Find the parameter values satisfying conditions for all roots of an equation

Question

Hi I'm very new at Mathematica.

My real problems concerns a much more difficult equation with many parameters, but I want to check that the method is good, so I made up a test equation.

My test equation and roots:

equ = (5 x^3 + z + x z + x^2 (1 + z) == 0)  
sol = x /. Solve[equ, x]

Here is problem. I'm looking to the parameter z such that all the roots of the equation are less than zero (edit: real parts less than 0, where, z is a real number and the roots can be complex:

Reduce[Re[sol[[1]]] < 0 && Re[sol[[2]]] < 0 && Re[sol[[3]]] < 0 && Im[z] == 0, z]

The answer is (I think) z > 4, but the function Reduce doesn't calculate this in a reasonable time. How do I proceed?

---

EDIT after few days: In equation with more parameters:

equ=-11.8193 + 12.0482 b - 59.0964 x + 12.0482 a x - 11.8193 d x + 12.0482 b d x + 0.320482 x^2 + 12.0482 c x^2 - 59.0964 d x^2 + 12.0482 a d x^2 + 1. x^3 + 0.320482 d x^3 + 1. d x^4 
Reduce[ForAll[x, equ == 0 \[And] Element[{a, b, c, d}, Reals], Re[x] < 0], {a, b, c, d}]

It is good code? I waiting and waiting for a response from the Mathematica.

Maybe it would be faster if the program could write only one/few numerical answer?

Answer 1

Finding the whole solution set

Inspired by @Szabolcs idea, you can let Reduce solve this problem with the help of existential quantifiers:

Reduce[ ForAll[x, 5 x^3 + z + x z + x^2 (1 + z) == 0 \[And] Im[z] == 0, Re[x] < 0], z]

It immediately affirms your conjecture:

(* z > 4 *)

This problem formulation via ForAll can be read as: Reduce the following statement with respect to z over the complex numbers: "For all x that satisfy the condition that 5 x^3 + z + x z + x^2 (1 + z) == 0 where z is restricted to the real number line, the real part of x is negative."

Finding a numeric solution

If you are not so much interested in all possible solutions but just one that meets certain criteria, you can use NMinimize to help solve that optimization problem. For that to work smoothly it's useful to set up some helper functions first:

PolyRoots[poly_, paramrules : {(_ -> _?NumericQ) ..}] :=
  x /. {ToRules@NRoots[poly /. paramrules, x]}

This takes a polynomial and a list of rules which specify the parameters and their values and returns a list of all roots. The big obfuscated looking pattern match has the purpose of making sure that the function definition fires only when the right hand side of all rules (a -> 123) assumes numeric values.

To make it easier for NMinimize to get an idea how bad the current set of parameter values is, we introduce a function that assigns a penalty value to a set of parameter values based on the sum of the squares of the real parts of all roots if they are positive and zero if they are negative. That allows NMinimze to "sense" it's way to better solutions until it hits the promised land, where solutions have penalty value zero.

Penalty[poly_, paramrules : {(_ -> _?NumericQ) ..}] :=
  Total[HeavisideTheta[#] #^2 & @ Thread @ Re @ PolyRoots[poly, paramrules]]

The final step is setting up the polynomial (including a cosmetic change of variable names so that the variables in our final solution have the desired names) and letting NMinimize do the work:

Module[{poly},
  poly = -11.8193 + 12.0482 b - 59.0964 x + 12.0482 a x - 
    11.8193 d x + 12.0482 b d x + 0.320482 x^2 + 12.0482 c x^2 - 
    59.0964 d x^2 + 12.0482 a d x^2 + 1. x^3 + 0.320482 d x^3 + 
    1. d x^4 /. {a -> aa, b -> bb, c -> cc, d -> dd};
  {penalty, solution} = 
  NMinimize[
    {Penalty[poly, {aa -> a, bb -> b, cc -> c, dd -> d}], a > 0, b > 0, c > 0, d > 0},
    {a, b, c, d}]
 ]

(* {0., {a -> 13.2852, b -> 7.3045, c -> 2.38894, d -> 0.171584}} *)

Answer 2

RegionPlot[ And @@ Thread[(Re[sol /. z -> (a + b I)]) < 0], {a, 3, 10}, {b, -10, 10}]

Edit

As per your edit, z is Real (really bad choice for a name), so:

Plot[Re[sol], {z, -1, 10}, Evaluated -> True, PlotRange -> {-2, 2}]

You can see that from z==4 onwards, the two complex conjugate solutions are negative, as well as the real one.