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Boundary Value Problems

, 2013:94 | Cite as

On the Volterra property of a boundary problem with integral gluing condition for a mixed parabolic-hyperbolic equation

  • Abdumauvlen S Berdyshev
  • Alberto Cabada
  • Erkinjon T Karimov
  • Nazgul S Akhtaeva
Open Access
Research

Abstract

In the present work, we consider a boundary value problem with gluing conditions of an integral form for the parabolic-hyperbolic type equation. We prove that the considered problem has the Volterra property. The main tools used in the work are related to the method of the integral equations and functional analysis.

Keywords

Strong Solution Unique Solvability Mixed Type Equation Tricomi Problem Hyperbolic Part 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

Introduction

The theory of mixed type equations is one of the principal parts of the general theory of partial differential equations. The interest for these kinds of equations arises intensively because of both theoretical and practical uses of their applications. Many mathematical models of applied problems require investigations of this type of equations.

The actuality of the consideration of mixed type equations has been mentioned, for the first time, by S. A. Chaplygin in 1902 in his famous work ‘On gas streams’ [1]. The first fundamental results in this direction was obtained in 1920-1930 by Tricomi [2] and Gellerstedt [3]. The works of Lavrent’ev [4], Bitsadze [5, 6], Frankl [7], Protter [8, 9] and Morawetz [10], have had a great impact in this theory, where outstanding theoretical results were obtained and pointed out important practical values of them. Bibliography of the main fundamental results on this direction can be found, among others, in the monographs of Bitsadze [6], Berezansky [11], Bers [12], Salakhitdinov and Urinov [13] and Nakhushev [14].

In most of the works devoted to the study of mixed type equations, the object of study was mixed elliptic-hyperbolic type equations. Comparatively, few results have been obtained on the study of mixed parabolic-hyperbolic type equations. However, this last type of equations have also numerous applications in the real life processes (see [15] for an interesting example in mechanics). The reader can find a nice example given, for the first time, by Gelfand in [16], and connect with the movement of the gas in a channel surrounded by a porous environment. Inside the channel, the movement of gas was described by the wave equation and outside by the diffusion one. Mathematic models of this kind of problems arise in the study of electromagnetic fields, in a heterogeneous environment, consisting of dielectric and conductive environment for modeling the movement of a little compressible fluid in a channel surrounded by a porous medium [17]. Here, the wave equation describes the hydrodynamic pressure of the fluid in the channel, and the equation of filtration-pressure fluid in a porous medium. Similar problems arise in the study of the magnetic intensity of the electromagnetic field [17].

In the last few years, the investigations on local boundary value problems, for mixed equations in domains with non-characteristic boundary data, were intensively increased. We point out that the studies made on boundary value problems for equations of mixed type, in domains with deviation from the characteristics (with a non-characteristic boundary), have originated with the fundamental works of Bitsadze [5], where the generalized Tricomi problem (Problem M) for an equation of mixed type is discussed.

In the works [18] and [19], the analog to the Tricomi problem for a modeled parabolic-hyperbolic equation, was investigated in a domain with a non-characteristic boundary in a hyperbolic part. Moreover, the uniqueness of solution and the Volterra property of the formulated problem was proved. We also refer to the recent works devoted to the study of parabolic-hyperbolic equations [20, 21, 22, 23].

In the last years, the interest for considering boundary value problems of parabolic-hyperbolic type, with integral gluing condition on the line of type changing, is increasing [24, 25].

In the present work, we study the analog to the generalized Tricomi problem with an integral gluing condition on the line of type changing. We prove that the formulated problem has the Volterra property. The obtained result generalizes some previous ones from Sadybekov and Tajzhanova given in [28].

Formulation of the problem

Let Ω R 2 Open image in new window be a domain, bounded at y > 0 Open image in new window by segments A A 0 Open image in new window, A 0 B 0 Open image in new window, B B 0 Open image in new window of straight lines x = 0 Open image in new window, y = 1 Open image in new window, x = 1 Open image in new window, respectively, and at y < 0 Open image in new window by a monotone smooth curve A C : y = γ ( x ) Open image in new window, 0 < x < l Open image in new window, 1 / 2 < l < 1 Open image in new window, γ ( 0 ) = 0 Open image in new window, l + γ ( l ) = 1 Open image in new window and by the segment B C : x y = 1 Open image in new window, l x < 1 Open image in new window, which is the characteristic curve of the equation
L u = f ( x , y ) , Open image in new window
(1)
where
L u = { u x u y y , y > 0 , u x x u y y , y < 0 . Open image in new window
(2)

Now we state the problem that we will consider along the paper:

Problem B To find a solution of Eq. (1), satisfying boundary conditions
and gluing conditions
u x ( x , + 0 ) = u x ( x , 0 ) , u y ( x , + 0 ) = α u y ( x , 0 ) + β 0 x u y ( t , 0 ) Q ( x , t ) d t , 0 < x < 1 , Open image in new window
(5)

where Q is a given function such that Q C 1 ( [ 0 , 1 ] × [ 0 , 1 ] ) Open image in new window, and α , β R Open image in new window satisfy α 2 + β 2 > 0 Open image in new window.

When the curve AC coincides with the characteristic one x + y = 0 Open image in new window, α = 1 Open image in new window and β = 0 Open image in new window, the Problem B is just the Tricomi problem for parabolic-hyperbolic equation with a non-characteristic line of type changing, which has been studied in [26].

Regular solvability of the Problem B with continuous gluing conditions ( α = 1 Open image in new window, β = 0 Open image in new window) have been proved, for the first time, in [27], and strong solvability of this problem was proved in the work [28].

Several properties, including the Volterra property of boundary problems for mixed parabolic-hyperbolic equations, have been studied in the works [29, 30, 31, 32, 33].

We denote the parabolic part of the mixed domain Ω as Ω 0 Open image in new window and the hyperbolic part by  Ω 1 Open image in new window.

A regular solution of the Problem B in the domain Ω will be a function
u C ( Ω ¯ ) C 1 ( Ω 0 A B ) C 1 ( Ω 1 A C A B ) C 1 , 2 ( Ω 0 ) C 2 , 2 ( Ω 1 ) , Open image in new window

that satisfies Eq. (1) in the domains Ω 0 Open image in new window and Ω 1 Open image in new window, the boundary conditions (3)-(4), and the gluing condition (5).

Regarding the curve AC, we assume that x + γ ( x ) Open image in new window is monotonically increasing. Then, rewriting it by using the characteristic variables ξ = x + y Open image in new window and η = x y Open image in new window, we have that the equation of the curve AC can be expressed as ξ = λ ( η ) Open image in new window, 0 η 1 Open image in new window.

Main result

Theorem 1 Let γ C 1 [ 0 , l ] Open image in new window and Q C 1 ( [ 0 , 1 ] × [ 0 , 1 ] ) Open image in new window. Then for any function f C 1 ( Ω ¯ ) Open image in new window, there exists a unique regular solution of the Problem  B.

Proof By a regular solution of the Problem B in the domain Ω 1 Open image in new window we look for a function that fulfills the following expression:
u ( ξ , η ) = 1 2 [ τ ( ξ ) + τ ( η ) ξ η ν 1 ( t ) d t ] ξ η d ξ 1 ξ 1 η f 1 ( ξ 1 , η 1 ) d η 1 , Open image in new window
(6)
where
ξ = x + y , η = x y , f 1 ( ξ , η ) = 1 4 f ( ξ + η 2 , ξ η 2 ) , τ ( x ) = u ( x , 0 ) , ν 1 ( x ) = u y ( x , 0 ) . Open image in new window
(7)
Based on (4) from (6), using the expressions on (7), we deduce that
ν 1 ( η ) = τ ( η ) 2 λ ( η ) η f 1 ( ξ 1 , η ) d ξ 1 , 0 η 1 . Open image in new window
(8)
By virtue of the unique solvability of the first boundary problem for the heat equation (1) satisfying condition (3), and the fact that u ( x , 0 ) = τ ( x ) Open image in new window, its solution can be represented as
u ( x , y ) = 0 x d x 1 0 1 G ( x x 1 , y , y 1 ) f ( x 1 , y 1 ) d y 1 + 0 x G y 1 ( x x 1 , y , 0 ) τ ( x 1 ) d x 1 , Open image in new window
(9)
where τ ( 0 ) = 0 Open image in new window and G ( x , y , y 1 ) Open image in new window is the Green’s function related to the first boundary problem, for the heat equation in a rectangle A A 0 B 0 B Open image in new window, which has the form [34]
G ( x , y , y 1 ) = 1 2 π x n = + [ exp { ( y y 1 + 2 n ) 2 4 x } exp { ( y + y 1 + 2 n ) 2 4 x } ] . Open image in new window
(10)
Calculating the derivative u y Open image in new window in (9) and passing to the limit at y 0 Open image in new window, we get
u y ( x , + 0 ) = 0 x k ( x t ) u x ( t , + 0 ) d t + F 0 ( x ) , Open image in new window
where
k ( x ) = 1 π x n = + e n 2 x = 1 π x 1 2 + k ˜ ( x ) , Open image in new window
(11)
and
F 0 ( x ) = 0 x d x 1 0 1 G y ( x x 1 , y , y 1 ) | y = 0 f ( x 1 , y 1 ) d y 1 . Open image in new window
(12)
Thus, the main functional relation between τ ( x ) Open image in new window and ν 0 ( x ) = u y ( x , + 0 ) Open image in new window, reduced to the segment AB from the parabolic part of the domain, implies that
ν 0 ( x ) = 0 x k ( x t ) τ ( t ) d t + F 0 ( x ) . Open image in new window
(13)
Suppose, in a first moment, that α 0 Open image in new window. From (8) and (13), considering the gluing condition (5), we obtain the following integral equation regarding the function τ ( x ) Open image in new window:
τ ( x ) + 0 x k 1 ( x , t ) τ ( t ) d t = F 1 ( x ) . Open image in new window
(14)
Here,
k 1 ( x , t ) = 1 α [ k ( x t ) + β Q ( x , t ) ] , Open image in new window
(15)
and
F 1 ( x ) = 1 α F 0 ( x ) + 2 λ ( x ) x f 1 ( ξ 1 , x ) d ξ 1 + 2 β α 0 x Q ( x , t ) d t λ ( t ) t f ( ξ 1 , t ) d ξ 1 . Open image in new window
(16)

Hence, the Problem B is equivalent, in the sense of unique solvability, to the second kind Volterra integral equation (14).

The restrictions imposed on the functions γ, Q, and the right-hand side of Eq. (1) guarantees that, by virtue of (11) and (15), the kernel k 1 ( x , t ) Open image in new window is a kernel with weak singularity. So, we have that Eq. (14) has a unique solution and τ C 1 ( 0 , 1 ) Open image in new window. Since τ ( 0 ) = 0 Open image in new window, we deduce the uniqueness of the function τ. Equation (8) gives us the uniqueness of function ν 1 Open image in new window and, as consequence, we deduce, from Eq. (6), the uniqueness of solution of Problem B when α 0 Open image in new window.

Consider now the other case, i.e. α = 0 Open image in new window and β 0 Open image in new window.

From functional relations (8) and (13), and taking gluing condition (5) into account at α = 0 Open image in new window, we have
0 x k ( x t ) τ ( t ) d t + F 0 ( x ) = β 0 x [ τ ( t ) 2 λ ( t ) t f 1 ( ξ 1 , t ) d ξ 1 ] Q ( x , t ) d t Open image in new window
or, which is the same,
0 x τ ( t ) [ k ( x t ) + β Q ( x , t ) ] d t = F 0 ( x ) + 2 β 0 x d t λ ( t ) t Q ( x , t ) f 1 ( ξ 1 , t ) d ξ 1 . Open image in new window
Considering the representation of k ( x t ) Open image in new window, the previous equation can be rewritten as follows:
0 x τ ( t ) d t ( x t ) 1 / 2 = π [ F 0 ( x ) + 2 β 0 x d t λ ( t ) t Q ( x , t ) f 1 ( ξ 1 , t ) d ξ 1 0 x τ ( t ) ( k ˜ ( x t ) + β Q ( x , t ) ) d t ] . Open image in new window
(17)
Since Eq. (17) is the Abel’s equation, it can be solved and so we arrive at the following identity:
τ ( x ) = F 0 ( 0 ) π x + 1 π { 0 x F 0 ( t ) d t x t + 2 β 0 x d t x t t 0 t d z λ ( z ) z Q ( t , z ) f 1 ( ξ 1 , z ) d ξ 1 0 x d t x t t 0 t τ ( z ) [ k ˜ ( x t ) + β Q ( t , z ) ] d z } . Open image in new window
Considering F 0 ( 0 ) = 0 Open image in new window, after some simplifications, we get
τ ( x ) + 0 x K 0 ( x , z ) τ ( z ) d z = F 2 ( x ) , Open image in new window
(18)
Since the kernel K 0 ( x , z ) Open image in new window has a weak singularity, then Eq. (18) has a unique solution, and it can be represented as
τ ( x ) = F 2 ( x ) + 0 x R ( x , z ) F 2 ( z ) d z , Open image in new window
(20)

where R ( x , z ) Open image in new window is the resolvent kernel of (18).

As a consequence, arguing as in the case α 0 Open image in new window, we deduce, from Eq. (6), the uniqueness of solution of Problem B for α = 0 Open image in new window and β 0 Open image in new window, and the result is proved. □

In the sequel, we will deduce the exact expression of the integral kernel related to the unique solution of Problem B.

To this end, we suppose, at the beginning, that α 0 Open image in new window. Note that the unique solution of Eq. (14) can be represented as
τ ( x ) = 0 x Γ ( x , t ) F 1 ( t ) d t + F 1 ( x ) , Open image in new window
(21)
where Γ ( x , t ) Open image in new window is the resolvent kernel of Eq. (14), and it is given by the recurrence formula:
From (21), taking τ ( 0 ) = 0 Open image in new window into account, we have that
τ ( x ) = 0 x Γ 1 ( x , t ) F 1 ( t ) d t , Open image in new window

where Γ 1 ( x , t ) = 1 + t x Γ ( z , t ) d z Open image in new window.

From the formula (6), and considering (8), one can easily deduce that
u ( ξ , η ) = τ ( ξ ) + ξ η d η 1 λ ( η 1 ) η 1 f 1 ( ξ 1 , η 1 ) d ξ 1 . Open image in new window
(22)
Substituting the representation of τ ( x ) Open image in new window into (22) and considering (12) and (16), after some evaluations we get
u ( x , y ) = 1 α 0 ξ d x 1 0 1 G 1 ( ξ x 1 , y 1 ) f ( x 1 , y 1 ) d y 1 + 2 0 ξ d η 1 λ ( η 1 ) η 1 Γ 1 ( ξ , η 1 ) f 1 ( ξ 1 , η 1 ) d ξ 1 + 2 β α 0 ξ d η 1 λ ( η 1 ) η 1 G 0 ( ξ η 1 , η 1 ) f 1 ( ξ 1 , η 1 ) d ξ 1 + ξ η d η 1 λ ( η 1 ) η 1 f 1 ( ξ 1 , η 1 ) d ξ 1 , Open image in new window
(23)
where
G 1 ( x , y 1 ) = 0 x Γ 1 ( x , t ) G y ( t , y 1 , 0 ) d t , Open image in new window
and
G 0 ( x , η ) = 0 x Q ( z + η , η ) Γ 1 ( x , z ) d z . Open image in new window
In an analogous way, substituting the representation of τ ( x ) Open image in new window into (9), we have
u ( x , y ) = 0 x d x 1 0 1 G 2 ( x x 1 , y , y 1 ) f ( x 1 , y 1 ) d y 1 + 2 0 x d η 1 λ ( η 1 ) η 1 G 1 ( x η 1 , y ) f 1 ( ξ 1 , η 1 ) d ξ 1 + 2 β α 0 x d η 1 λ ( η 1 ) η 1 G 01 ( x η 1 , η 1 ) f 1 ( ξ 1 , η 1 ) d ξ 1 , Open image in new window
(24)
where
G 2 ( x , y , y 1 ) = G ( x , y , y 1 ) + 1 α 0 x G 1 ( t , y 1 ) G y ( x t , y , 0 ) d t Open image in new window
and
G 01 ( x , η 1 ) = 0 x G y ( x 1 , y 1 , 0 ) G 0 ( x x 1 , η 1 ) d x 1 . Open image in new window
From (23) and (24), we arrive at the following expression:
u ( x , y ) = Ω K α β ( x , y , x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 , Open image in new window
Here,
θ ( y ) = { 1 , y > 0 , 0 , y < 0 . Open image in new window
When α = 0 Open image in new window and β 0 Open image in new window, by using a similar algorithm, we conclude that
u ( x , y ) = Ω K 0 β ( x , y , x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 , Open image in new window
and
G 5 ( x , y , η 1 ) = 0 x G y ( x x 1 , y , 0 ) Q 1 ( x 1 , η 1 ) d x 1 . Open image in new window

Thus, we have partially proved the following lemma.

Lemma 1 The unique regular solution of Problem  B can be represented as follows:
u ( x , y ) = Ω K ( x , y , x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 ( x , y ) Ω , Open image in new window
(25)
where K ( x , y , x 1 , y 1 ) L 2 ( Ω × Ω ) Open image in new window and

Proof Expression (25) has been proved before. Let us see that K ( x , y ; x 1 , y 1 ) L 2 ( Ω × Ω ) Open image in new window.

Note that in the kernel defined in (25), all the items are bounded except the first one. So, we only need to prove that
θ ( y ) θ ( y 1 ) θ ( x x 1 ) G ( x x 1 , y , y 1 ) L 2 ( Ω × Ω ) . Open image in new window
From the representation of the Green’s function G ( x x 1 , y , y 1 ) Open image in new window given in (10), it follows that, for the aforementioned aim, it is enough to prove that (for n = 0 Open image in new window):
B ( x x 1 , y , y 1 ) = θ ( y ) θ ( y 1 ) θ ( x x 1 ) 1 2 π ( x x 1 ) [ exp { ( y y 1 ) 2 4 ( x x 1 ) } exp { ( y + y 1 ) 2 4 ( x x 1 ) } ] Open image in new window

is bounded.

First, note that
B ( x x 1 , y , y 1 ) 1 2 π ( x x 1 ) e ( y y 1 ) 2 4 ( x x 1 ) . Open image in new window
Using this fact, we deduce that
B L 2 ( Ω × Ω ) 2 = 0 1 d x 0 1 d y 0 x d x 1 0 1 | B ( x x 1 , y , y 1 ) | 2 d y 1 = 0 1 d y 0 1 d y 1 0 1 d x 0 x | B ( x , y , y 1 ) | 2 d x 1 0 1 d y 0 1 d y 1 0 1 | B ( x , y , y 1 ) | 2 d x 1 4 π 0 1 d y 0 1 d y 1 0 1 1 x e ( y y 1 ) 2 4 x d x = 1 4 π 0 1 d y 0 1 d x x 0 1 e ( y y 1 ) 2 4 x d y 1 . Open image in new window
By means of the change of variables y y 1 2 x = y 2 Open image in new window, we get that this last expression is less than or equals to the following one:
1 4 π 0 1 d y 0 1 d x x y 1 2 x y 2 x e y 2 2 2 x d y 2 1 2 π 0 1 d y 0 1 d x x + e y 2 d y 2 = 1 π . Open image in new window

As a consequence, K ( x , y ; x 1 , y 1 ) L 2 ( Ω × Ω ) Open image in new window and Lemma 1 is completely proved. □

Define now
F α β ( x ) = { F 1 ( x ) , α 0 , F 2 ( x ) , α = 0 . Open image in new window

We have the following regularity result for this function.

Lemma 2 If f C 1 ( Ω ¯ ) Open image in new window, f ( 0 , 0 ) = 0 Open image in new window and Q C 1 ( [ 0 , 1 ] × [ 0 , 1 ] ) Open image in new window, then F α β C 1 [ 0 , 1 ] Open image in new window and F α β ( 0 ) = 0 Open image in new window.

Proof Using the explicit form of the Green’s function given in (10), it is not complicated to prove that function F α β Open image in new window, defined by formulas (16) and (19), belongs to the class of functions C 1 [ 0 , 1 ] Open image in new window and F α β ( 0 ) = 0 Open image in new window.

Lemma 2 is proved. □

Lemma 3 Suppose that Q C 1 ( [ 0 , 1 ] × [ 0 , 1 ] ) Open image in new window and f L 2 ( Ω ) Open image in new window, then F α β L 2 ( Ω ) Open image in new window and
F α β L 2 ( 0 , 1 ) C f L 2 ( Ω ) . Open image in new window
(26)
Proof Consider the following problem in Ω 0 Open image in new window:
ω x ω y y = f ( x , y ) , ω | A A 0 A 0 B 0 A B = 0 . Open image in new window
(27)

Obviously, we have that F 0 ( x ) = lim y 0 ω y ( x , y ) Open image in new window.

First, note that it is known [35] that problem (27) has a unique solution ω W 2 1 , 2 ( Ω 0 ) Open image in new window, and it satisfies the following inequality:
ω L 2 ( Ω 0 ) 2 + ω x L 2 ( Ω 0 ) 2 + ω y L 2 ( Ω 0 ) 2 + ω y y L 2 ( Ω 0 ) 2 C f L 2 ( Ω 0 ) 2 . Open image in new window
(28)
Using now the obvious equality
ω y ( x , 0 ) = ω y ( x , y ) 0 y ω y y ( x , t ) d t , Open image in new window
we have that
ω y ( , 0 ) L 2 ( 0 , 1 ) 2 = 0 1 | ω y ( x , 0 ) | 2 d x = 0 1 d y 0 1 | ω y ( x , 0 ) | 2 d x C [ ω y L 2 ( Ω 0 ) 2 + ω y y L 2 ( Ω 0 ) 2 ] . Open image in new window
(29)
From (28) and (29), we obtain
F 0 L 2 ( 0 , 1 ) = ω y ( , 0 ) L 2 ( 0 , 1 ) C f L 2 ( Ω 0 ) . Open image in new window
(30)

Now, by virtue of the conditions of Lemma 3 and the representations (16) and (19), from expression (30) and the Cauchy-Bunjakovskii inequalities, we get the estimate (26) and conclude the proof. □

Denote now l Open image in new window as the norm of the Sobolev space H l ( Ω ) W 2 l ( Ω ) Open image in new window with W 2 0 ( Ω ) L 2 ( Ω ) Open image in new window.

Lemma 4 Let u be the unique regular solution of Problem  B. Then the following estimate holds:
u 1 c f 0 . Open image in new window
(31)

Here, c is a positive constant that does not depend on u.

Proof By virtue of Lemma 3, and from (20) and (21), we deduce that
τ L 2 ( 0 , 1 ) C F α β L 2 ( 0 , 1 ) C f 0 . Open image in new window

The result follows from expression (22). □

Definition 1 We define the set W as the set of all the regular solutions of Problem B.

A function u L 2 ( Ω ) Open image in new window is said to be a strong solution of Problem B, if there exists a functional sequence { u n } W Open image in new window, such that u n Open image in new window and L u n Open image in new window converge in L 2 ( Ω ) Open image in new window to u and f, respectively.

Define L Open image in new window as the closure of the differential operator L : W L 2 ( Ω ) Open image in new window, given by expression (2).

Note that, according to the definition of the strong solution, the function u will be a strong solution of Problem B if and only if u D ( L ) Open image in new window.

Now we are in a position to prove the following uniqueness result for strong solutions.

Theorem 2 For any function Q C 1 ( [ 0 , 1 ] × [ 0 , 1 ] ) Open image in new window and f L 2 ( Ω ) Open image in new window, there exists a unique strong solution u of Problem  B. Moreover, u W 2 1 ( Ω ) W x , y 1 , 2 ( Ω 1 ) C ( Ω ¯ ) Open image in new window, satisfies inequality (31) and it is given by the expression (25).

Proof Let C 0 1 ( Ω ¯ ) Open image in new window be the set of the C 1 ( Ω ¯ ) Open image in new window functions that vanish in a neighborhood of Ω ( Ω is a boundary of the domain Ω). Since C 0 1 ( Ω ¯ ) Open image in new window is dense in L 2 ( Ω ) Open image in new window, we have that for any function f L 2 ( Ω ) Open image in new window, there exist a functional sequence f n C 0 1 ( Ω ¯ ) Open image in new window, such that f n f 0 Open image in new window, as n Open image in new window.

It is not difficult to verify that if f n C 0 1 ( Ω ¯ ) Open image in new window then F α β n C 1 ( [ 0 , 1 ] ) Open image in new window (with obvious notation). Therefore, Eqs. (14) and (18) can be considered as a second kind Volterra integral equation in the space C 1 ( [ 0 , 1 ] ) Open image in new window. Consequently, we have that τ n ( x ) = u n x ( x , 0 ) C 1 [ 0 ; 1 ] Open image in new window. Due to the properties of the solutions of the boundary value problem for the heat equation in Ω 0 Open image in new window and the Darboux problem, by using the representations (6) and (9), we conclude that u n W Open image in new window for all f n C 0 1 ( Ω ¯ ) Open image in new window.

By virtue of the inequality (31), we get
u n u 1 c f n f 0 0 . Open image in new window

Consequently, { u n } Open image in new window is a sequence of strong solutions, hence, Problem B is strongly solvable for all right hand f L 2 ( Ω ) Open image in new window, and the strong solution belongs to the space W 2 1 ( Ω ) W x , y 1 , 2 ( Ω 1 ) C ( Ω ¯ ) Open image in new window. Thus, Theorem 2 is proved. □

Consider now, for all n = 2 , 3 , Open image in new window , the sequence of kernels given by the recurrence formula
K n ( x , y ; x 1 , y 1 ) = Ω K ( x , y ; x 2 , y 2 ) K ( n 1 ) ( x 2 , y 2 , x 1 , y 1 ) d x 2 d y 2 , Open image in new window
with
K 1 ( x , y ; x 1 , y 1 ) = K ( x , y ; x 1 , y 1 ) , Open image in new window

and K defined in Lemma 1.

Lemma 5 For the iterated kernels K n ( x , y ; x 1 , y 1 ) Open image in new window we have the following estimate:
| K n ( x , y ; x 1 , y 1 ) | ( π M ) n ( 3 2 ) n 1 ( x x 1 ) n 2 1 Γ ( n 2 ) , n = 1 , 2 , 3 , , Open image in new window
(32)

where Open image in new window and Γ is the Gamma-function of Euler.

Proof The proof will be done by induction in n.

Taking the representation of the Green’s function given in (10) into account, and from the representation of the kernel K ( x , y ; x 1 , y 1 ) Open image in new window at n = 1 Open image in new window, the inequality (32)
| K 1 ( x , y ; x 1 , y 1 ) | M ( x x 1 ) 1 2 Open image in new window

is automatically deduced.

Let (32) be valid for n = k 1 Open image in new window. We will prove the validity of this formula for n = k Open image in new window. To this end, by using inequality (32), at n = 1 Open image in new window and n = k 1 Open image in new window, we have that
Evaluating the previous integrals, we have that
| K k ( x , y ; x 1 , y 1 ) | M k ( π ) k 1 ( 3 2 ) k 1 ( x x 1 ) k 2 1 Γ ( k 1 2 ) 0 1 σ 1 2 ( 1 σ ) k 2 3 2 d σ = ( π M ) k ( 3 2 ) k 1 ( x x 1 ) k 2 1 Γ ( k 2 ) , Open image in new window

which proves Lemma 5. □

Now we are in a position to prove the final result of this paper, which gives us the Volterra property for the inverse of operator L Open image in new window.

Theorem 3 The integral operator defined in the right hand of (25), i.e.
L 1 f ( x , y ) = Ω K ( x , y ; x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 , Open image in new window
(33)

has the Volterra property (it is almost continuous and quasi-nilpotent) in L 2 ( Ω ) Open image in new window.

Proof Since the continuity of this operator follows from the fact that K L 2 ( Ω × Ω ) Open image in new window. To prove this theorem, we only need to verify that operator L 1 Open image in new window, defined by (33), is quasi-nilpotent, i.e.
lim n L n 0 1 n = 0 , Open image in new window
(34)
where
L n = L 1 [ L ( n 1 ) ] , n = 1 , 2 , 3 , . Open image in new window
From (33), and by direct calculations, one can easily arrive at the following expression:
L n f ( x , y ) = Ω K n ( x , y ; x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 . Open image in new window
(35)
Consequently, using the inequality of Schwarz and expression (32), from the representation (35) we obtain that
L n f 0 2 = Ω | L n f | 2 d x d y = Ω [ Ω K n ( x , y ; x 1 , y 1 ) f ( x 1 , y 1 ) d x 1 d y 1 ] 2 d x d y Ω [ ( Ω | f ( x 1 , y 1 ) | 2 d x 1 d y 1 ) ( Ω | K n ( x , y ; x 1 , y 1 ) | 2 d x 1 d y 1 ) ] d x d y ( 3 2 π M ) 2 n 1 n ( n 1 ) Γ 2 ( n 2 ) f 0 2 . Open image in new window
From here, we get
L n 0 ( 3 2 π M ) n 1 Γ ( 1 + n 2 ) . Open image in new window

From the last equality, one can state the validity of the equality (34) and Theorem 3 is proved. □

Consequence 1 Problem B has the Volterra property.

Consequence 2 For any complex number λ, the equation
L u λ u = f Open image in new window
(36)

is uniquely solvable for all f L 2 ( Ω ) Open image in new window.

Due to the invertibility of the operator L Open image in new window, the unique solvability of Eq. (36) is equivalent to the uniqueness of solution of the equation
u λ L 1 u = L 1 f , Open image in new window

which is a second kind of Volterra equation. This proves Consequence 2 of Theorem 3.

Notes

Acknowledgements

This research was partially supported by Ministerio de Ciencia e Innovación-SPAIN, and FEDER, project MTM2010-15314 and KazNPU Rector’s grant for 2013.

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© Berdyshev et al.; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Authors and Affiliations

  • Abdumauvlen S Berdyshev
    • 1
  • Alberto Cabada
    • 2
  • Erkinjon T Karimov
    • 3
  • Nazgul S Akhtaeva
    • 1
  1. 1.Kazakh National Pedagogical University named after AbaiAlmatyKazakhstan
  2. 2.University of Santiago de CompostelaSantiago de CompostelaSpain
  3. 3.Institute of Mathematics, National University of Uzbekistan named after Mirzo UlughbekTashkentUzbekistan

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