5
$\begingroup$

I'm plotting complex-valued functions using a common technique: plot the magnitude (modulus) on the vertical axis and fill the area below with color hues to indicate the phase (argument). Here is an example:

f[x_] := Exp[-x^2] Exp[10 I x];
Plot[Abs[f[x]], {x, -3, 3}, 
  Filling -> Axis, 
  ColorFunction -> Function[x, Hue[Arg[f[x]]/(2 Pi)]], 
  ColorFunctionScaling -> False, 
  PlotPoints -> 500]

This works fine, but for aesthetic reasons I'd also like to add a black outline curve, like this:

enter image description here

I made this image by plotting the outline in a separate Plot, then combining the two plots using Show. But that's pretty awkward, so I'm wondering if there's a better way to do it, preferably with a single Plot. The problem seems to be that the ColorFunction setting controls both the curve and the filling underneath it. Ideas?

$\endgroup$
1
  • $\begingroup$ (The PlotPoints setting isn't irrelevant here: a setting greater than the default is needed for color hues to show correctly.) $\endgroup$
    – dvs
    Jun 6, 2017 at 18:17

2 Answers 2

4
$\begingroup$

I'm not sure if you can do this with the options to Plot but you can "fix" it afterwards:

Plot[Abs[f[x]], {x, -3, 3}, 
  Filling -> Axis, 
  ColorFunction -> Function[{x, y}, Hue[Arg[f[x]]/(2 Pi)]], 
  ColorFunctionScaling -> False] /. 
  Line[pts_, _] :> {Thick, Black, Line[pts]}

Filling with plot line

(We are removing the VertexColors -> Automatic option from the plot line.)

$\endgroup$
1
$\begingroup$

You can use the options MeshFunctions, Mesh and MeshShading:

Plot[Abs[f[x]], {x, -3, 3}, Filling -> Axis, 
 ColorFunction -> Function[{x, y}, Hue[Arg[f[x]]/(2 Pi)]], 
 ColorFunctionScaling -> False,
 Mesh -> {{-3}}, MeshFunctions -> {#1 &}, 
 MeshShading -> {Directive[Thick, Black], Directive[Thick, Black]}]

Mathematica graphics

Notes: We can not use PlotStyle because as stated in Documentation >> Plot:

Mathematica graphics

However, (see Documentation >> ColorFunction), we can use MeshShading since

Mathematica graphics

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.