# Why does the number of fixed points changes when I change the dimension of the system?

we define the following three dimensional system:

dyn3={b + r Subscript[\[Gamma], r] - s (b + Subscript[\[Gamma], s]) +
i s (-\[Beta] + \[Delta]), -b + s \[Beta] +
r Subscript[\[Beta],
r] - \[Gamma] + (-1 + i) \[Delta], -i r Subscript[\[Beta], r] +
i \[Gamma] - r (b + Subscript[\[Gamma], r]) +
s Subscript[\[Gamma], s] + i r \[Delta]}


which can be reduced to the following two dimensional system

dyn2={b - (-1 + i + s) Subscript[\[Gamma], r] -
s (b + Subscript[\[Gamma], s]) + i s (-\[Beta] + \[Delta]), -b +
s (\[Beta] - Subscript[\[Beta], r]) + Subscript[\[Beta], r] -
i Subscript[\[Beta], r] - \[Gamma] + (-1 + i) \[Delta]}


Now, solving the two systems yields a different number of solutions:

Solve[Thread[(dyn3)==0],{s,i,r}]

• How do you know that dyn3 reduces to dyn2 ? Did you solve for r to get there ? If so how ? Nov 30, 2022 at 11:30
• If you are sure that dyn3 is equivalent to dyn2 then you might want to use Reduce and include all assumptions you have about the variables Nov 30, 2022 at 11:36
• I did not try but you might also want to use LinearSolve. The problem seems linear. Nov 30, 2022 at 11:47
• The reason is that one of the three solutions does not satisfy the relation r+i+s==1. So when you impose that condition, you knock out a solution. Nov 30, 2022 at 15:21