Let us consider the dynamical system
\begin{aligned} \dot{x}&=\frac{(a-x)}{b} -z-u, \\ \dot{y}&=x-c y, \\ \dot{z}&=x-y, \\ \dot{u}&=x-1 \end{aligned}, where $a,b,c$ are parameters We can observe that the equilibrium point of the system is of the form $(x, y,z, u) = (1,1,\frac{a-1}{b}-t,t)$, where $t \in \Bbb{R}$. So here we have infinite number of equilibrium points depending on $t$. The equilibrium points exists only when $c=1$.
For the stability of the equilibrium points, I got the characteristic equation to be $P(\lambda) = \lambda^{4} + \lambda^{3}(c + \frac{1}{b}) + \lambda^{2}(2-\frac{c}{b})+\lambda(2c-1) =0$ implying one eigenvalue is zero.
For stability we need all real parts of four eigenvalues to be less than zero. I am thinking how to proceed from here?
I am thinking if there exists any stable equilibrium point or they are all unstable?
The $\dot y = x - cy$ implies that $c = 1$, then the characteristic polynomial is
$$\lambda (\lambda ^3+\left(1+\frac{1}{b}\right)\,\lambda ^2+\left(2+\frac{1}{b}\right)\,\lambda +1)$$
Since there is a single root equal $0$ the system can only be at most marginally stable
Using the Routh table:
$$ \begin{array}{|c|c|c|c|} \hline 1 & 2+1/b & 0\\ \hline 1+1/b& 1& 0\\ \hline (1/b^2+3/b + 1) b/(b+1)& 0& 0\\ \hline 1 & 0 & 0\\ \hline \end{array} $$ The condition for the system to be marginally stable is $1+1/b \geq 0$, $(1/b^2+3/b + 1) b/(b+1) \geq 0$. Consequently $b\in [\frac{-3-\sqrt{5}}{2},-1]$. Then you need to check $b$ at the boundaries to make sure that there all imaginary roots are distinct