156 M. Reissig
Step 5. Conclusion After choosing s
ξ
= st
ξ
= exp−t
1q ξ
for large t
ξ
and taking account of w
t
t, ξ =
1 2
τ
t
tτ t
−
1 2
v st, ξ
+ τ t
1 2
v
s
st, ξ we obtain |wt s, ξ| + |w
t
t s, ξ |
≤ τ t s
1 2
1 +
τ
t
t s 2τ t s
|vs, ξ| + τ t s
−
1 2
|v
s
s, ξ |
≤ 2τ t s
1 2
|vs, ξ| + τ t s
−
1 2
|v
s
s, ξ |
for large ξ . Finally, we use τ t s ∼ |ξ|. This follows from the definition λt
ξ
, ξ =
λ and lim
t
ξ
→∞
λ
2
t
ξ
= 0. Thus we have shown |ξ| | ˆus
ξ
, ξ | + | ˆu
s
s
ξ
, ξ | ≥ C
1
|ξ|
1 2
expC
2
η t
ξ
. The function ηt
= t
α
satisfies 18 if α
q −1
q
. The function t
ξ
behaves as ln |ξ|
q
. Together these relations give
|ξ| | ˆus
ξ
, ξ | + | ˆu
s
s
ξ
, ξ | ≥ C
1
|ξ|
1 2
expC
2
ln |ξ|
qα
≥ C
1
|ξ|
1 2
expC
2
ln |ξ|
γ
, where
γ ∈ 1, q − 1 .
From this inequality we conclude the statement of Theorem 10.
R
EMARK
6. The idea to apply Floquet’s theory to construct a counter-example goes back to [25] to study C
∞
well-posedness for weakly hyperbolic equations. This idea was employed in connection to L
p
− L
q
decay estimates for solutions of wave equations with time-dependent coefficients in [24]. The merit of [14] is the application
of Floquet’s theory to strictly hyperbolic Cauchy problems with non-Lipschitz coef- ficients. We underline that the assumed regularity b
∈ C
2
comes from statements of Floquet’s theory itself. An attempt to consider non-Lipschitz theory, weakly hyperbolic
theory and theory of L
p
− L
q
decay estimates for solutions of wave equations with a time-dependent coefficient is presented in [23].
5. How to weaken C
2
regularity to keep the classification of oscillations
There arises after the results of [6] and [7] the question whether there is something between the conditions
• a ∈ L
∞
[0, T ] ∩ C
1
0, T ] , |t
γ
a
′
t | ≤ C for t ∈ 0, T ] ;
21 • a ∈ L
∞
[0, T ] ∩ C
2
0, T ] , |a
k
t | ≤ C
k
1 t
ln 1
t
γ k
22 for
t ∈ 0, T ], k = 1, 2.
Hyperbolic equations with non-Lipschitz coeffi cients 157
The paper [15] is devoted to the model Cauchy problem u
t t
− at, x △ u = 0 , uT, x = ϕx, u
t
T , x = ψx ,
23 where a
= at, x ∈ L
∞
[0, T ], B
∞
R
n
and a ≤ at, x with a positive constant
a .
D
EFINITION
3. Definition of admissible space of coefficients. Let T be a positive small constant, and let γ
∈ [0, 1] and β ∈ [1, 2] be real numbers. We define the weighted spaces of H ¨older differentiable functions 3
β γ
= 3
β γ
0, T ] in the following way:
3
β γ
0, T ] = {a = at, x ∈ L
∞
[0, T ], B
k
R
n
: sup
t ∈0,T ]
katk
B
k
R
n
+ sup
t ∈0,T]
k∂
t
at k
B
k
R
n
t
−1
ln t
−1 γ
+ sup
t ∈0,T ]
k∂
t
a k
M
β− 1
[t,T ],B
k
R
n
t
−1
ln t
−1 γ
β
for all k ≥ 0} ,
where kFk
M
β− 1
I
with a closed interval I is defined by kFk
M
β− 1
I
= sup
s
1
, s
2
∈I,s
1
6=s
2
|Fs
1
− Fs
2
| |s
1
− s
2
|
β −1
. • If a satisfies 21 with γ = 1, then a ∈ 3
1
. • If a satisfies 22 with γ ∈ [0, 1], then a ∈ 3
2 γ
. D
EFINITION
4. Space of solutions. Let σ and γ be non-negative real numbers. We define the exponential-logarithmic scale H
γ ,σ
by the set of all functions f ∈ L
2
R
n
satisfying k f k
H
γ ,σ
: =
Z
R
n
| expσ lnhξi
γ
ˆ f ξ
|
2
dξ
12
∞ . In particular, we denote H
γ
= S
σ
H
γ ,σ
. T
HEOREM
11. Let γ ∈ [0, 1] and β ∈ 1, 2]. If a ∈ 3
β γ
0, T ], then the Cauchy problem 23 is well-posed in H
γ
on [0, T ], that is, there exist positive constants C
γ ,β
, σ and σ
′
with σ ≤ σ
′
such that k∇ut, u
t
t k
H
γ ,σ
≤ C
γ ,β
k∇ϕ, ψk
H
γ ,σ ′
for all t ∈ [0, T ].
R
EMARK
7. In the Cauchy problem 23 we prescribe data ϕ and ψ on the hy- perplane t
= T . It is clear from Theorem 4, that a unique solution of the backward Cauchy problem 23 exists for t
∈ 0, T ]. The statement of Theorem 11 tells us that in the case of very slow, slow or fast oscillations γ
∈ [0, 1], the solution possesses a continuous extension to t
= 0.
158 M. Reissig
O
PEN PROBLEM
5. Try to prove the next statement: If a
= at, x ∈ 3
β γ
0, T ] with γ 1 and β ∈ 1, 2, then these oscillations are
very fast oscillations The energy inequality from Theorem 11 yields the same connection between the
type of oscillations and the loss of derivatives as Theorem 8. T
HEOREM
12. Let us consider the Cauchy problem 23, where a ∈ 3
β γ
0, T ] with γ
∈ [0, 1] and β ∈ 1, 2]. The data ϕ, ψ belong to H
s +1
, H
s
, respectively. Then the following energy inequality holds:
Eut
H
s−s0
≤ CT Eu0
H
s
for all t ∈ [0, T ] ,
where • s
= 0 if γ = 0 very slow oscillations, • s
is an arbitrary small positive constant if γ ∈ 0, 1 slow oscillations,
• s is a positive constant if γ
= 1 fast oscillations. Proof of Theorem 11. The proof follows that for Theorem 8. But now the coefficient
depends on spatial variables, too. Our main goal is to present modifications to the proof of Theorem 8.
To Step 2. Symbols To given real numbers m
1
, m
2
≥ 0, we define S{m
1
, m
2
} and T
m
1
as follows: S
{m
1
, m
2
} = {a = at, x, ξ ∈ L
∞ loc
0, T , C
∞
R
2n
: |∂
τ x
∂
η ξ
at, x , ξ | ≤ C
τ,η
hξi
m
1
−|η|
1 t
ln 1
t
γ m
2
in Z
hyp
N };
T
m
1
= {a = at, x, ξ ∈ L
∞
0, T , C
∞
R
2n
: |∂
τ x
∂
η ξ
at, x , ξ | ≤ C
τ,η
hξi
m
1
−|η|
in Z
pd
N }.
Regularization Our goal is to carry out the first two steps of the diagonalization procedure because only
two steps allow us to understand a refined classification of oscillations. But the coeffi- cient a
= at, x doesn’t belong to C
2
with respect to t. For this reason we introduce a regularization a
ρ
of a. Let χ = χs ∈ B
∞
R be an even non-negative function
having its support on −1, 1. Let this function satisfy
R χ
sds = 1. Moreover, let
the function µ = µr ∈ B
∞
[0, ∞ satisfy 0 ≤ µr ≤ 1, µr = 1 for r ≥ 2 and
µ r
= 0 for r ≤ 1. We define the pseudo-differential operator a
ρ
= a
ρ
t, x , D
x
with the symbol
a
ρ
t, x , ξ = µ
t hξi
N ln hξi
γ
b
ρ
t, x , ξ |
{z }
Z
hy p
N
+ 1
− µ t
hξi N ln
hξi
γ
a |{z}
Z
pd
N
,
Hyperbolic equations with non-Lipschitz coeffi cients 159
where b
ρ
t, x , ξ = hξi
Z
R
as, x χ t − shξi
| {z
} regularization of a
ds .
L
EMMA
13. The regularization a
ρ
has the following properties: • a
ρ
t, x , ξ ≥ a
; • a
ρ
t, x , ξ ∈ S
1,0
; • ∂
t
a
ρ
t, x , ξ ∈ S{0, 1} ∩ T
−∞
; • ∂
2 t
a
ρ
t, x , ξ ∈ S{−β + 2, β} ∩ T
−∞
; • at, x − a
ρ
t, x , ξ ∈ S{−β, β} ∩ T
. To Step 4. Two steps of diagonalization procedure
We start with u
t t
− at, x △ u = 0. The vector-valued function U
= √
a
ρ
hD
x
iu, D
t
u
T
is a solution of the first order system D
t
− A − B
− R U
= 0, A
: =
√ a
ρ
hD
x
i √
a
ρ
hD
x
i ,
B :
= Op
h
D
t
a
ρ
2a
ρ
i a
− a
ρ
hD
x
i√a
ρ ♯
, where R
∈ S uniformly for all t
∈ [0, T ], that is, R = R
t, x , ξ ∈
L
∞
[0, T ], S .
First step of diagonalization, diagonalization modulo L
∞
[0, T ], S {0, 1} ∩ T
1
. Using the same diagonalizer in the form of a constant matrix we obtain from the above
system D
t
− A
1
− B
1
− R
1
U
1
= 0 , A
1
: =
√ a
ρ
hD
x
i 1
−1 ,
B
1
∈ L
∞
[0, T ], S {0, 1} ∩ T
1
, R
1
∈ L
∞
[0, T ], S .
R
EMARK
8. We can split B
1
into two parts B
10
: = Op
D
t
a
ρ
4a
ρ
1 1
1 1
, B
11
: =
1 2
a − a
ρ
hD
x
i √
a
ρ ♯
1 1
−1 −1 .
160 M. Reissig
The second part B
11
belongs to S {−β +1, β}∩ T
1
for all t ∈ [0, T ]. If β 1, then this
class is better than S {0, 1} ∩ T
1
. We need β 1 later, to understand that the influence of B
11
is not essential. This is the reason we exclude in Theorem 11 the value β = 1.
Second step of diagonalization, diagonalization modulo L
∞
[0, T ], S {−β + 1, β} ∩ T
1
+ L
∞
[0, T ], S .
We define the diagonalizer M
2
= M
2
t, x , D
x
: =
I − p
p I
, where p =
pt, x , ξ =
D
t
a
ρ
8a
ρ
√ a
ρ
hξ i
. Then a suitable transformation U
2
: = M
2
U
1
changes the above system to
D
t
− A
1
− A
2
− B
2
− R
2
U
2
= 0 , A
2
: = Op
D
t
a
ρ
4a
ρ
1 1
, B
2
∈ L
∞
[0, T ], S {−β + 1, β} ∩ T
1
, R
2
∈ L
∞
[0, T ], S .
Transformation by an elliptic pseudo-differential operator. We define M
3
= M
3
t, x , ξ : = exp
−
T
R
t D
s
a
ρ
4a
ρ
ds 1
1 . The transformation
U
2
: = M
3
U
3
gives from the last system D
t
− A
1
− B
3
− R
3
U
3
= 0, where B
3
, R
3
belong to the same symbol classes as B
2
, R
2
, respectively. R
EMARK
9. The last step corresponds to the fact from the proof of Theorem 8, that
t
R
r ∂
s
as 4as
ds depends only on a. Application of sharp G ˚arding’s inequality for matrix-valued operators.
We generalize an idea from [2] to our model problem. G
OAL
. Let us find a pseudo-differential operator θ = θt, D
x
in such a way that after transformation V t, x :
= e
−
T
R
t
θ s,D
x
ds
U
3
t, x the operator equation D
t
− A
1
− B
3
− R
3
U
3
= 0 is transformed to ∂
t
− P − P
1
V = 0, where we can show that for the
solution V of the Cauchy problem an energy estimate without loss of derivatives holds. A simple computation leads to
P + P
1
= iA
1
+ B
3
+ R
3
+ θt, D
x
I + i
e
−
T
R
t
θ s,D
x
ds
, A
+ B + R
e
T
R
t
θ s,D
x
ds
. The matrix-valued operator A
1
brings no loss of derivatives, here we feel the strict hyperbolicity. Taking account of the symbol classes for B
3
, R
3
and our strategy due to G˚arding’s inequality that θ
= θt, ξ should majorize iB
3
t, x , ξ + R
3
t, x , ξ the symbol of θ should consist at least of two parts:
Hyperbolic equations with non-Lipschitz coeffi cients 161
• a positive constant K , due to R
3
∈ L
∞
[0, T ], S ;
• K θ t, ξ :
= K µ
t hξ i
N ln hξ i
γ
1 hξ i
β− 1
1 t
ln
1 t
γ β
+K 1
− µ
t hξ i
N ln hξ i
γ
hξi, due to B
3
∈ L
∞
[0, T ], S {−β + 1, β} ∩ T
1
. It turns out that the symbol of the commutator doesn’t belong to one of these symbol
classes. For this reason we introduce a third part • K θ
1
t, ξ : = K µ
t hξ i
N ln hξ i
γ
ln
1 t
γ
+ K 1
− µ
t hξ i
N ln hξ i
γ
ln
1 t
ξ
γ
. Defining
• P = iA
1
+ B
3
+ R
3
+ K 1 + θ t, D
x
I , • P
1
= K θ
1
t, D
x
I + i
e
−
T
R
t
θ s,D
x
ds
, A
1
+ B
3
+ R
3
e
T
R
t
θ s,D
x
ds
one can show det
P + P
∗
2 t, x , ξ
≥ θ t, ξ
∈ L
∞
[0, T ], S
1 1,0
, det
P
1
+ P
∗ 1
2 t, x , ξ
≥ θ
1
t, ξ ∈ L
∞
[0, T ], S
ε ε,
. We use the sharp G˚arding’s inequality with see [19] with
• c = 0 , m = 1 , ρ = 1 , δ = 0
for P ,
• c = 0 , m = ε , ρ = ε , δ = 0
for P
1
, thus ReP
k
u, u ≥ −C
k
kuk
2 L
2
for k = 1, 2. These are the main inequalities for
proving the energy estimate kV t, ·k
2 L
2
≤ e
C T
kV T, ·k
2 L
2
for t ∈ [0, T ] .
It remains to estimate
T
R θ
s, ξ ds. This is more or less an exercise. A careful calcula- tion brings
T
R θ
s, ξ ds ≤ C lnhξi
γ
. The statements of Theorem 11 are proved.
162 M. Reissig
6. Construction of parametrix
