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How can I solve $-2K\alpha^2\alpha^*+2\epsilon\alpha^*-i\frac{\kappa}{2}\alpha=0$ for $\alpha$?

The solution is supposed to be $\alpha=0$ and $\alpha=r_0e^{i\theta_0}$, where $r_0=\Big(\frac{4\epsilon^2-\kappa^2/4}{4K^2}\Big)^{1/4}$ and $\theta_0=\frac{1}{2}\tan^{-1}\Big(\frac{\kappa}{\sqrt{16\epsilon^2-\kappa^2}}\Big)$

I've tried to solve it using Reduce, Solve and NSolve but I only get the output

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

I don't know what is wrong? Here is my input.

Solve[-2 k \[Alpha]^2 Conjugate[\[Alpha]] + 
2 \[Epsilon] Conjugate[\[Alpha]] - (I \[Kappa] \[Alpha])/2 == 0, \[Alpha]]
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  • $\begingroup$ Do you mean SuperStar[[Alpha]] as Conjugate[[Alpha]] ? $\endgroup$
    – user64494
    Commented Jan 31, 2018 at 12:19
  • $\begingroup$ Yes I mean SuperStar[[Alpha]] as Conjugate[[Alpha]] $\endgroup$ Commented Jan 31, 2018 at 12:23

1 Answer 1

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One way is to use the polar form and split the equation into real and imaginary part:

    aux = -2 k \[Alpha]^2 Conjugate[\[Alpha]] + 2 \[Epsilon] Conjugate[\[Alpha]] - (I \[Kappa] \[Alpha])/2;
    expr = ComplexExpand[aux /. \[Alpha] -> a Exp[I t], 
  TargetFunctions -> {Re, Im}]

(* -2 a^3 k Cos[t] + 2 a \[Epsilon] Cos[t] + 1/2 a \[Kappa] Sin[t] +  I (-1/2) a \[Kappa] Cos[t] - 2 a^3 k Sin[t] - 2 a \[Epsilon] Sin[t]) *)

    eq = {Simplify[Re[expr], 
   Assumptions -> {(k | \[Epsilon] | \[Kappa] | a | t) \[Element] 
      Reals}], 
  Simplify[Im[expr], 
   Assumptions -> {(k | \[Epsilon] | \[Kappa] | a | t) \[Element] 
      Reals}]}
(* {(-2 a^3 k + 2 a \[Epsilon]) Cos[t] + 
  1/2 a \[Kappa] Sin[t], -(1/2)
    a (\[Kappa] Cos[t] + 4 (a^2 k + \[Epsilon]) Sin[t])} *)

The solutions are:

sol = Reduce[eq == {0, 0}, {a, t}]
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  • $\begingroup$ Hmm, if I follow your proposal I get a very long and strange answer. $\endgroup$ Commented Jan 31, 2018 at 14:01
  • $\begingroup$ You can try to add additional conditions, for instance Reduce[{eq == {0, 0}, \[Epsilon] > 0, k > 0, \[Kappa] > 0, a>=0, -Pi <= t < Pi}, {a, t}] and see if you get what you expected. I should have added the conditions a>=0, -Pi<=t<Pi since the start. $\endgroup$ Commented Jan 31, 2018 at 15:29

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