程序代写代做代考 MAT332 – cscodehelp代写
MAT332
Introduction to Nonlinear Dynamical Systems and Chaos
Lecture 3
Qun WANG
University of Toronto Mississauga
September 16, 2021
(UTM)
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Review
Find out a fundamental set of solution for the following system 1 0
x ̇=Ax,A= 0 0 (1)
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Learning Objective
For today’s lecture, students are expected to:
Objective
Make the phase portrait for 1D linear system;
Make the phase portrait for 2D linear system;
Understand the notion of stable space and unstable space; Introduce the notation of the matrix exponential (If time permits)
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Linear System in 1D: Simple yet Essential
Solve the linear equation
There are three cases: k>0
k<0 k=0
x ̇ = kx,k ∈ R
(2)
If x(0) ̸= 0, will the solution touch 0 in finite time?
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Linear System in 2D: Classification
We would like to see if systems in 2D will exhibit similar behaviours. Natural idea: Decompose the space into two 1-dim subspaces...
Natural question: Which decomposition is most indicative? – After all it is an evolution!
Mission
Given a 2D autonomous linear dynamical system, we want to switch the system into a simple one s.t.
If possible, we can decompose it into two invariant directions.
In either case, we want to know the bavaiour of the solution in those
directions.
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Linear System in 2D: Classification
In linear algebra you have already seen the idea: decompose into invariant subspaces ⇐ eigenvalue + eigenspace
Quiz
Let A be a matrix and M ∈ Rn×n be an invertible matrix, and B = M−1AM
Relation between eigenvalues of A and those of B?
Relation between eigenvectors of A and those of B?
Now apply this idea to linear systems: How will a solution change under change of coordinates?
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Linear System in 2D: Classification Theorem
Normal Form of Real 2 × 2 matrix For A ∈ R2×2, there exists an invertible matrix P s.t. B = P−1AP becomes to one of the following:
1
2
3
λ1 0 B=0λ.
2
a b B= −b a .
λ 1 B=0λ.
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CASE I: The Diagonal Matrix
eigenvalues: λ1, λ2 ∈ R Fundamental Set of Solution. e
Subcases:
1 λ1 ̸= λ2
2
λ1t 1 0
. e
λ2t 1 0 .
λ1 > λ2 > 0 ( )
0>λ1 >λ2 (NodalSink)
λ1 >0>λ2 (Saddle)
λ1 > 0 = λ2 (Degenerate Source)
λ1 = 0 > λ2 (Degenerate Sink)
λ1 0 B=0λ.
2 λ1 = λ2 = λ
λ > 0 (Star Source)
λ < 0 (Star Sink)
λ = 0 (Guess What? You will not have this case in your final exam)
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CASE II: The Complex Eigenvalue Matrix
a b B= −b a .
eigenvalues: λ1 =a+ib,λ2 =a−ib
fundamental set of solutions: e
Subcases:
at cosbt − sin bt
, e
at sinbt cos bt
a > 0 (Spiral Source) a < 0 (Spiral Sink)
a = 0 (Center)
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CASE III: Sub-Diagonal Matrix
λ 1 B=0λ.
eigenvalues: λ1 = λ2 = λ 1 0
eigenvectors: 0 and 1 (augmented)
λ1 λ 1 0
Fundamental Set of Solutions: e 0 , and e (t 0 + 1 )
Subcases:
λ > 0 (degenerate source) λ < 0 (degenerate sink)
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Poincar ́e Det-Tr Diagram
Let’s try to classify the different situations by looking at determinant and trace.
Quiz
Write down the characteristic function in terms of determinant and trace
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Poincar ́e Det-Tr Diagram
Figure: Poincar ́e Det-Tr Diagram
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