Open Access
Issue
Wuhan Univ. J. Nat. Sci.
Volume 29, Number 5, October 2024
Page(s) 397 - 402
DOI https://doi.org/10.1051/wujns/2024295397
Published online 20 November 2024

© Wuhan University 2024

Licence Creative CommonsThis is an Open Access article distributed under the terms of the Creative Commons Attribution License (https://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

0 Introduction

In Ref.[1], the authors studied the structural stability for the Moore-Gibson-Thompson (M-G-T) heat equation. They obtained the results of the continuous and convergence for different coefficients in a bounded domain. These results gave measures of the structural stability between the solutions for different theories. The so-called M-G-T heat equation has the form:

C ( τ α + α ¨ ) = k * Δ α + k Δ α ˙ , (1)

where α denotes the thermal displacement, k is the thermal conductivity, k* denotes the conductivity rate, C and τ denote the thermal capacity and the relaxation parameter, respectively. In Refs.[2] and [3], Green and Nagdhi proposed that the motivation for considering (1) as a heat equation came from the heat conduction of type III.

The M-G-T heat equation and similar equations have deserved much interest in the last years, see Refs. [4-8]. In the present paper, equation (1) is defined in a semi-infinite cylindrical pipe R with boundary R. The pipe has arbitrary cross section denotes by D and the boundary D and the generators of the pope are parallel to the x3 axis. We introduce the notations:

R z = { ( x 1 , x 2 , x 3 ) | ( x 1 , x 2 ) D , x 3 > z 0 } ,   D z = { ( x 1 , x 2 , x 3 ) | ( x 1 , x 2 ) D , x 3 = z 0 } ,

where z is a running variable along the x3 axis. Clearly, R0=R, D0=D.

The lateral sides of the cylinder are constrained to have zero thermal displacement. Thus, we have

α = 0   o n   D × ( 0 , + ) = D . (2)

To have a well-determined problem, we impose boundary conditions on the finite end of the cylinder. Thus, we take as assumption

α ( x 1 , x 2 , 0 , t ) = f i ( x 1 , x 2 , t )   o n   ( x 1 , x 2 ) D , t > 0 . (3)

We give the initial conditions:

α ( x 1 , x 2 , x 3 , 0 ) = α ˙ ( x 1 , x 2 , x 3 , 0 ) = α ¨ ( x 1 , x 2 , x 3 , 0 ) = 0 . (4)

The classical Phragmén-Lindelöf theorem, without some a priori asymptotical decay assumptions for solution, states that harmonic function which vanishes on the cylindrical surface must either grow or decay exponentially with the distance from the finite end of the cylinder. In this paper, we add some a priori asymptotical decay assumptions for solution at the infinity. We can get the spatial decay estimates result. It can be seen as a version of Saint-Venant principle. For a review of results of the Saint-Venant principle have been published recently (in several situations), one could see Refs.[9-14]. In the present article, the comma is used to indicate partial differentiation and the differentiation with respect to the direction xi is denoted as ",i", thus α,i denotes αxi, and α˙ denotes αt. The usual summation convection is employed with repeated Latin subscripts i summed from 1 to 3. Hence, α,iα,i=α,12+α,22+α,32.

1 The Function Expression φ(z,t)

In order to obtain the decay estimates result, we must define a function φ(z,t) for the solutions. The function will be constructed by the following Lemmas.

Lemma 1  

Let α be classical solution of equation (1) and satisfy the initial boundary value problems (2)-(4), we define a function

φ 1 ( z , t ) = k * 0   t z D ξ e x p ( - ω η ) α ˙ α , 3 d x d η + k 0 t z D ξ e x p ( - ω η ) α ˙ α ˙ , 3 d x d η . (5)

φ 1 ( z , t ) can also be expressed as

φ 1 ( z , t ) = - C τ 0   t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ ω 2 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x d η + C τ z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ α ¨ d x + C ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x d η + C 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ˙ ) 2 d x + k * ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α , i α , i d x d η + k * 2 z D ξ e x p ( - ω t ) ( ξ - z ) α , i α , i d x + k * ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η . (6)

Proof  

Multiplying (1) by α˙ and integrating, we have

0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ [ C ( τ α + α ¨ ) - k * Δ α + k Δ α ˙ ] d x d η = - C τ 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ ω 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ α ¨ d x d η + C τ z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ α ¨ d x + C ω 2 0   t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x d η + C 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ˙ ) 2 d x + k * 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α , i d x d η - k * 0 t z D ξ e x p ( - ω η ) α ˙ α , 3 d x d η + k 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i   d x d η - k 0 t z D ξ e x p ( - ω η ) α ˙ α ˙ , 3 d x d η = - C τ 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ ω 2 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x d η + C τ ω 2 z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x + C τ z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ α ¨ d x + C ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ˙ ) 2 d x d η

+ C 2 z D ξ e x p ( - ω t ) ( ξ - z )   ( α ˙ ) 2 d x + k * ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α , i α , i d x d η + k * 2 z D ξ e x p ( - ω t ) ( ξ - z ) α , i α , i d x - k * 0 t z D ξ e x p ( - ω η ) α ˙ α , 3 d x d η + k 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i   α ˙ , i   d x d η - k 0 t z D ξ e x p ( - ω η ) α ˙ α ˙ , 3 d x d η = 0 . (7)

From (5) and (7), we can obtain (6).

Lemma 2  

Let α be classical solution of equation (1) and satisfy the initial boundary value problems (2)-(4), we define a function

φ 2 ( z , t ) = k * 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η + k 0 t z D ξ e x p ( - ω η ) α ¨ α ˙ , 3 d x d η . (8)

φ 2 ( z , t ) can also be expressed as

φ 2 ( z , t ) = ( C τ ω 2 + C ) 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ˙ ) 2 d x + ( k ω 2 - k * ) 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η + k * ω 2 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α , i α , i d x d η + k * ω 2 z D ξ e x p ( - ω t ) ( ξ - z ) α , i α , i d x + k * z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α , i d x + k 2 z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α ˙ , i d x . (9)

Proof  

Multiplying (1) by α¨ and integrating, we have

0   t z D ξ e x p ( - ω η ) ( ξ - z ) α ¨ [ C ( τ α + α ¨ ) - k * Δ α + k Δ α ˙ ] d x d η = C τ ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C 0   t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ¨ ) 2 d x + k * 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ¨ , i α , i d x d η - k * 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η + k z D ξ e x p ( - ω t )   ( ξ - z ) α ¨ , i   α , i d x - k 0 t z D ξ e x p ( - ω η )   α ¨ α , 3 d x d η = ( C τ ω 2 + C ) 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ¨ ) 2 d x - k * 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η + k * ω 0 t z D ξ e x p ( - ω η ) ( ξ - z )   α ˙ , i α , i d x d η + k * z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α , i d x - k * 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η + k ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η + k 2 z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α ˙ , i d x - k 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η = ( C τ ω 2 + C ) 0 t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ¨ ) 2 d x - k * 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η + k * ω 2 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α , i α , i d x d η + k * ω 2 z D ξ e x p ( - ω t ) ( ξ - z ) α , i α , i d x + k * z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α , i d x - k * 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η + k ω 2 0 t z D ξ e x p ( - ω η ) ( ξ - z ) α ˙ , i α ˙ , i d x d η + k 2 z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α ˙ , i d x - k 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η = 0 . (10)

From (8) and (10), we get the desired results (9).

We define a new function φ(z,t):

φ ( z , t ) = k ˜ φ 1 ( z , t ) + φ 2 ( z , t ) , (11)

where k˜ is an arbitrary positive constant which will be defined later.

From (5) and (8), we have

φ ( z , t ) = k * k ˜ 0   t z D ξ e x p ( - ω η ) α ˙ α , 3 d x d η - k k ˜ 2 0 t D z e x p ( - ω η ) ( α ˙ ) 2 d A d η + k * 0 t z D ξ e x p ( - ω η ) α ¨ α , 3 d x d η + k 0 t z D ξ e x p ( - ω η ) α ¨ α ˙ , 3 d x d η . (12)

From (6) and (9), φ(z,t) can also be rewritten as

φ ( z , t ) = ( C τ ω 2 + C - C τ k ˜ ) t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 2 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ¨ ) 2 d x + ( k ω 2 - k * + k k ˜ ) 0 t z D ξ e x p ( - ω η )   ( ξ - z ) α ˙ , i α ˙ , i d x d η + ( k * ω 2 2 + k * ω k ˜ 2 ) 0 t z D ξ e x p ( - ω η ) ( ξ - z )   α , i α , i d x d η + ( k * ω 2 + k * k ˜ 2 ) z D ξ e x p ( - ω t ) ( ξ - z ) α , i α , i d x + k 2 z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α ˙ , i d x + ( C τ ω 2 2 + C ω 2 ) k ˜ 0 t z D ξ e x p ( - ω η )   ( ξ - z ) ( α ˙ ) 2 d x d η + C τ k ˜ z D ξ e x p ( - ω t ) ( ξ - z )   α ˙ α ¨ d x + C k ˜ 2 z D ξ e x p ( - ω t ) ( ξ - z )   ( α ˙ ) 2 d x + k * z D ξ e x p ( - ω t ) ( ξ - z )   α ˙ , i α , i d x (13)

We will use the function φ(z,t) to obtain our main result in the next section.

2 Spatial Decay Estimates

Differentiating (13) and using the Schwarz's inequality, we have

- φ ( z , t ) z ( C τ ω 2 + C - C τ k ˜ ) 0 t z D ξ e x p ( - ω η ) ( α ¨ ) 2 d x d η + ( C τ 2 - C τ 2 k ˜ ) z D ξ e x p ( - ω t ) ( α ¨ ) 2 d x + ( k ω 2 - k * + k k ˜ ) 0 t z D ξ e x p ( - ω η )   α ˙ , i α ˙ , i d x d η + ( k * ω 2 2 + k * ω k ˜ 2 ) 0 t D z e x p ( - ω η ) α , i α , i d A d η + [ k * ω 2 + k * k ˜ 2 - ( k * ) 2 k ] z D ξ e x p ( - ω t )   α , i α , i d x + k 4 z D ξ e x p ( - ω t ) α ˙ , i α ˙ , i d x + C k ˜ 4 z D ξ e x p ( - ω t )   ( α ˙ ) 2 d x + ( C τ ω 2 2 + C ω 2 ) k ˜ 0 t z D ξ e x p ( - ω η )   ( α ˙ ) 2 d x d η . (14)

If we choose k˜=14τ, ω=4k*k, we have

( C τ ω 2 + C - C τ k ˜ ) > 0 ,   ( C τ 2 - C τ 2 k ˜ ) > 0 ,   k ω 2 - k * + k k ˜ > 0 ,   k * ω 2 + k * k ˜ 2 - ( k * ) 2 k > 0 .

Using the same method, we have

2 φ ( z , t ) z 2 C τ ω 4 0 t D z e x p ( - ω η ) ( α ¨ ) 2 d A d η + C τ 4 D z e x p ( - ω t ) ( α ¨ ) 2 d A + k ω 4 0 t D z e x p ( - ω η )   α ˙ , i α ˙ , i d A d η + ( k * ω 2 2 + k * ω k ˜ 2 ) 0 t D z e x p ( - ω η ) α , i α , i d A d η + k * ω 4 D z e x p ( - ω t )   α , i α , i d A + k 4 D z e x p ( - ω t ) α ˙ , i α ˙ , i d A + C k ˜ 4 D z e x p ( - ω t )   ( α ˙ ) 2 d A + ( C τ ω 2 2 + C ω 2 ) k ˜ 0   t D z e x p ( - ω η )   ( α ˙ ) 2 d A d η . (15)

We can easily get

φ ( z , t ) C τ ω 4 0   t z D ξ e x p ( - ω η ) ( ξ - z ) ( α ¨ ) 2 d x d η + C τ 4 z D ξ e x p ( - ω t ) ( ξ - z ) ( α ¨ ) 2 d x + k ω 4 0 t z D ξ e x p ( - ω η )   ( ξ - z ) α ˙ , i α ˙ , i d x d η + ( k * ω 2 2 + k * ω k ˜ 2 ) 0 t z D z e x p ( - ω η ) ( ξ - z ) α , i α , i d x d η + k * ω 4 z D ξ e x p ( - ω t )   ( ξ - z ) α , i α , i d x + k 4 z D ξ e x p ( - ω t ) ( ξ - z ) α ˙ , i α ˙ , i d x + C k ˜ 4 z D ξ e x p ( - ω t )   ( ξ - z ) ( α ˙ ) 2 d x + ( C τ ω 2 2 + C ω 2 ) k ˜ 0   t z D ξ e x p ( - ω η )   ( ξ - z ) ( α ˙ ) 2 d x d η = E ( z , t ) . (16)

Combining (12), (14) and (15), we have

φ ( z , t ) k 1 ( - φ z ) + k 2 ( 2 φ z 2 ) , (17)

where k1, k2 are positive computable constants.

We can rewrite (17) as

2 φ z 2 - k 1 k 2 φ z - 1 k 2 0 . (18)

We want to obtain the following result

( z - a ) ( φ z + b ) 0 . (19)

Combining (18), (19) and (15), we obtain

b - a = - k 1 k 2 , a b = 1 k 2 .

We obtain

a = ( k 1 k 2 ) 2 + 4 k 2 + k 1 k 2 2 ,   b = ( k 1 k 2 ) 2 + 4 k 2 - k 1 k 2 2 .

From (19), we have

z { e - a   z ( φ z + b φ ) } 0 . (20)

Integrating (20) with respect to z from z to , we have

φ ( z , t ) z + b φ ( z , t ) 0 . (21)

Solving inequality (21), we easily get

φ ( z , t ) φ ( 0 , t )   e - b z . (22)

Inserting (16) into (22), we obtain

E ( z , t ) φ ( 0 , t )   e - b z . (23)

Inequality (23) implies that the energy can decay as z.We thus have proved the following theorem:

Theorem 1   Let α be classical solution of equation (1) and satisfy the initial boundary value problems (2)-(4), we can get the decay estimates E(z,t)φ(0,t) e-bz.

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