Journal of Mathematical Sciences, Vol. 108, No. 4, 2002

LINEAR SECOND-ORDER PARTIAL DIFFERENTIAL EQUATIONSOF THE PARABOLIC TYPE

A. M. Ilyin, A. S. Kalashnikov, and O. A. Oleynik UDC 517.9

The article is a review of the theory of linear partial differential equations of the parabolic type, wherethe classical facts of this theory are described in detail.

Bibliography: 128 titles.

CONTENTS

Introduction 436

1 Maximum Principle. Uniqueness of Solutions of the Main Boundary-Value Problems 437

2 A priori Estimates 448

3 Solving of Boundary-Value Problems by the Rothe Method. Cauchy Problem 463

4 The Fundamental Solution of a Linear Parabolic Equation. Green’s Function.Method of Integral Equations for Solving Boundary-Value Problems 480

5 Generalized Solutions of Boundary-Value Problems. Uniqueness Theorems.Some Auxiliary Propositions 501

6 Method of Finite Differences 507

7 Some Functional Methods of Solving Boundary-Value Problems 513

8 Solution of the First Boundary-Value Problemby the Method of Continuation with Respect to a Parameter 517

9 Application of Galerkin’s Methodto the Construction of a Solution of the First Boundary-Value Method 518

10 Generalized Solution for the Cauchy Problem 520

11 On the Smoothness of Generalized Solutions 527

12 Behavior of Solutions with Infinite Increasing of Time 533

References 537

Translated from Trudy Seminara imeni I. G. Petrovskogo, No. 21, pp. 9–193, 2001. The Russian version of thisarticle is a reprint from the journal “Uspekhi Matematicheskikh Nauk” (1962, Vol. 17, issue 3 (105), pp. 3–146) withthe consent of the editorial board.

1072–3374/02/1084–0435 $ 27.00 c© 2002 Plenum Publishing Corporation 435

INTRODUCTION

In the general theory of partial differential equations, that of second-order equations occupies a special place.At present, the theory of second-order equations is the best developed. Many methods invented to study second-order equations turned out to be useful also for the study of higher-order equations and systems of partial differentialequations. The strong interest in the study of second-order equations, and parabolic equations in particular, is alsomotivated by their importance in mechanics, heat conduction, probability theory, and other fields of mathematicsand physics. These applications constantly produce new problems for partial differential equations. In particular,one may read in E. B. Dynkin’s article [1] about new problems for second-order partial differential equations arisingin connection with probability theory.

While there are monographs and large reviews on the theory of elliptic and hyperbolic equations (amongthem, such a detailed review as the work of C. Miranda [2]), up to now there has been no survey of the mostimportant facts concerning parabolic second-order equations.

The present article was written mainly owing to the influence of numerous questions that people have askedthe authors and the fact that we had difficulty in indicating the corresponding literature.

This article is a review of the theory of second-order parabolic partial differential equations with detailedproofs of many basic facts of this theory. We pay special attention to the methods of construction of classical andgeneralized solutions for boundary-value problems and the Cauchy problem. The majority of these methods havea larger range of applicability and can be used in the study of higher-order equations and systems of equationsof different types. In each section, we indicate papers where the method we discuss finds application in differentquestions connected with the theory of partial differential equations.

We do not consider methods applicable only to equations of some special form. Thus, we do not considerthe application of the Fourier method or the Laplace transform to solving boundary-value problems; neither do weconsider the application of the Fourier transform to solving the Cauchy problem. These questions are considered in[3–5]. Neither do we describe methods of solving boundary-value problems connected with the theory of semigroups:to do this it is necessary to use some special knowledge in the fields of semigroup theory and functional analysis.Recently, semigroup theory has been actively developed, and there are a large number of papers in this field,e.g., [6–8].

We have tried to give our article an elementary character and to use only well-known facts of functionalanalysis and function theory. As we did not want to enlarge the article, we have omitted many special ques-tions concerning the parabolic second-order equations, as, for example, degenerated parabolic equations, equationswith a small parameter in highest terms, the qualitative behavior of solutions in unbounded domains, equationswith piecewise-continuous coefficients and the corresponding conjunction conditions on the discontinuity surfaces,the uniqueness of the Cauchy problem with initial conditions given on different manifolds, and many other ques-tions. Some of these questions are treated in [9–17]. We also do not consider nonlinear differential equations andtheir numerous applications. These questions are considered in many papers (e.g., in review [18]).

Let us briefly indicate the contents of every section of this article. In Sect. 1, we give various forms ofthe maximum principle for homogeneous and nonhomogeneous equations in bounded and unbounded domains; wepose the main boundary-value problems and the Cauchy problem and prove the uniqueness theorems for theseproblems. In Sect. 2, we formulate the main a priori estimates for solutions of boundary-value problems in terms ofthe data corresponding to each problem (initial or boundary conditions, coefficients of the equation, the boundaryof the domain); then we give the proof of S. N. Bernstein’s estimates for closed or nonclosed domains and the proofof the main a priori estimates for the solutions, the latter also in integral norms. In Sect. 3, the solutions of the firstand second boundary-value problems are constructed by means of the Rothe method; then we investigate howthe smoothness of the solution of the first boundary-value problem depends on the smoothness of the problem data.After that, the first boundary-value problem is investigated in domains of various types with weak restrictions onthe boundary (conditions for the existence of barrier functions). In Sect. 4, we construct the fundamental solution forthe second-order parabolic equation of general form under the condition that the coefficients are Holder continuous.By means of the fundamental solution we construct the solution of the Cauchy problem and the Green function forthe solution of the first boundary-value problem. The application of integral equations to solving boundary-valueproblems is given in outline. In Sect. 5, we give the definition of the generalized solution of the first boundary-value problem; then we prove the corresponding uniqueness theorems and give the proofs of numerous auxiliarypropositions that are used in the following sections. In Sects. 6–9, the generalized solution of the first boundary-value problem is constructed by different methods (in Sect. 6 by the method of finite differences, in Sect. 7 by

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some functional methods, in Sect. 8 by the method of continuation with respect to a parameter, and in Sect. 9 byGalerkin method). In Sect. 10, we consider the problem of existence and uniqueness for the generalized solutionsof the Cauchy problem in classes of rapidly growing functions. In Sect. 11, we give a method for the investigationof the smoothness properties of the solution in terms of the smoothness of the data. In Sect. 12, we prove somesimple theorems concerning the behavior of the solutions of parabolic equations as t→∞.

This article is an extended version of the special course of lectures that O. A. Oleynik gave at the Departmentof Mechanics and Mathematics of the Moscow State University in 1960.

Note also that all of the main notations and definitions are given in Sects. 1 and 5.We discussed some sections with V. P. Mikhaylov; he also wrote Sect. 7. We are grateful for his help and

useful suggestions. We also thank E. B. Dynkin for his advice and useful comments.

1 MAXIMUM PRINCIPLE.UNIQUENESS OF SOLUTIONS OF THE MAIN BOUNDARY-VALUE PROBLEMS

1. Some Notation. By En we denote the Euclidean space of dimension n, the point coordinates being(x1, . . . , xn); (x, t) will denote an arbitrary point of an (n+ 1)-dimensional space Rn+1 = En × (−∞,+∞); D willdenote a bounded domain in Rn+1 lying between the planes t = 0 and t = T , and D is the closure of D. We supposethat the intersection D0 of D and the plane t = 0 is not empty. By S we denote the closure of the set of boundarypoints of D having time coordinates t = 0 and t = T . The points of S not lying on the plane t = 0 we shall denoteby S. Γ denotes the union of S and D0.

In the special case where D is a cylinder Ω × (0, T ) with elements parallel to the t-axis and a domain Ωof the plane t = 0 as a base, we shall use the notation Q instead of D. If σ denotes the boundary of Ω, we haveD0 = Ω ∪ σ and Γ = S ∪Ω ∪ σ.

The domain consisting of points lying in Q and such that their distance from S is greater than δ will becalled Qδ; Ωδ will denote the domain of points in Ω such that their distance from σ is greater than δ. Let ν bethe direction of the outer normal to the surface S at the points where this normal exists. We also shall considerthe layer H of points in the space Rn+1 such that 0 < t T .

In this and all the following sections, we shall study the parabolic equation

L(u) ≡n∑

i,j=1

aij(x, t)∂2u

∂xi ∂xj+

n∑i=1

bi(x, t)∂u

∂xi+ c(x, t)u − ∂u

∂t= f(x, t), (1.1)

where the functions aij , bi, c, and f are real, with finite values, aij = aji andn∑

i,j=1

aij(x, t)αiαj > 0 forn∑i=1

α2i > 0.

We shall say that the function u(x, t) satisfies Eq. (1.1) at a point (x, t) if u(x, t) is continuous at this point togetherwith its derivatives ∂u

∂xi, ∂u∂t ,

∂2u∂xi ∂xj

(i, j = 1, . . . , n) and (1.1) is satisfied.Now let us define some classes of functions. We shall say that a function v(x, t) is of class Cm(E) (m = 0, 1, . . .)

on a set E if at the points of E this function has continuous derivatives up to the orderm. A function v(x, t) satisfieson E the Holder condition with exponent λ > 0 if the ratio |v(P1)−v(P2)|

[r(P1,P2)]λis bounded from above for any P1 and P2

belonging to E ; here r(P1, P2) is the distance between the points P1 and P2. The upper bound of this ratio is calledthe Holder coefficient of the function v on E . A function v(x, t) belongs to the class Cm+λ(E) if its derivativesof order m are Holder continuous on E with exponent λ, 0 < λ < 1. A function v(x, t) belongs to the class

C2m+λ,m+λ(E) if on E it has continuous derivatives ∂lu

∂xl11 ...∂xlnn ∂tl0

withn∑i=0

li = l andn∑i=1

li + 2l0 2m, and thesederivatives are Holder continuous with exponent λ.

A surface S in the space Rn+1 belongs to the class Am+λ, 0 < λ < 1, (Am) if for every point P of S there isa sphere ω with center P such that inside ω the surface S can be represented by the equation

xi = χ(x1, . . . , xi−1, xi+1, . . . , xn, t)

for certain i n and the function χ is of class Cm+λ (respectively Cm). Likewise, an (n− 1)-dimensional surface σin En belongs to the class Am+λ (Am) if any of its points has a neighborhood where the surface is represented bythe equation

xi = η(x1, . . . , xi−1, xi+1, . . . , xn) (1.2)

for a certain i n and the function η is of class Cm+λ (respectively of class Cm).

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2. Posing of Problems. Second-order linear parabolic equations retain many properties of the simplestequation of this type — the heat equation

n∑i=1

∂2u

∂x2i

− ∂u

∂t= 0. (1.3)

One of the most important properties of Eq. (1.3) is the maximum principle. In this section, we shall investigatethe forms the maximum principle takes for different classes of second-order parabolic equations. Just as for the heatequation, for general second-order parabolic equations the maximum principle implies the uniqueness of solutionsfor the main boundary-value problems and the Cauchy problem. These problems for general parabolic equationsare posed in the same way as for the heat equation (see [19]).

Now we shall pose the main problems for the parabolic equation (1.1).Consider the domain D described in subsection 1. The function u(x, t) is a solution of the first boundary-

value problem for Eq. (1.1) in the domain D with conditions

u|t=0 = ϕ(x), (1.4)u|S = ψ(x, t) (1.5)

if it is continuous in D and satisfies (1.1) in D \ Γ and conditions (1.4) and (1.5) on Γ; here ϕ(x) and ψ(x, t) aregiven continuous functions.

Sometimes we shall consider also solutions of this problem having discontinuities at the points of σ (seesubsection 6).

The function u(x, t) is called a solution of the oblique derivative problem for (1.1) in the domain D withconditions

u|t=0 = ϕ(x),(∂u

∂γ+ a(x, t)u

)S

= ψ(x, t) (1.6)

if it is continuous in D and satisfies Eq. (1.1) in D \ Γ and conditions (1.6) on Γ; here ϕ(x), a(x, t), and ψ(x, t) aregiven continuous functions. It is supposed that S at every point has a tangent plane and γ is some direction notparallel to the t axis and such that the angle between γ and the inner normal to S is acute.

If the domainD is a cylinderQ, γ coincides with the direction conormal to S(∂u∂γ ≡

n∑i,j=1

aij(x, t) cos(ν, xj) ∂u∂xi

),

and a(x, t) ≡ 0, the above-mentioned problem is called the second boundary-value problem for Eq. (1.1).The Cauchy problem for Eq. (1.1) in the layer H = Rn × (0, T ] with the condition

u|t=0 = ϕ(x) (1.7)

has the function u(x, t) for solution if this function is continuous in H and satisfies Eq. (1.1) inH and condition (1.7),where ϕ(x) is a given continuous function.

Here we have given the definition of classical solutions of the main problems for Eq. (1.1). In Sects. 5 and 10we shall define the generalized solutions of the same problems.

3. Maximum Principle in a Bounded Domain. Now we shall study some properties of the solutionsu(x, t) of (1.1) in the domain D. These properties are generalizations of the well-known maximum principle forthe heat equation (see [19]).

Let us note that the transformation u = veαt, where α > 0 is a constant, changes (1.1) to an equation alsohaving the form (1.1), and the coefficient in v of the new equation is c(x, t)−α. If the coefficient c(x, t) is boundedfrom above (c(x, t) M), then for sufficiently large α the coefficient of the transformed equation is strictly negative.In what follows, we shall repeatedly use this fact.

Theorem 1. Suppose the function u(x, t) is continuous in D and has continuous derivatives occurring inthe operator L; suppose that L(u) 0 in D \ Γ and that c(x, t) < M , M = const. Then if u(x, t) 0 on Γ, wehave u(x, t) 0 in D.

Proof. Consider first the case M < 0, that is, c(x, t) < 0. If there is a point (x, t) in D such that u(x, t) isnegative, then the minimum of u(x, t), attained at the point (x0, t0), also is negative. According to the suppositionsof the theorem, the point (x0, t0) can be either inside D or inside the upper base t = T .

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Therefore, at this point we have ∂u∂xi

= 0 (i = 1, . . . , n), ∂u∂t 0, cu > 0. Simultaneously at the point

(x0, t0) the inequalityn∑

i,j=1

aij∂2u

∂xi ∂xj 0 is fulfilled. Indeed, after the transformation of the independent variables

ξi =n∑

j=1

kijxj (i = 1, . . . , n) we obtain

n∑i,j=1

aij∂2u

∂xi ∂xj=

n∑i,j=1

bij∂2u

∂ξi ∂ξj,

where the matrices A = ‖aij‖, B = ‖bij‖, and K = ‖kij‖ are connected as follows: B = KAK∗. It is well-knownthat we can choose a nondegenerate matrix K so that B will be diagonal at the point (x0, t0). Because we have∂2u∂ξ2i 0 (i = 1, . . . , n), the inequality

n∑i,j=1

bij∂2u

∂ξi ∂ξj

∣∣∣∣x=x0

t=t0

0

and thereforen∑

i,j=1

aij∂2u

∂xi ∂xj

∣∣∣∣x=x0

t=t0

0.

As a result, we have L(u) 0 at the point (x0, t0); this is a contradiction proving the theorem for c(x, t) < 0.If the coefficient c(x, t) < M and M > 0 we can make the change of variables u(x, t) = v(x, t)eMt. The

function v(x, t) satisfies an equation of the form (1.1) with a negative coefficient in v and nonpositive right-handterm. As we have already proved, v(x, t) 0 in D, and the same is true for u(x, t) = v(x, t)eMt. The proof ofthe theorem is thereby completed.

Obviously, if L(u) 0 in D \ Γ and u(x, t) 0 on Γ, and the remaining suppositions of Theorem 1 arefulfilled, we have u(x, t) 0 everywhere in D.

Remark. It follows easily from the proof of Theorem 1 that the following assertion is valid. If L(u) < 0 andc 0 in D, then u(x, t) cannot assume a negative minimum in D \ Γ. We can also see that if L(u) > 0 and c 0in D, u(x, t) cannot assume a positive maximum in D \ Γ.

Theorem 2. Suppose the function u(x, t) is continuous in D and satisfies Eq. (1.1), where f(x, t) is a boundedfunction (|f | N) and the coefficient c(x, t) 0. Suppose that |u(x, t)|Γ m. Then everywhere in D we have

|u(x, t)| Nt+m. (1.8)

Proof. Consider in D the functions w±(x, t) = Nt+m± u(x, t). These functions are nonnegative on Γ, andin D \ Γ, taking into account that c 0, we get

L(w±) = −N +Nct+ cm± L(u) −N + |f | 0.

According to Theorem 1, both functions w±(x, t) are nonnegative in D, whence (1.8) follows.

Corollary. Let the suppositions of Theorem 2 be fulfilled and f(x, t) ≡ 0. Then everywhere in D

|u(x, t)| maxΓ|u(x, t)|.

Theorem 3. Suppose the function u(x, t) is continuous in D and satisfies in D\Γ Eq. (1.1) with |f(x, t)|N .Suppose that |u(x, t)|Γ m and c(x, t) −c0 < 0. Then everywhere in D

|u(x, t)| maxN

c0,m

.

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Proof. Denote N1 = maxNc0,mand consider the functions w±(x, t) = N1 ± u(x, t) in D. These functions

are nonnegative on Γ and in D \ Γ we have

L(w±) = N1c± f −N1c0 +N 0.

According to Theorem 1, the functions w±(x, t) are nonnegative everywhere in D and the assertion of the theoremfollows.

If c ≡ 0 and f ≡ 0, Theorem 2 leads to the following consequence, analogous to the well-known maximumand minimum principle for the heat equation.

Theorem 4. Suppose that the hypotheses of Theorem 2 are valid, f ≡ 0, and c ≡ 0. Then the solutionu(x, t) attains its maximal and minimal values on Γ, that is, everywhere in D we have the inequalities

m1 ≡ minΓu(x, t) u(x, t) max

Γu(x, t) ≡ m2.

To prove this theorem we shall note that the functions u(x, t) −m1 and m2 − u(x, t) are nonnegative on Γand satisfy Eq. (1.1). We can apply to them the corollary to Theorem 2 and get

u(x, t)−m1 maxΓ

(u(x, t)−m1) = m2 −m1,

m2 − u(x, t) maxΓ

(m2 − u(x, t)) = m2 −m1.

Theorem 5. Let us change in Theorem 2 the condition c 0 to the condition c(x, t) M , where M > 0.Then everywhere in D

|u(x, t)| eMt(Nt+m). (1.9)

To prove this theorem we shall make the change of variables u = veMt and apply Theorem 2 to the function v.From the estimate (1.9) we can deduce the uniqueness theorem for the first boundary-value problem (1.1),

(1.4), (1.5) if the coefficient c(x, t) is bounded from above.From the same estimate it follows that the solution of the first boundary-value problem depends continuously

on the right-hand term f(x, t), the initial function ϕ(x), and the boundary function ψ(x, t). Indeed, it followsfrom (1.9) that |u(x, t)| eMT (T + 1)ε if |f | < ε, |ϕ| < ε, and |ψ| < ε, that is, |u(x, t)| is arbitrarily small if |f |,|ϕ|, and |ψ| are sufficiently small.

The following theorem is called the strict maximum principle.

Theorem 6. Suppose the function u(x, t) is continuous in D and satisfies the equation L(u) = 0 in D \ Γ.We shall also suppose that all the coefficients of this equation are bounded in D, c(x, t) 0 and

n∑i,j=1

aijαiαj µn∑i=1

α2i

for all real αi; µ > 0 is some constant. Let us suppose that at some point (x0, t0) ∈ D \ Γ the function u(x, t)attains a positive maximum: u(x0, t0) = max

Du(x, t) = m > 0. Then u(x, t) = m at every point (x, t) ∈ D such that

t < t0 and that can be connected with the point (x0, t0) by a continuous curve of the form x = x(t), lying entirelyin D \ Γ.

Proof. Let us suppose the contrary. Suppose at some point P1(x1, t1) that can be connected with the pointP0(x0, t0), t1 < t0, by a continuous curve x = x(t) lying in D \ Γ the value of u(x, t) is less than m. We canconnect the points (x1, t1) and (x0, t0) by a broken line contained in D \ Γ with vertices P1(x1, t1), P2(x2, t2),. . . ,Pk(xk, tk), P0, where t1 < t2 < . . . < tk < t0. If we prove that the inequality u(xs, ts) < m leads to the inequalityu(xs+1, ts+1) < m (s = 1, 2, . . . , k − 1) then, passing from vertex Ps(xs, ts) to the vertex Ps+1(xs+1, ts+1) we shallbe able to establish the inequality u(x0, t0) < m. The contradiction to the supposition that u(x0, t0) = m will provethe theorem.

To simplify the notations we shall suppose that s = 1. So we suppose that u(x1, t1) < m1 < m, and we shallprove that u(x2, t2) < m.

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We can suppose that x1 = x2 = 0, because we can reduce the general case to this one by the lineartransformation of variables

τ = t, ξi = xi − x1i − (x2

i − x1i )t− t1t2 − t1 (i = 1, . . . , n).

The equation L(u) = 0 fulfills the conditions of the theorem also in the new variables (ξ, τ).

Let us consider the cylinder Q1r 6, t1 t t2, where r = n∑

i=1

x2i

1/2

, and the constant 6 < 1 is

chosen so that the cylinder Q1 lies in the domain D and on the lower base of the cylinder we have u(x, t1) < m1.Consider the function

w(x, t) = m− (m−m1)(62 − r2)2e−α(t−t1) − u(x, t)

in the cylinder Q1. The constant α > 0 can be chosen so that the inequality L(w) < 0 holds in Q1. In fact,

∂

∂xi(62 − r2)2 = −4xi(62 − r2),

∂2

∂xi ∂xj(62 − r2)2 = 8xixj − 4(62 − r2)δij ,

L(w) = cm− (m−m1)e−α(t−t1)

8

n∑i,j=1

aijxixj − 4(62−r2)n∑i=1

aii − 4(62−r2)n∑i=1

bixi + c(62−r2)2 + α(62−r2)2

−(m−m1)e−α(t−t1)8µr2 + (62 − r2)f1(x, t) + α(62 − r2)2,

where f1(x, t) is a bounded function. As we have µ > 0, the function f2(x, t) = 8µr2+(62−r2)f1(x, t)+α(62−r2)2

is positive in some neighborhood 61 < r 6 of the side of the cylinder Q1. In the rest of the cylinder, wherer 61, the function (62 − r2) is bounded from below by some positive constant and, therefore, f2(x, t) is positivefor sufficiently large α. For such α, L(w) < 0 in Q1.

It is easy to verify that the function w(x, t) is nonnegative on the side part of the boundary of Q1 and on itslower base. Indeed, for r = 6 we have w = m− u 0. For t = t1 and r 6 we get

w(x, t) = m− (m−m1)(62 − r2)2 − u m− (m−m1)64 −m1 m− (m−m1)−m1 = 0.

According to Theorem 1, we have w(x, t) 0 everywhere in the cylinder Q1. Therefore

w(0, t2) = m− (m−m1)64e−α(t2−t1) − u(0, t2) 0.

Henceu(0, t2) m− (m−m1)64e−α(t2−t1) < m.

The theorem is proved.Another proof of this theorem and some of its generalizations can be found in L. Nirenberg’s paper [20].4. Oblique Derivative Problem. For the parabolic Eq. (1.1) with f ≡ 0 and c 0 there is a theorem

about the derivative of a solution at a maximum point on the boundary analogous to the corresponding theorem foran elliptic equation (see [21,22]). We shall use this theorem to prove the uniqueness of the solution for the obliquederivative problem (1.1), (1.6).

Theorem 7. Suppose that the function u(x, t) is continuous in D and satisfies the equation L(u) = 0in D \Γ. Suppose the coefficients of the equation satisfy the suppositions of Theorem 6. We shall also suppose thatfor any point P of the surface S there exists a ball AP containing P , such that all its points except P lie in D \ Γand its radius directed to P is not parallel to the t axis.

Suppose the function u(x, t) attains its maximal value at a point P1(x1, t1) ∈ S. Then either the function

u(x, t) is constant in a neighborhood of the point P1 for t t1 or ∂u(P1)∂γ < 0, where γ is an arbitrary direction such

that the angle between this direction and the direction from the point P1 to the center of the ball AP1 is acute.(We suppose that ∂u

∂γ exists at the point P1.)

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Proof. Let us suppose that t1 < T . If u(x, t) is not constant in a neighborhood of P1 for t t1, then,according to Theorem 6, we have

u(P ) < u(P1) (1.10)

for all the inner points of the ball AP1 belonging to a sufficiently small neighborhood of P1. We shall suppose thatthe origin is at the center of the ball, at the point AP1 , and that the radius R of the ball is so small that everywherein the ball the inequality (1.10) holds and AP1 ∈ D. Consider the function

v(x, t) = e−α(r2+t2) − e−αR2, where r =

( n∑i=1

x2i

)1/2

.

It is easy to see that

L(v) =[−2α

( n∑i=1

aii − t+n∑i=1

bixi

)+ 4α2

n∑i,j=1

aijxixj + c

]e−α(r2+t2) − ce−αR2

> 0 (1.11)

for r > δ and for sufficiently large α > 0.Let us consider the domain ω bounded by the part of the boundary of the ball AP1 containing the point P1

and by the plane perpendicular to the radius OP1 and situated so near to the point P1 as to make (1.11) validin ω. Consider the auxiliary function w(P ) = u(P1) − u(P ) − βv(P ), where β > 0 is a constant; this function isnonnegative on the surface of the ball AP1 , because on this surface we have v(P ) 0 and u(P ) u(P1). On the restof the boundary of ω (1.10) holds; so for sufficiently small β we have w(P ) 0 on the whole boundary of ω. As wehave L(w) = cu(P1)− βL(v) < 0, it follows from Theorem 1 that w(P ) 0 everywhere in ω. Moreover, w(P1) = 0.Therefore ∂w(P1)

∂γ 0, whence

∂u(P1)∂γ

−β ∂v(P1)∂γ

= 2βαe−α(r2+t2)√r2 + t2 cos(

−−→OP1, γ) < 0.

The proof given above will need practically no changes if t1 = T : only in AP1 the points with t T have tobe considered. The theorem is proved.

The uniqueness of the solution for the oblique derivative problem follows from Theorem 7 if the coefficientsof Eq. (1.1) and the surface S satisfy the conditions of this theorem and a(x, t) 0. Indeed, it is easy to show thatunder the above-mentioned conditions on the surface S any point of the domain D can be connected by the curveof the form x = x(t) with any point (x, 0) ∈ D0. The difference u(x, t) between two solutions of the obliquederivative problem is equal to zero for t = 0 and satisfies the equation L(u) = 0 in D \ Γ, and the condition givenon the surface S is

∂u

∂γ+ a(x, t)u = 0 (a(x, t) 0). (1.12)

According to Theorem 6, the function u(x, t) cannot attain its maximal positive or its minimal negative valuein D \ Γ. If the maximal positive value m of u(x, t) is attained at a point P0 ∈ S, then, by virtue of Theorem 7and the condition (1.12), u(x, t) ≡ m in a neighborhood of the point P0. Then Theorem 6 leads to u|t=0 = m. Thecase of the negative minimum at the point P0 ∈ S can be reduced to that of the positive maximum by changingthe sign of u(x, t). The contradiction we obtain proves the theorem.

Remark. If the direction γ lies in a plane t = const, then it is possible to prove the uniqueness of the solutionfor the oblique derivative problem also without the supposition c(x, t) 0. In fact, in this case the transformationu = veαt, changing the equation to one with a negative coefficient in v, does not change the boundary condi-tion (1.12). What is more, when the direction γ lies in a plane t = const, for the uniqueness of the solution itis sufficient to require that the coefficient a(x, t) in the condition (1.12) be bounded from above. In this case,condition (1.12) can be reduced to one with a 0 by the transformation u = vz(x, t), where the smooth positivefunction z is defined on the surface S and satisfies the inequality ∂z

∂γ +az < 0. It is easy to construct such a function,for example, when S ∈ A2 and

cos(γ, ν) > β > 0. (1.13)

442

Under the supposition that γ lies in a plane t = const and (1.13) is fulfilled, we shall prove that the solutionof the oblique derivative problem depends continuously on the right-hand term of the equation and on the initialand boundary functions. We shall consider a domain D with side boundary S, belonging to the class A2.

Let the function u(x, t) satisfy Eq. (1.1) in D \ Γ and the condition (1.6) on Γ. Suppose that |f(x, t)| < ε,|ϕ(x)| < ε, and |ψ(x, t)| < ε. As we have remarked before, one can suppose that c(x, t) −c0 < −1 in D and thata(x, t) 0.

Let us construct a function v(x, t) ∈ C2 such that v|S = 0 and ∂v∂γ

∣∣S> 1. Suppose that |L(v)| < M1 and

|v| < M1. Consider the auxiliary functions w± = εv ± u. The function w+ cannot attain a positive maximal valueon the surface S, because we have ∂w+

∂γ + aw+ = ε ∂v∂γ + εav +(∂u∂γ + au

)> ε + ψ > 0 on S. If the function w+

attains its positive maximal value at a point P0 of D \ Γ, at this point we have

w+(P0) max |L(w+)|

c0 εmax |L(v)|+max |f |

c0< ε

M1 + 1c0

.

This inequality follows from Eq. (1.1) at the point P0. For t = 0, we have w+ ε|v|+ ε < ε(M1 + 1). As a resultwe get w+ < ε(M1 + 1) in the domain D. Similarly we get w− < ε(M1 + 1). Therefore

|u(x, t)| < ε(2M1 + 1),

that is, |u(x, t)| is arbitrarily small for sufficiently small ε.Theorems similar to those proved here can be found in [22].5. The Maximum Principle in an Unbounded Domain and the Cauchy Problem. Now we shall

prove that the solution of the Cauchy problem for Eq. (1.1) is unique and depends continuously on the right-handterm of (1.1) and on the initial function. First we shall prove for the unbounded domain H theorems similar toTheorem 1. To do this we have to impose some conditions on the growth of the coefficients and on the admissible

growth of the solution asn∑i=1

x2i →∞.

Theorem 8. Suppose the function u(x, t) in H is continuous and bounded from below: u(x, t) > −m,m > 0. Suppose the function u(x, t) in H has continuous derivatives, appearing in the operator L, and the inequalityL(u) 0 holds. Suppose that the coefficients aij , bi, and c satisfy the following conditions:

|aij(x, t)| < M(r2 + 1), |bi(x, t)| < M√r2 + 1, c(x, t) < M,

where r =( n∑i=1

x2i

)1/2

and M is a positive constant.

Under these suppositions, if u 0 for t = 0, then u(x, t) 0 everywhere in H .

Proof. Let us consider the auxiliary function

w(x, t) =m

r20

(r2 +Kt)eαt + u(x, t).

It is possible to choose the constants K > 0 and α > 0 so that for all r0 > 0 the value of L(w) will be negative.Indeed,

L(w) =m

r20

eαt(2

n∑i=1

aii + 2n∑i=1

bixi + cr2 +Kct−K − αr2 −Kαt)+ L(u)

and for r 1 we haveL(w) m

r20

eαt[M(4n+ 1)(r2 + 1)− αr2 + (M − α)Kt] < 0

if α > 2M(4n+ 1). For r < 1, taking into account that |aij | < 2M and |bi| < 2M , we get

L(w) m

r20

eαt[M(8n+ 1)−K] < 0

if K > M(8n+ 1).

443

Consider now w(x, t) in the cylinder Qr0r r0, 0 t T . For t = 0 we have w(x, 0) u(x, 0) and forr = r0 the function w(x, t) m+ u > 0, so, according to Theorem 1, everywhere in Qr0 the inequality w(x, t) 0holds.

Any fixed point from H is contained in Qr0 for all sufficiently large r0. At this point, according to what wasproved above, we have w(x, t) = m

r20(r2 +Kt)eαt+ u(x, t) 0. Passing to the limit in this inequality as r0 →∞, we

obtain the assertion of the theorem.

Remark. Instead of the condition u(x, t) > −m in Theorem 8 it is sufficient to require u(x, t) > −γ(r)r2−mwith lim

r→∞γ(r) = 0. Under this condition the proof undergoes practically no changes, but as the auxiliary function

w(x, t) we have to take γ(r0)(r2 +Kt)eαt + u(x, t).Moreover, it is possible to prove Theorem 8 under the supposition u(x, t) > −m(rq + 1), where m and q are

some real numbers. To do this, one has to take instead of mr20eαt(r2+Kt) another auxiliary function 2m

r2p−q0

(r2+Kt)peαt

with 2p > q; then we only need to repeat the proof of Theorem 8.

Theorem 9. Suppose the coefficients aij and bi are bounded in H :

|aij | < M, |bi| < M, |c| < M,

and the function u(x, t) is continuous in H and satisfies the inequalities

L(u) 0, u(x, t) −eβ(r2+1)

in H ; here β is a positive constant. Then if u(x, 0) 0, then u(x, t) 0 everywhere in H.

To prove this theorem, we have to take the auxiliary function

w(x, t) = e2β(r2+1)eαt−β(r20+1) + u(x, t)

and repeat the reasonings we used to prove Theorem 8 with respect to the new function. Doing this, we have toverify the inequality

L(e2β(r2+1)eαt) < 0

if α > 0 is chosen in a correct way. We have

L(e2β(r2+1)eαt) = e2β(r2+1)eαt+αt

[4β

n∑i=1

aii + 16β2eαtn∑

i,j=1

aijxixj + 4βn∑i=1

bixi + ce−αt − 2βα(r2 + 1)]< 0

for t 1α , if α =M

(1β +8n+48βn

). So we get u(x, t) 0 for 0 t 1

α . We can repeat this argument for the layer1α t

2α , then for 2

α t 3α and so on; in this way we establish that u(x, t) 0 everywhere in H .

Theorem 10. Suppose that u(x, t) is continuous and bounded in H and satisfies in H Eq. (1.1) andthat the inequalities |u(x, 0)| M1, |f(x, t)| M2, and c(x, t) M3 hold. The coefficients aij and bi satisfythe suppositions of Theorem 8. Then everywhere in H

|u(x, t)| eM3t(M1 +M2t). (1.14)

To prove this theorem we shall consider the auxiliary functions w± = eM3t(M1 +M2t) ± u. According tothe suppositions of the theorem, w±(x, 0) 0. We have

L(w±) = eM3t[(c−M3)(M1 +M2t)−M2]± f −M2eM3t ± f 0.

By virtue of Theorem 8, w± 0 everywhere in H and therefore (1.14) holds.If only bounded solutions of (1.1) are considered and the coefficients of (1.1) are subject to the suppositions

of Theorem 8, then the uniqueness of the solution of the Cauchy problem for (1.1) and the continuous dependenceof this solution on the initial function ϕ(x) and the right-hand term f(x, t) of the equation follow immediately fromTheorem 10. Moreover, under the above-mentioned conditions on the coefficients of the equation it follows from

444

the remark to Theorem 8 that the solution of the Cauchy problem is unique in the class of functions growing notfaster than a degree of r as r →∞.

When the coefficients of Eq. (1.1) are bounded, it follows from Theorem 8 that the solution of the Cauchyproblem is unique in the class of functions growing not faster than eβr

2, β > 0, as r →∞.

The results stated below show that there is no uniqueness for the solutions of the Cauchy problem in the widerclass of functions characterized by the inequality

|u(x, t)| < eβ(r2+ε+1), ε > 0. (1.15)

This means that there are solutions of Eq. (1.3) different from identical to zero and satisfying the condition u(x, 0)=0and the inequality (1.15).

It is impossible to weaken the conditions imposed on the coefficients of Eq. (1.1) in Theorem 8, remainingin classes of estimates of the coefficients characterized by degrees of r. Indeed, for any δ > 0 the function

u(x, t) =

+∞∫Fδ(x,t)

e−y2dy for 0 < t T,

0 for t = 0,

where the function

Fδ(x, t) =(√x2 + 1 + x)δ

2√t

is continuous and bounded in H , vanishes for t = 0 and for t > 0 satisfies the equation

1δ2

(x2 + 1)(√x2 + 1− x)2δ ∂

2u

∂x2+

1δ2

(x− δ√x2 + 1)(

√x2 + 1− x)2δ ∂u

∂x− ∂u

∂t= 0.

In this equation, the coefficient of ∂2u∂x2 does not grow faster than M |x|2+2δ and the coefficient of ∂u

∂x does not growfaster than M |x|1+2δ. Therefore, for coefficients having such growth there is no uniqueness for solutions in the classof bounded functions.

The questions concerning the uniqueness of solutions of the Cauchy problem for parabolic equations and forsystems parabolic according to I. G. Petrovskiy were studied in many papers (see [3]).

For the heat equation (1.3) there are stronger results than Theorem 9. These results are due to E. Holm-gren [23], A. N. Tikhonov [24], and S. Tacklind [25]. It is proved that the solution of the Cauchy problem forthe heat equation is unique in the class of functions satisfying the condition

|u(x, t)| < eδ|x|h(|x|) (|x| > 1), (1.16)

where δ is an arbitrary positive constant and h(r) is a positive nondecreasing function such that

∞∫1

dr

h(r)=∞. (1.17)

One can construct examples to show that if the integral (1.17) converges, then a solution of the Cauchy problemfor the heat equation is not unique in the class of functions satisfying (1.16) (see [24, 25]).

For the parabolic systems the uniqueness theorems for the Cauchy problem in classes of growing functionsis studied in [26, 27] and some others. For parabolic equations and systems with growing coefficients the questionsof the uniqueness were studied in [28].

Remark. For Eq. (1.1), one can also study boundary-value problems in unbounded domains. Let Q∞ =Ω∞ × (0, T ), where Ω∞ is a complement of a bounded domain Ω in the space En. The uniqueness of a boundedsolution for the first and second boundary-value problems, as well as for the oblique derivative problem for Eq. (1.1)in Q∞ can be proved similarly to the case of a bounded domain, but the choice of the auxiliary function, growingas r →∞, is different (see, for example, [29]).

445

6. Some Generalizations of Uniqueness Theorems. One can pose the first boundary-value problemand the Cauchy problem for Eq. (1.1) in more general terms than we did in Sect. 1. We can suppose that the initialcondition u(x, 0) = ϕ(x) holds for all x except a certain set E , lying inD0. The function ϕ(x) can have discontinuitiesat the points of E .

The following theorems state the conditions under which the uniqueness theorems hold.

Theorem 11. Suppose the set E ⊂ D0 has the following property: there exists a positive number λ < nsuch that for every given ε > 0 the set E can be covered by a finite number of n-dimensional balls K1, . . . ,KN withradii 61, . . . , 6N , respectively, and

N∑s=1

6λs < ε. (1.18)

Suppose that the function u(x, t) is defined, continuous, and bounded in D \ E , satisfies in D \ Γ the equation

L(u) = 0, (1.19)

and on the boundary of D satisfies the conditions

u(x, t) = 0 on S, (1.20)u(x, 0) = 0 for x ∈ D0 \ E . (1.21)

The coefficients bi and c of the Eq. (1.19) are assumed to be bounded in D, and the coefficients aij are assumed tobe continuous in D and satisfy the inequality

n∑i,j=1

aij(x, t)αiαj µ

n∑i=1

α2i , µ > 0. (1.22)

Under these suppositions, u(x, t) ≡ 0 everywhere in D \ E .

Proof. We can choose a point (x, 0) ∈ D0 and consider the cylinder Q′x,&|x − x| < 6, 0 < t < 62, where

|x − x| =[ n∑i=1

(xi − xi)2]1/2

and 6 > 0 is sufficiently small. Suppose that Qx,& = Q′x,& ∩D. Consider the auxiliary

function

v(x, t; x, 6) =1

(t+ 62)λ/2e−α

n∑i,j=1

aij(x,0)(xi−xi)(xj−xj)

4(t+2) . (1.23)

Here aij(x, t) denotes the element of a matrix inverse to ‖aij(x, t)‖; λ is the same as in (1.18); α is a fixed numbersatisfying the inequalities

λ

n< α < 1. (1.24)

We shall show that in a domain Qx,&∗ , where 6∗ > 0 and does not depend on x and 6, the inequality

L(v(x, t; x, 6)) < 0

holds.Define akl = akl(x, 0), akl = akl(x, 0), and ξk = xk − xk (k, l = 1, . . . , n). We have

∂v

∂xk= −

αn∑i=1

akiξi

2(t+ 62)v,

∂2v

∂xk ∂xl=[− αakl

2(t+ 62)+

α2

4(t+ 62)2

n∑i,j=1

akialjξiξj

]v,

∂v

∂t=[− λ

2(t+ 62)+

α

4(t+ 62)2

n∑i,j=1

aijξiξj

]v,

446

L(v) =n∑

k,l=1

akl∂2v

∂xk ∂xl− ∂v

∂t+

n∑k,l=1

(akl − akl)∂2v

∂xk ∂xl+

n∑k=1

bk∂v

∂xk+ cv

=αv

2(t+ 62)

[−

n∑k,l=1

aklakl +

λ

α+

α

2(t+ 62)

n∑k,l,i,j=1

aklakialjξiξj −

12(t+ 62)

n∑i,j=1

aijξiξj −n∑

k,l=1

(akl − akl)akl

+α

2(t+ 62)

n∑i,j,k,l=1

(akl − akl)akialjξiξj −n∑

i,k=1

bkakiξi +

2c(t+ 62)α

]. (1.25)

Without loss of generality, we can suppose that c(x, t) 0. Sincen∑

k=1

aklaki = δli and the quadratic form

n∑i,j=1

aijξiξj is positive (this follows from (1.22)), from (1.25) we deduce

L(v) =αv

2(t+ 62)

−(n− λ

α

)− (1− α)

2(t+ 62)

n∑i,j=1

aijξiξj −n∑

k,l=1

(akl − akl)akl

+α

2(t+ 62)

n∑i,j,k,l=1

(akl − akl)akialjξiξj −n∑

i,k=1

bkakiξi +

2c(t+ 62)α

αv

2(t+ 62)

−(n− λ

α

)− µ1(1 − α)

|x− x|2t+ 62

+M1ω(6∗) +M2ω(6∗)|x− x|2t+ 62

+M36∗. (1.26)

Here ω(6∗) = max |akl(x, t)−akl(x′, t′)| for (x, t) ∈ D, (x′, t′) ∈ D, |x−x′| 6∗, |t−t′| 6∗2 (k, l = 1, . . . , n),and the constants µ1 > 0 and Mi > 0 (i = 1, 2, 3) do not depend on 6∗, 6, and x. The inequalities (1.24) and (1.26)show that L(v(x, t; x, 6)) < 0 in Qx,&∗ , where 6∗ depends on the modulus of continuity ω(6) of the coefficients aijand on the upper bounds of |bi| in D but does not depend on x and 6.

In D \ Qx,&∗ the function v(x, t; x, 6) is bounded uniformly in x and 6 together with all its derivativesappearing in L(v). Therefore |L(v(x, t; x, 6))| < M4, M4 > 0 for (x, t) ∈ D \Qx,&∗ , and M4 does not depend on x.As a consequence, for the function

z(x, t; x, 6) = v(x, t; x, 6) +M4t (1.27)

we have everywhere in D the inequalityL(z(x, t; x, 6)) < 0. (1.28)

By virtue of (1.23) and (1.27), we can obtain an estimate from below for z(x, t; x, 6) in the domain Qx,&∗ :

z(x, t; x, 6) 1(262)λ/2

e−M5|x−x|2

2 µ2

6λ. (1.29)

Here M5 > 0 and µ2 > 0 do not depend on x and 6.Now suppose that M = sup

D

|u(x, t)| and (x0, t0) is an arbitrary point in D\Γ. Let us take an arbitrary small

number ε > 0 and prove that |u(x0, t0)| < ε.From (1.23) and (1.27) one can deduce that

z(x0, t0; x, 6) M6, (1.30)

where M6 does not depend on x and 6 (the point (x0, t0) remains fixed). Let us cover the set E by n-dimensionalballs K1, . . . ,KN such that

N∑s=1

6λs <εµ2

MM6; (1.31)

here 6s is the radius of the ball Ks and xs = (xs1, . . . , xsn) is its center, s = 1, . . . , N . We shall also assume that6s <

√t0 (s = 1, . . . , N).

447

Set D′ = D \N⋃s=1

Qxs,&s ; Γ′ will denote the set of boundary points of D′ such that t < T . Compare on

the set D′ the function u(x, t) with the function

w(x, t) =M

µ2

N∑s=1

6λs z(x, t;xs, 6s),

where z(x, t;xs, 6s) (s = 1, . . . , N) are defined as in (1.27) and (1.23). Everywhere on Γ′ we have

|u(x, t)| w(x, t). (1.32)

In fact, on S and on D0 \ E , (1.32) follows from the conditions (1.20), (1.21). If (x, t) ∈ Qxs1 ,&s1for a certain s1,

1 s1 N , then according to (1.29) we get

w(x, t) M

µ26λs1

z(x, t;xs1 , 6s1) M |u(x, t)|.

By virtue of (1.28) we have L(w±u) < 0 everywhere in D′. Applying Theorem 1 to w+u and w−u, we cansee that the inequality (1.32) is valid everywhere in D′, and particularly in the point (x0, t0). Taking into account(1.30) and (1.31), we get from (1.32)

|u(x0, t0)| w(x0, t0) M6M

µ2

N∑s=1

6λs < ε.

As the point (x0, t0) and the number ε are arbitrary, the statement of Theorem 11 follows.

Remark 1. If the boundary σ of the domain D0 is smooth, Theorem 11 is also valid for E = σ.

Remark 2. If L(u) 0 in D \ Γ, u(x, t) 0 everywhere in Γ \ E , and the other suppositions of Theorem 11hold too, then u(x, t) 0 everywhere in D \ E . The proof of this assertion is similar to that of Theorem 11.

Theorem 12. Suppose E is a set in the plane t = 0 having the following property: there exists a positivenumber λ < n such that for any given ε > 0 any bounded domain E1 ⊂ E can be covered by a finite numberof n-dimensional balls K1, . . . ,KN of radii 61, . . . , 6N , respectively, so that the inequality (1.18) holds. Supposethe function u(x, t) is defined, continuous, and bounded in H \ E and satisfies Eq. (1.19) in H and the initialcondition (1.21) for t = 0 outside the set E . Suppose also that all the coefficients of Eq. (1.19) are bounded in H ,the coefficients aij are uniformly continuous in H , and the inequality (1.22) holds in H .

Under these suppositions, u(x, t) ≡ 0 everywhere in H \ E .

The proof of this theorem is similar to that of Theorem 11. As in Theorem 8, we only need to use, besidesthe auxiliary function z(x, t; x, 6), another auxiliary function, growing as r →∞: w1(x, t) = (r2 +Kt)eαt.

2 A PRIORI ESTIMATES

1. S. N. Bernstein’s a priori Estimates. In the study of boundary-value problems and the properties ofsolutions of partial differential equations, a priori estimates of solutions in various norms are of great importance.

An estimate is called an a priori estimate if it is obtained under the supposition of the existence of the solutionfor the problem considered and depends on the data of the problem. That is, it depends on the coefficients ofthe equation and its right-hand term, on the domain D, and on the initial and boundary data. In Theorems 2 and 5of Sect. 1 (theorems on the generalization of the well-known maximum principle for the heat equation), a prioriestimates of the absolute value of a solution of the first boundary-value problem are established in terms of the dataof the problem considered. In this section, we shall establish the most important a priori estimates for the solutionsand their derivatives for the second-order parabolic equations.

S. N. Bernstein first pointed out the important part of a priori estimates in the study of solutions forboundary-value problems in his papers on the elliptic equations (see, e.g. [30,31]). In these papers, he gave methodsof obtaining a priori estimates for the solutions. Here the method of auxiliary functions is very important. Thismethod became widespread recently in the study of second-order elliptic and parabolic equations. S. N. Bernstein’smethods were used in many works (see, e.g., [18, 32–36]). The following theorems of Bernstein, giving an a prioriestimate for derivatives of a solution of Eq. (1.1), are proved by the method of auxiliary functions.

448

Theorem 1 (see [37]). Suppose the function u(x, t) satisfies Eq. (1.1) at the points of the cylinder Q =Ω× (0, T ), where Ω is a bounded domain in the space En with the boundary σ. Suppose that |u(x, t)| M in Q\Γ,the coefficients aij , bi, and c and the right-hand term f of Eq. (1.1) are continuous in Q \ Γ together with theirderivatives of the first order with respect to x1, . . . , xn, and

n∑i,j=1

aij(x, t)αiαj µ

n∑i=1

α2i , µ > 0. (2.1)

Suppose also that the function u(x, t) in Q \ Γ has continuous derivatives ∂3u∂xi∂xj∂xk

and ∂2u∂xk ∂t

(i, j, k = 1, . . . , n).

Then in any cylinder Qδ = Qδ ∩ δ t T , where δ > 0 is an arbitrarily small number, the inequality

n∑k=1

(∂u

∂xk

)2

M1 (2.2)

holds. Here M1 depends only on δ, M , and µ and on the maximum of the absolute values of the coefficients of (1.1),the right-hand term f , and their derivatives with respect to x1, . . . , xn in Qδ/2.

Proof. Assume that c(x, t) 0. As we have remarked in subsection 3 of Sect. 1, this assumption can bemade without loss of generality. It is sufficient to prove the inequality (2.2) for the cylinder Kδ/4 × [δ, T ], whereKδ/4 is an n-dimensional ball of radius δ/4, lying together with its boundary inside Ωδ/2, because the set Ωδ canbe covered by a finite number of such balls. Moreover, one can assume that the concentric to Kδ/4 ball Kδ/2 ofradius δ/2 also lies in Ωδ/2.

For simplicity we shall assume that the center of the ball Kδ/2 is at the origin. Consider in the cylinderGδ = Kδ/2 ×

(δ2 , T)the auxiliary function

w(x, t) = τΦ(x)n∑

k=1

(∂u

∂xk

)2

+Nu2 + eN1(T−τ), (2.3)

where

τ = t− δ

2, Φ(x) =

(δ2

4−

n∑i=1

x2i

)2

.

Note the following property of Φ(x):(∂Φ∂xi

)2

M2Φ for x ∈ Kδ/2 (i = 1, . . . , n); (2.4)

here M2 > 0 is a constant.We shall show that for a correct choice of the constants N and N1 the function w(x, t) in Gδ attains its

maximal value either on t = δ2 or on

n∑i=1

x2i = δ2

4 . Taking into account the remark to Theorem 1 of Sect. 1, we can

see that it is sufficient to prove that everywhere in Gδ we have

L(w) > 0. (2.5)

To calculate L(w) we use the formula

L(v1v2) = v1L(v2) + v2L(v1) + 2n∑

i,j=1

aij∂v1

∂xi

∂v2

∂xj− cv1v2.

We have

L(w) = τΦ(x)[2

n∑k=1

∂u

∂xkL

(∂u

∂xk

)+ 2

n∑i,j,k=1

aij∂2u

∂xk ∂xi

∂2u

∂xk ∂xj− c

n∑k=1

(∂u

∂xk

)2]

+ [τL(Φ)− Φ]n∑

k=1

(∂u

∂xk

)2

+ 4τn∑

i,j,k=1

aij∂Φ∂xi

∂u

∂xk

∂2u

∂xk ∂xj+ 2NuL(u)

+ 2Nn∑

i,j=1

aij∂u

∂xi

∂u

∂xj− c[τΦ(x)

n∑k=1

(∂u

∂xk

)2

+Nu2

]+ (N1 + c)eN1(T−τ). (2.6)

449

Differentiating Eq. (1.1) with respect to xk, we get

L

(∂u

∂xk

)= −

n∑i,j=1

∂aij∂xk

∂2u

∂xi ∂xj−

n∑i=1

∂bi∂xk

∂u

∂xi− ∂c

∂xku+

∂f

∂xk(k = 1, . . . , n).

In the right-hand side we can omit members that are obviously positive and obtain

L(w) −2τΦn∑

i,j,k=1

∂aij∂xk

∂u

∂xk

∂2u

∂xi ∂xj− 2τΦ

n∑i,k=1

∂bi∂xk

∂u

∂xk

∂u

∂xi− 2τΦ

n∑k=1

∂c

∂xku∂u

∂xk

+ 2τΦn∑

k=1

∂f

∂xk+ 2τΦ

n∑i,j,k=1

aij∂2u

∂xk ∂xi

∂2u

∂xk ∂xj+ [τL(Φ) − Φ]

n∑k=1

(∂u

∂xk

)2

+ 4τn∑

i,j,k=1

aij∂Φ∂xi

∂u

∂xk

∂2u

∂xk ∂xj+ 2Nuf + 2N

n∑i,j=1

aij∂u

∂xi

∂u

∂xj+ (N1 + c)eN1(T−τ). (2.7)

Then we use the elementary inequality

|ab| 12

(εa2 +

1εb2), (2.8)

where ε > 0 is an arbitrary positive number. One can estimate the first term in the right-hand side of (2.7) bymeans of this inequality in the following way:

−2τΦn∑

i,j,k=1

∂aij∂xk

∂u

∂xk

∂2u

∂xi ∂xj −2M3τΦ

n∑i,j,k=1

∣∣∣∣ ∂u∂xk∂2u

∂xi ∂xj

∣∣∣∣ −εnM3τΦ

n∑i,j=1

(∂2u

∂xi ∂xj

)2

− n2M3

ετΦ

n∑k=1

(∂u

∂xk

)2

.

We denote by Mi, i = 3, . . ., the constants depending on the same data as M1.In a similar way we can estimate the remaining terms in (2.7). Taking additionally into account the condi-

tion (2.1), we obtain the following inequality:

L(w) 2τΦ[−εM4 − εM5

1Φ

n∑k=1

(∂Φ∂xk

)2

+ µ

] n∑i,j=1

(∂2u

∂xi ∂xj

)2

+(2Nµ− M6

ε−M7

) n∑k=1

(∂u

∂xk

)2

+ N1 + c−M8N −M9. (2.9)

Let us take ε > 0 so small that the coefficient ofn∑

i,j=1

(∂2u

∂xi ∂xj

)2 in the right-hand side of (2.9) is positive

everywhere inGδ; this can be achieved by virtue of (2.4). Having fixed ε, we can chooseN so large that the coefficient

ofn∑

k=1

(∂u∂xk

)2 in (2.9) becomes positive. Then we choose N1 so large as to make positive the term in braces in (2.9).

We have proved the inequality (2.5). As we have already noted, it follows that w(x, t) attains its maximum

in Gδ either on t = δ2 or on

n∑l=1

x2l = δ2

4 . Therefore, w(x, t) NM2 + eN1T everywhere in Gδ. Moreover, in Gδ

the inequality (t− δ

2

)Φ(x)

n∑k=1

(∂u

∂xk

)2

NM2 + eN1T

holds. In particular, for the points of the cylinder Kδ/4 × [δ, T ] we obtain

n∑k=1

(∂u

∂xk

)2

NM2 + eN1T(t− δ

2

)Φ(x)

NM2 + eN1T

δ2

(3δ2

16

)2 =M1,

which was to be proved.

450

Remark. By the same method it is possible to estimate the higher-order derivatives of the function u(x, t).We only have to suppose that the function u(x, t) and the coefficients and the right-hand term of (1.1) have a higherdegree of smoothness. So to obtain estimates for the derivatives ∂2u

∂xi ∂xj(i, j = 1, . . . , n) in the cylinder Qδ it is

sufficient to suppose that in Q \ Γ differential operators ∂2

∂xk ∂xl(k, l = 1, . . . , n) can be applied to Eq. (1.1) and to

use instead of function (2.3) the auxiliary function

w1(x, t) = τΦ(x)n∑

k,l=1

(∂2u

∂xk ∂xl

)2

+N2

n∑k=1

(∂u

∂xk

)2

+ eN3(T−τ).

The rest of the reasoning undergoes no changes. To obtain estimates for the derivatives ∂pu∂x

p11 ...∂xpnn

it is sufficientto suppose that in Q \ Γ one can differentiate (1.1) p times with respect to x1, . . . , xn.

After we have obtained in Qδ estimates for the derivatives of u(x, t) of the first and second order withrespect to x1, . . . , xn, the estimate for ∂u

∂t in Qδ follows from Eq. (1.1). Similarly, if we intend to get estimatesfor the higher-order derivatives of u(x, t) containing the differentiation with respect to t, we have to use equationsobtained from (1.1) by differentiation. To obtain the estimates for the derivatives ∂p+qu

∂xp11 ...∂xpnn ∂tq

, where p+ 2q r

(r 2), it is sufficient to suppose that the differential operator ∂k+l

∂xk11 ...∂xknn ∂tl

with k + 2l r, l [r2

]− 1, can be

applied to (1.1) in Q \ Γ.Applying Bernstein’s method, we can establish a priori estimates also in a closed domain.

Theorem 2. Suppose the function u(x, t) is a solution of the first boundary-value problem for Eq. (1.1) inthe cylinder Q = Ω× (0, T ) with the initial and boundary conditions

u(x, 0) = ϕ(x), u(x, t)|S = ψ(x, t). (2.10)

The coefficients and the right-hand side of (1.1) are supposed to be continuous in Q together with their first-orderderivatives with respect to x1, . . . , xn, and the inequality (2.1) holds. The boundary σ of the domain Ω belongs tothe class A2. Suppose also that the derivatives ∂u

∂xk(k = 1, . . . , n) are continuous in Q, and in Q \ Γ continuous

derivatives ∂3u∂xi∂xj∂xk

, ∂2u∂xk ∂t

(i, j, k = 1, . . . , n) exist.

Under these suppositions, everywhere in Q the inequality (2.2) holds with M1 depending only on the do-main Q, the constant µ, on M = max

Q|u(x, t)|, and on the maximum of the absolute values of the following functions:

aij , bi, c, f , ϕ, their first-order derivatives with respect to x1, . . . , xn, the function ψ(x, t), and its first- and second-order derivatives.

Proof. Without loss of generality, we can assume that c(x, t) 0 (as in Theorem 1). We shall estimaten∑

k=1

(∂u∂xk

)2 for (x, t) ∈ S. According to our supposition, the boundary σ of the domain Ω can be covered by a finite

number of overlapping pieces σp such that each of them can be represented in the form

xi = χi(ξ1, . . . , ξn−1) (i = 1, . . . , n) (2.11)

with χi ∈ C2. Therefore, in σp one can introduce a new coordinate system by the formulas xi = ηi(ξ1, . . . , ξn),ξi = η∗i (x1, . . . , xn) (i = 1, . . . , n) so that σp in these coordinates is defined by the equation ξn = 0, for the pointsof Ω the inequality ξn > 0 holds, and ηi ∈ C2 and η∗i ∈ C2.

We shall construct functions ζp(ξ1, . . . , ξn−1), having the following properties: 12 ζp 1, ζp = 1

2 onthe piece σp and ζp ≡ 1 in a certain subdomain σp of σp. The subdomains σp are chosen so that they still cover σ.

In the cylinder Qp

(ξ1, . . . , ξn−1) ∈ σp, 0 ξn 1

N , 0 t Twe shall consider an auxiliary function

z(ξ, t) = z(ξ1, . . . , ξn, t) = u(ξ1, . . . , ξn−1, ξn, t)− ψ(ξ1, . . . , ξn−1, t) +N1e−Nξnζp(ξ1, . . . , ξn−1), (2.12)

where the constants N > 0 and N1 > 0 will be chosen later. Calculate L(z) in the variables ξ1, . . . , ξn, t:

L(z) =n∑

i,j,k=1

αij(ξ, t)∂2z

∂ξi ∂ξj+

n∑i=1

βi(ξ, t)∂z

∂ξi+ c(ξ, t)z − ∂z

∂t

= L(u)− L(ψ) +N1e−NξnL(ζp) + L(N1e

−Nξn)ζp − 2NN1e−Nξn

n∑j=1

αnj∂ζp∂ξj−N1ce

−Nξnζp.

451

From these formulas we can deduce that

L(z) = f − L(ψ) +N1e−Nξn

[L(ζp) +N2αnnζp −Nβnζp − 2N

n∑j=1

αnj∂ζp∂ξj

]. (2.13)

According to (2.1) we have αnn µ′ > 0 everywhere in Qp; moreover, ζp 12 . Therefore, for sufficiently

large N > 0 the expression in square brackets on the right-hand term of (2.13) becomes positive. Having fixedsuch N , we can choose N1 > 0 so large that the whole right-hand term of (2.13) becomes positive.

Because L(z) > 0, according to the remark to Theorem 1 of Sect. 1, the function z(ξ, t) cannot attain itsmaximal value in Qp in Q, neither inside of Qp nor on t = T . We shall show that for sufficiently large N1 > 0the function z(ξ, t) takes its maximal value at points belonging to Sp = σp × [0, T ].

It follows from (2.12) that z(ξ, t) = N1 at the points of Sp and z(ξ, t) = N1ζp N1 at the points of Sp =σp× [0, T ]. If 0 ξn 1

N and (ξ1, . . . , ξn−1) belongs to the boundary of the piece σp, then z(ξ, t) 2M + N12 < N1

if N1 > 4M . For ξn = 1N we have z(ξ, t) 2M + N1

e < N1 if N1 > 4M . Finally, also for t = 0 the function z(ξ, t)does not exceed N1 if N1 is sufficiently large. In fact,

∂z(ξ, 0)∂ξn

=∂ϕ(ξ)∂ξn

−NN1e−Nξnζp max

∣∣∣∣∂ϕ(ξ)∂ξn

∣∣∣∣− 12eNN1 < 0

for N1 > 0 sufficiently large.So for a correct choice of N and N1 the function z(ξ, t) takes its maximal value on Sp, and there it is

identically equal to N1. Therefore, at all the points of Sp we have the inequality ∂z∂ξn 0, and this leads to

∂u

∂ξn

∣∣∣∣Sp

N1N. (2.14)

Considering the auxiliary function

z1(ξ, t) = u(ξ1, . . . , ξn−1, ξn, t)− ψ(ξ1, . . . , ξn−1, t)−N1e−Nξnζp(ξ1, . . . , ξn−1)

and reasoning in a similar way, we can establish the inequality

∂u

∂ξn

∣∣∣∣Sp

−N1N. (2.15)

From (2.14) and (2.15) we can deduce that the estimate∣∣∣∣ ∂u∂ξn∣∣∣∣ N1N

holds on Sp. Since the derivatives of u with respect to ξ1, . . . , ξn−1 on Sp coincide with the corresponding derivativesof the function ψ, we have proved the validity of (2.2) on Sp and therefore everywhere on S.

Consider now in the cylinder Q the function

w(x, t) =n∑

k=1

(∂u

∂xk

)2

e−N2t + eN2(T−t),

where the constant N2 > 0 will be chosen later. Let us compute L(w):

L(w) = e−N2t

2

n∑i,j,k=1

aij∂2u

∂xk ∂xi

∂2u

∂xk ∂xj− 2

n∑i,j,k=1

∂aij∂xk

∂u

∂xk

∂2u

∂xi ∂xj− 2

n∑i,k=1

∂bi∂xk

∂u

∂xi

∂u

∂xk

− 2un∑

k=1

∂c

∂xk

∂u

∂xk+ 2

n∑k=1

∂f

∂xk

∂u

∂xk+ (N2 − c)

n∑k=1

(∂u

∂xk

)2

+ (N2 + c)eN2T

. (2.16)

452

Taking into account (2.8), from (2.16) we get

L(w) e−N2t

(2µ− εM2)

n∑i,k=1

(∂2u

∂xi ∂xk

)2

+(N2 −

M3

ε−M4

) n∑k=1

(∂u

∂xk

)2

+ (N2 −M5)eN2T −M5

. (2.17)

First we take ε < 2µM2

, then choose N2 >M3ε +M4 +M5. According to (2.17), we have L(w) > 0. Therefore w(x, t)

cannot have a maximum in Q \ Γ.This means that the maximum is attained on t = 0 or on S, and according to what was already proved, we

have w(x, t) M ′1, where M′1 depends on the same quantities as M1 in the conditions of the theorem.

Hencen∑

k=1

(∂u

∂xk

)2

M ′1eN2T =M1.

The theorem is proved.

Theorem 3. Suppose the function u(x, t) is a solution of the first boundary-value problem (1.1), (2.10) inthe cylinder Q. The coefficients aij , bi, and c and the function f are continuous in Q together with their derivativesof the first and second order with respect to x1, . . . , xn and the inequality (2.1) holds. The surface σ belongs to

the class A3. The derivatives ∂u∂t , ∂u

∂xk, ∂2u∂xk ∂xl

(k, l = 1, . . . , n) are continuous in Q, and in Q \ Γ the derivatives∂4u

∂xi∂xj∂xk∂xl, ∂3u∂xk∂xl∂t

(i, j, k, l = 1, . . . , n) exist.

Under these suppositions, everywhere in Q we have

n∑k,l=1

(∂2u

∂xk ∂xl

)2

M1, (2.18)

where M1 depends only on the domain Q, the constant µ, and the maximum of absolute values of the followingfunctions: ϕ, aij , bi, c, f , their first and second derivatives with respect to x1, . . . , xn, ψ, and its derivatives up tothe third order inclusively.

Proof. We shall use the notations

MQ = maxQ

n∑k,l=1

(∂2u

∂xk ∂xl

)21/2

, MΓ = maxΓ

n∑k,l=1

(∂2u

∂xk ∂xl

)21/2

.

Consider in Q the following auxiliary function:

w(x, t) = e−Ntn∑

k,l=1

(∂2u

∂xk ∂xl

)2

+ eN(T−t).

For the correct choice of N > 0 this function attains its maximal value on Γ. We have

L(w) = e−Nt

2

n∑i,j,k,l=1

aij∂3u

∂xk∂xl∂xi

∂3u

∂xk∂xl∂xj+ 2

n∑k,l=1

∂2u

∂xk ∂xl

[ n∑i,j=1

aij∂4u

∂xi∂xj∂xk∂xl

+n∑i=1

bi∂3u

∂xi∂xk∂xl− ∂3u

∂xk∂xl∂t+ c

n∑k,l=1

∂2u

∂xk ∂xl

]+N

n∑k,l=1

(∂2u

∂xk ∂xl

)2

+ (N + c)eNT

.

The expression in square brackets can be transformed by means of the equation obtained from (1.1) by differentiation

with respect to xk and xl. We shall also take into account the estimate we have obtained forn∑i=1

(∂u∂xi

)2 in Theorem 2and the inequality (2.8). We get

L(w) e−Nt

(2µ− εM2)

n∑i,k,l=1

(∂3u

∂xk∂xl∂xi

)2

+(N − M3

ε−M4

) n∑k,l=1

(∂2u

∂xk ∂xl

)2

+ (N −M5).

453

It follows from this inequality that for ε < 2µM2

and N > M3ε +M4+M5 the function w(x, t) cannot have a maximum

in Q \ Γ. Therefore maxQ

w M2Γ + eNT ; this leads to M2

Q eNT (M2Γ + eNT ), and hence

MQ M6MΓ +M7. (2.19)

Now we shall get an estimate for MΓ. As we did in the proof of Theorem 2, we can introduce in the space(x1, . . . , xn) in the vicinity of the piece σp a new coordinate system (ξ1, . . . , ξn), such that σp in these coordinateshas the equation ξn = 0 and for the points of Ω the inequality ξn > 0 holds. In the cylinder

Qp

(ξ1, . . . , ξn−1) ∈ σp, 0 ξn

1N2

, 0 t T

we shall consider the auxiliary function

zk(ξ, t) =∂u(ξ1, . . . , ξn−1, ξn, t)

∂ξn− ∂ψ(ξ1, . . . , ξn−1, t)

∂ξk+N1e

−N2ξnζp(ξ1, . . . , ξn−1), k = n,

where the function ζp(ξ1, . . . , ξn−1) is the same as in Theorem 2. Similarly to (2.13), we get

L(zk) = N1e−N2ξn

[N2

2αnnζp −N2βnζp − 2N2

n∑j=1

αnj∂ζp∂ξj

+ L(ζp)]

− L(∂ψ

∂ξk

)−

n∑i,j=1

∂αij∂ξk

∂2u

∂ξi ∂ξj−

n∑i=1

∂βi∂ξk

∂u

∂ξi− ∂c

∂ξku+

∂f

∂ξk,

hence

L(zk) > N1e−N2ξn

(µ′

3N2

2 −M8

)−M9MQ −M10. (2.20)

It follows from the inequality (2.20) that if

N2 =√M11MQ +M12, (2.21)

then L(zk) > 0 everywhere inside Qp and also for t = T . According to the remark to Theorem 1 of Sect. 1,the function zk(ξ, t) cannot attain its maximal value in Qp, neither inside Qp nor on t = T .

After that, as in Theorem 2, we can see that zk|Sp = N1,

zk|Sp N1, zk|ξn= 1N2 2M1 +

N1

e< N1 for N1 > 4M1,

and zk 2M1 + N12 < N1 if (ξ1, . . . , ξn−1) belongs to the boundary of the piece σp and N1 > 4M1. For t = 0 we

have∂zk∂ξn

∣∣∣∣t=0

=∂2ϕ

∂ξk ∂ξn−N1N2e

−N2ξnζp max∣∣∣∣ ∂2ϕ

∂ξk ∂ξn

∣∣∣∣− N1N2

2e< 0

if

N1 =2eN2

max∣∣∣∣ ∂2ϕ

∂ξk ∂ξn

∣∣∣∣+ 1 =M13. (2.22)

Therefore, the function zk(ξ, t) attains its maximal value on Qp at the points of Sp, where zk ≡ N1, and wehave

∂zk∂ξn

∣∣∣∣Sp

0,

hence∂2u

∂ξk ∂ξn

∣∣∣∣Sp

N1N2.

454

Similar arguments lead to∂2u

∂ξk ∂ξn

∣∣∣∣Sp

−N1N2,

and we get ∣∣∣∣ ∂2u

∂ξk ∂ξn

∣∣∣∣ N1N2 on Sp (k = 1, . . . , n− 1). (2.23)

From Eq. (1.1) (in the variables ξ1, . . . , ξn, t) we obtain

∂2u

∂ξ2n

= − 1αnn

( ∑i+j<2n

αij∂2u

∂ξi ∂ξj+

n∑i=1

βi∂u

∂ξi+ cu− ∂u

∂t− f). (2.24)

On Sp we have ∣∣∣∣∂u∂t∣∣∣∣ =∣∣∣∣∂ψ(ξ, t)∂t

∣∣∣∣ M14. (2.25)

From (2.21)–(2.25) one can obtain

n∑i,j=1

(∂2u

∂ξi ∂ξj

)21/2∣∣∣∣Sp

M15 +M16

√MQ.

Hence n∑i,j=1

(∂2u

∂ξi ∂ξj

)21/2∣∣∣∣S

M17 +M18

√MQ.

Taking into account that n∑

i,j=1

(∂2u

∂xi ∂xj

)21/2∣∣∣t=0

= n∑

i,j=1

(∂2ϕ

∂xi ∂xj

)21/2

M19, we get

MΓ M20 +M21

√MQ. (2.26)

Comparing (2.19) and (2.26) we arrive at the inequality

MQ M22

√MQ +M23,

which leads toMQ M24.

The theorem is proved.

Corollary. Under the suppositions of Theorem 3, the estimate∣∣∂u∂t

∣∣ M1 holds in Q; here M1 depends onthe same quantities as M1.

In fact, by virtue of Eq. (1.1) the derivative ∂u∂t can be expressed by means of the derivatives ∂u

∂xiand ∂2u

∂xi ∂xj

(i, j = 1, . . . , n) and for them we have already obtained an estimate.By similar arguments one can obtain estimates in a closed domain Q for the derivatives of higher order of

u(x, t) (assuming they exist); we only need to require a higher degree of smoothness of the functions aij , bi, c, f ,ϕ, and ψ. First, we have to estimate simultaneously the derivatives ∂3u

∂xk∂xl∂xmand ∂2u

∂xk ∂t(k, l,m = 1, . . . , n); then

the derivatives ∂4u∂xk∂xl∂xm∂xq

and ∂3u∂xk∂xl∂t

(k, l,m, q = 1, . . . , n), and so on.Note that from Bernstein’s estimates one can deduce that a family of uniformly bounded solutions of (1.1)

is compact in the sense of uniform convergence in any closed domain contained in Q \ Γ (see Theorem 1).Results similar to Theorems 1, 2, and 3 are valid also for elliptic second-order equations (see [31, 38]).2. A priori Estimates in the Norms of Holder Spaces. In the study of nonlinear equations, a priori

estimates independent of the smoothness of the coefficients of equations play an important part (see, for exam-ple, [18]). Such estimates for the solutions of linear equations were only recently established ([39,40,124]). Now weshall formulate a theorem giving an a priori estimate for the Holder constant for the solution of a parabolic equation.This theorem is a generalization of a theorem proved by J. Nash [39]. The proof can be found in [39] and [18].

455

Theorem 4. Suppose that u(x, t) is a solution of the equation

n∑i,j=1

∂

∂xi

(Aij(x, t)

∂u

∂xj

)+

n∑i=1

Bi(x, t)∂u

∂xi+ C(x, t)u − ∂u

∂t= F (x, t) (2.27)

in the domain Q, and in Q the following inequalities hold:

|u| M, µ1

n∑i=1

α2i

n∑i,j=1

Aij(x, t)αiαj µ2

n∑i=1

α2i (0 < µ2 < µ1),

|Bi(x, t)| B (i = 1, . . . , n), |F (x, t)| N, |C(x, t)| C1, N + C1M = N1.

Then for (x1, t1) and (x2, t2) belonging to Qδ, 0 < t1 < t2, we have the following inequality:

|u(x2, t2)− u(x1, t1)|

Amax[M+N1

δα, (M+N1)Bα,

M

min(√t1, 1)

]|x2 − x1|α +Amax

[M+N1

δ2β, (M+N1)B2β ,

M

min(√t1, 1)

](t2 − t1)β ;

here A, α, and β are constants, depending only on µ1, µ2, and n, and 0 < α < 12 and 0 < β < 1

4 .

For the solutions of linear second-order parabolic equations, a priori estimates similar to the well-knownestimates of J. Schauder for linear elliptic equations can be proved (see [41]). In the case of one space variable, suchestimates are due to C. Ciliberto [42]. For the equations with several space variables estimates of this type wereestablished by A. Friedman [43–46]. These estimates can be useful in the study of the properties of solutions of linearparabolic equations, such as the dependence of the smoothness of solutions on the smoothness of the coefficients,the compactness of families of solutions in various norms, and many other properties. One can use these estimatesto establish existence theorems for the solutions of the first boundary-value problem for Eq. (1.1) (see [43]) and alsofor quasilinear parabolic equations [18].

We shall use the following notations:

H(u) = supP,Q∈D

|u(P )− u(Q)|d(P,Q)α

, where d(P,Q) = (|x − x′|2 + |t− t′|)1/2.

Let us introduce the norms

|u|α = supD|u|+H(u),

|u|1+α = |u|α +n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣α

,

|u|2+α = |u|α +n∑

=1

∣∣∣∣ ∂u∂xi∣∣∣∣α

+n∑

i,j=1

∣∣∣∣ ∂2u

∂xi ∂xj

∣∣∣∣α

+∣∣∣∣∂u∂t∣∣∣∣α

.

The function u(x, t) belongs to the class Hj+α(D) if the norm |u|j+α is finite, j = 0, 1, 2.

Theorem 5. Suppose the coefficients of Eq. (1.1) in the domain D satisfy the conditions

|aij |α M1, |bi|α M1, |c|α M1,n∑

i,j=1

aijαiαj µn∑i=1

α2i ,

(2.28)

and a solution u(x, t) of Eq. (1.1) belongs to the class H2+α(D). Then

|u|2+α M(|f |α + |ψ|2+α),

where the function ψ is defined in D and u|Γ = ψ; the constant M depends only on M1, µ, and the domain D. Itis also supposed that S belongs to the class A2+α.

456

Let Dδ be a domain consisting of points of D such that their distance from the boundary Γ is greaterthan δ. The function u(x, t) belongs to the class Kj+α(Dδ) if u and its derivatives with respect to x1, . . . , xn up tothe order j belong to the class Hα(Dδ). Define

|u|Kj+α(Dδ) =j∑

l=0

∑l1+...+ln=l

∣∣∣∣∣ ∂lu

∂xl11 . . . ∂xlnn

∣∣∣∣∣α

.

Theorem 6. Suppose the function u(x, t) satisfies Eq. (1.1) and has in D Holder continuous (with expo-nent α) derivatives with respect to xi of the first and second order and also a continuous derivative with respectto t. Let γ be a nonnegative integer. Then, if

|aij |Kγ+α(Dδ/2) M1, |bi|Kγ+α(Dδ/2) M1, |c|Kγ+α(Dδ/2) M1,

and (2.28) holds, we have|u|K2+γ+α(Dδ) M(|f |Kγ+α(Dδ/2) +max

Dδ/2

|u|),

where M depends only on M1, δ, and µ and on the size of the domain D.

The proof of Theorems 5 and 6 (in more general form) are given in [43, 46]. Note that in [45] there isan estimate for |u|1+δ in D in terms of max

D|f | when u|Γ = 0; the estimate depends only on the Hα(D) norms of

the coefficients aij , on the smoothness of aij on S, and on the maximal absolute values of the other coefficients ofEq. (1.1) in D.

The proofs of these theorems are based upon the representation of the solutions of (1.1) by means of the fun-damental solution (see Sect. 4).

3. Integral Estimates. When functional methods for solving boundary-value problems are applied, a prioriestimates in integral norms become very important. We shall use these estimates in Sects. 7 and 8. Theorems 7 and 8can be considered as theorems on the continuous dependence for the solutions of the first boundary-value problemon the initial function and the right-hand term of the equation in corresponding integral norms.

Theorem 7. Suppose the function u(x, t) has continuous derivatives of the first and second order in Q andsatisfies Eq. (2.27) and the initial condition

u(x, 0) = ϕ(x).

Suppose u|S = 0, the coefficients Aij ∈ C1(Q) and Bi, C, and F are bounded in Q. We also suppose thatn∑

i,j=1

Aijαiαj µn∑i=1

α2i , µ > 0, and S ∈ A1. Then for any τ , 0 τ T , we have the inequalities

∫Ω

u2(x, τ) dx +

τ∫0

∫Ω

n∑i=1

(∂u

∂xi

)2

dx dt M1

τ∫0

∫Ω

F 2 dx dt+∫Ω

ϕ2(x) dx, (2.29)

∫Ω

n∑i=1

(∂u

∂xi

)2∣∣∣∣t=τ

dx+

τ∫0

∫Ω

(∂u

∂t

)2

dx dt M2

τ∫0

∫Ω

F 2 dx dt +∫Ω

[ϕ2 +

n∑i=1

(∂ϕ

∂xi

)2]dx

, (2.30)

where the constant M1 depends only on the domain Q, on µ, and on the maximal absolute values of Bi and C, andthe constant M2 depends on the same quantities and on the maximal absolute values of Aij and

∂Aij∂t .

Proof. Multiplying both sides of Eq. (2.27) by u(x, t)e−θt and integrating over the cylinder Qτ = Ω× [0, τ ],we get

∫∫Qτ

e−θtun∑

i,j=1

∂

∂xi

(Aij

∂u

∂xj

)dx dt +

∫∫Qτ

e−θtun∑i=1

Bi∂u

∂xidx dt

+∫∫Qτ

e−θtCu2 dx dt−∫∫Qτ

e−θtu∂u

∂tdx dt =

∫∫Qτ

e−θtFu dx dt. (2.31)

457

This equality needs some transformations. By M3 we shall denote an expression depending on the same

quantities asM1, and by N1 — an expression such that |N1| M3

∫∫Qτ

(F 2 + u2 + |u|

n∑i=1

∣∣ ∂u∂xi

∣∣)e−θt dx dt. The equal-ity (2.31) can be rewritten in the form

−∫∫Qτ

e−θtun∑

i,j=1

∂

∂xi

(Aij

∂u

∂xj

)dx dt+

∫∫Qτ

e−θtu∂u

∂tdx dt = N1. (2.32)

Let us transform the first integral:

−∫∫Qτ

e−θtun∑

i,j=1

∂

∂xi

(Aij

∂u

∂xj

)dx dt =

∫∫Qτ

e−θtn∑

i,j=1

Aij∂u

∂xi

∂u

∂xjdx dt µ

∫∫Qτ

n∑i=1

(∂u

∂xi

)2

e−θt dx dt.

The second integral is equal to∫∫Qτ

e−θtu∂u

∂tdx dt =

∫∫Qτ

12e−θt

∂(u2)∂t

dx dt =12

∫Ω

u2(x, τ)e−θτ dx− 12

∫Ω

u2(x, 0) dx+θ

2

∫∫Qτ

e−θtu2 dx dt.

So we get from (2.32) the inequality

µ

∫∫Qτ

n∑i=1

(∂u

∂xi

)2

e−θt dx dt+e−θT

2

∫Ω

u2(x, τ) dx − 12

∫Ω

ϕ2(x) dx

+θ

2

∫∫Qτ

e−θtu2 dx dt M3

∫∫Qτ

F 2e−θt dx dt+M3

∫∫Qτ

u2e−θt dx dt+M3

∫∫Qτ

|u|n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣ e−θt dx dt.

The last integral can be estimated in the following way:

M3

∫∫Qτ

|u|n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣ e−θt dx dt µ

2

∫∫Qτ

n∑i=1

(∂u

∂xi

)2

e−θt dx dt+nM2

3

2µ

∫∫Qτ

u2e−θt dx dt.

Choose θ = 2nM23

µ + 4M3. Then

µ

2

∫∫Qτ

n∑i=1

(∂u

∂xi

)2

e−θt dx dt+e−θT

2

∫Ω

u2(x, τ) dx +θ

4

∫∫Qτ

e−θtu2 dx dt 12

∫Ω

ϕ2 dx+M3

∫∫Qτ

F 2e−θt dx dt.

By changing e−θt in the left-hand side to e−θT and to the unity in the right-hand side, we obtain (2.29).Let us multiply Eq. (2.26) by ∂u

∂t and integrate over Qτ . We get

∫∫Qτ

[∂u

∂t

n∑i,j=1

∂

∂xi

(Aij

∂u

∂xj

)+

n∑i=1

Bi∂u

∂xi

∂u

∂t+ Cu

∂u

∂t−(∂u

∂t

)2]dx dt =

∫∫Qτ

F∂u

∂tdx dt. (2.33)

Since we have ∫∫Qτ

∂u

∂t

n∑i,j=1

∂

∂xi

(Aij

∂u

∂xj

)dx dt = −

∫∫Qτ

n∑i,j=1

Aij∂u

∂xj

∂2u

∂xi ∂tdx dt

= −12

∫∫Qτ

∂

∂t

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj

)dx dt+

12

∫∫Qτ

n∑i,j=1

∂Aij

∂t

∂u

∂xi

∂u

∂xjdx dt

= −12

∫Ω

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj

)t=τ

dx+12

∫Ω

n∑i,j=1

Aij(x, 0)∂ϕ

∂xi

∂ϕ

∂xjdx+

12

∫∫Qτ

n∑i,j=1

∂Aij

∂t

∂u

∂xi

∂u

∂xjdx dt,

458

we can obtain from (2.33) the inequality

12

∫Ω

( n∑i,j=1

A∂u

∂xi

∂u

∂xj

)t=τ

dx+∫∫Qτ

(∂u

∂t

)2

dx dt

M4

∫∫Qτ

[∣∣∣∣∂u∂t∣∣∣∣

n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣+∣∣∣∣u∂u∂t

∣∣∣∣+∣∣∣∣F ∂u∂t

∣∣∣∣+n∑i=1

(∂u

∂xi

)2]dx dt+

∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx

12

∫∫Qτ

(∂u

∂t

)2

dx dt+M5

∫∫Qτ

[u2 + F 2 +

n∑i=1

(∂u

∂xi

)2]dx dt+

∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx

,

where the constants M4 and M5 depend on the same quantities as M2. Therefore,

µ

2

∫Ω

n∑i=1

(∂u

∂xi

)2∣∣∣∣t=τ

dx+12

∫∫Qτ

(∂u

∂t

)2

dx dt

M5

∫∫Qτ

F 2 dx dt+∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx+∫∫Qτ

[u2 +

n∑i=1

(∂u

∂xi

)2]dx dt

. (2.34)

To estimate the last integral in the right-hand side of (2.34) we shall use the inequality (2.29). Then from (2.34)we get (2.30).

Theorem 8. Suppose that a function u(x, t) ∈ C3(Q) satisfies Eq. (1.1) in Q and the conditionsu(x, 0) = ϕ(x) and u|S = 0. Suppose also that the coefficients aij(x, t) are continuous, bi(x, t), c(x, t), and f(x, t)are bounded in Q, and the boundary of the domain Ω belongs to the class A2. Then the following inequality holds:

∫∫Q

[u2 +

(∂u

∂t

)2

+n∑i=1

(∂u

∂xi

)2

+n∑

i,j=1

(∂2u

∂xi ∂xj

)2]dx dt M

∫∫Q

f2 dx dt+∫Ω

[ϕ2 +

n∑i=1

(∂ϕ

∂xi

)2]dx

; (2.35)

here the constant M depends only on the modulus of continuity of the coefficients aij(x, t), on the least eigenvalueof the matrix ‖aij‖, on the supremums of |aij(x, t)|, |bi(x, t)|, and |c(x, t)|, and on the domain Q.

Proof. First we consider a particular case of this problem, when the coefficients aij are constant, bi = c = 0,and u(x, t) vanishes in a neighborhood of S. In this case, Eq. (1.1) has the form

L0(u) ≡n∑

i,j=1

aij∂2u

∂xi ∂xj− ∂u

∂t= f.

We shall square both sides of the equation, multiply by e−θt, and integrate over the domain Q. The result is

∫∫Q

[e−θt

n∑i,j,k,l=1

aijakl∂2u

∂xi ∂xj

∂2u

∂xk ∂xl− 2e−θt

∂u

∂t

n∑i,j=1

aij∂2u

∂xi ∂xj+ e−θt

(∂u

∂t

)2]dx dt =

∫∫Q

e−θtf2 dx dt.

We transform the first sum by twice integrating by parts and the second sum by integrating by parts once. Takinginto account that the function u(x, t) is zero in a neighborhood of S, we get

∫∫Q

e−θt[ n∑i,j,k,l=1

aijakl∂2u

∂xi ∂xk

∂2u

∂xj ∂xl+ 2

n∑i,j=1

aij∂u

∂xi

∂2u

∂xj ∂t+(∂u

∂t

)2]dx dt =

∫∫Q

e−θtf2 dx dt. (2.36)

459

The integral of the second sum can be transformed in the following way:

2∫∫Q

e−θtn∑

i,j=1

aij∂u

∂xi

∂2u

∂xj ∂tdx dt =

∫∫Q

∂

∂t

(e−θt

n∑i,j=1

aij∂u

∂xi

∂u

∂xj

)dx dt+

∫∫Q

θe−θtn∑

i,j=1

aij∂u

∂xi

∂u

∂xjdx dt

=∫Ω

e−θtn∑

i,j=1

aij∂u

∂xi

∂u

∂xj

∣∣∣∣t=T

t=0

dx+ θ

∫∫Q

e−θtn∑

i,j=1

aij∂u

∂xi

∂u

∂xjdx dt

µ

∫Ω

e−θtn∑i=1

(∂u

∂xi

)2∣∣∣∣t=T

dx−∫Ω

n∑i,j=1

aij∂ϕ

∂xi

∂ϕ

∂xjdx+ θµ

∫∫Q

e−θtn∑i=1

(∂u

∂xi

)2

dx dt; (2.37)

here µ is the least eigenvalue of the matrix ‖aij‖. Note that1

n∑i,j,k,l=1

aijakl∂2u

∂xi ∂xk

∂2u

∂xj ∂xl µ2

n∑i,j=1

(∂2u

∂xi ∂xj

)2

. (2.38)

From (2.36), (2.37), and (2.38) we deduce that

µ2

∫∫Q

n∑i,j=1

(∂2u

∂xi ∂xj

)2

e−θt dx dt+ µθ

∫∫Q

n∑i=1

(∂u

∂xi

)2

e−θt dx dt+∫∫Q

(∂u

∂t

)2

e−θt dx dt

∫∫Q

f2e−θt dx dt+M1

∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx, (2.39)

which is valid for all θ > 0. By Mi (i = 1, 2, . . .) here and below we denote constants depending on the quantitiescited for M in the statement of the theorem.

Now we shall prove inequality (2.39) in the case where the aij are constant, bi = c = 0 and the functionu(x, t) vanishes outside a domain with boundary that can coincide with S only on a plane piece. Without loss ofgenerality, we can assume that this part of S is a part of the plane xn = 0.

To be definite we shall suppose that the domain where u(x, t) does not vanish lies in the half-space xn > 0.As before, we have

∫∫Q

e−θtf2 dx dt =∫∫Q

e−θt(∂u

∂t

)2

dx dt+∫Ω

e−θtn∑

i,j=1

aij∂u

∂xi

∂u

∂xj

∣∣∣∣t=T

t=0

dx

+ θ

∫∫Q

e−θtn∑

i,j=1

aij∂u

∂xi

∂u

∂xjdx dt+

∫∫Q

e−θtn∑

i,j,k,l=1

aijakl∂2u

∂xi ∂xk

∂2u

∂xj ∂xldx dt

+

T∫0

e−θtn∑

i,j,k,l=1

aijakl

( ∫xn=0

∂u

∂xi

∂2u

∂xk ∂xl

dx

dxj−∫

xn=0

∂u

∂xi

∂2u

∂xl ∂xj

dx

dxk

)dt. (2.40)

1The inequality (2.38) follows from the fact that for any symmetric matrix A = ‖aij‖ having the least eigenvalue µ > 0 and anysymmetric matrix B = ‖βij‖ the inequality

n∑i,j,k,l=1

aijaklβikβjl µ2n∑

i,k=1

β2ik (∗)

holds. Inequality (∗) is obvious for diagonal A. Note that the left-hand side of (∗) is the trace of the matrix K = ABAB. Let C bean orthogonal matrix such that CAC−1 is diagonal. Let sp(K) denote the trace of K. Since we have

sp(K) = sp(CKC−1) = sp(CAC−1 · CBC−1 · CAC−1 · CBC−1)

and under an orthogonal transformation the right-hand side of (∗) does not change, (∗) is valid also in the general case.

460

It follows from u|xn=0 = 0 that for i = n, j = n the derivatives ∂u∂xi

and ∂2u∂xi ∂xj

vanish on xn = 0. Moreover,the integrals over the boundary xn = 0 are nonzero only for j = n and k = n. So the nonzero terms of the last sumhave the following form:

n∑k,l=1

annakl

∫xn=0

∂u

∂xn

∂2u

∂xk ∂xl

dx

dxn−

n∑j,l=1

anjanl

∫xn=0

∂u

∂xn

∂2u

∂xl ∂xj

dx

dxn

= 2n∑l=1

annanl

∫xn=0

∂u

∂xn

∂2u

∂xn ∂xl

dx

dxn− 2

n∑l=1

annanl

∫xn=0

∂u

∂xn

∂2u

∂xl ∂xn

dx

dxn= 0.

The remaining integrals in (2.40) can be estimated as in the first case, and we obtain (2.39) from (2.40).Let us prove the inequality (2.35) for the general case. The domain Ω can be covered by a finite number of

overlapping domains Ωl (l = 1, . . . , N1) such that in any of them there exists a smooth transformation of independentvariables xi transforming the part of the boundary of Ωl lying on σ to a plane. Under such a transformation, the formof the equation and the smoothness of the coefficients do not change. We shall call the new independent variables ξi,and the coefficients aij(ξ, t), bi(ξ, t), and c(ξ, t) correspondingly. The least eigenvalue of the matrix ‖aij(ξ, t)‖ isµ > 0, and the maximum modulus of all the coefficients of the transformed equation is denoted by M2.

The idea of the proof of estimate (2.35) in the general case is as follows. The function u(x, t) is represented asa sum of the functions uk(x, t); each of these functions is nonzero only in a small domain. The domain can be interioror lying at the boundary. In a small domain, the part of the boundary of Ωl belonging to S can be transformed toa plane and the equation with variable coefficients can be approximated by an equation with constant coefficients.For this approximation Eq. (2.35) is already obtained.

To realize this plan, we cover the domain Q by N overlapping small domains Qk such that in any two pointsof Qk the difference of the values of the coefficients aij(ξ, t) does not exceed µ

2(n+1) and each domain Qk belongs toone of the domains Ωl × (0, T ).

Take a so-called partition of unity (see [47, p. 162]). This means that the unity is represented in the form

1 ≡N∑k=1

hk(x, t), where hk(x, t) is a non-negative sufficiently smooth function that is nonzero only in the domain Qk

(k = 1, . . . , N). We have

u(x, t) =N∑k=1

u(x, t)hk(x, t) =n∑

k=1

uk(x, t).

Let L(uk) = fk(x, t). In the new coordinates ξ, t, this equation can be rewritten asn∑

i,j=1

aij(ξ0k, t0k)∂2uk∂ξi ∂ξj

− ∂uk∂t

= f +n∑

i,j=1

[aij(ξ0k, t0k)− aij(ξ, t)]∂2uk∂ξi ∂ξj

−n∑i=1

bi(ξ, t)∂uk∂ξi− c(ξ, t)uk ≡ gk(ξ, t),

where (ξ0k, t0k) is some point in Qk. We shall estimate the right-hand side of the equation we obtained:

g2k (n+ 1)2

[f2k +

µ2

4(n+ 1)2

n∑i,j=1

(∂2uk∂ξi ∂ξj

)2

+M22

n∑i=1

(∂uk∂ξi

)2

+M22u

2k

],

since |aij(ξ0k, t0k)− aij(ξ, t)| µ2(n+1) . The function uk(ξ, t) satisfies a parabolic equation with constant coefficients

and the right-hand side gk and is nonzero only in one of the domains Qk. Therefore, according to what was provedbefore, for the function uk(ξ, t) the inequality (2.39) holds:

µ2

∫∫Q

n∑i,j=1

(∂2uk∂ξi ∂ξj

)2

e−θt dξ dt+ µθ

∫∫Q

n∑i=1

(∂uk∂ξi

)2

e−θt dξ dt+∫∫Q

(∂uk∂t

)2

e−θt dξ dt

∫∫Q

g2ke−θt dξ dt+M1

∫Ω

n∑i=1

(∂uk(ξ, 0)∂ξi

)2

dξ (n+ 1)2∫∫Q

[f2k +M2

2

n∑i=1

(∂uk∂ξi

)2

+M22u

2k

]e−θt dξ dt

+µ2

4

∫∫Q

n∑i,j=1

(∂2uk∂ξi ∂ξj

)2

e−θt dξ dt+M1

∫Ω

n∑i=1

(∂uk(ξ, 0)∂ξi

)2

dξ.

461

Hence∫∫Q

[ n∑i,j=1

(∂2uk∂ξi ∂ξj

)2

+ θ

n∑i=1

(∂uk∂ξi

)2

+(∂uk∂t

)2]e−θt dξ dt

M3

∫∫Q

[f2k +

n∑i=1

(∂uk∂ξi

)2

+ u2k

]e−θt dξ dt+

∫Ω

n∑i=1

(∂uk(ξ, 0)∂ξi

)2

dξ

. (2.41)

Making the inverse transformation from ξi to xi, we can obtain the estimate (2.41) also in the old coordinates(perhaps, with another constant M3).

Because 0 hk(x, t) 1, we have

u2k(x, t) = u2(x, t)h2

k(x, t) u2(x, t). (2.42)

For the derivative ∂uk∂xi

we obtain the following estimate:

(∂uk∂xi

)2

=[∂(hku)∂xi

]2

=(hk

∂u

∂xi+ u

∂hk∂xi

)2

2h2k

(∂u

∂xi

)2

+M4u2 2

(∂u

∂xi

)2

+M4u2. (2.43)

We estimate also fk(x, t):

[fk(x, t)]2 = [L(uk)]2 = [L(hku)]2 =[hkL(u) + uL(hk) + 2

n∑i,j=1

aij∂u

∂xi

∂hk∂xj− cuhk

]2

M5

[L(u)]2 + u2 +

n∑i=1

(∂u

∂xi

)2=M5

[f2 + u2 +

n∑i=1

(∂u

∂xi

)2]. (2.44)

Using the inequality (2.41) in the old coordinates and the estimates (2.42), (2.43) and (2.44), we obtainthe following inequality:

∫∫Q

e−θt[ n∑i,j=1

(∂2uk∂xi ∂xj

)2

+ θ

n∑i=1

(∂uk∂xi

)2

+(∂uk∂t

)2]dx dt

M6

∫∫Q

e−θt[f2 +

n∑i=1

(∂u

∂xi

)2

+ u2

]dx dt+

∫Ω

[ n∑i=1

(∂u(x, 0)∂xi

)2

+ u2(x, 0)]dx

. (2.45)

We have u(x, t) =N∑k=1

uk(x, t); hence

u2(x, t) M7

N∑k=1

u2k(x, t),

(∂u

∂t

)2

M7

N∑k=1

(∂uk∂t

)2

,

(∂u

∂xi

)2

M7

N∑k=1

(∂uk∂xi

)2

,

(∂2u

∂xi ∂xj

)2

M7

N∑k=1

(∂2uk∂xi ∂xj

)2

.

Using these estimates and summing the inequalities (2.45) with respect to k from 1 to N , we come to

∫∫Q

e−θt[ n∑i,j=1

(∂2u

∂xi ∂xj

)2

+ θ

n∑i=1

(∂u

∂xi

)2

+(∂u

∂t

)2]dx dt

M8

∫∫Q

e−θt[f2 +

n∑i=1

(∂u

∂xi

)2

+ u2

]dx dt+

∫Ω

[ n∑i=1

(∂ϕ

∂xi

)2

+ ϕ2

]dx

. (2.46)

462

Since u|S = 0, u(x1, x2, . . . , xn, t) =x1∫x0

1

∂u∂x1

(ζ, x2, . . . , xn, t) dζ, where the point (x01, x2, . . . , xn, t) ∈ S. There-

fore,

u2(x, t) d

x1∫x0

1

(∂u

∂x1

)2

dζ,

where d is the diameter of the domain Ω. Integrating this inequality over the cylinder Q, we get

∫∫Q

u2(x, t) dx dt M6

∫∫Q

n∑i=1

(∂u

∂xi

)2

dx dt. (2.47)

It follows from the inequalities (2.46) and (2.47) that

∫∫Q

e−θt[u2 + θ

n∑i=1

(∂u

∂xi

)2

+n∑

i,j=1

(∂2u

∂xi ∂xj

)2

+(∂u

∂t

)2]dx dt

M10

∫∫Q

e−θt[f2 +

n∑i=1

(∂u

∂xi

)2]dx dt+

∫Ω

[ϕ2 +

n∑i=1

(∂ϕ

∂xi

)2]dx

.

If we take θ = 2M10, we obtain the required estimate (2.35).

Remark. The estimate (2.35) holds not only in the cylinder Q but also in a noncylindrical domain D ifthe boundary S belongs to the class A2.

An estimate similar to (2.35) was first obtained by S. N. Bernstein in 1908 for the solution of the Dirichletproblem for a general elliptic second-order equation with two independent variables ([48], see also [49]). Anotherproof of (2.35) for a parabolic second-order equation is given in [50, 51]. In these papers, the authors use similarestimates of the solution to the Dirichlet problem for second-order elliptic equations with several independentvariables; these last estimates are obtained in [52, 53].

Note that estimates similar to (2.35) in Lp (p > 1) norms and also estimates for elliptic and parabolicequations of higher orders require essentially new approaches. The approach connected with the “parametrix”(that is, almost an inverse operator) of a problem, can be found in [54–57].

In [58,59], to obtain estimates for elliptic equations and systems the approach is based on the use of Fouriertransform for the equations with constant coefficients; after that the equation with variable coefficients is comparedto an equation with constant coefficients in small domains, similarly to the reasoning we used in proving (2.35).

Estimates of the type (2.35) are widely used in proving the existence for boundary-value problems. Weshall use these estimates in Sects. 7 and 8. Similar estimates in Lp norms are used also in the study of nonlinearequations [55].

3 SOLVING OF BOUNDARY-VALUE PROBLEMS BY THE ROTHE METHOD. CAUCHY PROBLEM

In this section, we construct the solutions of the main problems for Eq. (1.1) with several space variablesby a method similar to the method of straight lines for the equation with one space variable. E. Rothe [60] usedthis method to solve the first boundary-value problem for a parabolic equation with two independent variables.A review of works where Rothe’s method is used for two or more independent variables to prove the solvability ofboundary-value problems and to construct their approximate solutions can be found in [61]. Note that in [62, 63]the solutions of the first and second boundary-value problems for quasilinear parabolic equations with several spacevariables are constructed by means of the Rothe method.

Using this method we shall construct solutions of the first and second boundary-value problems for Eq. (1.1)in the cylinder Q = Ω× (0, T ). The construction is based on theorems of existence of solutions to the correspondingproblems for elliptic equations. In this section, we also study the smoothness of a solution of the first boundary-value problem in accordance with the smoothness of the data (coefficients of the equation, the boundary σ ofthe domain Ω, the initial and boundary functions). The first and second boundary-value problems are studied for

463

a cylindrical domain Q and a noncylindrical domain D under very weak assumptions on the boundaries of thesedomains (existence of the barrier functions). The existence of the solution of the Cauchy problem for Eq. (1.1) ina layer H will be proved as a corollary to the existence theorem for the solution of the first boundary-value problem.For the case of the Cauchy problem we shall suppose only that c(x, t) M and |f(x, t)| M and any growth of

the coefficients aij(x, t) and bi(x, t) asn∑i=1

x2i → ∞ is admitted. Examples are given showing that the suppositions

about c and f are essential for the solvability of the Cauchy problem in the class of bounded functions.1. Construction of a Solution of the First Boundary-Value Problem, Smooth up to the Bound-

ary of the Domain. Consider the first boundary-value problem for Eq. (1.1) in the cylinder Q = Ω× (0, T ) withthe following conditions:

u(x, 0) = ϕ(x), u|S = ψ(x, t). (3.1)

Theorem 1. There exists in the cylinder Q a unique solution of the problem (1.1), (3.1), continuous in Qtogether with its derivatives, appearing in Eq. (1.1), if the following suppositions are fulfilled.

(1)n∑

i,j=1

aij(x, t)αiαj µn∑i=1

α2i for (x, t) ∈ Q and any α1, . . . , αn; here µ = const > 0.

(2) The coefficients aij , bi, and c of Eq. (1.1) and its right-hand side f in Q belong to the class C4+λ withrespect to x1, . . . , xn and are Lipschitz continuous with respect to t together with their derivatives of the first andthe second order in x1, . . . , xn.

(3) The boundary σ of the domain Ω belongs to the class A6+λ.(4) ϕ(x) ∈ C6+λ(Ω); ψ ∈ C6+λ as a function of local coordinates ξ1, . . . , ξn−1 on the surface σ. The

derivatives ∂ψ∂t , ∂2ψ

∂ξi ∂t(i = 1, . . . , n− 1) exist; these derivatives and also the derivatives of ψ up to the fourth order

with respect to ξ1, . . . , ξn−1 are Lipschitz continuous with respect to t.(5) Consistency conditions are fulfilled:

ψ(x, 0) = ϕ(x) for x ∈ σ, (3.2)[ n∑i,j=1

aij(x, 0)∂2ϕ

∂xi ∂xj+

n∑i=1

bi(x, 0)∂ϕ

∂xi+ c(x, 0)ϕ− f(x, 0)

]x∈σ

=∂ψ

∂t(x, 0)

∣∣∣∣x∈σ

. (3.3)

Proof. Without loss of generality, one can suppose that c(x, t) < 0 in Q (if this is not so, this can be achievedby the transformation u = veαt). We can decompose the interval [0, T ] into m equal parts and instead of Eq. (1.1)we can take the following system of elliptic equations:

L∆t(u) ≡n∑

i,j=1

aij(x, k∆t)∂2u(x, k∆t)∂xi ∂xj

+n∑i=1

bi(x, k∆t)∂u(x, k∆t)

∂xi

+ c(x, k∆t)u(x, k∆t) − u(x, k∆t)− u(x, (k − 1)∆t)∆t

= f(x, k∆t) (3.4)

with the conditionsu(x, k∆t) = ψ(x, k∆t) for x ∈ σ, (3.5)

where ∆t = Tm , k = 1, . . . ,m; when k = 0 we take u(x, 0) = ϕ(x). The functions u(x, k∆t) can be found

consecutively by solving the Dirichlet problem for a second-order elliptic equation. According to Schauder’s theorem[2, p. 150], under the suppositions of Theorem 1 on the coefficients of (1.1), the boundary σ, and the functions ϕ(x)and ψ(x, t), for all k there is a unique solution u(x, k∆t) of the problem (3.4), (3.5), belonging to the class C6+λ(Ω)as a function of x. We shall prove that for sufficiently small ∆t the families of functions

u(x, k∆t),∂u(x, k∆t)

∂xi,

∆u∆t≡ u(x, k∆t)− u(x, (k − 1)∆t)

∆t,

∂2u(x, k∆t)∂xi ∂xj

,∆∆t

(∂u

∂xi

)≡ 1

∆t

[∂u(x, k∆t)

∂xi− ∂u(x, (k − 1)∆t)

∂xi

],

∂3(x, k∆t)∂xi∂xj∂xl

,∆∆t

(∂2u

∂xi ∂xj

)≡ 1

∆t

[∂2u(x, k∆t)∂xi ∂xj

− ∂2u(x, (k − 1)∆t)∂xi ∂xj

],

∆2u

∆t2≡ u(x, k∆t)− 2u(x, (k − 1)∆t) + u(x, (k − 2)∆t)

∆t2(i, j, l = 1, . . . , n)

(3.6)

464

are uniformly bounded. For k∆t t (k + 1)∆t we linearly extend the functions u(x, k∆t); the resultingfunctions are called u∆t(x, t). Because the functions (3.6) are uniformly bounded, we can select from the familyu∆t(x, t) a subsequence that is uniformly converging in Q as ∆t → 0; the limit u(x, t) satisfies (1.1) in Q andthe conditions (3.1) on Γ.

First we shall show that in Q the functions u(x, k∆t) are uniformly bounded with respect to ∆t.Let ∆t be fixed; if u(x, k∆t) attains its positive maximum (as a function of x and k) at a point lying in Q\Γ,

then at this point we have, owing to (3.4),

−c(x, k∆t)u(x, k∆t) + f(x, k∆t) 0,

whence

u(x, k∆t) maxQ

∣∣∣∣fc∣∣∣∣ =M1.

It is possible to establish a similar estimate for u(x, k∆t) at a point of its negative minimum. If the maximum of|u(x, k∆t)| is attained at a point of Γ, then we have at this point

|u(x, k∆t)| maxΓ|ϕ(x)|, |ψ(x, t)| =M2.

Thus for any x and k we have the inequality

|u(x, k∆t)| M3,

with the constant M3 not depending on ∆t.Now we shall prove that ∂u(x,k∆t)

∂xl(l = 1, . . . , n) is uniformly bounded with respect to ∆t in Q.

The estimate forn∑l=1

[∂u(x,k∆t)∂xl

]2 on S is obtained in the same way as in Theorem 2 of Sect. 2. To find

the estimate for this sum in Q we take the following auxiliary function:

w(x, k) = e−Nk∆tn∑l=1

[∂u(x, k∆t)

∂xl

]2

+ eN(T−k∆t);

N will be chosen later.Set ∆w = w(x, k) − w(x, k − 1); we have

L∆t,N(w) ≡n∑

i,j=1

aij∂2w

∂xi ∂xj+

n∑i=1

bi∂w

∂xi+ cw − e−N∆t∆w

∆t.

It is easy to see (as in Theorem 1 of Sect. 1) that if L∆t,N(w) > 0 in Q \ Γ, the function w(x, k) cannot attaina maximum in Q \ Γ.

To calculate L∆t,N(w) we shall use the identities

∆(v1v2)∆t

= v1∆v2

∆t+ v2(x, k − 1)

∆v1

∆t,

∆(v2)∆t

= 2v∆v∆t− (∆v)2

∆t.

We have

L∆t,N(w) = e−Nk∆t

2

n∑i,j,l=1

aij∂2u

∂xl ∂xi

∂2u

∂xl ∂xj+ 2

n∑l=1

∂u

∂xl

[ n∑i,j=1

aij∂3u

∂xl∂xi∂xj+

n∑i=1

bi∂2u

∂xl ∂xi− 1

∆t∆(∂u

∂xl

)]

+1∆t

n∑l=1

[∆(∂u

∂xl

)2]+ c

n∑l=1

(∂u

∂xl

)2

− e−N∆teNk∆t∆(e−Nk∆t)∆t

n∑l=1

(∂u

∂xl

)2

+(c− e−N∆t − 1

∆t

)eNT

e−Nk∆t

(2µ− εM4)

n∑l,i=1

(∂2u

∂xl ∂xi

)2

+(1−e−N∆t

∆t− M5

ε−M5

) n∑l=1

(∂u

∂xl

)2

+(1−e−N∆t

∆t−M6

)eNT −M7

;

465

here the Mi are constants independent of N and ∆t. Here we have used equations obtained by differentiation ofboth sides of (3.4) with respect to xl (compare to (2.16) and (2.17)).

Take ε < 2µM4

, and after this N = 4(M5 + M5

ε +M6 +M7

). Then for all ∆t < 1

N everywhere in Q \ Γthe inequality L∆t,N(w) > 0 is valid, owing to 1−e−N∆t

∆t > Ne−1 > N4 .

Therefore,

w(x, k) maxΓ

w(x, k) maxΓ

n∑l=1

(∂u(x, k∆t)

∂xl

)2

+ eNT M8.

Hencen∑l=1

(∂u(x, k∆t)

∂xl

)2

M8eNT =M9,

where M9 does not depend on ∆t.Now we shall prove that the functions ∂2u(x,k∆t)

∂xl ∂xp(l, p = 1, . . . , n) are uniformly bounded in Q with respect

to ∆t.Consider in Q the auxiliary function

w1(x, k) = e−N1k∆tn∑

l,p=1

(∂2u(x, k∆t)∂xl ∂xp

)2

+ eN1(T−k∆t).

Using the equations obtained by twice differentiating both sides of (3.4) with respect to xl and xp, we obtain

L∆t,N1(w1) e−N1k∆t

(2µ− εM10)

n∑i,l,p=1

(∂3u

∂xi∂xl∂xp

)2

+(1− e−N1∆t

∆t− M10

ε−M10

) n∑l,p=1

(∂2u

∂xl ∂xp

)2

+1− e−N1∆t

∆t−M10

> 0,

if

ε <2µM10

, N1 > 4(M10

ε+M10

), and ∆t <

1N1

.

Therefore, if N1 is chosen as above, we have in Q the inequality

w1(x, k) maxΓ

w1.

Setting

MQ = maxQ

n∑l,p=1

(∂2u(x, k∆t)∂xl ∂xp

)21/2

, MΓ = maxΓ

n∑l,p=1

(∂2u(x, k∆t)∂xl ∂xp

)21/2

,

we getMQ M11(MΓ + 1). (3.7)

Then, as in Theorem 3 of Sect. 2, we can estimaten∑

l,p=1

(∂2u

∂xl ∂xp

)2 on S. As a result we obtain

MΓ M12

(√MQ + 1

). (3.8)

It follows from (3.7) and (3.8) that MQ M13.As a consequence of Eqs. (3.4), the difference quotients

∆u∆t

=u(x, k∆t)− u(x, (k − 1)∆t)

∆t

are also uniformly bounded in Q with respect to ∆t.

466

Let us estimate in Q the derivatives of the form ∂3u(x,k∆t)∂xl∂xp∂xq

and ∂∂xl

(∆u∆t

)(l, p, q = 1, . . . , n).

Set

M ′Q = maxQ

n∑l,p,q=1

(∂3u

∂xl∂xp∂xq

)21/2

, M ′′Q = maxQ

n∑i=1

[∂

∂xl

(∆u∆t

)]21/2

,

and denote by M ′Γ and M ′′Γ the maximum values of these sums on Γ correspondingly. Using the function

w2(x, k) = e−N2k∆tn∑

l,p,q=1

(∂3u

∂xl∂xp∂xq

)2

+ eN2(T−k∆t),

with N2 sufficiently large, and reasoning as we did above, we can obtain the estimate

M ′Q M14(M ′Γ + 1). (3.9)

Then, using the functions

z±lq(ξ, k) =∂2u(ξ1, . . . , ξn−1, ξn, k∆t)

∂ξl ∂ξq− ∂2ψ(ξ1, . . . , ξn−1, k∆t)

∂ξl ∂ξq±N3e

−N4ξnζp(ξ1, . . . , ξn−1) (l, q = 1, . . . , n− 1)

(the notations are the same as in Theorem 2 of Sect. 2), we show that on any piece Sp the inequality

n−1∑l,q=1

(∂3u

∂ξl∂ξq∂ξn

)21/2∣∣∣∣Sp

M15

(√M ′Q + 1

)(3.10)

holds.Differentiating Eqs. (3.4) in the vicinity of Sp with respect to ξ1, . . . , ξn, t, we get

αnn∂3u

∂ξl∂ξ2n

= −∑

i+j<2n

αij∂3u

∂ξi∂ξj∂ξl+

∂

∂ξl

(∆u∆t

)+Rl(u); (3.11)

here Rl(u) denotes the sum of terms containing derivatives of u with respect to ξ1, . . . , ξn of order not higher thanthe second. From (3.10) and (3.11) it follows that

M ′Γ M16

(√M ′Q +M ′′Γ + 1

). (3.12)

Comparing (3.9) and (3.12), we getM ′Q M12(M ′′Γ + 1). (3.13)

Now let us estimate M ′′Γ . We extend u(x, k∆t) to k = −1 and x ∈ Ω by setting

u(x,−∆t) = ϕ(x) −∆t n∑

i,j=1

aij(x, 0)∂2ϕ(x)∂xi ∂xj

+n∑i=1

bi(x, 0)∂ϕ(x)∂xi

+ c(x, 0)ϕ(x) − f(x, 0). (3.14)

In the vicinity of Sp for 0 k m we consider the auxiliary functions

z±(ξ, k) =∆u(ξ1, . . . , ξn−1, ξn, k∆t)

∆t− ∆ψ(ξ1, . . . , ξn−1, k∆t)

∆t±N5e

−N6ξnζp(ξ1, . . . , ξn).

Because of the smoothness of ψ(x, t) and the consistency conditions (3.2), (3.3) the value L∆t

(∆ψ∆t

)is uniformly

bounded with respect to ∆t. Using the equations obtained by applying to both sides of (3.4) the operator ∆∆t , we

arrive at L∆t(z+(ξ, k)) > 0 and L∆t(z−(ξ, k)) < 0 in the neighborhood of Sp, if N5 and N6 are chosen in a correctway. Then, as in Theorem 2 of Sect. 2, we can get∣∣∣∣ ∂∂ξn

(∆u∆t

)∣∣∣∣Sp

M18.

467

Because of the smoothness of ψ(x, t) we also have

∣∣∣∣ ∂∂ξi(∆u∆t

)∣∣∣∣Sp

M19 (i = 1, . . . , n− 1).

Finally, the smoothness of ϕ(x) and equality (3.14) lead to

∣∣∣∣ ∂∂ξi(∆u∆t

)∣∣∣∣t=0

M20 (i = 1, . . . , n).

Thus we haveM ′′Γ M21. (3.15)

It follows from (3.13) and (3.15) thatM ′Q M22,

and this means that the derivatives ∂3u∂xl∂xp∂xq

are bounded uniformly in Q with respect to ∆t. It follows fromequations obtained by differentiating (3.4) with respect to xl that the derivatives ∂

∂xl

(∆u∆t

)also are uniformly

bounded in Q.Then by the same arguments we used to obtain bounds for ∂3u(x,k∆t)

∂xl∂xp∂xqand ∂

∂xl

(∆u(x,k∆t)∆t

)we can prove

the uniform boundedness of ∂4u(x,k∆t)∂xl∂xp∂xq∂xr

and ∂2

∂xl ∂xp

(∆u(x,k∆t)∆t

)(l, p, q, r = 1, . . . , n). The uniform boundedness of

the difference quotients ∆2u∆t2 = ∆

∆t

(∆u(x,k∆t)∆t

)follows from equations obtained by applying the operator ∆

∆t to bothsides of (3.4). So we can prove in Q the uniform boundedness of all functions (3.6) with respect to ∆t.

Now we shall consider a family of functions u∆t(x, t); they are continuous in Q. Because the functionsu(x, k∆t), their derivatives ∂u(x,k∆t)

∂xi(i = 1, . . . , n), and difference quotients ∆u(x,k∆t)

∆t are uniformly bounded in Q,the functions u∆t(x, t) are equicontinuous in Q. According to the Arcela theorem, one can single out from the familyu∆t(x, t) a subsequence um(x, t), converging uniformly in Q to some function u(x, t).

The families ∂u∆t(x,t)∂xi

, ∂2u∆t(x,t)∂xi ∂xj

, and ∆u∆t(x,t)∆t are also uniformly bounded and equicontinuous in Q; this

follows from the uniform boundedness of functions (3.6). Therefore, a new subsequence can be chosen in the sequenceum(x, t) (it will also be called um(x, t)), such that the corresponding subsequences ∂um(x,t)

∂xi, ∂2um(x,t)

∂xi ∂xj, and ∆um(x,t)

∆t

converge uniformly in Q to some continuous functions ui(x, t), uij(x, t), and u0(x, t) respectively.

It is easy to show that ui(x, t) = ∂u(x,t)∂xi

, uij(x, t) = ∂2u(x,t)∂xi ∂xj

, and u0(x, t) = ∂u(x,t)∂t . (The proof is similar

to one given for the heat equation in [19, p. 361–362].) Then for the chosen subsequence we pass to the limit asm → ∞ in Eqs. (3.4) and the result is that the function u(x, t) everywhere in Q satisfies Eq. (1.1). Because ofthe uniform convergence of um(x, t), the function u(x, t) satisfies also the boundary conditions (3.1) on Γ.

The uniqueness of the solution of the problem (1.1), (3.1) is proved in Sect. 1.

Remark. If greater smoothness is required from the data of problem (1.1), (3.1), then the function u(x, t)constructed in Theorem 1 will also be more smooth than is stated in this theorem. Namely, to prove the existencein Q \ S of continuous derivatives of the form

∂p+qu(x, t)∂xp1

1 . . . ∂xpnn ∂tq, where p+ 2q r (r > 2) (3.16)

it is sufficient to suppose that ϕ(x) ∈ Cr+4+λ(Ω) and that in any closed domain Q∗ ⊂ Q \ S there exist continuousderivatives that are Holder continuous in x:

∂p+qaij(x, t)∂xp1

1 . . . ∂xpnn ∂tq, where p+ 2q r + 2, q

[r

2

];

the same is supposed about the derivatives of bi(x, t), c(x, t), and f(x, t).

468

In fact, under such suppositions according to a theorem by E. Hopf [2, p. 152] the solution u(x, k∆t) ofproblem (3.4), (3.5) for any ∆t and for k = 1, . . . ,m = T

∆t in any closed domain Q∗ ⊂ Ω belongs to the classCr+4+λ as a function of x1, . . . , xn. In the cylinder

Rδ

n∑i=1

(xi − x0i )

2 δ2, 0 t T,

consider the auxiliary function

w(x, k) = Φδ(x)n∑l=1

(∂u(x, k∆t)

∂xl

)2

+N7u2(x, k∆t) + eN8(T−k∆t); (3.17)

here x0 = (x01, . . . , x

0n) ∈ Ω, Φδ(x) =

[δ2 −

n∑i=1

(xi − x0i )

2]2, δ > 0 is sufficiently small, and the constants N7 and N8

are sufficiently large. As in Theorem 1 of Sect. 2, we can show that L∆t(w) > 0 in the cylinder Rδ and therefore

in Rδ/2 the inequalityn∑l=1

(∂u(x,k∆t)∂xl

)2 M23 holds; M23 does not depend on ∆t. Similarly, one can prove that

the families of derivatives and difference quotients

∂p

∂xp11 . . . ∂xpnn

∆qu(x,∆t)∆tq

, where p+ 2q r + 2, (3.18)

are uniformly bounded with respect to ∆t in Rδ1 (δ1 > 0) (see the remark to Theorem 1 of Sect. 2). It is easy todeduce from the boundedness of these families that the limit function u(x, t) in Q \ S has continuous derivatives ofthe form (3.16).

To prove the existence of continuous derivatives (3.16) in the closed cylinder Q we need only additionalsuppositions that the boundary σ of the domain Ω belongs to the class Ar+4+λ, ϕ(x) ∈ Cr+4+λ(Ω) and ψ belongsto the class Cr+4+λ as a function of local coordinates ξ1, . . . , ξn−1 on σ and has on S continuous derivatives ofthe form

∂p+qψ

∂ξp11 . . . ∂ξ

pn−1n−1 ∂t

q, where p+ 2q r + 4, q

[r2

]+ 1.

Besides this, the functions ϕ and ψ on σ must satisfy some new consistency conditions. These conditions arededuced from the equation obtained by applying differential operators ∂p+q

∂xp11 ...∂xpnn ∂tq

with p + 2q r − 2 to (1.1)and then expressing the derivatives of u(x, t) in terms of ϕ and ψ.

In fact, under such suppositions one can show that in Q all the derivatives and difference quotients ofthe form (3.18) are uniformly bounded with respect to ∆t. This is done by the methods used in Theorem 1. Then,using the same arguments as above, we can prove the existence of continuous derivatives (3.16) in Q.

2. First Boundary-Value Problem for Noncylindrical Domains. In this section, a solution tothe first boundary-value problem for Eq. (1.1) is constructed under more general suppositions about the boundaryof the domain as well as about the initial and boundary functions.

Consider Eq. (1.1) in the domain D described in Sect. 1. A function vP0(x, t) is called a barrier at a pointP0(x0, t0) ∈ Γ if it has the following properties:

(A) vP0 (x, t) is defined and continuous in the intersectionGP0 of the closed domain D with some neighborhoodof the point P0.

(B) vP0(x0, t0) = 0, vP0(x, t) > 0 at all the points of GP0 different from P0.

(C) L(vP0) ≡n∑

i,j=1

aij(x, t)∂2vP0∂xi ∂xj

+n∑i=1

bi(x, t)∂vP0∂xi

+ c(x, t)vP0 −∂vP0∂t < 0 everywhere inside GP0 for all

continuous functions aij(x, t), bi(x, t), and c(x, t) such that the absolute values of the differences between them andaij(x, t), bi(x, t), and c(x, t) respectively are sufficiently small in GP0 .

The following theorem is an auxiliary one.

Theorem 2. In the cylinder Q = Ω×(0, T ), there exists a unique solution u(x, t) of the problem (1.1), (1.3),bounded and continuous in Q \ σ and satisfying Eq. (1.1) in Q \ Γ and the boundary conditions (3.1) on Γ \ σ, ifthe following conditions are fulfilled:

469

(1) Supposition (1) of Theorem 1.(2) The functions aij , bi, c, and f are continuous in Q, and in Q \ Γ the derivatives of the form

∂p+qaij∂xp1

1 . . . ∂xpnn ∂tq, where p+ 2q 4, q 1, (3.19)

exist and are Holder continuous with respect to x in any closed domain contained in Q \ Γ. The same is supposedabout the derivatives of this form of bi, c, and f .

(3) The boundary σ of the domain Ω has the following property: there exists a number γ, 0 < γ < n, suchthat for any ε > 0 the boundary σ can be covered by a finite number N of n-dimensional balls of radii 61, . . . , 6N

such thatN∑l=1

6γl < ε.

(4) The function ϕ(x) is bounded and continuous inside Ω and the function ψ(x, t) is bounded and continuouson S.

(5) There exists a barrier for every point P0 ∈ Γ.If at some point P1 ∈ σ the consistency condition ϕ(P1) = ψ(P1) is fulfilled, then the function u(x, t) is

continuous at the point P1 and satisfies the condition u(P1) = ϕ(P1).If in Q \ Γ there exist continuous derivatives of the form

∂p+qaij∂xp1

1 . . . ∂xpnn ∂tq, p+ 2q r + 2, q

[r2

], (3.20)

and the same derivatives of functions bi, c, and f , then the function u(x, t) in Q \ Γ has continuous derivatives ofthe form

∂p+qu

∂xp11 . . . ∂xpnn ∂tq

, p+ 2q r. (3.21)

If besides this in a domain Ω′ contained inside Ω the function ϕ(x) belongs to the class Cr+2 and the deriva-tives of the form (3.20) of the functions aij , bi, c, and f are continuous on the set t = 0, x ∈ Ω′, then the deriva-tives (3.21) also are continuous on this set.

Proof. We shall construct sequences of uniformly bounded and equicontinuous in Q functions amij , bmi , cm,

and fm that converge uniformly to aij , bi, c, and f , respectively, together with derivatives of the form (3.19) in anyclosed domain contained in Q \ Γ. Functions fm are constructed so that fm(x, t) = 0 at all the points situated ata distance not greater than 1

m (m = 1, 2, . . .) from σ. We can extend the functions ϕ(x) and ψ(x, t), defined on Γ, toall of the cylinder Q so that the resulting function U(x, t) will be bounded and continuous in Q\σ. Then we constructa sequence of uniformly bounded infinitely differentiable functions Um(x, t) that converge uniformly to U(x, t) inany closed domain contained in Q \ σ, and Um(x, t) = 0 at all the points situated at a distance from σ not greaterthan 1

m (m = 1, 2, . . .). Then we approximate the domain Ω from inside by a sequence of expanding domains Ωm

having the following properties: the boundary of Ωm belongs to the class A6+λ, Ωm ⊂ Ωm+1,∞⋃

m=1Ωm = Ω, and all

the points of Ω at a distance greater than 12m from σ belong to the domain Ωm (m = 1, 2, . . .).

Denote by σm the boundary of Ωm, by Qm the cylinder Ωm×(0, T ), and by Sm its side boundary σm×(0, T ].According to Theorem 1, for each m in Qm there exists a unique solution um(x, t) of the equation

Lm(u) ≡n∑

i,j=1

amij (x, t)∂2u

∂xi ∂xj+

n∑i=1

bmi (x, t)∂u

∂xi+ cm(x, t)u − ∂u

∂t= fm(x, t), (3.22)

satisfying the conditions

um(x, 0) = Um(x, 0) for x ∈ Ωm, um(x, t) = Um(x, t) on Sm; (3.23)

note that the consistency conditions (3.2), (3.3) are fulfilled because the functions Um and fm vanish in a neigh-borhood of σm.

According to Theorem 5 of Sect. 1, the functions um are uniformly bounded. Applying S. N. Bernstein’sestimates (see Theorem 1 of Sect. 2 and the remark to this theorem), we can show that in any closed domain

470

Q∗ ⊂ Q \ Γ the derivatives ∂um

∂xi, ∂2um

∂xi ∂xj, ∂um

∂t , ∂2um

∂xi ∂t, ∂3um

∂xi∂xj∂xl, ∂3um

∂xi∂xj∂t, and ∂2um

∂t2 (i, j, l = 1, . . . , n; m = 1, 2, . . .)are uniformly bounded. Therefore, one can choose in the sequence um(x, t) a subsequence that converges uniformlyin Q∗ together with the derivatives appearing in Eq. (1.1). Considering a sequence of domains Q∗1 ⊂ Q∗2⊂ . . . ⊂Q∗m ⊂ . . . such that

∞⋃m=1

Q∗m = Q \ Γ and using the diagonal process, we obtain a subsequence umk(x, t) that

converges uniformly together with ∂umk∂xi

, ∂2umk∂xi ∂xj

, and ∂umk∂t in any closed subdomain of Q \ Γ. The limit function

u(x, t) = limk→∞

umk(x, t) obviously satisfies Eq. (1.1) in Q \ Γ.As was done for the heat equation (see [19, pp. 362–365]), one can show that also for the general equation (1.1)

from the existence of a barrier for a point P0 ∈ Γ it follows that the boundary condition is satisfied at this point:lim

P→P0u(P ) = U(P0) if lim

m→∞Um(P ) = U(P ) uniformly in some neighborhood of the point P0. Therefore, according

to supposition (5) of this theorem, the function u(x, t) satisfies condition (3.1) everywhere on Γ \ σ.The uniqueness of the solution follows from Theorem 11 of Sect. 1. Note that because the solution is unique,

not only the subsequence umk(x, t) but the whole sequence um(x, t) converges to u(x, t) in Q \ σ.Suppose now that P1 ∈ σ, the functions ϕ and ψ are continuous at the point P1, and ϕ(P1) = ψ(P1) = κ. Let

us show that limP→P1

u(P ) = κ. First consider the case of κ = 0. Then we can define the function U(x, t) so that it is

continuous at the point P1 and construct the functions Um(x, t) so that they have the following additional property:for every ε > 0, there exists a number m0(ε) so large and a neighborhood GP1 of the point P1 so small that in GP1

the inequality |Um(x, t)| < ε holds for all m > m0(ε). From (3.23) it follows that |um(x, t)| < ε for m > m0(ε) onthe part of the surface Sm belonging to GP1 . Because it was supposed that a barrier exists (supposition 5), we cometo lim

P→P1u(P ) = 0 in the same way as was done in [19]. This means that the boundary condition (3.1) is satisfied

at the point P1.Now let ϕ(P1) = ψ(P1) = κ = 0. We can change the variable u in (1.1) and (3.1): u = v + κ. The func-

tion v(x, t) in Q \ Γ satisfies the equation

L(v) = f(x, t)− κc(x, t)

and the boundary conditions v|Γ\σ = U |Γ\σ−κ. As we have proved, the solution of this problem exists and is unique,and the function v(x, t) is continuous at the point P1 and v(P1) = 0. Obviously the function u(x, t) = v(x, t) + κis a solution of Eqs. (1.1), (3.1). Because of the uniqueness this function coincides with the solution constructedabove. Therefore, the latter at the point P1 satisfies the boundary condition u(P1) = κ, which was to be proved.

Now suppose that in Q\Γ the coefficients aij , bi, and c and the right-hand side f of Eq. (1.1) have continuousderivatives of the form (3.20). In this case one can additionally require that the derivatives (3.20) of functions amij ,bmi , cm, and fm converge uniformly (as m→∞) in any closed domain Q∗ ⊂ Q \Γ to the corresponding derivativesof aij , bi, c, and f . It follows from S. N. Bernstein’s estimates that in any closed domain Q∗ ⊂ Q\Γ the derivatives

∂p+qum

∂xp11 . . . ∂xpnn ∂tq

, p+ 2q r + 2, (3.24)

are uniformly bounded. This leads to the compactness of the derivatives of the form (3.21) of um(x, t), and we candeduce that the derivatives (3.21) of u(x, t) exist in Q∗ ⊂ Q \ Γ.

Finally, suppose that in a subdomain Ω′ ⊂ Ω the function ϕ(x) ∈ Cr+2 and the derivatives of the form (3.20)of aij , bi, c, and f are continuous on the set t = 0, x ∈ Ω′. Then one can construct the sequence Um(x, t) so thatin any closed domain contained in Ω′ the derivatives ∂pUm(x,0)

∂xp11 ...∂xpnn

, p r+2, converge uniformly to the correspondingderivatives of ϕ(x). We can also suppose that the derivatives of the form (3.20) of amij , b

mi , cm, and fm converge as

m→∞ uniformly in all of the cylinder Ω∗×[0, T ]; here Ω∗ is any closed subdomain of Ω′. Let x0 ∈ Ω′ be an arbitrary

point and let δ > 0 be so small that the n-dimensional balln∑i=1

(xi − x0i )

2 δ2 lies in Ω′. As we did in the remark

to Theorem 1 of this section, we can show that in the cylinder Rδ′ = n∑

i=1

(xi − x0i )

2 δ′2, 0 t T, where

0 < δ′ < δ, the derivatives (3.24) of the functions um(x, t) are uniformly bounded with respect to m. Therefore

the function u(x, t) has continuous derivatives (3.21) in Rδ′ and in particular on the sett = 0,

n∑i=1

(xi−x0i )

2 δ′2.

As the point x0 ∈ Ω′ is arbitrary, this means that the derivatives (3.21) of u(x, t) are continuous everywhere onthe set t = 0, x ∈ Ω′.

471

Remark 1. Supposition (5) of Theorem 1 is, for example, fulfilled when for every point x0 = (x01, . . . , x

0n) ∈ σ

there exists an n-dimensional sphere with center y0 = (y01 , . . . , y

0n) that contains x

0 and has no more common pointswith Ω. In this case the barrier at the point P0(x0, t0) ∈ S is the function

vP0(x, t) = 6−K0 − 6−K , (3.25)

with 6 = 6(x, t) = n∑

i=1

(xi − y0i )

2 + (t − t0)21/2

, 60 = 6(x0, t0), and K > 0 a sufficiently large number. The re-

quirements (A) and (B) for a barrier are obviously fulfilled. To verify (C) we calculate L(vP0):

L(vP0 ) = K6−K−4

62

[ n∑i=1

aii +n∑i=1

bi(xi − y0i )− (t− t0)

]− (K + 2)

n∑i,j=1

aij(xi − y0i )(xj − y0

j )+ cvP0

< K6−K−4[M1 − (K + 2)

µ

262

0

]< 0,

if c(x, t) < 0 (this does not lead to loss of generality) and K is sufficiently large. As a barrier for the point P1(x1, 0),where x1 = (x1

1, . . . , x1n) ∈ Ω, we can take the function

vP1(x, t) =n∑i=1

(xi − x1i )

2 +K1t (3.26)

with K1 > 0 sufficiently large. This function also obviously fulfills conditions (A) and (B). Let us verify (C):

L(vP1 ) = 2n∑i=1

[aii + bi(xi − x1i )] + cvP1 −K1 < 0,

ifn∑i=1

(xi − x1i )

2 < 1 and K1 > 0 is sufficiently large.

Remark 2. If in Theorem 2 we suppose in addition to (2) that in Q \ Γ the derivatives

∂p+qaij∂xp1

1 . . . ∂xpnn ∂tq, where p+ 2q 6, q 2,

exist in any closed domain Q∗ ⊂ Q\Γ, and they are Holder continuous with respect to x, and the same is supposedon the derivatives of bi and c, then condition (5) can be replaced by a weaker one:

(5′) For any point P0 ∈ Γ, there exists a function VP0(x, t), having the properties (A) and (B) of a barrierand the following property (C′):

(C′) L(VP0) ≡n∑

i,j=1

aij(x, t)∂2VP0

∂xi ∂xj+

n∑i=1

bi(x, t)∂VP0

∂xi+ c(x, t)VP0 −

∂VP0

∂t< 0 everywhere inside GP0 .

The function VP0 (x, t) is also called a barrier.Under the suppositions we have made, the existence of the solution of problem (1.1), (3.1) in the cylinder Q

is proved as in Theorem 2; we only have to take amij = aij , bmi = bi, and cm = c.

Remark 3. When the coefficients of Eq. (1.1) do not depend on t, condition (5′) is fulfilled if all the points

of σ are regular with respect to the Laplace equationn∑i=1

∂2u∂x2

i= 0 (see [64] or [19, p. 266]). This means that

the boundary σ of the domain Ω must be such that the Dirichlet problem for the Laplace equation is solvable forany continuous boundary function defined on σ.

In this case, we can again take for the barrier at the point P1(x1, 0), x1 ∈ Ω, the function (3.26). To constructa barrier for a point P0(x0, t0) ∈ S we first note that for such a point it is sufficient to make sure that (A), (B),and (C′) are fulfilled only for t t0. This can be proved for Eq. (1.1) in the same way as for the heat equation(see [19, pp. 364–365]).

472

Let us suppose that in Eq. (1.1) we have c(x) < −c0 < 0. Let w(x) be the solution of the elliptic equation

n∑i,j=1

aij(x)∂2w

∂xi ∂xj+

n∑i=1

bi(x)∂w

∂xi+ c(x)w = 0

in the domain Ω with the boundary condition

w(x)|σ = F (x),

where F (x) is a continuous function such that F (x0) = 1c0

and F (x) < 1c0

for x = x0. The function w(x) existsbecause of the regularity of the points of σ (see [65]). The maximum principle gives w(x) < 1

c0everywhere in Ω

except the point x = x0, where we have w = 1c0.

We shall demonstrate that at the point P0(x0, t0) we can take as a barrier the function

VP0(x, t) =1c0− w(x) + (t0 − t).

In fact, for t t0 we have

VP0 (x, t) > 0 forn∑i=1

(xi − x0i )

2 + (t0 − t) > 0,

VP0(x0, t0) = 0, L(VP0) =

c(x)c0

+ c(x)(t0 − t) + 1 < 0,

and thus VP0 is a barrier.Consider now Eq. (1.1) with conditions (3.1) in the domain D bounded by the planes t = 0 and t = T and

in general not cylindrical. We shall suppose that the intersection D0 of D and the plane t = 0 is a closure of somebounded n-dimensional domain Ω with boundary σ.

Theorem 3. In the domain D there exists a unique solution of the problem (1.1), (3.1), continuous in Dand satisfying (1.1) in D \ Γ if the following conditions are fulfilled:

(1) Supposition (1) of Theorem 1.(2) Supposition (2) of Theorem 2 (with Q changed to D).(3) ϕ(x) ∈ C(D0), ψ(x, t) ∈ C(S).(4) The consistency condition (3.2) holds.(5) For any point P0(x0, t0) ∈ S there exists an (n+1)-dimensional sphere having only one common point P0

with D, and the radius of this sphere directed to P0 is not parallel to the t axis.

Proof. Let U(x, t) be an arbitrary function, continuous in D and satisfying (3.1). We shall approximatethe coefficients aij , bi, c, and f of (1.1) by sequences of uniformly bounded infinitely differentiable functions amij ,bmi , cm, and fm converging uniformly in any closed domain D∗ ⊂ D \Γ together with the corresponding derivativesof the form (3.19). Let us construct the sequence of “step-solids” Dm (m = 1, 2, . . .), consisting of the cylindersQml = Ωm

l ×[ (l−1)T

2m , lT2m]. Here the boundaries of n-dimensional domains Ωm

l belong to the class A6+λ for allm and l, l = 1, 2, . . . , 2m; we have Dm+1 ⊃ Dm and

∞⋃m=1

Dm = D.

Let us denote by Γml the part of the boundary of the cylinder Qml , consisting of its lower base t = (l−1)T

2m andits side boundary. According to Theorem 2, for any m in the cylinder Qm

1 there exists a unique solution um1 (x, t) ofEq. (3.22) satisfying the condition

um1 (x, t)|Γm1 = U(x, t)|Γm1 .

The function um1 (x, t) is infinitely differentiable and satisfies Eq. (3.22) everywhere in Qm1 \ Γm1 , in particular for

t = T2m , x ∈ Ωm

1 .Now consider Eq. (3.22) in the cylinder Qm

2 with conditions um2 (x, t) = um1 (x, t) on the part of Γm2 wherethe function um1 is defined and um2 (x, t) = U(x, t) on the rest of the boundary Γm2 . According to Theorem 2, in Qm

2

there exists a unique solution um2 (x, t) of Eq. (3.22) satisfying these conditions. This function um2 (x, t) is infinitely

473

differentiable; it satisfies Eq. (3.22) in Qm2 \ Γm2 , and also for t = T

2m , x ∈ Ωm2 ∩ Ωm

1 . In this way the solutionsum1 and um2 are “glued” to each other continuously together with their derivatives of all orders with respect to xfor t = T

2m . It follows from Eq. (3.22) that their derivatives containing differentiation with respect to t are alsocontinuous for t = T

2m .In the same way we construct functions uml (x, t) in the cylinders Qm

l (l = 3, . . . , 2m). Set um(x, t) = uml (x, t)in Qm

l (l = 1, 2, . . . , 2m); we obtain a function um(x, t), that is infinitely differentiable in Dm, coincides with U(x, t)at the points of the boundary of Dm that belong to

⋃Γml , and satisfies Eq. (3.22) at all other points of Dm.

We can perform this for m = 1, 2, . . . Using S. N. Bernstein’s estimates, we can establish the compactness inthe sense of uniform convergence of the sequence um(x, t) and its derivatives ∂um

∂xi, ∂2um

∂xi ∂xj, and ∂um

∂t (i, j = 1, . . . , n)in any closed subdomain D∗ ⊂ D \ Γ. Using the diagonal process, we can select a subsequence ums(x, t) thatconverges as ms →∞ to a function u(x, t) satisfying Eq. (1.1) in D \ Γ.

Using Remark 2 to Theorem 11 of Sect. 1 and the barrier functions, we can prove that u(x, t) satisfiesconditions (3.1) on Γ. For a point P1(x1, 0) ∈ D0 we can again use the barrier (3.26). If the point P0(x0, t0) ∈ S,then, according to condition (5) of this theorem there exists a sphere with center (x∗, t∗), x∗ = x0, having onlyone common point P0 with D. Let us take a neighborhood of P0 so small that for all points in this neighborhood

the inequalityn∑i=1

(x∗i − xi)2 β > 0 holds. Then in this neighborhood the function similar to (3.25)

vP0(x, t) = 6−K0 − 6−K , (3.27)

where 6 = 6(x, t) = n∑

i=1

(xi−x∗i )2+(t−t∗)21/2

, 60 = 6(x0, t0), andK > 0 is sufficiently large, has all the properties

of a barrier. In fact, forn∑i=1

(x∗i − xi)2 β > 0 and c(x, t) < 0 (this does not restrict generality) we have

L(vP0) < K6−K−4[M1 − (K + 2)

µ

2β2]< 0

for sufficiently large K. Conditions (A) and (B) are obviously fulfilled.The uniqueness of the solution is proved in Sect. 1.

Remark 1. As in the previous theorem, in Theorem 3 one can omit the consistency condition (3.2), supposinginstead that the functions ϕ and ψ are bounded and continuous in Ω and on S respectively.

Remark 2. The suppositions of Theorems 2 and 3 concerning the smoothness of coefficients aij , bi, and c ofEq. (1.1) and its right-hand side f are too strong. This is connected with the method of proof, namely with the useof S. N. Bernstein’s estimates for the derivatives of the solution to the parabolic equation. Using A. Friedman’sestimates (see Sect. 2) it is possible to prove the existence of a solution of problem (1.1), (3.1) under the suppositionthat the functions aij , bi, c, and f are continuous in D and Holder continuous in any closed domain D∗ ⊂ D \ Γand the surface S has the same properties as in Theorem 3 (see [43]).

3. Cauchy problem. In this section, we shall consider the Cauchy problem in the layerH0<tT, x∈Enwith the initial condition

u(x, 0) = ϕ(x), x ∈ En. (3.28)

The solution of problem (1.1), (3.28) will be constructed as the limit of the solutions of the first boundary-valueproblems in a sequence of infinitely expanding cylinders. We shall require that the absolute value of the right-handside f(x, t) of Eq. (1.1) be bounded in H and the coefficient c(x, t) be bounded from above in H . The coefficients

aij(x, t) and bi(x, t) can grow in an arbitrary way asn∑i=1

x2i →∞. The uniqueness theorems for the Cauchy problem

are proved in subsection 5 of Sect. 1.

Theorem 4. The function u(x, t), continuous and bounded in H , satisfying for t = 0 the initial condition(3.28) and satisfying for t > 0 Eq. (1.1) exists under the following suppositions:

(1)n∑

i,j=1

aij(x, t)αiαj µn∑i=1

α2i for (x, t) ∈ H and all α1, . . . , αn (µ = const > 0).

(2) In the layer H there exist continuous derivatives of the form (3.19), and the same derivatives of func-tions bi, c, and f ; the coefficients aij , bi, c, and f are continuous in H .

(3) The function f(x, t) is bounded in the layer H .(4) c(x, t) M <∞ for (x, t) ∈ H .(5) The function ϕ(x) is continuous and bounded in the whole space En.

474

Proof. We shall approximate the coefficients aij , bi, c, and f of Eq. (1.1) by the sequences of infinitely differ-entiable functions amij , b

mi , cm, and fm, converging uniformly with the corresponding derivatives of the form (3.19)

in any closed bounded domain contained in H . We shall suppose that |fm(x, t)| M1 and cm(x, t) M2, where

M1 and M2 do not depend on m. Denote by Ωm the balln∑i=1

x2i < m2; by σm denote its boundary, and by Qm

denote the cylinder Ωm × (0, T ). According to Theorem 2, for all m = 1, 2, . . . there exists a solution um(x, t) ofEq. (3.22) in the cylinder Qm satisfying the conditions

um(x, 0) = ϕ(x) for x ∈ Ωm,

um(x, t) = ϕ(x) for x ∈ σm, 0 t T.

As follows from estimate (1.9) of Sect. 1, the absolute values |um(x, t)| are bounded by a constant not dependingon m.

Let us fix an arbitrary R > 0 and consider in the cylinder QR the functions um(x, t) for m > R. Accordingto Theorem 1 of Sect. 2 and the remark to this theorem, the families um,

∂um

∂xi

,

∂2um

∂xi ∂xj

, and

∂um

∂t

are

compact in the sense of uniform convergence in the cylinder QR ∩t δ for all δ > 0. Let R tend to infinity and δtend to zero. By the “diagonal process” we can select a subsequence umk(x, t) converging at any point (x, t) ∈ H toa bounded function u(x, t) that satisfies Eq. (1.1) in H . Using barriers of type (3.26), we can show that the functionu(x, t) is continuous for t = 0 and satisfies the initial condition (3.28).

Remark. Under the suppositions of Theorem 4 the solution of problem (1.1), (3.28) that we have justconstructed is not in general unique even in the class of bounded functions. To have uniqueness in this class we

have to impose additional restrictions on the growth of the coefficients of Eq. (1.1) asn∑i=1

x2i →∞ (see Theorem 10

of Sect. 1).

Let us give examples showing that conditions (3) and (4) of Theorem 4 cannot be omitted.Consider the equation

∂2u

∂x2− ∂u

∂t= −x (3.29)

in the strip H10 < t T1, −∞ < x < +∞ with the initial condition

u(x, 0) = 0, −∞ < x < +∞. (3.30)

It is easy to verify that the unbounded function u(x, t) = xt is a solution of problem (3.29), (3.30). If we supposethat there is another solution v(x, t) of this problem and it is bounded, then w(x, t) = u(x, t) − v(x, t) will satisfythe heat equation ∂2w

∂x2 − ∂w∂t = 0 in the strip H1 with zero as the initial condition. This contradicts the uniqueness

theorem for the solution of the Cauchy problem in the class of functions growing not faster than a linear functionas |x| → ∞ (Theorem 9 of Sect. 1). So Theorem 4 does not hold without condition (3).

Consider now in H1 the equation∂2u

∂x2+ c(x)u − ∂u

∂t= 0 (3.31)

with the initial conditionu(x, 0) = 1, −∞ < x < +∞. (3.32)

The function c(x) will be constructed in the following way. Take an integer k0 > 0 such that for all k > k0

the inequality4√T (2k + 1) < ek(e− 1)

holds. Then for k > k0 we shall have

ek + 4√Tk < ek+1 − 4

√T (k + 1).

For ek − 4√Tk x ek + 4

√Tk, set c(x) = ek (k = k0 + 1, k0 + 2, . . .). For the remaining x define c(x) so that it

is an infinitely differentiable positive even function, nondecreasing for x 0.

475

For ek |x| ek+1 (k = k0 + 1, k0 + 2, . . .) we have

0 c(x) ek+1 e|x|. (3.33)

We shall show that problem (3.31), (3.32) has no bounded solution. Suppose the contrary: let u(x, t) bea solution of (3.31), (3.32) such that |u(x, t)| M . Setting u = vete

k

we deduce from (3.31) and (3.32) that

∂v

∂t− ∂2v

∂x2= [c(x) − ek]v, v(x, 0) = 1. (3.34)

We shall consider Eq. (3.34) as a heat equation with the right-hand side [c(x)−ek]v. Since u(x, t) is assumedto be bounded and estimate (3.33) holds, the function [c(x) − ek]v grows as |x| → ∞ not faster than a linearfunction. Therefore, the solution of Cauchy problem (3.34) is given by the well-known formula

v(x, t) = 1 +

t∫0

dτ

+∞∫−∞

Z(x, t; ξ, τ)[c(ξ) − ek]v(ξ, τ) dξ,

whereZ(x, t; ξ, τ) = [4π(t− τ)]− 1

2 e−(x−ξ)24(t−τ)

is a fundamental solution of the heat equation. Hence we have

u(x, t) = etek

1 +

t∫0

dτ

+∞∫−∞

Z(x, t; ξ, τ)[c(ξ) − ek]e−τeku(ξ, τ) dξ. (3.35)

Thus, the bounded solution of problem (3.31), (3.32) must satisfy (3.35). From (3.35) we can deducethe inequality

u(x, t) etek

1−M

t∫0

dτ

+∞∫−∞

Z(x, t; ξ, τ)|c(ξ) − ek| dξ. (3.36)

We have constructed c(x) so that for |ξ − ek| 4√Tk we have c(ξ) − ek = 0. Setting x = ek in (3.36) and

using estimate (3.33), we obtain

u(ek, t) etek1−M

t∫0

dτ

( ek−4√Tk∫

−∞

+

+∞∫ek+4

√Tk

)Z(ek, t; ξ, τ)(e|ξ| + ek) dξ

= etek

1− M√

π

t∫0

dτ

( −2k√

Tt−τ∫

−∞

+

+∞∫2k√

Tt−τ

)(e|ek + 2

√t− τζ|+ ek)e−ζ

2dζ

etek1− 2M√

π

t∫0

dτ

+∞∫2k√

Tt−τ

[ek(e+ 1) + 2e√t− τζ]e−ζ2

dζ

. (3.37)

The quantity in braces in the right-hand side of the inequality (3.37) tends to unity as k → +∞ for any tfrom the interval (0, T ). In fact,

+∞∫2k√

Tt−τ

ek(e+ 1)e−ζ2dζ < ek(e+ 1)

+∞∫2k

e−ζ dζ = (e+ 1)e−k → 0 (k → +∞),

476

+∞∫2k√

Tt−τ

2e√t− τζe−ζ

2dζ =

+∞∫4k2Tt−τ

e√t− τe−s ds < e

√T

+∞∫2k2

e−s ds = e√Te−4k2

→ 0 (k → +∞).

Therefore, for sufficiently large k we have

u(ek, t) >12ete

k

,

and this contradicts the supposition that u(x, t) is bounded.4. Second Boundary-Value Problem. One can use the Rothe method also to construct the solution

of the second boundary-value problem for Eq. (1.1) in the cylinder Q = Ω × (0, T ). Boundary conditions for thisproblem have the form

u(x, 0) = ϕ(x), (3.38)n∑

i,j=1

aij(x, t) cos(ν, xi)∂u

∂xj+ a(x, t)u = 0 on S, (3.39)

where ν is the direction of the inner normal to S.

Theorem 5. There exists in the cylinder Q a unique function u(x, t), continuous in Q together with itsfirst-order derivatives with respect to x1, . . . , xn, satisfying boundary conditions (3.38), (3.39) on Γ having inQ \ Γ continuous second-order derivatives with respect to x1, . . . , xn, first-order derivatives with respect to t, andsatisfying Eq. (1.1), if the following suppositions are fulfilled:

(1) Supposition (1) of Theorem 1.(2) The functions aij , bi, c, and f belong to the class C8+λ,4+λ(Q) (see Sect. 1).(3) The boundary σ of domain Ω belongs to the class A6+λ.(4) ϕ(x) ∈ C6+λ(Ω); a(x, t) ∈ C8+λ,4+λ as a function of local coordinates ξ1, . . . , ξn−1, t on the surface S.(5) The consistency condition

[ n∑i,j=1

aij(x, 0) cos(ν, xi)∂ϕ(x)∂xj

+ a(x, 0)ϕ(x)]x∈σ

= 0 (3.40)

is fulfilled.

Proof. As we have already noted when we proved Theorem 1, one can suppose without loss of generalitythat c(x, t) < 0 in Q. We can also suppose that a(x, t) < 0 (see the remark to Theorem 7 of Sect. 1). Considerthe system of equations (3.4) with the boundary conditions

n∑i,j=1

aij(x, k∆t) cos(ν, xi)∂u(x, k∆t)

∂xj+ a(x, k∆t)u(x, k∆t) = 0 (k = 1, . . . ,m). (3.41)

For k = 0 we set u(x, 0) = ϕ(x). It is well known that for every k there exists a unique solution u(x, k∆t) ofthe elliptic equation (3.4) satisfying the condition (3.41), and the solution u(x, k∆t) ∈ C6+λ(Ω) (see, for example,[66], [2] and Theorem 7 from [63]).

We shall show that the functions u(x, k∆t) are uniformly bounded in the cylinder Q with respect to ∆t.As in Theorem 1, it is possible to estimate |u(x, k∆t)| at an inner point where |u| attains its maximum. It isobvious that |u(x, k∆t)| is uniformly bounded for k = 0. Finally it follows from (3.41) that u(x, k∆t) as a functionof x cannot attain its positive maximum or negative minimum on S. In fact, if the positive maximum or negativeminimum was attained on S, according to the well-known property of solutions of elliptic equations (see [21]) wewould have at the point where the positive maximum of u(x, k∆t) is attained

u(x, k∆t) > 0,n∑

i,j=1

aij(x, k∆t) cos(ν, xi)∂u(x, k∆t)

∂xj< 0,

477

and this is impossible because of (3.41). Similarly, one can prove that u(x, k∆t) cannot have a negative minimumon S.

Let us estimate the derivatives ∂u(x,k∆t)∂xl

(l = 1, . . . , n). By S. N. Bernstein’s method one can prove the uni-

form boundedness ofn∑l=1

[∂u(x,k∆t)∂xl

]2 in any closed domain Q∗ ⊂ Q \ S. To do this we have to use an auxiliary

function of the form (3.17) (see the remark to Theorem 1 of this section).Now we shall prove that the derivatives ∂u(x,k∆t)

∂xlare uniformly bounded in a neighborhood of S. Let σp be

a piece of σ allowing a parametric representation

xi = xip(ξ1, . . . , ξn−1) (i = 1, . . . , n), where xip ∈ C6+λ.

In the cylinder Qp = Gp × (0, T ), where Gp is an intersection of a neighborhood of σp with the domain Ω, we shallintroduce a new coordinate system (ξ1, . . . , ξn, t) given by formulas

xi = xip(ξ1, . . . , ξn−1) + ξnhi(ξ1, . . . , ξn−1, t) (i = 1, . . . , n);

here

hi(ξ1, . . . , ξn−1, t) =1

A(x, t)

n∑j=1

aij(x, t) cos(ν, xj),

A(x, t) = n∑

i=1

[ n∑j=1

aij(x, t) cos(ν, xj)]21/2

.

The boundary condition (3.41) will have the following form on the piece Sp = σp × [0, T ]:

[∂u(ξ, k∆t)

∂ξn+a(ξ, k∆t)A(ξ, k∆t)

u(ξ, k∆t)]ξn=0

= 0.

Set u(ξ, k∆t) = v(ξ, k∆t)κ(ξ, k∆t), where

κ(ξ, t) = 1− ξn[a(ξ, t)A(ξ, t)

∣∣∣∣ξn=0

+ β

](β > 0)

for 0 ξn δ (δ > 0 is sufficiently small); for ξn δ the function κ(ξ, t) is extended in a smooth way. The functionsv(ξ, k∆t) in the cylinder Qp satisfy the equations of the form

Lk(v(ξ, k∆t)) =n∑

i,j=1

aij(ξ, k∆t)∂2v

∂ξi ∂ξj+

n∑i=1

bi(ξ, k∆t)∂v

∂ξi

+ c(ξ, k∆t)v − v(ξ, k∆t)− v(ξ, (k − 1)∆t)∆t

= f(ξ, k∆t) (3.42)

and the boundary condition [∂v(ξ, k∆t)

∂ξn− βv(ξ, k∆t)

]ξn=0

= 0. (3.43)

Let us take an arbitrary point ξ1 = (ξ11 , . . . , ξ

1n−1, 0) and consider in the cylinder Qδ

p

0 ξn δ,

n∑i=1

(ξi − ξ1i )

2 δ2, 0 t Tan auxiliary function

z(ξ, k) = Φ1(ξ) n−1∑

l=1

(∂v

∂ξl

)2

+(∂v

∂ξn− βv

)2+Nv2 + eN1(T−k∆t) + εξn,

478

where Φ1(ξ) =[δ2 −

n−1∑i=1

(ξi − ξ1i )

2]2, and N , N1, and ε are positive constants. Using Eq. (3.42) and equations

obtained by differentiating (3.42) with respect to ξl, it is possible to prove (as was done in Theorem 1) that forsufficiently large N and N1 inside Qδ

p (and for t = T ) the inequality Lk(z(ξ, k)) < 0 holds for any ∆t and k. Itfollows that the function z(ξ, k) cannot attain its maximal value in Qδ

p inside Qδp (or for t = T ).

So we have proved that the function z(ξ, k) is uniformly bounded with respect to ∆t for ξn = δ and forn−1∑i=1

(ξi − ξ1i )

2 = δ2. Obviously z(ξ, k) is uniformly bounded also for t = 0.

Suppose that z(ξ, k) attains its maximal value at a point (ξ1, . . . , ξn, t), with ξn = 0. Then the function

z∗(ξ, k) = Φ1(ξ)n−1∑l=1

(∂v

∂ξl

)2

+Nv2 + eN1(T−k∆t) + εξn

also attains its maximal value at this point, because it follows from (3.43) that ∂v∂ξn− βv = 0 for ξn = 0. Therefore

at this point we have ∂z∗

∂ξn 0. On the other hand, by differentiating (3.43) with respect to ξl, we obtain

∂2v

∂ξl ∂ξn

∣∣∣∣ξn=0

= β∂v

∂ξl

∣∣∣∣ξn=0

(l = 1, . . . , n− 1). (3.44)

Now, taking into account (3.42) and (3.44), we have

∂z∗

∂ξn

∣∣∣∣ξn=0

= 2βΦ1(ξ)n−1∑l=1

(∂v

∂ξl

)2∣∣∣∣ξn=0

+ 2βNv2|ξn=0 + ε > 0.

This contradiction proves that the function z(ξ, k) cannot attain its maximum for ξn = 0.So we have proved that z(ξ, k) is uniformly bounded in the cylinder Qδ

p. The boundedness of the derivatives∂u(x,k∆t)

∂x1,. . . , ∂u(x,k∆t)

∂xnin the cylinder Qδ′

p , where 0 < δ′ < δ, follows. Since the piece σp and the point ξ1 arearbitrary, we have proved that these derivatives are uniformly bounded everywhere in Q.

Now we have to estimate in Q the derivatives ∂2u(x,k∆t)∂xl ∂xq

(l, q = 1, . . . , n) and the difference quotient∆u(x,k∆t)

∆t = u(x,k∆t)−u(x,(k−1)∆t)∆t . Using S. N. Bernstein’s method and the function

w(x, k) = Φ2(x)n∑

l,q=1

[∂2u(x, k∆t)∂xl ∂xq

]2

+N2

n∑l=1

[∂u(x, k∆t)

∂xl

]2

+ eN3(T−k∆t),

where Φ2(x) =[δ2 −

n∑i=1

(xi − x0i )

2]2, x0 ∈ Ω, we shall prove that

n∑l,q=1

[∂2u(x,k∆t)∂xl ∂xq

]2 is uniformly bounded in any

closed domain Q∗ ⊂ Q \ S. As follows from Eqs. (3.4), thus we have proved also the uniform boundedness in Q∗ ofthe difference quotient ∆u(x,k∆t)

∆t . Then we extend the function u(x, k∆t) to k = −1 by formula (3.14) and considerin the cylinder Qδ

p the auxiliary function

z1(ξ, k) = Φ1(ξ) n−1∑

l,q=1

(∂2v

∂ξl ∂ξq

)2

+n−1∑l=1

(∂2v

∂ξl ∂ξn− β ∂v

∂ξl

)2

+(∆v∆t

)2

+N4

[ n−1∑l=1

(∂v

∂ξl

)2

+(∂v

∂ξn− βv

)2]+ eN5(T−k∆t) + εξn.

Using this function and taking into account the consistency condition (3.40), we shall show as we did before thatin Qδ′

p , 0 < δ′ < δ, the derivatives ∂2u∂ξi ∂ξj

(i = 1, . . . , n; j = 1, . . . , n−1) and the difference quotient ∆u∆t are uniformly

bounded. It follows from Eqs. (3.14) that the derivative ∂2u∂ξ2nin Qδ′

p also is bounded uniformly with respect to ∆t. In

this way we have proved that in Q the functions ∂2u(x,k∆t)∂xi ∂xj

(i, j = 1, . . . , n) and ∆u(x,k∆t)∆t are uniformly bounded.

479

Then, using the function

w1(x, k) = Φ2(x)n∑

l,q,r=1

[∂3u(x, k∆t)∂xl∂xq∂xr

]2

+N6

n∑l,q=1

[∂2u(x, k∆t)∂xl ∂xq

]2

+ eN7(T−k∆t),

we shall estimate the derivatives ∂3u(x,k∆t)∂xl∂xq∂xr

(l, q, r = 1, . . . , n) in Q∗ ⊂ Q \ S. The equations obtained by differenti-ating (3.4) with respect to xl allow us to obtain an estimate in Q∗ for the functions ∂

∂xl

[∆u(x,k∆t)∆t

](l = 1, . . . , n).

Using the auxiliary function

z2(ξ, k) = Φ1(ξ) n−1∑

l,q,r=1

(∂3v

∂ξl∂ξq∂ξr

)2

+n−1∑l,q=1

(∂3v

∂ξl∂ξq∂ξn− β ∂2v

∂ξl ∂ξq

)2

+n−1∑l=1

[∂

∂ξl

(∆v∆t

)]2

+[∂

∂ξn

(∆v∆t

)− β∆v

∆t

]2+N8

[ n−1∑l,q=1

(∂2v

∂ξl ∂ξq

)2

+n−1∑l=1

(∂2v

∂ξl ∂ξn− β ∂v

∂ξl

)2

+(∆v∆t

)2]+ eN9(T−k∆t) + εξn,

we can show in the same way as we did before that the functions ∂3u(x,k∆t)∂xl∂xq∂xr

and ∂∂xl

[∆u(x,k∆t)∆t

](l, q, r = 1, . . . , n)

are uniformly bounded in Q.After that, using Bernstein’s method, we can prove that the derivatives ∂4u(x,k∆t)

∂xl∂xq∂xr∂xs(l, q, r, s = 1, . . . , n) are

uniformly bounded in any closed domain Q∗ ⊂ Q\S. Equations obtained by twice differentiating (3.4) with respectto x1, . . . , xn give the uniform boundedness of functions ∂2

∂xl ∂xq

[∆u(x,k∆t)∆t

](l, q = 1, . . . , n) in any closed domain

Q∗ ⊂ Q \ S.Let us extend the functions u(x, k∆t) by linear interpolation for k∆t t (k + 1)∆t. As we have proved,

the families obtained in this way, u∆t(x, t) and∂u∆t(x,t)

∂xl

(l = 1, . . . , n), are compact in Q and the families∂2u∆t(x,t)

∂xl ∂xq

(l, q = 1, . . . , n) and ∆u∆t(x,t)

∆t are compact in any closed domain contained in Q \ S together with itsboundary. As in the proof of Theorem 1 we can see that the sequence u∆tm(x, t) uniformly converges for ∆tm → 0in Q to a function u(x, t). This limit function satisfies Eq. (1.1) in Q \ Γ and conditions (3.38) and (3.39) on Γ.

The uniqueness of the solution of problem (1.1), (3.38), (3,39) is proved in Sect. 1.Note that one can obtain estimates for the solution of the second boundary-value problem (1.1), (3.38),

(3.39) in the same way as we did for the functions u(x, k∆t) and their derivatives and difference quotients.

4 THE FUNDAMENTAL SOLUTION OF A LINEAR PARABOLIC EQUATION. GREEN’S FUNCTION.METHOD OF INTEGRAL EQUATIONS FOR SOLVING BOUNDARY-VALUE PROBLEMS

In the study of linear partial differential equations, the so-called fundamental solutions, that is, solutionshaving singularities of a prescribed type, are of great importance. Many works are dedicated to the constructionof fundamental solutions for parabolic equations and systems and to their applications for solving boundary-valueproblems and the Cauchy problem. A fundamental solution for a general second-order parabolic equation with manyindependent variables and smooth coefficients was first constructed in [67, 68]. For parabolic equations of specialform, fundamental solutions were considered in [69–71] and others. In W. Pogorzelski’s papers [72,73] a fundamentalsolution is constructed for a second-order parabolic equation with only Holder continuous coefficients. For systemsof equations parabolic in the sense of I. G. Petrovskiy, fundamental solutions are constructed in [27, 74, 75]. In allthese papers the classical method of E. E. Levi [76] is used for the construction of fundamental solutions.

In this section, we construct a fundamental solution for the equation L(u) = 0 and establish some of itsimportant properties. With the help of fundamental solutions we obtain solutions for the Cauchy problem andthe first boundary-value problem for Eq. (1.1) under the supposition that the coefficients and the function f(x, t)are only Holder continuous.

1. Construction of a Fundamental Solution. As is known, the function

w(x, t; ξ, τ) =1

2√π(t− τ)e

− (x−ξ)24(t−τ)

480

satisfies the heat equation∂2u

∂x2− ∂u

∂t= 0 (4.1)

for t > τ and has the following property: for any bounded and continuous function ϕ(x) defined for all x we have

limt→τ+0

+∞∫−∞

w(x, t; ξ, τ)ϕ(ξ) dξ = ϕ(x), (4.2)

uniformly on any finite interval. For a more general equation

n∑i,j=1

aij∂2u

∂xi ∂xj− ∂u

∂t= 0, (4.3)

with coefficients aij independent of x and t, a solution with similar properties also can be constructed explicitly.In what follows, we shall suppose that the coefficients aij in Eq. (4.3) are functions of parameters ξ and τ ,

where ξ = (ξ1, . . . , ξn) ∈ En, 0 τ T . Denote by aij(ξ, τ) elements of the matrix inverse to ‖aij(ξ, τ)‖; A(ξ, τ)will denote the determinant of the matrix ‖aij(ξ, τ)‖. Then let us set

σ(ζ, θ; s) = σ(ζ, θ; s1, . . . , sn) =n∑

i,j=1

aij(ζ, θ)sisj ,

Wζ,θ(x, t; ξ, τ) = [4π(t− τ)]− n2 [A(ζ, θ)]−

12 e−

σ(ζ,θ;x−ξ)4(t−τ) . (4.4)

By direct calculation one can show that Wξ,τ (x, t; ξ, τ) as a function of variables x, t satisfies Eq. (4.3) for t > τ .Suppose the functions aij(x, t) and ϕ(x, t) are continuous and bounded in the layer H0 t T, x ∈ En, and

n∑i,j=1

aijαiαj µn∑i=1

α2i , µ > 0. (4.5)

We shall demonstrate that under these suppositions for Wξ,τ (x, t; ξ, τ) a formula similar to (4.2) holds:

limt→τ+0

∫En

Wξ,τ (x, t; ξ, τ)ϕ(ξ, t) dξ = ϕ(x, τ) (0 τ < T ). (4.6)

First suppose that aij and ϕ are constant. The change of variables

ξi = xi +n∑

j=1

γijηj (i = 1, . . . , n) (4.7)

transforms the quadratic form σ(x − ξ) =n∑

i,j=1

aij(xi − ξi)(xj − ξj) to the canonical form σ =n∑i=1

η2i . The determi-

nant of transformation (4.7) obviously is (det ‖aij‖)− 12 =√A. Hence

I(x, t, τ) ≡∫En

Wξ,τ (x, t; ξ, τ)ϕdξ = ϕ

∫En

[4π(t− τ)]− n2 e−

n∑i=1

η2i

4(t−τ) dη.

Passing to polar coordinates, we obtain

I(x, t, τ) = (4π)−n2 ϕωn

+∞∫0

(t− τ)−n2 rn−1e−

r24(t−τ) dr,

481

where ωn = 2πn2

Γ(n2 )is the area of the n-dimensional space. Setting r2

4(t−τ) = q, we obtain

I(x, t, τ) =4−

n2 ϕ

Γ(n2

) +∞∫0

4n2 q

n2−1e−q dq = ϕ

for all t and τ , t > τ . Thus we have proved (4.6) for constant aij and ϕ. Now let aij and ϕ not be constant. Wehave

I(x, t, τ) = ϕ(x, t)∫En

Wx,τ (x, t; ξ, τ) dξ + ϕ(x, t)∫En

[Wξ,τ (x, t; ξ, τ) −Wx,τ (x, t; ξ, τ)] dξ

+∫En

Wξ,τ (x, t; ξ, τ)[ϕ(ξ, τ) − ϕ(x, t)] dξ = I1 + I2 + I3. (4.8)

As we have proved,lim

t→τ+0I1 = ϕ(x, τ);

in fact, according to (4.4), the coefficients aij appearing in the formula forWx,τ (x, t; ξ, τ) are taken at a point (x, τ).To estimate the function under the integral sign in I2 we shall use the mean-value theorem. We obtain

|Wξ,τ (x, t; ξ, τ) −Wx,τ (x, t; ξ, τ)| (t− τ)− n2 e−

µ′|x−ξ|2t−τ

×M1|x− ξ|2t− τ max

i,j|aij(ξ, τ)− aij(x, τ)| +M2|A(ξ, τ) −A(x, τ)|

,

where µ′ > 0 and |x− ξ| = n∑

i=1

(xi − ξi)2 1

2. Using the elementary inequality

qνe−µ′q Kνe

−µ1q (0 q < +∞, 0 < µ1 < µ′), (4.9)

which holds for all ν 0, we obtain

|Wξ,τ (x, t; ξ, τ)−Wx,τ (x, t; ξ, τ)| (t−τ)− n2 e−

µ1|x−ξ|2t−τ M3 max

i,j|aij(ξ, τ)−aij(x, τ)|+M2|A(ξ, τ)−A(x, τ)|. (4.10)

Let us represent the integral I2 as a sum I(1)2 +I(2)

2 , where I(1)2 is taken over the ball of radius 6 and with center ξ = x.

The integral I(2)2 is taken over the remaining part of the space ξ1, . . . , ξn. Taking into account the estimate (4.10)

and the continuity of the functions aij(x, t) and A(x, t), one can see that the absolute value of the integral I(1)2 is

arbitrarily small for sufficiently small 6. If 6 is fixed and t− τ > 0 is sufficiently small, the integral I(2)2 also can be

made arbitrarily small in absolute value, because according to (4.10) it is uniformly converging with respect to xand the function under the integral sign uniformly tends to zero as t→ τ +0. Thus we have lim

t→τ+0I2 = 0. Similarly

we obtain limt→τ+0

I3 = 0. This proves (4.6).

Using similar arguments it is also possible to prove that

limτ→t−0

∫En

Wξ,τ (x, t; ξ, τ)ϕ(ξ, τ) dξ = ϕ(x, t) (4.11)

under the same suppositions on aij and ϕ.Now we shall consider a general linear second-order homogeneous parabolic equation in the layer H :

Lx,tu ≡n∑

i,j=1

aij(x, t)∂2u

∂xi ∂xj+

n∑i=1

bi(x, t)∂u

∂xi+ c(x, t)u − ∂u

∂t= 0. (4.12)

482

A function Z(x, t; ξ, τ) is called a fundamental solution of Eq. (4.12) in the layer H if it has the followingproperties:

(1) The function Z(x, t; ξ, τ) in the domain 0 τ < t T, x ∈ En, ξ ∈ En is continuous jointly in x,t, ξ, and τ together with its derivatives ∂2Z

∂xi ∂xj(i, j = 1, 2, . . . , n) and ∂Z

∂t and satisfies Eq. (4.12) with respect tovariables x and t. The function Z(x, t; ξ, τ) is bounded in any domain t− τ + |x− ξ| δ, where δ > 0.

(2) For any continuous and bounded function ϕ(ξ) in En and for all x ∈ En, τ ∈ [0, T ) the relation

limt→τ+0

∫En

Z(x, t; ξ, τ)ϕ(ξ) dξ = ϕ(x) (4.13)

holds, and the integral tends to its limit uniformly with respect to x in any bounded domain of the space En.

Theorem 1. Suppose all the coefficients of Eq. (4.12) are bounded and continuous in H jointly in x and tand are Holder continuous with respect to x:

|aij(x′, t)− aij(x, t)| M4|x′ − x|λ,|bi(x′, t)− bi(x, t)| M4|x′ − x|λ,|c(x′, t)− c(x, t)| M4|x′ − x|λ

(i, j = 1, . . . , n; λ > 0). (4.14)

Suppose additionally that the coefficients aij in H are Holder continuous with respect to t:

|aij(x, t′)− aij(x, t)| M4|t′ − t|λ. (4.15)

Let us also suppose that the inequality (4.5) holds.Then there exists a unique fundamental solution Z(x, t; ξ, τ) of Eq. (4.12) in the layer H . For Z(x, t; ξ, τ)

the estimates

|Z(x, t; ξ, τ)| < M5(t− τ)−n2 e−

µ1|x−ξ|2t−τ ,∣∣∣∣∂Z(x, t; ξ, τ)∂xi

∣∣∣∣ < M5(t− τ)−n+1

2 e−µ1|x−ξ|2

t−τ ,∣∣∣∣∂2Z(x, t; ξ, τ)∂xi ∂xj

∣∣∣∣ < M5(t− τ)−n2−1e−

µ1|x−ξ|2t−τ ,∣∣∣∣∂Z(x, t; ξ, τ)∂t

∣∣∣∣ < M5(t− τ)−n2−1e−

µ1|x−ξ|2t−τ

(4.16)

hold; here M5 and µ1 are positive constants. The function Z(x, t; ξ, τ) is positive everywhere for t > τ . If in H there

exist bounded and continuous derivatives∂aij∂xj

,∂2aij∂xi ∂xj

, and ∂bi∂xi

(i, j = 1, . . . , n), Holder continuous with respect

to x, then for t > τ the function Z(x, t; ξ, τ) as a function of variables ξ, τ satisfies the equation conjugate to (4.12):

L∗ξ,τ (Z) ≡n∑

i,j=1

∂2[aij(ξ, τ)Z]∂ξi ∂ξj

−n∑i=1

∂[bi(ξ, τ)Z]∂ξi

+ c(ξ, τ)Z +∂Z

∂τ= 0. (4.17)

Proof. We shall try to find the fundamental solution in the form

Z(x, t; ξ, τ) =Wξ,τ (x, t; ξ, τ) +

t∫τ

dθ

∫En

Wζ,θ(x, t; ζ, θ)Φ(ζ, θ; ξ, τ) dζ, (4.18)

where the function Wξ,τ (x, t; ξ, τ) is given by (4.4) and the function Φ(x, t; ξ, τ) is to be found.First we shall study the properties of the improper integral

V (x, t; ξ, τ) =

t∫τ

dθ

∫En

Wζ,θ(x, t; ζ, θ)Φ(ζ, θ; ξ, τ) dζ. (4.19)

483

According to (4.4), for the function Wζ,θ(x, t; ξ, τ) we have the following estimate:

|Wζ,θ(x, t; ξ, τ)| < M6(t− τ)−n2 e−

µ1|x−ξ|2t−τ , µ1 > 0. (4.20)

Using the inequality (4.9) we obtain estimates for the derivatives of Wζ,θ(x, t; ξ, τ):∣∣∣∣∂Wζ,θ(x, t; ξ, τ)∂xi

∣∣∣∣ < M6(t− τ)−n+1

2 e−µ2|x−ξ|2

t−τ , (4.21)∣∣∣∣∂2Wζ,θ(x, t; ξ, τ)∂xi ∂xj

∣∣∣∣ < M6(t− τ)−n2−1e−

µ2|x−ξ|2t−τ , (4.22)∣∣∣∣∂Wζ,θ(x, t; ξ, τ)

∂t

∣∣∣∣ < M6(t− τ)−n2−1e−

µ2|x−ξ|2t−τ ; (4.23)

here µ1 > µ2 > 0.Suppose that the function Φ(x, t; ξ, τ) is continuous jointly in x, t, ξ, τ for t > τ and for any x, ξ and can be

estimated in the following way:

|Φ(x, t; ξ, τ)| < M7(t− τ)−n2−1+λ1e−

µ3|x−ξ|2t−τ , (4.24)

where λ1 > 0 and µ3 > 0. Moreover, we shall suppose that for |x′ − x|2 < a(t− τ), where a > 0 is a constant, wehave

|Φ(x′, t; ξ, τ)− Φ(x, t; ξ, τ)| M7|x′ − x|λ2(t− τ)− n2−1+λ3e−

µ3|x−ξ|2t−τ , (4.25)

where λ2 > 0 and λ3 > 0. Let us show that under such suppositions the function (4.19) is continuous and hascontinuous derivatives ∂V

∂xi, ∂2V∂xi ∂xj

(i, j = 1, . . . , n), and ∂V∂t for t > τ and any x and ξ.

According to our suppositions, the function

J(x, t, θ; ξ, τ) =∫En

Wζ,θ(x, t; ζ, θ)Φ(ζ, θ; ξ, τ) dζ (4.26)

is continuous with respect to all its arguments and has derivatives of any order with respect to x and t for τ < θ < tand any x and ξ. Taking into account (4.20) and (4.24), we obtain

|J(x, t, θ; ξ, τ)| < M8

∫En

(t− θ)−n2 e−

µ1|x−ζ|2t−θ (θ − τ)− n

2−1+λ1e−µ3|ζ−ξ|2θ−τ dζ

< M8(θ − τ)λ1−1n∏i=1

∞∫−∞

[(t− θ)(θ − τ)]− 12 e−µ4

[(xi−ξi)

2

t−θ +(ζi−ξi)

2

θ−τ

]dζi

=M8(θ − τ)λ1−1n∏i=1

π

12 [µ4(t− τ)]−

12 e−µ4

(xi−ξi)2t−τ

=M9(θ − τ)λ1−1(t− τ)− n

2 e−µ4|x−ξ|2

t−τ . (4.27)

Therefore, for t− τ δ > 0, where δ > 0 is arbitrary, and for any x and ξ the inequality

|J(x, t, θ; ξ, τ)| < M(1)δ (θ − τ)λ1−1 (4.28)

holds. This allows us to establish the uniform convergence of the integralt∫τ

J(x, t, θ; ξ, τ) dθ with respect to x, t, ξ,

and τ . Thus, we have established that the function V (x, t; ξ, τ) is continuous jointly in all its arguments for t > τand any x and ξ. According to (4.27), we have

|V (x, t; ξ, τ)| t∫

τ

|J(x, t, θ; ξ, τ)| dθ < M10(t− τ)−n2 +λ1e−

µ4|x−ξ|2t−τ . (4.29)

484

Using (4.21) and (4.24), with the help of similar arguments we obtain the estimate

∣∣∣∣∂J(x, t, θ; ξ, τ)∂xi

∣∣∣∣ ∫En

∣∣∣∣∂Wζ,θ(x, t; ζ, θ)∂xi

∣∣∣∣ · |Φ(ζ, θ; ξ, τ)| dζ < M11(t− θ)−12 (θ − τ)λ1−1(t− τ)− n

2 e−µ4|x−ξ|2

t−τ . (4.30)

Hence ∣∣∣∣∂J(x, t, θ; ξ, τ)∂xi

∣∣∣∣ < M(2)δ (t− θ)− 1

2 (θ − τ)λ1−1

for t− τ δ > 0 and any x and ξ. Therefore the integral

t∫τ

∂J(x, t, θ; ξ, τ)∂xi

dθ

converges uniformly with respect to x, t, ξ, and τ in the above-mentioned domain.According to the well-known theorem (see [77]), it follows that for t > τ and any x and ξ the derivatives

∂V

∂xi=

t∫τ

dθ

∫En

∂Wζ,θ(x, t; ζ, θ)∂xi

Φ(ζ, θ; ξ, τ) dζ (i = 1, . . . , n) (4.31)

exist and are continuous. According to the inequality (4.30), for the derivatives ∂V∂xi

we have the estimate

∣∣∣∣∂V (x, t; ξ, τ)∂xi

∣∣∣∣ < M12(t− τ)−n+1

2 +λ1e−µ4|x−ξ|2

t−τ . (4.32)

To prove the existence of the derivatives ∂2V∂xi ∂xj

(i, j = 1, . . . , n) consider for t− τ δ > 0 the integral

F (x, t; ξ, τ) =

t∫τ

dθ

∫En

∂2Wζ,θ(x, t; ζ, θ)∂xi ∂xj

Φ(ζ, θ; ξ, τ) dζ =

t∫τ

∂2J(x, t, θ; ξ, τ)∂xi ∂xj

dθ.

It can be represented as a sum:

F (x, t; ξ, τ) =

τ+ δ2∫

τ

∂2J(x, t, θ; ξ, τ)∂xi ∂xj

dθ +

t∫τ+ δ

2

∂2J(x, t, θ; ξ, τ)∂xi ∂xj

dθ = F1(x, t; ξ, τ) + F2(x, t; ξ, τ).

Using the estimates (4.22) and (4.24), we get

∣∣∣∣∂2J(x, t, θ; ξ, τ)∂xi ∂xj

∣∣∣∣ < M13

∫En

(t− θ)−n2−1(θ − τ)− n

2−1+λ1e−µ2|x−ζ|2

t−θ −µ3|ζ−ξ|2θ−τ dζ

< M14(t− θ)−1(θ − τ)λ1−1(t− τ)− n2 e−

µ4|x−ξ|2t−τ . (4.33)

Therefore, for τ < θ < τ + δ2 the inequality

∣∣∣∣∂2J(x, t, θ; ξ, τ)∂xi ∂xj

∣∣∣∣ < M(3)δ (θ − τ)λ1−1

holds, and it follows that the integral F1 converges uniformly in the domain t− τ δ > 0, x ∈ En, ξ ∈ En.

485

To estimate the function under the integral sign in F2, we shall first transform ∂J(x,t,θ;ξ,τ)∂xi

in the followingway:

∂J

∂xi=∫En

∂Wζ,θ(x, t; ζ, θ)∂xi

Φ(ζ, θ; ξ, τ) dζ = Φ(y, θ; ξ, τ)∫En

∂Wy,θ(x, t; ζ, θ)∂xi

dζ

+Φ(y, θ; ξ, τ)∫En

[∂Wζ,θ(x, t; ζ, θ)

∂xi− ∂Wy,θ(x, t; ζ, θ)

∂xi

]dζ +

∫En

[Φ(ζ, θ; ξ, τ) − Φ(y, θ; ξ, τ)]∂Wζ,θ(x, t; ζ, θ)

∂xidζ;

(4.34)

here y is an arbitrary fixed point. Let x be located inside the ball K of radius 12 with center at an arbitrary

point x0. In the first term of the right-hand side of (4.34) we can single out the integral over a ball K1 of radius 1,concentric to the ball K, and apply to this integral the Gauss–Ostrogradski formula, taking into account that∂Wy,θ(x,t;ζ,θ)

∂xi= −∂Wy,θ(x,t;ζ,θ)

∂ζi. We obtain

∂J

∂xi= −Φ(y, θ; ξ, τ)

∫Σ1

Wy,θ(x, t; ζ, θ) cos(ν, ζi) dSζ +Φ(y, θ; ξ, τ)∫

En\K1

∂Wy,θ(x, t; ζ, θ)∂xi

dζ

+Φ(y, θ; ξ, τ)∫En

[∂Wζ,θ(x, t; ζ, θ)

∂xi− ∂Wy,θ(x, t; ζ, θ)

∂xi

]dζ +

∫En

[Φ(ζ, θ; ξ, τ) − Φ(y, θ; ξ, τ)]∂Wζ,θ(x, t; ζ, θ)

∂xidζ;

(4.35)

here Σ1 is the boundary of the ball K1 and ν is the outer normal to Σ1. Differentiating (4.35) with respect to xiand then setting y = x, we get

∂2J

∂xi ∂xj= −Φ(x, θ; ξ, τ)

∫Σ1

[∂Wy,θ(x, t; ζ, θ)

∂xj

]y=x

cos(ν, ζi) dSζ +Φ(x, θ; ξ, τ)∫

En\K1

[∂2Wy,θ(x, t; ζ, θ)

∂xi ∂xj

]y=x

dζ

+Φ(x, θ; ξ, τ)∫En

∂2Wζ,θ(x, t; ζ, θ)

∂xi ∂xj−[∂2Wy,θ(x, t; ζ, θ)

∂xi ∂xj

]y=x

dζ

+∫En

[Φ(ζ, θ; ξ, τ) − Φ(x, θ; ξ, τ)]∂2Wζ,θ(x, t; ζ, θ)

∂xi ∂xjdζ = I1 + I2 + I3 + I4. (4.36)

Since the inequality (4.21) holds, we have for the function under the integral sign in I1 for x ∈ K, ζ ∈ Σ1

the estimate ∣∣∣∣∣[∂Wy,θ(x, t; ζ, θ)

∂xj

]y=x

cos(ν, ζi)

∣∣∣∣∣ < M15(t− θ)−n+1

2 e−µ2

4(t−θ) < M16.

Taking into account additionally (4.24), we get

|I1| < M16ωn ·M7(θ − τ)−n2−1+λ1e−

µ3|x−ξ|2θ−τ < M17(θ − τ)−

n2−1+λ1e−

µ3|x−ξ|2θ−τ . (4.37)

(Here ωn is the area of the unit sphere in the space En.) Hence

|I1| < M(4)δ for τ +

δ

2< θ < t, x ∈ K, ξ ∈ En. (4.38)

Then with the help of the inequalities (4.22), (4.24), and (4.9), we can estimate |I2|:

486

|I2| < M18(θ − τ)−n2−1+λ1e−

µ3|x−ξ|2θ−τ

∫|x−ζ|1

2

(t− θ)−n2−1e−

µ2|x−ζ|2t−θ dζ

M19(θ − τ)−n2−1+λ1e−

µ3|x−ξ|2t−τ

∫|x−ζ|1

2

(t− θ)−n2 |x− ζ|−2e−

µ22|x−ζ|2t−θ dζ

4M19ωn(θ − τ)−n2−1+λ1e−

µ3|x−ξ|2t−τ

+∞∫12

(t− θ)−n2 rn−1e−

µ22

r2t−θ dr

< M20(θ − τ)−n2−1+λ1e−

µ3|x−ξ|2t−τ

+∞∫0

qn2−1e−

µ22 q dq =M21(θ − τ)−

n2−1+λ1e−

µ3|x−ξ|2t−τ . (4.39)

It follows that

|I2| < M(5)δ for τ +

δ

2< θ < t, x ∈ K, ξ ∈ En. (4.40)

Applying the mean-value theorem and using the inequalities (4.9), (4.14), and (4.22), we come to the followingestimate for the function under the integral sign in I3:

∣∣∣∣∣∂2Wζ,θ(x, t; ζ, θ)

∂xi ∂xj−[∂2Wy,θ(x, t; ζ, θ)

∂xi ∂xj

]y=x

∣∣∣∣∣ < M22(t− θ)−n2−1+λ

2 e−µ4|x−ζ|2

t−θ . (4.41)

From (4.24) and (4.41) we obtain

|I3| < M23(θ − τ)−n2−1+λ1(t− θ)λ2−1e−

µ3|x−ξ|2t−τ . (4.42)

It follows that

|I3| < M(6)δ (t− θ)λ2−1 for τ +

δ

2< θ < t, x ∈ K, ξ ∈ En. (4.43)

The integral I4 can be represented as the sum I4 = I(1)4 +I(2)

4 , where the integral I(1)4 is taken over the domain

defined by the inequality |ζ − x|2 < a(θ − τ). Taking into account (4.22) and (4.25), we obtain

|I(1)4 | < M24

∫|ζ−x|2<a(θ−τ)

|ζ − x|λ2(θ − τ)− n2−1+λ3e−

µ3|ζ−ξ|2θ−τ (t− θ)− n

2−1e−µ2|x−ζ|2

t−θ dζ

< M25

∫En

(t− θ)−n2−1+

λ22 (θ − τ)− n

2−1+λ3e−µ4

(|x−ζ|2t−θ + |ζ−ξ|

2

θ−τ

)dζ.

Using arguments similar to those used when we deduced the inequality (4.27), we get

|I(1)4 | < M26(t− θ)

λ22 −1(θ − τ)λ3−1(t− τ)− n

2 e−µ4|x−ξ|2

t−τ . (4.44)

Therefore, for τ + δ2 < θ < t, x ∈ K, ξ ∈ En the estimate

|I(1)4 | < M

(7)δ (t− θ)

λ22 −1 (4.45)

is valid. The integral I(2)4 can be estimated with the help of inequalities (4.22) and (4.24):

487

|I(2)4 | < M27

∫|ζ−x|2a(θ−τ)

(θ − τ)− n2−1+λ1e−

µ3|ζ−ξ|2θ−τ (t− θ)−n

2−1e−µ2|x−ζ|2

t−θ dζ

< M27

∫En

(θ − τ)− n2−1+

λ12 e−

µ3|ζ−ξ|2θ−τ a−

λ12 |ζ − x|λ1(t− θ)−n

2−1e−µ2|x−ζ|2

t−θ dζ

< M28

∫En

(θ − τ)− n2−1+

λ12 (t− θ)−n

2−1+λ12 e−µ4

(|x−ζ|2t−θ + |ζ−ξ|

2

θ−τ

)dζ

< M29(t− θ)λ12 −1(θ − τ)

λ12 −1(t− τ)− n

2 e−µ4|x−ξ|2

t−τ . (4.46)

If τ + δ2 < θ < t, x ∈ K, ξ ∈ En this leads to

|I(2)4 | < M

(8)δ (t− θ)

λ12 −1. (4.47)

It follows from relations (4.38), (4.40), (4.43), (4.45), and (4.47) that the integral

F2(x, t; ξ, τ) =

t∫τ+δ

2

∂2J(x, t, θ; ξ, τ)∂xi ∂xj

dθ

converges uniformly in the domain t− τ δ, x ∈ K, ξ ∈ En. We have shown before that the integral

F1(x, t; ξ, τ) =

τ+ δ2∫

τ

∂2J(x, t, θ; ξ, τ)∂xi ∂xj

dθ

also converges uniformly in this domain. Thus, we have proved the uniform convergence of the integral F (x, t; ξ, τ) =F1(x, t; ξ, τ) + F2(x, t; ξ, τ) in this domain. Therefore in this domain there exist continuous derivatives

∂2V (x, t; ξ, τ)∂xi ∂xj

=

t∫τ

dθ

∫En

∂2Wζ,θ(x, t; ζ, θ)∂xi ∂xj

Φ(ζ, θ; ξ, τ) dζ (i, j = 1, . . . , n). (4.48)

Since the number δ > 0 and the point x0 (the center of the ball K) are arbitrary, equality (4.48) holds everywherein the domain 0 τ < t T, x ∈ En, ξ ∈ En.

Because of inequalities (4.33), (4.37), (4.39), (4.42), (4.44), and (4.46), we get

∣∣∣∣∂2V (x, t; ξ, τ)∂xi ∂xj

∣∣∣∣ τ+ δ

2∫τ

∣∣∣∣∂2J(x, t, θ; ξ, τ)∂xi ∂xj

∣∣∣∣ dθ +t∫

τ+δ2

∣∣∣∣∂2J(x, t, θ; ξ, τ)∂xi ∂xj

∣∣∣∣ dθ < M30(t− τ)−n2−1+λ4e−

µ4|x−ξ|2t−τ , (4.49)

where λ4 > 0.Now we shall show that for t > τ and any x and ξ the function V (x, t; ξ, τ) has a continuous derivative ∂V

∂t ,which can be calculated by the following formula:

∂V (x, t; ξ, τ)∂t

= Φ(x, t; ξ, τ) +

t∫τ

dθ

∫En

∂Wζ,θ(x, t; ζ, θ)∂t

Φ(ζ, θ; ξ, τ) dζ = Φ(x, t; ξ, τ) +

t∫τ

∂J(x, t, θ; ξ, τ)∂t

dθ. (4.50)

According to Eq. (4.3), for Wξ,τ (x, t; ξ, τ) we have

∂J(x, t, θ; ξ, τ)∂t

=∫En

n∑i,j=1

aij(ζ, θ)∂2Wζ,θ(x, t; ζ, θ)

∂xj ∂xiΦ(ζ, θ; ξ, τ) dζ (4.51)

488

if τ < θ < t. Therefore the integralt∫

τ

∂J(x, t, θ; ξ, τ)∂t

dθ (4.52)

is similar to the integral (4.48) we have studied before, and as before, we can see that the integral (4.52) convergesabsolutely and uniformly in the domain t− τ δ > 0, x ∈ K, ξ ∈ En.

Let us consider the difference

V (x, t+∆t; ξ, τ)− V (x, t; ξ, τ)∆t

− Φ(x, t; ξ, τ) −t∫

τ

∂J(x, t, θ; ξ, τ)∂t

dθ

=∫τ

[J(x, t+∆t, θ; ξ, τ) − J(x, t, θ; ξ, τ)

∆t− ∂J(x, t, θ; ξ, τ)

∂t

]dθ +

1∆t

t+∆t∫t

J(x, t+∆t, θ; ξ, τ) dθ − Φ(x, t; ξ, τ)

= J(x, t+∆t, t′; ξ, τ)− Φ(x, t; ξ, τ) −( τ+η1∫

τ

+

t∫t−η2

)∂J

∂t(x, t, θ; ξ, τ) dθ

+( τ+η1∫

τ

+

t∫t−η2

)∂J

∂t(x, t′′, θ; ξ, τ) dθ +

t−η2∫τ+η1

[∂J

∂t(x, t′′, θ; ξ, τ) − ∂J

∂t(x, t, θ; ξ, τ)

]dθ, (4.53)

where t′ and t′′ are some values between t and t+∆t. To be definite we shall suppose that ∆t > 0. According torelation (4.6) we get

lim∆t→0

J(x, t+∆t, t′; ξ, τ)− Φ(x, t; ξ, τ) = 0.

Since the integral (4.52) converges uniformly with respect to t, for sufficiently small η1 > 0 and η2 > 0 the firstthree integrals in the right-hand side of (4.53) can be made arbitrarily small in absolute value independently of ∆t.The fourth integral can be estimated in the following way:

∣∣∣∣t∫

t−η2

∂J

∂t(x, t′′, θ; ξ, τ) dθ

∣∣∣∣ t′′∫

t−η2

∣∣∣∣∂J∂t (x, t′′, θ; ξ, τ)∣∣∣∣ dθ.

It follows that this integral can be made arbitrarily small for sufficiently small η2 > 0 and ∆t. Finally for fixedη1 and η2 the last integral in the right-hand side of (4.53) tends to zero as ∆t → 0, because the function underthe integral sign tends to zero and is continuous in θ and t′′ in the closed domain τ + η1 θ t− η2, t t′′ T .

Thus, the left-hand side of the equality (4.53) tends to zero as ∆t → +0. In a similar way we can considerthe case where ∆t→ −0.

The equality (4.50) is proved.Now we pass to the immediate construction of the fundamental solution Z(x, t; ξ, τ) in the form (4.18).Suppose that the function Φ(x, t; ξ, τ) satisfies the conditions (4.24) and (4.25). Using formulas (4.18), (4.31),

(4.48), and (4.50), we get

Lx,t(Z) = Lx,t(Wξ,τ (x, t; ξ, τ)) +

t∫τ

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ − Φ(x, t; ξ, τ).

The requirement that Lx,t(Z) be zero leads to the following equation for Φ:

Φ(x, t; ξ, τ) = Lx,t(Wξ,τ (x, t; ξ, τ)) +

t∫τ

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ. (4.54)

489

We seek a solution of the integral equation (4.54) as a series

Φ(x, t; ξ, τ) =∞∑

m=1

Φm(x, t; ξ, τ), (4.55)

where

Φ1(x, t; ξ, τ) = Lx,t(Wξ,τ (x, t; ξ, τ)),

Φm+1(x, t; ξ, τ) =

t∫τ

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))Φm(ζ, θ; ξ, τ) dζ (m = 1, 2, . . .). (4.56)

Let us show that the series (4.55) converges uniformly for t > τ . Taking into account (4.3), we have forWξ,τ (x, t; ξ, τ)

Φ1(x, t; ξ, τ) =n∑

i,j=1

[aij(x, t) − aij(ξ, τ)]∂2Wξ,τ (x, t; ξ, τ)

∂xi ∂xj+

n∑i=1

bi(x, t)∂Wξ,τ (x, t; ξ, τ)

∂xi+ c(x, t)Wξ,τ (x, t; ξ, τ).

By inequalities (4.14), (4.15), and (4.20)–(4.22) we get the estimate

|Φ1(x, t; ξ, τ)| = |Lx,t(Wξ,τ (x, t; ξ, τ))| < M31(t− τ)−n2−1+λe−

µ5|x−ξ|2t−τ , (4.57)

where µ5 > 0. Then we have

|Φ2(x, t; ξ, τ)| < M231

t∫τ

dθ

∫En

(t− θ)−n2−1+λe−

µ5|x−ζ|2t−θ (θ − τ)− n

2−1+λe−µ5|ζ−ξ|2θ−τ dζ

=M231

t∫τ

(t− θ)λ−1(θ − τ)λ−1 dθ

n∏i=1

+∞∫−∞

[(t− θ)(θ − τ)]− 12 e−µ5

[(xi−ζi)

2

t−θ +(ζi−ξi)

2

θ−τ

]dζi

=M231π

n2 [µ5(t− τ)]−

n2 e−

µ5|x−ξ|2t−τ

t∫τ

(t− θ)λ−1(θ − τ)λ−1 dθ

=M231π

n2 [µ5(t− τ)]−

n2 e−

µ5|x−ξ|2t−τ (t− τ)2λ−1

1∫0

sλ−1(1− s)λ−1 ds =Γ2(λ)Γ(2λ)

M231

(π

µ5

)n2

(t− τ)−n2−1+2λe−

µ5|x−ξ|2t−τ ,

where Γ(h) is the gamma-function.It is easy to show by induction with respect to m that

|Φm(x, t; ξ, τ)| < Γm(λ)Γ(mλ)

Mm31

(π

µ5

) (m−1)n2

(t− τ)mλ− n2−1e−

µ5|x−ξ|2t−τ (m = 1, 2, . . .). (4.58)

Since we have Γ(mλ) [mλ− 1]! for mλ > 2, it follows from (4.58) that the series (4.55) converges absolutely anduniformly for t > τ and the inequality (4.24) holds for the function Φ(x, t; ξ, τ). From the uniform convergenceof the series (4.55) it follows that the function Φ(x, t; ξ, τ) is continuous jointly in all its arguments in the domain0 τ < t T, x ∈ En, ξ ∈ En.

Let us show that the function Φ(x, t; ξ, τ) we have constructed satisfies the inequality (4.25) for |x′ − x|2 <a(t− τ). It is sufficient to prove that both terms of the right-hand side of Eq. (4.54) satisfy the inequality (4.25).

Consider first the difference L(Wξ,τ (x′, t; ξ, τ))− L(Wξ,τ (x, t; ξ, τ)). We shall concentrate on finding boundsonly for the terms most difficult to estimate, that is, terms containing second derivatives of the function Wξ,τ .Using the inequalities (4.14), (4.15), and (4.22), we obtain

490

∣∣∣∣[aij(x′, t)− aij(ξ, τ)]∂2Wξ,τ (x′, t; ξ, τ)∂xi ∂xj

− [aij(x, t)− aij(ξ, τ)]∂2Wξ,τ (x, t; ξ, τ)

∂xi ∂xj

∣∣∣∣∣∣∣∣[aij(x′, t)− aij(x, t)]∂2Wξ,τ (x′, t; ξ, τ)

∂xi ∂xj

∣∣∣∣+∣∣∣∣[aij(x, t)− aij(ξ, τ)]

[∂2Wξ,τ (x′, t; ξ, τ)

∂xi ∂xj− ∂2Wξ,τ (x, t; ξ, τ)

∂xi ∂xj

]∣∣∣∣< M32|x′ − x|λ(t− τ)−

n2−1e−

µ2|x′−ξ|2t−τ

+M33|x′ − x|[|x − ξ|λ + (t− τ)λ](|x − ξ|+ |x′ − ξ|)(t− τ)− n2−2

e−

µ2|x−ξ|2t−τ +

(1 +|x− ξ|2t− τ

)e−

µ4|x−ξ|2t−τ

,

(4.59)

where x belongs to the interval joining x and x′. We consider first such ξ that satisfy the condition |ξ−x| 2|x−x′|,and then the remaining values of ξ. Using the inequality (4.9) we can see that for |x′−x|2 < 2a(t−τ) the right-handside of (4.59) is not greater than

M34|x′ − x|λ2(t− τ)− n2−1+λ3e−

µ5|x−ξ|2t−τ ,

where λ2 > 0 and λ3 > 0. Similarly we can estimate the remaining terms of the difference L(Wξ,τ (x′, t; ξ, τ)) −L(Wξ,τ (x, t; ξ, τ)). Thus we have

|L(Wξ,τ (x′, t; ξ, τ))− L(Wξ,τ (x, t; ξ, τ))|

< M35|x′ − x|λ2 (t− τ)− n2−1+λ3e−

µ5|x−ξ|2t−τ for |x′ − x|2 < 2a(t− τ). (4.60)

Now we pass to the estimate of the term Ψ(x, t; ξ, τ) of the right-hand side of (4.54). We have

Ψ(x′, t; ξ, τ)− Ψ(x, t; ξ, τ)

=

t∫t− 1

2a |x′−x|2

dθ

∫En

L(Wζ,θ(x′, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ −t∫

t− 12a |x′−x|2

dθ

∫En

L(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ

+

t− 12a |x

′−x|2∫τ

dθ

∫En

L(Wζ,θ(x′, t; ζ, θ))− L(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ = J1 + J2 + J3. (4.61)

By (4.24) and (4.57), for |x′ − x|2 < a(t− τ) we get

|J2| < M36

t∫t− 1

a |x′−x|2

dθ

∫En

(t− θ)− n2−1+λ(θ − τ)− n

2−1+λe−µ5

(|x−ζ|2t−θ + |ζ−ξ|

2

θ−τ

)dζ

=M37(t− τ)−n2 e−

µ5|x−ξ|2t−τ

t∫t− 1

a |x′−x|2

(t− θ)λ−1(θ − τ)λ−1 dθ

< M37

(t− τ − 1

2a|x′ − x|2

)λ−1

(t− τ)− n2 e−

µ5|x−ξ|2t−τ

t∫t− 1

a |x′−x|2

(t− θ)λ−1 dθ

< M38|x′ − x|2λ(t− τ)−n2−1+λe−

µ5|x−ξ|2t−τ . (4.62)

In a similar way we can estimate |J1|. Then, taking into account (4.60) and (4.24), we find

491

|J3| < M39|x′ − x|λ2

t− 12a |x

′−x|2∫τ

dθ

∫En

(t− θ)−n2−1+λ3(θ − τ)− n

2−1+λe−µ5

(|x−ζ|2t−θ + |ζ−ξ|

2

θ−τ

)dζ

=M40|x′ − x|λ2(t− τ)− n2 e−

µ5|x−ξ|2t−τ

t− 12a |x

′−x|2∫τ

(t− θ)λ3−1(θ − τ)λ−1 dθ

< M40|x′ − x|λ2(t− τ)− n2−1+λ+λ3e−

µ5|x−ξ|2t−τ

1− |x′−x|2

2a(t−τ)∫0

sλ−1(1− s)λ3−1 ds

< M41|x′ − x|λ2(t− τ)− n2−1+λ+λ3e−

µ5|x−ξ|2t−τ . (4.63)

Comparing the inequalities (4.60)–(4.63), we obtain the required estimate (4.25) for the right-hand side ofEq. (4.54) and, therefore, for the function Φ(x, t; ξ, τ). Thus we have proved that the function Z(x, t; ξ, τ), givenby formula (4.18), for t > τ is continuous jointly in the variables x, t, ξ, and τ , together with its derivatives ∂Z

∂xi,

∂2Z∂xi ∂xj

(i, j = 1, . . . , n), and ∂Z∂t . The function Z(x, t; ξ, τ) satisfies Eq. (4.12) with respect to the variables x and t.

It follows from estimate (4.20) for the function Wξ,τ (x, t; ξ, τ) and estimate (4.29) for the function (4.19)that Z(x, t; ξ, τ) is bounded for t− τ + |x− ξ| δ > 0.

Now let us show that for Z(x, t; ξ, τ) the relation (4.13) holds.Let ϕ(ξ) be a continuous and bounded function in En. Suppose that x is an arbitrary point of the space En

and 0 τ < T . Then we have∫En

Z(x, t; ξ, τ)ϕ(ξ) dξ =∫En

Wξ,τ (x, t; ξ, τ)ϕ(ξ) dξ +∫En

V (x, t; ξ, τ)ϕ(ξ) dξ,

where the function V (x, t; ξ, τ) is given by (4.19). Taking into account (4.29), we can see that

∣∣∣∣∫En

V (x, t; ξ, τ)ϕ(ξ) dξ∣∣∣∣ < M42 sup |ϕ|(t− τ)λ1 ;

therefore,

limt→τ+0

∫En

V (x, t; ξ, τ)ϕ(ξ) dξ = 0 (4.64)

uniformly with respect to x ∈ En. From (4.6) and (4.64) the relation (4.13) follows.Thus we have constructed the function Z(x, t; ξ, τ) that is a fundamental solution of Eq. (4.12) in the layerH .Let us prove the uniqueness of the fundamental solution of Eq. (4.12).From the properties of the fundamental solution it follows that the function

u(x, t) =∫En

Z(x, t; ξ, τ)ϕ(ξ) dξ (4.65)

is a bounded solution of the Cauchy problem for Eq. (4.12) in the layer Hτ0 τ < t T, x ∈ En with the initialcondition

u(x, τ) = ϕ(x), (4.66)

where ϕ(x) is an arbitrary continuous function vanishing outside of some bounded domain. In fact, let ϕ(x) = 0for |x| R; then

u(x, t) =∫|ξ|R

Z(x, t; ξ, τ)ϕ(ξ) dξ.

492

For t > τ , we have

L(u) =∫|ξ|R

Lx,t(Z(x, t; ξ, τ))ϕ(ξ) dξ = 0.

According to (4.13) the function (4.65) for t = τ satisfies the initial condition (4.66). From (4.13) we can alsodeduce that the function (4.65) is bounded in the domain τ < t T, |x| 2R. Finally in the domainτ < t T, |x| 2R this function also is bounded because in this domain we have

|u(x, t)| sup|x−ξ|R

|Z(x, t; ξ, τ)|∫|ξ|R

|ϕ(ξ)| dξ MR;

this follows from the boundedness of Z(x, t; ξ, τ) for |x− ξ| R.According to Theorem 10 of Sect. 1, a bounded solution of the Cauchy problem (4.12), (4.66) is unique in

the layer Hτ .Suppose now that besides the fundamental solution Z(x, t; ξ, τ) of Eq. (4.12) that we have constructed

above, there exists another fundamental solution Z1(x, t; ξ, τ) of the same equation. Then, taking into accountthe uniqueness of the solution for the Cauchy problem (4.12), (4.66), we obtain∫

En

[Z(x, t; ξ, τ) − Z1(x, t; ξ, τ)]ϕ(ξ) dξ = 0

for any x ∈ En and 0 τ < t T and for any continuous function ϕ(ξ) vanishing outside some bounded domain.Since Z and Z1 are continuous functions for t > τ , it follows that Z1(x, t; ξ, τ) ≡ Z(x, t; ξ, τ) for t > τ , x ∈ En, andξ ∈ En.

Taking into account estimates (4.20)–(4.22) for Wξ,τ (x, t; ξ, τ), and also the inequalities (4.29), (4.32), and(4.49) for the function (4.19), we can deduce that for the functions

Z(x, t; ξ, τ),∂Z(x, t; ξ, τ)

∂xi,∂2Z(x, t; ξ, τ)

∂xi ∂xj(i, j = 1, . . . , n)

the estimates (4.16) hold. It follows from Eq. (4.12) that the estimate (4.16) holds for ∂Z(x,t;ξ,τ)∂t .

Now we shall prove that Z(x, t; ξ, τ) > 0 for t > τ .Suppose first that Z(x, t; ξ, τ) < 0 for some x0 ∈ En, ξ0 ∈ En, t = t0, and τ , t0 > τ . Then, because

Z(x0, t0; ξ, τ) is continuous, we have Z(x0, t0; ξ, τ) < 0 for |ξ − ξ0| < ε, where ε > 0. Consider Eq. (4.12) withinitial condition (4.66) in the layer Hτ , where ϕ(x) > 0 for |x − ξ0| < ε and ϕ(x) ≡ 0 for |x − ξ0| ε. Accordingto Theorem 8 of Sect. 1, the solution u(x, t) of the Cauchy problem (4.12), (4.66) must be nonnegative everywherein Hτ . On the other hand, we find from (4.65) that

u(x0, t0) =∫

|ξ−ξ0|<ε

Z(x0, t0; ξ, τ)ϕ(ξ) dξ < 0.

This contradiction shows that Z(x, t; ξ, τ) 0 everywhere for t > τ .Suppose now that Z(x, t; ξ, τ) = 0 for x = x1, ξ = ξ1, t = t1, and τ = τ1, where t1 > τ1.Consider Z(x, t; ξ1, τ1) in the cylinder Rτ1 + ε t t1, |x − x1| N, where 0 < ε < t1 − τ1 and

N > 0 is an arbitrary number. Since Z(x, t; ξ1, τ1) 0 in R and Z(x1, t1; ξ1, τ1) = 0, we have Z(x, t; ξ1, τ1) = 0everywhere in R. This follows from Theorem 6 of Sect. 1, which in the case of zero maximum or minimum holdsnot only for c(x, t) 0 but also for c(x, t) M , where M > 0. If ε tends to zero and N tends to infinity, we getZ(x, t; ξ1, τ1) = 0 everywhere in the layer τ1 < t t1, x ∈ En, and this is in contradiction with the formulas(4.4), (4.18), and (4.19) and the estimate (4.29). Thus we can see that the function Z(x, t; ξ, τ) is strictly positivein the domain 0 τ < t T, x ∈ En, ξ ∈ En.

Now we shall prove that Z(x, t; ξ, τ) satisfies Eq. (4.17) for t > τ as a function of the variables ξ and τ , if in Hthere exist continuous and bounded derivatives ∂aij(x,t)

∂xj, ∂2aij(x,t)

∂xi ∂xj, and ∂bi(x,t)

∂xi(i, j = 1, . . . , n), Holder continuous

with respect to x.

493

Transforming the left-hand side of Eq. (4.17) and changing ξ to x and τ to t, we get

L∗x,t(v) =n∑

i,j=1

aij(x, t)∂2v

∂xi ∂xj+

n∑i=1

[2

n∑j=1

∂aij(x, t)∂xj

− bi(x, t)]∂v

∂xi

+[ n∑i,j=1

∂2aij(x, t)∂xi ∂xj

−n∑i=1

∂bi(x, t)∂xi

+ c(x, t)]v − ∂v

∂(−t) = 0. (4.67)

According to the additional suppositions on the smoothness of aij and bi, we can see that the coefficients of Eq. (4.67)satisfy conditions (4.14) and (4.15). As we have proved, in the domain 0 t < τ T, x ∈ En, ξ ∈ En thereexists a unique fundamental solution Z∗(x, t; ξ, τ) of Eq. (4.67). We shall show that Z(x, t; ξ, τ) = Z∗(x, t; ξ, τ).

Consider the following easily verified identity:

vL(u)− uL∗(v) =n∑i=1

∂

∂xi

[ n∑j=1

aij

(v∂u

∂xj− u ∂v

∂xj

)+(bi −

n∑j=1

∂aij∂xj

)uv

]− ∂(uv)

∂t. (4.68)

Here we set u(x, t) = Z(x, t; ξ, τ) and v(x, t) = Z∗(x, t; ξ∗, τ∗), where ξ ∈ En, ξ∗ ∈ En, 0 τ < τ∗ T . For x ∈ En

and τ < t < τ∗ we have

Z∗(x, t; ξ∗, τ∗)L(Z(x, t; ξ, τ)) − Z(x, t; ξ, τ)L∗(Z∗(x, t; ξ∗, τ∗)) = 0. (4.69)

Let us integrate the identity (4.68) over the cylinder R′τ < t′ t t′′ < τ∗, |x| N1, where N1 > 0 isan arbitrary number, and after that use the Gauss–Ostrogradski formula. Taking into account (4.69), we get

0 =∫S′

n∑i=1

[ n∑j=1

aij

(Z∗

∂Z

∂xj− Z ∂Z

∗

∂xj

)+(bi −

n∑j=1

∂aij∂xj

)ZZ∗

]cos(ν, xi) dS

−∫

|x|N1

Z(x, t′′; ξ, τ)Z∗(x, t′′; ξ∗, τ∗) dx+∫

|x|N1

Z(x, t′; ξ, τ)Z∗(x, t′; ξ∗, τ∗) dx,

where S′ is the side boundary (|x| = N1) of the cylinder R′ and ν is a normal to S′. Let N1 tend to infinity.According to the estimates (4.16) for Z and similar estimates for Z∗, the integral over S′ tends to zero for N1 →∞,and we arrive at the following equality:∫

En

Z(x, t′′; ξ, τ)Z∗(x, t′′; ξ∗, τ∗) dx =∫En

Z(x, t′; ξ, τ)Z∗(x, t′; ξ∗, τ∗) dx.

Since the values t′ and t′′ are arbitrary, this means that the integral∫En

Z(x, t; ξ, τ)Z∗(x, t; ξ∗, τ∗) dx does not dependon t for τ < t < τ∗: ∫

En

Z(x, t; ξ, τ)Z∗(x, t; ξ∗, τ∗) dx = χ(ξ, τ ; ξ∗, τ∗). (4.70)

In the same way as we have proved the relation (4.13) it is easy to prove that

limt→τ+0

∫En

Z(x, t; ξ, τ)ϕ(x, t) dx = ϕ(ξ, τ) (4.71)

for any function ϕ(x, t) continuous and bounded in the layer τ t τ + δ, x ∈ En (δ > 0). Let us pass tothe limit in the equality (4.70) as t→ τ + 0; according to (4.71) we get

χ(ξ, τ ; ξ∗, τ∗) = Z∗(ξ, τ ; ξ∗, τ∗). (4.72)

On the other hand, taking into account the property of the function Z∗(x, t; ξ∗, τ∗) similar to (4.71), we can deducefrom (4.70) passing to the limit as t→ τ∗ − 0 that

χ(ξ, τ ; ξ∗, τ∗) = Z(ξ∗, τ∗; ξ, τ). (4.73)

From (4.72) and (4.73) it follows that Z(ξ∗, τ∗; ξ, τ) = Z∗(ξ, τ ; ξ∗, τ∗). Since ξ∗ and τ∗ were chosen in an arbitraryway, this means that Z(x, t; ξ, τ) = Z∗(x, t; ξ, τ) everywhere in the domain 0 τ < t T, x ∈ En, ξ ∈ En. Itfollows that Z(x, t; ξ, τ) satisfies (4.17) everywhere in this domain with respect to the variables ξ and τ .

Theorem 1 is proved.

494

Remark. We can get from relations (4.4), (4.18), and (4.19) the following estimate from below for Z(x, t; ξ, τ):

Z(x, t; ξ, τ) M43(t− τ)−n2 e−M44

|x−ξ|2t−τ −M45(t− τ)−

n2 +λ1e−µ4

|x−ξ|2t−τ . (4.74)

In particular, for |x− ξ|2 < a(t− τ), where a > 0, we obtain

Z(x, t; ξ, τ) M46(t− τ)−n2 . (4.75)

2. Cauchy Problem. The fundamental solution of Eq. (4.12) constructed above can be applied to the studyof the Cauchy problem for a linear second-order parabolic equation. We shall prove an existence theorem forthe solution of the Cauchy problem under very weak suppositions about the coefficients of the equation and itsright-hand side. In Sect. 3 we have proved the existence theorem by another method and under much strongersuppositions.

Theorem 2. In the layer H there exists a unique bounded solution of the Cauchy problem for Eq. (1.1)with the boundary condition

u(x, 0) = ϕ(x), x ∈ En, (4.76)

if the following suppositions are fulfilled:(1) The coefficients aij , bi, and c in the layer H are bounded, continuous, and satisfy the relations (4.14)

and (4.15), and the inequality (4.5) holds.(2) The function f(x, t) in the layer H is bounded, continuous, and Holder continuous with respect to x.(3) The function ϕ(x) is continuous and bounded in En.The solution of problem (1.1), (4.76) is given by the formula

u(x, t) = −t∫

0

dτ

∫En

Z(x, t; ξ, τ)f(ξ, τ) dξ +∫En

Z(x, t; ξ, 0)ϕ(ξ) dξ = U1(x, t) + U2(x, t). (4.77)

Proof. It follows from the properties of the fundamental solution Z(x, t; ξ, τ) and from estimates (4.16) thatthe function U2(x, t) in the layer H is bounded and for t > 0 satisfies the homogeneous equation (4.12). Accordingto (4.13), this function satisfies also the initial condition (4.76).

The first of the estimates (4.16) gives

|U1(x, t)| M47 sup |f |t∫

0

dτ

∫En

(t− τ)− n2 e−

µ1|x−ξ|2t−τ dξ M48t;

hence the function U1(x, t) is bounded in H and for t = 0 it satisfies the zero initial condition:

U1(x, 0) = 0.

It remains to prove that for t > 0 the function U1(x, t) satisfies Eq. (1.1). To do this, we represent U1(x, t)as a sum:

U1(x, t) = −t∫

0

dτ

∫En

Wξ,τ (x, t; ξ, τ)f(ξ, τ) dξ −t∫

0

dτ

∫En

V (x, t; ξ, τ)f(ξ, τ) dξ = U(1)1 (x, t) + U

(2)1 (x, t), (4.78)

where

V (x, t; ξ, τ) =

t∫τ

dθ

∫En

Wζ,θ(x, t; ζ, θ)Φ(ζ, θ; ξ, τ) dζ.

495

The integral U (1)1 (x, t) is similar to the integral (4.19) that we have studied when we constructed the funda-

mental solution. Taking into account condition (2) of this theorem, we deduce (as in the proof of Theorem 1) thatthe function U (1)

1 (x, t) for t > 0 satisfies the equation

L(U (1)1 (x, t)) = f(x, t)−

t∫0

dτ

∫En

Lx,t(Wξ,τ (x, t; ξ, τ))f(ξ, τ) dξ. (4.79)

In the integral U (2)1 (x, t) we can change the order of integration:

U(2)1 (x, t) = −

t∫0

dθ

∫En

Wζ,θ(x, t; ζ, θ)χ(ζ, θ) dζ, (4.80)

where

χ(ζ, θ) =

θ∫0

dξ

∫En

Φ(ζ, θ; ξ, τ)f(ξ, τ) dξ. (4.81)

We can do this because of estimates (4.20) and (4.24). By arguments similar to those used above we can deducethat the function (4.80) for t > 0 satisfies the following equation:

L(U (2)1 (x, t)) = χ(x, t)−

t∫0

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))χ(ζ, θ) dζ. (4.82)

Because of relations (4.14), (4.15), and (4.20)–(4.32) we have

|Lx,t(Wζ,θ(x, t; ζ, θ))| < M49(t− θ)−n2−1+λ

2 e−µ4|x−ζ|2

t−τ .

This inequality allows the change of integration order in the right-hand side of (4.82); we obtain

L(U (2)1 (x, t)) =

t∫0

dτ

∫En

[Φ(x, t; ξ, τ)−

t∫τ

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ]f(ξ, τ) dξ. (4.83)

Comparing (4.79) and (4.83), we get

L(U1(x, t)) = f(x, t) +

t∫0

dτ

∫En

f(ξ, τ)[Φ(x, t; ξ, τ) − Lx,t(Wξ,τ (x, t; ξ, τ))

−t∫

τ

dθ

∫En

Lx,t(Wζ,θ(x, t; ζ, θ))Φ(ζ, θ; ξ, τ) dζ]dξ = f(x, t),

because the function Φ(x, t; ξ, τ) for 0 τ < t T satisfies Eq. (4.54) and we deduce that the function U1(x, t) fort > 0 satisfies Eq. (1.1).

The uniqueness of the solution of problem (1.1), (4.76) is proved in Sect. 1.

Remark 1. With the help of inequalities (4.16) it is easy to show that the problem (1.1), (4.76) has a solutionalso in case where the function ϕ(x) is continuous for all x and satisfies the inequality

|ϕ(x)| < M50eM51|x|2−ε , ε > 0. (4.84)

496

In this case, the solution is again given by formula (4.77). It follows from this formula that

sup0tT

|u(x, t)| < M52eM53|x|2−ε + sup

H

|f(x, t)| · t. (4.85)

The uniqueness of the solutions of Cauchy problem in the class of functions satisfying (4.85) is proved in Sect. 1.It is interesting to note that one cannot prove the existence of the solution for the Cauchy problem for all

t > 0 if one allows ε = 0 in the estimate (4.84). In fact, the function u(x, t) = (1 − 4At)−12 e

Ax2

e1−4At is a solution ofthe equation ∂2u

∂x2 − ∂u∂t = 0 with the initial condition u(x, 0) = eAx

2(A > 0). This function is not defined for t 1

4A ,and lim

t→ 14A−0

u(x, t) =∞. According to the uniqueness theorem, for t < 14A there is no other solution of the Cauchy

problem growing not faster than eMx2.

Remark 2. If in Theorem 2 we suppose additionally that ϕ(x) ∈ C2+λ, then, using the methods of [43] wecan prove that the solution of problem (1.1), (4.76) belongs to the class C2,1 in the closed domain H .

3. First Boundary-Value Problem. Green’s Function. Consider now for Eq. (1.1) in the cylinderQ = Ω× (0, T ) the first boundary-value problem with conditions

u(x, 0) = ϕ(x) for x ∈ Ω, u|S = 0. (4.86)

It is well known that the first boundary-value problem for the heat equation can be solved with the help of the so-called double-layer heat potentials, similar to the double-layer potentials for the Laplace equation [78]. For Eq. (1.1)with coefficients satisfying the suppositions (4.14) and (4.15) it is not possible in general to construct a double-layerpotential, because the coefficients aij(ξ, τ) and, consequently, also the function Z(x, t; ξ, τ) are not assumed to bedifferentiable in ξ1, . . . , ξn. Therefore we shall use another method to construct the solution of problem (1.1), (4.86);it is due to W. Pogorzelski [79]. Note that we do not suppose the continuity of the solution of problem (1.1), (4.86)for t = 0 on the boundary σ of the domain Ω; therefore ϕ(x) on σ may be different from zero or even not defined.

Theorem 3. There exists in the cylinder Q a unique solution of the problem (1.1), (4.86) if the followingsuppositions are fulfilled:

(1) The coefficients aij , bi, and c in Q satisfy the inequalities (4.5), (4.14), and (4.15).(2) The function f(x, t) is continuous in Q jointly in the variables x and t and is Holder continuous with

respect to x.

(3) The domain Ω is bounded and belongs to the class A1+λ0 .(4) The function ϕ(x) is continuous and bounded in Ω.

Proof. A function G(x, t; ξ, τ) will be called a Green’s function of the boundary-value problem (1.1), (4.86)if it is defined and continuous jointly in the variables for x ∈ Ω, ξ ∈ Ω, 0 τ < t T and has the form

G(x, t; ξ, τ) = Z(x, t; ξ, τ) −Ψ(x, t; ξ, τ),

where Ψ(x, t; ξ, τ) has the following properties

(1) Lx,t(Ψ(x, t; ξ, τ)) = 0 for x ∈ Ω, ξ ∈ Ω, 0 τ < t T ; (4.87)(2) Ψ(x, t; ξ, τ) = Z(x, t; ξ, τ) for ξ ∈ Ω, (x, t) ∈ S, 0 τ < t T ; (4.88)(3) lim

t→τ+0Ψ(x, t; ξ, τ) = 0 for x ∈ Ω, ξ ∈ Ω. (4.89)

To construct a Green’s function it is sufficient to find the function Ψ(x, t; ξ, τ), satisfying the relations (4.87)–(4.89).We shall construct this function for ξ ∈ Ω and any x ∈ En.

Let us extend the coefficients aij , bi, c, and f of Eq. (1.1) to all of the layer H0 t T, x ∈ En so thatthey remain bounded and satisfy the conditions of this theorem. For x ∈ En \ Ω we set

Ψ(x, t; ξ, τ) = Z(x, t; ξ, τ),

497

where Z(x, t; ξ, τ) is a fundamental solution of Eq. (4.12) in the layer H . For a fixed ξ ∈ Ω, the function Ψ(x, t; ξ, τ)so defined is bounded in the domain Rτ0 τ < t T, x ∈ En \Ω because of the estimate (4.16). Inside Rτ thisfunction satisfies Eq. (4.12) and the conditions

∂Ψ(x1, t1; ξ, τ)∂γx1

=∂Z(x1, t; ξ, τ)

∂γx1on S,

lim|x|→∞

Ψ(x, t; ξ, τ) = 0, limt→τ

Ψ(x, t; ξ, τ) = 0.

(4.90)

Here∂F (x1, t)∂γx1

≡ limx→x1∈σx∈En\Ω

n∑i,j=1

aij(x, t) cos(νx1 , xi)∂F (x, t)∂xj

,

and νx1 is the direction of the inside normal (inside with respect to the domain En \ Ω) to σ at the point x1.According to the remark at the end of subsection 5 of Sect. 1, conditions (4.90) define a unique bounded

solution of Eq. (4.12) in the domain Rτ .On the other hand, the solution of problem (4.12), (4.90) in the domain Rτ can be found in the form of

an integral similar to the simple layer potential for the Laplace equation, that is, in the form

Ψ(x, t; ξ, τ) =

t∫τ

dθ

∫σ

Z(x, t; ζ, θ)ω(ζ, θ; ξ, τ) dσζ , (4.91)

where the function ω(x, t; ξ, τ) is to be defined. One can show (see [80]) that for integrals having the form (4.91)with a continuous function ω(x, t; ξ, τ) the formula

∂Ψ(x, t; ξ, τ)∂γx

=12ω(x, t; ξ, τ) +

t∫τ

dθ

∫σ

∂Z(x, t; ζ, θ)∂γx

ω(ζ, θ; ξ, τ) dσζ

holds for (x, t) ∈ S and t > τ , similar to the “jump” formula for the normal derivative for ordinary potentials. Bythe first of conditions (4.90) we obtain for ω(x, t; ξ, τ) the integral equation

ω(x, t; ξ, τ) = 2∂Z(x, t; ξ, τ)

∂γx−

t∫τ

dθ

∫σ

2∂Z(x, t; ζ, θ)

∂γxω(ζ, θ; ξ, τ) dσζ , x ∈ σ, τ < t T. (4.92)

The solution ω(x, t; ξ, τ) of (4.92) can be found by successive approximations, as was done above for the so-lution of Eq. (4.54). The solution ω(x, t; ξ, τ) is continuous for τ < t T and satisfies the inequality

|ω(x, t; ξ, τ)| < M54(t− τ)−n+1−λ0

2 e−µ4|x−ξ|2

t−τ ; (4.93)

here λ0 > 0 is the same number as in condition (3) of the theorem. Setting ω(x, τ ; ξ, τ) = 0, we obtain a functionthat is continuous everywhere for x ∈ σ and τ t T (recall that ξ ∈ Ω is a fixed point).

From (4.16) and (4.93) we can deduce that the function (4.91) is continuous and bounded in the closeddomain Rτ . Since the solution of the second boundary-value problem (4.12), (4.90) is unique, it follows that

Ψ(x, t; ξ, τ) = Z(x, t; ξ, τ)

for x ∈ En \ Ω, 0 τ < t T .The function (4.91) is defined and continuous not only outside Q but in all of the layer 0 τ < t T ,

x ∈ En. In particular, for x1 ∈ σ and τ < t we have

limx→x1

x∈Ω

Ψ(x, t; ξ, τ) = limx→x1

x∈En\Ω

Ψ(x, t; ξ, τ) = Z(x1, t; ξ, τ);

498

therefore, the function Ψ(x, t; ξ, τ) considered for x ∈ Ω, satisfies the relation (4.88). The relations (4.87) and (4.89)obviously hold.

Setting G(x, t; ξ, τ) = Z(x, t; ξ, τ) − Ψ(x, t; ξ, τ), we obtain the required Green’s function.Taking into account inequalities (4.16) and (4.93), it is easy to show that for Ψ(x, t; ξ, τ) the following

estimate holds:|Ψ(x, t; ξ, τ)| < M55(t− τ)−

n−λ02 e−

µ4|x−ξ|2t−τ . (4.94)

Therefore, for the Green’s function G(x, t; ξ, τ) we have the same estimate as for Z(x, t; ξ, τ):

|G(x, t; ξ, τ)| < M56(t− τ)−n2 e−

µ4|x−ξ|2t−τ . (4.95)

Now let us prove that the solution of the first boundary-value problem (1.1), (4.86) in the cylinderQ = Ω× (0, T ) is given by the formula

u(x, t) = −t∫

0

dτ

∫Ω

G(x, t; ξ, τ)f(ξ, τ) dξ +∫Ω

G(x, t; ξ, 0)ϕ(ξ) dξ = u1(x, t) + u2(x, t). (4.96)

The boundedness of the function (4.96) follows from the estimate (4.95) for G(x, t; ξ, τ). According to the sameestimate, |u1(x, t)| M57t; therefore u1(x, 0) = 0. The function u2(x, t) can be represented as a sum:

u2(x, t) =∫Ω

Z(x, t; ξ, 0)ϕ(ξ) dξ +∫Ω

−Ψ(x, t; ξ, 0)ϕ(ξ) dξ = u(1)2 (x, t) + u

(2)2 (x, t).

Similarly to (4.13) it is easy to show that if the function ϕ(x) is continuous and bounded in Ω we have

limt→+0

∫Ω

Z(x, t; ξ, 0)ϕ(ξ) dξ = ϕ(x) for x ∈ Ω; (4.97)

here the convergence is uniform in any closed domain lying strictly inside Ω. It follows that u(1)2 (x, 0) = ϕ(x) for

x ∈ Ω. As a consequence of (4.94) we have |u(2)2 (x, t)| M58t

λ0/2, hence u(2)2 (x, 0) = 0. Thus, the function (4.96)

satisfies the first of the conditions (4.86).Let us show that the function (4.96) satisfies Eq. (1.1) in Q \ Γ. The properties of G(x, t; ξ, τ) allow us to

assert that the function u2(x, t) in Q \ Γ is a solution of the homogeneous equation (4.12). The function u1(x, t)can be represented as follows:

u1(x, t) = −t∫

0

dτ

∫Ω

Z(x, t; ξ, τ)f(ξ, τ) dξ +

t∫0

dθ

∫σ

Z(x, t; ζ, θ)h(ζ, θ) dσζ = u(1)1 (x, t) + u

(2)1 (x, t),

where

h(ζ, θ) =

θ∫0

dτ

∫Ω

ω(ζ, θ; ξ, τ)f(ξ, τ) dξ;

the change of integration order effectuated in the second term is legitimate because of the estimate (4.93).It is obvious that u(2)

1 (x, t) in Q \Γ satisfies Eq. (4.12). In the same way as in Theorem 2, we can show thatu

(1)1 (x, t) in Q \ Γ is a solution of Eq. (1.1). Therefore, the function (4.96) in Q \ Γ satisfies Eq. (1.1).

It remains to verify that the function (4.96) satisfies the second of conditions (4.86).Let x1 ∈ σ. Denote byK& the intersection of the domain Ω and the n-dimensional ball of radius 6 with center

at the point x1. Consider the function u2(x, t) for x ∈ K& and 0 < δ t T , where δ > 0 is an arbitrarily smallnumber. The integral over Ω can be represented as a sum of the integral over K2& and the integral over Ω \K2&.In accordance with such a representation we get

u2(x, t) = u∗2(x, t) + u∗∗2 (x, t).

499

Using the estimate (4.95) for G(x, t; ξ, τ), we find

|u∗2(x, t)| < M59 sup |ϕ|∫

|x−ξ|3&

|G(x, t; ξ, 0)| dξ < M60t− 1

2

&∫0

(r√t

)n−1

e−µ4r

2

t dr < M61δ− 1

2 6 <ε

2(4.98)

if 0 < 6 < 60(ε, δ). Let us fix such a 6 and then choose 61(ε, δ) > 0 so small as to have the inequality

|u∗∗2 | <ε

2(4.99)

for |x− x1| < 61 and 0 < t T . This is possible because of the following property of the function G(x, t; ξ, τ):

limx→x1

G(x, t; ξ, τ) = 0

uniformly with respect to ξ ∈ Ω \K2& and 0 < t T . From (4.98) and (4.99) we get

|u2(x, t)| < ε for 0 < δ t T and |x− x1| < 61(ε, δ),

and it follows that u2(x, t)|S = 0 for 0 < t T .Similarly one can prove that u1(x, t)|S = 0 for 0 < t T .Thus, the function (4.96) is the required solution of the problem (1.1), (4.86) in the cylinder Q. The unique-

ness of the solution follows from Theorem 11 of Sect. 1.

Remark. The solution of the first boundary-value problem (1.1), (4.86) constructed in Theorem 3 turns outto be continuous everywhere in Q (including the points of the set x ∈ σ, t = 0) if in addition to the suppositionsof Theorem 3 we suppose that the consistency condition

ϕ(x) = 0 for x ∈ σ

holds. In fact, let us set ϕ(x) = 0 for x lying outside of Ω. Then the function ϕ(x) becomes continuous and boundedin all of the space En. According to (4.13) the relation (4.97) holds for any x ∈ En, including x ∈ σ, and the tendingto the limit is uniform in any closed domain of the space En, in particular, in Ω. Therefore

limt→+0

x→x1∈σ

∫Ω

Z(x, t; ξ, 0)ϕ(ξ) dξ = 0.

As was proved in Theorem 3, we have |u1(x, t)| M57t and |u(2)2 (x, t)| M58t

λ0/2 for t > 0, and it follows thatlimt→+0

x→x1∈σ

u(x, t) = 0.

4. For a second-order parabolic equation (1.1) it is possible to construct a potential theory similar tothe classic potential theory for second-order elliptic equations [2]. However, the suppositions on the smoothness ofthe coefficients of Eq. (1.1) have to be more strict than they were in previous sections. Here we give a sketch ofanother possible approach to solving the first boundary-value problem by means of potential theory (see [81]).

Suppose that the coefficients aij(x, t) in Q have continuous second derivatives with respect to x1, . . . , xn,the derivatives of bi(x, t) in x1, . . . , xn and the function c(x, t) are also continuous in Q, and the boundary σ ofthe domain Ω belongs to the class A1+λ. In this case, one can construct a double-layer potential with the help ofthe fundamental solution Z(x, t; ξ, τ):

v(x, t) =

t∫0

dτ

∫σ

P (Z(x, t; ξ, τ))ω(ξ, τ) dσξ ,

where ω(ξ, τ) is a function continuous on S that is called the density of the potential,

P (Z(x, t; ξ, τ)) ≡n∑

j=1

n∑i=1

∂

∂ξi[aij(ξ, τ)Z(x, t; ξ, τ)] − bj(ξ, τ)Z(x, t; ξ, τ)

cos(ν, ξj),

500

and ν is the direction of the inner normal to σ. We shall seek the solution of the first boundary-value problem forEq. (4.12) with the conditions

u(x, 0) = 0, u|S = ψ(x, t) (4.100)

as a potential with an unknown density ω(ξ, τ). There is a theorem about a jump of the double-layer potentialon S, that is, the relation

limx→x1∈σ

v(x, t) =12ω(x1, t) + v(x1, t), (4.101)

where

v(x1, t) =

t∫0

dτ

∫σ

P (Z(x1, t; ξ, τ))ω(ξ, τ) dσξ .

Taking into account conditions (4.100) and formula (4.101), we obtain an integral equation for the determinationof ω(ξ, τ):

12ω(x, t) +

t∫0

dτ

∫σ

P (Z(x, t; ξ, τ))ω(ξ, τ) dσξ = ψ(x, t). (4.102)

According to the suppositions on the smoothness of σ, we can see that the kernel of integral Eq. (4.102) has a weaksingularity

|P (Z(x, t; ξ, τ))| < M0(t− τ)−n+1

2 +λ0e−µ0|x−ξ|2

t−τ ,

whereM0, λ0, and µ0 are positive constants. Equation (4.102) can be solved by means of successive approximations.

5 GENERALIZED SOLUTIONS OF BOUNDARY-VALUE PROBLEMS.UNIQUENESS THEOREMS. SOME AUXILIARY PROPOSITIONS

In modern investigations on partial differential equations, the concept of generalized solution plays an im-portant role. The most frequent way to introduce a generalized solution is to define it either as a function satisfyingan integral identity that is in some sense equivalent to the differential equation or as a limit of a sequence of smoothsolutions of the equation (they are also called classical). Depending on the class of functions the generalized solutionbelongs to and on the integral identity that is taken to define it, we get different ways of posing problems. To showthat a certain way to define a generalized solution is reasonable it is necessary to prove the existence and uniquenessof this solution under some suppositions on the data. The question of the smoothness of this generalized solutionalso arises. That is, we have to investigate whether the generalized solution is sufficiently smooth if the coefficientsof the equation, the boundary data, and the boundary itself have a certain degree of smoothness and whether itwill satisfy the equation and the boundary conditions in the ordinary sense, if it is smooth.

In this section, we shall consider as an example two different definitions of the generalized solution of the firstboundary problem for a linear second-order parabolic equation and we shall prove the uniqueness of these solutions.In Sects. 6–9 this problem will serve as a model to demonstrate various methods to prove the existence of generalizedsolutions. In Sect. 11 we investigate the question of smoothness for generalized solutions of Eq. (1.1)

In what follows, we shall need some functional spaces and some auxiliary propositions.1. Some Auxiliary Propositions and Notations. As before we shall denote by Q the cylinder Ω×(0, T ),

where Ω is a domain in the space En. We shall suppose that σ ∈ A1.Denote by Ωδ the points of Ω that are situated at a distance greater than δ > 0 from the boundary σ of

the domain Ω, and Qδ will denote the cylinder Ωδ × (0, T ). The class of functions that are square integrable in Qwill be denoted by L2(Q). The scalar product of elements v1 and v2 from L2(Q) will be defined by the formula

(v1, v2) =∫∫Q

v1v2 dx dt.

The class of infinitely differentiable (k times continuously differentiable) functions in Q, vanishing in Q \Qδ

for some δ > 0 will be denoted by0

C∞(Q) (respectively0

Ck(Q)). In a similar way, we define classes0

C∞(Ω) and0

Ck(Ω).

501

A function w(x, t) integrable in Q is called a generalized derivative of the integrable in xk function v(x, t)

with respect to xk if for any Φ(x, t) ∈0

C∞(Q) the equality∫∫Q

(v ∂Φ∂xk

+wΦ)dx dt = 0 holds. In this case we shall write

w = ∂v∂xk

. In a similar way, we can define the generalized derivative with respect to t and higher-order generalizedderivatives (see [82, pp. 39, 40]). Note the following property: if a sequence of functions vm(x, t) converges weaklyas m→∞ to v(x, t) in the space L2(Q) and the norms of ∂vm

∂xkare bounded in L2(Q) uniformly with respect to m,

then v(x, t) has a generalized derivative ∂v∂xk∈ L2(Q) and ∂vm

∂xkconverge weakly to ∂v

∂xk([82, pp. 42, 43]).

Denote by W 1(Ω) the Hilbert space of functions ϕ(x) such that ϕ(x) ∈ L2(Ω) and ∂ϕ∂xi∈ L2(Ω) (i, . . . , n),

with the scalar product

(ϕ1, ϕ2)1 =∫Ω

ϕ1(x)ϕ2(x) dx +∫Ω

n∑i=1

∂ϕ1

∂xi

∂ϕ2

∂xidx.

By W 1,1(Q) we shall denote the Hilbert space of functions v(x, t) such that v(x, t) ∈ L2(Q), ∂v(x,t)∂xi

∈ L2(Q)

(i = 1, . . . , n), and ∂v(x,t)∂t ∈ L2(Q), with the scalar product

(v1, v2)1,1 =∫∫Q

v1(x, t)v2(x, t) dx dt +∫∫Q

( n∑i=1

∂v1

∂xi

∂v2

∂xi+∂v1

∂t

∂v2

∂t

)dx dt.

It is well known [82, p. 76] that the spaces W 1(Ω) and W 1,1(Q) are complete. The norms in spaces L2(Q), W 1(Ω),and W 1,1(Q) will be denoted by ‖v‖0, ‖v‖1, and ‖v‖1,1 respectively.

From S. L. Sobolev’s embedding theorems it follows that a function from W 1,1(Q) can be changed on a setof measure zero so that it is square integrable on an intersection of the cylinder Q and any n-dimensional planeor n-dimensional surface of class A1. In particular, such a function is square integrable on the surface S and onthe intersection of the cylinder Q and any plane t = const.

What is more, since the embedding operator is completely continuous, the values of the functionv(x, t) ∈ W 1,1(Q) on the n-dimensional planes or surfaces close to each other also are close in the sense of meanvalues [82, pp. 94, 96]. In particular, if v(x, t) ∈ W 1,1(Q) and v(x, 0) = ϕ(x), then

∫Ω

[v(x, t) − ϕ(x)]2 dx→ 0 fort→ 0.

By0

W 1(Ω) denote the subset of functions ϕ(x) ∈ W 1(Ω) vanishing on the surface σ. By0

W 1,1(Q) denotethe subspace of functions v(x, t) ∈W 1,1(Q) vanishing on the surface S.

An important auxiliary tool that will be used below is the averaging of functions. Let the functionv(x, t) ∈ L2(Q). We shall extend this function to the whole space by defining it as zero outside the cylinder Q.The function

vh(x, t) =

+∞∫−∞

∫En

v(ξ, τ)ωh(x − ξ, t− τ) dξ dτ

will be called an average function of radius h, where ωh(x−ξ, t−τ) is called the averaging kernel and has the followingproperties: ωh is infinitely differentiable; ωh = 0 for |x−ξ|2+(t−τ)2 h2 (h > 0); ωh > 0 for |x−ξ|2+(t−τ)2 < h2;+∞∫−∞

∫En

ωh(x− ξ, t− τ) dξ dτ = 1.

It is easy to show [82, pp. 19, 20, 41] that vh(x, t) → v(x, t) as h → 0 in the norm of L2(Q), and forv(x, t) ∈ W 1,1(Q) the equalities

(∂v∂xi

)h = ∂vh

∂xi(i = 1, . . . , n) and

(∂v∂t

)h= ∂vh

∂t hold for sufficiently small h in anyclosed domain lying in the interior of the cylinder Q.

Similarly it is possible to define the averaging operator in the domain Ω.

Lemma 1. The functions of class0

C∞(Ω) are everywhere dense in0

W 1(Ω), and the functions of class0

C∞(Q)

are everywhere dense in0

W 1,1(Q).

In [82, p. 102] it is proved that for any function ϕ(x) ∈0

W 1(Ω) there exists a sequence ϕm(x) ∈0

C∞(Ω) thatconverges weakly in W 1(Ω) to ϕ(x). The functions ϕm(x) are obtained by multiplying ϕ(x) by smooth compactly

502

supported in Ω functions χm(x), equal to unity in Ω1/m, and then averaging of this product ϕ(x)χm(x). Carrying outwith some more precision the proof of a lemma from [82, p. 102], it is possible to show that ϕm(x) converges stronglyto ϕ(x) in the norm ofW 1(Ω). Besides that, the first assertion of Lemma 1 can be obtained as a corollary to a well-known theorem of functional analysis (see [83, p. 203]). According to this theorem there exist linear combinations

ψN =N∑

m=1CNmϕm(x) of the functions ϕm(x), weakly converging to ϕ(x), such that ‖ψN (x)−ϕ(x)‖1 → 0 for N →∞.

Obviously ψN (x) also belongs to0

C∞(Ω) (N = 1, 2, . . .).The second assertion of the lemma can be proved in a similar way.

For the functions ϕ(x) ∈0

W 1(Ω) the following inequality holds:

∫Ω

ϕ2(x) dx M∫Ω

n∑i=1

(∂ϕ

∂xi

)dx, (5.1)

and for the functions v(x, t) ∈0

W 1,1(Q) we have the inequality

∫∫Q

v2(x, t) dx dt M∫∫Q

n∑i=1

(∂v

∂xi

)2

dx dt, (5.2)

where the constant M depends only on the size of the domain Ω.The inequalities (5.1) and (5.2) were established in Sect. 2 for smooth functions and in particular for the func-

tions belonging to classes0

C∞(Ω) and0

C∞(Q). Since the set0

C∞(Ω) is dense in0

W 1(Ω), the inequality (5.1) holds

also for ϕ(x) ∈0

W 1(Ω). The inequality (5.2) for v(x, t) ∈0

W 1,1(Q) can be proved in the same way.Note the following important proposition.

Theorem 1. From any sequence of functions vm(x, t) that is uniformly bounded in the norm W 1,1(Q) it ispossible to select a subsequence converging in the norm L2(Q) to a function v(x, t), belonging to the space W 1,1(Q).If vm|S = 0 and vm(x, 0) converge to the function ϕ(x) in the norm L2(Ω), then the limit function v(x, t) belongs

to0

W 1,1(Q) and is equal to ϕ(x) for t = 0.

The proof of this theorem can be found in [82, pp. 83–94].

Lemma 2. In the domain Ω with boundary belonging to class A2, there exists an orthonormal in L2(Ω)system of functions Xk(x) having the following properties:

(1) Xk(x) are twice continuously differentiable in Ω and vanish on the boundary σ of the domain Ω.

(2) For any function ϕ(x) ∈0

W 1(Ω) and any number ε > 0 there exists a linear combination ϕN (x) =N∑k=1

ckXk(x) of functions Xk(x) such that

‖ϕ− ϕN‖1 < ε. (5.3)

(3) For any function v(x, t) ∈0

C∞(Q), vanishing in a neighborhood of the boundary Q, and any ε > 0 there

exists a linear combination vN (x, t) =N∑k=1

ck(t)Xk(x) such that

∫∫Q

(v − vN )2 +

n∑i=1

[∂(v − vN )

∂xi

]2dx dt < ε.

Proof. It is sufficient to prove the inequality (5.3) only for ϕ(x) ∈0

C∞(Ω), because the set of these functions

is dense in0

W 1(Ω). Since σ ∈ A2, there exists a function ζ(x) ∈ C2(Ω), vanishing on σ and positive in the domain Ω.

Since ϕ(x)ζ(x) ∈

0

C2(Ω), we have[ϕ(x)ζ(x)

]h ∈ 0

C∞(Ω) and∥∥ϕζ −

(ϕζ

)h∥∥1< ε

2 for a sufficiently small h > 0.

503

Without loss of generality, one can assume that Ω lies inside the cube 0 xi π (i = 1, . . . , n). SetXk(x) = sink1x1 × sin k2x2 × . . . × sin knxn, where kj (j = 1, . . . , n) are some integers and k = (k1, k2, . . . , kn).

It is well known that a function from0

C∞(Ω) can be expanded in a Fourier series with respect to Xk(x) and thatthis series converges uniformly in Ω together with its derivatives. Therefore, there exists a linear combination∑1kjN

ckXk(x) of functions Xk(x) such that

∥∥∥∥(ϕ

ζ

)h−

∑1kjN

ckXk(x)∥∥∥∥

0

+n∑i=1

∥∥∥∥∥∂(ϕζ

)h∂xi

−∑

1kjNck∂Xk(x)∂xi

∥∥∥∥∥0

<ε

2.

Therefore, ∥∥∥∥ϕ(x)ζ(x)−

∑1kjN

ckXk(x)∥∥∥∥

0

< ε (5.4)

and ∥∥∥∥ ∂

∂xi

(ϕ(x)ζ(x)

)−

∑1kjN

ck∂Xk(x)∂xi

∥∥∥∥0

< ε (i = 1, . . . , n). (5.5)

Set X∗k (x) = ζ(x)Xk(x). It follows from the inequality (5.4) that∥∥∥∥ϕ− ∑

1kjNckX

∗k

∥∥∥∥0

=∥∥∥∥ζ(ϕ

ζ−

∑1kjN

ckXk

)∥∥∥∥0

< M1ε;

then ∥∥∥∥ ∂ϕ∂xi −∑

1kjNck∂X∗k∂xi

∥∥∥∥0

=∥∥∥∥ζ(1ζ

∂ϕ

∂xi−

∑1kjN

ck∂Xk

∂xi− 1ζ

∂ζ

∂xi

∑1kjN

ckXk

)∥∥∥∥0

=∥∥∥∥ζ[∂

∂xi

(ϕ

ζ

)+ϕ

ζ2

∂ζ

∂xi−

∑1kjN

ck∂Xk

∂xi− 1ζ

∂ζ

∂xi

∑1kjN

ckXk

]∥∥∥∥0

M2

∥∥∥∥ ∂

∂xi

(ϕ

ζ

)−

∑1kjN

ck∂Xk

∂xi

∥∥∥∥0

+M2

∥∥∥∥ϕζ −∑

1kjNckXk

∥∥∥∥0

< M3ε,

where the constants M1 and M2 do not depend on ε. Therefore,∥∥∥∥ϕ− ∑1kjN

ckX∗k

∥∥∥∥1

< (M1 +M3)ε.

To complete the proof of assertion (2) of the lemma it is sufficient to orthogonalize and normalize the functionsX∗k (x) in the space L2(Ω) and replace Xk(x) by these newly obtained functions.

Let us prove assertion (3). Consider the Fourier-series expansion of v(x, t) in Q with respect to the functionssin mπ

T t (m = 1, 2, . . .):

v(x, t) =∞∑

m=1

αm(x) sinmπ

Tt.

This series converges absolutely and uniformly in Q together with its derivatives (see [84, p. 507]). Denote

vl(x, t) =l∑

m=1αm(x) sin mπ

T t.

For a sufficiently large l we have

∫∫Q

(v − vl)2 +

n∑i=1

[∂(v − vl)∂xi

]2dx dt < ε. (5.6)

504

On the other hand, according to the assertion (2) that we have already proved, one can choose N such that

∥∥∥∥αm(x) −N∑k=1

ckmXk(x)∥∥∥∥

2

1

<ε

l, (m = 1, . . . , l).

From this inequality and from (5.6) it follows that∫∫Q

(v − vN )2 +

n∑i=1

[∂(v − vN )

∂xi

]2dx dt < M4ε,

where

vN (x, t) =l∑

m=1

sinmπ

Tt

N∑k=1

ckmXk(x) =N∑k=1

Xk(x)l∑

m=1

ckm sinmπ

Tt =

N∑k=1

ck(t)Xk(x),

and the constant M4 does not depend on ε and N .Lemma 2 is proved. It will be used in Sect. 9.In what follows, besides the spaces introduced above we shall consider the space W k(Ω) of the functions

ϕ(x) such that their derivatives up to order k belong to L2(Ω). By W 2k,k(Q) we shall denote the space of functionsv(x, t) such that

∂l+mv

∂xl11 . . . ∂xlnn ∂tm∈ L2(Q), where l + 2m 2k.

The functions fromW 2k,k(Q) have in the cylinder Q generalized derivatives belonging to L2(Q) up to order 2k withrespect to the space variables x1, . . . , xn and up to order k with respect to the time t and the corresponding mixedderivatives. The norm ‖v‖2k,k of the function v(x, t) in the space W 2k,k(Q) will be defined by the formula

‖v‖2k,k =∫∫

Q

∑0l+2m2k

(∂l+mv

∂xl11 . . . ∂xlnn ∂tm

)2

dx dt

1/2

.

Consider the space of functions obtained as a closure of the functions from the class C∞(Q) vanishing in Γin the norm of W 2,1(Q). Denote this space by W 2,1(Q). Obviously W 2,1(Q) is a subspace of the Hilbert spaceW 2,1(Q) with the scalar product

(v1, v2)2,1 =∫∫Q

(v1v2 +

n∑i=1

∂v1

∂xi

∂v2

∂xi+∂v1

∂t

∂v2

∂t+

n∑i,j=1

∂2v1

∂xi ∂xj

∂2v2

∂xi ∂xj

)dx dt

for any elements of v1 and v2.2. Generalized Solutions and Uniqueness Theorems. Here we consider generalized solutions of

the first boundary-value problem for a parabolic equation. We shall give two definitions of the generalized solutionof this problem and prove the corresponding uniqueness theorems.

Definition 1. A function u(x, t) is called a weak solution of the first boundary-value problem for the equation

L(u) ≡n∑

i,j=1

∂

∂xi

(Aij(x, t)

∂u

∂xj

)+

n∑i=1

Bi(x, t)∂u

∂xi+ C(x, t)u − ∂u

∂t= F (x, t) (5.7)

in the cylinder Q with conditionsu(x, 0) = ϕ(x), u|S = 0, (5.8)

if u(x, t) ∈0

W 1,1(Q), u(x, 0) = ϕ(x), and for any function Φ(x, t) ∈0

W 1,1(Q) the equality∫∫Q

[ n∑i,j=1

Aij∂u

∂xj

∂Φ∂xi−( n∑

i=1

Bi∂u

∂xi+ Cu− ∂u

∂t− F

)Φ]dx dt = 0 (5.9)

holds. We suppose that Aij(x, t), Bi(x, t), and C(x, t) are bounded and measurable and F (x, t) ∈ L2(Q).

505

We note immediately that the validity of the identity (5.9) for all Φ(x, t) ∈0

W 1,1(Q) is equivalent to its

validity for all Φ(x, t) ∈0

C∞(Q). This follows from the density of the set of functions0

C∞(Q) in0

W 1,1(Q).

Theorem 2. A weak solution of the problem (5.7), (5.8) is unique.

Proof. Denote by u(x, t) a difference between two solutions of the problem. This function obviously satisfiesthe equality ∫∫

Q

[ n∑i,j=1

Aij∂u

∂xj

∂Φ∂xi−( n∑

i=1

Bi∂u

∂xi+ Cu− ∂u

∂t

)Φ]dx dt = 0 (5.10)

for any Φ(x, t) ∈0

W 1,1(Q) and the condition u(x, 0) = 0.In the identity (5.10) we can set Φ(x, t) = u(x, t)e−θt and then transform the equality thus obtained, taking

into account that u(x, t) ∈0

W 1,1(Q) and u(x, 0) = 0. We have

0 =∫∫Q

[u∂u

∂te−θt + e−θt

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj−

n∑i=1

Bi∂u

∂xiu− Cu2

)]dx dt

=12

∫t=T

u2e−θT dx+∫∫Q

e−θt(θ

2u2 +

n∑i,j=1

Aij∂u

∂xi

∂u

∂xj−

n∑i=1

Bi∂u

∂xiu− Cu2

)dx dt.

Since the coefficients Bi(x, t) and C(x, t) are bounded and the matrix ‖Aij(x, t)‖ is positive definite, wearrive at the inequality

0 ∫∫Q

e−θt[θ

2u2 + µ

n∑i=1

(∂u

∂xi

)2

−Mu2 −Mn∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣ |u|

]dx dt

∫∫Q

e−θt[θ

2u2 + µ

n∑i=1

(∂u

∂xi

)2

−Mu2 − µn∑i=1

(∂u

∂xi

)2

−M1u2

]dx dt =

∫∫Q

e−θt(θ

2−M −M1

)u2 dx dt,

(5.11)

whereM andM1 are constants depending only on the coefficients of Eq. (5.7) and on the dimension n of the spaceEn.If we choose θ > 2(M +M1), then it follows from the inequality (5.11) that u(x, t) ≡ 0 in the cylinder Q.

Definition 2. Suppose that the functions um(x, t) ∈ C3(Q) satisfy in Q the equation

L(um) ≡n∑

i,j=1

aij(x, t)∂2um∂xi ∂xj

+n∑i=1

bi(x, t)∂um∂xi

+ c(x, t)um −∂um∂t

= fm(x, t).

Suppose thatum(x, 0) = ϕm(x), um|S = 0.

If the fm(x, t) converge in the norm of L2(Q) to the function f(x, t) as m → ∞, the ϕm(x) converge in the normof W 1(Ω) to the function ϕ(x), and the um(x, t) converge in the norm of L2(Q) to the function u(x, t), then u(x, t)is called a strong solution of the equation

L(u) ≡n∑

i,j=1

aij(x, t)∂2u

∂xi ∂xj+

n∑i=1

bi(x, t)∂u

∂xi+ c(x, t)u − ∂u

∂t= f(x, t) (5.12)

in the cylinder Q with the condition (5.8).

Note that it follows from the inequality (2.35) of Sect. 2 that the um converge to u in the norm of W 2,1(Q).We shall suppose that aij(x, t) are continuous in Q, bi(x, t) and c(x, t) are bounded, and σ ∈ A2.

506

Theorem 3. A strong solution of the problem (5.12), (5.8) is unique.

Proof. Suppose there are two solutions of this problem: u(x, t) and u(x, t). This means that there existsequences fm, ϕm, um and fm, ϕm, um, converging in the norm of L2 respectively to f , ϕ, u and f , ϕ, u.

Set vm = um − um. According to the a priori estimate (2.35) we have

‖vm‖2,1 M2(‖fm − fm‖0 + ‖ϕm − ϕm‖1). (5.13)

Since the right-hand side of inequality (5.13) tends to zero as m→∞, we also have vm → 0 as m→∞; therefore,u ≡ u.

Passing to the limit as m→∞ in the equation L(um) = fm, we can see that the strong generalized solutionu(x, t) satisfies Eq. (5.12) almost everywhere in Q and belongs to the space W 2,1(Q). Condition (5.8) is satisfied inthe mean, in the sense that was defined above for the functions of class W 1,1(Q).

The following proposition is also valid. Let the function v(x, t) ∈ W 2,1(Q), v|S = 0, v(x, 0) = ϕ(x), andlet v(x, t) satisfy Eq. (5.12) almost everywhere in Q. Let u(x, t) be a strong generalized solution of the problem(5.12), (5.8), σ ∈ A2, and aij(x, t) ∈ C1(Q). Then u(x, t) ≡ v(x, t).

In fact, multiplying (5.12) by a function Φ(x, t) ∈0

C∞(Q) and integrating over Q, we obtain

∫∫Q

n∑i,j=1

aij∂v

∂xi

∂Φ∂xj−[ n∑i=1

(bi −

n∑j=1

∂aij∂xj

)∂v

∂xi+ cv − ∂v

∂t− f]Φdx dt = 0.

The same identity is valid for the function u(x, t). Therefore u(x, t) and v(x, t) are weak solutions of the firstboundary-value problem for the equation

n∑i,j=1

∂

∂xi

(aij

∂u

∂xj

)+

n∑i=1

(bi −

n∑j=1

∂aij∂xj

)∂u

∂xi+ cu− ∂u

∂t= f

in the cylinder Q with the conditions u(x, 0) = ϕ(x) and u|S = 0. Since the weak solution of the first boundary-value problem is unique, we have u(x, t) ≡ v(x, t).

6 METHOD OF FINITE DIFFERENCES

Consider the first boundary-value problem for Eq. (5.7) in the cylinder Q = Ω× (0, T ), where Ω is a boundeddomain in the space En with the boundary σ ∈ A1 with the conditions (5.8). We suppose that the coefficients ofEq. (5.7) are continuous in Q, ∂Aij(x,t)

∂t are bounded, and the right-hand side F (x, t) ∈ L2(Q); the matrix ‖Aij(x, t)‖

is assumed to be symmetric and positive definite:n∑

i,j=1

Aijαiαj µn∑i=1

α2i , µ > 0. We shall prove the existence

of a weak generalized solution of the boundary-value problem (5.7), (5.8) by the method of finite differences (see

Sect. 5, Definition 1). We shall suppose that the function ϕ(x) belongs to the class0

W 1(Ω).First we prove the existence of a generalized solution of the problem (5.7), (5.8) under the supposition that

ϕ(x) ∈0

C1(Ω) and F (x, t) ∈ C(Q).Let us cover the domain 0 t T, x ∈ En by a rectangular grid consisting of the lines t = rh and

xi = pih (i = 1, . . . , n), where the pi take all integer values and r takes all integer values on the interval[0, Th]. We

suppose that T is a multiple of h. The node points of the grid are points with coordinates that are multiples of h.At the node points we shall define a function uh. Denote by Sh the node points belonging to Q \ Γ that are

neighbors to a node point lying outside Q \ S. The remaining nodes of the grid lying in Q \ Γ will be called innernodes: the set of inner nodes is denoted by Qh. Define the function uh as zero at the nodes Sh and at the nodes lyingoutside Q \ S. On the base of the cylinder Q set at the node points uh(x, 0) = ϕ(x). The values of the function uh

at the inner node points Qh are determined by a system of linear equations obtained by substituting the derivativesin Eq. (5.7) by corresponding difference quotients.

507

Let (x1, . . . , xn, t) be a node point of the grid. For simplicity we shall use the following notations:

v(x1, . . . , xn, t) = v, v(x1, . . . , xk−1, xk ± h, xk+1, . . . , xn, t) = v(xk ± h), v(x1, . . . , xn, t− h) = v(t− h),

v(xk + h)− vh

=∆v∆xk

,v − v(xk − h)

h=

∆v∆xk

,∆(

∆v∆xk

)∆xi

=∆2v

∆xi∆xk;

similar notations will be used for difference quotients with respect to t. To be concise we shall omit the index h ofthe function uh.

For every point of Qh the following linear equation can be written:

n∑i,j=1

∆∆xi

(Aij

∆u∆xj

)+

n∑i=1

Bi∆u∆xi

+ Cu− ∆u∆t

= F. (6.1)

Besides the value of the function u at the chosen point this equation contains the values of u in the neighboringnode points. The values of the coefficients and the right-hand side are taken at the chosen point. Thus we obtainm linear equations with m unknowns, wherem is the number of inner node points. The fact that u = 0 at the pointsof Sh and u|t=0 = ϕ has to be taken into account. The solvability of this system will be shown below.

Suppose that a solution of the system (6.1) exists. Under this supposition we can obtain some estimatesfor the solution and its difference quotients, independent of h. To do this we multiply both sides of Eq. (6.1) by∆u∆t e

−θt and then transform the equality thus obtained. For u and its difference quotients ∆u∆xi

and ∆u∆t we can

obtain an estimate similar to the a priori estimate (2.30) for the solutions of Eq. (5.7). The calculations includingdifference quotients are similar to those carried out when we deduced estimate (2.30). We also use the obviousidentities

v∆w∆xi

=∆(vw)∆xi

− w(xi + h)∆v∆xi

=∆(vw(xi + h))

∆xi− w ∆v

∆xi,

v∆w∆xi

=∆(vw)∆xi

− w(xi − h)∆v∆xi

,

v∆w∆t

=∆(vw)∆t

− w(t− h)∆v∆t

.

(6.2)

By Ni we shall denote quantities not greater in absolute value than

M

∣∣∣∣∆u∆t∣∣∣∣(|u|+

n∑i=1

∣∣∣∣ ∆u∆xi

∣∣∣∣)+

n∑i=1

(∆u(t− h)

∆xi

)2e−θt,

where the constant M depends only on the coefficients of Eq. (5.7).Thus, multiplying (6.1) by ∆u

∆t e−θt, we obtain

−F∆u∆t

e−θt =(∆u∆t

)2

e−θt −n∑

i,j=1

∆u∆t

∆∆xi

(Aij

∆u∆xj

)e−θt +N1

=(∆u∆t

)2

e−θt +n∑

i,j=1

Aij∆u∆xj

∆2u

∆xi∆te−θt −

n∑i,j=1

∆∆xi

[∆u∆t

Aij(xi + h)∆u(xi + h)

∆xje−θt

]+N1. (6.3)

This last relation is also valid for the points of Sh, because at these points we have ∆u∆t = 0, and this means

that both sides of equality (6.3) vanish. Recall that in accordance with (6.2), we have the identity

∆∆t

(Aij

∆u∆xi

∆u∆xj

e−θt)

=∆(Aije

−θt)∆t

∆u(t− h)∆xi

∆(t− h)∆xj

+Aije−θt ∆

∆t

(∆u∆xi

∆u∆xj

)

=(∆Aij

∆te−θt +Aij(t− h)

∆e−θt

∆t

)∆u(t− h)

∆xj∆u(t− h)

∆xj+Aije

−θt ∆∆t

(∆u∆xi

∆u∆xj

).

508

We transform the first sum in the right-hand side of (6.3):

n∑i,j=1

Aij∆2u

∆xi∆t∆u∆xj

e−θt =12

n∑i,j=1

∆∆t

(Aij

∆u∆xi

∆u∆xj

e−θt)− 1

2

n∑i,j=1

∆Aij

∆te−θt

∆u(t− h)∆xi

∆u(t− h)∆xj

− 12

n∑i,j=1

∆e−θt

∆tAij(t− h)

∆u(t− h)∆xi

∆u(t− h)∆xj

− 12

n∑i,j=1

Aije−θt ∆

∆t

(∆u∆xi

∆u∆xj

)+

n∑i,j=1

Aij∆2u

∆xi∆xj∆u∆xj

e−θt

=12

n∑i,j=1

∆∆t

(Aij

∆u∆xi

∆u∆xj

e−θt)+N2 −

12

n∑i,j=1

∆e−θt

∆tAij(t− h)

∆u(t− h)∆xi

∆u(t− h)∆xj

+12

n∑i,j=1

Aij

[2

∆2u

∆xi∆t∆u∆xj

− ∆2u

∆xi∆t

(∆u∆xj

+∆u(t− h)

∆xj

)]e−θt. (6.4)

The last sum is equal to

12

n∑i,j=1

Aij

(∆u∆xj

− ∆u(t− h)∆xj

)∆2u

∆xi∆t=h

2

n∑i,j=1

Aij∆2u

∆xi∆t∆2u

∆xj∆t 0. (6.5)

Moreover, we have∆e−θt

∆t=e−θt − e−θ(t−h)

h= −e−θt e

θh − 1h

< −θe−θt,

and therefore

−12

n∑i,j=1

∆e−θt

∆tAij(t− h)

∆u(t− h)∆xi

∆u(t− h)∆xj

θµe−θt

2

n∑i=1

(∆u(t− h)

∆xi

)2

. (6.6)

From (6.3) and (6.4), taking into account estimates (6.5) and (6.6), we obtain the inequality

−F∆u∆t

e−θt (∆u∆t

)2

e−θt +12

n∑i,j=1

∆∆t

(Aij

∆u∆xi

∆u∆xj

e−θt)

+θµe−θt

2

n∑i=1

(∆u(t− h)

∆xi

)2

−n∑

i,j=1

∆∆xi

[∆u∆t

Aij(xi + h)∆u(xi + h)

∆xie−θt

]+N3. (6.7)

Let us sum this inequality over all the points of Sh and Qh. The sum over all these points will be denotedby∑Q

, and the sum over the points lying on the plane t = t0, where 0 t0 T , will be denoted by∑t=t0

. Since

∆u∆t = 0 at the points Sh, the sum ∆

∆xi

[∆u∆tAij(xi + h)∆u(xi+h)

∆xj

], taken over the points Qh and Sh, is equal to zero.

Moreover, it is obvious that

∑Q

n∑i,j=1

∆∆t

(Aij

∆u∆xi

∆u∆xj

e−θt)

=1h

∑t=T

n∑i,j=1

Aij∆u∆xi

∆u∆xj

e−θT − 1h

∑t=0

n∑i,j=1

Aij∆u∆xi

∆u∆xj

.

Thus, summing (6.7) over Qh and Sh, we have

∑Q

[(∆u∆t

)2

+θµ

2

n∑i=1

(∆u(t− h)

∆xi

)2]e−θt +

12h

∑t=T

n∑i,j=1

Aij∆u∆xi

∆u∆xj

e−θT − 12h

∑t=0

Aij∆u∆xi

∆u∆xj

−∑Q

F∆u∆t

e−θt +M1

∑Q

∣∣∣∣∆u∆t∣∣∣∣(|u|+

n∑i=1

∣∣∣∣ ∆u∆xi

∣∣∣∣)+

n∑i=1

(∆u(t− h)

∆xi

)2e−θt

∑Q

[F 2 +

14

(∆u∆t

)2]e−θt +

14

∑Q

(∆u∆t

)2

e−θt +M2

∑Q

[u2 +

n∑i=1

(∆u∆xi

)2]e−θt +M1

∑t=0

n∑i=1

(∆u∆xi

)2

.

Here and below the constants Mi depend only on the coefficients of Eq. (5.7) and on the domain Q.

509

Therefore,

12

∑Q

(∆u∆t

)2

e−θt +θµ

2

n∑i=1

(∆u(t− h)

∆t

)2

e−θt +12h

∑t=T

n∑i,j=1

Aij∆u∆xi

∆u∆xj

e−θT

∑Q

F 2e−θt +M2

∑Q

[u2 +

n∑i=1

(∆u∆xi

)2]e−θt +M1

∑t=0

n∑i=1

(∆u∆xi

)2

+12h

∑t=0

n∑i,j=1

Aij∆u∆xi

∆u∆xj

.

Sincen∑

i,j=1

Aij∆u∆xi

∆u∆xj µ

n∑i=1

(∆u∆xi

)2 and u(x, 0) = ϕ(x), we have

∑Q

(∆u∆t

)2

e−θt + θµ∑Q

n∑i=1

(∆u(t− h)

∆xi

)2

e−θt +µ

h

∑t=T

n∑i=1

(∆u∆xi

)2

e−θT

2∑Q

F 2e−θt +M3

∑Q

[u2 +

n∑i=1

(∆u∆xi

)2]e−θt +

1h

∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2, (6.8)

where∑Ω

denotes the sum over all the node points of Ω.

Estimate now the sum∑Q

u2e−θt in terms of∑Q

n∑i=1

(∆u∆xi

)2e−θt. Let d0 denote the diameter of the domain Ω.

One can express the value of the function u at a chosen node point P (x1, . . . , xn, t) in terms of difference quotientsof this function with respect to the space variables. Let the point

Pi(x1, . . . , xi−1, xi − pih, xi+1, . . . , xn, t)

belong to Sh. Then

u(P ) = u(P )− u(Pi) = h

[∆u∆xi

+∆u(xi − h)

∆xi+ . . .+

∆u(xi − (pi − 1)h)∆xi

].

Therefore,

u2(P ) pih2

pi−1∑l=0

(∆u(xi − lh)

∆xi

)2

d0h

n∑i=1

pi−1∑l=0

(∆u(xi − lh)

∆xi

)2

.

Multiplying this inequality by e−θt and summing over all the points of Qh ∪ Sh, we get

∑Q

u2e−θt d20

∑Q

n∑i=1

(∆u∆xi

)2

e−θt. (6.9)

Hence, taking into account the inequality (6.8), we have

∑Q

(∆u∆t

)2

e−θt + θµ∑Q

n∑i=1

(∆u(t− h)

∆xi

)2

e−θt +µ

h

∑t=T

n∑i=1

(∆u∆xi

)2

e−θt

2∑Q

F 2e−θt +M4

∑Q

n∑i=1

(∆u∆xi

)2

e−θt +M3

h

∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2

. (6.10)

Suppose that θ > 2M4µ and h < 1

θ . Then from the inequality (6.10) it follows that

∑Q

(∆u∆t

)2

e−θt +θµ

2

∑Q

n∑i=1

(∆u∆xi

)2

e−θt 2∑Q

F 2e−θt +M3

h

∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2

.

510

Taking into account that 1 e−θt e−θT , we get

∑Q

[(∆u∆t

)2

+n∑i=1

(∆u∆xi

)2]M5

[∑Q

F 2 +1h

∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2].

With the help of (6.9), we arrive at the final inequality

∑Q

[u2 +

(∆u∆t

)2

+n∑i=1

(∆u∆xi

)2]M6

[∑Q

F 2 +1h

∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2]. (6.11)

From inequality (6.11) follows the solvability of the system (6.1). In fact, the solution of the homogeneoussystem (6.1) corresponds to the case where F ≡ 0 and ϕ ≡ 0. It follows from the inequality (6.11) that in this caseuh = 0 at all the points of Qh. The solvability of the system (6.1) follows from the uniqueness of its solution.

Let (x, t) be a point with coordinates satisfying the inequalities (pi − 1)h < xi pih (i = 1, . . . , n),(r − 1)h < t rh. We set uh(x, t) = uh(p1h, . . . , pnh, rh). In this way we define a piecewise-constant functionuh(x, t) (that in general is not continuous) everywhere in the layer H0 t T, x ∈ En. This function is equalto zero at all the points outside Q.

In this way we can extend to the whole layer H also the functions ∆uh

∆xiand ∆uh

∆t , obtaining the piecewise-constant functions uhi (x, t) and u

h0(x, t) respectively. Multiplying (6.11) by hn+1, we come to the following relation:

∫∫Q

[(uh)2 +

n∑i=1

(uhi )2 + (uh0)

2

]dx dt M6

[∑Q

F 2hn+1 +∑Ω

n∑i=1

(∆ϕ(x)∆xi

)2

hn]. (6.12)

Since ∆ϕ(x)∆xi

= ∂ϕ∂xi

(x1, . . . , xi−1, ξi, xi+1, . . . , xn), where xi − h < ξi < xi, in the right-hand side of the in-

equality (6.12) we have integral sums for the functions F 2(x, t) in Q andn∑i=1

(∂ϕ(x)∂xi

)2in Ω. These sums are uniformly

bounded with respect to h. Therefore, the functions uh, uhi , and uh0 are uniformly bounded in the L2(Q)-norm with

respect to h, and one can select a subsequence hk → 0 such that uhk , uhki , and uhk0 converge weakly in L2(Q) tothe functions u, ui, and u0 respectively. Since the norm of a weak limit of a sequence of functions does not exceedthe lower limit of their norms, we obtain from (6.12)

∫∫Q

(u2 +

n∑i=1

u2i + u2

0

)dx dt M6

[∫∫Q

F 2 dx dt+∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx

]. (6.13)

Let us prove that ui = ∂u∂xi

, u0 = ∂u∂t , and that the function u satisfies conditions (5.8). Suppose that the func-

tion Φ(x, t) ∈0

C∞(Q) and vanishes for t = T . We have the following obvious relation:

∑Q

(Φ(t− h)∆u

h

∆t+ uh

∆Φ∆t

)=∑Q

∆(Φuh)∆t

= − 1h

∑t=0

Φuh. (6.14)

Let us define piecewise-constant functions Φh(x, t) and Ψh0(x, t), setting at the node points Φh(x, t) = Φ(x, t − h)

and Ψh0 (x, t) =

∆Φ(x,t)∆t and extending the functions Φh and Ψh

0 to the whole layer H in the same way as was donefor the functions uh. We can multiply (6.14) by hn+1, and since uh(x, 0) = ϕ(x), we obtain∫∫

Q

(Φhuh0 + uhΨh0 ) dx dt = −

∫Ω

Φh(x, h)ϕh(x) dx, (6.15)

where ϕh(x) is a piecewise-constant function, equal to ϕ(x) at the nodes of the grid and extended to all x ∈ En

in the same way as uh. Since for h → 0 the functions Φh(x, t) converge uniformly in Q to Φ(x, t), the functionsΦh(x, h) converge uniformly in Ω to Φ(x, 0), and the functions Ψh

0(x, h) converge uniformly in Q to ∂Φ(x,t)∂t , we can

511

pass to the limit in the relation (6.15) as hk → 0. Taking into account that uhk and uhk0 converge weakly to u andu0 respectively, we get ∫∫

Q

(Φu0 + u

∂Φ∂t

)dx dt = −

∫Ω

Φ(x, 0)ϕ(x) dx. (6.16)

If the function Φ(x, t) has compact support in Q, then the relation (6.16) means that u(x, t) has a generalizedderivative ∂u

∂t in Q equal to u0(x, t) and that ∂u∂t ∈ L2(Q).

In the same way, one can show that ∂u∂xi

= ui, and this means that the function u(x, t) belongs to W 1,1(Q).Therefore, for u(x, t) we have the relation

∫∫Q

(Φ∂u

∂t+ u

∂Φ∂t

)dx dt = −

∫Ω

Φ(x, 0)u(x, 0) dx. (6.17)

From (6.16) and (6.17) it follows that the initial condition u|t=0 = ϕ(x) is satisfied.Since the functions uh are equal to zero at the nodes of Sh and outside Q\S, we can assume that the integral

in the left-hand side of (6.13) is taken over the whole layer H0 t T, x ∈ En. Therefore, u(x, t) ∈ W 1,1(H).

Since u(x, t) = 0 outside of Q, the function u(x, t) vanishes on S and therefore belongs to0

W 1,1.To show that u(x, t) is a generalized solution of the problem (5.7), (5.8), we only need to establish the rela-

tion (5.9).Suppose that Φ(x, t) ∈

0

C∞(Q). Let us multiply Eq. (6.1) at every inner node point (x, t) of the grid byΦ(x, t) and then transform the equality thus obtained. We get

FΦ = −Φ∆u∆t

+n∑

i,j=1

Φ∆∆xi

(Aij

∆u∆xj

)+

n∑i=1

ΦBi∆u∆xi

+ΦCu

= −Φ∆u∆t

+n∑

i,j=1

∆∆xi

[ΦAij(xi + h)

∆u(xi + h)∆xj

]−

n∑i,j=1

Aij∆Φ∆xi

∆u∆xj

+n∑i=1

ΦBi∆u∆xi

+ΦCu. (6.18)

Since Φ(x, t) = 0 at the points of Sh (for sufficiently small h), the last relation holds also at these points.Summing the equality (6.18) over all inner and boundary points, we obtain

∑Q

FΦ =∑Q

(−Φ∆u

∆t−

n∑i,j=1

Aij∆Φ∆xi

∆u∆xj

+n∑i=1

ΦBi∆u∆xi

+ΦCu). (6.19)

Define the piecewise-constant functions Fh(x, t), Φh(x, t), Ψhi (x, t), A

hij(x, t), B

hi (x, t), and Ch(x, t) as we

have defined uh(x, t), Φh(x, t), and Ψh0 (x, t). Multiplying (6.19) by hn+1, we can bring it to the form

∫∫Q

FhΦh dx dt =∫∫Q

(−Φhuh0 −

n∑i,j=1

AhijΨ

hi u

hj +

n∑i=1

ΦhBhi u

hi +ΦhChuh

)dx dt. (6.20)

In (6.20) we can pass to the limit as h = hk → 0. Taking into account that the functions Fh, Φh, AhijΨ

hi ,

ΦhBhi , and ΦhCh converge in L2(Q) as h → 0 to the functions F , Φ, Aij

∂Ψ∂xi

, ΦBi, and ΦC respectively, andthe functions uh, uhi , and u

h0 converge weakly as h = hk → 0 to u, ∂u

∂xi, and ∂u

∂t respectively, we obtain the iden-tity (5.9).

Thus we have proved the existence of the generalized solution of the problem (5.7), (5.8) under the supposition

that ϕ(x) ∈0

C1(Ω) and F (x, t) ∈ C(Q). The solution we have constructed by virtue of (6.13) satisfies the inequality

∫∫Q

[u2 +

n∑i=1

(∂u

∂xi

)2

+(∂u

∂t

)2]dx dt M6

[ ∫∫Q

F 2 dx dt+∫Ω

n∑i=1

(∂ϕ

∂xi

)2

dx

]. (6.21)

512

Suppose now that ϕ(x) ∈0

W 1(Ω) and F (x, t) ∈ L2(Q). We can select subsequences ϕm(x) ∈0

C∞(Ω) andFm(x, t) ∈ C(Q) such that ϕm → ϕ in the norm of W 1(Ω), and Fm → F in the norm of L2(Q). As we haveproved, for any pair of functions ϕm(x) and Fm(x, t) in the cylinder Q there exists a generalized solution um(x, t)of the equation

n∑i,j=1

∂

∂xi

(Aij(x, t)

∂um∂xj

)+

n∑i=1

Bi(x, t)∂um∂xi

+ C(x, t)um −∂um∂t

= Fm(x, t)

with the conditionsum|t=0 = ϕm(x), um|S = 0.

Taking into account the estimate (6.21), we can see that the functions um(x, t) are bounded in the norm ofW 1,1(Q)uniformly in m. Therefore, one can select a subsequence umk

(x, t) converging weakly in W 1,1(Q) to the functionu(x, t). In the same way as was done above, we can prove that u|t=0 = ϕ(x) and u|S = 0.

Passing to the limit in the identity (5.9), written for the solutions um(x, t), we can see that the limit functionu(x, t) also satisfies this identity.

The existence of a solution of the problem (5.7), (5.8) is proved.Since the generalized solution of the problem (5.7). (5.8) is unique, the whole sequence uh converges to u(x, t)

as h→ 0. The functions uh are approximate solutions of this problem.The method of finite differences is widely used in the theory of partial differential equations to prove the ex-

istence of solutions of boundary-value problems and also to find approximate solutions of these problems. The ele-ments of the method of finite differences can be found in [19]. For the first time the method of finite differences wasused to prove the existence of the solution of the Dirichlet problem for the Laplace equation in L. A. Lusternik’spaper [85]. Then this method was used to prove the existence of solutions of problems for equations of various typesby R. Courant, C. Friedrichs, and H. Levy [86].

A review of the literature concerning the use of the method of finite differences in proofs of the existencetheorems is given in [61]. Questions concerning numerical solution of parabolic equations by the method of finitedifferences are studied also in [87].

7 SOME FUNCTIONAL METHODS OF SOLVING BOUNDARY-VALUE PROBLEMS

1. Recently the so-called functional methods of solving boundary-value problems became widespread.The essence of these methods is to consider a boundary-value problem as a problem of solving some operatorequation in a specially chosen functional space.

The origin of functional methods is connected with the names of S. L. Sobolev [82,92] and K. Friedrichs [93].Later these methods were used for the solution of boundary-value problems for elliptic and hyperbolic equationsand systems in [88–91, 95–97] and so on. Boundary-value problems for parabolic equations and systems were alsostudied by functional methods (see, e.g., [91, 99–101,125–128]).

In what follows, we shall consider the first boundary-value problem as an example of the use of functionalmethods for the proof of existence theorems for generalized solutions of boundary-value problems. This methodcoincides in its most important features with the method used by M. I. Vishik in [90,99] for the proof of the solvabilityof the boundary-value problems for parabolic and hyperbolic systems.

2. Consider in the cylinder Q = Ω× (0, T ) Eq. (5.7) with conditions

u(x, 0) = 0, u|S = 0. (7.1)

Suppose thatn∑

i,j=1

Aij(x, t)αiαj µn∑i=1

α2i for all the points (x, t) ∈ Q, where µ > 0. We shall construct a generalized

solution of the problem (5.7), (7.1) in the sense of Definition 1 of Sect. 5.Denote by W 1,1(Q) the Hilbert space of functions belonging to W 1,1(Q) and vanishing on Γ. In the space

W 1,1(Q) the scalar product u1, u2 =∫∫Q

(∂u1∂t

∂u2∂t +

n∑i=1

∂u1∂xi

∂u2∂xi

)dx dt is introduced. The corresponding norm will

be denoted by ‖u‖01,1. As before, (u1, u2) =∫∫Q

u1u2 dx dt denotes the scalar product in L2(Q). By V (Q) we shall

513

denote the set of functions v(x, t) belonging to W 1,1(Q) and having mixed derivatives ∂2v∂xi ∂t

(i = 1, . . . , n) squareintegrable in Q. Obviously V (Q) is dense in W 1,1(Q).

According to Sect. 5, u(x, t) is a weak generalized solution of the problem (5.7), (7.1) if u(x, t) ∈ W 1,1(Q)and satisfies the integral identity

(∂u

∂t,Φ)+

n∑i,j=1

(Aij

∂u

∂xi,∂Φ∂xj

)−

n∑i=1

(Bi

∂u

∂xi,Φ)− (Cu,Φ) = −(F,Φ) (7.2)

for any function Φ(x, t) of class0

W 1,1(Q). Obviously, to prove the existence of a weak generalized solution ofthe problem (5.7), (7.1) it is sufficient to show that there exists a function u(x, t) ∈ W 1,1(Q), satisfying the integralidentity

(∂u

∂t,∂v

∂te−θt

)+

n∑i,j=1

(Aij

∂u

∂xi,∂2v

∂xj ∂te−θt

)−

n∑i=1

(Bi

∂u

∂xi,∂v

∂te−θt

)−(Cu,

∂v

∂te−θt

)= −

(F,∂v

∂te−θt

)(7.3)

for any function v(x, t) ∈ V (Q) and any θ > 0. In fact, if Φ(x, t) ∈0

W 1,1(Q), then v(x, t) =t∫

0

Φ(x, τ)e−θτ dτ

belongs to V (Q). Substituting this v(x, t) in (7.3), we get (7.2). The uniqueness of the weak generalized solutionof the problem (5.7), (7.1) is proved in Sect. 5.

Theorem. If Aij , Bi, C, and∂Aij∂t are bounded and measurable in Q and F ∈ L2(Q), then there exists

a weak generalized solution of the problem (5.7), (7.1).

The proof of this theorem is based on a lemma about operators in a Hilbert space. Let A be an operator ina Hilbert space H. Denote by DA (DA ⊂ H) the domain of operator A and by RA its range.

Lemma. If DA is a dense set in H and the operator A has a bounded inverse A−1, then the range RA∗ ofthe operator A∗ conjugate to A coincides with the whole space H.

To prove the lemma it is sufficient to show that for any element h ∈ H an element y can be found such thatA∗y = h.

Let us take an arbitrary element v ∈ DA. In view of the suppositions of the lemma, there exists an element wof RA such that Av = w and v = A−1w. The linear functional

l(w) = (h, v) = (h,A−1w)

is bounded in RA, since|l(w)| ‖h‖ ‖A−1w‖ ‖A−1‖ ‖h‖ ‖w‖.

Because the operator A−1 is bounded, the domain RA is closed, and this means that it is a subspace ofH. Accordingto Riesz’s theorem on the form of general linear functionals inH (see, e.g., [83, p. 180]) there exists a unique element ysuch that l(w) = (y, w). This last equality can be rewritten in the following form: (h, v) = (y,Av) for any v ∈ DA.This means that y ∈ DA∗ and A∗y = h, which was to be proved.

Proof of the theorem. We shall use the following notation:

l1(u, v) =(∂u

∂t,∂v

∂te−θt

)+

n∑i,j=1

(Aij

∂u

∂xi,∂2v

∂xj ∂te−θt

)−

n∑i=1

(Bi

∂u

∂xi,∂v

∂te−θt

)−(Cu,

∂v

∂te−θt

), (7.4)

l2(v) = −(F,∂v

∂te−θt

).

It is easy to verify that if v ∈ V (Q), then l1(u, v) is a bounded linear functional with respect to u, defined in W 1,1(Q).In fact,

|l1(u, v)| M1

(‖v‖01,1 +

n∑j=1

∥∥∥∥ ∂2v

∂xj ∂t

∥∥∥∥0

)‖u‖01,1,

514

where M1 depends only on the upper bound of the absolute values of Aij , Bi, C, and ∂Aij∂t in Q; this upper

bound will be called M . According to Riesz’s theorem, there exists an operator A, defined on V (Q) and such thatl1(u, v) = u,Av.

The functional l2(v) is defined for v ∈ W 1,1(Q) and it is bounded, because∣∣∣∣(F,∂v

∂te−θt

)∣∣∣∣ ‖F‖0∥∥∥∥∂v∂t

∥∥∥∥0

‖F‖0 ‖v‖01,1.

Therefore, there exists an element F1 ∈ W 1,1(Q) such that

l2(v) = F1, v. (7.5)

Now one can rewrite the equality (7.3) in the form u,Av = F1, v.Let us prove that the operator A defined on V (Q) ∈ W 1,1(Q) has a bounded inverse A−1. To do this we

shall show thatv, v M2v,Av,

where M2 does not depend on v.Now we shall estimate each term on the right-hand side of (7.4), having made the substitution u = v. We

have (∂v

∂t,∂v

∂te−θt

)=∥∥∥∥∂v∂t e− θ

2 t

∥∥∥∥2

0

, (7.6)

−n∑i=1

(Bi

∂v

∂xi,∂v

∂te−θt

) −

n∑i=1

∣∣∣∣(Bi

∂v

∂xi,∂v

∂te−θt

)∣∣∣∣ −M

n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥0

∥∥∥∥∂v∂t e− θ2 t

∥∥∥∥0

−Mnε

2

∥∥∥∥∂v∂t e− θ2 t

∥∥∥∥2

0

− M

2ε

n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

, (7.7)

where ε > 0 is any positive number. Then(−Cv, ∂v

∂te−θt

) −

∣∣∣∣(Cv,

∂v

∂te−θt

)∣∣∣∣ −M∥∥∥ve− θ

2 t∥∥∥

0

∥∥∥∥∂v∂t e− θ2 t

∥∥∥∥0

−Mε

2

∥∥∥∥∂v∂t e−θ2t

∥∥∥∥2

0

− M

2ε

∥∥∥ve− θ2 t∣∣∣20 −Mε

2

∥∥∥∥∂v∂t e− θ2 t

∥∥∥∥2

0

− MM3

2ε

n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

. (7.8)

In the last estimate we have used the inequality∥∥∥ve− θ

2 t∥∥∥2

0M3

n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

,

which is valid for functions belonging to W 1,1(Q) vanishing on S (see Sect. 5, inequality (5.2)). Estimate nown∑

i,j=1

(Aij

∂v∂xi

, ∂2v∂xj ∂t

e−θt):

n∑i,j=1

(Aij

∂v

∂xi,∂2v

∂xj ∂te−θt

)=

12

n∑i,j=1

∫∫Q

∂

∂t

[Aij

∂v

∂xi

∂v

∂xje−θt

]dx dt

− 12

n∑i,j=1

(∂Aij

∂t

∂v

∂xi,∂v

∂xje−θt

)+θ

2

n∑i,j=1

(Aij

∂v

∂xi,∂v

∂xje−θt

)

=12

∫t=T

n∑i,j=1

Aij∂v

∂xi

∂v

∂xje−θT dx− 1

2

n∑i,j=1

(∂Aij

∂t

∂v

∂xi,∂v

∂xje−θt

)+θ

2

n∑i,j=1

(Aij

∂v

∂xi,∂v

∂xje−θt

)

θµ

2

n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

− M

2

n∑i,j=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥0

∥∥∥∥ ∂v∂xj e−θ2 t

∥∥∥∥0

(θµ

2− Mn

2

) n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

. (7.9)

515

Taking into account (7.6)–(7.9), we obtain

v,Av = l1(v, v) (1− Mnε

2− Mε

2

)∥∥∥∥∂v∂t e− θ2 t

∥∥∥∥2

0

+(θµ

2− Mn

2− M

2ε− MM3

2ε

) n∑i=1

∥∥∥∥ ∂v∂xi e−θ2 t

∥∥∥∥2

0

. (7.10)

Let us choose ε > 0 sufficiently small to have

1− Mnε

2− Mε

2>

12,

and then choose θ sufficiently large to have

θµ

2− Mn

2− M

2ε− MM3

2ε>

12.

It follows from the inequality (7.10) that

v,Av 12e−θT

(∥∥∥∥∂v∂t∥∥∥∥

2

0

+n∑i=1

∥∥∥∥ ∂v∂xi∥∥∥∥

2

0

)=

12e−θT v, v.

Thus we get ‖A−1‖ 2eθT . According to the lemma the operator A∗ is defined on the whole space W 1,1(Q).Therefore, there exists an element u ∈ W 1,1(Q) such that

A∗u = F1,

where F1 is defined by formula (7.5).So for any v ∈ V (Q) the equality

F1, v = A∗u, vholds, and, consequently,

F1, v = u,Av.This means that u is a generalized solution of the problem (5.7), (7.1).

Note that for elliptic equations of any order the solution of the first boundary-value problem is constructedin a similar way by L. Garding in [88].

3. The method we shall describe below was used in its various forms for finding solutions of the boundary-value problems for elliptic, hyperbolic, and parabolic equations and systems and also for solving the Cauchy problemfor higher-order hyperbolic equations and systems (see [89, 95–97,101]). The description of various applications ofthis method can be found in [98]. It seems that the original idea of this method is due to K. Friedrichs. He appliedit to the first boundary-value problem for Eq. (5.12) with conditions (7.1). The idea is as follows. The operator Ldefined by the left-hand side of Eq. (5.12) is considered as defined on functions from the Hilbert space W 2,1(Q)(see Sect. 5). This operator maps W 2,1(Q) into L2(Q). If we show that the set L(u), where u is any element ofW 2,1(Q), coincides with the set L2(Q), then we shall have proved that the problem (5.12), (7.1) is solvable for anyf ∈ L2(Q). If the set L(u) is closed, it is sufficient to show that in L2(Q) there is no nonzero element orthogonalto all the elements of the form L(u). To prove this is the same as to prove the uniqueness of a generalized solution(defined in a certain sense) of the problem conjugate to (5.12), (7.1).

Now consider the problem (5.12), (7.1) and let us find a strong generalized solution of this problem (seeSect. 5) for some function f ∈ L2(Q). As we have already noted, for such generalized solutions of the problem(5.12), (7.1) the inequality (2.35) of Sect. 2 holds. It follows easily from this inequality that the range of opera-tor L(u), defined for the functions u(x, t), belonging to W 2,1(Q), is closed in L2(Q). We shall show that this rangecoincides with the whole space L2(Q) if we prove the following proposition: from the equality

(L(u), v) = 0, (7.11)

where u is any element of W 2,1(Q) and v is some element of L2(Q), it follows that v = 0. This proposition is true ifaij , bi, c, and

∂aij∂xk

and their first-order derivatives in t are bounded in Q and the boundary of the domain Ω belongsto the class A2. The proof of this proposition can be found in [101]. It is based on the substitution of a functionu ∈ W 2,1(Q), defined in a special way in terms of v, in the equality (7.11).

516

8 SOLUTION OF THE FIRST BOUNDARY-VALUE PROBLEM BY THE METHOD OF CONTINUATIONWITH RESPECT TO A PARAMETER

The method of solution of boundary-value problems by the method of continuation with respect to a param-eter takes its origin in S. N. Bernstein’s works on solutions of the Dirichlet problem for elliptic equations [48]. Thenthis method was substantially developed in Leray and Schauder’s work [102]. Now this method is widely used toprove the existence of solutions of boundary-value problems for various types of linear and nonlinear equations andsystems (see, e.g., [2, 18, 55, 103]).

For the demonstration of this method we shall take as an example the first boundary-value problem forEq. (5.12). For simplicity we consider this equation in the cylinder Q with conditions (7.1). In a similar way, it ispossible to consider the case of a noncylindrical domain D and nonzero boundary conditions.

Consider the strong generalized solution of the problem (5.12), (7.1) in the sense of Definition 2 of Sect. 5.In other words, u(x, t) is a solution of the problem (5.12), (7.1) if there exists a sequence of functions um(x, t)of class C3(Q), converging in the mean to u(x, t) such that um|Γ = 0 and the functions L(um) = fm converge inthe mean to f . It follows from the estimate (2.35) that the um converge to u in the norm of W 2,1(Q). Note thatthe inequality (2.35) holds also for a generalized solution of the problem (5.12), (7.1). We have already noted inSect. 5 that a strong generalized solution satisfies Eq. (5.12) almost everywhere and satisfies the condition (7.1) inthe mean. Thus, we have to find a function u(x, t) ∈ W 2,1(Q) such that L(u) = f , where f ∈ L2(Q) is a givenfunction. We make the same suppositions on the coefficients of (5.12) and the boundary of the domain Ω as inTheorem 8 of Sect. 2, when we obtained the inequality (2.35).

The following construction is based on the estimate (2.35) and on the existence in Q of the solution ofthe problem for some equation of type (5.12), e.g., for the heat equation:

L0(u) ≡n∑i=1

∂2u

∂x2i

− ∂u

∂t= f. (8.1)

The solution of Eq. (8.1) with conditions (7.1) can be obtained by many methods (it is possible to usethe Fourier method, the Laplace transform, or any of the methods considered in previous sections). It is sufficientto construct a solution only for functions f that are infinitely differentiable in Q and have compact support in thiscylinder (that is, vanish in a neighborhood of the boundary of Q). The set of these functions is dense in the spaceL2(Q), and we can use the inequality (2.35) to obtain a solution of the problem (8.1), (7.1) in the space W 2,1(Q)for any f ∈ L2(Q). The inequality (2.35) for the solution of problem (8.1), (7.1) means that there exists an inverseoperator L−1

0 for L0(u). This inverse operator is bounded and it maps the whole space L2(Q) into W 2,1(Q).To prove the existence of a strong generalized solution of problem (5.12), (7.1) consider the set of parabolic

equationsLα(u) ≡ αL(u) + (1 − α)L0(u) = f, (8.2)

depending on a parameter α, 0 α 1. For α = 0 the operator Lα(u) coincides with L0(u), and for α = 1 itcoincides with the operator L(u). The function u belongs to the space W 2,1(Q). Note that the estimate (2.35) isvalid for all the operators Lα(u) with a constant M not depending on α, since it is possible to estimate uniformlyin α the moduli of continuity of the coefficients in the higher derivatives, the least eigenvalues of the matrix of thesecoefficients, and the upper bound of absolute values of all the coefficients of the operators Lα(u), 0 α 1.

Rewrite equations (8.2) in the form

Lα(u) ≡ L0(u) + α(L − L0)u = f (8.3)

and then apply to both sides of (8.3) the operator L−10 . We obtain

u+ αL−10 (L− L0)u = L−1

0 f. (8.4)

Denote by A the operator L−10 (L − L0) mapping the functions from W 2,1(Q) into functions of the same

space. The operator A is bounded, since ‖(L− L0)u‖0 M1‖u‖2,1 and ‖L−10 v‖2,1 M2‖v‖0. Equation (8.4) has

the formu+ αAu = L−1

0 f, (8.5)

517

where ‖αA‖ < 1 for sufficiently small α, 0 < α α1(M). Therefore, Eq. (8.5) for 0 α α1 can be solved inthe space W 2,1(Q) for any function f ∈ L2(Q) by method of successive approximations (see [83, p. 47]). We canapply the operator L0 to both sides of (8.5) and obtain as a result that the solution of Eq. (8.5) belongs to W 2,1(Q)and satisfies Eq. (8.3) for 0 α α1 almost everywhere in Q.

The fact that the solution u(x, t) of Eq. (8.3) for α = α1 belongs to W 2,1(Q) means that there existsa sequence of functions um(x, t) ∈ C∞(Q) such that um|Γ = 0 and um → u in the norm of W 2,1(Q) for m → ∞.Therefore, for all solutions u(x, t) of Eq. (8.3) for α = α1 belonging to the class W 2,1(Q) the inequality (2.35) holds.Hence it follows that there exists an operator L−1

α1and its norm is bounded by a constant depending only on M .

Rewrite Eq. (8.2) in the formLα(u) ≡ Lα1(u) + (α − α1)(L− L0)u = f (8.6)

and carry out for (8.6) the same calculations we did for (8.3). The part of the operator L0 in Eq. (8.6) will playthe operator Lα1 . Thus we can show that Eq. (8.2) with condition (7.1) has a solution belonging to W 2,1(Q) for0 α − α1 α1, that is, for α 2α1, because the norm of the operator L−1

α1(L − L0) in W 2,1(Q) is bounded

uniformly with respect to α. Repeating this reasoning, after a finite number of steps equal to[

1α1

]+1 we establish

the solvability of Eq. (8.3) with condition (7.1) in the space W 2,1(Q) for α = 1, and this was to be proved.Note that we have obtained the solution of the problem (5.12), (7.1) under weak suppositions about the co-

efficients of the equation L(u) = f . Namely, we supposed only that the coefficients aij are continuous and bi and care bounded in Q. A similar existence theorem can be proved for any domain D such that for this domain the esti-mate (2.35) holds and there exists a smooth solution of the first boundary-value problem for the heat equation (8.1).

9 APPLICATION OF GALERKIN’S METHODTO THE CONSTRUCTION OF A SOLUTION OF THE FIRST BOUNDARY-VALUE METHOD

In this section, we construct a weak (see Sect. 5) generalized solution of the first boundary-value problemin the cylinder Q = Ω × (0, T ) for Eq. (5.7) with conditions (5.8). We shall suppose that the coefficients Aij are

of class C1(Q); Bi, C, and F are continuous in Q; the initial function ϕ(x) ∈0

W 1(Ω),n∑

i,j=1

Aij(x, t)αiαj µn∑i=1

α2i

(µ > 0), and the boundary σ of the domain Ω belongs to the class A2. We shall use Galerkin’s method in the formused by E. Hopf to construct a solution of the boundary-value problem for the Navier–Stokes system [104]. Laterthis method was widely used for the construction of solutions of boundary-value problems for linear equations ofvarious types, and also for linear elliptic, parabolic, and hyperbolic systems (see [99, 105]).

According to Lemma 2 of Sect. 5, in the domain Ω there exists a system of functions X1(x), X2(x),. . . ,Xk(x),. . . , with properties (1), (2), and (3), of Lemma 2. We shall seek the approximate solution uN (x, t) ofthe problem (5.7), (5.8) as a sum

uN(x, t) =N∑k=1

αNk (t)Xk(x). (9.1)

The functions αNk (t) are defined by the conditions∫Ω

[L(uN )− f ]Xk(x) dx = 0 for 0 t T (k = 1, . . . , N). (9.2)

This means that∫Ω

[−

n∑i,j=1

Aij∂uN∂xj

∂Xk

∂xi+( n∑

i=1

Bi∂uN∂xi

+ CuN −∂uN∂t− F

)Xk

]dx = 0 (k = 1, . . . , N). (9.3)

Taking into account the fact that the system Xk(x) is orthogonal in L2(Ω), we can obtain from equality (9.3)a system of ordinary linear first-order differential equations with continuous coefficients:

dαNk (t)dt

=N∑s=1

Aks(t)αNs (t) + Bk(t) (k = 1, . . . , N); (9.4)

this system allows us to define the functions αNk (t).

518

The initial function ϕ(x) belongs to0

W 1(Ω). Therefore, according to Lemma 2 of Sect. 5 there exists

a sequence of functions ϕN (x) =N∑k=1

cNk Xk(x), converging in the norm of W 1(Ω) to the function ϕ(x). DefineαNk (0) from the condition

uN(x, 0) ≡N∑k=1

αNk (0)Xk(x) = ϕN (x) ≡N∑k=1

cNk Xk(x). (9.5)

This means thatαNk (0) = cNk (k = 1, . . . , N). (9.6)

The solution of the system (9.4) with conditions (9.6) is constructed on the interval 0 t T .We shall show that the function u(x, t) = lim

N→∞uN(x, t) is a generalized solution of the problem (5.7), (5.8)

in the sense of Definition 1 of Sect. 5.First we show that uN (x, t) are bounded in the norm of W 1,1(Q) uniformly in N . To do this we multiply

Eqs. (9.3) by αNk (t)e−θt and take the sum of the equalities thus obtained with respect to k from 1 to N . Then weintegrate with respect to t from 0 to T . We arrive at the following relation:

∫Q

e−θt[−

n∑i,j=1

Aij∂uN∂xi

∂uN∂xj

+n∑i=1

BiuN∂uN∂xi

+ (CuN − F )uN − uN∂uN∂t

]dx dt = 0. (9.7)

We can apply the integration by parts to the last term in the left-hand side of (9.7). In the same way as wedid when we deduced the estimate (2.29) we shall obtain that for sufficiently large θ > 0 the inequality

12

∫∫Q

θe−θtu2N dx dt+

∫∫Q

e−θtn∑

i,j=1

Aij∂uN∂xi

∂uN∂xj

dx dt M1

(∫∫Q

F 2 dx dt+∫Ω

ϕ2N dx

)

holds. Here and below, Mi will be constants not depending on N . It follows from the last inequality that

∫∫Q

u2N dx dt M2 and

∫∫Q

n∑i=1

(∂uN∂xi

)2

dx dt M2.

Now multiply Eqs. (9.3) by dαNk (t)dt , take the sum with respect to k from 1 to N , and integrate with respect

to t from 0 to T . We get

∫∫Q

[−

n∑i,j=1

Aij∂uN∂xj

∂2uN∂xi ∂t

+n∑i=1

Bi∂uN∂xi

∂uN∂t

+ (CuN − F )∂uN∂t−(∂uN∂t

)2]dx dt = 0.

Transform the first sum under the integral sign using integration by parts, as we did when we deduced (2.30). Weobtain the inequality

∫t=T

n∑i=1

(∂uN∂xi

)2

dx+∫∫Q

(∂uN∂t

)2

dx dt M3

∫∫Q

F 2 dx dt+∫Ω

[ϕ2N +

n∑i=1

(∂ϕN∂xi

)2]dx

, (9.8)

and it follows that ∫∫Q

(∂uN∂t

)2

dx dt M4.

When we proved the inequality we used the fact that∫

t=0

n∑i,j=1

Aij∂uN∂xi

∂uN∂xjM5

∫Ω

n∑i=1

(∂ϕN∂xi

)2dx and the last integral

is bounded uniformly in N because of the convergence of ∂ϕN∂xi

to ∂ϕ∂xi

in the norm of L2(Ω).

519

Thus we have ‖uN‖1,1 M6. According to Theorem 1 of Sect. 5, it is possible to select from the sequenceuN (x, t) a subsequence uNk

(x, t) converging in the mean to the function u(x, t) ∈ W 1,1(Q). The derivatives ∂uNk∂xi

and ∂uNk∂t will converge weakly to the corresponding generalized derivatives ∂u

∂xiand ∂u

∂t as Nk →∞. Conditions (5.8)are satisfied for u(x, t); we have to take into account equalities (9.5), the condition uN |S = 0, and Theorem 1 ofSect. 5.

Now we show that u(x, t) satisfies the integral identity (5.9).Note that it is sufficient to prove (5.9) for the functions Φ(x, t) infinitely differentiable in Q and vanishing

in a neighborhood of the boundary of Q. In fact, we can insert into (5.9) a function Φ(x, t) = χm(t)Ψ(x, t), where

Ψ(x, t) ∈0

C∞(Q) and χm(t) is an infinitely differentiable function with the following properties: 0 χm(t) 1;χm(t) = 0 for 0 t 1

m and for T − 1m t T ; χm(t) = 1 for 2

m t T − 2m (m = 1, 2, . . .). Passing to

the limit in (5.9) as m → ∞, we see that the integral identity (5.9) is satisfied for any function Ψ(x, t) ∈0

C∞(Q),

and therefore for any function Ψ(x, t) ∈0

W 1,1(Q).

According to Lemma 2 of Sect. 5, there exists a sequence Φm(x, t) =n∑

k=1

cmk (t)Xk(x) such that the Φm

converge to Φ in the mean together with their derivatives in x1, . . . , xn. First we shall prove that the integralidentity (5.9) is satisfied for the functions Φm(x, t). Multiply the equality (9.3) by cmk (t), take the sum with respectto k from 1 to m, and integrate with respect to t from 0 to T . We obtain∫∫

Q

[−

n∑i,j=1

Aij∂uN∂xj

∂Φm

∂xi+( n∑

i=1

Bi∂uN∂xi

+ CuN − F −∂uN∂t

)Φm

]dx dt = 0.

In this relation, we can pass to the limit with fixed m as N tends to infinity along the subsequence Nk chosen above.We get ∫∫

Q

[−

n∑i,j=1

Aij∂u

∂xj

∂Φm

∂xi+( n∑

i=1

Bi∂u

∂xi+ Cu− F − ∂u

∂t

)Φm

]dx dt = 0.

In this equality, we can pass to the limit as m→∞ and obtain the integral identity (5.9). Thus we have proved thatu(x, t) is a weak generalized solution of problem (5.7), (5.8). The uniqueness of this solution is proved in Sect. 5.

Because we have constructed a unique solution u(x, t), the whole sequence uN (x, t) converges to it. Thereforethe construction used to prove the existence of a solution for problem (5.7), (5.8) also gives an easy way to obtainan approximate solution of this problem. Various modifications of this method are often used in practice.

Note that without major changes it is possible to prove the convergence of the functions uN (x, t) to the so-lution u(x, t) under weaker suppositions on the smoothness of the coefficients of the equation and the boundary ofthe domain.

10 GENERALIZED SOLUTION FOR THE CAUCHY PROBLEM

We shall consider a Cauchy problem in the domain H0 < t T1, x ∈ En for Eq. (5.7) with the initialcondition

u(x, 0) = ϕ(x), x ∈ En. (10.1)

The functions Aij(x, t),∂Aij(x,t)

∂t , Bi(x, t), and C(x, t) are assumed to be continuous and bounded in H . Let

the absolute values of these functions not exceed M and let the inequalityn∑

i,j=1

Aijαiαj n∑i=1

α2i , µ > 0 hold.

In this section, we shall define a generalized solution of the Cauchy problem (5.7), (10.1) and prove the ex-istence and uniqueness of such a solution in a class of functions that can grow in a certain manner as |x| ≡( n∑i=1

x2i

) 12 → ∞. We shall suppose that in any bounded domain ϕ(x) ∈ W 1, F (x, t) ∈ L2 and for any α > 0

the integrals ∫En

e−α|x|2[ϕ2(x) +

n∑k=1

(∂ϕ(x)∂xk

)2]dx,

∫∫H

e−α|x|2F 2(x, t) dx dt (10.2)

converge.

520

The function v(x, t) is said to belong to Vα if in any bounded domain v(x, t) ∈ W 1,1 and the integral∫En

e−α|x|2v2(x, t) dx

converges uniformly in t for 0 t T1 and, moreover, that

∫∫H

e−α|x|2[v2 +

n∑k=1

(∂v

∂xk

)2

+(∂v

∂t

)2]dx dt <∞

(for some α > 0). If v(x, t) ∈ Vα for all α > 0, then we shall say that v(x, t) ∈ V .

Definition. The function u(x, t) will be called a generalized solution of the problem (5.7), (10.1) if u(x, t)belongs to the class Vα for some α > 0, is equal to ϕ(x) for t = 0, and satisfies the equality

T∫0

∫En

[ n∑i,j=1

Aij∂u

∂xj

∂Φ∂xi−( n∑

i=1

Bi∂u

∂xi+ Cu− du

dt− F

)Φ]dx dt = 0, (10.3)

where 0 < T T1, and Φ(x, t) is an arbitrary function with compact support with respect to x, belonging tothe class W 1,1.

We shall prove the existence of a generalized solution of the Cauchy problem in the class V and the uniquenessof the generalized solution in a wider class of functions. That is, in proving the uniqueness of the solution of Cauchyproblem we shall assume that a solution u(x, t) belongs to the class Vα1 for some α1 > 0, where the numbers α1 ingeneral are different for different solutions.

To prove the uniqueness of the generalized solution it is sufficient to prove that the generalized solutionu(x, t) of class Vα1 for a certain α1 > 0, is identically zero if ϕ(x) ≡ 0 and F (x, t) ≡ 0.

1. Proof of the Uniqueness Theorem. Denote by γR(x) a smooth function equal to unity for |x| R

and equal to zero for |x| R + 2, such that 0 γR(x) 1 andn∑

k=1

(∂γR(x)∂xk

)2 1. Insert into the identity (10.3)

instead of Φ(x, t) the functione−2α1|x|2−β|x|2tγR(x)u(x, t),

where the constants β > 0 and R > 0 will be chosen later. We shall obtain

T∫0

∫En

e−2α1|x|2−β|x|2t n∑

i,j=1

Aij∂u

∂xj

[∂(uγR)∂xi

− 2(2α1 + βt)xiγRu]+ γRu

∂u

∂t− CγRu2 − γRu

n∑i=1

Bi∂u

∂xi

dx dt = 0.

(10.4)It follows from the equality (10.4) that

T∫0

∫|x|R

e−2α1|x|2−β|x|2t n∑

i,j=1

[Aij

∂u

∂xi

∂u

∂xj− 2(2α1 + βt)xiAiju

∂u

∂xj

]+ u

∂u

∂t

dx dt

M1

T∫0

∫En

e−2α1|x|2−β|x|2t(u2 + |u|

n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣)dx dt

+M1

T∫0

∫R|x|R+2

e−2α1|x|2−β|x|2t[u2 +

n∑i=1

(∂u

∂xi

)2

+(∂u

∂t

)2](1 + |x|) dx dt, (10.5)

where M1 depends only on M and n.

521

Let us transform the integral I1 =T∫0

∫|x|R

e−2α1|x|2−β|x|2tu∂u∂t dx dt. It is obvious that

I1 =12

T∫0

∫|x|R

e−2α1|x|2−β|x|2t ∂(u2)

∂tdx dt

=12

∫|x|R

u2(x, T )e−2α1|x|2−βT |x|2dx+β

2

T∫0

∫|x|R

|x|2u2e−2α1|x|2−β|x|2tdx dt,

because u(x, 0) = 0.Now we estimate the integral

I2 =

T∫0

∫|x|R

e−2α1|x|2−β|x|2tn∑

i,j=1

2(2α1 + βt)xiAiju∂u

∂xjdx dt.

We have

|I2| M2(βT + 2α1)

T∫0

∫|x|R

e−2α1|x|2−β|x|2t|x|n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣ |u| dx dt

µ

2

T∫0

∫|x|R

e−2α1|x|2−β|x|2tn∑i=1

(∂u

∂xi

)2

dx dt+ (βT + 2α1)2M3

T∫0

∫|x|R

|x|2u2e−2α1|x|2−β|x|2tdx dt,

where the constants M2 and M3 depend only on µ, M , and n.These transformations bring the inequality (10.5) to the following form:

T∫0

∫|x|R

e−2α1|x|2−β|x|2t(2

n∑i,j=1

Aij∂u

∂xi

∂u

∂xj+ β|x|2u2

)dx dt+

∫|x|R

e−2α1|x|2−βT |x|2u2(x, T )dx

2M1

T∫0

∫En

e−2α1|x|2−β|x|2t(u2 + |u|

n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣)dx dt+ µ

T∫0

∫|x|R

e−2α1|x|2−β|x|2tn∑i=1

(∂u

∂xi

)2

dx dt

+ 2(βT + 2α1)2M3

T∫0

∫|x|R

|x|2u2e−2α1|x|2−β|x|2tdx dt

+ 2M1

T∫0

∫R|x|R+2

e−2α1|x|2−β|x|2t(1 + |x|)[u2 +

n∑i=1

(∂u

∂xi

)2

+(∂u

∂t

)2]dx dt. (10.6)

The last of the integrals tends to zero as R → ∞, because e−2α1|x|2(1 + |x|) < e−α1|x|2 and u belongs to

the class Vα1 . Taking into account thatn∑

i,j=1

Aij∂u∂xi

∂u∂xj µ

n∑i=1

(∂u∂xi

)2 and passing to the limit in (10.6) as R→∞,we get

522

T∫0

∫En

e−2α1|x|2−β|x|2t[2µ

n∑i=1

(∂u

∂xi

)2

+ β|x|2u2

]dx dt +

∫En

e−2α1|x|2−βT |x|2u2(x, T ) dx

2M1

T∫0

∫En

e−2α1|x|2−β|x|2t(u2 + |u|

n∑i=1

∣∣∣∣ ∂u∂xi∣∣∣∣)dx dt

+ µ

T∫0

∫En

e−2α1|x|2−β|x|2tn∑i=1

(∂u

∂xi

)2

dx dt+ 2(βT + 2α1)2M3

T∫0

∫En

|x|2u2e−2α1|x|2−β|x|2tdx dt

M4

T∫0

∫En

e−2α1|x|2−β|x|2tu2 dx dt+ 2µ

T∫0

∫En

e−2α1|x|2−β|x|2tn∑i=1

(∂u

∂xi

)2

dx dt

+ 2(βT + 2α1)2M3

T∫0

∫En

e−2α1|x|2−β|x|2t|x|2u2 dx dt, (10.7)

where M4 depends only on µ, M , and n.Choose β > 0 so large and T0 > 0 so small as to have the inequality 2M3(βT + 2α1)2 < β for all T T0.

Then it follows from (10.7) that

∫En

e−2α1|x|2−βT |x|2u2(x, T ) dx M4

T∫0

∫En

e−2α1|x|2−βt|x|2u2(x, t) dx dt for 0 < T T0. (10.8)

Let us integrate the inequality (10.8) with respect to T from 0 to δ T0:

δ∫0

∫En

e−2α1|x|2−βT |x|2u2(x, T ) dx dT

M4

δ∫0

[ T∫0

∫En

e−2α1|x|2−βt|x|2u2(x, t) dx dt]dT M4δ

δ∫0

∫En

e−2α1|x|2−βt|x|2u2(x, t) dx dt.

If we choose δ = 12M4

, then it follows from the last inequality that u(x, t) ≡ 0 for 0 t δ. Note that δdepends only on µ,M , and n. Apply the same reasonings to the domain δ t 2δ, then to the domain 2δ t 3δ,and so on; after a finite number of steps we shall obtain u(x, t) ≡ 0 in the whole layer 0 t T . The uniquenessof the generalized solution of the Cauchy problem (5.7), (10.1) is proved.

A similar uniqueness theorem is proved in [106].2. Construction of the Generalized Solution. Now we shall show that under the conditions (10.2) on

the initial function ϕ(x) and the right-hand side F (x, t), there exists a generalized solution of the Cauchy problem(5.7), (10.1) belonging to the class V .

Consider the cylinders QN having as bases balls of radius N with center at the origin; the height of the cylin-ders is T1. Denote by uN (x, t) the generalized solution of the first boundary-value problem for Eq. (5.7) in the cylin-der QN with the initial condition uN (x, 0) = ϕ(x)γN−2(x) and the boundary condition uN |S = 0. In Sect. 6, it isproved that under the suppositions we have made on the coefficients of (5.7) and the functions ϕ(x) and F (x, t)

such a solution uN(x, t) ∈0

W 1,1(QN ) exists and by definition satisfies the relation

∫∫QN

[ n∑i,j=1

Aij∂uN∂xj

∂Φ∂xi−( n∑

i=1

Bi∂uN∂xi

+ CuN −∂uN∂t− F

)Φ]dx dt = 0 (10.9)

for Φ(x, t) ∈0

W 1,1(QN ).

523

Let us show that for any α > 0 the integrals

∫∫QN

e−α|x|2[ n∑i=1

(∂uN∂xi

)2

+(∂uN∂t

)2]dx dt,

∫|x|N

e−α|x|2u2N (x, T ) dx

are bounded uniformly in N and in T T1. Insert into (10.9) the function Φ(x, t) = e−α|x|2(t+1)uN (x, t). Repeating

the calculations we made above when we proved the uniqueness of the solution of the Cauchy problem (having nowβ = 2α1 = α), we obtain that for α α0 and all T , 0 < T T1, the following inequality holds:

T∫0

∫|x|N

e−α|x|2(t+1)

n∑i=1

(∂uN∂

xi

)2

dx dt+∫

|x|N

e−α|x|2(T+1)u2

N (x, T ) dx

M5

T∫0

∫|x|N

e−α|x|2(t+1)(u2

N + F 2) dx dt+∫

|x|N

e−α|x|2u2N (x, 0) dx

,

where α0 is sufficiently small (the constant α0 depends only on µ, M , and n). Hence, because of the suppositionswe made on the functions ϕ(x) and F (x, t) we can deduce that the integrals

T∫0

∫|x|N

e−α|x|2

n∑i=1

(∂uN∂xi

)2

dx dt,

∫|x|N

e−α|x|2u2N (x, T ) dx (10.10)

are bounded uniformly in N and T for α α0, and therefore for all positive α.The proof that the integral containing the derivative ∂uN

∂t is also uniformly bounded is a little morecomplicated. To prove this we insert the function Φ(x, t) = e−α|x|

2(t+1)uNτ (x, t) into the equality (10.9); herevτ (x, t) ≡ v(x,t)−v(x,t−τ)

τ (α is some fixed positive number). We can suppose that the function uN(x, t) is extendedto the domain t < 0: uN (x, t) ≡ uN (x, 0) if t < 0. The equality (10.9) takes the following form:

∫∫QN

[ n∑i,j=1

Aij∂uN∂xj

∂(uNτ e−α|x|2(t+1))∂xi

+ e−α|x|2(t+1)uNτ

(∂uN∂t− CuN −

n∑i=1

Bi∂uN∂xi

+ F

)]dx dt = 0. (10.11)

For simplicity we omit the index N in uN and we denote by Ki quantities bounded in absolute value by

M6

1 +[∫∫QN

e−α|x|2(t+1)u2

τ dx dt

]1/2,

where the constant M6 does not depend on N and τ . Doing this we take into account the uniform boundedness ofintegrals (10.10) for all α > 0, and, therefore, the uniform in N and T boundedness of integrals

∫∫QN

|x|me−α|x|2n∑i=1

(∂uN∂xi

)2

dx dt,

∫|x|N

|x|me−α|x|2u2N(x, T )dx

(m > 0 is an arbitrary number). It follows from (10.11) that

∫∫QN

n∑i,j=1

Aij∂u

∂xj

∂uτ∂xi

e−α|x|2(t+1)dx dt+

∫∫QN

uτ∂u

∂te−α|x|

2(t+1)dx dt = K1. (10.12)

Let us estimate the first integral in the left-hand side of the equality (10.12). To do this we use the relation

(vw)τ = vτw(x, t− τ) + vwτ .

524

We obtain ∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)

)τ

dx dt

=∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj

)τ

e−α|x|2(t+1) − α|x|2

n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)

dx dt

=∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj

)τ

e−α|x|2(t+1)dx dt +K2 =

∫∫QN

n∑i,j=1

Aij

(∂u

∂xi

∂u

∂xj

)τ

e−α|x|2(t+1)dx dt +K3; (10.13)

here0 < τ < τ.

The expression under the last integral sign can be transformed in the following way:n∑

i,j=1

Aij

(∂u

∂xi

∂u

∂xj

)τ

=n∑

i,j=1

Aij

(∂u

∂xi

∂uτ∂xj

+∂u(x, t− τ)

∂xi

∂uτ∂xj

)

= 2n∑

i,j=1

Aij∂u

∂xi

∂uτ∂xj−

n∑i,j=1

Aij

(∂u

∂xi− ∂u(x, t− τ)

∂xi

)∂uτ∂xj

= 2n∑

i,j=1

Aij∂u

∂xi

∂uτ∂xj− τ

n∑i,j=1

Aij∂uτ∂xi

∂uτ∂xj

.

Hence and from (10.13) we obtain that∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)

)τ

dx dt

= 2∫∫QN

n∑i,j=1

Aij∂u

∂xi

∂uτ∂xj

e−α|x|2(t+1)dx dt− τ

∫∫QN

n∑i,j=1

Aij∂uτ∂xi

∂uτ∂xj

e−α|x|2(t+1)dx dt+K3.

Therefore ∫∫QN

n∑i,j=1

Aij∂u

∂xi

∂uτ∂xj

e−α|x|2(t+1)dx dt =

12

∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)

)τ

dx dt

+τ

2

∫∫QN

n∑i,j=1

Aij∂uτ∂xi

∂uτ∂xj

e−α|x|2(t+1)dx dt +K4

12

∫∫QN

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)

)τ

dx dt +K4. (10.14)

It is obvious that

∫∫QN

vτ dx dt =1τ

T1∫0

∫|x|N

[v(x, t) − v(x, t− τ)] dx dt

=1τ

T1∫0

∫|x|N

v(x, t) dx dt − 1τ

T1−τ∫−τ

∫|x|N

v(x, ζ) dx dζ =1τ

T1∫T1−τ

∫|x|N

v(x, t) dx dt − 1τ

0∫−τ

∫|x|N

v(x, t) dx dt.

Therefore the inequality (10.14) can be rewritten in the following form:∫∫QN

n∑i,j=1

Aij∂u

∂xi

∂uτ∂xj

e−α|x|2(t+1)dx dt

12τ

T1∫T1−τ

∫|x|N

n∑i,j=1

Aij∂u

∂xi

∂u

∂xje−α|x|

2(t+1+τ)dx dt− 12

∫|x|N

( n∑i,j=1

Aij∂u

∂xi

∂u

∂xj

)t=0

e−α|x|2(1+τ)dx+K4 K5.

525

From this inequality and from (10.12) it follows that

∫∫QN

uτ∂u

∂te−α|x|

2(t+1)dx dt M7

1 +[ ∫∫QN

u2τe−α|x|2(t+1)dx dt

]1/2.

Passing to the limit as τ → 0 (M7 does not depend on τ), we obtain

∫∫QN

(∂uN∂t

)2

e−α|x|2(t+1)dx dt M7

1 +[ ∫∫QN

(∂uN∂t

)2

e−α|x|2(t+1)dx dt

]1/2.

From the latter inequality it follows that the integral

∫∫QN

(∂uN∂t

)2

e−α|x|2(t+1)dx dt (10.15)

is uniformly bounded with respect to N .Since in any bounded domain the functions uN(x, t) and their first-order derivatives are uniformly bounded in

the L2-norm, we can select a subsequence uNk(x, t) that in any bounded domain converges weakly to the function

u(x, t). The function u(x, t) in any bounded domain belongs to the class W 1,1 and for t = 0 satisfies the initialcondition (10.1). Since the integrals (10.10) and (10.15) are uniformly bounded, the integrals

T1∫0

∫En

e−α|x|2[ n∑i=1

(∂u

∂xi

)2

+(∂u

∂t

)2]dx dt and

∫En

e−α|x|2u2(x, T ) dx (10.16)

converge for any α>0. Moreover, since the integrals (10.15) and (10.16) are bounded, the integral∫En

e−α|x|2u2(x, t)dx

converges uniformly in t for 0 t T1. Therefore, u(x, t) ∈ V .Passing to the limit as N = Nk →∞ in the equality (10.9), where the function Φ(x, t) has compact support

with respect to x, we can see that the function u(x, t) satisfies the condition (10.3).Thus we have proved that the function u(x, t) is a generalized solution of the Cauchy problem (5.7), (10.1).From the estimates of the integrals (10.16) that we have established above it follows that there is a continuous

dependence of the solution on the right-hand side F (x, t) and on the initial function ϕ(x), that is, the integrals (10.16)are small if the integrals

T1∫0

∫En

F 2(x, t)e−α|x|2dx dt

and ∫En

e−α|x|2[ϕ2(x) +

n∑i=1

(∂ϕ(x)∂xi

)2]dx

are small.Let us note that the suppositions on the growth of the initial function ϕ(x) and the right-hand side F (x, t)

that we made when we proved the existence of the generalized solution of the Cauchy problem (5.7), (10.1), arefairly exact. In fact, we can consider as an example the Cauchy problem for the heat equation

∂2u

∂x2− ∂u

∂t= 0 (10.17)

with the initial conditionu(x, 0) = eAx

2, A > 0 (10.18)

526

(see subsection 2 of Sect. 4). The function (10.18) does not fulfill the condition of the convergence of the firstintegral of (10.2). The solution

u(x, t) = (1− 4At)−12 e

Ax21−4At (10.19)

of the problem (10.17), (10.18) in the strip 0 < t T1 < 14A belongs to the class Vα for some α = α(T1).

According to the uniqueness theorem, there is no other generalized solution of the problem (10.17), (10.18). Butthe function (10.19) tends to infinity as t→ 1

4A . Therefore the problem (10.17), (10.18) has no generalized solutiondefined for all t > 0. If we take as an initial function instead of (10.18) the function eAx

2−ε, where ε > 0, then

the integrals (10.2) turn out to be converging for all α > 0, and therefore there exists a generalized solution ofthe Cauchy problem for Eq. (10.17) with this initial function for all t > 0.

11 ON THE SMOOTHNESS OF GENERALIZED SOLUTIONS

In the previous sections we have proved, using various methods, the existence of the generalized solutions ofthe first boundary-value problem for Eqs. (5.7) or (5.12). The generalized solutions were understood as in Definitions1 or 2 of Sect. 5.

In this section, we shall show that these generalized solutions have derivatives up to a certain order ifthe right-hand side of the equation, its coefficients, and the boundary of the domain are sufficiently smooth.

In Theorem 1 we show that a weak generalized solution of the problem (5.7), (5.8) under some additional

suppositions on the data belongs to0

W 2,1(Q) and satisfies the equation almost everywhere. We shall prove thisusing finite differences; the essence of the proof is as follows. Let us consider the difference quotients

∆u∆xi

=u(x1, . . . , xn, t)− u(x1, . . . , xi−1, xi − h, xi+1, . . . , xn, t)

h.

It is possible to establish that they satisfy an “energy” inequality similar to the inequality (2.29). Hence wecan see that the difference quotients corresponding to the derivatives of the generalized solution with respect tothe space variables are bounded in L2(Q) uniformly in h. By Lemma 3, which is formulated below, we get thatthe solution u(x, t) has generalized derivatives of the second order with respect to the space variables. One canconsider difference quotients for the derivatives of the generalized solution and establish for them estimates that areuniform in h. This allows us to prove the existence of higher-order derivatives. The estimates near S are obtainedin the same way as in Theorem 1. But these estimates are rather complicated technically. Therefore, in Theorem 2we prove the smoothness of generalized solutions only inside the cylinder Q.

This theorem states that for sufficiently smooth coefficients and the right-hand side F of Eq. (5.7)u(x, t) ∈ W 2k,k(Q∗δ) in any interior subdomain Q∗δ of the cylinder Q. For sufficiently large k by embedding theo-rems [82] we obtain that u(x, t) has continuous derivatives of higher orders inside Q. The method described herewas used in [107] to prove the smoothness of generalized solutions for strongly elliptic systems of equations.

We have to note that the proof of Theorem 1 has as a principal goal the demonstration of the method. Infact, under the supposition that the coefficients in the higher derivatives in Eqs. (5.7) and (5.12) have continuousderivatives, it is clear that the forms (5.7) and (5.12) are equivalent. It follows from Theorem 2 of Sect. 5 and

the existence theorem of Sect. 8 that the weak generalized solution belongs to0

W 2,1(Q).First we prove three lemmas that establish the connection between the difference quotients of a function and

its generalized partial derivatives.

Lemma 1. Let the function u(x, t) ∈ W 1,1(Q) vanish in a neighborhood of S. Consider its differencequotients in the direction of some axis xk:

ukh(x1, . . . , xn, t) ≡1h[u(x1, . . . , xn, t)− u(x1, . . . , xk−1, xk − h, xk+1, . . . , xn, t)].

Then

‖ukh‖0 ∥∥∥∥ ∂u∂xk

∥∥∥∥0

(11.1)

and ∥∥∥∥ukh − ∂u

∂xk

∥∥∥∥0

→ 0 for h→ 0 (k = 1, . . . , n). (11.2)

527

Proof. First we establish the inequality (11.1) for an infinitely differentiable compact support function u(x)of one variable. We have

uh(x) =1h[u(x)− u(x− h)] = 1

h

x∫x−h

du

dξdξ.

Therefore,

[uh(x)]2 1h2

[ x∫x−h

∣∣∣∣dudξ∣∣∣∣dξ]2

1h

x∫x−h

(du

dξ

)2

dξ,

+∞∫−∞

u2h(x) dx

1h

+∞∫−∞

[ x∫x−h

(du

dξ

)2

dξ

]dx =

1h

+∞∫−∞

dξ

ξ+h∫ξ

(du

dξ

)2

dx =

+∞∫−∞

(du

dξ

)2

dξ.

We have proved inequality (11.1) for the case of one variable.Now let u(x, t) be a continuously differentiable function of (n+1) variables. For this function inequality (11.1)

can be written along any line parallel to the xk axis. Integrating by the remaining variables, we get the requiredinequality (11.1). If u(x, t) ∈ W 1,1(Q) vanishes in a neighborhood of S, we can consider its averaged functions (seeSect. 5) u&(x, t). For these averaged functions (11.1) holds. We can pass to the limit in (11.1) as 6 → 0, and thusestablish the validity of the first statement of the lemma.

Note that it follows from the estimate (11.1) that as h→ 0 the functions ukh converge weakly to the derivative∂u∂xk

(see [82]).To prove the convergence of ukh to ∂u

∂xkin the mean we shall first estimate uh for the functions of one variable:

uh(x)−du(x)dx

=1h

x∫x−h

du(ξ)dξ

dξ − du(x)dx

=1h

x∫x−h

[du(ξ)dξ

− du(x)dx

]dξ.

Therefore,+∞∫−∞

[uh(x) −

du(x)dx

]2

dx +∞∫−∞

1h

x∫x−h

[du(ξ)dξ

− du(x)dx

]2

dξ dx

=1h

+∞∫−∞

h∫0

[du(x)dx

− du(x− ξ)dx

]2

dξdx =1h

h∫0

dξ

+∞∫−∞

[du(x)dx

− du(x− ξ)dx

]2

dx.

Writing such an inequality for each line parallel to the xk axis and integrating with respect to the remainingvariables, we arrive at the following inequality:

∫∫Q

[ukh(x, t)−

∂u(x, t)∂xk

]2

dx dt 1h

h∫0

dξ

∫∫Q

[∂u(x, t)∂xk

− ∂u(x− ξ, t)∂xk

]2

dx dt. (11.3)

Since the estimate (11.3) is valid for any smooth functions with compact support in Q, it is also valid for functionsbelonging to W 1,1(Q) and vanishing in a neighborhood of S.

The right-hand side of (11.3) tends to zero as h → 0, because ∂u∂xk

belongs to L2(Q) and the theorem onthe continuity in L2 is valid ([82, p. 16]). Relation (11.2) is proved.

Similar assertions are true also for the differences with respect to t. It is clear that both assertions remaintrue if ukh is changed to uk−h ≡ 1

h [u(x1, . . . , xk−1, xk + h, xk+1, . . . , xn, t)− u(x1, . . . , xn, t)], that is,

‖uk−h‖0 ∥∥∥∥ ∂u∂xk

∥∥∥∥0

and ∥∥∥∥uk−h − ∂u

∂xk

∥∥∥∥0

→ 0 for h→ 0. (11.4)

528

Lemma 2. Let u(x, t) ∈ W 1,1(Q) vanish in a neighborhood of S \ S1, where S1 is a plane piece of S lyingin the plane xn = 0. Then relations (11.1) and (11.2) of the previous lemma hold for ukh if k = n.

Proof. For simplicity, let us suppose that the cylinder Q lies in the half-space xn > 0. Consider Qδ thatis a part of the cylinder Q such that xn δ. Denote by u&(x, t) averaged functions for u(x, t). If the averagingradius 6 is sufficiently small, the averaged functions are defined in Qδ. For the functions u&(x, t) in the domain Qδ

the inequality (11.1) holds if k = n. The functions u&(x, t) and their derivatives converge in the mean in Qδ tothe function u(x, t) and its derivatives. Passing to the limit as 6→ 0 in the equality (11.1), written for u& and Qδ

(for k = n), and then making δ tend to zero, we get the estimate (11.1) for the function u(x, t). The relation (11.2)can be proved in a similar way.

Lemma 3. Let the function u(x, t) belong to L2(Q) and either (1) vanish in a neighborhood of S or (2) vanishin a neighborhood of S \ S1, where S1 is a plane piece of the boundary S lying in the plane xn = 0. Considerthe differences ukh, where k = 1, . . . , n in the first case and k = n in the second one. Suppose that ‖ukh(x, t)‖0 M ,where the constant M does not depend on h.

Then there exist the corresponding derivatives ∂u(x,t)∂xk

∈ L2(Q) and their norms are bounded by a constantM .

Proof. Since the functions ukh are bounded in L2(Q), it is possible to select out of these sequences a weaklyconverging subsequence. Let the subsequence ukhm weakly converge to v(x, t) as m→∞. Then ‖v(x, t)‖0 M . It

remains only to prove that v(x, t) = ∂u(x,t)∂xk

. To do this let us consider the function Ψ(x, t) ∈0

C∞(Q). It is easy toverify that for sufficiently small hm

(ukhm,Ψ) = −(u,Ψk−hm). (11.5)

By Lemma 1, the Ψ−h converge in L2(Q) as h→ 0 to the function ∂Ψ∂xk

(see (11.4)). Passing to the limit as m→∞in (11.5), we get

(v,Ψ) = −(u,∂Ψ∂xk

).

Therefore, v = ∂u∂xk

, and the lemma is proved.

Theorem 1. Suppose u(x, t) is a weak generalized solution of the first boundary-value problem for Eq. (5.7)in the cylinder Q with conditions

u(x, 0) = 0, u|S = 0. (11.6)

Suppose the coefficients Aij(x, t) ∈ C1(Q), Bi(x, t), and C(x, t) are bounded, F (x, t) ∈ L2(Q), and S ∈ A2. Thenthe solution u(x, t) ∈W 2,1(Q) and satisfies Eq. (5.7) almost everywhere in Q.

Proof. By definition, the function u(x, t)

∫∫Q

[ n∑i,j=1

Aij∂Ψ∂xi

∂u

∂xj−( n∑

i=1

Bi∂u

∂xi+ Cu− ∂u

∂t

)Ψ]dx dt = −

∫∫Q

FΨ dx dt (11.7)

for any Ψ(x, t) ∈0

W 1,1(Q).We shall use the following notations:

(f, g) =∫∫Q

f(x, t)g(x, t) dx dt,

Af, g =∫∫Q

n∑i,j=1

Aij(x, t)∂f

∂xi

∂g

∂xjdx dt.

In what follows, the expression Ru will denote a linear differential operator of order not higher than one withbounded coefficients. The equality (11.7) can be rewritten in the following way:

Au,Ψ+ (Ru,Ψ)+(∂u

∂t,Ψ)= −(F,Ψ). (11.8)

529

Let us consider the differences

fh =1h[f(x1, . . . , xn, t)− f(x1, . . . , xk−1, xk − h, xk+1, . . . , xn, t)],

where xk is one of the space variables. Let ζ(x) be a smooth function in Ω, vanishing in a δ-neighborhood ofthe boundary σ of the domain Ω and equal to zero outside a 2δ-neighborhood of σ. Then for sufficiently small h

the function (Φζ)h ∈0

W 1,1(Q) if Φ ∈ W 1,1(Q). Therefore, one can insert Ψ = (Φζ)h into the equality (11.8) andget

Au, (Φζ)h+ (Ru, (Φζ)h) +(∂u

∂t, (Φζ)h

)= −(F, (Φζ)h). (11.9)

Let us denote by Mi constants not depending on h and the functions Φ(x, t). Since F (x, t) ∈ L2(Q) and

u(x, t) ∈0

W 1,1(Q), we have|(F, (Φζ)h)|+ |(Ru, (Φζ)h)| M1‖(Φζ)h‖0.

By Lemma 1, ‖(Φζ)h‖0 ‖Φζ‖1 M2‖Φ‖1. Therefore

|(F, (Φζ)h)|+ |(Ru, (Φζ)h)| M3‖Φ‖1. (11.10)

Taking into account the equality (f, gh) = −(f−h, g), which is valid for f and g from L2(Q) if one of thesefunctions vanishes in an h-neighborhood of S, we get(

∂u

∂t, (Φζ)h

)= −

((∂u

∂t

)−h

,Φζ)

= −(ζ∂u−h∂t

,Φ). (11.11)

Let us denote by Ni quantities not exceeding in absolute valueMi‖Φ‖1. Then by (11.9), (11.10), and (11.11)we have

−(ζ∂u−h∂t

,Φ)+Au, (Φζ)h = N1. (11.12)

Since (ζu)−h = ζu−h + u(xk + h)ζ−h, we have(ζ∂u−h∂t

,Φ)

=(∂(ζu)−h

∂t,Φ)−(ζ−h

∂u(xk + h)∂t

,Φ)

=(∂(ζu)−h

∂t,Φ)+N2. (11.13)

In a similar way, we get

Au, (Φζ)h =n∑

i,j=1

(Aij

∂u

∂xi,∂[ζΦh + ζhΦ(xk − h)]

∂xj

)

=n∑

i,j=1

(Aij

∂u

∂xi, ζ∂Φn

∂xj

)+

n∑i,j=1

(Aij

∂u

∂xi,Φh

∂ζ

∂xj+∂ζh∂xj

Φ(xk − h) + ζh∂Φ(xk − h)

∂xj

). (11.14)

Since ∥∥∥∥Φh∂ζ

∂xj+∂ζh∂xj

Φ(xk − h) + ζh∂Φ(xk − h)

∂xj

∥∥∥∥0

M4‖Φh‖0 +M5‖Φ‖0 +M6‖Φ‖1,

it follows from (11.14) that

Au, (Φζ)h =n∑

i,j=1

(Aijζ

∂u

∂xi,∂Φh

∂xj

)+N3.

Let us continue the transformation of the right-hand side of this equality:

Au, (Φζ)h =n∑

i,j=1

(Aijζ

∂u

∂xi,

(∂Φ∂xj

)h

)+N3

= −n∑

i,j=1

([Aij

∂(ζu)∂xi

−Aiju∂ζ

∂xi

]−h

,∂Φ∂xj

)+N3 = −

n∑i,j=1

(Aij

∂(ζu)−h∂xi

,∂Φ∂xj

)+N4.

530

Finally we arrive at the relationAu, (Φζ)h = −A(ζu)−h,Φ+N4. (11.15)

Combining equalities (11.12), (11.13), and (11.15), we get(∂(ζu)−h

∂t,Φ)+A(ζu)−h,Φ = N5.

Therefore ∣∣∣∣(∂(ζu)−h

∂t,Φ)+A(ζu)−h,Φ

∣∣∣∣ M7‖Φ‖1.

Denote Φ = (ζu)−h. Then we have∣∣∣∣(∂(ζu)−h

∂t, (ζu)−h

)+A(ζu)−h, (ζu)−h

∣∣∣∣ M7‖(ζu)−h‖. (11.16)

Since(f, ∂f∂t

)= 1

2

∫Ω

f2(x, T ) dx 0 for functions f(x, t) vanishing for t = 0 and belonging to W 1,1(Q) and

the quadratic form Ag, g µ‖g‖21 (µ > 0) for the functions g(x, t) ∈0

W 1,1(Q), it follows from (11.16) that

µ‖(ζu)−h‖21 M7‖(ζu)−h‖1.

Since ζ(x) ≡ 1 in the cylinder Q2δ and all the points of this cylinder are at a distance not less than 2δ from S, wehave ‖u−h‖1 M8 in Q2δ. This means that the difference quotients of the derivatives of u(x, t) with respect tothe space variables are bounded in the mean uniformly in h:∥∥∥∥

(∂u

∂xi

)−h

∥∥∥∥0

M8 in Q2δ (i = 1, . . . , n). (11.17)

Therefore, by Lemma 3 applied to ∂u∂xi

, there exist second-order generalized derivatives ∂∂xk

(∂u∂xi

), square integrable

in Q2δ. Since the numbers k and δ > 0 can be chosen arbitrarily, it follows that in Q \ S there exist second-ordergeneralized derivatives of the function u(x, t) with respect to the space variables.

Therefore, the relation (11.7), written for Ψ(x, t) ∈0

C∞(Q), can be integrated by parts. The result will be(L(u) − F,Ψ) = 0, and it follows that almost everywhere inside Q the function u(x, t) satisfies Eq. (5.7). We haveshown that ∂2u

∂xi ∂xk∈ L2(Q2δ) for any δ > 0 and for i, k = 1, . . . , n. To complete the proof it is sufficient to prove

that all these derivatives are integrable in the neighborhood of the boundary S.Let us consider a subdomain of the base Ω of the cylinder Q, adjacent to a part of the boundary σ, that can

be transformed to a piece of the plane xn = 0 by a change of the independent variables. Denote this subdomainby Ω. Then the domain Ω× (0, T ) is also transformed to a cylinder with a plane piece of boundary S, and this planepiece has the equation xn = 0. According to the suppositions made about the smoothness of S, the coefficients ofthe transformed equation will satisfy the same smoothness conditions as the coefficients of the original equation (5.7).

In the neighborhood of the plane piece S of the boundary S we shall estimate the difference quotients of ∂u∂xi

in the same way as was done for strictly interior domains Qδ. To do this we shall consider difference quotients onlyalong directions tangent to S, that is, for k = n, and the cutting-off smooth function ζ(x) we shall choose so that itvanishes outside Ω× (0, T ) and is equal to unity in a neighborhood of some piece of S lying strictly inside S. Thenin this neighborhood the estimates (11.17) hold, and by Lemma 2 all the second-order derivatives with respect tothe space variables, besides ∂2u

∂x2n, are square integrable in this domain.

By Eq. (5.7), which is satisfied almost everywhere, the derivative ∂2u∂x2

nalso is square integrable. Since

the neighborhood of the boundary can be divided into a finite number of parts similar to the one considered above,we have u(x, t) ∈W 2,1(Q). Theorem 1 is proved.

Theorem 2. Suppose that u(x, t) is a weak generalized solution of the first boundary-value problem (5.7),(11.6) in the cylinder Q. Suppose the coefficients Aij(x, t), Bi(x, t), and C(x, t) belong to the class C2k−2,k−1(Q),F (x, t) ∈ W 2k−2,k−1(Q) and the coefficients Aij(x, t) have in Q bounded derivatives up to order 2k−1 with respectto the space variables. Then u(x, t) ∈W 2k,k in the cylinder Q∗, contained together with its boundary in Q \ Γ.

531

Proof. Let us denote by Dq the operator of differentiation of order q with respect to the space variables.First we prove that D2ku and D2k−2 ∂u

∂t belong to L2 in any cylinder Q∗. The proof is conducted by induction.Suppose that Dqu ∈ L2 in any cylinder Q∗ and Dq−2 ∂u

∂t ∈ L2. (For q = 2 such as inclusion is proved in Theorem 1.)Let us prove that in any Q∗ also Dq+1u and Dq−1 ∂u

∂t are square integrable if q + 1 2k.Let us differentiate Eq. (5.7) (q−2) times with respect to the space variables (this is possible by the inductive

hypothesis). We have

n∑i,j=1

∂

∂xi

(Aij

∂

∂xjDq−2u

)− ∂

∂tDq−2u

=n∑

i,j=1

[∂

∂xi

(Aij

∂

∂xjDq−2u

)−Dq−2 ∂

∂xi

(Aij

∂u

∂xj

)]−Dq−2

( n∑i=1

Bi∂u

∂xi

)−Dq−2(Cu) +Dq−2F. (11.18)

Denote the right-hand side of (11.18) by F1. It contains the derivatives of u with respect to the space variables upto order q− 1, derivatives of Aij in x1, . . . , xn up to order q − 1, and derivatives of Bi, C, and F up to order q − 2.Therefore, by the inductive hypothesis and the suppositions on the smoothness of the coefficients of Eq. (5.7), wehave DF1 ∈ L2 in any cylinder Q∗. Let Φ be an arbitrary sufficiently smooth function with compact support in Q.Then, multiplying (11.18) by Φ−h, integrating over Q, and then integrating by parts, we obtain

ADq−2u,Φ−h+(∂

∂tDq−2u,Φ−h

)= −(F1,Φ−h).

(The notations are the same as in the previous theorem.)Since (F1,Φ−h) = −(F1h,Φ), we have∣∣∣∣ADq−2u,Φ−h+

(∂

∂tDq−2u,Φ−h

)∣∣∣∣ M1‖Φ‖0; (11.19)

by Mi here and below are denoted constants not depending on h and on the function Φ. Obviously (11.19) is validalso for any function Φ with compact support in Q belonging to W 1,1(Q).

Let ζ(x, t) be a smooth function vanishing in a neighborhood of Γ and equal to unity in some cylinderQ∗ ⊂ Q \Γ. Insert into the inequality (11.19) Φ = (ζ2Dq−2uhh1)−h1 . Let us denote by Nj quantities not exceeding

in absolute value M(1 +∑q‖ζDquh‖0

), where

∑q‖ζDquh‖0 is the sum taken over all the derivatives with respect

to the space variables of order q and the constantM depends only on the coefficients of the equation and the normsζDmu and ζDm−2 ∂u

∂t in L2(Q) for m q.Let us transform the left-hand side of the inequality (11.19):(

∂

∂tDq−2u,Φ−h

)= −

(∂

∂tDq−2uh,Φ

)=(∂

∂tDq−2uhh1 , ζ

2Dq−2uhh1

)

=(∂

∂t(ζDq−2uhh1), ζD

q−2uhh1

)−(∂ζ

∂tDq−2uhh1 , ζD

q−2uhh1

)=

T∫0

∂

∂t

∫Ω

[ζDq−2uhh1 ]2 dx

dt+N1 N1,

(11.20)

since ‖Dq−2uhh1‖0 ‖Dqu‖0 by Lemma 1.Transform in a similar way the first term of the inequality (11.19):

ADq−2u,Φ−h =n∑

i,j=1

([Aij

∂

∂xi(Dq−2u)

]hh1

,∂

∂xj[ζ2Dq−2uhh1 ]

)

=n∑

i,j=1

(Aij

∂

∂xiDq−2uhh1 ,

∂

∂xj[ζ2Dq−2uhh1 ]

)+N2

=n∑

i,j=1

(Aij

∂

∂xi[ζDq−2uhh1 ],

∂

∂xj[ζDq−2uhh1 ]

)+N3 = AζDq−2uhh1 , ζD

q−2uhh1+N3. (11.21)

532

From (11.19), (11.20), and (11.21), we can deduce that

AζDq−2uhh1 , ζDq−2uhh1 M1‖(ζ2Dq−2uhh1)−h1‖0 +N4.

Passing to the limit in this inequality as h1 → 0 and using Lemma 1, we get

AζDq−1uh, ζDq−1uh M1‖D(ζ2Dq−1uh)‖0 +N4 = N5.

Since this inequality is valid for any derivative Dq−1uh and Af, f µn∑i=1

(∂f∂xi

, ∂f∂xi

), we have

∑q

(ζDquh, ζDquh) M2

(1 +∑q

‖ζDquh‖0). (11.22)

As above, the sum is taken over all the derivatives with respect to the space variables of order q.It follows from the inequality (11.22) that ‖Dquh‖0 M3 in the domain where ζ ≡ 1. Therefore, by

Lemma 3 in this domain there exist derivatives Dq+1u. Differentiating Eq. (5.7) (q − 1) times with respect tothe space variables and expressing ∂

∂t (Dq−1u) from the resulting equality, we can see that the derivative ∂

∂t (Dq−1u)

also is square integrable in Q.The induction statement is proved. This means that D2ku ∈ L2(Q∗) and ∂

∂t (D2k−2u) ∈ L2(Q∗) for any

interior subdomain Q∗. Differentiating Eq. (5.7), it is easy to see that other derivatives of the solution u(x, t) alsoexist and belong to L2(Q∗). It follows that u(x, t) ∈W 2k,k(Q∗). The theorem is proved.

In some papers, another method is used to study the smoothness of generalized solutions; it is based onaveraged functions. The main idea of this method is as follows: instead of the function Φ(x, t) the derivative of itsaverage is inserted into the integral identity (5.9) (and not a difference quotient, as was done above). In this way,one can get an estimate for the derivatives of the averages of the generalized solution. This estimate is uniform withrespect to the radius of averaging. In the clearest form this method was presented by K. Friedrichs [108], wherehe proved the smoothness of generalized solutions of strongly elliptic systems at the interior points of the domain.The use of this method to prove the smoothness up to the boundary meets with considerable difficulties.

12 BEHAVIOR OF SOLUTIONS WITH INFINITE INCREASING OF TIME

In this section, we prove some simple theorems on the behavior of solutions of the Cauchy problem andboundary-value problems with infinite increasing of time. We suppose that the coefficients and the right-hand sideof Eq. (1.1) are bounded in the domains considered and c(x, t) 0. The solutions are also assumed to be bounded.

Theorem 1. Suppose the function u(x, t) is a solution of the Cauchy problem for Eq. (1.1) in the half-spacet > 0 or of a boundary-value problem for Eq. (1.1) in the cylinder Q = Ω × (0,+∞) with one of the boundaryconditions

u|S = 0 (12.1)

or

l(u) ≡(∂u

∂γ+ au

)S

= 0, (12.2)

where a 0 and γ is a direction in the space (x1, . . . , xn), making an acute angle with the inner normal tothe boundary of the domain Ω. Suppose that c(x, t) < −c0 < 0, where c0 is a constant, and f(x, t)→ 0 uniformlyin x as t→ +∞. Then u(x, t)→ 0 uniformly in x as t→ +∞.

Proof. To prove this theorem we first note that the solution u(x, t) is bounded for all x and t 0, that is,|u(x, t)| M . This follows from Theorem 3 of Sect. 1 for the problem (1.1), (12.1) and from similar estimates forthe remaining problems (see Theorems 7 and 8 of Sect. 1). Let us fix an arbitrary ε > 0 and choose T such that|f(x, t)| < εc0 for t T .

Consider the functions w±(x, t) = Me−c0(t−T ) + ε ± u(x, t). These functions are nonnegative for t = T . Ifcondition (12.1) is fulfilled on S, then w±(x, t)|S > 0 for t T . If we have (12.2) on S, then l(w±) 0 for t T .For t T we have L(w±) =ML(e−c0(t−T ))+εc±f Mec0(T−t)(c0+c)−εc0+ |f | < 0. It follows from Theorems 1,7 or 8 of Sect. 1 that w± 0 for t T , that is,

|u(x, t)| ε+Mec0(T−t) < 2ε

for t > T1(ε). The theorem is proved.

533

Theorem 2. Suppose the function u(x, t) satisfies Eq. (1.1) in the cylinder Q, u|S = 0, and f(x, t) → 0uniformly in x as t→∞. Then u(x, t)→ 0 uniformly in x ∈ Ω as t→∞.

Proof. We shall make the following change of the unknown function in (1.1): u(x, t) = v(x, t)ϕ(x), wherethe function ϕ(x) > 0 will be chosen later.

For the new unknown function v(x, t) we obtain an equation where the coefficient in v(x, t) is L(ϕ). Therefore,if we choose the function ϕ(x) so as to have L(ϕ) < −c0 < 0 for all values of x considered, we shall be able to applyTheorem 1 that we have proved above to the function v(x, t). According to Theorem 1, v(x, t)→ 0 as t→∞. Thismeans that also u(x, t)→ 0 uniformly with respect to x ∈ Ω as t→∞.

For example, we can take ϕ(x) = 1 − eαx1 . Without loss of generality, we can suppose that x1 < 0 forthe points of the domain Ω. Then we have in Ω

L(ϕ) = −a11α2eαx1 − b1αeαx1 + c(1− eαx1) < 0

for sufficiently large α > 0.

Remark 1. Theorem 1 is valid also for a noncylindrical domainD, bounded by the plane t = 0 and a surface S,and Theorem 2 is valid for a noncylindrical domain D under the additional supposition that the projection of Sto the plane t = 0 is situated in a bounded part of this plane. The condition u|S = 0 can be changed to u|S → 0uniformly in x as t→∞.

Remark 2. It follows easily from the proof of Theorems 1 and 2 that |u(x, t)| < Me−γt, γ > 0, if f(x, t) ≡ 0and u|S = 0.

Theorem 3. Suppose that the function u(x, t) satisfies Eq. (1.1) in the cylinder Q,(∂u∂γ + a(x, t)u

)S= 0,

where a(x, t) < −a0 < 0, and f(x, t)→ 0 uniformly in x as t→∞. Then u(x, t)→ 0 uniformly in x ∈ Ω as t→∞.

Proof. As above, we shall make a change of the unknown function in Eq. (1.1), u = vψ(x), where ψ(x) > 0.

The boundary condition will take the following form:(∂v∂γ +

aψ+ ∂ψ∂γ

ψ v)S= 0. Therefore, if we intend to reduce

the proof of the theorem to the use of Theorem 1, it is sufficient to construct a smooth function ψ(x) > 0 such thatL(ψ) < 0 and

(∂ψ∂γ + aψ

)S 0.

Set ψ(x) = ϕ(x)+A, where ϕ(x) is a function constructed in the previous theorem and A > 0. The operatorL(ψ) = L(ϕ) +Ac < 0. The constant A > 0 will be chosen so large that

∂ψ

∂γ+ aψ =

∂ϕ

∂γ+ aϕ+Aa < 0.

Remark 1. The remarks to Theorem 2 are also valid for Theorem 3. In Theorems 2 and 3, one can changethe supposition that u(x, t) satisfies Eq. (1.1) and conditions (12.1) or (12.2) can be changed to the suppositionthat the following inequalities are valid: L(u) 0 in Q and u|S 0 or

(∂u∂γ + au

)S 0. Then u(x, t) > −ε if t is

sufficiently large, that is, limt→∞

infxu(x, t) 0.

Remark 2. In Theorem 3, the condition a < −a0 < 0 cannot be omitted, because under the condition that∂u∂γ

∣∣∣S= 0 the solution does not necessarily tend to zero (e.g., u ≡ 1 for the equation ∂2u

∂x2 − ∂u∂t = 0). However,

if under the condition a 0 there exists a smooth function ψ(x) such that(∂ψ∂γ + aψ

)S 0 and L(ψ) < 0,

the statement of Theorem 3 remains valid.Such a function is easy to construct, for example, in the case of one dimension (n = 1) for the interval [0, X ]

under the boundary conditions(∂u∂x − a1(t)u

)x=0

= 0,(∂u∂x + a2(t)u

)x=X

= 0, if a1(t) 0 and a2(t) a0 > 0. Inthis case, we can take ψ(x) = A− α coshx for sufficiently large A > 0 and α > 0.

Theorem 4. Suppose that u(x, t) is a solution of the Cauchy problem for the equation L(u) = 0 witha bounded initial function u(x, 0). Suppose there exists a positive function v(x) such that L(v) 0 and v(x)→∞as |x| → ∞, where |x| =

∑x2i

1/2. Then the solution u(x, t) → 0 as t → ∞ uniformly with respect to x in any

bounded domain of the space En.

534

Proof. Let us take any ε > 0 and consider an n-dimensional ball of radius R with center at the origin. Toprove the theorem it is sufficient to show that |u(x, t)| ε if |x| R and t T1, where T1 is a sufficiently largenumber.

According to Theorem 10 of Sect. 1 (for M2 =M3 = 0) the solution u(x, t) is bounded: |u(x, t)| M1.ChooseR1 such that v(x) > 2M1

ε VR for |x| = R1 > R, where VR = max|x|R

v. Then w±(x, t) =εv(x)2VR±u(x, t) 0

for |x| = R1. The operator L(w±) 0 in the cylinder |x| R1, 0 < t < ∞. By Remark 1 to the previoustheorem, w±(x, t) − ε

2 for t T1 and for all x such that |x| R1. Therefore, |u(x, t)| ε2 + εv(x)

2VR ε for |x| R

and t T1, and this is what was to be proved.

Remark. In Theorem 1 of Sect. 1, the supposition L(u) 0 can be weakened. Namely, it is sufficient tosuppose that the function u(x, t) satisfies the inequality L(u) 0 everywhere in the domain D except for a finitenumber of smooth surfaces Sk (k = 1, . . . , N). The function u(x, t) has to be continuous in the closed domain Dand have derivatives along the normal on both sides of the surfaces Sk. On the surfaces Sk, the inequalities

∂u

∂ν+<

∂u

∂ν−(12.3)

must hold; here ∂u∂ν+

means a derivative in the normal direction taken from the side of the surface Sk wherethe normal is directed and ∂u

∂ν−is the derivative in the normal direction taken from the other side of the surface.

The proof of Theorem 1 of Sect. 1 under these suppositions remains the same; we only need to note that by virtueof (12.3) the function u(x, t) cannot attain its minimal value on the surfaces Sk.

In a similar way, it is possible to weaken the suppositions of Theorems 8 and 9 of Sect. 1. In Theorem 4of this section it is also possible to admit as v(x) a continuous function that satisfies condition (12.3) on a finitenumber of smooth surfaces.

Theorem 5. Suppose that u(x, t) is a bounded solution of the Cauchy problem for the equation L(u) = 0.Suppose that c(x, t) −c0 < 0 for all t 0 and x belonging to some domain in the space x1, . . . , xn. Suppose alsothat outside some ball (that is, for |x| R) there exists a positive function ψ(r) such that ψ(r) → ∞ as r → ∞and L(ψ) 0 (r = |x|). Then u(x, t)→ 0 as t→∞ uniformly in x in any bounded domain of the space x1, . . . , xn.

Proof. According to Theorem 4 and the remark to this theorem, it is sufficient to construct a function v(r)such that v(r) → ∞ as r → ∞ and L(v) 0 everywhere except for a finite number of surfaces where the firstderivatives of v have discontinuities and the condition (12.3) is fulfilled: ∂v

∂ν+< ∂v

∂ν−.

Without loss of generality, we can assume that c(x, t) −c0 < 0 for |x| 6, where 6 < R. Set v(r) = eαr2

for r 6 (α > 0). We have

L(v) =(4α2

n∑i,j=1

aijxixj + 2αn∑i=1

aii + 2αn∑i=1

bixi + c

)v < 0

for r < 6 and a sufficiently small α > 0.In the ring 6 r R we set v(r) = A(1 − e−βr2

) + B, where A, B, and β are positive constants. Let uscalculate L(v) for 6 < r < R:

L(v) = Ae−βr2(−4β2

n∑i,j=1

aijxixj + 2βn∑i=1

aii + 2βn∑i=1

bixi

)+ cv < 0

for sufficiently large β and any A and B. We can choose the constant A > 0 so small that the condition (12.3)is fulfilled for r = 6: 2Aβ6e−β&

2< 2α6eα&

2. Then we choose the constant B > 0 so that the function v(r) is

continuous for r = 6.Finally, for r R set v(r) = A1ψ(r) + B1, where A1 > 0 and B1 > 0 are again chosen so that the condi-

tion (12.3) is fulfilled for r = R and the function v(r) is continuous.The theorem is proved.

535

It is easy to see that the divergence of the integral∞∫R

re−

r∫r0

b(ζ)ζ dζ

dr, where

b(ζ) = sup|x|=ζ

n∑i=1

(aii + bixi)

n∑i,j=1

aijxixjζ2

is a sufficient condition for the existence of the function ψ(r). In the case where the integral diverges we can takeas ψ(r) the function

r∫R

6e−

∫r0

b(ζ)ζ dζ

d6.

Theorem 6. Suppose that u(x, t) is a solution of the Cauchy problem for the equation L(u) = 0 andthe initial function u(x, 0) tends to zero as |x| → ∞. If

n∑i=1

(aii + bixi) > α > 0 for all x and t 0, (12.4)

then u(x, t)→ 0 as t→∞ uniformly in x.

Proof. It is sufficient to prove this for the initial function u(x, 0) vanishing outside some ball. The function

v(x, t) = (t+1)−γeβr2t+1 (r = |x|) tends to zero as t→∞, and one can choose β > 0 and γ > 0 so that the inequality

L(v) < 0 is fulfilled. In fact,

L(v) = v

1

(t+ 1)2

[4β2

n∑i,j=1

aijxixj − βr2

]+

1t+ 1

[γ − 2β

n∑i=1

(aii + bixi)]+ c

. (12.5)

First we take β so small that the quantity in the first square brackets of the right-hand side of (12.5) becomes non-positive; then, using (12.4), we choose γ so small that the quantity in the second square brackets becomes negative.If M is sufficiently large, the functions Mv ± u are positive for t = 0 and satisfy the inequality L(Mv ± u) < 0for t > 0. According to Theorem 8 of Sect. 1, the functions Mv ± u are nonnegative for all t 0. Therefore,|u(x, t)| Mv(x, t) and u(x, t)→ 0 uniformly in x as t→∞.

Remark 1. Theorem 6 is valid also when the condition (12.4) is fulfilled only outside some cylinder, that is,for |x| R and t 0. One can find the proof in [109].

Remark 2. The theorems we have proved above concerned the cases where u(x, t) tends to zero with the in-creasing of t under some suppositions on the coefficients of Eq. (1.1). It is clear that these theorems can beextended to the case where the coefficients of Eq. (1.1) and the right-hand side f(x, t) tend uniformly in x tofunctions depending only on x as t→∞. Suppose that the limit elliptic equation

n∑i,j=1

aij(x)∂2v

∂xi ∂xj+

n∑i=1

bi(x)∂v

∂xi+ c(x)v = f(x) (12.6)

has a solution v(x), defined for all x from the domain considered. Suppose this solution is bounded together with itsderivatives occurring in the equation. To investigate the problem on when the solution u(x, t) of Eq. (1.1) tends tothe corresponding solution v(x) of Eq. (12.6) as t→∞ we have to apply the theorem proved above to the differenceu− v. We shall not give here the statements of theorems thus obtained.

If the coefficients of Eq. (1.1) do not depend on t, the behavior of solutions as t → ∞ is studied in detailin [110]. Some extensions of these results to equations with coefficients depending on x and t are given in [111].

In [112], the rate of decrease is studied for solutions of a second-order parabolic equation in the domain0 < x < X(t); this rate depends on the behavior of the function X(t) as t → ∞. It is supposed that the solutionu(x, t) is equal to zero for x = 0 and for x = X(t).

536

The question on the stabilization of the solutions of boundary-value problems for parabolic systems ofthe form ∂u

∂t + Au = f , where A is a strongly elliptic operator, is studied in [99]. The asymptotic behavior ofsolutions of the Cauchy problem for parabolic systems is studied in [113,114].

Conditions sufficient to assure the fact that solutions of the Cauchy problem for parabolic equations andsystems tend to the mean value of the initial function u(x, 0) as t→∞ are described in [114–116].

The asymptotic behavior of the solutions of boundary-value problems for parabolic equations and systemswhen the equations considered degenerate as t→∞ is studied in [117,118].

Questions on the asymptotic behavior of the solutions of the Cauchy problem and boundary-value problemsfor parabolic second-order equations were studied also in [119–123] and in other papers.

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