Ann Univ Ferrara (2012) 58:251–289DOI 10.1007/s11565-012-0148-6

On the asymptotic boundedness of the energyof solutions of the wave equation ut t − a(t)�u = 0

Chiara Boiti · Renato Manfrin

Received: 10 January 2012 / Accepted: 18 January 2012 / Published online: 1 February 2012© Università degli Studi di Ferrara 2012

Abstract Applying a class of quadratic forms introduced by Manfrin (Port. Math.65(4):447–484, 2008), we study the asymptotic behavior, as t → +∞, of the energyof the solutions to the Cauchy problem for wave equations with time dependentpropagation speed.

Keywords Wave equations · Energy estimates

Mathematics Subject Classification (2000) 35L15 · 35B05 · 35B35

1 Introduction

Let us consider the Cauchy problem for time dependent wave equation:

utt − a(t)�u = 0, (x, t) ∈ Rn × R

+, (1.1)

u(x, 0) = u0(x), ut (x, 0) = u1(x), (1.2)

where R+ = [0,∞), u0 ∈ H1(Rn), u1 ∈ L2(Rn) and a : R

+ → (0,∞) issufficiently regular.

It is well known (cf. Remark 1.9) that if a(t) is locally absolutely continuous anda′ a−1 ∈ L1(R+) then for every solution u =u(x, t) of (1.1)–(1.2) the energy

C. Boiti (B)Dipartimento di Matematica, Università di Ferrara, Via Machiavelli, 35, 44121 Ferrara, Italye-mail: chiara.boiti@unife.it

R. ManfrinUniversità IUAV di Venezia, Tolentini, S. Croce, 191, 30135 Venice, Italye-mail: manfrin@iuav.it

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252 Ann Univ Ferrara (2012) 58:251–289

Eu(t)def= √

a(t) ‖∇u(·, t)‖2L2 + ‖ut (·, t)‖2

L2√a(t)

, (1.3)

remains bounded as t → ∞. More precisely, there exists � ≥ 1 such that:

�−1 Eu(0) ≤ Eu(t) ≤ � Eu(0), t ∈ R+. (1.4)

The integrability of a′ a−1 in R+ is, in general, only a sufficient condition for the

validity of (1.4). Under the assumption that δ ≤ a(t) ≤ μ, with δ, μ > 0, Reissigand Smith [8] proved (1.4) when a ∈ C2(R+) satisfies |a(k)(t)| ≤ C (1 + t)−k fork = 1, 2 . Hirosawa [1] (cf. Remark 4.2) extended this result showing that (1.4) stillholds for a ∈ Cm(R+),m ≥ 2, satisfying

|a(k)(t)| ≤ C

(1 + t)kq∀ t ≥ 0, 1 ≤ k ≤ m, (1.5)

t∫

0

|a(s)− a∞|ds ≤ C (1 + t)p ∀ t ≥ 0, (1.6)

for some C ≥ 0, a∞ > 0 and p, q ∈ [0, 1) such that q ≥ p + 1−pm , with q > 1

mif p = 0.

Afterwards, Hirosawa and Wirth [2] obtained similar results in the case a(t) → ∞as t → ∞. More precisely, under a rather restrictive set of hypotheses on the structureof the coefficient a(t) (see the assumptions (A1)–(A5), (A1+) of [2, pp. 57–58]), they

proved that there exists � > 0 such that Eu(t) ≤ �(‖u0‖2

H1 + ‖u1‖2L2

)for all

t ∈ R+.

Our aim here is to obtain (1.4), or similar inequalities, under weaker assumptionson the behavior of a(t) as t → ∞. To this purpose we use a different argument. Moreprecisely, we apply some estimates derived from a quadratic form introduced in [4](see Sect. 2) for the Liouville equation wt t + λ2a(t)w = 0.

Indeed, considering (1.1), by Fourier transform with respect to the space variable,we are led to study the Liouville-type equation

vt t + |ξ |2a(t)v = 0 for t ∈ R+, (1.7)

with ξ ∈ Rn . Setting, for v : R

nξ × R

+t → C differentiable with respect to t ,

Ev def= a12 |ξ |2|v|2 + a− 1

2 |vt |2 , E0v

def= Ev∣∣∣t=0, (1.8)

(1.4) is certainly verified if one can show that there exists � ≥ 1 such that

�−1E0v ≤ Ev ≤ � E0

v for t ∈ R+, |ξ | > 0, (1.9)

for every v = v(ξ, t) solution of (1.7).

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Ann Univ Ferrara (2012) 58:251–289 253

To estimate Ev , we distinguish two cases: (i) |ξ | ≥ M , (ii) 0 < |ξ | ≤ M , fora suitable M > 0. In the first case, see Sect. 3, it is enough to apply the energyestimates derived from the quadratic form introduced in Sect. 2. To treat the secondcase, see Sect. 4, we need also suitable perturbation arguments developed in Appen-dix A. Namely, for 0 < |ξ | ≤ M , we consider (1.7) as linear perturbation of a givenLiouville-type equation,

vt t + |ξ |2a0(t)v = 0 for t ∈ R+, |ξ | > 0, (1.10)

with a0 satisfying suitable conditions: a0 of general type (cf. Sect. A.1) (Theorems 1.1,1.4); a0 regularly varying (cf. Sect. A.2) (Theorem 1.5); a0 almost nondecreasing(cf. Sect. A.3) (Theorem 1.7).

To begin with, we set:

Evdef= a0

12 |ξ |2|v|2 + a

− 12

0 |vt |2 , E0v

def= Ev∣∣∣t=0. (1.11)

Theorem 1.1 Let a, a0 : R+ → (0,∞) be continuous functions and m ≥ 2 integer

such that:

(i) a(t), a0(t) ≥ δ, μ−1 ≤ a0(t) a(t)−1 ≤ μ for some δ, μ ∈ (0,∞);

(i i) a ∈ Cm(R+), (a− 12 )(m) ∈ L1(R+);

(i i i) ∃ �, M > 0 s.t. �−1 E0v ≤ Ev ≤ �E0

v ∀ t ≥ 0, 0 < |ξ | ≤ M for everysolution of (1.10);

(iv) supt≥0

(∫ t0 |a − a0 | a0

− 12 ds

)m−1 ‖a− 12 ‖m−2

L∞(t,∞)

∫∞t |(a− 1

2 )(m)|ds < ∞.

Then there exists � ≥ 1 s.t. (1.9) holds for all solutions v of (1.7).

If a is bounded and a(t) ≥ δ > 0 , by Lemma B.4 of Appendix B∫∞

t |(a− 12 )(m)|ds ≤ C(δ,m) ‖a‖m−1

L∞(t,∞)

∫∞t |a(m)|ds for all t ∈ R

+. Hence,Theorem 1.1 gives:

Corollary 1.2 Let a, a0 : R+ → (0,∞) be bounded, continuous functions and m ≥ 2

integer s.t. (i), (i i i)hold, a ∈ Cm(R+), a(m) ∈ L1(R+), supt≥0

(∫ t0 |a − a0 |ds

)m−1

∫∞t |a(m)|ds < ∞. Then there exists� ≥ 1 s.t. (1.9) holds for all solutions v of (1.7).

In proving Theorem 1.1 we require (i i i), (iv) only to treat the case |ξ | ≤ M of (1.9).In fact, (i), (i i) are sufficient to prove that for every � > 0 there exists �� ≥ 1 suchthat�−1

� E0v ≤ Ev ≤ �� E0

v for all t ≥ 0 and |ξ | ≥ �; see Lemma 3.1. Further, if onlythe second inequality of (i i i) holds, i.e. Ev ≤ �E0

v for t ≥ 0, 0 < |ξ | ≤ M, than thesecond inequality of (1.9) is still valid.

By using [6, Theorem 1.2], in Lemma 5.3 we show that Ev ≤ � E0v , for t ≥ 0 , 0 <

|ξ | ≤ 1, does not hold for all solutions of

vt t + (1 + t)k |ξ |2v = 0 , t ∈ R+, (1.12)

if k > 0. Thus, we have the following negative result:

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254 Ann Univ Ferrara (2012) 58:251–289

Corollary 1.3 Let a0 :R+ → (0,∞) be a continuous functions s.t. μ−1 ≤ a0(t) (1+t)−k ≤ μ for some μ ≥ 1, k > 0. Then, independently of �,M > 0, the estimateEv ≤ �E0

v cannot be satisfied uniformly on {0 < |ξ | ≤ M} × R+t for every solution

v of (1.10).

This means that, in general, we cannot expect an estimate like (1.9) when a(t) isunbounded. On the other hand setting , for ε ≥ 0,

Iε,v def= |ξ |2−2ε |v|2 + |vt |2, I0ε,v

def= Iε,v∣∣∣t=0, (1.13)

the solutions of (1.12) satisfy (cf. Lemma 5.4):

Ev ≤ C I0k

2(k+2) ,von {0 < |ξ | ≤ 1} × R

+t , (1.14)

for a suitable C > 0. This leads us to state the following variant of Theorem 1.1.

Theorem 1.4 Let us suppose that assumptions (i), (i i) hold. Besides, assume that

(v) ∃�, M > 0, ε ∈ [0, 1) s.t. Ev ≤ � I0ε,v ∀ t ≥ 0, 0 < |ξ | ≤ M , for

every solution of (1.10).

(vi) supt≥0

(∫ t0 |a − a0 | a0

− 12 ds

)m−11−ε ‖a− 1

2 ‖m−2L∞(t,∞)

∫∞t |(a− 1

2 )(m)|ds < ∞.

Then there exists� ≥ 1 s.t. Ev ≤ �(I0ε,v + E0

v

),∀ t ∈ R

+, |ξ | > 0, for all solutionsv of (1.7).

In particular, under the assumptions of Theorem 1.4 on the coefficients a(t), a0(t),there exists C > 0 such that the solutions of (1.1)–(1.2) satisfy

Eu(t) ≤ C(‖u0‖2

H1 + ‖u1‖2L2

)∀ t ∈ R

+. (1.15)

We can prove that (1.15) still holds under the assumptions of the following:

Theorem 1.5 Let us suppose that (i), (ii), (iv) hold and that a0 ∈ C2 satisfies thefollowing conditions: there exists N ≥ 1 s.t.

a0(s) ≤ N a0(t) for s, t ∈ R+, s ≤ t, (1.16)

4√

a0

t∫

0

4√

a0 ds ≤ N

t∫

0

√a0 ds for t ∈ R

+; (1.17)

for all T > 0,

∣∣∣∣

d

dt

1√a0

∣∣∣∣ ≤ N

∫ T0

√a0 ds

for t ∈ [T,∞), (1.18)

∞∫

T

∣∣∣∣

d2

dt2

1√a0

∣∣∣∣ ds ≤ N

∫ T0

√a0 ds

. (1.19)

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Ann Univ Ferrara (2012) 58:251–289 255

Then ∃� ≥ 1 s.t. Ev ≤ �(|v(ξ, 0)|2 + E0

v

),∀ t ∈ R

+, |ξ | > 0, for all solutions vof (1.7).

Remark 1.6 From this we can easily generalizes the result proved in [2]. Indeed, thehypotheses (i), (1.16), (1.17), (1.18), (1.19) of Theorem 1.5 are immediately verifiedby using the assumptions (A1), (A2), (A1+) of [2, pp. 57–58]) and Lemma 3 of [2,p. 62]. Besides, by direct inspection of the proof of Theorem 1.5, one can show that theconclusion still holds true under slight weaker assumptions than (i i), (iv). Namely,instead of (i i), (iv), we may assume the conditions (1.22) – (1.23) of Sect. 1.1 below.These last conditions are easily verified under the assumptions (A1), (A2), (A4), (A5)of [2, p. 57]). See Remark 4.3 below.

Finally, a result that requires a different set of hypotheses on a0(t) is the following:

Theorem 1.7 Let us suppose that (i), (ii), (iv) hold and that a0 ∈ C2 be non-decreas-ing, or, more generally, that (a′

0)−

a−10

∈ L1(R+) where (a′0)− = min{a′

0, 0}. Then the

following hold:

(1) If a′0a0

−1 ∈ L∞(R+) anda′′

0

a0

− 3

2

(a′

0

a0

)2

∈ L1(R+) , then there exists � ≥ 1

s.t. Ev ≤ � E0v

(1 + |ξ |−1

)for all t ∈ R

+, |ξ | > 0 .

(2) If a′0a−5/4

0∈ L∞(R+) and

a′′0

a5/40

− 3

2

a′0

2

a9/40

∈ L1(R+), then there exists � ≥ 1

s.t. Ev ≤ � E0v

(1 + |ξ |−2

)for all t ∈ R

+, |ξ | > 0.

Considering equation (1.7) with a(t) ≥ δ > 0 s.t. (i i) holds, by [6, Theorem 1.3]one can also show that for every ξ = 0 there exists finite limit

limt→∞ Ev(ξ, t)

def= E∞v (ξ) , with E∞

v (ξ) > 0 if v(ξ, ·) ≡ 0. (1.20)

Thus, we immediately get:

Corollary 1.8 Under the assumptions as in Theorems 1.1, 1.4, 1.5, for every solution

u of (1.1)–(1.2) there exists finite limit limt→∞ Eu(t)def= E∞

u with E∞u > 0 if u ≡ 0.

Remark 1.9 If a ∈ ACloc(R+) , a(t) > 0 for t ∈ R

+, and a′a−1 ∈ L1(R+), then(1.9) is satisfied. Indeed, for every solution v of (1.10), we have

|E ′v| = 1

2

∣∣∣∣a′

a

(√a |ξ |2|v|2 − 1√

a|vt |2

)∣∣∣∣ ≤ |a′|

2aEv. (1.21)

Then, integrating (1.21), we easily obtain the inequalities �−1 E0v ≤ Ev ≤ � E0

v for

t ≥ 0, |ξ | > 0, with a constant � = exp(

12

∫ +∞0

|a′|a ds

).

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256 Ann Univ Ferrara (2012) 58:251–289

1.1 Some weak forms of assumptions (i i), (iv)

In the statements of Theorems 1.1, 1.5, 1.7 one can substitute assumptions (i i), (iv)with some slight weaker conditions.

(1) Instead of assumptions (i i) and (iv), by direct inspection of the proofs (fromSect. 2 to Sect. 4), one can see that it is sufficient to suppose the following:a ∈ Cm(R+),m ≥ 2,

supt≥0

⎛

⎝t∫

0

|a − a0 |√a0

ds

⎞

⎠(

1√a

)1− 1h∣∣∣∣

dh

dth

1√a

∣∣∣∣

1h

<∞ , 1 ≤ h ≤ m − 1 ,

(1.22)

and

supt≥0

⎛

⎝t∫

0

|a − a0 |√a0

ds

⎞

⎠

m−1 ∞∫

t

∣∣∣∣

(1√a

)η0(

d

dt

1√a

)η1

· · ·(

dm

dtm

1√a

)ηm∣∣∣∣ ds<∞

(1.23)

for all integers η0 , . . . , ηm ≥ 0 s.t.

∑

0≤h≤m

ηh = m − 1 ,∑

1≤h≤m

h ηh = m.

The conditions (1.22), (1.23) are easily verified if assumptions (i i), (iv) hold,and, in same cases, they are less restrictive. In particular, (1.22), (1.23) are verifiedunder the assumptions (A1), (A2), (A4), (A5) of [2, p. 57]). See Remark 4.3.

(2) One can also replace (i i), (iv) with the following:

there exists m ≥ 2 integer s.t.(a− 1

2)(m−1) ∈ BV (R+), and

supt≥0

⎛

⎝t∫

0

|a − a0 |a− 12

0 ds

⎞

⎠

m−1

‖a− 12 ‖m−2

L∞(t,∞) Var∞t((a− 1

2 )(m−1))< ∞.

(1.24)

Similar extensions can be obtained for Corollary 1.2 and Theorem 1.4.

1.2 Main notation

Throughout these notes R+ = [0,∞). For ξ = (ξ1, . . . ξn) ∈ R

n, |ξ |2 = ∑ni=1 ξ

2i .

We often indicate by C (or C0,C1, . . .) various positive constants independent of t ≥ 0and ξ ∈ R

n .

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Ann Univ Ferrara (2012) 58:251–289 257

Given an interval I ⊂ R and f : I → R, we write f ∈ BV (I ) if f is ofbounded variation on I ; f ∈ ACloc(I ) if f is locally absolutely continuous on I . Fora, b ∈ I , we denote with Varb

a( f ) the total variation of f in [a, b] . Besides, we alsoset ( f )

− = min{ f, 0}.

2 Quadratic forms for Liouville-type equations

Let us consider, for ξ ∈ Rn, |ξ | > 0, the Liouville-type equation

vt t + |ξ |2a(t)v = 0 , t ∈ R+, (2.1)

with a ∈ Cm(R+),m ≥ 2 integer, a(t) > 0 for all t ∈ R+.

Given a solutions v = v(ξ, t) of (2.1), we define quadratic form Q = Q(ξ, t) as

Qdef=

∑

0≤i≤[m2

]−1

αi |ξ |−2i (a|ξ |2|v|2 + |vt |2)

+∑

0≤i≤[m2

]−1

βi |ξ |−2i Re(vvt )+∑

0≤i<m2 −1

γi |ξ |−2i−2|vt |2, (2.2)

where αi , βi , γi : R+ → R satisfy the system of linear conditions:

γ−1 ≡ 0,

⎧⎨

⎩

(aαi )′ − aβi = 0

α′i + βi = −γ ′

i−1β ′

i − 2aγi = 0for 0 ≤ i ≤

[m

2

]− 1. (2.3)

The following result was proved in [4, Theorem 1.1].

Theorem 2.1 System (2.3) is solvable in R+. If {αi , βi , γi }0≤i≤[m

2

]−1 is a solution of

(2.3), then αi ∈ Cm−2i (R+), βi ∈ Cm−2i−1(R+), γi ∈ Cm−2i−2(R+), for 0 ≤ i ≤[m2

]− 1.Moreover, for any complex-valued solution v(ξ, t) of (2.1), we have:

d Q

dt=⎧⎨

⎩

β ′[m2

]−1|ξ |−m+2 Re(vvt ) f or m even,

γ ′[m2

]−1|ξ |−m+1|vt |2 f or m odd.

(2.4)

To give the structure of αi , βi , γi , we state first the following:

Definition 2.2 Given j, l ≥ 0 integers, we denote by H jl the set of homogeneous

polynomials P in the variables ζ0, . . . , ζ j of the form

P(ζ0, . . . , ζ j ) =∑

cη0,...,η j ζη00 ζ

η11 · · · ζ η j

j ,

where cη0,...,η j ∈ R, η0, . . . , η j ≥ 0 are integers such that∑

0≤h≤ j ηh = l and∑

0≤h≤ j hηh = j . Besides, given P ∈ H jl and a function ϕ ∈ Cm with m ≥ j , we

denote by Pϕ the function

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258 Ann Univ Ferrara (2012) 58:251–289

Pϕ(t) =∑

cη0,...,η jϕ(t)η0ϕ′(t)η1 · · ·ϕ( j)(t)η j . (2.5)

Setting, for the coefficient a(t) of Eq. (2.1),

ω(t)def= 1

2√

a(t), (2.6)

the following result was proved in [4, Theorem 1.2].

Theorem 2.3 There exist polynomials Pi ∈ H2i2i and Qi ∈ H2i+1

2i+1, for 0 ≤ i ≤[m2

]− 1, with P0 ≡ 1, such that {αi , βi , γi }0≤i≤[m2

]−1 satisfies (2.3) if and only if

αi = ω

i∑

h=0

ci−h Phω, βi =i∑

h=0

ci−h Qhω, γi = 2ω2β ′i (2.7)

for 0 ≤ i ≤ [m2

] − 1, where c0, . . . , c[m2

]−1 ∈ R. Moreover, Pi , Qi are independentof m and ω.

Finally, for 1 ≤ j ≤ m and t ∈ R+, we define

φ j (t)def= max

1≤h≤ jω(t)1− 1

h |ω(h)(t)| 1h . (2.8)

Applying Theorem 2.3, one can prove the following estimates [5, Lemma 2.6].

Proposition 2.4 Assume a ∈ Cm(R+) with m ≥ 2, a(t) > 0. There exists thenC = C(m) > 0 such that, for all t ∈ R

+:

|Piω(t)| ≤ Cφ2i (t)2i 1 ≤ i ≤

[m

2

]− 1 ,

|Qiω(t)| ≤ Cφ2i+1(t)2i+1 0 ≤ i ≤

[m

2

]− 1 ,

ω(t)|(Qiω)′(t)| ≤ Cφ2i+2(t)

2i+2 0 ≤ i <m

2− 1.

2.1 Some preliminary estimates

Applying Theorem 2.3, from now on we choose

αi = 2ωPiω, βi = 2Qiω, γi = 4ω2(Qiω)′ (2.9)

for 0 ≤ i ≤ [m2 ] − 1. Since P0 ≡ 1 and ω = 2−1a− 1

2 , we find that

Q(ξ, t) = Ev(ξ, t)+ F(ξ, t), (2.10)

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Ann Univ Ferrara (2012) 58:251–289 259

where Ev(ξ, t) is the energy introduced in (1.8) and

F def=∑

1≤i≤[m2

]−1

αi

|ξ |2i(a|ξ |2|v|2 + |vt |2)+

∑

0≤i≤[m2

]−1

βi

|ξ |2iRe(vvt )

+∑

0≤i<m2 −1

γi

|ξ |2i+2 |vt |2, ∀ t ≥ 0, ξ ∈ Rn \ {0}. (2.11)

Let us set

ei = αi

|ξ |2i(a(t)|ξ |2|v|2 + |vt |2), 0 ≤ i ≤

[m

2

]− 1

fi = βi

|ξ |2iRe(vvt ), 0 ≤ i ≤

[m

2

]− 1

gi = γi

|ξ |2i+2 |vt |2, 0 ≤ i <m

2− 1

so that

F =∑

1≤i≤[m2

]−1

ei +∑

0≤i≤[m2

]−1

fi +∑

0≤i<m2 −1

gi . (2.12)

Let us give an estimate of ei , fi , gi . For all t ≥ 0 and |ξ | > 0 we have:

|ei | = 2ω|Piω||ξ |2i

(a(t)|ξ |2|v|2 + |vt |2) = |Piω||ξ |2i

Ev ,

| fi | = 2|Qiω||ξ |2i

| Re(vvt )| ≤ |Qiω||ξ |2i+1 Ev ,

|gi | = 4ω2|(Qiω)′|

|ξ |2i+2 |vt |2 ≤ 2ω

|ξ |2i+2 |(Qiω)′| Ev ,

since | Re(vvt )| ≤ |v||vt | ≤ 12|ξ | Ev . Then, applying Proposition 2.4, there exists a

constant C1 = C1(m) > 0 such that, for all t ≥ 0 and |ξ | > 0,

|ei | ≤ C1

|ξ |2iφ2i

2i Ev , 1 ≤ i ≤[m

2

]− 1, (2.13)

| fi | ≤ C1

|ξ |2i+1 φ2i+12i+1 Ev , 0 ≤ i ≤

[m

2

]− 1, (2.14)

|gi | ≤ 2C1

|ξ |2i+2 φ2i+22i+2 Ev , 0 ≤ i <

m

2− 1 (2.15)

and hence, from (2.12), we easily have

|F | ≤ 3C1Ev∑

1≤i≤m−1

(φi

|ξ |)i

. (2.16)

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260 Ann Univ Ferrara (2012) 58:251–289

Lemma 2.5 Assume a ∈ Cm(R+) with m ≥ 2, a(t) ≥ δ > 0, ω(m) ∈ L1(R+) . Thenfor all � > 0 there exists τ� ≥ 0 such that

|F(ξ, t)| ≤ 2−1 Ev(ξ, t) , ∀ t ≥ τ� , |ξ | ≥ �. (2.17)

Proof By (2.6), a(t) ≥ δ ⇒ |ω| ≤ 2−1δ− 12 . Then, ω ∈ L∞(R+) and ω(m) ∈

L1(R+) ⇒ limt→+∞ ω(m−1)(t) = 0. Thus, we have

ω(m−1)(t) = −+∞∫

t

ω(m)(s)ds , ∀ t ≥ 0.

Applying the Gagliardo–Nirenberg inequality (B.3), with m − 1 instead of m, wededuce that for all τ ≥ 0, 1 ≤ h ≤ m − 1:

‖ω(h)‖L∞(τ,∞) ≤ 4h(m−1−h)‖ω(m−1)‖h

m−1L∞(τ,∞) ‖ω‖

m−1−hm−1

L∞(τ,∞)

≤ 4h(m−1−h)‖ω‖m−1−h

m−1L∞(τ,∞)

⎛

⎝∞∫

τ

|ω(m)|ds

⎞

⎠

hm−1

.

Therefore, for all t ≥ τ ≥ 0 and 1 ≤ i ≤ m − 1:

φi (t) ≤ max1≤h≤i

‖ω‖1− 1h

L∞(τ,∞) ‖ω(h)‖1hL∞(τ,∞)

≤ 4m−2‖ω‖m−2m−1L∞(τ,∞)

⎛

⎝∞∫

τ

|ω(m)|ds

⎞

⎠

1m−1

. (2.18)

In particular, for any ρ > 0 there exists τρ ≥ 0 such that, for 1 ≤ i ≤ m − 1,

φi (t) ≤ ρ ∀ t ≥ τρ. (2.19)

Hence, from (2.16) with |ξ | > ρ :

|F | ≤ 3C1Ev∑

1≤i≤m−1

ρi

|ξ |i ≤ 3C1ρ

|ξ | − ρEv ≤ 1

2Ev ,

if t ≥ τρ and 3C1ρ/(|ξ | − ρ) ≤ 12 , i.e.

|ξ | ≥ ρ (6C1 + 1) . (2.20)

To obtain (2.17) it is now sufficient to take ρ = �(6C1 + 1)−1 and τ� = τρ . ��

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Ann Univ Ferrara (2012) 58:251–289 261

To continue in the estimates, let us set

kmdef={ |β ′[m

2

]−1| for m even,

ω−1 |γ ′[m2

]−1| for m odd.

(2.21)

Lemma 2.6 Assume a ∈ Cm(R+) with m ≥ 2, a(t) ≥ δ > 0, ω(m) ∈ L1(R+) . Thenthere exists C = C(m) > 0 such that, for all t ∈ R

+,

∞∫

t

|km | ds ≤ C ‖ω‖m−2L∞(τ,∞)

∞∫

t

|ω(m)| ds. (2.22)

Proof Let m ≥ 2 even. From (2.9), we have |β ′[m2

]−1| = 2|(Q[m

2

]−1ω)′|, with 2([m

2 ]−1)+ 1 = m − 1. Hence Q[m

2

]−1 ∈ Hm−1m−1 and, by Definition 2.2,

β ′[m2

]−1 =∑

qη ωη0(ω′)η1 · · · (ω(m))ηm ,

where qη = qη0,...,ηm ∈ R,∑

0≤h≤m ηh = m − 1 and∑

1≤h≤m hηh = m. Applyingthe inequality (B.5) with l = m − 1, we get

‖β ′[m2

]−1‖L1(τ,∞) ≤ C ‖ω‖m−2L∞(τ,∞)‖ω(m)‖L1(τ,∞) ∀ τ ≥ 0, (2.23)

with C = C(m) > 0. Next, let m ≥ 2 odd. In this case 2([m2 ] − 1)+ 1 = m − 2, so

that Q[m2

]−1 ∈ Hm−2m−2. Then, from (2.9) and Definition 2.2, we have

ω−1γ ′[m2

]−1 = 4ω−1(ω2(

Q[m2

]−1ω)′)′

=∑

rη ωη0(ω′)η1 · · · (ω(m))ηm ,

with rη = rη0,...,ηm ∈ R,∑

0≤h≤m ηh = m − 1 and∑

1≤h≤m hηh = m. Applying(B.5), we get

‖ω−1γ ′[m2

]−1‖L1(τ,∞) ≤ C ‖ω‖m−2L∞(τ,∞)‖ω(m)‖L1(τ,∞) ∀ τ ≥ 0, (2.24)

with C = C(m) > 0. From (2.23), (2.24) we obtain (2.22). ��

3 Estimates for |ξ | bounded away from 0

Lemma 3.1 Let us suppose a ∈ Cm(R+) with m ≥ 2, a(t) ≥ δ > 0, ω(m) ∈L1(R+) . Then, given an arbitrary � > 0, there exists �� ≥ 1 such that

�−1�

Ev(ξ, 0) ≤ Ev(ξ, t) ≤ ��Ev(ξ, 0) ∀ t ≥ 0, |ξ | ≥ �. (3.1)

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262 Ann Univ Ferrara (2012) 58:251–289

Proof To begin with, we set

ψmdef={β ′[m

2

]−1|ξ |−m+2 Re(vvt ) for m even,

γ ′[m2

]−1|ξ |−m+1|vt |2 for m odd.

(3.2)

From Theorem 2.1, (2.10) and Lemma 2.5, for t ≥ τ� , |ξ | ≥ � we get

Ev(ξ, t) = −F(ξ, t)+ Ev(ξ, τ� )+ F(ξ, τ� )+t∫

τ�

ψm(ξ, s)ds,

≤ 1

2Ev(ξ, t)+ 3

2Ev(ξ, τ� )+

t∫

τ�

|ψm(ξ, s)|ds. (3.3)

Now, we observe that (2.21) and (3.2) give

|ψm(ξ, t)| ≤ km(t)

2|ξ |m−1 Ev(ξ, t), t ≥ 0, |ξ | > 0. (3.4)

Then, we have

Ev(ξ, t) ≤ 3 Ev(ξ, τ� )+ 2

t∫

τ�

|ψm(ξ, s)|ds

≤ 3 Ev(ξ, τ� )+ 1

|ξ |m−1

t∫

τ�

km(s) Ev(ξ, s)ds ∀t ≥ τ� , |ξ | ≥ �. (3.5)

We can now apply Gronwall’s inequality and Lemma 2.6 to (3.5). We obtain that

Ev(ξ, t) ≤ 3 Ev(ξ, τ� ) exp

⎧⎪⎨

⎪⎩

1

|ξ |m−1

t∫

τ�

km ds

⎫⎪⎬

⎪⎭

≤ 3 exp

{C

|ξ |m−1 ‖ω‖m−2L∞(τ

�,∞) ‖ω(m)‖L1(τ

�,∞)

}Ev(ξ, τ� )

≤ 3 exp

{C δ1− m

2

2m−2 �m−1 ‖ω(m)‖L1(τ�,∞)

}

Ev(ξ, τ� ), (3.6)

for all t ≥ τ� and |ξ | ≥ �.On the other hand, since a(t) ≥ δ and a′ ∈ L1(0, τ� ), by the same argument of

Remark 1.9 we easily have

C−1�

Ev(ξ, 0) ≤ Ev(ξ, τ� ) ≤ C� Ev(ξ, 0), (3.7)

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Ann Univ Ferrara (2012) 58:251–289 263

with C� = exp(

12

∫ τ�

0|a′|a ds

). Then, from (3.6) and (3.7), we deduce the second

inequality of (3.1). The proof of the first inequality of (3.1) is similar. ��Remark 3.2 In proving the first inequality of (3.1), i.e. �−1

�Ev(ξ, 0) ≤ Ev(x, t) for

t ≥ τ� , |ξ | ≥ �, one can merely perform the previous estimates proceeding backwardin time.

4 Estimates for |ξ | > 0 bounded

4.1 Proof of Theorem 1.1

To estimate Ev for |ξ | > 0 small we apply Lemma A.1 of Appendix A. First of allwe set

λ = |ξ |, b = a − a0 ,

and define B(t) as in (A.6). By (i i i) and (A.7) we obtain

Ev ≤ 4�E0v exp {4�|ξ |B(t)} ∀t ≥ 0, 0 < |ξ | ≤ M, (4.1)

for every solution v of (1.7). Moreover, assumption (i) implies that

Ev√μ

≤ Ev ≤ √μEv, ∀ t ≥ 0, |ξ | > 0. (4.2)

Thus (4.1) and (4.2) give

Ev ≤ 4μ� E0v exp {4�|ξ |B(t)} ∀t ≥ 0, 0 < |ξ | ≤ M. (4.3)

To continue, for y ≥ 0, we set

γ (y)def={∞ if B(t) ≤ y ∀ t ≥ 0,

max {t ≥ 0 : B(t) ≤ y} otherwise.(4.4)

The function γ : R+ → [0,∞] is well defined, because B : R

+ → R is continuous,non-decreasing and B(0) = 0. For τ ∈ R

+ we have

γ (y) = τ ⇔ B(τ ) = y, B(t) > y for t > τ. (4.5)

By assumption (iv) there exists L ≥ 0 such that

B(t)m−1 ‖ω‖m−2L∞(t,∞)

+∞∫

t

|ω(m)| ds ≤ L , ∀ t ≥ 0. (4.6)

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264 Ann Univ Ferrara (2012) 58:251–289

Thus, given χ ∈ (0,∞) and 0 < |ξ | ≤ M , such that γ(χ|ξ |)< ∞, from (4.6) we

obtain that

‖ω‖m−2L∞(t,∞)

+∞∫

t

|ω(m)|ds ≤ L

B(γ(χ|ξ |))m−1 = L

χm−1 |ξ |m−1, (4.7)

for all t ≥ γ(χ|ξ |)

, since B(γ(χ|ξ |))

= χ|ξ | > 0 and B is non-decreasing. This

permits to estimate φi (t), for t ≥ γ (χ|ξ | ). In fact, for 1 ≤ i ≤ m − 1, from (2.18) we

have

φi (t) ≤ 4m−2‖ω‖m−2m−1L∞(t,+∞)

⎛

⎝+∞∫

t

|ω(m)|ds

⎞

⎠

1m−1

≤ C2 χ−1|ξ |, (4.8)

with C2 = C2(m, L) > 0 . Then, coming back to (2.16), for t ≥ γ (χ|ξ | ) we obtain:

|F | ≤ 3C1Ev∑

1≤i≤m−1

(C2

χ

)i

= 3C1C2

χ − C2Ev ≤ 1

2Ev, (4.9)

provided

χ ≥ χdef= C2(6C1 + 1). (4.10)

Taking (4.9), (4.10) into account, we now distinguish the two cases:

(1) 0 < |ξ | ≤ M, γ(χ|ξ |)

= ∞ ; (2) 0 < |ξ | ≤ M, γ(χ|ξ |)< ∞.

Case (1) We have B(t) ≤ χ|ξ | ∀ t ≥ 0, because of (4.4). Therefore, from (4.3) we

have that

Ev ≤ 4μ� exp {4�χ} E0v ∀ t ≥ 0. (4.11)

Case (2) For simplicity, let us set

γ

(χ

|ξ |)

def= T . (4.12)

From (2.10), (3.2), (4.9)–(4.10) and Theorem 2.1 we get, for all t ∈ [T ,∞),

Ev = −F + Ev∣∣∣t=T

+ F∣∣∣t=T

+t∫

T

ψmds ≤ 1

2Ev + 3

2Ev∣∣∣t=T

+t∫

T

|ψm | ds.

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Ann Univ Ferrara (2012) 58:251–289 265

Hence, from (3.4),

Ev ≤ 3 Ev∣∣∣t=T

+ 1

|ξ |m−1

t∫

T

km Ev ds. (4.13)

From (2.22) and (4.7) with χ = χ , we obtain that

‖km‖L1(T ,∞) ≤ C ‖ω‖m−2L∞(T ,∞)

‖ω(m)‖L1(T ,∞) ≤ C L

χm−1 |ξ |m−1.

We can therefore apply Gronwall’s Lemma to (4.13) and obtain that

Ev(ξ, t) ≤ 3 exp

{C L

χm−1

}Ev(ξ, T ) = C3 Ev(ξ, T ) ∀ t ∈ [T ,∞)

, (4.14)

with C3 = 3 eC Lχm−1 ≥ 1. To investigate what happens for 0 ≤ t ≤ T , we now remark

that B(t) ≤ χ|ξ | for 0 ≤ t ≤ T . Therefore, from (4.3) we have that

Ev ≤ 4μ� exp {4�χ} E0v for 0 ≤ t ≤ T . (4.15)

In particular, from (4.14) and (4.15), there exists C4 ≥ 1 such that:

Ev ≤ C4 E0v ∀ t ≥ 0. (4.16)

Conclusion of the proof of Theorem 1.1 From the inequalities (4.11) and (4.16) wededuce that there exists � ≥ 1 such that the second inequality of (1.9) holds fort ≥ 0, 0 < |ξ | < M.

Setting � = M, by Lemma 3.1, we also know that there exists �M such that (3.1)holds for |ξ | ≥ M. Summarizing up, we have the second inequality of (1.9).

The first inequality of (1.9), i.e. �−1E(ξ, 0) ≤ E(ξ, t) for all t ≥ 0, |ξ | > 0, canbe proved applying the last part of Lemma A.1 and arguing almost in the same way.

��Corollary 4.1 Let a ∈ Cm(R+), with m ≥ 2, s.t.

∫∞0 |a(m)|ds < +∞, δ ≤ a(t) ≤ μ

for some δ, μ > 0, and assume that there exists a constant a∞ > 0 such that

supt≥0

⎛

⎝t∫

0

|a − a∞|ds

⎞

⎠

m−1 +∞∫

t

|a(m)|ds < ∞. (4.17)

Then there exists � ≥ 1 s.t. every solution v of (1.7) satisfies (1.9).

Proof Setting a0 ≡ a∞ , and applying Remark 1.9, assumptions (i), (i i i) are easilyverified. Lemma B.4 and (4.17) imply (i i), (iv). Thus, the hypotheses of Theorem 1.1are satisfied. ��

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266 Ann Univ Ferrara (2012) 58:251–289

Remark 4.2 In [1] Hirosawa proved the energy estimate (1.4) for a ∈ Cm(R+), withm ≥ 2, satisfying 0 < δ ≤ a(t) ≤ μ and conditions (1.5), (1.6). This result can beeasily derived from Corollary 4.1, because assumptions (1.5) and (1.6) imply (4.17).Indeed, for t ≥ 0 one has:

⎛

⎝t∫

0

|a − a∞|ds

⎞

⎠

m−1 +∞∫

t

|a(m)|ds ≤ Cm

mq − 1

(1 + t)p(m−1)

(1 + t)mq−1 ≤ const.

because the conditions on m, p, q imply that mq > 1 and p(m − 1) ≤ mq − 1.

4.2 Proof of Corollary 1.3

We set a(t) = (1+ t)k . Then, since k > 0, assumption (i i) of Theorem 1.1 is satisfiedfor all m ≥ 2. Besides, the condition μ−1 ≤ a0(t) (1 + t)−k ≤ μ gives (i) and

⎛

⎝t∫

0

|a − a0 | a− 1

20 ds

⎞

⎠

m−1

‖a− 12 ‖m−2

L∞(t,∞)

+∞∫

t

|(a− 12 )(m)|ds

≤ C(1 + t)(k2 +1)(m−1)− k

2 (m−2)−m− k2 +1 = C (4.18)

for all t ∈ R+, for a suitable C > 0. Thus, assumption (iv) is verified.

Now, by Example 5.2, we know that (1.9) does not hold when a(t) = (1 + t)k withk > 0. This means that (i i i) of Theorem 1.1 cannot be verified. A detailed inspectionof the proofs of Example 5.2 and Theorem 1.1 reveals that, independently of �,M > 0,the estimate Ev ≤ �E0

v, t ≥ 0, cannot be satisfied, uniformly for 0 < |ξ | ≤ M, forevery solution of (1.10).

4.3 Proof of Theorem 1.4

We can follow almost the same lines of the proof of Theorem 1.1 in Sect. 4.1.To estimate Ev , for |ξ | > 0 small, we apply Lemma A.2. with λ = |ξ |, b = a − a0

and B(t) as in (A.6). To begin with, from assumptions (i), (v) and (A.18), we have:

Ev ≤ 4√μ� I0

ε,v exp{

4�|ξ |1−ε B(t)}

∀t ≥ 0, 0 < |ξ | ≤ M, (4.19)

for every solution v of (1.7). Then, as in the proof of Theorem 1.1, for y ≥ 0 we set

γ (y)def={∞ if B(t) ≤ y ∀ t ≥ 0,

max {t ≥ 0 : B(t) ≤ y} otherwise.(4.20)

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Ann Univ Ferrara (2012) 58:251–289 267

By assumption (vi) there exists L ≥ 0 such that

B(t)m−11−ε ‖ω‖m−2

L∞(t,∞)

+∞∫

t

|ω(m)| ds ≤ L , ∀ t ≥ 0. (4.21)

Thus, given χ ∈ (0,∞) and 0 < |ξ | ≤ M , such that γ(χ1−ε|ξ |1−ε

)< ∞, from (4.21) we

find

‖ω‖m−2L∞(t,∞)

+∞∫

t

|ω(m)|ds ≤ L

B(γ(χ1−ε|ξ |1−ε

))m−11−ε

= L

χm−1 |ξ |m−1, (4.22)

for all t ≥ γ(χ1−ε|ξ |1−ε

), since B

(γ(χ1−ε|ξ |1−ε

))= χ1−ε

|ξ |1−ε > 0. This fact gives an estimate

of the terms φi (t), for t ≥ γ(χ1−ε|ξ |1−ε

), because of (2.18). Namely, for 1 ≤ i ≤ m − 1,

we find

φi (t) ≤ 4m−2‖ω‖m−2m−1L∞(t,+∞)

⎛

⎝+∞∫

t

|ω(m)|ds

⎞

⎠

1m−1

≤ C2 χ−1|ξ |, (4.23)

with C2 = C2(m, L) > 0 . Then, coming back to (2.16), for t ≥ γ(χ1−ε|ξ |1−ε

)we have

|F | ≤ 12 Ev provided χ ≥ χ

def= C2(6 C1 + 1) . Hence, we can now distinguish thetwo cases:

(I) 0 < |ξ | ≤ M, γ(χ1−ε|ξ |1−ε

)= ∞ ; (II) 0 < |ξ | ≤ M, γ

(χ1−ε|ξ |1−ε

)< ∞.

Case (I) We have B(t) ≤ χ1−ε|ξ |1−ε ∀ t ≥ 0, because of the definition (4.20). Therefore,

from (4.19) we have that

Ev ≤ 4√μ� exp

{4�χ1−ε} I0

ε,v ∀ t ≥ 0. (4.24)

Case (II) Setting γ(χ1−ε|ξ |1−ε

)def= T , and proceeding as in Case (2) of the proof of

Theorem 1.1 (see Sect. 4.1), we obtain the inequalities: Ev(ξ, t) ≤ C3 Ev(ξ, T ) for allt ∈ [T ,∞)

; Ev ≤ 4√μ� exp

{4�χ1−ε} I0

ε,v for 0 ≤ t ≤ T .From these estimates, we may conclude the proof as in the case of Theorem 1.1.

4.4 Proof of Theorem 1.5

The proof is similar to the previous ones. First of all, we set λ = |ξ |, b = a − a0 anddefine Ev,B(t) as in (A.3), (A.6). Assumption (i) and (1.16), (1.17), (1.18), (1.19)

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268 Ann Univ Ferrara (2012) 58:251–289

permit us to apply Lemma A.5. Hence, fixed an arbitrary M > 0, there exists � =�(δ,M, μ, N ) > 0 s.t.

Ev ≤ �(|v(ξ, 0)|2+|v′(ξ, 0)|2

)exp{�|ξ |B(t)}, for t ≥ 0, 0 < |ξ | ≤ M.

(4.25)

Noting that Ev ≤ √μEv , we can proceed as in the proof of Theorem 1.1. More

precisely, for y ≥ 0, we set

γ (y)def={∞ if B(t) ≤ y ∀ t ≥ 0,

max {t ≥ 0 : B(t) ≤ y} otherwise.(4.26)

By (iv) there exists L ≥ 0 such that

B(t)m−1 ‖ω‖m−2L∞(t,∞)

+∞∫

t

|ω(m)| ds ≤ L , ∀ t ≥ 0. (4.27)

Thus, given χ > 0, |ξ | > 0, such that γ(χ|ξ |)< +∞, from (4.27) we obtain that

‖ω‖m−2L∞(t,∞)

+∞∫

t

|ω(m)|ds ≤ L

χm−1 |ξ |m−1, ∀ t ≥ γ

(χ

|ξ |). (4.28)

As in (4.8), this gives an estimate of φi (t), for t ≥ γ (χ|ξ | ), because of (2.18). Namely,

φi (t) ≤ C2 χ−1|ξ |, for 1 ≤ i ≤ m − 1. Then, coming back to (2.16), for t ≥ γ (

χ|ξ | )

we have:

|F | ≤ 3C1Ev∑

1≤i≤m−1

(C2

χ

)i

= 3C1C2

χ − C2Ev ≤ 1

2Ev, (4.29)

provided

χ ≥ χdef= C2(6 C1 + 1). (4.30)

We now distinguish two cases:

(J) 0 < |ξ | ≤ M, γ(χ|ξ |)

= ∞, (JJ) 0 < |ξ | ≤ M, γ(χ|ξ |)< ∞.

Case (J) We have B(t) ≤ χ|ξ | ∀ t ≥ 0. Therefore, from (4.25) we obtain

Ev ≤ √μ�

(|v(ξ, 0)|2 + |v′(ξ, 0)|2

)exp {�χ} ∀ t ≥ 0. (4.31)

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Ann Univ Ferrara (2012) 58:251–289 269

Case (JJ) For simplicity, we set γ(χ|ξ |)

= T as in (4.12). From (2.10), (3.2),

(4.29)–(4.30) and Theorem 2.1 we get, for all t ∈ [T ,∞),

Ev ≤ 3 Ev∣∣∣t=T

+ 1

|ξ |m−1

t∫

T

km Ev ds. (4.32)

From (2.22) and (4.28) with χ = χ , we obtain that

‖km‖L1(T ,∞) ≤ C L

χm−1 |ξ |m−1.

We can therefore apply Gronwall’s Lemma to (4.32) and obtain that

Ev(ξ, t) ≤ C3 Ev(ξ, T ) ∀ t ∈ [T ,∞), (4.33)

with C3 = 3 eC Lχm−1 ≥ 1. Finally, since B(t) ≤ χ

|ξ | for 0 ≤ t ≤ T , from (4.25) wehave

Ev ≤ √μ�

(|v(ξ, 0)|2 + |v′(ξ, 0)|2

)exp {�χ} for 0 ≤ t ≤ T . (4.34)

In particular, from (4.33) and (4.34) we deduce that:

Ev ≤ C(|v(ξ, 0)|2 + |v′(ξ, 0)|2

)∀ t ≥ 0, (4.35)

for as suitable constant C ≥ 1.

Conclusion of the proof of Theorem 1.5 Having proved the estimates (4.31), (4.35)for 0 < |ξ | ≤ M , it only remains to apply Lemma 3.1. Setting � = M we also knowthat there exists �M such that (3.1) holds for |ξ | ≥ M. Thus, summarizing up, wededuce that there exists C ≥ 1 such that Ev ≤ C(|v(ξ, 0)|2+E0

v ) for all t ≥ 0, |ξ | > 0.��

Remark 4.3 As observed in Sect. 1.1, Theorem 1.5 continues to holds if (i), (1.16),(1.17), (1.18), (1.19) and (1.22), (1.23) are verified. All these conditions are satis-fied if assumptions (A1), (A2), (A4), (A5), (A1+) of [2, pp. 57–58]) hold. Indeed,the assumptions of [2] imply that there exist α, β ∈ Cm(R+) (m ≥ 2) s.t.

√a0 =

α,√

a = αβ, α > 0, α′ ≥ 0, 0 < c1 ≤ β ≤ c2 ,

lim supt→∞

(1 + ∫ t

0 α ds)α′

α2 < 2, (4.36)

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270 Ann Univ Ferrara (2012) 58:251–289

and c > 0 constant, �, � : R+ → (0,∞) nondecreasing, continuous functions s.t.

|α(k)| ≤ c αk+1

(1 + ∫ t

0 α ds)k, |β(k)| ≤ c�−k (1 ≤ k ≤ m), (4.37)

t∫

0

α|β − 1| ds ≤ c�,

∞∫

t

α1−m �−mds ≤ c�1−m, (4.38)

�(t) ≤ c α(t)�(t), α(t)�(t) ≤ c

⎛

⎝1 +t∫

0

α ds

⎞

⎠ . (4.39)

From these hypotheses, it is clear that a0 is nondecreasing and that a0 ≥ a0(0) > 0 .Thus (i), (1.16) hold. Condition (4.36) and Lemma 3 of [2, p. 62] give (1.17). Condi-tion (1.18) follows from the first of (4.37) with k = 1. Besides, the first of (4.37) (withk = 1, 2) also give

∣∣∣∣

d2

dt2

1√a0

∣∣∣∣ ≤ C α

(1 + ∫ t

0 α ds)2 . (4.40)

Thus (1.19) holds. To verify the conditions (1.22), (1.23), we first observe that

∣∣∣∣

dk

dtk

1√a

∣∣∣∣ ≤ C α−1�−k (0 ≤ k ≤ m). (4.41)

This can be proved, by induction, using (4.37) and the second of (4.39). Then, applying(4.41), we easily obtain:

(1√a

)1− 1h∣∣∣∣

dh

dth

1√a

∣∣∣∣

1h

≤ C α−1�−1 (1 ≤ h ≤ m) ; (4.42)

∣∣∣∣

(1√a

)η0(

d

dt

1√a

)η1

· · ·(

dm

dtm

1√a

)ηm∣∣∣∣ ≤ C α1−m �−m, (4.43)

for all integers η0 , . . . , ηm ≥ 0 s.t.∑

0≤h≤m ηh = m − 1,∑

1≤h≤m h ηh = m .Finally, (4.42) and the first inequalities of (4.38), (4.39) imply (1.22), because

∫ t0

|a−a0 |√a0ds ≤ (c2 + 1)

∫ t0 α|β − 1|ds . While, (4.43) and the second of (4.38) give

(1.23).

4.5 Proof of Theorem 1.7

We sketch here the proof of part (1) of Theorem 1.7, since the proof of (2) is similar.

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Ann Univ Ferrara (2012) 58:251–289 271

We set λ = |ξ |, b = a − a0 and define Jv, J0v,B(t),A1(t) respectively as in

(A.5), (A.6) and (A.59). The assumptions of part (1) permit us to apply Lemma A.11.Estimate (A.70) gives

Ev ≤ C[E0v + |ξ |J0

v

]exp {2|ξ |B(t)} ∀ t ≥ 0, |ξ | > 0, (4.44)

for every solution v of (1.7). By definition (A.5) and assumption (i), we have

Ev√μ

≤ Ev ≤ √μEv, (4.45)

Jv ≤ δ−12 |ξ |−2Ev, (4.46)

for all t ≥ 0, |ξ | > 0 . Thus, from (4.44) and (4.45)–(4.46), we obtain

Ev ≤ C E0v

(1 + |ξ |−1

)exp {2|ξ |B(t)} ∀ t ≥ 0, |ξ | > 0. (4.47)

From this estimate we can proceed as in the proof of Theorem 1.1 in Sect. 4.1.

5 Examples

Example 5.1 Let us consider Eq. (1.7) for

a(t) = 2 + 1

log(e + t)+ sin(1 + t)γ

(1 + t)β, t ∈ R

+, (5.1)

with γ, β ∈ (0, 1). Then 1 ≤ a(t) ≤ 4 and for every integer k ≥ 1 there exists ck > 0s.t.

|a(k)(t)| ≤ ck

(e + t)k log2(e + t)+ ck

(1 + t)β+k(1−γ ) , ∀t ≥ 0. (5.2)

Besides, one can easily see that a(k) ∈ L1(R+) if and only if

β + k(1 − γ ) > 1. (5.3)

In order to apply Corollary 1.2, let us then choose

a0(t) = 2 + 1

log(e + t), t ≥ 0. (5.4)

Since a′0

∈ L1(R+), (i i i) holds. Assuming β + m(1 − γ ) > 1,m ≥ 2, from (5.2) wehave:

⎛

⎝t∫

0

|a − a0 |ds

⎞

⎠

m−1 +∞∫

t

|a(m)|ds ≤ C t (m−1)(1−β)[t1−m + t1−β−m(1−γ )],

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272 Ann Univ Ferrara (2012) 58:251–289

for all t ≥ 1, for a suitable C = C(m) > 0. The right-hand side of this estimate isuniformly bounded from above if γ ≤ β and, in this case, a(m) ∈ L1(R+) for allm ≥ 2.

In conclusion, for a(t) defined by (5.1), the assumptions of Corollary 1.2 are sat-isfied if γ ≤ β and if γ = β we are not in a trivial case, i.e. a′ ∈ L1(R+) whenγ = β.

On the contrary, such an a(t) never satisfies the conditions of [1], while it satisfiesthe conditions of [8] only under the more restrictive condition γ ≤ β

2 , which falls into

the trivial case, because γ ≤ β2 ⇒ a′ ∈ L1(R+). Indeed, if the assumption (1.6) of

Hirosawa is satisfied, then a∞ = 2, since limt→∞ a(t) = 2. But then, for any p < 1,

limt→∞

1

(1 + t)p

t∫

0

|a−2|ds = limt→∞

1

(1 + t)p

t∫

0

∣∣∣∣

1

log(e + s)+ sin(1 + s)γ

(1 + s)β

∣∣∣∣ ds = ∞.

This proves that (1.6) cannot hold for any p < 1.On the other hand, Reissig and Smith proved in Theorem 1.1 of [8] (cf. also Theorem

1 of [1]) the energy estimate (L2 energy estimate) for a ∈ C2(R+) satisfying

|a(k)(t)| ≤ C

(1 + t)k∀ t ≥ 0, k = 1, 2 (5.5)

for some C ≥ 0. From (5.2), with k = 1, 2, we have that (5.5) is satisfied if and onlyif

{β + 1 − γ ≥ 1β + 2(1 − γ ) ≥ 2

⇔{β ≥ γ

β ≥ 2γ⇔ γ ≤ β

2.

However, for γ ≤ β2 , a′ ∈ L1(R+), as one can see from (5.2) with k = 1. ��

Example 5.2 In this example we want to consider the Liouville type equation (1.12).For sake of simplicity, we shall shift R

+ to [1,∞) and consider the equation

vt t + tk |ξ |2v = 0, t ∈ [1,∞), (5.6)

with |ξ |, k > 0. First of all, we shall see that

Lemma 5.3 Independently of C > 0, the estimate

Ev(ξ, t) ≤ C Ev(ξ, 1), t ∈ [1,∞), 0 < |ξ | ≤ 1, (5.7)

cannot be satisfied by all solutions of (5.6).

Proof To this aim we introduce the solution φ of

{φ′′ + tkφ = 0, t ∈ [0,∞)

φ(0) = 1, φ′(0) = 0(5.8)

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Ann Univ Ferrara (2012) 58:251–289 273

and look for a solution w of (5.6) of the form w = φ(ϑ t) , with ϑ = ϑ(ξ) > 0 to bedetermined. Substituting in (5.6), we impose that ϑ2φ′′(ϑ t)+ tk |ξ |2φ(ϑ t) = 0. Since

φ′′(ϑ t)+ (ϑ t)kφ(ϑ t) = 0, we choose ϑ such that tk |ξ |2ϑ2 = ϑk tk , i.e. ϑ(ξ) = |ξ | 2

k+2 .Hence, from now on, we assume

w(ξ, t)def= φ(|ξ | 2

k+2 t) for t ∈ [1,∞). (5.9)

Then, according to (1.8), we have the following expression for the energy:

Ew(ξ, t) = tk/2|ξ |2 |w(ξ, t)|2 + t−k/2 |wt (ξ, t)|2= tk/2 |ξ |2 |φ(|ξ | 2

k+2 t)|2 + t−k/2 |ξ | 4k+2 |φ′(|ξ | 2

k+2 t)|2= |ξ | k+4

k+2

[(|ξ | 2

k+2 t)k2 |φ(|ξ | 2

k+2 t)|2 + (|ξ | 2k+2 t)−

k2 |φ′(|ξ | 2

k+2 t)|2]

= |ξ | k+4k+2 E(φ, |ξ | 2

k+2 t), (5.10)

where, using the notation of [6], E(φ, t) is the energy

E(φ, t) = tk/2|φ(t)|2 + t−k/2|φ′(t)|2

of the solution φ of (5.8). By Theorem 1.2 of [6], there exists finite limit

limt→∞ E(φ, t)

def= Eφ, (5.11)

with Eφ > 0. Therefore

limt→+∞ Ew(ξ, t) = Eφ |ξ | k+4

k+2 for |ξ | > 0. (5.12)

In the following we shall see that Ew(ξ, 1) ≤ 2 |ξ |2 . Thus (5.7) cannot hold uniformlyfor 0 < |ξ | ≤ 1, because k+4

k+2 < 2. To prove this we study the behavior of φ(t) for0 ≤ t ≤ 1.

Let t∗ = inf{t > 0 : φ(t) = 0} and note that 0 < t∗ < ∞ and φ′′(t) = −tkφ(t) <0 in (0, t∗) . Thus φ′(t) ≤ 0 and 0 ≤ φ(t) ≤ 1 for all t ∈ [0, t∗]. Integrating, it easilyfollows that:

φ′(t) ≥ − tk+1

k + 1, φ(t) ≥ 1 − tk+2

(k + 1)(k + 2)in [0, t∗]. (5.13)

Therefore 1 − tk+2

2 ≤ φ(t) ≤ 1 for all t ∈ [0, t∗]. Hence t∗ ≥ 21

k+2 > 1 and12 ≤ φ(t) ≤ 1 ∀ t ∈ [0, 1] . It follows that for all t ∈ [0, 1]

φ′(t) = −t∫

0

skφ(s)ds ≤ −1

2

t∫

0

skds = −1

2

tk+1

k + 1. (5.14)

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274 Ann Univ Ferrara (2012) 58:251–289

Hence 12

tk+1

k+1 ≤ |φ′(t)| ≤ tk+1

k+1 for all t ∈ [0, 1]. Then, since w(ξ, t) = φ(|ξ | 2k+2 t), it

follows that

{ 12 ≤ w(ξ, 1) ≤ 1

12(k+1) |ξ |2 ≤ wt (ξ, 1) ≤ |ξ |2 for 0 < |ξ | ≤ 1. (5.15)

Hence, for 0 ≤ |ξ | ≤ 1, we obtain

|ξ |24

≤ Ew(ξ, 1) ≤ 2 |ξ |2. (5.16)

��Next, let us consider the solution ψ = ψ(t) of

{ψ ′′ + tkψ = 0, t ∈ [0,∞)

ψ(0) = 0, ψ ′(0) = 1(5.17)

As above, we can see that

z(ξ, t)def= |ξ |− 2

k+2 ψ(|ξ | 2k+2 t) (5.18)

is a solution of (5.6) and that

Ez(ξ, t) = |ξ | kk+2 E(ψ, |ξ | 2

k+2 t), (5.19)

where E(ψ, t)def= tk/2|ψ(t)|2 + t−k/2|ψ ′(t)|2 is the energy of the solutions of (5.17).

By Theorem 1.2 of [6], the exists limt→∞ E(ψ, t)def= Eψ, with Eψ ∈ (0,∞). Hence,

we have

limt→+∞ Ez(ξ, t) = Eψ |ξ | k

k+2 for |ξ | > 0. (5.20)

To study the behavior of ψ(t) for 0 ≤ t ≤ 1, let t = inf{t > 0 : ψ(t) = 0}. Then0 < t < ∞ and ψ ′′(t) ≤ 0, ψ ′(t) ≤ 1 in [0, t ] . Since ψ(0) = 0, we have ψ(t) ≤ tand ψ ′′(t) ≥ −tk+1 in [0, t ] . Hence, integrating, we find that

ψ ′(t) ≥ 1 − tk+2

k + 2, ψ(t) ≥ t − tk+3

(k + 2)(k + 3)∀t ∈ [0, t ]. (5.21)

This implies that t ≥ 61

k+2 > 1. Thus we may conclude that

5

6t ≤ ψ(t) ≤ t,

1

2≤ ψ ′(t) ≤ 1 for t ∈ [0, 1]. (5.22)

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Ann Univ Ferrara (2012) 58:251–289 275

In particular, since z(ξ, t) = |ξ |− 2k+2 ψ(|ξ | 2

k+2 t), it follows that

{56 ≤ z(ξ, 1) ≤ 112 ≤ zt (ξ, 1) ≤ 1

for 0 < |ξ | ≤ 1. (5.23)

Summarizing up the previous estimates for w, z, we have:

Lemma 5.4 Given M > 0 there exists C = C(M) > 0 such that

Ev ≤ C(|ξ |2− k

k+2 |v(ξ, 1)|2 + |vt (ξ, 1)|2)

∀ t ≥ 1, 0 < |ξ | ≤ M , (5.24)

for every solutions v of (5.6).

Proof Since limt→∞ E(φ, t) = Eφ < ∞, limt→∞ E(ψ, t) = Eψ < ∞, there existsK > 0 s.t.

E(φ, t) ≤ K , E(ψ, t) ≤ K max{t− k2 , 1}, ∀ t > 0. (5.25)

Thus, from (5.10) and (5.19) we find that

Ew ≤ K |ξ | k+4k+2 , Ez ≤ K max{1, |ξ | k

k+2 }, ∀ t ≥ 1. (5.26)

Given v, solution of (5.6), we look for α = α(ξ), β = β(ξ) such that v = αw + βz.To this end, it will be sufficient that α, β satisfy the system

{αw(ξ, 1)+ βz(ξ, 1) = v(ξ, 1)αwt (ξ, 1)+ βzt (ξ, 1) = vt (ξ, 1)

(5.27)

That is

α = v(ξ, 1)zt (ξ, 1)− vt (ξ, 1)z(ξ, 1)

W (ξ, 1), β= w(ξ, 1)vt (ξ, 1)− wt (ξ, 1)v(ξ, 1)

W (ξ, 1),

(5.28)

where W (ξ, 1) = w(ξ, 1)zt (ξ, 1) − wt (ξ, 1)z(ξ, 1). By (5.15), (5.23) we haveW (ξ, 1) ≥ 1

4 − |ξ |2 . Hence, for 0 < |ξ | ≤ 12√

2, we find

|α| ≤ 8 (|v(ξ, 1)| + |vt (ξ, 1)|) , |β| ≤ 8(|ξ |2 |v(ξ, 1)| + |vt (ξ, 1)|

). (5.29)

Noting that k+4k+2 = 2 − k

k+2 < 2, for t ≥ 1 and 0 < |ξ | ≤ 12√

2we easily have

Ev ≤ 2 |α|2Ew + 2 |β|2Ez

≤ 27 K |ξ | k+4k+2 (|v(ξ, 1)| + |vt (ξ, 1)|)2 + 27 K

(|ξ |2 |v(ξ, 1)| + |vt (ξ, 1)|

)2

≤ C(|ξ |2− k

k+2 |v(ξ, 1)|2 + |vt (ξ, 1)|2), (5.30)

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276 Ann Univ Ferrara (2012) 58:251–289

for a suitable C > 0. This gives (5.24) for |ξ | ≤ 12√

2. To conclude it is now sufficient

to observe that, by Lemma 3.1 (in [1,∞) instead of R+), Ev ≤ � 1

2√

2Ev(ξ, 1) for

|ξ | ≥ 12√

2and that Ev(ξ, 1) can be estimated with the right hand-side of (5.24) for

12√

2≤ |ξ | ≤ M . ��

Appendix A: some perturbation lemmas

Given a0 : R+ → (0,∞) continuous and b : R

+ → C locally integrable, let usconsider the second order equations with parameter λ ∈ (0,∞)

vt t + λ2a0(t)v = 0, t ∈ R+, (A.1)

vt t + λ2 [a0(t)+ b(t)]v = 0, t ∈ R+. (A.2)

Given v : (0,∞)× R+t → C , differentiable with respect to t , we set:

Evdef= a

12

0 λ2|v|2 + a

− 12

0 |v′|2, E0v = Ev

∣∣t=0, (A.3)

Iε,v def= λ2−2ε |v|2 + |v′|2, I0ε,v

def= Iε,v∣∣t=0, (A.4)

Jvdef= |v|2 + a−1

0λ−2|v′|2, J0

v = Jv∣∣t=0 , (A.5)

for t ≥ 0, λ > 0, ε ≥ 0 (with ∂tv = v′ ). Finally, for t ≥ 0, we also define:

B(t)def=

t∫

0

|b|√a0

ds. (A.6)

A.1 a0 of general type

Lemma A.1 Assume that there exist � ≥ 1,M ∈ (0,∞] s.t.every solution v of (A.1)satisfies Ev ≤ �E0

v for t ≥ 0, λ ∈ (0,M]. Then every solution v of (A.2) verifies:

Ev ≤ 4�E0v exp {4�λB(t)} , ∀ t ≥ 0, λ ∈ (0,M]. (A.7)

If, in addition, the solutions of (A.1) satisfy E0v ≤ �Ev for t ≥ 0, λ ∈ (0,M], then

E0v ≤ 4�2 Ev exp

{4�2λB(t)

}for t ≥ 0, λ ∈ (0,M] for every solution v of (A.2).

Proof For λ > 0, let w(λ, t), z(λ, t) be the solutions of (A.1) satisfying

{w(λ, 0) = a0(0)

− 14 λ−1

w′(λ, 0) = 0

{z(λ, 0) = 0

z′(λ, 0) = a0(0)14 .

(A.8)

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Ann Univ Ferrara (2012) 58:251–289 277

The Wronskian W = wzt − wt z satisfies: W (λ, t) = W (λ, 0) = λ−1 for all t ≥ 0.Since E0

w = E0z = 1, the assumptions give

Ew ≤ �, Ez ≤ �, t ≥ 0, 0 < λ ≤ M. (A.9)

In particular,

|w|2, |z|2 ≤ � a0− 1

2 λ−2, t ≥ 0, 0 < λ ≤ M. (A.10)

Given a solution v = v(λ, t) of (A.2), we proceed as in the method of variation ofparameters, looking for α, β : (0,∞)× R

+t → C such that

v = αw + βz , v′ = αw′ + βz′. (A.11)

Substituting in (A.2), we find that α, β must satisfy the linear system

{αt = αλ3bwz + βλ3b z2

βt = −αλ3bw2 − βλ3bwz(A.12)

with the initial conditions α(λ, 0) = v(λ, 0) a0(0)14 λ, β(λ, 0) = v′(λ, 0) a0(0)

− 14 .

Therefore, integrating (A.12) with respect to t , we find:

α = v(λ, 0)a0(0)14 λ+ λ3

t∫

0

(αbwz + βb z2)ds, (A.13)

β = v′(λ, 0)a0(0)− 1

4 − λ3

t∫

0

(αbw2 + βbwz)ds. (A.14)

From (A.10), for 0 < λ ≤ M we have that

|α| ≤ |v(λ, 0)| a0(0)14 λ+�λ

t∫

0

(|α| + |β|) |b|√a0

ds,

|β| ≤ |v′(λ, 0)| a0(0)− 1

4 +�λ

t∫

0

(|α| + |β|) |b|√a0

ds.

Hence, applying Gronwall’s Lemma:

|α| + |β| ≤ (|v(λ, 0)| a0(0)14 λ+ |v′(λ, 0)| a0(0)

− 14 ) exp

⎧⎨

⎩2�λ

t∫

0

|b|√a0

ds

⎫⎬

⎭.

(A.15)

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278 Ann Univ Ferrara (2012) 58:251–289

From this, (A.9) and (A.11) it follows that

Ev ≤ √a0 λ

2(|α||w| + |β||z|)2 + 1√a0

(|α||w′| + |β||z′|)2

≤ 2|α|2Ew + 2|β|2Ez ≤ 2�(|α|2 + |β|2) ≤ 2�(|α| + |β|)2

≤ 2�[|v(λ, 0)| a0(0)

14 λ+ |v′(λ, 0)| a0(0)

− 14

]2e4�λ

∫ t0 |b|a− 1

20 ds

≤ 4�Ev(λ, 0) e4�λ∫ t

0 |b|a− 12

0 ds, (A.16)

for all t ≥ 0, 0 < λ ≤ M, which is the desired estimate (A.7).Finally to prove the last statement of Lemma A.1, note that �−1 E0

v ≤ Ev ≤ �E0v

for t ≥ 0, 0 ≤ λ ≤ M , implies

Ev(λ, s) ≤ �2 Ev(λ, t) for s, t ≥ 0, 0 ≤ λ ≤ M , (A.17)

for every solution v of (A.1). Hence, for t > 0, performing the previous estimatesbackward in time in the interval [0, t], we obtain: E0

v ≤ 4�2 Ev exp{4�2λB(t)

}, for

all t ≥ 0, λ ∈ (0,M]. ��Lemma A.2 Assume that there exist �,M ∈ (0,∞), ε ≥ 0 s.t. every solution v of(A.1) satisfies Ev ≤ � I0

ε,v for all t ≥ 0, λ ∈ (0,M]. Then every solution v of (A.2)verifies:

Ev ≤ 4� I0ε,v exp

{4�λ1−ε B(t)

}, ∀ t ≥ 0, λ ∈ (0,M]. (A.18)

Proof For λ > 0, let w(λ, t), z(λ, t) be solutions of (A.1) satisfying the initial condi-tions

{w(λ, 0) = λ−1+εw′(λ, 0) = 0

{z(λ, 0) = 0z′(λ, 0) = 1.

(A.19)

In this case W (λ, t) = λ−1+ε for all t ≥ 0. Moreover, since I0ε,w = I0

ε,z = 1, we haveEw ≤ �,Ez ≤ � for t ≥ 0, 0 < λ ≤ M. Proceeding as in the proof of Lemma A.1,we obtain that

|α| + |β| ≤ (|v(λ, 0)| λ1−ε + |v′(λ, 0)| ) exp

⎧⎨

⎩2�λ1−ε

t∫

0

|b|√a0

ds

⎫⎬

⎭. (A.20)

From this, we easily deduce (A.18). ��

A.2 a0 regularly varying

Here we require that a0 : R+ → (0,∞) satisfies the following: there exists

N1 ≥ 1 s.t.

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Ann Univ Ferrara (2012) 58:251–289 279

a0(s) ≤ N1 a0(t) for s, t ∈ R+, s ≤ t, (A.21)

4√

a0

t∫

0

4√

a0 ds ≤ N1

t∫

0

√a0 ds, ∀ t ∈ R

+. (A.22)

Besides, given N2 > 0, for λ > 0 we define Tλ = Tλ(N2) by the identity:

λ

Tλ∫

0

√a0 ds

def= N2. (A.23)

Lemma A.3 Assume (A.21), (A.22), a0(t) ≥ δ > 0. Let v be a solution (A.2) s.t.v(λ, 0) = 0. Then there exists C = C(δ, N1, N2) > 0 s.t.

Ev ≤ C |v′(λ, 0)|2 exp{CλB(t)}, ∀ λ > 0, t ∈ [0, Tλ]. (A.24)

Proof From (A.2), we derive that v′ satisfies the integral equation:

v′(λ, t) = v′(λ, 0)− λ2

t∫

0

⎛

⎝t∫

z

(a0 + b) ds

⎞

⎠ v′dz, (A.25)

for all λ > 0, t ∈ R+. Hence

|v′|4√

a0

≤ |v′(λ, 0)|4√

a0

+t∫

0

k1(λ, t, z)|v′|4√

a0

dz +t∫

0

k2(λ, t, z)|v′|4√

a0

dz (A.26)

where

k1(λ, t, z)def= λ2

4√

a0(z)4√

a0(t)

t∫

z

a0 ds, (A.27)

k2(λ, t, z)def= λ2

4√

a0(z)4√

a0(t)

t∫

z

|b| ds. (A.28)

For 0 ≤ z ≤ t , by (A.21) we have∫ t

z a0 ds ≤ √N1√

a0(t)∫ t

0√

a0 ds. Thus, we easilyfind

k1(λ, t, z) ≤ √N1 λ

2 4√

a0(z)4√

a0(t)

t∫

0

√a0 ds

≤ N1 N2 λ4√

a0(Tλ)4√

a0(z), (A.29)

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280 Ann Univ Ferrara (2012) 58:251–289

for t ∈ [0, Tλ]. In the same way, for t ∈ [0, Tλ], 0 ≤ z ≤ t , we obtain

k2(λ, t, z) ≤ √N1 λ

2 4√

a0(z)4√

a0(t)

t∫

0

|b|√a0

ds

≤ N1 λ2 4√

a0(Tλ)

⎛

⎝t∫

0

|b|√a0

ds

⎞

⎠ 4√

a0(z). (A.30)

Hence, we get

|v′|4√

a0

≤ |v′(λ, 0)|4√

a0(0)+N1

⎛

⎝N2+λt∫

0

|b|√a0

ds

⎞

⎠ λ 4√

a0(Tλ)

t∫

0

4√

a0

|v′|4√

a0

dz, (A.31)

for t ∈ [0, Tλ]. Noting that∫ t

0|b|√a0

ds = B(t) is non decreasing, applying Gronwall’s

lemma and (A.22), (A.23) we find that

|v′|4√

a0

≤ |v′(λ, 0)|4√

a0(0)exp

⎧⎨

⎩N1

⎛

⎝N2 + λ

t∫

0

|b|√a0

ds

⎞

⎠ λ 4√

a0(Tλ)

t∫

0

4√

a0 dz

⎫⎬

⎭

≤ |v′(λ, 0)|4√

a0(0)exp

{N 2

1 N2 (N2 + λB(t))}

(A.32)

for t ∈ [0, Tλ] , i.e. |v′| ≤ N3(t) |v′(λ, 0)| 4√

a0 with

N3(t)def= 1

4√

a0(0)exp

{N 2

1 N2(N2 + λB(t)

)}. (A.33)

Finally, for |v(λ, t)| we find

|v| ≤t∫

0

|v′|ds ≤ N3(t) |v′(λ, 0)|t∫

0

4√

a0 ds

≤ N1 N3(t) |v′(λ, 0)|∫ t

0√

a0 ds4√

a0

. (A.34)

We can now estimate Ev for t ∈ [0, Tλ]. In fact, from (A.32) and (A.34), we find that

Ev(λ, t) = a12

0 λ2|v|2 + a

− 12

0 |v′|2 ≤ |v′(λ, 0)|2(

N 21 N 2

2 + 1)

N3(t)2 , (A.35)

for all λ > 0 and t ∈ [0, Tλ]. Thanks to (A.33), this gives (A.24). ��

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Ann Univ Ferrara (2012) 58:251–289 281

Lemma A.4 Assume (A.21), (A.22), a0(t) ≥ δ > 0. Let |b| ≤ μ1 a0 with μ1 ≥ 0.Let v(λ, t) be a solution of (A.2) s.t. v′(λ, 0) = 0, then ∀M > 0 there exists C =C(δ,M, μ1, N1, N2) > 0 s.t.

Ev ≤ C |v(λ, 0)|2, ∀ λ ∈ (0,M], t ∈ [0, Tλ]. (A.36)

Proof In this case v′(λ, t) satisfies the integral equation

v′(λ, t) = −λ2 v(λ, 0)

t∫

0

(a0 + b)ds − λ2

t∫

0

⎛

⎝t∫

z

(a0 + b) ds

⎞

⎠ v′dz, (A.37)

for all λ > 0, t ∈ R+. Setting α0(t)

def= (1 + μ1)∫ t

0 a0 ds, for all λ, t > 0 we maywrite:

|v′|α0

≤ λ2 |v(λ, 0)| +t∫

0

k3(λ, t, z)|v′|α0

dz, (A.38)

where

k3(λ, t, z)def= λ2 α0(z)

α0(t)

t∫

z

(a0 + |b|) ds. (A.39)

The function t → |v′|α0

may be assumed continuous for t = 0 , because

limt→0+ |v′|(∫ t

0 a0 ds)−1 = λ2|v(λ, 0)| |1 + b(0)a0(0)

−1| . Besides, k3(λ, t, z) ≤λ2α0(z) for 0 ≤ z ≤ t . Hence, we can apply Gronwall’s lemma to (A.38) and obtain

|v′| ≤ λ2 |v(λ, 0)|α0 exp

⎧⎨

⎩λ2

t∫

0

α0 ds

⎫⎬

⎭. (A.40)

Further, by (A.21), α0 ≤ √N1(1+μ1)

√a0

∫ t0

√a0 ds . Hence, we find that

∫ t0 α0 ds ≤

√N1(1 + μ1)

(∫ t0

√a0 ds

)2. Thus, for t ∈ [0, Tλ], we also have

|v′| ≤ λ√

N1(1 + μ1) N2 |v(λ, 0)| √a0 exp{√

N1(1 + μ1) N 22

}, (A.41)

i.e. |v′| ≤ N4 λ |v(λ, 0)|√a0 , with N4def= √

N1(1 +μ1) N2 exp{√

N1(1 + μ1) N 22

}.

Then, for t ∈ [0, Tλ], we obtain

|v| ≤ |v(λ, 0)| +t∫

0

|v′| ds

≤ |v(λ, 0)|⎛

⎝1 + N4λ

t∫

0

√a0 ds

⎞

⎠ ≤ |v(λ, 0)| (1 + N2 N4) (A.42)

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282 Ann Univ Ferrara (2012) 58:251–289

and thus the energy Ev satisfies

Ev ≤ C |v(λ, 0)|2 λ2 √a0 , ∀ λ > 0, t ∈ [0, Tλ], (A.43)

for a suitable C = C(N2, N4) > 0. Now it remains to estimate λ2√a0 . From (A.22),we find

λ2√a0 ≤ λ2 N 21

⎛

⎝t∫

0

√a0 ds

⎞

⎠

2 ⎛

⎝t∫

0

4√

a0 ds

⎞

⎠

− 2

≤ N 21 N 2

2

⎛

⎝t∫

0

4√

a0 ds

⎞

⎠

− 2

, (A.44)

for all t ∈ (0, Tλ]. Finally, by (A.21), it follows that

λ2√

a0(t) ≤ λ2√

N1√

a0(Tλ) ≤ N32

1 N 22

⎛

⎝Tλ∫

0

4√

a0 ds

⎞

⎠

− 2

, (A.45)

for all t ∈ [0, Tλ]. From (A.23) we know that Tλ > 0 depends continuously on λand that Tλ → ∞ as λ → 0+. It follows that for all M > 0 there exists δ1 =δ1(δ,M, N2) > 0 such that

∫ Tλ0

4√

a0 ds ≥ δ1 for all λ ∈ (0,M]. Hence, (A.36) holds.��

Lemma A.5 Assume (A.21), (A.22), a0(t) ≥ δ > 0, and let |b| ≤ μ1a0 for someμ1 ≥ 0. Besides, let us suppose that there exists N > 0 such that, for all T > 0,

|a′0|

a3/20

≤ N∫ T

0√

a0 ds, ∀ t ∈ [T,∞), (A.46)

∞∫

T

∣∣∣∣∣

(a′

0

a3/20

)′∣∣∣∣∣ds ≤ N

∫ T0

√a0 ds

. (A.47)

Let v(λ, t) be a solution (A.2), then for all M > 0 there exists � = �(δ,M, μ1 , N ,N1) > 0 s.t.

Ev≤�(|v(λ, 0)|2+|v′(λ, 0)|2

)exp{�λB(t)}, ∀ λ ∈ (0,M], t ∈ R

+. (A.48)

Proof Let v be a solution of (A.2). Deriving with respect to t we find

d

dt

(

Ev + a′0

2a3/20

Re(vv′))

= 1

2

(a′

0

a3/20

)′Re(vv′)− 2λ2 1√

a0

Re(bv v′)− a′0

2a3/20

λ2 Re(bv v). (A.49)

123

Ann Univ Ferrara (2012) 58:251–289 283

From now on we set N2 = N in (A.23), i.e.we define Tλ = Tλ(N ) from the equation

λ∫ Tλ

0√

a0 dsdef= N . Then conditions (A.46), (A.47) give

|a′0|

2 a3/20

∣∣Re(vv′)

∣∣ ≤ |a′

0|

4λ a3/20

Ev ≤ 1

4Ev, (A.50)

|a′0|

a3/20

λ2 |Re(bv v)| ≤ |a′0|

λ a3/20

λ|b|√

a0

Ev ≤ λ|b|√

a0

Ev, (A.51)

1

λ

∞∫

t

∣∣∣∣∣

(a′

0

a3/20

)′∣∣∣∣∣

ds ≤ 1, (A.52)

for all λ > 0 and t ∈ [Tλ,∞). Noting also that λ2√a0

∣∣Re(bv v′)

∣∣ ≤ λ

|b|√a0Ev , integrat-

ing (A.49) on [Tλ, t) we derive the integral inequality

Ev = Ev(λ, Tλ)− a′0

2a3/20

Re(vv′)∣∣∣t

Tλ+ 1

2

t∫

Tλ

(a′

0

a3/20

)′Re(vv′) ds

−t∫

Tλ

[2λ2

√a0

Re(bv v′)+ a′0

2a3/20

λ2 Re(bv v)

]

ds

≤ Ev(λ, Tλ)+ 1

4(Ev(λ, Tλ)+Ev)+ 1

4λ

t∫

Tλ

∣∣∣∣∣

(a′

0

a3/20

)′∣∣∣∣∣Ev ds+3λ

t∫

Tλ

|b|√a0

Ev ds

(A.53)

for all t ≥ Tλ . Hence, Gronwall’s lemma and (A.52) give

Ev(λ, t) ≤ 4Ev(λ, Tλ) exp{4λ (B(t)− B(Tλ))} ∀ t ≥ Tλ. (A.54)

To conclude it is now sufficient to observe that Lemma A.3 and A.4 imply that

Ev ≤ C(|v(λ, 0)|2 + |v′(λ, 0)|2

)exp{4λB(t)}, ∀ t ∈ [0, Tλ], (A.55)

for all λ ∈ (0,M] , for a suitable constant C = C(δ,M, μ1 , N , N1) > 0. ��

A.3 a0 almost nondecreasing

Now we suppose that a0 : R+ → (0,∞) satisfies the following assumption: a0 ∈

ACloc(R+) with a′

0≥ 0, or, more generally,

(a′0)−

a−10

∈ L1(R+), (A.56)

where (a′0)− = min{a′

0, 0}. The following propositions hold:

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284 Ann Univ Ferrara (2012) 58:251–289

Proposition A.6 Suppose that a0 ∈ ACloc(R+) verifies (A.56). Then every solution v

of (A.1) satisfies: Jv≤C0 J0v for all t ≥ 0, λ > 0, with C0

def= exp{− ∫∞0 (a′

0)−

a−10

dt}.Proof Let v be a solution of (A.1). For λ > 0, deriving with respect to t , we easilyhave:

Jv′ = − a′

0

λ2a20

|v′|2 ≤ − (a′0)−

a0

|v′|2λ2a0

≤ − (a′0)−

a0

Jv. (A.57)

Hence, Jv ≤ C0 J0v holds for all t ≥ 0. ��

Now, for brevity, we set:

A(t)def=

t∫

0

∣∣∣

a′′0

a5/40

− 3

2

a′0

2

a9/40

∣∣∣ ds , (A.58)

A1(t)def=

t∫

0

∣∣∣a′′

0

a0

− 3

2

a′0

2

a20

∣∣∣ ds. (A.59)

Proposition A.7 Assume (A.56), a′0

∈ ACloc(R+) and |a′

0|2 a

− 52

0 ≤ K < ∞.Then every solution v of (A.1) satisfies Ev ≤ 4

(E0v + C0 KJ0

v

) + C0J0v A(t)

2 fort ≥ 0, λ > 0.

Proof Let v be a solution of (A.1). For λ > 0, deriving with respect to t , we find that

d

dt

(

Ev + a′0

2a3/20

Re(vv′))

= 1

2

(a′

0

a3/20

)′Re(vv′). (A.60)

Hence, integrating we obtain

Ev = E0v − a′

0

2a3/20

Re(vv′)∣∣∣t

0+ 1

2

t∫

0

(a′

0

a3/20

)′Re(vv′) ds. (A.61)

Since |a′0|2 a

− 52

0 ≤ K and |v| |v′| ≤ a14

0 (Jv Ev)12 , the following estimates hold:

∣∣∣

a′0

a3/20

Re(vv′)∣∣∣ ≤ |a′

0|

a5/40

(Jv Ev)1/2 ≤ 1

2(Ev + KJv) ,

∣∣∣( a′

0

a3/20

)′Re(vv′)

∣∣∣ ≤

∣∣∣

a′′0

a3/20

− 3

2

a′0

2

a5/20

∣∣∣ |v| |v′|

≤∣∣∣

a′′0

a5/40

− 3

2

a′0

2

a9/40

∣∣∣ (Jv Ev)1/2 .

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Ann Univ Ferrara (2012) 58:251–289 285

Now, by Proposition A.6, we know that Jv ≤ C0 J0v for all t ≥ 0, with C0 ≥ 1.

Hence introducing the previous inequalities into (A.61), we easily have:

Ev ≤ 2[E0v + C0 KJ0

v

]+(

C0J0v

)1/2t∫

0

∣∣∣

a′′0

a5/40

− 3

2

a′0

2

a9/40

∣∣∣√Ev ds.

It is now sufficient to apply Lemma B.5 to deduce the desired estimate. ��Proposition A.8 Assume (A.56), a′

0∈ ACloc(R

+) and |a′0| a−1

0≤ K1 < ∞. Then

every solution v of (A.1) satisfies Ev ≤ E0v + C0 λJ

0v (K1 + A1(t)) for t ≥ 0, λ > 0.

Proof Let v be a solution of (A.1). Let λ > 0, noting that 2|v| |v′| ≤ √a0 λJv the

following inequalities are easily verified:

∣∣∣

a′0

a3/20

Re(vv′)∣∣∣ ≤ |a′

0|

2a0

λJv ≤ K1

2λJv, (A.62)

∣∣∣( a′

0

a3/20

)′Re(vv′)

∣∣∣ ≤

∣∣∣

a′′0

a3/20

− 3

2

a′0

2

a5/20

∣∣∣ |v| |v′|

≤ 1

2

∣∣∣a′′

0

a0

− 3

2

a′0

2

a20

∣∣∣ λJv. (A.63)

By Proposition A.6, we know that Jv ≤ C0 J0v for all t ≥ 0. Thus, from (A.61), we

easily obtain:

Ev ≤ E0v + 1 + C0

4K1λJ

0v + C0

4λJ0

v

t∫

0

∣∣∣a′′

0

a0

− 3

2

a′0

2

a20

∣∣∣ ds. (A.64)

Since C0 ≥ 1, we immediately get the desired estimate. ��Lemma A.9 Assume that a0 ∈ ACloc(R

+) verifies (A.56), b ∈ L1loc(R

+t ). Then every

solution v of (A.2) satisfies Jv ≤ C0 J0v exp{λB(t)} for all t ≥ 0, λ > 0.

Proof Let v be a solution of (A.2). For λ > 0 we have

Jv′ = − 2

a0

Re(bv v′)− a′0

λ2a20

|v′|2

≤ 2|b|a0

|v||v′| − (a′0)−

a0

|v′|2λ2a0

≤(

λ|b|√

a0

− (a′0)−

a0

)

Jv.

Hence, integrating the above inequality, we deduce that Jv ≤ C0 J0v exp{λB(t)} for

all t ≥ 0, λ > 0, where C0 is the constant introduced in Proposition A.6. ��

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286 Ann Univ Ferrara (2012) 58:251–289

Lemma A.10 Under the assumptions of Proposition A.7, every solution v of (A.2)satisfies

Ev ≤ 5[E0v + C0J

0v

(K + A(t)2

)]exp{2λB(t)}, ∀ t ≥ 0, λ > 0. (A.65)

Proof Let v be a solution of (A.2). Deriving with respect to t , we find

d

dt

(

Ev + a′0

2a3/20

Re(vv′))

= 1

2

(a′

0

a3/20

)′Re(vv′)− 2λ2 1√

a0

Re(bv v′)− a′0

2a3/20

λ2 Re(bv v). (A.66)

The last two terms can be estimated as follows: noting that 2λ|v| |v′| ≤ Ev , we have

λ2

√a0

∣∣Re(bv v′)

∣∣ ≤ λ

|b|2√

a0

Ev ; (A.67)

besides, for λ > 0, it is easily verified that

|a′0|

a3/20

λ2 |Re(bv v)| ≤ |a′0|

a5/40

λ|b|√

a0

(Jv Ev)1/2

≤ (KJv)1/2 λ

|b|√a0

√Ev , (A.68)

because of the inequalities |a′0|a− 5

40 ≤ √

K and λ|v|2 = a− 1

40 (a

14

0 λ|v|) |v| ≤ a− 1

40√

Ev√Jv .

Further, by Lemma A.9,

Jv ≤ C0 J0v exp{λB(t)} ∀ t ≥ 0,

where B : R+ → R

+ is non-decreasing. Hence, integrating (A.66) with respect to t ,and using the previous inequalities, we derive the following:

Ev = E0v − a′

0

2a3/20

Re(vv′)∣∣∣t

0+ 1

2

t∫

0

(a′

0

a3/20

)′Re(vv′) ds

−t∫

0

[2λ2

√a0

Re(bv v′)+ a′0

2a3/20

λ2 Re(bv v)

]

ds

≤ E0v + 1

4

(E0v + KJ0

v

)+ 1

4

(Ev + K C0J

0v exp{λB(t)}

)

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Ann Univ Ferrara (2012) 58:251–289 287

+1

2

(C0J

0v exp {λB(t)}

)1/2t∫

0

(∣∣∣

a′′0

a5/40

− 3

2

a′0

2

a9/40

∣∣∣+

√Kλ

|b|√a0

)√Ev ds

+t∫

0

λ|b|√

a0

Ev ds. (A.69)

Next, applying Lemma B.5, we find that (A.65) holds. ��Lemma A.11 Under the assumptions of Proposition A.8, every solution v of (A.2)satisfies

Ev ≤[E0v + C0λJ

0v (K1 + A1(t))

]exp{2λB(t)}, ∀ t ≥ 0, λ > 0. (A.70)

Proof It is enough to follow the same arguments of the proof of Proposition A.8. Letv be a solution of (A.2). Starting from (A.66), and assuming λ > 0, we estimate theterms by applying the inequalities (A.62), (A.63) and (A.67). We estimate the last termof (A.66) by

|a′0|

a3/20

λ2 |Re(bv v)| ≤ |a′0|

a0

λ|b|√

a0

λJv ≤ K1 λ|b|√

a0

λJv. (A.71)

Then, we use Lemma A.9 to estimateJv . In this way we easily get an integral inequalityto which we apply Gronwall’s lemma. ��

Appendix B: some inequalities

We recall here the Gagliardo–Nirenberg inequality in one space-dimension (see [3]).Then we give some estimates we need in the proofs. Below, I denotes an unboundedopen interval.

Theorem B.1 (Gagliardo–Nirenberg) Let 1 ≤ r, q ≤ +∞ and m an integer ≥ 1. Ifϕ ∈ Lq(I ) and ϕ(m) ∈ Lr (I ), then for all integers h, 0 ≤ h ≤ m, ϕ(h) ∈ L ph (I ) if

m

ph= m − h

q+ h

r. (B.1)

Moreover,

‖ϕ(h)‖L ph (I ) ≤ 4h(m−h)‖ϕ(m)‖hmLr (I )‖ϕ‖

m−hm

Lq (I ). (B.2)

Remark B.2 In particular, taking ph = q = r = +∞, for 0 ≤ h ≤ m we get from(B.2):

‖ϕ(h)‖L∞(I ) ≤ 4h(m−h)‖ϕ(m)‖hmL∞(I )‖ϕ‖

m−hm

L∞(I ). (B.3)

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288 Ann Univ Ferrara (2012) 58:251–289

Taking in (B.2) q = +∞, r = 1 and hence ph = m/h, for 1 ≤ h ≤ m, we get

‖ϕ(h)‖L

mh (I )

≤ 4h(m−h)‖ϕ(m)‖hmL1(I )

‖ϕ‖m−h

mL∞(I ). (B.4)

Recalling Definition 2.2, we can prove the following:

Lemma B.3 Let P ∈ Hml and ϕ ∈ Cm(I ) ∩ L∞(I ) with ϕ(m) ∈ L1(I ). Then, there

exists C > 0, independent of I , s.t. the following estimate holds:

‖Pϕ‖L1(I ) ≤ C ‖ϕ‖l−1L∞(I )‖ϕ(m)‖L1(I ). (B.5)

Proof For simplicity, let us set cη = cη0,...,η j in (2.5). Then, applying the generalizedHölder’s inequality and then (B.4), we find:

‖Pϕ‖L1(I ) ≤∑

|cη| ‖ϕ‖η0L∞(I )

m∏

h=1

‖ϕ(h)‖ηh

Lmh (I )

≤∑

|cη| ‖ϕ‖η0L∞(I )

m∏

h=1

4h(m−h)ηh ‖ϕ(m)‖hm ηh

L1(I )‖ϕ‖

m−hm ηh

L∞(I )

≤ c ‖ϕ‖η0+η1+...+ηm− η1+2η2+...+mηmm

L∞(I ) ‖ϕ(m)‖η1+2η2+...+mηm

mL1(I )

= c ‖ϕ‖l−1L∞(I ) ‖ϕ(m)‖L1(I ).

��

Lemma B.4 Assume a(t) ≥ δ > 0, a ∈ L∞(R+), a(m) ∈ L1(R+). Then there existsC = C(δ,m) such that, for any unbounded interval I ⊂ R

+,

‖(a− 12 )(m)‖L1(I ) ≤ C ‖a‖m−1

L∞(I ) ‖a(m)‖L1(I ). (B.6)

Proof To prove the inequality, we first prove by induction that

(a− 12 )( j) = R j a

a j+1/2 for 1 ≤ j ≤ m, (B.7)

where R j is a suitable polynomial of H jj , i.e.

R j a =∑

rη aη0(a′)η1 · · · (a( j))η j (B.8)

with rη = rη0 ,...η j ∈ R and η0 , . . . , η j ≥ 0 integers s.t.∑

0≤h≤ j ηh = j,∑

0≤h≤ j

hηh = j (see Definition 2.2). Indeed, for j = 1, (a− 12 )′ = − 1

2a′

a1+1/2 . If (B.7) holdsfor some j ≤ m − 1, then it also holds for j + 1:

((a− 1

2 )( j))′ =

(R j a

a j+1/2

)′= (R j a)′a − ( j + 1/2)a′ R j a

a j+1+1/2

and (R j a)′a − ( j + 1/2)a′ R j a = R j+1a, with R j+1 ∈ H j+1j+1 if R j ∈ H j

j .

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Ann Univ Ferrara (2012) 58:251–289 289

Therefore, from (B.7), (B.8) and the inequality (B.5), with l = m, we obtain that

∥∥∥(a− 1

2 )(m)∥∥∥

L1(I )≤ δ−m−1/2‖Pm(a(t))‖L1(I )

≤ δ−m−1/2 C ‖a‖m−1L∞(I )‖a(m)‖L1(I ), (B.9)

for all b ≥ 0, for a suitable C > 0, which proves (B.6). ��

B.1 A Gronwall’s type inequality

The following lemma, due to A. I. Perov, is proved in [7].

Lemma B.5 Let y : R+ → R

+ be a continuous function that satisfies the integralinequality y(t) ≤ c + ∫ t

0

[h(s) y(s) + k(s) yα(s)

]ds , where c ≥ 0, 0 ≤ α < 1;

h(t), k(t) are nonnegative locally integrable functions in R+. Then, for t ≥ 0, y(t)

satisfies

y(t) ≤{

c1−αe(1−α) ∫ t0 h(s) ds + (1 − α)

t∫

0

k(s)e(1−α) ∫ ts h(r) dr ds

} 11−α. (B.10)

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3. Hörmander, L.: Lectures on Nonlinear Hyperbolic Differential Equations. Springer, Berlin (1997)4. Manfrin, R.: Quadratic forms for the Liouville equation wt t + |ξ |2a(t)w = 0 with applications to

Kirchoff equation. Port. Math. 65(4), 447–484 (2008)5. Manfrin, R.: On the boundedness of solutions of the equation u′′ + (1 + f (t))u = 0. Discret. Contin.

Dyn. Syst. 23(3), 991–1008 (2009)6. Manfrin, R.: L p solutions of second order differential equations. Funkcial. Ekvac. 52, 255–279 (2009)7. Mitrinovic, D.S., Pecaric, J.E., Fink, A.M.: Inequalities for Functions and Their Integrals and

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