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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.

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1 Answer 1

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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]
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