C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 67, N umb er 1, Pages 286–292 (2018) D O I: 10.1501/C om mua1_ 0000000850 ISSN 1303–5991
http://com munications.science.ankara.edu.tr/index.php?series= A 1
ENERGY DECAY RATE OF THE SOLUTIONS OF A MARINE RISER EQUATION WITH A VARIABLE COEFFICIENT
MUGE MEYVACI
Abstract. In this work the initial boundary value problem for a fourth order non linear equation which describes the marine riser is studied :
utt+ kuxxxx [a(x)ux]x+ utx+ b(t)utjutjp= 0; x 2 [0; l]; t > 0;
Under appropriate conditions on a(x) and b(t), we prove that the energy of the problem tends to zero as t ! 1.
1. INTRODUCTION
We work on the decay properties of solutions to the initial boundary value prob-lem of the marine riser equation:
utt+ kuxxxx [a(x)ux]x+ utx+ b(t)utjutjp= 0; x 2 [0; l]; t > 0; (1.1)
u(0; t) = uxx(0; t) = u(l; t) = uxx(l; t) = 0; t > 0; (1.2)
where k; p; are given positive numbers, a; b are given functions. This equation without the variable damping coe¢ cient is studied in [1] and [2]. This problem about the o¤shore drilling operations which done by a long slender vertical pipe that is including a drilling string and drilling mud, which is so called Marine riser. The problem of riser stability, that is the stability of pipes conveying ‡uid has caught the attention of many authors (see e.g. [1]-[11]).
Since our equation includes a variable coe¢ cient b(t) the techniques used in above articles is not applicable to our problem. Therefore we adapt the study of Martinez [8], in this article a new weighted integral inequality method was used to estimate the decay rate of solutions of the wave equation. This method is originated a result of Haraux [3] .
In [9], the following simplest equation that can be used in modeling of marine riser:
utt+ uxxxx N uxx= 0 x 2 (0; 1); t > 0;
Received by the editors: September 06, 2016; Accepted: March 24, 2017. 2010 Mathematics Subject Classi…cation. 35B30, 35B35.
Key words and phrases. Energy decay rate, marine riser, fourth order wave equation, asymp-totic behaviour.
c 2 0 1 8 A n ka ra U n ive rsity C o m m u n ic a tio n s d e la Fa c u lté d e s S c ie n c e s d e l’U n ive rs ité d ’A n ka ra . S é rie s A 1 . M a th e m a t ic s a n d S t a tis t ic s .
under the homogeneous boundary conditions (1.2) is considered. Where N is a positive number. Lyapunov’s direct method is used in detail.
In [7], the following nonlinear marine riser equation:
mutt+ EIuxxxx (N ux)x+ auxt+ butjutj = 0 x 2 (0; l); t > 0;
under the boundary conditions (1.2) is studied and the stability of zero solution of this problem is established.
In [11], the initial boundary value problem for the fourth order equation mutt+ (EIuxx)xx+ P (t)uxx= 0; x 2 (0; l); t > 0;
under the boundary conditions (1.2) is considered. The necessary conditions on P (t) for the stability of solutions are obtained. In [6], the initial boundary value problem for the marine riser equation:
mutt+ kuxxxx (a(x)ux)x+ utx+ butjutjp= 0 x 2 (0; l); t > 0;
under the boundary conditions (1.2) is considered. The global asymptotic stability of solutions and the estimates for the rate of decay of the solutions were obtained. In [1], the globally asymptotically stability of the zero solution to the problem for multidimensional marine riser equation:
utt+ k 2u + a u + ~g:rut+ butjutjp= 0; x 2 ; t > 0;
under the initial boundary conditions
u(x; 0) = u0(x); ut(x; 0) = u1(x); x 2 ;
u(x; t) = @u(x; t)
@ = 0; x 2 @ ; t > 0;
where RN, N 3 is a bounded domain with su¢ ciently smooth boundary
@ , is the unit outward normal vector to the boundary, k > 0, p 1, b > 0 and a 2 R are given numbers and ~g = (g1; g2; :::; gN) 2 RN, is studied. Furthermore,
continuous dependence of the weak and the strong solutions of the problem on the coe¢ cients a, b and g were proved.
There are many articles devoted to the study of boundary control of initial boundary value problems for marine riser type equations (see,e.g.[4],[5], [10]). In what follows, we will use the following notations:
ku(t)k := Z l 0 u2(x; t)dx !1 2 ; ku(t)kq := Z l 0 uq(x; t)dx !1 q :
The proof of our main result will be based on the following pre mentioned Lemma. Lemma 1.1. (Martinez,[8]) Let E : R+ ! R+ be a non increasing function and
: R+! R+ a strictly increasing function of class C1 such that
Assume that there exist 0 and ! > 0 such that Z +1
S
E(t)1+ 0(t)dt 1
!E(0) E(S): (1.4)
Then E(t) has the following decay property:
if = 0; then E(t) E(0)e1 ! (t); 8t 0; (1.5)
if > 0; then E(t) E(0) 1 + 1 + ! (t)
1
; 8t 0: (1.6)
2. Asymptotic behavior
Theorem 2.1. Suppose that b(t) is a nonincreasing function of class C1 on R+ satisfyingR0tb(s)ds ! 1 as t ! 1 and there exists a positive number a0 such that
a(x) a0:
Then each solution of the problem (1.1)-(1.2) satis…es the following energy decay property: E(t) E(0) p + 2 2 + !pR0tb(s)ds !2 p ; 8t > 0; where ! 1=2max ( c ; (2 3p+2 p pl) p 2 (p + 2)( (p + 2))p2Eq(0) ; (p + 1)(4(d2 ) p+2)p+11 (p + 2)( (p + 2))p+11 E q p 1 (0) ) ; and =1 2 c 2 k12 > 0:
Proof. Suppose that u is a solution to the problem (1.1)-(1.2). Multiplying equation (1.1) by utand integrating over (0; l) we get
d dtE(t) = 2 Z l 0 b(t)ju t(x; t)jp+2dx; (2.1) where E(t) := kut(t)k2+ k kuxx(t)k2+ Z l 0 a(x)u2x(x; t)dx: (2.2)
Now, multiplying equation (1.1) by 0qu and integrating over (0; l) (S; T ) and using boundary conditions we get
0(t)Eq Z l 0 (u(x; t)ut(x; t)) jTSdx Z T S 0(t)Eq (t) kut(t)k2dt Z T S Z l 0
[ 00(t)Eq(t) + q 0(t)Eq 1(t)E0(t)]u(x; t)ut(x; t)dxdt
+ Z T S 0(t)Eq(t)[E(t) kut(t)k2]dt + Z T S 0(t)Eq(t) Z l 0 u(x; t)utx(x; t)dxdt + Z T S 0(t)Eq(t) Z l 0 b(t)u(x; t)ju t(x; t)jp+1dxdt = 0: So we have Z T S 0(t)Eq+1(t)dt = 0(t)Eq(t) Z l 0 (u(x; t)ut(x; t)) jTSdx +2 Z T S 0(t)Eq (t) kut(t)k2dt Z T S 0(t)Eq(t) Z l 0 u(x; t)utx(x; t)dxdt + Z T S Z l 0 h
00(t)Eq(t) + q 0(t)Eq(t)E0(t)iu(x; t)u
t(x; t)dxdt Z T S 0(t)Eq(t) Z l 0 b(t)u(x; t)ju t(x; t)jp+1dxdt: (2.3)
Using Cauchy inequality, Hölder’s inequality, de…nition of E(t) and 0(t) we get Z l 0 u(x; t)ut(x; t)dx Z l 0 u2(x; t)dx ! Z l 0 u2t(x; t)dx ! cE(t); (2.4) 0(t)Eq Z l 0 u(x; t)ut(x; t)dxjTS c Eq+1(S); (2.5) 2 Z T S 0(t)Eq (t)jjut(t)jj2dt 2 Z T S 0(t)Eq(t)lp+2p Z l 0 ju t(x; t)jp+2dx ! 2 p+2 dt p(2 1) p+2 p p + 2 Z T S 0(t)l Eq(p+2)p dt + 2 (p + 2) p+2 2 1 Z T S ( E0(t) 2 )dt; (2.6)
Z T S
Z l 0
[ 00(t)Eq(t) + q 0(t)Eq 1(t)E0(t)]u(x; t)ut(x; t)dxdt
Z T S
00(t)Eq(t) + q 0(t)Eq 1(t)E0(t) cE(t)dt
qc q + 1E q+1(S); (2.7) Z T S 0(t)Eq Z l 0 u(x; t)utx(x; t)dxdt d1 2 Z T S 0q+1(t)dt; (2.8) Z T S 0(t)Eq(t) Z l 0 b(t)u(x; t)ju t(x; t)jp+1dxdt ( d2 2)p+2 p + 2 Z T S 0(t)Eq(q+1 2)(p+2)(t)dt + p + 1 (p + 2) p+2 p+1 2 E(S); (2.9) where c = l 2 2k12 ; d1= 1 + 2l2 2k; d2= l32+p+21 k122 p+1 p+2 : Thanks to Sobolev inequality ku(t)kp+2 l
1 2+
1
p+2ku
x(t)k (ref. [6]) and the
de…ni-tion of E(t) we have
ku(t)kp+2
l12+ 1 p+2
E(t)12:
Employing the inequalities 2.5–2.9 and 2.3 we get Z T
S
0(t)Eq+1(t)dt c E(0)qE(S) +pl(2 1)
p+2 p p + 2 Z T S 0(t)Eq(p+2)2 dt + 1 (p + 2) p+2 2 1 Z T S ( E0(t))dt + qc q + 1E q+1(S) +d1 2 Z T S 0(t)Eq+1(t)dt +( d2 2) p+2 p + 2 Z T S 0(t)E(q+1 2)(p+2)(t)dt + p + 1 (p + 2) p+2 p+1 2 E(S); If we choose q = p2 we get Z T S
0(t)Eq+1(t)dt c E(0)qE(S) +pl(2 1)
p+2 p p + 2 Z T S 0(t)Eq(p+2)p (t)dt + 1 (p + 2) p+2 2 1 Z T S ( E0(t))dt + qc q + 1E q+1(S) +d1 2 Z T S 0(t)Eq+1(t)dt +( d2 2) p+2 p + 2 Z T S 0(t)Eq+1+ (t)dt + p + 1 (p + 2) p+2 p+1 2 E(S);
Here =p(p+2)2 . Choosing k > 2l222 we get =12
c 2
k12 > 0,
Z T S
0(t)Eq+1(t)dt c E(0)qE(S) + pl(21)
p+2 p p + 2 Z T S 0(t)Eq(p+2)p (t)dt + 1 (p + 2) p+2 2 1 Z T S ( E0(t))dt + qc q + 1E q(0)E(S) +( d2 2) p+2 p + 2 Z T S 0(t)Eq+1+ (t)dt + p + 1 (p + 2) p+2p+1 E(S); If we choose p+22 =4(d (p+2) 2 )p+2E (0) and p+2 p 1 = (p+2) 2 3p+2 p pl we get 2 Z T S 0(t)Eq+1(t)dt c Eq(0)E(S) + qc q + 1E q(0)E(S) + (2 3p+2 p pl)p2 (p + 2)( (p + 2))p2 E(S) +(p + 1)(4(d2 ) p+2)p+11 Ep+1(0) (p + 2)( (p + 2))p+11 E(S); Thus we obtain Z T S 0(t)Eq+1(t)dt 1 !E q(0)E(S):
Now, using Lemma 1.1 we get
E(t) E(0) p + 2 2 + !pR0tb(s)ds !2 p ; 8t > 0; Here ! 1= 2max ( c ; (2 3p+2 p pl)p2 (p + 2)( (p + 2))p2Eq(0) ; (p + 1)(4(d2 ) p+2)p+11 (p + 2)( (p + 2))p+11 E q p 1 (0) ) : References
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Current address : Department of Mathematics, Mimar Sinan Fine Art University 34380,¸Si¸sli, Istanbul, Turkey
E-mail address : muge.meyvaci@msgsu.edu.tr ORCID: http://orcid.org/0000-0002-4772-3869
