Is a one-dimensional unstable manifold of an ODE a union of the associated equilibrium point and two full orbits? [closed]

Consider an ordinary differential equation (ODE) system \begin{align} \frac{dx}{dt} = f(x) \end{align} where $x \in \mathbb{R}^n$ ($n \geq 2$) and the vector field $f$ is defined on an open subset $X$ of $\mathbb{R}^n$. Furthermore, we let $f$ be smooth enough so that each $x_0 \in X$ corresponds to a unique solution of the system. This would guarantee a flow $\phi_t$ (see Perko's book "Differential Equations and Dynamical Systems").

Now, we suppose that $x^*$ is a hyperbolic saddle point. That is:

1. $f(x^*) = 0$; and,
2. The Jacobian matrix $J(x^*)$ has exactly one eigenvalue with positive real part and $n - 1$ eigenvalues with negative real part.

Then, there is a one-dimensional unstable manifold $x^*$, given by \begin{align} W^U(x^*) := \{x_0 \in X : \text{$\phi_t(x_0) \rightarrow x^*$ as $t \rightarrow -\infty$}\}. \end{align} A local version $W^U_\text{loc}(x^*)$ exists by the (un)stable manifold theorem, from which we have \begin{align} W^U(x^*) = \bigcup_{t \geq 0} \phi_t\left(W^U_\text{loc}(x^*)\right). \end{align} May I ask if the following is true?

Claim. There exists two full orbits $\gamma_1$ and $\gamma_2$ are full orbits that start from $x^*$ (i.e. with alpha-limit set $\{x^*\}$), such that $W^U(x^*) = \{x^*\} \cup \gamma_1 \cup \gamma_2$.

After searching through the internet, I think the claim is true. However, I wish to know if this is indeed true, and how it might be proven, or if there are some reliable sources. Otherwise, may I ask if $W^U(x^*)$ can still be expressed as the union of distinct orbits?

closed as off-topic by Loïc Teyssier, Wolfgang, Michael Renardy, Sebastian Goette, Ryan BudneyApr 14 '16 at 19:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about research level mathematics within the scope defined in the help center." – Loïc Teyssier, Wolfgang, Michael Renardy, Sebastian Goette, Ryan Budney
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• It's possible for the stable and unstable manifolds to intersect, in what's called a "homoclinic point". This would violate #2, if I understand what you mean. – Martin M. W. Apr 14 '16 at 18:06
• Thanks. I agree with you that condition #2 does not hold when the equilibrium point $x^*$ is connected by a homoclinic orbit. This would mean that one of the full orbits $\gamma_k$ connects $x^*$ to itself. I have removed that condition in the post. – Alexis Erich Almocera Apr 15 '16 at 4:30