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I have this equation but it takes so long and no answer comes out. Could you please suggest me what to do or should I just wait more. Even if you could help me understand different numerical solves, that would be great. Note that want it to be solved with respect to q and other variables are positive and from 0 to infinity.

Sqrt[1 + q^2 + w^2] ((q^2 + w^2) Subscript[\[Alpha], 
 1] - (l^2 + q^2 + w^2) (-1 + Subscript[\[Alpha], 2] - 
   Subscript[\[Alpha], 3])) (Subscript[C, 
 3333] (l^2 + (q^2 + w^2) (1 + Subscript[\[Alpha], 1]) - (1 + 
      q^2 + w^2) Subscript[\[Alpha], 2]) - (q^2 + w^2) Subscript[
 C, 1133] Subscript[\[Alpha], 3]) - Sqrt[l^2 + q^2 + w^2] (Subscript[C, 
 3333] ((q^2 + w^2) Subscript[\[Alpha], 
    1] - (l^2 + q^2 + w^2) (-1 + Subscript[\[Alpha], 2])) - (q^2 +
    w^2) Subscript[C, 1133] Subscript[\[Alpha], 
 3]) (l^2 + (q^2 + w^2) (1 + Subscript[\[Alpha], 1]) + (1 + q^2 + 
   w^2) (-Subscript[\[Alpha], 2] + Subscript[\[Alpha], 3]))
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  • $\begingroup$ Welcome to Mathematica StackExchange! Do you want to have a general symbolic result, or do you know the values of the parameters and just want a numerical result? $\endgroup$
    – Domen
    Mar 11, 2023 at 16:42
  • 3
    $\begingroup$ The symbols C and K are used by Mathematica and should be avoided. It is also better to avoid using subscripts. Indexed variables are better (you can display indexed variables as subscripts using Format). $\endgroup$
    – Bob Hanlon
    Mar 11, 2023 at 16:56
  • $\begingroup$ If you look carefully you should find that your expression is not just up to q^4, it looks like it has has even powers up to q^6. I saw that by using Expand on your expression. It might make your problem slightly easier if you replace all q^(2n) by newq^n and solve for newq and then look at the two square roots of newq. But none of that is going to help deal with the Sqrt[l^2+q^2+w^2] and Sqrt[1+q^2+w^2] that will complicate finding any "simple" solution in q.. $\endgroup$
    – Bill
    Mar 12, 2023 at 3:58
  • $\begingroup$ @Bill thank you for your asnwer. I get what you mean. yes it's q^6. i used Expand and Factor and tried even Collect for some parts but as you said, Sqrt[l^2+q^2+w^2] and Sqrt[1+q^2+w^2] are my real problems which doesn't let me have an answer. I don't know it can help but for Alpha 1 to 3 (which are constants), I have them and w changes from 0 to infinity and q is what i want. This expression has real and imaginary parts as well but it can not be shown in what i wrote. i want to try numerical ways to calculate it in order to see where it can make roots. could you help me on that? $\endgroup$
    – Ali Taher
    Mar 12, 2023 at 9:34
  • $\begingroup$ @Bill maybe even using the absolute value of the function helps better to solve numerically. however, if you can help me with an example or a link of video or anything, i would really apreciate $\endgroup$
    – Ali Taher
    Mar 12, 2023 at 9:35

1 Answer 1

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Use Solve.

expr = -Sqrt[
      l^2 + q^2 + 
       w^2] (Subscript[c, 
        3333] ((q^2 + w^2) Subscript[α, 
           1] - (l^2 + q^2 + w^2) (-1 + 
            Subscript[k, d]^2 Subscript[α, 2])) - (q^2 + 
         w^2) Subscript[c, 1133] Subscript[α, 
        3]) (l^2 + (q^2 + w^2) (1 + Subscript[α, 1]) + (1 + 
         q^2 + w^2) Subscript[k, d]^2 (-Subscript[α, 2] + 
         Subscript[α, 3])) + 
   Sqrt[1 + q^2 + 
      w^2] (Subscript[c, 
        3333] (l^2 + (q^2 + w^2) (1 + Subscript[α, 1]) - (1 + 
            q^2 + w^2) Subscript[k, d]^2 Subscript[α, 
           2]) - (q^2 + w^2) Subscript[c, 1133] Subscript[α, 
        3]) (l^2 + (q^2 + w^2) (l^2 + q^2 + w^2) Subscript[k, 
         d]^2 (1 + Subscript[α, 1]) (-Subscript[α, 2] + 
         Subscript[α, 3]));
params = {Subscript[α, 1] -> 3.5, 
   Subscript[α, 2] -> 1,  Subscript[α, 3] -> 1.75, 
   l -> 1.25, Subscript[c, 1133] ->  0.75, 
   Subscript[c, 3333] -> 0.167, Subscript[k, d] -> 1};

sol = q /. Solve[expr == 0 /. params, q, Reals] // Quiet;

Plot[sol, {w, 0, 1}, AspectRatio -> Automatic, 
 AxesLabel -> {w, q}] // Quiet

Plot

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  • $\begingroup$ thank you very much. Is it possible to plot it from 0<w<infinity as well? $\endgroup$
    – Ali Taher
    Mar 12, 2023 at 18:40
  • $\begingroup$ Can you please write down the value of parameters ($\alpha_1, \alpha_2, \alpha_3, C_{3333}, K_d, \dots$) so that I can get a realistic plot? $\endgroup$
    – Domen
    Mar 12, 2023 at 19:13
  • $\begingroup$ Yes. Subscript[[Alpha], 1] -> 3.5, Subscript[[Alpha], 2] -> 1, \ Subscript[[Alpha], 3] -> 1.75, l -> 1.25, Subscript[C, 1133] -> \ 0.75, Subscript[C, 3333] -> 0.167 and let me make the expr a bit simple without K_d. I also have 0<w<infinity but seems this condition is not entirely true. I have updated the question part. $\endgroup$
    – Ali Taher
    Mar 13, 2023 at 6:24
  • $\begingroup$ @AliTaher, is solution q supposed to be real? Because given your parameters, I only get complex solutions ... $\endgroup$
    – Domen
    Mar 13, 2023 at 11:18
  • $\begingroup$ yes it is complex. it came from fourier integrals. i just wrote the results here which i wanted to evaluate numerically. $\endgroup$
    – Ali Taher
    Mar 14, 2023 at 9:22

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