# Stable and unstable manifolds

I would like to know how to finish this problem, and if what I have done so far is correct.

Problem: Determine the stable and unstable manifolds for the rest point of the system $$\dot{x}=2x-(2+y)e^y, \dot{y}=-y.$$

Attempt and outline: The rest point of the system is $(1,0).$ Now, I changed the coordinates of the system to be centered at the origin, giving me the system $$\dot{x}=2(x+1)-(2+y)e^y, \dot{y}=-y.$$

I have found the solutions of the (shifted) system to be $$x=e^{{c_1}e^{-t}}+c_2e^{2t}-1, y=c_1e^{-t}$$

Then I computed $Df(x_0)$ and obtained the eigenvalues of the associated linear system as $\lambda_1 = 2, \lambda_2 = -1$.

My question is: Now that I have the stable and unstable eigenspaces (manifolds) for the linear system, how can I find them explicitly for the nonlinear system?

Thanks for any help.

Here's what I think:

Based on the (presumably correct) given form for the solutions:

$x=e^{{c_1}e^{-t}}+c_2e^{2t}-1, \tag{1}$

$y=c_1e^{-t}, \tag{2}$

we can obtain the stable and unstable manifolds of the system by arguing as follows: the stable manifold is, by definition, the set of points $p$ such that $\phi_t(p) \to (0, 0)$ as $t \to \infty$, where $\phi_t$ is the flow of the given system. Since for nonzero $c_2$, $c_2e^{2t} \to \text{sign}(c_2) \infty$ as $t \to \infty$, $c_2 = 0$ on the stable manifold. It is thus given by

$x=e^{{c_1}e^{-t}}-1, \tag{3}$

$y=c_1e^{-t}, \tag{4}$

or if you like

$x = e^y - 1, \tag{5}$

or

$y = \ln (x + 1), x > -1. \tag{6}$

Note that from (3), we have $x > -1$ for any combination of finite $c_1$ and $t$, so (6) is well-defined. Likewise, the unstable manifold consists of those $p$ such that $\phi_t(p) \to (0, 0)$ as $t \to -\infty$. This condition forces $c_1 = 0$ so that the unstable manifold is given by

$x = c_2e^{2t}, \tag{7}$

$y = 0, \tag{8}$

i. e., it is the entire $x$-axis. For a further analysis of this system, see my answer to the OP's other question relating to it, here.

Nota Bene: Not to put too fine a point on it, but it appears to me that there is an erroneous assertion in Evgeny's answer to this question. Though it is true that $y$ is decoupled from $x$, since $\dot y = -y$, the $y$-axis is not invariant under the flow. To see this, consider $\dot x$ at some point on the $y$-axis, $(0, y_0)$. Then $\dot x = 2 - (2 + y_0)e^{y_0} = 2(1 - e^{y_0}) - y_0e^{y_0} <0$ for $y_0 > 0$. This shows that at least some points on the $y$-axis are moved off of it by the flow, so it cannot be invariant. End of Note.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

• I realized that my assertion wasn't true immediately after I've posted the answer. I didn't know what's better: to delete this part or to remark it as not true in comments. But still thanks for correction :) – Evgeny Mar 4 '14 at 6:45
• @Evgeny: Probably best to edit your post with the correct answer and an explanation. But don't worry, we all make mistakes! Best Wishes, RKL. – Robert Lewis Mar 4 '14 at 6:48
• @Robert Lewis Why are this the formula for stable/ unstable manifold for the NON-linear system? – Rhjg Jan 15 '16 at 13:54

First, you may noticed that $Oy$ axis is an invariant manifold and moreover it's a stable manifold for your steady point. (I've decided to mark it wrong explicitly; for stable manifold part see my first comment) So the only question is equation of unstable manifold. Since you know exact solution, you may try to find constants $c_1$ and $c_2$ from two following conditions:

• $x(t) \rightarrow 0$, $y(t) \rightarrow 0$ when $t \rightarrow - \infty$;

• $(\dot{x}(t), \dot{y}(t))$ tends to the direction of unstable eigenspace.

• I've been mistaken in the first sentence, $Oy$ is not an invariant manifold. But the general idea remains the same: point of stable manifold must tend to steady point when $t \rightarrow +\infty$ and direction of $(\dot{x}(t), \dot{y}(t))$ must tend to the direction of stable eigenspace. – Evgeny Mar 4 '14 at 5:46
• $O$ is ususally stands for "origin", i.e. point $(0, 0)$. By $Oy$ I meant a set $\lbrace (x, y) \colon x = 0 \rbrace$ in $(x, y)$ coordinate system. – Evgeny Mar 4 '14 at 5:52
• Yep, this sounds correct. Then you have to find stable manifold: the same procedure, but with $t \rightarrow + \infty$. Also I realized that one condition (the first one) is enough to find stable/unstable manifolds (because steady point is a saddle point on plane). – Evgeny Mar 4 '14 at 6:08
• The $y$-axis is neither invariant nor the stable manifold of the critical point. See my answer for details. Regards. – Robert Lewis Mar 4 '14 at 6:44