0
$\begingroup$

I'm new to Mathematica and would like to use it to solve a system of polynomials. In this system, every variable (and every solution I'm interested in) is a positive real number.

I tried this:

Solve[k1*A*N - k2*C == 0 && k3*B*N - k4*D == 0 && 
  N0 - C - D - N == 0 && A0 - C - A == 0 && B0 - D - B == 0, {A, B, C,
   D, N}]

which generates an enormous expression and isn't constrained to positive reals. I am using Mathematica 11.3.0.0 which I believe doesn't support PositiveReals as an argument to Solve. When I try to encode the constraints as additional equations, Mathematica hangs:

Solve[k1*A*N - k2*C == 0 &&
      k3*B*N - k4*D == 0 && 
      N0 - C - D - N == 0 && 
      A0 - C - A == 0 && 
      B0 - D - B == 0 &&
      k1 >= 0 && k2 >= 0 && k3 >= 0 && k4 >= 0 && A >= 0 && B >= 0 && N >= 0 && C >= 0 && D >= 0 && A0 >= 0 && B0 >= 0 && N0 >= 0,
      {A, B, C, D, N}, Reals]

Trying to encode the constraints with Assuming also hangs:

Assuming[k1 >= 0 && k2 >= 0 && k3 >= 0 && k4 >= 0 && A >= 0 && 
  B >= 0 && N >= 0 && C >= 0 && D >= 0 && A0 >= 0 && B0 >= 0 && 
  N0 >= 0,
 Solve[k1*A*N - k2*C == 0 &&
       k3*B*N - k4*D == 0 && 
       N0 - C - D - N == 0 && 
       A0 - C - A == 0 && 
       B0 - D - B == 0,
       {A, B, C, D, N}, Reals]]

any ideas on how to solve this system? I don't expect it to have any enormously complicated solutions.

$\endgroup$

1 Answer 1

0
$\begingroup$
Clear["`*"];
Solve[k1*A*N - k2*C == 0 && k3*B*N - k4*D == 0 && 
  N0 - C - D - N == 0 && A0 - C - A == 0 && B0 - D - B == 0 && 
  k1 >= 0 && k2 >= 0 && k3 >= 0 && k4 >= 0 && A >= 0 && B >= 0 && 
  N >= 0 && C >= 0 && D >= 0 && A0 >= 0 && B0 >= 0 && N0 >= 0, {A, B, 
  C, D, N}, Reals, Method -> Reduce]
$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.