Given a system: $\dot{x_1} = x_1(x_n - \phi(x))$, $\dot{x_n} = x_n(x_(n-1)-\phi(x))$ where $\phi(x) = x_1x_2 + x_2x_3 + \ldots + x_nx_1$. Let $S_n = \left\{x\in R^n: \sum_{i=1}^{n} x_i = 1, x_i\geq 0\ \forall\ i = 1,\ldots, n\right\}$.
(a) For $n=2$, find all the stable, unstable and center manifolds for each equilibrium of the system on $S_2$, provided they exist.
(b) For $n=3$, find all the stable, unstable and center manifolds for each equilibrium of the system on $S_2$, provided they exist.
My attempt: For part (a), I was able to solve for the equilibrium points on $S_2$, which are: $(x_1, x_2) = (0,1), (1,0), (\frac{1}{2}, \frac{1}{2})$. Now, at the equilibrium point $(1,0)$ or $(0,1)$, the matrix $Df((1,0))$ and $Df((0,1))$ of the corresponding linearized system all have unique eigenvalue = 0 (double multiplicity). So $(1,0)$ or $(0,1)$ is not hyperbolic, which means the center manifold don't exist at these points. So we're left with finding the stable/unstable manifold. But since the eigenvalue is just $0$, based on definition of stable/unstable subspace, I think neither stable manifold or unstable manifold at $(1,0)$ or $(0,1)$ exists? But this really contradicts to the center manifold theorem:P
Also, based on the definition, in this particular case, I think the stable manifold = center manifold must happen, as otherwise I don't know how to solve directly for both stable manifold and center manifold (both requires us to find the polynomial $h(y)$ such that $x = h(y)$ satisfies the DEs system, so we should always have the same $h(y)$ for center manifold and stable manifold via this method). Can someone please help clarify for me the method for finding BOTH stable and center manifold?
Assume that there exists such a stable/unstable manifold. Then we have: $\frac{dx_2}{dx_1} = \frac{1-2x_2}{1-2x_1}$. Thus, by separation of variables, we get: $\frac{ln|1-2x_2|}{2} = \frac{ln|1-2x_2|}{2} + C$ where C is some constant. This implies $|1-2x_2| = D|1-2x_1| $. At the point $(0,1)$, we would have: $2x_2 - 1 = D(1-2x_1)$, so $x_2 = \frac{D(1-2x_1)+1}{2}$ with $x_2(0) = 0$ and $x_2'(0) = 0$. Plugging the first condition in, we get $D=-1$ (contradiction since $D>0$), so there doesn't exist a stable manifold at $(0,1)$. Similarly, at $(1,0)$, we would have: $1-2x_2 = D(2x_1-1)$ with $x_1(0) = 0$ and $x_1'(0) = 0$. The first condition implies $D=-1$, which is wrong. Thus there also doesn't exist a stable manifold at $(1,0)$.