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A study of the multitime evolution equation with
time-nonlocal conditions
A. Benrabah, F. Rebbani and N. Boussetila

Abstract. The aim of this paper is to prove existence, uniqueness, and
continuous dependence upon the data of solutions to the multitime evolution equations with nonlocal initial conditions. The proofs are based on
a priori estimates established in non-classical function spaces and on the
density of the range of the operator generated by the studied problem.
M.S.C. 2010: 35R20, 34B10.
Key words: Multidimensional time problem; nonlocal conditions; a priori estimates;
strong generalized solution.

1

Introduction

The classical time-dependent partial differential equations (PDEs) of mathematical
physics involve evolution in one-dimensional time. Space can be multidimensional, but
time stayed one dimensional until 1932, then the adjective multitime was introduced
for the first time in physics by Dirac (1932) where he considered the multitime wave
functions via m-time evolution equation and later it was used in mathematics.

Multitime evolution equation arise for example in Brownian motion [1], Transport theory (Fokker-plank-type equations ), biology (age structured population dynamics)[18],
wave and maxell’s equations [3], [17], mechanics, physics and cosmology([24], [31]).
The important step in the theoretical study of multidimensional time problems
was made by Friedman and Littman ([13], [20]) where they have proved the existence and uniqueness of the following mixed problem with two-dimensional time
ut1 ,t2 − Lu = F (x, t), u |t1 =0 = u |t2 =0 = u |x∈∂D , where L is a second order elliptic
self-adjoint differential operator. The further development of the theory was elaborated in the series of papers by Brish and Yurchuk ([4], [5], [6]) and Rebbani, Zouyed
and Boussetila ([22], [35], [23]) for the Mixed and Goursat problems, hyperbolic equations and recently by (Rebbani, Zouyed and Boussetila) for the multitime evolution
equations with nonlocal initial conditions, all these works were studied by the energy
inequality method. In [11], Dezin showed for the first time that, for the description
of all solvable extensions of differential operators generated by a general differential

Balkan
Journal of Geometry and Its Applications, Vol.16, No.2, 2011, pp. 13-24.
c Balkan Society of Geometers, Geometry Balkan Press 2011.
°

14

A. Benrabah, F. Rebbani and N. Boussetila


equation with constant coefficients, one should use not only local but also nonlocal
conditions.
The problems with nonlocal conditions in the time variable for some classes
of PDEs depending on one time variable have attracted much interest in recent
years, and have been studied extensively by many authors, see for instance [21],
[25], Fardigola,[12], Chesalin and Yurchuk [7], [8], [9] Gordeziani and Avalishvili [15],
[16], and Shakhmurov and al. [26]. However, the case of multitime equations with
time-nonlocal conditions does not seem to have been widely investigated and few results are available, see, e.g., the articles by the authors Rebbani and al [22, 23, 35].
In the present paper, we consider time-nonlocal problems for a class of multitime
evolution equations, and we will apply the same technic used in [35]. We will prove
the existence and the uniqueness of the strong solution, this results are proved by the
energy inequality method.

2

Assumptions and statement of the problem

Throughout this paper H will represent a complex Hilbert space, endowed with the
inner product (., .) and the norm |.|, and L(H) denote the Banach algebra of bounded
linear operators on H. Let T1 , T2 > 0, Ω = ]0, T1 [ × ]0, T2 [ = Q1 × Q2 be a bounded

rectangle in the plane R2 . We consider the following problem: Given the data f ,ϕ, ψ
and H, find a function u(t1 , t2 ) satisfying the multitime evolution equation
¸
·
∂2u
∂u
∂u
(2.1)
Lu ≡
+ A(t)u = f (t), t ∈ Ω,
+B
+
∂t1 ∂t2
∂t1
∂t2
(2.2)

ℓλ1 u ≡ u |t1 =0 − λ1 u |t1 =T1 = ϕ(t2 ),

t 2 ∈ Q2 ,


ℓλ2 u ≡ u |t2 =0 − λ2 u |t2 =T2 = ψ(t1 ),

t 1 ∈ Q1 ,

where u and f are H-valued functions on Ω, ϕ (resp. ψ) is H-valued function on Q2
(resp. Q1 ) and satisfy the compatibility condition
(2.3)

ϕ(0) − λ2 ϕ(T2 ) = ψ(0) − λ1 ψ(T1 ),

λ1 and λ2 are two complex parameters, A(t) is an unbounded linear operator in H,
with domain of definition D(A) densely defined and independent of t and B ∈ L(H).
We require the following assumptions
1. The operator A(t) is self-adjoint for every t ∈ Ω and verifies
(2.4)

2

(A(t)u, u) ≥ c0 |u| ,


∀u ∈ D(A),

(2.5)

A(0, t2 ) = A(T1 , t2 ),

t2 ∈ Q2 ,

(2.6)

A(t1 , 0) = A(t1 , T2 ),

t1 ∈ Q1 .

where c0 is a positive constant not depending on u and t.
2. λi 6= 0 (i = 1, 2) such that,
(2.7)

αi = |λi |2 exp(3C(T1 + T2 )) < 1,


where C is a positive constant depending on B, A(t) and its derivatives and λ1 , λ2
are tow complex parameters belonging to M, (C, M will be defined later).

15

Multitime evolution equation

3
3.1

Spaces and auxiliary inequalities
Abstract formulation

Let us reformulate problem ((2.1) − (2.2) = P) as the problem of solving the operator
equation
Lu = F = (f, ϕ, ψ),

(3.1)


where L = (L, ℓλ1 , ℓλ2 ) is generated by (P), with domain of definition D(L), the
operator L is considered from the Banach space E into the Hilbert space F, which
will be defined later. For this operator we establish an energy inequality
kukE ≤ kkLukF .

(3.2)

If the operator L is closable then we denote by L the closure of L and by D(L) its
domain.
Definition 3.1. A solution of the abstract equation Lu = F is called a strongly
generalized solution of problem (P).
Inequality (3.2) can be extended to u ∈ D(L), that is,
kukE ≤ kkLukF ,

(3.3)

∀u ∈ D(L).

From this inequality, we obtain the uniqueness of a strong solution, if it exists, and
the equality of the sets R(L) and R(L). Thus, to prove the existence of a strong

solution for any F ∈ F, it remains to prove that the set R(L) is dense in F.

3.2

Function spaces

In this subsection, we introduce and study certain fundamental function spaces. For
this purpose, let us denote by W r = D(Ar (0)), 0 ≤ r ≤ 1, the space W r endowed
with the inner product (x, y)r = (Ar (0)x, Ar (0)x) and the norm |x|r = |Ar (0)x| is a
1
Hilbert space. We show that the operator A(t) (resp. A 2 (t)) is bounded from W 1
1
1
1
(resp. W 2 ) into H, i.e., A(t) (resp. A 2 (t)) ∈ L(W 1 ; H) (resp. L(W 2 ; H)) (see [19]).
Thus, we have the following results
Proposition 3.1. [6] If the function Ω ∋ t 7−→ A(t) ∈ L(W 1 ; H) is continuous with
respect to the topology of L(W 1 ; H), then there exist positive constants c1 and c2 such
that
c1 |u|1 ≤ |A(t)u| ≤ c2 |u|1 ,


(3.4)
(3.5)



1

c1 |u| 21 ≤ |A 2 (t)u| ≤



∀u ∈ W 1 ,

c2 |u| 21 ,

1

∀u ∈ W 2 .


Lemma 3.2. If the function Ω ∋ t 7−→ A(t) ∈ L(W 1 ; H) admits bounded derivatives
with respect to t1 and t2 with respect to the simple topology in L(W 1 ; H), then we
have the estimates
°
°
°
°
°
° ∂A(t) 12
°
° ∂A(t)
°
− 12 °
−1 °
°
≤ δ°
, (i = 1, 2),
(3.6)
A(t) °
A(t) °

°
°
° ∂ti
∂ti
L(H)
L(H)

16

A. Benrabah, F. Rebbani and N. Boussetila

where δ =

R∞
0



s
ds. (see [19], Lemma 1.9, p. 186).
(1 + s)2
1

Proposition 3.3. The operators
1

i.e.,

1
∂A(t) 2
∂A(t)
A(t)−1 ,
A(t)− 2 are uniformly bounded,
∂ti
∂ti

1
∂A(t) 2
∂A(t)
A(t)−1 ,
A(t)− 2 ∈ L∞ (Ω; L(H)), (i = 1, 2).
∂ti
∂ti

To show the estimate (3.2) we introduce the following spaces
H 1,1 (Ω; W 1 ) is the space obtained by completing C ∞ (Ω; W 1 ) with respect to the norm
!
¯
¯
¯
Z ï 2 ¯ 2 ¯
¯ ∂u ¯2 ¯ ∂u ¯2
¯ ∂ u ¯
2
2
¯
¯
¯
¯
¯
¯
kuk1,1 =
¯ ∂t1 ∂t2 ¯ + ¯ ∂t1 ¯ + ¯ ∂t2 ¯ + |u|1 dt.
1
1
1


1
2

1

Let H 1 (Qi ; W ) be the obtained space by completing C ∞ (Qi ; W 2 ), (i = 1, 2) with
µ
µ


T
T
R1 ¯¯ ′ ¯¯2
R2 ¯¯ ′ ¯¯2
2
2
2
2
respect to the norms kϕk1 =
¯ψ ¯ + |ψ| 1 dt1 .
¯ϕ ¯ + |ϕ| 1 dt2 , kψk1 =
2

0



2

0

1

By completing C (Ω; W ) with respect to the norm
¡
¢
kuk2E = (J(λ))2 sup ku(τ1 , .)k21 + ku(., τ2 )k21 ,
τ ∈Ω

where J(λ) =

(1−α1 )(1−α2 )
(1+α1 )(1+α2 ) ,

we obtain the space E.
1

1

Denoting by F the Hilbert space L2 (Ω; H) × V 1 (Q2 ; W 2 ) × V 1 (Q1 ; W 2 ), consisting
of vector-valued functions F = (f, ϕ, ψ) for which the norm kFk2F = kf k2 + kϕk21 +
1
1
1
kψk21 , is finite. V 1 (Q2 ; W 2 ) × V 1 (Q1 ; W 2 ) is the closed subspace of H 1 (Q2 ; W 2 ) ×
1
H 1 (Q1 ; W 2 ) composed of elements (ϕ, ψ) satisfying (2.3).
To prove the existence of the strong generalized solution we need the following
Hilbert structure.
Let H 1,1 (Ω; H) be the Hilbert
space obtained by completion!of C ∞ (Ω; H) with respect
Ã
¯
¯
R ¯¯ ∂ 2 u ¯¯2 ¯¯ ∂u ¯¯2 ¯ ∂u ¯2
2
¯
¯
+
+
to the norm kuk21,1 =
dt.
¯ ∂t1 ¯
¯ ∂t1 ∂t2 ¯
¯ ∂t2 ¯ + |u|

Let H 1 (Q2 ; H) be the Hilbert space obtained by completion of the space C ∞ (Q2 ; H)
with respect to the norm kϕk21 = kϕk2 + kϕ′ k2 .
We construct H 1 (Q1 ; H) in a similar manner.
Denoting by E the Hilbert space L2 (Ω; H) × V 1 (Q2 ; H) × V 1 (Q1 ; H) composed of
elements F = (f, ϕ, ψ) such that the norm kFk2E = kf k2 +kϕk21 +kψk21 is finite, where
V 1 (Q2 ; H) × V 1 (Q1 ; H) is the closed subspace of H 1 (Q2 ; H) × H 1 (Q1 ; H) composed
of elements (ϕ, ψ)n such that λ2 ϕ(0) − ϕ(T2 ) = λ1 ψ(0) − o
ψ(T1 ).
1,1
1
1,1
1
is the closed subspace of
H0 (Ω; W ) = u ∈ H (Ω; W ) : ℓλ1 u = 0, ℓλ2 u = 0
H 1,1 (Ω; W 1 ). n
o
is the closed subspace of
H01,1 (Ω; H) = u ∈ H 1,1 (Ω; H) : ℓλ1 u = 0, ℓλ2 u = 0
H 1,1 (Ω; H)

o
n
1,1
H 0 (Ω; H) = u ∈ H 1,1 (Ω; H) : λ1 u |t1 =0 −u |t1 =T1 = 0, λ2 u |t2 =0 −u |t2 =T2 = 0 ,

17

Multitime evolution equation
is the closed subspace of H 1,1 (Ω, H). We further denote
©
ª
M = (λ1 , λ2 ) ∈ C2 : λi 6= 0 and αi < 1, (i = 1, 2) .

4

Uniqueness and continuous dependence

We are now in a position to state and to prove the main theorem of this section
for the operator L = (L, ℓλ1 , ℓλ2 ) acting from E into F with domain of definition
D(L) = H 1,1 (Ω; W 1 ) ⊂ E, from which we conclude the uniqueness and continuous
dependence of the solution with respect to the data.
Theorem 4.1. Let the function Ω ∋ t 7−→ A(t) ∈ L(W 1 ; H) have bounded derivatives
with respect to t1 and t2 with respect to the simple convergence topology of L(W 1 ; H)
and the conditions (2.4), (2.5), (2.6) and (2.7) be fulfilled. Then we have
(4.1)

kuk2E ≤ SkLuk2F ,

∀u ∈ H 1,1 (Ω; W 1 ),

where S is a positive constant independent of λ1 , λ2 and u.
Lemma 4.2. (generalized Gronwall’s lemma)
(GV1) Let v(t1 , t2 ) and F (t1 , t2 ) be two non negative integrable functions on Ω such
that the function F (t1 , t2 ) is non-decreasing
with respect to the variables
t1 and t2 .
¾
½ t1
Rt2
R
v(τ1 , t2 ) dτ1 + v(t1 , τ2 ) dτ2 + F (t1 , t2 ) ,
Then the inequality v(t1 , t2 ) ≤ c3
0
0
³
´
(c3 ≥ 0), gives v(t1 , t2 ) ≤ exp 2c3 (t1 + t2 ) F (t1 , t2 ).
(GV2) Let v(t1 , t2 ) and G(t1 , t2 ) be two non negative integrable functions on Ω such
that the function G(t1 , t2 ) is non-increasing
with respect to the variables
t1 and t2 .
)
(
T
T
2
1
R
R
v(τ1 , t2 ) dτ1 + v(t1 , τ2 ) dτ2 + G (t1 , t2 ) ,
Then the inequality v(t1 , t2 ) ≤ c4
t2
³ t1
´
(c4 ≥ 0), yields v(t1 , t2 ) ≤ exp 2c4 (T1 + T2 − t1 − t2 ) G(t1 , t2 ).
Proof. see [35].

¤

1
, 1), g be a function of variable
2
m
t∈ [0, T ] in W , and let hi = g(0) − λi g(T ), (i = 1, 2). Then, if the condition (2.7)
(1+αi )
holds, we have θ3 |g(0)|2m − 12 (1 + αi )|g(T )|2m ≤ θ3 (1−α
|hi |2m , θ3 = |λαii|2 (i = 1, 2).
i)
Lemma 4.3. Let |.|m be the norm in W m (m =

Proof. It sufficient to use the ε inequality with ε =

(1−αi )
2αi , (i

= 1, 2) .

¤

Lemma 4.4. [The method of continuity]
Let X1 , X2 be two Banach spaces and L0 , L1 be bounded operators from X1 into X2 .
For each r ∈ [0, 1], set Lr = (1 − r)L0 + rL1 and suppose that there is a constant k
such that kukX1 ≤ kkLr ukX2 for r ∈ [0, 1]. Then L1 maps X1 onto X2 if and only if
L0 maps X1 onto X2 . (see [14], Th. 5.2, p. 75).

We also need the ε-inequality: 2(a, b) ≤ ε|a|2 + ε−1 |b|2 , ε > 0. Let us return now to
the demonstration of the theorem 4.1.

18

A. Benrabah, F. Rebbani and N. Boussetila

Proof. The proof is based on detailed analysis of the forms

R

Ωτ

2Re(Lu, Mu) dt1 dt2 ,

∂u
∂u
+
and Ω ⊃ Ωτ = (0, τ1 ) × (0, τ2 ), (0, τ1 ) × (0, T2 ), (0, T1 ) ×
∂t1
∂t2
(0, τ2 ), (τ1 , T1 ) × (τ2 , T2 ) and by making use some technical elementary estimates,
propositions (3.1, 3.3) and lemmas (3.6, 4.2, 4.3) we get kuk2E ≤ SkLuk2F , ∀u ∈ D(L).
¤
where Mu =

It follows from estimation (4.1), that there is a bounded inverse operator L−1 on
the range R(L) of L. However, since we have no information concerning R(L), except
that R(L) ⊂ F, we must extend L so that the estimation (4.1), holds for the extension
and its range is the whole space. We first show that L : E −→ F, with the domain
D(L), has a closure.
Proposition 4.5. If the conditions of theorem 4.1 are satisfied, then the operator L
admits a closure L with domain of definition denoted by D(L).
The solution of the equation
Lu = F,

(4.2)

F ∈ F,

is called a strong generalized solution of problem (P). Passing to the limit, we extend
the inequality (4.1) to the strong generalized solution, we obtain
(4.3)
from which we deduce

° °2
2
kukE ≤ S °Lu°F ,

∀u ∈ D(L),

Corollary 4.6. From the inequality (4.3) we deduce that, if the strong generalized solution exists, then this solution is unique and it depends continuously on F = (f, ϕ, ψ).
Corollary 4.7. The set of values R(L) of the operator L is equal to the closure R(L)
of R(L) and (L)−1 = L−1 .
This corollary allows us to claim that, to establish the existence of the strong generalized solution to problem (P) it suffices to prove the density of the set R(L) in
F.

5

Existence of a solution

To show the existence, we need the following condition
Condition (H)

Ω ∋ t 7−→ A(t) ∈ L(Ω; W 1 ) admits mixed derivatives

∂2A
∂A
∂A
∂2A
,
with
A−1 ,
A−1 ∈ L2 (Ω; L(H)).
∂t1 ∂t2 ∂t2 ∂t1
∂t1 ∂t2
∂t2 ∂t1
We are now, in a position to state and to prove the main result of this paper i.e., establish the density of R(L) in F, who is equivalent to show that, R(L)⊥ = {(0, 0, 0)} for
this purpose, we meet some difficulties (derivation), and to surmount these difficulties
we introduces the regularization operators, so we use the regularization technique.

19

Multitime evolution equation

−1
Definition 5.1. We put Aε (t) = (I + εA(t)), Jε (t) = A−1
,
ε (t) = (I + εA(t))
1
−1
Rε (t) = A(t)(I + εA(t)) = ε (I − Jε (t)), ε > 0, and call Rε (t) the Yosida approximation of A(t).

Theorem 5.1. Under the conditions of the Theorem (4.1) and the condition (H),
the set R(L) is dense in F.
Proof. We use the method of continuity given in the book [14]. We introduce the
family of operators Lω = (Lω , ℓλ1 , ℓλ2 ), ω ∈ [0, 1], where
¸
·
∂2


Lω =
.
+ ωB + A(t) = (1 − ω)L0 + ωL1 , B = B
+
∂t1 ∂t2
∂t1
∂t2
Let’s start with showing the result (i.e., R(L0 )⊥ = {(0, 0, 0)}) in the case ω = 0, and
by means of the method of continuity, we establish the general case.
ω = 0. Let V = (v, v1 , v2 ) be an orthogonal element to R(L0 ). Then

First step
we have
(5.1)

hL0 u, V iF = hL0 u, vi + hℓλ1 u, v1 i + hℓλ2 u, v2 i = 0,

∀u ∈ H 1,1 (Ω, W 1 ).

Let’s show that V = (0, 0, 0), but ℓλ1 , ℓλ2 are independent and their ranges are dense,
then it is sufficient to prove the following proposition
Proposition 5.2. If for every v ∈ L2 (Ω; H) we have
¿ 2
À
∂ u
(5.2)
hL0 u, vi =
+ A(t)u, v = 0, ∀u ∈ H01,1 (Ω; W 1 ), then v = 0.
∂t1 ∂2
Proof. Let w = A−1
ε v and h = Aε u, then the relation (5.2) becomes

(5.3)

¿



∂2h


(B ∗ h) −
(B ∗ h) + B3ε
h, w
∂t1 ∂t2
∂t2 1ε
∂t1 2ε

À

= −hh, Awi,

∂A −1


Aε −
∈ L(H), (i = 1, 2, 3) are given by B1ε
= ε ∂t
h ∈ H01,1 (Ω; H) and Biε
1
2

∂A −1
A
∂A −1
∂A −1


B,
B2ε
= ε ∂t
= ε ∂t∂1 ∂t
Aε − B,B3ε
A−1
ε − εB ∂t1 Aε − εB ∂t2 Aε , (∗) denotes
2
2
the symbol of the adjoint.
Since the equation (5.3) is true for all function h ∈ H01,1 (Ω; H) , it remains true for
h ∈ C0∞ (Ω; H), what gives to the sense of distributions
À
¿
∂w
∂w
∂2w
+ B1ε
+ B2ε
= −hh, (A + B3ε )wi, ∀h ∈ C0∞ (Ω; H).
(5.4)
h,
∂t1 ∂t2
∂t2
∂t1 D′

We define the following operators
(5.5)

(5.6)

2
e = H 1,1 (Ω; H), Lu
e = ∂ u + B2ε ∂u + B1ε ∂u ,
D(L)
0
∂t1 ∂t2
∂t1
∂t2


∂2u

D(Le′ ) = H01,1 (Ω; H), Le′ u =

(B ∗ u) −
(B ∗ u).
∂t1 ∂t2
∂t1 2ε
∂t2 1ε

20

A. Benrabah, F. Rebbani and N. Boussetila

e ∗ . Let’s come back to the
According to (5.3) and (5.4), we can show that Le′ = (L)
equation (5.4), we have, for each ε 6= 0, w is the weak solution to the problem

2

e ≡ ∂ w + B1ε ∂w + B2ε ∂w = −(B3ε A−1 ε + AA−1 ε)v,

 Lw

∂t1 ∂t2
∂t2
∂t1

(5.7)
e
ℓλ1 w ≡ λ1 w |t1 =0 − w |t1 =T1 = 0,




 e
ℓλ2 w ≡ λ2 w |t2 =0 − w |t2 =T2 = 0,

where v ∈ L2 (Ω; H), Bjε ∈ L(H), (j = 1, 2, 3).
We are going to show that, w is a solution in the strong sense of the problem (5.7)
and that it verifies an a priori estimate, then we shows that v = 0.
e = (L,
ee
To establish these results, we can show that the operator L
l1µ , e
l2µ ) acting from
1,1
H (Ω; H) into E is isomorphism
e is isomorphism from H 1,1 (Ω; H) into E.
Proposition 5.3. The operator L
e = E and
Proof. We must show that R(L)
(5.8)

(i)

(5.9)

(ii)

2

e 2E ≤ d1 kuk ,
kLuk
1,1

2
e 2E ,
kuk1,1 ≤ d2 kLuk

∀u ∈ H 1,1 (Ω; H),
∀u ∈ H 1,1 (Ω; H),

where d1 and d2 are positive constants independent of u.
(i) It is easy to show
(5.10)

2

e 2 ≤ 4 max(1, C 2 ) kuk ,
kLuk
1,1

∀u ∈ H 1,1 (Ω; H).

By virtue of the continuity of the operators ℓeλ1 , ℓeλ2 from H 1,1 (Ω; H) into H 1 (Q2 ; H),
H 1 (Q1 ; H) respectively and the inequality (5.10), we obtain the estimate (i).
(ii) We use the same techniques to those used to establish the estimate (4.1) in
Theorem (4.1), then we establish the estimate (5.9).
e and the inequality (5.10), we conclude that
From the continuity of the operator L
e is an isomorphism from H 1,1 (Ω; H) into the closed subspace R(L)
e =
the operator L
¡ 1,1
¢
e H (Ω; H) .
L
e = E, for this purpose, we introduce the family of
It remainsnto oshow that R(L)

operators L
defined by
η∈[0, 1]

(5.11)


e η = (Leη , ℓeλ , ℓeλ ), η ∈ [0, 1] ,

L

2
1



2
∂ u
∂u
∂u
Leη u =
+ ηBε u , with Bε u = B2ε
+ B1ε
,

∂t
∂t
∂t
∂t

1
2
1
2



e η ) = H 1,1 (Ω, H).
D(L

e 1 ) = R(L)
e = E. This
We proceed by the method of continuity, we can show that R(L
proves the proposition (5.3).
¤

Multitime evolution equation

21

e = Le is closed
Proposition 5.4. The operator L

¤

Proof. The proof is similar to the proof of Proposition 4.6 in [35].
From the properties of the operators with closed range, it follows
e ′ ) = R(L)
e ⊥ = L2 (Ω; H)⊥ = {0} ,
N (L

e ′ ) = N (L)
e ⊥ = {0}⊥ = L2 (Ω; H).
e ) = R(L
R(L

e is an isomorphism from H 1,1 (Ω; H) into L2 (Ω; H) and it is closed in the
Hence L
0
topology of L2 (Ω; H).
³ ´∗
Definition 5.2. We Denote by L̂ = Le′ the weak extension of the operator Le
defined by
E D
E
D
(5.12) Le′ u, v = u, L̂ v = hu, f i , ∀u ∈ H01,1 (Ω, H) and L̂v = f ∈ L2 (Ω, H)


Proposition 5.5. The weak extension L̂ of the operator Le coincides with its strong
³ ´′

extension L̂ = Le .
Proof. see[35].

¤

From the proposition (5.5), we deduce that the weak solution to problem (5.7)
coincides with its strong solution. Hence w ∈ H 1,1 (Ω; H) ∩ L2 (Ω, W 1 ) and satisfies
the problem (5.7) in the strong sense, i.e.,

1,1

 D(L) = H 0 (Ω; H),
(5.13)
2

 Lw = ∂ w + B2ε ∂w + B1ε ∂w + Aw = −B3ε w = f.
∂t1 ∂t2
∂t1
∂t2

By a similar calculations to those used to establish theorem (4.1), we show

Proposition 5.6. Under the assumptions of the theorem (4.1) we have the estimate
1

kA 2 wk2 ≤ d6 kB3ε wk2 ,

(5.14)

from (5.14) and (2.4) it follows kwk2 ≤
A−1
ε v in the last inequality, we obtain

2
kA−1
ε vk ≤

(5.15)
We have
∗ ∗ −1
(B3ε
) Aε v


+
+

1,1

∀w ∈ H 0 (Ω; H),

1
d6
1
kA 2 wk2 ≤
kB3ε wk2 , replacing w by
c0
c0

d6
2
kB3ε A−1
ε vk .
c0

° °
¶∗
¶∗ °
µ 2
µ 2

°¡
°
¢° °
¢
∂ A −1 ¡ −1
∂ A −1
−1
° I − A−1
°
°
A
Aε v − v ° + °(I − Aε )
A

ε
°
°
∂t2 ∂t1
∂t1 ∂t2
°
°
µ 2
¶∗
1
°¡
¢
¢°
∂ A −1 ¡ −1
2
° I − A−1
A
Aε v − v °
2 kBkL(H)
ε
°
°
∂t2 ∂t1
° °
°o
°
°
°
∂A −1 ∗ °
∂A −1 ∗ °
−1 ∗

°
°
°(I − A−1
A ) v ° + °(I − Aε ) (B
A ) v°
ε ) (B
° → 0, ε → 0.
°
∂t1
∂t2

22

A. Benrabah, F. Rebbani and N. Boussetila

While taking account of the last inequality, and while passing to the limit in (5.15),
when ε −→ 0 and applying the properties of A−1
ε , we obtain v = 0. This completes
the proof of proposition (5.2).
¤
Let us go back now to (5.1), by virtue of proposition (5.2), we obtain hℓλ1 u, v1 i0 +
hℓλ2 u, v2 i0 = 0. Since ℓλ1 , ℓλ2 are independent and the ranges of the operators ℓλ1 , ℓλ2
are dense in the corresponding spaces, we obtain v1 = v2 = 0. Hence V = (0, 0, 0),
therefor R(Lω ) = F for ω = 0.
Second step ω 6= 0. We need the following lemma
Lemma 5.7. The operator (L1 − L0 ) is bounded, and we have
(5.16)

k(L1 − L0 )ukE ≤ kkukE ,

where the constant k does not depend on u .
Proof. The equation Lω u = F can be written as
u + (ω − ω0 )(Lω0 )−1 (L1 − L0 )u = (Lω0 )−1 F.

From (4.3) and (5.16)√
we have k(Lω0 )−1 FkE ≤ SkFkE , and k(Lω0 )−1 (L1 − L0 )ukE ≤
1
, putting Λ = (ω −ω0 )(Lω0 ) (L1 − L0 )
mkukE , where m = k S. Let |ω −ω0 | ≤ ρ < m
−1
and N = (Lω0 ) F, (5.17) can be written as u + Λu = N.
P∞
||Λu||1
Observing that ||Λ|| = sup
< 1. The Neumann series u = n=0 (−Λ)n N
||u||
1
u∈D(Lλ )
(5.17)

is then a solution to equation (5.17). We have thus proved that if R(Lω0 ) = F and
1
|ω − ω0 | ≤ ρ < , then R(Lω ) = F. Proceeding step by step in this way we establish
m
that R(Lω ) = F for every ω ∈ [0, 1]. For the case ω = 1, we have R(L) = F. The
proof of theorem (5.1) is achieved.
¤

Theorem 5.8. For every element F = (f, ϕ, ψ) ∈ F there exists a unique strong
generalized solution u = (L)−1 F = (L−1 )F to problem (P) satisfying the estimate
kuk2E ≤ SkLuk2F ,

∀u ∈ H 1,1 (Ω; W 1 ),

where S is a positive constant independent of λ1 , λ2 and u.

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Author’s address:
Abderafik Benrabah, Nadjib Boussetila
Department of Mathematics, 08 Mai 1945 - Guelma University,
P.O.Box.401, Guelma, 24000, Algeria.
Applied Math Lab, Badji Mokhtar-Annaba University,
P.O.Box. 12, Annaba, 23000, Algeria.
E-mail: [email protected] , [email protected]
Faouzia Rebbani
Applied Math Lab, Badji Mokhtar-Annaba University,
P.O.Box. 12, Annaba, 23000, Algeria.
E-mail: [email protected]