Finding the stable and unstable manifold of this system

Question

Consider the system $$\begin{cases}\dot{x} = x \\ \dot{y} = -y + x^2\end{cases}$$

This has fixed point $\overline{X} = (0,0)$, which is a saddle point. The aim is to find the equation of the stable and unstable manifold for the system. My lecture notes finds the unstable manifold in the following way:

Use the trick: $\dot{x}\frac{dy}{dx} = \dot{y}$. For the system this gives $x\frac{dy}{dx}=-y+x^2$. We insert our power series $y(x) = \sum^\infty_{k=2}a_kx^k$ into the equation to give $\sum^\infty_{k=2}a_kkx^k = \sum^\infty_{k=2}(-a_k)x^k + x^2$. Comparing coefficients gives $a_2 = \frac{1}{3}$ and $a_k = 0$, for all $k > 2$. Hence, $y(x) = \frac{x^2}{3}$.

This is correct and I'm fine working through this. But I don't know how to use this to find the stable manifold (which turns out to be $x = 0$). I've tried interchanging $x$ and $y$ to try and end up with an equation in $x$ but I can't actually get anywhere with it.

How would I go about finding the stable manifold?

Answer 1

I might be missing something, but it looks like you can just solve this. $x=Ae^{t}$ so you just solve the other equation, $\dot{y}+y = A^2e^{2t}$. Get your integrating factor as $e^t$ and solve

$$d(y\cdot e^t)= A^2e^{3t}$$

integrate to get

$$y\cdot e^{t}={A^2\over 3}e^{3t}+C$$

so $y= {A^2\over 3}e^{2t}+Ce^{-t}$ i.e.

$$y= {1\over 3}x^2+{C'\over x}, x>0.$$

As for the stable/unstable manifolds, note you can see that if you are not at the origin, you cannot possibly approach it with this flow, the solutions being $y={1\over 3} x^2+O(1)$, so it pushes you away unless you are already at the origin, in which place you need $A=C=0$. Similarly for going to $-\infty$ along the flow lines. If $x=0$ this will also give you stability since then $y′=−y$ and so $y=Ae^{−t}+C$ and you can easily see that the flow leads to $C$ as $t\to-\infty$ for all points on the $x$-axis and diverges as $t\to-\infty$ unless this new $A=0$.

Answer 2

It is (clearly) $x=0$ (this is due to the fact that you can show that there is a solution of the form $(0,y(t))$, where $y'=-y$, and since this "curve" is tangent to the stable space it will be the stable amnifold). Since we know that it is unique, nothing else is needed.

Anyways, you can proceed as for the unstable manifold, nothing changes other than interchanging the variables: you get $$ \sum^\infty_{k=2}a_kky^k\left(-y+\left(\sum^\infty_{k=2}a_ky^k\right)^2\right) = \sum^\infty_{k=2}a_ky^k $$ and after some computations you will inevitably get zero coefficients. But I should note that many times it is really complicated to check that all will be zero (or something else). If you really must do it, only making the computations you'll know.