New York Journal of Mathematics
New York J. Math. 4 (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
2. Notations and Main Result
3. Solvability of strong and weak solutions
4. Asymptotic Behaviour
References
1.
167
168
169
173
176
Introduction
Let be a bounded domain of Rn with C 2 boundary ; and let (;0 ;1 ) be a
partition of ;both parts with positive measure.
We consider the following hyperbolic problem:
(1.1)
@ 2 y ; r (a (x)ry) ; @ r (b (x)ry) = f in Q = (0 1)
ij
ij
@t2
@t
y = 0 on 1 = ;1 (0 1) @y
@y @ @y
@a + @t @b + (x) @t ; g = 0 on 0 = ;0 (0 1)
1
y(0) = y0 @y
@t (0) = y in @y ) is the outer cornormal derivative with respect to the matrix
where @@ya (resp. @
b
(aij ) (resp. (bij )) dened by
@y
@y
@a = i aij @xj
and = (1 n ) denotes the exterior unit normal at ;.
Received June 30, 1997.
Mathematics Subject Classication. 35B40, 35L80.
Key words and phrases. elasticity, viscosity, boundary stabilization.
167
c 1998 State University of New York
ISSN 1076-9803/98
168
M. M. Cavalcanti
In physical terms the entries aij and bij are related to coecients of elasticity and
viscosity, respectively. Without viscosity, i.e., if bij = 0 and assuming that aij = 1,
f = 0 and g = 0 the 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 difculties 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) y (t))L2 () and (g(t) y (t))L2 (;0 ) on the energy
Z
Z
@y (x t) @y (x t) dx
(1.2)
E (t) = 21 jy (x t)j2 dx + 12 aij (x) @x
@xj
i
or about the sign of its derivative E (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.
0
0
0
0
2.
Notations and Main Result
We dene
(u v) =
(u v);0 =
and let V be
Z
Z
u(x)v(x) dx juj2 =
Z
;0
u(x)v(x) dx juj =
2
;0
ju(x)j2 dx
Z
;0
ju(x)j2 dx
V = fv 2 H 1 (
) v = 0 on ;1 g
which, equipped with the topology jr j is a Hilbert subspace of H 1 (
).
In order to establish our main result, we make the following assumptions on the
coecients:
(2.1)
aij bij 2 C 1 (
):
There exist positive constants a0 and b0 such that
(2.2)
(2.3)
aij = aji and for 2 Rn bij = bji and for 2 Rn n
X
ij =1
n
X
ij =1
aij i j a0 j j2
bij i j b0 j j2 2 L (
) (x) 0 a.e. on ;0 (x) = 0 8x 2 ;0 \ ;1
and (x) ! 0 whenever x tends to a point x 2 ;0 \ ;1 :
This choice of the function was done in order to avoid eventual singularities.
(2.4)
1
Boundary Stabilization of a Hyperbolic Equation
Dening
(2.5)
a(u v) =
(2.6)
b(u v) =
n
X
Z
ij =1 n Z
X
ij =1
169
@u @v dx 8u v 2 V
aij (x) @x
@x
i
j
@u @v dx 8u v 2 V
bij (x) @x
@x
i
j
from (2.1), (2.2) and (2.3) there exist positive constants a0 a1 b0 and b1 such that
(2.7)
a0 jruj2 a(u u) a1 jruj2 8u 2 V
(2.8)
b0jruj2 b(u u) b1 jruj2 8u 2 V:
Now, we are in position to state our main result.
Theorem 2.1. Let
y0 y1 f g 2 V L2 (
) L2 (0 1 L2(
)) H 1 (0 1 L2 (;0 )):
Under the assumptions (2.1)-(2.3) and assuming that g(0) = 0, the problem (1.1)
possesses a unique weak solution y : (0 1) ! R such that
(2.9)
y 2 L (0 1 V ) y 2 L (0 1 L2(
)):
Moreover, provided that for large t, the inequality
1
(2.10)
Z
0
t
0
1
exp( 2" s) jf (s)j2 + j g(s)j2;0 ds tr
p
holds for some positive constants ", , and r, we obtain the following energy decay
E (t) C exp(; 2" t) 8t 0 and 8" 2 (0 "0 ]
where C and "0 are positive constants.
Remark 1. The hypothesis (2.10) means that the map
t 7;!
t
Z
0
exp( 2" s) jf (s)j2 + j g(s)j2;0 ds
p
must be bounded by a polinomial P (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 existence of solutions may be proven either by the Galerkin method or using semigroup
arguments. We employ the Galerkin method.
We dene the Hilbert space
(3.1)
H = u 2 V Au Bu 2 L2 (
)
where A and B are the operators dened by
A = ; @x@ aij (x) @x@
i
j
@ b (x) @
and B = ; @x
ij @x
i
j
and H is endowed by the natural inner product
(u v)H = (u v)V + (Au Av) + (Bu Bv):
170
M. M. Cavalcanti
Let us consider
(3.2)
y0 y1 2 H
satisfying the compatibility condition
@y0 + @y1 + (x) ;y1 ; g(0) = 0:
@a @b
(3.3)
In addition, let us assume that
(3.4)
f 2 H 1 (0 1 L2(
)) g 2 H 2 (0 1 L2(;0 )):
The variational formulation associated with problem (1.1) is given by
(y (t) w) + a(y(t) w) + b(y(t) w) + (y (t) w);0
= (f (t) w) + (g(t) w);0 8w 2 V:
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
(y (t) w);0 , we need to derive the above expression with respect to t: But it
lead us to technical diculties when we estimate y (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
(3.5)
v(x t) = y(x t) ; (x t)
where
(x t) = y0 (x) + ty1 (x) (x t) 2 0 1)
we obtain the equivalent problem for v :
00
0
0
00
v ; r (aij (x)rv) ; r (bij (x)rv ) = F in Q
v = 0 on 1
@v + @v + v = G on 1
@a @b
v(0) = v (0) = 0
00
(3.6)
0
0
0
0
where
(3.7)
F = f + r (aij (x)r
) + r (bij (x)r
)
@
; @
; :
(3.8)
G = g ; @
a @b
If v is a solution of (3.6) in 0,T], then y = v + is a solution of (1.1) in the same
0
0
0
interval. However, after two estimates we are going to prove later, we have that
(3.9)
jAv(t)j2 + jrv (t)j2 C (T ) 8t 2 0 T ] and 8T > 0:
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 0 1) using the standard
argument
;
2
2
Tmax = 1 or if Tmax < 1 then t lim
j
Av
(
t
)
j
+
jr
v
(
t
)
j
= 1:
T
0
0
!
max
Hence, it is sucient to prove that (3.6) has a solution in 0,T], which will be
done by the Galerkin method.
Boundary Stabilization of a Hyperbolic Equation
171
We represent by (! ) a basis in H which is orthonormal in L2 (
), by Vm the
subspace of H generated by the m-rst vectors !1 !m and dene
vm (t) =
(3.10)
m
X
i=1
gim (t)!i
where vm (t) is the solution of the folowing Cauchy problem:
(3.11) (vm (t) !j ) + a(vm (t) !j ) + b(vm (t) !j ) + (vm (t) !j );0
= (F (t) !j ) + (G(t) !j );0 00
0
0
vm (0) = vm (0) = 0 j = 1 m:
0
The aproximate system is a normal one of ordinary dierential equations. It has
a solution in 0 tm): 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) by gjm (t) summing
over 1 j m and considering (2.2), we obtain
(3.12)
1 d njv (t)j2 + a(v (t) v (t))o + b(v (t) v (t)) + pv (t) 2
m
m
m
m
m
2 dt m
0
0
0
0
0
;0
d (G(t) v (t)) ; (G (t) v (t))
= (F (t) vm (t)) + dt
m ;0
m ;0
2
12 jF (t)j2 + 12 jvm (t)j2 + dtd (G(t) vm (t));0 + C20 jG (t)j2;0 + 12 jrvm (t)j2
where C0 is a positive constant such that jvj;0 C0 jrvj , 8v 2 V:
Integrating (3.12) over (0,t) 0 < t < tm , taking into consideration (2.7) and (2.8)
and noting that vm (0) = vm (0) = 0, we get
Z t
Z t
p
2
jrv (s)j2 ds +
(3.13) 1 jv (t)j2 + a0 jrv (t)j2 + b
v (s) ds
0
0
0
0
0
2 m
0
m
0
m
m ;0
2
0
0
2
12 kF k2L2(0T L2()) + C20 kG k2L2(0T L2 (;0 )) + (G(t) vm (t));0
Z tn
o
1
jvm (s)j2 + jrvm (s)j2 ds:
+2
0
0
0
0
0
On the other hand, for ar arbitrary > 0 we have
(G(t) vm (t));0 C40 jG(t)j2;0 + jrvm (t)j2 :
2
(3.14)
Combining (3.13), (3.14) and choosing > 0 small enough, we obtain the rst
estimate
(3.15)
jvm (t)j2 + jrvm (t)j2 +
Z
0
0
t
jrvm (s)j2 ds +
t
Z
0
0
p
2
vm (s) ; ds L1
0
0
where L1 is a positive constant independent of m 2 N and t 2 0 T ]:
172
M. M. Cavalcanti
3.1.2. The Second Estimate. First of all we are going to estimate vm (0) in L2 (
)
norm. Taking t = 0 in (3.11) and taking into account that vm (0) = vm (0) = 0, it
follows that
(3.16)
(vm (0) !j ) = (F (0) !j ) + (G(0) !j );0 :
From (3.3) and (3.8) we have that G(0) = 0, and from (3.16) we conclude
jvm (0)j2 = (F (0) vm (0))
which implies that
(3.17)
jvm (0)j L 8m 2 N
where L is a positive constant independent of m 2 N:
Now, taking the derivative of (3.11) with respect to t, we can write
00
0
00
00
00
00
(3.18) (vm (t) !j ) + a(vm (t) !j ) + b(vm (t) !j ) + (vm (t) !j );0
= (F (t) !j ) + (G (t) !j );0 :
Multiplying both sides of (3.18) by gjm (t) and summing over 1 j m we
obtain
d njv (t)j2 + a(v (t) v (t))o + b(v (t) v (t)) + pv (t) 2
(3.19) 12 dt
m ;0
m
m
m
m
m
d (G (t) v (t)) ; (G (t) v (t)) :
= (F (t) vm (t)) + dt
m ;0
m ;0
000
0
00
00
0
0
00
00
0
0
0
00
00
0
00
00
0
00
0
Using arguments analogous to those considered in the rst estimate, observing
that vm (0) = 0, and taking into account (3.17), from (3.19) we obtain the second
estimate
0
(3.20)
jvm (t)j + jrvm (t)j +
2
00
2
0
Z
0
t
jrvm (s)j ds +
00
2
t
Z
0
p
2
vm (s) ; ds L2
00
0
where L2 is a positive constant independent of m 2 N and t 2 0 T ]:
Due to estimates (3.15) and (3.20) we can extract a subsequence fv g of fvm g
such that
(3.21)
v * v weak star in L (0 T V )
(3.22)
v * v weak star in L (0 T V )
(3.23)
v * v weak in L2 (0 T V )
(3.24)
v * v weak star in L (0 T L2(
))
(3.25)
v * v weak in L2 (0 T L2(;0 )):
The above convergences are sucient to pass to the limit.
1
0
0
1
00
00
00
00
0
1
0
3.2. Uniqueness. Supose we have two solutions y and y^ of problem (1.1). Then
z = y ; y^ satises
(3.26)
(z (t) w) + a(z (t) w) + b(z (t) w) + (z (t) w);0 = 0 8w 2 V
z (0) = z (0) = 0:
00
0
0
0
173
Boundary Stabilization of a Hyperbolic Equation
Taking w = 2z (t) in (3.26) we obtain
0
d jz (t)j2 + a(z (t) z (t))o + 2b(z (t) z (t)) + 2
dt
n
(3.27)
0
0
Integrating (3.27) over (0,t) we obtain
t
Z
0
Z
t
p
p
2
z (t) ; = 0:
0
0
2
z (s) ; ds = 0
jz (t)j + a0 jrz (t)j + 2b0 jrz (s)j ds + 2
0
0
0
which implies that jrz (t)j = jz (t)j = 0: This completes the proof.
3.3. Solvability of weak solutions. We have just proved the existence of solu2
0
2
2
0
0
0
tions of the problem (1.1) when y0 and y1 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 since g(0) = 0 given y0 y1 2 V L2(
) it is sucient
to consider
@u = 0 on ;
y0 2 D(A) = u 2 V Au 2 L2 (
) and @
0
a
such that
y0 ! y0 in V
and
y1 2 D(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
p
2
E (t) = ;b(y (t) y (t)) ; y (t) ; + (f (t) y (t)) + (g(t) y (t));0 :
0
Let and be positive constants such that
(4.2)
jvj2 jrvj2 8v 2 V
(4.1)
and
(4.3)
0
0
0
p
0
0
2
v ; jrvj2 8v 2 V:
0
For an arbitrary " > 0 we dene the perturbed energy
(4.4)
E" (t) = E (t) + "(t)
where
Z
(t) = y ydx:
(4.5)
0
0
174
M. M. Cavalcanti
Proposition 4.1. There exists C1 > 0 such that
jE" (t) ; E (t)j "C1 E (t) 8t 0 and 8" > 0:
Proof. From (2.7), (4.2) and (4.5) we obtain
;
j(t)j 12 jy j2 + 21 a0 1 a(y y) a0 1 + 1 E (t):
(4.6)
;
0
;
If we dene C1 = a0 1 + 1then from (4.4) and (4.6) we can write
jE" (t) ; E (t)j = "j(t)j "C1 E (t):
This concludes the proof.
Proposition 4.2. There exist C2 = C2 (") and "1 positive constants such that
p
E" (t) ;"E (t) + C2 jf (t)j2 + j g(t)j2;0 8t 0 and 8" 2 (0 "1 ]:
;
0
Proof. First of all we must estimate (t) in terms of E (t): Taking the derivative
of (t) given in (4.5) and replacing y by r (aij (x)ry) + r (bij ry ) + f in the
0
00
0
expression obtained it follows that
(4.7)
Z
Z
Z
Z
(t) = r (aij (x)ry) y dx + r (bij ry ) y dx + fy dx + jy j2 dx:
0
0
0
On the other hand, from Gauss' theorem and taking into account that
@y @y
@a + @b = (g ; y ) on ;0
0
0
we obtain
(4.8)
Z
Z
r (aij (x)ry) y dx + r (bij ry ) y dx
0
= ;a(y(t) y(t)) ; b(y (t) y(t)) ;
Z
0
y y d; +
Z
gy d;:
0
;0
;0
Replacing (4.8) in (4.7) adding and subtracting the term jy j2 dx in (4.7) we get
(4.9)
Z
Z
Z
Z
2
(t) = ;2E (t) ; b(y (t) y(t)) + 2 jy j dx ; y y d; + fy dx + gy d;:
R
0
0
0
0
0
;0
;0
Now, from (2.7), (4.2), (4.3) and (4.9), using for an arbitrary > 0 the inequality
ab a42 + b2 we can write
(4.10)
2a 1 jj
b
jj
(t) ; (2 ; 8)E (t) + 2 + 4 0 jry (t)j2
0
1
+ a40
;
where
;
0
p
1
y (t) ; + a04 jf (t)j2 +
0
jjbjj =
2
0
n
X
ij =1
;
jjbij jjL1() :
p
2
g(t) ;
0
175
Boundary Stabilization of a Hyperbolic Equation
Choosing = 18 from (4.10) we have
(4.11)
p
2
2
(t) ;E (t) + M1 jry (t)j + M2 y (t) + M3 jf (t)j2 +
0
0
where
p
0
;0
2
g(t) ;
0
M1 = 2 + jjbjj2 a0 1 M2 = 2a0 1 and M3 = 2a0 1:
;
;
;
;
Thus, combining (2.8), (4.1), (4.4) and (4.11), we conclude
E" (t) = E (t) + " (t)
0
0
0
p
2
;b0 jry (t)j2 ; y (t) ; + (f (t) y (t)) + (g(t) y (t));0 ; "E (t)
0
0
+ "M1 jry (t)j2 + "M2
0
0
p
Dening
0
2
y (t) ; + "M3 jf (t)j2 +
0
2
p
g(t) ;
0
which implies
(4.12)
0
0
p
2
E" (t) ; (b0 ; "(M1 + 1)) jry (t)j2 ; (1 ; "(M2 + 1)) y (t) ;
0
1
2
2
; "E (t) + 4" + "M3 jf (t)j + 4" + "M3 jg(t)j;0 :
0
0
"1 = min M b0+ 1 M 1+ 1
1
2
and choosing " 2 (0 "1 ] it follows that
E" (t) ;"E (t) + C2 (") jf (t)j2 +
0
which concludes the proof.
p
2
g(t) ;
0
Proof of the exponential decay. We dene
"0 = minf"1 2C1 g
1
and let us consider " 2 (0 "0 ]: From Proposition 4.1 we have
(4.13)
(1 ; C1 ")E (t) E" (t) (1 + C1 ")E (t)
Since " 1=2C1 ,
1 E (t) E (t) 3 E (t) 2E (t) 8t 0
(4.14)
"
2
2
and therefore
(4.15)
;"E (t) ; 2" E" (t):
Hence, from (4.15) and considering Proposition 4.2 we obtain
p
E" (t) ; 2" E" (t) + C2 jf (t)j2 + j g(t)j2;0 :
Consequently
0
d E (t) exp( " t) C jf (t)j2 + jpg(t)j2 exp( " t):
2
;0
dt "
2
2
0
176
M. M. Cavalcanti
Integrating the above inequality over 0,t] we get
Z t
p
E" (t) exp(; 2" t)E" (0) + C2 exp(; 2" t) exp( 2" s) jf (s)j2 + j g(s)j2;0
0
and taking into consideration (4.14) we see that
(4.16)
E (t) 3E (0) + 2C2
Z
0
t
exp( 2" s) jf (t)j2 + j g(t)j2;0
p
exp(; 2" t):
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, Maringa - PR, Brasil.
marcelo@gauss.dma.uem.br
This paper is available via http://nyjm.albany.edu:8000/j/1998/4-11.html.