A mixed problem for the wave equation in coordinate domains. Ii. Obtaining of a priori estimates protect in mixed problems of protect multidimensional wave equation Текст научной статьи по специальности «Математика»

Вычислительные технологии

Том 1, № 2, 1996

A MIXED PROBLEM FOR THE WAVE EQUATION IN COORDINATE DOMAINS.

II. OBTAINING OF A PRIORI ESTIMATES PROTECT IN MIXED PROBLEMS OF PROTECT

MULTIDIMENSIONAL WAVE EQUATION*

A. M. Blokhin Institute of Mathematics SB RAS, Novosibirsk, Russia

D. L. Tkachev Novosibirsk State University, Russia

In the second part of the article (the first one has been published in the previous issue of journal) there is suggested a new and simple method to obtain the known a priori estimation of solution from W2 of the mixed problem for the multidimensional wave equation in a half plane with boundary conditions of an oblique derivative type. The method combines elements of the Fourier-Laplace transform technique as well as the energy integrals technique. For the case of the mixed problem for the multidimensional wave equation in a coordinate corner, a domain of values of the boundary conditions parameters is selected out where the a priori estimation of solution from without loss of generality is valid.

Total list of the literature has been adduced in the first part of the article.

1. Mixed problem for the wave equation (the real coefficients case)

In this section, the following mixed problem for the wave equation in the domain t > 0, (x,y, z) G R++2, n > 1 is considered:

L(r,£,n,(i ,---,(n)u = utt - uxx - Uyy - Az u = 0, t> 0, x > 0, (1.1)

ut — aux — buy — (c, (u) = 0, x = 0, (1.2)

u = <^(x, y, z), ut = ^(x,y, z), t = 0. (1.3)

Here

d d d z =(zi, ...,zn), T = dt, e = dx n = dy,

d _

Z = (Zi, ..., Zn)T, Zk = dr, k = i,n,

CJZk

*© A. M. Blokhin, D. L. Tkachev, 1996.

A,

(ci,

(Z,Z ) = £ Zk2 = E

k=1

k=1

dzk;

du

l)T' (c'Zu)^CkdZk;

k=1

a, b, ck, k = 1,n are real numbers. Without loss of generality we will assume that cn = 0, R++2 = {(x, y, z); x > 0, (y, z) G Rn+1}.

Remark 1.1. We will assume that for the problem (1.1)-(1.3) the uniform Lopatinski condition (ULC) holds. Mixed problem (1.1)-(1.3) is said to satisfy ULC on a boundary if:

n

f + a^f2 + IyI2 - ibYo - i CkYk = 0

k=i

when Ref > 0, |f|2 + |y|2 = 0 (for more details about ULC see [17, 27]). Here f is a complex number.

Y = (7o,7i, •••> Yn), |Y|2 = (7,Y),

a = 0, n are real numbers. It can be shown that in the case when ULC fails there exist the examples of ill-posedness of Hadamard's type for problem (1.1)-(1.3) or the problems close to it. In terms of the boundary condition coefficients from (1.2), ULC can be written as (see [27]):

a > 0,

b2 + |c|2 < 1.

(1.4)

Remark 1.2. By straightfoward manipulations problem (1.1)-(1.3) can be put in a more simple way. The essence of these manipulations is the following. Let T be a real orthogonal matrix of order (n + 1). Then by replacement of the initial differential operators (i.e., by passing from the initial differential operators to their linear combinations):

ß

( \

ßi \ßn )

t *

(1.5)

we obtain the following relations:

t 2 - e2 - n2-£ Zk2 = t 2 - e2ß2

k=1

/ y fai

a=0

T - ae - bn - Ck Zk = T - ae - ( T

k=1

ß

t — ae — bß0

The latter relation is true since there exists such an orthogonal matrix T that

(b,C1, ..., cn) ■ T = (b, 0, ..., 0),

where b = sign(cra^^/b2 + |c|2. With the above-mentioned relations taken into account, initial problem (1.1)-(1.3) can be reduced to the following so called canonical form.

c

c

*

Problem I. We seek the solution of the wave equation

utt - uxx - Uyy - Azu = 0, t > 0, (x, y, z) G R++2.

(1.1')

which satisfies at x = 0 the boundary condition

ut - aux - buy = 0, t > 0, (y, z) G Rn+1,

(1.2')

and at t = 0 the initial data (1.3) (while formulating Problem I, we return to the former notations). The uniform Lopatinski condition for Problem I is formulated as

a > 0, b2 < 1.

(1.4')

The study of Problem I is the matter the present section.

In Problem I we carry out the Fourier transform with respect to the variables zk, k = 1, n. Then Problem I looks like

utt - uxx - uiyy + 4n |£| u = 0, t > 0, (x, y) G R+, ut - aux - buy = 0, t > 0, y G R1, x = 0, u = ^(x,y,£), ut = V^y^), (x,y) G R

Here

u = u(t,x,y,C) = / e-2ni(z>i)u(t,x,y, z)dz

(1.1'')

(1.2'') (1.3')

is Fourier transform of the function u, £ = • the functions ^ and ^ (see (1.3)). Following [5, 17, 27], for the vector

. , G Rn, (/3, V are Fourier transforms of

U

Ut u/ x

u/

y

we write the symmetric system (its validity on the solutions of (1.1") is easily verified):

{AoT - Bo£ - Con}U + 4n2|£|2uF = 0

(1.6)

where

Ao

0 =

Co =

k I

I k

m in

m —in k

I

k

k in I m

-in m -1

m -in k

in -m /

k / m

rp* T0

rp* T0

To*

H O2 O2 H

•To

O2 -H -H O2

-H O2 O2 H

•To

To

F

k I

m H

Tn

1

72

( 1 0 -1 \ 0

0 1

10 10 01

k — m —l — in —I + in k + m

O2 is the zero matrix of order 2; k, l, m, n are certain real constants. A0, B0, C0, H are the Hermitian matrices. System (1.6) can also be rewritten in the form

{AnT — — Cnn}U + 4n2|£| 2UF = 0.

(1.6')

Here the bar means the complex conjugation.

Let us multiply system (1.6) scalarly by U and system (1.6') by U and sum up the expressions. Finally we obtain the following identity:

(U, AoU)t - (U, BoU)x - (U, CoU)y+

+ 4n2|£|2(k(|«|2)t + l(|u|2)x + m(|u|2)y} = 0. Here |u|2 = u ■ u. Let integrate (1.7) over the domain R+, assuming that

|U|2 = (U, U) —> 0, |u| —> 0 as r —> to,

where r = \Jx2 + y2. As a consequence, we obtain

d.Ut) + j {(U ,BnU) — 4n2|£|2/|U|2}

R1

x=0

dy = 0.

Here

J (t)

{(U, AnU) + 4n2|^|2 k|U|2}dxdy

R+

Now let us consider the forms (U, A0U) and (U, B0U)|x=0. The form

(U -AnU'= (V •( H H2 )' V

(V1 ,H • V1 ) + (V11 ,H • V11 ) > 0

Tll\

if H > 0, i.e. k > 0, k 2 - m— I- n2 > 0. Here

V = Tn • U

VI

VII

(1.7)

(1.8)

V1 = 7

V2

ut — u

y

V11 =

-Ux

-Ux

V2 \ Uit + Uly

Preparatory to considering the form (U,B0U)|x_0, let us rewrite the boundary conditions (1.2") in the way

(1.2'")

V1 = S ■ V11, x = 0,

where

2a 1 - b

S = | 1 + b 1 + b 1 0

Note that the matrix S is the Hurwitz one if ULC is true (i.e. its eigen-values lie on the left complex semi-plane). The form

' O2 — H

(UU)U = |VA rr „ )• V

-H O

2

x=0

= — (V1, H ■ V11) |x=o — (V11, H ■ V1) |x=o = — (V11, [S*H + HS] ■ V11) |x=o•

Let

S *H + HS = —G, where G = G* > 0 is a certain matrix. Then

(U,BoU)|x=o = ("V11, G ■ V11 )|x=o > 0. (1.9)

Remark 1.3. If S is the Hurwitz matrix, then the Lapunov matrix equation

S *H + HS = —G (1.10)

is uniquely solvable with respect to H for any Hermitian matrix G = G*. As this takes place (if G > 0), then H = H* > 0 (about the solution of the Lapunov matrix equation see, for example, [25]).

Presenting (1.10) as

—si k — m —l — in \ f k — m —l — in \ / — si — s2

—s2 0/y —l + in k + m J y —l + in k + m J y 1 0

—g2 — ig3l, gi4 > 0, gig4 — g2 — g2 > 0,

— g2 + ig3 g4 7 '4 4 2 3

we easily find that

1 + b

l = 2(1—b) < 0 (111)

Here si =1205, S2 = .

In view of (1.9), (1.11), it follows from (1.8):

d -

-Ji(t) < 0. (1.12)

Since

d

dt , , , |U|2dxd^ < Jo(t), (1.13)

summing up (1.12) and (1.13), we can obtain d

— I Ji(t) + // |«|2dxdy} < J°(t) < Mi • {j/1(t) + / |U|2dxdy I. (1.14)

Here

f?2 d2

j/°(t) JJ {|U|2 + (4n2|C|2 + 1)|u|2}dxdy

M1 = max <i 1

k — V/2 + m2 + n2 (1.14) yields:

Jo(t) < M1M2j/o(0)eMli, t> 0, (1.15)

where M2 = max {1, k + v/lT + m2 + n2. In view of the Parceval equality, we obtain the desired a priori estimate for the solutions of Problem I at last:

J J0(t)d£ = J {u2(t,x,y, z) + z) +

Rn R++2

(t, x, y, z) + ^(t, x, y, z) + |Z«(t, x, y, z)|2}dxdydz < < MiM2eMlt J {p2(x,y, z) + ^X(x,y, z) +

R++2

+^2(x, y, z) + (t, x, y, z) + |(^(x, y, z) |2}dxdydz, t > 0

or

llU(t)llW21(R++2) + lUt(t)!L2(R++2) < < MiM2eMlt{|^!W2i(R++2) + lHl!2(R++2)}, t > 0 (1.16)

Here W2 is the Sobolev space (see [5]).

Remark 1.4. Replacement (1.5) and Fourier transform are just an auxiliary expedient. Therefore estimate (1.16) also holds true for initial problem (1.1)-(1.3) provided that ULC (1.4) is fulfilled.

2. Mixed problem for the wave equation the complex coefficients case

Let in boundary condition (1.2) the coefficients a,b,ci, ..., cn be complex numbers: a = a' + ¿a", b = b' + ¿b'', ck = Ck + ¿c'k', k = 1, n. Without loss of generality, we will assume that < = 0. In terms of coefficients of boundary condition (1.2) ULC can be formulated in rather sophisticated way (see [27]), and this situation is not discussed here.

Once again we will try to simplify initial problem (1.1)-(1.3) by a replacement of operators. To this end, first we turn the vector (b'',cf, ... , ^) to make all its components, except for

the first one, equal to zero and then turn the vector (c'1, ..., c^) (its components are defined below). As a consequence, problem (1.1)-(1.3) can be reduced to the following canonical form.

Problem II. We seek the solution of the wave equation

utt - uxx - Uyy - Azu = 0, t > 0, (x, y, z) 6 R++2, (1.1) which satisfy at x = 0 the boundary condition

ut - aux - buy - c 1uz1 = 0, t > 0, (y, z) 6 Rra+1, (2.1) and at t = 0 the initial data (1.3'). Here b = b' + ib'', and

,_ b'b'' + (c' c'')

b'' = sign(cnnMb'')2 + |c''|2, b' = bb +c('C,c ),

b''

C' =(c1, ...,Cn), c'' =(c1', ..., cnn),

ci = sign(c;)|c'1, c' = (4 4),

4 = sign(Cnn )

r'' r' — r' r"

H\ rn-1rn rn-1rn

\! (rii-i)2 + (rn)2

n n

rk-1 S (rjrj0 — rk-1 S (rj0

j=k j=k j "j-7"

c k =-, =-, k = 1, n — 1,

nn

E (rj')^ j j=k-1 j=k

(r0 = b). In what follows we will again denote b, r1 by b and r1. Note that u, ^ are complex-valued functions.

Remark 2.1. Following [27], we formulate ULC for Problem II as the requirement of the positive definiteness for the matrix made up from the coefficients of boundary condition (2.1):

> 0,

i.e.

a' > 0,

(a')2 — (b'')2 — (Re(ab))2 > 0, (2.2)

(1 — c1)[(a')2 — (b'')2] — (Re(ab))2 > 0.

It follows from (2.2), for example, that c1 < 1.

Remark 2.2. In Problem II we make the following replacement of operators t, Z1:

t = {t' + nd }, Z1 = {C1T ' + Z1},

( a' 0 —Re(ab) — a'r1 \

0 a' ib'' 0

—Re(ab) —ib'' a' 0

V — a'r1 0 0 a' /

>/1—' y/T— r

1

or

t' {t — r^}, Z1 = {Z1 — r1T},

1 — r21 1 — r21

Then the relations

T2 - e2 - n2 - £ z2 = (r')2 - e2 - n2 - (ci)2 - £ z2,

fc=i

k=2

t — — bn — ciCi = y 1 — cir' — — bn

are valid. With these relations taken into account, Problem II can be formulated as

n

{(t')2 — £2 — n2 — (Z1 )2 — £ Ck> = 0, t > 0, (x, y, z) G R++2,

k=2

Here

{t' - ae - bn}u = 0, t > 0, x = 0, (y, z) G Rn+1.

a i b

a =—, , b =—, .

(2.1')

In Problem II we carry out the Fourier transform with respect to the variables zk, k = 2, n. Then we obtain the problem

{(t ')2 - e2 - n2 - (zi )2 + 4n2ie'i2}u = o,

t > 0, (x,y,z1) G R+, {t' - ae - bn}u = 0, t > 0, x = 0, (y, z1) G R2, U = ^(x,y,zi,e'), Ut = V(x,y,zi,e'), (x,y,z) G R

(1.1'')

(2.1'') (1.3'')

Here

U = U(t, x, y, z1,e ')

are Fourier transform of the function u, £ = (£2, transforms of the functions ^ and ^ (see (1.3)). Following [5, 17, 27], for the vector

e-2ni(z,'«,)u(t,x,y,zi, z')dz'

, en), z' = (z2, ..., zn) J are Fourier

U

/ t'U \

eu

nu V Ziu y

we write the corresponding symmetric system (by direct calculations one can easily obtain the validity of this system on the solutions (1.1)):

or

where

A

{AT' - Be - Cn - CiZi}U + 4n2|e'|2UF = 0 {AT - Be - Cn - CiZi}U + 4n2|e'|2UF = 0,

C =

(2.3) (2.3')

Ao 0>| B = f Bo 0

0 kj I 0 -l

Co 0

0 - m

3

C1

O3 F

F * 0

T

F 0

the matrices A0, B0, C0, the vector F are described above (see the section 1).

l

A

Vl-cf

(ciCi + A} > 0 if H > 0,

i.e., k > 0, k2 - /2 - m2 - n2 > 0 (c2 < 11);

Ci

= {ciA + Ci}.

Cf

System (2.3') can also be rewritten as follows:

{AT - B£ - Cn - Ci<i}U + 4n2|£'|2uF = 0, (2.3'')

Multiply system (2.3') scalarly by U and system (2.3'') by U and sum up the expressions: (U, AU)t - (U, BU)x - (U, CU)y - (U, CiU)^! +

+4n2ici2

k (|U|2)t - (|U|2)z! + l(|u|2)x + m(|u|2)4 = 0.

x/l-^f

(2.4)

We integrate (2.4) over the domain R+, assuming that |U|2 = (U, U) ^ 0, |U|2 ^ 0 as r ^ to where r = \Jx2 + y2 + z2. In the end, we obtain

d dt

Jt) + ((UU) - 4n2|C'|2/|u|2}|x=0dydzi = 0.

(2.5)

R2

Here

Ji(t)= /((U ,AU) +

R+

vT-

4n2|C'|2l|u|2}dxdydzi.

The form (U, AU) > 0, if H > 0. The form

(U,BU) Ix=0 =(U/,B0U')L=0 - lKi

x=0 "l^lu|

where

And the form where

Ul

(Ul,B0Ul)|x=0 = -(V11, [S*H + HS]V7/),x=0>

S

2a

1 + b 1

1 -b

l + b 0

V

VI

VII

T0U1.

1

k

2

c

1

Note that if ULC (2.2) is true, the matrix S is the Hurwitz one. Hence the Lyapunov matrix equation

S *H + HS = - G

(2.6)

is uniquely solvable with respect to H for any Hermitian matrix G = G* > 0, and H = H* > 0. Then

(U/,BcU/)|x=0 = -(V11 ,GV//) > 0. (2.7)

Remark 2.3. As in the section 1, we can show that l > 0. Let

H

-i

hi h2 h-2 h3

Then equation (2.6) can be rewritten as

H-1S* + SH-1 = -H-1GH-1.

Since

H-1S * + SH-1 =

—2h1Res1 — 2Re(h2s2) h1 — s1h2 — s2h3

hi — s1h2 — «2h3 h2 + h,2

< 0,

where s1

Reh2

2a 1 — b

-, s2 = -r, the inequality 2Reh2 < 0 is true. It is easy to show that

1 + 6 l

k^ l2 — m2 — n2

1+ 6 , i.e. I < 0.

In view of (2.7) and Remark 2.3, (2.5) yields

d

dt

J1(t) < 0.

(2.8)

Further reasonings are analogous to that presented at the end of section 1. That leads us to estimate (1.16) again. Remark 1.4 from section 1 remains also true. In fact, since

(U ,AU) = (W, A W),

where

U = rw, r

/ 1 0 0 — C1 \

1 0 d 0 0

d 0 0 d 0

V — C1 0 0 1 /

w

( TU \

nU

V Z1 u /

d

1 — c1, A = rAr;

then

Amin(A)(WV, W) < (U,AU) < Amax(A)(WV, W). Here Amin(A), Amax(A ) are minimal and maximal eigen-values of matrix A. Since

dtiJL

|u|2dxdydz^ < J0(t),

(2.9)

(2.10)

summing up (2.8) and (2.10), in view of (2.9), we finally obtain

^ jji(i) + if Î |U|2dxdyd^j < J0(t) <

dt

R+

< M J J1(t) + / / / |U|2dxdydz1

(2.11)

R+

Here

Jo(t) =11J (|W|2 + (4n |£'|2 + 1)|ûf}dxdydzi, M3 = max 11,

R+

It follows from (2.11):

Amin (A)

Jo(t) < M3M4 Jo(0)eM3t, t > 0,

(2.12)

where M4 = max(1,Amax(A)}. Regarding the Parceval equality, from (2.12) we obtain the desired a priori estimate for the solutions of Problem II in the form of (1.6). Remark 2.4. The matrix A looks like

A

( k l m — kc

l k in 0

m —in k 0

\ —kc1 0 0 k

7 k _ n , ^ v

k = - , n = - (A > 0). d d

Then where

Amin(A) = k - Q, Amax(A) = k + Q,

Q = \

qo + ^qO - 4n2k2c1

2 2 2 k2 2 q0 = m + n + l +kc1

3. Mixed problem for the wave equation in a domain with a corner

In this section, the mixed problem for the wave equation in the domain

t> 0, (x,y, z) G R++2, n > 1

is studied (the case n = 0 has been considered in [4, 5]):

utt — uxx — uyy — Azu = 0 at t > 0, x > 0, y > 0,

ut — aux — buy — (c, Zu) = 0 at x = 0, ut — auy — — (d, Zu) = 0 at y = 0,

(3.1)

(3.2)

(3.3)

2

u = <^(x, y, z), ut = ^(x, y, z) at t = 0. (3.4)

Here d = (di, ... , dn)T; a, b, ck, a, dk, k = 1, n are real numbers. Without loss of generality, we will assume cn = 0; = {(x, y, z) : x > 0, y > 0, z G Rn}.

Following the idea of section 1, we simplify problem (3.1)-(3.4) by an appropriate replacement of the initial differential operators. For this purpose let us first turn the vector c, making all its components equal to zero except for the first one. Then we make the analogous turn of the vector (d2, ... , dn) (the components of this vector will be defined below). As a consequence, the problem (3.1)-(3.4) can be put in the following canonical form.

Problem III. We seek the solution of the wave equation

utt - uTx - uyy - Azu = 0, t > 0, (x, y, z) G Rn+2, (3.1')

satisfying at x = 0 and y = 0 the boundary conditions

ut - aux - buy - c1uz1 = 0, t > 0, (y, z) G R+1, (3.2')

ut - auy - - d1uZl - d2uZ2 =0, t > 0, (y, z) G R+1, (3.3')

and at t = 0 the initial data (3.1). Here

~ . / \ii T /\(C' d) 7 • / \ dncn-1 dn-1On

Cl = sign(cn)|c|, Oil = sign(cn) . I , dn = sign(cn)-

|c| ' — '' ycn^ c2

Ok-i E (cj dj ) - dk-i E c2

j=k j=k

Ec2• E c2 j=k j=k-1

k = 2, n — 1,

(¿2 = sign(d„)|d'|, d' = (¿2, ... , dn).

In the subsequent discussion c,d1,d2 will be again denoted as ci,di,d2. In addition, we will assume d2 = 0.

Remark 3.1. ULC on the boundary x = 0 for Problem III can be written as (see (1.4)):

a > 0, b2 + c2 < 1, (3.5)

and for the boundary y = 0 as follows:

a > 0, + d2 + d2 < 1. (3.6)

Remark 3.2. In Problem III we make the following replacement of the operators t, Z1)2:

ci I = T ( t1

Z2) v Z2

where T = (tkj), k,j = 1, 3 is the matrix of real coefficients tkj. We will assume that the following relations hold:

t2 - Z? - Z22 = (t')2 - (Z1 )2 - (Z2)2,

k

T-CiCl = rr'- rid, T - diZi - ^2(2 = rr' - r2<1,

where r, r1>2 are certain constants which will be defined below. Then for the coefficients of matrix T and the quantities r, r1)2 we obtain the following system:

t2 t2 t2 = 1 , ni l21 l31 = 1,

Solving this system, we find:

t2 t2 t2 = 1 ,

t12 - t22 - t32 = -1, t2 t2 t2 = 1 ,

t13 - t23 - t33 = -1,

t11t12 - t21t22 - t31132 = 0, t11t13 - t21t23 - t31t33 = 0 t12t13 - t22t23 - t32t33 = 0,

t11 - C1t21 = r, t12 - c1t22 = -r1, t13 - c1123 = 0, t11 - d1t21 - d2t31 = r t12 - d1t22 - d2t32 = -r2, t13 - d1t23 - d2t33 =

t13 = c1t23, t33 = Pt23, t23 = A , A = 1 - c1

22

P =

c1 - d1

c1r

112 = —;—, 111 = r- . 12 A 11 A

t21 = 0, t22 = -

C1rp 2 t31 = —— , r

r1 = -

C1pg

The matrix T 1 looks like

T

-1

C1

At

gP

A

t32 =

g

g =

r2 - r1

A

C1

23

At

23

g

1 + p2' 22

C1P

1

g

At

23

Due to all the above stated, Problem III can be reformulated as

{(r')2 - C2 - n2 - (Z1 )2 - (Z2)2 ^ Z2}u = 0, t> 0, (x,y, z) G R++2;

k=3

(3.1'')

2

2

2

2

2

2

2

2

r

r

0

1

P

(rr' - aC - bn - rid}u = 0, t > 0, x = 0, (y, z) G R++1; (rr' - aC - ßn - r2(i }u = 0, t > 0, y = 0, (x, z) G R++ 1.

(3.2") (3.3")

We carry out the Fourier transform with respect to the variables zk, k = 3, n in Problem III. Then we obtain

((r')2 - C2 - n2 - (Z ')2 - (Z2)2 + 4n2|C''|2}U = 0, t> 0, (x,y,z 1,Z2) G R++;

(rr' - aC - bn - r 1Zi}u = 0, t > 0, x -(rr' - an - ßn - r2Z '}U = 0, t > 0, y

0, (y,z 1,z2) G R+; - 0, (x, zbz2) G R+.

U = <^(x,y,z 1,Z2,C ), Ut = ^^(x,y, z 1,Z2, C ) , (x, y, Z 1, Z2) G R++.

(3.1''')

(3.2''') (3.3''') (3.4')

Here

u = u(t,x,y,zi,z2,£")= J e-2ni(Z''^,)M(i,x,y,Zi,Z2, z")dz';

Rn-2

is Fourier transform of the function u, £'' = (£3, ... , £n), z'' = (Z3, ..., zn). For the vector

( T'u \

zu

U = nu

Z' u Z2 u y

we write down the following symmetric system (its validity can be easily verified by the straightforward calculations):

or

where

A

(AT' - BC - Cn - C1Z' - C2Z2}U + 4n2|C''|2UF = 0

(Ar - BC - Cn - C1Z1 - C2Z2}U + 4n2|C''|2UF = 0,

B

(Ar' - BC

(Ar - BC

/ An 0

= 1 0 k

(3.7) (3.7')

Bn 0 ^ , C = Cn 0

0 -l J 0 -m

C1

An

Do 0 C2

0 -n

(k l m n

l k 0 0

m 0 k 0

n 0 0 k

O4 F

F * 0

F

Bn

F 0

0

Cn

( m 0 k 0 \

0 -ml 0

k l m n

y 0 0 n m y

( k \

I

D

F

m n

n=

1

( I k 0

k l m n 0 m -l 0

\ 0 n 0 -i y

( n 0 0 k \

0 -n 0 l

0 0 -n m

y k l m n y

C1

A = -A +^C2, r At23

(i = /1 i 2\ A - P^^i + "a— C2, C2 = 2s A + -(7i + -T^C2;

r(1 + P2) g At23 r(1+ P2) g At23

k,1,m, n, are real constants. Note that the matrix A > 0 if k > 0, k2 - /2 -

-m2 - n2 > 0 and r = , / ^ 2, r > 0. Rewrite system (3.7') as

{AT - BC - Cn - CiZi - C2C2}U + 4n2|£''|2UF = 0. (3.7'')

Multiply scalarly system (3.7') by U, and system (3.7'') by U, and sum up the expressions:

(U, AU)t - (U, BU)x - (U, CU)y - (U, CiU)z! - (U, C2U)z2 + +4n2|C''|2{k [(N2)t - ^(N2)z! - ^(|U|2)z2] +

+1(|U|2)x + m(|u|2)y + n [-P(|u|2)z! + (|U|2)z2 ]} = 0. (3.8)

g

Integrate (3.8) over the domain R++, assuming that |U|2 ^ 0, |U|2 ^ 0 at r —► 00, where r = \Jx2 + y2 + z2 + z|. In the end, we obtain

ddtJi(t) + //\{(U,BU) - 4n2|e''|2/|u|2}|x=0dydzidz2+ + II/{(UC, CU) - 4n2|C''|2m|U|2}|y=0dxdzidz2 = 0. (3.9)

Here

k

Ji(t)= / {(U,AU) + -4n2|£T|«|2}|x=0dxdydzidz2

R4

R++

The form (U, AU) > 0 if k > 0, k2 - /2 - m2 - n2 > 0. The form

(U ,B U) I x=0 = (U l,B0Ul)L=0 - 11Z2 U |2 I x=0.

Analogously, the form

(U,CU) Iy=0 = (Ul,C0Ul)|y=0 - m|Z2u|2|y=0.

Here

Ul

( t1« \

n«

V ciu)

In view of (3.2111), (3.3111), the conditions

(U 1,B0U1)L=0 > 0,

(U 1,C0Ul)|y=0 > 0

can be reformulated as the requirement of the positive definiteness for the following matrices:

2ka + /(a2 + 1) b(/a + k) + m c(/a + k) + n

b (la + k) + m -/(1 — b2) c (/a + k) + n / be

a b b

c =

/ bc —/(1 — c2)

ri

> 0

(3.10)

—m(1 — ß2) ß (ma + k) + / mße

ß (ma + k) + / 2ka + m(a2 + 1) e (ma + k) + n mße e (ma + k) + n —m(1 — e2)

- a ß = ß - = r r r

>0

(3.11)

a

Besides, we require

k > 0 , m < 0 , / < 0 , k2 — m2 - /2 - n2 > 0.

(3.12)

Then with regard to (3.10), (3.11), (3.12) it follows from (3.9) that

d

dJi(t) < 0. (3.13)

Further reasonings are analogous to that at the end of the sections 1, 2. Actually, since

(U, AU) = (W, A W),

where

U = rW, A = r*Ar,

r

1 0 0 Ci Cip

r r(1 + p2) r(1 + p

0 1 0 0 0

0 0 1 0 0

0 0 0 p 1

g g

Ci 0 0 1 p

At23 At23 At23

W

( tu \

nu

Ciu V Z2U )

then Since

Amin(A)(W, W) < (U, AU) < Amax(A)(WV, W).

~dt \ J Iu12< J0(t),

R+ +

(3.14)

(3.15)

r

summing up (3.13), (3.15) and using (3.14), we finally obtain

— i Ji(t) + J |U|2dxdydzidz2 ^ < J0(t) <

R4

R++

< MA Ji(t) + |U|2dxdydzidz2 >. (3.16)

R++

Here

Jo(t) = / {|W|2 + 4n2|«|2|£T + |U|2}dxdydzidz2

R++

M5 = max 11,

1 k

Amin(A ) r

From (3.16) it follows that

Jo(t) < M5M6 Jo(Q)eMBi, t> 0, (3.17)

where M6 = max{1,Amax(A ),k}. Regarding the Parceval equality, from (3.17) we obtain in the end the desired a priori estimate for the solutions of Problem III from (1.16). Remark 1.4 from section 1 remains true.

Remark 3.3. A priori estimate (3.17) was obtained under conditions that matrices (3.10), (3.11) are positive definite and unequalities (3.12) hold. It has not been shown yet that ULC (3.5), (3.6) are sufficient for the existance of such real numbers k, l, m, n that (3.12) is true and the matrices (3.10), (3.11) are positive definite. The special examples have been considered in the diploma thesis of A. A. Beljaev, a student of the Novosibirsk University.

4. Mixed problem for the vector wave equation

In this section, the mixed problem for the wave equation in the domain t > 0, (x, y, z) G R++2, n > 1 is briefly discussed (the case n = 0 has been considered in [18]):

L(r,e,n,Ci, ..., Zn)U = Utt - Uxx - Uyy - AzU = 0 at t> 0, x> 0; (4.1)

JiUt - AiUx - BiUy = 0, at x = 0; (4.2)

U = $(x,y, z), Ut = ^(x,y, z) at t = 0. (4.3)

Here Ji, Ai, Bi are the constant complex matrices of order N.

Remark 4.1. We will assume that for problem (4.1)-(4.3) ULC holds. In terms of the coefficients of boundary conditions (4.2) ULC can be written as follows (see [18]):

a) the matrix Ji + Bi is not degenerate,

b) all the eigen-values of the matrix

S--

-Si -S2

In On

lie strictly in the left semi-plane, i.e.

ReAi(S) < 0, i = 1, 2N.

Here Si = 2(Ji + Bi)-1Ai, S2 = (Ji + Bi)-1(Ji — Bi). Note that the characteristic equation for S

det(S — A/2N) = 0

can be written as

det(A2/N + ASi + S2) = 0.

In problem (4.1)-(4.3) we make the Fourier transform with respect to the variables zk, k = 1,n. In the end, we obtain:

Utt - Uxx - Uyy + 4n2|f|2U = 0, t> 0, (x,y) G R+, JiU - AiUUx - BiUuy = 0, t> 0, y G R1, x = 0,

U = $(x,y,f), Ut = *(x,y,f), (x,y) G R

For the vector

W

/U t\

U x

V U y /

we write the symmetric system

(4.1')

(4.2') (4.3')

(AqT - BQC - Con}W + 4n2|f|2FU = 0

(4.4)

Here

A0

Bo

Co

K L M

L K iN

m -iN k

L K iN

K L M

-iN M -L

M -iN K

iN -M L

K L M

To*

To*

T0*

H °2N

°2N H

02N - H

- H O2N

- H 02N

O2N H

T

T

T

( In On -In

1 On -In On

72 On -In On

V In On In )

M -L - iN \

iN K + M

K

F = I L

M

H

K, L,M, N are some arbitrary for the present Hermitian matrices of order N. System (4.4) can also be written in the form

{Aot - Bof - (Con}W + 4n2|f|2FU = 0.

(4.4')

2

u

Let multiply scalarly system (4.4) by W and system (4.4') by W. Sum up the expressions obtained. In the end, we have:

(W, AoW)t - (W, BoW)x - (W, CoW)y+

+4n2|£|2{(U, KU)t + (U, LU)x + (U, MU)y} = 0. (4.5)

Integrate (4.5) over R+, assuming that |W|2 = (W, W) ^ 0, |U|2 ^ 0 as r ^ to where r = \Jx2 + y2. Thus, we obtain

-dJi(t)+ /{(W,BoW) - 4n2|£|2(U,LU)} dy = 0. (4.6)

dt J x=o

R1

Here

Ji(t) = JJ{(W, AoW) + 4n2|£|2(U, KU)}dxdy. R+

Now we consider the forms (W, A0W) and (W, B0W)|x=0. The form

H O2N

oW

'2N

(W.A„W) = jv,^ -j.v

= (V1, H ■ V1 ) + (V11, H ■ V11 ) > 0,

if H > 0. Here

V1 V

V = Tn ■ W = ,

V1 = W Ut - Uy V11 = W -Ux

V2 \ -U y ' v^ V Ut + U

We rewrite boundary conditions (4.2') as:

V1 = S ■ V11, x = 0, (4.2'')

The form

(W,BnW)U = (V,( -2H -£) ■ v)U =

= -(V1, H ■ V11 )|x=n - ("V11, H ■ V1 )|x=n = — ("V11, [S*H + HS] ■ V11 )|x=n.

Let

S *H + HS = - G, where G = G* > 0 is a certain matrix. Then

(W,BnW)|x=n = (V11, G ■ V11 )|x=n > 0. (4.7)

Now we assume that

L < 0. (4.8)

Then from (4.6), in view of (4.7), (4.8), we obtain

d

J < 0. («)

Since

d

dt , , , (U, U)dxdy }> < J0(t), (4.10)

R+2

summing up (4.9) and (4.10), in the end we obtain

dft< Ji(t) + / I Udxd^ < J0(t) < M7 ^ Ji(t) + 11 |U|adxdy }>. (4.11)

Here

22 R+ R+

J0(t) = JJ (|W|2 + (4n2|£|2 + 1)|U|2}dxdy

R+2

M7 = max < 1

(4.11) yields:

Amin (A0) Amin (K)

J0(i) < M7M8J0(0)eM7t, t> 0, (4.12)

where Ms = max {1, Amax(A0)}. With regard to the Parceval equality, from (4.12) we obtain the desired a priori estimate for the solutions of problem (4.1)-(4.3):

II^IIwkr^ +11^)11^+2) < < M7MseM7i{|$|W2i(R++2) + ii^tHL2(R++2)>, t >

Now we discuss the question of validity of equality (4.8). By virtue of the Lapunov theorem (see [25]), ULC being true (see the Remark 4.1), the matrix equation

S *H + HS = -G (4.13)

has the unique solution H = H* > 0 for any right-hand side G = G* > 0. Here

G = ( Gi G2 H = ( Hi H2

G = ^ G2 G3 J , H = ^ H2* H3 Gi,3 = Gi,3 > 0, Hi,3 = Hi*3 > 0,

and

k = k * = 2(Hi + H3), M = M* = 2(H3 - Hi),

L = L* = - 2(H2 + H2*),

1

1

N = N * = - - H2).

Rewrite the matrix equation (4.13) as

H1S1 + S*Hi — Gi + H2 + H*

Я* C 1 c* u _

2 S2 + S2 H2 — G3,

H3 — H1S2 — S* H2 — —G2,

ЯC* U U* Q _

3 — S2 Hi — H, Si — —G2.

(4.13')

On the whole, the question of validity of (4.8) remains open. However, it is possible to point out such boundary conditions (4.2) for which this question can be easily solved. Indeed, let B = ON. Then S2 = , and second subsystem (4.13') can be written as

H2 + H2 = G3,

i.e.

L — — 1(H2 + H*) — — 2G3 < 0.

Received for publication Oktober 13, 1996