A Boundary Value Problem for a Hyperbolic Equation with a Third-Order Wave Operator

Differential Equations, Vol. 40, No. 2, 2004, pp. 218–226. Translated from Differentsial’nye Uravneniya, Vol. 40, No. 2, 2004, pp. 208–215. c 2004 by Korzyuk. Original Russian Text Copyright

PARTIAL DIFFERENTIAL EQUATIONS

A Boundary Value Problem for a Hyperbolic Equation with a Third-Order Wave Operator V. I. Korzyuk Institute for Mathematics, National Academy of Sciences, Minsk, Belarus Received July 10, 2001

This paper is actually a continuation of [1], where the energy inequality kukE ≤ ckAukL2 (Ω)

(1)

was proved for a boundary value problem for the third-order linear hyperbolic equation Au = A1 u + A2 u = f (x),

(2)



 ∂2u ∂ n 2 x = (x1 , . . . , xn ) ∈ Ω ⊂ R , A1 u = − a ∆u , ∂x1 ∂x21   X ∂ ∂ α A2 u = , aα (x)D u, D= ,..., ∂x1 ∂xn

∆=

n X i=2

∂2 , ∂x2i

|α|≤2

where α = (α1 , . . . , αn ) is a multi-index, |α| = α1 + · · · + αn , Ω is a bounded domain in Rn with piecewise smooth boundary ∂Ω, and L2 (Ω) is the space of square integrable functions in Ω. In the present paper, we use variable-step averaging operators to prove the existence and uniqueness theorem for a strong solution of the boundary value problem for Eq. (2). Let us state this problem using the notation in [1]. By ν(x) = (ν1 (x), . . . , νn (x)) we denote the unit normal on the hypersurface ∂Ω at a point x ∈ ∂Ω. Let δ be a sufficiently small positive number. We assume that the boundary ∂Ω can consist of hypersurfaces of seven kinds, which can be defined as follows in terms of δ and ν(x) :   n X 3 2 2 S0 = x ∈ ∂Ω A1 (ν) = ν1 − a ν1 νi > 0, ν1 > 0 , i=2

S1 S2 S3 S4 S5 S6

= {x ∈ ∂Ω = {x ∈ ∂Ω = {x ∈ ∂Ω = {x ∈ ∂Ω = {x ∈ ∂Ω = {x ∈ ∂Ω

| | | | | |

A1 (ν) = 0, ν1 > 0} , A1 (ν) ≤ −δ, ν1 ≥ δ} , A1 (ν) = 0, ν1 = 0} , A1 (ν) ≥ δ, ν1 ≤ −δ} , A1 (ν) = 0, ν1 < 0} , A1 (ν) < 0, ν1 < 0} .

Other hypersurfaces are excluded from ∂Ω. The simplest example is the deformed (stretched) ball of radius R shown in the figure. We consider this domain in a Cartesian coordinate system x1 , . . . , xn in which the center of the ball coincides with the origin. Here all above-mentioned parts of the hypersurfaces Si (i = 0, . . . , 6) of the boundary ∂Ω are present. They can be described with the use of the conical surfaces n o 2 2 Zm = x | x21 − (tan ϕm ) |x0 | = 0 (m = 0, 1, 2), 2

where 0 < ϕ0 < ϕ1 < ϕ2 < ϕ3 = π/2, tan ϕ0 = |a|, and |x0 | =

Pn i=2

x2i .

c 2004 MAIK “Nauka/Interperiodica” 0012-2661/04/4002-0218

A BOUNDARY VALUE PROBLEM FOR A HYPERBOLIC EQUATION

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Figure.

Here S0 is the upper part of the surface of the sphere |x| = R lying inside the cone n o 2 Z0 = x | x21 − a2 |x0 | = 0 ; i.e., S0 =

  n X x |x| = R, ν12 − a2 νi2 > 0, ν1 > 0 , i=2

where ν = (ν1 , . . . , νn ) is the unit outward normal vector. The surface S1 lies between the cones Z0 and Z1 . More precisely,   n X R 2 2 2 S 1 = x ν1 − a νi = 0, ν1 > 0; |x| = R for x ∈ Z0 , |x| = for x ∈ Z1 , cos (ϕ1 − ϕ0 ) i=2 S2 = {x | |x| = R/ cos (ϕ1 − ϕ0 ) , |x0 | cot ϕ2 < x1 < |x0 | cot ϕ1 } , S3 = {x | |x0 | = R cos (ϕ3 − ϕ2 ) / cos (ϕ1 − ϕ0 ) , − |x0 | cot ϕ2 ≤ x1 ≤ |x0 | cot ϕ2 } , S4 = {x | |x| = R cos (ϕ1 − ϕ0 ) , − |x0 | cot ϕ1 < x1 < − |x0 | cot ϕ2 } ,   n X R S5 = x ν12 − a2 νi2 = 0, ν1 < 0, |x| = R for x ∈ Z0 , |x| = for x ∈ Z1 , cos (ϕ1 − ϕ0 ) i=2 S6 = {x | |x| = R, − R ≤ x1 < − |x0 | /|a|} . The boundary value problem consists of Eq. (2) in Ω and the homogeneous boundary conditions 2 u ∂u ∂ = 0, u|S6i=2 Si = = (3) ∂ν S6 Si ∂ν 2 S6 i=4

where ∂/∂ν is the derivative along the normal ν and ∪ stands for the union. It is of interest to consider problem (2), (3) for n > 2, since such problems were already considered, e.g., in [3–5] in the case of two independent variables on the plane (n = 2) for hyperbolic equations and systems. The proof of the energy inequality (1) in [1] shows that, for n > 3, there are new essential difficulties in the study of boundary value problems for hyperbolic equations defined in noncylindrical domains. Therefore, the methods used for plane problems in [3–5] cannot be DIFFERENTIAL EQUATIONS

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directly applied to our case. If the energy inequality (1) is valid, then the variable-step averaging operators [2] permit one to show that problem (2), (3) is well posed and prove the existence of a strong solution under some conditions on the problem data. Similar results for second-order hyperbolic equations of rather general form can be found in [6]. The statement of problems and the proof of their solvability, especially for the case in which the main equation is defined in a noncylindrical domain, are topical for higher-order hyperbolic equations. We also note the paper [7], where boundary value problems were considered for a hyperbolic equation in divergent form for n = 3 in very complicated domains.  ¯ | x1 = t be a cross-section of Ω ¯ = Ω ∪ ∂Ω perpendicular to the x1 -axis. Let S(t) = x ∈ Ω
3 ¯ ¯ we introduce the norm For functions u ∈ C Ω (three times continuously differentiable in Ω), kukE = sup

X

¯ x∈Ω |α|≤2

kD α ukL2 (S(t)) ,

(4)

where k·kL2 (S(t)) is the norm of the space of
Lebesgue square integrable functions on S(t). By D (A) ¯ satisfying condition (3). The energy inequality (1) with we denote the set of functions u ∈ C 3 Ω constant c > 0 independent of u was proved in [1] for every function u ∈ D (A) and for the operator A under some conditions on the boundary ∂Ω. By E we denote the closure of D (A) in the norm (4). Then the operator A treated as a mapping ¯ This can be justified by verifying the of E into L2 (Ω) with domain D (A) admits the closure A. closability criterion: Auk → 0 in the norm of L2 (Ω) as k → ∞ whenever kuk kE → 0 as k → ∞ [6].
¯ = f with f ∈ L2 (Ω) Definition 1. A function u ∈ D A¯ satisfying the operator equation Au is called a strong solution of problem (2), (3). By passing to the limit in inequality (1), we obtain the
same energy inequality for the operator ¯ ¯ A with the same constant c for each element u ∈ D A . To prove the existence Pn of a strong solution, we introduce some more additional conditions on Ω. Let P (ξ) = ξ12 − a2 i=2 ξi2 = 0 be the characteristic cone for the wave operator in Ω. We define a vector field R of elements r(x) = (1, r2 (x), . . . , rn (x)) as follows. Let b(i) (x) (i = 1, 2) be ¯ and let b(i) (x) ≥ δ1 for some arbitrarily small but nonzero δ1 > 0. some positive functions in Ω, We assume that r(x) = b(1) (x)r (1) (x) + b(2) (x)r (2) (x), where the vectors r(x), r (i) (x) (i = 1, 2), and (1, 0, . . . , 0) lie in some two-dimensional plane γ(x) passing through the vertex of the characteristic cone and the r (i) (x) are perpendicular to the generators ξ (i) (x) obtained in the intersection of γ(x) with P (ξ) = 0. ¯ generates a The vector field R , whose elements are uniquely determined at each point x ∈ Ω, family {%} of lines to which it is tangent. Definition 2. A subdomain Qj ⊂ Ω is said to be line % to which R is tangent is connected.

R -convex

if the intersection of Qj with each

Condition 1. The domain Ω can be partitioned by cross-sections S(t) into finitely subdomains Qj (j = 1, . . . , j0 ) satisfying the following conditions: (i) for each j, there exists a vector field R such that Qj is R -convex;
¯ j , k = 2, . . . , n; (ii) rk (x) ∈ C 3 Q (iii) (r(x), ν(x)) = 0 for each x ∈ ((S2 ∪ S3 ∪ S4 ) ∩ ∂Qj ). Condition 2. Each subdomain Qj ⊂ Ω (j = 1, . . . , j0 ) is convex with respect to straight lines parallel to the x1 -axis. Theorem. Suppose that Conditions 1 and 2, as well as the assumptions of the theorem in [1] providing the validity of the energy inequality, be valid, S0 ∪ S1 6= ∅, S5 = ∅, and S6 6= ∅ (where ∅ is the empty set). Then for each function f ∈ L2 (Ω), there exists a unique strong solution of problem (2), (3). DIFFERENTIAL EQUATIONS

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Proof. Since the problem is linear, it follows that the uniqueness of a strong solution is an ¯ immediate consequence of the energy inequality for A. By the general theory of closable operators, to prove the existence of a strong solution of problem (2), (3) for every f ∈ L2 (Ω), it suffices [6] to prove that the range R(A) of A is dense in L2 (Ω). If we also use the method of continuation with respect to a parameter (see the proof of the theorem in [8]), then it suffices to prove that the set R (A1 ), where the domain D (A1 ) of the operator A1 coincides with D (A), is dense in L2 (Ω). Let v ∈ L2 (Ω) be a function such that the inequality (A1 u, v)L2 (Ω) = 0

(5)

o n
¯ | u satisfies condition (3) . Here is valid for all u ∈ D (A1 ) = u ∈ C 3 Ω  A1 =

∂2 − a2 ∆ ∂x21



∂ . ∂x1


¯ and satisfies the boundary If u ∈ D (A1 ), then the function w = ∂u/∂x1 belongs to the class C 2 Ω conditions ∂w S w| 6i=3 Si = = 0. (6) ∂ν S6 Relation (5) can be restated as (L w, v)L2 (Ω) = 0

(7)

o n
¯ | w satisfies condition (6) , where w ∈ C2 Ω

∂2 − a2 ∆ and ∂x21 v ∈ L2 (Ω). If one uses the variable-step averaging operators Eδ in which the partition of unity is such that the boundary conditions are preserved, then Eδ w ∈ D (L ) for each w ∈ D (L ). Here Eδ is a variable-step averaging operator [2], where the parameters δ = (δ0 , δ1 , . . . , δk , . . .) depend on w, δk > 0 (k = 0, 1, . . .), and δk → 0 as k → ∞. The partition of unity in the construction of Eδ is such that the subdomains accumulate near the boundary ∂Ω to the effect that the boundary conditions for w are preserved. Therefore, Eδ w ∈ D (L ) provided that w ∈ D (L ). For this choice of Eδ w, relation (7) can be rewritten in the form for all w ∈

D (L )

=

L

=

0 = (L Eδ w, v)L2 (Ω) = (w, L Eδ∗ w)L2 (Ω) + (L Eδ w − Eδ L w, v)L2 (Ω) ! Z n X ∂w ∂w 2 + ν1 − a νi Eδ∗ v ds ∂x1 ∂x i i=2 ∂Ω ! Z n X ∂Eδ∗ v ∂Eδ∗ v 2 − w ν1 − a νi ds, ∂x1 ∂xi i=2

(8)

∂Ω

where Eδ∗ is the adjoint of Eδ , for each w ∈ that relation (8) is valid provided that

S

Eδ∗ v| Consider the commutator

4 i=0

D (L ).

Si

By varying w within

D (L ),

one can show

∂ ∗ = E v = 0. ∂ν δ S0

(9)

L Eδ w − Eδ L w, which can be represented in detailed form as

(L Eδ w − Eδ L w) (x) = (K0 w) (x) +

n  X i=1

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∂w Ki ∂xi

 (x),

(10)

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KORZYUK

where

" # ∞ n 2 X X ∂ 2 ψk (x) ∂ ψ (x) k (K0 w) (x) = (Aδ w) (x) − a2 (Aδ w) (x) , 2 2 ∂x ∂x 1 i i=2 k=1     ∞ X ∂w ∂ψk (x) ∂w K1 (x) = Aδ (x), ∂x1 ∂x1 ∂x1 k=1     ∞ X ∂w ∂ψk (x) ∂w 2 Ki (x) = −a Aδ (x), i = 2, . . . , n, ∂xi ∂xi ∂xi k=1 ∞

{ψk (x)}k=1 is a partition of unity, and the Aδk are the Sobolev averaging operators. The operators Ki (i = 0, . . . , n) are also variable-step averaging operators preserving the boundary conditions on ∂Ω. By taking account of (9) and (10), one can represent relation (8) in the form

0 = (w, L

Eδ∗ v)L2 (Ω)

n X ∂ − Ki∗ v ∂x i i=1

w, K0∗ v

+

! +

n Z X i=1

L2 (Ω)

w (Ki∗ v) νi ds

(11)

∂Ω

for each w ∈ D (L ). The operator Ki∗ (i = 0, . . . , n) is the adjoint of Ki . It follows from (11) that Ki∗ v = 0

(12)

on S0 ∪ S1 , i = 0, . . . , n. Therefore, (w, L

Eδ∗ v)L2 (Ω)

+

w, K0∗ v

n X ∂ − Ki∗ v ∂x i i=1

! = 0.

(13)

L2 (Ω)

Since D (L ) is dense in L2 (Ω), it follows that relation (13) is valid for each w ∈ L2 (Ω). Let the subdomain Q1 ⊂ Ω (see Condition 1) lie “above” the other Qj (j = 2, . . . , j0 ); i.e., Sj0 ˜ 1 be ˜ 1 and Q1 \Q x1 > y1 for each x = (x1 , . . . , xn ) ∈ Q1 and each y = (y1 , . . . , yn ) ∈ j=2 Qj . Let Q
˜ 1. ˜ 1 lies “above” Q1 \Q the subdomains of Q1 obtained from Q1 by the cross-section S t˜ , where Q In (13), we set

  w(x) =

I Jv =

x

˜1 Eδ∗ v ds for x ∈ Q

(14) ˜ 1. for x ∈ Ω\Q 
 Hx ˜ 1 ∩ S6 to a point x belonging Here x˜ is a curvilinear integral from a point x ˜ ∈ S˜− = S t˜ ∪ ∂ Q ˜ 1 ∩ (S0 ∪ S1 ) along a line % to which the vector field R is tangent. to ∂ Q We substitute the function w given by (14) into (13) and transform relation with  the resulting  ∂ the use of the boundary conditions (9) and (12) and the relation Jv (x) = (Eδ∗ v) (x) for ∂r ˜ 1 . Then we obtain x∈Q 

∂ ∂2 Jv, Jv ∂x1 ∂x1 ∂r





x ˜

0

 n  X ∂ ∂2 −a Jv, Jv ∂xi ∂xi ∂r ˜1) ˜1) L 2 (Q L 2 (Q i=2   n X ∂ ∗ ∗ = (Jv, K0 v)L2 (Q˜ 1 ) + Jv, Ki v . ∂xi ˜1) L 2 (Q i=1 2

DIFFERENTIAL EQUATIONS

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Further, Z 

1 2

∂ Jv ∂x1

2

˜1 ∂Q

2 n Z  ∂ a2 X rν ds − Jv rν ds 2 i=2 ∂xi ˜1 ∂Q

=

(Jv, K0∗ v)L2 (Q˜ 1 )

 Z n  X ∂ ∗ + Jv, Ki v + ∂xi ˜1) L 2 (Q i=1

(16) (Jv)ds,

F

˜1 Q

where rν = ν1 +

Pn

νi ri , ν = (ν1 (x), . . . , νn (x)), r = (1, r2 (x), . . . , rn (x)), and ∂ quadratic form in the first derivatives Jv, i = 1, . . . , n. We have ∂xi i=2

X ∂ ∂ Jv = Eδ∗ v − ri (x) Jv. ∂x1 ∂x i i=2

F

(Jv) is a

n

(17)

It follows from (17) that

∂ x0 = (x2 , . . . , xn ) , Jv t˜, x0 = Eδ∗ v t˜, x0 , ∂x1 n X ∂ ∂ ˜1 ∩ Jv(x) = − ri (x) Jv(x), x ∈ ∂Q ∂x1 ∂x i i=2

(18) 4 [

! Si

.

(19)

i=0

˜ 1 ∩ (S3 ∪ S4 ∪ S6 ). By virtue of the definition of J, we have rν ≤ 0 and Jv(x) = 0 on S˜˜ − = ∂ Q Therefore, 2 2 Z  Z  n ∂ ∂ a2 X − Jv rν ds + Jv rν ds ≥ 0. (20) ∂x1 2 i=2 ∂xi ˜ ˜− S

˜ ˜− S

Relations (17)–(20), together with (16), imply the inequality Z S(t˜)

2 (Eδ∗ v)

Z n
0 X  2  ∂ ∂ 0 ˜ t, x dx + a δij − ri (x)rj (x) Jv Jvrν ds ∂x ∂x i j i,j=2 ˜+ S

 Z n  X ∂ ≤ 2 (Jv, K0∗ v)L2 (Q˜ 1 ) + Jv, Ki∗ v + F (Jv)dx ∂x i ˜ L2 (Q1 ) i=1 ˜1 Q !

2 n n X X

∂ 2 2

≤ c1 kEδ∗ vkL2 (Q˜ 1 ) + kJvk2L2 (Q˜ 1 ) + kKi∗ vkL2 (Q˜ 1 ) ,

∂xi Jv ˜ + L2 (Q1 ) i=2 i=0

(21)

˜ 1 ∩ (S0 ∪ S1 ), δij is the Kronecker delta, and c1 is a positive constant. where S˜+ = ∂ Q To apply
the Gronwall H x¯ ∗ inequality to (21), along with the function Jv(x), we introduce the ˜ function J v (x) = x Eδ v ds, where the integration is also performed along the lines % to which R is tangent and z ∈ S˜+ . It follows from the definition of J and J˜ that     (Jv)(x) + J˜v (x) = J˜v (˜ x) . (22) Note that the points on the integration line % are functionally related. In particular, the zi = zi (˜ x) − ˜ (i = 1, . . . , n) thus defined can be treated as continuous functions with domain D (zi ) = S and range R (zi ) = S˜+ . DIFFERENTIAL EQUATIONS

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˜ Then we obtain In inequality (21), we use relation (22) and replace the function Jv by Jv. Z n X






a δij − ri rj z(x) 2

i,j=2 ˜− S

 ∂ ˜ ∂ ˜ Jv (˜ x) Jv (˜ x) βi (˜ x) βj (˜ x) (rν ) (z (˜ x)) β (˜ x) ds ∂xi ∂xj

+ kEδ∗ vkL2 (S(t˜)) 2

2

2 ˜ ≤ c1 (ε0 ) kEδ∗ vkL2 (Q˜ 1 ) + J˜v (˜ x) − Jv(x)

(23)

˜1) L 2 (Q

!

2 n n X X

∂

∂ 2 ˜ x) − ˜

Jv(x) + kKi∗ vkL2 (Q˜ 1 ) ,

∂xi Jv (˜

˜ + ε0 ∂x i L2 (Q1 ) i=2 i=0 where the functions βi (˜ x) 6= 0 can be chosen on the basis of the vector field R , β (˜ x) ≥ c2 > 0, and c1 (ε0 ) is inversely proportional to ε0 > 0 as ε → 0. By the definition of the Pn vector field R , for each Pn vector ξ(x) = ¯(ξ1 (x), . . . , ξn (x)) orthogonal to r(x), we have ξ12 − a2 i=2 ξi2 ≤ −c3 |ξ|2 = −c3 i=1 ξi2 for all x ∈ Ω with some constant c3 > 0. As such a vector, we take ! n X ∂ ˜ ∂ ˜ ∂ ˜ ξ(˜ x) = − Jv (˜ x) , β2 (˜ Jv (˜ x) , . . . , βn (˜ Jv (˜ x) . ri (z (˜ x)) β (˜ x) x) x) ∂x ∂x ∂x i 2 n i=2 In particular, it is obviously orthogonal to the vector r (z (˜ x)) = (1, r2 (z (˜ x)) , . . . , rn (z (˜ x))) for − ˜ each x ˜ ∈ S . Since βi (˜ x) 6= 0 (i = 2, . . . , n), we have Z n X






a δij − ri rj z(x) 2

i,j=2 ˜− S

 ∂ ˜ ∂ ˜ Jv (˜ x) J v (˜ x) βi (˜ x) βj (˜ x) (rν ) (z (˜ x)) β (˜ x) ds ∂xi ∂xj

n X

∂

˜

≥ c4

∂xi Jv i=2

Further,

(24) ,

c4 > 0.

˜− ) L 2 (S

2 n X

∂

∂

˜ (˜ Jv x) − J˜v(x) +

˜ ˜1) ∂xi ∂xi L 2 (Q L2 (Q1 ) i=2 !

n   2 X ∂

2

˜1 ˜ ˜ x) ≤ ε1 Q +

Jv (25)

∂xi Jv (˜

˜− ˜− ) L 2 (S L 2 (S ) i=2 !

2 n

2 X

∂

˜

˜ + c5 J v , +

∂xi Jv ˜ ˜1) L 2 (Q L2 (Q1 ) i=2
˜ 1 can be chosen sufficiently small if we appropriately diminish the where the positive constant ε Q ˜ 1 . By the properties of averaging operators, we have domain Q

2

˜

˜ x) − Jv(x)

J v (˜

n X

kKi∗ vkL2 (Q˜ 1 ) ≤ c6 kvk2L2 (Q˜ 1 ) . 2

(26)

i=0

Along with (23), consider the inequality

2

˜

Jv

˜− ) L 2 (S

2

˜ 2 ≤ c7 kEδ∗ vkL2 (Q˜ 1 ) + Jv

˜1) L 2 (Q

,

DIFFERENTIAL EQUATIONS

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which is obtained from the relation Z Z h i 1  ˜ 2 ˜ (˜ Jv (˜ x) rν (z (˜ x)) β (˜ x) ds = Eδ∗ v(x) Jv x) − J˜v(x) dx; 2 ˜− S

˜1 Q

in turn, the last formula can be derived from the relation 1 ∂ (Jv)(x) = Eδ∗ v(x)Jv(x) 2 ∂r ˜ 1 . By adding inequalities (23) and (27), by taking account of (24)–(26), and by by integration over Q choosing an appropriate value of ε1 , we obtain the inequality # Z " 2 X n  ∂ 2 ˜ kEδ∗ v(x)kL2 (S(t˜)) + J˜v + Jv (˜ x) ds ∂x i i=2 ˜− S   (28) # Z " 2 X n  ∂   2 ˜ ˜ ≤ c8 (ε) kEδ∗ v(x)kL2 (S(t˜)) + Jv Jv (x)dx + ε0 c9 kvk2L2 (Q˜ 1 ) . + ∂x i i=2 ˜1 Q

It follows from (28) that tZmax




Φ t˜ ≤ c10 (ε0 )

Φ(t)dt + ε0 c11 kvkL2 (Q˜ 1 ) ,

(29)

t˜

where




˜ 2 Φ t˜ = kEδ∗ v(x)kL2 (S(t˜)) + Jv

n X

∂

˜

Jv +

∂xi L2 (S(t˜)) i=2

,

tmax = max x1 , ¯ x∈Ω

L2 (S(t˜))

with a positive constant c10 (ε0 ) inversely proportional to ε0 . Inequality (29) can be obtained in a τ τ ˜ ˜ ˜ ˜ similar way for any subdomain Q ⊂ Q1 ⊂ Ω, where Q = x ∈ Ω | t ≤ τ < x1 < tmax . Therefore, tZmax

Φ(τ ) ≤ c10 (ε0 )

Φ(t)dt + ε0 c11 kvkL2 (Q˜ τ ) ,

(30)

τ

where Φ(t) =

kEδ∗ vkL2 (S(τ ))



˜ + Jv



∂ ˜

Jv + ∂xi L2 (S(τ ))

. L2 (S(τ ))

The Gronwall inequality, together with (30), implies that Φ(τ ) ≤ ε0 c11 exp {c10 (ε0 ) (tmax − τ )} kvkL2 (Q˜ 1 ) , 
whence it follows that kEδ∗ vkL2 (S(τ )) ≤ ε0 c11 exp c10 (ε0 ) tmax − t˜ kvkL2 (Q˜ 1 ) . On the left-hand side of the last inequality, we pass to the least upper bound with respect to τ , where τ varies from t˜
2 2 ∗ to tmax . Further, obviously, kEδ vkL2 (Q˜ 1 ) ≤ tmax − t˜ supt˜