I have the following coupled non-linear ODE system, which describes a biological system:
$$ \begin{cases} \dfrac{dp}{dt} = -\gamma p f,\\ \\ \dfrac{df}{dt} = \gamma p f,\\ \\ \dfrac{dT}{dt} = \left( 1 - \dfrac{k}{T} \right) \left\{ b (1 - T) \dfrac{r^n}{f^n + r^n} - m \dfrac{f^n}{f^n + r^n} \right\}, \end{cases} $$ where the parameters are: $\gamma = 0.5432$, $k = 0.0026$, $b = 0.0885$, $n = 14.9832$, $r = 0.8265$, and $m = 0.9780$.
I want to investigate the stability of the system. To this end, first, one observes that the first two equations imply that $p + f$ must be a constant, let's call it $w$, and in our case, it must be non-negative (for the chosen parameters' values of the system in the above, it's $0.801$). Now, by excluding $p$ from the above system, we can write:
$$ \begin{cases} \dfrac{df}{dt} = \gamma (w - f) f,\\ \\ \dfrac{dT}{dt} = \left( 1 - \dfrac{k}{T} \right) \left\{ b (1 - T) \dfrac{r^n}{f^n + r^n} - m \dfrac{f^n}{f^n + r^n} \right\}. \end{cases} $$
The second system has four equilibria, $(f, T)$:
$$(0, k), (0, 1), (w, k), \left( w, 1- \frac{m}{b} (w / r)^n \right).$$
Now, by calculating the eigenvalues of the Jacobian, it turns out that the first two equilibria are unstable, and the last two are stable equilibria.
My question is as follows: If we simulate our original, i.e., the first system, using Mathematica for some initial conditions:
s = NDSolve[{p'[t] == -gamma p[t] f[t], f'[t] == gamma p[t] f[t],
T'[t] == b (1 - T[t]) (r^n/(f[t]^n + r^n)) (1 - 0.0026/T[t]) - m (f[t]^n/(f[t]^n + r^n)) (1 - 0.0026/T[t]),
p[0] == p0, f[0] == 0.001, T[0] == T0}/.{gamma -> 0.5432, b -> 0.0885, m -> 0.9780, r -> 0.8265, n -> 14.9832, p0 -> 0.8, T0 -> 0.8}, {p, f, T}, {t, 0, 720}];
Plot[Evaluate[{f[t], T[t]}/.s], {t, 0, 720}, PlotStyle -> {Blue, Red}, AxesOrigin -> {-1, 0}, PlotRange -> All]
we obtain:
From the above plot it's clear that solutions are converging to $(w, k)$ in the long run, which here $w$ is around $0.8$. However, it seems that the system never converges to $\left( w, 1- \frac{m}{b} (w / r)^n \right)$, for any initial conditions. What is the reason behind that? Can a system never converge to one of its stable equilibria?
EDIT
The reason for transforming the original system: If one wants to do the stability analysis for the original system, one runs into a problem. It has two equilibria: $(f \to 0, T \to k)$ and $(f \to 0, T \to 1)$, regardless of the value of $p$. Since $p$ can be anything, it isn't clear how one can calculate the Jacobian. However, by transforming the system, we can eliminate this issue.
The direct answer to your question is "no". At least if by "stable" you mean all eigenvalues of the Jacobian at $x^*$ have negative real part. (See, e.g., Section 2.9 Theorems 1 and 2 from Differential Equations and Dynamical Systems, L. Perko 3rd edition).
How then do we square this with this particular example? The issue comes when you simplify the system. You are correct in saying $w=f+p$ is constant, but you have to understand what this means: This means that as time evolves, $w$ is constant for a given trajectory it does not mean that all trajectories have the same constant $w$. Thus to successfully transform the system we would need to include
$$\frac{dw}{dt} = 0$$
Note that from the original system it is clear that the last two fixed points you give cannot be fixed points of the original system since the first two equations imply right away that $f=0$ or $p=0$ for any fixed point.
Edit: I decided to add some information about the system since this is an interesting case. We have that either $p=0$ or $f=0$. Start with $f=0$. Then we have a fixed point for any value of $p$ and $T=k$ or $T=1$. If instead $p=0$ then either $T=k$ (and $f$ is any value) or $T=1-\frac{mf^n}{br^n}$. Thus we get four curves which fully describe all the fixed points of the system. We can still analyze these fixed points as we would anyway. The issue is that these fixed points are what is called "nonhyperbolic"--you will see that the Jacobian of each has a zero eigenvalue. As the book I referenced puts it, "The question as to whether a nonhyperbolic equilibrium point is stable, asymptotically stable, or unstable is a delicate question." The reason is that linearization "doesn't work" in the nonhyperbolic case.