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.
Best Answer
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!!!