1 Introduction

In recent years, there have been extensive studies on well-posedness of the following nonlinear variational wave equation with general data:

t 2 u - c ( u ) x c ( u ) x u = 0 in ( 0 , ) × R , u | t = 0 = u 0 on R , t u | t = 0 = u 1 on R ,
(1.1)

where c(·) is given smooth, bounded, and positive function with c'(·) ≥ 0 and c'(u0) > 0,u0H1(R),u1(x) ∈ L2(R). Equation (1.1) appears naturally in the study for liquid crystals [14]. In addition, Chang et al. [5], Su [6] and Kian [7] discussed globally Lipschitz continuous solutions to a class one dimension quasilinear wave equations

u t t - p ρ ( x ) , u x x = ρ ( x ) h ρ ( x ) , u , u x , u ( x , 0 ) = u 0 ( x ) , u t ( x , 0 ) = ω 0 ( x ) ,
(1.2)

where (x,t) ∈ R × R+, u0(x),ω0(x) ∈ R. Furthermore, Nishihara [8] and Hayashi [9] obtained the global solution to one dimension semilinear damped wave equation

u t t + u t - u x x = f ( u ) , ( t , x ) R + × R + ( u , u t ) ( 0 , x ) = ( u 0 , u 1 ) ( x ) .
(1.3)

Ikehata [10] and Vitillaro [11] proved global existence of solutions for semilinear damped wave equations in RNwith noncompactly supported initial data or in the energy space, in where the nonlinear term f(u) = |u|por f(u) = 0 is too special; some authors [1214] discussed the regularity of invariant sets in semilinear wave equation, but they didn't refer to any the initial value condition of it. Unfortunately, it is difficulty to classify a class wave operator equations, since the differential operator structure is too complex to identify whether have variational property. Our aim is to classify a class of nonlinear damped wave operator equations in order to research them more extensively and go beyond the results of [12].

In this article, we are interested in the existence of global solutions of the following nonlinear damped wave operator equations:

d 2 u d t 2 + k d u d t = G ( u ) , k > 0 u ( x , 0 ) = φ ( x ) , u t ( x , 0 ) = ψ ( x ) ,
(1.4)

where G : X 2 × R + X 1 * is a mapping, X2X1, X1, X2 are Banach spaces and X 1 * is the dual spaces of X1, R+ = [0, ∞), u = u(x,t). If k > 0, (1.4) is called damped wave equation. We obtain the existence, uniformly bounded and regularity of solutions by dividing the differential operator G(u) into two parts, variational and non-variational structure.

2 Preliminaries

First we introduce a sequence of function spaces:

X H 2 X 2 X 1 H , X 2 H 1 H ,
(2.1)

where H, H1, H2 are Hilbert spaces, X is a linear space, X1, X2 are Banach spaces and all inclusions are dense embeddings. Suppose that

L : X X 1 is one to one dense linear operator , L u , v H = u , v H , u , v X .
(2.2)

In addition, the operator L has an eigenvalue sequence

L e k = λ k e k , ( k = 1 , 2 , . . . )
(2.3)

such that {e k } ⊂ X is the common orthogonal basis of H and H2. We investigate the existence of global solutions of the Equation (1.4), so we need define its solution. Firstly, in Banach space X, introduce

L p ( ( 0 , T ) , X ) = u : ( 0 , T ) X | 0 T u p d t < ,

where p = (p1, p2,..., p m ),p i ≥ 1(1 ≤ im),

u p = k = 1 m u k p k ,

where | · | k is semi-norm in X, and X = i = 1 m i . Similarily, we can define

W 1 , p ( ( 0 , T ) , X ) = u : ( 0 , T ) X | u , u L p ( ( 0 , T ) , X ) .

Let L loc p ( ( 0 , ) , X ) = u ( t ) X | u L p ( ( 0 , T ) , X ) , T > 0 ..

Definition 2.1. Set (φ, ψ) ∈ X2 × H1, u W loc 1 , 0 , , H 1 L loc ( ( 0 , ) , X 2 ) is called a globally weak solution of (1.4), if for ∀vX1, it has

u t , v H + k u , v H = 0 t G u , v d t + k φ , v H + ψ , v H .
(2.4)

Definition 2.2. Let Y1,Y2 be Banach spaces, the solution u(t, φ, ψ) of (1.4) is called uniformly bounded in Y1 × Y2, if for any bounded domain Ω1 × Ω2Y1 × Y2, there exists a constant C which only depends the domain Ω1 × Ω2, such that

u Y 1 + u t Y 2 C , ( φ , ψ ) Ω 1 × Ω 2 and t 0 .

Definition 2.3. A mapping G: X 2 X 1 * is called weakly continuous, if for any sequence {u n } ⊂ X2, u n u0 in X2,

lim n G ( u n ) , v = G ( u 0 ) , v , v X 1 .

Lemma 2.1. [15]Let H2, H be Hilbert spaces, and H2H be a continuous embedding. Then there exists a orthonormal basis {e k } of H, and also is one orthogonal basis of H2.

Proof. Let I : H2H be imbedded. According to assume I is a linear compact operator, we define the mapping A : H2H as follows

A u , v H 2 = I u , v H = u , v H , v H 2 .

obviously, A : H2H2 is linear symmetrical compact operator and positive definite. Therefore, A has a complete eigenvalue sequence {λ k } and eigenvector sequence k H 2 such that

A k = λ k k , k = 1 , 2 , . . . ,

and k is orthogonal basis of H2. Hence

i , j H = A i , j H 2 = λ i i , j H 2 = 0 , if i j .

it implies i is also orthogonal sequence of H. Since H2H is dense, i is also orthogonal sequence of H, so { e i } = i / i H is norm orthogonal basis of H. The proof is completed.

Now, we introduce an important inequality

Lemma 2.2. [16] (Gronwall inequality) Let x(t), y(t), z(t) be real function on [a, b], where x(t) ≥ 0,∀atb, z(t) ∈ C[a, b], y(t) is differentiable on [a, b]. If the inequality as follows is hold

z ( t ) y ( t ) + a t x ( τ ) z ( τ ) d τ , a t b ,
(2.5)

then

z ( t ) y ( a ) e a t x ( s ) d s + a t e a t x ( τ ) d y d s d s .
(2.6)

3 Main results

Suppose that G=A+B: X 2 × R + X 1 * . Throughout of this article, we assume that

  1. (i)

    There exists a function FC 1 : X 2R 1 such that

    A u , L v = - D F ( u ) , v , u , v X
    (3.1)
  2. (ii)

    Function F is coercive, if

    F ( u ) u X 2
    (3.2)
  3. (iii)

    B as follows

    B u , L v C 1 F ( u ) + C 2 v H 1 2 , u , v X
    (3.3)

for some g L loc 1 ( 0 , ) .

Theorem 3.1. Set G : X 2 × R + X 1 * is weakly continuous, (φ, ψ) ∈ X2 × H1, then we obtain the results as follows:

(1) If G = A satisfies the assumption (i) and (ii), then there exists a globally weak solution of (1.4)

u W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 )

and u is uniformly bounded in X2 × H1;

(2) If G = A + B satisfies the assumption (i), (ii) and (iii), then there exists a globally weak solution of (1.4)

u W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) ;

(3) Furthermore, if G = A + B satisfies

G u , v 1 2 v H 2 + C F ( u ) + g ( t )
(3.4)

for someg L loc 1 ( 0 , ) , thenu W loc 2 , 2 ( 0 , ) , H .

Proof. Let {e k } ⊂ X be the public orthogonal basis of H and H2, satisfies (2.3).

Note

X n = i = 1 n α i e i | α i R 1 , X ̃ n = j = 1 n β j ( t ) e j | β j C 2 0 , .
(3.5)

From the assumption, we know L X n = X n ,L X n ̃ = X n ̃ , apply the Galerkin method to make truncate in X n ̃ :

d 2 u i d t 2 + k d u i d t = G ( u n ) , e i , 1 i n u i ( x , 0 ) = φ , e i H , u i ( x , 0 ) = ψ , e i H
(3.6)

there exists u n = i = 1 n u i ( t ) e i C 2 ( 0 , ) , X n for any v X n ̃ satisfies

0 t d 2 u n d t 2 + k d u n d t , v H d t = 0 t G u n , v d t
(3.7)

for any vX n , it yields that

d u n d t , v H + k u n , v H = 0 t G u n , v d t + k φ , v H + ψ , v H
(3.8)
  1. (1)

    If G=A, u n X n ̃ substitute v= d d t L u n into (3.7), we get

    0 t d 2 u n d t 2 + k d u n d t , d d t L u n H 1 d t = 0 t G u n , d d t L u n d t

combine condition (2.2) with (3.1), we get

0 t Ω d 2 u n d t 2 d u n d t d x d t + 0 t Ω k d u n d t d u n d t d x d t + 0 t D F ( u n ) d u n d t d x d t = 0 0 t 1 2 d d t d u n d t H 1 2 d t + k 0 t d u n d t H 1 2 d t + 0 t d d t F ( u n ) d t = 0 1 2 d u n d t H 1 2 - 1 2 ψ n H 1 2 + k 0 t d u n d t H 1 2 d t + F ( u n ) - F ( φ n ) = 0

consequently, we get

F ( u n ) + 1 2 u n H 1 2 + k 0 t u n H 1 2 d t = F ( u n ) + 1 2 ψ n H 1 2 .
(3.9)

Assume φH2, combine(2.2)with(2.3), we know {e n } is also the orthogonal basis of H1, then φ n φ in H2, ψ n ψ in H1, owing to H2X2 is embedded, so

φ n φ i n X 2 ψ n ψ i n X 1
(3.10)

due to the condition (3.6), from (3.9)and (3.10) we easily know

{ u n } W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) i s b o u n d e d .

consequently, assume that

u n u 0 i n W loc 1 , ( ( 0 , ) , H 1 ) L loc ( ( 0 , ) , X 2 ) a . e . t > 0

i.e. u n u0in X2a.e. t > 0, and G is weakly continuous, so

lim n G u n , v = G u 0 , v .

By (3.8), we have

lim n d u n d t H + k u n , v H = lim n 0 t G u n , v d t + k φ , v H + ψ , v H d u 0 d t , v H + k u 0 , v H = 0 t G u 0 , v d t + k φ , v H + ψ , v H

it indicates for any v n = 1 X n X 2 , it holds. Hence, for any vX2, we have

d u 0 d t , v H + k u 0 , v H = 0 t G u 0 , v d t + k φ , v H + ψ , v H .
(3.11)

Consequently, u0 is a globally weak solution of (1.4).

Furthermore, by (3.9) and (3.10), for any R > 0, there exists a constant C such that if

φ X 2 + ψ H 1 R
(3.12)

then the weak solution u(t, φ, ψ) of (1.4) satisfies

u ( t , φ , ψ ) X 2 + u t ( t , φ , ψ ) H 1 C . t 0
(3.13)

Assume (φ,ψ) ∈ X2 × H1 satisfies (3.12), by H2X2 is dense. May fix φ n H2 such that

φ n X 2 + ψ H 1 R , lim n φ n = φ in X 2

by (3.13), the solution {u(t, φ n , ψ)} of (1.4) is bounded in W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 ) a.e. t > 0.

Therefore, assume u(t, φ n , ψ) ⇀ u in W loc 1 , ( 0 , ) , H 1 L loc ( ( 0 , ) , X 2 ) then u(t) is a weak solution of (1.4), it satisfies uniformly bounded of (3.13). So the conclusion (1) is proved.

  1. (2)

    If G=A+B, u n X n ̃ , substitute v= d d t L u n into (3.7), we get

    0 t d 2 u n d t 2 , d d t L u n H 1 + k d u n d t , d d t L u n 1 H 1 d t = 0 t A u n , d u n d t + B u n , d u n d t d t

combine the condition (2.2) and (3.1), we have

0 t Ω d 2 u n d t 2 d u n d t d x d t + k 0 t Ω d u n d t 2 d u n d t d x d t + 0 t D F ( u n ) d u n d t d t = 0 t B u n , d u n d t d t 0 t 1 2 d d t d u n d t H 1 2 d t + k 0 t d u n d t H 1 2 d t + 0 t d d t F ( u n ) d t = 0 t B u n , d u n d t d t 1 2 u n H 1 2 - 1 2 ψ n H 1 2 + k 0 t u n H 1 2 d t + F ( u n ) + F ( φ n ) = 0 t B u n , d u n d t d t

consequently, we have

F ( u n ) + 1 2 u n H 1 2 + k 0 t u n H 1 2 d t = 0 t B u n , d u n d t d t + F ( φ n ) + 1 2 ψ n H 1 2
(3.14)

by the condition (3.3),(3.14)implies

F ( u n ) + 1 2 u n H 1 2 C 0 t F ( u n ) + 1 2 u n H 1 2 d t + f ( t )
(3.15)

where f ( t ) = 0 t g ( τ ) dt+ 1 2 ψ H 1 2 + sup n F ( φ n ) ..

by Gronwall inequality [Lemma(2.2)], from (3.15) we easily know:

F ( u n ) + 1 2 u n H 1 2 f ( 0 ) e C t + 0 t f ( τ ) e C ( t - τ ) d τ
(3.16)

it implies that, for any 0 < T < ∞

{ u n } W 1 , ( 0 , T ) , X 2 L 0 , T , X 2 is bounded .

now, use the same way as (1), we can obtain the result (2).

  1. (3)

    If the condition (3.4) is hold, u n X n ̃ , substitute v= d 2 u d t 2 into (3.7), we can get

    0 t d 2 u n d t 2 , d 2 u n d t 2 H + k d u n d t , d 2 u n d t 2 H d t = 0 t G u n , d 2 u n d t 2 d t

then

0 t d 2 u n d t 2 , d 2 u n d t 2 H d t + k 2 0 t d d t u n ( t ) H 2 d t 0 t 1 2 u n ( t ) H 2 + C F ( u n ) + g ( t ) d t 0 t d 2 u n d t 2 , d 2 u n d t 2 H d t + k 2 u n H 2 k 2 ψ n H 2 + 0 t 1 2 d 2 u n d t 2 H 2 + C F ( u n ) + g ( τ ) d τ

by (3.16), it implies that

0 t d 2 u n d t 2 H 2 d τ C , ( C > 0 )

consequently, for any 0 < T < ∞

{ u n } W 2 , 2 ( ( 0 , T ) , H ) is bounded .

it implies that uW2,2((0,T), H), the main theorem (3.1) has been proved.