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

In this paper we study the first initial boundary problem for semilinear hyperbolic equations in nonsmooth cylinders, where is a nonsmooth domain in Rn, n >=2. We established the existence and uniqueness of a global solution in time.

Mathematical and Physical Sci., 2013, Vol 58, No 7, pp 39-49

This paper is available online at http://stdb.hnue.edu.vn

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 inRn , 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

Let Ω⊂ R n be a bounded domain with non-smooth boundary ∂Ω, set Ω T = Ω×

(0, T ), with 0 < T < + ∞ We use the notations H1(Ω), H1

0(Ω) as the usual Sobolev

spaces and H −1 (Ω) as the dual space of H1

0(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 L p (0, ∞; X) consists of all measurable functions u : [0, ∞) −→ X with norm

∥u∥ L p (0,T ;X) =





∞

∫

0

∥u(t)∥ p

X dt





1

p

< + ∞ for 1 ≤ p < +∞.

We consider the partial differential operator

Lu = −

n

∑

i,j=1

∂

∂x j

(

a ij (x, t) ∂u

∂x i

) +

n

∑

i=1

b i (x, t) ∂u

∂x i + c(x, t)u (1.1)

where (x, t) ∈ Q = Ω × (0, ∞); a ij , b i , c ∈ C1(Q) (i, j = 1, · · · , n)

Received March 12, 2013 Accepted June 5, 2013

Contact Nguyen Thanh Tung, e-mail address: thanhtung70tbu@gmail.com

a ij (x, t) = a ji (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 ∈ R n , ∀(x, t) ∈ Q

such that

n

∑

i,j=1

In this paper, we consider the initial-boundary value problem in the cylinder Q for

semilinear PDE’s:

u tt + Lu + f (x, t, u, Du) = h(x, t), (x, t) ∈ Q, (1.4)

u(x, 0) = u0(x), u t (x, 0) = u1(x), x ∈ Ω, (1.5)

u(x, t) = 0, (x, t) ∈ ∂Ω × (0, ∞), (1.6)

where u0 ∈ H1

0(Ω), u1 ∈ L2(Ω), h ∈ L2(0, ∞; L2(Ω)) and f : Q × R × R n −→ 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)

∫

Ω

(

f (x, t, u, Du) − f(x, t, v, Dv))(u − v)dx ≥ 0, a.e t ∈ [0, +∞) (1.8)

We introduce the Sobolev space H ∗ 1,1 (Q) which consists of all functions u defined on Q such that u ∈ L2(0, ∞; H1

0(Ω)), u t ∈ L2(0, ∞; L2(Ω)), and u tt ∈

L2(0, ∞; H −1(Ω)) with the norm

∥u∥2

H ∗ 1,1 (Q)=∥u∥2

L2(0, ∞;H1 (Ω))+∥u t ∥2

L2(0, ∞;L2 (Ω))+∥u tt ∥2

L2(0, ∞;H −1(Ω)).

By⟨·, ·⟩ we denote pairs of elements in H −1 (Ω) and H1

0(Ω); By the notation (·, ·)

we mean the inner product in L2(Ω) Let

B[u, v; t] =

∫

Ω

[∑n i,j=1

a ij (x, t)u x i v x j +

n

∑

i=1

b i (x, t)u x i v + c(x, t)uv

]

dx

which is a bilinear form defined on H1(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:

- ⟨u tt (t), v ⟩ + B[u(t), v; t] +(f ( ·, t, u, Du), v)=(

h( ·, t), v) with each v ∈ H1

0(Ω) and a.e t ∈ [0, ∞).

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

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).

To confirm this, we first see that if{ω i (x) } ∞

i=1 is a basis in H1

0(Ω)∩ L2(Ω) and N

is a positive integer chosen then existence

u N (x, t) =

N

∑

i=1

such that

(u N tt , ω i ) + B[u N , ω i ; t] +(

f (u N , Du N ), ω i)

= (h, ω i) 0≤ t ≤ T, i = 1, · · · , N (2.2)

in which g i (t) are defined on [0, ∞) such that with i = 1, · · · , N

{

Since (2.1) - (2.4), applying the Caratheodory theorem, the functions g i (i =

1, 2 · · · , N) always exist in [0, T ].

Theorem 2.1 For u N (x, t) defined by (2.1) we obtain:

(

∥u N

t ∥2

L2 (Ω)+∥u N ∥2

H1 (Ω)

) +∥u N

tt ∥2

L2(0,T ;H −1(Ω))

≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥ L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

) (2.5)

for all t ∈ [0, T ].

Proof (i) From (2.2), we multiply both sides by g i ′ , sum i = 1, · · · , N, we obtain:

(u N tt , u N t ) + B[u N , u N t ; t] = −(f (u N , Du N ), u N t )

+ (h, u N t ). (2.6)

We have:

(u N tt , u N t ) = d

dt

( 1

2∥u N

t ∥2

L2 (Ω)

)

B[u N , u N t ; t] =

∫

Ω

( n

∑

i,j=1

a ij (x, t)u N x

i u N t,x

j

)

dx +

∫

Ω

( n

∑

i=1

b i (x, t)u N x

i u N t + c(x, t)u N u N t

)

dx

=: B1+ B2

and if setting A[u N , u N ; t] =∫

Ω

n

∑

i,j=1

a ij (x, t)u N

x i u N

x j dx, then

B1 ≥ d dt

( 1

2A[u

N , u N ; t]

)

− C∥u N ∥2

|B2| ≤ C(∥u N ∥2

H1 (Ω)+∥u N

t ∥2

L2 (Ω)

)

Using (1.7), we get:

∥f∥2

L2 (Ω) ≤ C(∥k∥2

L2 (Ω)+∥u∥2

H1 (Ω)

)

.

Therefore,

|(f (u N , Du N ), u N t )

| ≤ 1

2

(

∥f∥2

L2 (Ω)+∥u N

t ∥2

L2 (Ω)

)

≤ C(∥k∥2

L2 (Ω)+∥u N ∥2

H1 (Ω)+∥u N

t ∥ L2 (Ω)

)

(2.10)

Continuing, we have

|(h, u N

t )| ≤ 1

2

(

∥h∥2

L2 (Ω)+∥u N

t ∥2

L2 (Ω)

)

(2.11) Combining (2.6) - (2.11) gives

d

dt

(

∥u N

t ∥2+ A[u N , u N ; t])

≤ C(∥k∥2

L2 (Ω)+∥u N ∥2

H1 (Ω)+∥h∥2

L2 (Ω)+∥u N

t ∥2

L2 (Ω)

)

.

(2.12)

We have:

θ

∫

Ω

|Du N |2dx ≤

∫

Ω

n

∑

i,j=1

a ij (x, t)u N x i u N x j = A[u N , u N ; t] (2.13)

by (1.3), and applying the Friedrichs theorem, we get

∥u N ∥2

H1 (Ω) ≤ C.A[u N , u N ; t].

So (2.12) becomes

d

dt

(

∥u N

t ∥2

L2 (Ω)+ A[u N , u N ; t])

≤ C(∥k∥2

L2 (Ω)+∥u N

t ∥2

L2 (Ω)+ A[u N , u N ; t] + ∥h∥2

L2 (Ω)

)

(2.14)

Now setting η(t) = ∥u N

t ∥2

L2 (Ω) + A[u N , u N ; t], ψ(t) = ∥h∥2

L2 (Ω)+∥k∥2

L2 (Ω), then (2.14)

will be written in the form η ′ (t) ≤ C1η(t) + C2ψ(t) for 0 ≤ t ≤ T, and C1, C2 are constants Thus, applying Gronwall’s inequality, we deduce that

η(t) ≤ e C1t



η(0) + C2

t

∫

0

ψ(s)ds



It is found that η(0) = ∥u N

t (0)∥2 + A[u N (0), u N (0); 0] By (2.4) then ∥u N

t (0)∥2

L2 (Ω) ≤

C ∥u1∥2

L2 (Ω) and A[u N (0), u N(0); 0] ≤ C∥u N (x, 0) ∥2

H1 (Ω) ≤ C∥u0∥2

H1 (Ω), which is

obtained from (2.3) Therefore

η(0) ≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)

)

On the other hand we see that

t

∫

0

ψ(s)ds ≤

T

∫

0

(

∥h∥2

L2 (Ω)+∥k∥2

L2 (Ω)

)

ds ≤ C(∥h∥2

L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

)

.

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

∥u N

t ∥2

L2 (Ω)+A[u N , u N ; t] ≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥2

L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

)

.

Applying (2.13) we get:

(

∥u N

t ∥2

L2 (Ω)+∥u N ∥2

H1 (Ω)

)

≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥2

L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

) (2.18) for∀t ∈ [0, T ].

(ii) For each v ∈ H1

0(Ω) selected, there is∥v∥ H1 (Ω)≤ 1 Then for each positive integer N

there exists v1 ∈ span{ωk}N

k=1 and v2 ∈ (span{ωk}N

k=1)⊥ such that v = v1+ v2 Inferred

∥v1∥ H1 (Ω) ≤ 1 and (v2, ω k ) = 0 for all k = 1, 2, · · · , N.

From (2.1) and (2.2), we have:

⟨u N

tt , v ⟩ = (u N

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

By estimates

|(f, v1)| ≤ ∥f∥ L2 (Ω)∥v1∥ H1 (Ω) ≤ ∥f∥ L2 (Ω) ≤ C(∥k∥ L2 (Ω)+∥u N ∥ H1 (Ω)

)

,

|(h, v1)| ≤

∫

Ω

|hv1|dx ≤ ∥h∥ L2 (Ω),

|B[u N , v1; t] | ≤ C∥u N ∥ H1 (Ω),

then we obtain:

|⟨u N

tt , v ⟩| ≤ C(∥h∥ L2 (Ω)+∥u N ∥ H1 (Ω)+∥k∥ L2 (Ω)

)

, ∀ v ∈ H1

0(Ω), ∥v∥ ≤ 1.

It follows readily (2.18) that

∥u N

tt ∥2

L2(0,T ;H −1(Ω))

≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥2

L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

)

(2.19)

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∥2

H ∗ 1,1(ΩT)≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥2

L2(0,T ;L2 (Ω))+∥k∥2

L2(0,T ;L2 (Ω))

)

Proof. (i) From (2.5) we obtain sequences {u N } ∞

N =1 , {u N

t } ∞

N =1 , {u N

tt } ∞

N =1 respectively

bounded in L2(0, T ; H1

0(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 {u N } ∞

N =1 and u ∈

H ∗ 1,1(ΩT) such that











u N ⇀ u weak in L2(0, T ; H1

0(Ω))

u N

t ⇀ u t weak in L2(0, T ; L2(Ω))

u N

tt ⇀ u tt weak in L2(0, T ; H −1(Ω))

(2.20)

(ii) We prove that⟨u tt , v ⟩ + B[u, v; t] + (f, v) = (h, v) for all v ∈ H1

0(Ω) Indeed, for a fixed positive integer N arbitrary and we choose a function v ∈ C1([0, T ]; H1(Ω)) of the

form v(x, t) =

N

∑

k=1

d k (t)ω(x), where {d k } N

k=1 are smooth functions We select m ≥ N,

multiply (2.2) by d i (t), the sum i = 1, · · · , N, then we obtain (u N

tt , v) + B[u N , v; t] +

(f, v) = (h, v) Setting (F (u N ), v) := (u N

tt , v) + B[u N , v; t] − (h, v), then

(F (u N ), v) = −(f (x, t, u N , Du N ), v)

By

∥f(x, t, u N , Du N)∥2

L2 (Ω) ≤ C(∥k∥2

L2 (Ω)+∥u N ∥2

H1 (Ω)

)

,

and also by (2.5) we have f (x, t, u N , Du N ) which is bounded in L2(Ω) a.e 0 ≤ t < T.

Therefore, there exists ξ ∈ Ł2(Ω) to f (x, t, u N , Du N ) ⇀ ξ which is weak in L2(Ω) when

N −→ ∞ and hence (f (x, t, u N , Du N ), v)

−→ (ξ, v) So when N −→ ∞ then (2.21)

becomes

On the other hand, from (1.8) then(

f (x, t, u N , Du N)− f(x, t, ω, Dω), u N − ω)≥ 0 for

∀ω ∈ H1

0(Ω) Integrating Ω and N −→ ∞, we have:

∫

Ω

[F (u)u − ξω − f(x, t, ω, Dω)(u − ω)]dx ≥ 0.

By (2.22) we also have (F (u), u) = −(ξ, u) and so∫

Ω

[−ξu + ξω − f(x, t, ω, Dω)(u − ω)]dx ≥ 0, that is

∫

Ω

[ξ − f(x, t, ω, Dω)](u − ω)dx ≤ 0 (2.23)

Putting ω = u −λv for λ > 0, into (2.23), we get∫

Ω

[ξ −f(x, t, u−λv, Du−λDv)]vdx ≤ 0.

Sending λ −→ 0, yields∫

Ω

[ξ − f(x, t, u, Du)]vdx ≤ 0 By an argument analogous to the

above with ω = u + λv for λ > 0, we obtain∫

Ω

[ξ − f(x, t, u, Du)]vdx ≥ 0 From this we

deduce

(

f (x, t, u, Du), v)

Combining (2.21), (2.22) and (2.24) yields

⟨u tt , v ⟩ + B[u, v; t] +(f (u, Du), v)

(iii) We have to prove u(x, 0) = u0(x), u t (x, 0) = u1(x), x ∈ Ω To prove this, we

choose any function v ∈ C2([0, T ]; H1

0(Ω)) with v(T ) = v t (T ) = 0 With this function v, integrating by t in (2.25) on [0, T ] and by

T

∫

0

(u tt , v)dt = −(u t (0), v(0))

+(

u(0), v t(0))

+

T

∫

0

(v tt , u)dt so

T

∫

0

{(v tt , u) + B[u, v; t] }dt +

T

∫

0

(f, v)dt =

T

∫

0

(h, v)dt +(

u t (0), v(0))

−(u(0), v t(0))

(2.26) Similarly, we deduce

T

∫

0

{(v tt , u N ) + B[u N , v; t] }dt+

T

∫

0

(f, v)dt =

T

∫

0

(h, v)dt +(

u N t (0), v(0))

−(u N (0), v t(0))

When N −→ ∞ then

T

∫

0

{(v tt , u) + B[u, v; t] }dt +

T

∫

0

(f, v)dt =

T

∫

0

(h, v)dt +(

u1(0), v(0))

−(u0(0), v t(0))

.

(2.27)

From (2.26) and (2.27) we obtain u(x, 0) = u0(x), u t (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 ∈ H1

0(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:











u tt + Lu + F (ω1, ω2, Dω1, Dω2) = 0 in ΩT

Next, we will prove u ≡ 0 on Ω T

(i) For each 0≤ s < T that is fixed, we set

v(t) =







s

∫

t

u(τ )dτ if 0≤ t ≤ s

for each t ∈ [0, T ] then v(t) ∈ H1

0(Ω) From the definition of weak solution, we have

⟨u tt , v ⟩ + B[u, v; t] + (F, v) = 0, (2.29)

integrating by t in (2.29) on [0, s] with s ∈ (0, T ), it follows further that

s

∫

0

(

⟨u tt , v ⟩ + B[u, v; t])dt +

s

∫

0

(F, v)dt = 0.

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

s

∫

0

{

− (u t , v t ) + B[u, v; t]}

dt +

s

∫

0

(F, v)dt = 0.

Now v t=−u (0 ≤ t ≤ s), and so

s

∫

0

{

(u t , u) − B[v t , v; t]

}

dt +

s

∫

0

(F, v)dt = 0.

Thus

s

∫

0

d

dt

(1

2∥u∥2

L2 (Ω)− 1

2B[v, v; t]

)

dt +

s

∫

0

(F, v)dt = −

s

∫

0

{

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

(2.30) where

C[u, v; t] =

∫

Ω

(

−

n

∑

i=1

b i (x, t)uv x i −

n

∑

i=1

b i (x, t)vv t,x i

)

dx

D[v, v; t] =

∫

Ω

(∑n i,j=1

a ij t (x, t)v x i v x j +

n

∑

i=1

b i,t (x, t)v x i v + c t (x, t)vv

)

dx

(2.30) is written into

1

2∥u(s)∥2

L2 (Ω)+1

2B[v(0), v(0); 0] +

s

∫

0

(F, v)dt = −

s

∫

0

{

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

(2.31)

By using the Cauchy inequality and the Lipschitz condition (2.28), we obtain the following estimates:

s

∫

0

C[u, v; t] dt ≤∫s

0

(

∥v∥2

H1 (Ω)+∥u∥2

L2 (Ω)

)

dt,

s

∫

0

D[v, v; t] dt ≤∫s

0

∥v∥2

H1 (Ω)dt,

s

∫

0

(F, v) dt ≤ C∫s

0

∥v∥2

H1 (Ω)dt.

Employing the estimates above and inequality (1.3), we get from (2.31) that

∥u(s)∥2

L2 (Ω)+∥v(0)∥2

H1 (Ω) ≤ C





s

∫

0

(

∥v∥2

H1 (Ω)+∥u∥2

L2 (Ω)

)

dt + ∥v(0)∥2

L2 (Ω)





(2.32)

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

t

∫

0

u(τ )dτ (0≤ t ≤ T ), then (2.32) become

∥u(s)∥2

L2 (Ω)+∥w(s)∥2

H1 (Ω)

≤ C





s

∫

0

(

∥w(s) − w(t)∥2

H1 (Ω)+∥u(t)∥2

L2 (Ω)

)

dt + ∥w(s)∥2

L2 (Ω)



 (2.33)

Since

∥w(s) − w(t)∥2

H1 (Ω)≤ 2∥w(s)∥2

H1 (Ω)+ 2∥w(t)∥2

H1 (Ω)

and

∥w(s)∥2

L2 (Ω) ≤ C

s

∫

0

∥u(t)∥2

L2 (Ω)dt,

we conclude from (2.33) that

∥u(s)∥2

L2 (Ω)+ (1− 2C1s) ∥w(s)∥2

H1 (Ω) ≤ C1

s

∫

0

(

∥u∥2

L2 (Ω)+∥w∥2

H1 (Ω)

)

dt.

We choose T1 which is so small that 1− 2T1C1 ≤ 1

2 Then if 0 ≤ s ≤ T1, we will have

∥u(s)∥2

L2 (Ω)+∥w(s)∥2

H1 (Ω)≤ C

s

∫

0

(

∥u∥2

L2 (Ω)+∥w∥2

H1 (Ω)

)

dt.

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 Ω T

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∥2

H ∗ 1,1 (Q) ≤ C(∥u1∥2

L2 (Ω)+∥u0∥2

H1 (Ω)+∥h∥2

L2(0, ∞;L2 (Ω))+∥k∥2

L2(0, ∞;L2 (Ω))

)

.

[1] Evans L.C., 1997 Partial differential equations AMS.

[2] Robert A.Adam, 1975 Sobolev spaces Academic press New York San Francisco,

London

[3] Alexander Doktor, 1973 Mixed problem for semilinear hyperbolic equation of

second order with Dirichlet boundary condition Czechoslovak Mathematical

Journal, Vol.23, No.1, pp 95-122

[4] E.A.codington and N.Levison, 1955 Theory of odinary differential equations.

McGraw-Hill

[5] N.M Hung, 1999 Boundary problem for nonstationary systems in domains with a

non-smooth boundry Doctor dissertation, Mech Math.Department MSU, Moscow.

[6] Hung, N M, Luong, V T, 2008 Unique solvability of initial boundary-value

problems problems for hyperbolic systems in cylinders whose base is a cups domain,

Electron J Diff Eqns., Vol 2008, No 138, pp 1-10

[7] Hung, N M, Luong, V T, 2010 The L p −Unique solvability of the first initial boundary-value problem for hyperbolic systems Taiwanese Journal of Mathematics,

Vol 14, No 6, pp 2365-2381

problems problems for hyperbolic systems in cylinders whose base is a cups domain,

Electron J Diff Eqns.,... Hung, N M, Luong, V T, 2010 The L p −Unique solvability of the first initial boundary- value problem for hyperbolic systems Taiwanese Journal of Mathematics,

Vol 14,... E.A.codington and N.Levison, 1955 Theory of odinary differential equations.

McGraw-Hill

[5] N.M Hung, 1999 Boundary problem for nonstationary systems in domains with a