Journal of Mathematical Sciences, Vol. 190, No. 1, April, 2013

FIRST BOUNDARY-VALUE PROBLEMFOR SECOND-ORDER ELLIPTIC-PARABOLIC EQUATIONS WITHDISCONTINUOUS COEFFICIENTS

I. T. Mamedov∗ UDC 517.9

Abstract. In this paper, the first boundary value problem for the second-order degenerated elliptic-parabolic equations in nondivergent form is considered. Unique strong (almost everywhere) solvabilityof this problem in the corresponding weight Sobolev space is proved.

CONTENTS

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1042. Auxiliary Statements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1053. Main Coercive Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1144. Solvability of the First Boundary Value Problem for the Equation M0u = f . . . . . . . . 1225. Local Solvability of the First Boundary Value Problem . . . . . . . . . . . . . . . . . . . . 1246. General Solvability of the First Boundary Value Problem . . . . . . . . . . . . . . . . . . 1277. On the Accuracy of the Cordes Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

1. Introduction

Let En be n-dimensional Euclidean space of points x = (x1, . . . , xn) , Ω be a bounded domain in En

with the boundary ∂Ω, ∂Ω ∈ C2, and QT be a cylinder Ω× (0, T ) , T ∈ (0,∞) . Consider in QT a firstboundary value problem

Lu =n∑

i,j=1

aij (x, t)uxixj +n∑

i=1

bi (x, t)uxi + c (x, t)u+ϕ (T − t)utt − ut = f (x, t) , (x, t) ∈ QT ,

(1.1)

u|Γ(QT ) = 0, (1.2)

where Γ (QT ) = (Ω× {(x, t) : t = 0}) ∪ (∂Ω× [0, T ]) is a parabolic boundary of domain QT , andsuppose that the following conditions relative to the coefficients of the operator L are fulfilled:

γ |ξ|2 ≤n∑

i,j=1

aij (x, t) ξiξj ≤ γ−1 |ξ|2 ; (x, t) ∈ QT , ξ ∈ En (1.3)

σ = supQT

⎡

⎣n∑

i,j=1

a2ij (x, t) /

(n∑

i=1

aii (x, t)

)2⎤

⎦− 1

n− λ2< 0; (1.4)

∗Deceased

Translated from Sovremennaya Matematika. Fundamental’nye Napravleniya (Contemporary Mathematics.Fundamental Directions), Vol. 39, Partial Differential Equations, 2011.

104 1072–3374/13/1901–0104 c© 2013 Springer Science+Business Media New York

ϕ (z) ∈ C1 [0, T ] , ϕ (z) ≥ 0, ϕ′ (z) ≥ 0, ϕ (0) = 0, ϕ′ (0) = 0,

ϕ (z) ≥ β zϕ′ (z) , β > 0− const,(1.5)

bi (x, t) ∈ Ln+2 (QT ) ; i = 1, . . . , n, c (x, t) ∈ LS (QT ) . (1.6)

Here

λ =

infQT

n∑i=1

aii(x, t)

supQT

n∑i=1

aii(x, t)

,

s =

{max

(2, n+2

2

)if n �= 2,

2 + ν if n = 2

with some constant ν = 0. Moreover, we assume that the elements of the matrix ‖aij (x, t)‖ aremeasurable in QT functions. Condition (1.4) is called the parabolic Cordes condition. It is understoodup to a nonsingular linear transformation, i.e., QT may be covered by a finite number of domainsQ1,. . . ,QM such that in each Qi there exists a nonsingular linear transformation of coordinates underwhich the matrix of principal coefficients relative to arbitrary spatial variables of image of the operatorL satisfies Condition (1.4) in the image of Qi, i = 1, . . . ,M. Note that Condition (1.4) coincides withthe classical Cordes condition (see [4, 22]) appearing during the investigation of second-order ellipticequations of nondivergent structure for matrices with constant trace, i.e., for λ = 1. However, for theabove-mentioned equations, the constancy condition for the trace of the matrix of the principal part,without loss of generality, can always be assumed as satisfied. As to elliptic-parabolic and parabolicequations of the form (1.1), the constancy requirement for the trace of the matrix ‖aij (x, t)‖ is asubsidiary condition imposed on the coefficients of the operator L.

The aim of the present paper is to prove the unique strong (almost everywhere) solvability of thefirst boundary value Problem (1.1)-(1.2) in the corresponding weight Sobolev space for any f (x, t) ∈L2 (QT ) .

Investigations in the theory of degenerate elliptic-parabolic equations go back to the classical workof Keldysh [11], in which correct statements of boundary value problems for equations of the form (1.1)with one spatial variable were found. In [6], Fichera established the unique weak solvability of thefirst boundary value problem for wide class of second-order equations with nonnegative characteristicform (see also [20]). As to the strong solvability of the first boundary value problem for some classesof elliptic-parabolic equations in nondivergent form, note the papers [3, 9, 12, 17]. We also note thepapers [22] and [1, 2, 7, 16, 23, 24], in which unique strong solvability of the first boundary valueproblem of second-order elliptic and parabolic equations in nondivergent form with discontinuouscoefficients under the Cordes type conditions was proved. We mention the papers [8, 21], wherequestions of weak solvability of the first boundary value problem for elliptic-parabolic equations indivergent form were investigated.

2. Auxiliary Statements

First, we agree on some notations. By ui, uij , and uit, denote the derivatives uxi , uxixj , and uxit,

respectively, i, j = 1, . . . , n. Let W 1,02 (QT ) , W

2,02 (QT ) and W 2,1

2 (QT ) , W2,22,ϕ (QT ) be Banach spaces

of functions u (x, t) given on QT with finite norms

‖u‖W 1,0

2 (QT )=

⎛

⎜⎝∫

QT

(u2 +

n∑

i=1

u2i

)dxdt

⎞

⎟⎠

1/2

,

105

‖u‖W 2,0

2 (QT )=

⎛

⎜⎝∫

QT

⎛

⎝u2 +n∑

i=1

u2i +

n∑

i,j=1

u2ij

⎞

⎠ dxdt

⎞

⎟⎠

1/2

,

‖u‖W 2,1

2 (QT )=

⎛

⎜⎝∫

QT

⎛

⎝u2 +n∑

i,j=1

u2i +n∑

i,j=1

u2ij + u2t

⎞

⎠ dxdt

⎞

⎟⎠

1/2

,

and

‖u‖W 2,2

2,ϕ(QT )=

( ∫

QT

(u2 +

n∑

i=1

u2i +n∑

i,j=1

u2ij + u2t + ϕ2(T − t)u2tt

+ϕ(T − t)n∑

i=1

u2it

)dxdt

)1/2(2.1)

respectively, where W 2,22,ϕ (QT ) is a subspace ofW

2,22,ϕ (QT ) in which the set of all functions from C∞ (

QT

)

vanishing on Γ (QT ) is a dense set. A function u (x, t) ∈ W 2,22,ϕ (QT ) is called a strong solution of

boundary value Problem (1.1)-(1.2) if it satisfies Eq. (1.1) almost everywhere on QT .For R > 0, x0 ∈ En, we denote by BR

(x0)a ball

{x :

∣∣x− x0∣∣ < R

}and denote by QR

T

(x0)a cylin-

der BR

(x0)× (0, T ) . Let BR

(x0) ⊂ Ω. We say the u (x, t) ∈ A (

QRT

(x0))

if u (x, t) ∈ C∞ (QR

T

(x0)),

u|t=0 = 0, and supp u ⊂ QρT

(x0)for some ρ ∈ (0, R) .

In the sequel, the notation C (· · · ) means that the positive constant C depends only on what is inthe parentheses:

Consider the following model operator

L0 = Δ+ ϕ (T − t)∂2

∂t2− ∂

∂t,

where Δ =n∑

i=1

∂2

∂x2iis the Laplace operator.

Lemma 2.1. If a function ϕ (z) satisfies Conditions (1.5), then there exists T1 (ϕ, n) such that forT ≤ T1 and any function u (x, t) ∈ A (

QRT

(x0))

the following estimate is valid:

∫

QRT (x0)

( n∑

i,j=1

u2ij + u2t + ϕ2(T − t)u2tt + ϕ(T − t)n∑

i=1

u2it

)dxdt≤ (1 + 2 (n+ 1) q (T ))

∫

QRT (x0)

(L0u)2 dxdt,

(2.2)where q (T ) = sup

[0,T ]ϕ′ (z) .

Proof. Here and later on, for the sake of simplicity, by Q and B we denote BR

(x0)and QR

T

(x0)

respectively. We have

106

I =

∫

Q

(L0u)2 dxdt =

∫

Q

(Δu)2 dxdt+

∫

Q

ϕ2 (T − t)u2ttdxdt

+

∫

Q

u2tdxdt+ 2

∫

Q

ϕ (T − t)uttΔudxdt− 2

∫

Q

utΔudxdt

−∫

Q

ϕ (T − t)uttutdxdt = i1 +

∫

Q

ϕ2 (T − t) u2ttdxdt

+

∫

Q

u2tdxdt+ i2 + i3 + i4.

(2.3)

Obviously,

i1 =

∫

Q

n∑

i,j=1

u2ijdxdt; (2.4)

i2 = 2

∫

Q

ϕ (T − t)n∑

i=1

uiiuttdxdt = −2

∫

Q

n∑

i=1

(ϕ (T − t)uii)t utdxdt

(since ϕ (T − t) |t=T = 0, uii|t=0 = 0) = 2

∫

Q

ϕ′ (T − t)n∑

i=1

uiiutdxdt

−∫

Q

ϕ (T − t)n∑

i=1

uiitutdxdt ≥ −∫

Q

ϕ′ (T − t)n∑

i=1

u2iidxdt

−n∫

Q

ϕ′ (T − t)u2tdxdt+ 2

∫

Q

ϕ (T − t)n∑

i=1

u2itdxdt

≥ −∫

Q

ϕ′ (T − t)

n∑

i,j=1

u2ijdxdt− n

∫

Q

ϕ′ (T − t)u2tdxdt

+2

∫

Q

ϕ (T − t)n∑

i=1

u2itdxdt;

(2.5)

i3 = 2

∫

Q

n∑

i=1

uiuitdxdt =

∫

Q

n∑

i=1

(u2i)tdxdt =

∫

B

n∑

i=1

u2i (x, T ) dx ≥ 0; (2.6)

i4 = −∫

Q

ϕ (T − t)(u2t)tdxdt = −

∫

Q

ϕ′ (T − t)u2tdxdt+ϕ (T )

∫

B

u2t (x, 0) dx ≥ −∫

Q

ϕ′ (T − t)u2tdxdt.

(2.7)Taking into account (2.4)-(2.7) in (2.3), we obtain the inquality

I ≥ (1− q (T ))

∫

Q

n∑

i,j=1

u2ijdxdt+

∫

Q

ϕ2 (T − t)u2ttdxdt

+(1− (n+ 1) q (T ))

∫

Q

u2tdxdt+ 2

∫

Q

ϕ (T − t)n∑

i=1

u2itdxdt.

(2.8)

107

According to Conditions (1.5), q (T ) → 0 as T → 0. We choose T1 so small that (n+ 2) q (T1) ≤ 1/2.Then for T ≤ T1,

1

1− (n+ 1) q (T )≤ 1 +

(n+ 1) q (T )

1− (n+ 2) q (T )≤ 1 + 2 (n+ 1) q (T ) . (2.9)

Now the required estimate (2.2) follows from (2.8)-(2.9). The lemma is proved.

Remark. If the function ϕ (z) satisfies conditions (1.5), then for T ≤ T1, τ ∈ [0, 1] , and any functionu (x, t) ∈ A (

QRT

(x0)), the following estimate is valid:∫

QRT (x0)

( n∑

i,j=1

u2ij + u2t + ϕ2 (T − t)u2tt + ϕ (T − t)n∑

i=1

u2it

)dxdt

≤ (1 + 2 (n+ 2) q (T ))

∫

QRT (x0)

(L0u− τ

Tu)2dxdt.

Obviously, it suffices to consider the case where τ > 0. By μ′ denote τ/T. We have

I1 =

∫

Q

(L0 − μ′u

)2dxdt ≥ (1− q (T ))

∫

Q

n∑

i,j=1

u2ijdxdt+

∫

Q

ϕ2 (T − t)u2ttdxdt

+(1− (n+ 1) q (T ))

∫

Q

u2tdxdt+ 2

∫

Q

ϕ (T − t)n∑

i=1

u2itdxdt

+(μ′)2∫

Q

u2dxdt− 2μ′∫

Q

uΔudxdt+ 2μ′∫

Q

u utdxdt− 2μ′∫

Q

ϕ (T − t)uttudxdt.

However, on the other hand,

−2μ′∫

Q

uΔudxdt = 2μ′∫

Q

n∑

i=1

u2i dxdt ≥ 0;

2μ′∫

Q

u utdxdt = μ′∫

B

u2 (x, T ) dx ≥ 0;

Moreover, for any α > 0, the inequality

−2μ′∫

Q

ϕ (T − t)uttudxdt = 2μ′∫

Q

ϕ (T − t)u2i dxdt

−2μ′∫

Q

ϕ′ (T − t)u utdxdt ≥ −2μ′q (T )∫

Q

|u| |ut| dxdt≥ −μ′αq (T )∫

Q

u2tdxdt− μ′α−1q (T )

∫

Q

u2dxdt

is satisfied.Choosing α = (μ′)−1 and taking into account that q (T ) ≤ 1, we obtain that

I1 ≥ (1− q (T ))

∫

Q

n∑

i,j=1

u2ijdxdt+

∫

Q

ϕ2 (T − t)u2ttdxdt

+(1− (n+ 2) q (T ))

∫

Q

u2tdxdt+ 2

∫

Q

ϕ (T − t)n∑

i=1

u2itdxdt,

108

which implies the required inequality.

In the sequel, we restrict ourselves to consideration of the most interesting case where ϕ (z) > 0for z > 0. If ϕ (z) ≡ 0, then equation (1) is parabolic, and the corresponding solvability result forthe first boundary value problem was obtained in [1]. If ϕ (z) = 0 for z ∈ [

0, z0], then a solution of

Problem (1.1)-(1.2) is obtained by splicing solution u (x, t) of this problem in the cylinder Qz0 andthe solution ϑ (x, t) of the first boundary value problem for the parabolic equation in the cylinderΩ× (

z0, T)with the boundary data ϑ

(x, z0

)= u(x, z0), ϑ|∂Ω×[z0,T ] = 0.

We fix an arbitrary ε ∈ (0, T ) and introduce a function ϕε (z) in the following form:

ϕε (z) = ϕ (ε)− ϕ′ (ε) εm

+ϕ′ (ε)mεm−1

zm for z ∈ [0, ε)

and ϕε (z) = ϕ (z) for z ∈ [ε, T ] , where m = 2β−1. Obviously, ϕε (z) ∈ C1 [0, T ] . Show that forz ∈ [0, T ] ,

ϕε (z) ≥ 1

2ϕ (z) . (2.10)

It suffices to prove inequality (2.10) for z ∈ [0, ε) . Obviously, by virtue of the monotonicity of ϕ (z) ,inequality (2.10) will be satisfied if

ϕ (ε)− ϕ′ (ε) · εm

≥ 1

2ϕ (ε)

or

ϕ (ε) ≥ 2

mϕ′ (ε) · ε.

However, the last estimate is valid by virtue of (1.5). Then inequality (2.10) is proved.Without loss of generality, we will assume that m > 1. Then

qε (T ) = sup[0,T ]

ϕ′ε (z) ≤ q (T ) . (2.11)

In fact,

qε (T ) ≤ max

(sup[0,ε]

ϕ′ (ε)(zε

)m−1, q (T )

)= max

(ϕ′ (ε) , q (T )

)= q (T ) .

We will say that u (x, t) ∈ B (QR

T

(x0))

if u (x, t) ∈ A (QR

T

(x0))

and, moreover, u|t=T = ut|t=T = 0.Now we define an operator

Lε = Δ+ ϕε (T − t)∂2

∂t2− ∂

∂t.

Lemma 2.2. If ϕ (z) satisfies the Conditions (1.5), then for any function u (x, t) ∈ B (QR

T

(x0)), the

following estimate holds for T ≤ T1:

∫

QRT (x0)

⎛

⎝n∑

i,j=1

u2ij + u2t + ϕ2ε (T − t)u2tt + ϕε (T − t)

n∑

i,j=1

u2it

⎞

⎠ dxdt

≤ (1 + 2 (n+ 1) q (T ))

∫

QRT (x0)

(Lεu)2 dxdt.

(2.12)

109

Proof. We have

∫

Q

(Lεu)2 dxdt ≥

∫

Q

n∑

i,j=1

u2ijdxdt+

∫

Q

ϕ2ε (T − t)u2ttdxdt+

∫

Q

u2tdxdt

+2

∫

Q

ϕε (T − t)uttΔudxdt− 2

∫

Q

ϕε (T − t)uttutdxdt.

(2.13)

However, on the other hand,

2

∫

Q

ϕε (T − t)uttΔudxdt = −2

∫

Q

n∑

i=1

(ϕε (T − t)uii)t utdxdt

= (since uii|t=0 = uii|t=T = 0) = 2

∫

Q

ϕ′ε (T − t)

n∑

i=1

uiiutdxdt

−2

∫

Q

ϕε (T − t)n∑

i=1

uiitutdxdt ≥ −qε (T )∫

Q

n∑

i,j=1

u2ijdxdt

−nqε (T )∫

Q

u2tdxdt+ 2

∫

Q

ϕε (T − t)n∑

i=1

u2itdxdt

(2.14)

and, analogously,

−2

∫

Q

ϕε (T − t)uttutdxdt = −∫

Q

ϕ′ε (T − t)u2tdxdt

+ϕε (T )

∫

B

u2t (x, 0) dx (since ut|t=T = 0) ≥ −qε (T )∫

Q

u2tdxdt.

(2.15)

Now, using (2.9) and (2.11) and arguing in the same way as in the proof of Lemma 2.1, from (2.13)-(2.15), we obtain the required estimate (2.12). The lemma is proved.

We introduce into consideration an elliptic-parabolic operator

L′ =n∑

i,j=1

aij (x, t)∂2

∂xi∂xj+ ϕ (T − t)

∂2

∂t2− ∂

∂t.

Also, let

δ = supQT

∣∣∣∣g (x, t)

m− 1

∣∣∣∣+ supQT

√√√√2∑

i<j

a2ij (x, t) +n∑

i=1

(aii (x, t)− g (x, t)

m

)2

,

where g (x, t) =n∑

i=1aii (x, t) and the constant m > 0 will be chosen later.

Lemma 2.3. If the coefficients of the operator L′ satisfy Conditions (1.3), (1.5) and δ < 1, thenthere exists T2 (ϕ, δ, n) such that if T ≤ T2, then for any function u (x, t) ∈ A

(QR

T

(x0))

the following

110

estimate holds:

J =

∫

QRT (x0)

( n∑

i,j=1

u2ij + u2t + ϕ2 (T − t)u2tt + ϕ (T − t)n∑

i=1

u2it

)dxdt

≤ C2.1 (ϕ, δ, n)

∫

QRT (x0)

(L′u)2dxdt.

(2.16)

Proof. Assuming that T ≤ T1, from Lemma 2.1 we obtain that

J12 ≤ C2.2 ‖L0u‖L2(Q) ≤ C2.2

∥∥L′u∥∥L2(Q)

+ C2.2

∥∥(L′ − L0

)u∥∥L2(Q)

, (2.17)

where C2.2 = (1 + 2 (n+ 1) q (T ))1/2 . On the other hand (δij is the Kronecker symbol),

L′u =g (x, t)

mΔu+

n∑

i,j=1

(aij (x, t)− δij

g (x, t)

m

)uij + ϕ (T − t)utt − ut.

Therefore,

∥∥(L′ − L0

)u∥∥L2(Q)

≤ supQT

∣∣∣∣g (x, t)

m− 1

∣∣∣∣ ‖Δu‖′L2(Q)

+supQT

√√√√2∑

i<j

a2ij (x, t) +n∑

i=1

(aii (x, t)− g (x, t)

m

)2

⎛

⎜⎝∫

Q

n∑

i,j=1

u2ijdxdt

⎞

⎟⎠

12

.

Using (2.4), from the last inequality and (2.17), we conclude that

J12 ≤ C2.2

∥∥L′u∥∥L2(Q)

+ δC2.2

⎛

⎜⎝∫

Q

n∑

i,j=1

u2ijdxdt

⎞

⎟⎠

12

. (2.18)

Since δ < 1 and q (T ) → 0 as T → 0, it follows that there exists T ′ (ϕ, δ, n) such that

δ(1 + 2 (n+ 2) q

(T ′)) 1

2 ≤ 1 + δ

2< 1.

Now we fix T2 = min (T1, T′) . Then for T ≤ T2, the required estimate (2.16) follows from (2.18).

The lemma is proved.

Remark (Remark 1). From the remark to Lemma 2.1, it follows that if the coefficients of the operatorL′ satisfy conditions (1.3) and (1.5) and δ < 1, then for T ≤ T2, τ ∈ [0, 1], and any function u (x, t) ∈A(QR

T

(x0))

the following estimate holds:

J ≤ C2.1

∫

QRT (x0)

(L′u− τ

Tu)2dxdt.

Remark (Remark 2). Condition (1.3) was nowhere used in the proof. Actually, it follows from thecondition δ < 1. In fact, let δ < 1, (x, t) ∈ QT and ξ ∈ En. We have

111

n∑i,j=1

aij (x, t) ξiξj =n∑

i,j=1

(aij (x, t)− δij

g (x, t)m

)ξiξj +

(g (x, t)m − 1

)|ξ|2 + |ξ|2

≥ − supQT

√2∑i<j

a2ij (x, t) +n∑

i=1

(aii (x, t)− g (x, t)

m

)2(

n∑i,j=1

ξ2i ξ2j

) 12

− supQT

∣∣∣∣g (x, t)m − 1

∣∣∣∣ |ξ|2 + |ξ|2 = (1− δ) |ξ|2 .

The inequality

n∑

i,j=1

aij (x, t) ξiξj ≤ (1 + δ) |ξ|2

is proved quite analogously.

Obviously, for n = 1 Conditions (1.3) and δ < 1 are equivalent up to a nonsingular linear transfor-mation. The analogous fact is also valid for n = 2 if the trace of the matrix ‖aij (x, t)‖ is constant inQT .

Now, we will replace Condition (1.3) by a weaker condition; exactly, we suppose that

infQT

n∑

i=1

aii (x, t) = γ′ > 0. (2.19)

Lemma 2.4. If (2.19) is fulfilled, then the condition δ < 1 (up to a nonsingular linear transformation)follows form the Cordes Condition (1.4).

Proof. Denote supQT

g (x, t) by p. We make a linear transformation of coordinates:

x′i =

xi√p; i = 1, . . . , n; t′ = t.

Then if a′ij (x′, t′) are coefficients of the image of the spatial gradient of the operator L′, then

a′ij (x′, t′) = p−1aij (x, t) . Let λ have the same meaning as in Condition (1.4) and g′ (x′, t′) = p−1g (x, t) .

Suppose that m = 1 for n = 1 and m = n− λ2 for n > 1. Obviously, if n = 1 and Condition (2.19) issatisfied, then both conditions δ < 1 and (1.4) are equivalent. If n > 1, then g′ (x′, t′) ≤ m. Therefore,

supQ

∣∣∣∣g′ (x′, t′)

m− 1

∣∣∣∣ = 1− 1

minfQg′(x′, t′

), (2.20)

where QT is the image of cylinder Q (T ) . Moreover,

supQT

√√√√2∑

i<j

(a′ij(x′, t′))2 +n∑

i=1

(a′ii(x′, t′)−

g′(x′, t′)m

)2

≤ supQT

g′(x′, t′) supQT

√√√√√n∑

i,j=1(a′i,j(x′, t′))2

(g′(x′, t′))2+n− 2m

m2

(2.21)

From (2.20) and (2.21), we conclude that condition δ < 1 will be satisfied if

112

1− 1

minfQT

g′(x′, t′) + supQT

g′(x′, t′) supQT

√√√√√n∑

i,j=1(a′i,j(x′, t′))2

(g′(x′, t′))2+n− 2m

m2< 1,

i.e.,

supQT

n∑i,j=1

a2ij(x, t)

g2 (x, t)<λ2 + 2m− n

m2. (2.22)

However, (2.22) is equivalent to Condition (1.4). The lemma is proved.

Remark. It easy to show that Condition (1.4) implies either uniform positive or uniform negativedefiniteness of the matrix ‖aij (x, t)‖ in QT .We impose additional Condition (2.19) in order to providethe positive definiteness of the matrix ‖aij (x, t)‖ in QT .

In the sequel, without loss of generality, we will assume that R ≤ 1.

Lemma 2.5. If the coefficients of the operator L′ satisfy Conditions (2.19), (1.4) and (1.5), then forany T ≤ T2 and any function u (x, t) ∈ A

(QR

T

(x0))

the following estimate holds:

‖u‖W 2,2

2,ϕ(QRT (x0)) ≤ C2.3 (ϕ, σ, n)

∥∥L′u∥∥L2(QR

T (x0)) . (2.23)

Proof. We will fix an arbitrary t′ ∈ (0, T ) and continue the function u (x, t′) to get zero into thecomplement BR

(x0)of the n-dimensional parallelepiped P = {x :

∣∣xi − x0i∣∣ < R; i = 1, . . . , n}.

Denote the continued function by u (x, t′) again. Let x′ = (x2, . . . , xn) and x ∈ (x01 −R, x01 +R

). We

have

u(x1, x

′, t′)=

x01∫

x01−R

u1(z, x′, t′

)dz,

i.e.,

u2(x1, x

′, t′) ≤ 2R

x01+R∫

x01−R

u21(z, x′, t′

)dz.

Thus,

∫

P

u2(x, t′

)dx ≤ 4R2

∫

P

u21(x, t′

)dx ≤ 4

∫

P

n∑

i=1

u2i(x, t′

)dx.

If we recall that u (x, t′) = 0 in P\ BR

(x0)and integrate the last inequality with respect to t′ from 0

to T, then we obtain that

∫

Q

u2dxdt ≤ 4

∫

Q

n∑

i=1

u2i dxdt. (2.24)

One can analogously prove that

∫

Q

n∑

i=1

u2i dxdt ≤ 4

∫

Q

n∑

i,j=1

u2ijdxdt. (2.25)

Now the required estimate (2.23) follows from (2.24), (2.25), and Lemma 2.3. The lemma is proved.

113

Remark. Obviously, if the low-order coefficients of the operator L are bounded, i.e.,

|bi (x, t)| ≤ b0, i = 1, . . . , n, |c (x, t)| ≤ b0, (2.26)

then for any T ≤ T3 (ϕ, σ, n, bo) and any function u (x, t) ∈ A(QR

T

(x0))

the estimate

‖u‖W 2,2

2,ϕ(QRT (x0)) ≤ C2.4 (ϕ, σ, n) ‖Lu‖L2(QR

T (x0)) (2.27)

holds.In fact, according to the imbedding theorem proved in [13], for any function u (x, t) ∈ W 2,1

2 (QT )the following inequalities hold:

n∑

i=1

‖ui‖Lp(QT ) ≤ C2.5 (p, n) ‖u‖W 2,12 (QT )

if p ≥ 2,

(1

2− 1

p

)(n+ 2) ≤ 1; (2.28)

‖u‖Lq(QT ) ≤ C2.6 (q, n) ‖u‖W 2,12 (QT )

if q ≥ 2,

(1

2− 1

q

)(n+ 2) ≤ 2; (2.29)

‖u‖L∞(QT ) ≤ C2.7 ‖u‖W 2,12 (QT )

if n = 1. (2.30)

On the other hand,

∥∥L′u∥∥L2(Q)

≤ ‖Lu‖L2(Q) +∥∥(L′ − L)u∥∥

L2(Q)≤ ‖Lu‖L2(Q) +

n∑

i=1

‖biui‖L2(Q) + ‖cu‖L2(Q)

≤ ‖Lu‖L2(Q) + b0

(n∑

i=1

‖ui‖L2(Q) + ‖u‖L2(Q)

)≤ ‖Lu‖L2(Q)+b0 (mesQ)

1n+2 (2.31)

×(

n∑

i=1

‖ui‖L 2(n+2)n

(Q) + ‖u‖L 2(n+2)n

(Q)

).

Applying inequalities (2.26) and (2.29) for p = q =2 (n+ 2)

n , from (2.31), we obtain that

∥∥L′u∥∥L2(Q)

≤ ‖Lu‖L2(Q) + C2.8 (n) (ωnT )1

n+2 ‖u‖W 2,1

2 (Q), (2.32)

where ωn is the volume of the unit n-dimensional ball.Then from (2.23) and (2.24) for T3 ≤ T2, it follows that

‖u‖W 2,2

2,ϕ(Q)≤ C2.3 ‖Lu‖L2(Q) + C2.3C2.8 (ωnT )

1n+2 ‖u‖

W 2,22,ϕ(Q)

.

Now, it suffices to fix T3 = min

(T2,

1ωn (2C2.3C2.8)

n+2

), and the required estimate (2.27) is

proved.

3. Main Coercive Inequality

We say that u (x, t) ∈ A1

(QR

T

(x0))

if u (x, t) ∈ C∞ (QR

T

(x0)), u|t=0 = 0.

Lemma 3.1. If the coefficients of the operator L satisfy Conditions (2.19), (1.4), (1.5), and (2.25),then for any T ≤ T3, any ε > 0, and any function u (x, t) ∈ A1

(QR

T

(x0)), the following estimate

holds:‖u‖

W 2,22,ϕ

(Q

R2T (x0)

) ≤ C2.4 ‖Lu‖L2(QRT (x0)) + ε ‖u‖

W 2,22,ϕ(QR

T (x0))

+ε−1R−2C3.1 (ϕ, σ, n, b0) ‖u‖L2(QRT (x0)) .

(3.1)

114

Proof. Denote BR2

(x0)× (0, T ) by Q1. Consider a function η (x) ∈ C∞

0

(BR

(x0))

such that η (x) = 1

for x ∈ BR2

(x0), 0 ≤ η (x) ≤ 1, and

|ηi| ≤ C3.2 (n)R−1, |ηij | ≤ C3.2(n)R

−2; i, j = 1, . . . , n. (3.2)

Applying estimate (2.28) to the function u (x, t) η (x), we obtain that

‖u‖W 2,2

2,ϕ(Q1)= ‖uη‖

W 2,22,ϕ(Q1)

≤ ‖uη‖W 2,2

2,ϕ(Q)≤ C2.4 ‖L (uη)‖L2(Q) . (3.3)

On the other hand,

L (uη) = ηLu+ uL′′η + 2n∑

i,j=1

aij (x, t)uiηj ,

where L′′ = L − c (x, t) . Thus, taking into account (3.2) in (3.3), we conclude that

‖u‖W 2,2

2,ϕ(Q1)≤ C2.4 ‖Lu‖L2(Q) +

C3.3 (ϕ, σ, n, b0)

R2‖u‖L2(Q)

+C3.4 (σ, n)

R

n∑

i=1

‖ui‖L2(Q) ≤ C2.4 ‖Lu‖L2(Q) +C3.5 (ϕ, σ, n, b0)

R2‖u‖

W 1,02 (Q)

.

(3.4)

According to the interpolation inequality [13], for any ε′ > 0, we have

‖u‖W 1,0

2 (Q)≤ ε′ ‖u‖

W 2,02 (Q)

+C3.6 (n)

ε′‖u‖L2(Q) . (3.5)

Fix any arbitrary ε′ > 0. Let ε′ = εR2

C3.5. Then (3.4) and (3.5) imply the required estimate (3.1). The

lemma is proved.For ρ > 0, denote the set {x : x ∈ Ω, dist (x, ∂Ω) > ρ} by Ωρ. Let QT (ρ) = Ωρ × (0, T ) .

Corollary. If the coefficients of the operator L′ satisfy Conditions (2.19), (1.4), and (1.5), then forany ρ ≤ 1, T ≤ T2, and ε > 0 and any function u (x, t) ∈ C∞ (

QT

), u|t=0 = 0, the following estimate

holds:

‖u‖W 2,2

2,ϕ(QT (ρ))≤ C3.7 (ϕ, σ, n, ρ,Ω)

∥∥L′u∥∥L2(QT )

+ ε ‖u‖W 2,2

2,ϕ(QT )

+C3.8 (ϕ, σ, n, ρ,Ω)

ε‖u‖L2(QT ) .

(3.6)

Lemma 3.2. Let Conditions (2.19), (1.4), and (1.5) be fulfilled for the coefficients of the operatorL′. Then there exists ρ1 (σ, n,Ω) such that if T ≤ T3, then for any ε > 0 and any function u (x, t) ∈C∞ (

QT

), u|Γ(QT ) = 0. the following estimate holds:

‖u‖W 2,2

2,ϕ(Q′T (ρ1)) ≤ C3.9 (ϕ, σ, n,Ω)

∥∥L′u∥∥L2(QT )

+ ε ‖u‖W 2,2

2,ϕ(QT )+C3.10 (ϕ, σ, n,Ω)

ε‖u‖L2(QT ), (3.7)

where Q′T (ρ1) = QT \QT (ρ1) .

Proof. It suffices to consider the case where the coefficients of the operator L′ are infinitely dif-ferentiable in QT . We fix arbitrary t ∈ (0, T ) , x0 ∈ ∂Ω, and ε > 0. Consider the domain Ωt =Ω × {(y, τ) : τ = tn} . There exists an orthogonal transformation of coordinates x → y such that the

hyperplane tangent to ∂Ωt at the point(y0, t

)is perpendicular to the yn axis. Here Ωt and

(y0, tn

)

are the images of the domain Ωt and point(x0, t

)respectively. Denote by L′ the image of the operator

L′. For simplicity we will assume that the intersection of ∂Ωt′ with some ball Br = Br(y0)

(y0, tn

)

is given by the equation yn = ψ (y1, . . . , yn−1) and ψ ∈ C2, which is the part of Ωt adjacent to

115

∂Ωt ∩ Br, lies on the set {(y, t′) : yn > ψ (y1, . . . , yn−1)} . Now we give an equivalent form of Con-dition (1.4). Let, μi (x, t) i = 1, . . . , n be eigenvalues of the matrix A (x, t) = ‖aij (x, t)‖ . ThenTr (A (x, t)) =

n∑i=1

aii (x, t) =n∑

i=1μi (x, t) ,

n∑i,j=1

a2ij (x, t) = Tr(A2 (x, t)

)=

n∑i=1

μi (x, t) , where μi (x, t)

are eigenvalues of the matrix A2 (x, t) ; i = 1, . . . , n. However, it is clear that

μi (x, t) = μ2i (x, t) , i = 1, . . . , n.

This implies that Condition (1.4) can be written in the following equivalent form:

σ = supQT

n∑i=1

μ2i (x, t)

(n∑

i=1μi (x, t)

)2 − 1

n− λ2< 0,

where

λ =

infQT

n∑i=1

μi (x, t)

supQT

n∑i=1

μi (x, t)

.

Thus, if the matrixA (x, t) satisfies Condition (1.4) with the constant σ, then the image of this matrixafter an orthogonal transformation also satisfies a condition of form (1.4) with the same constant σ(as well as a condition of the form (2.19) with the constant γ′). We make one more transformation ofcoordinates y → ϑ as follows: ϑ1 = y1, . . . , ϑn−1 = yn−1, ϑn = yn −ψ (y1, . . . , yn−1) . Denote by L theimage of operator L′ and by aij (ϑ, t) the coefficients of the operator L for the high-order derivativeswith respect to spatial variables, i, j = 1, . . . , n. Obviously, L is an operator of the form L and absolutevalue of its low-order coefficients are bounded by a constant depending only on n and Ω. Moreover,

aij (ϑ, t) =n∑

k,l=1

akl (y, t)∂ϑi∂yk

∂ϑj∂, yl

; i, j = 1, . . . , n,

where akl (y, t) are coefficients of the operator L′ at high-order higher derivatives with respect tospatial variables, k, l = 1, . . . , n. Thus,

aij (ϑ, t) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

aij (y, t) if 1 ≤ i, j ≤ n− 1,

anj (y, t)−n−1∑k=1

akj (y, t)ψyk if i = n, 1 ≤ j ≤ n− 1,

ann (y, t) +n−1∑k,l=1

akl (y, t)ψykψyl − 2n−1∑k,l=1

akn (y, t)ψyk if i = j = n.

Let Ω′t and

(ϑ0, t

)be the images of the domain Ωt and the point

(y0, t

)respectively and y = (y1, . . . ,

yn−1). Obviously, Ω′t, being an n-dimensional domain, does not depend on t, i.e., Ω′

t = Ω′. Now, takinginto account the arbitrariness of t and the fact that ψyi

(y0)= 0 for i = 1, . . . , n− 1, we conclude that

there exists a small ν(ϑ0, σ, n,Ω

)such that if Conditions (2.19) and (1.4) are fulfilled for the matrix

‖aij (y, t)‖ with constants γ′ and σ, then the same conditions are fulfilled for the matrix ‖aij (ϑ, t)‖with constants γ′ ≥ γ′/2 and σ ≤ σ/2, respectively, if (ϑ, t) ∈ C

+ν(ϑ0)

=(Ω′ ∩Bν(ϑ0)

(ϑ0)) × (0, T ) .

For brevity we denote here ν(ϑ0, σ, n,Ω

)by ν

(ϑ0).

Let u (ϑ, t) be the image of the function u (x, t) under the transformations of spatial coordinatesx→ y and y → ϑ. Obviously, C+

ν(ϑ0)is a cylinder whose base is a half-ball

B+ν(ϑ0)

={ϑ :

∣∣ϑ− ϑ0∣∣ < ν

(ϑ0), ϑn > 0

}. For each t ∈ (0, T ) we continue the function u (ϑ, t) to get

116

an odd function (and the coefficients of the operator L to get an even function) through the hyperplaneϑn = 0 into the half-ball

B−ν(ϑ0)

={ϑ :

∣∣ϑ− ϑ0∣∣ < ν

(ϑ0), ϑn < 0

}

and denote the continued function and operator again by u (ϑ, t) and L, respectively. If Cν(ϑ0) =(B+

ν(ϑ0)∪ B−

ν(ϑ0)

)× (0, T ) , then u (z, t) ∈W 2,2

2,ϕ

(Cν(ϑ0)

). According to Lemma 3.1, for any ε′ > 0,

‖u‖W 2,2

2,ϕ

(C ν(ϑ0)

2

) ≤ C3.11 (ϕ, σ, n,Ω)∥∥Lu∥∥

L2

(Cν(ϑ0)

) + ε′ ‖u‖W 2,2

2,ϕ

(Cν(ϑ0)

)

+C3.12 (ϕ, σ, n,Ω)

ε′ν2 (ϑ0)‖u‖L2(Cν(ϑ0))

, (3.8)

where C ν(ϑ0)2

= B ν(ϑ0)2

(ϑ0)× (0, T ) . Let C+

ν(ϑ0)2

= B+ν(ϑ0)

2

× (0× T ) .

Taking into account the rule to continue u (z, t) and the coefficients of the operator L, we obtainthe relation

‖u‖2W 2,2

2,ϕ

(Cν(ϑ0)

) = ‖u‖2W 2,2

2,ϕ

(C+

ν(ϑ0)

) + ‖u‖2W 2,2

2,ϕ

(B−

ν(ϑ0)×(0,T )

)

= 2 ‖u‖2W 2,2

2,ϕ

(C+

ν(ϑ0)

) .

Obviously, similar relations hold for the norms ‖u‖W 2,2

2,ϕ

(C ν(ϑ0)

2

) , ‖u‖L2

(Cν(ϑ0)

) and ∥∥Lu∥∥L2(Cν(ϑ0))

.

Therefore, from (3.8), we derive the inequality

‖u‖W 2,2

2,ϕ

⎛⎝C

+

ν(ϑ0)2

⎞⎠

≤ C3.11

∥∥Lu∥∥L2(C+

ν (ϑ0)) + ε′ ‖u‖W 2,2

2,ϕ

(C+

ν(ϑ0)

)+ C3.12

ε′ν2 (ϑ0)‖u‖

L2

(C+

ν(ϑ0)

) .

Returning to the variables (x, t) , we conclude that

‖u‖2W 2,2

2,ϕ(C1(x0)) ≤ C3.13(ϕ, σ, n,Ω)

(∥∥L′u∥∥2L2(C(x0))+

(ε′)2 ‖u‖2

W 2,22,ϕ(C(x0)) +

1

(ε′)2 ν4 (ϑ0)‖u‖2

L2(C(x0))

),

(3.9)where C

1(x0) and C(x0) are the inverse images of the cylinders C+ν(ϑ0)

2

and C+ν(ϑ0)

, respectively. By

the definition of the boundary of the class C2 (see [14]), there exists ρ1 (σ, n,Ω) such that Q′T (ρ1) can

be covered by a finite number l1 of domains of the form C1(xi), i = 0, . . . , l1 − 1, and l1 depends only

on σ, n, and Ω. Denote mini=0,...,l1−1

ν(ϑi)by ν0. Applying inequalities of form (3.9) to the sets C

1(xi)

and C(xi), respectively (changing ν(ϑi)for ν0), and then summing them over i from 0 to l1 − 1, we

come to estimate

‖u‖2W 2,2

2,ϕ(Q′T (ρ1))

≤ C3.13l1

(∥∥L′u∥∥2L2(QT )

+(ε′)2 ‖u‖2

W 2,22,ϕ(QT )

+1

(ε′)2 ν40‖u‖2L2(QT )

).

Now it suffices to choose ε′ = ε√C3.13l1

, and the required estimate (3.7) is proved.

Lemma 3.3. Let the conditions of the previous lemma be satisfied. Then for any T ≤ T3 and anyfunction u (x, t) ∈ C∞ (

QT

), u|Γ(QT ) = 0, the following estimate holds:

‖u‖W 2,2

2,ϕ(QT )≤ C3.14 (ϕ, σ, n,Ω)

(∥∥L′u∥∥L2(QT )

+ ‖u‖L2(QT )

). (3.10)

117

Proof. From (3.6) and (3.7), for any ε > 0, we have

‖u‖2W 2,2

2,ϕ(QT (ρ1))≤ 2ε2 ‖u‖2

W 2,22,ϕ(QT )

+ 2

(C3.7

∥∥L′u∥∥L2(QT )

+C3.8

ε‖u‖L2(QT )

)2

,

‖u‖2W 2,2

2,ϕ(Q′T (ρ1))

≤ 2ε2 ‖u‖2W 2,2

2,ϕ(QT )+ 2

(C3.9

∥∥L′u∥∥L2(QT )

+C3.10

ε‖u‖L2(QT )

)2

.

Now, it suffices to sum these inequalities, assigning ε = 12√2, and we come to the required esti-

mate (3.10).Let h (x, t) ∈ Lp (QT ) , p ∈ [1,∞) . The value

Hh;p (ν) = supe⊂QT , mes e≤ν

⎛

⎝∫

e

|h (x, t)|p dxdt⎞

⎠

1p

is called the AC-modulus of the function |h (x, t)|p . Everywhere below the notation C (L) means thatthe positive constant C depends only on ϕ, σ,Hb1;n+2,. . . ,Hbn;n+2,Hc;s, ‖b1‖Ln+2(QT ) ,. . . ,‖bn‖Ln+2(QT ) ,

and ‖c‖Ls(QT ) .

Theorem 3.1. Let the coefficients of the operator L satisfy Conditions (2.19), and (1.4)–(1.6) be

satisfied. Then at T ≤ T0 (L, n,Ω) for any function u (x, t) ∈ W 2,22,ϕ (QT ) the following estimate holds:

‖u‖W 2,2

2,ϕ(QT )≤ C3.15 (ϕ, σ, n,Ω) ‖Lu‖L2(QT ) . (3.11)

Proof. It is obvious that it suffices to consider the case where u (x, t) ∈ C∞ (QT

), u|Γ(QT ) = 0.

According to inequality (3.10), we have the estimate

‖u‖W 2,2

2,ϕ(QT )

≤ C3.14

(‖Lu‖L2(QT ) +

n∑

i=1

‖biui‖L2(QT ) + ‖cu‖L2(QT ) + ‖u‖L2(QT )

).

(3.12)

Using estimate (2.28) for p = 2 (n+ 2) /n, we obtain the inequality

n∑

i=1

‖biui‖L2(QT ) ≤n∑

i=1

‖bi‖Ln+2(QT ) ‖ui‖L 2(n+2)n (QT )

≤n∑

i=1

Hbi;n+2 (T mesΩ) ‖ui‖L 2(n+2)n (QT )

≤ 1

4C3.14‖u‖

W 2,22,ϕ(QT )

(3.13)

if T ≤ T4 and T4 ≤ T3 is so small that

Hbi;n+2 (T4mesΩ) ≤ 1

4C3.14C2.5, i = 1, . . . , n.

Suppose now that n ≥ 3. Using estimate (2.29) for q =2 (n+ 2)n− 2 , we conclude that

‖cu‖L2(QT ) ≤ ‖c‖Ln+22

(QT ) ‖u‖L 2(n+2)n−2

(QT )≤ Hc;n+22

(T mesΩ) ‖u‖L 2(n+2)n−2

(QT ) ≤1

4C3.14‖u‖

W 2,22,ϕ(QT )

(3.14)if T ≤ T5 and T5 ≤ T3 is so small that

Hc;n+22

(T5mesΩ) ≤ 1

4C3.14C2.6.

118

Let now n = 2. For q = 2 (2 + ν) /ν, estimate (2.29) implies that

‖cu‖L2(QT ) ≤ ‖c‖L2+ν(QT ) ‖u‖L 2(2+ν)ν

(QT )≤ Hc;2+ν (T mesΩ) ‖u‖L 2(2+ν)ν

(QT ) ≤1

4C3.14‖u‖

W 2,22,ϕ(QT )

(3.15)if T ≤ T6 and T6 ≤ T3 is so small that

Hc;2+ν (T6mesΩ) ≤ 1

4C3.14C2.6.

Finally, let n = 1. From (2.30), we obtain the inequality

‖cu‖L2(QT ) ≤ ‖c‖L2(QT ) ‖u‖L∞(QT ) ≤ Hc;2 (T mesΩ) ‖u‖L∞(QT )≤1

4C3.14‖u‖

W 2,22,ϕ(QT )

(3.16)

if T ≤ T7 and T7 ≤ T3 is so small that

Hc;2 (T7mesΩ) ≤ 1

4C3.14C2.7.

On the other hand, according to the Friedrichs inequality (see the proof of Lemma 2.5),

‖u‖L2(QT ) ≤ T ‖ut‖L2(QT ) ≤ T ‖u‖W 2,2

2,ϕ(QT )≤ 1

4C3.14‖u‖

W 2,22,ϕ(QT )

(3.17)

if T ≤ T8 = min(T3,

14C3.14

).

Now we fix T0 = min (T4, T5, T6, T7, T8) . If T ≤ T0, then, using (3.13)–(3.17) in (3.12), we concludethat

‖u‖W 2,2

2,ϕ(QT )≤ C3.14 ‖Lu‖L2(QT ) +

3

4‖u‖

W 2,22,ϕ(QT )

,

which implies the required estimate (3.11). The theorem is proved.

Remark. If the conditions of the theorem are satisfied, then for any T ≤ T0, τ ∈ [0, 1], and u (x, t) ∈W 2,2

2,ϕ (QT ) , the following estimate holds:

‖u‖W 2,2

2,ϕ(QT )≤ C3.15

∥∥∥Lu− τ

Tu∥∥∥L2(QT )

. (3.18)

In order to prove it, it suffices to use Remark 1 to Lemma 2.3.

Theorem 3.2. Let the function ϕ (z) satisfy Conditions (1.5) and the operator Lε, ε > 0, be the same

as in Lemma 2.2. Then for any T ≤ T9 (ϕ, n,Ω) and any function u (x, t) ∈ W 2,22,ϕε

(QT ) , the estimate

‖u‖W 2,2

2,ϕε(QT )

≤ C3.16 (ϕ, n,Ω) ‖Lεu− μu‖L2(QT ) (3.19)

holds. Here μ = T−1, W 2,22,ϕε

(QT ) is the Banach space of functions u (x, t) defined on QT with the

finite norm defined (2.1), where the function ϕ (T − t) is change by ϕε (T − t) , and W 2,22,ϕε

(QT ) is

the completion of the set of all functions from C∞ (QT

)vanishing on ∂QT by the norm of the space

W 2,22,ϕε

(QT ) .

Proof. It suffices to prove the theorem for functions u (x, t) ∈ C∞ (QT

), u|∂QT

= 0. Note thatq (T1) ≤ 1 according to (2.9). Then, reasoning in the same way as in the proofs of Lemmas 2.5 and3.1–3.3, from (2.12), we conclude that there exists T10 (ϕ, n,Ω) ≤ T1 such that if T ≤ T10, then forany function v (x, t) ∈ C∞ (

QT

), v|Γ(QT ) = 0, v|t=T = vt|t=T = 0, the estimate

119

‖v‖W 2,2

2,ϕ(QT )≤ C3.17 (ϕ, n,Ω)

(‖Lεv‖L2(QT ) + ‖v‖L2(QT )

)(3.20)

holds. We assume that T ≤ T10/2. Suppose that R = T/4 and u (x, t) ∈ C∞ (QT

), u|∂QT

= 0.Consider a function ζ (t) ∈ C∞[0, T ] such that ζ (t) = 1 for t ∈ [0, T −R], ζ (t) = 0 for t ∈ [T −R/2, T ],0 ≤ ζ (t) ≤ 1, and

∣∣ζ ′ (t)∣∣ ≤ C3.18

R,∣∣ζ ′′ (t)

∣∣ ≤ C3.18

R2. (3.21)

Assuming in (3.20) that v (x, t) = u (x, t) ζ (t) and taking (3.21) into account, we obtain the inequality

‖v‖W 2,2

2,ϕε(QT−R)

≤ C3.17

(‖Lε (uζ)‖L2(QT ) + ‖u‖L2(QT )

)

≤ C3.17

(‖Lεu‖L2(QT )

+(C3.18R + 1

) ‖u‖L2(QT )

(3.22)

+2C3.18

R‖ϕεut‖L2(QT ) +

C3.18

R2‖ϕεu‖L2(QT )

).

From Conditions (1.5), it follows that sup[0,T ]

ϕ (z) ≤ C3.19 (ϕ)T. Therefore, taking into account that

sup[0,T ]

ϕε (z) = sup[0,T ]

ϕ (z) , we conclude that

‖ϕεu‖L2(QT ) ≤ C3.19T ‖u‖L2(QT ) . (3.23)

On the other hand, for any α′ > 0, the following interpolation inequality holds:

‖ϕεut‖L2(QT ) ≤ C3.19Tα′ ‖ϕεutt‖L2(QT ) + 3(α′)−1 ‖u‖L2(QT ) . (3.24)

In fact, we fix arbitrary α′ and consider for ν > 0 the integral

K =

∫

QT

(νϕ2

ε (T − t)utt +1

νu

)2

dxdt.

Obviously, K ≥ 0. At the same time

K = ν2∫

QT

ϕ4ε (T − t)u2ttdxdt+

1

ν2

∫

QT

u2dxdt+ 2

∫

QT

ϕ2ε (T − t)uttudxdt

≤ C23.19T

2ν2∫

QT

ϕ2ε (T − t)u2ttdxdt+

1

ν2

∫

QT

u2dxdt− 2

∫

QT

ϕ2ε (T − t)u2tdxdt

+4

∫

QT

ϕε (T − t)ϕ′ε (T − t)u utdxdt.

(3.25)

Moreover, using the fact that q (T ) ≤ 1 and taking into account (2.11), we obtain the inequality

4

∫

QT

ϕε (T − t)ϕ′ε (T − t)u utdxdt ≤

∫

QT

ϕ2ε (T − t)u2tdxdt

+4

∫

QT

(ϕ′ε (T − t)

)2u2dxdt ≤

∫

QT

ϕ2ε (T − t)u2tdxdt+ 4

∫

QT

u2dxdt.

(3.26)

From (3.25)-(3.26), it follows that

120

∫

QT

ϕ2ε (T − t)u2tdxdt ≤ C2

3.19T2ν

∫

QT

ϕ2ε (T − t)u2ttdxdt+

(ν−2 + 4

) ∫

QT

u2dxdt.

Now it suffices to assume ν = min (α′, 1) , and inequality (3.24) is proved.Using (3.23) and (3.24) in (3.22), we conclude that the inequality

‖u‖W 2,2

2,ϕ(QT−R)≤ C3.17 ‖Lεu‖L2(QT )+8α′C3.17C3.18C3.19 ‖u‖W 2,2

2,ϕε(QT )

+ (α′)−1R−1C3.20 ‖u‖L2(QT )

(3.27)holds for any α′ > 0.

We fix arbitrary α > 0 and choose α′ = α8C3.17C3.18C3.19

. Then (3.27) implies that

‖u‖W 2,2

2,ϕε(QT−R)

≤ C3.17 ‖Lεu‖L2(QT ) + α ‖u‖W 2,2

2,ϕε(QT )

+α−1T−1C3.21 (ϕ, n,Ω) ‖u‖L2(QT ) . (3.28)

In the same way, one can show that if Q′ = Ω× (T − 2R, T + 2R) , Q′′ = Ω× (T −R, T +R) , andS (Q′) = ∂Ω× [T − 2R, T + 2R] , then for any function w (x, t) ∈ C∞ (

Q′) , w|S(Q′) = 0, and any α > 0the following estimate holds:

‖w‖W 2,2

2,ϕε(Q′′) ≤ C3.17 ‖Lεw‖L2(Q′) + α ‖w‖

W 2,22,ϕε

(Q′) +C3.22 (ϕ, n,Ω)

α T‖w‖L2(Q′) . (3.29)

Let Q′+ = Ω × (T − 2R, T ) , Q′− = Ω × (T, T + 2R) , and Q′′

+ = Ω × (T −R, T ) . We extend thefunctions u (x, t) and ϕε (T − t) through the hyperplane t = T from Q′

+ to Q′− and denote the extendedfunctions again by u (x, t) and ϕε (T − t) , respectively. Assuming that w = u in (3.28) and taking intoaccount the relation ‖u‖

W 2,22,ϕε

(Q′′) =√2 ‖u‖

W 2,22,ϕε(Q

′′+)

and similar relations for the norms ‖u‖W 2,2

2,ϕε(Q′) ,

‖u‖L2(Q′) , and ‖Lεu‖L2(Q′) , we obtain the estimate

‖u‖W 2,2

2,ϕε(Q′′+)

≤ C3.17 ‖Lεu‖L2(Q′+)

+ α ‖u‖W 2,2

2,ϕε(Q′+)

+C3.22

αT‖u‖L2(Q′

+). (3.30)

Combining (3.28) and (3.30) and selecting an appropriate α, we conclude that

‖u‖2W 2,2

2,ϕε(QT )

≤ C3.23 (ϕ, n,Ω)(‖Lεu‖2L2(QT ) + T−2 ‖u‖2L2(QT )

). (3.31)

On the other hand, if we recall that μ = T−1, then we have the relation∫

QT

(Lεu− μu)2 dxdt = ‖Lεu‖2L2(QT ) + μ2 ‖u‖2L2(QT )

−2μ

∫

QT

uLεudxdt = ‖Lεu‖2L2(QT ) + μ2 ‖u‖2L2(QT ) +K1.

(3.32)

Moreover,

K1 = −2μ

∫

QT

u (Δu+ ϕε (T − t)utt − ut) dxdt = 2μ

∫

QT

n∑

i=1

u2i dxdt

−2μ

∫

QT

ϕε (T − t)u uttdxdt+ μ

∫

QT

(u2)tdxdt ≥ 2μ

∫

QT

ϕε (T − t)u2tdxdt

−2μ

∫

QT

ϕ′ε (T − t)u utdxdt.

(3.33)

121

Now we will show that the inequality

ϕε (z) ≥ β zϕ′ε (z) (3.34)

is satisfied for z ∈ (0, T ). By virtue of Conditions (1.5), it suffices to justify the validity of (3.34) onlyfor z ∈ (0, ε) . However, for such z, (3.34) is equivalent to the inequality

ϕ (ε)− ϕ′ (ε) εm

≥ ϕ′ (ε) zm

m εm−1,

where m = 2β−1. However, the last inequality will be valid if the following estimate holds:

ϕ (ε) ≥ 2

mϕ′ (ε) ε. (3.35)

Now it suffices to note that (3.35) is satisfied by virtue of (1.5). Thus, from (3.32), (3.33), and (2.11),we obtain the inequality

K1 ≥ −μ2

∫

QT

[ϕ′ε (T − t)]2

ϕε (T − t)u2dxdt ≥ − μ

2β

∫

QT

ϕ′ε (T − t)

T − tu2dxdt≥ −μ q (T )T

2β

∫

QT

u2

(T − t)2dxdt. (3.36)

Now we apply Hardy’s inequality (see [10]), according to which

∫

QT

u2

(T − t)2dxdt ≤ 4

∫

QT

u2tdxdt. (3.37)

Then, from (3.32), (3.36), and (3.37) we conclude that

‖Lεu‖2L2(QT ) + μ2 ‖u‖2L2(QT ) ≤ ‖Lεu− μu‖2L2(QT ) + q (T )β−1 ‖u‖2W 2,2

2,ϕε(QT )

. (3.38)

Now we choose a small T11 such that q (T11) ≤ β4C3.23

and fix T9 = min(T102 , T11

). Then (3.31)

and (3.38) imply the required estimate (3.19). The theorem is proved.In the sequel, we will denote by M0 and M1 the operators L0 − μ and Lε − μ respectively.

4. Solvability of the First Boundary Value Problem for the Equation M0u = f

Denote min (T0, T9) by T0.

Theorem 4.1. Let the function ϕ (z) satisfy Conditions (1.5). Then for T ≤ T 0, the first boundaryvalue problem

M0u = f (x, t) , (x, t) ∈ QT , u|Γ(QT ) = 0 (4.1)

is uniquely strongly solvable in the space W 2,22,ϕ (QT ) for any f (x, t) ∈ L2 (QT ) .

Proof. First, suppose that f (x, t) ∈ C∞ (QT

). Let ϑ (x, t) be a classical solution of the first

boundary value problem

Δϑ− ϑt = f(x, t), (x, t) ∈ QT , ϑ|Γ(QT ) = 0.

Obviously, this solution exists. According to [13], ϑ (x, t) belongs to W 2,22 (QT ) and

‖ϑ‖W 2,2

2 (QT )≤ C4.1 (n,Ω, f) . (4.2)

Here W 2,22 (QT ) is the Banach space of functions defined on QT with the finite norm of form (2.1)

for ϕ ≡ 1. By virtue of the fact that ϕε (z) ≤ 1 for ε ∈ (0, T ) , from (4.2), we conclude that

122

‖ϑ‖W 2,2

2,ϕε(QT )

≤ C4.1. (4.3)

Denote by W 2,22 (QT ) the completion by the norm of spaceW 2,2

2 (QT ) of the set of all functions fromC∞ (

QT

)vanishing on ∂QT . Let u

ε (x, t) be a strong (almost everywhere) solution of the Dirichletproblem

Mεuε = f (x, t) , (x, t) ∈ QT , (uε (x, t)− ϑ (x, t)) ∈ W 2,2

2 (QT )

According to [10] such solution exists for any ε > 0. Obviously, (uε (x, t)− ϑ (x, t)) ∈ W 2,22,ϕε

(QT ) .

Taking into account the relation ϑ|Γ(QT ) = 0 and also inequality (2.10), we obtain that uε (x, t) ∈W 2,2

2,ϕ (QT ) . Besides, if Fε (x, t) = Mεϑ, then taking into account (4.3), we see that

‖Fε‖L2(QT ) ≤ C4.2 (n,Ω, T, f) . (4.4)

From Theorem 3.2, it follows that

‖uε − ϑ‖W 2,2

2,ϕε(QT )

≤ C3.16

(‖f‖L2(QT ) + ‖Fε‖L2(QT )

).

Then, from (4.3), (4.4), and (2.10) we conclude that

‖uε‖W 2,2

2,ϕ(QT )≤ C4.3 ‖u‖W 2,2

2,ϕ(QT )≤ C4.4 (ϕ, n,Ω, T, f) . (4.5)

Thus, the set of functions {uε (x, t)} is uniformly bounded with respect to ε by the norm of the

space W 2,22,ϕ (QT ) . Then this set is weakly compact in W 2,2

2,ϕ (QT ) . The latter, in particular, means that

there exists a sequence of positive numbers {εk} , limk→∞

εk = 0 and a function u0 (x, t) ∈ W 2,22,ϕ (QT )

such that for any ψ (x, t) ∈ C∞ (QT

)we have

limk→∞

(M0uεk , ψ) = (M0u0, ψ) . (4.6)

Here (f1, f2) =∫

QT

f1f2dxdt. On the other hand,

(M0uεk , ψ) = ((M0 −Mεk)u

εk , ψ) + (Mεkuεk , ψ) = ((M0 −Mεk)u

εk , ψ) + (f, ψ) . (4.7)

Moreover, taking into account (2.10) and (4.5), we see that

J (k) = |(M0 −Mεk)uεk , ψ| ≤ ‖(ϕ− ϕεk)u

εktt ‖L2(Q(εk))

‖ψ‖L2(Q(εk))

≤ 3 ‖uεk‖W 2,2

2,ϕεk(QT )

‖ψ‖L2(Q(εk))≤ 3C4.4 ‖ψ‖L2(Q(εk))

,

where Q (ε) = Ω× (T − ε, T ) . This implies that

limk→∞

J (k) = 0. (4.8)

From (4.6)–(4.8) we obtain that (M0u0ψ) = (f, ϕ) , i.e., M0u0 = f (x, t) almost everywhere in QT .Now, let f (x, t) ∈ L2 (QT ) . Then there exists a sequence {fm (x, t)} , m = 1, 2, . . . , such that

fm (x, t) ∈ C∞ (QT

)and lim

m→∞ ‖fm − f‖L2(QT ) = 0. For a positive integer m, consider the sequence

{um (x, t)} of strong solutions of first boundary value problems

M0um = fm (x, t) , (x, t) ∈ QT , um|Γ(QT ) = 0.

As proved above, a function um (x, t) exists for each m. Applying estimate (3.18) for L = L0 andτ = 1, we conclude that

‖um‖W 2,2

2,ϕ(QT )≤ C3.15 ‖fm‖L2(QT ) ≤ C4.5 (ϕ, n,Ω, f) . (4.9)

123

Thus, sequence {um (x, t)} is weakly compact in W 2,22,ϕ (QT ) , i.e., there exists a subsequence of

positive integers {mk} , limk→∞

mk = ∞, and a function u (x, t) ∈ W 2,22,ϕ (QT ) such that for any ψ (x, t) ∈

C∞ (QT

)

limk→∞

(M0umk, ψ) = (M0u, ψ) .

However,

limk→∞

(M0umk, ψ) = lim

k→∞(fmk

, ψ) = (f, ψ) .

Therefore, (M0u, ψ) = (f, ψ) , i.e., M0u = f (x, t) almost everywhere in QT . Thus, the existence ofa strong solution of Problem (4.1) is proved. Its uniqueness follows from estimate (3.18). The theoremis proved.

5. Local Solvability of the First Boundary Value Problem

Theorem 5.1. If the coefficients of the operator L satisfy Conditions (2.19) and (1.4)–(1.6), then at

T ≤ T 0 the first boundary value Problem (1.1)-(1.2) is uniquely strongly solvable in the space W 2,22,ϕ (QT )

for any f (x, t) ∈ L2 (QT ) , and the following estimate holds for solution u (x, t):

‖u‖W 2,2

2,ϕ(QT )≤ C3.15 ‖f‖L2(QT ) . (5.1)

Proof. The uniqueness of the solution and estimate (5.1) follow from the Theorem 3.1. Prove the

existence of solution. To do this, for τ ∈ [0, 1], we introduce the set of operators L(τ) = (1− τ)M0+τL.We will show that the set E of points τ for which the problem

L(τ)u = f (x, t) , (x, t) ∈ QT , u|Γ(QT ) = 0 (5.2)

is uniquely strongly solvable in W 2,22,ϕ (QT ) for any f (x, t) ∈ L2 (QT ) is nonempty, open, and closed

with respect to the segment [0, 1] . From here it follows that E = [0, 1] and, in particular, Problem (5.2)

is solvable for τ = 1 if L(1) = L.The nonemptiness of the set E follows from Theorem 4.1. Prove its openness. Let τ0 ∈ E and ε > 0

be numbers that will be chosen later. Show that Problem (5.2) is solvable for all τ ∈ [0, 1] such that|τ − τ0| < ε. We represent Problem (5.2) in the equivalent form

L(τ0)u = f (x, t)−(L(τ) − L(τ0)

)u, (x, t) ∈ QT ; u(x, t) ∈ W 2,2

2,ϕ (QT ) . (5.3)

Consider an arbitrary function v (x, t) ∈ W 2,22,ϕ (QT ) and the first boundary value problem

L(τ0)u = f (x, t)−(L(τ) − L(τ0)

)v (x, t) , (x, t) ∈ QT ; u(x, t) ∈ W 2,2

2,ϕ (QT ) . (5.4)

Obviously,(L(τ) − L(τ0)

)v (x, t) ∈ L2 (QT ) . Further, note that Conditions (2.19) and (1.4) are satisfied

for all operators L(τ) with constants γ′ (τ) ≥ min(γ′, n) and σ(τ) ≤ σ, respectively. The first is obvious.

Prove the second fact. Denote by a(τ)ij (x, t) , i, j = 1, . . . , n, the coefficients at high-order derivatives

with respect to spatial variables of operator L(τ). Let

r = supQT

n∑i,j=1

a2ij (x, t)

g2 (x, t), r(τ)(x, t) =

n∑i,j=1

(a(τ)ij (x, t)

)2

(n∑

i=1a(τ)ii (x, t)

)2 , r(τ) = supQT

r(τ) (x, t) ,

124

where g (x, t) =n∑

i,j=1aii (x, t) . Taking into account (1.4) and the fact that the inequality r ≥ n−1 holds

for any operator of the form L, we obtain the inequality

r(τ) (x, t) =

n (1− τ)2 + 2τ (1− τ) g (x, t) + τ2n∑

i,j=1a2ij (x, t)

n2 (1− τ)2 + 2τ (1− τ)ng (x, t) + τ2g2 (x, t)

≤ 1

n+

τ2(r − 1

n

)g2 (x, t)

n2 (1− τ)2 + 2τ (1− τ)ng (x, t) + τ2g (x, t)≤ 1

n+τ2

(r − 1

n

)g2 (x, t)

τ2g2 (x, t)= r.

(5.5)

Further, let

λ− = infQT

g (x, t) , λ+ = supQT

g (x, t) , λ (τ) =

infQT

n∑i=1

a(τ)ii (x, t)

supQT

n∑i=1

a(τ)ii (x, t)

.

Obviously,

λ (τ) =(1− τ)n+ τ λ−

(1− τ)n+ τλ+.

On the other hand,

λ′ (τ) =λ− − λ+

[(1− τ)n+ τλ+]2≤ 0

Therefore,

λ (τ) ≥ λ (1) = λ. (5.6)

From (5.5) and (5.6), we conclude that

σ(τ) = r(τ) − 1

n− λ2 (τ)≤ r − 1

n− λ2= σ,

and the required statement is proved. Finally, note that if b(τ)i (x, t) and c(τ) (x, t) are the corresponding

coefficients of operator L(τ), then the conditions of form (1.6) are satisfied. Moreover, the followinginequalities hold:

Hb(τ)i ; n+2

(ν) = Hbi; n+2 (ν) ; i = 1, . . . , n; Hc(τ); s (ν) ≤ Hc; s (ν) .

From the aforesaid and (3.18), it follows that at T ≤ T 0, τ ∈ [0, 1] for any function u (x, t) ∈W 2,2

2,ϕ (QT ) , the following estimate holds:

‖u‖W 2,2

2,ϕ(QT )≤ C3.15

∥∥∥L(τ)u∥∥∥L2(QT )

. (5.7)

By supposition, the first boundary value Problem (5.4) has a strong solution u (x, t) for each v (x, t) ∈W 2,2

2,ϕ (QT ) . In other words, an operator P from W 2,22,ϕ (QT ) to W 2,2

2,ϕ (QT ) is defined such that u =

Pv. We will show that for ε chosen appropriately the operator P is a contraction. Let v(i) (x, t) ∈W 2,2

2,ϕ (QT ) , u(i) = Pv(i); i = 1, 2. Then taking into account that L(τ) −L(τ0) = (τ − τ0) (L −M0) , we

conclude that u(1) (x, t)− u(2) (x, t) is a strong solution of the first boundary value problem

L(τ0)(u(1) − u(2)

)= (τ − τ0) (L −M0)

(v(1) − v(2)

);(u(1) − u(2)

)∈ W 2,2

2,ϕ (QT ) .

125

Applying (5.7), we obtain the inequality∥∥∥u(1) − u(2)

∥∥∥W 2,2

2,ϕ(QT )≤ C3.15 |τ − τ0|

∥∥∥(L −M0)(v(1) − v(2)

)∥∥∥L2(QT )

.

On the other hand,∥∥∥(L −M0)

(v(1) − v(2)

)∥∥∥L2(QT )

≤ C5.1 (L, n,Ω, T )∥∥∥v(1) − v(2)

∥∥∥W 2,2

2,ϕ(QT ). (5.8)

Thus, ∥∥∥u(1) − u(2)∥∥∥W 2,2

2,ϕ(QT )≤ C3.15C5.1ε

∥∥∥v(1) − v(2)∥∥∥W 2,2

2,ϕ(QT ).

Choosing now ε = (2C3.15C5.1)−1, we prove that the operator P is a contraction. Hence, it has a fixed

point u = Pu, which is a strong solution of the boundary value Problem (5.3) and, consequently, (5.2).Thus, the openness of the set E is proved. Now, let us show that the set E is closed. Let τk ∈ E,k = 1, 2, . . . , lim

k→∞τk = τ. For positive integer k, denote by u[k] (x, t) the strong solution of the first

boundary value problem

L(τk)u[k] = f (x, t) , (x, t) ∈ QT ; u[k]|Γ(QT ) = 0.

According to (5.7), we have ∥∥u[k]∥∥W 2,2

2,ϕ(QT )≤ C3.15 ‖f‖L2(QT ) . (5.9)

Thus, the set of functions{u[k] (x, t)

}is weakly compact in W 2,2

2,ϕ (QT ) , i.e., there exists a subsequence

of positive integers {kl} , liml→∞

kl = ∞, and a function u (x, t) ∈ W 2,22,ϕ (QT ) such that for any ψ (x, t) ∈

C∞ (QT

)

liml→∞

(L(τkl)u[kl], ψ

)=(L(τ)u, ψ

). (5.10)

However,

(L(τ)u[kl], ψ

)=((

L(τ) − L(τkl))u[kl], ψ

)+ (f, ψ) = J1 (l) + (f, ψ) . (5.11)

Moreover, taking into account (5.8) and (5.9), we obtain that

|J1 (l)| ≤ |τ − τkl |∣∣((L −M0)u[kl], ψ

)∣∣

≤ |τ − τkl |C5.1

∥∥u[kl]∥∥W 2,2

2,ϕ(QT )‖ψ‖L2(QT )≤ C3.15C5.1 |τ − τkl1 | ‖f‖L2(QT ) ‖ψ‖L2(QT ) .

(5.12)

From (5.12), it follows that liml→∞

J1 (l) = 0. Now, from (5.10) and (5.11) we conclude that(L(τ)u, ψ

)=

(f, ψ) , i.e., almost everywhere in QT . Thus, we proved that α(r)u = f(x, t), which means that τ ∈ E,i.e., the set E is closed. The theorem is proved.

Remark. Obviously, if Φ (x, t) ∈W 2,22,ϕ (QT ) is given and the conditions of Theorem 5.1 are satisfied,

then for any T ≤ T 0, the first boundary value problem

Lu = f(x, t), (x, t) ∈ QT , (u(x, t)− Φ(x, t)) ∈ W 2,22,ϕ (QT )

is uniquely strongly solvable. For its solution, the following estimate holds:

‖u‖W 2,2

2,ϕ(QT )≤ C5.2 (L, n,Ω)

(‖f‖L2(QT ) + ‖Φ‖

W 2,22,ϕ(QT )

).

126

6. General Solvability of the First Boundary Value Problem

In order to prove the unique strong solvability of the first boundary value Problem (1.1)-(1.2) forany T ∈ (0,∞) , we have to impose stronger conditions on the functions ϕ (z) and f (x, t) and tolow-order coefficients of the operator L. Exactly, we assume that f (x, t) ∈ Ln+1 (QT ) , the functionϕ (z) , unlike (1.5), satisfies the additional condition

T∫

0

(ϕ (z)

z2

)n+1

dz <∞, (6.1)

and, instead of (1.6), conditions

bi (x, t) ∈ L2(n+1) (QT ) , i = 1, . . . , n, c (x, t) ∈ Ln+1 (QT ) , (6.2)

are satisfied such that there exists a constant H > 0 such that for (x, t) ∈ QT we have∣∣B (x, t)

∣∣2 +Hc (x, t) ≤ 0. (6.3)

Here B (x, t) = (b1 (x, t) , . . . , bn (x, t)) , and∣∣B (x, t)

∣∣ =

(n∑

i=1b2i (x, t)

) 12

. In the proof, we use the

following analogue of the Aleksandrov–Bakel’man inequality established in [15, 19]: if the coefficients

of the operator L satisfy Conditions (1.3), (1.5) and (6.1)–(6.3) and u (x, t) ∈ W 2,22,ϕ (QT ) is such that

Lu (x, t) ∈ Ln+1 (QT ) for T ∈ (0,∞) , then the following estimate holds:

‖u‖L∞(QT ) ≤ C6.1 (γ,H, n,Ω, T ) ‖Lu‖Ln+1(QT ) . (6.4)

Theorem 6.1. If the coefficients of the operator L satisfy Conditions (2.19), (1.4), (1.5), and (6.1)–(6.3) and T ∈ (0,∞) , then the first boundary value Problem (1.1)-(1.2) is uniquely strongly solvable

in the space W 2,22,ϕ (QT ) for any f (x, t) ∈ Ln+1 (QT ) and the following estimate holds for the solution

u (x, t)‖u‖

W 2,22,ϕ(QT )

≤ C6.2 (L, H, n,Ω, T ) ‖f‖Ln+1(QT ) . (6.5)

Proof. It is obvious that it suffices to prove the theorem in the cylinder QT with T ∗ = 3T 0/2. Further,we can solve the first boundary value Problem (1.1)-(1.2) for any T ∈ (0,∞) in a finite number ofsteps since on each step the height of the cylinder in which the solution is constructed increases onT 0/2.

Let Rn+1 = En×(−∞,∞) .We will assume that the coefficients aij(x, t), bi(x, t), i, j = 1, . . . , n, andc(x, t) of the operator L are continued to CQT ∗ = Rn+1\QT ∗ with preservation of Conditions (2.19),

(1.4), (6.2), and (6.3). For this, it suffices to assume that aij(x, t) =γ′n δij , bi(x, t) = 0; i, j = 1, . . . , n;

c(x, t) = 0 if (x, t) ∈ CQT ∗ . Moreover, we continue function f (x, t) into CQT ∗ to get zero. Denote byB1 the unit (n+ 1)-dimensional ball. Let w1(x, t) ∈ C∞

0

(B1

), w1(x, t) ≥ 0 in Rn+1 and

∫

Rn+1

w1(x, t)dxdt = 1, wh(x, t) = h−n−1w1

(x

h,t

h

).

If F (x, t) ∈ Lloc1 (Rn+1) , then the function

F h (x, t) =

∫

Rn+1

F (y, τ)wh(x− y, t− τ)dxdt

is called a function averaging F (x, t) with parameter h. Further, let

Lh =n∑

i,j=1

ahij(x, t)∂2

∂xi∂xj+

n∑

i=1

bhi (x, t)∂

∂xi+ ch(x, t) + ϕ (T − t)

∂2

∂t2− ∂

∂t.

127

First, for sufficiently small h, we will solve the problem

Lhuh = fh(x, t), (x, t) ∈ QT ∗ , uh|Γ(QT∗ ) = 0, (6.6)

applying the alternating Schwartz method (see [5]). Denote Ω× (T 0/2, 3T 0/2

)and Ω× (

0, T 0)by Q1

and Q2, respectively. Let u′h,1 (x, t) be a strong solution of the boundary value problem

Lhu′h,1 = fh(x, t), (x, t) ∈ Q1; u′h,1|Γ(Q1) = 0. (6.7)

Obviously, Conditions (2.19) and (1.4) are satisfied for the operator Lh with the constants γ′ andσ respectively. Moreover, the functions Hbhi , n+2 (ν) and Hch; s (ν) are majorized by the functions

2Hbi; n+2 (ν) and 2Hc; s (ν) ; i = 1, . . . , n, respectively, if h is sufficiently small. According to The-orem 5.1, a strong solution of Problem (6.7) exists and is unique. This solution is actually classic,since the coefficients of the operator Lh are continuously differentiable in Q1. Denote by u′′h,1 (x, t) aclassical solution of the Dirichlet problem

Lhu′′h,1 = fh (x, t) , (x, t) ∈ Q2; u′′h,1

∣∣Γ(Q2)

= 0, u′′h,1∣∣t=T 0 = u′h,1

∣∣t=T 0 . (6.8)

A solution of Problem (6.8) exists and is unique according to [14]. Now, let u′h,2(x, t) and u′′h,2(x, t)

be solutions of the first boundary value problems

Lhu′h,2 = fh(x, t), (x, t) ∈ Q1; u′h,2|S(Q1) = 0, u′h,2|t=T0

2

= u′′h,1|t=T0

2

(6.9)

andLhu

′′h,2 = fh(x, t), (x, t) ∈ Q2; u′′h,2|Γ(Q2) = 0, u′′h,2|t=T 0 = u′h,2|t=0,

respectively. Here S(Q1

)is the lateral surface of the cylinder Q1. A strong solution of Problem (6.9)

exists and is unique according to the remark to Theorem 5.1 since u′′h,1|t=T0

2

is a trace of some function

from W 2,22,ϕ

(Q1

)on the lower base Q1. More exactly, let Q3 = Q × (

T 0, 3T 0/2). Then, by virtue of

the smoothness of the boundary ∂Ω and the coefficients of the operator Lh, the function u′′h,1 (x, t) istwice continuously differentiable in Q2 (see [22]) and is extendable into Q3 such that the continued

function belongs to W 2,22,ϕ (QT ∗) . The same smoothness considerations for ∂Ω and the coefficients of

Lh allow us to conclude that the above-stated strong solution of Problem (6.9) is its classical solution.Continue the process analogously. On the m-th step we will obtain classical solutions u′h,m (x, t)

and u′′h,m (x, t) of the first boundary value problems

Lhu′h,m = fh(x, t), (x, t) ∈ Q1; u′h,m|S(Q1) = 0, u′h,m|

t=T0

2

= u′′h,m−1|t=T0

2

andLhu

′′h,m = fh(x, t), (x, t) ∈ Q2; u′′h,m|Γ(Q2) = 0, u′′h,m|t=T 0 = u′h,m|t=T 0 ,

respectively. Then, reasoning in the same way as in [5], we obtain that there exist limits

u′h (x, t) = limm→∞u′h,m(x, t) for (x, t) ∈ Q1, u′′h(x, t) = lim

m→∞u′′h,m(x, t) for (x, t) ∈ Q2.

Moreover, u′h(x, t) = u′′h(x, t) for (x, t) ∈ Q1 ∩Q2 and the function

uh (x, t) =

{u′h (x, t) , (x, t) ∈ Q1

u′′h (x, t) , (x, t) ∈ QT ∗\Q1

is a classical solution of the boundary value Problem (6.6).Let now α ∈ (0, 3/2) be a number that will be chosen later and

Q4 = Ω× ((3/2− α)T 0, 3T 0/2

), Q5 = Ω× (

(3− α)T 0/2, 3T 0/2).

We introduce a function ξ (t) = C∞ [(3/2− α)T 0, 3T 0/2

]such that ξ (t) = 1 for t ∈ [

(3− α)T 0/2, 3T 0/2],

ξ (t) = 0 for t ∈ [(3/2− α)T 0, (3/2− 3α/4)T 0

], 0 ≤ ξ (t) ≤ 1, and

128

∣∣ξ′ (t)∣∣ ≤ (αT 0)−1C6.3,

∣∣ξ′′ (t)∣∣ ≤ (αT 0)−2C6.3. (6.10)

Applying Theorem 3.1 to the function uh (x, t) ξ (t) in the cylinder Q4 and using estimates (6.10),we obtain the inequality

‖uh‖2W 2,22,ϕ(Q

5)≤ 2C2

3.15 ‖Lhuh‖L2(Q4) + C6.4

(ϕ, σ, n,Ω, α, T 0

) ‖uh‖2L2(Q4)

+(αT 0)−2C6.5 (ϕ, σ, n,Ω) ‖ϕ (uh)t‖2L2(Q4) .(6.11)

On the other hand, according to Conditions (1.5),

if q1 (T ) = sup[0,T ]

ϕ (z) , then limT→0

q1(T )T−1 = 0.

We will choose and fix a small α < 3/2 such that q1(αT0)(αT 0)−1C

1/26.5 ≤ 1/2. Obviously, α depends

only on ϕ, σ, n, Ω, and T 0. Then, from (6.11), we conclude that

‖uh‖2W 2,22,ϕ(Q

5)≤ 2C2

3.15

∥∥∥fh∥∥∥2

L2(Q4)+ C6.4

(ϕ, σ, n,Ω, T 0

) ‖uh‖2L2(Q4)

+1

4

∥∥∥(uh)2t∥∥∥2

L2(Q4)≤ 2C2

3.15

∥∥∥fh∥∥∥2

L2(Q4)+ C6.4 ‖uh‖2L2(Q4) +

1

4‖u‖2

W 2,22,ϕ(Q

4).

(6.12)

Further, let

Q6 = Ω× (0, (3/2− α/4)T 0

), Q7 = Ω× (

0, (3− α)T 0/2).

We introduce a function ξ (t) = C∞ [0, (3/2− α/4)T 0

]such that ξ (t) = 1 for t ∈ [

0, (3− α)T 0/2],

ξ (t) = 0 for t ∈ [(3/2− 3α/8)T 0, (3/2− α/4)T 0

], 0 ≤ ξ (t) ≤ 1, and

∣∣ξ′ (t)∣∣ ≤ (αT 0)−1C6.6,

∣∣ξ′′ (t)∣∣ ≤ (αT 0)−2C6.6. (6.13)

Obviously, the operator Lh is uniformly elliptic in the cylinder Q6 and its ellipticity constant dependsonly on σ, α, T 0, and ϕ. According to the coercive estimate of the paper [16], for any functionu (x, t) ∈ C2

(Q6

)vanishing on ∂Q6, the following inequality holds:

‖u‖W 2,2

2 (Q6)≤ C6.7

(ϕ, σ, n,Ω, T 0

) (‖Lhu‖2L2(Q6) + ‖u‖2L2(Q6)

). (6.14)

Applying inequality (6.14) to the function u (x, t) ξ (t) and taking into account the inequality‖u‖

W 2,22,ϕ(Q

6)≤ C6.8

(ϕ, T 0

) ‖u‖W 2,2

2 (Q6)and estimates (6.13), we obtain the inequlaity

‖u‖2W 2,2

2,ϕ(Q6)≤ C6.8

(ϕ, σ, n,Ω, T 0

) (‖Lhuh‖2L2(Q6) + ‖uh‖2L2(Q6) + ‖(uh)t‖2L2(Q6)

). (6.15)

According to the interpolation inequality [13], for any ε > 0, we have the inequality

‖(uh)t‖2L2(Q6) ≤ ε ‖(uh)tt‖2L2(Q6) + ε−1C6.10 ‖uh‖2L2(Q6) . (6.16)

On the other hand, since

inf[0,(3/2−α/4)T 0]

ϕ (T ∗ − t) = ϕ(αT 0/4),

it follows that

‖(uh)tt‖2L2(Q6) ≤ εC6.11

(ϕ, σ, n,Ω, T 0

) ‖ϕ (uh)tt‖2L2(Q6) . (6.17)

We choose and fix ε = (4C6.9C6.11)−1. Then, from (6.15)–(6.17), we conclude that

‖uh‖2W 2,22,ϕ(Q

7)≤ C6.12

(ϕ, σ, n,Ω, T 0

)(∥∥∥fh∥∥∥2

L2(Q6)+ ‖uh‖2L2(Q6)

)+

1

4‖uh‖2W 2,2

2,ϕ(Q6). (6.18)

Now, summing inequalities (6.12) and (6.18), we come to the estimate

129

‖uh‖W 2,22,ϕ(QT∗ ) ≤ C6.13

(ϕ, σ, n,Ω, T 0

)(∥∥∥fh∥∥∥L2(QT∗ )

+ ‖uh‖L2(QT∗ )

). (6.19)

Using (6.4) in (6.19), we have

‖uh‖W 2,22,ϕ(QT∗ ) ≤ C6.14

(ϕ, σ,H, n,Ω, T 0

) ∥∥∥fh∥∥∥Ln+1(QT∗ )

. (6.20)

By virtue of the fact that∥∥fh

∥∥Ln+1(QT∗ ) ≤ 2 ‖f‖Ln+1(QT∗ ) for sufficiently small h, inequality (6.20)

implies that

‖uh‖W 2,22,ϕ(QT∗ ) ≤ 2C6.14 ‖f‖Ln+1(QT∗ ) . (6.21)

Thus, the set of functions {uh (x, t)} is weakly compact in W 2,22,ϕ (QT ∗) . Hence, there exist a sequence

{hk} , limk→∞

hk = 0, and a function u (x, t) ∈ W 2,22,ϕ (QT ∗) such that for any ψ (x, t) ∈ C∞ (

QT ∗)

limk→∞

(Luhk, ψ) = (Lu, ψ) . (6.22)

On the other hand,

|J2 (k)| = |((L − Lhk)uhk

, ψ)| ≤ C6.15

×( n∑

i,j=1

∥∥∥aij − ahkij

∥∥∥L2(QT∗ )

+n∑

i=1

∥∥∥bi − bhki

∥∥∥L2(QT∗ )

+∥∥∥c− chk

∥∥∥L2(QT∗ )

)‖uhk

‖W 2,2

2,ϕ(QT∗ ) ,

i.e., (6.21) is satisfied; then

limk→∞

J2 (k) = 0. (6.23)

From (6.22) and (6.23), we obtain (see the proof of Theorem 4.1) that Lu = f (x, t) almost every-where in QT ∗ . Thus, the existence of a strong solution of the first boundary value Problem (1.1)-(1.2) isproved. Its uniqueness follows from inequality (6.4). It remains to prove the validity of estimate (6.5).Form (6.21), we conclude that one can extract a convergent subsequence from the number sequence‖uhk

‖W 2,2

2,ϕ(QT ). Without loss of generality, we can assume that this sequence coincides with {hk} and

limk→∞

‖uhk‖W 2,2

2,ϕ(QT∗ ) ≤ 2C6.14 ‖f‖Ln+1(QT∗ ) . (6.24)

Further, from the weak convergence in W 2,22,ϕ (QT ∗) of the sequence {uhk

} as k → ∞ to the function

u (x, t) , it follows, in particular, that for any ψ (x, t) ∈ L2 (QT ∗) and any i, 1 ≤ i ≤ n,

limk→∞

((uhk

)i , ψ)=(ui, ψ

). (6.25)

We fix arbitrary i, 1 ≤ i ≤ n, and suppose in (6.25) that ψ (x, t) = ui (x, t) . We obtain the inequality

‖ui‖2L2(QT∗ ) = limk→∞

((uhk

)i , ui) ≤ ‖ui‖L2(QT∗ ) lim

k→∞∥∥(uhk

)i∥∥L2(QT∗ )

≤ ‖ui‖L2(QT∗ ) limk→∞

‖uhk‖W 2,ϕ

2,2 (QT∗ ) ,

i.e., if (6.24) is satisfied, then

‖ui‖L2(QT∗ ) ≤ 2C6.14 ‖f‖Ln+1(QT∗ ) .

One can similarly prove that the norms

‖uij‖L2(QT∗ ) , ‖√ϕuit‖L2(QT∗ ) , i, j = 1, . . . , n, ‖ut‖L2(QT∗ ) , ‖ϕutt‖L2(QT∗ )

130

are majorized by 2C6.14 ‖f‖Ln+1(QT∗ ) . Now, it suffices to recall (6.4), and the required estimate (6.5)

is proved.

Remark. From the proof, it is not obvious at first sight why the constant C6.2 depends on all theparameters included in the notation C (L) . This constant depends on the number of steps necessaryto construct a strong solution of the first boundary value Problem (1.1)-(1.2) in the cylinder QT witharbitrary T ∈ (0,∞) by means of the process suggested in the proof. Obviously, the number of stepsis in proportion to the relation T/T 0. Now it suffices to note that T 0 = T 0 (L, n,Ω) .

7. On the Accuracy of the Cordes Condition

Now we give an example showing that if Condition (1.4) is violated, then the unique strong solv-

ability of the first boundary value Problem (1.1)-(1.2) in the space W 2,22,ϕ (QT ) , generally speaking,

does not occur. Note that the idea of the example is taken form the paper [2] (see also [18]).Let QT 0 = B1/2 (0)×

(0, T 0

), n ≥ 3, ϕ (z) be an arbitrary function satisfying Conditions (1.5), and

L = Δ+n

n− 2

n∑

i,j=1

xixj

|x|2∂2

∂xi∂xj+ ϕ

(T 0 − t

) ∂2∂t2

− ∂

∂t.

Obviously, the constant σ (see (1.4)) equals zero for the the operator L. Consider the first boundaryvalue problem

Lu = f (x, t) , (x, t) ∈ QT 0 , u|Γ(QT0) = 0, (7.1)

where f(x, t) ∈ L2 (QT 0) will be chosen later. For ε > 0, construct a set of operators{L(ε)

}with

smoothed coefficients tending to the corresponding coefficients of the operator L almost everywherein QT 0 as ε → 0 and also the set of Lipschitz continuous functions {fε (x, t)} , lim

ε→0fε (x, t) = f (x, t)

almost everywhere in QT 0 such that if u(ε) (x, t) is a classical solution of the boundary value problem

L(ε)u(ε) = fε (x, t) , (x, t) ∈ QT 0 , u(ε)∣∣Γ(QT0)

= 0, (7.2)

then∥∥u(ε)

∥∥W 2,2

2,ϕ(QT0)→ ∞ for ε→ 0.

To do this, we consider the auxiliary Dirichlet problem

Nεzε = Δzε + gε (r)n∑

i,j=1

xixjr2

(zε)ij = Fε (r) , x ∈ B1/2 (0) ; zε|∂B1/2(0) = 0. (7.3)

Here

r = |x| , gε (r) =n

n− 2for r > ε, gε (r) =

n

n− 2

r

εfor r ≤ ε,

Fε (r) = r−n2

(log

1

r

)− 32

for r > ε, Fε (r) = ε−n2

(log

1

r

)− 32

for r ≤ ε.

We seek a solution of Problem (7.3) in the form zε (x) = vε (r) . Then (7.3) implies that

(1 + gε (r)) v′′ε +

n− 1

rv′ε = Fε (r) . (7.4)

We reduce the solution of Eq. (7.4) satisfying the boundary condition vε (1/2) = 0 to the form

vε (r) =

1/2∫

r

exp

((n− 1)

1/2∫

τ

dν

ν (1 + gε (ν))

)×

τ∫

0

Fε (λ) exp

(− (n− 1)

1/2∫

λ

dτ

τ (1 + gε (τ))

)dλ dτ. (7.5)

131

On the other hand,n∑

i,j=1

(zε)2ij = (v′′ε )

2 +n− 1

r2(v′ε)

2 ≥ (v′ε)2

r2.

Therefore,

I (ε) =

∫

B1/2(0)

n∑

i,j=1

(zε)2ij dx ≥ nωn

1/2∫

0

rn−3(v′ε)2dr

= nωn

1/2∫

0

rn−3 exp

(2 (n− 1)

1/2∫

r

dτ

τ (1 + gε (τ))

)

×( r∫

0

Fε (λ) exp

(− (n− 1)

1/2∫

λ

dτ

τ (1 + gε (τ))

)dλ

)2

dr

≥ nωn

1/2∫

ε

rn−3 exp

((n− 2)

1/2∫

r

dτ

τ

)

×( r∫

ε

λ−n/2(log λ−1

)−3/2exp

(−n− 2

2

1/2∫

λ

dτ

τ

)dλ

)2

dr

= C7.1(n)

1/2∫

ε

r−1

( r∫

ε

dλ

λ (log λ−1)3/2

)2

dr

= 4C7.1

( 1/2∫

ε

dr

r log 1r

+1

log 1ε

1/2∫

ε

dr

r− 2(log 1

ε

)1/2

1/2∫

ε

dr

r(log 1

r

)1/2

)

= 4C7.1 (p1 (ε) + p2 (ε) + p3 (ε)) . (7.6)

Obviously,

limε→0

p1 (ε) = ∞, limε→0

p2 (ε) = 1, limε→0

p3 (ε) = −4. (7.7)

Taking into account (7.7) in (7.6), we conclude that

limε→0

I (ε) = ∞. (7.8)

From (7.5), it follows that limε→0

vε (r) = v (r) almost everywhere in B1/2 (0) , where

v (r) =

1/2∫

r

exp

(n− 2

2log

1

2τ

) τ∫

0

λ−n/2

(log

1

λ

)−3/2

exp

(−n− 2

2log

1

2λ

)dλdτ

= C7.2 (n)

1∫

r

τ−(n−2)/2

τ∫

0

dλ

λ(log 1

λ

)3/2dτ = C7.3 (n)

1∫

r

τ−(n−2)/2

(log

1

τ

)−1/2

dτ

≤ C7.4 (n)

1∫

r

τ−(n−2)/2dτ.

132

Thus, if we denote z (x) = v (r) , then

∫

B1/2(0)

z2dx = nωn

1/2∫

0

rn−1v2dr <∞. (7.9)

Suppose now that

L(ε) = Nε + ϕ(T 0 − t

) ∂2∂t2

− ∂

∂t,

u(ε) (x, t) = tvε (|x|) , fε (x, t) = tFε (|x|) − vε (|x|) in (7.2), and f (x, t) = tF (|x|) − z (x) in (7.1),

where F (|x|) = |x|−n/2(log 1

|x|)−3/2

. From (7.9), it follows that f (x, t) ∈ L2 (QT 0) . At the same

time, if (7.8) is satisfied, then limε→0

∥∥u(ε)∥∥W 2,2

2,ϕ(QT0)= ∞. The required example is constructed.

REFERENCES

1. Yu.A. Alkhutov and I. T. Mamedov, “Some properties of the solutions of the first boundary valueproblem for parabolic equations with discontinuous coefficients,” Sov. Math., Dokl., 32, No. 2,343–347 (1985).

2. Yu.A. Alkhutov and I. T. Mamedov, “The first boundary value problem for second-order nondi-vergence parabolic equations with discontinuous coefficients,”Math. USSR Sb., 59, No. 2, 471–495(1988).

3. A. Canfora, “Esistenza ed unicita delle soluzioni di un problema al contorno relativo ad un’equazione elliptico-parabolica di ordine 2m,” Recerche Mat., 25, 247–304 (1976).

4. H.O. Cordes, “Uber die erste randwertaufgabe bei quasilinearen differentialgleichungen zweiterornung in mehr als zwei variablen,” Math. Ann., 131, 278–312 (1956).

5. R. Courant and D. Hilbert, Methods of Mathematical Physics, Vol. II, Interscience, New York(1962).

6. G. Fichera, “On a unified theory of boundary value problems of elliptic-parabolic equationsof second-order,” Boundary Probl. in Differ. Equations (Proc. Sympos., Madison, April 20–22,1959), 97–120 (1960).

7. G. Fiorito, “Un contributo alla risoluzione del problema di Cauchy–Dirichlet,” Matematiche, 35,53–70 (1980).

8. M. Franciosi, “Sul di un’equazione ellittico-parabolica a coefficienti discontinui,” Boll. Un. Mat.Ital., VI. Ser., C, Anal. Funz. Appl., 2, 63–75 (1983).

9. M. Franciosi, “Un teorema di esistenza ed unicita per la soluzione di un’equazione ellittico-parabolica, a coefficienti discontinui, in forma non divergenza,” Boll. Un. Mat. Ital., VI. Ser., B,4, 253–263 (1985).

10. G.H. Hardy, J. E. Littlewood, and G. Polya, Inequalities, Cambridge Univ. Press, Cambridge(1952).

11. M.V. Keldysh, “On some cases of degeneration of elliptic type equations on the boundary of thedomain,” Dokl. Akad. Nauk, 77, No. 2, 181–183 (1951).

12. J. J. Kohn and L. Nirenberg, “Degenerate elliptic-parabolic equations of second-order,” Commun.Pure Appl. Math., 20, 797–872 (1967).

13. O.A. Ladyzhenskaya, V.A. Solonnikov, and N.N. Ural’tseva, Linear and Quasilinear Equationsof Parabolic Type, AMS, Providence (1968).

14. O.A. Ladyzhenskaya and N.N. Ural’tseva, Linear and Quasilinear Elliptic Equations, AcademicPress, New York–London (1968).

15. I. T. Mamedov, “An inequality of A.D. Aleksandrov type for degenerate elliptic-parabolic oper-ators of second-order,” Proc. Inst. Math. Mech. Natl. Acad. Sci. Azerb., 13, 92–96 (2000).

133

16. I. T. Mamedov and R.A. Agayeva, “The first boundary value problem for nondivergent linearsecond-order elliptic equations of Cordes type,” Trans. Natl. Acad. Sci. Azerb., 1, 150–167 (2002).

17. I. T. Mamedov and G.T. Guseynov, “On solvability of the first boundary value problem fordegenerating elliptic-parabolic equations of second-order,” Dokl. Nats. Akad. Nauk Azerb., 1,No. 3, 30–39 (2000).

18. I. T. Mamedov and T.R. Muradov, “On the Dirichlet problem for Gilbarg–Serrin equation,” Proc.Inst. Math. Mech. Natl. Acad. Sci. Azerb., 16, 81–93 (2002).

19. I. T. Mamedov and Sh.Yu. Salmanova, “The A.D. Aleksandrov type inequality for a class ofsecond-order equations with non-negative characteristic form,” Trans. Natl. Acad. Sci. Azerb., 4,108–114 (2001).

20. O.A. Oleinik and E.V. Radkevich, Second-Order Equations with Non-Negative CharacteristicForm, VINITI, Moscow (1971).

21. Sh.Yu. Salmanova, “On solvability of the first boundary value problem fro the second-orderdegenerate elliptic-parabolic equations,” Proc. Inst. Math. Mech. Natl. Acad. Sci. Azerb., 15,132–145 (2001).

22. G. Talenti, “Sopra una classe di equazioni ellittiche a coefficienti misurableili,” Ann. Mat. PuraAppl. (4), 69, 285–304 (1965).

23. G.C. Wen, “Initial-mixed boundary value problems for parabolic equations of second-order withmeasurable coefficients in a higher dimensional domain,” Proceedings of the Second ISAAK Con-gress, Vol. I, Kluwer Acadmic Publishers, Dordrecht–Boston–London (2000).

24. G.C. Wen and T. Maoying, “Initial-oblique derivative problems for nonlinear parabolic equationswith measurable coefficients,” Commun. Nonlinear Sci. Numer. Simul., 2, 109–113 (1998).

I. T. MamedovInstitute of Mathematics and Mechanicsof Azerbaijan National Academy of Sciences, Baku, AzerbaijanE-mail: ilham@lan.ab.az

134