Skip to main content

plotting - StreamPoints for all of the domain (how to color basins of attraction)


Although that title is probably as confusing as I'll get out, here is the actual question. In StreamPlot, the StreamPoints will help trace out a 'trajectory'. So this nice, but I'm interested in areas throughout the domain that will converge to a certain location.


So for the following function, (often call the unforced Duffing system) I would like to have a StreamPlot and in the background almost like a contour plot of two colors (I know a contour plot isn't right at all I'm just trying to help you see what I'm interested in). So if it is one color then it will converge at one hole and another color for the other hole. Not sure if this is possible, but any ideas would be awesome!



StreamPlot[{y, -.15 y + x - x^3}, {x, -2, 2}, {y, -2, 2}, 
StreamPoints -> {{{{1, 1}, Red}, {{1, 1.6}, Green}, Automatic}}]

enter image description here



Answer



Horribly slow, but you probably get the idea. If this system can be solved analytically, then one could implement a far better approach, but I haven't looked into this.


What you are after is to find out which point in your plane converges to which sink. You can do this by solving the differential equation where your direction vector is the speed. You provide a starting point and start integrating. If you have done this, you can look what point is reached at the end of the integration and use the specific color for the sink.


A function that does exactly this is the following. It additionally stores the path starting from the initial point to the end.


line[{x0_, y0_}] :=
Module[{x, y, t, res, col},

res = NDSolveValue[{x'[t] == y[t],
y'[t] == -.15 y[t] + x[t] - x[t]^3, x[0] == x0, y[0] == y0}, {x,
y}, {t, 0, 100}];
col = If[Norm[{-1, 0} - Through[res[100]]] < .01, Green, Red];
{col, Line[Table[Through[res[t]], {t, 0, 100, .1}]]}
]

Now you can sample your region and for each starting point you create a colored line (this may take a while)


lines = ParallelTable[line[{x, y}], {x, -2, 2, .1}, {y, -2, 2, .1}];


After that, you can combine all lines into one graphics. This is again very slow due to Opacity and the sheer amount of line points


Show[
Graphics[{Opacity[.01], Thickness[0.01], lines}],
StreamPlot[{y, -.15 y + x - x^3}, {x, -2, 2}, {y, -2, 2}],
Frame -> True, FrameTicks -> True, PlotRange -> {{-2, 2}, {-2, 2}},
PlotRangeClipping -> True
]

but the result looks somewhat nice and I believe it is what you are after


Mathematica graphics



Please be aware that there is much room for improvement.


Line integral convolution


Another idea is to simply provide a test-function that tells you to which sink the starting point converges. This is close to the line function but doesn't create so many line-points.


test[{x0_?NumericQ, y0_?NumericQ}] :=
Module[{x, y, t, res},
res = NDSolveValue[{x'[t] == y[t],
y'[t] == -.15 y[t] + x[t] - x[t]^3, x[0] == x0, y[0] == y0}, {x,
y}, {t, 0, 100}];
Norm[{-1, 0} - Through[res[100]]] < .01
]


Now, you could create a colored image that shows you the regions of convergence in different colors. We will split a color scheme into two halves and introduce some noise which will be an excellent starting point for a LIC


mat = ParallelTable[
List @@
ColorData["TemperatureMap",
Boole[test[{x, y}]]/2 + RandomReal[1/2]], {y, -2,
2, .01}, {x, -2, 2, .01}];

bg = LineIntegralConvolutionPlot[{{y, -.15 y + x - x^3},
Image[Reverse@mat]}, {x, -2, 2}, {y, -2, 2}, RasterSize -> 512]


and then


Show[
bg,
StreamPlot[{y, -.15 y + x - x^3}, {x, -2, 2}, {y, -2, 2},
StreamStyle -> Gray]
]

Mathematica graphics


and finally, note that you can use this test-function with RegionPlot or you can employ it to color your StreamPlot accordingly:



stP = StreamPlot[{y, -.15 y + x - x^3}, {x, -2, 2}, {y, -2, 2}, 
StreamColorFunction ->
Function[{x, y},
If[test[{x, y}], ColorData["TemperatureMap", .9],
ColorData["TemperatureMap", .1]]],
StreamColorFunctionScaling -> False]

Show[bg, stP]

Mathematica graphics



Speed


As mentioned in the comments, the real system has no analytical solution. The bottlenecks in the last LIC example are (1) the creation of the domain image and (2) the calculation of the LIC itself. As for (2) there is nothing much we can do. However, one can speed up (1) by implementing a simple numerical scheme that can be compiled. Here is a hack of a simple Runge-Kutta solver that can be called in parallel. If you don't have a C compiler then you need to remove the compile to C option:


f[{x_, y_}] := {y, -.15 y + x - x^3}
step[yn_, h_] := Module[{k1, k2, k3},
k1 = f[yn];
k2 = f[yn + h/2.0*k1];
k3 = f[yn - h*k1 + 2.0 h*k2];
yn + h (1./6 k1 + 4./6 k2 + 1./6 k3)
]


fc = Compile[{{x0, _Real, 0}, {y0, _Real, 0}},
Module[{eps = 0.001, xn = x0, yn = y0, h = .1, maxSteps = 5000,
sink = -1, xtmp = 0., ytmp = 0.},
Do[
{xtmp, ytmp} = #;
If[
Norm[{xtmp, ytmp} - {xn, yn}] < eps,
Break[]
];
{xn, yn} = {xtmp, ytmp},

maxSteps
];
If[Norm[{-1, 0} - {xn, yn}] < .1, 1, 0]
], Parallelization -> True, RuntimeAttributes -> {Listable},
CompilationTarget -> "C"
] &[FullSimplify[step[{xn, yn}, h]]]

For a region sampling of 512x512 points, the runtime of the call to NDSolve within the test function is about 100s, the same compution with fc only needs 4s.


mat = fc @@ Transpose[
Table[{x, y}, {y, -2, 2, 4/511}, {x, -2, 2, 4/511}], {2, 3,

1}]; // AbsoluteTiming
Image[Reverse[mat]]

Mathematica graphics


I promise it will be the last graphics, but I quite like the style. It's in remembrance of a good friend of mine, who had a fast line integral convolution for Mathematica long before Wolfram could even spell the word. He always liked flow fishes in vector fields (which are called "Drop" in Mathematica) and often combined them:


lic = LineIntegralConvolutionPlot[{{y, -.15 y + x - x^3}, 
Colorize[ImageAdjust@Image[
GaussianFilter[Reverse[mat + RandomReal[{-1, 1}, {512, 512}]],
4]]]}, {x, -2, 2}, {y, -2, 2}, RasterSize -> 512];
Show[

lic,
StreamPlot[{y, -.15 y + x - x^3}, {x, -2, 2}, {y, -2, 2},
StreamStyle -> White, StreamMarkers -> "Drop"]
]

Mathematica graphics


Comments

Popular posts from this blog

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....