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Symbolic computation using "Reduce" does not give a result

Posted 8 months ago

Firstly, I have proved [Gamma] = [Beta]^2/12 is one of the solutions of (A12 P312 + A13 P213 + A23 P123)/P123123 + A12 A13 A23==0

In[1]:= \[Gamma] = \[Beta]^2/12;
l1 = 1/6 (Sqrt[\[Beta]^2 \[Omega]1^2 - 12 \[Gamma] \[Omega]1^2 - 
      12 k1^4 - 12 k1 \[Omega]1] - \[Beta] \[Omega]1);
l2 = 1/6 (Sqrt[\[Beta]^2 \[Omega]2^2 - 12 \[Gamma] \[Omega]2^2 - 
      12 k2^4 - 12 k2 \[Omega]2] - \[Beta] \[Omega]2);
l3 = 1/6 (Sqrt[\[Beta]^2 \[Omega]3^2 - 12 \[Gamma] \[Omega]3^2 - 
      12 k3^4 - 12 k3 \[Omega]3] - \[Beta] \[Omega]3);
A12 = (2 \[Gamma] \[Omega]1 \[Omega]2 + k2 (4 k1^3 + \[Omega]1) - 
   6 k1^2 k2^2 + 
   k1 (4 k2^3 + \[Omega]2) + \[Beta] (l1 \[Omega]2 + l2 \[Omega]1) + 
   6 l1 l2)/(
  2 \[Gamma] \[Omega]1 \[Omega]2 + k2 (4 k1^3 + \[Omega]1) + 
   6 k1^2 k2^2 + 
   k1 (4 k2^3 + \[Omega]2) + \[Beta] (l1 \[Omega]2 + l2 \[Omega]1) + 
   6 l1 l2);
A13 = (2 \[Gamma] \[Omega]1 \[Omega]3 + k3 (4 k1^3 + \[Omega]1) - 
   6 k1^2 k3^2 + 
   k1 (4 k3^3 + \[Omega]3) + \[Beta] (l1 \[Omega]3 + l3 \[Omega]1) + 
   6 l1 l3)/(
  2 \[Gamma] \[Omega]1 \[Omega]3 + k3 (4 k1^3 + \[Omega]1) + 
   6 k1^2 k3^2 + 
   k1 (4 k3^3 + \[Omega]3) + \[Beta] (l1 \[Omega]3 + l3 \[Omega]1) + 
   6 l1 l3);
A23 = (2 \[Gamma] \[Omega]2 \[Omega]3 + k3 (4 k2^3 + \[Omega]2) - 
   6 k2^2 k3^2 + 
   k2 (4 k3^3 + \[Omega]3) + \[Beta] (l2 \[Omega]3 + l3 \[Omega]2) + 
   6 l2 l3)/(
  2 \[Gamma] \[Omega]2 \[Omega]3 + k3 (4 k2^3 + \[Omega]2) + 
   6 k2^2 k3^2 + 
   k2 (4 k3^3 + \[Omega]3) + \[Beta] (l2 \[Omega]3 + l3 \[Omega]2) + 
   6 l2 l3);
P312 = \[Gamma] (-\[Omega]1 - \[Omega]2 + \[Omega]3)^2 + (-k1 - k2 + 
      k3) (-\[Omega]1 - \[Omega]2 + \[Omega]3) + (-k1 - k2 + 
     k3)^4 + \[Beta] (-l1 - l2 + 
      l3) (-\[Omega]1 - \[Omega]2 + \[Omega]3) + 3 (-l1 - l2 + l3)^2;
P213 = \[Gamma] (-\[Omega]1 + \[Omega]2 - \[Omega]3)^2 + (-k1 + k2 - 
      k3) (-\[Omega]1 + \[Omega]2 - \[Omega]3) + (-k1 + k2 - 
     k3)^4 + \[Beta] (-l1 + l2 - 
      l3) (-\[Omega]1 + \[Omega]2 - \[Omega]3) + 3 (-l1 + l2 - l3)^2;
P123 = \[Gamma] (\[Omega]1 - \[Omega]2 - \[Omega]3)^2 + (k1 - k2 - 
      k3) (\[Omega]1 - \[Omega]2 - \[Omega]3) + (k1 - k2 - 
     k3)^4 + \[Beta] (l1 - l2 - 
      l3) (\[Omega]1 - \[Omega]2 - \[Omega]3) + 3 (l1 - l2 - l3)^2;
P123123 = \[Gamma] (\[Omega]1 + \[Omega]2 + \[Omega]3)^2 + (k1 + k2 + 
      k3) (\[Omega]1 + \[Omega]2 + \[Omega]3) + (k1 + k2 + 
     k3)^4 + \[Beta] (l1 + l2 + 
      l3) (\[Omega]1 + \[Omega]2 + \[Omega]3) + 3 (l1 + l2 + l3)^2;
Simplify[(A12 P312 + A13 P213 + A23 P123)/P123123 + A12 A13 A23 == 0]

Out[12]= True

Next, I want to solve the relation between [Beta] and [Gamma] through ``Reduce". However, it took too much time, and failed to give the output in finite time. How can I get the relation between [Beta] and [Gamma]?

ClearAll[\[Beta], \[Gamma]]
l1 = 1/6 (Sqrt[\[Beta]^2 \[Omega]1^2 - 12 \[Gamma] \[Omega]1^2 - 
      12 k1^4 - 12 k1 \[Omega]1] - \[Beta] \[Omega]1);
l2 = 1/6 (Sqrt[\[Beta]^2 \[Omega]2^2 - 12 \[Gamma] \[Omega]2^2 - 
      12 k2^4 - 12 k2 \[Omega]2] - \[Beta] \[Omega]2);
l3 = 1/6 (Sqrt[\[Beta]^2 \[Omega]3^2 - 12 \[Gamma] \[Omega]3^2 - 
      12 k3^4 - 12 k3 \[Omega]3] - \[Beta] \[Omega]3);
A12 = (2 \[Gamma] \[Omega]1 \[Omega]2 + k2 (4 k1^3 + \[Omega]1) - 
   6 k1^2 k2^2 + 
   k1 (4 k2^3 + \[Omega]2) + \[Beta] (l1 \[Omega]2 + l2 \[Omega]1) + 
   6 l1 l2)/(
  2 \[Gamma] \[Omega]1 \[Omega]2 + k2 (4 k1^3 + \[Omega]1) + 
   6 k1^2 k2^2 + 
   k1 (4 k2^3 + \[Omega]2) + \[Beta] (l1 \[Omega]2 + l2 \[Omega]1) + 
   6 l1 l2);
A13 = (2 \[Gamma] \[Omega]1 \[Omega]3 + k3 (4 k1^3 + \[Omega]1) - 
   6 k1^2 k3^2 + 
   k1 (4 k3^3 + \[Omega]3) + \[Beta] (l1 \[Omega]3 + l3 \[Omega]1) + 
   6 l1 l3)/(
  2 \[Gamma] \[Omega]1 \[Omega]3 + k3 (4 k1^3 + \[Omega]1) + 
   6 k1^2 k3^2 + 
   k1 (4 k3^3 + \[Omega]3) + \[Beta] (l1 \[Omega]3 + l3 \[Omega]1) + 
   6 l1 l3);
A23 = (2 \[Gamma] \[Omega]2 \[Omega]3 + k3 (4 k2^3 + \[Omega]2) - 
   6 k2^2 k3^2 + 
   k2 (4 k3^3 + \[Omega]3) + \[Beta] (l2 \[Omega]3 + l3 \[Omega]2) + 
   6 l2 l3)/(
  2 \[Gamma] \[Omega]2 \[Omega]3 + k3 (4 k2^3 + \[Omega]2) + 
   6 k2^2 k3^2 + 
   k2 (4 k3^3 + \[Omega]3) + \[Beta] (l2 \[Omega]3 + l3 \[Omega]2) + 
   6 l2 l3);
P312 = \[Gamma] (-\[Omega]1 - \[Omega]2 + \[Omega]3)^2 + (-k1 - k2 + 
      k3) (-\[Omega]1 - \[Omega]2 + \[Omega]3) + (-k1 - k2 + 
     k3)^4 + \[Beta] (-l1 - l2 + 
      l3) (-\[Omega]1 - \[Omega]2 + \[Omega]3) + 3 (-l1 - l2 + l3)^2;
P213 = \[Gamma] (-\[Omega]1 + \[Omega]2 - \[Omega]3)^2 + (-k1 + k2 - 
      k3) (-\[Omega]1 + \[Omega]2 - \[Omega]3) + (-k1 + k2 - 
     k3)^4 + \[Beta] (-l1 + l2 - 
      l3) (-\[Omega]1 + \[Omega]2 - \[Omega]3) + 3 (-l1 + l2 - l3)^2;
P123 = \[Gamma] (\[Omega]1 - \[Omega]2 - \[Omega]3)^2 + (k1 - k2 - 
      k3) (\[Omega]1 - \[Omega]2 - \[Omega]3) + (k1 - k2 - 
     k3)^4 + \[Beta] (l1 - l2 - 
      l3) (\[Omega]1 - \[Omega]2 - \[Omega]3) + 3 (l1 - l2 - l3)^2;
P123123 = \[Gamma] (\[Omega]1 + \[Omega]2 + \[Omega]3)^2 + (k1 + k2 + 
      k3) (\[Omega]1 + \[Omega]2 + \[Omega]3) + (k1 + k2 + 
     k3)^4 + \[Beta] (l1 + l2 + 
      l3) (\[Omega]1 + \[Omega]2 + \[Omega]3) + 3 (l1 + l2 + l3)^2;
Reduce[(A12 P312 + A13 P213 + A23 P123)/P123123 + A12 A13 A23 == 
  0, {\[Beta], \[Gamma]}, Reals]

This question is also posted at https://mathematica.stackexchange.com/questions/289814/symbolic-computation-using-reduce-does-not-give-a-result

Attachments:
POSTED BY: Lingfei Li
2 Replies
Posted 8 months ago

Thanks. It is worth to try.

POSTED BY: Lingfei Li
Posted 8 months ago

If you

Simplify[(A12 P312 + A13 P213 + A23 P123)/P123123 + A12 A13 A23 ==   0]

you should see it is a very large numerator/denominator==0. If you replace that with numerator==0 then the problem is about half as big and may be easier to solve. If it can do that then you can check denominator != 0 afterwards.

Sometimes not asking Reduce to solve over the Reals seems faster for large problems. If it can do that then you can check for Real solutions afterwards.

Sometimes Reduce[singleexpr==0,singlevariable] seems more likely to succeed for large problems. If it can do that then perhaps you can find the relation between Beta and Gamma afterwards.

But your problem is still large and complicated enough that these may not be enough to give a solution.

POSTED BY: Bill Nelson
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