Here we obtain the unique solvability of the problem in the Sobolev function spaces in the casep &gt

SOLVABILITY IN SOBOLEV SPACES OF

A PROBLEM FOR A SECOND ORDER PARABOLIC EQUATION WITH TIME DERIVATIVE IN THE BOUNDARY CONDITION

E. Frolova

1 – Introduction

In this paper we consider the initial boundary value problem for a second order parabolic equation with both time and spatial derivatives in the boundary condition. The unique solvability in H¨older function spaces, in weighted H¨older function spaces, and in Sobolev function spaces in the case p = 2 for problems with such a boundary condition in domains with smooth boundaries was obtained in the papers [1–5]. Here we obtain the unique solvability of the problem in the Sobolev function spaces in the casep > n+ 2. Condition (20) in Theorem 5 is a usual assumption on the combination of signs in the boundary condition. In the case of another combination the problem is ill-posed, it was shown in [6].

In §2 we consider the model problem for the heat equation in a half-space and obtainL_p-estimates for its solution using some results on Fourier multipliers [7, 8]. This allows us to prove the solvability of the problem in a domain with smooth boundary by the construction of a regularizer in the same way as in [9, 3].

In§3 we make this for a periodic case. Periodic problem (19) arose in studying the periodic one-phase Stefan problem [10].

In§4 we reduce the assumptions on the smoothness of the boundary which is usually made in the construction of a regularizer. Assuming that the boundary belongs to W_p^2−1/p, we prove the solvability of the problem with time derivative in the boundary condition.

We also consider in§2 a model problem with time derivative in theconjugation condition which is useful in studying the two-phase Stefan problem in the Sobolev function spaces.

Received: December 12, 1997; Revised: May 29, 1998.

2 – Model problems(^∗)

We introduce the following notations:

x= (x⁰, xn)∈Rⁿ ,

R1 =Rⁿ−=ⁿx∈Rⁿ|xn<0^o, R2=Rⁿ+=ⁿx∈Rⁿ|xn>0^o , Γ =ⁿx∈Rⁿ|x_n= 0^o, R_i,T =R_i×[0, T), i= 1,2,

Γ_T = Γ×[0, T) .

We consider the following model problem in the half-space:

(1)

v_t−∆v=f , (x, t)∈R_2,T ,

·

β v_t− ∂v

∂x_n +

n−1X

l=1

b_l ∂v

∂x_l

¸¯¯¯¯

ΓT

=ϕ(x⁰, t), v|t=0= 0 , whereβ >0,b_l∈R.

The specific character of this problem consists in the fact that the bound- ary condition contains both time and normal derivatives. Boundary operator in problem (1) does not satisfy the complementing condition for n > 1, therefore this problem can not be included into the general theory of parabolic problems for second order equations constructed in [9].

Problem (1) arises in studying the one-phase Stefan problem and some prob- lems of the heat conduction theory. If we consider the two-phase Stefan problem then it is useful to investigate the following model problem with time derivative in the conjugation condition:

(2)

∂v_k

∂t −a²_k∆v_k= 0, (x, t)∈R_k,T ,

∂v₁

∂t +b· ∇v₁−c· ∇v₂|Γ_T =ψ(x⁰, t) , v₁−v₂|ΓT = 0, v_k|t=0= 0 ,

wherek= 1,2;b,c∈Rⁿ;a₁, a₂∈R;b_n>0,c_n>0.

Problems (1) and (2) were studied in H¨older function spaces [1, 2, 13], in weighted H¨older spaces [3, 14], and in Sobolev spaces with p = 2 [4, 15] (only

(^∗) This part of the paper has been published in Russian [11]; the Lp-estimates for such model problems also has been obtained by H. Koch [12].

problem (1)). Here we obtainL_p-estimates of the solutions for problems (1) and (2) in the casep >1.

Let us putT = +∞and make the Fourier transform with respect to the space variablesx⁰ and the Laplace transform with respect to t:

(3) F v(ξ, x_n, s) =v(ξ, x_{e n}, s) = 1 (√

2π)ⁿ⁻¹ Z _+∞

0

dt Z

Rⁿ⁻¹v(x⁰, x_n, t)e^{−ts+ix0·ξ}dx⁰ , ξ ∈Rⁿ⁻¹, Res >0 .

Then, from problem (1) with f = 0 and problem (2) we pass to the following problems for ordinary differential equations

(4)

− d²ve

dx²_n + (ξ²+s)ev= 0, x_n>0,

·

− dve

dx_n +^³β s−i

n−1X

j=1

ξjbj

´ve^¸¯¯¯_¯

xn=0

=ϕ(ξ, s)e and

(5)

−a²_kd²ev_k

dx²_n + (a²_kξ²+s)ve_k= 0 , e

v₁|_xn=0=ve₂|_xn=0 ,

sev₁−i(b⁰−c⁰)·ξ+b_n dve₁

dx_n −c_n dve₂ dx_n

¯¯

¯¯

xn=0

=ψ(ξ, s)^e , whereb⁰ = (b₁, ..., b_n−1)^T,c⁰ = (c₁, ..., c_n−1)^T.

Solutions of problems (4) and (5) can be found in the explicit form:

(6) v(ξ, x_{e n}, s) = 1

r+β s−i

n−1X

j=1

ξjbj

e^−rxnϕ(ξ, s)_e ,

wherer=^ps+ξ², Rer >0, and (7) ev_k(ξ, x_n, s) = 1

s+^b_aⁿ

1 r₁+^c_aⁿ

2 r₂−i(b⁰−c⁰)·ξe^−akrk|xn|ψ(ξ, s)^e , wherer_k=^qs+a²_kξ², Rer_k>0,k= 1,2.

To obtain the solutions of problems (1) with f = 0 and (2) we have to use the integral transform inverse to (3).

Let us begin the formulation of the results by reminding some definitions.

The norm in Lp(R_k,T) is defined by the formula kukp,R_k,T =

µZ _T

0

dt Z

Rk

|u(x, t)|^pdx

¶1/p

.

ByWp^m,m/2(R_k,T) (m/2∈N) we mean the closure of the set of smooth functions in the norm

kuk^(m)p,R_k,T = ^X

0≤|α|+2a≤m

kD^atDx^αukp,Rk,T , α= (α₁, ..., α_n), |α|=

Xn i=1

α_i, α_i ∈N∪ {0}. The norm in the space of traces Wm−1/p, m/2−1/2p

p (Γ_T) is defined by the for- mula

kuk^(m−1/p)p,Γ_T = ^X

0≤|α|+2a<m

kD^atDx^αukp,ΓT +hhuii^(m−1/p)p,Γ_T , where

hhuii^(m−1/p)_p,ΓT =

= ^X

|α|+2a=m−1

µZ _T

0

dt Z

Γ

dx Z

Γ

¯¯

¯Dt^aDx^αu(x, t)−Dt^aD^αyu(y, t)^¯¯_¯^p dy

|x−y|n−1+p(1−1/p)

¶1/p

+

+ ^X

0<m−1/p−2a−|α|<2

µZ

Γ

dx Z _T

0

Z _T

0

¯¯

¯Dt^aD^αxu(x, t)−D^aτDx^αu(x, τ)^¯¯_¯^p dt dτ

|t−τ|^1+p(1−1/p)

¶1/p

,

hh·ii^(m)p and hh·ii^(m−1/p)p are the main parts of the norms k · k^(m)p and k · k^(m−1/p)p respectively.

The subspace of Wp^m,m/2(R_k,T) (Wm−1/p, m/2−1/2p

p (Γ_T)) which consists of functions satisfying zero initial conditions we denote by W

◦ m,m/2 p (R_k,T) (W

◦

m−1/p, m/2−1/2p

p (Γ_T)).

Theorem 1. For any ϕ∈W

◦

m+1−1/p, m/2+1/2−1/2p

p (Γ_T),f ∈W

◦ m,m/2

p (R_2,T), there exists a unique solution of problem (1)v ∈ W

◦

m+2, m/2+1

p (R_2,T), and there holds the estimate

(8) hhvii^(m+2)_p,R2,T ≤c^³hhϕii^(m+1−1/p)_p,ΓT +hhfii^(m)_p,R2,T^´, m

2 ∈N∪ {0}.

Theorem 2. For any ψ ∈ W

◦

m+1−1/p, m/2+1/2−1/2p

p there exists a unique

solution of problem (2)v_k∈W

◦

m+2, m/2+1

p (R_k,T)and there holds the estimate X

k=1,2

hhv_kii^(m+2)_p,Rk,T ≤chhψii^(m+1−1/p)_p,ΓT .

We prove theorems 1 and 2 by using the following results on Fourier multipliers [7, 8, 16].

Definition. Function g is called Fourier multiplier of class (p, p) if for any v∈C₀^∞ the estimate

kF⁻¹g F vkp≤ckvkp, 1≤p <∞. holds.

We set

M^pp(g) = sup

v∈C₀^∞

kF⁻¹g F vkp

kvkp

.

Theorem 3. Let ξ ∈ Rⁿ. Assume that function g(ξ) and all its mixed derivatives

∂^lg(ξ)

∂ξ_k1· · ·∂ξ_kl, where l≥1, ki6=kj fori6=j, i, j= 1, ..., l ,

are continuous for|ξ_j|>0,j = 1, ..., n, and that there exists a positive constant M such that

(9)

¯¯

¯¯ξ_k1· · ·ξ_kl ∂^lg(ξ)

∂ξ_k1· · ·∂ξ_kl

¯¯

¯¯≤M .

Then g is a Fourier multiplier of class (p, p)and M^pp(g)≤c M.

Theorem 4. Let functions H(x⁰, x_n, t) and η(x⁰, x_n, t) be defined in Rⁿ+,∞, η(x, t) ∈ Lp(Rⁿ+,∞), let F H(ξ, xn, s) be a Fourier multiplier of class (p, p), and M^pp(H(·, x_n,·))≤ _x^c_n. Let

Hη= Z _+∞

0

F^−1hH(ξ, x_n+y, s)F η(ξ, y, s)ⁱdy . ThenHη∈Lp(Rⁿ+,∞) and there holds the estimate

kHηkp,Rⁿ_+,∞ ≤ckηkp,Rⁿ_+,∞ .

Proof of Theorem 1: We consider the following auxiliary problem:

w_t−∆w=f(x, t), (x, t)∈R_2,T , w|Γ_T = 0, w|t=0 = 0 .

There exists the unique solution w∈W

◦

m+2, m/2+1

p (R_2,T) of this problem [9], and there holds the estimate

(10) hhwii^(m+2)_p,R2,T ≤chhfii^(m)_p,R2,T .

Looking for the solution of problem (1) in the form v = w+u, for the new unknown functionuwe have problem (1) with f = 0:

(1a)

u_t−∆u= 0, (x, t)∈R_2,T , µ

β ut− ∂u

∂x_n +

n−1X

l=1

b_l ∂u

∂x_l

¶¯¯¯¯

Γ_T

=ϕ(x⁰, t) + ∂w

∂x_n =ϕ^∗(x⁰, t), u|_t=0= 0 .

At first we find ϕ₁ ∈W

◦

m+1−1/p, m/2+1/2−1/2p

p (Γ_∞) such that

(11)

ϕ₁|_t≤T =ϕ^∗, ϕ₁|_t≥T+ε= 0, ε >0 ,

hhϕ1ii^(m+1−1/p)_p,Γ∞ ≤chhϕ^∗ii^(m+1−1/p)_p,ΓT ≤c^³hhϕii^(m+1−1/p)_p,ΓT +hhfii^(m)_p,R2,T^´,

and, making the Fourier–Laplace transform (3), we arrive at problem (4).

Then we extend the function ϕ₁ into the half-space R_2,∞, finding $ ∈ W◦

m+1, m/2+1/2

p (R_2,∞) such that$|_Γ∞ =ϕ₁ and the estimate

(12) hh$ii^(m+1)_p,R2,∞ ≤chhϕ1ii^(m+1−1/p)_p,Γ∞

holds. With the help of the Newton–Leibnitz formula we obtain from (6) the

following integral representation for the solutionu_e of problem (4) with ϕ=ϕ₁

e

u(ξ, x_n, s) =− Z _+∞

0

∂

∂y









1 r+β s−i

n−1X

j=1

b_jξ_j

e^−r(xn+y)$(ξ, y, s)e







 dy

= Z _+∞

0

r r+β s−i

n−1X

j=1

ξjbj

e^−r(xn+y)$(ξ, y, s)_e dy

− Z _+∞

0

1 r+β s−i

n−1X

j=1

ξ_jb_j

e^−r(xn+y) ∂

∂y$(ξ, y, s)^e dy .

Hence, for j= 2, ..., m+ 2,

(13)

F⁻¹(r^jF u) =

= Z _+∞

0

F⁻¹

·

H(ξ, x_n+y, s) µ

r^j−1$(ξ, y, s)_e −r^j−2 ∂$(ξ, y, s)_e

∂y

¶¸

dy , where

H(ξ, x_n, s) = r² r+β s−i

n−1X

j=1

ξ_jb_j

e^−rxn .

Let us show that H(ξ, xn, s) is a Fourier multiplier of class (p, p) and M^pp

³H(·, x_n,·)^´≤ c x_n .

It is obvious that the functionH and its mixed derivatives are continuous for Res > 0, |ξ_j| >0, j = 1, ..., n−1. Let us check that for H and its derivatives the estimate (9) holds withM= _x^c_n.

It is easy to see that

(14) Rer

¯¯

¯¯r+β s−i

n−1X

l=1

ξ_lβ_l

¯¯

¯¯

≤c .

So,

¯¯

¯H(ξ, x_n, s)^¯¯_¯≤ |r|²

Rere^{−Rer xn} , (15)

¯¯

¯¯ξj ∂H(ξ, xn, s)

∂ξ_j

¯¯

¯¯=

¯¯

¯¯

¯¯

¯¯

¯¯

¯

2ξ²_j −xnξ_j²r r+β s−i

n−1X

l=1

ξ_lb_l

e^−rxn− ξ_j²r−i bjξjr²

³r+β s−i

n−1X

l=1

ξ_lb_l^´2 e^−rxn

¯¯

¯¯

¯¯

¯¯

¯¯

¯

≤c^³|r|+x_n|r²|^´e^{−Rer xn} . Using the inequality

(16) y^je^−yz≤ c

z^j , ∀y >0, z >0, j∈N, withj= 1,2, we obtain

(17) Rer e^{−Rer xn} ≤ c

x_n, (Rer)²xne^{−Rer xn} ≤ c x_n . So, (15) and (17) imply

¯¯

¯H(ξ, x_n, s)^¯¯_¯≤ |r|² (Rer)² · 1

x_n ≤ c x_n ,

¯¯

¯¯ξ_j ∂H(ξ, x_n, s)

∂ξ_j

¯¯

¯¯≤c µ |r|

Rer + |r|² (Rer)²

¶ 1 x_n ≤ c

x_n .

Here we take into account the fact that the function r = ^pξ²+s with Rer >0, Re(ξ²+s)>0 satisfies the inequality _Re^|r|_r ≤c.

Further, we have

¯¯

¯¯s∂H(ξ, xn, s)

∂s

¯¯

¯¯=

¯¯

¯¯

¯¯

¯¯

¯¯

¯

s(1−¹₂r x_n) r+β s−i

n−1X

l=1

b_lξ_l

e^−rxn− s(¹₂r+β r)

³r+β s−i

n−1X

l=1

b_lξ_l^´2 e^−rxn

¯¯

¯¯

¯¯

¯¯

¯¯

¯

≤c









1 + |s|

¯¯

¯r+β s−i

n−1X

l=1

ξ_lb_l^¯¯_¯







 1 xn

.

Let us prove that

(18) P = |s|

¯¯

¯r+β s−i

n−1X

l=1

ξ_lb_l^¯¯_¯

≤c .

If|^Pn−1l=1 ξ_lb_l| ≤ ¹₂(β|Ims|+|Imr|) then, taking into account thatβ >0, Res >0 and sign(Ims) = sign(Imr), we have

P ≤ |s|

q

(Rer+βRes)²+^β₄²(Ims)² ≤c . If |^Pn−1_l=1 ξ_lb_l| ≥ ¹₂(β|Ims|+|Imr|) then|Ims| ≤c|ξ|and

P ≤c Res+|ξ| Re^pξ²+s ≤c .

So, for the all possible values of ξ and s, the inequality (18) holds and

|s^∂H(ξ,x_∂s^n,s)|can be estimated by _x^c

n.

Thus, we obtain the required estimates (9) for the function H and its deriva- tives of the first order. The estimates (9) for the higher order derivatives of the functionH are obtained in a similar way.

Taking into account (14) and (18), we have

¯¯

¯¯ξ_k1· · · ξ_kl∂^lH(ξ, x_n, s)

∂ξ_k1· · · ∂ξ_kl

¯¯

¯¯+

¯¯

¯¯s ξ_k1· · ·ξ_kl−1 ∂^lH(ξ, x_n, s)

∂ξ_k1· · ·∂ξ_kl−1∂s

¯¯

¯¯ ≤

≤ c^³|r|+x_n|r|²+...+x^l_n|r|^l+1´e^{−Rer xn} , and, with the help of inequality (16) with j = 1,2, ..., l+ 1, we arrive at the required estimates.

Hence, due to Theorem 3, we make a conclusion that H(ξ, xn, s) is a Fourier multiplier of class (p, p).

According to Theorem 4, it follows from the representation formula (13) that kF⁻¹(r^ju)e kp ≤c^µ°°_°F⁻¹(r^j−1F $)^°°_°

p,R2,∞

+^°°_°F^−1³r^j−2 ∂

∂x_nF $^´°°_°

p.R2,∞

¶ , j= 2, ..., m+2. Making the integral transform inverse to (3), we find the solution of problem (1a)u=F⁻¹(u).e

From the explicit form (6), for this solution we obtain

∂^ju

∂x^jn

=F⁻¹ µ ∂^j

∂x^jn

F u

¶

=F^−1³(−1)^jr^jue^´ . Thus,

hhuii^(m+2)_p,R2,T ≤ hhuii^(m+2)_p,R2,∞≤c µ

kF⁻¹r^m+2F ukp,R2,∞

°°

°°

∂^m+2u

∂x^m+2n

°°

°°

ρ,R2,∞

¶

≤c^µ°°_°F⁻¹(r^m+1F $)^°°_°

p,R2,∞

+^°°_°F^−1³r^m ∂

∂x_nF $^´°°_°

p,R2,∞

¶

≤chh$ e^tResii^(m+1)_p,R2,∞ .

From here, taking into account estimates (11) and (12), we obtain the desired estimate

hhuii^(m+2)_p,R2,T ≤c^³hhϕii^(m+1−1/p)_p,ΓT +hhfii^(m)_p,R2,T^´ .

Together with estimate (10) this estimate implies (8) for the solution of prob- lem (1)v=w+u.

Remark. There holds also the estimate

kvk^(m+2)p,R_2,T ≤c^³kϕk^(m+1−1/p)p,Γ_T +kfk^(m)p,R_2,T

´.

The uniqueness of the solution follows from the fact that if we consider ho- mogeneous problem corresponding to problem (1) and make the Laplace–Fourier transform (3) we arrive at the homogeneous problem (4) which has only zero solution.

Proof of Theorem 2: At first we extend the functionψfrom Γ_T intoR_k,∞, constructing

ψ_k ∈W

◦

m+1,^m₂+¹₂

p (R_k,∞), k= 1,2 , such that

ψ_k^¯¯_¯xn=0

t≤T

=ψ , ψ_k^¯¯_{¯ xn=0}

t≥T+ε

= 0, ε >0 , and

hhψ_kii^(m+1)_p,Rk,∞ ≤chhψii^(m+1−1/p)_p,ΓT .

Forv_ek which is given by formula (7) we use an integral representation similar to (13). As in the proof of Theorem 1, we should check that the function

H_k(ξ, x_n, s) = r_k²e^−|xn|rk s+^b_aⁿ

1 r1+^c_aⁿ

2 r2−i(b⁰−c⁰)·ξ

is a Fourier multiplier of class (p, p) and that M^pp

³H_k(·, xn,·)^´≤ c x_n .

Calculating the derivatives of the function H_k and using the same arguments as in the proof of the estimates for the mixed derivatives of the function H, we make sure that estimates (9) withM = _x^c_n hold true.

For example,

¯¯

¯¯s∂H_k

∂s

¯¯

¯¯=

=

¯¯

¯¯

¯¯

¯

s−¹₂r_k|x_n|s s+^b_aⁿ

1 r₁+^c_aⁿ

2 r₂−i(b⁰−c⁰)·ξ − s r_k^2³1 +_2a^bk

1r1 +_2a^cn

2r2

´

³s+^b_aⁿ

1 r₁+^c_aⁿ

2r₂−i(b⁰−c⁰)·ξ^´2

¯¯

¯¯

¯¯

¯

·^¯¯¯e^{−rk|xn|¯¯}_¯

≤



_¯ |s r_kx_n|

¯¯s+^b_aⁿ

1r₁+^c_aⁿ

2 −i(b⁰−c⁰)·ξ^¯¯_¯+ |s| |r_k|²

¯¯

¯s+^b_aⁿ

1 r₁+^c_aⁿ

2 r₂−i(b⁰−c⁰)·ξ^¯¯_¯²



 e^−Rerk|xn|

≤ c

|x_n|

|r_k| Rer_k ≤ c

|x_n| .

Then, we use Theorem 4 and complete the proof of Theorem 2 in the same way as it has been done above in the proof of Theorem 1.

3 – Periodic problem

In this section we assume that all the given functions and surfaces are periodic with respect to the space variablesxi (i= 1, ..., n−1) with the period 1 and call such functions simply periodic.

Let Ω ⊂ Rⁿ be a domain which lies between two mutually disjoint smooth periodic surfacesS₁ and S₂.

We denote by B1(a),a∈Rⁿ⁻¹, the cube in the space Rⁿ⁻¹ B₁(a) =

½

x⁰∈Rⁿ⁻¹| |x_i−a_i|< 1

2, i= 1, ..., n−1

¾ . By ^Q_1,a we denote the stripB₁(a)×R={x∈Rⁿ|x⁰ ∈B₁(a)}. For an arbitrary set E⊂Rⁿ we set E⁰(a) =E∩^Q1,a.

By L^e_p(Ω_T) we mean the space of periodic functions with a finite norm in L_p(Ω⁰_T(a)), ∀a∈Rⁿ⁻¹.

We denote by W^f_p^2,1(Ω_T) the space of periodic functions which are defined in Ω_T = Ω×[0, T), have generalized derivativesDx^αu(x, t),|α| ≤2,α= (α1, ..., αn), Dtu(x, t) and a finite norm

kuk⁽²⁾_p,ΩT =kuk_Wep^2,1_(ΩT)=kuk_Wp^2,1_(Ω⁰

T(a)), ∀a∈Rⁿ⁻¹ .

Wfp^{2−1/p,1−1/2p}(S_T) is the space of traces of functions from W^f_p^2,1(Ω_T) on the periodic surfaceS ∈Ω. The norm in this space is determined by the formula

kuk^(2−1/p)_p,ST =kuk_Wep^{2−1/p,1−1/2p}_(ST)=kuk_W2−1/p,1−1/2p

p (S_T⁰(a)), ∀a∈Rⁿ⁻¹ . Wf

◦ 2,1

p (Ω_T) (Wf

◦

2−1/p,1−1/2p

p (S_T)) is a subspace of the space W^f_p^2,1(Ω_T) (W^fp^{2−1/p,1−1/2p}(S_T)) containing the functions satisfying zero initial conditions:

u|_t=0 = 0 .

By W^f_p^2−γ(Rⁿ⁻¹),γ ∈(0,1), we denote the space of periodic functions having first order generalized derivatives and a finite norm

kuk_Wep^2−γ_(Rn−1) =kuk_Wp^2−γ_(B1(a)), ∀a∈Rⁿ⁻¹ . We consider the following problem

(19)

Lu≡u_t− Xn i,j=1

a_ij(x, t)u_xixj−

n−1X

i=1

a_i(x, t)u_xi −a(x, t)u=f, (x, t)∈Ω_T , Bu=µ u_t+

Xn i=1

b_i(x, t)u_xi|x∈S1 =ϕ₁ ,

Cu= Xn i=1

ci(x, t)uxi +c(x, t)u|_x∈S2 =ϕ2 , u|_t=0= 0 ,

where

γ1ξ² ≤aij(x, t)ξiξj ≤γ2ξ², γ1, γ2>0, ∀ξ∈Rⁿ, t∈[0, T) .

We assume that the coefficients a_ij,a_i,a,b_i,c_i, cand the given functions f, ϕ₁,ϕ₂ are periodic.

The surfaces S₁ and S₂ are supposed from the class C². It means that there exists such a positive number dthat in the ball K_d with the radius d and with the center in anyx₀ ∈S_j (j= 1,2), the equation of the surface S_j∩K_d in local coordinates with the center in the point x0 has the form yn = F(y1, ..., yn−1), where the functionF belongs to C²(Rⁿ⁻¹).

Theorem 5. Let in problem (19) a_ij ∈ C(Ω_T), a_i, a ∈ L^e_p(Ω_T), bi∈W^fp^{1−1/p,1/2−1/2p}(S_1,T), ci ∈W^fp^{1−1/p,1/2−1/2p}(S_2,T) with a certain p > n+2, and there hold the inequalities:

(20) µ >0, Xn j=1

b_j(x, t)ν_j ≥δ >0, (x, t)∈Ω_T , whereν = (ν₁, ..., ν_n)^T is the outward normal to the surface S₁.

Then, for any f ∈L^e_p(D_T), ϕ1 ∈W^f

◦

1−1/p,1/2−1/2p

p (S_1,T), ϕ2 ∈W^f

◦

1−1/p,1/2−1/2p

p (S_2,T) ,

problem (19) has a unique solutionu∈W^f

◦ 2,1

p (Ω_T), this solution has the additional smoothness on the surfaceS₁: u_t|x∈S1 ∈W^f

◦

1−1/p,1/2−1/2p

p (S_1,T) and the estimate (21) kuk⁽²⁾_p,ΩT +kutk^(1−1/p)_p,S1,T ≤c^³kfkp,Ω_T +kϕ1k^(1−1/p)_p,S1,T +kϕ2k^(1−1/p)_p,S2,T ^´

holds.

Proof: We are going to prove this result with the help of constructing a regularizer in the same way as it is usually done in a nonperiodic case [9, 17].

We rewrite problem (19) in the form

(22) Au=h ,

where

Au=^hLu, Bu, Cu^iT , h= [f, ϕ₁, ϕ₂]^T .

For any smallλ >0 we construct the systems of subdomainsω^(k)⊂Ω^(k)⊂Ω with the following properties:

1) ^S_kω^(k)=^S_kΩ^(k)= Ω.

2) For any point x ∈ Ω there can be found such ω^(k) that x ∈ ω^(k) and dist(x,Ω\ω^(k))

≥dλ,d >0.

3) There existsN₀ ∈Nsuch that^TN_j=1⁰⁺¹Ω^(kj)=∅for anyk₁, ..., k_N0+1,k_i6=k_j fori6=j.

4) The systems of subdomainsω^(k), Ω^(k)(k∈N) are periodic with respect to the space variablesx_i,i= 1, ..., n−1, with the period 1, diam Ω^k< ¹₂. We identify the domainsω^(k), Ω^(k) which can be obtained from each other by displacement by the period. Then, we obtain a finite system of periodic setsω_e^(j) andΩ^e(j) (j = 1, ..., M).

Let ξ^(j)(x) (j= 1, ..., M) be smooth periodic functions such that ξ^(j)(x) =

½1, x∈ωe^(j), 0, x∈Ω\Ω^e(j) , 0≤ξ^(j) ≤1, |D^s_xξ^(j)| ≤ c

λ^|s| . By virtue of property 3 of the domains Ω^(k)

1≤ XM k=1

ξ^k(x)≤N₀ for any x∈Ω, and, consequently, the periodic functions

η^(j)(x) = ξ^(j)(x) X

j

ξ^(j)2(x) have the following properties:

η^(j)(x) = 0, x∈Ω\Ω^e(j), |D^s_xη^(j)(x)| ≤ c λ^|s|

and XM

j=1

η^(j)(x)ξ^(j)(x) = 1 for any x∈Ω .

We fix in every ω_e^(k) the set of points x_ek which can be obtained from each other by displacement by the period. Without loss of generality we can assume that ifΩ^e(k)∩S_j 6=∅thenωe^(k)∩S_j 6=∅, in this case we fix pointsxe_kon ωe^(k)∩S_j. If we freeze the coefficients in problem (19) at the point x_k ∈ xe_k, take the principal part of the operator A, and multiply the given functions onξ^(k), then,

by virtue of the periodicity of the coefficients and the given functions, we arrive at one of the following periodic problems:

1) in the case whenΩ^e(k)⊂Ω:

(23) Lu^(k)=u_t− Xn i,j=1

a_ij(x_k,0)u_xixj =f ξ^(k)=f^(k), u|t=0= 0, k= 1, ..., M₁ ;

2) in the case whenΩ^e(k)∩S₂ 6=∅: (24) L^(k)u=f^(k), C^(k)u=

Xn l=1

C_l(x_k,0)u_xl|x∈S2 =ϕ₂ξ^(k)=ϕ^(k)₂ ,

u|t=0= 0, k=M₁+1, ..., M₂ ; 3) in the case whenΩ^e(k)∩S₁ 6=∅:

(25)

L^(k)u=f^(k) , B^(k)u=µ ut+

Xn l=1

bl(xk,0)ux_l|_x∈S1 =ϕ1ξ^(k)=ϕ^(k)₁ , u|_t=0 = 0, k=M₂+ 1, ..., M .

Let us choose the domain Ω^k from the periodic set Ω^ek (k = 1, ..., M1) and consider problem (23) in this domain. Making the zero extension of the given functions from Ω^k to Rⁿ we arrive at the Cauchy problem in Rⁿ. The unique solvability of the Cauchy problem for second order parabolic equations with con- stant coefficients in Sobolev function spaces was established in [9]. We multiply the solution by the function η^(k)|Ω^k and do the periodic extension from Ω^k to Ωe^(k). By virtue of the periodicity of the given functions and uniqueness of the solution of the Cauchy problem, the obtained periodic function does not depend on the choice of the domain Ω^(k) from the setΩ^ek.

If ωe^(k) is closed to the boundary and in neighbourhoods of the points xe^(k) the boundary surface can be given in a local coordinate system by the equation y_n=F^(k)(y⁰), we make in every Ω^(k)∈Ω^e(k) the coordinate transformationZ^(k):

z_i =y_i, i= 1, ..., n−1, z_n=y_n−F^(k)(y⁰) ,

which straightens the boundary. After this transformation we arrive at a periodic problem of type (24) or (25) in a half-space in which the operatorsZ^(k),C^(k)and B^(k) are recalculated in the new coordinates and take the form Z^ek,C^ek,B^ek.

If ω_e^(k) is closed toS₂ we have (26)

Ze^(k)u^(k)=f^e(k), x∈Rⁿ+, t∈[0, T) , Ce^(k)u^(k)|xn=0 =ϕ_e^(k)₂ , u^(k)|t=0 = 0 , where

fe^(k)=P_k⁻¹f^(k), ϕ_e^(k)₂ =P_k⁻¹ϕ^(k)₂ ,

P_k is an operator which makes correspond to a function in local coordinates the same function in the original coordinates.

We consider problem (26) in the image of the domain Ω^k∈Ω^ek (k =M₁+1, ..., M₂) under the transformZ^(k)and do the zero extension of the given functions into Rⁿ+. We obtain the second boundary value problem in a half-space, the unique solvability of which in Sobolev function spaces is known [9]. We do the transform inverse to Z^(k), multiply the solution by η^(k)|_Ω^k, and do the periodic extension toΩ^ek.

If ωe^(k) is closed to S1 then, after the transformation Z^(k), we arrive at the problem

(27)

Ze^(k)u^(k)=f^e(k), x∈Rⁿ−, t∈[0, T) , Be^(k)u^(k)|_xn=0=ϕe^(k)₁ , u^(k)|_t=0= 0 ,

e

ϕ^(k)₁ =P_k⁻¹ϕ^(k)₁ .

Let us consider the problem in a half-space for the second order parabolic equation with constant coefficients:

(28)

ut− Xn i,j=1

aijux_ix_j =g , x∈Rⁿ−, t∈[0, T), µ u_t+

Xn i=1

b_iu_xi|_xn=0=ϕ , u|_t=0= 0 ,

where

aij, bi ∈R, bn<0, µ >0 .

We make the orthogonal coordinate transform which brings the matrix of the coefficients in the equation to a diagonal form. Under this transform the boundary condition in problem (28) transforms to the boundary condition of the same type. After additional linear change of variables we obtain the problem for the heat equation. Making once more orthogonal coordinate transform we

have in the new variablesz_i the boundary condition on the planez_n= 0. So, we can reduce problem (28) to the corresponding problem for the heat equation, the unique solvability for this problem in the Sobolev function spaces is established by Theorem 1.

We consider problem (27) in the image of the domain Ω^k∈Ω^e(k) (k=M₂+1, ..., M) under the transformZ^(k) and do the zero extension of the given functions intoRⁿ+, then we arrive at the problem of type (28). We do the transform inverse to Z^(k), multiply the solution of the obtained problem by η^(k)|_Ω^k, and do the periodic extension toΩ^e(k).

Note, that by virtue of the periodicity of the given functions and uniqueness of the solution of the second boundary value problem and of problem (28), the obtained periodic functions are independent on the choice of the domain Ω^kfrom the setΩ^e(k).

We put

Rh=

M1

X

k=1

uη^(k)(x)R^(k)(ξ^(k)f) +

M2

X

k=M1+1

uη^(k)(x)P_kR^(k)(P_k⁻¹ξ^(k)f, P_k⁻¹ξ^(k)ϕ₂) +

XM k=M2+1

uη^(k)(x)P_kR^k(P_k⁻¹ξ^(k)f, P_k⁻¹ξ^(k)ϕ₁),

where by R^k we mean the operator which assigns to the given functions the solution of

1) problem (23) for k= 1, ..., M₁, 2) problem (26) for k=M₁+ 1, ..., M₂, 3) problem (27) for k=M₂+ 1, ..., M.

in the domain Ω^k ∈ Ω^ek, by u we mean the operator which makes the periodic extension.

In just the same way as it was done in [9] for nonperiodic case, making all the estimates in the norms of periodic Sobolev spaces, it can be shown that for a sufficiently smallτ >0

kAR−Ik<1 and kRA−Ik<1.

The termµ u_tin the boundary condition on the surfaceS₁does not complicate the estimates because it is presented also in problems (25), µis a constant, and

the functions ξ^(k), η^(k) and F^(k) do not depend on the variable t. So in the operators AR−I and RA−I there is no additional terms in comparison with the case of the second boundary condition.

Consequently, the operator R is a regularizer and, for the sufficiently small τ the operator A in problem (22) has a bounded inverse operator. It means that problem (19) has a unique solutionu∈W^f

◦ 2,1

p (Ω_τ), which has the additional smoothness onS_1τ and for which there holds estimate (21). To prove the existence of the solution on the whole time interval [0, T) we use the same reasoning as in [15]. Namely, we chooset₀ ∈(^3τ₄ , τ) and extendu|Ω×[0,t0] to Ω×[0,2t₀] using the Hestence–Whitney method:

b

u(x, t) =







u(x, t), t∈[0, t₀], XN

j=1

u µ

x,2t0−t j

¶

λj, t∈[t0,2t0], XN

j=1

λ_j µ

−1 j

¶_s

= 1, s= 0,1, ..., N−1. For a sufficiently large N we have

ub∈W^f

◦ 2,1

p (Ω_2,t0), u_et|S1 ∈W^f

◦

1−1/p,1/2−1/2p

p (S_1,2t0) .

We find a solution w(x, t) of the problem

(29)

Lw=f−Lbu=f ,^b x∈Ω, t∈[t0,2t0), Cw|_x∈S2 =ϕ₂−Cub|_x∈S2 =ϕb₂ ,

Bw|_x∈S1 =ϕ₁−µu_bt− Xn i=1

b_i(x, t)u_bxi|_x∈S1 =ϕ_b1 , w|t=t0 = 0 .

Note that by construction a regularizer one can prove the existence of the solution of problem (29) on the small time interval [t₀, t₀+τ) in the same way as for problem (19). Because of uniform parabolicity of the operatorLand condition (20) for the operatorB,τ can be chosen one and the same for anyt₀ ∈[0, T).

Thus, we find w ∈ W^f_p^2,1(Ω×[t0, t0 +τ)) such that wt|_{x∈S1×[t0,t0+τ)} ∈ Wfp^{1−1/p,1/2−1/2p}(S1×[t0, t0+τ)), w|_t=t0 = 0, wt|x∈S1

t=t0

= 0. We make the zero extensionw_b of the function w to Ω_t0. Obviously,

b w∈W^f

◦ 2,1

p (Ω_t0+τ), wb_t|_x∈S1,t

0+τ ∈W^f

◦

1−1/p,1/2−1/2p

p (S_1,t0+τ).