Stability analysis of equilibrium point of non-linear ODE system with Jacobian going to infinity

Question

Let's say I have a nonlinear system of ODEs, where one of equations looks like:

$$ \frac{dX_i}{dt} = a_1X_0+\dotsb+a_j\frac{X_j^{0.5}}{X_j^{0.5}+b_j^{0.5}}+\dotsb. $$

And equilibrium point is 0. I want to investigate the stability of a given point depending on the initial conditions ($X_k(0) \ge 0$, $k=0,\dotsc,n$). The usual procedure: linearization of the system does not work, since the Jacobian will tend to infinity for a term with power of 0.5.

Is there any way to get around this problem?

Is there any adequate numerical algorithm for investigating the stability of a point (if system contains, for example, up to 10 equations)?

For instance, let's take the model from "Mathematical modeling and stability analysis of macrophage activation in left ventricular remodeling post-myocardial infarction" by Yunji Wang et al.

And here instead of $\frac{T_\alpha}{T_\alpha +c_{T_\alpha}}$ we'll use $\frac{T_\alpha^{0.5}}{T_\alpha^{0.5} +c_{T_\alpha}^{0.5}}$.

There will be one (not only) stationary point at $\mathbf0$. I want to investigate stability of this particular point.

Answer 1

This is what (in my opinion) the general setup is and what should happen and why. Note that I haven't proved anything yet. What I offer is just a back of envelope computation at the physicist level of rigor (which is not much from the mathematical point of view, but you can check it against numerical simulations, nevertheless, and see if it makes sense). I'll update it when I understand things better.

You have a system of the kind $\dot X=-DX+AX+e_{i_0}\psi(\sqrt{X_1})+O(\|X\|^2)$ where $\psi(x)$ is a smooth function behaving like $x$ near $0$, $e_i$ is the vector with $i_0$-th coordinate $1$ and the rest $0$ for $i_0\ne 1$, $D$ is a diagonal matrix with positive elements on the diagonal (degradation matrix in your example), $A$ is a matrix with non-negative entries that has $0$ on the diagonal, and $O(\|X\|^2)$ is some quadratic and higher order nonsense.

We say that $k$ feeds upon $\ell\ne k$ if $A_{k,\ell}>0$. We say that $i$ is at the level $m\ge 0$ in the food chain starting from $i_0$ if there is a sequence of indices $i=i_m,i_{m-1},\dots, i_0$ in which each index feeds on the next one and $m$ is the smallest length of any such sequence (which is the same as if we consider an oriented graph with the adjacency matrix induced by $A$ and measure the number of steps needed to reach $i$ from $i_0$).

Case 1. $1$ is at some finite level $m$. Then what should happen is the following. We have a solution $X=X(t)$ in which $X_1$ is of order $t^{2p-2}$, if $i$ is at level $\ell$, then $X_i$ is of order $t^{p+\ell}$, and to have it consistent with the statement about $X_1$ (to close the loop), we must have $2p-2=p+m$, so $p=m+2$. That solution can start at arbitrarily small $t$ and grow to fixed size at $t=1$, so the system is always unstable in this case.

Case 2: $1$ is not in the above food chain. Then, if we run the food chain from $1$ in the other direction (indices $1$ feeds on, indices indices $1$ feeds on feed on, etc.), we'll get some set if indices $I$ containing $1$ but not $I_0$ that feed only on each other. Then one needs to investigate the linear stability of the corresponding submatrix. If it is stable, one should investigate the linear stability of the submatrix corresponding to the complementary set $I^c$ of indices. If that one is stable too, the whole system is stable.

As I said, I have no proofs yet, just an educated guess. The person who voted to close the question has, apparently, been able to figure everything out in under 5 minutes and could explain the full solution in under 10, but he/she decided not to condescend to poor idiots like myself and left no remark, so we'll have to investigate this problem using our inferior brains. :-)