0
|
4532 Views
|
8 Replies
|
1 Total Likes
View groups...
Share
GROUPS:

# How to plot implicit function with intermediate variable

Posted 2 years 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:
8 Replies
Sort By:
Posted 2 years ago
 Cross-posted in https://mathematica.stackexchange.com/q/261815/1871
Posted 2 years 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 2 years ago
Posted 2 years 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 2 years 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.