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?
Best Answer
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.
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 $A=0$.