本次新西兰作业案例分享主要为一个assignment数学代写
Maths 761: Dynamical SystemsAssignment 3
一季度。考虑微分方程组
8
<
：
_ x = y x z
_ y = x y3
_ z = z 2 x y 2 x4
+ x2
定义为 (x; y; z) 2 R3
.
(a) 证明原点是一个非双曲均衡。
(b) 导出原点中心流形的幂级数近似。
(c) 使用你的中心流形方程来寻找动力学的近似值
中心歧管。
(d) 绘制中心流形上动力学的局部相图和相图
局部靠近原点的三维相空间。
注意：中心流形上的平衡解的稳定性分析可能是最好的
使用 Glendinning（第 38 页）的定理 2.11 和 Lyapunov 函数完成
形式为V（x1; x2）= x2
1 + x2
2.
[30 分]
Q2。再次考虑微分方程的单参数族
_ x = x + 4 x3
(1 x2
)
从第四季度开始。作业 2。
(a) 回想一下，你为这个系统确定了三个分支。对于每个分叉
点，验证是否满足适当的通用性条件。
（b）注意，对于相同的参数值= 1，发生两个分叉。
你会或不会期望这是一个巧合。
[10 分]
Q3。二参数族
_ x = x + 4 x3
(1 x2
)
与; 2 R 包含来自上述 Q2. 的族，在 = 0 时获得。
(a) 首先，x = 0:1 仅略大于 0。
证明如果 = 0:1，则 x = 1 是该系统的均衡。
使用 XPPAUT 计算分岔图并交出这张分岔图
与你的任务。选择适当的范围，以便您的图表显示所有
这个系统发生的分岔。颜色和/或标记曲线（手动）
清楚地表明所有平衡点的稳定性和分岔点的位置。
(b) 接下来，x = 1 并用 XPPAUT 以同样的方式计算分岔图
至于(a)部分。
(c) 证明（用手）对于 =
和 =
也存在于 =
和
=
.证明这种平衡的任何分岔也发生在 ( ; ) = (
;有的
)
和 ( ; ) = (
;有的
）。
(d) 使用 XPPAUT 计算具有两者和变化的双参数分岔集。
标识在两个参数中找到的= 0，= 0：1和= 1的分叉
分岔集。
两参数分岔曲线在几个点相交。解释会发生什么
在这样的交叉点。
Q1. Consider the system of di erential equations
8
<
:
_ x = y x z
_ y = x y3
_ z = z 2 x y 2 x4
+ x2
de ned for (x; y; z) 2 R3
.
(a) Show that that the origin is a non-hyperbolic equilibrium.
(b) Derive a power series approximation for the centre manifold of the origin.
(c) Use your equation for the centre manifold to nd an approximation to the dynamics on
the centre manifold.
(d) Draw a local phase portrait of the dynamics on the centre manifold and a phase portait
of the three-dimensional phase space locally near the origin.
Note: Stability analysis of the equilibrium solution on the centre manifold is perhaps best
done using Theorem 2.11 from Glendinning (page 38) with a Lyapunov function
of the form V (x1; x2) = x2
1 + x2
2.
[30 marks]
Q2. Consider again the one-parameter family of di erential equations
_ x = x + 4 x3
(1 x2
)
from Q4. of Assignment 2.
(a) Recall that you identi ed three bifurcations for this system. For each of the bifurcation
points, verify that the appropriate genericity conditions are satis ed.
(b) Note that two of the bifurcations occur for the same parameter value = 1. Explain why
you would or would not expect this to be a coincidence.
[10 marks]
Q3. The two-parameter family
_ x = x + 4 x3
(1 x2
)
with ; 2 R contains the family from the above Q2., obtained when = 0.
(a) First, x = 0:1 only slightly larger than 0.
Show that x = 1 are equilibria of this system if = 0:1.
Use XPPAUT to compute the bifurcation diagram and hand in this bifurcation diagram
with your assignment. Choose an appropriate range for so that your diagram shows all
the bifurcations that occur for this system. Colour and/or label the curves (by hand) to
indicate clearly the stability of all equilibria and the locations of the bifurcation points.
(b) Next, x = 1 and use XPPAUT to compute the bifurcation diagram in the same way
as for part (a).
(c) Show (by hand) that any equilibrium for =
and =
also exists for =
and
=
. Show that any bifurcation for such equilibria also occur for both (; ) = (
;
)
and (; ) = (
;
).
(d) Use XPPAUT to compute a two-parameter bifurcation set with both and varying.
Identify the bifurcations you found for = 0, = 0:1, and = 1 in the two-parameter
bifurcation set.
The two-parameter bifurcation curves intersect in several points. Explain what happens
at such intersection points.