Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2009, Article ID 182371, 15 pages
doi:10.1155/2009/182371
Review Article
Well-Posedness of the Cauchy Problem for
Hyperbolic Equations with Non-Lipschitz
Coefficients
Akbar B. Aliev and Gulnara D. Shukurova
Baku State University, Academic Zahid Xalilov str., 23, AZ 1148 Baku, Azerbaijan
Correspondence should be addressed to Akbar B. Aliev, alievagil@yahoo.com
Received 12 March 2009; Accepted 16 May 2009
Recommended by Pavel Sobolevskii
We consider hyperbolic equations with anisotropic elliptic part and some non-Lipschitz
coefficients. We prove well-posedness of the corresponding Cauchy problem in some functional
spaces. These functional spaces have finite smoothness with respect to variables corresponding to
regular coefficients and infinite smoothness with respect to variables corresponding to singular
coefficients.
Copyright q 2009 A. B. Aliev and G. D. Shukurova. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Let us consider the Cauchy problem for a second-order hyperbolic equation:
ü −
n
aij tuxi xj i,j1
n
bj tuxj ctu 0,
t, x ∈ 0, T × Rn ,
1.1
j1
u0, x u0 x,
ut 0, x u1 x,
x ∈ Rn ,
1.2
where the matrix aij t is real and symmetric for all t ∈ 0, T , ü utt .
Suppose that 1.1 is strictly hyperbolic, that is, there exists λ0 > 0 such that
at, ξ ≡
n
aij t
i,j1
for all t, ξ ∈ 0, T × Rn \ {0}.
ξi ξj
|ξ|2
≥ λ0 > 0,
1.3
2
Abstract and Applied Analysis
It is known that if at, ξ satisfies the Lipschitz condition and bj t, ct ∈ L∞ 0, T , j 1, 2, . . . , n, then for any u0 ∈ H s Rn , u1 ∈ H s−1 Rn the problem 1.1, 1.2 has a unique
solution
u· ∈ C0, T ; H s Rn ∩ C1 0, T , H s−1 Rn ,
1.4
where s ≥ 1 see 1, Chapter 5 and 2, Chapter 3 .
If we reject the Lipschitz condition, this result, generally speaking, stops to be valid
see 3.
In the paper 4 it is proved that if at, ξ ∈ LLω 0, T , that is, if at, ξ satisfies the
logarithmic Lipschitz condition:
|at τ, ξ − at, ξ| ≤ c|τ| · log|τ| · ω|τ|,
1.5
where ω|τ| monotonically decreasing tends to zero, and log |τ| · ω|τ| tends to infinity, then
there exists δ > 0 such that, for all u0 ∈ H s Rn , u1 ∈ H s−1 Rn the problem 1.1, 1.2 has
a unique solution u ∈ C0, T ; H s−δ Rn ∩ C1 0, T , H s−1−δ Rn this behavior goes under
the name of loss of derivatives.
j, a part of coefficients
In the paper 5 it is considered the case when ai,j t 0, i /
belongs to the class LLω 0, T , and another part of coefficients satisfies the Lipschitz
condition. It is proved that the loss of derivatives occurs in those variables xk for which
appropriate coefficient akk t belongs to the class LLω 0, T .
It is interesting to investigate the Cauchy problem for 1.1, with singular coefficients.
Many interesting results have been obtained in this direction. For example, in the paper 6 it
is supposed that for each ξ ∈ Rn \ {0}at, ξ ∈ C1 0, T and
tq |ȧt, ξ| ≤ c,
t, ξ ∈ 0, T × Rn \ {0},
1.6
where q ≥ 1, c > 0. It is proved that if q 1, the problem 1.1, 1.2 is well-posed in C∞ Rn .
If q > 1 and
tp |at, ξ| ≤ c,
t, ξ ∈ 0, T × Rn \ {0},
1.7
where p ∈ 0, 1 ∩ 0, q − 1, then the problem 1.1, 1.2 is well-posed in the Geverey class
γ s Rn , s < q − p/q − 1 see 6. If the coefficients aij t satisfy only Holder conditions of
order α < 1 then in 3 it is established that the problem 1.1, 1.2 is γ s well-posed for all
s < 1/1 − α. In this direction see also the results obtained in the papers 6–13.
In this paper we consider the Cauchy problem for a higher-order hyperbolic equation
with anisotropic elliptic part:
ü n
bα tDxα u 0,
−1lk ak tDx2lkk u k1
t, x ∈ 0, T × Rn ,
|α:l|≤1
u0, x u0 x,
ut 0, x u1 x,
x ∈ Rn ,
where lk ∈ N, {1, 2, . . . , }, αk ∈ N ∪ {0}, k 1, 2, . . . , n, |α : l| α1 /l1 · · · αn /ln .
1.8
Abstract and Applied Analysis
3
Here the coefficients ak t satisfy different conditions of type 1.6 and 1.7, so that qk
and pk corresponding to different k are different. The smoothness of the solution depending
on smoothness on initial data with respect to each variable xk depends not only on lk but also
on qk and pk .
2. Statement of the Problem and Results
We considered the Cauchy problem 1.8. Suppose that ak t and bα t satisfy the following
conditions:
ak t ≥ a > 0,
t ∈ 0, T , k 1, 2, . . . , n,
2.1
tqk |ȧk t| ≤ c,
t ∈ 0, T , k 1, 2, . . . , n,
2.2
bα t ∈ L∞ 0, T ,
|α : l| ≤ 1.
2.3
In order to formulate the basic results we introduce some denotation. Let H be some
Hilbert space. By W2λ,L Rm , H we will denote a functional space with the norm
uW λ,L Rm ,H
2
⎡
⎣
Rm
m
1 ηk2Lk
k1
λ
⎤1/2
2
u
η H dη⎦ ,
2.4
where L L1 , . . . , Lm , Lj ∈ N, j 1, 2, . . . , m, λ ≥ 0, and u
η Fx uη; Fx is a Fourier
n
transformation with respect to variable x ∈ R .
For s ≥ 1 by γβs,L Rm , H we will denote a functional space with the norm
⎧ ⎡
⎤1/2
1/s ⎫
m
⎨ ⎬ 2
u
η H dη⎦ .
⎣ exp β ηk2 ⎩ k1 ⎭
uγ s,L Rm ,H
β
2.5
Denote W2λ,L Rm , R W2λ,L Rm , γβs,L Rm , R γβs,L Rm ,
C∞ Rm ; H W2λ,L Rm ; H,
λ≥0
γ s Rm ; H s
γβ Rm ; H.
2.6
β≥0
If L 1, . . . , 1 then W2λ,L Rm , H H λ Rm ; H, γβs,L Rm , H γβs Rm , H, and
s
s
γβs,L Rm , R γβ , where γβ
W2λ,L Rm , H
is the Geverey space of order s see 12, 13. If λ ∈ H then
λL ,...,λL m
is Hilbert-valued anisotropic Sobolev space W2 1
Rm ; H. For the read
valued functions the anisotropic Sobolev spaces are stated in 14. The basic results led in
14 are also valid for abstract-valued functions.
4
Abstract and Applied Analysis
We introduce also the following denotation:
|ξ| x
x1 , . . . , xn1 ,
x
xn1 1 , . . . , xn ,
ξ
ξ1 , . . . , ξn1 ,
ξ
ξn1 1 , . . . , ξn ,
l
l1 , . . . , ln1 ,
l
ln1 1 , . . . , ln ,
n
ξk2lk ,
k1
n1
ξ ξk2lk ,
l
k1
n
ξ ξk2lk ,
l
2.7
n2 n − n1 .
kn1 1
The main results are the following theorems.
Theorem 2.1. Let the conditions 2.1–2.3 be satisfied, where
qk ∈ 0, 1,
qk 1,
for k 1, 2, . . . , n1 ,
for k n1 1, . . . , n.
2.8
2.9
Then for any λ
≥ 0, λ
≥ 0 the energy estimates
E t, λ
, λ
≤ ME 0, λ
, λ
λ0 ,
2.10
hold, where M and λ0 are some constants indepent of t ∈ 0, T ,
λ
λ
λ 1 ξ
l
1 ξ
l
|v̇t, ξ|2 1 |ξ|l |vt, ξ|2 dξ,
E t, λ , λ λ Rn
λ ≥ 0,
∂vt, ξ
.
v̇t, ξ ∂t
2.11
Theorem 2.2. Let the conditions 2.1–2.3 be satisfied, where
qk ∈ 0, 1,
qk q > 1,
for k 1, 2, . . . , n1 ,
2.12
for k n1 1, . . . , n.
2.13
Additionally, let the conditions
tp |ak t| ≤ c,
t ∈ 0, T ,
for k n1 1, . . . , n.
2.14
be satisfied, where p ∈ 0, 1 ∩ 0, q − 1. Then for any β > 0, λ
≥ 0, and 1 ≤ s < q − p/q − 1 the
energy estimates,
E t, β, s, λ
≤ ME 0, β δ, s, λ
,
2.15
Abstract and Applied Analysis
5
hold, where M and δ are some constants independent of t ∈ 0, T ,
E t, β, s, λ Rn
λ 1/s
exp βξ
l
1 ξ
l
|v̇t, ξ|2 1 |ξ|l |vt, ξ|2 dξ.
2.16
Remark 2.3. It is clear by our notation that
E t, λ
, λ
≤ u̇t, ·W λ
,l
Rn2 ;W λ
1,l
Rn1 ut, ·W λ
,l
Rn2 ;W λ
1,l
Rn1 x
2
x
2
2
x
2
x
ut, ·W λ
1,l
Rn2 ;W λ
,l
Rn1 2
≤ 2E λ
, λ
, t ,
x
2.17
x
2
and we can write
E t, β, s, λ
ut, ·γ s,l
Rn2 ;W λ
,l
Rn1 .
β
x
2
2.18
x
Remark 2.4. It is possible to replace the conditions a1 t, . . . , an1 t ∈ C1 0, T and 2.8 or
2.12 by Lipschitz conditions.
The following theorems are obtained from Theorems 2.1 and 2.2.
Theorem 2.5. Let condition 2.1–2.9 be satisfied. Then for any s ≥ 0, u0 ∈ C∞ Rnx2
; W2s1,l Rnx1
,
u1 ∈ C∞ Rnx2
; W2s,l Rnx1
the problem 1.1, 1.2 admits a unique solution
∩ C1 0, T ; C∞ Rnx2
; W2s,l Rnx1
.
u ∈ C 0, T ; C∞ Rnx2
; W2s1,l Rnx1
2.19
Theorem 2.6. Let conditions 2.1–2.3 and 2.12–2.14 be satisfied. Then for any s
≥ 0, 1 ≤
s
< q − p/q − 1, u0 ∈ γ s Rnx2
; W2s 1,l Rnx1
, u1 ∈ γ s Rnx2
; W2s ,l Rnx1
the problem 1.1, 1.2
admits a unique solution
∩ C1 0, T ; γ s Rnx2
; W2s ,l Rnx1
.
u ∈ C 0, T ; γ s Rnx2
; W2s 1,l Rnx1
2.20
In particular it follows from Theorem 2.1 that if the conditions 2.1–2.3 are satisfied,
then the problem 1.1, 1.2 is well-posed in C∞ Rn , and if the conditions 2.1–2.3 and
2.12–2.14 are satisfied then the problem 1.1, 1.2 is well-posed in the Geverey class γ s .
3. Proof of Theorems
At first we reduce some auxiliary statements.
We denote vt, ξ Fx ut, ξ and define the weighted energetic function in the
following way:
Φt Φ t, ξ, λ
, λ
, β, r |v̇t, ξ|2 1 ξ
l
d t, ξ
|vt, ξ|2 · Ht, ξ,
3.1
6
Abstract and Applied Analysis
where
Ht, ξ H t, ξ, λ
, λ
, β, r
λ λ
1 ξ
l
1 ξ
l
⎤
⎡
t
q−1/r
⎦,
× exp⎣− α τ, ξ dτ βξ l
0
r
⎧ ⎨s q − 1 ,
for q > 1,
⎩1,
for q 1,
⎧ n
⎪
⎪
⎪
ak T ξk2lk ,
⎪
⎪
⎪
⎪
kn1 1
⎪
⎪
⎪ n
⎨ ξk2lk ,
ak |ξ
|−1/r
d t, ξ
l
⎪
⎪
kn1 1
⎪
⎪
⎪
n
⎪
⎪
⎪ a tξ2lk ,
⎪
⎪
k
k
⎩
kn1 1
λ
≥ 0, λ
≥ 0, β > 0,
for T r |ξ
|l
≤ 1,
3.2
for T r |ξ
|l
> 1, tr |ξ
|l
≤ 1,
for tr |ξ
|l
> 1,
⎧
n
⎪
2l
⎪
k
⎪
a
−
dt,
ξ
tξ
,
k
⎪
k
⎪
⎪
kn1 1
⎨
α t, ξ
n
2l ⎪
⎪
kn1 1 ȧk tξk k ⎪
⎪
⎪
,
⎪
⎩ n
2lk
kn1 1 ak tξk
for tr |ξ
|l
≤ 1,
for tr |ξ
|l
> 1.
The following auxiliary lemmas are proved similar to the paper 6. The proofs of the
lemmas are in appendix.
Lemma 3.1. If qk 1, k n1 1, . . . , n, then there exits such c1 > 0, c2 > 0, that
aξ
l
≤ d t, ξ
≤ c1 c2 ln 1 ξ
l
ξ
l
.
3.3
If qk > 1, k n1 1, . . . , n, then there exits such c1 > 0, c2 > 0, that
p/r aξ
l
≤ d t, ξ
≤ c1 c2 ξ
l
ξ
.
l
3.4
Lemma 3.2. If qk 1, k 1, 2, . . . , n1 , then there exits such constant c3 > 0, γ > 0, that
t
ατ, ξdτ ≤ c3 c4 ln1 |ξ
|l
.
0
If qk > 1, k 1, 2, . . . , n1 then there exits such c3 > 0, c4 > 0, that
t
0
q−1/r
ατ, ξdξ ≤ c3 c4 ξ
l
.
3.5
Abstract and Applied Analysis
7
By the definition of Φt Φt, ξ, λ
, λ
, β, r we have
dΦt
2Re v̈t, ξv̇t, ξ 1 ξ
d t, ξ
vt, ξv̇t, ξ Ht, ξ
dt
ḋ t, ξ
|vt, ξ|2 Ht, ξ − αt, ξΦt.
3.6
On the other hand from 1.8 we have
v̈t, ξ n
ak t ξk2k vt, ξ bα tiξα vt, ξ 0,
|α:|≤1
k1
v0, ξ v0 ξ,
v̇0, ξ v1 ξ,
3.7
3.8
where v0 ξ Fu0 ξ, v1 ξ Fu1 ξ, v̈t, ξ ∂2 vt, ξ/∂t2 .
From 3.6 and 3.7 we obtain
"
! n
n
1
dΦt
2k
2k
2Re −
ak t ξk 1 ξ d t, ξ −
ak t ξk
dt
k1
kn 1
× vt, ξv̇t, ξHt, ξ − 2Re
1
bα tiξα vt, ξv̇t, ξHt, ξ
3.9
|α:|≤1
ḋ t, ξ |vt, ξ|2 Ht, ξ − αt, ξΦt.
If tr |ξ
| < 1, then by definition of dt, ξ and αt, ξ
we have
"
! n
1
dΦt
2k
2Re − ak t ξk 1 ξ αt, ξ vt, ξv̇t, ξHt, ξ
dt
k1
− 2Re
3.10
α
bα tiξ vt, ξv̇t, ξHt, ξ − αt, ξΦt.
|α:|≤1
By our supposition qk < 1 for k 1, 2, . . . , n1 . Therefore we can easily see that
a ≤ ak t ≤ aT ,
with some constant aT > a.
k 1, 2, . . . , n1
3.11
8
Abstract and Applied Analysis
Using the Cauchy inequality, definition of αt, ξ, Ht, ξ, and ϕt we have
3.12
2Reαt, ξvt, ξv̇t, ξHt, ξ − αt, ξΦt ≤ 0,
bα tiξα vt, ξv̇t, ξHt, ξ
2Re
|α:|≤1
≤ 2bT
|ξα ||vt, ξ| · |v̇t, ξ| · Ht, ξ
3.13
|α:l|≤1
!
n
1 |ξk |2lk
≤ 2bT c5
"
2
2
|vt, ξ| |v̇t, ξ| Ht, ξ,
k1
where bT sup|α:l|≤1 bα tL∞ 0,T , c5 supξ∈Rn |α:|≤1 |ξα |2 / nk1 |ξk |2lk 1.
From 3.10–3.13 we get that when tr |ξ
|l
< 1, then there exists such a constant
M1 > 0, that
dΦt
≤ M1 Φt.
dt
3.14
If tr |ξ
|l
≥ 1 then by definition of dt, ξ and αt, ξ
from 3.9 we have that
⎤
⎡
n1
dΦt
2
α
k
2Re⎣− ak tξk −
bα tiξ vt, ξv̇t, ξ⎦Ht, ξ
dt
k1
|α:|≤1
n
2 kn1 1 ȧk tξk k ȧk t ξk2k |vt, ξ|2 Ht, ξ − n
Φt.
2k
kn1 1
kn1 1 ak tξk
3.15
n
On the other hand
n
2 kn1 1 ȧk tξk k ȧk t ξk2k |vt, ξ|2 Ht, ξ − n
Φt
2k
kn1 1
kn1 1 ak tξk
n
2 n
kn1 1 ȧk tξk k ȧk tξk2k |vt, ξ|2 Ht, ξ − n
2k
kn1 1
kn1 1 ak tξk
"
!
n
2
2k
2
2
× |v̇t, ξ| 1 ξ ak tξ
|vt, ξ| Ht, ξ ≤ 0.
n
l
kn1 1
3.16
k
From 3.13, 3.15, and 3.16 we again get inequality 3.14.
It follows from 3.14 that
Φt ≤ MΦ0,
where M M1 eT .
t ∈ 0, T ,
3.17
Abstract and Applied Analysis
9
Proof of Theorem 2.1. Let qk q 1, k n1 1, . . . , n, then r 1. From Lemmas 3.1 and 3.2 we
have
λ λ
Φ t, ξ, λ
, λ
, 0, 1 dξ ≤
1 ξ
1 ξ
Rn
Rn
2 × |v̇t, ξ|2 1 ξ
l
ξ
l
c1 c2 ln 1 ξ
l
|vt, ξ|2
× exp c3 c4 ln 1 ξ
l
dξ
λ λ
c2 ≤ c6
1 ξ
1 ξ
|v̇t, ξ|2 1 |ξ| |vt, ξ|2 dξ
Rn
c6 E t, λ
, λ
λ0 ,
3.18
where λ0 c4 1, c6 max{1, c1 ec3 , c2 ec3 }.
Thus,
Φ t, ξ, λ
, λ
, 0, 1 dξ ≤ c6 E t, λ
, λ
λ0 .
3.19
Rn
On the other hand from the definition of Φ and E we have
Φ t, ξ, λ
, λ
, 0, 1 dξ ≥ c7 E t, λ
, λ
.
3.20
Rn
It follows from 3.17–3.20 that
E t, λ
, λ
≤ c8 E 0, λ
, λ
d .
3.21
Proof of Theorem 2.2. Let qk q > 1, k n1 1, . . . , n, then r q − 1s. Taking into account
Lemmas 3.1 and 3.2 and Theorem 2.5 we have
Φ t, ξ, λ
, 0, β, r dξ
Rn
p/21 λ 1 ξ
|vt, ξ|2 1 ξ
l
c1 ξ
l
c2 ξ
≤
Rn
3.22
× |vt, ξ|2 exp c3 β c4 |ξ|q−1/2 dξ.
Further using the inequality ηp/21 ≤ c9 expcη1/s we obtain
Φ t, ξ, λ
, 0, β, r dξ ≤ c10 E t, λ
, s, β δ ,
Rn
where δ c4 c.
3.23
10
Abstract and Applied Analysis
On the other hand from the definition of φ and E we have
Φ t, ξ, λ
, 0, β, r dξ ≥ c11 E t, λ
, s, β .
3.24
Rn
From inequalities 3.17, 3.23, and 3.24 it follows that
E t, λ
, s, β ≤ c12 E 0, λ
, s, β d .
3.25
Proof of Theorem 2.5. For any ξ ∈ Rn the problem 3.7, 3.8 has a unique solution vt, ξ ∈
C1 0, T see 15, Chapter I.
Let u0 ∈ C∞ Rnx2
; W2λ 1,l Rnx1
, u1 ∈ C∞ Rnx2
; W2λ ,l Rnx1
, then for any s ≥ 0, λ
≥ 0,
E 0, λ
, s λ0 ≤ cs,λ
,
3.26
where cs,λ > 0 is some constant dependent on s ≥ 0 and λ
≥ 0.
Taking into account Theorem 2.1 it follows from 3.20 that
E t, λ
, s ≤ Mcλ
,s ,
t ∈ 0, T ,
3.27
that is,
u̇t, ·W s,l
Rn2 ;W λ
,l
Rn1 ut, ·W s1,l
Rn2 ;W λ
,l
Rn1 2
x
2
x
2
x
ut, ·W s,l
Rn2 ;W λ
1,l
Rn1 ≤ Mcλ
,s ,
2
x
2
x
2
x
t ∈ 0, T .
3.28
It follows from 3.28 that
,
u ∈ C 0, T ; C∞ Rnx2
; W2λ 1,l Rnx1
.
u̇ ∈ C 0, T ; C∞ Rnx2
; W2λ ,l Rnx1
1.8.
3.29
By the expression of ut, x it follows that the function ut, x is the solution of problem
The uniqueness of the solution is proved by standard method.
The proof of Theorem 2.6 is carried out in the similar way.
Abstract and Applied Analysis
11
Appendices
A. Proof of Lemmas
Proof of Lemma 3.1. Let qk 1, k n1 1, . . . , n. Then from 2.2 we have
ak t ≤ ak T |ak t − ak T |
T
≤ ak T |ȧk s|ds
t
≤ ak T c ln
A.1
T
t
$
#
1
.
≤ c1 c2 ln 1 t
It follows from 2.1 and 2.14 that
aξ
l
≤ d t, ξ
.
A.2
By definition of dt, ξ
for T |ξ
|l
≤ 1 we have
n
ak T ξk2lk ≤ c1 ξ
l
.
d t, ξ
A.3
kn1
If T |ξ
|l
> 1, and t|ξ
|l
< 1, then from A.1 we have
n
−1
d t, ξ
ak ξ
l
ξk2lk
kn1
!
≤ c1 c2 ln 1 "
1
|ξ
|−1
l
n
kn1
ξk2lk
A.4
c1 c2 ln 1 ξ
l
ξ
l
.
If t|ξ
|l
> 1, then using A.1 we get
n
ak tξk2lk
d t, ξ
kn1 1
$&
#
1 ξ l
≤ c1 c2 ln 1 t
≤ c1 c2 ln 1 ξ
ξ
.
%
l
l
Consequently if q 1, the statement of the lemma follows from A.2–A.5.
A.5
12
Abstract and Applied Analysis
Let qk > 1, k n1 1, . . . , n. By definition of dt, ξ
for T r |ξ
|l
< 1 we have
d t, ξ
≤ c1 ξ
l
.
A.6
If T r |ξ
|l
> 1 and tr |ξ
| < 1 then
n
−1/r
d t, ξ
ξk2lk
ak |ξ
|l
kn1 1
≤
n
M
kn1 1
|ξ
|−1/r
l
A.7
2l
p ξk k
1p/r
Mξ
l
.
If tr |ξ
| > 1 then
n
ak tξk2lk
d t, ξ
kn1 1
≤
n
M 2lk
ξ
tp k
kn 1
A.8
1
p/r
Mξ
l
· ξ
1p/r
Mξ
.
Thus if qk > 1, k n1 1, . . . , n then the statement of the lemma follows from A.2,
A.6, and A.8.
The lemma is proved.
Proof of Lemma 3.2. At first we consider the case when qk 1, k n1 1, . . . , n. If T |ξ
| ≤ 1,
then
t
0
α τ, ξ
dτ ≤
T
α τ, ξ
dτ
0
T n
n
2k
2k ak T ξk −
ak τξk dτ
≤ 0 kn
kn
1
≤
n
kn1
≤T·
ξk2k
T
1
A.9
|ak T − ak τ|dτ
0
max ak T ξ
ξk
kn1 1,...,n
T
ak τdτ
0
≤ aT ,
where aT maxkn1 1,...,n ak τ 1/T maxkn1 1,...,n
T
a τdτ
0 k
< ∞.
Abstract and Applied Analysis
13
If T |ξ
| > 1, then
t
α τ, ξ
ds ≤
|ξ
|−1
l
0
α s, ξ
dτ 0
T
|ξ
|−1
α s, ξ
ds
T n ȧk τξ2k |ξ
|−1
n
l
kn1
k
≤
ak τξk2lk dτ dτ
d τ, ξ −
n
−1
0
|ξ | ak τξ2k
kn
1
≤
|ξ
|−1
l
0
≤
kn1
0
k
T
|ξ
|−1
n
n
l
dτ
c
d τ, ξ
dτ ξk2lk ·
αk τdτ ξk2k
−1 τ
a
0
|ξ | kn
kn
1
|ξ
|−1
1
A.10
c1 c2 n 1 ξ
ξ
dτ
n
kn1
ξk2k · c
|ξ
|−1
0
n
c
T
n dτ τ
a kn
T
1
|ξ
|−1
dτ
τ
|ξ
|−1
−1 c T dτ
T
n dτ c1 c2 n 1 ξ c ξ ·
τ
a |ξ
|−1
τ
0
≤ c3 c4 n 1 ξ
.
Now let us consider the case qk > 1, k n1 1, . . . , n. In this case r q − 1s. If
T r |ξ
| ≤ 1, then
t
α τ, ξ
dτ ≤
0
T
α τ, ξ
dτ
0
≤
T
n
kn1 1
≤
0
|ak T − ak τ|ξk2k dτ
max ak T T ξ
kn1 1,...,n
≤ aT · T 1−r c ·
T
0
A.11
cτ −p dτ ξ
1
c
T 1−p ξ
≤ aT T 1−r T 1−p−r .
1−p
1−p
14
Abstract and Applied Analysis
If T r |ξ
| > 1, then
t
α τ, ξ
dτ ≤
|ξ
|−1/r
0
ατ, ξdτ T
|ξ
|−1/r
0
α τ, ξ
dτ
T
|ξ
|−1/r
n
2lk ak τξk dτ α τ, ξ
dτ
≤
dτ, ξ −
0
|ξ
|−1/r
kn 1
1
≤
n
−1/r
ak ξ
ξk2k
|ξ
|−1/r
l
dτ 0
kn1 1
n
ξk2k
|ξ
|−1/r
kn1 1
ak τdτ
0
n
2 kn1 1 ȧk τξk k dτ
n
2k
|ξ
|−1/r
a
τξ
k
kn1 1
k
T
≤
c
|ξ
|−1/r
p · ξ
·
|ξ
|−1/r
0
dτ cξ
·
p/r1 −1/r
≤ cξ
· ξ cξ
·
A.12
|ξ
|−1/r
0
dτ
c
dτ p
τ
a
T
|ξ
|−1/r
dτ
τq
1 −1/r 1−p
ξ
1−p
#
$
c 1
−1/r 1−q
T 1−q − ξ
a 1−q
1−1−p/r < cξ
q−1 /r
c 1−1−p/r c
ξ ξ
.
1−p
a q−1
As r q − 1s, and s < q − p/q − 1, it follows that 1 − 1 − p/s < 1/s and
q − 1/r 1/s. Then according to the Young inequality there exists such δ > 0 that
1−1−p/r 1/s
ξ ≤ c1 δ δ1 ξ
.
A.13
Thus, by A.9–A.13 the following inequality is valid:
t
α τ, ξ
dτ ≤ δ|ξ|1/s cδ ,
0
where δ δ1 a2 p/1 − p c/aq − 1cδ c1δ c2 δ/1 − p.
A.14
Abstract and Applied Analysis
15
B. Example
Let us consider the Cauchy problem in 0, T × R2 :
'
3
utt − 1 t2 uxx − 1 t2 uyy 0,
u 0, x, y ϕ1 xψ1 y ,
ut o, x, y ϕ2 xψ2 y ,
B.1
where ϕ1 x, ϕ2 x ∈ C∞ R ∩s≥0 W2s R, ψ1 y ∈ W22 R, ψ2 y ∈ W21 R, u ut, x, y.
It follows from Theorem 2.5 that the problem B.1 has a unique solution
u ∈ C 0, T ; C∞ R; W22 R ∩ C1 0, T ; C∞ R; W21 R .
B.2
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