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I wrote the simple following code to plot a given vector field:

rr := Sqrt[x^2 + y^2];
a := 0.5;
xlim = 1;
splot = StreamPlot[{
    rr^a (x (x rr + y rr + y^2))/rr^3,
    rr^a (y (x rr + y rr - x^2))/rr^3},
   {x, -xlim, xlim}, {y, -xlim, xlim},
   StreamColorFunction -> "Heat", AxesLabel -> {"x", "y"}];
Show[splot]

enter image description here

I would like to color the trajectories which enter and leave the origin in different colors. For example:

enter image description here

Is there a simple way to do this? Thanks very much in advance!

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1 Answer 1

Reset to default
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This should get you started:

First, make a color function that uses the point $(x,y)$ and vector $(v_x,v_y)$ at that point to get the $\cos$ of the angle between the two:

Clear[color]
color[x_, v_] := Module[{q},
  q = x.v/(Norm[x] Norm[v]);
  Return[Blend[{Blue, Black, Red}, (q + 1)/2]]
  ]

Edit Another color Function:

Clear[color]
color[x_, v_] := Module[{q, c},
Which[
  x[[1]] > 0 && x[[2]] <= 0,
  c = GrayLevel[.2]
 ,
 True,
  q = x.v/(Norm[x] Norm[v]);
 c = Blend[{Blue, Black, Red}, (q + 1)/2]
 ];
 Return[c]
 ]

Then use the function in your plot:

rr := Sqrt[x^2 + y^2];
a := 0.5;
xlim = 1;
splot = StreamPlot[{rr^a (x (x rr + y rr + y^2))/rr^3, 
rr^a (y (x rr + y rr - x^2))/rr^3},
     {x, -xlim, xlim}, {y, -xlim, xlim},
StreamColorFunctionScaling -> False, 
StreamColorFunction -> (color[{#1, #2}, {#3, #4}] &), 
AxesLabel -> {"x", "y"}];
Show[splot]

enter image description here

You can play with the colors in the Blend[] function to get the look what you want.

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6
  • $\begingroup$ Thank you -- this is almost exactly what I needed! One small question, how can it be modified so that ll curves under the x-axis are blue. I didn't describe this well, but I only want to color the exiting curves from the origin red (as in my hand-drawn picture) $\endgroup$
    – J. Doee
    Commented Nov 1, 2017 at 23:49
  • $\begingroup$ You almost never need to use Return[], and Normalize[] is built-in. Thus: color[x_, v_] := Blend[{Blue, Black, Red}, (1 + Normalize[x].Normalize[v])/2]. Alternatively: color[x_, v_] := Blend[{Blue, Black, Red}, 1 - CosineDistance[x, v]/2]. $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Nov 2, 2017 at 4:12
  • $\begingroup$ Thanks! Is there a way to modify the color function to make it piecewise defined so that it follows this rule above the x-axis and below it colors everything blue? $\endgroup$
    – J. Doee
    Commented Nov 2, 2017 at 11:43
  • $\begingroup$ Actually, I'd like all the vectors in the fourth quadrant to be colored gray. $\endgroup$
    – J. Doee
    Commented Nov 2, 2017 at 14:38
  • $\begingroup$ I was able to do this simply by "showing" two plots, where I excluded the relevant regions. Thanks! $\endgroup$
    – J. Doee
    Commented Nov 2, 2017 at 15:24

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