Continuing from this question: How do I simplify the equilibrium point to this simplified expression?
We found our equilibrium points.
We also define $$\mathcal{R}_0= \frac{[\beta_1 b_3(\epsilon p+b_1(1-p))+\beta_2p_1(\epsilon q +b_1(1-q))]\frac{\nu}{\mu}}{b_1b_2b_3-(\epsilon\xi_1+b_1 \xi_2)p_1}$$
Component $S$ of the second equilibrium point can be simplified as
$$ S^*=\frac{\nu}{\mu \mathcal{R}_0}$$
Component $J$ of the second equilibrium point can be simplified as
$$J^*=\frac{p_1 \mu}{\beta_1 b_3 +\beta_2 p_1}(\mathcal{R}_0-1) $$
Component $I_1$ of the second equilibrium point can be simplified as
$$ \frac{1}{b_1}\left[p\beta_1 \frac{\nu b_3}{(\beta_1 b_3+\beta_2 p_1)J^* +\mu p_1}+q \beta_2 \frac{\nu p_1}{(\beta_1 b_3+\beta_2 p_1)J^* +\mu p_1}+\xi_1 \right]J^*$$
My question therefore is; how can we do it in Mathematica without knowing the simplified expressions beforehand?
Following Bob's solution:
LeafCount /@ {eqPts, eqPts2 = eqPts // Simplify}
vars = Variables[Level[eqPts2, {-1}]]
eqns = eqPts2 /. Rule :> Equal;
sol4 = Solve[#, s, {a, i2, i1, b1}] & /@ eqns
Resulting in:
s -> (p1 \[Nu])/(b3 j \[Beta]1 + j p1 \[Beta]2 + p1 \[Mu])
This is good, it gives the same expression as the one described above for $S^*$ however there is a simpler expression, namely the one described above. How can I achieve this for $S^*$, $J^*$ and $I_1^*$? The other components i.e. $I_2^*$ and $A^*$ follow directly from the initial ODE system.
