Dynamical system 4

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Created this as a copy of Dynamical system 3.

Name	Status	Author	Last Modified
Dynamical system 4	Ready to use	Bill Foster	16/11/2020 10:02
Dynamical system 6:Centre.	draft	Bill Foster	16/11/2020 10:02
Dynamical system 7: Stable spiral.	Ready to use	Bill Foster	16/11/2020 10:02
Dynamical system 9: Saddle.	Ready to use	Bill Foster	16/11/2020 10:02
Bill's copy of Dynamical system 6:Centre.	Ready to use	Bill Foster	16/11/2020 10:02

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Question statement

Consider the two-dimensional dynamical system. Complex eigenvalues case , so centres or spirals.

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

{ode1()}

The above diagram shows the plot of $(x(t),y(t))$ .

At $t=0$ we have initally $x=-5,\;\;y=5$ . Moving the point gives phase diagrams for the following initial values at $t=0$ :

$x=\;$ $y=\;$

Dotted lines give points at which $\dot{x}=0,\;\dot{y}=0$ .

The plots below are for $x(t),\;x(0)=-1$ and $y(t),\;y(0)=1$ , in black and red respectively where the horizontal axis is $t$ .

{ode(xr,yr)}

Give any introductory information the student needs.

			Name	Type	Generated Value

lam1	number	2.9 + 2.5i
r	integer	1
be	integer	5
c	number	-12.52
m	number	-6.26
al	number	-0.2
delta	number	14.66
natures	list	List of 5 items
vec2	number	-0.2 - 5i
disc	number	-25
vec1	number	-0.2 + 5i
d	number	2.8
lam2	number	2.9 - 2.5i
a	number	3
b	number	0.5
tau	number	5.8
revec1	number	-0.2
yr	integer	100
rorc	integer	1
s	integer	1
naturetext	string	because $\mathrm{Im}(\lambda_1
xr	integer	10
revec2	number	-0.2
nature	string	unstable spiral

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xxxxxxxxxx
 
b*vec1+a

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Generated value: `number`

2.9 + 2.5i

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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 $\mathsf{A}$.

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Advice

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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Dynamical system 4

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Taxonomy: mathcentre

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Checkpoint description

Bill Foster 8 years, 1 month ago

Bill Foster 8 years, 1 month ago

Bill Foster 11 years, 4 months ago

Error in variable testing condition

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