Classify fixed points of a 2D dynamical system

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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Question statement

Consider the two-dimensional dynamical system

$\begin{align}\dot{x}&=\simplify[std]{{a}*x+{b}*y},\\\dot{y}&=\simplify[std]{{c}*x+{d}*y}.\end{align}$

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			Name	Type	Generated Value

a	number	6
tau	number	12
c	number	4
b	integer	1
d	number	6
nature	string	unstable node
vec1	number	2
lam2	number	4
lam1	number	8
vec2	number	-2
naturetext	string	because $\lambda_1>0$ and $\la
delta	number	32
r	integer	4
disc	number	16
natures	list	List of 5 items

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tau-d

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2. Gap Determinant $\delta$ Number entry
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1. Gap $\lambda_1$ Mathematical expression
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2. Gap $\lambda_2$ Mathematical expression
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2. Gap Corresponding to $\lambda_2$ Mathematical expression
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The system can be written in the form $\dot{\boldsymbol{x}}=\mathsf{A}\boldsymbol{x}$ , where $\boldsymbol{x}=\pmatrix{x,y}^\mathsf{T}$ .

Identify the components of the matrix $\mathsf{A}$ .

$\mathsf{A}=$ Unnamed gap

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The elements of the matrix $\mathsf{A}$ are the coefficients of $x$ and $y$ in the system of equations, hence

$\mathsf{A}=\pmatrix{\var{a} & \var{b}\\\var{c} & \var{d}}.$

The trace $\tau$ and determinant $\delta$ of a matrix

$\mathsf{A}=\pmatrix{a & b\\c & d}$

are given by $\tau=a+d$ and $\delta=ad-bc$ .

A straight forward substitution for $a=\var{a}$ , $b=\var{b}$ , $c=\var{c}$ , and $d=\var{d}$ reveals that $\tau=\var{tau}$ and $\delta=\var{delta}$ .

The eigenvalues $\lambda$ of a matrix $\mathsf{A}$ can be calculated from the expression

$\mathsf{A}\boldsymbol{x}=\lambda\boldsymbol{x},$

which is equivalent to

$\left(\mathsf{A}-\lambda\mathsf{I}\right)\boldsymbol{x}=\boldsymbol{0},$

where $\mathsf{I}$ is the identity matrix.

This linear system has a solution only when

$\det\left(\mathsf{A}-\lambda\mathsf{I}\right)=0,$

i.e. when $(a-\lambda)(d-\lambda)-bc=0$ . This leads to $\lambda^2-(a+d)\lambda+(ad-bc)=0$ , which is equivalent to

$\lambda^2-\tau\lambda+\delta=0.$

The eigenvalues $\lambda_1$ and $\lambda_2$ of the matrix $\mathsf{A}$ are therefore given by

$\lambda_{1,2}=\frac{\tau\pm\sqrt{\tau^2-4\delta}}{2}.$

Following through this calculation for the specific values of $a$ , $b$ , $c$ , and $d$ in this question leads to $\lambda_1=\var{lam1}$ and $\lambda_2=\var{lam2}$ .

The eigenvectors corresponding to the eigenvalues $\lambda_1$ and $\lambda_2$ can be found by solving the linear system of equations given by

$\mathsf{A}\boldsymbol{x}=\lambda\boldsymbol{x},$

$\begin{align}ax+by&=\lambda x,\\cx+dy&=\lambda y,\end{align}$

where $\lambda$ is either $\lambda_1$ or $\lambda_2$ .

Because this is an eigenvalue problem, we can arbitrarily choose one of the eigenvector components to be $1$ , the $x$ -component in this case. The $y$ -component can then be found from

$y=\frac{x(\lambda-a)}{b}\quad\text{or}\quad y=\frac{cx}{\lambda-d},$

with $x=1$ . Both expressions are equivalent, and will lead to the same value for $\lambda$ .

Making the necessary substitutions reveals that the $y$ -component of the eigenvector corresponding to $\lambda_1$ is $\var{vec1}$ , and the $y$ -component of the eigenvector corresponding to $\lambda_2$ is $\var{vec2}$ .

The nature of the fixed point at the origin can be determined by examining the eigenvalues. If the eigenvalues are real, then the sign of each eigenvalue determines the nature of the fixed point. If the eigenvalues are complex, then the sign of the real part of each eigenvalue determines the nature of the fixed point.

In this case, {naturetext}.

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This question is used in the following exam:

Chaos theory by Newcastle University Mathematics and Statistics in Content created by Newcastle University.

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Classify fixed points of a 2D dynamical system

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Classify fixed points of a 2D dynamical system

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Newcastle University Mathematics and Statistics 9 years, 3 months ago

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