轴向力作用下具有记忆项的热弹耦合梁方程组的初边值问题
The Initial Boundary ValueProblem of the Thermoelastic Coupled Beam Equations with Memory Term under Axial Force

作者: 刘帅博 , 张建文 :太原理工大学,数学学院,山西 太原;

关键词: 轴向力记忆项热弹耦合Axial Force Memory Terms Thermoelastic Beam

摘要: 本文研究了轴向力作用下具有记忆项的热弹耦合梁方程的初边值的问题,通过先验估计证明了系统弱解和正则解的存在唯一性。

Abstract: In this paper, we study the initial boundary value of a thermally coupled beam equation with a memory term under the action of an axial force.

1. 引言

1968年,Gurtin Monton E等 [1] 提出了具有记忆项的热传导理论。之后关于含记忆性的各类非线性梁方程的初边值问题的研究有若干的进展 [2] - [8],比如2009年,Wang Junmin等 [2] 研究了如下具有记忆项的热传导方程

θ t + 0 t g ( t τ ) θ x x ( x , τ ) d τ = 0

解的存在性。2011年,Lazo P P D [3] 运用Galerkin方法证明了如下具有记忆项的波动方程

u t t M ( | u x | 2 ) u x x + 0 t g ( t τ ) u x x ( x , τ ) d τ = 0

整体解存在唯一性和稳定性。2011年,Santos M L [4] 考虑热记忆项,研究了如下热弹耦合杆方程组

{ u t t u x x + α θ x = 0 θ t θ x x 0 t g ( t τ ) θ x x ( x , τ ) d τ + β u t x = 0

解的存在唯一性及指数衰减率。

本文在前人的基础上考虑如下轴向力作用下具有记忆项热弹耦合梁方程组

{ u t t + u x x x x 0 t k 1 ( t τ ) u x x x x ( x , τ ) d τ M ( 0 L | u x | 2 d x ) u x x + α θ x x = q ( x ) θ t θ x x 0 t k 2 ( t τ ) θ x x ( x , τ ) d τ α u x x t = 0 (1)

在边界条件

u ( 0 , t ) = u ( L , t ) = u x x ( 0 , t ) = u x x ( L , t ) = 0 , θ ( 0 , t ) = θ ( L , t ) = 0 (2)

和初始条件

, u t ( x , 0 ) = u 1 ( x ) , θ ( x , 0 ) = θ 0 ( x ) (3)

下解的存在唯一性。

2. 预备知识

本文的讨论是基于以下的Hilbert空间:

S 1 = { u H 4 ( 0 , L ) | u , u x x H 0 1 ( 0 , L ) }

S 2 = H 0 1 ( 0 , L ) H 2 ( 0 , L )

S 1 S 2 L 2

方程中的函数满足如下基本假设

(h1) 对函数M,假设 M C 1 ( [ 0 , ) ) ,且M为非负函数,有

M ( s ) a + b s ( a , b > 0 ) , | M ( s ) | < σ (4)

其中 M ^ ( s ) = 0 s M ( z ) d z ,其中 σ 为大于0的正常数。

(h2) 假设 k 1 ( t ) C 1 [ [ 0 , ) , [ 0 , ) ] , k 1 ( t ) 0 , k 1 ( t ) 0 , 0 k 1 ( t ) d t < a

(h3) 假设 k 2 ( t ) C 1 [ [ 0 , ) , [ 0 , ) ] , k 2 ( 0 ) = 0 ,且 1 0 k 2 ( t ) d t = l > 0 , ξ > 0 ,使得 k 2 ( t ) ξ k 2 ( t )

(h4) 假设 q ( x ) L 2 ( 0 , L )

3. 主要结论

3.1. 弱解的存在性

定理1若 u 0 S 2 , u 1 , θ 0 H 0 1 ( 0 , L ) ,且假设(h1)~(h4)成立,则存在使得 u L ( 0 , T ; S 2 ) , u t L ( 0 , T ; L 2 ) , θ L ( 0 , T ; L 2 ) , T > 0 。且在下述定义下满足方程(1),即对几乎处处的 t ( 0 , T )

{ d d t ( u t , ω ) + ( u x x , ω x x ) 0 t k 1 ( t τ ) ( u x x , ω x x ) d τ + M ( u x ( t ) 2 2 ) ( u x , ω x ) + α ( θ x x , ω ) = ( q ( x ) , ω ) d d t ( θ , ω ˜ ) + ( θ x , ω ˜ x ) + 0 t k 2 ( t τ ) ( θ x , ω ˜ x ) d τ α ( u x x t , ω ˜ ) = 0 , ω S 2 , ω ˜ H 0 1 (5)

证明:1) 构造近似解:设 { ω j } j = 1 分别是 S 2 H 0 1 的标准正交基,使得 u 0 , u 1 s p a n { ω 1 , ω 2 } θ 0 = ω ˜ 1 ,记

u m ( x , t ) = j = 1 m r j ( t ) ω j ( x ) , θ m ( t ) = j = 1 m h j ( t ) ω ˜ j ( x )

其中 r j ( t ) , h j ( t ) 为未知函数,仿(5)作如下关于 u m ( x , t ) 的方程组

{ ( u t t m , ω j ) + ( u x x m , ω j x x ) 0 t k 1 ( t τ ) ( u x x m , ω j x x ) d τ + M ( u x m ( t ) 2 2 ) ( u x m , ω j x ) + α ( θ x x m , ω j ) = ( q ( x ) , ω j ) ( θ t m , ω ˜ j ) + ( θ x m , ω ˜ j x ) + 0 t k 2 ( t τ ) ( θ x m , ω ˜ j x ) d τ α ( u x x t m , ω ˜ j ) = 0 (6)

另外有

{ u m ( x , 0 ) = u 0 m u 0 S 2 u t m ( x , 0 ) = u 1 m u 1 L 2 θ m ( x , 0 ) = θ 0 m θ 0 L 2 (7)

根据常微分的解的存在唯一性理论可知,方程(6) (7)存在唯一解,下面进行先验估计。

2) 先验估计

以下讨论中C表示与 m , T 无关的常数,它在不同的地方代表不同的表达式中可能有不同的值。

( k φ ) ( t ) = 0 t k ( t τ ) 0 L | φ ( t ) φ ( τ ) | d x d τ (8)

令方程组(6)中两式分别与 r j t h j 做积,并对 j = 1 , 2 , 3 , , m 求和,得

( u t t m , u t m ) + ( u x x m , u t x x m ) 0 t k 1 ( t τ ) ( u x x m ( τ ) , u t x x m ( t ) ) d τ + M ( u x m ( t ) 2 2 ) ( u x m , u t x m ) + α ( θ x x m , u t m ) = ( q ( x ) , u t m ) (9)

( θ t m , θ m ) + ( θ x m , θ x m ) + 0 t k 2 ( t τ ) ( θ x m ( τ ) , θ x m ( t ) ) d τ α ( u x x t m , θ x m ) = 0 (10)

两式相加,并从0到t积分所得两式相加,由假设(h1)及(8)式,得

F m ( t ) + ( k 1 u x x m ) ( t ) + 0 t [ k 1 ( s ) u x x m 2 2 ( k 1 u x x m ) ( s ) ] d s + 2 0 t [ 0 s k 2 ( s τ ) ( θ x m ( τ ) , θ x m ( s ) ) d τ ] d s + 2 0 t θ x m ( s ) 2 2 d s = F m ( 0 ) + 0 t ( q ( x ) , u t m ) d s (11)

其中

F m ( t ) = u t m 2 2 + u x x m 2 2 + M ^ ( u x m 2 2 ) + θ m 2 2 0 t k 1 ( s ) d s u x x m 2 2

F m ( 0 ) = u 1 m 2 2 + u 0 x x m 2 2 + M ^ ( u 0 x m 2 2 ) + θ 0 m 2 2 0 t k 1 ( s ) d s u 0 x x m 2 2

结合假设(h3),由Schwarz不等式和Young不等,得到

| 0 s k 2 ( s τ ) ( θ x m ( τ ) , θ x m ( s ) ) d τ | = | 0 L 0 s k 2 ( s τ ) θ x m ( τ ) θ x m ( s ) d τ d x | η θ x m ( s ) 2 2 + 1 4 η 0 L [ 0 s k 2 ( s τ ) θ x m ( τ ) d τ ] 2 d x η θ x m ( s ) 2 2 + 1 4 η 0 s k 2 ( τ ) d τ 0 s k 2 ( s τ ) 0 L | θ x m ( τ ) | 2 d x d τ η θ x m ( s ) 2 2 + ( 1 l ) k 2 ( 0 ) 4 η 0 s θ x m ( τ ) 2 2 d τ (12)

其中 0 < η < 1 。则

| 2 0 t [ 0 s k 2 ( s τ ) ( θ x m ( τ ) , θ x m ( s ) ) d τ ] d s | 2 η 0 t θ x m ( s ) 2 2 d s (13)

由假设(h4),得

2 0 t ( q ( x ) , u t m ( s ) ) d s 2 0 t [ q ( x ) u t m ( s ) ] d s q ( x ) 2 2 + 0 t F m ( s ) d s (14)

由(12)~(14),得

F m ( t ) + 2 ( 1 η ) 0 t θ x m ( s ) 2 2 d s F m ( 0 ) + q ( x ) 2 2 + C 0 t F m ( s ) d s (15)

故由Gronwall引理可得

F m ( t ) + 2 ( 1 η ) 0 t θ x m ( s ) 2 d s ( F m ( 0 ) + q ( x ) 2 2 ) e C t

因此得到

u t m C , u x x m C , θ m C , 0 t θ x m ( s ) 2 2 d s < C .

由于上述关于t一致有界,故对 T , t ( 0 , T ) 内均成立。由Poincare不等式可得 u m C

3) 收敛:

u m u L ( 0 , T ; S 2 ) 弱*收敛

u t m u t L ( 0 , T ; L 2 ) 弱*收敛

θ m θ L ( 0 , T ; L 2 ) 弱*收敛

0 t θ x m ( s ) 2 2 d s L ( 0 , T ; L 2 ) 弱*收敛

u m u L 2 ( 0 , T ) 中强收敛且几乎处处收敛

φ L 1 ( 0 , T ) ,有 ω φ L 1 ( 0 , T ; L 2 ) ,那么有

0 T ( u x x m , ω x x ) φ d t = 0 T 0 L u x x m ω x x φ d x d t = 0 T ( u m , ω x x x x φ ) d t 0 T ( u , ω x x x x φ ) d t = 0 T ( u x x , ω x x ) φ d t

( u x x m , ω x x ) ( u x x , ω x x ) L ( 0 , T ) 弱*收敛。

0 T ( θ x m , ω ˜ x ) φ d t = 0 T 0 L θ x m ω ˜ x φ d x d t = 0 T ( θ m , ω ˜ x x φ ) d t 0 T ( θ , ω ˜ x x φ ) d t = 0 T ( θ x , ω ˜ x ) φ d t

( θ x m , ω x ) ( θ x , ω x ) L ( 0 , T ) 弱*收敛。

同理可得

0 t k 1 ( t τ ) ( u x x m , ω x x ) d τ 0 t k 1 ( t τ ) ( u x x , ω x x ) d τ L ( 0 , T ) 弱*收敛。

M ( u t m ( t ) 2 2 ) ( u x m , ω x ) M ( u x ( t ) 2 2 ) ( u x , ω x ) L ( 0 , T ) 弱*收敛。

由以上可知

{ ( u t t , ω ) + ( u x x , ω x x ) 0 t k 1 ( t τ ) ( u x x , ω x x ) d τ + M ( u x ( t ) 2 2 ) ( u x , ω x ) = q ( L ) ( θ t , ω ˜ ) + ( θ x , ω ˜ x ) 0 t k 2 ( t τ ) ( θ x , ω ˜ x ) d τ = 0 (16)

成立。定理1证毕。

定理2设 ( u , θ ) , ( v , η ) 是方程(5)的两个解,且 u , v L ( 0 , T ; S 2 ) , θ , η L ( 0 , T ; L 2 ) , u t , v t L ( 0 , T ; L 2 ) ,以及假设 ( u , θ ) , ( v , η ) 满足初始条件: u ( x , 0 ) = u 0 , u t ( x , 0 ) = u 1 , , v ( x , 0 ) = v 0 , v t ( x , 0 ) = v 1 , η ( x , 0 ) = η 0 u 0 , v 0 S 2 , θ 0 , η 0 L 2 u 1 , v 1 L 2 , u 0 , v 0 S 2 ,令 ( z , λ ) = ( u v , θ η ) ,则有

{ ( z t t , ω j ) + ( z x x , ω j x x ) 0 t k 1 ( t τ ) ( z x x , ω j x x ) d τ + M ( z x ( t ) 2 2 ) ( z x , ω j x ) + α ( γ x x , ω j ) = 0 ( γ t , ω ˜ j ) + ( γ x , ω ˜ j x ) + 0 t k 2 ( t τ ) ( γ x , ω ˜ j x ) d τ α ( z t x x , ω ˜ j ) = 0

证明:与定理1先验估计类似,可得到

z t 2 2 + z x x 2 2 + λ 2 2 { u 1 v 1 2 2 + u 0 x x v 0 x x 2 2 + θ 0 η 0 2 2 } e C t (17)

其中C是关于 u 1 , v 1 , u 0 x x , v 0 x x , θ 0 , η 0 的连续函数。

推论方程(1)-(3)的解是唯一的。

证明:由(17)得,当 u 0 = v 0 , u 1 = v 1 , θ 0 = η 0 时,得到 z t 2 2 + z x x 2 2 + λ 2 2 0 ,则有 z t = 0 , z x x = 0 , λ = 0 ,从而有 u = v , θ = η 。解的唯一性得证。

3.2. 正则性

本节我们假设

(h5) k 1 ( 0 ) = 0 , k 1 ( 0 ) = 0 , k 2 ( t ) 0 , k 2 ( t ) 0

定理3若 u 0 S 1 , u 1 , θ 0 S 2 ,且假设(h1)—(h5)成立,则存在唯一的函数 u u ( x , t ) , θ θ ( x , t ) 使得 u L ( 0 , T ; S 1 ) , u t L ( 0 , T ; S 2 ) , u t t L ( 0 , T ; L 2 ( 0 , L ) ) , θ L ( 0 , T ; S 2 ) , θ t L ( 0 , T ; L 2 ( 0 , L ) ) , T > 0 。且在下述定义下满足方程(1),即对几乎处处的 t ( 0 , t )

{ ( u t t , ω ) + ( u x x , ω x x ) 0 t k 1 ( t τ ) ( u x x , ω x x ) d τ + M ( u x ( t ) 2 2 ) ( u x , ω x ) + α ( θ x x , ω ) = ( q ( x ) , ω ) ( θ t , ω ˜ ) + ( θ x , ω ˜ x ) + 0 t k 2 ( t τ ) ( θ x , ω ˜ x ) d τ α ( u x x t , ω ˜ ) = 0 ω S 1 , ω ˜ S 2 (18)

证明:1) 构建近似解:设 { ω j } j = 1 { ω ˜ j } j = 1 分别是 S 1 S 2 的标准正交基,使得 u 0 , u 1 s p a n { ω 1 , ω 2 } , θ 0 = ω ˜ 1 ,记

u m ( x , t ) = j = 1 m r j ( t ) ω j ( x ) , θ m ( t ) = j = 1 m h j ( t ) ω ˜ j ( x )

其中 r j ( t ) , h j ( t ) 为未知函数,仿(5)作如下关于 u m ( x , t ) 的方程组

{ ( u t t m , ω j ) + ( u x x x x m , ω j ) 0 t k 1 ( t τ ) ( u x x x x m , ω j ) d τ M ( u x m ( t ) 2 2 ) ( u x x m , ω j ) + α ( θ x x m , ω j ) = ( q ( x ) , ω j ) ( θ t m , ω ˜ j ) ( θ x x m , ω ˜ j ) 0 t k 2 ( t τ ) ( θ x x m , ω ˜ j ) d τ α ( u x x t m , ω ˜ j ) = 0 (19)

另外有

(20)

根据常微分的解的存在唯一性理论可知,方程(19) (20)存在唯一解,下面进行先验估计。

2) 先验估计

与之前一样可得 u t m C , u x x m C , θ m C

r j = r j t t ( 0 ) , h j = h j t ( 0 ) 两式分别与方程组(19)中两式做积,对 j = 1 , 2 , 3 , , m 求和后,得

u t t m ( 0 ) 2 2 + ( u x x x x m ( 0 ) , u t t m ( 0 ) ) M ( u x m ( 0 ) 2 2 ) ( u x x m ( 0 ) , u t t m ( 0 ) ) + α ( θ x x m ( 0 ) , u t t m ( 0 ) ) = ( q ( x ) , u t t m ( 0 ) ) (21)

θ t m ( 0 ) 2 2 + ( θ x x m ( 0 ) , θ t m ( 0 ) ) α ( u x x t m ( 0 ) , θ t m ( 0 ) ) = 0 (22)

结合假设(h4)应用Schwarz不等式,得

u t t m ( 0 ) 2 2 [ u 0 x x x x m + M ( u 0 x m 2 2 ) u 0 x x m + α θ 0 x x + q ( x ) ] u t t m ( 0 ) (23)

θ t m ( 0 ) 2 2 [ θ 0 x x m + α u 1 x x m ] θ t m ( 0 ) (24)

因此得到

u t t t m + u x x x x t m [ M ( u x m ( t ) 2 2 ) + 2 M ( u x m , u x t m ) ] u x x m M ( u x ( t ) 2 2 ) u x x t m 0 t k 1 ( t τ ) u x x x x m ( τ ) d τ k 1 ( 0 ) u x x x x t m + α θ x x t m = 0 (25)

θ t t m θ x x t m 0 t k 2 ( t τ ) θ x x m ( τ ) d τ k 2 ( 0 ) θ x x m α u x x t m = 0 (26)

(25) (26)两式分别与 u t t m θ t m 做积,并从0到t积分,所得两式相加,结合假设(h1)及(8)式,得

F m ( t ) + 2 0 t 0 s k 1 ( s τ ) ( u x x x m ( τ ) , u t t x m ( s ) ) d τ d s ( k 2 θ x m ) ( t ) + 0 t [ k 2 ( s ) θ x m 2 2 + ( k 2 θ x m ) ( s ) ] d s + 2 0 t θ x t m 2 2 d s = F m ( 0 ) + 0 t M ( || u x ( t ) || 2 2 ) ( u x x t m , u t t m ) + [ M ( u x ( t ) 2 2 ) + 2 M ( u x m , u x t m ) ] ( u x x m , u t t m ) d s (27)

其中 F m ( t ) = u t t m 2 2 + u x x t m 2 2 + θ t m 2 2 + θ x m 2 2 0 t k 2 ( τ ) d τ

由假设(h1)及定理1结论,并由Schwarz不等式和Young不等式,得到

M ( u x m 2 2 ) ( u x x t m , u t t m ) + [ M ( u x m 2 2 ) + 2 M ( u x m , u x t m ) ] ( u x x m , u t t m ) 1 2 σ ( u x x m 2 2 + u t t m 2 2 ) + 2 M u x m 2 2 u x t m 2 2 1 2 ( u x x m 2 2 + u t t m 2 2 ) + M ( u x m 2 2 ) 1 2 ( u x x t m 2 2 + u t t m 2 2 ) C ( u x x m 2 2 + u t t m 2 2 + u x m 2 2 + u x t m 2 2 + u x x t m 2 2 ) C F m ( t ) (28)

由假设(h5),及并结合Schwarz不等式和Young不等式,得到

| 0 s k 1 ( s τ ) ( u x x x m ( τ ) , u x t t m ( s ) ) d τ | | 0 L 0 s k 1 ( s τ ) u x x x m ( τ ) u x t t m ( s ) d τ d x | η u x t t m ( s ) 2 2 + 1 4 η 0 L ( 0 s k 1 ( s τ ) u x x x m ( τ ) d τ ) 2 d x η u x t t m ( s ) 2 2 + 1 4 η 0 s k 1 ( τ ) d τ 0 s k 1 ( s τ ) 0 L | u x x x m ( τ ) | 2 d x η u x t t m ( s ) 2 2 + ( 1 l ) k 1 ( 0 ) 4 η 0 s u x x x m ( τ ) 2 2 d τ (29)

其中 0 < η < 1 。故

| 0 t 0 s k 1 ( s τ ) ( u x x x x m ( τ ) , u x t t m ( s ) ) d s d τ | 2 η 0 t u x t t m ( s ) 2 2 d s 2 η 0 t F m ( s ) 2 2 d s (30)

由(28)~(31),可得

F m ( t ) + 2 0 t θ x t m 2 2 d s F m ( 0 ) + C 0 t F m ( s ) d s (31)

由Gronwall引理,得

F m ( t ) + 2 0 t θ x t m 2 2 d s F m ( 0 ) e C t

因此得到

u t t m C , u t x x m C , θ t m C , 0 t θ x t m 2 2 d s C

又因为 u x x m ( 0 , t ) = u x x m ( L , t ) = 0 ,故由罗尔中值定理,有

ξ 1 ( 0 , L ) ,使得 u x x x m ( ξ 1 , t ) = 0

由Poincare不等式,得

u x x x m 1 2 u x x x x m < C

又因为 θ m ( 0 , t ) = θ m ( L , t ) = 0 ,故由罗尔中值定理,得

ξ 2 ( 0 , 1 ) ,使得 θ x m ( ξ 1 , t ) = 0

由Poincare不等式,得

θ x m 1 2 θ x x m < C

3) 收敛性:

u m u L ( 0 , T ; S 1 ) 弱*收敛。

u t m u t L ( 0 , T ; S 2 ) 弱*收敛。

θ m θ L ( 0 , T ; S 2 ) 弱*收敛。

u m u H 1 ( 0 , T ) 中强收敛且几乎处处收敛。

u t t m u t t L ( 0 , T ; L 2 ) 弱*收敛。

θ t m θ t L ( 0 , T ; L 2 ) 弱*收敛。

0 t θ x t m d s 0 t θ x t d s L ( 0 , T ; L 2 ) 弱*收敛

θ m θ L 2 ( 0 , T ) 中强收敛且几乎处处收敛。

由定理2的证明可证 ( u , θ ) 满足初始条件 u ( 0 ) = u 0 , θ ( 0 ) = θ 0 , u t ( 0 ) = u 1 。进而可证解是唯一的。证毕。

4. 结论

本文主要讨论的轴向力作用下具有记忆项的热弹耦合梁方程组的初边值问题具有如下结论:

1) 在满足初始条件的情况下,系统(1)~(3)弱解存在且唯一。

2) 在满足初始条件的情况下,系统(1) ~ (3)正则解存在且唯一。

基金项目

国家自然科学基金(11872264)。

参考文献

NOTES

*通讯作者。

文章引用: 刘帅博 , 张建文 (2020) 轴向力作用下具有记忆项的热弹耦合梁方程组的初边值问题。 应用数学进展, 9, 492-500. doi: 10.12677/AAM.2020.94060

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