Consider the system
$$\begin{cases}\dot{x}=x+y\cos(y)\\
\dot{y}=-y
\end{cases}$$
which has the unique equilibrium point $(0,0)$. I want to find the invariant manifolds for this system. The Jacobian at the origin is
$$\begin{pmatrix}
1&1\\
0&-1
\end{pmatrix}$$
which has eigenvalues $\lambda_1=1$ and $\lambda_2=-1$. An eigenvector for $\lambda_1$ is $v_1=(1,0)^\intercal$ and for $\lambda_2$, an eigenvector is $v_2=(1,-2)^\intercal$. It is clear that the unstable manifold is the unstable subspace $\operatorname{span}(v_1)$ aka the $x$-axis. However, I am completely lost on how to find the stable manifold. How does one go about finding this?
Find the invariant manifolds of the equilibrium
dynamical systemsordinary differential equationsset-invariance
Best Answer
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: