Question: Consider the following autonomous vector field on the plane: $$\dot x = −x $$ $$\dot y = −x^2$$ $$(x,y) \in \mathbb{R}^2$$

$\bullet$ Compute the flow generated by this vector field.

$\bullet$ Can this vector field have periodic orbits? You must justify your answer.

$\bullet$ Compute the centre manifold of the origin.

$\bullet$ Compute the stable manifold of the origin.

$\bullet$ Show that the origin is Lyapunov stable.

I will be showing my working throughout this question then pointing out my questions and queries with my answers that I am unsure on.


$\bullet$ $\dot x = -x \Rightarrow \bar x = 0,$

$\dot y = -x^2 \Rightarrow \bar y = 0 \Rightarrow (0,0) $ being the fixed point of the system

Calculating the flow:

$\dot x = -x \Rightarrow \frac{\dot x}{x} = -1$

Integrating: $\int \frac{\dot x}{x} dt$ = $\int \frac{dx}{x}$ = $-\int dt$

$\Rightarrow log(x) = -t + c_x$, $c_x$ is a constant

$\Rightarrow x = e^{-t + c_x} = Ce^{-t}$ where $C = e^{c_x}$

Taking an initial condition $x(0) = x_0$

$\Rightarrow x_0 = C \rightarrow x = x_0 e^{-t}$

We have that $\dot y = -x^2 = -(x_0 e^{-t})^2 = -x_0 ^2 e^{-2t}$

$\Rightarrow \int \dot y dt = \int dy = -x_0 ^2\int e^{-2t}dt$

$\Rightarrow y = \frac{x_0 ^2}{2}e^{-2t} + c_y$, $c_y$ is a constant.

Taking $y(0) = y_0$ as an initial condition:

$y_0 = \frac{x_0 ^2}{2} + c_y$

$\Rightarrow c_y = y_0 - \frac{x_0 ^2}{2}$

Thus the flow is:

$y = \frac{x_0 ^2}{2}e^{-2t} + y_0 - \frac{x_0 ^2}{2} = \frac{x_0 ^2}{2}(e^{-2t}-1) + y_0$

$\bullet$ As $t \rightarrow \infty $, $x \rightarrow 0$ regardless of the choices of $x_0$, $x_0$ is free. For $y \rightarrow 0$ as $t \rightarrow \infty$, we must have $ y_0 = \frac{x_0 ^2}{2} $

$\Rightarrow y = \frac{x^2}{2}$ is the stable manifold

$\bullet$ To analyse the fixed point, we take the Jacobian of the system.

$J = \begin{pmatrix} -1 & 0 \\ -2x & 0 \end{pmatrix}$

$\Rightarrow J(0,0) = \begin{pmatrix} -1 & 0 \\ 0 & 0 \end{pmatrix}$ which is nonhyperbolic, thus linearization is not sufficient, however the eigenvalues of this matrix are $0$ and $-1$ which correspond to a centre subspace $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ and a stable subspace $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$

I think that the curve $y = \frac{x^2}{2}$ is tangent to the stable subspace spanned by $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ at the origin, and it is invariant thus it is stable manifold? As the choice of $x_0$ is free, the whole y axis (x=0) is the centre manifold. Someone please confirm or verify this.

The vector field cannot have periodic orbits due to both component of the field having exponential decay solutions which are strictly decreasing.

$\bullet$ For lyapunov stability:

We have $|y(t_0)| = |\frac{x_0 ^2}{2}| \lt x_0 ^2$ and $|y(t)| = |\frac{x_0 ^2 (e^{-2t} - 1)}{2} + y_0| \leq |\frac{x_0 ^ 2}{2} + y_0| \lt |\frac{x_0 ^2}{2}+\frac{x_0 ^2}{2}| = x_0^2$ choosing $\delta = \epsilon = x_0^2$

I am not sure about this last bit regarding the centre manifold of the origin, and thus showing it is Lyapunov stable? Is calculating the eigenvalues of the linearized matrix not enough to show this is if the eignevalues aren't positive? The order of which these questions are asked makes me think that I should be doing it another way but I am unsure of how to go about this. Center manifold theory comes to mind but if I were to do this, I am unsure about the degree of the polynomial that I would use to approximate the center manifold.

Any help and guidance would be appreciated.

  • 1
    $\begingroup$ Look at your $y$ calculation again. $\endgroup$ – Moo Aug 22 '18 at 12:33
  • $\begingroup$ @Moo thank you, just double checked my calculation and realized I made an amateur integration mistake, I followed it through however it seems to have made no difference to the end result of the stable manifold? $\endgroup$ – A. Bharj Aug 22 '18 at 13:11
  • $\begingroup$ If I understand you correctly, in your proof of Lyapunov stability you have shown only that if $(x_0,y_0)$ is close to $(0,0)$ and lies on $y=x^2/2$ then its positive orbit is close to $(0,0)$. This is not enough: you have to consider all initial values $(x_0,y_0)$ close to $(0,0)$. $\endgroup$ – user539887 Aug 24 '18 at 9:16
  • $\begingroup$ @user539887 The definition that I am using is: $y= 0$ is said to be Lyapunov stable at $t_0$ given $\epsilon > 0$ there exists $\delta = \delta(t_0,\epsilon) $s.t $$|y(t_0)| \lt \delta \Rightarrow |y(t)|\lt\epsilon, \forall t \gt t_0$$ As this appears to apply for all $t \gt t_0$ I think it holds? I should have added that to my answer, thank you for pointing it out $\endgroup$ – A. Bharj Aug 24 '18 at 14:16
  • $\begingroup$ @A.Bharj You do not need to use $t_0$: the system is autonomous, so it suffices to find, given $\epsilon>0$, $\delta=\delta(\epsilon)$ such that $\lvert y(0)\rvert<\delta$ implies $\lvert y(t)\rvert<\epsilon$ for all $t>0$. $\endgroup$ – user539887 Aug 24 '18 at 17:40

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