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copied!<h2>An Image based solution</h2> <p>I don't claim this one neither robust nor general. But it's quick and fun. It uses Image Transformations to find the edges (possible because the heavy oscillatory character of your function): </p> <p>Function: </p> <pre><code>envelope[plot_] := Module[{boundary, Pr, rescaled}, (* "rasterize" the plot, identify the lower edge and isolate pixels*) boundary = Transpose@ImageData@Binarize@plot /. {x___, 0, 1, y___} :> Join[Array[1 &, Length[{x}]], {0}, Array[1 &, Length[{y}] + 1]]; (* and now rescale *) Pr = PlotRange /. Options[plot, PlotRange]; rescaled = Position[boundary, 0] /. {x_, y_} :> { Rescale[x, {1, Dimensions[boundary][[1]]}, Pr[[1]]], Rescale[y, {1, Dimensions[boundary][[2]]}, Reverse[Pr[[2]]]] }; (* Finally, return a rescaled and slightly smoothed plot *) Return[ListLinePlot@ Transpose@{( Transpose[rescaled][[1]])[[1 ;; -2]], MovingAverage[Transpose[rescaled][[2]], 2]}] ] </code></pre> <p>Testing code: </p> <pre><code>tk0 = phi'[t] phi'[t] - phi[t] phi''[t]; tk1 = phi''[t] phi''[t] - phi'[t] phi'''[t]; a = tk0/Sqrt[tk1]; f = Sqrt[tk1/tk0]; s = NDSolve[{ phi''[t] + phi[t] - 0.167 phi[t]^3 == 0.005 Cos[t - 0.5*0.00009*t^2], phi[0] == 0, phi'[0] == 0}, phi, {t, 0, 1000}]; plot = Plot[Evaluate[f /. s], {t, 0, 1000}, Axes -> False]; Show[envelope[plot]] </code></pre> <p><img src="https://i.stack.imgur.com/qwY7H.png" alt="alt text"></p> <p><b>Edit</b> </p> <p>Fixing a bug in the code above, the results are more accurate: </p> <pre><code>envelope[plot_] := Module[{boundary, Pr, rescaled}, (*"rasterize" the plot, identify the lower edge and isolate pixels*) boundary = Transpose@ImageData@Binarize@plot /. {x___, 0, 1, y___} :> Join[Array[1 &, Length[{x}]], {0}, Array[1 &, Length[{y}] + 1]]; (*and now rescale*) Pr = PlotRange /. Options[plot, PlotRange]; rescaled = Position[boundary, 0] /. {x_, y_} :> {Rescale[ x, {(Min /@ Transpose@Position[boundary, 0])[[1]], (Max /@ Transpose@Position[boundary, 0])[[1]]}, Pr[[1]]], Rescale[y, {(Min /@ Transpose@Position[boundary, 0])[[2]], (Max /@ Transpose@Position[boundary, 0])[[2]]}, Reverse[Pr[[2]]]]}; (*Finally,return a rescaled and slightly smoothed plot*) Return[ListLinePlot[ Transpose@{(Transpose[rescaled][[1]])[[1 ;; -2]], MovingAverage[Transpose[rescaled][[2]], 2]}, PlotStyle -> {Thickness[0.01]}]]] </code></pre> <p><img src="https://i.stack.imgur.com/yejb9.png" alt="enter image description here"> . .</p>

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