# Compare Mathematica with MATLAB by plotting the same equation

Posted 19 days ago
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 Firstly, the MATLAB programme is t = 0; A = 2; c1 = 1; c2 = 1; c = 1/25; omega = 2; [xi,eta]=meshgrid(-5:0.1:5,-4:0.1:6); P = ((2*sinh(xi+omega*t))./(3*cosh(xi+omega*t)))+((5*sinh(xi+omega*t))./(6*cosh(xi+omega*t)).^3)+0.9; Q = -(sinh(eta))./(cosh(eta)); x = xi-2.5*tanh(xi+omega*t); y = eta; Px = -(sech(xi+omega*t)).^2; Qy = -(sech(eta)).^2; u = ((A-c1.*c2).*Px.*Qy)./(1+c1.*P+c2.*Q+c.*P.*Q).^2; view([-45 70]); surf(x,y,u); xlabel('x','FontSize',26,'FontName','Times New Roman','FontAngle', 'italic') ylabel('y','FontSize',26,'FontName','Times New Roman','FontAngle', 'italic') zlabel('$u$','interpreter','latex','FontSize',24,'FontName','Times New Roman','FontAngle', 'italic') and it comesWith the same equation, the Mathematica program is \[Omega] = 2; A = 2; Subscript[c, 1] = Subscript[c, 2] = 1; Subscript[c, 3] = 1/25; px = -Sech[\[Xi] + \[Omega] t]^2; p = (2 Sinh[\[Xi] + \[Omega] t])/( 3 Cosh[\[Xi] + \[Omega] t]) + (5 Sinh[\[Xi] + \[Omega] t])/( 6 Cosh[\[Xi] + \[Omega] t]^3) + 0.9; qy = -Sech[\[Eta]]^2; q = -(Sinh[\[Eta]]/Cosh[\[Eta]]); func[\[Xi]_, \[Eta]_, t_] = ((A - Subscript[c, 1] Subscript[c, 2]) px qy)/(1 + Subscript[c, 1] p + Subscript[c, 2] q + Subscript[c, 3] p q)^2 // Rationalize // Simplify; xyToXiEta[x_, y_, t_] := NSolve[{x == \[Xi] - 2.5 Tanh[\[Xi] + 2 t], y == \[Eta]}, {\[Xi], \[Eta]}, Reals]; With[{t = 0}, ListPlot3D[ Flatten[Table[{x, y, func[\[Xi], \[Eta], t]} /. xyToXiEta[x, y, t], {x, -2, 2, 0.04}, {y, -1, 5, 0.04}], 2], Axes -> True, PlotRange -> {All, All, {0, 2}}, AxesLabel -> {x, y, z}, ColorFunction -> "TemperatureMap"]] and it comesThanks Gianluca Gorni for providing the Mathematica program.It is obvious that the MATLAB picture forms like a bridge with empty space under it. However, the Mathematica picture is solid. How can I improve the Mathematica program to make it like MATLAB's result? Attachments:
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Posted 18 days ago
 A direct translation of the code is not possible as that function does not exist (ListParametricPlot3D or so). But we can recreate the graphic "manually" by recreating the polygons, or rewrite the code to use ParametricPlot3D (preferred and nicer), I included both:
Posted 18 days ago
Posted 18 days ago
 Cross-posted in https://mathematica.stackexchange.com/q/261847/1871 Please include the link(s) of other version(s) of your question when cross-posting to avoid repetive work.
Posted 18 days ago
 Thank you for your notification, I will pay attention to this matter.
 It is much easier to bypass x,y and plot directly with respect to xi,eta: With[{t = 0}, With[{x = \[Xi] - 2.5 Tanh[\[Xi] + 2 t], y = \[Eta]}, ParametricPlot3D[{x, y, func[\[Xi], \[Eta], t]}, {\[Xi], -1, 2}, {\[Eta], -1, 4}, PlotRange -> All, ColorFunction -> "TemperatureMap"]]]