非线性振动第1章 Ritz-Galerkin法

1.9 Ritz-Galerkin法伽辽金法基本思想：伽辽金法是一种变分方法，亦称 伽辽金法基本思想 里兹(Ritz)平均法。基本思想是假设一含待定系数 的近似解，代入控制方程后产生偏差(残值),为使 偏差最小，用一权函数(变分)乘以该偏差，并使其 在一周期内积分为零。从而得到确定待定系数的代数 方程组，解此方程组求出待定系数，即得所求近似解。

自治系统

+ f ( x ) = 0 x 看成静力平衡方程 表示惯性力 表示转动力和约束反力

ω

为待求的圆频率

+ f ( x ) = 0 x x f (x) 由虚位移原理： 由虚位移原理：

[ + f ( x )] δ x = 0 x

设解

x ( t ) = ∑ ai wi ( t )i =1

N

代入原方程，由于近似解一般不会刚好等于真解， 所以会产生不等于零的残值 N & & R(ai ) = ∑ ai wi + f ∑ ai wi ≠ 0 i =1 i =1 N

近似解的变分δ x = ∑ wi ( t )δ aii =1 N

为使偏差最小，取这个残值与近似解的变分的乘积，在 一周期内积分(也即使偏差在一个周期内平均分布)为零：

∫

T

0

R ( ai ) δ xdt = 0T 0

∑∫i =1

N

N & & ∑ ai wi + i =1

N f ∑ ai wi wiδ ai dt = 0 i =1

由于δ ai 任意，则：

∫

T

0

N & & ∑ ai wi + i =1

N f ∑ ai wi wi dt = 0 i =1

i=0,1,2...

解此代数方程组，求出N个待定系数 ai ,代回原方程即得近似解

例1 Duffing方程的周期 方程的周期

+ bx + cx x

3

f ( x ) = bx + cx

= 03

设 x = a1 1 (t ) = a1 cos ωt = a1 cos T

∫ [∑0

n

i =1

a i i ( t ) + f ( ∑ a i i ( t ) )] j ( t ) dt = 0i =1 T

n

j = 1, 2 ,

∫ [a0

1

1 ( t ) + f [ a 1 1 ( t ))] 1 ( t ) dt = 0

T

[ a 1ω 2 co s ω t + b a 1 co s ω t + c ( a 1 co s ω t ) 3 ] co s ω td t = 0 ∫0

T

[ a1ω 2 cos pt + ba1 cos ω t + c ( a1 cos ω t ) 3 ] cos ω tdt = 0 ∫0 T

a 1 ( b ω 2 ) co s ω t + ca 1 3 co s 3 ω t ] co s ω td t = 0 ∫0

ωt = φ2π

∫0

a 1 ( b ω 2 ) co s 2 φ + ca 1 3 co s 4 φ td φ = 0

1π 3 1 π 2 4 a1 ( b ω ) + 4 ca 1 3 = 0 2 2 4 2 2

ω

2

3 = b + ca 1 2 4

例1 Duffing方程的周期 方程的周期

+ bx + cx x

3

= 03

f ( x ) = bx + cxx = a1 1 (t ) + a3 3 (t ) = a1 cos ωt + a3 cos3ωtT

= a1 cos + a3 cos3 1

∫ [a0

1 ( t ) + f [ a 1 1 ( t ))] 1 ( t ) dt = 0n

T

∫ [∑0

n

i =1

a i i ( t ) + f ( ∑ a i i ( t ) )] j ( t ) dt = 0i =1

j = 1, 2

T

∫ [∑01

n

i =1

a i i ( t ) + f ( ∑ a i i ( t ) )] j ( t ) dt = 0i =1

n

j = 1, 2

T

∫ [ a0

cos 9 + a 3 cos 3 + f [ a 1 cos + a 3 cos 3 ] 1 ( t ) dt = 0 cos 9 + a 3 cos 3 + f [ a 1 cos + a 3 cos 3 ] 3 ( t ) dt = 0

T

∫ [ a0

1

T

[ a1ω 2 c

os pt 9 ω 2 a 3 cos 3ω t + f [ a1 cos ω t + a 3 cos 3ω t ] cos ω tdt = 0. ∫0

T

[ a1ω 2 co s ω t 9 ω 2 a 3 cos 3ω t + f [ a1 cos ω t + a 3 co s 3ω t ] co s 3ω td t = 0 ∫0

2π

∫0

F1 ( ) d =F2 ( ) d =

1 (4 ba1 4 ω 2 a1 + 3 ca13 + 3 ca1 2 a 3 + 6 ca1 a 3 2 ) 2π = 0 81 (4 ba 3 36 ω 2 a 3 + ca13 + 6 ca1 2 a 3 + 3 ca 3 3 ) 2 π = 0 8

2π

∫0

4 ba1 4 ω 2 a1 + 3 ca13 + 3 ca1 2 a 3 + 6 ca1 a 3 2 = 04 ba 3 36 ω 2 a 3 + ca13 + 6 ca1 2 a 3 + 3 ca 3 3 = 0

3 3 3 ( b ω 2 ) a1 + ca13 + ca1 2 a 3 + ca1 a 3 2 = 0 4 4 2 ( b 9 ω 2 ) a + 1 ca 3 + 3 ca 2 a + 3 ca 3 = 0 3 1 1 3 3 4 2 4

初始条件

x ( 0 ) = 1, x ( 0 ) = 0

3 3 3 ( b ω 2 ) a1 + ca1 3 + ca1 2 a 3 + ca1 a 3 2 = 0 4 4 2 ( b 9 ω 2 ) a + 1 ca 3 + 3 ca 2 a + 3 ca 3 = 0 3 1 1 3 3 4 2 4

a 1 = 0 .9 9 7 0 9 3 , a 2 = 0 .0 0 2 9 0 7 , ω = 1 .0 3 6 7 2

x = a 1 cos pt + a 3 cos 3 pt = 0 . 997093 cos 1 . 03672 t + 0 . 002907 cos 3 . 11036 t

例2 用伽辽金法求Duffing方程的周期解

+ x + εx 3 = 0 x设解 x = acosωt = acos (仅取一项权函数cos ) 代入Duffing方程，并令与权函数乘积在一周期内积分=0x ∫ ( + x + ε x ) cos dt = ∫ ( aω + a + ε a cos )cos cos dtT 3 0 T 2 3 2 0

3 1 = a ∫ [( ω 2 + 1 + ε a 2 )cos + ε a 2 cos3 ]cos dt = 0 0 4 4T

x ∫ ( + x + ε x ) cos dt = ∫ ( aω + a + ε a cos )cos cos dtT 3 0 T 2 3 2 0

3 1 = a ∫ [( ω 2 + 1 + ε a 2 )cos + ε a 2 cos3 ]cos dt = 0 0 4 4T

3 2 1 + cos 2 1 2 T ∫ ( ω + 1 + ε a ) dt + ε a ∫ cos 3 cos dt = 0 0 4 2 4 0T 2

3 ω 2 + 1 + ε a 2 = 0 4

ω 2 = 1 + ε a2

3 4

设解：x = a1 cos ωt + a3 cos 3ωt = a1 cos + a3 cos 3 & & x = - a1ω 2 cos - 9a3ω 2 cos 3 x 3 = (a1 cos + a3 cos )3 = a13 cos3 + a33 cos3 3 + 3a12 a3 cos 2 cos 3 + 3a1a32 cos cos 2 3 3 1 3 1 = a13 ( cos + cos 3 ) + a33 ( cos 3 + cos 9 ) + 4 4 4 4 3 2 3 a1 a3 (1 + cos 2 ) cos 3 + a1a32 cos (1 + cos 6 ) 2 2 3 3 3 1 3 3 = ( a13 + a1a32 + a12 a3 ) cos + ( a13 + a33 + a12 a3 ) cos 3 + D ( ) 4 2 4 4 4 2

x ∫ ( + x + ε x ) cos dt = 0 x ∫ ( + x + ε x ) cos 3 dt = 0T 3 0 T 3 0

3 3 3 3 2 2 ò0 {[- a1ω + a1 + ε ( 4 a1 + 2 a1a3 + 4 a1 a3 )]cos φ}cos φ dt = 0 T 1 3 3 2 3 3 2 ò0 {[- 9ω a3 + a3 + ε ( 4 a1 + 2 a1 a3 + 4 a3 )]cos φ}cos 3φ dt = 0T 2

3 3 2 3 (1 ω 2 )a1 + ε a13 + a12 a3 + a1a3 = 0 4 2 4 (1 9ω 2 )a + ε 1 a 3 + 3 a 2 a + 3 a 3 = 0 3 1 3 3 1 2 4 4

3 3 2 3 (1 ω 2 ) + ε a12 + a1a3 + a3 = 0 4 2 4 (1 9ω 2 )a + ε 1 a 3 + 3 a 2 a + 3 a 3 = 0 3 1 3 3 1 2 4 4

给定适当参数后，采用C++语言编程求解。

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