Vol 3, No 1 (2012)

(1)

SEPARATION OF THE GENERAL TRICOMI DIFFERENTIAL OPERATOR

IN HILBERT SPACE AND ITS APPLICATION

Saleh Omran*, Khaled Gepreel, and Sayed Khalil

1

Math. Depart., Faculty of Science, Taif University, Kingdom of Saudi Arabia. ²Math. Depart., Faculty of Science, South valley University, Egypt.

³Math. Depart., Faculty of Science, Zagazig University, Egypt. Math. Depart., Faculty of Science, Minofia University, Egypt.

E-mail: [email protected], [email protected]

(Received on: 17-10-11; Accepted on: 01-11-11)

________________________________________________________________________________________________

ABSTAACT

In this article, we separate the differential operator A of the form

Au

(

x,y

)

= Pu(x,y)+Vu(x,y) for all x y, ∈R, in the Hilbert spaceH =L₂

(

R2,H₁

)

with the operator potential V

(

x,y

)

, where L H( ₁)is the space of all bounded

operators on an arbitrary Hilbert spaceH and1 =− − ∂

∂ ∂

∂

2 2 4

4 2 2 2 1

y x x

P is the general tricomi operator onR . 2

Moreover, we study the existence and uniqueness of the solution of the differential equation

Au= Pu+Vu=f,in the Hilbert space H , wheref∈H , as an application of this separation.

Keywords: Separation; Tricomi differential operator; Hilbert space; Coercive estimate.

________________________________________________________________________________________________

1. INTRODUCTION:

The concept of separation for differential operators was first introduced by Everitt and Giertz

[

6, 7 .

]

They have obtained

the separation results for the Sturm Liouville differential operator

( )

x y

( )

x V

( ) ( )

x y x, x R,

Ay =− ′′ + ∈ (1)

in the spaceL₂

( )

R .They have studied the following question: What are the conditions on V x

( )

such that if

( )

x L

( )

R

y ∈ 2 and Ay

( )

x ∈L2

( )

R imply that both of y′′

( )

x and V

( ) ( )

x y x ∈L2

( )

R?More fundamental results of

separation of differential operators were obtained by Everitt and Giertz [8,9]. A number of results concerning the property referred to the separation of differential operators was discussed by Biomatov [2], Otelbaev [16], Zettle [21] , Mohamed et al [10-15] and Zayed et al [17-19]. The separation for the differential operators with the matrix potential was first studied by Bergbaev [1]. Brown [3] has shown that certain properties of positive solutions of disconjugate second order differential expressions imply the separation. Some separation criteria and inequalities associated with linear second order differential operators have been studied by many authers, see for examples [4, 5, 11, 13, 14, 15]. Recently, Zayed [20] has studied the separation for the following biharmonic differential operator in the Hilbert spaceH. In this article, we study the separation of the differential operator A of the form

(

)

(

)

(

)

V

(

x,y

) (

ux,y

)

. y

y , x u x

x y , x u y

, x

Au +

∂ ∂ + ∂ ∂ −

= ₂

2 4

2 2

4 2

1

(2)

Simply we denote the differential operator A by

(

x,y

)

Pu V

(

x,y

) (

u x,y

)

,

Au = + (3)

(2)

application$ % & ' (

) * & ! * ! + ,

where =− −

∂ ∂ ∂

∂

2 2 4

4 2 2 2 1

y x x

P is the general Tricomi differential operator. We construct the coercive estimate for the

differential operator (2) . The existence and uniqueness theorem for the solution of the differential equation

(

x,y

)

Pu V

(

x,y

) (

u x,y

)

f

(

x,y

)

,

Au = + =

in the Hilbert space H=L₂

(

R2,H₁

)

is also given, where f

(

x,y

)

∈H.

2. SOME NOTATIONS:

In this section we introduce the definitions that will be used in the subsequent section. We consider the Hilbert space

1

H with the norm .₁ and the inner product space u,v₁. we introduce the Hilbert space H =L2

(

Rn,H1

)

of all

functions u(x), x∈Rnequipped with the norm

. dx ) x ( u u

n R

= ₁2

2

(5)

The symbol u,v where u,v∈H denotes the scalar product inH which is defined by

. dx v , u v

,

u _Rn

1

= (6)

Let W22

(

Rn,H1

)

be the space of all functionsu

( )

x , x∈R which have generalized derivatives Dαu

( )

x ,α≤2 such

that u

( )

x and Dαu

( )

x belong toH. We say that the function u

( )

x for allx∈Rnbelongs to W₂2_,_loc

(

Rn,H₁

)

if for all functions Q

( )

x ∈C₀∞

( )

Rn the functions Q

( ) ( )

x y x ∈W22,loc

(

Rn,H1

)

.

3. MAIN RESULTS:

Definition: The operator A of the form Au

(

x,y

)

=−Pu

(

x,y

)

+V

(

x,y

) (

u x,y

)

,x,y∈R is said to be separated in the Hilbert space Hif the following statement holds:

If u(x,y)∈H∩W₂2_,_loc

(

R2,H₁

)

and Au

( )

x ∈Himply that both −Pu

(

x,y

)

and V

(

x,y

) (

ux,y

)

∈H.

Theorem: 1 If the following inequalities hold for all x,y∈R,

, Vu Vu

V x V

V2 1 ₁

1

0 _∂ ≤σ

∂ ₋

(7)

, Vu Vu

V y V x

V 2 2 1 ₂

1

0 _∂ ≤σ

∂ −

−

(8)

where σ1 and σ2are positive constants satisfying σ1+σ2<4 , then the coercive estimate

, Au N y u x V x u V Pu

Vu ≤

∂ ∂ + ∂ ∂ +

+ 2 2

1

0 2

1

0 (9)

is valid, where V₀=Re(V) and

(

)

(

)

(

)

,

N

1

2 1 2

1

2

1

2 1 2

1

1 1

2 1

4 16

1 1 16

8 1

4 16

1 1 4

2 1 4

16 1 1 2 1

− −

−

σ + σ β − σ

β − +

σ + σ β − σ β − + σ + σ β − + =

(3)

) * & ! * ! +

is a constant independent on u(x,y) while is given by

. 2 1 2

1 4

16 4

σ + σ < β < σ + σ

Then the operator Agiven by (4) is separated in the Hilbert spaceH.

Proof: From the definition of the inner product in the Hilbert spaceH, we can obtain

n ,..., , , i , , u v ,

i x

v i

x

u ₌₋ ₌₁₂₃

∂ ∂ ∂

∂ _{for all}_u_,_v_∈_C∞

( )

_Rn_.

0

Hence

2 2

4

2 2

2 4 2

4 4

1

1 ,

,

2

8

1 (

)

1 ,

,

2

8

1

1 ,

,

.

2

8

8 u

u

Au Vu

Vu

x

Vu

Vu Vu

x

y

u

Vu

u

x Vu

Vu Vu

x

y

u

V

u

V

u

x V

x u

Vu Vu

x

y

∂

= −

−

+

∂

=

−

+

∂

=

+

∂

(11)

Equating the real parts of the two sides of (11), we get

, Vu , Vu Vu V y V x V , y u V x Re

y u V x , y u V x Vu V x V V , x u V Re x

u V , x u V Vu , Au Re

+ ∂

∂ ∂

∂ +

∂ ∂ ∂

∂ +

∂ ∂ ∂

∂ +

∂ ∂ ∂ ∂ =

− −

1 2 2 1

0 2 1

0 2

2 1

0 2 2 1

0 2 1

2 1

0 2 1

0 2

1

0 2 1

0

8 1

8 1 2

1

(12)

Since for any complex number Z,_{we have}

−Z ≤ReZ≤ Z, (13)

then by the Cauchy- Schwarz inequality, we obtain

Re Au,Vu ≤ Au,Vu ≤ Au Vu. (14)

With the help of (14) and conditions (7), (8), we have

, Vu x u V Vu

V x V V x u V Vu V x V V , x u V Re

∂ ∂ σ − ≥ ∂

∂ ∂ ∂ − ≥ ∂

∂ ∂

∂ − − 2

1

0 1 1

2 1

0 2 1

0 1

2 1

0 2 1

0 (15)

, Vu y u V x Vu

V y V V x y u V x y u V x , y u V x Re

∂ ∂ σ

− ≥ ∂

∂ ∂

∂ − ≥ ∂ ∂ ∂

∂ − − − 2

1

0 2 2 1

2 1

0 2 2 1

0 2 2

1

0 2 2 1

0

2 ₍₁₆₎

It is easy to see that for any β>0 and y1,y2∈R2, then we have the inequality

. y y y y

β + β ≤

2 2

2 2 2 1 2

(4)

) * & ! * ! + (

Applying (17) to (15) and (16), we have

, Vu x

u V Vu

V x V V , x u V

Re 1 2

2 2 1

0 1 1

2 1

0 2 1

0

2

2 β

σ − ∂ ∂ βσ − ≥ ∂

∂ ∂

∂ − ₍₁₈₎

, Vu y

u V x Vu

V y V V x , y u V x

Re 2 2

2 2 1

0 2 2 1

2 1

0 2 2 1

0 2

2

2 β

σ − ∂ ∂ βσ

− ≥ ∂

∂ ∂

∂ − − ₍₁₉₎

From (13),(14) and (18) - (19), we conclude that

. Au Vu Vu ) (

x u y V y

u V

) σ +σ ≤

β − + ∂ ∂ σ

β

− 1 2 2

2

2 1 2 1

0 2 2

2 1

0 1

2 1 1 2

1 2

1 (20)

Choosing

16 2 1 4σ +σ

, 2 1

4 σ + σ

< β

< we obtain from (20) that

, Au ) (

Vu

1

2 1

4 16

1 1

−

σ + σ β −

≤ (21)

, Au ) (

x u

V 2

1

2 1 2

1

1 2

1

0 ₁₆ 4

1 1 4

2

1 − −

σ + σ β − σ

β − ≤ ∂ ∂

(22)

. Au ) (

y u x

V 2

1

2 1 2

1

2 2

2 1

0 4

16 1 1 16

8

1 − −

σ + σ β − σ

β − ≤ ∂ ∂

(23)

From (3) and (21) we get

. Au ) (

Au Vu Pu

1

2 1

4 16

1 1

−

σ + σ β − ≤ +

≤ (24)

Consequently, we deduce from (21)-(24) that

, Au N x u x V x u V Pu

Vu ≤

∂ ∂ +

+ 2 2

1

0 2

1

0 (25)

where

(

)

(

)

(

)

.

N

1

2 1 2

1

2

1

2 1 2

1

1 1

2 1

16 1 1 16

8 1

16 1 1 4

2 1 4

16 1 1 2 1

− −

−

σ + σ β − σ

β − +

σ + σ β − σ β − + σ + σ β − + =

This completes the proof..

Theorem: 2 If the operator Agiven by (3) is separated in the Hilbert space H and if there are positive functions

) y , x (

ψ and t(x,y)∈C1(R2)satisfying

, V

y t t x , V

x t

t 2 ₂

1

0 1 2 1

2 1

0

1 ₂ _≤₂ _σ

∂ ∂ σ

≤ ∂

∂ − ₋ −

− ₍₂₆₎

where ,

2 2 1

0<σ +σ <β while βis defined in theorem1. Then the differential equationAu=−Pu+Vu= f, for all

H

(5)

) * & ! * ! +

-Proof: First, we prove the differential equation

Au=−Pu+Vu=0, (27)

has the only zero solution u(x,y)=0 for all x,y∈R. To this end, we assume that t(x,y) and ψ(x,y) are

two positive functions belonging to C1(R2).

. y t u x , y u y tu x , y u y u t x , y u x t u , x u x tu , x u x u t , x u u t x , y u x ) u t ( , x u u t , y u x u t , x u u t , y u x x u u t , Pu u t , Vu ∂ ∂ ψ ∂ ∂ − ∂ ψ ∂ ∂ ∂ − ∂ ∂ ψ ∂ ∂ − ∂ ∂ ψ ∂ ∂ − ∂ ψ ∂ ∂ ∂ − ∂ ∂ ψ ∂ ∂ − = ψ ∂ ∂ + ∂ ψ ∂ ∂ ∂ − = ψ ∂ ∂ + ψ ∂ ∂ = ψ ∂ ∂ + ∂ ∂ = ψ = ψ 4 4 4 4 2 2 2 2 4 2 2 2 2 4 2 2 8 1 8 1 8 1 2 1 2 1 2 1 8 1 2 1 8 1 2 1 4 2 1 (28)

Equating the real parts of both sides of (28), we have

. y t u x , y u Re y tu x , y u Re y u x t , y u x t x t u , x u Re x tu , x u Re x u t , x u t u V t , u V t u t , u V ∂ ∂ ψ ∂ ∂ − ∂ ψ ∂ ∂ ∂ − ∂ ∂ ψ ∂ ∂ ψ − ∂ ∂ ψ ∂ ∂ − ∂ ψ ∂ ∂ ∂ − ∂ ∂ ψ ∂ ∂ ψ − = ψ ψ = ψ 4 4 2 2 1 2 1 2 2 1 2 1 2 1 2 1 2 1 2 1 2 1 0 2 1 2 1 2 1 0 2 1 2 1 0 8 1 8 1 8 1 2 1 2 1 2 1 (29)

By using Cauchy- Schwarz inequality, we obtain

, u ) x t x x t ( u ) x t x x t ( , u ) x t x u x tu ( , u x t x u , u x tu , u x u x t , u x t x u , u x u x t , u x tu , u x u x t , u ) x tu ( x , u x u x t , u x tu , x u x tu , x u x tu , x u x tu , x u Re 2 2 1 2 2 2 2 2 2 2 2 2 2 2 ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ − ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ∂ ∂ ψ ∂ − ∂ ∂ ∂ ψ ∂ − ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ ∂ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ ∂ ∂ = ∂ ψ ∂ ∂ ∂ + ∂ ψ ∂ ∂ ∂ = ∂ ψ ∂ ∂ ∂ (30)

On the other hand, we have

(6)

) * & ! * ! + .

, u ) y t y y t ( x u ) y t y y t ( x , u y t y u x y u tx , u y u y tx , u y t y u x , u y u y tx , u y u tx , u y u y tx , u ) y u tx ( y , u y u y tx , u y u tx , y u y u tx , y u y u tx , y u y u tx , y u Re 2 2 1 2 2 2 2 2 4 4 2 2 4 4 4 4 2 2 4 4 4 4 4 4 4 4 2 ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ∂ ∂ ψ ∂ − ∂ ∂ ∂ ψ ∂ − ∂ ψ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ ∂ ∂ − = ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ ∂ ∂ = ∂ ψ ∂ ∂ ∂ + ∂ ψ ∂ ∂ ∂ = ∂ ψ ∂ ∂ ∂ (32) . u V t y u t x u V V y t t x t y u t x u V V y t t x t y u t x u V V y t t x t , y u t x Re u V V y t t t x , y u t x Re y t u x , y u Re 2 2 1 0 2 1 2 1 2 2 2 1 2 1 2 2 2 1 0 2 1 0 1 2 2 1 2 1 2 2 1 2 1 2 2 1 0 2 1 0 1 2 2 1 2 1 2 1 2 1 2 2 1 0 2 1 0 1 2 2 1 2 1 2 1 2 1 2 2 1 0 2 1 0 1 2 1 2 1 2 2 1 2 1 2 4 2 2 2 1 2 ψ β σ − ∂ ∂ ψ β − ≥ ∂ ∂ ψ β − ∂ ∂ ψ β − ≥ ∂ ∂ ψ ∂ ∂ ψ − ≥ ∂ ∂ ψ ∂ ∂ ψ = ∂ ∂ ψ ∂ ∂ ψ = ∂ ∂ ψ ∂ ∂ − − − − − − − − (33)

Putting β=2, in the above inequalities and using Eqs (30)-(33), we have:

[

]

) u .

y t y y t ( x u ) x t x x t ( u V t 2 2 1 2 2 2 2 2 1 2 2 2 2 1 0 2 1 2 1 2 1 2 1 2 1 1 ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ + ∂ ∂ ∂ ψ ∂ + ∂ ψ ∂ ≤ ψ σ − σ

− (34)

Choosing ψ(x,y)=1,t(x,y)=1 for all x,y∈R , then if 0<σ₁+σ₂<1 , we deduce from (34) that

[

1

]

t V u 0.

2 2 1 0 2 1 2 1 2

1−σ ψ ≤

σ

− (35)

From (5) and (35), we deduce that

[

]

t V u dxdy .

R 0 1 0 2 1 2 1 0 2 1 2 1 2 2

1−σ ψ ≤

σ −

< (36)

The inequality (36) holds only for u(x,y)=0. This proves that u(x,y)=0 is the only solution of Eq.(27).

Second, we know that the linear manifold M={f : Au=f for all u∈C₀∞(R2)} is dense everywhere in H.

So there exist a sequence of functions

{ }

ur ∈C0∞(R2) , such that for all f H, we have Aur−f →0, as

.

r

→ ∞

By applying the coercive estimate (25), we find that

, ) u u ( A N y ) u u ( x V x ) u u ( V ) u u ( p ) u u (

V _p _r _p _r p r p r ≤ _p− _r

∂ − ∂ + ∂ − ∂ + − +

− 2 2

1

0 2

1

0 (37)

where u=up−urfor all p,r=1,2,,.... As p→∞,r→∞. It follows that the sequences

{ }{

Vur ,Pur

}

, _∂

(7)

) * & ! * ! + (

and ∂_∂

y r u x V2 2

1

0 converge in H. Then there exist a vector functions w0,w1, w2 and w3 in H such

that V(u_r−w₀),P(u_r−w₁)

x ) w r u ( V ,

∂ −

∂ 2

2 1

0 and y

) w r u ( x V

∂ −

∂ 2

2 2 1

0 are convergent to zero, as r →∞.

This implies that u_r →V−1w₀=u, Pu_r →Pu

y u x

r u

V V

, ∂_∂ → 2 _∂∂ 1

0 2

1

0 and y

u y

r u

x V x

V ∂_∂ → 2 2_∂∂ 1

0 2

2 1

0 asr →∞ . Hence for

any f∈Hthere exist u∈H∩W22,loc(R2,H1), such that Au= f.

Suppose that u is another solution of the equationAu=f , then A(u−u)=0.But Au=0 has only the zero solution,

then u=uand the uniqueness is proved. Hence, the proof of theorem is completed.

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( )

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(8)

) * & ! * ! + (

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