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I am trying to solve these equations and want to plot the value of $a^{\dagger}a$ versus P0. I have tried this code but this gives me an error like "Conjunction::argr: Conjunction called with 1 argument; 2 arguments are expected. >>". If there is another method to solve this I don't know. If anybody can solve this will be appreciated.

delta = 4;
g0 = 1;
Pb = 2;
KL = 1;
kb = 0.2;
ome1 = 1;
gma = 0.02;
nth = 2;
Lf = Conjugate[a1]*b1;
Bf = a1*b1;
NL = Conjugate[a1]*a1;
sol = NSolve[{I*P0*(Conjugate[a1] - a1) + 
      I*Pb*(Lf - Conjugate[Lf] + Conjugate[Bf] - Bf) - KL*NL - 
      kb*(Lf*a1 + Conjugate[Lf]*Conjugate[a1] + Bf*Conjugate[a1] + 
         Conjugate[Bf]*
          a1 + (NL - 2*Abs[a1]^2)*(b1 + Conjugate[b1])) == 0, 
    I*g0*((NL - 2*Abs[a1]^2)*(Conjugate[b1] - b1) + 
         Conjugate[Lf]*Conjugate[a1] - Lf*a1 + Conjugate[Bf]*a1 - 
         Bf*Conjugate[a1]) + 
      I*Pb*(Conjugate[Bf] - Lf + Conjugate[Lf] - Bf) - 
      gma*Conjugate[b1]*b1 + gma*nth == 
     0, -I*delta*Lf - I*ome1*Lf + 
      I*g0*((2*NL - Abs[a1]^2 - Conjugate[b1]*b1 + 2*Abs[b1]^2 + 
            b1^2)*Conjugate[a1] - (Conjugate[Bf] + 2*Lf)*b1 - 
         Lf*Conjugate[b1]) - I*P0*b1 == 0, 
    I*delta*Bf - I*ome1*Bf + 
      I*g0*((2*NL - Abs[a1]^2 + Conjugate[b1]*b1 - 2*Abs[b1]^2)*
          a1 + (2*Bf - a1*b1 + Conjugate[Lf])*b1 + Bf*Conjugate[b1]) +
       I*P0*b1 + 
      I*Pb*(NL + Conjugate[b1]*b1 + a1^2 + b1^2 + 1) - (KL + gma)*
       Bf/2 - kb/
        2*((2*Bf - a1*b1 + Conjugate[Lf])*
          b1 + (Conjugate[b1]*b1 - 2*Abs[b1]^2)*a1 + 
         Bf*Conjugate[b1]) == 0, 
    I*delta*a1 + I*delta*a1 + I*g0*(Conjugate[Lf] + Bf) + I*P0 + 
      I*Pb*(b1 + Conjugate[b1]) - KL*a1/2 - 
      kb*(Conjugate[Lf] + Bf)/2 == 
     0, -I*ome1*b1 + I*g0*NL + I*Pb*(a1 + Conjugate[a1]) - gma*b1/2 ==
      0}, {a1, b1}];
Plot[{Evaluate[Conjugate[a1]*a1] /. sol}, {P0, 0, 30}]
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    $\begingroup$ You have one Conjunction[b] in your code and since you have dozens of Conjugate perhaps you just misspelled the first one. As an experiment I changed that to Conjugate and ran the code. It grinds for a while. I don't know whether it is going to finish or not. Maybe you can let it run for a while, minutes or an hour, and report back what it gives you. $\endgroup$
    – Bill
    Jun 29, 2020 at 4:21
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    $\begingroup$ The equations in NSolve have the undefined variables {b, ig0, Kb, kL, ome1}. Please provide their values. $\endgroup$
    – Bob Hanlon
    Jun 29, 2020 at 4:51
  • $\begingroup$ There are what seem to me multiple typos. For example, you have ig0 which should probably be I * g0, you have Conjunction[b] which should be Conjugate[b1] etc. $\endgroup$
    – Randy Marsh
    Jun 29, 2020 at 4:58
  • $\begingroup$ Thanks for selecting these errors but sir, still it is running for more than one hour. $\endgroup$
    – vidya
    Jun 29, 2020 at 10:12
  • $\begingroup$ Do you have any reason to believe that it should be faster? You have a non-linear system with 6 equations, 2 unknowns and their conjugates, and a generic constant. You could try some special cases, e.g. $|a_1|=|b_1|=1$ that will simplify the equations to some extent, and see if the system can be solved then. $\endgroup$
    – Randy Marsh
    Jun 29, 2020 at 11:06

1 Answer 1

Reset to default
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Clear["Global`*"]

Use exact values for the constants to facilitate simplification.

delta = 4;
g0 = 1;
Pb = 2;
KL = 1;
kb = 2/10;
ome1 = 1;
gma = 2/100;
nth = 2;
Lf = Conjugate[a1]*b1;
Bf = a1*b1;
NL = Conjugate[a1]*a1;

eqns = And @@ 
   Simplify[{I*P0*(Conjugate[a1] - a1) + 
       I*Pb*(Lf - Conjugate[Lf] + Conjugate[Bf] - Bf) - KL*NL - 
       kb*(Lf*a1 + Conjugate[Lf]*Conjugate[a1] + Bf*Conjugate[a1] + 
          Conjugate[Bf]*a1 + (NL - 2*Abs[a1]^2)*(b1 + Conjugate[b1])) == 0, 
     I*g0*((NL - 2*Abs[a1]^2)*(Conjugate[b1] - b1) + 
          Conjugate[Lf]*Conjugate[a1] - Lf*a1 + Conjugate[Bf]*a1 - 
          Bf*Conjugate[a1]) + I*Pb*(Conjugate[Bf] - Lf + Conjugate[Lf] - Bf) -
        gma*Conjugate[b1]*b1 + gma*nth == 
      0, -I*delta*Lf - I*ome1*Lf + 
       I*g0*((2*NL - Abs[a1]^2 - Conjugate[b1]*b1 + 2*Abs[b1]^2 + b1^2)*
           Conjugate[a1] - (Conjugate[Bf] + 2*Lf)*b1 - Lf*Conjugate[b1]) - 
       I*P0*b1 == 0, 
     I*delta*Bf - I*ome1*Bf + 
       I*g0*((2*NL - Abs[a1]^2 + Conjugate[b1]*b1 - 2*Abs[b1]^2)*
           a1 + (2*Bf - a1*b1 + Conjugate[Lf])*b1 + Bf*Conjugate[b1]) + 
       I*P0*b1 + 
       I*Pb*(NL + Conjugate[b1]*b1 + a1^2 + b1^2 + 1) - (KL + gma)*Bf/2 - 
       kb/2*((2*Bf - a1*b1 + Conjugate[Lf])*
           b1 + (Conjugate[b1]*b1 - 2*Abs[b1]^2)*a1 + Bf*Conjugate[b1]) == 0, 
     I*delta*a1 + I*delta*a1 + I*g0*(Conjugate[Lf] + Bf) + I*P0 + 
       I*Pb*(b1 + Conjugate[b1]) - KL*a1/2 - kb*(Conjugate[Lf] + Bf)/2 == 
      0, -I*ome1*b1 + I*g0*NL + I*Pb*(a1 + Conjugate[a1]) - gma*b1/2 == 0}];

Convert a1 and b1 to Cartesian form so that all unknowns are real.

eqns2 = Assuming[Element[{x1, y1, x2, y2, P0}, Reals],
  (eqns /. {a1 -> x1 + I*y1, b1 -> x2 + I*y2}) // 
   ComplexExpand // FullSimplify]

(* False *)

This indicates that the equations are inconsistent. Further, there are too many equations for the number of unknowns, so the system is overdetermined.

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  • $\begingroup$ Sir, thankyou for your response. After your code, I write $Plot[{Evaluate[x1^2 + y1^2] /. eqns2}, {P0, 0, 30}]$ and then run the program but it is showing this error: {False} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing. >> and General::stop: Further output of ReplaceAll::reps will be suppressed during this calculation. >> $\endgroup$
    – vidya
    Jun 29, 2020 at 16:36
  • $\begingroup$ eqns2 is not a Rule (LHS -> RHS) or a set of rules so it makes no sense to place it after ReplaceAll. You should determine why your equations are inconsistent. $\endgroup$
    – Bob Hanlon
    Jun 29, 2020 at 16:48

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