3
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Here is my todo list and what i've done:

  1. Using FrameTicks on specific points only (failed).
  2. Plotting six graphs with Table on ParametricPlot (Success).
  3. Marking each graph with Mesh "independently" (have no idea, but have the related topic, See 8).
  4. Coloring each graph (failed).
  5. Legending each graph (works but only the first graph).

I will explain the point $(3)$ with this sketch (simplification drawing):

enter image description here

The distance between two marks based on the codomain should have the same length.

Here is my attempt without different marks:

M = 5;
markers = {{"*", 25}, {"@", 10}, {"$", 15}, {"#", 10}, {"&", 
    10}, {"+", 25}};
mesh = {Most@Subdivide[-2, 2.8, 5]};
eqn = Table[
   2 M/t (\[Theta] Cot[\[Theta]] + I  \[Theta]), {t, {9, 19, 24, 29, 
     39, 49}}];
ticks = {{{-5, -2}, -4, -3, -2, -1, 0, 1}, {-Pi, Pi}, {}, {}};
graph = ParametricPlot[ReIm@eqn, {\[Theta], -3.14, 3.14}, 
  AspectRatio -> 1, 
  FrameLabel -> {"Re(s(\[Theta]))", "Im(s(\[Theta]))"}, 
  Frame -> True, FrameTicks -> ticks, PlotStyle -> Thick, 
  Axes -> False, MeshFunctions -> {#2 &}, Mesh -> mesh, 
  MeshStyle -> PointSize[Large], 
  PlotLegends -> 
   Placed[LineLegend[{"t=9", "t=19", "t=24", "t=29", "t=39", "t=49"}, 
     LegendFunction -> "Frame"], {.85, .8}]]

The Output:

enter image description here

The markings still sharing the same scale and not evaluated on each graph, has no coloring, FrameTicks doesn't work, and the Legend only shows $1$ graph.

Meanwhile, when i try to add the different markings, it becomes messy and there's an error message

M = 5;
markers = {{"*", 25}, {"@", 10}, {"$", 15}, {"#", 10}, {"&", 
    10}, {"+", 25}};
mesh = {Most@Subdivide[-2, 2.8, 5]};
eqn = Table[
   2 M/t (\[Theta] Cot[\[Theta]] + I  \[Theta]), {t, {9, 19, 24, 29, 
     39, 49}}];
ticks = {{{-5, -2}, -4, -3, -2, -1, 0, 1}, {-Pi, Pi}, {}, {}};
graph = ParametricPlot[ReIm@eqn, {\[Theta], -3.14, 3.14}, 
   AspectRatio -> 1, 
   FrameLabel -> {"Re(s(\[Theta]))", "Im(s(\[Theta]))"}, 
   Frame -> True, FrameTicks -> ticks, PlotStyle -> Thick, 
   Axes -> False, MeshFunctions -> {#2 &}, Mesh -> mesh, 
   MeshStyle -> PointSize[Large], 
   PlotLegends -> 
    Placed[LineLegend[{"t=9", "t=19", "t=24", "t=29", "t=39", "t=49"},
       LegendFunction -> "Frame"], {.85, .8}]];
meshStyles = 
  Association[
   Join @@ Cases[
     graph, {___, Directive[___, c_?ColorQ, ___], Line[x_]} :> 
      Thread[x -> c], All]];
styleToMarkers = 
  AssociationThread[ColorData[97] /@ Range[3], Style @@@ markers];
graph /. Point[
   x_] :> ({meshStyles@#, Text[styleToMarkers[meshStyles@#], #]} & /@ 
    x)

Output:

enter image description here

And i wish i have something like this:

enter image description here

Hope my question isn't too much and you want to help me. Thanks in advance!

Here is my references:

  1. parametricplot-table-of-complex-functions-in-several-graphs
  2. coloring-parametricplot-of-several-complex-functions
  3. color-parametric-plot-by-parameter
  4. marking-a-continuous-plot-with-the-same-subinterval-based-on-the-range-of-the-co
  5. how-do-i-reverse-the-axis-in-parametricplot
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1 Answer 1

2
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1. Wrap the first argument of ParametricPlot with Evaluate to get the 6 curves treated as separate curves so that each has its own style end legend entry.

2. Add LegendMarkers -> markers in LineLegend[...]

3. Replace 3 in the definition of styleToMarkers with Length @ markers.

4. Define ticks as ticks = {{{-Pi, Pi}, Automatic}, {Range[-5, 1], Automatic}};

5. Add the option Exclusions -> None: This makes each curve a single line rather than two lines, and ensures that meshStyles works without modification.

6. Use the options MeshFunctions -> {"ArcLength"} and Mesh -> 5 to add 5 markers that divide each curve into 6 equal-length pieces.

7. Add the option RegionFunction -> (-5 <= # <= 2 &): This ensures that the underlying curves produced by the kernel do not extend to the left of -5. (PlotRange does not guarantee this.) Without this option, MeshFunctions -> {"ArcLength"} will not give the desired result.

With all these changes:

graph = ParametricPlot[Evaluate[ReIm@eqn], {θ, -3.14, 3.14},
   PlotStyle -> Thick, AspectRatio -> 1, 
   FrameLabel -> {"Re(s(θ))", "Im(s(θ))"},
   Frame -> True, FrameTicks -> ticks, Axes -> False, 
   ImageSize -> Large,
   MeshStyle -> PointSize[Large],
   MeshFunctions -> {"ArcLength"},
   Mesh -> 5,
   Exclusions -> None,
   RegionFunction -> (-5 <= # <= 2 &),
   PlotRange -> {{-5, 2}, Automatic},
   PlotRangePadding -> Scaled[.02],
   PlotLegends -> Placed[LineLegend[{"t=9", "t=19", "t=24", "t=29", "t=39", "t=49"},
       LegendFunction -> "Frame", LegendMarkers -> markers], {.85, .8}]];

meshStyles = Association[Join @@ Cases[graph, 
   {___, Directive[___, c_?ColorQ, ___], Line[x_]} :> Thread[x -> c], All]];

styleToMarkers = AssociationThread[ColorData[97] /@ Range[Length @ markers], 
   Style @@@ markers];

ticks = {{{-Pi, Pi}, Automatic}, {Range[-5, 1], Automatic}};

graph /. Point[x_] :> ({meshStyles@#, Text[styleToMarkers[meshStyles@#], #]} & /@ x)

enter image description here To add markers at the start and end of curves use

graph /. {Point[x_] :> ({meshStyles@#, Text[styleToMarkers[meshStyles@#], #]} & /@ x), 
 Line[x_] :> {Line[x], 
    {meshStyles@#, Text[styleToMarkers[meshStyles@#], #]} & /@ x[[{1, -1}]]}}

enter image description here

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7
  • $\begingroup$ Thank you so much for the help! $\endgroup$
    – user516076
    Commented Mar 9, 2021 at 2:54
  • $\begingroup$ Hi. How to my third problem? The marking still have the same common place. How to work with pattern since the scale of each graph is different? Thanks. $\endgroup$
    – user516076
    Commented Mar 9, 2021 at 3:39
  • $\begingroup$ @user516076, do you get what you meed with MeshFunctions -> {#3 &} and Mesh -> 5? $\endgroup$
    – kglr
    Commented Mar 9, 2021 at 4:12
  • $\begingroup$ error sir. Isn't #3 & for 3d? $\endgroup$
    – user516076
    Commented Mar 9, 2021 at 4:36
  • 1
    $\begingroup$ @user516076, please see the substantially revised version. $\endgroup$
    – kglr
    Commented Mar 9, 2021 at 7:10

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