The pointwise estimates to solutions for 1-dimensional linear thermo-visco-elastic system

Acta Mathematica Scientia 2011,31B(4):1259–1271 http://actams.wipm.ac.cn

THE POINTWISE ESTIMATES TO SOLUTIONS FOR 1-DIMENSIONAL LINEAR THERMO-VISCO-ELASTIC SYSTEM∗ Hao Xingwen (

)

Mathematics Department, Shanghai Jiaotong University, Shanghai 200240, China School of Mathematics and Information Sciences, Weifang University, Weifang 261061, China E-mail: [email protected]


)

Peng Shaoyu (

School of Mathematics and Information, Ludong University, Yantai 264000, China E-mail: [email protected]

Abstract In this paper, we study the linear thermo-visco-elastic system in one-dimensional space variable. The mathematical model is a hyperbolic-parabolic partial differential system. The solutions of the system show some decay property due to the parabolicity. Based on detailed analysis on the Green function of the system, the pointwise estimates of the solutions are obtained, from which the generalized Huygens’ principle is shown. Key words thermo-visco-elastic system; Fourier transform; Green function; pointwise estimate; Huygens’ principle 2000 MR Subject Classification

1

53A08; 52A20

Mathematical Model and Main Result

In this paper, we study the initial value problem of linear thermo-visco-elasticity system as follows ⎧ ⎪ ut − αvx = 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ vt − αux + βθx − μvxx = 0, ⎪ ⎪ ⎪ ⎪ ⎨ θ − κθ + βv = 0, t xx x (1.1) ⎪ ⎪ u(0, x) = u0 (x), ⎪ ⎪ ⎪ ⎪ ⎪ v(0, x) = v0 (x), ⎪ ⎪ ⎪ ⎪ ⎩ θ(0, x) = θ (x), 0 ∗ Received

March 16, 2010. Xingwen Hao’s research was supported in part by National Natural Science Foundation of China (10571120 and 10971135), Shanghai Shuguang Project (06SG11) and the Program for New Century Excellent Talents of Chinese Ministry of Education (NCET-07-0546) and Doctorial Foundation of Weifang University (2011BS11).

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where u, v and θ denote the deformation gradient, velocity and temperature respectively, α, β, κ and μ are all constants with α, μ, κ > 0, β = 0. This system is the linearization of the balance law of mass, momentum and energy with density ρ = 1 in Lagrangian coordinate in the framework of the thermo-mechanics for one dimension. Thermo-visco-elastic system describes the elastic and thermal behavior of elastic heat conductive media, see [4] for details. It is easy to verify that this system is a hyperbolic-parabolic coupled system. There were many documents and researches devoted to it such as [1–11, 15] etc. In [10], by using energy method, the authors got the existence of smooth solutions in 1-D nonlinear case with second sound. The global existence of smooth solutions in H¨ older space for the 1-D nonlinear thermo-visco-elastic equations were obtained in [4] in which fixed point method was used. For some other existence results, see [5, 7] and reference therein. For system (1.1), there are many literatures about the properties of the solutions. Such as in [8], the authors studied the asymptotic behavior of the solution in 2-D and 3-D cases. Green function method was widely used in studying the decay property for the hyperbolic and parabolic systems. In [18], by Green function method and frequency analysis, the authors got the pointwise estimates of the solutions. In this paper, we will use this method to give the pointwise decay estimate of the solutions to (1.1). About this method, there are many documents such as [12–20]. Denote U (t, x) = (u(t, x), v(t, x), θ(t, x)) and U0 (x) = (u0 (x), v0 (x), θ0 (x)) . The following theorem is the main result of this paper. Theorem 1.1 For any given nonnegative integer α, assume that U0 ∈ L1 (R) ∩ C α (R) and U0 has compact support. Then, for any positive integer N , the solution U (x, t) to problem (1.1) satisfies the following estimate: |Dα U (x, t)| ≤ Ct−

1+α 2

√ √ [BN (x, t) + BN (x + t 1 + δ, t) + BN (x − t 1 + δ, t)]||U0 (x)||L1 (R) , (1.2)

  x2 −N and C depends only on α, N and the support of U0 (x). where BN (x, t) = 1 + 1+t Remark If initial data U0 (x) has no compact support, then the decay estimate can still be obtained if U0 (x) has some decay property. See the details after the proof of the theorem. The rest of this paper is organized as follows. In Section 2, we take Fourier transform to system (1.1) and obtain the approximate expression of Green function. Some important properties of Fourier transform of the Green function are also given. In Section 3, pointwise estimate is obtained to the Green function. Finally in Section 4, we complete the proof of Theorem 1.1.

2

Fourier Analysis We use 1 fˆ(ξ) = √ 2π



∞

e−ixξ f (x)dx

−∞

to denote the Fourier transform of f (x) and use 1 fˇ(x) = √ 2π



∞

−∞

eixξ f (ξ)dξ

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to denote the inverse Fourier transform of f (x). Here and in the sequel the notation i denotes the imaginary unit satisfying i2 = −1. Taking Fourier transform in system (1.1) with respect to x, we obtain ⎧ ⎪ u ˆt − iαξˆ v = 0, ⎪ ⎪ ⎨ (2.1) vˆt − iαξ u ˆ + iβξ θˆ + μξ 2 vˆ = 0, ⎪ ⎪ ⎪ ⎩ θˆ + κξ 2 θˆ + iβξˆ v = 0. t

In order to study the Green function of Cauchy problem (1.1), we consider initial data U0 (x) = (u0 (x), v0 (x), θ0 (x)) = δ(x)I, where δ(x) is the Dirac function and I is the unit matrix. After Fourier transformation, we get (ˆ u0 (ξ), vˆ0 (ξ), θˆ0 (ξ)) = I.

(2.2)

Denote ˆ ξ)) ˆ ξ) = (ˆ G(t, u(t, ξ), vˆ(t, ξ), θ(t, and ˆ ξ) = (ˆ G(0, u0 (ξ), vˆ0 (ξ), θˆ0 (ξ)) , then the problem (2.1)–(2.2) can be rewritten as ⎧ ⎨G ˆ = 0, ˆ t + A(ξ)G

(2.3)

⎩ G(0, ˆ ξ) = I, where

⎛

0

⎜ A(ξ) = ⎜ ⎝ −iαξ 0

− iαξ

0

⎞

μξ 2

⎟ iβξ ⎟ ⎠.

iβξ

κξ 2

Next, we try to calculate the eigenvalues for the matrix A(ξ). From |λI − A(ξ)| = 0, the eigenvalues of the matrix A(ξ) satisfy λ3 − (μ + κ)ξ 2 λ2 + (α2 + β 2 + μκξ 2 )ξ 2 λ − α2 κξ 4 = 0.

(2.4)

It is difficult to get explicit expressions of the eigenvalues from (2.4). In fact, in order to obtain ˆ ξ), we need only to give the approximate values of λ near ξ = 0 and ξ = ∞. the estimate of G(t, First, we calculate the approximate eigenvalues near ξ = 0. By some tedious calculation, we can get α2 κ 2 ξ + O(1)ξ 3 , + β2  μ(α2 + β 2 ) + κβ 2 λ2 = i α2 + β 2 ξ + + O(1)ξ 3 , 2(α2 + β 2 )  μ(α2 + β 2 ) + κβ 2 λ3 = −i α2 + β 2 ξ + + O(1)ξ 3 . 2(α2 + β 2 )

λ1 =

α2

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ˆ ξ) near ξ = 0 as After some more tedious but direct calculation, we can write G(t, ⎛ ⎞ e−λ1 t 0 0 ⎟ −1 ⎜ ˆ ξ) = P ⎜ 0 G(t, e−λ2 t 0 ⎟ ⎠P , ⎝ −λ3 t 0 0 e where P is a 3 × 3 matrix, the column vectors of which are the eigenvectors of A(ξ) near ξ = 0. After neglecting the higher order terms, P can be approximately written as ⎛ ⎞ iαβξ iαξ iαξ ⎜ ⎟   P ⎜ i α2 + β 2 ξ −i α2 + β 2 ξ ⎟ ⎝ 0 ⎠. −iα2 ξ iβξ iβξ ˆ can be written as Thus G ˆ=G ˆ1 + G ˆ2 + G ˆ3, G where ⎛ ⎜ ˆ1 = P ⎜ G ⎝

e−λ1 t

0 0

2κ ⎟ −1 2 − α2α+β 2ξ t P = A (ξ)e (1 + O(1)ξ 3 t), 0 0⎟ 1 ⎠ 0 0

0 0

⎛

0

0

0

0 0

⎞

√ μ(α2 +β 2 )+κβ 2 2 ⎟ −1 −i α2 +β 2 ξt − 2(α2 +β 2 ) ξ t P = A (ξ)e e (1 + O(1)ξ 3 t), 0⎟ 2 ⎠

⎜ ˆ 2 = P ⎜ 0 e−λ2 t G ⎝ 0 0 ⎛

⎞

0 0

⎞

√ μ(α2 +β 2 )+κβ 2 2 ⎜ ⎟ −1 i α2 +β 2 ξt − 2(α2 +β 2 ) ξ t ⎟ ˆ3 = P ⎜ 0 0 = A (ξ)e e (1 + O(1)ξ 3 t), P G 0 ⎠ 3 ⎝ 0 0 e−λ3 t here Ak (ξ), k = 1, 2, 3, are 3 × 3 matrices which are bounded near ξ = 0. We omit their detail expressions since they are of no importance. ˆk (ξ) (k = 1, 2, 3) near ξ = ∞. We will continue to give the approximate expressions of λ They can be obtained by similar procedure as near ξ = 0. We give their expressions as follows. (i) If κ = μ, then   2 ˆ 1 (ξ) = α + O 1 , λ μ ξ   2 ˆ 2 (ξ) = μξ 2 + iβξ − α + O 1 , λ 2μ ξ   2 ˆ 3 (ξ) = μξ 2 − iβξ − α + O 1 . λ 2μ ξ (ii)

If κ = μ, then   2 ˆ 1 (ξ) = α + O 1 , λ μ ξ

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1  2 2  ˆ 2 (ξ) = μξ 2 − α + β +O , λ μ μ−κ ξ 1 2 ˆ 3 (ξ) = μξ 2 − β λ +O . μ−κ ξ ˆ k (ξ), we know that if |ξ| 1 then Re(λ ˆ k (ξ)) > 0. Thus, we can From the expression of λ ˆ ξ) near ξ = ∞, get the following approximate formula of G(t, ˆ 2 (t, ξ) + G ˆ 3 (t, ξ). ˆ ξ) = G ˆ 1 (t, ξ) + G G(t, ˆ k (t, ξ) (k = 1, 2, 3) can be written as If κ = μ, then G   2 ˆ 1 = C1 (ξ) exp − α t + O(1) t , G μ ξ     2 ˆ 2 = C2 (ξ) exp − μξ 2 + iβξ − α t + O(1) t , G 2μ ξ     2 t α 3 2 ˆ t + O(1) . G = C3 (ξ) exp − μξ − iβξ − 2μ ξ If κ = μ, then they have the following expressions   2 ˆ 1 = C1 (ξ) exp − α t + O(1) t , G μ ξ      2 β2 t α 2 2 ˆ + t + O(1) , G = C2 (ξ) exp − μξ − μ μ−κ ξ     2 t ˆ 3 = C3 (ξ) exp − μξ 2 − β t + O(1) , G μ−κ ξ here Ck (ξ) (k = 1, 2, 3) are 3 × 3 bounded matrices near ξ = ∞, the detail expressions are omitted. Lemma 2.1 For any ξ ∈ (−∞, +∞), the real parts of the eigenvalues of the matrix A(ξ) are positive. Proof From expressions of the eigenvalues obtained above, the conclusion is obvious near ξ = 0 and ξ = ∞. We only need to prove that A(ξ) has no pure imaginary eigenvalue, i.e., (2.4) has no pure imaginary solution. Assume that λ = ia (a ∈ R) is one of the eigenvalues of A(ξ) for some ξ = ξ1 , i.e., λ(ξ1 ) = ia(ξ1 ), a(ξ1 ) ∈ R, a(ξ1 ) = 0. Taking λ = ia in λ3 − (μ + κ)ξ 2 λ2 + (α2 + β 2 + μκξ 2 )ξ 2 λ − α2 κξ 4 = 0, we get −ia3 + (κ + μ)ξ 2 a2 + i(α2 + β 2 + μκξ 2 )ξ 2 a − α2 κξ 4 = 0. Thus we have

⎧ ⎨ (α2 + β 2 + μκξ 2 )ξ 2 − a2 = 0, ⎩ (μ + κ)a2 = α2 κξ 2 .

By direct and simple calculation, we get (κ + μ)β 2 + μα2 +

μ(κ + μ)2 2 a = 0, α2

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which means that (κ + μ)β 2 = 0. 2

This is obviously a contradiction to the assumption in problem (1.1).

Next, we will give several lemmas which are useful later, simple proofs of which are given. Also see [18]. Lemma 2.2 Let f (x, t) and g(x, t) be two given bounded functions. Suppose the Fourier transform of f and g have relation fˆ(ξ, t) = eiaξ g(ξ, t), a ∈ R being a constant. If g(x, t) ≤ p(x, t), then we have f (x, t) ≤ p(x + at, t). We can get this lemma from the properties of Fourier transform easily or directly prove it by the formulae of Fourier transform. This lemma has important use in the following text. Lemma 2.3

 1 ∨ ξ

1 = √ 2π

 1 π H(x), eixξ dξ = i ξ 2 −∞



∞

(2.5)

where H(x) is the Heaviside function. In fact, let f (x) =

 1 ∨ ξ

1 =√ 2π

and it can be rewritten as 1 f (x) = − √ 2π





∞

−∞

∞

1 eixξ dξ, ξ −∞

1 e−ixξ dξ. ξ

Adding the above two formulae together, we get   sin xξ π iH(x) ∞ sin ξ dξ = √ dξ = i H(x), ξ 2 2π −∞ ξ −∞ ∞ here we have used the well known result −∞ sinx x dx = π. i f (x) = √ 2π



∞

Lemma 2.4 For any nonnegative integers α, β and real number b > 0, we have     2   β α −bξ2 t   ) ≤ C ξ α−β (1 + ξ 2 t)β e−bξ t , Dξ (ξ e

(2.6)

where C is a positive constant. Proof First, by induction we know that the following estimate holds ⎧ ⎨ O(1)ξ α−β (1 + ξ 2 t)β e−bξ2 t , α ≥ β, 2 β α −bξ t )= Dξ (ξ e α+β 2 2 −bξ t ⎩ O(1)t β−α 2 (1 + ξ t) 2 e , α < β.

(2.7)

Since when β > α, t

β−α 2

(1 + ξ 2 t)

α+β 2

e−bξ

2

t

= ξ α−β (ξ 2 t)

β−α 2

(1 + ξ 2 t)

α+β 2

e−bξ

2

t

2

= O(1)ξ α−β (1 + ξ 2 t)β e−bξ t , (2.6) follows immediately.

2

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Pointwise Estimate on Green Function

ˆ t) It is well known that the decay of the solution is mainly related to the properties of G(ξ, near ξ = 0 in the frequency space. To do this, we can use a cut-off function to divide the frequency space into low frequency part, middle frequency part and high frequency part. We ˆ t) in these different regions and give asymptotic decay behavior of will carefully analyze G(ξ, U (x, t). Let ⎧ ⎧ ⎨ 1, |ξ| < ε, ⎨ 1, |ξ| > R + 1, χ1 (ξ) = χ3 (ξ) = ⎩ 0, |ξ| ≥ 2ε, ⎩ 0, |ξ| < R be two smooth cut-off functions and 2ε < R. Denote χ2 (ξ) = 1 − χ1 (ξ) − χ3 (ξ) and ˆ k = χj G ˆ k , j = 1, 2, 3; k = 1, 2, 3. G j ˆ t) as Then we rewrite G(ξ, ˆ ξ) = G(t,

3  3 

ˆ k (t, ξ). G j

(3.1)

k=1 j=1

For simplicity, in the following we denote ˆ 2 (t, ξ) = χ1 (ξ)A2 (ξ)e− L ˆ 3 (t, ξ) = χ1 (ξ)A3 (ξ)e L

μ(α2 +β 2 )+κβ 2 2(α2 +β 2 )

2 +β 2 )+κβ 2 − μ(α2(α 2 +β 2 )

ξ2 t 2

ξ t

(1 + O(1)ξ 3 t),

(3.2)

(1 + O(1)ξ 3 t).

(3.3)

Lemma 3.1 If ε is small enough and |ξ| < ε, then we have α2 κ

2

ˆ 11 (t, ξ))| ≤ Cξ γ−η (1 + ξ 2 t)η+1 e− α2 +β2 ξ t , |Dξη (ξ γ G ˆ 2 (t, ξ))| ≤ Cξ γ−η (1 + ξ 2 t)η+1 e |Dξη (ξ γ L ˆ 3 (t, ξ))| ≤ Cξ γ−η (1 + ξ 2 t)η+1 e |Dξη (ξ γ L

(3.4)

2 +β 2 )+κβ 2 − μ(α2(α 2 +β 2 )

ξ t

2 +β 2 )+κβ 2 − μ(α2(α 2 +β 2 )

2

2

,

ξ t

for some constant C. Proof Direct calculation gives ˆ 1 (ξ, t)) ˆ 1 (t, ξ)) = Dη (ξ γ χ1 (ξ)G Dξη (ξ γ G 1 ξ 2  − α κ ξ2 t = O(1) Dξl (A1 (ξ))χ1 (ξ))Dξm (ξ γ e α2 +β2 (1 + O(1)ξ 3 t)). l+m=η

Since |Dξl (A1 (ξ)χ1 (ξ))| ≤ C and ξ −m < ξ −η (m < η), from Lemma 2.4 we have |Dξm (ξ γ e

2

2 κ − α2α+β 2ξ t

)| ≤ O(1)|ξ γ−m (1 + ξ 2 t)m e ≤ O(1)|ξ γ−η (1 + ξ 2 t)β e 2

=

2

2 κ − α2α+β 2ξ t

2κ − α2α+β 2

− α κ ξ2 t O(1)|Dξη (ξ γ e α2 +β2 |.

ξ2 t

|

|

(3.5) (3.6)

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Thus we get ˆ 1 (t, ξ))| ≤ O(1)|Dη (ξ γ e |Dξη (ξ γ G 1 ξ

2

2 κ − α2α+β 2ξ t

≤ C|ξ γ−η (1 + ξ 2 t)η e

| + O(1)|Dξη (ξ γ+2 te

2κ − α2α+β 2

ξ2 t

2

e

)|

| + C|ξ γ+2−η (1 + ξ 2 t)η e

2 α κ η+1 − α2 +β 2 ξ t

≤ Cξ γ−η (1 + ξ 2 t)

2

2 κ − α2α+β 2ξ t

2

2 κ − α2α+β 2ξ t

|

.

The proof of (3.4) is completed. (3.5)–(3.6) can be obtained similarly. 2 Proposition 3.1 Assume that Gk1 (k = 1, 2, 3) are the inverse Fourier transforms of ˆ k1 (k = 1, 2, 3), respectively and α ≥ 0. Then when ε is small enough, we have the following G estimates γ+1 2

(3.7)

− γ+1 2

(3.8)

|Dxγ (G11 (t, x))| ≤ C(1 + t)−

BN (x, t),  |Dxγ (G21 (t, x))| ≤ C(1 + t) BN (x − α2 + β 2 t, t),  γ+1 |Dxγ (G31 (t, x))| ≤ C(1 + t)− 2 BN (x + α2 + β 2 t, t),

(3.9)

2

x where BN (x, t) = (1 + 1+t )−N and C depends only on α and N . Proof From the property of Fourier transform and Lemma 3.1, we know that    1  ∞ ixξ η γ ˆ 1 |xη Dxγ (G11 (t, x))| = √  e Dξ (ξ G1 (t, ξ))dξ  2π ∞  ∞ 2 − α κ ξ2 t ≤C ξ γ−η (1 + ξ 2 t)η+1 e α2 +β2 dξ 0  ∞ 2 η−γ−1 γ−η−1 − α κ ξ2 t ≤ Ct 2 ξ 2 (1 + ξ)η+1 e α2 +β2 dξ 0

η−γ−1 2

≤ Ct

.

Therefore, |Dxγ (G11 (t, x))| ≤ Ct

γ−1 2

η

η

·

γ−1 γ−1 t2 1 + t2 1+t η ≤ Ct 2 · 2 η ≤ Ct 2 · ( 2 ) 2 . η x x (x ) 2

On the other hand, |Dxγ (G11 (t, x))|

   1  ∞ ixξ γ ˆ 1 = √  e ξ G1 (t, ξ)dξ  2π ∞  ∞  2 − α κ ξ2 t ≤C ξ γ e α2 +β2 dξ + C ≤ Ct

0 − γ+1 2

+ Ct · t−

γ+3 2

≤ Ct−

∞

ξ γ+2 te

2

2 κ − α2α+β 2ξ t

dξ

0

γ+1 2

.

Thus we get |Dxα G11 (t, x)| ≤ Ct− Since

α+1 2

  1 + t N  . min 1, x2

⎧ ⎨ 2, x 2 ≤ 1+ 1 + t ⎩ 2x , 1+t 2

x2 ≤ t + 1, x2 > t + 1,

(3.10)

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we have

  1 + t N  2N ≤ min 1, = 2N BN (x, t). N x2 (1 + 1+t x2 )

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(3.11)

Then inequality (3.7) follows. By the same discussion as above, we know that |Dxγ L2 (t, x)| ≤ C(1 + t)− Since ˆ 2 (t, ξ) = e−i G 1

√

γ+1 2

BN (x, t).

α2 +β 2 ξt ˆ

L(t, ξ)

using Lemma 2.2, we know that inequalities (3.8) and (3.9) hold. 2 Next, we will give the estimates of Gk2 (t, x) (k = 1, 2, 3). For this, we need the following lemma. Lemma 3.2 For any positive numbers γ and η, there exist a positive constant b0 > 0 such that ˆ k2 (t, ξ))| ≤ C(1 + t)η e−b0 t . |Dξη (ξ γ G (3.12) Proof From Lemma 2.1, we know that the real part of eigenvalue in region ε ≤ |ξ| ≤ R ˆ k (ξ, t)| ≤ Ce−b0 t . It is obvious is positive. Therefore we can find a number b0 > 0 such that |G 2 that (3.12) holds when η = 0. Suppose that (3.12) holds for β ≤ l − 1, we will prove that it is true for β = l. From equation (2.3), we obtain ∂   η! ˆ k2 (t, ξ)) = − ˆ k2 (ξ, t)). + A(ξ) Dξη (ξ γ G Dl A(ξ)Dξη−l (ξ γ G ∂t (η − l)!l! ξ 1≤l<η

Thus, when β = l, ˆ k2 (t, ξ))| ≤ |Dξη (ξ γ G

 1≤l<η



≤C

t

η! (η − l)!l!



t 0

ˆ k2 (t − s, ξ)Dξl (A(ξ))Dη−l (ξ γ G ˆ k2 (s, ξ))|ds |G ξ

e−b0 (t−s) e−b0 s (1 + s)η−1 ds

0

≤ C(1 + t)η e−b0 t . 2 Gk2 .

In fact, we have Next, we’ll give the estimate of Proposition 3.2 For any fixed ε and R, where |ε| 1 and |R| 1, there exist positive constants b and C such that |Dxγ (G12 + G22 + G32 )(t, x)| ≤ Ce−bt BN (x, t).

(3.13)

Proof From the proof of Lemma 3.2, we know that there exists a number b0 > 0 such ˆ k (ξ, t)| ≤ Ce−b0 t , k = 1, 2, 3. Besides, from Lemma 3.2, we have that |G 2   ˆ k (t, ξ)dξ ≤ C(1 + t)η e−b0 t Dξη ξ γ G dξ ≤ C(1 + t)η e−b0 t . |xη Dxγ Gk2 (t, x)| = 2 R

ε≤ξ≤R

We can choose a number b ∈ (0, b0 ), such that η

|xη Dxγ Gk2 (t, x)| ≤ C(1 + t) 2 e−bt .

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Therefore, |Dxγ Gk2 (t, x)| ≤ C

 1 + t η x2

Vol.31 Ser.B

e−bt .

By direct calculation, we get         ixξ γ ˆ k  ˆ k2 (t, ξ)dξ  =  e ξ (t, ξ)dξ G |Dxγ Gk2 (t, x)| =  eixξ ξ γ G 2    R ε≤|ξ|≤R  ≤ Ce−b0 t ξ γ dξ ≤ Ce−bt . ε≤|ξ|≤R

With the help of inequality (3.11), we know that (3.13) holds. 2 Finally, we give the estimates of Gk3 (k = 1, 2, 3). For convenience, we first give the approximate Taylor expansion of χ3 (ξ)Ck (ξ) near ξ = ∞: χ3 (ξ)Ck (ξ) = Ck0 + Ck1

1 1 + O(1) 2 , ξ ξ

where Ckj (j = 0, 1; k = 1, 2, 3) are constant matrices. The estimates on Green Functions corresponding to the high frequency part is the following proposition. Proposition 3.3 If R is large enough, then for any nonnegative integer γ, there exist positive numbers C and b such that   3    γ  k  Dx G3 (t, x) − Pγ (t, x) ≤ Ce−bt BN (x, t), 

(3.14)

k=1

where Pγ (t, x) is a polynomial which has the following form Pγ (t, x) = O(1)

3 

e−bk t Fγ (t, x)

(3.15)

Dxl δ(x),

(3.16)

k=1

with Fγ (t, x) =

γ  l=0

and bk is some positive real number. Proof First, we give the estimate of G13 (t, x). By direct calculation, we have  (i)γ ˆ 13 (t, ξ)dξ eixξ ξ γ G Dxγ G13 (t, x) = √ 2π R  α2 t (i)γ = √ eixξ ξ γ χ3 (ξ)C1 (ξ)e− μ t+O(1) ξ dξ 2π R  t (i)γ − αμ2 t = √ e eixξ ξ γ χ3 (ξ)C1 (ξ)eO(1) ξ dξ 2π R t (i)γ − αμ2 t 1 = √ e eixξ ξ γ (Ck0 + O(1) )(1 + o(1) )dξ ξ ξ 2π |ξ|>R   γ γ  2 2 α α (i) 1 = √ e− μ t O(1) eixξ ξ l dξ + Ce− μ t eixξ (1 + t) dξ ξ 2π R |ξ|>R l=0 = I1 + I2 .

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From the property of Fourier transform, we have I1 = e−

α2 μ

t



eixξ

O(1) R

γ 

ξ l dξ = O(1)e−

α2 μ

t

l=0

γ 

Dl δ(x).

l=0

From Lemma 2.3, for some real number b, 0 < b <

a2 μ ,

we have

       1   − αμ2 t eixξ (1 + t) dξ  ≤ Ce−bt . − |I2 | = Ce   ξ R |ξ|
  γ 2    γ 1 l Dx G3 (t, x) − O(1)e− aμ t D δ(x) ≤ Ce−bt . 

(3.17)

l=0

On the other hand,    α2 t (i)γ eixξ Dξη ξ γ χ3 (ξ)C1 (ξ)e− μ t+O(1) ξ dξ xη Dxγ G13 (t, x) = √ 2π R  (i)γ − αμ2 t t 1 = √ e eixξ Dξη (ξ γ (Ck0 + O(1) )(1 + o(1) ))dξ ξ ξ 2π |ξ|>R   (γ−η) +  α2 α2 (i)γ 1 = √ e− μ t O(1) eixξ ξ l dξ + Ce− μ t eixξ (1 + t) dξ ξ 2π R |ξ|>R l=0 = I1 + I2 .

As the above calculation, we can obtain I1 = O(1)e−

a2 μ

t

(γ−η)+



a2

I2 = O(1)(1 + t)e− μ t

Dxl δ(x),

l=0

with the help of the following estimate η

x

γ  l=0

Dxl δ(x)

 =

e R

ixξ

Dξη

γ 



l

ξ dξ =

e R

l=0

ixξ

(γ−η)+

 l=0

l

ξ dξ =

(γ−η)+



Dxl δ(x).

l=0

Thus we can choose some positive number b > 0 such that   N  γ 2   γ 1  l D G (t, x) − O(1)e− aμ t  ≤C 1+t D δ(x) e−bt .  x 3  x2

(3.18)

l=0

From (3.17) and (3.18), and using (3.11) again, we can get   γ 2   γ 1  l Dx G3 (t, x) − O(1)e− aμ t  ≤ Ce−bt BN (x, t). D δ(x)   l=0

Now we estimate G23 (t, x) and G33 (t, x). Whether μ = κ or μ = κ, from Lemma 2.1 we know that there exists a positive number b0 < Reλi (i = 1, 2). By the same discussion, we can get the above relations for G23 (t, x) and G33 (t, x). Adding these inequalities, we get (3.14). 2

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4

ACTA MATHEMATICA SCIENTIA

Vol.31 Ser.B

Proof of Theorem 1.1

In this section, we will use the estimates obtained in Section 3 to prove Theorem 1.1. From Duhumal’s principle, we know that the solution of problem (1.1) can be written as U (x, t) = G(x, t) ∗ U0 (x). Differentiating both sides of this formula with respect to x and using equation (3.1), we get Dxα U (x, t) =

3  3 

  Dxα Glk (x, t) ∗ U0 (x) .

k=1 l=1

First, from Proposition 3.1, we have     3   γ k   ≤ C(1 + t)− γ+1 2 B D G (x, t) (x, t) + B (x − α2 + β 2 t, t) N N x 1   k=1   +BN (x + α2 + β 2 t, t) ||U0 ||L1 , where C depends on the support of U0 . Secondly, from Proposition 3.2, we know that   3  γ  k D  ≤ Ce−bt BN (x, t)||U0 ||L1 . G (t, x) 2  x 

(4.1)

(4.2)

k=1

Since 3 

Dxγ Gk3 (t, x) ∗ U0 (x) =

k=1

 3

 Dxγ Gk3 (t, x) − Pγ (t, x) ∗ U0 (x) + Pγ (t, x) ∗ U0 (x)

k=1

= R1 + R2

from Proposition 3.3, we know that |R1 | ≤ Ce−bt BN (t, x)||U0 ||L1 .

(4.3)

However,     γ γ 3 3         −bk t l −bk t l    |R2 | = O(1) e Dx δ(x) ∗ U0 (x) = O(1) e Dx U0 (x). k=1

l=0

Choosing b = min(b1 , b2 , b3 ), we get |R2 | ≤ Ce

k=1

l=0

   γ  l Dx U0 (x) ≤ Ce−bt ||U0 (x)||C γ . 

−bt 

(4.4)

l=0

Because U0 (x) has compact support, we have |Dxl U0 (x)| ≤ C(1 + x2 )−s , l = 0, 1, · · ·

(4.5)

for any positive number s. Thus we have  |R2 | ≤ Ce−bt (1 + x2 )−s ≤ Ce−bt 1 +

x2 −s ≤ Ce−bt Bs (x, t). 1 + t2

(4.6)

If we choose s ≥ N , then by combining (4.1)–(4.6), we get (1.2). Thus the proof of Theorem 1.1 is completed. 2

No.4

X.W. Hao & S.Y. Peng: THE POINTWISE ESTIMATES TO SOLUTIONS

Acknowledgments and valuable suggestions.

1271

The authors would like to thank professor Yachun Li for her help

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