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

3.2 Decay of Solutions

We can now state the asymptotic behavior of the solution of (P).

Theorem 3.2.1 Suppose that (G1) , (G2) and (2.3) hold. Assume further that u0 2 W and ui 2 1¹₀ (Q) satisfying (3.12) . Then the global solution satisfies

E(t) < E (0) exp (--At) , Vt > 0 if m = 2, (3.25)

or

E(t) < (E(0)^' + Kort)³ , Vt > 0 if m > 2, (3.26)

where A and K0 are constants independent of t, r = m 1 and s =

2

2 2 -- m.

The following Lemma will play a decisive role in the proof of our result. The proof of this lemma was given in Nakao [34].

Lemma 3.2.1 ( [37]) Let cp(t) be a nonincreasing and nonnegative function defined on [0, T] , T > 1, satisfying

cp^l#177;r(t) < k( (cp (t) -- cp (t + 1)) , t 2 [0, T] ,

for ko > 1 and r > 0. Then we have, for each t 2 [0, T] ,

cp (t) < cp (0) exp (--k [t -- 1]⁺) , r = 0

c° (t) ~{c (0)^--r + k0r [t -- ¹]⁺1

{

_1 ^r ,r > 0

,

(3.27)

where [t -- 1]⁺ = max ft -- 1,0} , and k = ln ( k0 k0 1 ₁) .

Proof of Theorem 3.2.1.

Multiplying the first equation in (P), by ut and integrate over ft to obtain d dt E (t) + w 11out 1²2 + a Mud 2 : = (g' 2

Vu) (t) -- g (t) 11V u(t)11²₂

Then, integrate the last equality over [t, t + 1] to get

t+1 t+1

E(t 1) -- E(t) + w kbut1²₂ ds + a Mud: ds

t t

=

t+1

t

1 (g^' 0 V u) (s)ds --

²

t+1

₂g(s) 1Vu(t) 1 2ds

t

Therefore,

1

E(t) -- E(t 1) = fi^m(t) -- ₂

1

(g' o VU) (s)ds +

2

t+1

g(t) 1Vu(t)k²₂ ds, (3.28)

t

t+1

t

where

t+1 t+1

F^m(t) = a Mud: ds w 1Vutk²₂ ds (3.29)

t t

Using Poincaré's inequality to find

t+1 t+1

Iutk²₂ ds < C (12) Mutem ds (3.30)

t t

Exploiting Holder's inequality, we obtain

_{0 1}

_Zt + 1

kutk2 _m ds ~ @ ds _A

t

m-2

_m 0

@

₁

2

~kutk2 m A

m

t+1

t

t+1

t

2
m

ds

0 ~ @

t+1 t

1 2 ~kutk2 m A m

2 m

ds. (3.31)

Combining (3.29), (3.30), and (3.31), we obtain, for a constant C1, depending on ~

t+1

Iutk²₂ ds < F²(t), C1 > 0. (3.32)

t

~ ~

By applying the mean value theorem, ( Theorem 1.3.3. in chapter1), we get for some t1 2 t_; t + 1 ;

4

~ ~

t + 3

t2 2 _4; t + 1

Ilut(ti)112 < 2c (n)

1 ^2 F(t), = 1, 2. (3.33)

Hence, by (G2) and since

t+1

1Vut1²₂ ds < C2F (t)², C2 > 0 (3.34)

t

1 3

, there exist t1 2 [t' 4 t + 1 , 4 t2 2 [t + t + 11 such that

11Vut(ti)11²₂ < 4C (Q)F(t)², i = 1, 2. (3.35)

_Zt 2

tl

Next, we multiply the first equation in (P) by u and integrate over Q x [t1, t2] to obtain

_{2 0 1 3}

_Zt

_{4 @}1 ~ g(~)d~ _A kru(t)k2 ₂ ds ~ b kuk^p ₅ ds

0

p

= ~

Zt 2 tl

I

u.uttdxds -- w

Zt 2 tl

I

Vu.Vutdxds -- a

Zt 2 tl

Z

u: jutj^m-2utdxds

t2 +f tl

Zs 0

g(s ^-T) I

Vu(s). [Vu(T) -- Vu(s)] dxdrds.

Obviously, t 2 Z tl

I(s)ds = --

Zt 2 tl

Z

u.uttdxds -- w

Zt2 tl

Z

Vu.Vutdxds -- a

Zt 2 tl

Z

u.lutr^-2 utdxds

t 2 + f tl

Zs 0

g(s ^-T) I

Vu(s). [Vu(T) -- Vu(s)] dxdrds

~~~~~~

=

~~~~~~

u.uttdxds

~_~~Zt2

~~~

tl

I

_{2 3}

_{Z Z}t 2 _Z

₄ utudx ₅ ~ utdxds

t2

~ t1 t1 ~

=

~~~~~~

Z

ut(t2)u(t2)dx -- I ut(ti)u(ti)dx --

Zt 2 tl

kutk^2 2ds

~~~~~~

;

t 2

+ f (g o Vu) (s)ds. (3.36)

tl

Note that by integrating by parts, to obtain

Using Hölder's and Poincaré's inequalities, we get

~~~~~~

2

_Z

u.uttdxds

< C2 ~

_X
i=i

~_~~Zt 2

~~~

tl

1Vutk²₂ dt. (3.37)

Iout(ti)12 Ilvu(ti)112 + C2 ~ _Zt2

tl

By using Hölder's inequality once again, we have

~~~Zt 2 ~~~ tl

I

Vu.Vutdxds

~~~~~~

t2 I~ tl

1V7k2 1Vuth ds (3.38)

Furthermore, by (3.35) and (3.16), we have

1 1

kVtt(ti)1₂ 1Vt(ti)1₂ < C3 (C(Q)) 2 F(t) sup E(s) 2 ; (3.39)

h<s<t2

where, C3 = 2 (1 _(p ²--^p 2))

1

2

:

From (3.34) we have by Hölder's inequality

Zt 2 tl

1V7k2 1Vuth dt <

Zt 2 tl

E(s)

1 ~ 2p ~~ 1 2 l p ~ 2

1 2

11VUt112 ds

< 1 2C3 sup E(s) ti<8<t2

1 2

Zt 2 tl

1Vuth ds,

which implies

1

1

2

1 krutk2 ₂ dt _A

0

11Vut112 dt < _@1dt

_Zt2

tl

t1

₁

_A

t2

₂ 0

@

_Zt 2

tl

Then,

~~~Zt 2 ~~~ tl

Z

u.uttdxds

~~~~~~

< 2C^2~C3F(t) sup h<s<t2

1 E(s)^2 + C!C2F(t)^2. (3.41)

-- 2

N/3C2 F(t).

C3 3C2

where C4 =

4

1

E(s) 2 (3.40)

_Zt 2

tl

1Vu12 1Vuth dt < C4F(t) sup h<s<t2

. Therefore (3.37) , becomes

We then exploit Young's inequality to estimate

Zt 2 ,

I i st 0

g(s - r)Vu(t). [Vu(s) - Vu(t)1 drdxdt

(3.42)

< 8

Zt 2 h

t I 0

g(s - 7^-)11Vu11^22c/rdt+ 4^16.

, I h

(g o Vu) (t)dt, VS > 0.

Now, the third term in the right-hand side of (3.36), can be estimated as follows

Zt 2 h

t 2 f 17/11^m-2 ut.udxds < I ~ tl

I

lutl^m-1 . lul dads.

By Holder's inequality, we find

Zt 2 h

I

lutr^n-1 . lul dxds <

Zt 2 h

2 664

0 1 Z @jutj^m dx A ~

m1 m 0 1 Z @ juj^m dx A ~

3 ^1 ^m 5 7 7

ds

kutk

=

_Zt 2

h

^m~1 m ll'all_m ds.

By Sobolev-Poincare's inequality, we have

, I ti

kutk^m~1 m Mullm ds < C(S2)

Zt 2 ti

kutk^m~1 m 1Vu12 ds,

for 2 < m ~ 2n n - 2

if n > 3, or 2 < m < 1 if n = 1, 2.

Using Holder's inequality, and since ti, t2 2 [t, t + 1] and E(t) decreasing in time, we conclude from the last inequality, (3.16) and (3.29) , that

, I ti.

Ilut1C^-1 Mullm ds < C (Q) (l (p ^2--^p 2))

1 2

Zt 2 ti.

Ilut1C^-1 (J (u))^I2 - ds

< C (Q) (l (p ^2! 2))

1 2

Zt 2 ti

1 Ilut1C^-1 (E(u))^2 ds

1

< C (Q) (l (p ^2! 2))	2		(E(u))	1
		sup h<8<t2		2 x
0 Zt2 @t1					1 kutkm m ds A		m-1	0 Zt 2 @t1	1 ds A	1 M
											(3.43)

m-1 1

( 1a )

M

C (Q) sup h<8<t2

(E(t))

21 ( 2p ) l (p -- 2))

2

F (t)^m-1

Then, taking into account (3.41) -- (3.43), estimate (3.36) takes the form

_Zt 2

ti.

1

I(t)dt < (2C! + ^3C2 w) C3F(t) sup E(s)² + C:C2F(t)²

4 ti <8<t2

2 C3C (Q) sup (E(t))

1

₂ F(t)m-1

1

M

a

+

h<8<t2

(3.44)

+8

, I ti

t I 0

g(t -- s) 11V u112 dsdt + (1 48 + 1)

, I h

(g 0 Vu) (t)dt.

Moreover, from (3.4) and (3.10), we see that

E(t) = 2 kutk2

1 2 dt + J(t)

t

=

2 2p

1 Ilutg + (p 2) 1 -- I g(s)ds) 11V u112

0

(3.45)

1p 2p-- 2 \

+ ) (g 0 Vu) (t) + ¹ I (t).

By integrating (3.45) over [t1, t2] , we obtain

, I ,

E(t)dt = 1 2

Zt 2 h

11utg dt + 132p2/

, I ,

0 @1 --

t I 0

1 g(s)ds A 11Vu11^22 dt

+ (p -- 2) 2p

Zt 2 ,

(g o Vu)(t)dt + 1 P

Zt 2 ,

I(t)dt, (3.46)

which implies by exploiting (3.32)

t 2 I .

E(t)dt < c 1 (F (t))^2 + 2 1 P

Zt 2 ,

/(t)dt + (P -- 2) 2p

Zt 2 ,

(g o Vu) (t)dt

(3.47)

+ (p -- 2) 2p

Zt 2 h

0 @1 --

t I 0

1 g(s)ds A 11Vu11^22 dt.

By using (3.11), Lemma 3.1.5, we see that

0 @1 --

t I 0

)

g(s)ds 11V ug < ¹1(t). (3.48)

⁷1

Therefore, (3.47), takes the form

t 2 I

Q)( E(t)dt < C 2 (F(t))^2 + (P -- 2) 2p

, I ,

(g o Vu) (t)dt

+ (1 + p -- 2) .75, 2pii )

Zt 2 ,

I(t)dt. (3.49)

Again an integration of (3.14) over [s, t2] , s E [0, t2] gives

E(s) = E(t2) + a

Zt 2 8

1 11ut(t)11: dr + 2

, I 8

g(T) 11Vu(t)11^22 dr

--

1 2

Zt 2 8

(g o Vu) (t)dr + w

Zt 2 8

11Vut(t)11^22 dr (3.50)

Zt 2 ,

E(s)ds >

t 2 I

1 E(t2)ds > 2E(t2) · (3.51)

1

'

By using the fact that t2 -- ti > 2 we have

The fourth term in (3.44), can be handled as

t I 0

g(t -- s) 1Vu1^22 ds = 1Vu1^22

t I 0

g(t -- s)ds

(3.52)

< 2p (1 -- l) E (t). -- l (p -- 2)

Thus,

t2 I t1

t I 0

g(t -- s) 1Vuk^22 dsdt < 2p (1 ~ l) l (p -- 2)

t2 I t1

E(t)dt

p (1 ~ l)

~ l (p -- 2)E(ti)

~ p (1 ~ l) l (p -- ₂₎E(t). (3.53)

Hence, by (3.53) , we obtain from (3.44)

t2

_I

t1

1

I(t)dt < ( 20, + ^3C2w) C3F(t) sup E(s)² + C!C2F(t)²

4 ti<8<t2

2 C3C (Q) sup (E(t))

1

₂ F(t)m-1

1

m

a

+

ti<8<t2

E(t) + ( 1 + 1) +8 l (p -- 2) 48

t2 I t1

(g o Vu) (t)dt. (3.54)

E(t) < 2

t2 I t1

E(s)ds + a

t + 1 I t

1 Iut(t)1^mm dr + 2

t + 1 I t

g(r) 1Vu(t)k^22 dr

~

1 2

t + 1 I t

(g^' o Vu) (t)dr + w

t + 1 I t

1Vut(t)k^22 dr. (3.55)

From (3.50) and (3.51) we have

Obviously, (3.49) and (3.55) give us

( c (Q) )

f /(t)dt E(t) < 2 (F(t))² + (P 2 ² ) t 2 t 2

I (g ° Vu) (t)dt + (¹ + p -- 2)

2 p 2pn

tl tl

+a

Zt 2 ,

1 kut(t)km m dt + 2

, I ,

g (t) 11V u(t)g dt -- 12

, I ,

(g^' o Vu) (t)dt

_Zt 2

,

+w

1Vut(t)k²₂ dt.

Consequently, plugging the estimate (3.54) into the above estimate, we conclude

E(t) < C (Q) (F(t))^2 + (P 2) P

, I ,

(g o Vu) (t)dt

.V3C2 1

+2 1¹ + P-- ²) [(2C: + ₄ w) C3F(t) sup E(s) 2 + C!C2F(t)²1

P 2pn ti<8<t2

+

(1 + P 2) _am¹ C3C (Q) sup (E(t)) ² F(t)^m-1 p 2pi
t

i

<

8

<t2

+2 (1 p + 2pn p -- 2) [Sp1 (p -- 2) E (t) + (_4S + 1 ) I (g 0 V u) (t)dt)1

1

l

t

t2

(3.56)

+F^m(t) -- 1 2

, I ,

(g^' o Vu) (t)dt + 2

, I ,

g (t) 11V u(t)g dt.

We also have, by the Poincaré's inequality

I I u(s) I I 2 < C I I vu(s)II2

1

< C (1 (p --^p 2))

2

E(t)

1 2

, (3.57)

Choosing 8 small enough so that

1 -- 2 (¹ + p -- 2) _Sp (1 -- 1)

(3.58)

p 2pi 1 (p -- 2) > 0'

we deduce, from (3.56) that there exists K > 0 such that

E(t) < K [ F(t)² + E(t) 2 F(t) + E(t) 2 F(t)^m-1 + F(t)^m_]

+

1 2

t + 1 I t

g(s) 1Vu(s)k^22 ds -- 1 2

t + 1 I t

(g^' o Vu) (s)ds

(3.59)

#177;[(p ;2) #177; 2 ( 15. #177; 0 (p i_ #177; p 2-13712)1

t + 1 I t

(g o Vu) (s)ds

,

_I

,

Using (G2) again we can write

, I ,

(g o Vu) (t)dt < --

(g^' 0 Vu) (t)dt, > 0.

Then, we obtain, from (3.59),

1 1 ,

E(t) < K [ F(t)² + E(t) ² F(t) + E(t) ² F(t)^m-1 + F(t)^m_]

(3.60)

g(s)11Vu(s)gds -- ( 6+ 2)

1

+ 2

t + 1

_I

t

t + 1

_I

t

(g^' o Vu) (s)ds.

where 6_. = [(p ;2) + 2 (_.731 #177; p 2p--n) ( 45+ 1 )J

An appropriate use of Young's inequality in (3.60), we can find K1 > 0 such that

E(t) < K1 [F(t)² + F(t)^2(m-1) + F(t)^m] (3.61)

g(s)11Vu(s)gds -- ( 6+ 2)

₃

(g^' o Vu) (s)ds 5 ,

t + 1

_I

t

t + 1

_I

t

[

1

2

for K1 a positive constant.

Using (G2) again to get

E(t) < K1 [F(t)² + F(t)^2(m-1) + F(t)^m]

+ [ (1 + 2 f

g (s) 11V u(s) g ds -- (61 + 2)

t + 1

_I

t

t+ 1

_I

t

(g^' o Vu) (s)ds

< K1 [F(t)² + F(t)^2(m-1) + F(t)^m]

(3.62)

t + 1 t + 1

+ (1 + gi)

[2

1 I

I g(s) 11V u(s)112 ² 2 ds -- (g^' o Vu) (s)ds

t t

At this end we distinguish two cases:

Case 1. m = 2. In this case we use (3.28) and (3.62), we can find K2 > 0 such that

E(t) < K1 F(t)²

t+ 1 t + 1

+ (1 + gi)

[2

I g(s) 11V u(s)11₂ ds -- 2 II (g^' o Vu) (s)ds

t 1

t

< K2 [E(t) -- E(t + 1)] . (3.63)

Since E(t) is nonincreasing and nonnegative function, an application of Lemma 3.2.1 yields

E(t) < K2 [E(t) -- E(t + 1)] , t > 0, (3.64)

which implies that

E(t) < E (0) exp (--A [t -- 1]1 , on [0, oo) , (3.65)

where A = ln

K2 -- 1

( K2

Case 2. m > 2. In this case we, again use (3.28) and (3.62) to arrive at

2

m

. (3.66)

t + 1 t+ 1t t

F(t)² = (E(t) -- E(t + 1)) -- 2 1 I g(s) 11V u(s) g ds + 21 I (g^' o Vu) (s)ds

m

2 < 2

We then use the algebraic inequality

(a + b)

m ( m m

2 _a2 + b²), m > 2. (3.67)

To infer from (3.62), and by using (3.67), that

[E(t)]

m [1 + F (t)^2(m_2) + F (t)^m_2] 2 < K3

m 2 F(t)m

	m 2 (1 + 21)	m	2 4	ft + 1 1 g(s) MVu(s)M2 2 ds _ 1 2 2 t	Zt + 1 t	3 (g^' o Vu) (s)ds 5		m 2	+2

[1 + F (t)^2(m_2) + F (t)^m_2]^m

< K3 2 x [E(t) - E(t + 1)]

(3.68)

	m 2 (1 + 21)	m	2 4	ft + 1 1 g(s) MVu(s)M2 2 ds _ 1 2 2 t	Zt + 1 t	3 (g^' o Vu) (s)ds 5		m 2	+2

m

where K3 = 2

2 K1. We use (3.28) to obtain

0 @

1 2

Zt + 1 t

g(s) VuM2 2 ds - 1 2

Zt + 1 t

1 (g^' o Vu) (s)ds A

m 2

m

< (E(t) - E(t + 1)) A combination of (3.68), (3.69) yields

2 (3.69)

[E(t)]

m 2 < K3 [1 + F(t)^2(m_2) + F(t)^m_2]

m 2 x (E(t) - E(t + 1))

+2

m 2 (1 + 21)

m 2 [E(t) -- E(t + 1)]

m 2 -1 [E(t) - E(t + 1)]

[ ]

m m m

< K3 [1 + F (t)^2(m_2) + F (t)m_2]m ² + 2 2 ~1

2 (1 + 21) ² [E(t) - E(t + 1)] x

[E(t) - E(t + 1)] (3.70)

By using (3.62), the estimate (3.70) takes the form

[E(t)]

m 2 <

{ m m m m K32m [1 + E(0)^(m_2) + (E(0)) 2 ~1i 2 _1} + 2 ^2 (1 + 21) ^2 (E(0)) ~

(E(t) - E(t + 1))

< K0 (E(t) - E(t + 1)). (3.71)

Again, using Lemma 3.2.1, we conclude

E(t) < [E(0)_^r + K0r [t -- 11⁺1⁸ , (3.72)

Tn

with r =

2

2

1 > 0, s = and K0 is some given positive constant.

2 -- n-i

This completes the proof.