Global Existence and Exponential Stability of Solutions to ... - IM-UFRJ

Journal of Elasticity 75: 125–145, 2004. © 2004 Kluwer Academic Publishers. Printed in the Netherlands.

125

Global Existence and Exponential Stability of Solutions to Thermoelastic Equations of Hyperbolic Type YUMING QIN1 and JAIME E. MUÑOZ RIVERA2 1 Department of Applied Mathematics, College of Sciences, Donghua University, Shanghai 200051,

P. R. China; and Department of Mathematics, Henan University, Kaifeng 475001, P. R. China. E-mail: [email protected] 2 National Laboratory for Scientific Computation (LNCC), Rua Getulio Vargas 333, Quitandinha 25651-070, Petropolis-RJ, Brazil. E-mail: [email protected] Received 1 May 2003 Abstract. In this paper we prove the global existence and exponential stability of solutions to thermoelastic equations of hyperbolic type provided that the initial data are close to the equilibrium and the relaxation kernel is strongly positive definite and decays exponentially. Moreover, the global solution, together with its the third-order full energy, is exponentially stable for any t > 0. Mathematics Subject Classifications (2000): 35B40, 35Q99, 74A15. Key words: nonlinear hyperbolic thermoelasticity, global existence, exponential stability, strongly positive definite kernel.

1. Introduction This paper is concerned with the global existence, uniqueness and exponential stability of solutions to the following thermoelastic equations of hyperbolic type ut t − σ (ux )x + αθx = 0, θt − k ∗ θxx + βuxt = 0,

in [0, 1] × [0, +∞), in [0, 1] × [0, +∞),

(1.1) (1.2)

subject to the initial conditions u(x, 0) = u0 (x),

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

θ(x, 0) = θ0 (x),

∀x ∈ [0, 1],(1.3)

and the boundary conditions u(0, t) = u(1, t) = 0,

θx (0, t) = θx (1, t) = 0,

∀t 0.

(1.4)

Here by u = u(x, t) and θ = θ(x, t) we denote the displacement and the temperature difference respectively. By σ = σ (s) we denote a nonlinear function and

126

Y. QIN AND J.E.M. RIVERA

k = k(t) is the t relaxation kernel. The sign ∗ denotes the convolution product, i.e., k ∗y(·, t) = 0 k(t −τ )y(·, τ ) dτ . Finally, α and β are constants satisfying αβ > 0. Since the pioneer work of Dafermos [3] on the existence, differentiability and asymptotic stability of solutions to the system of linear thermoelasticity, significant progress has been made on the mathematical aspects in this direction. Rivera [28] established the decay rate of energy in one-dimensional linear thermoelasticity obeying Fourier’s law without memory effect. Concerning the nonlinear one-dimensional thermoelastic model obeying Fourier’s law without thermal memory, Slemrod [31] proved the global existence and asymptotic stability of small solutions with Neumann–Dirichlet (ux |x=0,1 = θ|x=0,1 = 0) or Dirichlet–Neumann (i.e., (1.4)) boundary conditions. Racke and Shibata [25] proved the global existence and polynomial decay of small smooth solutions with Dirichlet–Dirichlet (u|x=0,1 = θ|x=0,1 = 0) boundary conditions, and later for this type of boundary conditions, Racke et al. [26] showed the exponential stability of small global smooth solutions by using the similar idea as in [28]. Rivera and Barreto [29] improved the results in [26] for small initial data (u0 , u1 ) in the H 2 × H 1 norm. Recently, the authors of the present paper [30] established the global existence and exponential stability of small solutions to a nonlinear one-dimensional thermoelastic model obeying Fourier’s law with thermal memory subject to Dirichlet and mixed boundary conditions at the endpoints. In this direction, we also refer to the works by Burns et al. [1], Gibson et al. [15], Hansen [18] and Kim [19]. When the heat flux takes the form: ∞ t µ(s) θx (x, τ ) dτ ds, K0 > 0 constant, q(t) = −K0 θx (x, t) − 0

t −s

Giorgi and Naso [16] proved the exponential stability of C0 -semigroup associated with the corresponding system. When the heat flux obeys the theory of Gurtin and Pipkin [17], that is, q(t) = −k ∗ θx , Fatori and Rivera [14] established the energy decay for a linear hyperbolic thermoelastic system provided the relaxation kernel k(t) is a strongly positive definite and decays exponentially. To the authors’ knowledge, there is no result on nonlinear models when the heat flux follows the Gurtin and Pipkin’s law above. For this reason we study this topic here. The aim of this paper is to establish the global existence and exponential stability of “small” solutions to problem (1.1)–(1.4). Now let us explain some difficulties in deriving our results. When deriving exponential decay (or stability) of solution (or energy), we usually strive to construct a functional L(t), equivalent to the energy, satisfying (t) + λ L(t) g(t), L

(1.5)

where λ > 0 is a constant and g(t) is an exponential function. But in our case, we conclude that the energy is not necessarily a decreasing function which in particular means that inequality (1.5) is not possible to achive. To overcome this difficulty, ˜ := eδt k(t), and we in advance make the following exponential transforms in t: k(t) v(x, t) = eδt u(x, t), φ(x, t) = eδt θ(x, t) with a small parameter δ > 0 (see (2.5)

127

GLOBAL EXISTENCE AND EXPONENTIAL STABILITY

and (2.9)). Then, we study the new transformed problem (see (2.13), (2.14)) and we prove that it admits a unique global solution which is uniformly bounded (the bounded constants are independent of any units of time). This implies the global existence and exponential stability of solutions to the original problem (1.1)–(1.4). To show the uniform boundedness of the new system, we use some multiplicative techniques and the fact that the relaxation kernel is a strongly positive definite. In the directions of viscoelasticity and thermoviscoelasticity, we would like to refer to the works by Dafermos [4–8], Dafermos and Hsiao [9, 10], Dafermos and Nohel [11, 12], Fabrizio and Lazzari [13], Lagnese [20], Liu and Zheng [22], the first author of this paper [23, 24], Renardy et al. [27] and Zheng [33] and references cited therein. Throughout this paper we assume that σ = σ (s) is a C 3 function in a neighbourhood of s = 0, say, O = {s ∈ R: |s| < 1}, satisfying σ (0) > 0

(1.6)

and concerning the kernel we assume that k(t) ∈ C 1 (R+ ) and that k(t) is a strongly positive definite kernel. Additionally, we assume that there exist positive constants c0 c1 , and c2 such that k(t) > 0,

k (t) + c0 k(t) 0 k (t) + c1 k(t),

∀t 0.

(1.7)

By u2 and θ1 we mean u2 := [σ (ux )x − αθx ]|t =0 = σ (u0x )x − αθ0x ,

θ1 := −βuxt |t =0 = −βu1,x ,

which are supposed to satisfy the compatibility conditions u0 = u1 = u2 = θ0x = θ1x = 0

at x = 0, x = 1.

(1.8)

For the initial data we assume that (u0 , u1 , u2 ) ∈ H 3 (0, 1) × H 2 (0, 1) × H 1 (0, 1), (θ0 , θ1 ) ∈ H 2 (0, 1) × H 1 (0, 1) and

1

θ0 (x) dx = 0.

(1.9)

(1.10)

0

The notation in this paper is standard and follows Lions and Magenes’s book [21]. We put · = · L2 (0,1). We use C (sometimes C1 , C2 , . . .) to denote the generic positive constant independent of time t > 0. Our main results of this paper read as follows. THEOREM 1.1. Under assumptions (1.6)–(1.10), there exists a small constant 0 < 0 < 1 such that for any ∈ (0, 0 ) and for any initial data satisfying u0 2H 2 (0,1) + u1 2H 2 (0,1) + u2 2H 1 (0,1) + θ0 2H 2 (0,1) + θ1 2H 1 (0,1) < 2 , (1.11)

128


problem (1.1)–(1.4) admits a unique global solution (u(t), θ(t)) satisfying 3 u(t) ∈ C j [0, +∞), H 3−j , j =0

(k ∗ θ)(t), θ(t) ∈

1

(1.12)

C [0, +∞), H 2−j , j

j =0

(k ∗ θ)(t), θ(t) ∈ C 2 [0, +∞), L2 , j (k ∗ ∂ti θx )(t), (k ∗ ∂t θxx )(t) ∈ L2 [0, +∞), L2 ,

(1.13)

i = 0, 1, 2; j = 0, 1. (1.14) Moreover, there exist positive constants C1 , C2 such that for any t > 0, u(t)2H 3 + ut (t)2H 2 + ut t (t)2H 1 + ut t t (t)2 + θ(t)2H 2 + θt (t)2H 1 1 1 + θt t (t)2 + (k ∗ ∂ti θx )(t)2 + ∂ti (k ∗ θ)(t)2H 2−i i=0

+ (k ∗ θxx )(t)2 C1 e−C2 t .

i=0

(1.15)

REMARK 1.1. If (1.1) is replaced by the more general version ut t −S(ux , θ)x = 0, we can rewrite this equation as ut t − auxx + bθx = f,

(1.16)

where S = S(ux , θ) is the Piola–Kirchhoff stress and ∂S ∂S (ux , θ) + b θx , (ux , θ) − a uxx + f = ∂ux ∂θ ∂S ∂S (0, 0), b = − (0, 0). a= ∂ux ∂θ

(1.17) (1.18)

Assume that S = S(ux , θ) ∈ C 3 in a neighbourhood of (0, 0), say, |ux | 1, |θ| 1, and a > 0, b = 0 and (1.7)–(1.10) hold. Then the conclusion in Theorem 1.1 also holds. 2. Proof of Theorem 1.1 In this section we will prove Theorem 1.1. The proof of Theorem 1.1 is based on a priori estimates which we use to continue a local solution globally in time. The existence of a local solution to problem (1.1)–(1.4) under the assumptions in Theorem 1.1 can be established by a standard contraction mapping argument and we omit details here. THEOREM 2.1. Under the assumptions in Theorem 1.1, problem (1.1)–(1.4) admits a unique local solution (u(t), θ(t)) such that u(t) ∈

3 j =0

C j [0, Tm ), H 3−j ,

(k ∗ θ)(t), θ(t) ∈

1 j =0

C j [0, Tm ), H 2−j ,


129

2

(k ∗ θ)(t), θ(t) ∈ C [0, Tm ), L2 , (k ∗ ∂ti θx )(t) ∈ L2 [0, Tm ), L2 , i = 0, 1, 2, j 2 2 (k ∗ ∂t θxx )(t) ∈ L [0, Tm ), L , j = 0, 1, where [0, Tm ) is the maximal existence interval of solution (u(t), θ(t)). Moreover, if

3 1 j j ∂t u(t)2H 3−j + ∂t θ(t)2H 2−j < ∞, sup t ∈[0,Tm )

j =0

j =0

then Tm = +∞. The next lemma concerns with the property of a strongly positive definite kernel. ˆ LEMMA 2.2. Assume that k(t) ∈ L1 (R+ ) is a strongly positive definite kernel 1 + satisfying kˆ (t) ∈ L (R ), then for any y(t) ∈ L1loc (R+ ), it follows that t t 2 ˆ |k ∗ y(τ )| dτ β0 k1 y(τ ) kˆ ∗ y(τ ) dτ, 0

+∞

0

+∞ ˆ dt)2 + 4( 0 |kˆ (t)| dt)2 and β0 > 0 is a constant such where k1 = ( 0 |k(t)| ˆ − β0 e−t is a positive definite kernel. that the function k(t) Proof. See, e.g., [32]. 2 Without loss of generality, we suppose that σ (0) = 1 and α > 0, β > 0. In order to obtain global solution we need to show that u(t)H 3 (0,1) + ut (t)H 2 (0,1) + θ(t)H 2 (0,1) + θt (t)H 1 (0,1) C,

∀t 0. (2.1)

To this end we reduces system (1.1), (1.2) to ut t − uxx − αθx = ηuxx θt − k ∗ θxx + βut x = 0

in (0, 1) × [0, Tm ), in (0, 1) × [0, Tm ),

(2.2) (2.3)

where η = σ (ux ) − σ (0) = σ (ux ) − 1. It follows from (1.7) that the kernel k(t) satisfies that for any t 0, k(0)e−c1 t k(t) k(0)e−c0 t k(0).

(2.4)

Thus we can choose δ ∈ δ0 ≡ (0, min(1, c0 /2)) such that for any t 0, ˜ := eδt k(t) k(0)e−(c0 /2)t k(0)e−c1 t k(t)

(2.5)

130


and ˜ > 0, k(t)

c0 ˜ ˜ 0 k˜ (t) + c1 k(t), k˜ (t) + k(t) 2

∀t 0.

(2.6)

By the Paley and Wiener theorem (e.g., [27, pp. 149–150]) and (2.5), (2.6), there ˜ is a strongly positive definite exists δ1 ∈ (0, δ0 ] such that for any δ ∈ (0, δ1 ), k(t) ˆ ˜ kernel, and hence applying Lemma 2.2 to k(t) = k(t), using (2.5), (2.6), there is a ∗ positive constant k1 , independent of δ, ∞

2 ∞

2 ∗ ˜ ˜ |k(t)| dt + 4 |k (t)| dt , (2.7) k1 k1 = k1 (δ) = 0

0

such that for any δ ∈ (0, δ1 ] and for any y(t) ∈ L1loc (R+ ), t t 2 ∗ ˜ |k ∗ y(τ )| dτ β0 k1 y(τ )k˜ ∗ y(τ ) dτ. 0

(2.8)

0

Let us set v(x, t) = eδt u(x, t),

φ(x, t) = eδt θ(x, t).

(2.9)

Then v(x, t) and φ(x, t) satisfy ut eδt = vt − δv, θt eδt = φt − δφ, k˜ ∗ φxx = eδt k ∗ θxx .

ut t eδt = vt t − 2δvt + δ 2 v, θt t eδt = φt t − 2δφt + δ 2 φ,

(2.10) (2.11) (2.12)

Then, using (2.10)–(2.12), we transfer equations (2.2), (2.3) into (2.13) vt t − vxx + αφx = f in (0, 1) × [0, Tm ), (2.14) φt − k˜ ∗ φxx + βvt x = g in (0, 1) × [0, Tm ), v(0, t) = v(1, t) = 0, φx (0, t) = φx (1, t) = 0 in (0, 1) × [0, Tm ), where

1 0

1

φ0 (x) dx =

θ0 (x) dx = 0

(2.15)

0

and f = ηuxx eδt + 2δvt − δ 2 v,

g = δφ + δβvx .

(2.16)

To prove (2.1), it suffices to show that the solution (v(t), φ(t)) is bounded in H 3 × H 2 . We easily get from (2.3) and (2.14), (2.15) 1 1 θ(x, t) dx = φ(x, t) dx = 0, ∀t 0, (2.17) 0

0

131


which together with (2.16) gives 1 g(x, t) dx = 0, ∀t 0.

(2.18)

0

To continue our analysis let us introduce the linear problem Vt t − Vxx + α x = F

in (0, 1) × [0, Tm ),

(2.19)

t − k˜ ∗ xx + βVt x = G in (0, 1) × [0, Tm ),

(2.20)

V (x, 0) = V0 ,

Vt (x, 0) = V1 ,

(x, 0) = 0 ,

V (0, t) = V (1, t) = x (0, t) = x (1, t) = 0, with

1

0 (x) dx = 0.

(2.21)

0

In fact, it is obvious from (1.8), (2.9), (2.14), (2.15) and (2.18) that (2.21) is satisfied ˜ if ( , G) = (φ, g) or ( , G) = (φt , gt + k(t)φ 0xx ). Thus it follows from (2.14), (2.16), (2.18) and (2.21) that when (V , ) = (v, φ), (F, G) = (f, g) or (V , ) = ˜ (vt , φt ), (F, G) = (ft , gt + k(t)φ 0xx ), 1 1 1 t (x, t) dx = G(x, t) dx = 0, t (x, t) dx = 0, ∀t 0. 0

0

0

(2.22) In the sequel we are going to study the linearized system (2.19), (2.20). To this end we define the following energy functions 1 1 2 Vt + Vx2 + αβ −1 2 dx, (2.23) E1 (t; V , ) = 2 0 1 1 2 Vt t + Vt2x + αβ −1 2t dx, (2.24) E2 (t; V , ) = 2 0 1 1 2 2 + αβ −1 2x dx. Vt x + Vxx (2.25) E3 (t; V , ) = 2 0 Multiplying (2.19) and (2.20) by Vt and αβ −1 , respectively, and summing the results, we have 1 1 d −1 ˜ E1 (t; V , ) = −αβ x k ∗ x dx + (F Vt + αβ −1 G ) dx. dt 0 0 (2.26) Assuming regular initial data and noting that Vt and t satisfy the same boundary conditions, we get

132


1 1 d ˜ E2 (t; V , ) = −αβ −1 t x k˜ ∗ t x dx − αβ −1 k(t) 0x t x dx dt 0 0 1 (Ft Vt t + αβ −1 Gt t ) dx + 0

1 1 −1 −1 d ˜ ˜ k(t) t x k ∗ t x dx − αβ 0x x dx = −αβ dt 0 0 1 1 −1 ˜ + αβ k (t) 0x x dx + (Ft Vt t + αβ −1 Gt t ) dx. 0

0

(2.27) respectively, and

−1

Similarly, multiplying (2.19) and (2.20) by −Vxxt and −αβ xx summing the results, we obtain 1 1 d −1 ˜ E3 (t; V , ) = −αβ xx k ∗ xx dx − (F Vxxt + αβ −1 G xx ) dx dt 0 0 1 d 1 −1 ˜ xx k ∗ xx dx − F Vxx dx = −αβ dt 0 0 1 (Ft Vxx − αβ −1 G xx ) dx. (2.28) + 0

Now we introduce the following functionals: 1 x x dyVt t dx, E4 (t; V , ) = − 0 0 1 1 Vt x dx, E6 (t; V , ) = Vt x Vx dx, E5 (t; V , ) = 0 0 1 1 ˜ t k ∗ t dx, E8 (t; V , ) = − x k˜ ∗ x dx. E7 (t; V , ) = − 0

0

Thus, integrating (2.20) over (0, x) and using the boundary conditions, we derive x x ˜ t dy − k ∗ x + βVt = G dy. (2.29) 0

0

By (2.19) and (2.29), we easily get

β β 2 1 d 2 2 E4 (t; V , ) − Vt t + Vt x + α + t 2 + k 2 (0) x 2 dt 2 8 β β

x 1 x 2 ˜ + k ∗ x + Gt dyVt t + t dyFt dx. 0

0

0

(2.30) Now, define

1

n(t; V , ) = 0

2 + 2t + 2x (t) dx Vt2t + Vt2x + Vxx

133


and

1 −1 ˜ 0x x dx L(t; V , ) = N E1 (t; V , ) + E2 (t; V , ) + αβ k 0 + E3 (t; V , ) + E4 (t; V , ) + E5 (t; V , ) +

β E6 (t; V , ) + a1 E7 (t; V , ) + a2 E8 (t; V , ), 4

where N > 0 is a parameter to be specified later on and

2 4 + k 2 (0) α 2 β 4 4 α+ , a2 = α+ + + a1 . a1 = k(0) β k(0) β 8 Under the above notations, we have: LEMMA 2.3. There exists positive constants β1 , β2 , β3 , C3 , C4 sufficiently large constant N > N0 := 2β0 k1∗ β(C4 + a12 /8)/α such that, for any t > 0, L(t; V , ) satisfies the following inequalities: d L(t; V , ) −C3 n(t; V , ) + C4 k˜ ∗ x 2 + k˜ ∗ t x 2 + k˜ ∗ xx 2 dt 1 ( x k˜ ∗ x , + t x k˜ ∗ t x + xx k˜ ∗ xx ) dx −αNβ −1 0

+R(t; V , )

(2.31)

and (2.32) L(t; V , ) β2 n(t; V , ) + k˜ ∗ t 2 + k˜ ∗ x 2 + k˜ 2 (t) 0x 2 , 2 2 2 2 ˜ ˜ ˜ L(t; V , ) β1 n(t; V , ) − β3 k ∗ t + k ∗ x + k (t) 0x , (2.33) where

1

(F Vt + αβ −1 G + Ft Vt t + αβ −1 Gt t + Ft Vxx 0 d 1 −1 − αβ G xx ) dx − N F Vxx dx dt 0 1 1 −1 ˜ 0x x dx − a1 Gt k˜ ∗ t dx + αNβ k (t) 0 0

x 1 1 x Gt dyVt t + t dyFt dx + a2 Gk˜ ∗ xx dx + 0 0 0 0

1 β Vt x G − F x − F Vxx dx. (2.34) + 4 0

R(t; V , ) = N

134


Proof. By (2.19), (2.20) and integration by parts, we get 1 d 2 2 (Vxx + F ) x E5 (t; V , ) = −βVt x + α x − dt 0 1 + Vt x (k˜ ∗ xx + G) dx 0

β 4 β 2 2 x 2 − Vt x + Vxx + α + 2 16 β 1 1 ˜ 2 k ∗ xx + + (Vt x G − F x ) dx 2β 0 and

(2.35)

1 1 d x Vxx dx − F Vxx dx E6 (t; V , ) = Vt x 2 − Vxx 2 + α dt 0 0 1 1 α2 − Vxx 2 + x 2 + Vt x 2 − F Vxx dx. (2.36) 2 2 0

Thus, it follows from (2.30) and (2.35), (2.36) that

β d E4 (t; V , ) + E5 (t; V , ) + E6 (t; V , ) dt 4

β β 2 β 2 2 2 t 2 − Vt t − Vt x − Vxx + α + 2 8 16 β

α2β k 2 (0) 4 1 1 ˜ + x 2 + k˜ ∗ x 2 + k ∗ xx 2 + α+ + β 8 β β 2β

1 x x + Gt dyVt t + t dyFt dx 0 0 0

1 β Vt x G − F x − F Vxx dx. (2.37) + 4 0 On the other hand, differentiating (2.20) with respect to t, multiplying the resulting equation by k˜ ∗ t and integrating by parts, we deduce d E7 (t; V , ) dt

1

( t k˜ ∗ t + k(0) x k˜ ∗ t x + k˜ ∗ x k˜ ∗ t x ) dx = −k(0) t + 0 1 (βVt t k˜ ∗ t x + Gt k˜ ∗ t ) dx − 2

0

β 1 k(0) t 2 + k˜ ∗ t 2 Vt t 2 + x 2 + − 2 4a1 2k(0)

135


1 k 2 (0) ˜ k ∗ t x 2 + k˜ ∗ x 2 + k˜ ∗ t x 2 4 2 1 Gk˜ ∗ t dx. +βa1 k˜ ∗ t x 2 − +

(2.38)

0

Similarly, differentiating (2.20) with respect to x, multiplying the resulting equation by k˜ ∗ x and integrating by parts, we infer 1 d 2 2 E8 (t; V , ) = −k(0) x + k˜ ∗ xx + Gk˜ ∗ xx dx dt 0 1 ( x k˜ ∗ x + βVt x k˜ ∗ xx ) dx − 0

β 1 k(0) x 2 + k˜ ∗ x 2 Vt x 2 + − 2 4a2 2k(0) 1 2 ˜ + (1 + βa2 )k ∗ xx + Gk˜ ∗ xx dx.

(2.39)

0

Combining (2.38) and (2.39) with (2.37) gives β d E4 (t; V , ) + E5 (t; V , ) + E6 (t; V , ) dt 4 + a1 E7 (t; V , ) + a2 E8 (t; V , )

a1 a2 1 + + k˜ ∗ x 2 −C3 n(t; V , ) + β 2 2k(0) 1 a1 ˜ + (1 + βa2 )a2 k˜ ∗ xx 2 + k ∗ t 2 + β 2k(0)

a1 a1 + βa12 k˜ ∗ t x 2 + R1 (t; V , ), + k 2 (0) + 4 2 where C3 = min[β/16, k(0)a1 /4, k(0)a2 /4]. In view of (1.7), (2.22) and Poincare’s inequality, we have 1 k˜ ∗ t Ck˜ ∗ t x Ck˜ ∗ t x . k˜ ∗ t dx = 0,

(2.40)

(2.41)

0

Thus, it follows from (2.26)–(2.28) and (2.40), (2.41) that there is a constant C4 > 0 such that inequality (2.31) holds. From the definition of L(t; V , ), we easily know that there exist constants β1 , β2 , β3 > 0 and a sufficiently large constant N > N0 such that (2.32) and (2.33) hold. The proof is complete. 2 Now we define M(t; v, φ) = n(t; v, φ) + n(t; vt , φt ) + φxx 2 .

136


Differentiating (2.20) with respect to t, we arrive at φt t − k(0)φxx − k˜ ∗ φxx + βvt t x = gt , which, combined with (2.6) and (2.20), yields φxx 2 C φt t 2 + vt t x 2 + k˜ ∗ φxx 2 + gt 2 C φt t 2 + vt t x 2 + φt 2 + vt x 2 + g2 + gt 2 C5 (n(t; v, φ) + n(t; vt , φt )).

(2.42)

Thus, n(t; v, φ) + n(t; vt , φt ) M(t; v, φ) C6 (n(t; v, φ) + n(t; vt , φt )). (2.43) By (2.6) and noting that (k˜ ∗ φt )t = k(0)φt + k˜ ∗ φt , we easily obtain d ˜ k ∗ φt 2 + k˜ ∗ φt t 2 + k˜ ∗ φx 2 + k˜ ∗ φt x 2 dt C7 k˜ ∗ φt x 2 + k˜ ∗ φt t x 2 + k˜ ∗ φx 2 C3 φt t 2 + φx 2 + φt x 2 . + 8C6 β3

(2.44)

By the smallness condition of initial data (1.11) and (1.1), (1.2), there is a constant α1 > 1, independent of δ, such that M(0; u, θ) < α1 2 .

(2.45)

Using equations (2.13), (2.14) and (2.9)–(2.11), there exists a constant α2 > 1, independent of δ, such that n(0; v, φ) + n(0; vt , φt ) M(0; v, φ) α2 M(0; u, θ) < α1 α2 2 ,

(2.46)

which leads to v0 2H 2 + v1 2H 2 + v2 2H 1 + φ0 2H 2 + φ1 2H 1 α1 α2 2 .

(2.47)

We infer from (2.5), (2.6) and (2.47) that there exists a constant η0 > 0, independent of δ, such that +∞ λ3 (k˜ (t))2 φ0x 2 + φ1x 2 + λ1 (k˜ (t))2 + λ2 k˜ 2 (t) φ0xx 2 dt 0

< η0 2 ,

(2.48)

where 2C6 4N 2 α 2 C6 + , C3 C3 β 2 2N 2 α 2 C6 . λ3 = C3 β 2 λ1 = 1 +

λ2 =

2C6 2N 2 α 2 C6 + , C3 C3 β 2

and


137

Using the continuity of the solution it follows that there exists some t0 ∈ [0, Tm ) such that M(t; v, φ) α0 2 ,

∀t ∈ [0, t0 ),

(2.49)

where 4C6 α3 , β1 α1 α2 (β1 + β2 ) N 2 α 2 C5 2 α3 = + k (0)α1 α2 2(β2 + β3 ) + + η0 . 2 β1 β 2

α0 = α1 α2 +

Define

t1 = sup τ1 > 0; M(t; v, φ) α0 2 in [0, τ1 ) .

(2.50)

Then we have either t1 = Tm or t1 < Tm . In the former case, (v(t), φ(t)) with its corresponding derivatives are bounded in the L2 -norm for any t ∈ [0, Tm ). Thus, by Theorem 2.1, Tm = +∞. We will show that the latter case will not happen. To this end, let us assume that t1 < Tm . By Sobolev’s embedding theorem and (2.50), we obtain that for any (x, t) ∈ [0, 1] × [0, t1 ), |vx (x, t)| + |φ(x, t)| + |φx (x, t)| + |φt (x, t)| C9 ,

(2.51)

which implies that for any (x, t) ∈ [0, 1] × [0, t1 ), |ux (x, t)| + |θ(x, t)| + |θx (x, t)| + |θt (x, t)| C10 e−δt .

(2.52)

Thus, if is small enough, we have that for any (x, t) ∈ [0, 1] × [0, t1 ), |ux (x, t)| < 1,

ux (x, t) ∈ O.

Define ν = sup {|∂ ρ η(s)|; 0 ρ 2}, |s|1

where ∂ ρ denotes the derivatives of order |ρ|. Recalling the definition of η, we deduce |η| C11 ,

(2.53)

with C11 = C11 (ν) > 0 being a constant. By (2.50)–(2.53), we easily derive that for any (x, t) ∈ [0, 1] × [0, t1 ), |vt (x, t)| + |vt x (x, t)| + |vt t (x, t)| C12 , which, together with (2.10)–(2.12), implies that for any (x, t) ∈ [0, 1] × [0, t1 ), |ut (x, t)| + |ut x (x, t)| + |ut t (x, t)| C13 e−δt .

(2.54)

138


By (2.13), (2.51) and (2.53), (2.54), we get |vxx (x, t)| C + C |vxx (x, t)|, which gives |vxx (x, t)| C14 ,

|uxx (x, t)| C14 e−δt ,

∀(x, t) ∈ [0, 1] × [0.t1 ). (2.55) Similarly, differentiating (2.13) with respect to x, we conclude that vxxx 2 C M(t; v, φ) + fx 2 CM(t; v, φ) + C 2 vxxx 2 , which gives that for any t ∈ [0, t1 ), vxxx 2 CM(t; v, φ) C15 2 ,

uxxx 2 C15 2 e−δt ,

(2.56)

provided is small enough. In the next two lemmas we will estimate each term in R(t; V , ) for the cases of both (V , ) = (v, φ), (F, G) = (f, g) and (V , ) = (vt , φt ), (F, G) = (ft , gt + ˜ k(t)φ 0xx ), respectively. LEMMA 2.4. Under the assumptions in Theorem 1.1, the following estimates hold for any t ∈ [0, t1 ): R(t; v, φ) C( + δ)M(t; v, φ) + Cδ k˜ ∗ φt x 2 + k˜ ∗ φxx 2 2N 2 α 2 C6 ˜ d 1 C3 2 2 2 φx + (k (t)) φ0x − N f vxx dx, + 8C6 C3 β 2 dt 0 1 f vxx dx C( + δ)M(t; v, φ). N 0

Proof. The estimates in the lemma are easily proved from the definition of M(t; v, φ) and (2.16). 2 LEMMA 2.5. Under the assumptions in Theorem 1.1, the following estimates hold for any t ∈ [0, t1 ): R(vt , φt )

a12 ˜ k ∗ φt x 2 + Cδk˜ ∗ φt xx 2 C( + δ)M(t; v, φ) + Cδ + 4 C3 + vt t t 2 + vt t x 2 + φt 2 + φt t 2 + φt x 2 + φxx 2 8C6 + λ1 (k˜ (t))2 + λ2 k 2 (t) φ0xx 2 + λ3 (k˜ (t))2 φ1x 2 d t N 2 2 −1 ˜ (ηvt xx − vt t x ) − Nαβ k(t)φ0xx φxx − Nft vt xx dx, + dt 0 2 (2.57)


139

1

ft vt xx dx C M(t; v, φ), 1 −1 ˜ φ0xx φxx dx −Nαβ k(t) N

(2.58)

0

0

N 2 α 2 C5 ˜ 2 β1 (n(t; v, φ) + n(t; vt , φt )) + k (t)φ0xx 2 . 2 2β1 β 2

(2.59)

Proof. For the proof of (2.57), we need only prove that the following estimate holds for some terms in R(t; vt , φt ), i.e., 1 ˜ ft t vt xx − αβ −1 (gt + k(t)φ N 0xx )φt xx dx 0

1 d ˜ C3 φxx 2 − Nαβ −1 φ0xx φxx dx k(t) C( + δ)M(t; v, φ) + 8C6 dt 0 1 2 2 N d 2N α C 6 ˜ + ηvt2xx dx + (k (t))2 φ0xx 2 ; (2.60) 2 dt 0 C3 β 2 the other terms in R(t; vt , φt ) can be proved following the same arguments. In fact, it is obvious that 1 ˜ ft t vt xx − αβ −1 (gt + k(t)φ N 0xx )φt xx dx 0 1 1 −1 ft t vt xx dx + Nαβ φt x gt x dx =N 0 0

1 1 −1 d −1 ˜ ˜ k(t) φ0xx φxx dx + Nαβ k (t) φ0xx φxx dx. − Nαβ dt 0 0 (2.61) By virtue of (2.10)–(2.12), (2.16), (2.52)–(2.56) and integration by parts, we get 1 ft t vt xx dx N 0 1 ηt t vxx + 2ηt (vt xx − δvxx ) + 2δηt vxx + η(vt t xx − 2δvt xx + δ 2 vxx ) =N 0 + 2δη(vt xx − δvxx ) + δ 2 ηvxx + 2δvt t t − δ 2 vt t vt xx dx N d 1 2 ηvt xx dx. (2.62) C( + δ)M(t; v, φ) + 2 dt 0 Similarly, Nαβ

−1

0

1

φt x gt x dx C( + δ)M(t; v, φ),

(2.63)

140


Nαβ −1 k˜ (t)

1

φ0xx φxx dx

0

C3 2N 2 α 2 C6 ˜ φxx 2 + (k (t))2 φ0xx 2 . 8C6 C3 β 2 (2.64)

Thus, (2.60) follows from (2.62)–(2.64). The proof is complete.

2

Let us introduce the following function:

1 (f vxx + ft vt xx ) dx L1 (t) = L(t; v, φ) + L(t; vt , φt ) + N 0 1 N 1 2 2 −1 ˜ + η(vt t x − vt xx ) dx + Nαβ k(t) φ0xx φxx dx. 2 0 0 Then it follows from (2.43), (2.52)–(2.56) and Lemmas 2.3–2.5 that if + δ is small enough,

β1 (n(t; v, φ) + n(t; vt , φt )) + C( + δ)M(t; v, φ) L1 (t) β2 + 2 N 2 α 2 C5 ˜ 2 k (t)φ0xx 2 + β2 k˜ ∗ φt 2 + k˜ ∗ φt t 2 + k˜ ∗ φx 2 + 2β1 β2 + k˜ ∗ φt x 2 + k˜ 2 (t)φ0x 2 + k˜ 2 (t)φ1x 2 n(t; v, φ) + n(t; vt , φt ) (β2 + β1 ) 2

N 2 α 2 C5 2 2 2 ˜2 φ0xx k (t) + β2 φ0x + β2 φ1x + 2β1 β 2 (2.65) + β2 k˜ ∗ φt 2 + k˜ ∗ φt t 2 + k˜ ∗ φx 2 + k˜ ∗ φt x 2 and β1 (n(t; v, φ) + n(t; vt , φt )) − C( + δ)M(t; v, φ) 2 N 2 α 2 C5 ˜ 2 2 (t)φ − β k k˜ ∗ φt 2 + k˜ ∗ φt t 2 − 0xx 3 2β1 β 2 + k˜ ∗ φx 2 + k˜ ∗ φt x 2 + k˜ 2 (t)φ0x 2 + k˜ 2 (t)φ1x 2 β1 (n(t; v, φ) + n(t; vt , φt )) 4

N 2 α 2 C5 2 ˜2 φ k (t) − β3 φ0x 2 + β3 φ1x 2 + 0xx 2β1 β 2 − β3 k˜ ∗ φt 2 + k˜ ∗ φt t 2 + k˜ ∗ φx 2 + k˜ ∗ φt x 2 .

L1 (t)

Define

(2.66)

N 2 α 2 C5 2 ˜2 φ0xx k (t) L(t) = L1 (t) + β3 φ0x + β3 φ1x + 2β1 β 2 + β3 k˜ ∗ φt 2 + k˜ ∗ φt t 2 + k˜ ∗ φx 2 + k˜ ∗ φt x 2 . 2

2

(2.67)


141

Then, it follows from (2.66), (2.6) and (2.43), (2.44) that if + δ is small enough, β1 β1 (n(t; v, φ) + n(t; vt , φt )) M(t; v, φ), 4 4C6 d d C3 L(t) L1 (t) + (n(t; v, φ) + n(t; vt , φt )) dt dt 8 + C7 β3 k˜ ∗ φt x 2 + k˜ ∗ φt t x 2 + k˜ ∗ φx 2 . L(t)

(2.68)

(2.69)

Proof of Theorem 1.1. We will assume that the initial data u0 , u1 and θ0 belong to H 4 × H 3 × H 4 . Our result will follow using the standard density argument. By virtue of Lemmas 2.4, 2.5, we easily obtain d L1 (t) −Nαβ −1 dt

1

(φx k˜ ∗ φx + 2φt x k˜ ∗ φt x + φt t x k˜ ∗ φt t x

0

+ φxx k˜ ∗ φxx + φt xx k˜ ∗ φt xx ) dx C3 M(t; v, φ) − C3 (n(t; v, φ) + n(t; vt , φt )) + 8C6 + C16 ( + δ)M(t; v, φ) + C4 k˜ ∗ φx 2 + 2k˜ ∗ φt x 2 a2 + k˜ ∗ φt t x 2 + k˜ ∗ φxx 2 + k˜ ∗ φt xx 2 + 1 k˜ ∗ φt x 2 4 2 2 2 + C16 δ k˜ ∗ φt x + k˜ ∗ φxx + k˜ ∗ φt xx 2 + λ3 (k˜ (t))2 φ0x 2 + φ1x 2 + λ1 k˜ (t) + λ2 k˜ 2 (t) φ0xx 2 , which together with (2.68), (2.69) and (2.44) yields that if + δ is small enough, d L(t) −Nαβ −1 dt

1

(φx k˜ ∗ φx + 2φt x k˜ ∗ φt x + φt t x k˜ ∗ φt t x

0

+ φxx k˜ ∗ φxx + φt xx k˜ ∗ φt xx ) dx C3 (n(t; v, φ) + n(t; vt , φt )) 2

a12 + C7 β3 + C16 δ k˜ ∗ φx 2 + k˜ ∗ φt x 2 + 2C4 + 4 + k˜ ∗ φt t x 2 + k˜ ∗ φxx 2 + k˜ ∗ φt xx 2 2 + λ3 (k˜ (t))2 φ0x 2 + φ1x 2 + λ1 k˜ (t) + λ2 k˜ 2 (t) φ0xx 2 . (2.70) −

Integrating (2.70) with respect to t, using (2.7), (2.8), (2.65), (2.67), (2.68) and (2.46), (2.47), and taking δ( δ1 ) and small enough, we deduce

142


C3 L(t) + 2

t

(n(τ ; v, φ) + n(τ ; vt , φt )) dτ t 2 1 i 2 i 2 + C8 k˜ ∗ ∂t φx + k˜ ∗ ∂t φxx dτ 0

0

i=0

i=0

α1 α2 2 N 2 α 2 C5 2 α1 α2 2 + η0 2 + k (0) 2(β2 + β3 ) + (β1 + β2 ) 2 β1 β 2 =: α3 2 , (2.71) where C8 = [Nα/(ββ0 k1∗ ) − 2C4 − a12 /4]/2 > 0 (see Lemma 2.3). Thus, it follows from (2.43), (2.68) and (2.71) that for any t ∈ [0, t1 ), 2C3 t M(τ, v, φ) dτ M(t; v, φ) + β1 0 2 1 4C6 C8 t ˜ k ∗ ∂ti φx 2 + k˜ ∗ ∂ti φxx 2 dτ + β1 0 i=0 i=0

4C6 α3 2 = (α0 − α1 α2 ) 2 . β1

(2.72)

Letting t → t1 in (2.72), we have M(t1 ; v, φ) (α0 − α1 α2 ) 2 < α0 2 , which is contradictory to the definition of t1 , (2.50). Thus we conclude that t1 = Tm = +∞ and all the estimates above are valid for any t > 0. Note that M(t; v, φ) is equivalent to the third order full energy E(t; v, φ) := E1 (t; v, φ) + E2 (t; v, φ) + E3 (t; v, φ) + E2 (t; vt , φt ) + E3 (t; vt , φt ), that is, −1 M(t; v, φ) E(t; v, φ) C17 M(t; v, φ), C17

∀t > 0.

(2.73)

On the other hand, it is easy to verify that −1 M(t; u, θ)e2δt M(t; v, φ) C18 M(t; u, θ)e2δt , C18

∀t > 0.

(2.74)

In fact, note that {E1 (t; v, φ), E2 (t; v, φ), E3 (t; v, φ), E2 (t; vt , φt ), E3 (t; vt , φt )} is equivalent to {e2δt E1 (t; u, θ), e2δt E2 (t; u, θ), e2δt E3 (t; u, θ), e2δt E2 (t; ut , θt ), e2δt E3 (t; ut , θt )}. Thus, (2.74) follows from (2.73). By (2.72) and (2.56), we have M(t; u, θ) CM(t; v, φ)e−2δt Ce−2δt ,

∀t > 0, (2.75)

uxxx 2 Cvxxx 2 e−2δt CM(t; v, φ)e−2δt Ce−2δt ,

∀t > 0.

(2.76)

143


By (2.6), we deduce

1 d ˜ (k ∗ ∂ti φx )(t)2 + (k˜ ∗ φxx )(t)2 dt i=0

1 (k˜ ∗ ∂ti φx )(t)2 + (∂ti φx )(t)2 C i=0

+ (k˜ ∗ φxx )(t)2 + φxx (t)2

C

1

(k˜ ∗ ∂ti φx )(t)2 + (k˜ ∗ φxx )(t)2 + M(t; v, φ) .

(2.77)

i=0

Integrating (2.77) with respect to t, and exploiting (2.72), we finally obtain 1

(k˜ ∗ ∂ti φx )(t)2 + (k˜ ∗ φxx )(t)2 C,

i=0

which, together with (2.10)–(2.12), implies 1

(k˜ ∗ ∂ti θx )(t)2 + (k˜ ∗ θxx )(t)2 Ce−2δt .

(2.78)

i=0

Differentiating (1.1), (1.2) with respect to t respectively and using (2.75) and (2.78), we have ut t t (t)2 CM(t; u, θ) Ce−2δt , θt t (t)2 C M(t; u, θ) + k ∗ θxx 2 Ce−2δt .

(2.79) (2.80)

Similarly, by (2.75), (2.76) and (2.78)–(2.80),

1 1 ∂ti (k ∗ θ)(t)2H 2−i C (k ∗ ∂ti θx )(t)2 + (k ∗ θxx )(t)2 + k 2 (t) i=0

i=0 −2δt

Ce

.

(2.81)

Thus, (u(t), θ(t)) and (v(t), φ(t)) are uniformly bounded in H 3 × H 2 . Therefore, problem (1.1)–(1.4) admits a unique global solution (u(t), θ(t)) in H 3 × H 2 and the estimate (1.15) follows from (2.75), (2.76) and (2.78)–(2.81) with C2 = 2δ. The proof of Theorem 1.1 is now complete. 2

Acknowledgements Yuming Qin is partially supported by the grants of Prominent Youth (0412000100) and the Young Key Teachers from Henan Province.

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