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T< ∞ , S isthesetofpositiveintegers, f :Ω × CB (Ω) 3 ( t,x,s ) → f ( t,x,s ) ∈ where id istheidentityoperator,( t,x ) ∈ Ω:= { ( t,x ): t ∈ (0 ,T ) ,x ∈ R } , ∂x ∂x F := −A , A := a ( t,x ) + b ( t,x ) + c ( t,x ) id, ∂∂t ∂ ∂∂x X X Here u (0 ,x )= ϕ ( x )f

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Podíl "T< ∞ , S isthesetofpositiveintegers, f :Ω × CB (Ω) 3 ( t,x,s ) → f ( t,x,s ) ∈ where id istheidentityoperator,( t,x ) ∈ Ω:= { ( t,x ): t ∈ (0 ,T ) ,x ∈ R } , ∂x ∂x F := −A , A := a ( t,x ) + b ( t,x ) + c ( t,x ) id, ∂∂t ∂ ∂∂x X X Here u (0 ,x )= ϕ ( x )f"

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EXISTENCE OF SOLUTIONS OF THE CAUCHY PROBLEM FOR SEMILINEAR INFINITE SYSTEMS OF PARABOLIC

DIFFERENTIAL–FUNCTIONAL EQUATIONS

by Anna Pude lko

Abstract. We consider the Cauchy problem for a countable system of weakly coupled semilinear differential–functional equations of parabolic type. The right-hand sides of the system are functionals of unknowns.

The object of this paper is to transform the system of parabolic differential equations into the associated system of integral equation in order to prove the existence of the solution of the latter problem with use of the Schauder fixed point theorem.

1. Introduction. We consider a countable system of weakly coupled semi- linear differential–functional equations of the form

(1) Fi[ui](t, x) =fi(t, x, u), i∈S, with the initial condition

(2) u(0, x) =ϕ(x) for x∈Rm.

Here Fi:= ∂

∂t − Ai,Ai:=

m

X

j,k=1

aijk(t, x) ∂2

∂xj∂xk +

m

X

j=1

bij(t, x) ∂

∂xj

+ci(t, x)id, where id is the identity operator, (t, x)∈Ω :={(t, x) :t∈(0, T), x∈Rm}, T <∞,Sis the set of positive integers,fi: Ω×CBS (Ω)3(t, x, s)→fi(t, x, s)∈

2000Mathematics Subject Classification. 35R10, 35K55.

Key words and phrases. Infinite systems, parabolic differential–functional equations, Schauder fixed point theorem, Cauchy problem.

Part of this work is supported by local Grant No.11.420.04.

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R, i∈S, u stands for the mapping

u:S×Ω3(i, t, x)→ui(t, x)∈R, composed of unknowns ui and ϕ= (ϕ1, ϕ2, . . .).

A functionuis said to be aC–solutionof differential problem (1), (2) in Ω if u ∈CBS(Ω) and satisfies the system of integral equations associated with the differential problem in Ω.

This paper can be considered as a continuation of the author’s study of the certain infinite systems of parabolic differential–functional equations. Now the author considers a more general form of operator with lower order of terms with respect to the derivatives in x–variables. There have appeared papers devoted to an analysis of the initial-boundary value problem, e.g. [2], [3]. The infinite systems of parabolic type and properties of their solutions are treated, e.g., in [7] and [14].

The goal of this paper is to prove the existence of a solution of Cauchy problem (1), (2) for a countable systems using the Schauder fixed point theorem under weaker assumptions than in [11]. In paper [11], to solve the above problem the Banach fixed point theorem was used. This approach allowed us to prove the existence and uniqueness of the solution under rather restrictive assumptions which were previously imposed on the initial data (namely equi- boundeness of all components) and the right-hand sides (among other things, equi-boundeness of all components and the Lipschitz condition with respect to functional argument).

Now, resigning from that and applying the Schauder fixed point theorem (see [4], [10] or [15]), we will get the existence of the considered problem in a layer [0, τ)×Rm,where τ is a sufficiently small number. Unfortunately, this theorem does not help us to obtain the uniqueness of the problem (1), (2).

In order to apply the Schauder fixed point theorem, we have to check, among other things, that certain operator is compact. We obtained it by using the well-known Fr´echet compactness theorem. It is in some way an extension of the idea we have first come across in [10].

Let us stress that the result is obtained without using the weighted-spaces.

Instead, we assume that the truncated operator FN of the operator F (ge- nerated by right-hand sides) uniformly smothers the functions at the infinity.

An infinite system of equations was first considered by M. Smoluchowski as a model for coagulation of colloids moving according to a Brownian mo- tion. A system of infinite number of reaction–diffusion equations related to the system of ODE derived by Smoluchowski is investigated in [1]. A nonlocal discrete model of cluster coagulation and fragmentation expressed in terms of an infinite system of integro–differential bilinear equations was studied in [8].

The coagulation–fragmentation local model which add spatial diffusion to the

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classical coagulation equations is considered in [12]. For a treatment of infinite systems of parabolic differential–functional equations with the initial-boundary value problem, where Schauder’s method of fixed point is also applied, we refer the reader to [2].

This paper is organized as follows. In the next section the necessary no- tations and definitions are introduced. In section 3 we impose the necessary assumptions and formulate the main result of this paper. In section 4 we state and prove the auxiliary lemmas. The last section contains a proof of the main theorem.

2. Notations and definitions. Throughout the paper, we use the fol- lowing notation. By Ωτ we denote the set (0, τ)×Rm for eachτ ∈(0, T].The set ΩT is denoted for short by Ω. The norm in RN we note by | · |N. B(x, δ) denotes an open ball with center at x and radiusδ >0.

Here, by CB(Ω) we unusually denote the space of functionsh ∈C(Ω), such that h vanishes uniformly at infinity, i.e.

∀ >0 ∃R >0∀t∈[0, T]∀x∈Rm\B(0, R) : |h(t, x)|< . The author hopes that it causes no misunderstanding.

The space CBS(Ω) comprises all functions h = (h1, h2, . . .) such that hi ∈CB(Ω), i∈S with the finite norm

khkΣ =

X

i=1

1

Qikhik, where khik= sup

(t,x)∈Ω

|hi(t, x)|fori∈S,

and Q is an arbitrary real number. Obviously, the space CBS(Ω) endowed with the norm k · kΣ is a Banach space.

For fixedN ∈ N letCBN(Ω) be the space of all functionsh= (h1, . . . , hN) such that hi ∈ CB(Ω) for i = 1, . . . , N. We endow this space with the following norm: khkN = max

i=1,...,N sup

(t,x)∈Ω

|hi(t, x)|.

We understand the spacesCB(Rm), CBS(Rm), CBN(Rm) analogously.

Now we introduce the following (?)–condition.

The set K⊂CBN(Ω) is said to satisfy(?)–conditionif and only if

∀ >0 ∃R,N >0 ∀h∈K ∀t∈[0, T]∀x∈Rm\B(0, R,N)

|hi(t, x)|< fori= 1, . . . , N.

Let η∈CBS(Ω). We define the following operatorF= (F1,F2, ...) F:η→F[η],

setting

Fi[η](t, x) :=fi(t, x, η), i∈S.

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3. Assumptions and the main result. Now we formulate the assump- tions necessary for obtaining the result which is given at the end of this section.

We will assume that:

(H) : the coefficients aijk(t, x), bij(t, x), ci(t, x), i∈ S, j, k = 1, . . . , m are bounded continuous functions such that aijk(t, x) =aikj(t, x) and satisfy the followingH¨older continuous condition with exponent α (0< α61) in Ω :

∃H >0 ∀i∈S ∀ j, k= 1, . . . , m∀ t∈[0, T]∀ x, x0 ∈Rm :

|aijk(t, x)−aijk(t, x0)|6H|x−x0|α

|bij(t, x)−bij(t, x0)|6H|x−x0|α

|ci(t, x)−ci(t, x0)|6H|x−x0|α.

We suppose as well that the operators Fi, i∈ S,are uniformly parabolic in Ω (the operators Ai are uniformly elliptic in Ω), i.e.

∃ µ1, µ2>0∀ ξ= (ξ1, . . . , ξm)∈Rm ∀ (t, x)∈Ω∀ i∈S: µ1

m

X

j=1

ξj2

m

X

j,k=1

aijk(t, x)ξjξk2

m

X

j=1

ξj2.

The crucial assumptions related to the initial data ϕ= (ϕ1, ϕ2, . . .) are:

1) : ϕi ∈CB(Rm) for each i∈S (i.e. ϕ∈CBS(Rm))

2) : there exist Q0 ∈(0, Q) and ¯M > 0 such that kϕikRm 6 M(Q¯ 0)i−1 for each i∈S.

Our main requirements concerning the right-hand sides are as follows: Let the function f = (f1, f2, . . .) generating the operator Fbe such that for each τ ∈(0, T]

(F1) : F: CBS(Ωτ)→CBS(Ωτ) is continuous;

(F2) : if K is a closed, bounded set inCBS (Ωτ) then there exist Q0 ∈(0, Q) and ¯M >0 such that sup

w∈K

kFi[w]kτ 6M(Q¯ 0)i−1 for each i∈S;

(F3) : for eachN ∈ N,the mapping FN := (F1, . . . ,FN), FN :CBS(Ωτ) → CBN(Ωτ) transforms a bounded set in CBS(Ωτ) into a set which sat- isfies(?)–condition;

(V) : functions fi satisfy the Volterra condition, i.e. for arbitrary (t, x) ∈ Ω and arbitrary η,η˜ ∈ CBS(Ω) such that ηj(¯t, x) = η˜j(¯t, x), for 06¯t6t, j ∈S there isfi(t, x, η) =fi(t, x,η), i˜ ∈S.

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The examples of the F operator which satisfy the imposed conditions in- clude:

Fi[u](t, x) = ( Pi

n=1bn(t, x)un(t, x) for (t, x)∈[0, T]×B(0,1)

Pi

n=1bn(t,x)un(t,x)

|x|1/i for (t, x)∈[0, T]×(Rm\B(0,1)), where bn∈C(Ω) satisfykbnk < M

Q0 Q

i−1

for someM >0, Q0 ∈(1, Q) or Fi[u](t, x) =

Z t

0

bi(τ, x)ui(τ, x)ui−1(τ, x) dτ,

where bi∈C(Ω) satisfykbik < M

Q0 Q2

i−1

for someM >0, Q0∈(1, Q).

Now, let us state the main result of the paper.

Theorem. Let all the above assumptions hold and τ ∈ (0, T] be a suf- ficiently small number. Then the Cauchy problem (1), (2) has at least one C–solution u∈CBS(Ωτ),where 0< τ 6τ 6T.

4. Auxiliary lemmas. To get the proof of the foregoing theorem, we first formulate the following lemmas.

Lemma 1. ([5], Th. 2.1, p. 71 [6], Th. 10, p. 23)

If the operators Fi (i ∈ S) are uniformly parabolic in Ω with the constants µ1, µ2 and the coefficients aijk(t, x), bij(t, x), ci(t, x), i∈S, j, k= 1, . . . , m sa- tisfy the condition (H) in Ω then there exist the fundamental solutions Γi(t, x;τ, ξ) of the equations

Fi[ui](t, x) = 0, i∈S and the following inequalities hold

i(t, x;τ, ξ)|6C(t−τ)m2 exp

− µ|x−ξ|2 4(t−τ)

, i∈S

for any µ < µ, where µ depends on µ1, µ2, H and C depends on µ1, µ2, α, T and the nature of continuity aijk(t, x) in t.

Owing to the fact that Fi (i ∈ S) are uniformly parabolic in Ω and the coefficients aijk(t, x) (i ∈ S, j, k = 1, . . . , m), satisfy condition (Ha) in Ω, we notice that R

Rmi(t, x;τ, ξ)|dξ, i∈S are equi-bounded. ByC we denote the infimum of their upper bounds.

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Using the fundamental solutions Γi(t, x;τ, ξ), i ∈ S, for the equations Fi[ui](t, x) = 0, we consider the following system of integral equations as- sociated with differential problem (1), (2)

(3) ui(t, x) = Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ+

t

Z

0

Z

Rm

Γi(t, x;τ, ξ)Fi[u](τ, ξ)dξdτ i∈S fort >0, x∈Rm. Now, we define T:CBS(Ω)−→CBS(Ω) in the following way

T[z] := (T1[z1],T2[z2], . . .), where

Ti[zi] = Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ+ Zt

0

Z

Rm

Γi(t, x;τ, ξ)zi(τ, ξ)dξdτ.

By TN we denote the operatorTN = (T1, . . . ,TN).

We will now prove the following lemmas.

Lemma 2. Let N ∈ N be fixed, ϕi ∈ CB(Rm), i = 1, . . . , N. If G ⊂ CBN(Ω)is a bounded set which satisfies (?)–condition, then

TN(G)⊂CBN(Ω) is a family of equicontinuous functions.

Proof. To begin with, we notice that ϕi, i = 1, . . . , N, are uniformly continuous. We denote M0 := sup

z∈G

kzkN.

Fix η > 0. Let (t, x), (t0, x0) ∈ Ω. First of all, we consider the easiest case t=t0 = 0.Since ϕi, i= 1, . . . N are uniformly continuous, there exists δ1 >0 such that

(2.1)

Ti[zi](0, x)−Ti[zi](0, x0)

=|ϕi(x)−ϕi(x0)|< η if only |x−x0|m < δ1.Now lett > t0 = 0.It is obvious that

Ti[zi](t, x)−Ti[zi](0, x0)

=

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ+

t

Z

0

Z

Rm

Γi(t, x;τ, ξ)zi(τ, ξ)dξdτ −ϕi(x0)

6

t

Z

0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ+

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x0) .

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We estimate the first term in the following way (2.2)

t

Z

0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ 6M0

t

Z

0

Z

Rm

i(t, x;τ, ξ)|dξdτ =CM0t < η 2 if only t < δ2 := 2CMη 0.In order to estimate the second one, we select R1η >0 such that for x∈Rm\B(0, R1η)

(2.3) |ϕi(x)|< η

6C fori= 1, . . . , N.

Now, we choose R1η > R1η such that for all x ∈ Rm\B(0, R1η), ξ ∈B(0, R1η), t∈[0, T]

(2.4) |Γi(t, x; 0, ξ)|< η

6kϕikRmm(B(0, R1η)), where m stands for the Lebesgue measure.

Let us remark that it is enough to considerx, x0 such that|x−x0|m< δ2.Thus, if x∈B(0, R1η2),thenx, x0 ∈B(0, R1η+ 2δ2).According to the definition of the fundamental solution,

t&0lim Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ=ϕi(x)

follows for each x ∈ Rm. Since B(0, R1η + 2δ2) is a compact set and ϕi, i = 1, . . . , N are continuous functions, R

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ converges to ϕi(x) uniformly on B(0, R1η + 2δ2), that is, there exists δ3 > 0 independent of x such that fort∈(0, δ3), x∈B(0, R1η + 2δ2)

(2.5)

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x)

< η 4

follows. Next, we choose 0< δ4 < δ3 such that for all|x−x0|m < δ4

(2.6)

ϕi(x)−ϕi(x0)

< η

4 fori= 1, . . . , N.

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Then, (2.5) and (2.6) yield the estimate

(2.7)

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ −ϕi(x0)

6

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x)

+|ϕi(x)−ϕi(x0)|< η 2 for t∈(0, δ4), x, x0∈B(0, R1η+ 2δ2), |x−x0|< δ4.

If x /∈B(0, R1η2) thenx, x0 ∈/ B(0, R1η).Certainly,

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x0) 6

Z

Rm

i(t, x; 0, ξ)||ϕi(ξ)|dξ+|ϕi(x0)|.

We see from (2.3) that the second term is less than η6, (as it is less than 6Cη ) and we present the first one as follows

Z

Rm

i(t, x; 0, ξ)||ϕi(ξ)|dξ

= Z

Rm\B(0,R1η)

i(t, x; 0, ξ)||ϕi(ξ)|dξ+ Z

B(0,R1η)

i(t, x; 0, ξ)||ϕi(ξ)|dξ.

To estimate the first integral, we employ (2.3):

(2.8)

Z

Rm\B(0,R1η)

i(t, x; 0, ξ)||ϕi(ξ)|dξ < η 6C

Z

Rm

i(t, x; 0, ξ)|dξ < η 6, and the second one is a directly estimated by (2.4) as follows

(2.9) Z

B(0,R1η)

i(t, x; 0, ξ)||ϕi(ξ)|dξ <

Z

B(0,R1η)

η

6kϕikRmm(B(0, Rη1))kϕikRmdξ= η 6. Hence (2.2), (2.7), (2.8) and (2.9) lead to the assertion that

if max{t, |x−x0|m}< δ5 := min{δ1, δ2, δ4}then (2.10)

Ti[zi](t, x)−Ti[zi](0, x0)

< η.

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In the next step of our proof we consider the case t, t0 > 0 the sake of simplicity let us now assume that min{t, t0}=t0.We estimate

Ti[zi](t, x)−Ti[zi](t0, x0)

6

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ− Z

Rm

Γi(t0, x0; 0, ξ)ϕi(ξ)dξ

+

t

Z

0

Z

Rm

Γi(t, x;τ, ξ)zi(τ, ξ)dξdτ−

t0

Z

0

Z

Rm

Γi(t0, x0;τ, ξ)zi(τ, ξ)dξdτ

=:I1+I2. To estimate I1,we select R2η >0 such that, for eachx∈Rm\B(0, Rη2) (2.11) |ϕi(x)|< η

8C fori= 1, . . . , N

takes place. Next, we choose R2η > R2η such that, for all x ∈ Rm\B(0, R2η), ξ ∈B(0, R2η), t∈(0, T]

(2.12) |Γi(t, x; 0, ξ)|< η

8kϕikRmm(B(0, R2η)) fori= 1, . . . , N

follows. As previously, let us remark that it is enough to consider x, x0 such that |x−x0|m< δ5.

Obviously, if x ∈ B(0, R2η5), then x, x0 ∈ B(0, R2η + 2δ5). By the same argument as before, there exists 0 < δ6 < δ5 independent of x such that for t∈(0, δ6), x∈B(0, R2η+ 2δ5),the following inequality holds

(2.13)

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x)

< η

8 fori= 1, . . . , N.

If we select 0< δ7< δ6 such that (2.14) |ϕi(x)−ϕi(x0)|< η

4 fori= 1, . . . , N,

takes place for all |x−x0|m < δ7 then for t < δ6, x, x0 ∈B(0, R2η) such that

|x−x0|m < δ7,using (2.13) and (2.14) we immediately obtain

(2.15) I1 6

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ−ϕi(x)

+

Z

Rm

Γi(t0, x0; 0, ξ)ϕi(ξ)dξ−ϕi(x0)

+|ϕi(x)−ϕi(x0)|< η 2.

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Now, let t ∈ [δ6, T]. We remark that it is enough to consider t0 such that

|t−t0|< δ7. Then certainly t, t0 ∈[δ6−δ7, T].Since Γi(·,·,0, ξ), i= 1, . . . , N are uniformly continuous on [δ6−δ7, T]×B(0, R2η+2δ5) there exists 0< δ8 < δ7 such that, if max{|t−t0|, |x−x0|m}< δ8,then

(2.16) |Γi(t, x; 0, ξ)−Γi(t0, x0; 0, ξ)|< η

4kϕikRmm(B(0, R2η)). We estimate the term I1 in the following way

(2.17)

I1 6 Z

Rm

i(t, x; 0, ξ)−Γi(t0, x0; 0, ξ)||ϕi(ξ)|dξ

= Z

B(0,R2η)

i(t, x; 0, ξ)−Γi(t0, x0; 0, ξ)||ϕi(ξ)|dξ

+ Z

Rm\B(0,R2η)

i(t, x; 0, ξ)−Γi(t0, x0; 0, ξ)||ϕi(ξ)|dξ=:I11+I12.

Hence, applying (2.16) and (2.11), we easily see that for (t, x), (t0, x0)∈[δ6− δ7, T]×B(0, R2η+ 2δ5) such that max{|t−t0|, |x−x0|m}< δ8 there is

(2.18) I11<

Z

B(0,R2η)

η

4kϕikRmm(B(0, R2η))kϕikRmdξ= η 4 as well as

(2.19) I12< η 8C

Z

Rm

i(t, x; 0, ξ)|+|Γi(t0, x0; 0, ξ)|dξ < η 4.

If, however, x /∈ B(0, R2η5), then x, x0 ∈/ B(0, R2η). Consequently, using (2.17), (2.11) and (2.12) we estimate I1 as follows

(2.20)

I1 6I11+I12<kϕikRm Z

B(0,R2η)

i(t, x; 0, ξ)|+|Γi(t0, x0; 0, ξ)|dξ

+ η 8C

Z

Rm\B(0,R2η)

i(t, x; 0, ξ)|+|Γi(t0, x0; 0, ξ)|dξ6 η 2.

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Let us now recall (2.15), (2.18), (2.19) and (2.20) to state that for all (t, x), (t0, x0) ∈ Ω such that max{|t−t0|, |x−x0|m} < δ8 the following es- timate follows

(2.21) I1=

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ− Z

Rm

Γi(t0, x0; 0, ξ)ϕi(ξ)dξ

< η 2. Next, let us demonstrate thatI2< η2 as well. To confirm this assertion we note that

I2 6

t0

Z

0

Z

Rm

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)||zi(τ, ξ)|dξdτ

+ Zt

t0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ =:I21+I22. There exists δ9 := min{4CMη 0, δ8}, where M0 := sup

z∈G

kzkN, such that for all x∈Rm and t, t0 such thatt−t0 < δ9

(2.22) I226CM0(t−t0)< CM0δ9 < η 4.

Since Gsatisfies(?)–condition, there existsR3η >0 such that for all t∈[0, T], x∈Rm\B(0, R3η) the inequality

(2.23) |zi(t, x)|< η

16CT for i= 1, . . . , N.

holds. We present I21 as follows I21 =

t0

Z

0

Z

Rm\B(0,R3η)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)||zi(τ, ξ)|dξdτ

+

t0

Z

0

Z

B(0,R3η)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)||zi(τ, ξ)|dξdτ :=I211+I212.

Applying (2.23) we can establish the following estimate (2.24) I211 < η

16CT

t0

Z

0

Z

Rm

i(t, x;τ, ξ)|+|Γi(t0, x0;τ, ξ)|dξdτ 6 η 8.

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Obviously

(2.25) I212 6M0

t0

Z

0

Z

B(0,Rη3)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξdτ.

Let us notice that in view of the integrability of fundamental solution, there exists 0< δ10< δ9 such that if (t, x),(t0, x0)∈Ω and t0 6t, τ ∈(0, t0) then for each x∈Rm

(2.26)

t0

Z

0

Z

B(x,δ10)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξ < η

16M0 fori= 1, . . . , N follows. In order to estimate (2.25) we present it as follows

(2.27)

t0

Z

0

Z

B(0,R3η)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξdτ

=

t0

Z

0

Z

B(0,R3η)∩B(x+x2 010)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξdτ

+

t0

Z

0

Z

B(0,R3η)\B(x+x0

2 10)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξdτ.

(2.26) yields that the first term of (2.27) is less than 16Mη 0. Now, we assume that max{|t−t0|, |x−x0|m} < δ11 < δ10 for some δ11 > 0. By virtue of the mean-value theorem there exists (t0, x0) belonging to the segment connecting (t, x), (t0, x0),such that

(2.28)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|

=

(t−t0)∂Γi

∂t (t0, x0, τ, ξ) +

m

X

k=1

(xk−x0k)∂Γi

∂xk

(t0, x0, τ, ξ)

< δ11

∂Γi

∂t (t0, x0, τ, ξ)

+

m

X

k=1

∂Γi

∂xk(t0, x0, τ, ξ)

.

(13)

If|x0−ξ| ≥ δ102 ,then the derivatives appearing in (2.28) are esimated as below (cf. [9], p. 427)

(2.29)

∂Γi

∂t (t0, x0, τ, ξ)

< C0(t0−τ)m+22 exp

−C00|x0−ξ|2 t0−τ

6C0(t0−τ)m+22 exp

−C00 δ102 4(t0−τ)

6Mt,

∂Γi

∂xk(t0, x0, τ, ξ)

< C0(t0−τ)m+12 exp

−C00|x0−ξ|2 t0−τ

6C0(t0−τ)m+12 exp

−C00 δ102 4(t0−τ)

6Mx fork= 1, . . . , m.

Owing to (2.29), we discover that for all (t, x), (t0, x0) ∈ Ω such that max{|t−t0|, |x−x0|m}< δ11 we obtain

(2.30)

t0

Z

0

Z

B(0,R3η)\B(x+x2 010)

i(t, x;τ, ξ)−Γi(t0, x0;τ, ξ)|dξdτ

< δ11(Mt+kMx)m(B(0, R3η))T < η 16M0 if only δ11< η

16(Mt+kMx)m(B(0,R3η))M0T.

Thus, combining (2.22), (2.24), (2.25), (2.26) and (2.30), we see that for all (t, x), (t0, x0)∈Ω such that max{|t−t0|, |x−x0|m}< δ11,

(2.31) I2 =

t

Z

0

Z

Rm

Γi(t, x;τ, ξ)zi(τ, ξ)dξdτ −

t0

Z

0

Z

Rm

Γi(t0, x0;τ, ξ)zi(τ, ξ)dξdτ

< η 2. And finally, owing to (2.1), (2.10), (2.21) and (2.31) we conclude that for all (t, x), (t0, x0)∈Ω, z ∈G, if max{|t−t0|, |x−x0|m}< δ11,then

Ti[zi](t, x)−Ti[zi](t0, x0)

< η, which completes the proof of Lemma 2.

Next we will show a certain property of the operatorTN.Roughly speaking, we will check that the operator TN transfers the(?)–condition.

Lemma 3. Let N ∈ N be fixed. If ϕi ∈CB(Rm), i= 1, . . . N, then the operator TN transforms a bounded set in CBN(Ω), satisfying (?)–condition into a set satisfying (?)–condition in CBN(Ω).

(14)

Proof. Let K ⊂ CBN(Ω) be a bounded set which satisfies the (?)–

condition. Fix >0.There existsR1 >0 such that forx∈Rm\B(0, R1)

(3.1) |ϕi(x)|<

4C fori= 1, . . . , N

follows. We selectR1 > R1 such that forx∈Rm\B(0, R1), ξ∈B(0, R1), t∈ [0, T] we can obtain the inequality

(3.2) |Γi(t, x; 0, ξ)|<

4kϕikRmm(B(0, R1)) fori= 1, . . . , N.

It is obvious that

Z

Rm

i(t, x; 0, ξ)||ϕi(ξ)|dξ=

= Z

B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ+ Z

Rm\B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ.

Therefore, by (3.1) and (3.2) for x ∈ Rm \ B(0, R1), we easily obtain the following estimates

(3.3)

Z

Rm\B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ <

4C Z

Rm\B(0,R1)

i(t, x; 0, ξ)|dξ6 4, (3.4)

Z

B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ <

Z

B(0,R1)

4kϕikRmm(B(0, R1))kϕikRmdξ= 4. From the (?)–conditionforK,it follows that there existsR2 >0 such that for each function z∈K for all t∈[0, T], x∈Rm\B(0, R2),we can assert (3.5) |zi(t, x)|<

4T C fori= 1, . . . , N.

Now consider

Z t

0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ

= Z t

0

Z

Rm\B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ+

Z t

0

Z

B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ.

(15)

Making use of (3.5) we estimate the first term as follows (3.6)

Z t

0

Z

Rm\B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ <

4T C Z t

0

Z

Rm

i(t, x;τ, ξ)|dξdτ 6 4. In order to estimate the second one, we choose R2 > R2 such that for all x∈Rm\B(0, R2), ξ∈B(0, R2), t, τ ∈[0, T] : τ < t, we can get the estimate (3.7) |Γi(t, x;τ, ξ)|<

4T M0m(B(0, R2)) fori= 1, . . . , N, where M0 := sup

z∈K

kzkN.Then using (3.7) we readily assert (3.8)

Z t

0

Z

B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ <

4.

Thus, for R = max{R1, R2}, z ∈K, t∈[0, T], x∈Rm\B(0, R) combining (3.3), (3.4), (3.6) and (3.8), we find

|(Li)−1[zi](t, x)|6 Z

Rm

i(t, x; 0, ξ)||ϕi(ξ)|dξ+ Z t

0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ

= Z

B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ+ Z

Rm\B(0,R1)

i(t, x; 0, ξ)||ϕi(ξ)|dξ

+ Z t

0

Z

B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ

+ Z t

0

Z

Rm\B(0,R2)

i(t, x;τ, ξ)||zi(τ, ξ)|dξdτ < fori= 1, . . . , N.

The Lemma 3 is proved.

Let us now prove the next auxiliary lemma. We adopt the Arzela–Ascoli compactness theorem, developing it for the case of functions defined on an unbounded domain but satisfying(?)–condition. The diagonal method will be our main tool in a proof of this lemma, like in the proof of the Arzela–Ascoli compactness criterion.

Lemma 4. Let Y ⊂ CBN(Ω). If Y is a closed set of equicontinuous, uniformly bounded functions andY satisfies(?)–condition, thenY is a compact set in CBN(Ω).

(16)

Proof. Let {vn}n∈N be a sequence in Y. Let1= 1.

From the (?)–condition it follows that there exists a constant R1 > 0 such that |vn(t, x)|N < 21 follows for each n∈ N, t∈[0, T], x∈Rm\B(0, R1).

Thus, for each t∈[0, T], x∈Rm\B(0, R1), n, k∈ N,there is

|vn(t, x)−vk(t, x)|N 6|vn(t, x)|N +|vk(t, x)|N 6.

Now, let us remark that [0, T]× B(0, R1) =: B1 is separable. Let Ξ1 :={(t1, x1),(t2, x2), . . .} be a countable and dense subset in B1.

In view of the boundness of Y,the sequence{vn(ti, xi)}n∈N is bounded in RN for each (ti, xi)∈Ξ1.

In particular,{vn(t1, x1)}n∈N is a bounded sequence inRN.Therefore there exists a convergent subsequence {vn,1(t1, x1)}n∈N. Obviously, {vn,1}n∈N is a subsequence of the sequence {vn}n∈N.

Since {vn,1(t2, x2)}n∈N is a bounded sequence in RN, then there exists a convergent subsequence {vn,2(t2, x2)}n∈N and {vn,2}n∈N is a subsequence of the sequence{vn,1}n∈N.

Proceeding in this way, we will define an infinite matrix as follows:

v1,1(t1, x1) v2,1(t1, x1) v3,1(t1, x1) . . . v1,2(t2, x2) v2,2(t2, x2) v3,2(t2, x2) . . . v1,3(t3, x3) v2,3(t3, x3) v3,3(t3, x3) . . . . . . .

From the Cantor diagonal procedure it follows that the sequence{vk,k}k∈N is characterized by the fact that the sequence {vk,k(ti, xi)}k∈N converges in every point (ti, xi)∈Ξ1

Hereafter we denote vk1:=vk,k.

Now, we take 2 = 12. There exists a constant R2 > 0 ≥ R1 such that

|vk1(t, x)|N < 22 for each k ∈ N, t ∈ [0, T], x ∈ Rm \B(0, R2). In [0, T]× B(0, R2) =:B2,we take a dense and countable subset Ξ2 such that Ξ2∩B1 = Ξ1.Proceeding as above, we choose such a subsequence{vk2}k∈N of the sequence {v1k}k∈N which converges in every point of Ξ2.

Repeating the previous procedure with respect to i = 1i for i= 3,4, . . . , we obtain an infinite matrix of functions

v11 v12 v31 . . . v12 v22 v32 . . . v13 v32 v33 . . . . . . .

Using the Cantor procedure once again, we choose a subsequence {vkk}k∈N of the sequence {vn}n∈N at every point of S

n∈N Ξn.

(17)

Now, we prove that the sequence {vkk}k∈N is a Cauchy sequence in CN(Ω).

Let >0 be arbitrary. By virtue of an assumption that{vn}n∈N are equicon- tinuous, there exists δ >0 such that for all k = 1,2, . . .

|vkk(t, x)−vkk(s, y)|N <

3 if only max{|t−s|, |x−y|m}< δ.

There exists j0 such that j0 = j1

0 < .

From the (?)–condition it follows that there exists a constant Rj

0 >0 such that |vn(t, x)|N < 21 follows for each n∈ N, t∈[0, T], x∈Rm\Bj

0. Thus, for each t∈[0, T], x∈Rm\Bj

0, p, q > k0,there is

(4.1) |vpp(t, x)−vqq(t, x)|N 6|vpp(t, x)|N +|vqq(t, x)|N 6j0 < .

ByBi we denote an open ball with center at (ti, xi)∈Ξj0 and radiusδ >0.

Then Bj0 ⊂S

i=1Biδ and, moreover

(4.2) |vkk(t, x)−vkk(ti, xi)|N 6

3 for (t, x)∈Biδ.

By the Borel-Lebesque theorem there exists i0 ∈ N such thatBj0 ⊂Si0

i=1Biδ. Since the sequence {vkk(ti, xi)}k∈N converges for i= 1,2, . . . , i0,it follows that there exists a constant k0 such that

(4.3) |vpp(ti, xi)−vqq(ti, xi)|N 6

3 forp, q > k0, i= 1, . . . , i0.

If x ∈ B(0, Rj0), then there exists i 6 i0, such that (t, x) ∈ Biδ. Hence, applying (4.2) and (4.3), we get

(4.4) |vpp(t, x)−vqq(t, x)|N 6

|vpp(t, x)−vpp(ti, xi)|N +|vpp(ti, xi)−vqq(ti, xi)|N +|vqq(ti, xi)−vqq(t, x)|N <

for p, q > k0,where k0 is independent of (t, x).

Then owing to (4.1) and (4.4), we discover that {vkk}k∈N is a Cauchy sequence in CN(Ω), which in view of the completeness of the space implies the convergence of {vkk}k∈N inCBN(Ω).Thus, since Y is closed, the limit of {vkk}k∈N belongs to Y,which completes the proof of Lemma 4.

5. Proof of the theorem. In this section let us finally give a proof of the main result of the paper.

To begin with, we remark that proving that there exists a C–solution of problem (1), (2) is equivalent to showing that the operator TF:CBS(Ωτ)→ CBS(Ωτ) has a fixed point. We will prove it on the basis of the Schauder fixed point theorem.

(18)

For fixed M > 0 in CBS(Ωτ) we consider a closed ball Bτ(0, M) =:Bτ with center at 0 and radius M. Obviously, Bτ is a closed and convex set τ ∈(0, T].

Let us first demonstrate that TF(Bτ) ⊂Bτ for a sufficiently smallτ. Let w∈Bτ. It is easy to see that

|TiFi[w](t, x)|=

Z

Rm

Γi(t, x; 0, ξ)ϕi(ξ)dξ+

t

Z

0

Z

Rm

Γi(t, x;τ0, ξ)Fi[w](τ0, ξ)dξdτ0

6kϕikRm Z

Rm

i(t, x; 0, ξ)|dξ+kFi[w]kτ

t

Z

0

Z

Rm

i(t, x;τ0, ξ)|dξdτ0 6CkϕikRm+CτkFi[w]kτ.

This immediately implies

kTiFi[w]kτ 6CkϕikRm+CτkFi[w]kτ. Next, we note that kϕkΣ

Rm <∞ and kF[w]kΣτ <∞ for eachw∈Bτ.This is a direct consequence of assumptions (F1), (F2), (ϕ1) and (ϕ2).These facts instantly lead to the following estimate

kTF[w]kΣτ =

X

i=1

1

QikTiFi[w]kτ

6C

X

i=1

1

QiikRm+Cτ

X

i=1

1

QikFi[w]kτ =CkϕkΣ

Rm+CτkF[w]kΣτ. If

0< τ 6min

M−CkϕkΣ

Rm

CM1

, T

:=τ, where M1 := sup

w∈BT

kF[w]kΣτ (due to assumption (F2),we see that M1 <∞), then we readily obtain kTF[w]kΣτ 6M.

Now, let us check thatTF:Bτ →Bτ is the continuous operator. SinceF is a continuous operator in Bτ, it is enough to prove thatT is continuous on F(Bτ). To verify this assertion, fix > 0 and z = (z1, z2, . . .) ∈ F(Bτ). Let z= (z1, z2, . . .)∈F(Bτ) be such thatkz−zkΣτ < δ:= .Then

|Ti[zi](t, x)−Ti[zi](t, x)|

6

t

Z

0

Z

Rm

i(t, x;τ, ξ)||zi(τ, ξ)−zi(τ, ξ)|dξdτ 6Cτkzi−zikτ

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