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I have this plot but it does not do what I want it to do. I have made A MOCK-UP OUTPUT of what I want.

 ClearAll[x, xi, σ, n]
 xi = 2; k0 = 5; σ = 2;
 Show[
  Table[
   sol = NDSolve[{D[x[t], t] == (Sinh[(σ^2) x[t] (t - xi)])/((Cos[2 k0 x[t]]))), x[0] == n}, x[t], {t, 0, 4}];
  ParametricPlot[{x[t], t} /. sol, {t, 0, 4}, 
   PlotRange -> All, 
   BaseStyle -> Thick, AxesStyle -> Thickness[.001], 
   LabelStyle -> {Black, Medium}, ColorFunctionScaling -> False, 
   ColorFunction -> 
     Function[{x, y, t}, 
      Directive[
       Opacity[0.02 + (e^(-1/2 (x[t] - xi)^2 σ^2) sqrt (π/
               2) σ (1 + e^(2 x[t] (-xi) σ^2) + 
               2 e^(x[t] (-xi) σ^2) cos (2 k0 x))) /. sol[[1]] //
          First], Blue]]], {n, -3, 3 - 1/4, 1/4}]]

This is what it should output with each iteration, the curves are faded based upon the function - such that curves close to the center fade. Notice that the curve holds it opacity all through its progress.

enter image description here

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3
  • $\begingroup$ Betty,there ar esime syntax errors in your code. For instance e^ should probably be E^ (uppercase) or Exp[...]. Similarly sqrt should be Sqrt[...]. Does it get any better if you fix those problems? I also think that your code should probably read sol = Table[...], rather than Table[sol =...]. $\endgroup$
    – MarcoB
    Jun 4, 2018 at 3:24
  • $\begingroup$ Marco not really - the output above is not something that comes from the code I'd written, it what I want to code to produce. $\endgroup$
    – Betty
    Jun 4, 2018 at 5:55
  • $\begingroup$ Did you notice that your NDSolve expression does not actually work though? If you try to evaluate it, it returns NDSolve::ndsz: At t == 9.421270636234184*^-14, step size is effectively zero; singularity or stiff system suspected.`. Until you solve this issue, you won't be able to obtain a plot no matter what. $\endgroup$
    – MarcoB
    Jun 6, 2018 at 15:37

1 Answer 1

Reset to default
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functions = Table[n x^4 + n/2 x^2 + n/5, {n, -4, 4}];

styled = MapThread[
      Style[#1, Opacity@#2] &,
      {#, Abs@Standardize[#, Mean, 2 StandardDeviation[#]&]& @Range@Length@#}
   ]& @functions;

Plot[
   Evaluate@styled, {x, -0.7, 0.7},
   PlotRange -> All, PlotStyle -> Directive[Thick, Darker@Red]
]

Mathematica graphics

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3
  • $\begingroup$ This looks right but what if I wanted to use my orginal equation - as I may want to vary it: sol = NDSolve[{D[x[t], t] == (Sinh[(σ^2) x[t] (t - xi)])/((Cos[2 k0 x[t]]))), x[0] == n}, x[t], {t, 0, 4}] $\endgroup$
    – Betty
    Jun 4, 2018 at 23:05
  • $\begingroup$ thanks for that, but in terms of my function I do not see how it can work, in fact being new to Mathematica it looks like it is only designed for this pattern. Can you please tell me more? $\endgroup$
    – Betty
    Jun 5, 2018 at 3:22
  • $\begingroup$ @Betty Did you notice that your NDSolve expression does not actually work though? If you try to evaluate it, it returns NDSolve::ndsz: At t == 9.421270636234184*^-14, step size is effectively zero; singularity or stiff system suspected.`. Until you solve this issue, you won't be able to obtain a plot no matter what. $\endgroup$
    – MarcoB
    Jun 6, 2018 at 15:37

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