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Problem Statement

Pplane produces this phase plane portrait:

pplane raw

Note the yellow line which represents the locus where $\dot{y}=0$.

My attempts to recreate this plot in Mathematica have failed.

mm

The vexatious piece is boxed below.

vex


Question

How to plot the $y$ nullcline?


Best effort

The differential equations are

xdot=4y^3+(x+1)-(x+1)((x+1)^2+y^4);

ydot=-2(x+1)+2y^3-2y^3 ((x+1)^2+y^4);

The nullcline components are computed using

Solve[ydot == 0, y];
nully = y /. %

{Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 1], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 2], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 3], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 4], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 5], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 6], 
 Root[1 + x + (2 x + x^2) #1^3 + #1^7 &, 7]}

The problematic component is the first element:

trouble

The other elements were successful:

gya = Plot[nully[[2]], {x, -\[Lambda], \[Lambda]},PlotStyle -> {{Red, Thick}}, PlotPoints -> 1500];

gyc = Plot[nully[[1]], {x, -1.265, \[Lambda]},PlotStyle -> {{Red, Thick}}, PlotPoints -> 1500];

Show[{gya, gyc}, PlotRange -> All]

aandc

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  • 3
    $\begingroup$ ContourPlot[-2 (x + 1) + 2 y^3 - 2 y^3 ((x + 1)^2 + y^4) == 0, {x, -4, 4}, {y, -4, 4}] $\endgroup$ – Alan Jun 10 '17 at 0:13
  • $\begingroup$ @Alan. Wonderful. Intuitive and much simpler. Please promote this to an answer for up voting. $\endgroup$ – dantopa Jun 10 '17 at 0:16
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    $\begingroup$ The answer in the proposed duplicate uses ContourPlot to plot the nullclines. This is what Alan also suggested here. I have not tried it, but you indicated that it also works for your, less well behaved, system. If the solution is the same, then I think it is fair to say that they are duplicates. If you disagree, please state why you are dissatisfied with ContourPlot. $\endgroup$ – C. E. Jun 10 '17 at 10:53
  • $\begingroup$ @ C. E.: You are entirely correct. An up vote to you. $\endgroup$ – dantopa Jun 10 '17 at 10:56