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Stable and unstable manifolds 2

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Consider the system $$x' = -x,$$ $$y' = y + g(x),$$ where $g$ is a class $C^1$ function with $g(0) = 0$.

Compute the stable manifold $W^s (\mathbf{0}).$
Using $g(x) = x^n (n \geq 1)$, compute $W^s (\mathbf{0})$ and $W^u (\mathbf{0})$.

The other was an exercise I found, this is an actual exercise from the test. How would I compute the stable/unstable manifold using the Taylor approximation method or successive approximations? :confused:
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Writing $g(x)$ as a power series,
$$y'= y+ ax+ bx^2+ cx^3.$$

From the first equation, $x'= -x$, we can see that $x$ decreases rapidly as the independent variable (which I will call $t$) increases (in fact, $x= Ce^{-t}$).

So for large $t$, this can be approximated by
$y'= y+ aCe^{-t}$. The general solution to the associated homogeneous equation is $y= De^{t}$. Looking for a specific solution of the form $y= Ae^{-t}$, $y'= -Ae^{-t}$ so the equation becomes $-Ae^{-t}= Ae^{-t}+ aCe^{-t}$. $-2Ae^{-t}= aCe^{-t}$, so $A= -aC/2$. The general solutions to the equations are $x= Ce^{-t}$ and $y= De^{t}-(aC/2)e^{-t}$. If $D$ is not $0$, $y$ will increase without bound so the stable manifold will be $x= Ce^{-t}$, $y= -(aC/2)e^{-t}$, which can be written $y/x= -(aC/2)e^{-t}/Ce^{-t}= -a/2$ so that $y= -(a/2)x$ or, equivalently, $ax+ 2y= 0$.

That is not the general method of finding the stable manifold but it works for this simple situation.
($a$ is the coefficient of $x$ in the linearization of $x$.)
 
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Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Thanks HoI! What if I wanted a third order approximation? Something that perhaps yields a third degree polynomial. Would it proceed as $$y' = y + aCe^{-t} +bCe^{-2t} + cC e^{-3t},$$ followed by finding $y$ and expliciting it as a function of $x$?