0
$\begingroup$

I've been struggling to solve equations in two variables containing constant symbolic parameters. Take for example the function in two variables defined by:

F[x_,y_]:=Sqrt[2 t1^2 + t3^2 + 4 t1 t3 Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 2 t1^2 Cos[Sqrt[3] y]]

My goal is to solve $$F(x,y)=0$$ so I want to find the coordinates of the points in the 2D xy plane where the function is 0. These coordinates will be given by an expression containing the parameters $t_{1}$, $t_{3}$. I can find some solutions by guessing, for example: $$(x,y)=(0,\frac{2}{\sqrt{3}}\arccos{(-\frac{t_{3}}{2t_{1}})}) $$

or $$(x,y)=(0,-\frac{2}{\sqrt{3}}\arccos{(-\frac{t_{3}}{2t_{1}})}) $$

I would like to find these solutions (and the others) in Mathematica,so I've tried:

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

or

Reduce[F[x, y] == 0, {x, y}]

However my problem is that Mathematica gives me an expression where the solution for x contains y or viceversa, but I would like the solutions separate (so the solution for x shall not contain y and viceversa, as in the two I wrote above). Also, sometimes the parameters $t_{1}$,$t_{3}$ are treated as variables by Mathematica, and a solution is provided also for them, which I don't want.

This is a problem I had in general for these kind of equations where I solve for 2 variables and I also have additional constant parameters. So how can I get the solution for x and y (without them been coupled) in terms of the constant parameters?

$\endgroup$
4
  • $\begingroup$ Have a look at Solve. $\endgroup$ Aug 26, 2022 at 13:08
  • $\begingroup$ I've used that, but it gives me the problem I explained. $\endgroup$
    – Mathew
    Aug 26, 2022 at 14:04
  • 1
    $\begingroup$ There are quit a lot of solutions as Reduce shows: Reduce[F[x, y] == 0, {x, y}, Reals] $\endgroup$ Aug 26, 2022 at 15:04
  • $\begingroup$ With Solve use Solve[F[x, y] == 0, {x, y}, Reals, Method -> Reduce] $\endgroup$
    – Bob Hanlon
    Aug 26, 2022 at 15:38

1 Answer 1

1
$\begingroup$

Solve for Reals with Method->Reduce like @BobHanlon proposes.

Plot solutions for x and y in dependence of t1 and t3, or get values at definite t1,t3.

F[x_, y_] = 
  Sqrt[2 t1^2 + t3^2 + 4 t1 t3 Cos[(3 x)/2] Cos[(Sqrt[3] y)/2] + 
    2 t1^2 Cos[Sqrt[3] y]]

sol3[c1_, c2_, t1_, t3_] = 
 Solve[F[x, y] == 0, {x, y}, Reals, Method -> Reduce] /. {C[1] -> c1, 
   C[2] -> c2}

Plot3D[Evaluate[y /. sol3[0, 0, t1, t3] // Simplify // Union], {t1, 0,
   1}, {t3, 0, 1}, PlotStyle -> {Red, Green, Blue, Magenta}, 
 PlotRange -> All]

{x, y} /. sol3[0, 0, 1/3, 1/2] // TableForm
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.