TransWikia.com

Stable and unstable manifolds of a linear system

Mathematics Asked on February 25, 2021

Consider the linear system of the form $dot{x}=Ax$, with $x in R^2$.

(a) Solve the system and find stable and unstable manifolds in the case of

$A=begin{pmatrix}
1 & 2 \
1 & 0
end{pmatrix}$

(b) Sketch the phase portrait of the above system.

To begin with, I found eigenvalues to be 2 and -1. Then I calculated the unstable subspace to be $E^u=text{span} left{begin{pmatrix} 2\ 1end{pmatrix}right}$ and

stable subspace to be $E^s=text{span} left{begin{pmatrix} 1\ -1end{pmatrix}right}$.

Now the system of equations can be written as $dot{x}=x+2y$ and $dot{y}=x$.

To find stable and unstable manifolds, I need to solve the differential equation which is where I am having difficulty.

Taking the first equation $(dot{x}-x=2y)$, we can write it as $(-xe^{-t})’=xe^{-t}-dot{x}e^{-t}=(x-dot{x})e^{-t}=2ye^{-t}$.

Now I don’t know how to proceed after that. Can the second equation be used to solve the first one? How do we find stable and unstable manifolds using the above calculations and find their phase portraits?

3 Answers

$$x'=x+2y$$ $$y'=x$$ You can also sum both Differential equations: $$(x+y)'=2(x+y)$$ And integrate: $$(ln (x+y))'=2$$ $$ln (x+y)=2t+C$$ $$x+y=C_1e^{2t}$$ Then solve the DE: $$y'=x=C_1e^{2t}-y$$ $$y'+y=C_1e^{2t}$$

Answered by Aryadeva on February 25, 2021

You can also go the slow way, from $dot x-x=2y$ it follows that $$ ddot x-dot x=2dot y=2x $$ which you can now solve as a scalar equation to get $$ x=c_1e^{2t}+c_2e^{-t} $$ and consequently $$ y=frac12(dot x-x)=frac12(c_1(2-1)e^{2t}+c_2(-1-1)e^{-t}) $$ so that $$ pmatrix{x(t)\y(t)}=c_1pmatrix{1\frac12}e^{2t}+c_2pmatrix{1\-1}e^{-t} $$

Answered by Lutz Lehmann on February 25, 2021

In the case of distinct real eigenvalues, the general solution to the linear system is given by

$$ x(t) = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2,$$

where $v_1, v_2$ are eigenvalues with respect to eigenvectors $lambda_1, lambda_2$ respectively, and $c_1, c_2$ are any real constants.

In our situation $lambda_1 =-1 <0$ and $lambda_2 = 2 >0$. Let $sin mathbb R^2$. Then there are unique $c_1, c_2$ so that $c_1 v_1 + c_2 v_2 = s$. Then $phi_t(s) = x(t)$ (with that unique $c_1, c_2$). Using $e^{lambda_1 t}to 0$ and $e^{lambda_2 t} to infty$ as $tto infty$,

$$ x(t) = c_1 e^{lambda_1 t} v_1 + c_2 e^{lambda_2 t} v_2 to vec 0$$

if and only if $c_2 = 0$. Thus $s$ lies in the stable manifold if and only if $s = c_1v_1$ for some $c_1$. That is, precisely when $s$ is an eigenvector of $A$ with eigenvalues $lambda_1 = -1$. Similarly one shows that the unstable manifolds are given by the second eigenspaces.

Answered by Arctic Char on February 25, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP