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I am trying to solve $$ \left(\eta ^2-1\right) \cosh ^2(\beta \eta )+2 J^2=2 \eta (J \sinh (\beta \eta ) \sinh (\beta J)+\eta \cosh (\beta \eta ) \cosh (\beta J)) $$ for $\eta$. Where, $\beta, J$ are positive and reals, and $B$ is real. I used

Reduce[2 J^2 + (η^2 - 1) Cosh[ β η]^2 == 
   2 η (η Cosh[J β] Cosh[β η] + 
      J Sinh[J β] Sinh[β η]) && η > 0 && 
  J > 0 && β > 0, η, Reals]

with no answers.

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ContourPlot3D shows the solution in parameterspace:

ContourPlot3D[2 J^2 + (\[Eta]^2 - 1) Cosh[\[Beta] \[Eta]]^2 ==2 \[Eta] (\[Eta] Cosh[J \[Beta]] Cosh[\[Beta] \[Eta]] +J Sinh[J \[Beta]] Sinh[\[Beta] \[Eta]]), {J, 0, 3}, {\[Beta], 0, 5}, {\[Eta], 0, 5}, AxesLabel -> Automatic ]

There are two solutions for J>1/Sqrt[2] and one for 0<J<1/Sqrt[2] enter image description here

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  • $\begingroup$ @yarchik The parameterspace {J, \[Beta], \[Eta]} is threedimansional, that's why I prefer ContourPlot3D . $\endgroup$ – Ulrich Neumann Jul 3 at 14:39
  • $\begingroup$ You are right. However, I was wondering how many solution branches are there. ContourPlot[(2 J^2 + (η^2 - 1) Cosh[ β η]^2 == 2 η (η Cosh[J β] Cosh[β η] + J Sinh[J β] Sinh[β η]))/.{J->1/Sqrt[2]}, {η,-100,100},{β,-100,100}] does not give me any. $\endgroup$ – yarchik Jul 3 at 14:45
  • $\begingroup$ It's necessary to evaluate the argument of ContourPlot. Try ContourPlot[ Evaluate[2 J^2 + (\[Eta]^2 - 1) Cosh[\[Beta] \[Eta]]^2 == 2 \[Eta] (\[Eta] Cosh[J \[Beta]] Cosh[\[Beta] \[Eta]] + J Sinh[J \[Beta]] Sinh[\[Beta] \[Eta]]) /. J -> 1/Sqrt[2]] , {\[Eta], 0, 10}, {\[Beta], 0, 5}, FrameLabel -> Automatic] , this gives one branch. $\endgroup$ – Ulrich Neumann Jul 3 at 15:07

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