Group Abstract

Message Boards

WOLFRAM COMMUNITY

4.8K Views

6 Replies

1 Total Like

View groups...

Follow this post

Share this post:

GROUPS:

Fading curves based on a function?

Betty Boo

Posted 8 years ago

I have some code that outputs what I want, although I may change the function later. However, I want the output to be like a screenshot I have provided, such that it fades as with the properties of another function. Here is the code: ClearAll["Global`*"]; xi = 2; k0 = 9; Print["k0=", k0, " xi=", xi]; \ Show[Table[ sol = NDSolve[{D[xtraj[t], t] == (Sinh[ xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[ xtraj[t] (t - xi)])), xtraj[0] == n}, xtraj[t], {t, 0, 4}]; ParametricPlot[{t, xtraj[t]} /. sol, {t, 0, 4}, PlotRange -> All, PlotStyle -> {Blue, Full, Medium}, AxesStyle -> Thickness[.003], LabelStyle -> {Black, Medium}, AxesLabel -> {xtraj, t}], {n, -4, 4 - 0.5, 0.5}]] Here is the output - or a simplified version of it: And now this is what I would like - the fading is based upon another function - given below. Here is the fading function: (E^(-2 (-2 + t + x)^2) Sqrt[2 \[Pi]] (1 + E^(8 (-2 + t) x) + 2 E^(4 (-2 + t) x) Cos[10 x])) It is a tricky one that I really would like to solve.

I have some code that outputs what I want, although I may change the function later. However, I want the output to be like a screenshot I have provided, such that it fades as with the properties of another function.

Here is the code:

ClearAll["Global`*"]; xi = 2; k0 = 9; Print["k0=", k0, " xi=", xi]; \
Show[Table[
  sol = NDSolve[{D[xtraj[t], 
       t] == (Sinh[
         xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[ 
           xtraj[t] (t - xi)])), xtraj[0] == n}, xtraj[t], {t, 0, 4}];
  ParametricPlot[{t, xtraj[t]} /. sol, {t, 0, 4}, PlotRange -> All, 
   PlotStyle -> {Blue, Full, Medium}, AxesStyle -> Thickness[.003], 
   LabelStyle -> {Black, Medium}, AxesLabel -> {xtraj, t}], {n, -4, 
   4 - 0.5, 0.5}]]

Here is the output - or a simplified version of it:

enter image description here

And now this is what I would like - the fading is based upon another function - given below. enter image description here

Here is the fading function:

(E^(-2 (-2 + t + x)^2) Sqrt[2 \[Pi]] (1 + E^(8 (-2 + t) x) + 2 E^(4 (-2 + t) x) Cos[10 x]))

It is a tricky one that I really would like to solve.

POSTED BY: Betty Boo

6 Replies

Sort By:

Betty Boo

Posted 8 years ago

Thanks that is very helpful.

POSTED BY: Betty Boo

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 8 years ago

This may be closer to what you want: xi = 2; k0 = 9; fadingFunction[t_, x_] = (E^(-2 (-2 + t + x)^2) Sqrt[ 2 \[Pi]] (1 + E^(8 (-2 + t) x) + 2 E^(4 (-2 + t) x) Cos[10 x])); sol = ParametricNDSolveValue[{D[xtraj[t], t] == (Sinh[ xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[ xtraj[t] (t - xi)])), xtraj[0] == n}, xtraj, {t, 0, 4}, {n}]; Plot[Table[sol[n][t], {n, -4, 4 - 0.5, 0.5}], {t, 0, 4}, PlotRange -> All, ColorFunction -> Function[{t, x}, Blend[{Blue, White}, Rescale[ArcTan[fadingFunction[t, x]], {0, Pi/2}]]], ColorFunctionScaling -> False]

This may be closer to what you want:

xi = 2; k0 = 9;
fadingFunction[t_, 
   x_] = (E^(-2 (-2 + t + x)^2) Sqrt[
     2 \[Pi]] (1 + E^(8 (-2 + t) x) + 2 E^(4 (-2 + t) x) Cos[10 x]));
sol = ParametricNDSolveValue[{D[xtraj[t], 
      t] == (Sinh[
        xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[
          xtraj[t] (t - xi)])), xtraj[0] == n}, xtraj, {t, 0, 4}, {n}];
Plot[Table[sol[n][t], {n, -4, 4 - 0.5, 0.5}], {t, 0, 4}, 
 PlotRange -> All, 
 ColorFunction -> 
  Function[{t, x}, 
   Blend[{Blue, White}, 
    Rescale[ArcTan[fadingFunction[t, x]], {0, Pi/2}]]], 
 ColorFunctionScaling -> False]

POSTED BY: Gianluca Gorni

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 8 years ago

Mine was just a hint in a general direction. I leave it to you to fine-tune it.

POSTED BY: Gianluca Gorni

Betty Boo

Posted 8 years ago

I understand but to make the function fade this will not work, I've already tried it but thanks.

POSTED BY: Betty Boo

Betty Boo

Posted 8 years ago

Gianluca, this is a color function on y, if you look closely at my second diagram you will see that the fade is the same along each curve, whereas this code fades in the direction of y evenly.

POSTED BY: Betty Boo

Gianluca Gorni

Gianluca Gorni, University of Udine

Posted 8 years ago

Have you tried `ColorFunction` and `ColorFunctionScaling`? For example xi = 2; k0 = 9; Show[ Table[sol = NDSolve[{D[xtraj[t], t] == (Sinh[ xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[ xtraj[t] (t - xi)])), xtraj[0] == n}, xtraj[t], {t, 0, 4}]; ParametricPlot[{t, xtraj[t]} /. sol, {t, 0, 4}, PlotRange -> All, ColorFunction -> Function[{x, y}, Directive[Opacity[Abs[y]/4], Blue]], ColorFunctionScaling -> False], {n, -4, 4 - 0.5, 0.5}]]

Have you tried ColorFunction and ColorFunctionScaling? For example

xi = 2; k0 = 9; Show[
 Table[sol = 
   NDSolve[{D[xtraj[t], 
       t] == (Sinh[
         xtraj[t] (t - xi)])/((Cos[xtraj[t]]) + (Cosh[
           xtraj[t] (t - xi)])), xtraj[0] == n}, 
    xtraj[t], {t, 0, 4}];
  ParametricPlot[{t, xtraj[t]} /. sol, {t, 0, 4}, PlotRange -> All, 
   ColorFunction -> 
    Function[{x, y}, Directive[Opacity[Abs[y]/4], Blue]], 
   ColorFunctionScaling -> False], {n, -4, 4 - 0.5, 0.5}]]

POSTED BY: Gianluca Gorni

Reply to this discussion

Reply Preview

Attachments

Remove Add a file to this post

Follow this discussion

or Discard

Feedback