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ISSN Online: 2333-9721 ISSN Print: 2333-9705

DOI: 10.4236/oalib.1104631 Jun. 26, 2018 1 Open Access Library Journal

Solvability of a Class of Operator-Differential

Equations of Third Order with Complicated

Characteristic on the Whole Real Axis

Abdel Baset I. Ahmed, Mohamed A. Labeeb

Egyptian Russian University,Badr City, Egypt

Abstract

On the whole real axis, we demonstrate sufficient conditions of regular solva-bility of third order operator-differential equations with complicated charac-teristics. These conditions were formulated only by the operator coefficients of the equation. In addition, by the principal part of the equation, the norms of the operators of intermediate derivative were estimated.

Subject Areas

Ordinary Differential Equation

Keywords

Operator-Differential Equation, Hilbert Space, Self-Adjoint Operator, Intermediate Derivative Operator

1. Introduction

In a separable Hilbert space H, we have the following equation:

( )

0

( )

1

( )

( )

, ,

Pu xp u x +p u x = f x x R∈ (1)

where

( )

( )

( )

( )

2 0

3 2

1 3

1

d d .

d d

d

, d

s

s s

s

p u x A A u x

t t

u x

p u x A

x

− =

  

= +

  

=

A is a self-adjoint positive-definite operator, and A ss, =1,2 are generally linear unbounded operators. All derivatives are understood in the sense of

distribu-How to cite this paper: Ahmed, A.B.I. and Labeeb, M.A. (2018) Solvability of a Class of Operator-Differential Equations of Third Order with Complicated Characteristic on the Whole Real Axis. Open Access Library Journal, 5: e4631.

https://doi.org/10.4236/oalib.1104631

Received: May 2, 2018 Accepted: June 23, 2018 Published: June 26, 2018

Copyright © 2018 by authors and Open Access Library Inc.

This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).

http://creativecommons.org/licenses/by/4.0/

(2)

DOI: 10.4236/oalib.1104631 2 Open Access Library Journal

tions theory.

We consider f x

( )

L R H2

(

;

)

, where

(

)

( )

( )

( )

(

( )

)

2

1

2 2 2

2 ; : L R H; Hd

L R H =f x f x −∞+∞ f x t < +∞

 

(see [1] [2]), and

( )

3

(

)

2 ;

u x W R H,which are determined as follows:

(

)

( )

3

( )

(

)

( )

(

)

3 3

2 3 2 2

d

; : ; , ; .

d u x

W R H u x L R H A u x L R H

x +

 

 

= ∈ ∈

 

 

With the norm

( )

( ) ( )

3

2 2

2

1

2 2

3 2

3 3

; ;

;

d ,

d

W R H L R H

L R H

u

u A u

x

+

 

 

= +

 

 

See [2].

Notice that the principal part of the investigated equation possesses compli-cated characteristic, not multiple characteristics as in [3].

Definition 1. If for any f x

( )

L R H2

(

;

)

there exists a vector function

( )

2

(

)

2 ;

u x W R H that satisfies Equation (1) almost everywhere in R, then it is

called a regular solution of Equation (1)

Definition 2. If for any f x

( )

L R H2

(

;

)

there exists a regular solution of

Equation (1), and satisfies the inequality

( ) ( )

3

2 ; 2 ; ,

W R H L R H

uconst f (2)

then Equation (1) is called regularly solvable.

It is known that if 3

(

)

(

)

1: 2 ; 2 ;

p W R HL R H , then 3

( )

(

)

2

d

; d

s s

s

u x

A L R H

x

,

1,2

s= .

And the following inequalities are valid (see [2]).

( )

( ) ( )

2 2 2

3

; ;

d

, 1,2. d

s s

s

s W R H

L R H

u x

A c u s

x

= (3)

Definition 3. Parseval’s equality

( )

2 1

( )

2

d d

f x x f ζ ζ

+∞ +∞

−∞ = −∞

where,

( )

( )

e d .ix

f ζ +∞f x − ζ x

−∞

=

2. Main Results

Theorem 1. The operator P0 is an isomorphism from the space W R H23

(

;

)

to

the space L R H2

(

;

)

.

Proof. From (2), it is easy to prove that the operator P0 acts from

(

)

3

2 ;

W R H to L R H2

(

;

)

be bounded. Using Fourier transforms for the

(3)

DOI: 10.4236/oalib.1104631 3 Open Access Library Journal

(

)(

) ( )

2

( )

.

i E A

ξ

i E A u

ξ

ξ

f

ξ

− − − +  =  (4)

(E is the unit operator), where u

( ) ( )

ξ

,f

ξ

are Fourier transform for the func-tions u x f x

( ) ( )

, , respectively. The operator pencil

(

i E A

ξ

)(

i E A

ξ

+

)

2 is

invertible and moreover

( ) (

) (

1

)

2

( )

= ,

u

ξ

i E A

ξ

i E A

ξ

+f

ξ

 (5)

Hence,

( )

1

(

) (

1

)

2

( )

e d .

i x

u x +∞ i E Aξi E Aξf ξ ζ ζ

−∞

=

− − − +  (6)

We show that

( )

3

(

)

2 ;

u x W R H. By using the Parseval equality and (3), we

obtain:

( )

( ) ( )

( )

( )

( )

( )

(

) (

)

( )

( )

(

) (

)

( )

( )

(

) (

)

( )

( )

(

) (

)

( )

( )

3

2 2 2 2

2

2

2

2

2 2

3 2 2 2

2 3 3 3

3

; ; ; ;

;

2

1 2

3

; 2

1 2

3

;

2 2

1 2

3

;

2 2

1 2

3

; d

d

sup sup

W R H L R H L R H L R H

L R H

L R H

L R H

L R H H H

R

L R H H H

R

u

u A u i u A u

t

i i E A i E A f

A i E A i E A f

i i E A i E A f

A i E A i E A f

ζ

ζ

ζ ξ ξ

ζ ξ ξ ξ

ξ ξ ξ

ζ ζ ζ ζ

ζ ζ ζ

− −

− −

− −

→ ∈

− −

→ ∈

= + = − +

= − − − − +

+ − − − +

≤ − − − − +

+ − − − +

 

(7)

If σ

( )

A is a spectrum of the operator A, then we consider

(

) (

)

( )

(

) (

)

(

2 2 2

)

3

1 2

3

1 2

3 3 sup

sup sup

sup 1,

H H R

R A

R

i i E A i E A

i i i

ξ

ξ σ σ

ξ ζ σ

ζ ζ ζ

ζ ζ σ ζ σ

ζ

− −

→ ∈

− −

∈ ∈

+

− − − − +

≤ − − − − +

= ≤

(8)

(

) (

)

( )

(

) (

)

( )

(

2 2 2

)

3

1 2

3

1 2

3 3

sup

sup sup

sup 1

H H R

R A

A

A i E A i E A

i i

ξ

ξ σ σ

σ σ ζ σ

ζ ζ

σ ζ σ ζ σ

σ

− −

→ ∈

− −

∈ ∈

+

− − +

≤ − − − +

= ≤

(9)

Taking into account (5) and (6) into (4) we obtain:

( )

( )

( )

( )

( )

3

2 2 2

2 2

2

; 2 ; = 2 ; .

W R H L R H L R H

uf

ζ

f x (10)

Consequently,

( )

3

(

)

2 ;

u x W R H.

Applying Banach theorem on the inverse operator, we get that the operator

0

P is an isomorphism from 3

(

)

2 ;

W R H to L R H2

(

;

)

.

(4)

sol-DOI: 10.4236/oalib.1104631 4 Open Access Library Journal

vability of the given equation, expressed only by its operator coefficients. From theorem 1, we have that the norms p u0 L R H2( ; ) and 3( )

2 ;

W R H

u are

equivalent in the space 3

(

)

2 ;

W R H . Therefore by the norm p u0 L R H2( ; ) the

theorem on intermediate derivatives is valid as well. Theorem 2. Let

( )

3

(

)

2 ;

u x W R H. Then there hold the following

inequali-ties:

( )

( ) 2( )

2 3

0 ;

; d

, 1,2. d

s s

s

s L R H

L R H

u x

A a p u s

x

= (11)

where 1 2 2

3 3

a =a = .

Proof. To establish the validity of inequality (11) we make change

( )

( )

0

p u x = f x and apply the Fourier transformation. We get

(

) (

) (

)

( )

( )

(

) (

) (

)

( )

( )

2

2

1 2

3

;

1 2

3

;

sup , 1,2.

s s

L R H s

s

L R H H H

R

A i i E A i E A f

A i i E A i E A f s

ξ

ζ ζ ζ ζ

ζ ζ ζ ζ

− −

− −

→ ∈

− − − − +

≤ − − − − + = (12)

For

ζ

R we estimate the following norms:

(

) (

) (

)

( )

(

) (

) (

)

( )

(

)

(

)

(

)

( )

2 2

1 2

3

1 2

3

1 2

2

3 2 2

3 2 0

sup

sup 1 1

1

sup 3 , 1,2.

3 3 1

s s

H H s

s A

s s A

s

s s

A i i E A i E A

i i i

i i i

s s s

σ σ

σ σ

ξ µ

σ

ζ ζ ζ

σ ζ ζ σ ζ σ

ζ ζ

σ ζ

σ σ

µ µ

− −

− −

− −

= ≥

− − − − +

≤ − − − +

   

= − − −  − +

   

≤ = − =

+

(13)

Finally, from (12), we have

(

) (

) (

)

( )

( )

( )

2( )

2

1 2

3

;

; , 1,2.

s s

s L R H

L R H

Aiζ i E Aζ i E Aζ +f ζ a f ξ s=

(14) Lemma. The operator P1 continuously acts from W R H23

(

;

)

to L R H2

(

;

)

provided that the operators s, 1,2

s

A A s= are bounded in H.

Taking into account the results found up [4] to now we get possibility to es-tablish regular solvability conditions of Equation (1).

Theorem 3. Let the operators s, 1,2

s

A A s= be bounded in H and it holds

the inequality 2 ( )3 3

1 1

s

s s H H

s a A A

− −

=

 , where the numbers a ss, =1,2 are de-termined in theorem 2. Then the Equation (1) is regularly solvable.

Proof. By theorem 1, provided that the operator P0 has a bounded inverse

operator 1 0

P acting from

(

)

2 ;

L R H to 3

(

)

2 ;

W R H , then after replacing

( ) ( )

0

p u x =v x in Equation (1) can be written as

(

1

)

( )

( )

1 0

E p p v x+= f x .

(5)

DOI: 10.4236/oalib.1104631 5 Open Access Library Journal

( ) ( )

2 2

1

1 0 L R H; L R H; 1.

p p

→ <

By theorem (2), we have:

( ) ( )

( )

( )

( )

( )

2 2

2

2

2

2 3 2 1

1 0 ; 1 ; 1 3

; 3

2

3

1 ;

2

3 0 ;

1 2

3 ;

1

d d d d

s s

s L R H

L R H s

L R H s

s s

s H H s

s L R H

s

s s H H L R H s

s

s s H H L R H s

u

p p v p u A

x u

A A A

x

a A A p u

a A A v

− −

− =

− −

− →

=

=

=

= ≤

=

Consequently,

( ) ( )

2 2

2 1

1 0 ; ; 3

1 1.

s s s

L R H L R H s H H

p pa A A

→ ≤

= → 

Thus, the operator 1

1 0

E p p+ is invertible in

(

)

2 ;

L R H and hence u x

( )

can be determined by

( )

1

(

1

)

1

( )

0 1 0

u x = p E p p+ − − f x , moreover

( ) ( ) ( )

(

(

)

)

( ) ( ) ( )

( )

3 3

2 2 2 2

2 2

2

1

1 1

0 1 0

; ; ; ;

; ;

; .

W R H L R H W R H L R H L R H L R H

L R H

u p E p p f

const f

− −

≤ +

≤ (15)

The theorem is proved.

3. Conclusion

We formulated exact conditions on regular solvability of Equation (1), expressed only by its operator coefficients. We estimated the norms of intermediate deriv-ative operators participating in the principle part of the given equation. In the case when in the perturbed part of Equation (1), the participant variable opera-tor coefficients, i.e. A x ss

( )

, =1,2 are linear operators, which determined for

all x R∈ , are investigated in a similar way.

References

[1] Hille, E. and Phillips, R. (1962) Functional Analysis and Semi-Groups. IL, Moscow, 829 p. (In Russian)

[2] Lions, J.L. and Majenes, E. (1971) Inhomogeneous Boundary Value Problems and Their Applications. Mir, Moscow, 371 p. (In Russian)

[3] Aliev, A.R. and Elbably, A.L. (2012) On the Solvability in a Weight Space of a Third-Order Operator-Differential Equation with Multiple Characteristic. Doklady Mathematics, 85, 233-235. https://doi.org/10.1134/S106456241201036X

[4] Gasymov, M.G. and Mirzoev, S.S. (1992) On Solvability Boundary Value Problems for Elliptic Type Operator Differential Equations of Second Order. Different Urav-nenie, 28, 651-666. (In Russian)

[5] Mirzoyev, S.S. (2003) On the Norms of Operators of Intermediate Derivatives,

References

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