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I have the following variables

p = (Γ + κ1 + κ2)/2;
u1 = 36 g1^2 (-2 p + 3 κ2) + (36 g2^2 + (2 p - 3 κ1) (2 p - 3 κ2)) (4 p - 3 (κ1 + κ2));
u2 = 2^(2/3)*(12 g1^2 + 12 g2^2 - 4 p^2 + 6 p (κ1 + κ2) - 3 (κ1^2 + κ1 κ2 + κ2^2));
sig = ((u1 + Sqrt[u1^2 + u2^3])/16)^(1/3);

and I define my functions as

o1 = 1/3*(p + sig - u2/(2^(8/3)*sig) - 3*I*ω);
o2 = 1/3*(p - 1/2*(sig - u2/(2^(8/3)*sig)) + (I*Sqrt[3])/2*(sig + u2/(2^(8/3)*sig)) - 3*I*ω);
o3 = 1/3*(p - 1/2*(sig - u2/(2^(8/3)*sig)) - (I*Sqrt[3])/2*(sig + u2/(2^(8/3)*sig)) - 3*I*ω);

Now, I wish to find the common value of g1 for which 'o1 = o2 = o3'. I do the following

Solve[o1==o2==o3,g1]//Simplify

And I'm returned with two values of such g1

{{g1 -> -(Sqrt[-12 g2^2 + Γ^2 + κ1^2 - κ1κ2 + κ2^2 - Γ (κ1 + κ2)]/(2 Sqrt[3]))}, 
{g1 -> Sqrt[-12 g2^2 + Γ^2 + κ1^2 - κ1κ2 + κ2^2 - Γ (κ1 + κ2)]/(2 Sqrt[3])}}

But upon checking by substituting them back into o1, o2 and o3 (for the first common g1), they do not yield the same result

tei1 = (o1 /. {g1 -> -(Sqrt[-12 g2^2 + Γ^2 + κ1^2 - κ1κ2 + κ2^2 - Γ (κ1 + κ2)]/(2 Sqrt[3]))}) // FullSimplify
tei2 = o2 /. {g1 -> -(Sqrt[-12 g2^2 + Γ^2 + κ1^2 - κ1κ2 + κ2^2 - Γ (κ1 + κ2)]/(2 Sqrt[3]))} // FullSimplify

Doing

(tei1-tei2)//FullSimplify

Gives

((3 - I Sqrt[3]) (-(Γ + κ1 - 2 κ2)^3 +108 g2^2 (Γ - κ2) + Sqrt[((Γ + κ1 - 2κ2)^3 + 108 g2^2 (-Γ + κ2))^2])^(1/3))/(12 2^(1/3))

Which is not 0. Clearly if o1 and o2 are common at that g1 value, subtracting them from one another should be zero. But clearly, that's not the case. I've also tried this for o2 and o3, and o1 and o3. But subtracting them from one another do not yield zero. What is the issue here?

Thanks!

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When I check Reduce[o1==o2==o3], I get False. I think equations are equal to each other only for possibly isolated values.

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  • $\begingroup$ Thanks but do you have any means of finding those values? What do you mean by isolated? $\endgroup$ – kowalski Sep 4 '18 at 20:17
  • $\begingroup$ I meant that it seems like there is no "generic" formula for g1, hence it may not be possible to write it as a function in terms of the remaining independent variables $g2, \Gamma, \kappa1, \kappa2, \Omega$. But this does not mean that there are no set of numbers for $g1, g2, \Gamma, \kappa1, \kappa2, \Omega$ for which the equations are satisfied; that is what I meant by isolated values. The straightforward method that comes to my mind is to use FindInstance. $\endgroup$ – Soner Sep 4 '18 at 22:07
  • $\begingroup$ Still, I am not sure such isolated solutions exist either! Naively I would expect Reduce to capture them, so the fact that Reduce yields False means they should not exist either! Another way to see this is to use MaxExtraConditions -> All option in Solve: You see that you no longer get any solution set! This generic/nongeneric solution issue is discussed in Possible Issues section of the documentation of Solve, you can get more information there. $\endgroup$ – Soner Sep 4 '18 at 22:21

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