Open Access

Global existence and asymptotic behavior of solutions to a class of fourth-order wave equations

Boundary Value Problems20132013:168

DOI: 10.1186/1687-2770-2013-168

Received: 9 May 2013

Accepted: 27 June 2013

Published: 16 July 2013

Abstract

This paper is concerned with the Cauchy problem for a class of fourth-order wave equations in an n-dimensional space. Based on the decay estimate of solutions to the corresponding linear equation, a solution space is defined. We prove the global existence and optimal decay estimate of the solution in the corresponding Sobolev spaces by the contraction mapping principle provided that the initial value is suitably small.

MSC:35L30, 35L75.

Keywords

fourth-order wave equation global existence decay estimate

1 Introduction

We investigate the Cauchy problem for a class of fourth-order wave equations
a u t t + Δ 2 u + u t = Δ f ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ1_HTML.gif
(1.1)
with the initial value
t = 0 : u = u 0 ( x ) , u t = u 1 ( x ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ2_HTML.gif
(1.2)

Here u = u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq1_HTML.gif is the unknown function of x = ( x 1 , , x n ) R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq2_HTML.gif and t > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq3_HTML.gif, and a > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq4_HTML.gif is a constant. The nonlinear term f ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq5_HTML.gif is a smooth function with f ( u ) = O ( u 2 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq6_HTML.gif for u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq7_HTML.gif.

Equation (1.1) is reduced to the classical Cahn-Hilliard equation if a = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq8_HTML.gif (see [1]), which has been widely studied by many authors. Galenko et al. [25] proposed to add inertial term a u t t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq9_HTML.gif to the classical Cahn-Hilliard equation in order to model non-equilibrium decompositions caused by deep supercooling in certain glasses. For more background, we refer to [46] and references therein. It is obvious that (1.1) is a fourth-order wave equation. For global existence and asymptotic behavior of solutions to more higher order wave equations, we refer to [714] and references therein.

Very recently, global existence and asymptotic behavior of solutions to the problem (1.1), (1.2) were established by Wang and Wei [7] under smallness condition on the initial data. When u 0 H s + 2 L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq10_HTML.gif, u 1 H s L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq11_HTML.gif, they obtained the following decay estimate:
x k u ( t ) H s + 2 k C ( u 0 L 1 H s + 2 + u 1 L 1 H s ) ( 1 + t ) n 8 k 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ3_HTML.gif
(1.3)
for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq12_HTML.gif and s [ n / 2 ] + 5 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq13_HTML.gif. The main purpose of this paper is to refine the result in [7] and prove the following decay estimate for the solution to the problem (1.1), (1.2) for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq14_HTML.gif with L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq15_HTML.gif data,
x k u ( t ) H s + 2 k CE 0 ( 1 + t ) n 8 k 4 1 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ4_HTML.gif
(1.4)
for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq12_HTML.gif and s max { 0 , [ n / 2 ] 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq16_HTML.gif. Here E 0 : = u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq17_HTML.gif is assumed to be small. We also establish the decay estimate for the solution to the problem (1.1), (1.2) for n 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq14_HTML.gif with L 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq18_HTML.gif data,
x k u ( t ) H s + 2 k CE 1 ( 1 + t ) k 4 1 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ5_HTML.gif
(1.5)

for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq12_HTML.gif and s max { 0 , [ n / 2 ] 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq16_HTML.gif. Here E 1 : = u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq19_HTML.gif is assumed to be small. Compared to the result in [7], we obtain a better decay estimate of solutions for small initial data.

The paper is organized as follows. In Section 2, we study the decay property of the solution operators appearing in the solution formula. We prove global existence and asymptotic behavior of solutions for the Cauchy problem (1.1), (1.2) in Sections 3 and 4, respectively.

Notations We introduce some notations which are used in this paper. Let F [ u ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq20_HTML.gif denote the Fourier transform of u defined by
u ˆ ( ξ ) = F [ u ] ( ξ ) : = R n e i ξ x u ( x ) d x . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equa_HTML.gif

We denote its inverse transform by F 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq21_HTML.gif. For 1 p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq22_HTML.gif, L p = L p ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq23_HTML.gif denotes the usual Lebesgue space with the norm L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq24_HTML.gif. The usual Sobolev space of order s is defined by W s , p = ( I Δ ) s 2 L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq25_HTML.gif with the norm f W s , p = ( I Δ ) s 2 f L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq26_HTML.gif. The corresponding homogeneous Sobolev space of order s is defined by W ˙ s , p = ( Δ ) s 2 L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq27_HTML.gif with the norm f W ˙ s , p = ( Δ ) s 2 f L p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq28_HTML.gif; when p = 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq29_HTML.gif, we write H s = W s , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq30_HTML.gif and H ˙ s = W ˙ s , 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq31_HTML.gif. We note that W s , p = L p W ˙ s , p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq32_HTML.gif for s 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq33_HTML.gif.

For a nonnegative integer k, x k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq34_HTML.gif denotes the totality of all the k th order derivatives with respect to x R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq35_HTML.gif. Also, for an interval I and a Banach space X, C k ( I ; X ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq36_HTML.gif denotes the space of k-times continuously differential functions on I with values in X.

Throughout the paper, we denote every positive constant by the same symbol C or c without confusion. [ ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq37_HTML.gif is the Gauss symbol.

2 Decay property

The aim of this section is to derive the solution formula to the Cauchy problem (1.1), (1.2). Without loss of generality, we take a = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq38_HTML.gif. We first study the linearized equation of (1.1),
u t t + Δ 2 u + u t = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ6_HTML.gif
(2.1)
with the initial data in (1.2). Taking the Fourier transform, we have
u ˆ t t + u ˆ t + | ξ | 4 u ˆ = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ7_HTML.gif
(2.2)
The corresponding initial value are given as
t = 0 : u ˆ = u ˆ 0 ( ξ ) , u ˆ t = u ˆ 1 ( ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ8_HTML.gif
(2.3)
The characteristic equation of (2.2) is
λ 2 + λ + | ξ | 4 = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equb_HTML.gif
Let λ = λ ± ( ξ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq39_HTML.gif be the corresponding eigenvalues, we obtain
λ ± ( ξ ) = 1 ± 1 4 | ξ | 4 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ9_HTML.gif
(2.4)
The solution to the problem (2.2), (2.3) is given in the form
u ˆ ( ξ , t ) = G ˆ ( ξ , t ) u ˆ 1 ( ξ ) + H ˆ ( ξ , t ) u ˆ 0 ( ξ ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ10_HTML.gif
(2.5)
where
G ˆ ( ξ , t ) = 1 λ + ( ξ ) λ ( ξ ) ( e λ + ( ξ ) t e λ ( ξ ) t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ11_HTML.gif
(2.6)
and
H ˆ ( ξ , t ) = 1 λ + ( ξ ) λ ( ξ ) ( λ + ( ξ ) e λ ( ξ ) t λ ( ξ ) e λ + ( ξ ) t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ12_HTML.gif
(2.7)
We define G ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq40_HTML.gif and H ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq41_HTML.gif by G ( x , t ) = F 1 [ G ˆ ( ξ , t ) ] ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq42_HTML.gif and H ( x , t ) = F 1 [ H ˆ ( ξ , t ) ] ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq43_HTML.gif, respectively, where F 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq44_HTML.gif denotes the inverse Fourier transform. Then, applying F 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq21_HTML.gif to (2.5), we obtain
u ( t ) = G ( t ) u 1 + H ( t ) u 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ13_HTML.gif
(2.8)
By the Duhamel principle, we obtain the solution formula to (1.1), (1.2)
u ( t ) = G ( t ) u 1 + H ( t ) u 0 + 0 t G ( t τ ) Δ f ( u ) ( τ ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ14_HTML.gif
(2.9)

The aim of this section is to establish decay estimates of the solution operators G ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq45_HTML.gif and H ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq46_HTML.gif appearing in the solution formula (2.8).

Lemma 2.1 The solution of the problem (2.2), (2.3) verifies the estimate
( 1 + | ξ | 2 ) 2 | u ˆ ( ξ , t ) | 2 + | u ˆ t ( ξ , t ) | 2 C e c ω ( ξ ) t ( ( 1 + | ξ | 2 ) 2 | u ˆ 0 ( ξ ) | 2 + | u ˆ 1 ( ξ ) | 2 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ15_HTML.gif
(2.10)

for ξ R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq47_HTML.gif and t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq48_HTML.gif, where ω ( ξ ) = | ξ | 4 ( 1 + | ξ | 2 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq49_HTML.gif.

Proof We apply the energy method in the Fourier space to prove (2.10). Such an energy method was first developed in [15]. We multiply (2.2) by u ˆ ¯ t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq50_HTML.gif and take the real part. This yields
1 2 d d t { | u ˆ t | 2 + | ξ | 4 | u ˆ | 2 } + | u ˆ t | 2 = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ16_HTML.gif
(2.11)
Multiplying (2.2) by u ˆ ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq51_HTML.gif and taking the real part, we obtain
1 2 d d t { | u ˆ | 2 + 2 Re ( u ˆ t u ˆ ¯ ) } + | ξ | 4 | u ˆ | 2 | u ˆ t | 2 = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ17_HTML.gif
(2.12)
Combining (2.11) and (2.12) yields
d d t E + F = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ18_HTML.gif
(2.13)
where
E = | u ˆ t | 2 + [ 1 2 + | ξ | 4 ] | u ˆ | 2 + Re ( u ˆ t u ˆ ¯ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equc_HTML.gif
and
F = | ξ | 4 | u ˆ | 2 + | u ˆ t | 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equd_HTML.gif
A simple computation implies that
CE 0 E CE 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ19_HTML.gif
(2.14)
where
E 0 = ( 1 + | ξ | 2 ) 2 | u ˆ | 2 + | u ˆ t | 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Eque_HTML.gif
Note that
F | ξ | 4 ( 1 + | ξ | 2 ) 2 E 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equf_HTML.gif
It follows from (2.14) that
F c ω ( ξ ) E , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ20_HTML.gif
(2.15)
where
ω ( ξ ) = | ξ | 4 ( 1 + | ξ | 2 ) 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equg_HTML.gif
Using (2.13) and (2.15), we get
d d t E + c ω ( ξ ) E 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equh_HTML.gif
Thus
E ( ξ , t ) e c w ( ξ ) t E ( ξ , 0 ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equi_HTML.gif

which together with (2.14) proves the desired estimates (2.10). Then we have completed the proof of the lemma. □

Lemma 2.2 Let G ˆ ( ξ , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq52_HTML.gif and H ˆ ( ξ , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq53_HTML.gif be the fundamental solutions to (2.1) in the Fourier space, which are given in (2.6) and (2.7), respectively. Then we have the estimates
( 1 + | ξ | 2 ) 2 | G ˆ ( ξ , t ) | 2 + | G ˆ t ( ξ , t ) | 2 C e c ω ( ξ ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ21_HTML.gif
(2.16)
and
( 1 + | ξ | 2 ) 2 | H ˆ ( ξ , t ) | 2 + | H ˆ t ( ξ , t ) | 2 C ( 1 + | ξ | 2 ) 2 e c ω ( ξ ) t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ22_HTML.gif
(2.17)

for ξ R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq47_HTML.gif and t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq48_HTML.gif, where ω ( ξ ) = | ξ | 4 ( 1 + | ξ | 2 ) 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq49_HTML.gif.

Proof If u ˆ 0 ( ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq54_HTML.gif, from (2.5), we obtain
u ˆ ( ξ , t ) = G ˆ ( ξ , t ) u ˆ 1 ( ξ ) , u ˆ t ( ξ , t ) = G ˆ t ( ξ , t ) u ˆ 1 ( ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equj_HTML.gif

Substituting the equalities into (2.10) with u ˆ 0 ( ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq54_HTML.gif, we get (2.16).

In what follows, we consider u ˆ 1 ( ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq55_HTML.gif, it follows from (2.5) that
u ˆ ( ξ , t ) = H ˆ ( ξ , t ) u ˆ 0 ( ξ ) , u ˆ t ( ξ , t ) = H ˆ t ( ξ , t ) u ˆ 0 ( ξ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equk_HTML.gif

Substituting the equalities into (2.10) with u ˆ 1 ( ξ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq55_HTML.gif, we get (2.17). The lemma is proved. □

Lemma 2.3 Let G ˆ ( ξ , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq52_HTML.gif and H ˆ ( ξ , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq53_HTML.gif be the fundamental solutions to (2.1) in the Fourier space, which are given in (2.6) and (2.7), respectively. Then there exists a small positive number R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq56_HTML.gif such that if | ξ | R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq57_HTML.gif and t 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq48_HTML.gif, we have the following estimate:
| G ˆ t ( ξ , t ) | C | ξ | 4 e c | ξ | 4 t + C e c t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ23_HTML.gif
(2.18)
and
| H ˆ t ( ξ , t ) | C | ξ | 4 e c | ξ | 4 t + C e c t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ24_HTML.gif
(2.19)
Proof For sufficiently small ξ, using the Taylor formula, we get
λ + ( ξ ) = | ξ | 4 + O ( | ξ | 8 ) , λ ( ξ ) = 1 + | ξ | 4 + O ( | ξ | 8 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ25_HTML.gif
(2.20)
and
1 λ + ( ξ ) λ ( ξ ) = 1 + 2 | ξ | 4 + O ( | ξ | 8 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ26_HTML.gif
(2.21)
It follows from (2.6) and (2.7) that
{ G ˆ t ( ξ , t ) = λ + ( ξ ) e λ + t λ ( ξ ) e λ t λ + ( ξ ) λ ( ξ ) , H ˆ t ( ξ , t ) = 1 λ + ( ξ ) λ ( ξ ) ( λ + ( ξ ) λ ( ξ ) e λ ( ξ ) t λ ( ξ ) λ + ( ξ ) e λ + ( ξ ) t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ27_HTML.gif
(2.22)

Equations (2.18) and (2.19) follow from (2.20)-(2.22). The proof of Lemma 2.3 is completed. □

Lemma 2.4 Let 1 p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq58_HTML.gif and k 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq59_HTML.gif. Then we have
x k G ( t ) ϕ L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k + l 4 ϕ W ˙ l , p + C e c t x ( k 2 ) + ϕ L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ28_HTML.gif
(2.23)
x k H ( t ) ϕ L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k + l 4 ϕ W ˙ l , p + C e c t x k ϕ L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ29_HTML.gif
(2.24)
x k G t ( t ) ϕ L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k + l 4 1 ϕ W ˙ l , p + C e c t x k ϕ L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ30_HTML.gif
(2.25)
and
x k H t ( t ) ϕ L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k + l 4 1 ϕ W ˙ l , p + C e c t x k + 2 ϕ L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ31_HTML.gif
(2.26)
x k G ( t ) Δ g L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k 4 1 2 g L p + C e c t x k g L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ32_HTML.gif
(2.27)
x k G t ( t ) Δ g L 2 C ( 1 + t ) n 4 ( 1 p 1 2 ) k 4 3 2 g L p + C e c t x k + 2 g L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ33_HTML.gif
(2.28)

where ( k 2 ) + = max { 0 , k 2 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq60_HTML.gif in (2.23).

Proof By the property of the Fourier transform and (2.16), we obtain
x k G ( t ) ϕ L 2 2 = R n | ξ | 2 k | G ˆ ( ξ , t ) | 2 | ϕ ˆ ( ξ ) | 2 d ξ | ξ | R 0 | ξ | 2 k ( 1 + | ξ | 2 ) 2 e c ω ( ξ ) t | ϕ ˆ ( ξ ) | 2 d ξ + C | ξ | R 0 | ξ | 2 k ( 1 + | ξ | 2 ) 2 e c ω ( ξ ) t | ϕ ˆ ( ξ ) | 2 d ξ C | ξ | R 0 | ξ | 2 k e c | ξ | 4 t | ϕ ˆ ( ξ ) | 2 d ξ + C | ξ | R 0 | ξ | 2 k | ξ | 4 e c t | ϕ ˆ ( ξ ) | 2 d ξ C | ξ | 2 k + 2 l e c | ξ | 4 t L q | ξ | l ϕ ˆ ( ξ ) L p 2 + C e c t x ( k 2 ) + ϕ L 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ34_HTML.gif
(2.29)

where R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq56_HTML.gif is a positive constant in Lemma 2.3, and 1 q + 2 p = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq61_HTML.gif and 1 p + 1 p = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq62_HTML.gif.

By a straight computation, we get
| ξ | 2 k + 2 l e c | ξ | 4 t L q ( | ξ | R 0 ) C ( 1 + t ) n 4 q k + l 2 C ( 1 + t ) n 4 ( 2 p 1 ) k + l 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ35_HTML.gif
(2.30)
It follows from the Hausdorff-Young inequality that
| ξ | l ϕ ˆ ( ξ ) L p C ( Δ ) l 2 ϕ L p C ϕ W ˙ l , p . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ36_HTML.gif
(2.31)

Combining (2.29)-(2.31) yields (2.23). Similarly, we can prove (2.24)-(2.28). Thus we have completed the proof of the lemma. □

3 Global existence and decay estimate (I)

The purpose of this section is to prove global existence and asymptotic behavior of solutions to the Cauchy problem (1.1), (1.2) with L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq15_HTML.gif data. We need the following lemma, which comes from [16] (see also [17]).

Lemma 3.1 Assume that f = f ( v ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq63_HTML.gif is smooth, where v = ( v 1 , , v n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq64_HTML.gif is a vector function. Suppose that f ( v ) = O ( | v | 1 + θ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq65_HTML.gif ( θ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq66_HTML.gif is an integer) when | v | ν 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq67_HTML.gif. Then, for the integer m 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq68_HTML.gif, if v , w W m , q ( R n ) L p ( R n ) L ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq69_HTML.gif and v L ν 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq70_HTML.gif, w L ν 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq71_HTML.gif, then f ( v ) f ( w ) W m , r ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq72_HTML.gif. Furthermore, the following inequalities hold:
x m f ( v ) L r C v L p x m v L q v L θ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ37_HTML.gif
(3.1)
and
x m ( f ( v ) f ( w ) ) L r C { ( x m v L q + x m w L q ) v w L p + + ( v L p + w L p x m ( v w ) L q ) } ( v L + w L ) θ 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ38_HTML.gif
(3.2)

where 1 r = 1 p + 1 q https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq73_HTML.gif, 1 p , q , r + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq74_HTML.gif.

Based on the decay estimates of solutions to the linear problem (2.1), (1.2), we define the following solution space:
X = { u C ( [ 0 , ) ; H s + 2 ( R n ) ) C 1 ( [ 0 , ) ; H s ( R n ) ) : u X < } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equl_HTML.gif
where
u X = sup t 0 { k s + 2 ( 1 + t ) n 8 + k 4 + 1 2 x k u ( t ) L 2 + h s ( 1 + t ) n 8 + h 4 + 3 2 x h u t ( t ) L 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equm_HTML.gif
For R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq75_HTML.gif, we define
X R = { u X : u X R } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equn_HTML.gif
The Gagliardo-Nirenberg inequality gives
u ( t ) L C u ( t ) L 2 1 n 2 s 0 x s 0 u ( t ) L 2 n 2 s 0 C ( 1 + t ) ( n 4 + 1 2 ) u X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ39_HTML.gif
(3.3)

where s 0 = [ n 2 ] + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq76_HTML.gif, s 0 s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq77_HTML.gif (i.e., s [ n 2 ] 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq78_HTML.gif).

Theorem 3.1 Assume that u 0 W ˙ 2 , 1 ( R n ) H s + 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq79_HTML.gif, u 1 W ˙ 2 , 1 ( R n ) H s ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq80_HTML.gif ( s max { 0 , [ n / 2 ] 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq81_HTML.gif). Put
E 0 = u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equo_HTML.gif
If E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq82_HTML.gif is suitably small, the Cauchy problem (1.1), (1.2) has a unique global solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq83_HTML.gif satisfying
u C ( [ 0 , ) ; H s + 2 ( R n ) ) C 1 ( [ 0 , ) ; H s ( R n ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equp_HTML.gif
Moreover, the solution satisfies the decay estimate
x k u ( t ) L 2 CE 0 ( 1 + t ) n 8 k 4 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ40_HTML.gif
(3.4)
and
x h u t ( t ) L 2 CE 0 ( 1 + t ) n 8 h 4 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ41_HTML.gif
(3.5)

for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq84_HTML.gif and 0 h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq85_HTML.gif.

Proof Let us define the mapping
Φ ( u ) = G ( t ) u 1 + H ( t ) u 0 + 0 t G ( t τ ) Δ f ( u ( τ ) ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ42_HTML.gif
(3.6)
Using (2.23), (2.24), (2.27), (3.1) and (3.3), for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq12_HTML.gif, we obtain
x k Φ ( u ) L 2 x k G ( t ) u 1 L 2 + C x k H ( t ) u 0 L 2 + C 0 t x k G ( t τ ) Δ f ( u ( τ ) ) L 2 d τ C ( 1 + t ) n 8 k 4 1 2 ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 f ( u ) L 1 d τ + C t 2 t ( 1 + t τ ) 1 2 x k f ( u ) L 2 d τ + C 0 t e c ( t τ ) x k f ( u ) L 2 d τ C ( 1 + t ) n 8 k 4 1 2 ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 u L 2 2 d τ + C t 2 t ( 1 + t τ ) 1 2 x k u L 2 u L d τ + C 0 t e c ( t τ ) x k u L 2 u L d τ C ( 1 + t ) n 8 k 4 1 2 ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + CR 2 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 ( 1 + τ ) n 4 1 d τ + CR 2 t 2 t ( 1 + t τ ) 1 2 ( 1 + τ ) 3 n 8 k 4 1 d τ + CR 2 0 t e c ( t τ ) ( 1 + τ ) 3 n 8 k 4 1 d τ C ( 1 + t ) n 8 k 4 1 2 { ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + R 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equq_HTML.gif
Thus
( 1 + t ) n 8 + k 4 + 1 2 x k Φ ( u ) L 2 CE 0 + CR 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ43_HTML.gif
(3.7)
It follows from (3.6) that
Φ ( u ) t = G t ( t ) u 1 + H t ( t ) u 0 + 0 t G t ( t τ ) Δ f ( u ( τ ) ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ44_HTML.gif
(3.8)
By exploiting (3.8), (2.25), (2.26), (2.28), (3.1) and (3.3), for h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq86_HTML.gif, we have
x h Φ ( u ) t L 2 C x h G t ( t ) u 1 L 2 + C x h H t ( t ) u 0 L 2 + C 0 t x h G t ( t τ ) Δ f ( u ( τ ) ) L 2 d τ C ( 1 + t ) n 8 h 4 3 2 ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + C 0 t 2 ( 1 + t τ ) n 8 h 4 3 2 f ( u ) L 1 d τ + C t 2 t ( 1 + t τ ) 3 2 x h f ( u ) L 2 d τ + C 0 t e c ( t τ ) x h f ( u ) L 2 d τ C ( 1 + t ) n 8 h 4 3 2 { ( u 0 W ˙ 2 , 1 H s + 2 + u 1 W ˙ 2 , 1 H s ) + R 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equr_HTML.gif
The above inequality implies
( 1 + t ) n 8 + h 4 + 3 2 x h Φ ( u ) t L 2 CE 0 + CR 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ45_HTML.gif
(3.9)
Combining (3.7) and (3.9) and taking E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq82_HTML.gif and R suitably small, we get
Φ ( u ) X R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ46_HTML.gif
(3.10)
For u ˜ , u ¯ X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq87_HTML.gif, (3.6) gives
Φ ( u ˜ ) Φ ( u ¯ ) = 0 t G ( t τ ) Δ [ f ( u ˜ ) f ( u ¯ ) ] d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ47_HTML.gif
(3.11)
By (2.27), (3.2) and (3.3), for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq88_HTML.gif, we infer that
x k ( Φ ( u ˜ ) Φ ( u ¯ ) ) L 2 0 t x k G ( t τ ) Δ ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 f ( u ˜ ) f ( u ¯ ) L 1 d τ + C t 2 t ( 1 + t τ ) 1 2 x k ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ + C 0 t e c ( t τ ) x k ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 ( u ˜ L 2 + u ¯ L 2 ) u ˜ u ¯ L 2 d τ + C t 2 t ( 1 + t τ ) 1 2 { ( x k u ˜ L 2 + x k u ˜ L 2 ) u ˜ u ¯ L + ( u ˜ L 2 + u ¯ L 2 ) x k ( u ˜ u ¯ ) L 2 } d τ + C 0 t e c ( t τ ) { ( x k u ˜ L 2 + x k u ˜ L 2 ) u ˜ u ¯ L + ( u ˜ L + u ¯ L ) x k ( u ˜ u ¯ ) L 2 } d τ CR u ˜ u ¯ X 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 ( 1 + τ ) n 4 1 2 d τ + CR u ˜ u ¯ X t 2 t ( 1 + t τ ) n 8 1 2 ( 1 + τ ) 3 n 8 k 4 1 d τ + CR | u ˜ u ¯ X 0 t e c ( t τ ) ( 1 + τ ) 3 n 8 k 4 1 d τ CR ( 1 + t ) n 8 k 4 1 2 u ˜ u ¯ X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equs_HTML.gif
which implies
( 1 + t ) n 8 + k 4 + 1 2 x k ( Φ ( u ˜ ) Φ ( u ¯ ) ) L 2 CR u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ48_HTML.gif
(3.12)
Similarly, for 0 h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq89_HTML.gif, from (3.11), (2.28) and (3.2), (3.3), we deduce that
x h ( Φ ( u ˜ ) Φ ( u ¯ ) ) t L 2 0 t x h G t ( t τ ) Δ [ f ( u ˜ ) f ( u ¯ ) ] L 2 d τ C 0 t 2 ( 1 + t τ ) n 8 h 4 3 2 ( f ( u ˜ ) f ( u ¯ ) ) L 1 d τ + C t 2 t ( 1 + t τ ) 3 2 x h ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ + C 0 t e c ( t τ ) x h ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ CR ( 1 + t ) n 8 h 4 3 2 u ˜ u ¯ X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equt_HTML.gif
which gives
( 1 + t ) n 8 + h 4 + 3 2 x h ( Φ ( u ˜ ) Φ ( u ¯ ) ) t L 2 CR u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ49_HTML.gif
(3.13)
Combining (3.12) and (3.13) and taking R suitably small yields
Φ ( u ˜ ) Φ ( u ¯ ) X 1 2 u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ50_HTML.gif
(3.14)

From (3.10) and (3.14), we know that Φ is a strictly contracting mapping. Consequently, we conclude that there exists a fixed point u X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq90_HTML.gif of the mapping Φ, which is a solution to (1.1), (1.2). We have completed the proof of the theorem. □

4 Global existence and decay estimate (II)

In the previous section, we have proved global existence and asymptotic behavior of solutions to the Cauchy problem (1.1), (1.2) with L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq15_HTML.gif data. The purpose of this section is to establish global existence and asymptotic behavior of solutions to the Cauchy problem (1.1), (1.2) with L 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq18_HTML.gif data. Based on the decay estimates of solutions to the linear problem (2.1), (1.2), we define the following solution space:
X = { u C ( [ 0 , ) ; H s + 2 ( R n ) ) C 1 ( [ 0 , ) ; H s ( R n ) ) : u X < } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equu_HTML.gif
where
u X = sup t 0 { k s + 2 ( 1 + t ) k 4 + 1 2 x k u ( t ) L 2 + h s ( 1 + t ) h 4 + 3 2 x h u t ( t ) L 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equv_HTML.gif
For R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq75_HTML.gif, we define
X R = { u X : u X R } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equw_HTML.gif
Thanks to the Gagliardo-Nirenberg inequality, we get
u ( t ) L C ( 1 + t ) ( n 8 + 1 2 ) u X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ51_HTML.gif
(4.1)
Theorem 4.1 Suppose that u 0 H ˙ 2 ( R n ) H s + 2 ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq91_HTML.gif, u 1 H ˙ 2 ( R n ) H s ( R n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq92_HTML.gif ( s max { 0 , [ n / 2 ] 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq16_HTML.gif). Put
E 1 = u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equx_HTML.gif
If E 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq82_HTML.gif is suitably small, the Cauchy problem (1.1), (1.2) has a unique global solution u ( x , t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq83_HTML.gif satisfying
u C ( [ 0 , ) ; H s + 2 ( R n ) ) C 1 ( [ 0 , ) ; H s ( R n ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equy_HTML.gif
Moreover, the solution satisfies the decay estimate
x k u ( t ) L 2 CE 1 ( 1 + t ) k 4 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ52_HTML.gif
(4.2)
and
x h u t ( t ) L 2 CE 1 ( 1 + t ) h 4 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ53_HTML.gif
(4.3)

for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq93_HTML.gif and 0 h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq85_HTML.gif.

Proof Let the mapping Φ be defined in (3.6).

For k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq94_HTML.gif, (2.23), (2.24), (2.27), (3.1) and (4.1) give
x k Φ ( u ) L 2 x k G ( t ) u 1 L 2 + C x k H ( t ) u 0 L 2 + C 0 t x k G ( t τ ) Δ f ( u ( τ ) ) L 2 d τ C ( 1 + t ) k 4 1 2 ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 f ( u ) L 1 d τ + C t 2 t ( 1 + t τ ) 1 2 x k f ( u ) L 2 d τ + C 0 t e c ( t τ ) x k f ( u ) L 2 d τ C ( 1 + t ) k 4 1 2 ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 u L 2 2 d τ + C t 2 t ( 1 + t τ ) 1 2 x k u L 2 u L d τ + C 0 t e c ( t τ ) x k u L 2 u L d τ C ( 1 + t ) k 4 1 2 ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + CR 2 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 ( 1 + τ ) 1 d τ + CR 2 t 2 t ( 1 + t τ ) 1 2 ( 1 + τ ) n 8 k 4 1 d τ + CR θ + 1 0 t e c ( t τ ) ( 1 + τ ) n 8 k 4 1 d τ C ( 1 + t ) k 4 1 2 { ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + R 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equz_HTML.gif
Thus we get
( 1 + t ) k 4 + 1 2 x k Φ ( u ) L 2 CE 1 + CR 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ54_HTML.gif
(4.4)
Applying t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq95_HTML.gif to (3.6), we obtain
Φ ( u ) t = G t ( t ) u 1 + H t ( t ) u 0 + 0 t G t ( t τ ) Δ f ( u ( τ ) ) d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ55_HTML.gif
(4.5)
By using (2.25), (2.26), (2.28), (3.1), (4.1), for 0 h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq85_HTML.gif, we have
x h Φ ( u ) t L 2 C x h G t ( t ) u 1 L 2 + C x h H t ( t ) u 0 L 2 + C 0 t x h G t ( t τ ) Δ f ( u ( τ ) ) L 2 d τ C ( 1 + t ) h 4 3 2 ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + C 0 t 2 ( 1 + t τ ) n 8 h 4 3 2 f ( u ) L 1 d τ + C t 2 t ( 1 + t τ ) 3 2 x h f ( u ) L 2 d τ + C 0 t e c ( t τ ) x h f ( u ) L 2 d τ C ( 1 + t ) h 4 3 2 { ( u 0 H ˙ 2 H s + 2 + u 1 H ˙ 2 H s ) + R 2 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equaa_HTML.gif
This yields
( 1 + t ) h 4 + 3 2 x h Φ ( u ) t L 2 CE 1 + CR 2 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ56_HTML.gif
(4.6)
Combining (4.4) and (4.6) and taking E 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq96_HTML.gif and R suitably small, we obtain
Φ ( u ) X R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ57_HTML.gif
(4.7)
For u ˜ , u ¯ X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq87_HTML.gif, by using (3.6), we have
Φ ( u ˜ ) Φ ( u ¯ ) = 0 t G ( t τ ) Δ [ f ( u ˜ ) f ( u ¯ ) ] d τ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ58_HTML.gif
(4.8)
It follows from (2.27), (3.2) and (4.1) for 0 k s + 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq88_HTML.gif that
x k ( Φ ( u ˜ ) Φ ( u ¯ ) ) L 2 0 t x k G ( t τ ) Δ ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ C 0 t 2 ( 1 + t τ ) n 8 k 4 1 2 f ( u ˜ ) f ( u ¯ ) L 1 d τ + C t 2 t ( 1 + t τ ) 1 2 x k ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ + C 0 t e c ( t τ ) x k ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ CR ( 1 + t ) k 4 1 2 u ˜ u ¯ X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equab_HTML.gif
which implies
( 1 + t ) k 4 + 1 2 x k ( Φ ( u ˜ ) Φ ( u ¯ ) ) L 2 CR u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ59_HTML.gif
(4.9)
Similarly, for 0 h s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq89_HTML.gif, from (4.5), (2.28), (3.2) and (4.1), we infer that
x h ( Φ ( u ˜ ) Φ ( u ¯ ) ) t L 2 0 t x h G t ( t τ ) Δ [ f ( u ˜ ) f ( u ¯ ) ] L 2 d τ C 0 t 2 ( 1 + t τ ) n 8 h 4 3 2 ( f ( u ˜ ) f ( u ¯ ) ) L 1 d τ + C t 2 t ( 1 + t τ ) 3 2 x h ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ + C 0 t e c ( t τ ) x h ( f ( u ˜ ) f ( u ¯ ) ) L 2 d τ CR ( 1 + t ) h 4 3 2 u ˜ u ¯ X , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equac_HTML.gif
which implies
( 1 + t ) h 4 + 3 2 x h ( Φ ( u ˜ ) Φ ( u ¯ ) ) t L 2 CR u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ60_HTML.gif
(4.10)
Using (4.9) and (4.10) and taking R suitably small yields
Φ ( u ˜ ) Φ ( u ¯ ) X 1 2 u ˜ u ¯ X . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_Equ61_HTML.gif
(4.11)

It follows from (4.7) and (4.11) that Φ is a strictly contracting mapping. Consequently, we infer that there exists a fixed point u X R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-168/MediaObjects/13661_2013_Article_415_IEq90_HTML.gif of the mapping Φ, which is a solution to (1.1), (1.2). We have completed the proof of the theorem. □

Declarations

Authors’ Affiliations

(1)
School of Mathematics and Information Sciences, North China University of Water Resources and Electric Power

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© Zhuang and Zhang; licensee Springer. 2013

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