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# How to plot implicit function with intermediate variable

Posted 1 year ago
 Here, I intend to plot a 3D picture of with two implicit functions where x, y, t, \xi, \eta, are variables. Of course, these variables can be constrained to certain interval that will guarantee the function to be well-posed. How can I draw this one? Attachments:
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Posted 1 year ago
 Cross-posted in https://mathematica.stackexchange.com/q/261815/1871
Posted 1 year ago
 With : t=1; With[{t = 1}, ListPlot3D[ Flatten[Table[{x, y, func[\[Xi], \[Eta], t]} /. xyToXiEta[x, y, t], {x, -1, 1, 1/20}, {y, -1, 1, 1/20}], 2], BoxRatios -> {1, 1, 1}, Axes -> True, PlotRange -> {All, All, 2000 {-1, 1}}, AxesLabel -> {x, y, z}]] 
Posted 1 year ago
Posted 1 year ago
 Here is an attempt: func[\[Xi]_, \[Eta]_, t_] = -((1350. Sech[\[Eta]]^2 Tanh[\[Eta]] (1. Sech[ 0.1 t - \[Xi]]^2 Tanh[0.1 t - \[Xi]] + 0.6 Sech[0.3 t - \[Xi]]^2 Tanh[0.3 t - \[Xi]] + 1.6 Sech[0.1 t + \[Xi]]^2 Tanh[0.1 t + \[Xi]]))/((1 - 1.5 Sech[\[Eta]]^2) (1 - Sech[0.1 t - \[Xi]]^2 - 0.5 Sech[0.3 t - \[Xi]]^2 - 1.5 Sech[0.1 t + \[Xi]]^2) (0.3 Sech[\[Eta]]^2 + 0.5 Sech[0.1 t - \[Xi]]^2 + 0.3 Sech[0.3 t - \[Xi]]^2 - 0.8 Sech[0.1 t + \[Xi]]^2)^2)) // Rationalize // Simplify; xyToXiEta[x_?NumericQ, y_, t_] := NSolve[{x == \[Xi] - 1/2 Tanh[\[Xi] - 3/10 t] - Tanh[\[Xi] - 1/10 t] - 3/2 Tanh[\[Xi] + 1/10 t], y == \[Eta] - 3/2 Tanh[\[Eta]]}, {\[Xi], \[Eta]}, Reals]; With[{t = 1}, ListPointPlot3D[ Flatten[ Table[{x, y, func[\[Xi], \[Eta], t]} /. xyToXiEta[x, y, t], {x, -1, 1, 1/20}, {y, -1, 1, 1/20}], 2], BoxRatios -> {1, 1, 1}, Axes -> True, PlotRange -> {All, All, 2000 {-1, 1}}, AxesLabel -> {x, y, z}]] 
Posted 1 year ago
 Unfortunately, NSolve[{x == \[Xi] - 1/2 Tanh[\[Xi] - 3/10 t] - Tanh[\[Xi] - 1/10 t] - 3/2 Tanh[\[Xi] + 1/10 t], y == \[Eta] - 3/2 Tanh[\[Eta]]}, {\[Xi], \[Eta]}, Reals] does not function.
 When t=0` the function you want to plot is identically zero.