非线性分析作业第2次(硕士博士非线性分析)
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
1. For the following dynamical systems 1）''
30x x x ++=
2）'
'2(1),(1)3x
x x xy y y y xy =--=--
a) Find all fixed points and classify them. b) Sketch the phase space portrait. Solution for 1）：''
30x
x x ++=
Set 121
,y x y y '==. Then, the equation becomes to ， 1
23
2
11y y y y y '=⎧⎨'=--⎩ Set vector variable z, we can write
()z f z =, where 12y z y ⎡⎤
=⎢⎥⎣⎦ 213211()y y f z f y y y ⎡⎤⎡⎤==⎢⎥⎢⎥--⎣⎦⎣⎦
There is only fixed point 00z ⎡⎤
=⎢⎥⎣⎦
The Jacobian matrix 2
10
1310Df y ⎡⎤=⎢⎥--⎣⎦ Jacobian matrix for linearized system at the fixed point,
()
0110Df y ⎡⎤
=⎢⎥
-⎣⎦
Eigenvalues for this system are 12i λ=±, so they have zero real part and the method of linearization cannot decide about the stability.
Solution for 2）：'
'
2(1),(1)3x
x x xy y y y xy =--=--
Jacobian matrix ：243123x y x A y
y x ---⎡⎤
=⎢⎥---⎣⎦ Jacobian matrix for linearized system at the fixed point 00⎡⎤⎢⎥⎣⎦
is
2001⎡⎤⎢⎥⎣⎦
Eigenvalues for this system are 122,1λ= , repelling node, which is unstable.
2. Given the system
'''30x x x x +++=
Show that the equilibrium (0, 0) is globally asymptotically stable. Solution ：
Set 121
,y x y y '==. Then, the equation becomes to ， 1
23
2
211y y y y y y '=⎧⎨'=---⎩ Set vector variable z, we can write
()z f z =, where 12y z y ⎡⎤
=⎢⎥⎣⎦ 213
2211()y y f z f y y y y ⎡⎤⎡⎤==⎢⎥⎢⎥---⎣⎦⎣⎦ There is only fixed point 00z ⎡⎤
=⎢⎥⎣⎦
The Jacobian matrix 2
10
1311Df y ⎡⎤=⎢⎥---⎣⎦ Jacobian matrix for linearized system at the fixed point,
()
0111Df y ⎡⎤
=⎢⎥
--⎣⎦
Eigenvalues for this system are 120.50.866i λ=-± , so they have negative real parts. Thus, it is stable.
3. For a real number c, define the one-parameter family
()()(23),a f x x a x a x c =--++ for what values of c is there a
bifurcation in this family? Describe the bifurcations and list the bifurcation points (a, x), and Sketch the bifurcation diagram. Solution:
Suppose 0a =. Set ()212f x x =,()2f x x c =--.
()()()120a f x f x f x =⇒=
When 1
8c =, there is a bifurcation in ()a f x .
Bifurcation points: 10,4⎛
⎫- ⎪⎝
⎭
4. Show that the one parameter system
''2'2'()0x x x x x μ++-+=
undergoes a Hopf bifurcation at μ = 0. Plot the phase portraits and sketch the bifurcation diagram. Solution:
Set 12,x x x x '==, the corresponding state-space equations is
23
112221
22
x x x x x x x x μ⎧⎪⎨--+⎪⎩-'='= Solve the equations
23
1122220
x x x x x x μ⎧⎪⎨⎪⎩=--+=- Fixed points are obtained as (0,0). Jacobian matrix and Eigenvalues are
1,2,011λμ⎡⎤⎢
⎥⎢⎥⎣⎦
=
=-A
When 0μ= , there is an node center. The phase space portrait is shown next