T -t) -l+2 α-l ≤ E(t) ≤ C 1 (T -t) -l+2 α-l , ∀t ∈ [0, T ].
ii) if β(l+1)+l β+2 < α < β, we have β-α α+1 < β-l l+2 and (β + 2) α-l l+2 > β-l l+2 .

We choose

γ = β -α α + 1 and σ = 1 + β -α (β + 2)(α + 1)
.

From (4.23), we obtain

F (t) ≥ 4 c 3 (α, β)F (t) 1+ β-α (β+2)(α+1) , (4.26)
by (4.26), we have

d dt F (t) -β-α (β+2)(α+1) = - β -α (β + 2)(α + 1) d dt F (t)F (t) -1-β-α (β+2)(α+1) ≤ -c 6 ,
by integrating the above inequality from t to τ , we have

F (τ ) -β-α (β+2)(α+1) -F (t) -β-α (β+2)(α+1) ≤ -c 6 (τ -t), if τ → T, we obtain F (t) ≤ -(β+2)(α+1) β-α c 6 (T -t) -(β+2)(α+1) β-α , assuming C = -(β+2)(α+1) β-α c 6 E(t) ≤ C (T -t) -(β+2)(α+1) β-α .
The proof of Theorem 4.1 is now completed.

5 Oscillatory blow-up of solutions to (1.1) for α small

In this section, we establish the oscillatory blow-up of nontrivial solutions of (1.1). We can use the method from [START_REF] Aloui | Oscillatory behavior near blow-up of the solutions to some second-order nonlinear ODE[END_REF], we obtain the following result.

Theorem 5.1. Assume that

l < α < β(l + 1) + l β + 2 or α = β(l + 1) + l β + 2 , c < (β + 2) (β + 2)(l + 1) d(β + 1)(l + 2) β+1 β+2 ,
then, all nontrivial solutions of (1.1) have oscillatory blow-up at time T < ∞ and

lim sup t→T u(t) = lim sup t→T u (t) = +∞, lim inf t→T u(t) = lim inf t→T u (t) = -∞.
Proof. We proceed in 2 steps.

Step 1. For T > 0, u (t) has at least a zero on [0, T ]. Assume the contrary, which means that u (t) has a constant sign on [0, T ].

For t ∈ [0, T ], we introduce the polar coordinate as follows

d(l + 2) (β + 2)(l + 1) 1 2 |u| β 2 u = r(t) cos θ(t), |u | l 2 u = r(t) sin θ(t), (5.27)
where r and θ are two C 1 functions and r(t) = l+2 l+1 E(t)

1 2 > 0.
A simple calculations shows that θ satisfies at any non-singular point, the differential equation

θ = Ar 2(α-l) l+2 sin θ cos θ| sin θ| 2(α-l) l+2 -Br 2(β-l) (β+2)(l+2) | cos θ| β β+2 | sin θ| -l l+2 , (5.28)
where

A = c l + 2 2(l + 1) , B = (β + 2)(l + 1) d(l + 2) β+1 β+2 l + 2 2(l + 1) . Since l < α < β(l+1)+l β+2 , we have 2(α-l) l+2 < 2(β-l) (β+2)(l+2) and if t → T , r(t) ∼ C(T -t) -l+2 2(α-l) .
Then, if t → T, we have

r(t) 2(α-l) l+2 | sin θ| 2(α-l) l+2 +1 cos θ = r(t) 2(α-l) l+2 | sin θ| 2(α+1) l+2 -l l+2 cos θ ≤ r(t) 2(β-l) (β+2)(l+2) | sin θ| -l l+2 | cos θ| β β+2 , then θ ≤ -ξ(T -t) -γ | sin θ| -l l+2 | cos θ| β β+2 , if t → T,
where ξ > 0 and

γ = l + 2 α -l β -l (β + 2)(l + 2) > 1.
In the case α = β(l+1)+l β+2 , we have

θ = - l + 2 2(l + 1) r(t) 2(α-l) l+2 | sin θ| -l l+2 | cos θ| β β+2 (β + 2)(l + 1) d(l + 2) β+1 β+2 -c| sin θ| 2α-l l+2 +1 | cos θ| 1-β β+2 , since α = β(l+1)+l β+2 , we have β β+2 = 2α-l l+2 . Then θ ≤ - l + 2 2(l + 1) r(t) 2(α-l) l+2 | sin θ| -l l+2 | cos θ| β β+2 (β + 2)(l + 1) d(l + 2) β+1 β+2 -c| sin θ| 2α-l l+2 +1 | cos θ| 1-2α-l l+2 , assuming f (θ) = | sin θ| 2α-l l+2 +1 | cos θ| 1-2α-l l+2 , θ ∈ R. Then, we have max θ∈R f (θ) = 1 β + 2 1 β+2 β + 1 β + 2 β+1 β+2 .
(5.29)

Hence (β + 2)(l + 1) d(l + 2) β+1 β+2 -c| sin θ| 2α-l l+2 +1 | cos θ| 1-2α-l l+2 ≥ (β + 2)(l + 1) d(l + 2) β+1 β+2 -c 1 β + 2 1 β+2 β + 1 β + 2 β+1 β+2 (β + 2)(l + 1) d(l + 2) β+1 β+2 -c 1 β + 2 1 β+2 β + 1 β + 2 β+1 β+2 > 0 ⇔ c < (β + 2) (β + 2)(l + 1) d(β + 1)(l + 2) β+1 β+2 ,
then, we find in all cases for t → T,

θ ≤ -ξ(T -t) -1 | sin θ| -l l+2 | cos θ| β β+2 .
We introduce the function

H(s) = s a | sin u| l l+2 | cos u| β β+2
du, suppose that u does not vanish if t → T and for t ∈ [t 0 , T ], we may assume for instance θ(t) ∈ (-π 2 , π 2 ) and H(θ(t)) = F (t)

∀t 0 ≤ t ≤ T, F (t) ≤ -ξ(T -t) -1 ,
we integrate from t 0 to t

H(θ(t)) ≤ H(θ(t 0 )) -ξ log(T -t 0 ) + ξ log(T -t),
if t → T, we find H(θ(t)) → -∞. Or H(θ(t)) is non-negative, then, we obtain a contradiction. Therefore, u has a zero on each half-line.

Step 2. Applying Step 1, we know that u has an infinite sequence of zeroes tending to infinity.

We claim that between two successive zeroes of u there is a zero of u. Indeed let u Then we observe that when sin θ vanishes, the right hand side of the above equality is 0. Actually it is also continuous at points where sin θ vanishes, so that finally G • θ is C 1 everywhere. Now using Cauchy-Schwarz inequality, we obtain

Br(t) 2(β-l) (β+2)(l+2) - 2(l+1) l+2 | cos θ| β β+2 sin θ| 2α l+2 sin θ cos θ ≤ B 2 A r(t) 4(β-l) (β+2)(l+2) - 2(α-l) l+2 + Ar(t) 2(α-l) l+2 | sin θ| 4(α+1) l+2 + l l+2 cos 2 θ, then [G(θ(t)] ≥ -Cr(t) 4(β-l) (β+2)(l+2) - 2(α-l) l+2 . Since β > α > β(l+1)+l β+2 , from (4.14), we have r(t) ≤ C (T -t) -(β+2)(α+1) β-α for t close enough to T, then [G(θ(t)] ≥ -C (T -t) -λ , with λ = 4(β -l) (β + 2)(l + 2) - 2(α -l) l + 2 (α + 1)(β + 2) β -α = 1 + α β(l + 1) + l -α(β + 2) (β -α)(l + 2) < 1.
To finish the proof we shall use the following Lemma (cf. [START_REF] Haraux | Sharp decay estimates of the solutions to a class of nonlinear second order ODE's[END_REF] for proof).

Lemma 6.2. Let θ ∈ C 1 (a, T ) and G be a non constant τ -periodic function. We assume

G • θ ∈ C 1 (a, T ) and for some h ∈ L 1 (a, T ) [G(θ(t))] ≥ h(t), ∀t ∈ [a, T ].
Then, for t 1 ≤ t < T , θ(t) remains in some interval of length ≤ τ . In addition, if G has finite number of zeroes on [0, τ ], then θ(t) has a limit for t → T .

The proof of Theorem 6.1. From Lemma 6.2, θ(t) → Θ as t → T . We distinguish two cases: Case 1:

If Θ = π 2 mod [π], u ∼ Cr 2 β+2 > 0 if t → T , then u has a constant sign. Case 2: If Θ = π 2 mod [π], |u | ∼ r(t) 2 l+2 > 0 if t → T , then u (t)
does not vanish and u(t) has a constant sign if t → T . Let t 0 be such that u has a constant sign on (t 0 , T ), if u (t) has several zeroes in (T -, T ) for > 0 small enough, then (|u (t)| l u (t)) must have different signs at two successive zeroes |u (t)| l u (t). From equation (1.1) u must have different signs also, which is impossible. Thus, u (t) has a constant sign as t → T . E(t) is unbounded, then E(t) → ∞ as t tends to T . Then β + 1 β + 2 β+1 β+2 = 0 K(θ) > 0, so that θ is non-increasing. The distance of two consecutive zeroes of K(θ) other than π 2 (modπ) is not more than π, therefore we have two cases: Case 1: if θ(t) remains in an interval of length less than π, then θ is bounded from above and is non-increasing thus it converges to a limit as t → T and achieves at most one a value for which u vanishes. Case 2: if θ(t) coincides with one of these zeros for a finite value of t, due to existence and uniqueness for the ODE satisfies by θ(t) near the non-trivial equilibria, θ(t) is constant and u never vanishes.

ii) If c > c 0 , K(θ) < 0. We have two cases:

Case 1: if θ(t) = π 2 , then, it is bounded and since K(θ) < 0 near the trivial zeros, θ(t) is monotone, and therefore it is convergent as t → T . Case 2: if θ(t) = π 2 , then it remains constant and u never vanishes.

  β+2 and c < d(β + 2)( (β+2)(l+1) d(β+1)(l+2) ) β+1 β+2 (A 3 ) α < β(l+1)+l β+2 (A 4 ) α = β(l+1)+l β+2 and c ≥ d(β + 2)( (β+2)(l+1) d(β+1)(l+2) ) β+1 β+2

6 . 1 .

 61 (a) = u (b) = 0 with a < b and u = 0 in (a, b). If u has a constant sign in (a, b), by the equation (|u | l u ) has the same sign for t = a and t = b, which implies that (|u | l u ) have opposite signs on (a, a + η) and (b -η, b) for η > 0 small enough, a contradiction with u = 0 in (a, b). Finally, by (4.13) we have lim t→T u 2 (t) + u (t) = +∞. From the existence of infinitely many zeroes of u(t) and u (t) as t → T it is easy to deduce that lim sup t→T u(t) = lim sup t→T u (t) = +∞, and lim inf t→T u(t) = lim inf t→T u (t) = -∞. The proof of Theorem 5.1 is now completed. 6 Non-oscillatory blow-up of solutions to (1.1) for α large Theorem Assume l ≤ α and β(l+1)+l β+2 < α < β. Then any solution u(t) has a finite number of zeroes in (T -, T ), for some > 0 and blows-up as t → T , where T is the blow-up time. Proof. We introduce G(s) = s 0 | sin v| 2α+l l+2 sin v cos v dv. First we observe that G • θ is C 1 on any interval where u does not vanish. Indeed on such an interval, θ is C 1 and [G(θ(t)] = Ar(t) 2(α-l) l+2 cos 2 θ| sin θ| 4(α+1) l+2 + l l+2 -Br(t) 2(β-l) (β+2)(l+2) | cos θ| β β+2 | sin θ| 2α l+2 sin θ cos θ.

  lim t→T u(t) = lim t→T u (t) = ±∞, Since u(t) and u (t) have the same sign if t → T . Theorem 6.3. Assuming l ≤ α andα = β(1 + l) + l β + 2 , c ≥ c 0 = (β + 2) (β + 2)(l + 1) d(β + 1)(l + 2) β+1 β+2 ,then any solution u(t) of (1.1) blows-up in finite time T and has a finite number of zeroes in [0, T ].Proof. If α = β(1+l)+l β+2, then clearly β β+2 = 2α-l l+2 . In this case sin θ cos θ| sin θ|2α-l l+2 = -l + 2 2(l + 1) r 2(α-l) l+2 | sin θ| -l l+2 | cos θ| 2α-l l+2 (β + 2)(l + 1) d(l + 2) β+1 β+2 -c sin θ cos θ| sin θ| 2α-l l+2 | cos θ| -β β+2 .We setK(θ) = l + 2 2(l + 1) | sin θ| -l l+2 (β + 2)(l + 1) d(l + 2)β+1 β+2 | cos θ| 2α-l l+2 -c sin θ cos θ| sin θ| 2α-l l+2 i) If c = c 0 , using (5.29), we have (β + 2)(l + 1) d(l + 2) β+1 β+2 -c sin θ cos θ| sin θ| 0 | sin θ| 1+ 2α-l l+2 | cos θ| 1-

Acknowledgments We thank very much Professor Alain Haraux for suggesting us to study this delicate problem and for his interesting remarks and comments.