2
$\begingroup$

Today, the tutor of our dynamical system course said that in the exam one part will be to determine equilibria and to compute the local equations of stable and unstable manifold.

I do not know what is meant by giving the local equations for stable/ unstable manifold.

Can you give me a task/ example?

$\endgroup$
1
$\begingroup$

You need to be more specific, at least whether you have discrete or continuous time. But a (local) stable manifold is simply a graph over the stable space, so you can (with reasonable regularity assumptions) write it say with the first few terms of a Taylor series.

For example, take $x'=-x+xy$, $y'=y+x^2$. The stable manifold will be the graph of the function $\varphi(x)=ax^2+bx^3+\cdots$ (in this example you can go up to any order), that is, the set of points $(x,y)$ with $y=\varphi(x)$. Computing you get $$ y'=\varphi'(x)x'=(2ax+3bx^2+\cdots)(-x+x(ax^2+bx^3+\cdots)). $$ On the other hand, $$ y'=y+x^2=ax^2+bx^3+\cdots+x^2. $$ The two right-hand sides are equal (write the first few terms), which gives you equations to determine $a$, $b$, etc successively.

Added:

We are taking as definition of stable manifold any smooth invariant curve that is tangent to the stable space at the origin. That is, $$V^s=\{(x,\varphi(x)):x\in(-\delta,\delta)\}$$ for some $C^k$ function $\varphi\colon(-\delta,\delta)\to\mathbb R$ with $\varphi(0)=\varphi'(0)=0$ and with the property that any solution starting in $V^s$ remains in $V^s$ for all positive time. This means that if $(x(t),y(t))$ is a solution with $(x(t_0),y(t_0))\in V^s$ for some $t_0$, then $(x(t),y(t))\in V^s$ for all $t>t_0$. This is why we must have $y(t)=\varphi(x(t))$ (written above simply as $y=\varphi(x)$.

$\endgroup$
8
  • $\begingroup$ I am not sure what you mean with this function $\phi(x)$. To my computations, the equilibrium is $(0,0)$, the linearization matrix is $A=\begin{pmatrix}-1 & 0\\0 & 1\end{pmatrix}$ and $A$ has two eigenvalues $\lambda_1=+1$ and $\lambda_2=-1$. Hence, the solutions should be $x=x_0e^{t}, y=y_0e^{-t}$. So, $(0,0)$ is a saddle and locally, the $x$-axis should be the stable manifold where the $y$-axis is the unstable manifold? $\endgroup$
    – Rhjg
    Jan 15 '16 at 21:36
  • $\begingroup$ Fine if it is, the example applies to whatever nonlinear part you want, provided that you keep the linear part unchanged. $\endgroup$
    – John B
    Jan 15 '16 at 22:50
  • $\begingroup$ Sorry! I still do not understand you and what you are aiming at. $\endgroup$
    – Rhjg
    Jan 15 '16 at 22:54
  • $\begingroup$ You write "the $x$-axis should be the stable manifold where the $y$-axis is the unstable manifold". This might be true for some examples, but if fails in general. What then your definition of "stable manifold"? $\endgroup$
    – John B
    Jan 15 '16 at 22:56
  • $\begingroup$ Good question. We never had such situations. - Ok, so back to your idea. I cannot see where your function comes from. $\endgroup$
    – Rhjg
    Jan 15 '16 at 22:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.