EXISTENCE AND UNIQUENESS OF THE STRONG SOLUTION FOR AN NON-LOCAL BOUNDARY VALUE PROBLEM

Sciences

&

Technologie– N°17, Juin (2002), pp. 11-14.

© Université Mentouri, Constantine, Algérie, 2002.

EXISTENCE AND UNIQUENESS OF THE STRONG SOLUTION FOR AN NON-LOCAL BOUNDARY VALUE PROBLEM

Reçu le 27/03/2001 – Accepté le 13/03/2002

Abstract

In this work we are interested in a certain class of equations with operator coefficients and non-local boundary conditions. For the boundary value problem generated by these equations we establish the existence and uniqueness of the strong solution as well as its continuous dependence on the data. We use the method of energy inequality.

Keywords: Operator differential equation, Non-local boundary conditions, Energy inequality.

Résumé

Dans ce travail, on s'intéresse à une certaine classe d'équations à coefficients opérationnelles et des conditions aux limites non locales. Pour le problème aux limites engendré par ces équations, on établie l'existence et l'unicité de la solution forte, ainsi que la dépendance continue par rapport aux données initiales. La méthode utilisée est celle des inégalités énergétiques.

Mots clés: Equation différentielle opérationnelle, Conditions aux limites non-locales, Inégalités énergétiques.

his paper concerns the existence and uniqueness of a strong solution to a boundary value problem. More precisely, we search a function u solution of the problem:

) ( )

1 )(

1 (

) 1 ( )

1 (

2 2 12

2 2 1 2

2 2 1

1 2

t f Au µ

µ sign

t µ u t sign

µ u sign t B

t u u

 

 

  





 







 

 

 

 



 

L

);

(₂

1

0 _{1 1}

1

1u u_t  u_{t T} t

  _  _ 



) (₁

2

0 _{2 2}

2

2u u_t  u_{t T} t

  _  _ 

 where µ =(µ1, µ2) in C².

For  =0, we find the Goursat problem; this type of problem has been studied in [1, 2]. See also the recent results in [3-5]. The present work is an extension of the works [6] and [7], in two directions: first, the studied equation contains operator coefficients, and second, the variable t belongs to a bounded domain in R². We proceed as follows:

We considered the problem as a resolution of the operator equation F

L_ , where the operator Lµ is considered from the Hilbert space Eµ

into the Hilbert space E, which will be defined later. For this operator, we establish an a priori estimate, then we prove the density of the range of this operator in the space E.

POSITION OF THE PROBLEM

Let H be a Hilbert space with norm and inner product respectively denoted by | . | and (.,.). We suppose in the domain D =]0,T1[  ]0,T2[  R² the differential equation:

) ( )

1 )(

1 (

) 1 ( )

1 (

2 2 1 2

2 2 2 1

12 2

1 2

t f Au µ

µ sign

t µ u t sign

µ u sign t B

t u u

 

 

  





 







 

 

 

 



 

L

with the non-local boundary data:

T

A. GUEZANE-LAKOUD F. REBBANI

Département de Mathématiques Faculté des sciences

Université de Annaba 23000 Annaba, Algérie

صخلم فانصأ ضعب ةساردب ثحبلا اذه يف متهن

ريغ ةيدحلا اهطورش و ةرثؤم لماوع تاذ تلادعملا يوقلا لحلا ةيندحو و دوجو ىلع نهربن .ةيلحم تلادعملا هذهب ةدلوملا ةيدحلا ةلأسملل ،ةيلضافتلا

دمتعت .ةيدحلا طورشلاب رمتسملا طابترلاا كلذكو ةيوق اطلا تاحجارتملا ىلع نيهاربلا .

لا ةيحاتفم تاملك طورش .ةرثؤم ةيلضافت تلاداعم :

ةيوق اطلا تاحجارتملا .ةيلحمريغ ةيدح .

(1)

(2)

A. GUEZANE-LAKOUD and F. REBBANI

12 );

(₂

1

0 _{1 1}

1

1u u_t  u_{t T} t

  _  _ 



) (1 2

0 _{2 2}

2

2u ut  ut T t

  _  _ 



the functions u and f take values in H, the operators A and B are linear in H, with domains D(A) and D(B) everywhere denses in H. The parameter µ = (µ1, µ2) C² and satisfies

|µi|  1, i =1,2. For study this problem, we impose the following conditions:

C1: The operator A is self-adjoint and satisfies:

(Av,v)  c0|v|² ,  v D(A).

where c0 is a positive constant not depending on v.

C2: The operator B commutes with A and we have:

Re(Bv,v) > 0 ,  v D(B).

C3: the functions  and  take values in H and satisfy the compatibility conditions:

(0)  µ2(T2) = (0)  µ1(T1).

To study this problem, we introduce the following spaces. We consider on D(A) the hermitian norm |u|1=|Au|.

(D(A); | . |1) becomes a Hilbert space which we denoted by W¹, we then prove that A  L (W¹,H) .

In a similar way, we construct the Hilbert space W^1/2 on D(A^1/2) provided with the norm |u|1/2 =|A^1/2u| and we show that A^1/2 L(W^1/2,H).

We denoted by H¹([0,Ti], W^1/2) the completion of C^([0,Ti], W^1/2) with respect to the norms:

0 2

2 2 1 2 2

1

2 ' dt

T



^_^{ } / ^_^

  



0 1

2 2 1 2 2

1

2 ' dt

T



^_^{ } / ^_^

  



Let H^1,1(D,W) be the completion of the space C^(D,W) with respect to the norm:

















 

 



 





 

D

dt t u

u t

u t

t

u u ₁² .

2 21 2 11 2

21 1 2 2

1 , 1

Eµ designs the completion of the space C^(D,W) with respect to the norm:



 



 



















 

 



 

 



 

  





2 2 1 2 1 1 2

2 2 2

1

2 2 2 2 1

1 sup ( .) ( )

1 1

1 1

min  

 u u .,

µ µ

µ , µ

u _,

D

Let E be the Hilbert space L2(D,H)Hˆ¹((0,T2)W^1/2)

Hˆ¹((0,T1) W^1/2) whose elements F = (f,,) are such that

|| |F| ||² =||f||² +||||₁²+||||₁² is finite

where Hˆ¹((0,T2)W^1/2)Hˆ¹((0,T1)W^1/2) is the closed subspace of the Hilbert space Hˆ¹((0,T2)W^1/2) Hˆ¹((0,T1) W^1/2) whose elements  and  are such that the condition C3 is satisfies. We associate to problem (2)-(3) the operator



L (L,_1,_2)with domain of definition: D(Lµ) = H^1,1(D,W).

Theorem 1. Assume conditions C1 and C2 hold, then we have

u₁²C L_ ² for every uD(L_) (4) where C is a positive constant not depending on u and µ .

Proof . Multiply in H, the expression Lu by Mu = sign(1|µ1|)

t1

u



 + sign(1|µ2|²) . t2

u



 setting:

2 1 1

) ,

( ₁² ₂

2 2 2

1 u , i ,

t sign u

t t

F _/

i i

i 











 





 

 

 

this yields:

2Re(Lu,Mu) 2Re(B Mu, Mu) = t2



 F1(t1,t2) + t1



 F2(t1,t2)

We integrate the equality (5) with respect to (t1,t2) over the domain (0,1)  (0,2); we get:



0¹

 F1(t1,2)dt1 +



0²

 F2(1,t2)dt2 = 2Re

 

0 0^{1 2}

  ((Lu, Mu)  (BMu, Mu))dt +



0¹

 F1(t1,0)dt1 +



0²

 F2(0,t2)dt2 (Eq1) Similarly, integrating the equality (5) over respectively the domains:

(1,T1)  (2,T2) , (0, 1)  (2,T2) and (1,T1)  (0,2) We then get the equalities (Eq 2), (Eq 3) and (Eq 4), which we sum up as follows:

(Eq 1) + 

   2  2²

1 1

4 1

1   (Eq 2) +



 

 1 ₂²

12  (Eq 3) +  1 ₁²

12  (Eq 4)]

We find:











 





 

 





 

 

²

2 2

2 2 1 2 2 2 0 2 1 2 2

12 1 ( )

2 ) 1 ( 2 1

1 F ,t dt ^TF ,t dt



   

 +













 





 





 

 

 

¹

1 1

1 2 1 2 1 1 0

1 2 1 2 1

2 1 ( )

2 ) 1 ( 2 1

1 F t, dt ^TF t, dt



   



 

 

    



²

0

2 2 1 2 2 1 2

2 1 ( , )

21 ) (0,

 F t  F T t dt

 

 

 





 





 

  ²²



^{2 2 2 12 2 1 2 2}

2

) ( 2 1

) 1 0 ( 2 1

1 F t, F T t, dt

T







 

 

    



¹

0

1 2 1 2 1 2 1

1 1 ( , )

12 ) 0 , (

 F t  F t T dt

 

 

 





 





 

  ¹²



^{1 1 1 22 1 1 2 1}

1

) ( 2 1

) 1 0 ( 2 1

1 F t , F t ,T dt

T











 

 



 

 



 

 ^{1 2 12 22}

0 0

1 4 1

)) 1 (

) ((

2 ^{1 2}  

 

dt dt Mu , BMu Mu

, u

Re L





 

 



 

 1 2 1 2²

2 )) 1

( ) ((

2 ¹

1 2

2



 

dt dt Mu , BMu Mu

, u Re

T T

L (3)

(5)

Existence and uniqueness of the strong solution for an non-local boundary value problem.

13





 

 



 

 1 2 1²

0 0

2 1 )) 1

( ) ((

2 ^{1 2} 

 

dt dt Mu , BMu Mu

, u

Re L

2 1 0

1

1 2

)) (

) ((

2Re u,Mu BMu,Mu dtdt

T

 

^





L (6)

In order to majorize the right hand side of the equality (6), we need the following lemma:

Lemma 1. Let g be an arbitrary function g: D  H. We defined by h the function

i

i j ti T

t g

g

h _0 _ , i=1,2;

j=1,2, where µ =(µ1,µ2) is a complex parameter satisfies

|µj|  1. So, we have:

(i) if |µj| < 1, then:

.

| h |µ

| |µ

j j T

j t

t_{i i i}

2 2 2 2

2 2

0 (1 )

) 1

g ( 2 1

g 1



 



 

 



 _  _

(ii) if |µ| > 1 , then:

.

| h

|µ

| |µ

j j T

j t

t_{i i i}

2 2

2 2 2 2

0 ( 1)

) 1

g ( 2 1

g 1



 



 

 

 _

 

For the proof see [6].

Now, we use conditions C2, we then divided the obtained inequality by (1 +|µ1|²)(1 +|µ2|²), we find:

 

 

 

































 





























 













 

 

 













2 1 0 0

2 2 2

2 1 0 0

2 1 1

2 1 0 0 1 2 2 0

11 1 2

2 1 2 1

0

2 2

2 1 2 2 2

2 2 1

22 12

1 2 1 2

1 2 1

2 2

2

1 1

) (

) 1

)(

1 ( 4

1

; 1

dt t dt k u

dt t dt k u

dt dt u k

k dt t u

u

dt t u

u

| |µ

| |µ

| |µ

| |µ min

T T T T

T T T

t /

T

t /

L





) 1 1

min( ₁² ₂²

2 1 2

1 ²

1

|

|µ ,

|

|µ u

u µ

µ





 



(7)

where:

 

2 , ) 1

1 )(

1 ( 8

1

; 1

min

2 3 2 2

1

22 12

 





 



i T ,

| |µ

| |µ

| |µ

| k |µ

i i

Multiply (7) by

 

|

|µ

|

|µ

|

|µ

|

|µ

) 1 )(

1 ( 4

1

; 1 min

22 12

22 12







 , then by using some

other elementary estimates, we find:

 

 













 





 













 





















) ) ( 1 )(

1 ( 4

1

; 1 min

1

2 2

2

1 1

0

1 2

2 1 2 1

0 2 2

2 1 2 2 2

2 2 1

22 12

T

t /

T

t /

dt t u

u

dt t u

u

|

|µ

|

|µ

|

|µ

|

|µ





2 1 2

2 1 1 0 0

2 2

1 _{1 2}

1 2

8 )

(

32T T u dtdt _µu _µ u

T T



 





 

^L

Since the left hand side is not depending on , we take the supremum on   D, we find the inequality (4), this achieves the proof of the theorem.

Proposition 1. The operator Lµ admits a closure denoted

,

L with domain of definition D(L_ )=D(L_).

Proof. See [4]

Definition 1. The solution of the equation L_ u =F, where F = (f,,), is called a strong solution of problem (2)-(3).

By taking the limit we extend inequality (4) to strong

solutions: ^{2 2} ( )

1 Lu , u D L

u   

and we have: R(L_ )=R(L_) and (L_)^-1=(L^_µ¹).

From the inequality (8), we deduce the uniqueness of strong solution and the closure of R(L_ ). For the existence of the strong solution, it remains to prove that the range R(Lµ) is dense in E, which is equivalent to R(Lµ)^ = {(0,0,0)}.

EXISTENCE OF THE STRONG SOLUTION

Let V = (v,1,1)  R(Lµ)^, then for u  D(Lµ ) we have:

0 ) (

) (

) ( )

( _{( ) 11 11}

2 1

2   

 _  _ 

u,V u,v u, u,

L _E L _{L D,H}  

Choosing u such that _1u_2u, the equality (9) becomes:

( ) ( ) _{( )} ( ) 0

2  

 L D,H



D

E u,v u,v dt

V , u

L_ L L (10)

Let  be the solution of the problem:

t v

t 







2 1

2

; 0,

1 1 0 1

1  _ 

 t T

µt  0

2 2 0 2

2  _ 

 t T

µt 

Setting:  =Ag then v = A

2 1

2

t t

g







Replacing v by this expression in (10), integrating by part and using (11) this yields:

' 0

2 1

2 2

1

2  















 























D D

dt g t , t A u t dt

t A g ,

u L

L )dt (12)

where:



 







 

 

 

 



 

2 2 1 2

2 2 1

1 2

) 1 ( )

1

( t

µ g t sign

µ g sign t B

t

g g ^*

L'

Ag µ

µ

sign 

  

 (1 ₁²)(1 ₂²)

We denoted byL'_µthe differential operator generated by the problem:

 















 

 



 

 





 

 



 

 















 

 













 

 



 

0 0 0 0

1 1

1 1

2 2 2

2

1 1 1

1

2 2 2 1

2 2 1 2

2 2 1

1 2

2 1

T gt gt

µ g '

T ; gt gt

µ g '

Ag µ µ

sign

t µ g t sign

µ g

* sign t B t g g







 L'

(11) (9)

A. GUEZANE-LAKOUD and F. REBBANI

14 Since the set













2 1

2

t t

A u is dense in L2(D,H), from the (12), we deduce that L'g = 0. By techniques similar to those used to prove the estimate (4), we prove that:

2 2

1 C' L' g

g  _

So, since L’µ g = 0, we conclude that g = 0, then v = 0.

The equality (9) becomes:

) ( 0 ) (

)

(_1u;₁₁ _2u;₁₁ , uD L_u

As the domain of the operator (_1,_2)is dense in the product space Hˆ¹((0,T2) ,W^1/2)  Hˆ¹((0, T1) ,W^1/2) then 1

= 1 = 0 and consequently R(Lµ) is dense in E.

Theorem 4. Under conditions of theorem 1, the conditions C1 and C2, we have for all f  L2(D,H) ,  Hˆ¹((0,T2) , W^1/2),   Hˆ¹ ((0, T1) ,W^1/2), satisfying (0)  µ2 (T2) =

(0)  µ1(T1), there exists one and only one solution u  Eµ for the problem (2)-(3) and the following inequality holds:

)

( ² ₁² ₁²

2

1 C f     u

where C is not depending on u, f, , , µ1 and µ2.

CONTINUITY OF THE SOLUTION WITH RESPECT TO THE PARAMETERS

Let E¹ be the completion of the space C^(D,W¹) with respect to the norm:























 



 























 



 







 

2

1 1 1

2 2 1

0 2 2

2 / 1 2 2

0 2 1

2 / 1 2 1

2 sup

T

t T

D t E

dt t u

u

dt t u

u u



 

Theorem 5. If the conditions of theorem 4 hold, then for the convergent sequence



,



; 1 1,2

) ,

(µ_1n µ_2n  _n_ ₁ ₂ _i  i we have:

 

¹

) 1

(Lµ_n ^  _n_ L ^

in L(E,E¹) provided with the topology of simple convergence.

Proof. Since the constant in the inequality (8) is not depending on µ and the norm in the spaceE_nis minored by the norm of E¹ with a constant not depending on µn, then we get:

)

2 (

2

1 µn µn

E C L u u D L

u    (13)

where C is not depending on u and µn.

From this, we deduce that

) ( 1

) 1

(Lµn E,E

sup L

 is finite.

SinceR(L_) is dense in E, it suffices to establish that:

 

¹

) 1

(Lµ_n ^  _n_ L ^ in R(L_). For all F R(L_), we have:

 

^{( )}

)

(Lµn ^1F L ^1FD Lµ_n From (13), we deduce that:

 

^{1 2}

 

^{1 2}

1

) 1

(L F L F C F L L F

µn

n E

µ         (14)

and if (L_)^1Fh,then for sufficiently great n and for h

 H^1,1(D,W), we have:

2 0

2 2 1 2 2 2 2

2

1 0

2 2 1 2 1 1 2

1 2

1 1 2

2 2 1

dt t h

h

dt t h

h h L h L

t T

n /

t T

n / µ_n

































 







 











 









This, together with the inequality (14), implies:

 

⁰

) (

1 2 1

1

 ^

 n E

µ F L F

L  when(µ_1n,µ_2n) (₁,₂) for each F R(L_).This completes the proof.

REFERENCES

[1]- N. Brich N.I., Yurchuk N.I., "Goursat problem for a second order abstract linear differential equation''. Differ. Uravn., T.7, N° 6, (1971), pp. 1017-1030 (Russian).

[2]- Guezane-Lakoud A., Rebbani F., Yurchuk N.I., ''Problème aux limites pour une équation différentielle opérationnelle de second ordre'', Maghreb. Mathematical Review, Vol.6, N°1, (1997), pp. 39-48 (French).

[3]- Guezane-Lakoud A., Rebbani F., "A priori estimate for hyperbolic abstract equation with non local boundary conditions", Mathematical Maghreb Review, Vol.8, N°1&2, (1999), pp.111-118.

[4]- Guezane-Lakoud A., Rebbani F., "Strong solution for an abstract non local boundary value problem", Int. J. of Applied Math., Vol. 4, N°4, (2000), pp. 469-478.

[5]- Guezane-Lakoud A., "On a class of hyperbolic equation with abstract non-local boundary conditions", Int. J. of Applied Sc.

Computations, Vol.8, N°1, (2001).

[6]- Chesalyn V.I., Yurchuk N.I., ''Problem with non-local bundary conditions for an Liav's abstract equation'', I. zv.

Akad. Nauk. BSSR. Serie Phys. Mat., N° 6, (1973), pp. 30-35, (Russian).

[7]- Engelman U., "Boundary value problem for an operator hyperbolic partial differential equation of second order with non-local boundary conditions'', Differ. Uravn., T.11, N°9, (1975), pp. 1687-1963 (Russian).

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