I have a problem with making contour plots of the form $f(x,y,c)=0$ work well with Mathematica, where c defines the levelset. For example the lemniscate $(x^2 + y^2)^2 - c (x^2 - y^2) =0$. Obviously rewriting this as $\frac{(x^2 + y^2)^2}{(x^2 - y^2)} =c$ is going to cause some problems for anywhere $x^2=y^2$. If I try to plot multiple curves

ContourPlot[Evaluate@Table[(x^2 + y^2)^2 - m (x^2 - y^2) == 0, {m, 0,4}],
{x, -2, 2}, {y, -2, 2}]

this works but I lose the nice shading in between curves and hover over information about the level sets as seen in this example:

ContourPlot[x^2 + y^2, {x, -2, 2}, {y, -2, 2}]

The best I could do is the following:

lemniscates = 
  RegionPlot[{(x^2 + y^2)^2 - m (x^2 - y^2) < 0}, {x, -2, 
     2}, {y, -2, 2}, PerformanceGoal -> "Quality", 
    BoundaryStyle -> None, 
    PlotStyle -> {ColorData[{"RedBlueTones", {0, 7}}][m]}], {m, 0, 
 Prolog -> {ColorData[{"RedBlueTones", {0, 7}}][7], 
   Rectangle[Scaled[{0, 0}], Scaled[{1, 1}]]}]

but this looks awful compared to the default contour plotting and I still lose the level set information. How do I get my lemniscate plots to look like the ContourPlot of the circles example?

Thank you!

EDIT: It seems I used a bit too simple an example with the lemniscates to illustrate my problem. Here is another similar function:

 Evaluate@Table[(x^2 + m y^2)^2 - m (x^2 - y^2) == 0, {m, 0, 
    4}], {x, -2, 2}, {y, -2, 2}]

But now solving for m gives you two roots so this will not be easy to put in the required form. What can be done?


Generate separate ContourPlot for each m with contour 0 and change the tooltip label to m and combine all with Show:

Show @ Table[ContourPlot[Evaluate[(x^2 + y^2)^2 - m (x^2 - y^2)], {x, -2, 2}, {y, -2, 2}, 
    ContourStyle -> Directive[Thick, ColorData[63][m]], 
    ContourShading -> {Opacity[.2, ColorData[63][m]], White}, 
    Contours -> {{0}}] /. Tooltip[a_, _] :> Tooltip[a, m], {m, 0, 4}]

enter image description here

Generate the plots in the opposite order to get opaque colors without blending:

colors = {Cyan, Red, Blue, Orange, Green};
Show @ Table[ContourPlot[(x^2 + y^2)^2 - m (x^2 - y^2) , {x, -2, 2}, {y, -2, 2}, 
    ContourStyle -> Directive[Thick, Darker@Darker@colors[[m + 1]]], 
    ContourShading -> {Opacity[1, colors[[m + 1]]], None}, 
    Contours -> {{0}}] /. Tooltip[a_, _] :> Tooltip[a, m], 
  {m, 4, 0, -1}]

enter image description here

  • $\begingroup$ This looks the most promising! I did not know about Tooltips. It even works with the update to my question! The only problem is the hover over only works for the contour m=4 on my Mathematica. Do you know whats the matter? $\endgroup$ – Takoda Sep 2 at 16:16
  • $\begingroup$ @Takoda, I see all 4 tooltips when I hover over the contour lines both in v9 (windows10) and v12 (wolfram cloud). $\endgroup$ – kglr Sep 2 at 16:25
  • $\begingroup$ Ok its probably just a problem on my system. Thanks for the excellent answer! $\endgroup$ – Takoda Sep 2 at 16:31
  • $\begingroup$ @Takoda, my pleasure. Thank you for the accept. $\endgroup$ – kglr Sep 2 at 16:33

If you can tolerate solving for $1/m$ rather than $m$ (as $m=0$ doesn't given you real numbers), you could use the following:

ContourPlot[(x^2 - y^2)/(x^2 + y^2)^2, {x, -2, 2}, {y, -2, 2},
 Contours -> {1, 1/2, 1/3, 1/4}, PlotRange -> All,
 ContourShading -> {White, Red, Blue, Green, Cyan}]


Maybe a more interesting display might be with ContourPlot3D which also allows your actual form of $f(x,y,c)=0$:

ContourPlot3D[(x^2 + y^2)^2 - m (x^2 - y^2) == 0, {x, -2, 2}, {y, -2, 2}, {m, 0, 4}, 
  PlotRange -> All, PlotPoints -> 100]

ContourPlot3D example

For your more complex example, I think it becomes more important to consider ContourPlot3D:

ContourPlot3D for more complicated example

  • $\begingroup$ This is again a very smart answer and I like it, except that the problem I'm really interested in has a very complicated $f(x,y,c)=0$ that cannot be put in the form $g(x,y)=c$. I used the lemniscates as a minimum non-working example but obviously it was a bit too minimum non-working. $\endgroup$ – Takoda Sep 2 at 16:03
  • $\begingroup$ Even for your "a bit too minimal" example, wouldn't `ContourPlot3D be more interesting? $\endgroup$ – JimB Sep 2 at 17:23

To get ContourPlot to do what you seek you need to solve for the parameter in the form $m = F(x,y)$; then plot $F(x,y)$ with the desired contours $m$. However, in this OP's case, this causes some numerical problems, which can be alleviated with the following dodge:

 (x^2 + y^2)^2/($MachineEpsilon + Abs[x^2 - y^2]),
 {x, -2, 2}, {y, -2, 2},
 Contours -> Table[m, {m, 0, 4}], 
 RegionFunction -> 
  Function[{x, y, z}, x^2 != y^2 && 0 <= (x^2 + y^2)^2/(x^2 - y^2) < 5], 
 MaxRecursion -> 3, PlotPoints -> 25]

enter image description here

  • $\begingroup$ Thank you for your answer, unfortunately this dodge is very specific to the exact case of the lemniscate. My actual problem is much more complicated with basically no hope of putting it in the form $f(x,y)=c$. I put the leminscate down as a minimal non-working example. Is there any hope? $\endgroup$ – Takoda Sep 2 at 16:02

This feels somewhat old-school, perhaps because of the expansion of *Plot* functions, I haven't used it in a while. I used to see it used, and used it myself, more often. Basically we project a 3D contour plot of $f(x,y,m)=0$ and project it onto the $xy$-plane.

Like all solutions, it assumes $m$ is a function $m=F(x,y)$ of $x$ and $y$, although unlike my other answer, one does not need to solve for $m$. (If $m$ is not a function of $x$ and $y$, contour regions will overlap where there are multiple values of $m$ for a given $(x,y)$.)

 ContourPlot3D[(x^2 + y^2)^2 - m (x^2 - y^2) == 0,
  {x, -2, 2}, {y, -2, 2},
  {m, 0, 4},
  MeshFunctions -> {#3 &}, Mesh -> {Range@4}, 
  MeshShading -> 
   Table[ColorData["M10DefaultDensityGradient"][m/4], {m, 0, 4}],
  PlotPoints -> 35],
 {Graphics3D[dir_, opts___] :>
    dir /. {GraphicsComplex[pts_, g_, rest___] :>
        pts /. {x_Real, y_Real, z_Real} :> {x, y},
        g /. Line[p_] :> Tooltip[Line[p], pts[[First@p, -1]]],
    PlotRange -> 
     Replace[PlotRange /. 
       FilterRules[{opts}, {PlotRange}], {x_, y_, z_} :> {x, y}],
    Frame -> True, Axes -> False,
    FilterRules[{opts}, Options@Graphics]]}

enter image description here

  • $\begingroup$ This should work for a general function as well! Thanks! $\endgroup$ – Takoda Sep 3 at 14:44

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