The first initial boundary value problem for semilinear hyperbolic equations in nonsmooth domains

JOURNAL OF SCIENCE OF HNUE
Mathematical and Physical Sci., 2013, Vol. 58, No. 7, pp. 39-49
This paper is available online at

THE FIRST INITIAL-BOUNDARY VALUE PROBLEM
FOR SEMILINEAR HYPERBOLIC EQUATIONS IN NONSMOOTH DOMAINS

Vu Trong Luong and Nguyen Thanh Tung
Faculty of Mathematics, Tay Bac University
Abstract. In this paper we study the first initial boundary problem for semilinear
hyperbolic equations in nonsmooth cylinders Q = Ω × (0, ∞), where Ω is a
nonsmooth domain in Rn , n ≥ 2. We established the existence and uniqueness
of a global solution in time.
Keywords: Initial boundary value problem, semilinear hyperbolic equation, global
solution, non-smooth domain.

1. Introduction
Let Ω ⊂ Rn be a bounded domain with non-smooth boundary ∂Ω, set ΩT = Ω ×
(0, T ), with 0 < T < +∞. We use the notations H 1 (Ω), H01 (Ω) as the usual Sobolev
spaces and H −1 (Ω) as the dual space of H01 (Ω) is defined in [1]. We denote L2 (Ω) as the
space L2 (Ω) is defined in [2]. Suppose X is a Banach space with the norm ∥ · ∥X . The
space Lp (0, ∞; X) consists of all measurable functions u : [0, ∞) −→ X with norm
∥u∥Lp (0,T ;X)

 p1
∞
∫
=  ∥u(t)∥pX dt < +∞

for 1 ≤ p < +∞.

0

We consider the partial differential operator
(
) ∑
n
n
∑
∂u
∂u
∂
ij
a (x, t)
+
bi (x, t)
+ c(x, t)u
Lu = −
∂x
∂x
∂x
j
i
i
i=1
i,j=1
where (x, t) ∈ Q = Ω × (0, ∞); aij , bi , c ∈ C 1 (Q)

(1.1)

(i, j = 1, · · · , n)

Received March 12, 2013. Accepted June 5, 2013
Contact Nguyen Thanh Tung, e-mail address:

39


Vu Trong Luong and Nguyen Thanh Tung

aij (x, t) = aji (x, t)

for i, j = 1, 2, · · · , n; (x, t) ∈ Q.

(1.2)

The operator L is strongly elliptic. Then there exists θ > 0, ∀ξ ∈ Rn , ∀(x, t) ∈ Q
such that
n
∑
aij (x, t)ξi ξj ≥ θ|ξ|2 .
(1.3)
i,j=1

In this paper, we consider the initial-boundary value problem in the cylinder Q for
semilinear PDE’s:
utt + Lu + f (x, t, u, Du) = h(x, t),
u(x, 0) = u0 (x),

ut (x, 0) = u1 (x),
u(x, t) = 0,

(x, t) ∈ Q,
x ∈ Ω,

(x, t) ∈ ∂Ω × (0, ∞),

(1.4)
(1.5)
(1.6)

where u0 ∈ H01 (Ω), u1 ∈ L2 (Ω), h ∈ L2 (0, ∞; L2 (Ω)) and f : Q × R × Rn −→ R is
continuous and satisfies the following two conditions:
(
)
|f (x, t, u, Du)| ≤ C k(x, t) + |u| + |Du| , ∀(x, t) ∈ Q, k ∈ L2 (0, ∞; L2 (Ω)), (1.7)
∫
(
)(
)
f (x, t, u, Du) − f (x, t, v, Dv) u − v dx ≥ 0, a.e. t ∈ [0, +∞). (1.8)
Ω

We introduce the Sobolev space H∗1,1 (Q) which consists of all functions u
defined on Q such that u ∈ L2 (0, ∞; H01 (Ω)), ut ∈ L2 (0, ∞; L2 (Ω)), and utt ∈
L2 (0, ∞; H −1 (Ω)) with the norm
∥u∥2H 1,1 (Q) = ∥u∥2L2 (0,∞;H 1 (Ω)) + ∥ut ∥2L2 (0,∞;L2 (Ω)) + ∥utt ∥2L2 (0,∞;H −1 (Ω)) .
∗

0

By ⟨·, ·⟩ we denote pairs of elements in H −1 (Ω) and H01 (Ω); By the notation (·, ·)
we mean the inner product in L2 (Ω). Let
B[u, v; t] =

∫ [∑
n
Ω

ij

a (x, t)uxi vxj +

i,j=1

n
∑

]
bi (x, t)uxi v + c(x, t)uv dx

i=1

which is a bilinear form defined on H 1 (Ω).
Definition 1.1. A function u ∈ H∗1,1 (Q) is called a weak solution of the (1.4) - (1.6) if it
satisfies the following conditions:
(
) (
)
- ⟨utt (t), v⟩ + B[u(t), v; t] + f (·, t, u, Du), v = h(·, t), v
with each v ∈ H01 (Ω) and a.e. t ∈ [0, ∞).

- u(x, 0) = u0 (x), ut (x, 0) = u1 (x) with x ∈ Ω.
40


The first initial-boundary value problem for semilinear hyperbolic equations...

Normally we write f (u, Du) instead of f (x, t, u, Du). The problem (1.4)−(1.6) in
the case f −linear was considered in [5-7] in which authors proved the unique existence
and regularity of a weak solution on the domain with singularity points on the boundary. In
[3] of A. Doktor, problem (1.4) - (1.6) has been considered on smooth domains. By using
the results of the linear problem respectly, he proves the global solution of the problem. It
is noted that the method approaching is used in [3] can not be applied for the problem if
the domains is nonsmooth.
In the present paper, we consider problem (1.4) - (1.6) with domain Ω, which is
a non-smooth domain. The monotonic method is used to obtain the unique existence of
global solution in time.

2. The local existence and uniqueness of a weak solution
In this section, we use the monotonic method to prove the existence and uniqueness
of a weak solution of the problem (1.4)−(1.6).
1
To confirm this, we first see that if {ωi (x)}∞
i=1 is a basis in H0 (Ω) ∩ L2 (Ω) and N
is a positive integer chosen then existence
N

u (x, t) =

N
∑

gi (t)ωi (x)

(2.1)

i=1

such that
(
)
N
N
N
(uN
,
ω
)
+
B[u
,
ω
;
t]
+
f
(u
,
Du
),
ω

= (h, ωi ) 0 ≤ t ≤ T, i = 1, · · · , N (2.2)
i
i
i
tt
in which gi (t) are defined on [0, ∞) such that with i = 1, · · · , N
{
gi (0) = (u0 , ωi )
gi′ (0)

= (u1 , ωi )

(2.3)
(2.4)

Since (2.1) - (2.4), applying the Caratheodory theorem, the functions gi (i =
1, 2 · · · , N ) always exist in [0, T ].
Theorem 2.1. For uN (x, t) defined by (2.1) we obtain:
(
)
N 2
N 2
2
∥ut ∥L2 (Ω) + ∥u ∥H 1 (Ω) + ∥uN
tt ∥L2 (0,T ;H −1 (Ω))
0
)
(
≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H 1 (Ω) + ∥h∥L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) (2.5)
0

for all t ∈ [0, T ].
Proof. (i) From (2.2), we multiply both sides by gi′ , sum i = 1, · · · , N , we obtain:
(
)
N
N
N
N
N
N
(uN
+ (h, uN
(2.6)
tt , ut ) + B[u , ut ; t] = − f (u , Du ), ut
t ).
41


Vu Trong Luong and Nguyen Thanh Tung

We have:
N
(uN
tt , ut )

B[uN , uN
t ; t] =

∫ (∑

n

d
=
dt
)

N
aij (x, t)uN
xi ut,xj

i,j=1

Ω

(

)
1 N 2
∥u ∥
,
(2.7)
2 t L2 (Ω)
)
∫ (∑
n
N
N N
dx
dx +

bi (x, t)uN
xi ut + c(x, t)u ut
Ω

i=1

=: B1 + B2
and if setting A[uN , uN ; t] =

n
∫ ∑
Ω i,j=1

N
aij (x, t)uN
xi uxj dx, then

(
)
d 1
N
N
A[u , u ; t] − C∥uN ∥2H 1 (Ω)
B1 ≥
0
dt 2
(
)
2
|B2 | ≤ C ∥uN ∥2H 1 (Ω) + ∥uN

t ∥L2 (Ω) .
0

Using (1.7), we get:

(2.8)
(2.9)

(
)
∥f ∥2L2 (Ω) ≤ C ∥k∥2L2 (Ω) + ∥u∥2H 1 (Ω) .
0

Therefore,
(
)
)
1(
2
| f (uN , DuN ), uN
∥f ∥2L2 (Ω) + ∥uN
t | ≤
t ∥L2 (Ω)
2
(
)
≤ C ∥k∥2L2 (Ω) + ∥uN ∥2H 1 (Ω) + ∥uN
∥
t L2 (Ω) . (2.10)
0

Continuing, we have
)
1(
2
∥h∥2L2 (Ω) + ∥uN
t ∥L2 (Ω)
2

|(h, uN
t )| ≤

(2.11)

Combining (2.6) - (2.11) gives
)
(
)
d ( N 2
2
∥
∥ut ∥ + A[uN , uN ; t] ≤ C ∥k∥2L2 (Ω) + ∥uN ∥2H 1 (Ω) + ∥h∥2L2 (Ω) + ∥uN
t L2 (Ω) .
0
dt
(2.12)
We have:
∫
∫ ∑
n

N
N
N
N 2
(2.13)
θ |Du | dx ≤
aij (x, t)uN
xi uxj = A[u , u ; t]
Ω

Ω i,j=1

by (1.3), and applying the Friedrichs theorem, we get
∥uN ∥2H 1 (Ω) ≤ C.A[uN , uN ; t].
0

So (2.12) becomes
)
d ( N 2
∥ut ∥L2 (Ω) + A[uN , uN ; t]
dt
)
(
N
N
2
2
≤ C ∥k∥2L2 (Ω) + ∥uN
t ∥L2 (Ω) + A[u , u ; t] + ∥h∥L2 (Ω) . (2.14)
42

The first initial-boundary value problem for semilinear hyperbolic equations...
2
N
N
2
2
Now setting η(t) = ∥uN
t ∥L2 (Ω) + A[u , u ; t], ψ(t) = ∥h∥L2 (Ω) + ∥k∥L2 (Ω) , then (2.14)
will be written in the form η ′ (t) ≤ C1 η(t) + C2 ψ(t) for 0 ≤ t ≤ T, and C1 , C2 are
constants. Thus, applying Gronwall’s inequality, we deduce that


∫t
η(t) ≤ eC1 t η(0) + C2 ψ(s)ds , t ∈ [0, T ].
(2.15)
0
2
N
N
N
2
It is found that η(0) = ∥uN
t (0)∥ + A[u (0), u (0); 0]. By (2.4) then ∥ut (0)∥L2 (Ω) ≤
C∥u1 ∥2L2 (Ω) and A[uN (0), uN (0); 0] ≤ C∥uN (x, 0)∥2H 1 (Ω) ≤ C∥u0 ∥2H 1 (Ω) , which is
0
0
obtained from (2.3). Therefore
(

)
η(0) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H 1 (Ω) .
(2.16)
0

On the other hand we see that
∫t

∫T
ψ(s)ds ≤

0

(

)
(
)
∥h∥2L2 (Ω) + ∥k∥2L2 (Ω) ds ≤ C ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) .

0

(2.17)
Using (2.15), (2.16) and (2.17), for ∀t ∈ [0, T ], we obtain:
)
(
2
N
N
2

2
2
2
∥uN
∥
+A[u
,
u
;
t]
≤
C
∥u
∥
+
∥u
∥
+
∥h∥
+
∥k∥
1
1
0
t L2 (Ω)
L2 (Ω)
L2 (0,T ;L2 (Ω))
L2 (0,T ;L2 (Ω)) .
H (Ω)
0

Applying (2.13) we get:
)
(
2
N 2
∥uN
∥
+
∥u
∥
t L2 (Ω)
H01 (Ω)
)
(
2
2
2
2
≤ C ∥u1 ∥L2 (Ω) + ∥u0 ∥H 1 (Ω) + ∥h∥L2 (0,T ;L2 (Ω)) + ∥k∥L2 (0,T ;L2 (Ω)) (2.18)
0

for ∀t ∈ [0, T ].
(ii) For each v ∈ H01 (Ω) selected, there is ∥v∥H01 (Ω) ≤ 1. Then for each positive integer N
N
⊥
there exists v1 ∈ span{ωk }N
k=1 and v2 ∈ (span{ωk }k=1 ) such that v = v1 + v2 . Inferred
∥v1 ∥H01 (Ω) ≤ 1 and (v2 , ωk ) = 0 for all k = 1, 2, · · · , N.
From (2.1) and (2.2), we have:

N
N
N
⟨uN
tt , v⟩ = (utt , v) = (utt , v1 ) = −(f, v1 ) + (h, v1 ) − B[u , v1 ; t].

By estimates

(
)
|(f, v1 )| ≤ ∥f ∥L2 (Ω) ∥v1 ∥H01 (Ω) ≤ ∥f ∥L2 (Ω) ≤ C ∥k∥L2 (Ω) + ∥uN ∥H01 (Ω) ,
∫
|(h, v1 )| ≤ |hv1 |dx ≤ ∥h∥L2 (Ω) ,
Ω

|B[u , v1 ; t]| ≤ C∥uN ∥H01 (Ω) ,
N

43


Vu Trong Luong and Nguyen Thanh Tung

then we obtain:
|⟨uN
tt , v⟩|

(

)

≤ C ∥h∥L2 (Ω) + ∥u ∥
N

H01 (Ω)

+ ∥k∥L2 (Ω) , ∀ v ∈ H01 (Ω), ∥v∥ ≤ 1.

It follows readily (2.18) that
2
∥uN
tt ∥L2 (0,T ;H −1 (Ω))
)
(
≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H 1 (Ω) + ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) . (2.19)
0

Combining (2.18) and (2.19) we deduce (2.5).
Theorem 2.2. If conditions (1.7), (1.8) are satisfied, then the problem (1.4) - (1.6) has a
weak solution u ∈ H∗1,1 (ΩT ) such that
(
)
∥u∥2H 1,1 (Ω ) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H 1 (Ω) + ∥h∥2L2 (0,T ;L2 (Ω)) + ∥k∥2L2 (0,T ;L2 (Ω)) .
∗

T

0

N ∞

N ∞
Proof. (i) From (2.5) we obtain sequences {uN }∞
N =1 , {ut }N =1 , {utt }N =1 respectively
bounded in L2 (0, T ; H01 (Ω)), L2 (0, T ; L2 (Ω)), L2 (0, T ; H −1 (Ω)), therefore there exists
a subsequence, without the loss of generality we take the sequence {uN }∞
N =1 and u ∈
1,1
H∗ (ΩT ) such that

N

weak in L2 (0, T ; H01 (Ω))

u ⇀ u
(2.20)
uN
weak in L2 (0, T ; L2 (Ω))
t ⇀ ut


uN ⇀ u
weak in L (0, T ; H −1 (Ω))
tt

tt

2

(ii) We prove that ⟨utt , v⟩ + B[u, v; t] + (f, v) = (h, v) for all v ∈ H01 (Ω). Indeed, for a
fixed positive integer N arbitrary and we choose a function v ∈ C 1 ([0, T ]; H 1 (Ω)) of the

N
∑
form v(x, t) =
dk (t)ω(x), where {dk }N
k=1 are smooth functions. We select m ≥ N,
k=1

N
multiply (2.2) by di (t), the sum i = 1, · · · , N, then we obtain (uN
tt , v) + B[u , v; t] +
N
(f, v) = (h, v). Setting (F (uN ), v) := (uN
tt , v) + B[u , v; t] − (h, v), then
(
)
(F (uN ), v) = − f (x, t, uN , DuN ), v .
(2.21)

By
∥f (x, t, u , Du
N

N

)∥2L2 (Ω)

≤C

(

∥k∥2L2 (Ω)

+

∥uN ∥2H 1 (Ω)
0

)
,

and also by (2.5) we have f (x, t, uN , DuN ) which is bounded in L2 (Ω) a.e. 0 ≤ t < T.
Therefore, there exists (ξ ∈ Ł2 (Ω) to f (x, t, )uN , DuN ) ⇀ ξ which is weak in L2 (Ω) when
N −→ ∞ and hence f (x, t, uN , DuN ), v −→ (ξ, v). So when N −→ ∞ then (2.21)
becomes
(F (u), v) = −(ξ, v).
(2.22)
44


The first initial-boundary value problem for semilinear hyperbolic equations...

(
)
On the other hand, from (1.8) then f (x, t, uN , DuN ) − f (x, t, ω, Dω), uN − ω ≥ 0 for
∀ω ∈ H01 (Ω). Integrating Ω and N −→ ∞, we have:
∫
[F (u)u − ξω − f (x, t, ω, Dω)(u − ω)]dx ≥ 0.
Ω

∫

By (2.22) we also have (F (u), u) = −(ξ, u) and so [−ξu + ξω − f (x, t, ω, Dω)(u −
Ω

ω)]dx ≥ 0, that is
∫

[ξ − f (x, t, ω, Dω)](u − ω)dx ≤ 0

(2.23)

Ω

∫
Putting ω = u−λv for λ > 0, into (2.23), we get [ξ−f (x, t, u−λv, Du−λDv)]vdx ≤ 0.
Ω
∫
Sending λ −→ 0, yields [ξ − f (x, t, u, Du)]vdx ≤ 0. By an argument analogous to the
Ω
∫
above with ω = u + λv for λ > 0, we obtain [ξ − f (x, t, u, Du)]vdx ≥ 0. From this we
Ω

deduce
(

)
f (x, t, u, Du), v = (ξ, v).

(2.24)

Combining (2.21), (2.22) and (2.24) yields
(
)
⟨utt , v⟩ + B[u, v; t] + f (u, Du), v = (h, v).

(2.25)

(iii) We have to prove u(x, 0) = u0 (x), ut (x, 0) = u1 (x), x ∈ Ω. To prove this, we
choose any function v ∈ C 2 ([0, T ]; H01 (Ω)) with v(T ) = vt (T ) = 0. With this function v,
(
) (
)
∫T
integrating by t in (2.25) on [0, T ] and by (utt , v)dt = − ut (0), v(0) + u(0), vt (0) +
∫T

0

(vtt , u)dt so

0

∫T

∫T
{(vtt , u) + B[u, v; t]}dt +

∫T
(f, v)dt =
0

0

0

(
) (
)
(h, v)dt + ut (0), v(0) − u(0), vt (0)
(2.26)

Similarly, we deduce
∫T

∫T
{(vtt , u ) + B[u , v; t]}dt +
N

0

N

∫T
(f, v)dt =

0

) ( N
)
(

(h, v)dt + uN
t (0), v(0) − u (0), vt (0)

0

45


Vu Trong Luong and Nguyen Thanh Tung

When N −→ ∞ then
∫T

∫T
{(vtt , u) + B[u, v; t]}dt +

0

∫T
(f, v)dt =

0

(
) (
)
(h, v)dt + u1 (0), v(0) − u0 (0), vt (0) .

0

(2.27)
From (2.26) and (2.27) we obtain u(x, 0) = u0 (x), ut (x, 0) = u1 (x) x ∈ Ω.
In order to study the uniqueness of the weak solution of problem (1.4)−(1.6), we replace
condition (1.7) with the following condition
|f (x, t, u, Du) − f (x, t, v, Dv)| ≤ µ(|u − v| + |Du − Dv|)

(2.28)

∀u, v ∈ H01 (Ω), ∀(x, t) ∈ Q.
Theorem 2.3. If conditions (1.8), (2.28) are satisfied, then problem (1.4)−(1.6) has at
most one weak solution in H∗1,1 (ΩT ).
Proof. First, suppose ω1 , ω2 are the solutions of problem (1.4)−(1.6). If setting u = ω1 −
ω2 and F (ω1 , ω2 , Dω1 , Dω2 ) = f (x, t, ω1 , Dω1 ) − f (x, t, ω2 , Dω2 ), then u is a weak
solution of the following problem:



utt + Lu + F (ω1 , ω2 , Dω1 , Dω2 ) = 0 in ΩT
u=0
on ∂Ω × (0, T )


u = 0, u = 0
on Ω × {t = 0}
t

Next, we will prove u ≡ 0 on ΩT .
(i) For each 0 ≤ s < T that is fixed, we set

v(t) =

∫s
 u(τ )dτ


if 0 ≤ t ≤ s

t

0

if s ≤ t ≤ T

for each t ∈ [0, T ] then v(t) ∈ H01 (Ω). From the definition of weak solution, we have
⟨utt , v⟩ + B[u, v; t] + (F, v) = 0,
integrating by t in (2.29) on [0, s] with s ∈ (0, T ), it follows further that
∫s (
0

46

∫s
)
⟨utt , v⟩ + B[u, v; t] dt + (F, v)dt = 0.
0

(2.29)


The first initial-boundary value problem for semilinear hyperbolic equations...

Since ut (0) = 0, v(s) = 0 and integrating by parts twice with respect to t, we find
∫s

{

}
− (ut , vt ) + B[u, v; t] dt +

∫s
(F, v)dt = 0.
0

0

Now vt = −u (0 ≤ t ≤ s), and so
∫s {
∫s
}
(ut , u) − B[vt , v; t] dt + (F, v)dt = 0.
0

Thus
∫s

0

∫s
∫s
)

{
d (1
1
∥u∥2L2 (Ω) − B[v, v; t] dt + (F, v)dt = −
C[u, v; t] + D[v, v; t]}dt,
dt 2
2

0

0

0

(2.30)
where

∫ (
C[u, v; t] =

−

n
∑
i=1

Ω

∫ (∑
n

D[v, v; t] =

bi (x, t)uvxi −

n
∑
i=1

aij
t (x, t)vxi vxj

i,j=1

Ω

)
bi (x, t)vvt,xi dx

+

n
∑

)
bi,t (x, t)vxi v + ct (x, t)vv dx

i=1

(2.30) is written into

1
1
∥u(s)∥2L2 (Ω) + B[v(0), v(0); 0] +
2
2

∫s

∫s
(F, v)dt = −

0

{

C[u, v; t] + D[v, v; t]}dt

0

(2.31)
By using the Cauchy inequality and the Lipschitz condition (2.28), we obtain the following
estimates:
∫s
∫s
(
)
∥v∥2H 1 (Ω) + ∥u∥2L2 (Ω) dt,
C[u, v; t] dt ≤
0

0

0

∫s

∫s
D[v, v; t] dt ≤

0

∫s

∥v∥2H 1 (Ω) dt,
0

0

∫s

(F, v) dt ≤ C
0

∥v∥2H 1 (Ω) dt.
0

0

47

Vu Trong Luong and Nguyen Thanh Tung

Employing the estimates above and inequality (1.3), we get from (2.31) that
 s

∫ (
)
∥u(s)∥2L2 (Ω) + ∥v(0)∥2H 1 (Ω) ≤ C 
∥v∥2H 1 (Ω) + ∥u∥2L2 (Ω) dt + ∥v(0)∥2L2 (Ω)  .
0

0

0

(ii) Now we set w(t) :=

∫t

(2.32)
(0 ≤ t ≤ T ), then (2.32) become

u(τ )dτ

0

∥u(s)∥2L2 (Ω) + ∥w(s)∥2H 1 (Ω)
 s 0


∫ (
)
≤C
∥w(s) − w(t)∥2H 1 (Ω) + ∥u(t)∥2L2 (Ω) dt + ∥w(s)∥2L2 (Ω)  . (2.33)
0

0

Since
∥w(s) − w(t)∥2H 1 (Ω) ≤ 2∥w(s)∥2H 1 (Ω) + 2∥w(t)∥2H 1 (Ω)
0

0

and

0

∫s
∥w(s)∥2L2 (Ω)

≤C

∥u(t)∥2L2 (Ω) dt,
0

we conclude from (2.33) that
∥u(s)∥2L2 (Ω)

+ (1 −

2C1 s)∥w(s)∥2H 1 (Ω)
0

≤ C1

∫s (

∥u∥2L2 (Ω)

+

∥w∥2H 1 (Ω)
0

)
dt.

0

1
We choose T1 which is so small that 1 − 2T1 C1 ≤ . Then if 0 ≤ s ≤ T1 , we will have
2
∥u(s)∥2L2 (Ω) + ∥w(s)∥2H 1 (Ω) ≤ C

∫s (

)
∥u∥2L2 (Ω) + ∥w∥2H 1 (Ω) dt.
0

0

0

Applying Gronwall’s inequality, we obtain u ≡ 0 on [0, T1 ].
(iii) We apply the same argument for the intervals [T1 , 2T1 ], [2T1 , 3T1 ], . . . . Eventually we
deduce u ≡ 0 on ΩT .
With the result (2.5) and condition (2.28), we have the following result.
Theorem 2.4. If conditions (1.8), (2.28) are satisfied, then problem (1.4)−(1.6) has a
global unique solution u ∈ H∗1.1 (Q) that satisfies the condition
)
(
∥u∥2H 1,1 (Q) ≤ C ∥u1 ∥2L2 (Ω) + ∥u0 ∥2H 1 (Ω) + ∥h∥2L2 (0,∞;L2 (Ω)) + ∥k∥2L2 (0,∞;L2 (Ω)) .
∗

48

0


The first initial-boundary value problem for semilinear hyperbolic equations...

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