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arXiv:0805.0244v2 [math.AP] 22 Jan 2009
Derivative loss for Kirchhoff equations with
non-Lipschitz nonlinear term
Marina Ghisi
Universit` a degli Studi di Pisa
Dipartimento di Matematica “Leonida Tonelli”
PISA (Italy)
e-mail: ghisi@dm.unipi.it
Massimo Gobbino
Universit` a degli Studi di Pisa
Dipartimento di Matematica Applicata “Ulisse Dini”
PISA (Italy)
e-mail: m.gobbino@dma.unipi.it
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Abstract
In this paper we consider the Cauchy boundary value problem for the integro-differential
equation
??
with a continuous nonlinearity m : [0,+∞) → [0,+∞).
It is well known that a local solution exists provided that the initial data are regular
enough. The required regularity depends on the continuity modulus of m.
In this paper we present some counterexamples in order to show that the regularity
required in the existence results is sharp, at least if we want solutions with the same
space regularity of initial data. In these examples we construct indeed local solutions
which are regular at t = 0, but exhibit an instantaneous (often infinite) derivative loss
in the space variables.
utt− m
Ω
|∇u|2dx
?
∆u = 0in Ω × [0,T)
Mathematics Subject Classification 2000 (MSC2000): 35L70, 35L80, 35L90.
Key words: integro-differential hyperbolic equation, continuity modulus, Kirchhoff
equations, Gevrey spaces, derivative loss.
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1Introduction
In this paper we consider the hyperbolic partial differential equation
utt(t,x) − m
??
Ω
|∇u(t,x)|2dx
?
∆u(t,x) = 0∀(x,t) ∈ Ω × [0,T),(1.1)
where Ω ⊆ Rnis an open set, ∇u and ∆u denote the gradient and the Laplacian of
u with respect to the space variables, and m : [0,+∞) → [0,+∞). Equation (1.1) is
usually considered with initial conditions
u(x,0) = u0(x),ut(x,0) = u1(x)∀x ∈ Ω,(1.2)
and boundary conditions, for example of Dirichlet type (but the theory is more or less
the same also with Neumann or periodic boundary conditions)
u(x,t) = 0∀(x,t) ∈ ∂Ω × [0,T).(1.3)
From the mathematical point of view, (1.1) is probably the simplest example of quasi-
linear hyperbolic equation. From the mechanical point of view, this Cauchy boundary
value problem is a model for the small transversal vibrations of an elastic string (n = 1)
or membrane (n = 2). In the string context it was introduced by G. Kirchhoff in
[15].
A lot of papers have been devoted to existence of local or global solutions to (1.1),
(1.2), (1.3). The results can be divided into four main families.
(A) Local existence in Sobolev spaces. If m is a locally Lipschitz continuous function
such that m(σ) ≥ ν > 0 for every σ ≥ 0 (strict hyperbolicity), then (1.1), (1.2),
(1.3) admits a unique local solution for initial data in Sobolev spaces. This result
was first proved by S. Bernstein in the pioneering paper [3], and then extended
with increasing generality by many authors. The more general statement is prob-
ably contained in A. Arosio and S. Panizzi [1] (see also the references quoted
therein), where it is proved that the problem is locally well posed in the phase
space
Vβ:= Hβ+1(Ω) × Hβ(Ω) + boundary conditions
for every β ≥ 1/2.
(B) Global existence for analytic data. If m is a continuous function such that m(σ) ≥ 0
for every σ ≥ 0 (weak hyperbolicity), and u0(x) and u1(x) are analytic functions,
then (1.1), (1.2), (1.3) admits at least one global solution (T = +∞), which is
analytic in the space variables for any time.
This result was proved with increasing generality by S. Bernstein [3], S. I.
Pohozaev [19], A. Arosio and S. Spagnolo [2], P. D’Ancona and S. Spa-
gnolo [8, 9].
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(C) Local existence with assumptions between (A) and (B). More recently, H. Hiro-
sawa [13] considered the continuity modulus ω of the nonlinear term m. He proved
existence of local solutions in suitable classes of initial data, depending on ω, both
in the strictly hyperbolic and in the weakly hyperbolic case. From the point of
view of local solutions these results represent an interpolation between the results
of (A) and (B). The precise relations between ω and the regularity required on
initial data are stated in section 2. Roughly speaking, in the strictly hyperbolic
case the situation is summed up by the following scheme:
ω(σ) = o(1) → analytic data,
ω(σ) = σα(with α ∈ (0,1)) → Gevrey space Gs(Ω) with s = (1 − α)−1,
ω(σ) = σ|logσ| → C∞(Ω) (or Vβwith finite derivative loss),
ω(σ) = σ → V1/2(with no derivative loss).
More regularity is required in the weakly hyperbolic case, as shown in the following
scheme:
ω(σ) = o(1) → analytic data,
ω(σ) = σα(with α ∈ (0,1)) → Gevrey space Gs(Ω) with s = 1 + α/2,
ω(σ) = σ → Gevrey space G3/2(Ω).
(D) Global existence in special situations. Besides (B), there are four special cases in
which global solutions are known to exist. We refer the interested reader to the
quoted literature for the details. We just point out that in all these results the
nonlinearity m is assumed to be Lipschitz continuous and strictly positive.
• K. Nishihara [18] proved global existence for quasi-analytic initial data.
This class of functions strictly contains the space of analytic functions, but
it is strictly contained in every Gevrey class Gs(Ω) with s > 1.
• J. M. Greenberg and S. C. Hu [12], and then P. D’ancona and S.
Spagnolo [10] proved global existence for small initial data in Sobolev spaces
in the case Ω = Rn, where dispersion plays a crucial role. Later on this
dispersion-based approach was extended to external domains (see [22, 23]
and the references quoted therein).
• S. I. Pohozaev [20] proved global existence for initial data in the Sobolev
space V1in the case where m(σ) = (a + bσ)−2, with a > 0, b > 0. In this
particular case indeed equation (1.1) admits a second order invariant.
• In some recent papers R. Manfrin [16] (see also [17], [14], [11]) proved
global existence in a new class of nonregular initial data. Manfrin’s spaces
are small in the sense that they don’t contain any Gevrey class Gs(Ω) with
s > 1, but they are large in the sense that any initial condition (u0,u1) ∈ V1
is the sum of two initial conditions belonging to these spaces!
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Despite of the many positive results, as far as we know no paper has been devoted
to negative results. In particular in the literature we found no counterexample even
against the most optimistic conjecture, according to which a global solution to (1.1),
(1.2), (1.3) exists assuming only that m is a nonnegative continuous function, and the
initial data belong to the “energy space” V0.
In this paper we make a first step in the direction of counterexamples. We focus on
local solutions, and we prove that Hirosawa’s results [13] are sharp, both in the strictly
hyperbolic and in the weakly hyperbolic case.
In Theorem 3.1 and Theorem 3.4 we construct indeed local solutions u(x,t) of (1.1),
(1.2), (1.3) with very regular initial data, but such that for every t > 0 we have that
(u(x,t),ut(x,t)) belongs to the phase space Vβif and only if β ≤ 1/2. In a few words
these solutions exhibit an instantaneous derivative loss up to V1/2. In these examples
the maximal regularity admitted for the initial data depends on the continuity modulus
of m. Just to give some examples, in the strictly hyperbolic case we have the following
three situations (see Example 3.2).
• If m is continuous we can have derivative loss for quasi analytic initial data.
This proves in particular that in Nishihara’s result [18] the Lipschitz continuity
assumption on the nonlinear term cannot be relaxed to continuity, even when
looking for local solutions.
• If m is α-H¨ older continuous for some α ∈ (0,1), then we can have derivative loss
for initial data in any Gevrey space Gs(Ω) with s > (1 − α)−1.
• If m is α-H¨ older continuous for every α ∈ (0,1), then we can have derivative loss
for initial data in C∞(Ω). This proves in particular that in the results stated in (A)
(i.e., well posedness in Vβfor every β ≥ 1/2) the Lipschitz continuity assumption
on m cannot be relaxed to H¨ older continuity.
In the weakly hyperbolic case we have for example the following two situations (see
Example 3.5).
• If m is α-H¨ older continuous for some α ∈ (0,1), then we can have derivative loss
for initial data in any Gevrey space Gs(Ω) with s > 1 + α/2.
• If m is Lipschitz continuous we can have derivative loss for initial data in any
Gevrey space Gs(Ω) with s > 3/2. This proves in particular that in the results
stated in (A) the strict hyperbolicity cannot be relaxed to weak hyperbolicity.
Our examples don’t exclude that a local solution can always exist in V1/2. This
remains an open problem.
Our approach is based on the theory developed by F. Colombini, E. De Giorgi,
and S. Spagnolo [5], and then extended by F. Colombini and S. Spagnolo [6],
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and by F. Colombini, E. Jannelli, and S. Spagnolo [7]. They considered linear
equations with a time dependent coefficient such as
utt(t,x) − c(t)∆u(t,x) = 0.(1.4)
If c(t) is not Lipschitz continuous or vanishes for t = 0, they showed how to construct
“solutions” of (1.4) which are very regular at time t = 0, but very irregular (not even
distributions) for t > 0.
The construction of our counterexamples is divided into two steps. In the first step
we consider the linear equation (1.4), and we modify the parameters in the construction
described in [5, 6, 7] in order to obtain solutions of (1.4) which are very regular at
time t = 0, but with a prescribed minimal regularity (they belong to Vβ if and only
if β ≤ 1/2) for every t > 0. These counterexamples for the linear equation, stated in
Proposition 3.6 and Proposition 3.7, are maybe interesting in themselves because they
extend the theory developed in [5, 6, 7] to coefficients c(t) with arbitrary continuity
modulus. The assumptions on initial data in these counterexamples are sharp because
they are complementary to those required in the existence results (both for linear and
for Kirchhoff equations).
In the second step we show that these solutions, up to modifying one dominant
Fourier component, are solutions of (1.1) for a suitable choice of the nonlinearity m.
This paper is organized as follows. In section 2 we rephrase (1.1), (1.2), (1.3) as an
abstract evolution problem in a Hilbert space, and we restate Hirosawa’s local existence
results in the abstract setting. In section 3 we state our counterexamples, which we
prove in section 4. For the convenience of the reader in appendix A we sketch the main
point of the proof of the local existence results in the abstract setting.
2Preliminaries and local existence results
Continuity moduli
recalled throughout the paper.
Let ω : [0,+∞) → [0,+∞). The following assumptions are
(ω1) We have that ω ∈ C0([0,+∞)) is an increasing function such that ω(0) = 0, and
ω(a + b) ≤ ω(a) + ω(b) for every a ≥ 0 and b ≥ 0.
(ω2) The function σ → σ/ω(σ) is nondecreasing.
A function f : X → R (where X ⊆ R) is said to be ω-continuous if there exists a
constant L ∈ R such that
|f(a) − f(b)| ≤ Lω(|a − b|)
Two simple properties of continuity moduli are stated in Lemma A.2 of the appendix.
In particular from (A.4) it follows that the composition of a Lipschitz continuous function
and an ω-continuous function (in any order) is again an ω-continuous function.
∀a ∈ X, ∀b ∈ X. (2.1)
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It is not difficult to see that the set of ω-continuous functions only depends on the
values of ω in a right neighborhood of 0. For this reason, with a little abuse of notation,
we often consider continuity moduli which are defined, and satisfy (ω1) and/or (ω2), in
a right neighborhood of 0. For the same reason when needed we assume also that ω is
invertible.
Abstract setting for Kirchhoff equations
For every x and y in H, let |x| denote the norm of x, and let ?x,y? denote the scalar
product of x and y. Let A be an unbounded linear operator on H with dense domain
D(A). We always assume that A is self-adjoint and nonnegative, so that the power Aβ
is defined for every β ≥ 0 in a suitable domain D(Aβ).
We consider the second order evolution problem
Let H be a separable real Hilbert space.
u′′(t) + m(|A1/2u(t)|2)Au(t) = 0,∀ t ≥ 0, (2.2)
with initial data
u(0) = u0,u′(0) = u1. (2.3)
It is well known that (2.2), (2.3) is just an abstract setting of (1.1), (1.2), (1.3),
corresponding to the case where H = L2(Ω), and Au = −∆u, defined for every u in a
suitable domain D(A) depending on the boundary conditions (see [1]).
Functional spaces
able complete orthonormal system {ek}k≥1made by eigenvectors of A. We denote the
corresponding eigenvalues by λ2
Under this assumption (which in the concrete case corresponds to bounded domains)
we can work with Fourier series. However, any definition or statement of this section
can be easily extended to the general setting just by using the spectral theorem for
self-adjoint operators ([21, Theorem VIII.4, p. 260]).
By means of the orthonormal system every u ∈ H can be written in a unique way in
the form u =?∞
Aβu :=
?
so that we can consider the quantity
For the sake of simplicity, let us assume that H admits a count-
k, so that Aek= λ2
kekfor every k ≥ 1.
k=1ukek, where uk= ?u,ek? are the Fourier components of u. Moreover
for every β > 0 we have that
∞
k=1
λ2β
kukek,
?u?2
D(Aβ):=
∞
?
k=1
λ4β
ku2
k,
and characterize the spaces D(Aβ) and D(A∞) as follows
D(Aβ) :=
?
u ∈ H : ?u?2
D(Aβ)< +∞
?
,D(A∞) :=
?
β>0
D(Aβ).
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With these notations the phase spaces Vβare defined as Vβ:= D(A(β+1)/2)×D(Aβ/2).
Let now ϕ : [0,+∞) → [1,+∞) be any function. Then for every r > 0 and β > 0
we can set
∞
?
and then define the spaces
?u?2
ϕ,r,β:=
k=1
λ4β
ku2
kexp?rϕ(λk)?,(2.4)
Gϕ,r,β(A) :=?u ∈ H : ?u?2
If two continuous functions ϕ1(σ) and ϕ2(σ) coincide for every large enough σ, then
Gϕ1,r,β(A) = Gϕ2,r,β(A). For this reason, with a little abuse of notation, we consider these
spaces even if ϕ(σ) is defined, continuous, and greater than 1 only for large values of σ.
ϕ,r,β< +∞?,Gϕ,β(A) :=
?
r>0
Gϕ,r,β(A).
Hirosawa’s results
stated in the abstract setting. Since the abstract statements do not seem to follow
trivially from the ones in [13], we sketch the proofs in appendix A.
The first result concerns the strictly hyperbolic case (see Theorem 2.2 in [13]).
We are now ready to recall the main results proved in [13], re-
Theorem A (Strictly hyperbolic case) Let ω : [0,+∞) → [0,+∞) be a function
satisfying (ω1). Let m : [0,+∞) → (0,+∞) be an ω-continuous function satisfying the
strict hyperbolicity condition m(σ) ≥ ν > 0 for every σ ≥ 0.
Let ϕ : [0,+∞) → [1,+∞) be a function such that
σ
ϕ(σ)ω
limsup
σ→+∞
?1
σ
?
< +∞. (2.5)
Let us assume that
(u0,u1) ∈ Gϕ,r0,3/4(A) × Gϕ,r0,1/4(A)(2.6)
for some r0> 0.
Then there exist T > 0, and R > 0 with RT < r0 such that problem (2.2), (2.3)
admits at least one local solution
u ∈ C1?[0,T];Gϕ,r0−Rt,1/4(A)?∩ C0?[0,T];Gϕ,r0−Rt,3/4(A)?.
Remark 2.1 Condition (2.7), with the range space depending on time, simply means
that
u ∈ C1?[0,τ];Gϕ,r0−Rτ,1/4(A)?∩ C0?[0,τ];Gϕ,r0−Rτ,3/4(A)?
for every τ ∈ (0,T].
This amounts to say that scales of Hilbert spaces are the natural setting for this
problem.
(2.7)
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Example 2.2 Let us give some examples in order to clarify the interplay between as-
sumptions (2.5) and (2.6).
• If ω(σ) = o(1) as σ → 0+(which simply means that m is continuous), then (2.5)
holds true with ϕ(σ) = σ. In this case (2.6) means that u0and u1are analytic,
and one has re-obtained the classical local existence result for analytic data in the
strictly hyperbolic case. In (2.5) one can also take ϕ(σ) = σω(1/σ), thus obtaining
local existence for a larger class of initial data.
• If ω(σ) = σαfor some α ∈ (0,1) (which means that m is H¨ older continuous), then
(2.5) holds true with ϕ(σ) = σ1−α. In this case (2.6) means that one can take
initial data in the Gevrey space Gs(A) with s = (1 − α)−1. We recall that Gs(A)
is the space Gϕ,β(A) corresponding to ϕ(σ) = σ1/s(β is insignificant in this case).
• If ω(σ) = σ|logσ| (which means that m is log-Lipschitz continuous), then (2.5)
holds true with ϕ(σ) = logσ. In this case Gϕ,r,β(A) = D(Aβ+r/4). One can
therefore take initial data in D(A∞) and obtain a solution in the same space, or
even initial data in Vγ for some γ > 1/2 and obtain a solution with a possible
progressive derivative loss (due to the term r0− Rt in (2.7)).
• If ω(σ) = σ (which means that m is Lipschitz continuous), then (2.5) holds true
with ϕ(σ) ≡ 1. Therefore (2.6) means that one can take initial data in V1/2. This
is of course the classical existence result in Sobolev spaces.
Since we are interested in local solutions, Theorem A can be applied also in the
mildly degenerate case, namely when m may vanish but m(|A1/2u0|2) > 0.
The following result (see Theorem 2.1 of [13]) concerns the weakly hyperbolic case,
and it is essential to deal with the really degenerate case, i.e., when m(|A1/2u0|2) = 0.
Theorem B (Weakly hyperbolic case) Let ω : [0,+∞) → [0,+∞) be a function
satisfying (ω1). Let m : [0,+∞) → [0,+∞) be an ω-continuous function.
Let ϕ : [0,+∞) → [1,+∞) be a function such that
?
?ω(1/σ)
Let (u0,u1) and r0> 0 be such that (2.6) holds true.
Then there exist T > 0, and R > 0 with RT < r0 such that problem (2.2), (2.3)
admits at least one local solution u satisfying (2.7).
limsup
σ→+∞
σϕ
?
σ
??−1
< +∞.(2.8)
Example 2.3 Let us give some examples.
• If ω(σ) = o(1) as σ → 0+, then (2.8) holds true with ϕ(σ) = σ. Once again (2.6)
means that u0 and u1 are analytic, and one has re-obtained the classical local
existence result for analytic data in the weakly hyperbolic case.
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• If ω(σ) = σαfor some α ∈ (0,1], then (2.8) holds true with ϕ(σ) = σ2/(α+2).
Therefore (2.6) means that one can take initial data in the Gevrey space Gs(A)
with s = 1 + α/2. This is true in particular for α = 1 (i.e., when m is Lipschitz
continuous). In this case we have local existence for initial data in G3/2(A).
Remark 2.4 One cannot expect local solutions to be unique if m is not Lipschitz
continuous (see some examples in [2]). However the existence results rely on an a priori
estimate (see Proposition A.1) which implies in particular that under assumptions (2.5)
or (2.8) any local solution satisfying the minimal regularity requirement (A.1) also
satisfies the stronger condition (2.7). This is a sort of propagation of regularity: if the
space in (2.6) is strictly contained in D(A3/4) × D(A1/4), then also solutions lie in a
scale of spaces which is strictly contained in D(A3/4) × D(A1/4). We are going to see
that this is no more true when condition (2.5) or (2.8) are not satisfied.
3Statements of counterexamples
The first counterexample shows the optimality of Theorem A in the non-Lipschitz case.
Theorem 3.1 (Strictly hyperbolic case) Let A be a self adjoint linear operator on
a Hilbert space H. Let us assume that there exist a countable (not necessarily complete)
orthonormal system {ek}k≥1 in H, and an increasing unbounded sequence {λk}k≥1 of
positive real numbers such that Aek= λ2
Let ω : [0,+∞) → [0,+∞) be a function satisfying (ω1) and (ω2).
Let ϕ : [0,+∞) → [1,+∞) be a function such that
σ
ϕ(σ)ω
kekfor every k ≥ 1.
lim
σ→+∞
?1
σ
?
= +∞.(3.1)
Then there exist a function m : [0,+∞) → [1/2,3/2], a real number T0> 0, and a
function u : [0,T0] → H such that
(i) m is ω-continuous;
(ii) (u(0),u′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) for every r > 0;
(iii) u ∈ C1([0,T0];D(A1/4)) ∩ C0([0,T0];D(A3/4)) is a solution of (2.2);
(iv) for every t ∈ (0,T0] we have that (u(t),u′(t)) ?∈ Vβfor every β > 1/2.
Example 3.2 Let us consider some examples.
• The assumptions of Theorem 3.1 are satisfied if we take
ω(σ) =
1
|logσ|1/2,ϕ(σ) =
σ
logσ.
In this case m is continuous, the initial data are quasi analytic, and the solution
u has an instantaneous infinite derivative loss.
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• Let α ∈ (0,1). The assumptions of Theorem 3.1 are satisfied if we take
ω(σ) = σα,ϕ(σ) =σ1−α
logσ.
In this case m is α-H¨ older continuous, the initial data are in the Gevrey space
Gs(A) for every s > (1 − α)−1, and the solution u has an instantaneous infinite
derivative loss.
• The assumptions of Theorem 3.1 are satisfied if we take
ω(σ) = σ|logσ|3,ϕ(σ) = log2σ.
In this case m is α-H¨ older continuous for every α ∈ (0,1) (but not log-Lipschitz
continuous), the initial data are in D(A∞), and once again the solution u has an
instantaneous infinite derivative loss.
Remark 3.3 Theorem 3.1, as it is stated, is void in the log-Lipschitz case. Indeed when
ω(σ) = σ|logσ| and ϕ satisfies (3.1), then all initial data satisfying (ii) belong to V1/2
but not to Vβfor β > 1/2. In a certain sense they have no derivatives to loose!
On the other hand, a careful inspection of the proof reveals that Theorem 3.1 can be
improved as follows. Given any function ψ : [0,+∞) → [1,+∞) such that ψ(σ) → +∞
as σ → +∞ we can find a solution satisfying (i), (ii), (iii), and
(iv’) for every t ∈ (0,T0] we have that (u(t),u′(t)) ?∈ Gψ,r,3/4(A) × Gψ,r,1/4(A) for every
r > 0.
Thus for example we can take
ω(σ) = σ|logσ|,ϕ(σ) = log|logσ|,ψ(σ) = log|log|logσ||,
and obtain a solution with initial data in Gϕ,r,3/4(A)×Gϕ,r,1/4(A) which for every positive
time doesn’t belong even to the weaker scale Gψ,r,3/4(A)×Gψ,r,1/4(A). So also in the log-
Lipschitz case we have a (infinitesimally small) derivative loss.
We spare the reader from this generalization.
The second counterexample shows the optimality of Theorem B.
Theorem 3.4 (Weakly hyperbolic case) Let A, {ek}, {λk} be as in Theorem 3.1.
Let ω : [0,+∞) → [0,+∞) be a function satisfying (ω1) and (ω2).
Let ϕ : [0,+∞) → [1,+∞) be a function such that
?
?ω(1/σ)
Then there exist a function m : [0,+∞) → [0,3/2], a real number T0 > 0, and a
function u : [0,T0] → H such that
lim
σ→+∞σϕ
?
σ
??−1
= +∞. (3.2)
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(i) m is ω-continuous;
(ii) (u(0),u′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) for every r > 0;
(iii) u ∈ C1([0,T0];D(A1/4)) ∩ C0([0,T0];D(A3/4)) is a solution of (2.2);
(iv) for every t ∈ (0,T0] we have that (u(t),u′(t)) ?∈ Vβfor every β ≥ 1;
(v) there exists a sequence τk→ 0+such that |A(β+1)/2u(τk)| is unbounded for every
β > 1/2.
Example 3.5 Let α ∈ (0,1]. The assumptions of Theorem 3.1 are satisfied if we take
ω(σ) = σα,ϕ(σ) =σ2/(α+2)
logσ
.
In this case m is α-H¨ older continuous (Lipschitz continuous if α = 1), the initial
data are in the Gevrey space Gs(A) for every s > 1 + α/2, and the solution u has an
instantaneous infinite derivative loss.
In particular, in contrast to the strictly hyperbolic case (see Remark 3.3), The-
orem 3.4 provides examples of infinite derivative loss even in the Lipschitz (or log-
Lipschitz) case.
The counterexamples of Theorem 3.1 and Theorem 3.4 originate from two counterex-
amples for the linear equation
v′′+ c(t)Av = 0.(3.3)
This equation is well studied in mathematical literature. It is well known for example
that, if the coefficient c(t) is ω-continuous, then the Cauchy problem is well posed in
Gϕ,β(A) provided that ϕ and ω satisfy (2.5) in the strictly hyperbolic case, and (2.8)
in the weakly hyperbolic case. The main argument is that the approximated energy
estimates introduced in [5] can be extended word-by-word to arbitrary continuity moduli
as we do in the appendix below.
This result is sharp. Indeed if ϕ and ω do not satisfy (2.5) or (2.8), hence if they
satisfy (3.1) or (3.2) (see also Remark 3.9 below), then the Cauchy problem is not well
posed in Gϕ,β(A). In literature we found a lot of counterexamples for special choices of
ω and ϕ (see for example [4] and the references quoted therein), but we didn’t find the
general counterexamples under assumptions (3.1) or (3.2).
In the following two propositions we state them in the form which is needed in the
proof of Theorem 3.1 and Theorem 3.4.
Proposition 3.6 (Strictly hyperbolic case) Let A, {ek}, {λk}, ω, ϕ be as in The-
orem 3.1.
Then there exist c : [0,1] → [1/2,3/2], and v : [0,1] → H such that
(SH-1) c is ω-continuous;
10
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(SH-2) (v(0),v′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) for every r > 0;
(SH-3) v ∈ C1([0,1];D(A1/4)) ∩ C0([0,1];D(A3/4)) is a solution of (3.3);
(SH-4) for every t ∈ (0,1] we have that (v(t),v′(t)) ?∈ Vβfor every β > 1/2.
Proposition 3.7 (Weakly hyperbolic case) Let A, {ek}, {λk}, ω, ϕ be as in The-
orem 3.4.
Then there exist c : [0,1] → [0,3/2], and v : [0,1] → H such that
(WH-1) c is ω-continuous;
(WH-2) (v(0),v′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) for every r > 0;
(WH-3) v ∈ C1([0,1];D(A1/4)) ∩ C0([0,1];D(A3/4)) is a solution of (3.3);
(WH-4) for every t ∈ (0,1] we have that (v(t),v′(t)) ?∈ Vβfor every β ≥ 1;
(WH-5) there exists a sequence τk→ 0+such that |A(β+1)/2v(τk)| is unbounded for every
β > 1/2.
Remark 3.8 Proposition 3.6 and Proposition 3.7 are strongly based on the counterex-
amples shown in [5, 6, 7]. On the other hand, besides the fact that we deal with arbitrary
continuity moduli, and arbitrary sequences of eigenvalues, there are some differences we
would like to point out.
• In [5, 6, 7] the derivative loss is bigger because solutions instantaneously lie outside
the space of distributions. Here we need to be more careful since we want solutions
to lie in V1/2and nothing more.
• “Derivative loss” has a slightly different meaning in [5, 6, 7] and in this paper.
Losing the m-th derivative in [5, 6, 7] means that there exists a sequence τk→ 0+
such that the norm of (v(τk),v′(τk)) in Vm tends to +∞. This of course may
happen also if v(τk) and v′(τk) are in D(A∞) for every k. In other words, what is
actually lost is a uniform bound on the m-th derivative.
In statements (SH-4) and (WH-4), and in the corresponding statements of Theo-
rem 3.1 and Theorem 3.4, we lose the m-th derivative in a stronger sense, namely
(v(t),v′(t)) ?∈ Vmfor every t > 0. On the contrary in statement (WH-5), and in the
corresponding statement of Theorem 3.4, we are forced to lose the last derivatives
only in the weaker sense.
Remark 3.9 A careful inspection of the proofs shows that the same conclusions hold
true also if the limit in (3.1) and (3.2) is replaced by the corresponding limsup computed
along the sequence {λk}.
11
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4 Proofs
The proof is organized as follows. In section 4.1 we construct functions c(t) and v(t) de-
pending on several parameters. In Proposition 4.2, Proposition 4.3 and Corollary 4.4 we
relate the regularity properties of c(t) and v(t) to suitable conditions on the parameters.
Then in section 4.2 we choose the parameters in order to prove Proposition 3.6. In sec-
tion 4.3 we do the same for Proposition 3.7. Finally in section 4.4 we prove Theorem 3.1
and Theorem 3.4.
4.1General Construction
IngredientsThe starting point of the construction are the functions
b(ε,t) := 1 − 4εsin(2t) − ε2(1 − cos(2t))2,
w(ε,t) := sint · exp?ε?t −1
introduced in section 7 of [5].
Then the construction is based on seven sequences {tk}, {sk}, {τk}, {ηk}, {δk}, {εk},
{ak} satisfying the following assumptions.
(Hp-tk) The sequence {tk} is a decreasing sequence of positive real numbers (which
we think as times) such that t0= 1, and tk→ 0+as k → +∞.
(Hp-sk) The sequence {sk} is a sequence of positive real numbers (which we think as
times) such that tk< sk< tk−1for every k ≥ 1.
(Hp-τk) The sequence {τk} is a sequence of positive real numbers (which we think as
times) such that tk< τk< skfor every k ≥ 1.
(Hp-ηk) The sequence {ηk} is an increasing subsequence of the sequence {λk} of the
eigenvalues of A1/2.
(4.1)
(4.2)
2sin(2t)??,
(Hp-δk) The sequence {δk} is a nonincreasing sequence of positive real numbers with
δ0= 1. Moreover we require that√δkηktk/(2π) and√δkηksk/(2π) are inte-
gers, and 2√δkηkτk/π is an odd integer for every k ≥ 1.
(Hp-εk) The sequence {εk} is a sequence of positive real numbers such that εk≤ 1/16
for every k ≥ 1. Moreover we require that εkδk→ 0 and that {√δkεkηk} is
a nondecreasing sequence.
No special assumption is required on the sequence {ak}. We denote the limit of {δk}
by δ∞.
12
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Definition and properties of c(t)
and for every k ≥ 1
Let c : [0,1] → R be the function defined by c(0) = δ∞,
c(t) :=
δkb?εk,√δkηkt?
δk−1− δk
tk−1− sk(t − sk) + δk
if t ∈ [tk,sk],
if t ∈ [sk,tk−1].
(4.3)
Note that in the interval [sk,tk−1] the function c(t) is the affine interpolation between
δkand δk−1.
From the assumptions on the parameters we have that c(tk) = c(sk) = δk. Moreover
δk− 8εkδk≤ c(t) ≤ δk+ 8εkδk
δk≤ c(t) ≤ δk−1
∀t ∈ [tk,sk],(4.4)
∀t ∈ [sk,tk−1].(4.5)
Since εkδk→ 0, and δk→ δ∞, estimates (4.4) and (4.5) imply that c(t) is continuous
in [0,1]. Since εk≤ 1/16 we have also that
1
2δk≤ c(t) ≤3
2δk
∀t ∈ [tk,sk], (4.6)
and globally
1
2δ∞≤ c(t) ≤3
2
∀t ∈ [0,1]. (4.7)
Concerning the derivative we have that
|c′(t)| ≤ 16εkηkδ3/2
k
∀t ∈ (tk,sk),(4.8)
and of course
c′(t) =δk−1− δk
tk−1− sk
∀t ∈ (sk,tk−1).(4.9)
The ω-continuity of c(t) in the whole interval [0,1] can be deduced from the ω-
continuity in the intervals [tk,tk−1] provided that the following uniform estimates hold
true (we omit the easy and classical proof).
Lemma 4.1 Let c : [0,1] → R be any function. Let {tk} be any sequence satisfying
(Hp-tk).
Then c is ω-continuous in [0,1] if and only if there exists a constant L such that
(i) |c(ti) − c(tj)| ≤ Lω(|ti− tj|) for every pair of positive integers i and j;
(ii) |c(a) − c(b)| ≤ Lω(|a − b|) for every k ≥ 1, and every a and b in [tk,tk−1].
2
Applying the lemma we find the following sufficient condition for the for the ω-
continuity of c(t).
13
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Proposition 4.2 The function c(t) defined in (4.3) is ω-continuous in [0,1] if
sup
i<j
δi− δj
ω(ti− tj)+ sup
k≥1
δk−1− δk
ω(tk−1− sk)+ sup
k≥1
εkδk
ω?2π/(ηk
√δk)? < +∞. (4.10)
Proof. We apply Lemma 4.1. Assumption (i) is satisfied because the first supre-
mum in (4.10) is finite. In order to verify assumption (ii) we consider separately the
subintervals [tk,sk] and [sk,tk−1]. Thanks to (ω2) in [sk,tk−1] we have that
|c(a) − c(b)|
ω(|a − b|)
=δk−1− δk
tk−1− sk
·
|a − b|
ω(|a − b|)≤
δk−1− δk
ω(tk−1− sk).
This is bounded uniformly in k because the second supremum in (4.10) is finite.
In [tk,sk] the function c(t) is periodic, hence we can limit ourselves to consider
points a and b with |a − b| less or equal than a period 2π/(ηk
we therefore have that
√δk). By (4.8) and (ω2)
|c(a) − c(b)|
ω(|a − b|)
=
????
c(a) − c(b)
a − b
????·
|a − b|
ω(|a − b|)≤ 16εkηkδ3/2
k
·
2π
√δk
ηk
·
1
ω?2π/(ηk
√δk)?.
This is bounded uniformly in k because the third supremum in (4.10) is finite.
2
Definition and properties of vk(t)
ordinary differential equation
Let vk : [0,1] → R be the solution of the linear
v′′
k(t) + η2
kc(t)vk(t) = 0,(4.11)
with “initial” data
vk(tk) = 0,v′
k(tk) = ηk
?
δk.(4.12)
In order to estimate vkwe consider the usual “Kovalevskian” energy
Ek(t) := |v′
k(t)|2+ η2
k|vk(t)|2, (4.13)
and the usual hyperbolic energy
Fk(t) := |v′
k(t)|2+ η2
kc(t)|vk(t)|2.(4.14)
It is simple to show that their time derivatives are given by
E′
k(t) = 2η2
k(1 − c(t))vk(t)v′
k(t);(4.15)
F′
k(t) = η2
kc′(t)|vk(t)|2.(4.16)
Now we estimate Ek(t) in four cases.
14
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Case 1: t ∈ [tk,sk]In this interval we have the explicit expression
vk(t) = exp
?
−εkηk
?
δktk
?
· w
?
εk,ηk
?
δkt
?
,
which is indeed the main motivation of definitions (4.1) and (4.2). In particular we have
the following equalities
Ek(tk) = Fk(tk) = δkη2
k, (4.17)
Ek(sk) = Fk(sk) = δkη2
kexp
?
2εkηk
?
δk(sk− tk)
?
, (4.18)
and the estimate
Ek(t) ≤ 3η2
kexp
?
2εkηk
?
δk(sk− tk)
?
∀t ∈ [tk,sk]. (4.19)
From the explicit expression, and our assumption that 2√δkηkτk/π is an odd integer,
we have also that
|vk(τk)| = exp
?
εkηk
?
δk(τk− tk)
?
.(4.20)
Case 2: t ∈ [sk,tk−1]
that
In this interval c′(t) ≥ 0, and therefore from (4.16) we have
k(t) ≤c′(t)
F′
c(t)Fk(t)∀t ∈ [sk,tk−1].
Integrating in [sk,t] we find that
Fk(t) ≤ Fk(sk)exp
??t
sk
c′(s)
c(s)ds
?
= Fk(sk)exp
?
logc(t)
c(sk)
?
= Fk(sk) ·c(t)
δk
(4.21)
for every t ∈ [sk,tk−1]. Using (4.18), and the fact that c(t) ≤ δk−1in this interval, we
obtain in particular that
Fk(t) ≤ δk−1η2
kexp
?
2εkηk
?
δk(sk− tk)
?
∀t ∈ [sk,tk−1].(4.22)
On the other hand we have that F′
k(t) ≥ 0 in this interval, hence
?
Fk(t) ≥ Fk(sk) = δkη2
kexp
?
2εkηk
δk(sk− tk)
?
∀t ∈ [sk,tk−1].(4.23)
Since c(t) ≤ 3/2 we have also that
Ek(t) ≤3
2·
1
c(t)Fk(t)∀t ∈ (0,1],
which combined with (4.21) and (4.18) gives
Ek(t) ≤3
2η2
kexp
?
2εkηk
?
δk(sk− tk)
?
∀t ∈ [sk,tk−1].(4.24)
15
Page 18
Case 3: t ∈ [0,tk] From (4.15) and (4.7) we have that
|E′
k(t)| ≤ ηk|1 − c(t)|Ek(t) ≤ ηkEk(t)
Integrating in [t,tk] we obtain that
∀t ∈ [0,1].
Ek(tk)exp(−ηk(tk− t)) ≤ Ek(t) ≤ Ek(tk)exp(ηk(tk− t))
hence by (4.17) we conclude that
∀t ∈ [0,tk],
δkη2
kexp(−ηktk) ≤ Ek(t) ≤ δkη2
kexp(ηktk)∀t ∈ [0,tk].(4.25)
Case 4: t ∈ [tk−1,1]From (4.16) we have that
F′
k(t) =c′(t)
c(t)· η2
kc(t)|vk(t)|2≤|c′(t)|
c(t)Fk(t),
hence
Fk(t) ≤ Fk(tk−1)exp
??1
tk−1
|c′(s)|
c(s)
ds
?
∀t ∈ [tk−1,1].(4.26)
Now we observe that
?1
tk−1
|c′(s)|
c(s)
ds =
k−1
?
i=1
?si
ti
|c′(s)|
c(s)
ds +
k−1
?
i=1
?ti−1
si
|c′(s)|
c(s)
ds,(4.27)
and we estimate the two summands in the right hand side. From (4.8) and (4.6) we
have that
|c′(s)|
c(s)δi/2
√δkis nondecreasing. Therefore we have that
≤16εiηiδ3/2
i
= 32εiηi
?
δi
∀s ∈ [ti,si].
By (Hp-εk) the sequence εkηk
k−1
?
i=1
?si
ti
|c′(s)|
c(s)
ds ≤ 32εk−1ηk−1
?δk−1
k−1
?
i=1
|si− ti| ≤ 32εk−1ηk−1
?δk−1.
On the other hand, in the interval [si,ti−1] we have that c′(s) > 0, hence
?ti−1
si
|c′(s)|
c(s)
ds =
?ti−1
si
c′(s)
c(s)ds = logc(ti−1)
c(si)
= logδi−1
δi
,
and therefore (we recall that δ0= 1)
k−1
?
i=1
?ti−1
si
|c′(s)|
c(s)
ds =
k−1
?
i=1
logδi−1
δi
= log
?k−1
i=1
?
δi−1
δi
?
= log
1
δk−1.
16
Page 19
Coming back to (4.27) and (4.26) we have proved that
Fk(t) ≤ Fk(tk−1)
1
δk−1exp
?
32εk−1ηk−1
?δk−1
?
.
Taking (4.22) into account we finally obtain that
Fk(t) ≤ η2
kexp
?
2εkηk
?
δk(sk− tk) + 32εk−1ηk−1
?δk−1
?
∀t ∈ [tk−1,1]. (4.28)
In an analogous way we can prove that
Fk(t) ≥ Fk(tk−1)exp
?
−
?1
tk−1
|c′(s)|
c(s)
ds
?
≥ Fk(tk−1)δk−1exp
?
−32εk−1ηk−1
?δk−1
?
,
which by (4.23) gives
Fk(t) ≥ δkδk−1η2
kexp
?
2εkηk
?
δk(sk− tk) − 32εk−1ηk−1
?δk−1
?
∀t ∈ [tk−1,1].
(4.29)
In order to obtain estimates on Ek(t) we just remark that
δk−1
2
≤ c(t) ≤3
2
∀t ∈ [tk−1,1]
(here we used that the sequence δkis nonincreasing), so that
2
3Fk(t) ≤ Ek(t) ≤
3
δk−1Fk(t).
Therefore (4.28) and (4.29) imply the following estimates for every t ∈ [tk−1,1]
Ek(t) ≤3η2
k
δk−1exp
?
?
2εkηk
?
?
δk(sk− tk) + 32εk−1ηk−1
?δk−1
?δk−1
?
?
,(4.30)
Ek(t) ≥2
3δkδk−1η2
kexp2εkηk
δk(sk− tk) − 32εk−1ηk−1
.(4.31)
Definition and properties of v(t)
the eigenvalue η2
Let eikdenote the eigenvector of A corresponding to
k. We set
v(t) :=
∞
?
k=1
akvk(t)eik.(4.32)
At this level of generality this definition is purely formal, because we have no infor-
mation about the convergence of the series. All regularity properties of v(t) are related
17
Page 20
to convergence properties of suitable series. In particular for every t ∈ [0,1], and every
β ≥ 0, ϕ : [0,+∞) → [1,+∞), and r > 0 we have that
(v(t),v′(t)) ∈ Vβ
⇐⇒
∞
?
∞
?
k=1
a2
kη2β
kEk(t) < +∞, (4.33)
(v(t),v′(t)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) ⇐⇒
k=1
a2
kηkEk(t)exp?rϕ(ηk)?< +∞.(4.34)
Combining these implications with our estimates on Ek(t) we obtain sufficient con-
ditions in terms of the parameters for the regularity or non regularity of v(t).
Proposition 4.3 Let v(t) be the function defined in (4.32).
(1) For every r > 0 we have that (v(0),v′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) if
∞
?
k=1
a2
kδkη3
kexp?ηktk+ rϕ(ηk)?< +∞. (4.35)
(2) For every t > 0 and β ≥ 0 we have that (v(t),v′(t)) ∈ Vβif
a2
k
δk−1
∞
?
k=1
kη2β+2
exp
?
2εkηk
?
δk(sk− tk) + 32εk−1ηk−1
?δk−1
?
< +∞. (4.36)
(3) For every t > 0 and β ≥ 0 we have that (v(t),v′(t)) ?∈ Vβif
∞
?
k=1
a2
kδkδk−1η2β+2
k
exp
?
2εkηk
?
δk(sk− tk) − 32εk−1ηk−1
?δk−1
?
= +∞.(4.37)
(4) We have that v ∈ C1([0,1];D(A1/4))∩C0([0,1];D(A3/4)) if (4.35) holds true with
r = 0, and (4.36) holds true with β = 1/2.
(5) For every β ≥ 0 we have that the sequence
??A(β+1)/2v(τk)??is unbounded if
?
lim
k→+∞a2
kη2β+2
k
exp
?
2εkηk
δk(τk− tk)
?
= +∞.(4.38)
Proof. Statement (1) follows combining (4.34) and (4.25) with t = 0.
Let us examine statements (2) and (3). For t ∈ (0,1] we have that t ≥ tk−1for every
large enough k, and therefore the relevant estimates are (4.30) and (4.31). At this point
statement (2) follows combining (4.33) with (4.30), statement (3) follows by combining
(4.33) with (4.31).
18
Page 21
Let us consider now statement (4). A sufficient condition in order to prove that v(t)
has the required time regularity is that
∞
?
k=1
a2
kηk sup
t∈[0,1]Ek(t) < +∞. (4.39)
Now for every k ≥ 1 we have that
kexp(ηktk) +3η2
sup
t∈[0,1]Ek(t) ≤ δkη2
k
δk−1exp
?
2εkηk
?
δk(sk− tk) + 32εk−1ηk−1
?δk−1
?
.
This estimate indeed easily follows from (4.25) if t ≤ tk, from (4.19) or (4.24) if
tk≤ t ≤ tk−1, from (4.30) if tk−1≤ t ≤ 1.
Therefore (4.39) follows from (4.35) with r = 0 and (4.36) with β = 1/2.
Let us finally prove statement (5). By (4.20) we have that
??A(β+1)/2v(τk)??2≥ a2
so that (4.38) implies that the left hand side is unbounded.
kη2β+2
k
|vk(τk)|2= a2
kη2β+2
k
exp
?
2εkηk
?
δk(τk− tk)
?
,
2
Choice of ak
Let us set
a2
k:=δk−1
k2η3
k
exp
?
−2εkηk
?
δk(sk− tk) − 32εk−1ηk−1
?δk−1
?
. (4.40)
Conditions (4.35) through (4.38) can thus be restated in terms of the remaining
parameters. We obtain the following result (proof is trivial).
Corollary 4.4 Let v(t) be the function defined in (4.32), with akdefined by (4.40).
(1) For every r > 0 we have that (v(0),v′(0)) ∈ Gϕ,r,3/4(A) × Gϕ,r,1/4(A) if
∞
?
k=1
δkδk−1
k2
exp
?
ηktk+ rϕ(ηk) − 2εkηk
?
δk(sk− tk) − 32εk−1ηk−1
?δk−1
?
< +∞.
(4.41)
(2) For every t > 0 and β ≥ 0 we have that (v(t),v′(t)) ∈ Vβif
∞
?
k=1
1
k2η2β−1
k
< +∞. (4.42)
(3) For every t > 0 and β ≥ 0 we have that (v(t),v′(t)) ?∈ Vβif
∞
?
k=1
δkδ2
k2
k−1
η2β−1
k
exp
?
−64εk−1ηk−1
?δk−1
?
= +∞.(4.43)
19
Page 22
(4) We have that v ∈ C1([0,1];D(A1/4))∩C0([0,1];D(A3/4)) if (4.41) holds true with
r = 0 (note that (4.42) is trivial for β = 1/2).
(5) For every β ≥ 0 we have that the sequence
δk−1
k2η2β−1
??A(β+1)/2v(τk)??is unbounded if
δk(τk− sk) − 32εk−1ηk−1
lim
k→+∞
k
exp
?
2εkηk
?
?δk−1
?
= +∞.(4.44)
4.2Proof of Proposition 3.6
We define c(t) and v(t) according to the general construction. In order to choose the
parameters we first set t0= δ0= 1. Then for every k ≥ 1 we set
tk:=2π
ηk,
2tk
sk:=
?tk−1
?
tk,δk:= 1,εk:= ω
?1
ηk
?
, (4.45)
where the brackets denote the integer part, and the sequence ηkis defined as follows:
• η1is any element of the sequence {λk} which is greater than 4π and such that
ω(1/η1) ≤ 1/16;
• for every k ≥ 2 the term ηkis recursively defined as any element of the sequence
{λk} satisfying the following three inequalities
ηk
≥ 4ηk−1,
ηk
ϕ(ηk)ω
ηk
ηk
≥ exp(k + kεk−1ηk−1).
(4.46)
?1
?
≥
k
tk−1,(4.47)
(4.48)
We didn’t define τkbecause we don’t need it in this proof.
The first thing we have to verify is that ηkis well defined. The right hand sides of
(4.46) through (4.48) only depend on the values of the sequences with index k−1. Since
the sequence {λk} is unbounded, we can find ηkwith the required properties if we show
that the left hand sides tend to +∞ as ηk→ +∞. This is trivial in the case of (4.46)
and (4.48), and it follows from assumption (3.1) in the case of (4.47).
The second thing we need is that assumptions (Hp-tk), (Hp-sk), (Hp-ηk), (Hp-δk),
and (Hp-εk) are satisfied (they are required for the entire construction to work). By
(4.46) the sequence ηkis increasing and tends to +∞. These facts are enough to prove
all properties, except the monotonicity of εkηk
We are now ready to prove the conclusions of Proposition 3.6.
√δk, which however follows from (ω2).
20
Page 23
Conclusion (SH-1)
and our definition of δk.
In order to prove the ω-continuity of c(t) we apply Proposition 4.2. The first and
second supremum in (4.10) are trivially zero. Moreover by the monotonicity of ω we
have that
εkδk
ω?2π/(ηk
hence also the third supremum is finite.
First of all we remark that 1/2 ≤ c(t) ≤ 3/2. This is due to (4.7)
√δk)? =
εk
ω(tk)=ω(tk/(2π))
ω(tk)
≤ 1,
Conclusion (SH-2)
also for r = 0) if we show that the argument of the exponential function is bounded from
above. Since ηktkand εkηktkare bounded, it is enough to prove that rϕ(ηk) ≤ 2εkηksk
for every large enough k. In turn this is true if we show that
Let us examine (4.41). The series converges for every r > 0 (and
lim
k→+∞
εkηksk
ϕ(ηk)= +∞.(4.49)
To this end we first use the definition of tkand (4.46) to obtain that
sk
tk−1
=
?tk−1
2tk
?
tk
tk−1
≥1
2−
tk
tk−1
=1
2−ηk−1
ηk
≥1
4.
Then using the definition of skand εk, and estimate (4.47), we deduce that
εkηksk
ϕ(ηk)=
ηk
ϕ(ηk)ω
?1
ηk
?
sk≥ k ·
sk
tk−1
≥k
4,
which implies (4.49).
Conclusion (SH-3)
r = 0. We did this in the previous paragraph.
By statement (4) of Corollary 4.4 it is enough to prove (4.41) with
Conclusion (SH-4) Let us examine (4.43). By (4.48) we have that
1
k2η2β−1
k
exp(−64εk−1ηk−1) ≥
1
k2e(2β−1)kexp?[(2β − 1)k − 64]εk−1ηk−1
?.
It is therefore clear that the series cannot converge if β > 1/2.
2
21
Page 24
4.3
Let g : (0,+∞) → (0,+∞) be defined by g(σ) :=√σω−1(σ), where ω−1is the inverse
function of ω. It is easy to see that g is invertible. Its inverse function, which we denote
by h(σ), is increasing and vanishes as σ → 0+.
Now we define c(t) and v(t) according to the general construction. In order to choose
the parameters we first set t0= δ0= 1. Then for every k ≥ 1 we set
2π
ηk
2tk
where the brackets denote the integer part, and the sequence ηkis defined as follows:
Proof of Proposition 3.7
tk:=
√δk
,sk:=
?tk−1
?
tk,τk:= sk−tk
4,εk:=
1
16,δk:= h
?1
ηk
?
,(4.50)
• η1is any element of the sequence {λk} with h(1/η1) < 1, and η1
• for every k ≥ 2 the term ηkis recursively defined as any element of the sequence
{λk} such that ηk≥ ηk−1+ 1 and
?
ηkh
ηk
??1/2
ηkh
ηk
δ2
k−1
?
The first thing we need is that ηkis well defined. As in the proof of Proposition 3.6
it is enough to show that the left hand sides of (4.51) through (4.55) tend to +∞ as
ηk→ +∞. This is trivial in the case of (4.55). As for (4.51), (4.52), (4.54), with the
variable change z = 1/g(ω(σ)) we have that
??γ
By (ω2) the function ω(σ)/σ is bounded from below in a right neighborhood of 0.
It follows that the limit is +∞ for every γ < 3/2.
Let us finally consider the left hand side of (4.53). Thanks to assumption (3.2) and
the variable change z = 1/g(ω(1/σ)) we have that
?h(1/η1) > 4π;
ηk
h
?1
ηk
??1/2
?1
≥ 4ηk−1
?δk−1,(4.51)
?
≥
k
tk−1, (4.52)
ηk
ϕ(ηk)
?
h
?1
ηk
≥
k
tk−1,
k2
(4.53)
?1
logηk
?
≥
exp
?
?δk−1− logδk−1+ 2logk
64εk−1ηk−1
?δk−1
?
,(4.54)
≥ k2ηk−1
?
.(4.55)
lim
z→+∞z
?
h
?1
z
= lim
σ→0+
[ω(σ)]γ
g(ω(σ))= lim
σ→0+[ω(σ)]γ−3/2ω(σ)
σ
.
lim
z→+∞
z
ϕ(z)
?
h
?1
z
??1/2
=lim
σ→+∞
1
g(ω(1/σ))·
?
1
ϕ(1/g(ω(1/σ)))· [ω(1/σ)]1/2
??−1
=lim
σ→+∞σϕ
?
σ
?ω(1/σ)
= +∞.
22
Page 25
The second thing we need is that assumptions (Hp-tk), (Hp-sk), (Hp-τk), (Hp-ηk),
(Hp-δk), and (Hp-εk) are satisfied. Assumption (4.51) is equivalent to say that ηk
4ηk−1
?δk−1, hence also tk≤ tk−1/4. All the other properties are now almost trivial.
√δk≥
We are now ready to prove the conclusions of Proposition 3.7.
Conclusion (WH-1)
only if t = 0). This is due to (4.6), (4.7), and our definition of δk.
In order to prove the ω-continuity of c(t) we apply Proposition 4.2. Let us consider
the first supremum in (4.10). By definition of tkand δkwe have that δk= ω(tk/(2π)).
Since the function ω is itself ω-continuous, for j > i we have that
First of all we remark that 0 ≤ c(t) ≤ 3/2 (with c(t) = 0 if and
δi− δj= ω
?ti
2π
?
− ω
?tj
2π
?
≤ ω
?ti− tj
2π
?
≤ ω(ti− tj),
which implies that the first supremum in (4.10) is finite.
Since clearly sk/tk−1≤ 1/2, we have that tk−1− sk≥ tk−1/2, hence
δk−1− δk≤ δk−1= ω
?tk−1
2π
?
≤ ω
?tk−1
2
?
≤ ω (tk−1− sk),
which implies that the second supremum in (4.10) is finite. Finally
δk= ω
?tk
2π
?
≤ ω (tk) = ω
?
2π
√δk
ηk
?
,
hence also the third supremum is trivially finite.
Conclusion (WH-2)
we show that the argument of the exponential function is bounded from above. Since
εkηk
Let us examine (4.41). The series converges for every r > 0 if
√δktkis constant, it is enough to prove that
ηktk+ rϕ(ηk) ≤ 2εkηk
?
δksk
whenever k is large enough. Since εkis constant this is true if we show that
√δksk
tk
lim
k→+∞
= +∞,lim
k→+∞
ηk
√δksk
ϕ(ηk)
= +∞.(4.56)
Since we already observed that tk/tk−1≤ 1/4, we have also that
sk
tk−1
2tk
=
?tk−1
?
tk
tk−1
≥1
2−
tk
tk−1
≥1
4.(4.57)
Combining with (4.52) we obtain that
√δksk
tk
=δkηksk
2π
=
1
2π· h
?1
ηk
?
ηk·
sk
tk−1
· tk−1≥
k
8π,
23
Page 26
which implies the first limit in (4.56). Using (4.57) and (4.53) we have that
ηk
√δksk
ϕ(ηk)
=
ηk
ϕ(ηk)
?
h
?1
ηk
??1/2sk
tk−1
· tk−1≥k
4,
which implies the second limit in (4.56).
Conclusion (WH-3)
paragraph.
As in the strictly hyperbolic case it follows from the previous
Conclusion (WH-4)
term of the series is greater or equal than η2β−2
β ≥ 1.
Let us examine (4.43). Assumption (4.54) implies that the general
. Therefore the series diverges for every
k
Conclusion (WH-5)
is greater or equal than
Let us finally examine (4.44). By (4.55) each term in the sequence
exp
?
[(2β − 1)k − 1]
?
2ηk−1
?δk−1− logδk−1+ 2logk
2
?
−π
16
?
,
hence its limit is +∞ whenever β > 1/2.
4.4Proof of Theorem 3.1 and Theorem 3.4
For shortness’s sake we only prove Theorem 3.1 (the proof of Theorem 3.4 is completely
analogous).
Let c(t) and v(t) be as in Proposition 3.6. Since properties of v(t) don’t depend
on a single Fourier component, we can always assume that the component of v(t) with
respect to e1is identically zero.
Now we consider the solution w : [0,1] → R of the Cauchy problem
w′′(t) + λ2
1c(t)w(t) = 0,w(0) = 1, w′(0) = 1,
and for every ε > 0 we set
uε(t) := w(t)e1+ εv(t)∀t ∈ [0,1].
The function uε: [0,1] → H is again a solution of (3.3), and it has the same regularity
and non regularity properties of v(t).
Let ψε(t) := |A1/2uε(t)|2. Since the component of v(t) with respect to e1is zero we
have that
ψ′
ε(t) = 2?A3/4uε(t),A1/4u′
ε(t)? = 2λ2
1w(t)w′(t) + 2ε?A3/4v(t),A1/4v′(t)?.
24
Page 27
From (SH-3) we have that ?A3/4v(t),A1/4v′(t)? is bounded in [0,1]. Since w(0)w′(0) =
1 there exists T0∈ (0,1] and ε0> 0 such that
ε0(t) ≥1
2
ψ′
∀t ∈ [0,T0].
It follows that ψε0: [0,T0] → [ψε0(0),ψε0(T0)] is invertible, and therefore we can
define m : [0,+∞) → [1/2,3/2] by
The function m is ω-continuous because ψ−1
m(σ) :=
c(0)
c?ψ−1
c(T0)
if σ ≤ ψε0(0),
if ψε0(0) ≤ σ ≤ ψε0(T0),
if σ ≥ ψε0(T0).
ε0(σ)?
ε0is Lipschitz continuous. Moreover
c(t) = m(|A1/2uε0(t)|2)∀t ∈ [0,T0].
Therefore for this choice of the nonlinearity m the function u(t) := uε0(t) is a solution
of (2.2) in the interval [0,T0], and it exhibits the required derivative loss.
2
A Proof of Theorem A and Theorem B (sketch)
It is well known that local existence for Kirchhoff equations can be easily proved in
different ways (for example Galerkin approximations) provided that solutions are known
to satisfy an a priori estimate yielding the uniform continuity of the nonlinear term.
In this appendix we prove this estimate under the assumptions of Theorem A and
Theorem B. The statement is the following.
Proposition A.1 Let ω, m, ϕ, u0, u1, r0be as in Theorem A or Theorem B.
Then there exist positive real numbers T, H, R, with RT < r0, such that every
solution
u ∈ C0?[0,T];D(A3/4)?∩ C1?[0,T];D(A1/4)?
of problem (2.2), (2.3) satisfies
(A.1)
|A1/4u′(t)|2+ |A3/4u(t)|2≤ H∀t ∈ [0,T],(A.2)
and
u ∈ C1?[0,T];Gϕ,r0−Rt,1/4(A)?∩ C0?[0,T];Gϕ,r0−Rt,3/4(A)?.
The constants T, H, R depend only on ω, m, and on the norms of u0and u1in the
spaces Gϕ,r0,3/4(A) and Gϕ,r0,1/4(A), respectively.
In the proof of Proposition A.1 we need two simple properties of continuity moduli
and convolutions (see Lemma 4.1 and Lemma 4.2 in [11]).
(A.3)
25
Page 28
Lemma A.2 Let ω : [0,+∞) → [0,+∞) be a continuity modulus satisfying (ω1).
Then we have that
ω(λσ) ≤ (1 + λ)ω(σ)
1
ω(σ)≤
∀λ ≥ 0, ∀σ ≥ 0;
1 +1
σ
(A.4)
1 +
?
1 +
1
ω(1)
???
∀σ > 0. (A.5)
Lemma A.3 Let ρ : R → [0,+∞) be a function of class C∞, with support contained in
[−1,1], and integral equal to 1.
Let a > 0, and let f : [0,a] → R be a continuous function. Let us extend f(x) to the
whole real line by setting f(x) = f(0) for every x ≤ 0, and f(x) = f(a) for every x ≥ a.
For every ε > 0 let us set
fε(x) :=
?
R
f(x + εs)ρ(s)ds∀x ∈ R.
Then fε∈ C∞(R), and fεhas the following properties.
(1) If µ1≤ f(x) ≤ µ2for every x ∈ [0,a], then µ1≤ fε(x) ≤ µ2for every x ∈ R and
every ε > 0.
(2) Let ω be a continuity modulus. Let us assume that
|f(x) − f(y)| ≤ H0ω(|x − y|)
for some H0≥ 0. Then there exists a constant γ0(depending on ρ, but independent
on ε, H0, f) such that
∀x ∈ [0,a], ∀y ∈ [0,a],(A.6)
|fε(x) − f(x)| ≤ γ0H0ω(ε)∀x ∈ R, ∀ε > 0,
|f′
ε(x)| ≤ γ0H0ω(ε)
ε
∀x ∈ R, ∀ε > 0.
A priori estimates in the strictly hyperbolic case
stants. From the assumptions on m and ϕ we know that there exist L and Λ such
that
|m(σ1) − m(σ2)| ≤ Lω(|σ1− σ2|)
σω(1/σ) ≤ Λϕ(σ)
Let γ0be the constant appearing in Lemma A.3, and let
Let us introduce some con-
∀σ1≥ 0, ∀σ2≥ 0,
∀σ > 0.
(A.7)
(A.8)
γ1:=?m(|A1/2u0|2) + 1?· max?1,ν−1?,
H := γ1
??u1?2
26
ϕ,r0,1/4+ ?u0?2
ϕ,r0,3/4
?+ 1,
Page 29
R := γ0LΛ(H + 1)
?1
ν+
1
√ν
?
.
Let T > 0 be such that
ω(T) ≤
1
L(H + 1),T ≤r0
R.
(A.9)
We claim that (A.2) and (A.3) hold true for these values of T, H, and R. To this
end we introduce
S := sup?τ ≤ T : |A1/4u′(t)|2+ |A3/4u(t)|2≤ H ∀t ∈ [0,τ]?.
We have that S > 0 because γ1≥ 1 and
?u1?2
(A.10)
D(A1/4)+ ?u0?2
D(A3/4)≤ ?u1?2
ϕ,r0,1/4+ ?u0?2
ϕ,r0,3/4.
In order to prove (A.2) we only need to show that S = T. So let us assume by
contradiction that S < T. By the maximality of S we have that necessarily
|A1/4u′(S)|2+ |A3/4u(S)|2= H.(A.11)
Moreover for every t ∈ [0,S] we have that
????
Therefore if we set
d
dt|A1/2u(t)|2
????= 2???A3/4u(t),A1/4u′(t)???≤ H.
c(t) := m(|A1/2u(t)|2),
(A.12)
(A.13)
by (A.7), (A.12), and (A.4) we have that
|c(t) − c(s)| ≤ Lω???|A1/2u(t)|2− |A1/2u(s)|2???≤ L(H + 1)ω(|t − s|)
for every t and s in [0,S]. In particular from the strict hyperbolicity and the first
inequality in (A.9) we obtain that
ν ≤ c(t) ≤ c(0) + L(H + 1)ω(t) ≤ m(|A1/2u0|2) + 1
Let us extend c(t) outside the interval [0,S] as in Lemma A.3, and let us set
∀t ∈ [0,S].
cε(t) :=
?
R
c(t + εs)ρ(s)ds∀t ∈ R.(A.14)
Since estimate (A.6) holds true with H0:= L(H + 1), from statements (1) and (2)
of Lemma A.3 we deduce that
ν ≤ cε(t) ≤ m(|A1/2u0|2) + 1∀t ∈ R, ∀ε > 0,(A.15)
27
Page 30
|cε(t) − c(t)| ≤ γ0L(H + 1)ω(ε)
|c′
Let us consider the Fourier components uk(t) of u(t), and let us set
∀t ∈ R, ∀ε > 0,(A.16)
ε(t)| ≤ γ0L(H + 1)ω(ε)
ε
∀t ∈ R, ∀ε > 0. (A.17)
Ek,ε(t) := |u′
k(t)|2+ λ2
kcε(t)|uk(t)|2.(A.18)
An easy computation shows that
E′
k,ε(t) = c′
ε(t)λ2
|c′
k|uk(t)|2+ 2λ2
ε(t)|
k(cε(t) − c(t))uk(t)u′
k(t)
≤
cε(t)Ek,ε(t) + λk|cε(t) − c(t)|
?cε(t)
Ek,ε(t),
hence by (A.15), (A.16), and (A.17) we obtain that
E′
k,ε(t) ≤ γ0L(H + 1)
?1
ν
ω(ε)
ε
+
1
√νλkω(ε)
?
Ek,ε(t)∀t ∈ [0,S].(A.19)
Let us consider now the eigenvalues λk> 0, and let us set εk:= 1/λk. By (A.8) we
have that
ω(εk)
εk
hence by (A.19)
= λkω(εk) = λkω
?1
λk
?
≤ Λϕ(λk),
E′
k,εk(t) ≤ γ0L(H + 1)
?1
ν+
1
√ν
?
Λϕ(λk)Ek,εk(t) = Rϕ(λk)Ek,εk(t).
Integrating this differential inequality we find that
Ek,εk(t) ≤ Ek,εk(0)exp(Rϕ(λk)t)
Combining this inequality with (A.15) we obtain that
∀t ∈ [0,S].
|u′
k(t)|2+ λ2
k|uk(t)|2
≤ max?1,ν−1?Ek,εk(t)
= γ1
?|u1k|2+ λ2
≤ max?1,ν−1?· (c(0) + 1)?|u1k|2+ λ2
k|u0k|2?exp(Rtϕ(λk))
k|u0k|2?exp(Rtϕ(λk)), (A.20)
where u0kand u1kdenote the Fourier components of u0and u1, respectively.
Recalling that RT ≤ r0, and the definition of H, we have that
|A1/4u′(t)|2+ |A3/4u(t)|2
=
?
k
λk
?|u′
λk
k(t)|2+ λ2
k|uk(t)|2?
k|u0k|2?exp(r0ϕ(λk))
ϕ,r0,1/4+ ?u0?2
≤ γ1
?
??u1?2
28
k
?|u1k|2+ λ2
= γ1
ϕ,r0,3/4
?< H,
Page 31
for every t ∈ [0,S]. This inequality with t = S contradicts (A.11) and thus proves (A.2).
A sufficient condition in order to prove (A.3) is that
E(τ) :=
?
k
λkexp((r0− Rτ)ϕ(λk)) sup
t∈[0,τ]
?|u′
k(t)|2+ λ2
k|uk(t)|2?< +∞
for every τ ∈ (0,T]. On the other hand from (A.20) we have that
?
= γ1
??u1?2
A priori estimates in the weakly hyperbolic case
appearing in (A.7) and in Lemma A.3. Let Λ be such that
E(τ) ≤ γ1
k
λk
?|u1k|2+ λ2
ϕ,r0,1/4+ ?u0?2
k|u0k|2?exp(r0ϕ(λk))
ϕ,r0,3/4
?.
Let L and γ0be the constants
σ ≤ Λϕ
?
σ
?ω(1/σ)
?
∀σ > 0.(A.21)
Let us set
γ2:= 1 +
1
ω(1),γ3:= ω(1) + max
?
ω(σ) : 0 ≤ σ
?
ω(σ) ≤ 1
?
,
γ4:= 2γ2
?γ3+ m(|A1/2u0|2) + 1?,
???u1?2
γ6:= 2γ5max
H := γ4
?
1 +2Λ
r0
ϕ,r0,1/4+ ?u0?2
ϕ,r0,3/4
?+ 1,
?
γ5:= γ0L(H + 1) + 1,
?
Λ,1 +?ω(1),R := 2γ6.
Let T > 0 be such that
ω(T) ≤
1
L(H + 1),T ≤
r0
2γ6. (A.22)
We claim that in the weakly hyperbolic case (A.2) and (A.3) hold true with these
new values of T, H, and R. To this end we define once again S according to (A.10),
and we assume by contradiction that S < T, so that (A.11) holds true.
Let us consider the function c(t) defined according to (A.13), let us extend it outside
the interval [0,S] as in Lemma A.3, and let us set
cε(t) := ω(ε) +
?
R
c(t + εs)ρ(s)ds∀t ∈ R.(A.23)
Arguing as in the strictly hyperbolic case we find that
|c(t) − c(s)| ≤ L(H + 1)ω(|t − s|)
29
Page 32
for every t and s in [0,S]. This estimate, together with the first inequality in (A.22),
gives that
c(t) ≤ c(0) + L(H + 1)ω(t) ≤ m(|A1/2u0|2) + 1
Moreover from statement (2) of Lemma A.3 we deduce that
∀t ∈ [0,S].(A.24)
|cε(t) − c(t)| ≤ (1 + γ0L(H + 1))ω(ε) = γ5ω(ε),
ε(t)| ≤ γ0L(H + 1)ω(ε)
Let us consider the Fourier components uk(t) of u(t), and let us define Ek,ε(t) as
in (A.18). Computing the time derivative as in the strictly hyperbolic case, and using
(A.25), (A.26), and the fact that cε(t) ≥ ω(ε), we find that
?1
Now we choose ε as a function of k. If λk< 1 we set εk= 1, and we have that
(A.25)
|c′
ε
≤ γ5ω(ε)
ε
. (A.26)
E′
k,ε(t) ≤ γ5
ε+ λk
?
ω(ε)
?
Ek,ε(t)∀t ∈ [0,S]. (A.27)
γ5
?1
εk
+ λk
?
ω(εk)
?
≤ γ5
?
1 +
?
ω(1)
?
≤ γ6ϕ(λk),
where we exploited that ϕ(σ) ≥ 1 for all σ ≥ 0.
If λk≥ 1 we consider the function h(σ) = σ?ω(σ), which is invertible, and we set
ω(εk) =1
εk
εk:= h−1(1/λk). By (A.21) we have that
λk
?
≤ Λϕ
?
1
h(εk)
?
= Λϕ(λk),(A.28)
hence
γ5
?1
εk
+ λk
?
ω(εk)
?
≤ γ6ϕ(λk).
Going back to (A.27), in both cases we have that E′
t ∈ [0,S]. Integrating this differential inequality we obtain that
Ek,εk(t) ≤ Ek,εk(0)exp(γ6ϕ(λk)t)
In order to estimate Ek,εk(0) we need an estimate on cεk(0). To this end we first
observe that
ω(εk) ≤ γ3
Indeed this estimate is trivial if λk< 1, while for λk≥ 1 it follows from the fact that
h(εk) = 1/λk≤ 1. Thanks to (A.23), (A.24), and (A.30) we thus obtain that
cε(0) ≤ γ3+ m(|A1/2u0|2) + 1,
k,εk(t) ≤ γ6ϕ(λk)Ek,εk(t) for every
∀t ∈ [0,S].(A.29)
∀k ∈ N.(A.30)
30
Page 33
hence
Ek,εk(0) ≤?γ3+ m(|A1/2u0|2) + 1??|u1k|2+ λ2
Moreover from (A.5) and (A.28) it follows that
k|u0k|2?.(A.31)
max
?
1,
1
ω(εk)
?
≤ 1 +
1
ω(εk)≤ γ2
?
1 +1
εk
?
≤ 2γ2(1 + Λϕ(λk)), (A.32)
both for λk< 1 and for λk≥ 1. From (A.29), (A.31), and (A.32) it follows that
?
≤ γ4(1 + Λϕ(λk))?|u1k|2+ λ2
We are now ready to prove (A.2). We consider the inequality
|u′
k(t)|2+ λ2
k|uk(t)|2≤ max1,
1
ω(εk)
?
Ek,εk(t)
k|u0k|2?exp(γ6ϕ(λk)t). (A.33)
1 + Λx ≤
?
1 +Λ
M
?
exp(Mx)∀Λ ≥ 0, ∀x ≥ 0, ∀M > 0,(A.34)
and we apply it with x = ϕ(λk), M = r0/2. Combined with (A.33) and the fact that
γ6T ≤ r0/2 we obtain that
?
Multiplying by λk, and summing over all k’s, we obtain the same contradiction as in
the strictly hyperbolic case.
The proof of (A.3) is quite similar. We fix τ ∈ (0,T] and we apply inequality (A.34)
with x = ϕ(λk), M = γ6τ. Combined with (A.33) we obtain that
|u′
k(t)|2+ λ2
k|uk(t)|2≤ γ4
1 +2Λ
r0
??|u1k|2+ λ2
k|u0k|2?exp(r0ϕ(λk)).
exp((r0− Rτ)ϕ(λk)) sup
t∈[0,τ]
?
?|u′
Λ
γ6τ
k(t)|2+ λ2
k|uk(t)|2?
k|u0k|2?exp(r0ϕ(λk))
≤
≤ γ4
1 +
??|u1k|2+ λ2
for every t ∈ [0,τ]. Multiplying by λk, and summing over all k’s, we conclude as in the
strictly hyperbolic case.
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