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Existence et comportement asymptotique des solutions d'une équation de viscoélasticité non linéaire de type hyperbolique

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par Khaled ZENNIR
Université Badji Mokhtar Algérie - Magister en Mathématiques 2009
  

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1.2 Functional Spaces

1.2.1 The LP (Q) spaces

Definition 1.2.1 Let 1 < p < oo, and let Q be an open domain in Rn, n 2 N. Define the standard Lebesgue space LP(Q), by

LP(Q) =

8

<

:

f : Q -- >R : f is measurable and I

n

If(x)1P dx < 1

9

=

;

· (1.8)

Notation 1.2.1 For p 2 R and 1 < p < oo, denote by

1

P


·

(1.9)

0 Z

kfkp = @ 1f(x)r dx
n

If p = oo, we have

L°°(Q) = If : Q -- R : f is measurable and there exists a constant C

(1.10)

such that, 1f(x)1 < C a.e in Q1.

Also, we denote by

11/1100 = Inf {C, If(x)1 < C a.e in Q} . (1.11)

+

1
q

=1.

1

Notation 1.2.2 Let 1 ~ p ~ 1; we denote by q the conjugate of p i.e.

P

Theorem 1.2.1 ([48])

It is well known that LP(Q) supplied with the norm 11.11p is a Banach space, for all 1 < p < 00. Remark 1.2.1 In particularly, when p = 2, L2 (Q) equipped with the inner product

(f, 942(n) = f

n f(x)g(x)dx, (1.12)

is a Hilbert space.

Theorem 1.2.2 ([43], Corollary 3.2)

For 1 < p < oo, LP(Q) is reflexive space.

Some integral inequalities

We will give here some important integral inequalities. These inequalities play an important role in applied mathematics and also, it is very useful in our next chapters.

Theorem 1.2.3 ([48], Holder's inequality )

Let 1 < p < oo. Assume that f 2 LP(Q) and g 2 Lq(Q), then, fg 2 L1(Q) and

 

fgl dx < 11/4 mgmq (1.13)

Corollary 1.2.1 (Holder's inequality - general form)

Lemma 1.2.1 Let f1, 12, ...fk be k functions such that, fi 2 LA(Q), 1 < i < k, and

1

1
p

=

+

P1

1

+ ::: +

P2

1 < 1.

Pk

Then, the product f1f2
·
·
·fk 2 LP(1) and If1f2
·
·:fklp ~ kf1kp1
·
·
·11fklIpk
·

Lemma 1.2.2 ( [48], Young's inequality)

r

=

1
p

+

1
q

-- 1 > O. Then

Let f 2 LP(I18) and g 2 Lq(118) with 1 < p < oo, 1 < q < 1 and 1 f * g 2 UM and

MI * ql1L7-(R) < MIlli,p(R)11q1lifl(R)- (1.14)

Lemma 1.2.3 ([43], Minkowski inequality) For 1 < p < oo, we have

Mu +OLP < 117/11/,/, +11v1ILP (1.15)

Lemma 1.2.4 ([43])

1

Let 1 < p < r < q,

r

1--a

+

q

, and 1 < a < 1. Then

a
p

=

kukLr ~11u117,p kuk1~~

Lq (1.16)

Lemma 1.2.5 ([43])

If au (Q) < oo, 1 < p < q < oo, then Lq y LP, and

1
p

uhp < it(Q)

1

q kukLq :

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