THE DEPENDENCE OF SOLUTION UNIQUENESS CLASSES OF BOUNDARY VALUE PROBLEMS FOR GENERAL PARABOLIC SYSTEMS ON THE GEOMETRY OF AN UNBOUNDED DOMAIN

GEORGIAN MATHEMATICAL JOURNAL: Vol. 5, No. 4, 1998, 31-33 THE DEPENDENCE OF SOLUTION UNIQUENESS CLASSES OF BOUNDARY VALUE PROBLEMS FOR GENERAL PARABOLIC SYSTEMS ON THE GEOMETRY OF AN UNBOUNDED DOMAIN A. GAGNIDZE Abstract. General boundary value problems are considered for general parabolic (in the Douglas Nirenberg Solonnikov sense) systems. The dependence of solution uniqueness classes of these problems on the geometry of a nonbounded domain is established. The dependence of solution uniqueness classes of the first boundary value problem for a second-order parabolic equation in an unbounded domain on the domain geometry was considered in []. It was established there that the uniqueness class could be wider than the solution uniqueness class of the Cauchy problem for the above-mentioned equation. Analogous results were obtained in [] by the method of barrier functions. In [3] O. Oleinik constructed examples of second-order parabolic equations in the exponentially narrowing domains for which the solution uniqueness class for the second boundary value problem is the same as the solution uniqueness class of the Cauchy problem although the solution uniqueness class in this domain is wider. Later, in [4] E. Landis showed that if the domain narrows with a sufficient quickness at x, then the solution uniqueness class for the second boundary value problem can be wider than that of the Cauchy problem. In [5] the author considered degenerating parabolic equations of second order and obtained analogous uniqueness theorems for general boundary value problems. This paper deals with general boundary value problems for general parabolic systems in unbounded domains. Such problems were studied in [6], where solvability conditions similar to the Shapiro-Lopatinski conditions for 1991 Mathematics Subject Classification. 35K50. Key words and phrases. Parabolic systems, uniqueness classes, influence of domain geometry. 31 107-947X/98/0700-031$15.00/0 c 1998 Plenum Publishing Corporation

3 A. GAGNIDZE elliptic boundary value problems were found. The solvability conditions for analogous problems in unbounded domains were obtained in [7], [8]. Let ω be an unbounded domain in R n+1 contained between the planes {t = 0} and {t = T = const > 0} and the surface γ lying in-between these planes. Let us consider, in this domain, a linear system of differential equations with complex-valued coefficients of the form a αβ kj (x, β t)dα x t β u j(x, t) = 0, (1) α +bβ s k +t j where b, N are positive integers, s 1, s,..., s N, t 1, t,..., t N, β are integers, s j 0 and t j 0 for all j = 1,,..., N, a αβ kj (x, t) 0 if s k + t j < 0; (s j + t j ) = bm, m is a positive integer, α = (α 1,..., α n ) is a multiindex; α = α 1 + α + + α n ; D α x = D α1 x 1 is the imaginary unit). The matrix L 0 (x, t, ξ, σ) with elements a αβ kj (x, t)ξα σ β α +bβ s k +t j Dx α D α n x n, D xj = i x j (i (k, j = 1,,..., N), where ξ R n, σ is a complex-valued number, ξ α = ξ α 1 1 ξαn n, will be called the principal part of the symbol of system (1). It is assumed that system (1) is uniformly parabolic in the domain ω. Following [6] [9], we shall say that system (1) is parabolic in the domain ω if there exists a positive constant λ 0 called a parabolicity constant such that for any (x, t) ω and ξ R n the roots σ 1,..., σ m of the polynomial P (x, t, ξ, σ) = det L 0 (x, t, ξ, σ) with respect to σ satisfy the inequality where ξ = n ξi. Re σ s (x, t, ξ) λ 0 ξ b (s = 1,,..., m), () Denote by L 0 (x, t, ξ, σ) the matrix of algebraic complements to the elements of the matrix L 0 (x, t, ξ, σ). Then L 0 = p L 1 0. On the surface γ, we give general boundary conditions of the form b αβ ν j (x, t)d α β x t β u j(x, t) = 0 (ν = 1,,..., bm), (3) γ α +bβ q ν+t j where q ν are some integers.

BOUNDARY VALUE PROBLEMS FOR PARABOLIC SYSTEMS 33 Let ( x, t) γ. Consider the matrix B 0 ( x, t, ξ, σ) with ν j ( x, t)ξ α σ β. α +bβ=q ν+t j b αβ Let ν 0 = (ν 1,..., ν n ) be the unit vector of the external normal to the surface s t = γ {x, t : t = t} at the point ( x, t), and η( x, t) be any tangential vector to s t at the same point. It follows from the parabolicity condition (see [6] and [10]) that the polynomial P ( x, t, η( x, t) + τν 0 ( x, t), σ) with respect to τ has bm roots, τ s +( x, t, η, σ), (s = 1,,..., bm), with positive imaginary parts if there exists a constant λ 1 such that 0 < λ 1 < λ 0 and the inequalities Re σ λ 1 η( x, t) b, σ + η( x, t) 4b > 0 are fulfilled. We set M + ( x, t, η, σ, τ) = bm (τ τ+( x, s t, η, σ)). s=1 As to the boundary conditions (3) and system (1), it is assumed that the conditions for being complementary are fulfilled for any point ( x, t) γ. The condition for being complementary for system (1) and the boundary conditions (3) is fulfilled at the point ( x, t) if the rows of the matrix A( x, t, η( x, t) + τν 0 ( x, t), σ) B 0 ( x, t, η( x, t) + τν 0 ( x, t), σ ) L 0 ( x, t, η( x, t) + τν 0 ( x, t), σ ) are linearly independent modulo the polynomial M + ( x, t, η, σ, τ) with respect to τ provided that Re σ λ 1 η( x, t) b, σ + η( x, t) 4b > 0. We set ( bm 1 s=0 q s hj( x, t, η, σ)τ s) A ( x, t, η( x, t) + τν 0 ( x, t), σ ) ( mod M + ( x, t, η, σ, τ) ). Consider the matrix Q( x, t, η, σ) with elements q s b j ( x, t, η, σ), which has bm rows and bmn columns (h = 1,,..., bm; s = 0, 1,..., bm 1; j = 1,,..., N). Let k ( x, t, η, σ), (k = 1,,..., N 1 ) be the minors of order bm of the matrix Q and let ( x, t, η, σ) = max k k ( x, t, η, σ). According to [6] the condition for being complementary is fulfilled uniformly on the surface γ if γ inf ( x, t, η( x, t), σ ) > 0, (4) ( x, t) γ Re σ λ η( x, t) b, σ + η( x, t) 4b = 1.

34 A. GAGNIDZE On ω 0 = ω {x, t : t = 0} we give initial conditions of the form c αβ h j (x, t)d α β x t β u ω0 j(x, t) = 0 (h = 1,,..., m), (5) α +bβ r h +t j where r h are some negative integer numbers and the coefficients c αβ h j (x, t) 0, when r h + t j < 0. Consider the matrix C 0 (x, t, ξ, σ) with elements α +bβ=r h +t j c αβ h j (x, t)ξ α σ β. For system (1) and the boundary conditions (5) the condition for being complementary is fulfilled at the point ( x, t) ω if the rows of the matrix H( x, t, σ)=c 0 ( x, t, 0, σ) L 0 ( x, t, 0, σ) are linearly independent modulo σ m. Let ( m 1 s=0 d s h j ( x, t)σ s ) H( x, t, σ)(mod σ m ). Consider the matrix H( x, t) with elements d s h j ( x, t), which has m rows and mn columns (s = 0, 1,..., m 1; j = 1,,..., N). Let E k ( x, t), (k = 1,,..., L 1 ) be the minors of order m of the matrix H( x, t) and let ( x, t) = max E k ( x, t). k According to [6] the condition for being complementary is fulfilled uniformly in ω if 0 inf ω ( x, t) > 0. (6) We introduce a space of functions, where a solution of problem (1), (3), (5) will be considered. Let G R n x (0, T ) be a finite domain. Denote by u G p,0 the norm of the function u in the space L p (G), u G p,0 = ( G u p dx dt) 1/p. We define the norm u G p,bs = α +bβ bs Dx α β u G t p,0, where s is some β positive integer. Denote by Wp bs,s (G) the space of functions obtained by completing with respect to the norm u p,bs G the set of smooth functions in G. As to the surface γ, it is assumed that it satisfies the conditions of uniform local unbending. By [9] this means the following:

BOUNDARY VALUE PROBLEMS FOR PARABOLIC SYSTEMS 35 In the space R n+1 x,t we consider the sets H( x, t; ρ) = { x, t : x j x j < ρ, j = 1,,..., n; p b < t t < ρ b} ; H ( x, t; ρ) = { x, t : x j x j < ρ, j = 1,,..., n; p b < t t 0}; H + ( x, t; ρ) = { x, t : x j x j < ρ, j = 1,,..., n; 0 t t < ρ b} ; H + ( x, t; ρ 1, ρ ) = { x, t : x j x j < ρ 1, j = 1,,..., n; 0 t t < ρ b Analogous sets in the space R n+1 y,τ are denoted by s(ŷ, τ, ρ), s (ŷ, τ, ρ), s + (ŷ, τ, ρ), and s + (ŷ, τ, ρ 1, ρ ), respectively. The surface γ will be said to satisfy the condition of uniform local unbending with the constants d(ω), M(ω) and γ C s +t,b, where t = max{t j } and s > q = max(0, q 1,..., q bm ), if the following conditions are fulfilled: 1. For any point ( x, 0) ω 0 there exists a neighborhood such that O x,0 the set ω is homeomorphic under some nondegenerate transformation O x,0 of the coordinates = {y = f(x, t), τ = t : f( x, 0) = x} to the set Ψ x,0 s + (0, 0; κ 1 ) {y n 0}, 0 < κ 1, and the set γ is homeomorphic to O x,0 s + (0, 0; κ 1 ) {y n = 0}. It is assumed that the number κ 1 does not depend on the point ( x, 0) ω 0. Let M 0 = Ψ 1 x,0 ( x,0) ω 0 ( s +( 0, 0; κ 1 ) ) {y n 0}.. For any point ( x, T ) ω T, where ω T = ω {t = T }, there exists a neighborhood such that the set ω is homeomorphic under O x,t O x,t some nondegenerate transformation of the coordinates = {y = f(x, t), Ψ x,t τ = t T ; f( x, T ) = 0} to the sets s (0, 0; κ ) {y n 0}, and the set γ is homeomorphic to s O x,t (0, 0; κ ) {y n = 0}, 0 < κ 1. It is assumed that the number κ does not depend on the point ( x, T ) ω T. Let M T = ( Ψ 1 x,t ( x,0) ω T s ( 0, 0; κ ) ) {y n 0}. 3. The transformations,, Ψ Ψ x,0 Ψ x,t 1, x,0 Ψ 1 are given by the functions x,t whose norms in the space C s +t,t are bounded by the variable K 1 (γ) 1. 4. For any point ( x, t) γ 1, where { γ 1 = γ x, t : 1 ( κ1 ) b(k1 (γ)(n + 1)) 1 t T 1 ( κ ) b(k1 (γ)(n + 1)) 1} there exists a neighborhood such that the set ω is homeomorphic O x, t O x, t under some nondegenerate transformation of the coordinates Ψ x, t = {y = }.

36 A. GAGNIDZE f(x, t), τ = t t; f( x, t) = 0} to the set s(0, 0; κ 3 ) {y n 0}, 0 < κ 3 1, and the set O x, t γ is homeomorphic to s(0, 0; κ 3) {y n = 0}. Assume that κ 3 does not depend on the point ( x, t) γ 1. Let ( ( M γ1 = Ψ 1 s 0, 0; κ ) ) 3 {y n 0}. x, t ( x,wht) γ 1 5. The transformations and ψ 1 are given by the functions whose ψ x, t x, t norms in the space C s +t,t are bounded by the variable K (γ) 1. Let K(γ) = max{k 1 (γ), K (γ)}. Since the transformations ψ 1, x,0 ψ 1 x,t and ψ 1 shorten the distance between the points (n+1)k(γ) times at most, x, t the distance from any point of the set ω 1 = ω\{m 0 M T M γ1 } to γ is at least { 1 d 1 (γ) = min ( κ1 ) b(k(j)(n + 1)) 1, ( κ3 1 ) b(k(j)(n + 1)) 1 }. ( κ ) b(k(j)(n + 1)) 1, We set d(ω) = d 1 (γ)(n + 1) 1, M(ω) = K(γ)(n + 1). It is easy to check that d(ω) < 1, and for any point (x 0, t 0 ) ω 1 the parallelepiped H(x 0, t 0 ; ρ) {0 < t < T }) belongs to ω when ρ < d(ω). The following statement is proved in [9]. Lemma 1. Let system (1) be uniformly parabolic in ω with the parabolicity constant λ 0. Let system (1), and the boundary and initial conditions (3) and (5) satisfy the conditions for being complementary to (4) and (6), respectively. It is assumed that the surface γ satisfies the condition of a uniform local unbending with the constants d(ω), M(ω), and γ C s +t,b, where t = max(t 1,..., t N ), s > q = max(0, q b1,..., q bn ). Let the integer numbers t j, s k, q ν, and r h be divisible by b, and for the coefficients of problem (1), (3), (5) the condition a αβ kj ; s k,b Cs (ω) + b α β νj ; C s q ν,b (γ) + + sup c α β hj ; C s r h (ω {t = τ}) M, τ [0,T ] be fulfilled, where M = const > 0, k, j = 1,,..., N, ν = 1,,..., bm, h = 1,,..., m, α +bβ s k +t j, α +bβ q ν +t j, α +bβ r h +t j. Let l > 0, 0 < τ T, 1 p <, and 0 < ρ min(1, d(ω), τ 1/b ). Then for any solution u of the homogeneous problem (1), (3), (5) such that u j W s +t j,(s +t j)/b( ) p ω(l + M(ω)ρ; 0, τ) (j = 1,,..., N),

BOUNDARY VALUE PROBLEMS FOR PARABOLIC SYSTEMS 37 we have an estimate C 0 ρ s +t uj ; W s +t j,(s +t j )/b p (ω(l; 0, τ)) N ρ s t j uj ; L p (ω(l + M(ω)ρ; 0, τ)), (7) where ω(l; 0, τ) = ω {x, t : x j < l, j = 1,,..., n; 0 < t < τ}; the constant C 0 depends on the set of constants {K} = {λ 0, n, s, m, b, t 1,..., t N, s 1,..., s N, M, M(ω), d(ω)}. Let us introduce an additional independent variable x 0 and assume that the domain Ω = ω R 1 x 0. In the domain Ω we consider an additional system of the form ( a αβ β kj (x, t)dα x t β + p 1Dx b 0 )ν j (x, x 0, t) = 0 (8) α +bβ s k +t j (k = 1,,..., N) with the boundary conditions on Γ = γ R 1 x 0 ( b αβ β νj (x, t)dα x t β + p 1Dx b 0 )ν j (x, x 0, t) = 0 (9) α +bβ q ν +t j (ν = 1,,..., bm), and the initial conditions on Ω 0 = ω R 1 x 0 ( c αβ β hj (x, t)dα x t β + p 1Dx b 0 )ν j (x, x 0, t) = 0 (10) α +bβ r h +t j (h = 1,,..., m), where p 1 is a positive integer. The following statement is also proved in [9]. Lemma. (a) System (8) is uniformly parabolic in the domain Ω with the parabolicity constant λ(p 1 ); λ(p 1 ) λ 0 for p 1, where λ 0 is the parabolicity constant of system (1). (b) For system (8) and the initial conditions (10), the conditions for being complementary are fulfilled uniformly in Ω with the constant 0 defined by condition (6).

38 A. GAGNIDZE (c) There are positive constants A 0 and p 0 such that, for p 1 p 0, system (8) and the boundary conditions (9) satisfy the conditions for being complementary uniformly on Γ with the constant Γ A 0 γ, where γ is defined by condition (4). (d) The coefficients of problem (8) (10) satisfy the conditions of Lemma 1 with the constant M(p 1 ) > 0. (e) The function v(x, x 0, t) = exp{iµx 0 p 1 µ b t}u(x, t) is a solution of problem (8) (10) for any real parameter µ if only u is a solution of problem (1), (3), (5). Let u be a solution of problem (1), (3), (5). Consider two additional functions w(x, t) = exp{ p 1 µ b t}u(x, t) and v(x, x 0, t) = exp{iµx 0 }w(x, t). By Lemma the function v is a solution of problem (8) (10) and thus this problem satisfies the conditions of Lemma 1. Then for v the inequality ρ s +t vj ; W s +t j,(s +t j)/b (Ω(l; 0, τ)) C 0 N ρ s t j vj ; L p (Ω(l + M(ω)ρ; 0, τ)) (11) holds, where Ω(l; 0, τ) = ω(l; 0, τ) { x 0 < l} and the constant C 0 depends on the set of constant {K p1 } = {λ(p 1 ), n, s, t 1,..., t N, s 1,..., s N, b, m, M(p 1 ), s(ω), M(ω)}. Lemma 3. For the functions w and v, for any l > 0 and 0 < τ T the estimate vj ; L p (Ω(l; 0, τ)) C j wj 1 ; L p (ω(l; 0, τ)) (1) holds, where the constant C j 1 depends on l and p. Proof. It is easy to see that v j ; L p (Ω(l; 0, τ)) = exp{iµx0 }w j ; L p (Ω(l; 0, τ)) = ( ) 1/p ( 1/p = w j p dx dx 0 dt = e 1/p w j p dx dt). Ω(l;0,τ) Hence follows inequality (1). ω(l;0,τ) Lemma 4. For the functions v and w, for any l > 0, µ > 0, 0 < τ T the inequality C j (µ, l) wj ; L p (ω(l; 0, τ)) v j ; W s +t j,(s +t j )/b p (Ω(l; 0, τ)) (13) holds, where the constant C j depends only on l, m, p.

BOUNDARY VALUE PROBLEMS FOR PARABOLIC SYSTEMS 39 Proof. As in proving Lemma 3, it is easy to see that vj ; W s +t j,(s +t j )/b p (Ω(l; 0, τ)) = ( D s +t j x 0 v j Ω(l;0,r) p,0 = Ω(l;0,τ) α +bβ s +t j D α x,x 0 β p,0 t β v j Ω(l;0,τ) µ (s +t j)p w j p dx 0 dx dt) 1/p = = µ s +t j e 1/p w j ; L p (ω(l; 0, τ)). Hence follows inequality (13) for C j = µs +t j l 1 p. Lemma 5. For the function w, for any l > 0, 0 < τ T, and 0 < ρ min(1, d(ω), τ 1/b ) the inequality C j 3 (µ, ρ, t j wj l)ρt ; L p (ω(l; 0, τ)) ρ t t j w j ; L p (ω(l + M(ω)ρ; 0, τ)) (14) holds, where the constant C j 3 depends only on µ, ρ, l, p. Proof. With (1) and (13) taken into account, inequality (11) implies C 0 N ρ s +t C j (µ, l) w j ; L p (ω(l; 0, τ)) ρ t t j C j 1 (l + M(ω)ρ) wj ; L p (ω(l + M(ω)ρ; 0, τ)). Hence follows inequality (14) for C j 3 = e 1 p (l + M(ω)ρ) 1/p 1 C 0 (µρ) s +t j. Lemma 6. Let u be a solution of problem (1), (3), (5). Then for the function w, for any m 1 m 0 = const > 0 and 0 < τ T the inequality wj ; L p (ω( m1 ; 0, τ)) e k(m 1 ) w j ; L p (ω( m1+1, m 1 ; 0, τ)) (15) holds, where k(m 1 ) = [ λ b b 1 m1] and λ is a positive number, while ω(l, l 1 ; 0, τ) = ω(l ; 0, τ)\ω(l 1 ; 0, τ) for l > l 1.

330 A. GAGNIDZE Proof. Assume l 1 = m 1 and l = l 1. Let ρ(m 1 ) = m1 /(M(ω)k(m 1 )). Let m 0 be a sufficiently large number such that for m 1 m 0 the inequality m1 /(M(ω)K(m 1 )) min(1, d(ω), τ 1 b ) (16) is fulfilled. Let l 1 (n 1 ) = l 1 + n 1 M(ω)ρ and l 1 (0) = l 1. Then inequality (14) implies C j t j wj 3 ρt ; L p (ω(l 1 + n 1 M(ω)ρ; 0, τ)) ρ t t j wj ; L p (ω(l 1 + (n + 1)M(ω)ρ; 0, τ)). It is easy to see that ( C j 3 = m 1 + n 1 M(ω)ρ ) 1 p 1 m (µρ) s +t j 1 p 1 C 1 + 0 (n1 + 1)M(ω)ρ C (µρ(m 1)) s +t j. 0 Let µ = λb 0 m 1 b 1. Since ρ(m1 ) m 1 (M(ω)λ) 1 bm 1 m 1 b 1 =(M(ω)λ b 1 ) 1, we have C j 3 1 p C 1 +t j 0 bs 0. It is easy to check that if b 0 = ( p+1 p C0 e) 1 s, then C j 3 e for any j = 1,,..., N. In that case we obtain eρ t t j wj ; L p (ω(l 1 (n 1 ); 0, τ)) N ρ t t j wj ; L p (ω(l (n 1 ); 0, τ)), where l (n 1 ) = l 1 (n 1 + 1). Hence we find (e 1)ρ t t j wj ; L p (ω(l 1 (n 1 ); 0, τ)) ρ t t j wj ; L p (ω(l (n 1 ), l 1 (n 1 ); 0, τ)). But since e 1 > 0, we have e 1 ρ t t j wj ; L p (ω(l 1 (n 1 ); 0, τ)) N ρ t t j wj ; L p (ω(l (n 1 ), l 1 (n 1 ); 0, τ)). (17) If we now apply inequality (17) corresponding to n 1 = ν + 1 to estimate the right-hand side of this inequality corresponding to n 1 = ν and assume

BOUNDARY VALUE PROBLEMS FOR PARABOLIC SYSTEMS 331 successively that ν = 0, 1,,..., (k 1) and also take into account the fact that the domain ω(l (k(m 1 ) 1), l 1 (k(m 1 ) 1); 0, τ) is contained in the domain ω(l, l 1 ; 0, τ), then, since ρ 1, we obtain ρ t t)j wj ; L p (ω( m1 ; 0, τ)) e k(m 1) wj ; L p (ω( m1+1, m 1 ; 0, τ)). Since t t j 0 for any j, we have (ρ(m 1 )) t t j ( m1 /(M(ω)k(m 1 )) ) t t j (M(ω)k(m1 )) tj t. Since k(m 1 ) as m 1, the estimate ( M(ω)k(m 1 ) ) t t j exp(k(m 1 )/) holds for m 1 m 0, where m 0 is a sufficiently large number. Thus for m 1 m 0 the function w satisfies the inequality wj ; L p (ω( m 1 ; 0, τ)) } N e k(m 1 ) w j ; L p (ω( m1+1, m1 ; 0, τ)). Theorem. Let the domain ω be such that for a solution u of problem (1), (3), (5) in the domain ω the equality b lim exp{ σr b 1 } u; Lp (ω(r, R; 0, T )) = 0 (18) R holds, where 1 ρ < and σ is a positive constant. Then u 0 in ω. Proof. Since w(x, t) = exp{ p 1 µ b t}u(x, t), inequality (15) implies u j ; L p (ω( m1 ; 0, τ)) { exp 1 } k(m 1) + p 1 µ b (m 1 )τ uj ; L p (ω( m1 +1, m1 ; 0, τ)), where k(m 1 ) = [ bm 1 b 1 λ ], µ(m1 ) = λb 0 m 1 b 1. Let λ = 4σ and τ 0 = (4p 1 λ b 1 b b 0 ) 1. Then for any τ τ 0 we have u j ; L p (ω( m 1 ; 0, τ)) { exp σ b b 1 m1 } uj ; L p (ω( m+1, m1 ; 0, τ)).

33 A. GAGNIDZE Passing in the latter inequality to the limit as m 1 and taking into account condition (18), we find that u 0 in the domain ω {x, t : 0 t τ 0 }. Next we find that u 0 in the domains ω {x, t : τ 0 t τ 0 }, ω {x, t : τ 0 t 3τ 0 },.... Therefore u(x, t) 0 in ω. References 1. O. A. Oleinik and G. A. Iosifyan, An analog of Saint Venant s principle and the uniqueness of a solution of the boundary value problems in unbounded domains for parabolic equations. (Russian) Uspekhi Mat. Nauk 31(1976), No. 6, 14 166.. Ya. M. Tsoraev, On classes of uniqueness of solutions of the first boundary value problem in unbounded domains and of the Cauchy problem for nonuniformly parabolic equations. (Russian) Vestnik Mosk. Univ. Ser. I Mat. Mekh. 3(1970), 38 44. 3. O. A. Oleinik, On examples of nonuniqueness of a solution of the boundary value problem in an unbounded domain. (Russian) Uspekhi Mat. Nauk 38(1983), No. 1, 181 18. 4. E. M. Landis, On the dependence of the uniqueness class of the second initial-boundary problem of heat conductivity in an unbounded domain on the domain geometry. (Russian) Dokl. Akad. Nauk SSSR 75(1984), No. 4, 790 793. 5. A. G. Gagnidze, On the solution uniqueness classes of boundary value problems in unbounded domains for parabolic equations. (Russian) Trudy Sem. Petrovsk. 13(1988), 13 138. 6. V. F. Solonnikov, On the boundary value problem for linear parabolic systems of differential equations of general kind. (Russian) Trudy Mat. Inst. Steklov. 83(1963). 7. M. L. Marinov, A priori estimates of solutions of boundary value problems for general parabolic systems in unbounded domains. (Russian) Uspekhi Mat. Nauk 3(1977), No., 17 18. 8. M. L. Marinov, The existence of solutions of boundary value problems for general parabolic systems in an unbounded domain. (Russian) Vestnik Mosk. Univ. Ser. I Mat. Mekh. 6(1977), 56 63. 9. J. A. Oleinik and E. V. Radkevich, The method of introducing a parameter to investigate evolution equations. (Russian) Uspekhi Mat. Nauk 33(1978), No. 5, 7 76. (Received 18.04.1996) Author s address: I. Javakhishvili Tbilisi State University Faculty of Mechanics and Mathematics, University St., Tbilisi 380043 Georgia