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 ˙ , Mathematical equation(1)

where αMathematical equation denotes the thermal displacement, kMathematical equation is the thermal conductivity, k*Mathematical equation denotes the conductivity rate, CMathematical equation and τMathematical equation 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 RMathematical equation with boundary RMathematical equation. The pipe has arbitrary cross section denotes by DMathematical equation and the boundary DMathematical equation and the generators of the pope are parallel to the x3Mathematical equation 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 } , Mathematical equation

where zMathematical equation is a running variable along the x3Mathematical equation axis. Clearly, R0=RMathematical equation, D0=DMathematical equation.

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

α = 0   o n   D × ( 0 , + ) = D . Mathematical equation(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 . Mathematical equation(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 . Mathematical equation(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 xiMathematical equation is denoted as ",iMathematical equation", thus α,iMathematical equation denotes αxiMathematical equation, and α˙Mathematical equation denotes αtMathematical equation. The usual summation convection is employed with repeated Latin subscripts iMathematical equation summed from 1 to 3. Hence, α,iα,i=α,12+α,22+α,32.Mathematical equation

1 The Function Expression φ(z,t)Mathematical equation

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

Lemma 1  

Let αMathematical equation 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 . Mathematical equation(5)

φ 1 ( z , t ) Mathematical equation 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 . Mathematical equation(6)

Proof  

Multiplying (1) by α˙Mathematical equation 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 Mathematical equation

+ 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 . Mathematical equation(7)

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

Lemma 2  

Let αMathematical equation 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 . Mathematical equation(8)

φ 2 ( z , t ) Mathematical equation 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 . Mathematical equation(9)

Proof  

Multiplying (1) by α¨Mathematical equation 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 . Mathematical equation(10)

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

We define a new function φ(z,t)Mathematical equation:

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

where k˜Mathematical equation 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 . Mathematical equation(12)

From (6) and (9), φ(z,t)Mathematical equation 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 Mathematical equation(13)

We will use the function φ(z,t)Mathematical equation 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 . Mathematical equation(14)

If we choose k˜=14τMathematical equation, ω=4k*kMathematical equation, 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 . Mathematical equation

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 . Mathematical equation(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 ) . Mathematical equation(16)

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

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

where k1Mathematical equation, k2Mathematical equation are positive computable constants.

We can rewrite (17) as

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

We want to obtain the following result

( z - a ) ( φ z + b ) 0 . Mathematical equation(19)

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

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

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 . Mathematical equation

From (19), we have

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

Integrating (20) with respect to zMathematical equation from zMathematical equation to Mathematical equation, we have

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

Solving inequality (21), we easily get

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

Inserting (16) into (22), we obtain

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

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

Theorem 1   Let αMathematical equation 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.Mathematical equation

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