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I have the following two coupled non-linear equations

eqns = {0 == \[Alpha] (I*d - k/2 - I*g*\[Alpha] (b + Conjugate[b])) && 
   0 == I*w_m*b - I*g*Abs[\[Alpha]] - \[Gamma]/2*b}
Solve[eqns, {\[Alpha], b}]

Here, $\alpha$ and $b$ are my two complex variables I want to solve for. The other variables are constant, i.e. $d, k, g, \gamma$ are experimental constants.

Now, when I run this commmand, Solve gives the error message:

This system cannot be solved with the methods available to Solve.

Why is that?

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    $\begingroup$ You need a space between variable names. E.g.: "Ig[Alpha]". Further do not use capitalized variable names as they are reserved for the system. E.g. "D" is an derivative operator. $\endgroup$
    – Daniel Huber
    May 3, 2022 at 14:55
  • $\begingroup$ I edited the question accordingly. $\endgroup$
    – LionCereals
    May 3, 2022 at 14:57
  • $\begingroup$ Variable names should not include a pattern (_). Change w_m to wm. If you want it displayed as a subscript, use Format[wm] := Subscript[w, m]. Are the experimental constants assumed to be real? $\endgroup$
    – Bob Hanlon
    May 3, 2022 at 16:13
  • $\begingroup$ @BobHanlon Yes they are real. $\endgroup$
    – LionCereals
    May 3, 2022 at 16:35

1 Answer 1

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Let us somewhat rewrite eqns, making use of b + Conjugate[b]=2*Re[b] and Abs[\[Alpha]]=Sqrt[Re[\[Alpha]]^2 + Im[\[Alpha]]^2] and w=w_m.Then the following works.

eqns = {0 == \[Alpha] (I*d - k/2 - I*g*\[Alpha] *(2*Re[b])) && 
0 == I*w*b -  I*g*Sqrt[Re[\[Alpha]]^2 + Im[\[Alpha]]^2] - \[Gamma]/2*b};
f[a1_, a2_, a3_, a4_, a5_] := Solve[eqns /. {g -> a1, d -> a2, w -> a3, 
k -> a4, \[Gamma] -> a5}, {\[Alpha], b}]
f[1, Pi*I, 1, 2, -3]

{{\[Alpha] -> 0, b -> 0}, {\[Alpha] -> (I (169 + 338 \[Pi] + 169 \[Pi]^2)^(1/4))/( 2 Sqrt[2]), b -> ((2/13 + (3 I)/13) (169 + 338 \[Pi] + 169 \[Pi]^2)^(1/4))/Sqrt[ 2]}}

f[1.1,Pi*I,-I/2,2,-3]

{{\[Alpha] -> 0, b -> 0}}

It should be noticed Reduce is preferable, since Solve may use non-equivalent transforms.

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  • $\begingroup$ Thanks. As far as I understand, this gets you the solution for specific parameter values, which are provided as input to the function f. Is there a general solution as well? $\endgroup$
    – LionCereals
    May 4, 2022 at 9:01

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