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I have a function in 2 variables $F(x,y)$, which also contains the constant parameters $\epsilon_{1}$, $\epsilon_{2}$,$\beta$. It is defined as:

F[x_,y_]:=(3 - 3 beta (e1 + e2) + 3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) + (-1 + beta e1) (-4 + beta (e1 + 3 e2)) Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y])

My goal is to find the solutions to the equation $F(x,y)=0$, i.e. the coordinates of the points (x,y) on the plane (as a function of the parameters $\epsilon_{1}$, $\epsilon_{2}$,$\beta$) where the function is 0. I'm looking for an analytic expression of these coordinates.

I've tried with

FindInstance[F[x,y] == 0, {x,  y}, Reals]

or something similar.

The problem is that it either returns me an error because of the presence of the parameters $\epsilon_{1}$, $\epsilon_{2}$,$\beta$, or it gives me a solution for these parameters as well, treating them as variables just like x,y.

So how can I find the solutions?

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4 Answers 4

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Let us do the following:

Step 1:

F[x_, y_] := (3 - 3 beta (e1 + e2) + 
   3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) + (-1 + beta e1) (-4 + 
      beta (e1 + 3 e2)) Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 
   1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y])


eq1 = F[x, y] == 0

Step 2:

eq2 = eq1 /. Cos[(3 x)/2] -> z

(*   3 - 3 beta (e1 + e2) + 
  3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) + (-1 + beta e1) (-4 + 
     beta (e1 + 3 e2)) z Cos[(Sqrt[3] y)/2] + 
  1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y] == 0  *)

Step 3:

eq3 = Equal @@ Solve[eq2, z][[1, 1]] /. z -> Cos[(3 x)/2] 

(*  Cos[(3 x)/
  2] == ((-3 + 3 beta (e1 + e2) - 
    3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) - 
    1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]) Sec[(Sqrt[3] y)/
   2])/((-1 + beta e1) (-4 + beta (e1 + 3 e2)))  *)

Step 4:

eq4 = Map[ArcCos, eq3] /. ArcCos[Cos[t_]] :> t

(*  (3 x)/2 == 
 ArcCos[((-3 + 3 beta (e1 + e2) - 
     3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) - 
     1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]) Sec[(Sqrt[3] y)/
    2])/((-1 + beta e1) (-4 + beta (e1 + 3 e2)))]  *)

Last step:

Solve[eq4, x][[1, 1]]

(*   x -> 2/3 ArcCos[((-3 + 3 beta (e1 + e2) - 
      3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) - 
      1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]) Sec[(Sqrt[3] y)/
     2])/((-1 + beta e1) (-4 + beta (e1 + 3 e2)))]   *)

Done. Have fun!

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Solve[F[x, y] == 0, x]

Gives

{{x -> ConditionalExpression[
    2/3 (-ArcCos[((-3 + 3 beta (e1 + e2) - 
            3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) - 
            1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]) Sec[(
           Sqrt[3] y)/2])/((-1 + beta e1) (-4 + beta (e1 + 3 e2)))] + 
       2 \[Pi] ConditionalExpression[1, \[Placeholder]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] 
     Integers]}, {x -> 
   ConditionalExpression[
    2/3 (ArcCos[((-3 + 3 beta (e1 + e2) - 
           3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) - 
           1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]) Sec[(
          Sqrt[3] y)/2])/((-1 + beta e1) (-4 + beta (e1 + 3 e2)))] + 
       2 \[Pi] ConditionalExpression[1, \[Placeholder]]), 
    ConditionalExpression[1, \[Placeholder]] \[Element] Integers]}}

Notice the two solutions with + or - in front of the ArcCos[]
So your {x,y} points will be: enter image description here Where C\[1\] is an integer.

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sol = FindInstance[F[x, y] == 0, {x, y, e1, e2, beta}, Reals, 2]
F[x, y] == 0 /. sol // Simplify

{True, True}.

And the Reduce gave a complex expression.

(* Reduce[F[x, y] == 0, {x, y}, Reals] *)

Solve[F[x, y]== 0, {x, y}, Reals, Method -> Reduce]

period = FunctionPeriod[F[x, y], {x, y}]
sol = Solve[{F[x, y] == 0, 0 <= x <= period[[1]], 
   0 <= y <= period[[2]]}, {x, y}, Reals, Method -> Reduce]
F[x, y] == 0 /. sol // FullSimplify
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  • $\begingroup$ if I try this I get a solution also for the parameters e1, e2 and beta. But I only want to solve for the x,y variables and the solution for them will contain e1,e2,beta. $\endgroup$
    – Mathew
    Aug 25, 2022 at 16:19
  • $\begingroup$ Try: Solve[F[x, y] == 0, y] of Solve[F[x,y]==0,x] Do not specify "Reals" as without more knowledge about the parameters, real solutions can not be garanteed. $\endgroup$ Aug 25, 2022 at 16:47
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Regarding only real variables and parameters. Due to Cos multiple solutions represented by C1,C2

F[x_, y_] = (3 - 3 beta (e1 + e2) + 
    3/8 beta^2 (3 e1^2 + 2 e1 e2 + 3 e2^2) + (-1 + beta e1) (-4 + 
       beta (e1 + 3 e2)) Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 
    1/8 (-4 + beta (e1 + 3 e2))^2 Cos[Sqrt[3] y]);

red3 = Reduce[F[x, y] == 0, {x, y}, Reals];

StandardForm[
 red3 //. Or -> 
   Composition[(Column[#, Right, Background -> {{White, LightGray}}, 
       Frame -> All] &), List]]

enter image description here

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