New York J. Math. (1998) 167{176.
Boundary Stabilization of a Hyperbolic Equation
with Viscosity
M. M. Cavalcanti
Abstract. This paper is concerned with the solvability and uniform stability of a hyperbolic equation with spatially varying coecients of viscosity and elasticity and boundary damping.
Contents
1. Introduction 167
2. Notations and Main Result 168
3. Solvability of strong and weak solutions 169
4. Asymptotic Behaviour 173
References 176
1.
Introduction
Let be a bounded domain of
R
n with C2 boundary ; and let (;0;1) be a partition of ;both parts with positive measure.We consider the following hyperbolic problem:
@2y
@t2 ;r(aij(x)ry); @
@tr(bij(x)ry) =f in Q= (01) y= 0 on 1= ;1(01)
@@ya + @
@t @y
@b +(x)
@y
@t ;g= 0 on 0= ;0(01) y(0) =y0 @y
@t(0) =y1 in (1.1)
where @@ya (resp. @@yb) is the outer cornormal derivative with respect to the matrix (aij) (resp. (bij)) dened by
@@ya =iaij @y
@xj
and= (1n) denotes the exterior unit normal at ;.
Received June 30, 1997.
Mathematics Subject Classication. 35B40, 35L80.
Key words and phrases. elasticity, viscosity, boundary stabilization.
c1998StateUniversityofNewYork
ISSN1076-9803/98
167
In physical terms the entriesaijandbijare related to coecients of elasticity and viscosity, respectively. Without viscosity, i.e., ifbij = 0 and assuming thataij = 1, f = 0 andg= 0the problem (1.1) was studied by many authors: See J. P. Quinn and D. L. Russell 10], G. Chen 2, 3, 4], J. E. Lagnese 7, 8] and V. Komornik and E. Zuazua 6]. And when= 0, an asymptotic regularization procedure is proved by G. C. Hsiao and J. Sprekels 5]. Inspired by the above works we show solvability of strong and weak solutions to problem (1.1), and obtain boundary stabilization.
To obtain the existence of solutions we make use of Galerkin's aproximation.
However, as we are also interested in srong solutions we have some technical dif- culties wich lead us to transform the problem (1.1) into an equivalent one with zero initial data.
Stability problems with nonhomogeneous conditions require a special treatement because we don't have any information about the inuence of the inner products (f(t)y0(t))L2() and (g(t)y0(t))L2(;0
)on the energy E(t) = 12Z
jy0(xt)j2dx+ 12Z
aij(x)@y
@xi(xt) @y
@xj(xt)dx (1.2)
or about the sign of its derivativeE0(t):
To obtain the uniform decay we use the perturbed energy method. Our paper is organized as follows. In Section 2 we give notations and state our main result.
In Section 3 we prove solvability of strong and weak solutions of (1.1) while in Section 4 we obtain the boundary stabilization of solutions from Section 3.
2.
Notations and Main Result
We dene
(uv) =
Z
u(x)v(x)dx juj2=
Z
ju(x)j2dx (uv);0 =
Z
;0
u(x)v(x)dx juj2;0 =
Z
;0
ju(x)j2dx and letV be
V =fv2H1() v= 0 on ;1g
which, equipped with the topologyjrj is a Hilbert subspace ofH1().
In order to establish our main result, we make the following assumptions on the coecients:
aijbij 2C1(): (2.1)
There exist positive constantsa0 andb0 such that aij =aji and for 2
R
n Xnij=1aijij a0jj2 (2.2)
bij=bji and for 2
R
n Xnij=1bijij b0jj2 (2.3)
2L1() (x)0 a.e. on ;0 (x) = 0 8x2;0\;1 (2.4)
and (x)!0 wheneverx tends to a point x2;0\;1: This choice of the function was done in order to avoid eventual singularities.
Dening
a(uv) = Xn
ij=1
Z
aij(x)@u
@xi @v
@xjdx 8uv2V (2.5)
b(uv) = Xn
ij=1
Z
bij(x)@u
@xi @v
@xjdx 8uv2V (2.6)
from (2.1), (2.2) and (2.3) there exist positive constantsa0a1b0andb1such that a0jruj2a(uu)a1jruj2 8u2V
(2.7)
b0jruj2b(uu)b1jruj2 8u2V:
(2.8)
Now, we are in position to state our main result.
Theorem 2.1.
Lety0y1fg2V L2()L2(01L2())H1(01L2(;0)):
Under the assumptions (2.1)-(2.3) and assuming that g(0) = 0, the problem (1.1) possesses a unique weak solution y: (01)!
R
such thaty2L1(01V) y02L1(01L2()): (2.9)
Moreover, provided that for large t, the inequality
Z t
0
exp("
2s)jf(s)j2+jpg(s)j2;0ds tr (2.10)
holds for some positive constants", , and r, we obtain the following energy decay E(t)Cexp(;"
2t) 8t0 and 8"2(0"0] whereC and"0 are positive constants.
Remark 1.
The hypothesis (2.10) means that the map t7;!Z t0
exp("
2s)jf(s)j2+jpg(s)j2;0ds must be bounded by a polinomialP(t).
3.
Solvability of strong and weak solutions
In this section we are going to prove the existence of strong solutions of problem (1.1). Using density arguments we conclude the same for weak solutions. The exis- tence of solutions may be proven either by the Galerkin method or using semigroup arguments. We employ the Galerkin method.
We dene the Hilbert space
H=u2VAuBu2L2() (3.1)
whereAandB are the operators dened by A=; @
@xi
aij(x) @
@xj
and B=; @
@xi
bij(x) @
@xj
andH is endowed by the natural inner product
(uv)H= (uv)V + (AuAv) + (BuBv):
Let us consider
y0y12H (3.2)
satisfying the compatibility condition
@y0
@a +@y1
@b +(x);y1;g(0) = 0: (3.3)
In addition, let us assume that
f 2H1(01L2()) g2H2(01L2(;0)): (3.4)
The variational formulation associated with problem (1.1) is given by (y00(t)w) +a(y(t)w) +b(y(t)w) + (y0(t)w);0
= (f(t)w) + (g(t)w);0 8w2V:
In order to obtain strong solutions and since we can not use a `special basis' (for instance, one formed by eigenfunctions) because of the boundary condition (y0(t)w);0, we need to derive the above expression with respect to t: But it lead us to technical diculties when we estimate y00(0): To solve this problem we transform the boundary value problem (1.1) into an equivalent one with null initial data. In fact, considering the change of variables
v(xt) =y(xt);(xt) (3.5)
where
(xt) =y0(x) +ty1(x) (xt)201) we obtain the equivalent problem forv:
v0 0;r(aij(x)rv);r(bij(x)rv0) =F in Q v= 0 on 1
@@va +@v0
@b +v0=G on 1 v(0) =v0(0) = 0
(3.6) where
F =f+r(aij(x)r) +r(bij(x)r0) (3.7)
G=g; @
@a ;@0
@b ;0: (3.8)
If v is a solution of (3.6) in 0,T], theny=v+is a solution of (1.1) in the same interval. However, after two estimates we are going to prove later, we have that
jAv(t)j2+jrv0(t)j2C(T) 8t20T] and 8T >0: (3.9)
Thus, from (3.4) and (3.5) we have the same estimate obtained in (3.9) for the solution y: So, we can extend y to the whole interval 01) using the standard argument
Tmax=1 or if Tmax<1 then t lim
!Tmax
;
jAv(t)j2+jrv0(t)j2 =1: Hence, it is sucient to prove that (3.6) has a solution in 0,T], which will be done by the Galerkin method.
We represent by (!) a basis inH which is orthonormal in L (), by Vm the subspace ofH generated by the m-rst vectors!1!m and dene
vm(t) =Xm
i=1gim(t)!i
(3.10)
wherevm(t) is the solution of the folowing Cauchy problem:
(3.11) (v00m(t)!j) +a(vm(t)!j) +b(vm0 (t)!j) + (vm0 (t)!j);0
= (F(t)!j) + (G(t)!j);0 vm(0) =vm0 (0) = 0 j= 1m:
The aproximate system is a normal one of ordinary dierential equations. It has a solution in 0tm): The extension of the solution on the whole interval 0,T] is a consequence of the rst estimate we are going to obtain below.
3.1.
A Priori Estimates.
3.1.1. The First Estimate. Multiplying both sides of (3.11) byg0jm(t)summing over 1j mand considering (2.2), we obtain
12d dt
n
jvm0 (t)j2+a(vm(t)vm(t))o+b(v0m(t)vm0 (t)) + pvm0 (t)2
;0
(3.12)
= (F(t)v0m(t)) + d
dt(G(t)vm(t));0
;(G0(t)vm(t));0
12jF(t)j2+ 12jv0m(t)j2+ d
dt(G(t)vm(t));0+C02
2 jG0(t)j2;0+ 12jrvm(t)j2 whereC0 is a positive constant such thatjvj;0 C0jrvj ,8v2V:
Integrating (3.12) over (0,t) 0< t < tm, taking into consideration (2.7) and (2.8) and noting thatvm(0) =v0m(0) = 0, we get
12jv0m(t)j2+a0
2 jrvm(t)j2+b0Z t
0
jrv0m(s)j2ds+
Z t
0
pv0m(s)2
;0
(3.13) ds
12kFk2L2(0TL2())+C02
2 kG0k2L2(0TL2(;0))+ (G(t)vm(t));0 + 12
Z t
0 n
jv0m(s)j2+jrvm(s)j2ods:
On the other hand, for ar arbitrary >0 we have (G(t)vm(t));0 C02
4 jG(t)j2;0+jrvm(t)j2: (3.14)
Combining (3.13), (3.14) and choosing >0 small enough, we obtain the rst estimate
jv0m(t)j2+jrvm(t)j2+
Z t
0
jrvm0 (s)j2ds+
Z t
0
pv0m(s)2
;0
dsL1 (3.15)
whereL1is a positive constant independent ofm2
N
andt20T]:3.1.2. The Second Estimate. First of all we are going to estimatevm(0) inL () norm. Takingt = 0 in (3.11) and taking into account thatvm(0) =vm0 (0) = 0, it follows that
(vm00(0)!j) = (F(0)!j) + (G(0)!j);0: (3.16)
From (3.3) and (3.8) we have thatG(0) = 0, and from (3.16) we conclude
jv0 0m(0)j2= (F(0)vm00(0)) which implies that
jvm00(0)jL 8m2
N
(3.17)
whereLis a positive constant independent ofm2
N
:Now, taking the derivative of (3.11) with respect tot, we can write (3.18) (v000m(t)!j) +a(v0m(t)!j) +b(v00m(t)!j) + (v0 0m(t)!j);0
= (F0(t)!j) + (G0(t)!j);0: Multiplying both sides of (3.18) by gjm00 (t) and summing over 1 j m we obtain
(3.19) 12 d dt
n
jv00m(t)j2+a(vm0 (t)vm0 (t))o+b(vm00(t)vm0 0(t)) + pvm0 0(t)2
;
0
= (F0(t)vm0 0(t)) + d
dt(G0(t)v0m(t));0;(G00(t)vm0 (t));0: Using arguments analogous to those considered in the rst estimate, observing that vm0 (0) = 0, and taking into account (3.17), from (3.19) we obtain the second estimate
jv00m(t)j2+jrvm0 (t)j2+
Z t
0
jrvm00(s)j2ds+
Z t
0
pv00m(s)2
;
0
dsL2 (3.20)
whereL2is a positive constant independent ofm2
N
andt20T]:Due to estimates (3.15) and (3.20) we can extract a subsequencefvgoffvmg
such that
v* v weak star in L1(0TV) (3.21)
v0 * v0 weak star in L1(0TV) (3.22)
v00* v00 weak in L2(0TV) (3.23)
v0 0* v0 0 weak star in L1(0TL2()) (3.24)
v0* v0 weak in L2(0TL2(;0)): (3.25)
The above convergences are sucient to pass to the limit.
3.2.
Uniqueness.
Supose we have two solutions y and ^y of problem (1.1). Then z=y;y^satises(z00(t)w) +a(z(t)w) +b(z0(t)w) + (z0(t)w);0 = 0 8w2V (3.26)
z(0) =z0(0) = 0:
Takingw= 2z(t) in (3.26) we obtain dtd
n
jz0(t)j2+a(z(t)z(t))o+ 2b(z0(t)z0(t)) + 2 pz0(t)2
;0
= 0: (3.27)
Integrating (3.27) over (0,t) we obtain
jz0(t)j2+a0jrz(t)j2+ 2b0Z t
0
jrz0(s)j2ds+ 2
Z t
0
pz0(s)2
;0
ds= 0 which implies thatjrz(t)j=jz0(t)j= 0:This completes the proof.
3.3.
Solvability of weak solutions.
We have just proved the existence of solu- tions of the problem (1.1) when y0 andy1 are smooth. By density arguments we conclude the same for weak solutions. However, the principal diculty is due to the existence of a sequence of initial data which satises the hypothesis of compatibility (3.3). For this end and sinceg(0) = 0 giveny0y12 V L2() it is sucient to considery02D(A) =
u2V Au2L2() and @u
@a = 0 on ;0
such that
y0!y0 in V
and y12D(B)\H01() such that y1!y1 in L2():
The uniqueness of weak solutions requires a regularization procedure and can be obtained using the standard method of Visik-Ladyshenskaya, cf. J. L. Lions 9]
Chapter 3, Section 8.2.
4.
Asymptotic Behaviour
In this section we obtain the uniform decay of the energy given in (1.2) for strong solutions, since the same occurs for weak solutions using standard density arguments.
The derivative of the energy given by (1.2) is E0(t) =;b(y0(t)y0(t)); py0(t)2
;
0
+ (f(t)y0(t)) + (g(t)y0(t));0: (4.1)
Letand be positive constants such that
jvj2jrvj2 8v2V (4.2)
and
pv2
;
0
jrvj2 8v2V:
(4.3)
For an arbitrary" >0 we dene the perturbed energy E"(t) =E(t) +"(t) (4.4)
where
(t) =
Z
y0ydx:
(4.5)
Proposition 4.1.
There existsC1>0 such thatjE"(t);E(t)j"C1E(t) 8t0 and 8" >0:
Proof.
From (2.7), (4.2) and (4.5) we obtainj(t)j1
2jy0j2+ 12a;10 a(yy);a;10 + 1 E(t): (4.6)
If we deneC1=a;10 + 1then from (4.4) and (4.6) we can write
jE"(t);E(t)j="j(t)j"C1E(t): This concludes the proof.
Proposition 4.2.
There existC2=C2(") and "1 positive constants such that E"0(t);"E(t) +C2jf(t)j2+jpg(t)j2;08t0 and 8"2(0"1]:
Proof.
First of all we must estimate 0(t) in terms of E(t): Taking the derivative of(t) given in (4.5) and replacingy00 byr(aij(x)ry) +r(bijry0) +f in the expression obtained it follows that0(t) =
Z
r(aij(x)ry)y dx+
Z
r(bijry0)y dx+
Z
fy dx+
Z
jy0j2dx:
(4.7)
On the other hand, from Gauss' theorem and taking into account that
@@ya +@y0
@b =(g;y0) on ;0 we obtain
(4.8)
Z
r(aij(x)ry)y dx+
Z
r(bijry0)y dx
=;a(y(t)y(t));b(y0(t)y(t));
Z
;0
y0y d; +
Z
;0
gy d;: Replacing (4.8) in (4.7) adding and subtracting the termRjy0j2dxin (4.7) we get 0(t) =;2E(t);b(y0(t)y(t)) + 2Z
jy0j2dx;Z
;
0
y0y d; +Z
fy dx+Z
;
0
gy d;: (4.9)
Now, from (2.7), (4.2), (4.3) and (4.9), using for an arbitrary >0 the inequality ab a42 +b2 we can write
0(t);(2;8)E(t) +
2+jjbjj2a;10 4
j ry0(t)j2 (4.10)
+a;10 4
py0(t)2
;0
+a;10 4
jf(t)j2+ pg(t)2
;0
where
jjbjj= Xn
ij=1
jjbijjjL1():
Choosing= 8 from (4.10) we have
0(t);E(t) +M1jry0(t)j2+M2 py0(t)2
;0
+M3
jf(t)j2+ pg(t)2
;0
(4.11) where
M1= 2;+jjbjj2a;10 M2= 2a;10 and M3= 2a;10 : Thus, combining (2.8), (4.1), (4.4) and (4.11), we conclude
E0"(t) =E0(t) +"0(t)
;b0jry0(t)j2; py0(t)2
;0
+ (f(t)y0(t)) + (g(t)y0(t));0
;"E(t) +"M1jry0(t)j2+"M2 py0(t)2
;0
+"M3
jf(t)j2+ pg(t)2
;0
which implies
E"0(t);(b0;"(M1+ 1))jry0(t)j2;(1;"(M2+ 1)) py0(t)2
;0
(4.12)
;"E(t) +
4" +"M3jf(t)j2+
1
4"+"M3jg(t)j2;0: Dening
"1=min b0
M1+ 1M21+ 1
and choosing"2(0"1] it follows that E"0(t);"E(t) +C2(")
jf(t)j2+ pg(t)2
;
0
which concludes the proof.
Proof of the exponential decay.
We dene"0=minf"12C11g
and let us consider"2(0"0]:From Proposition 4.1 we have (1;C1")E(t)E"(t)(1 +C1")E(t) (4.13)
Since"1=2C1,
12E(t)E"(t)3
2E(t)2E(t) 8t0 (4.14)
and therefore
;"E(t);"
2E"(t): (4.15)
Hence, from (4.15) and considering Proposition 4.2 we obtain E"0(t);"
2E"(t) +C2jf(t)j2+jpg(t)j2;0: Consequently
dtd
E"(t)exp("
2t)C2jf(t)j2+jpg(t)j2;0exp("
2t):
Integrating the above inequality over 0,t] we get E"(t)exp(;"
2t)E"(0) +C2exp(;"
2t)
Z t
0
exp("
2s)jf(s)j2+jpg(s)j2;0 and taking into consideration (4.14) we see that
E(t)
3E(0) + 2C2Z t
0
exp("
2s)jf(t)j2+jpg(t)j2;0
exp(;"
2t): (4.16)
Combining (4.16) with the assumption (2.10) we prove the desired decay and nish the proof of Theorem 2.1.
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Departamento de Matematica, Universidade Estadual de Maringa, 87020-900, Mar- inga - PR, Brasil.
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