ON ANALYTIC SOLUTIONS TO THE NAVIER–STOKES EQUATION IN 3–D TORUS

ON ANALYTIC SOLUTIONS TO THE NAVIER–STOKES EQUATION IN 3–D TORUS

Oleg Zubelevich

(Received April 2004)

Abstract. We consider NSE with H

–initial conditions on the 3–dimensional torus and prove that there exists a solution that is analytic in all variables.

1. Introduction

The first existence and uniqueness theorems for weak solutions to Navier-Stokes equation were proven by Lere [7] in whole space R ² . Later these results were extended by Hopf [3]. In 1962 Ladyzhenskaya proved existence and uniqueness results for strong solutions for general two–dimensional domains [5].

Yudovich, Foias, Temam, Constantin, and others developed strong methods which provided deep insights into dynamics of NSE (see [19], [12], [13], [2]).

The regularity problems for NSE have been studied by many authors. J. Serrin showed that if a weak solution is from the space L ^s (0, T ), L ^r (D)
with n/r +2/s <

1, where n is the dimension of the space variables and external force is of the space L ¹ (0, T ), L ¹ (D)
(D is a domain in R ⁿ and the solution with zero boundary conditions is considered) then this solution is of C ^∞ in the spatial variables [16].

Under the same assumptions C. Kahane showed that the solution is analytic in the spatial variables [4].

It was shown by K. Masuda that if a solution is of C (0, T ), H ¹ (D)
then this solution is as regular as allowed by the external force (including C ^∞ regularity and analyticity) [8], [9].

Note that if external force and initial data are analytic in all variables, then the (local in time) existence of analytic solutions in all variables immediately follows from the result announced in the end of the paper [11].

All those regularity results study which extra hypotheses for the weak solution are sufficient for its regularity.

In the present paper we consider NSE on the 3–dimensional torus without ex- ternal force and show that under H ¹ –initial conditions there exists a solution that becomes analytic in time and spatial variables right after the beginning of the mo- tion. This theorem in the couple with some uniqueness result allows to conclude that the weak solution actually almost everywhere coincides with the analytic func- tion.

1991 Mathematics Subject Classification 76D03.

Key words and phrases: Navier–Stokes equation, Mathematical fluid, Fluid mechanics, Cauchy–

Kovalevskaya problem.

Partially supported by grants RFBR 02-01-00400, INTAS 00-221.

Note that in the case of dimension 2 the theory is clearly satisfactory and such a type results are known.

The main mathematical tools we use is the extension of the Cauchy–Weierstrass–

Kovalevskaya majorant method to the case of functional equations and Fourier series.

The author wishes to thank Ju. A. Dubinskii and A.L. Skubachevskii for useful discussions.

2. Main Theorem

Consider the Navier–Stokes equation with initial data and the condition of in- compressibility of a fluid:

v t + (v , ∇)v = −∇p + ν∆v , (1)

div v = 0, (2)

v | _t=0 = b v (x), (3)

where by v = v ¹ (t, x), . . . , v ³ (t, x)
we denote a vector–function, p(t, x) is a scalar function and ν is a positive constant. The differential operators are defined in the usual way: ∂ j = ∂/∂x j ,

∇p = (∂ 1 p, . . . , ∂ ₃ p), div v =

3

X

k=1

∂ _k v ^k ,

∆f = div∇f, (v , ∇)f =

3

X

k=1

v ^k ∂ k f.

Application of any scalar operator to a vector–function implies that this operator is applied to each component of the vector–function.

The variables x in problem (1)–(3) belong to the 3–dimensional torus: x ∈ T ³ = R ³ /(2πZ) ³ .

Due to the Galilean invariance of the equations (1)-(3) (they are not changed after the substitution x 7→ x + ct, v 7→ v + c, where c is a constant) we assume without loss of generality that the initial vector field v (x) is of zero mean value: b

Z

T

b v (x) dx = 0.

All the functions below are of zero mean value.

Let k, x ∈ C ³ , introduce some notations:

(k, x) = k 1 x 1 + . . . + k 3 x 3 , |x| = |x 1 | + . . . + |x 3 |, i ² = −1.

Sometimes we will use the Euclidian norm: |x| ² _e = |x ₁ | ² + . . . + |x ₃ | ² , and the norm

|x| m = max _k |x k |. These norms satisfy the inequalities:

|x| m ≤ |x| e ≤ |x|, |x| ≤ α|x| e ,

where a positive constant α does not depend on x. We also consider the following complex neighborhood of the torus:

T ³ r = {x | <e x ∈ T ³ , |=m x| m < r}.

So, define the set Z ³ 0 = Z ³ \{0}, and denote by H ^s (T ³ ) the Sobolev space of zero mean value functions with the norm:

f (x) = X

k∈Z

f k e ^i(k,x) , kf k ² _H

(T

) = X

k∈Z

|f k | ² |k| ^2s _e .

Drop the subscript for the H ¹ (T ³ )–norm:

k · k = k · k _H

(T

) . Let I T = [0, T ]. We use the norm

ku(t, x)k ^C = sup

t∈I

ku(t, ·)k in C I _T , H ¹ (T ³ )
.

The substitution p 7→ p + c(t) c(t) is an arbitrary function
in (1) also does not change the equation. So we will find the function p just up to an additional function of t.

Define the set

Q _r (T ) = {t ∈ C | |=m t| < <e t < T, r < <e t} × T ³

= n

(t, x) | |=m t| < <e t < T, 4α

ν |=m x| m < r < <e t, <e x ∈ T ³ o . The set Q r (T ) is open as a cross product of open sets. The set

Q(T ) = [

0<r<T

Q r (T )

= n

(t, x) | |=m t| < <e t < T, 4α

ν |=m x| m < <e t, <e x ∈ T ³ o is open as a union of open sets.

Assume that the initial vector field b v (x) belongs to the space H ¹ (T ³ ) and div v (x) = 0. We mean that the operator div consists of generalized derivatives. b Namely, for any function f (x) = P

k∈Z

f k e ^i(k,x) ∈ L ² (T ³ ) the series f _x

(x) = X

k∈Z

ik _j f _k e ^i(k,x)

possibly diverges but it defines a generalized function which value on a test function φ(x) ∈ C ^∞ (T ³ ) computes as follows:

(f x

, φ) = Z

T

f x

(x)φ(x) dx = X

k∈Z

ik j f k

Z

T

e ^i(k,x) φ(x) dx.

Denote by O(D) the space of holomorphic functions in a domain D.

Theorem 2.1. There exists a positive constant c that depends only on ν such that for

T = c

(1 + k v k) b ¹⁶ (1) initial problem (1)–(3) has a solution:

v (t, x) = X

k∈Z

v k (t)e ^i(k,x) ∈ O Q(T )
,

and

k v − v (t, ·)k → 0 b as t → 0.

Furthermore, there is a function V (t, x) = X

k∈Z

V k (t)e ^i(k,x) ∈ C I T , H ¹ (T ³ )
,

such that for |=m t| < <e t we have:

|v k (t)| ≤ V k (<e t) exp(−<e t|k| e ν/2).

(2) There exists a positive constant µ such that from k b v k ≤ µ it follows that the first part of this theorem remains valid for T = ∞.

As a corollary we have:

Proposition 2.2. Under the conditions of the second part of the theorem for any t ≥ 0 we have an estimate

kv (t, x)k ≤ ce ^−νt/2 , where a positive constant c does not depend on t.

As it was established by Leray [7] there exists a solution which belongs to L ² R + , H ¹ (T ³ )
\ L ^∞ R + , L ² (T ³ )
,

so called weak solution.

On the other hand by Serrin’s result [15] a regular solution, as long as it exists, is unique in the class of weak solutions.

Both of these theorems in conjunction with Theorem 4, imply that the weak solution is analytic in all variables except for the set of such singular values e t for which lim _{t→e t−0} kv (t, ·)k = ∞.

This set has been studied by many authors but nobody still knows whether it is empty or not. For example, Leray studied the possible occurrence of singularities and noticed that this set has Lebesgue measure 0 and even a 1/2–Hausdorff di- mension 0 in [0, T ]. Furthermore the complement of this set in [0, T ] is a countable union of semiclosed intervals [a i , b i ). In the special case we discuss, these results follow from Theorem 4.

As another consequence of Theorem 4 note that if a weak solution v (t, x) is equals to zero at a moment t for infinite number of points x ∈ T ³ then it equals to zero identically at least for all t between two singularities.

Before approaching to a proof of the theorem we must develop some technique.

3. Definitions and Technical Tools

We will denote inessential constants by c, C or by these letters with subscripts.

Provide the space O(D) with a collection of norms: let u ∈ O(D) and K be a subset of D, then kuk K = sup _z∈K |u(z)|. A sequence u k ∈ O(D) converges to u ∈ O(D) if for any compact K we have ku k − uk K → 0 as k → ∞. This kind of convergence is referred to as compact convergence. Equipped with such a convergence the space O(D) becomes a seminormed space.

The continuity of a function from O(D) implies the continuity with respect to

the compact convergence.

Particularly, a linear operator A : O(D) → O(D) is continuous iff for any com- pact set K ⊂ D there exists a compact set K ⁰ ⊂ D and a constant c such that for all u ∈ O(D) we have kAuk K ≤ ckuk K

and the constant c does not depend on u.

Evidently, the compact convergence in the space O Q(T )
follows from the con- vergence with respect to the norms k · k _Q

_{(T )} .

We say that a subset M of O(D) is bounded if for any compact set K ⊂ D there is a constant C K such that if u belongs to M then kuk K ≤ C K .

Other details on seminormed spaces topology see in [17], [18].

Theorem 3.1 (Montel). If M is a closed and bounded subset of O(D), then it is a compact set.

Theorem 3.2 ([1]). Let K be a convex compact subset of O(D). Then a contin- uous map f : K → K has a fixed point b x ∈ K i.e. f (ˆ x) = x. b

The Laplace operator ∆ is invertible in the space of zero mean value functions:

there exists a bounded operator ∆ ⁻¹ : L ² (T ³ ) → L ² (T ³ ) such that if u(x) = X

k∈Z

u k e ^i(k,x)

is the Fourier expansion of u then an explicit form of this operator is

∆ ⁻¹ u(x) = − X

k∈Z

u _k

|k| ² _e e ^i(k,x) . (4)

Formula (4) implies that for any function f ∈ O(T ³ R ) there is an estimate:

k∆ ⁻¹ f k _T

≤ c(r, δ)kf k _T

r+δ

, δ > 0, r + δ < R.

So, ∆ ⁻¹ is a continuous operator of the space O(T ³ R ) to itself.

Let

u(x) = X

k∈Z

u k e ^i(k,x) ∈ L ² (T ³ ).

Define semigroups by the formulas:

S ^t u(x) = X

k∈Z

u k e ^i(k,x)−|k|

^t ,

R ^t u(x) = X

k∈Z

u _k e ^i(k,x)−|k|

^t/2 ,

H ^t = S ^t R ^−t .

For any fixed t > 0 the operators S ^t , R ^t , H ^t are continuous on O(T ³ R ).

3.1. Majorant functions.

Let

v(t, x) = X

k∈Z

v k (t)e ^i(k,x) ∈ O Q(T )
,

V (τ, x) = X

k∈Z

V k (τ )e ^i(k,x) ∈ C I T , H ¹ (T ³ )
.

A notation v  V implies that the inequality |v k (t)| ≤ V k (<e t) holds for all

|=m t| < <e t < T and k ∈ Z ³ 0 .

If u, U are vector–functions, then a relation u  U means that each component of the vector U majorates corresponding component of the vector u.

Define a continuous linear operator of O(T ³ R ) to itself:

Du = X

k∈Z

|k| e u _k e ^i(k,x) .

The operator −D is an infinitesimal operator for the semigroup R ^2t and the operator

∆ is an infinitesimal operator for the semigroup S ^t . Define the contour L(t) as follows:

L(t) =



s + i =m t

<e t s | 0 ≤ s ≤ <e t,



, |= t| < < t < T.

Enumerate main properties of the relation ””. Suppose u(t, x)  U (τ, x) and v(t, x)  V (τ, x); then:

u + v  U + V, uv  U V,

λu  |λ|U, R

L(t)

u(s, x) ds  2

<e t

R

0

U (s, x) ds,

∂ l u  DU, S ^νt u  U,

D(uv)  U DV + V DU, ∆ ⁻¹ u  U.

In these formulas λ is a complex number.

Another property of ”” is as follows: there exists some positive constant c such that an estimate

∆ ⁻¹ ∂ j ∂ l u  c U (5)

holds for all functions u  U . Indeed, expanding the left- and the right-hand side of (5) to Fourier series by formula (4) we, see that the estimate follows from the inequality:

|k j k l | ≤ c|k| ² _e , k ∈ Z ³ 0 . We say that a map F : C I T , H ¹ (T ³ )

→ C I T , H ¹ (T ³ )

majorates a map f : O Q(T )
→ O Q(T )
(denoted by f  F ) if for any functions u, U the relation u  U implies f (u)  F (U ).

Lemma 3.3. Let u, v ∈ H ¹ (T ³ ) be zero mean functions, then

kH ^t D(u · v)k ≤ h(t)kuk · kvk, h(t) =





 c

t ^7/8 if 0 < t ≤ 1, ce ^−t/2 if t > 1,

positive constant c does not depend on u and v. (Note that the product u · v belongs

at least to L ¹ (T ³ ).)

Proof. Consider the estimate for 0 < t ≤ 1. The estimate for t > 1 is easy to obtain. Introduce some notations:

J _k = {(m, n) ∈ Z ³ 0 × Z ³ 0 | m + n = k 6= 0}, N k = {(m, n) ∈ J k | |m| e ≥ |k| e , |n| e ≥ |k| e }, M k = J k \N k .

The set M _k is finite and every pair (m, n) ∈ M _k is estimated as follows:

|m| e < 2|k| _e , |n| e < 2|k| _e . (6) Indeed, if |n| e ≥ |k| e then |m| e < |k| e and by the equality n = k − m we have

|n| e ≤ |k| e + |m| e < 2|k| e . Let

u(x) = X

k∈Z

u k e ^i(k,x) , v(x) = X

k∈Z

v k e ^i(k,x) , then

H ^t Du = X

k∈Z

|k| e e ^−|k|

t(1−1/(2|k|

))+i(k,x) u _k , and

H ^t D u(x) · v(x)

= X

k∈Z

|k| e e ^−|k|

t 1−1/(2|k|

)

+i(k,x) X

(m,n)∈J

u n · v m .

So, by the formula

e ^−|k|

t 1−1/(2|k|

)

≤ e ^−|k|

^t/2 we have:

kH ^t D(u · v)k ² ≤ X

k∈Z

e ^−|k|

^t · |k| ⁴ _e ·



 X

(m,n)∈J

|u _n | · |v _m |





2

. (7)

Substitute to (7) the following expansion:



 X

(m,n)∈J

|u n | · |v m |





2

=



 X

(m,n)∈N

|u n | · |v m |





2

+ 2 X

(m,n)∈N

|u n | · |v m | · X

(m,n)∈M

|u n | · |v m | +



 X

(m,n)∈M

|u n | · |v m |





2

, (8)

and estimate these sums separately. By the Cauchy inequality one has:

X

(m,n)∈N

|u n | · |v m | ≤ 1

|k| ² _e X

(m,n)∈N

|n| e |u n | · |m| e |v m |

≤ 1

|k| ² _e

s X

n∈Z

|n| ² _e |u n | ²

s X

n∈Z

|n| ² _e |v n | ² = 1

|k| ² _e kuk · kvk. (9)

Thus for the first sum of (7)–(8)we have:

X

k∈Z

e ^−|k|

^t |k| ⁴ _e  X

(m,n)∈N

|u _n | · |v _m |  2

≤ kuk ² kvk ² X

k∈Z

e ^−|k|

^t

≤ c

t ^3/2 kuk ² kvk ² . (10) The last inequality follows from the formula

X

k∈Z

e ^−|k|

^t ≤ c  Z ∞ 0

e ^−x

^t dx  3

≤ c

t ^3/2 . (11)

Denote the last sum of (7)–(8) by A and estimate it with the help of formulas (6):

A = X

k∈Z

e ^−|k|

^t |k| ⁴ _e



 X

(m,n)∈M

|u n | · |v m |





2

≤ X

k∈Z

e ^−|k|

^t/2 |k| e

|k| e



 X

(m,n)∈M

e ⁻

|u n |e ⁻

|v m |





2

.

Using the inequality

e ^−|k|

^t/2 |k| e

≤ c

t ^1/4 , k ∈ Z ³ 0

we obtain A ≤ c

t ^1/4 kS ^t/32 u · S ^t/32 vk ² _H

(T

) ≤ c

t ^1/4 kS ^t/32 uk ² _H

(T

) kS ^t/32 vk ² _H

(T

) . Estimate the term:

kS ^t/32 uk ² _H

(T

) = X

k∈Z

e ⁻

|k|2e t

32

|k| ^3/2 _e |k| ² _e |u k | ² ≤ c t ^3/4 kuk ² , where we use again the inequality

e ^−|k|

^t/32 |k| e

≤ c

t ^3/4 , k ∈ Z ³ 0 . Finally it follows that

A ≤ c

t ^7/4 kuk ² kvk ² . (12)

Estimate the middle sum of (7)–(8) by formula (9):

B = X

k∈Z

e ^−|k|

^t |k| ⁴ _e X

(m,n)∈N

|u n | · |v m | · X

(m,n)∈M

|u n | · |v m |

≤ kuk · kvk X

k∈Z

e ^−|k|

^t |k| ² _e X

(m,n)∈M

|u n | · |v m |.

Then using the Cauchy inequality and formulas (11), (12) one has:

B ≤ kuk · kvk

s X

k∈Z

e ^−|k|

^t

s X

k∈Z

e ^−|k|

^t |k| ⁴ _e ( X

(m,n)∈M

|u n | · |v m |) ²

≤ kuk · kvk c t ^3/4

√

A ≤ c

t ^13/8 kuk ² kvk ² .

Gathering this estimate with (10) and (12) we obtain the assertion of the Lemma.

Lemma is proved. 

3.2. Existence lemma.

Define a map

Φ(u, v) = Z t

0

H ^ν(t−ξ) D u(ξ, x)v(ξ, x)
dξ.

Lemma 3.4. The map Φ takes the space C I T , H ¹ (T ³ )
× C I T , H ¹ (T ³ )
to the space C I _T , H ¹ (T ³ )
, and

kΦ(u, v)k ^C ≤ ckuk ^C kvk ^C , (13)

positive constant c does not depend on u, v and T . Proof. Taking τ = t − ξ, by Lemma 3.3 we have:

kΦ(u, v)k ^C = kuk ^C kvk ^C sup

t≥0

Z t 0

h(ντ ) dτ.

Evidently, the last sup is lower than infinity. Lemma is proved.  Define a function b V (x) as follows:

V (x) = b X

k∈Z

| b v _k | m e ^i(k,x) , (14) where b v k is the Fourier coefficients of the initial data b v (x). Evidently we have k b V k = k v k. b

Lemma 3.5. Let the function b V ∈ H ¹ (T ³ ) be defined by (14). Then the equation V (t, x) = H ^νt V (x) + 2 b

Z t 0

H ^(t−ξ)ν D(V ² )(ξ, x) dξ (15) has a unique solution V (t, x) ∈ C I T , H ¹ (T ³ )

with the same constant T as in Theorem 4.

If sup _t≥0 kH ^νt V (x)k is sufficiently small then previous assertion remains valid b for T = ∞.

Proof. We shall prove that for sufficiently small T the map F (V ) = H ^νt V (x) + 2 b

Z t 0

H ^(t−ξ)ν D(V ² )(ξ, x) dξ is a contraction of the ball

B = {V ∈ C I _T , H ¹ (T ³ )
| kV − H ^νt V (x)k b ^C ≤ 1}.

For selected T a parameter ε = sup

t∈I

Z t 0

h(ντ ) dτ = cT ^1/8

is sufficiently small to operator F be a contraction of the ball B. Indeed, let V ∈ B it follows that

kV k ^C ≤ 1 + kH ^νt V k b ^C and take T such a small constant as

kF (V ) − H ^νt V k b ^C ≤ 2ε(kV k ^C ) ² ≤ 1 thus F (B) ⊆ B. If V ⁰ , V ⁰⁰ ∈ B then

kF (V ⁰ ) − F (V ⁰⁰ )k ^C ≤ 2εkV ⁰ + V ⁰⁰ k ^C · kV ⁰ − V ⁰⁰ k ^C . One checks that

2εkV ⁰ + V ⁰⁰ k ^C < 1 so F is contraction of the ball B.

The second assertion of the lemma follows from similar arguments.

Lemma is proved. 

4. Proof of Theorem

We shall prove Theorem 4 by the Majorant functions method. Namely, the original problem will be replaced with the so called majorant problem. Then we prove the existence theorem for the majorant problem and show that it implies the existence theorem for the original one.

The majorant functions method was originated by Cauchy and Weierstrass and applied by Kovalevskaya to prove an existence of analytic solutions to initial prob- lems for PDE. Further studies and applications of this technique are contained in [6], [14], [20], [21].

Now prove the first part of the theorem, the second one results in the same way due to the last assertion of Lemma 3.5.

Apply formally some standard procedure. The permissibility of such a procedure will be clear from the further context.

So, take the operator div from the right and the left sides of equation (1). Us- ing equation (2) we get ∂ i ∂ j (v ⁱ v ^j ) = −∆p, where we sum through the repeated subscripts. Thus

p = −∆ ⁻¹ ∂ _i ∂ _j (v ⁱ v ^j ). (16) Substituting this formula to equation (1) we obtain the following problem:

(v ^k ) _t = A ^k _l ∂ _j (v ^j v ^l ) + ν∆v ^k , A ^k _l = (∆ ⁻¹ ∂ _k ∂ _l − δ _kl ),

v ^k | t=0 = b v ^k , (17)

where δ kl = 1 for k = l and 0 otherwise.

The operator A ^k _l is continuous on the space O(T ³ R ):

kA ^k _l uk _T

r

≤ c _r,δ kuk _T

r+δ

. (18)

Present equation (17) in the form:

(v ^k )(t, x) = G ^k (v ) = S ^νt b v ^k (x) + Z

L(t)

S ^ν(t−ξ) A ^k _l ∂ j (v ^j v ^l )(ξ, x) dξ,

G(v ) = (G ¹ , . . . , G ³ ). (19)

Lemma 4.1. Let the function b V be defined by (14) and the function V (t, x) is the solution of equation (15) then the map G takes the set

W = {u ∈ O(Q(T )) | u  R ^{ν<e t} V, div u = 0}

into itself.

Proof. It is easy to check that the map G takes a solenoidal vector–field to a solenoidal vector–field, indeed it follows from the equality:

∂ k A ^k _l = 0.

For any

v ^k  R ^{ν<e t} V we have

G ^k (v )  S ^{ν<e t} V + 2 b Z <e t

0

S ν(<e t−ξ) D(R ^νξ V ) ² dξ.

Then using the inequality |j + k| _e ≤ |j| _e + |k| _e one checks that (R ^t V ) ²  R ^t (V ) ² .

Thus we have S ^ντ V + 2 b

Z τ 0

S ^{ν(τ −ξ)} D(R ^νξ V ) ² dξ

 R ^ντ (H ^ντ V + 2 b Z τ

0

H ^{ν(τ −ξ)} D(V ) ² dξ) = R ^ντ V.

Lemma is proved. 

Lemma 4.2. The set W is convex and it is compact in O Q(T )
.

Proof. The convexity is obvious.

According to the Montel theorem it is sufficient to prove that the set W is bounded. By the estimate |e ^i(j,x) | ≤ e ^{|j||=m x|}

for any (t, x) ∈ Q r (T ) we have

|v ^k (t, x)| ≤      

X

j∈Z

v _j ^k (t)e ^i(j,x)      

≤ X

j∈Z

|v ^k _j (t)|e ^|j|νr/(4α)

≤ X

j∈Z

V j (<e t)e ^−ν|j|

r/2+|j|νr/(4α) ≤ X

j∈Z

V j (<e t)e ^−νr|j|

^/4 . Thus

kv ^k k _Q

_{(T )} ≤ sup

t∈I

X

j∈Z

V _j (<e t)e ^−νr|j|

^/4 . (20) Since V ∈ C I T , H ¹ (T ³ )
the right–hand side of estimate (20) is bounded for all r > 0.

Lemma is proved.

The map G is continuous in W with respect to the seminormed topology of O Q(T )
.

Thus according to the Theorem 3.2 and Lemma 4.2 it has a fixed point v (t, x) ∈ W . This fixed point is a solution of equations (1), (2) for (t, x) ∈ Q(T ).

The Theorem is proved. 

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Oleg Zubelevich

Department (# 803) of Differential Equations Moscow State Aviation Institute

Volokolamskoe Shosse 4 125993

Moscow RUSSIA

[email protected]

Current Address:

2–nd Krestovskii Pereulok 12-179 129110

Moscow

RUSSIA