How many solutions do you get from simultaneous polynomial equations?

Question

I have the following four simultaneous polynomial equations

eqns = {4 a^2 k^2 (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) == 
    I1 I2 M1 M2 ω1^2 ω2^2, 
   4 c (a^2 + e^2) k (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) == 
    2 I1 I2 M1 M2 ζ2 ω1^2 ω2 + 
     2 I1 I2 M1 M2 ζ1 ω1 ω2^2, 
   2 (2 c^2 e^2 (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) + 
       k (a^2 I2 M1 M2 + I1 ((a^2 + b^2) M1 M2 + I2 (M1 + M2)))) == 
    I1 I2 M1 M2 ω1^2 + 
     4 I1 I2 M1 M2 ζ1 ζ2 ω1 ω2 + 
     I1 I2 M1 M2 ω2^2, 
   2 c (e^2 I2 M1 M2 + I1 ((b^2 + e^2) M1 M2 + I2 (M1 + M2))) == 
    2 I1 I2 M1 M2 ζ1 ω1 + 
     2 I1 I2 M1 M2 ζ2 ω2};

The values of a, e, k and c are unknowns and I have values for the other parameters as follows

values = {M1 -> 4.8`, M2 -> 2.1420000000000003`, I1 -> 0.0090397`, 
   I2 -> 0.0115032`, 
   b -> 0.0275`, ω1 -> 1212.851537626128`, ω2 -> 
    1640.0348023828728`, ζ1 -> 0.1848599246177704`, ζ2 -> 
    0.11006812856373586`};

I can use NSolve and table the results

solutions = NSolve[eqns /. values];
Style[TableForm[{a, e, k, c} /. solutions, 
  TableHeadings -> {Automatic, {"a", "e", "k", "c"}}], FontSize -> 8]

I have 24 solutions most of them complex. How can I determine the number of solutions to expect? Can finding the number of solutions be generalised for other similar polynomial equations I will be solving? Is there a function to do this?

Thanks

Answer 1

For generic values of the parameters, the (complex and real) solution count will indeed be 24. One can see this simply by using a random substitution. This is a general method, albeit probabilistic.

eqns = {4 a^2 k^2 (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) == 
    I1 I2 M1 M2 ω1^2 ω2^2, 
   4 c (a^2 + e^2) k (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) == 
    2 I1 I2 M1 M2 ζ2 ω1^2 ω2 + 
     2 I1 I2 M1 M2 ζ1 ω1 ω2^2, 
   2 (2 c^2 e^2 (b^2 M1 M2 + I1 (M1 + M2) + I2 (M1 + M2)) + 
       k (a^2 I2 M1 M2 + I1 ((a^2 + b^2) M1 M2 + I2 (M1 + M2)))) == 
    I1 I2 M1 M2 ω1^2 + 
     4 I1 I2 M1 M2 ζ1 ζ2 ω1 ω2 + 
     I1 I2 M1 M2 ω2^2, 
   2 c (e^2 I2 M1 M2 + I1 ((b^2 + e^2) M1 M2 + I2 (M1 + M2))) == 
    2 I1 I2 M1 M2 ζ1 ω1 + 
     2 I1 I2 M1 M2 ζ2 ω2};

values = {M1 -> 4.8`, M2 -> 2.1420000000000003`, I1 -> 0.0090397`, 
   I2 -> 0.0115032`, 
   b -> 0.0275`, ω1 -> 1212.851537626128`, ω2 -> 
    1640.0348023828728`, ζ1 -> 0.1848599246177704`, ζ2 -> 
    0.11006812856373586`};

From this we isolate the parameters and replace the values with random ones (in case the specific given values put us on a piece of the complex discriminant variety).

params = values[[All, 1]];
SeedRandom[12345];
randsubsts = 
  Thread[params -> RandomInteger[{100, 1000}, Length[params]]];

Now solve and count:

Timing[Length[NSolve[polys /. randsubsts]]]

(* Out[647]= {0.168101, 24} *)

The upshot is that for random sets of parameter values there will be 24 solutions. Moreover since the given set of values also gave the generic solution count, it follows that for an open neighborhood in parameter space, centered at those specific values, we still get 24 solutions (in other words, the count is stable with respect to small perturbations of the given set of values).