Gen. Math. Notes, Vol. 3, No. 1, March 2011, pp. 124-129 ISSN 2219-7184; Copyright © ICSRS Publication, 2011 www.i-csrs.org Available free online at http://www.geman.in
On the Generalized Solution for Composite Type Differential Equation Tarig. M. Elzaki Mathematics Department, Faculty of Sciences and Arts-Alkamil, King Abdulaziz University, Jeddah Saudi Arabia Mathematics Department, Sudan University of Science and Technology (www.sustech.edu) E-mail:
[email protected] and
[email protected] (Received: 29-11-10/ Accepted: 30-1-11) Abstract
In this paper, we study a boundary-value problem for a class of composite equation of a mixed–type problem in the space. The existence and uniqueness of the generalized solution is proved, the proof is based on an energy inequality and the density of the range of the operator generated by the problem. Keywords: Energy inequality, operator, generalized solution.
1
Introduction
In the rectangle Ω = (0,1) Χ (0, T ) we consider the boundary-value problem. lu = (
∂ ∂ 2 ∂ 2u ∂ 2u − 2 ) ( 2 − 2 ) = f (t , x) ∂t ∂x ∂t ∂x
The initial conditions: ∂u
∂2 u
u (0, x) = ∂t (0, x) = 2 (0, x) = 0 , ∀ x∈Ω ∂t
(1-1)
On the Generalized Solution…
125
and the boundary conditions:
Where s = ∂Ω , r is the exterior point.
∂ u ∂3u = =0 ∂r ∂r3
on s
Analogous to the problem (1 – 1), we consider its dual problem. We denote by l ∗ the formal dual of the operator l which is defined with respect to the inner product in the space L2 (Q) using
( lu, v ) = ( u, l ∗v ) ,
∀ u , v ∈ C0 3,4 (Q )
(1 – 2)
The dual equations: l *v = ( −
∂ ∂ 2 ∂ 2v ∂ 2v − 2 ) ( 2 − 2 ) = g (t , x) ∂t ∂x ∂t ∂x
With the final conditions:
v(T , x) =
∂v ∂t
(T , x) =
∂2 v ∂t 2
(T , x) = 0 , ∀ x ∈ Ω
(1 – 3)
and the boundary conditions: ∂ v ∂ 3v = =0 ∂r ∂r 3
2
on s
Functional Spaces
The domain
D (l )
of the operator l is D (l ) = H 03,4 (Ω) , the subspace of the
Sobolev space H 3,4 (Ω) ,which consists of all the functions u ∈ H 3,4 (Ω) satisfying the conditions of (1-1). The domain of l * is D (l * ) = H □3,4 (Ω ) which consists of all the functions v ∈ H 3,4 (Ω) satisfying the conditions of (1-3). Let H −3,−4 (Ω ) be the dual space of the space H 3,4 (Ω) . Definition The solution of (1-1) is called the generalized solution of the operational equation (2-1) lu = f , u ∈ D (l ) and the solution of the problem (1 – 3) is called the generalized solution of the operational equation l ∗v = g , v ∈ D(l ∗ ) (2-2) Where l , l * are extension of the operators L, L* Then we obtain:
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Tarig. M. Elzaki
3
(l *v, u ) = (v, Lu ),
∀u ∈ D(l ) , v ∈ H 02,3 (Ω)
(v, lu ) = ( L*v, u ),
∀u ∈ H 02,3 (Ω) v ∈ D(l * )
A Priori Estimates
Theorem For problems (1 – 1) and (1 – 3) we have the following a priori estimates: u v
2,3 2,3
≤ c Lu
−2, −3
≤ c∗ L∗v
−2, −3
, ∀u ∈ D( L)
(3 – 1)
, ∀v ∈ D ( L∗ )
(3 – 2)
Where the constants c > 0 and c∗ > 0 are independent of u and v . Proof. Firstly we prove the inequality (3 –1) for the function u ∈ D (l ) . For u ∈ D (l ) we define the operator: Mu = Φ (t ) ∂∂ t 2u − Φ(t ) 2
∂ 3u ∂x 2 ∂t
Where Φ (t ) = (t − T ) 2
Problem (1-1) can be written in the form: lu =
∂ 3u ∂ 3u ∂ 4u ∂ 4u − − + = f (t , x) ∂t 3 ∂t ∂x 2 ∂x 2∂t 2 ∂x 4
Then we have:
∫ luMudtdx = ∫ fMudtdx
Ω
Ω
We but 2 3 ∂ 3u ∂ 3u 2 ∂ u 2 ∂ u ( t − T ) dtdx , I = ( t − T ) dtdx 2 3 2 3 2 ∫ ∂ t ∂ t ∂ t ∂ x ∂ t Ω Ω
I1 = ∫
Integrating by parts and using the conditions we obtain: I1 = − ∫ (t − T )( Ω
∂ 2u 2 ∂ 2u 2 ∂ 2Ω ∂ 4u 2 ) dtdx , I = − ( ) dtdx − ( t − T ) dtdx 2 ∫Ρ ∂t∂x ∫ ∂u 2 ∂t 2 ∂x 2∂t 2 Ω
Also we but: 2 ∂ 3u ∂ 3u 2 2 ∂ u 2 J1 = ∫ (t − T ) dtdx , J 2 = ∫ (t − T ) ( 2 ) dtdx ∂t ∂x 2 ∂t 2 ∂x ∂t Ω Ω
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127
Integrating J1 by parts and using the conditions (1-1), then we have: ∂ 2u 2 , J1 = ∫ (t − T ) ( ) dtdx ∂x ∂t Ω Also we have: 2 3 ∂ 4u ∂ 4Ω 2 ∂ u 2 ∂ u ( t − T ) dtdx , K = ( t − T ) dtdx 2 2 2 2 2 2 2 ∫ ∂ x ∂ t ∂ t ∂ x ∂ t ∂ x ∂ t Ω Ω Integrating the last integral by parts and using conditions (1-1) we have:
K1 = ∫
2
K 2 = − ∫ (t − T ) ( Ω
∂ 3u 2 ) dtdx ∂x 2 ∂t
Finally we but: 2 ∂ 4u 2 ∂ u ( t − T ) dtdx ∂x 4 ∂t 2 Ω
L1 = ∫
3 ∂ 4u 2 ∂ u ( t − T ) dtdx ∂x 4 ∂x 2 ∂t Ω
, L2 = ∫
Integrating the last integral by parts and using conditions (1-1) we find that: 2 2 2 3 ∂ 2u ∂ 3u 2 ∂ u L1 = − ∫ (t − T ) + dtdx . L2 = ∫ (t − T ) 3 dtdx ∂t ∂x 2 ∂x 2 ∂x Ω Ω
Then we have: 2
2
2
2 3 ∂2u ∂2u ∂2u 2 ∂ u 2 ∂ u lu ,( t − T ) − ( t − T ) = ( T − t ) dxdt + ( T − t ) dxdt + 2 ∫Ρ ∫ ∂x2 dxdt ∂t 2 ∂x2∂t Ω∫ ∂t ∂t∂x Ω 2
2
2
∂2u ∂3u ∂3u +∫ dxdt + ∫ (T − t ) 2 dxdt + ∫ (T − t) 3 dxdt ∂x∂t ∂x ∂t ∂x Ω Ω Ρ
(3-3) We use the following Poincare estimates: ∂u ∫Ω u dxdt ≤ 4T Ω∫ ∂t dtdx , ∀u ∈ D(l ) 2
2
2
2
2
∂ u ∂ 2u dxdt ≤ 4 T ( T − t ) 2 dxdt , ∀u ∈ D(l ) ∫ ∂t ∫ ∂t Ω Ω 2
2
∂ u ∂ 2u dxdt ≤ 4 T ( t − t ) dxdt ∫ ∂x ∫ ∂x∂t Ω Ω
, ∀u ∈ D(l )
(3-4)
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Now apply ε − inequality to the left hand side of (3 – 3) and using inequalities (3 –4) we obtain (3 – 1) for u ∈ D (L ) .
4
Solvability of the Problem (Existence of Solution)
The uniqueness of the solution follows immediately from inequality (3-1). This inequality also ensures the closure of the range set R (L ) of the operator L. We prove the existence of solution or we prove that for all functions f ∈ H −2,−3 (Ω) there is a unique solution of the problem (1-1). To prove that R ( L ) equals the space H
− 2 ,− 3
(Ω ) ,
we obtain the inclusion
R ( L ) ⊆ R ( L) , and R ( L ) = H −2,−3 (Ω) .
Let { f k }k ∈ N be a Cauchy sequence in the space H −2,−3 (Ω ) , which consists
of element of set R (L) . Then it corresponds to a sequence {un }n∈N ⊆ D ( L ) such that: Lu n = f n ,
n∈ N
From the energy inequality (3 – 1) we have un − um ≤ c L ( un − um ) − c Lun − Lum = c f n − f m ≤ ε
Then {u n }n∈N is also a Cauchy sequence in the space H − 2 , − 3 ( Ω ) , and converges to an element u in H 2 ,3 ( Ω ) . Then we have: Lun = f n
, lim Lun = lim f n then Lu = f
and f ∈ R ( L )
this means that f ∈ R ( L ) and R ( L ) ⊆ R ( L) then we have that: R ( L ) = R ( L ) and R ( L ) is closed.
It remains to obtain the density of the set R ( L) in the space H −2,−3 (Ω ) when u ∈ D ( L) . Therefore we establish an equivalent result which amounts to proving that R ⊥ = {0} . Let v ∈ H −2,−3 (Ω) be such that ( L u , v) = 0, ∀u ∈ D( L) , that is ( L*v, u ) = 0
,
∀u ∈ D( L) . By the equality ( L*v, u ) = (v, Lu )
we have (v, Lu ) = 0
,
∀u ∈ D( L), v ∈ H
−2, −3
(Ω ) .
,
∀u ∈ D( L)
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129
From ( L*v, u ) = 0 we have that L*v = 0 and by virtue of the inequality (3-2) we conclude that v = 0 in the space H −2,−3 (Ω ) when u ∈ D ( L) .Then R ( L) is dense in H −2,−3 (Ω ) . The inequality (3-2) can be proved in a similar way by using the operator: M *v = t 2
∂ 2 v 2 ∂ 3v −t ∂t 2 ∂x 2 ∂t
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