Compute stable / unstable manifolds when eigenvectors have both non-zero x and y components

bifurcationdynamical systemsordinary differential equationsstability-in-odes

In class I was taught to compute stable/unstable manifolds as done in the following example:

We have the dynamical system:
$$
\dot{x} = x-xy
$$
$$
\dot{y} = -y+x^2
$$

We are interested in the manifolds for the fixed point $(0,0)$. We find the Jacobian to be:

$$
\begin{bmatrix}1-y&-x\\2x&-1\end{bmatrix}$$

Evaluating the Jacobian at the origin gives

$$
\begin{bmatrix}1&0\\0&-1\end{bmatrix}$$

Thus, we get the eigenvalue $\lambda_+ = 1$ with associated eigenvector $(1,0)$ and the eigenvalue $\lambda_+ = -1$ with associated eigenvector $(0,1)$.

Thus, we have $E^s = \text{span}${$(0,1)$} = { $x=0$} $= y-$axis and $E^u = \text{span}${$(1,0)$} = {$y = 0$} = x-axis.

Due to symmetry of the system, the stable manifold just be the y-axis. For the unstable manifold, we let $y = g(x)$ since $y \in E^s$ and $x \in E^u$. We then observe that $\dot{y} = \dot{x}g'(x)$. We know $\dot{x}$ and $\dot{y}$ from our original system and so we get

$$
-y+x^2 = (x-xy)g'(x)
$$

But on the manifold $y = g(x)$ and so we get

$$
-g(x) + x^2 = (x-xy)g'(x)
$$

We then try a taylor series expansion for $g(x)$ as $g(x) = a_0 + a_1 x + a_2 x^2 + \dots$ and substitute this into our equation. We can then match coefficients to whatever degree we want.

However I don't know how to do this if the eigenvectors aren't so nice. If we don't have eigenvectors of the form $(1,0)$ and $(0,1)$, then the stable/unstable manifolds will not just be on the x or y axis, and will instead have both x and y components. So then I can't let $y=g(x)$ as was done in the example above.

How do we deal with this?

A question I have come across where this occurs is:

$$
\dot{x} = -2x -3y -x^2
$$
$$
\dot{y}= x + 2y + xy -3y^2
$$

In this case I get $E^u = $ span{$(1,-1)$} and $E^u = $ span{$(3,-1)$}. I'm not sure how to continue from here.

Thanks

Best Answer

You can do exactly the same, set for the first direction $y(x)=x(-1+v(x))$ and insert into the quotient $$ \frac{dy}{dx}=\frac{x+2x(-1+v)+x^2(-1+v)-3x^2(-1+v)^2}{-2x-3x(-1+v)-x^2} \\ =\frac{-1+2v-x+xv-3x(1-2v+v^2)}{1-3v-x} \\ =-1+\frac{-v-2x+xv-3x(1-v)^2}{1-3v-x} \\ =-1+v+xv' $$ Isolating $v'$ gives $$ xv'=\frac{-2v-5x +3v^2+8xv-3xv^2}{1-3v-x} $$ Inserting a power series $v(x)=v_1x+v_2x^2+...$ gives $$ 1-(3v_1+1)x-3v_2x^2-...)(v_1+2v_2x+3v_3x^2+...) \\= -5-2(v_1+v_2xv_3x^2+...)+8x(v_1+v_2x+..) \\ +3x(v_1^2+2v_1v_2x+...)-3x^2(v_1^2+2v_1v_2x+...) $$ Comparing coefficients gets $v_1=-\frac53$ $$ 2v_2-v_1(3v_1+1)=-2v_2+8v_1+3v_1^2\\ 4v_2=v_1(9+6v_1)=-\frac53(9-10)=\frac53 $$ etc.

Related Solutions

[Math] Stable and unstable manifolds

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

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

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

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!!!

Find the invariant manifolds of the equilibrium

You can solve for the trajectories directly. Observe, \begin{align} \frac{dx}{dy} &= \frac{\dot{x}}{\dot{y}}\\ &= \frac{x + y\cos(y)}{-y} \end{align} By Mathematica, this has the general solution \begin{align} x &= \frac{C - \cos(y)}{y} - \sin(y) \,,\quad C \in \mathbb{R} \end{align} We need $\lim_{y \to 0} x(y) = 0$, so we infer that $C = 1$. Thus, we have the equation for the unstable manifold: \begin{align} \boxed{x = \frac{1 - \cos(y)}{y} - \sin(y) } \end{align}

To verify that this is the case, we can view a plot of the phase portrait:

If you want to play around with this, here's some Mathematica code:

sol = DSolve[{x'[y] == (x[y] + y*Cos[y])/-y, x[0] == 0}, x[y], 
    y][[1]] // FullSimplify
xmin = -4; xmax = 4; ymin = -3; ymax = 3;
cplot = ContourPlot[
   x == (x[y] /. sol), {x, xmin, xmax}, {y, ymin, ymax}, 
   ContourStyle -> {Black}];
splot = StreamPlot[{x + y*Cos[y], -y}, {x, xmin, xmax}, {y, ymin, 
    ymax}, StreamColorFunction -> "Rainbow"];
Manipulate[
 Show[splot, cplot, 
  ParametricPlot[
   Evaluate[
    First[{x[t], y[t]} /. 
      NDSolve[{x'[t] == x[t] + y[t]*Cos[y[t]], y'[t] == -y[t], 
        Thread[{x[0], y[0]} == point]}, {x, y}, {t, 0, T}]]], {t, 0, 
    T}, PlotStyle -> Red]], {{T, 20}, 1, 100}, {{point, {3, 0}}, 
  Locator}, SaveDefinitions -> True]

Best Answer

Related Solutions

[Math] Stable and unstable manifolds

Find the invariant manifolds of the equilibrium

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