Given the following system of nonlinear ODEs: $$\dfrac{dx}{dt} = Q-qxy + \ell z \\ \dfrac{dy}{dt} = (\ell+v)z - qxy \\ \dfrac{dz}{dt} = qxy - (\ell+v)z $$
and initial conditions: $x(0) = 100, y(0) = 1, z(0) = 0,$
I see some symmetry in the equations, but is it possible to solve this system analytically given the arbitrary (positive) parameters $Q,q,\ell, v$?
If necessary, I have values for $Q,q$ and $\ell$ as well: $Q=0.01$, $q = 0.02, \ell = 0.001.$
The goal is to identify the parameter regime for $v.$
I have it running in Maple with this code, but don't know how long it will take.
sys := diff(x(t), t) = 0.1e-1-0.2e-1*x(t)*y(t)+0.1e-2*z(t), diff(y(t), t) = (0.1e-2+v)*z(t)-0.2e-1*x(t)*y(t), diff(z(t), t) = 0.2e-1*x(t)*y(t)-(0.1e-2+v)*z(t);
f := {x(t), y(t), z(t)};
dsolve({sys, x(0) = 100, y(0) = 1, z(0) = 0}, f)
Thanks for any help.
Quick observation: $\frac{\text{d}y}{\text{d}t} + \frac{\text{d}z}{\text{d}t} = 0$, so $y+z$ is constant in time. Using the initial conditions, we see that $z(t) = 1-y(t)$. That allows us to reduce the system to \begin{align} \frac{\text{d} x}{\text{d} t} &= Q - q x y + \ell(1-y),\\ \frac{\text{d} y}{\text{d} t} &= (\ell + v)(1-y) - q x y. \end{align} It is convenient to shift $x$ and $y$ as to eliminate the constant terms, by introducing $u = x + \frac{Q(v+\ell)}{q(v-Q)}$, $w = y+1 - \frac{Q}{v}$, to yield \begin{align} \frac{\text{d} u}{\text{d} t} &= a u + b w -qu w ,\\ \frac{\text{d} w}{\text{d} t} &= a u + (b-v) w - q u w. \end{align} with $a = (1-Q/v)q$ and $b = \frac{Q(v+\ell)}{v-Q} - \ell$. You can then do a phase plane analysis, sketching null clines and determining the stability of the origin (which is the only equilibrium).