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Let me try to be a bit schematic because my own expression is a bit complicated and could be not useful for future questions.

x=f[y*a[x]]

My problem is how to Plot y=y[x]. In principle, y is not a function but a parameter; but I need the curve y vs x

EDIT: Let me show you my functions

H=Sqrt[2α(1+2Sinh[Fk[α,k,H]]^2)-α^2-1] 

with

Fk[α,k,H]=-(((1+Sqrt[1 + 4 k^2 (1+(-1+H^2+α^2)^2/(4 H^2))]) (EllipticK[1/2 (1+1/Sqrt[1+4 k^2 (1+(-1+H^2 +α^2)^2/(4 H^2))])]-EllipticPi[1/2-1/(2 Sqrt[1 + 4 k^2 (1+(-1+H^2 +α^2)^2/(4 H^2))]),1/2(1+1/Sqrt[1+4 k^2 (1+(-1+H^2 + α^2)^2/(4 H^2))])]))/(2(1+4 k^2 (1+(-1+H^2+α^2)^2/(4 H^2)))^(1/4) Abs[k Sqrt[1+(-1+H^2+α^2)^2/(4 H^2)]]))

Sorry, I know it looks horrible. Thus, what I want is a plot k=k[H].

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  • $\begingroup$ does the following example ClearAll[a, f]; f[x_] := Sin[x]; a[x_] := x Cos[x]; ContourPlot[x == f[y a[x]], {x, -Pi, Pi}, {y, 0, 20}, PlotPoints -> 100] give what you need? $\endgroup$ – kglr Mar 15 '19 at 23:34
  • $\begingroup$ @kglr Thanks. I'm running it now for my case but it is taking a lot of time. I edited my question to put my explicit example. $\endgroup$ – Patrick El Pollo Mar 15 '19 at 23:59
  • $\begingroup$ what is the correspondence between {H, \[Alpha], k} and {y,a,x,f}? $\endgroup$ – kglr Mar 16 '19 at 0:07
  • $\begingroup$ @kglr you can fix \[Alpha] (sorry for not mention it), say \[Alpha]=0.8, it is not a problem. The correspondence is x=H, the huge square root would be f, a` the combination of elliptic integrals; and y=k $\endgroup$ – Patrick El Pollo Mar 16 '19 at 0:27
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ClearAll[k, H, f]
α = 4/5;
f[k_, H_] := -(((1 + Sqrt[1 + 4 k^2 (1 + (-1 + H^2 + α^2)^2/(4 H^2))]) 
 (EllipticK[1/2 (1 + 1/Sqrt[1 + 4 k^2 (1 + (-1 + H^2 + α^2)^2/(4 H^2))])] - 
  EllipticPi[1/2 - 1/(2 Sqrt[1 + 4 k^2 (1 + (-1 + H^2 + α^2)^2/(4 H^2))]), 
      1/2 (1 + 1/Sqrt[1 + 4 k^2 (1 + (-1 +  H^2 + α^2)^2/(4 H^2))])]))/
  (2 (1 + 4 k^2 (1 + (-1 + H^2 + α^2)^2/(4 H^2)))^(1/4) 
   Abs[k Sqrt[1 + (-1 + H^2 + α^2)^2/(4 H^2)]]));


Quiet @ ContourPlot[H == f[k, H], {H, 0, 1}, {k, -2, 2}, 
  PlotPoints -> 100, FrameLabel -> {"H", "k"}]

enter image description here

|improve this answer|||||
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  • $\begingroup$ Thanks a lot. I had totally forgotten ContourPlot. I made a mistake when defined my functions, please check the correct definitions now. I was using ContourPlot in this case but it is taking infinite time. Is there a way to improve how ContourPlot works? Also, and didnt mention, sorry, that k>0 $\endgroup$ – Patrick El Pollo Mar 18 '19 at 14:40
  • $\begingroup$ @resanrom, try (1) Clear all definitions using ClearAll[H, Fk, h, \[Alpha], k], (2) Use Fk[\[Alpha]_, k_, H_] := ... when you define Fk, then (3) use \[Alpha] = 3/4; ContourPlot[ Sqrt[2 \[Alpha] (1 + 2 Sinh[Fk[\[Alpha], k, H]]^2) - \[Alpha]^2 - 1] == Fk[\[Alpha], k, H], {H, 0, 1}, {k, -2, 2}, PlotPoints -> 100, FrameLabel -> {"H", "k"}] It is still slow but it eliminates the recursion limit issue and gives some output. $\endgroup$ – kglr Mar 18 '19 at 15:01

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