IJMMS 25:1 (2001) 43–52
PII. S0161171201003994
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
THE INITIAL BOUNDARY VALUE PROBLEM OF A MIXED-TYPED
HEMIVARIATIONAL INEQUALITY
GUO XINGMING
(Received 20 May 1999)
Abstract. A mixed-typed differential inclusion with a weakly continuous nonlinear term
and a nonmonotone discontinuous nonlinear multi-valued term is studied, and the existence and decay of solutions are established.
2000 Mathematics Subject Classification. Primary 34G20, 35M10, 35M20.
1. Introduction. In the present paper, the following initial boundary value problem
of a degenerate multi-valued hyperbolic-parabolic inequality will be considered:
ü(t) + A(t)(u̇)(t) + B(u)(t) + ϕ u(x, t) f (t),
u(x, t) = 0,
a.e. t ∈ [0, T ],
a.e. (x, t) ∈ Σ = ∂Ω × [0, T ],
u(0) = u0 ,
(1.1)
u̇(0) = u1 ,
where A is a weakly continuous operator; B is a linear, continuous, and symmetric
operator; ϕ is a nonmonotonous, discontinuous, and nonlinear set-valued mapping.
Physical motivations for studying inequality (1.1) come partly from problems of continuum mechanics and optimal control problems, where nonmonotone, nonlinear, discontinuous, and multi-valued constitutive laws and boundary or external constraints
lead to various typed hemivariational inequalities, the mixed hyperbolic-parabolic
hemivariational inequality is one of those [11, 12, 14].
For inequality (1.1), its stationary problems have been studied by many researchers
(see [1, 2, 4, 13, 14, 15] and references therein). When ϕ degenerates into a class
of single-valued mappings, inequality (1.1) becomes an equation, and when A and B
were some special linear mappings and satisfy some conditions, equation (2.1) and
some of its evolution equations have been investigated and applied intensively (see
[5, 3, 6, 7, 8, 9, 10] and the references therein).
In this paper, we investigate the existence and decay of weak solution of the mixed
hyperbolic-parabolic inequality (1.1) with ϕ, A, and B satisfying some conditions. We
apply the Faedo-Galerkin method for the proof of existence of solutions.
2. Preliminaries. Let Ω be a bounded open set of Rn with regular boundary Γ . Let
T denote a positive real number, Q = Ω × [0, T ]. Suppose that b ∈ L∞
loc (R), for every
ρ > 0, set
b
inf b ξ1 ,
b̄ρ (ξ) = ess sup b ξ1 ,
(2.1)
- ρ (ξ) = |ξess
−ξ|<ρ
1
|ξ1 −ξ|<ρ
44
GUO XINGMING
they are all monotone for ρ > 0. Set
b
lim b
ρ (ξ),
- ρ (ξ) = ρ→0
+-
ϕ(ξ) = b(ξ),
b̄(ξ) .
-
b̄(ξ) = lim b̄ρ (ξ),
ρ→0+
(2.2)
ξ
Let J(ξ) = 0 b(t) dt, then ∂ C J(ξ) ⊆ ϕ(ξ), where ∂ C J(ξ) denotes the Clarkesubdifferential of J (see [2]). If b(ξ± ) exists for every ξ ∈ R, then ϕ(ξ) = ∂ C J(ξ).
If b is continuous at the point ξ, then ϕ(ξ) is single-valued at ξ, if J is convex, ϕ(ξ)
is maximal monotone (see [2]).
Let V = H01 (Ω), (·, ·) denotes the inner product of L2 (Ω), ·, · denotes the dual pair
between V and V = H −1 (Ω) which is compatible with the inner product of L2 (Ω). Let
|x|X denote the norm of the element x of the Banach space X.
Considering the following initial boundary value problem of a hyperbolic-parabolic
hemivariational inequality:
ü(t) + A(t)u̇(t) + Bu(t) + g(t) = f (t),
u(x, t) = 0,
a.e. t ∈ [0, T ],
a.e. (x, t) ∈ Σ = ∂Ω × [0, T ],
(2.3)
u̇(0) = u1 ,
u(0) = u0 ,
g(x, t) ∈ ϕ u(x, t) , a.e. (x, t) ∈ QT = Ω × [0, T ],
where f , u0 , and u1 are given.
First we list some assumptions:
(1) ∃c > 0, |b(ξ)| ≤ c(1 + |ξ|), a.e. ξ ∈ R.
(2) A : L2 (0, T ; L2 (Ω)) → L2 (0, T ; L2 (Ω)) is weakly continuous, and A(t) is nonnegative, that is, A(t)v, v ≥ 0, for a.e. t ≥ 0 and every v ∈ L2 (Ω).
(3) The function t → A(t)u, v is measurable on [0, T ] for all u, v ∈ L2 (Ω).
(4) B : H01 (Ω) → H −1 (Ω) is linear, continuous, symmetric, and semicoercive, that is,
∃c1 > 0, c2 > 0, c3 > 0
|Bv|H −1 (Ω) ≤ c1 |v|H 1 (Ω) ,
0
Bu, v = Bv, u,
Bv, v + c3 |v|2L2 (Ω) ≥ c2 |v|2H 1 (Ω) ,
0
∀u, v ∈ H01 (Ω),
Let β be any mollifier satisfying β ∈ C ∞ (R), β ≥ 0, supp β ⊂ (−1, 1), and
dξ = 1. Set
bε (ξ) =
1
ε
R
β
(2.4)
∀v ∈ H01 (Ω).
ξ −z
b(z) dz =
β(z)b(ξ − εz) dz,
ε
|z|≤1
for every ε > 0.
R β(ξ)
(2.5)
It is easy to see that bε is a smooth function, and also satisfies assumption (1) with
possible different constant c if b is agreeable with assumption (1). For convenience,
we denote b1/n by bn for any positive integer n.
THE INITIAL BOUNDARY VALUE PROBLEM OF A MIXED-TYPED . . .
45
3. Existence of solution
Theorem 3.1. Assume that f ∈ L2 (0, T ; L2 (Ω)), u0 ∈ H01 (Ω)∩Lp+1 (Ω), u1 ∈ L2 (Ω).
Then, under assumptions (1), (2), (3), and (4), there exists a function u defined in Ω ×
[0, T ] such that
u ∈ L∞ 0, T ; H01 (Ω)
C [0, T ]; L2 (Ω) ,
u̇ ∈ L∞ 0, T ; L2 (Ω)
C [0, T ]; H −1 (Ω) ,
ü ∈ L2 0, T ; H −1 (Ω) ,
(3.1)
and
ü(t) + A(t)u̇(t) + Bu(t) + g(t) = f (t), in L2 0, T ; H −1 (Ω) ,
g(t) ∈ ϕ u(x, t) , a.e. (x, t) ∈ Ω × [0, T ],
u(0) = u0 ,
(3.2)
u̇(0) = u1 .
Proof. Let {en }∞
be a subset of V = H01 (Ω) satisfying span{en } = V , (ei , ej ) =
n 1n=1
n
δij . Let xn = 1 ωi ei → u0 strongly in V and Lp+1 (Ω), yn = 1 ω2i ei → u1 strongly
in L2 (Ω).
Considering the following regularized equation of inequality (1.1)
ξ̈ n = M n + N n + h,
ξ n t=0 = ω1n ,
ξ̇ n t=0 = ω2n ,
(3.3)
where ξ n = {ξin }1×n , ω1n = {ω1i }1×n , ω2n = {ω2i }1×n , h = {f , ei }1×n , M n = {Min }1×n ,
n n n
n n n n n
n
Min =− A(t)
1 ξ̇j ej , ei , N = {Ni }1×n , Ni = − B
1 ξj ej , ei − bn
1 ξj ej , ei ,
where “·” denotes time derivate.
Equation (3.3) is a vector-valued ordinary differential equation and its local solution
n
ξ n exists on In = [0, Tn ], 0 < Tn ≤ T . Set un (t) = 1 ξjn ej (t ∈ In ). Equation (3.3) is
equal to
ün , ei = − A(t)u̇n , ei − Bun , ei − bn un , ei + f , ei ,
i = 1, 2, . . . , n.
(3.4)
Multiplying (3.4) by ξ̇in , summing over from i = 1 to i = n and integrating over [0, t]
(t ≤ In ), we get
u̇n (t)22
L
(Ω) + Bun (t), un (t) + 2
t
0
Au̇n , u̇n dτ + 2
=2
t
0
t
0
bn un , u̇n dτ
f , u̇n dτ + yn , yn + Bxn , xn ,
(3.5)
46
but
GUO XINGMING
t
0
t
J un (x, τ) 0 dx
un (x,t)
un (x,0)
=
bn (λ) dλ −
bn (λ) dλ dx,
bn un , u̇n dτ =
Ω
Ω
0
0
t u (x,t)
n
|λ| dλ
b u , u̇n dτ ≤ c
un (x, t) + un (x, 0) + 0 n n
Ω
0
un (x,0)
+
|λ| dλ dx,
0
1
un (x,t)
2
|λ| dλ = un (x, t) ,
2
0
un (x,0)
1
2
|λ|dλ = un (x, 0) ,
0
2
t c
2
2
1 + |Ω| un (t)L2 (Ω) + xn L2 (Ω) ,
bn un , u̇n dτ ≤
0
2
where |Ω| denotes the Lebesgue measure of the domain Ω.
From (3.5), it follows that there exists c4 > 0 such that
t
u̇n (t)22 +un (t)2 1 ≤ c4 + c3 + c 1+|Ω| un (t)22 +2
f , u̇n dτ.
L (Ω)
L (Ω)
H0 (Ω)
2
0
We note that
un (t) = un (0) +
t
0
t
0
(3.8)
u̇n L2 (Ω)
dτ,
using Hölder’s inequality, we get that there exists c5 , c6 > 0 such that
t
2
u̇n 2 dτ,
un (t)22
L (Ω) ≤ c5 + c6
L (Ω)
0
(3.7)
u̇n dτ,
un (t)22
L (Ω) ≤ un (0) L2 (Ω) +
t
(3.6)
0
f , u̇n dτ ≤ |f |L2 (0,T ;L2 (Ω)) · u̇n L2 (0,t;L2 (Ω))
(3.9)
≤
2
1 2
|f |L2 (0,T ;L2 (Ω)) + u̇n L2 (0,t;L2 (Ω)) .
2
From (3.7), (3.9), and (3.10), we obtain that there exists c7 , c8 > 0 such that
t
u̇n (τ)22 dτ
u̇n (t)22 + c2 un (t)2 1
≤
c
+
c
t ∈ In ,
7
8
L (Ω)
L (Ω)
H (Ω)
0
0
(3.10)
(3.11)
this implies that
u̇n (t)22
L (Ω)
≤ c7 + c 8
t
0
u̇n (τ)22
L (Ω) dτ
Using Gronwall’s inequality it follows that
u̇n (t)22
L (Ω) ≤ c7 exp c8 t
t ∈ In .
t ∈ In .
(3.12)
(3.13)
47
THE INITIAL BOUNDARY VALUE PROBLEM OF A MIXED-TYPED . . .
Therefore, from (3.9), (3.11), and (3.13), we get that there exists c9 > 0,
un (t) 2
un (t) 1
u̇n (t) 2
t ∈ In ,
L (Ω) ≤ c9 ,
L (Ω) ≤ c9 ,
H (Ω) ≤ c9 ,
(3.14)
0
where c4 , c5 , c6 , c7 , c8 , c9 are positive constants independent of n and Tn , from which
we can assert that In = [0, T ] (∀n).
For every η ∈ span{e1 , e2 , . . . , en }, from (3.4)
ün , η ≤ A(t) u̇n 2 · η 2 + f (t) 2 · η 2
L (Ω)
L (Ω)
L (Ω)
L (Ω)
(3.15)
+ bn un L2 (Ω) · ηL2 (Ω) + B · un H 1 (Ω) · ηH 1 (Ω) ,
0
0
where |B| is the norm of linear continuous operator B
ün (t) −1
H (Ω)
ün (t), η = sup ün (t), η =
sup
|η|V =1
η∈span{e1 ,...,en }
|η|V =1
≤ c10 A(t) u̇n L2 (Ω) +
f (t)
L2 (Ω) +
bn un L2 (Ω)
+ |B| · un (t)H 1 (Ω) ,
where c10 is the imbedding constant which H01 (Ω) imbeds in L2 (Ω)
2
2
2
bn un (t) dx ≤
bn un (t) 2
c 2 1 + un (x, t) dx
L (Ω) =
Ω
Ω
2 2
≤ 2c 2
1 + un (x, t) dx = 2c 2 Ω + un (t)L2 (Ω) ,
0
(3.16)
(3.17)
Ω
this shows that {bn (un )} is also a bounded subset of L∞ (0, T ; L2 (Ω)). Since A is weakly
continuous, it must be a bounded operator from L2 (0, T ; L2 (Ω)) to L2 (0, T ; L2 (Ω)).
But {u̇n } is a bounded subset of L2 (0, T ; L2 (Ω)), {A(t)(u̇n )} must be a bounded
subset of L2 (0, T ; L2 (Ω)). Inequality (3.16) implies that {ün } is a bounded subset of
L2 (0, T ; H −1 (Ω)).
Therefore, there exist a subsequence of {un }, still denoted by itself, and a function
u such that u ∈ L∞ (0, T ; H01 (Ω)), u̇ ∈ L∞ (0, T ; L2 (Ω)), ü ∈ L2 (0, T ; H −1 (Ω)) satisfying
un → u weakly-star in L∞ 0, T ; H01 (Ω) ,
u̇n → u̇ weakly-star in L∞ 0, T ; L2 (Ω) ,
(3.18)
ün → ü weakly in L2 0, T ; L−1 (Ω) ,
bn un → g weakly-star in L∞ 0, T ; L2 (Ω) .
Since, the space W (V ) defined by W (V ) = {u ∈ L2 (0, T ; V ), u̇ ∈ L2 (0, T ; V )} forms
a real Hilbert space with the norm |u|W = |u|L2 (0,T ;V ) +|u̇|L2 (0,T ;V ) and is continuously
imbedded in C([0, T]; L2 (Ω)), it is obvious that u ∈ C(0, T ; L2 (Ω)), u̇ ∈ C(0, T ; H −1 (Ω)).
Hence, u(0), u̇(0) make sense.
For λ ∈ L2 (0, T ), from (3.4) we have
T
T
T
ün , λei dt = −
A(t) u̇n , λei dt −
B un , λei dt
0
0
0
(3.19)
T
T
bn un , λei dt +
f (t), λei dt, i = 1, 2, . . . , n.
−
0
0
48
GUO XINGMING
For every given positive integer i, let n → ∞ in (3.19), it follows that
T
0
ü, λei dt = −
−
T
0
T
0
A(t)(u̇), λei dt −
g, λei dt +
T
T
0
B(u), λei dt
f (t), λei dt,
0
(3.20)
i = 1, 2, . . . , n.
Therefore, we have from (3.20)
ü(t) + A(t)(u̇) + B(u) + g(t) = f (t),
in L2 0, T ; H −1 (Ω) .
(3.21)
Next, we demonstrate that
g(x, t) ∈ ϕ u(x, t)
a.e. (x, t) ∈ QT = Ω × [0, T ].
(3.22)
Since, un (x, t) → u(x, t) a.e. (x, t) ∈ QT , by Eropoß’s theorem [9], for every δ > 0,
there exists a subset Qδ ⊆ QT = Ω × [0, T ], |Qδ | ≤ δ,
un (x, t) → u(x, t)
uniformly in QT \Qδ
(3.23)
that is, for every ε > 0, there exists a positive integer N, when n ≥ N,
un (x, t) − u(x, t) ≤ ε
∀(x, t) ∈ QT \Qδ .
(3.24)
It is obvious that, when 1/n ≤ ε and n ≥ N, for almost everywhere (x, t) ∈ QT \Qδ
b
- n un (x, t) = bn un (x, t) = b̄n un (x, t) ≤ b̄ε un (x, t) ≤ b̄2ε u(x, t) .
For every µ ∈ L1 (0, T ; L2 (Ω)), µ ≥ 0
g(x, t)µ(x, t) dx dt = lim
bn un (x, t) µ(x, t) dx dt
n→∞
QT \Qδ
QT \Qδ
b̄2ε u(x, t) µ(x, t) dx dt,
≤
QT \Qδ
g(x, t)µ(x, t) dx dt ≤ lim sup
b̄2ε u(x, t) µ(x, t) dx dt
QT \Qδ
≤
ε→0+
QT \Qδ
(3.25)
(3.26)
QT \Qδ
b̄ u(x, t) µ(x, t) dx dt.
Analogously, we can obtain
g(x, t)µ(x, t) dx dt ≥
QT \Qδ
QT \Qδ
b
- u(x, t) µ(x, t) dx dt.
(3.27)
Hence, (3.26) and (3.27) imply that
g(x, t) ∈ ϕ u(x, t)
a.e. (x, t) ∈ QT \Qδ .
(3.28)
a.e. (x, t) ∈ QT = Ω × [0, T ].
(3.29)
Finally, let δ → 0+ , we get
g(x, t) ∈ ϕ u(x, t)
THE INITIAL BOUNDARY VALUE PROBLEM OF A MIXED-TYPED . . .
49
Let λ ∈ C 1 [0, T ], λ(T ) = 0, integrating by parts the left-hand side of equations (3.19)
and (3.20) gives
T
T
t
u̇n , λ̇ei dt =−
A(t) u̇n , λei dt−
B un , λei dt
− u̇n (0), λ(0)ei −
0
0
0
(3.30)
T
T
bn un , λei dt −
f (t), λei dt,
−
− u̇(0), λ(0)ei −
T
0
0
u̇, λ̇ei dt = −
−
T
0
A(t) u̇ , λei dt −
0
T
0
g, λei dt −
T
0
t
0
B(u), λei dt
f (t), λei dt,
making comparison between (3.30) and (3.31) we get that
lim u̇n (0) − u̇(0), ei = 0, i = 1, 2, . . . , n
n→∞
(3.31)
(3.32)
therefore, this implies that
u̇n (0) → u̇(0) weakly in H −1 (Ω)
(3.33)
uniqueness of limit implies that u̇(0) = u1 (in H −1 (Ω)).
Let λ ∈ C 2 [0, T ], λ(T ) = 0, λ̇(T ) = 0. Analogously, integrating by parts the left-hand
side of equations (3.30) and (3.31), and making comparison with the obtained results
again gives: u(0) = u0 (in L2 (Ω)).
Theorem 3.2. Let f ∈ L2 (0, T ; L2 (Ω)), u0 ∈ H01 (Ω) ∩ L∞ (Ω), u1 ∈ L2 (Ω). Assume
that b satisfies
(1 ) b(ξ)ξ ≥ 0 for almost everywhere ξ ∈ R, and ∃c̄ > 0,
b(ξ) ≤ c̄ 1 + |ξ|p , a.e. ξ ∈ R, if n > 2, 0 < p ≤ 2n ; if n ≤ 2, 0 ≤ p < ∞. (3.34)
n−2
Then, under assumptions (2), (3), and (4), there exists a function v defined in Ω × [0, T ]
satisfying
v̇ ∈ L∞ 0, T ; L2 (Ω) ,
v ∈ L∞ 0, T ; H01 (Ω) ,
v̈ + A(t)(v̇) + B(v) + ḡ(t) = f (t) in L1 0, T ; H −1 (Ω) + L1 (Ω) ,
(3.35)
ḡ(x, t) ∈ ϕ v(x, t) a.e. (x, t) ∈ QT = Ω × [0, T ],
v(0) = u0 ,
v̇(0) = u1 .
Proof. It is also easy to see that bε satisfies assumption (1) with possible different
∞
constant c̄. Analogously to Theorem 3.1, we still may get (3.5), where en n=1 is a basis
of H01 (Ω) ∩ L∞ (Ω) satisfying (ei , ej ) = δij . Set
ξ
bn (t) dt,
(3.36)
Jn (ξ) =
0
then Jn (ξ) ≥ 0, ∀ξ ∈ R, and
t
bn un , u̇n dτ = Jn un (x, t) dx− Jn un (x, 0) dx ≥ − Jn un (x, t) dx
Ω
Ω
Ω
0
p
z bn (ξ) ≤ β(z)b ξ −
dz ≤ d1 + d2 |ξ|p ,
|z|≤1
n
(3.37)
50
GUO XINGMING
where d1 and d2 are positive constants independent of n.
xn
xn Jn xn = bn (t) dt
bn (t) dt ≤ sgn xn ·
0
0
p+1
d2 xn ≤ sgn xn ·
,
d1 + d2 |t|p dt = d1 |xn | +
(p + 1)
0
p+1
d x Jn xn dx ≤ d1 xn 1 + 2 n Lp+1 (Ω) .
Jn xn dt ≤
L (Ω)
(p + 1)
Ω
Ω
xn
(3.38)
Since Lp+1 (Ω) ⊂ L1 (Ω) and un (0) = xn → u0 strongly in Lp+1 (Ω), and |xn |L1 (Ω) are
bounded, and so is Ω Jn (xn (x)) dx. From (3.5) we have
2
u̇n (t)22 + c2 un (t)2 1
L (Ω)
H (Ω) ≤ c4 + c3 un (t) L2 (Ω) + 2
0
t
0
f , u̇n dτ.
(3.39)
It is easy to see that (3.9), (3.10), (3.11), (3.13), and (3.14) are still true and the solution
of (3.3) can be extended to interval [0, T ]. By Sobolev imbedding theorem, we have,
∗
for a.e. t ∈ [0, T ], if n > 2, then H01 (Ω) ⊂ Lp (Ω) ⊂ Lp (Ω), p ∗ = 2n/(n − 2), and
|un (t)|Lp (Ω) ≤ c10 |un (t)|H 1 (Ω) ≤ c10 c9 ; if n = 2, then H01 (Ω) ⊂ Lq (Ω), when 1 ≤ q < ∞,
0
so |un (t)|Lp (Ω) ≤ c10 |un (t)|H 1 (Ω) ≤ c10 c9 ; if n = 1, then H01 (Ω) ⊂ C(Ω̄) and ditto,
0
|un (t)|C(Ω̄) = maxx∈Ω̄ |un (x, t)| ≤ c10 c9 , where Ω̄ denotes the closure of Ω and c10 is
the imbedding constant which H01 (Ω) imbeds in Lp (Ω) or C(Ω̄). Note that, we always
have that bn (un ) ∈ L∞ (0, T ; Lp0 (Ω)), where p0 = (n + 1)/(n − 2) and {bn (un )} is a
bounded subset of L∞ (0, T ; Lp0 (Ω)). Therefore, there exist a subsequence of {un }, still
denoted by itself, and a function v such that v ∈ L∞ (0, T ; H01 (Ω)), v̇ ∈ L∞ (0, T ; L2 (Ω))
satisfying
un → v weakly-star in L∞ 0, T ; H01 (Ω) ,
u̇n → v̇ weakly-star in L∞ 0, T ; L2 (Ω) ,
(3.40)
∞
p0
bn un → g weakly-star in L 0, T ; L (Ω) .
Since, the dual of the space H01 (Ω) ∩ L∞ (Ω) is the space L1 (0, T ; H −1 (Ω) + L1 (Ω)), it is
easy to obtain from (3.4) that
v̈(t) + A(t)v̇ + B(v) + ḡ(t) = f (t) in L1 0, T ; H −1 (Ω) + L1 (Ω) .
(3.41)
Analogous to Theorem 3.1, we can complete the proof of this theorem.
Remark 3.3. If A(t) = A and A is linear, then the uniqueness of such solution will
be obtained in the same way as in [3].
4. Decay of the solution
Theorem 4.1. Let T = +∞, f ≡ 0. Suppose that for every t ≥ 0, the operator A(t)
satisfies
(4.1)
A(t)w, w ≥ δ0 |w|2L2 (Ω) , ∀w ∈ L2 (Ω).
2
≤ c2 , here c10 is an imbedding constant
Moreover, if Bw, w ≥ 0, ∀w ∈ H01 (Ω) or c3 c10
1
which H0 (Ω) imbeds in L2 (Ω). Then, under conditions of Theorem 3.2, the solution in
THE INITIAL BOUNDARY VALUE PROBLEM OF A MIXED-TYPED . . .
51
Theorem 3.2 obtained from the regularized equation (3.3) satisfies
u̇(t)22
L (Ω)
≤ µ1 exp − µ2 t ,
a.e. t ≥ 0,
(4.2)
where δ0 , µ1 , µ2 are positive constants.
Proof. Let un be a solution of (3.3), that is, satisfies (3.4) and (3.5). Since
Jn (un (x, t)) ≥ 0, by (3.5) we have
u̇(t)22
L
(Ω) + Bun (t), un (t) ≤ c11 − 2δ0
t
0
u̇n (τ)22
L (Ω) dτ,
t ∈ [0, +∞),
(4.3)
where c11 is a positive constant independent of n. If Bw, w ≥ 0, for every w ∈ H01 (Ω),
Bun (t), un (t) ≥ 0. Analogously to [7, Theorem 4], we obtain:
u̇n (t)22
L (Ω)
≤ c11 exp − 2δ0 t ,
a.e. t ≥ 0.
(4.4)
2
≤ c2 , we get from (4.3) that
If c3 c10
u̇n (t)22
L
2
2
(Ω) +c2 un (t) H 1 (Ω) ≤ c11 +c3 un (t) L2 (Ω) −2δ0
0
t
0
2
2 un (t)H 1 (Ω) −2δ0
≤ c11 +c3 c10
0
u̇n (τ)22
L (Ω) dτ
t
0
u̇n (τ)22 dτ,
L (Ω)
(4.5)
from which it is permitted to get inequality (4.4).
Since |u̇n (t)|L2 (Ω) ≤ c9 , u̇ → u̇ weakly-star in L∞ (0, ∞; L2 (Ω)), it is easy to obtain
that u̇(t) → u̇(t) weak in L2 (Ω) for a.e. t ≥ 0. But L2 (Ω) is a real Hilbert space,
therefore, |u̇(t)|L2 (Ω) ≤ limn→∞ |u̇n (t)|L2 (Ω) , a.e. t ≥ 0. Finally, we get |u̇(t)|2L2 (Ω) ≤
c11 exp(−2δ0 t), (a.e. t ≥ 0).
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Guo Xingming: Shanghai Institute of Applied Mathematics and Mechanics, Shanghai
University, Shanghai 200072, China
E-mail address: xmguo@yc.shu.edu.cn