Skip to main content

plotting - Constrain Locator to specified region


How can I constrain the locator to stay within the region defined by RegionPlot?


When the Locator remains within the region, NDSolve generates a periodic solution. (The boundary of the region represents a homoclinic orbit for the DE. When the Locator is outside the boundary of the region, NDSolve generates an unbounded orbit. Actually, the boundary of the region is the solution to the DE with initial condition $x(0) = -6,\; y(0) = 0$.)


Additionally a strange behavior occurs whenever the left mouse button is pressed: the right end of the region is truncated. Why?


Manipulate[

region = RegionPlot[y^2 < x^2*(1 + x/6), {x, -6, 0}, {y, -2.5, 2.5}];
sol = NDSolve[{x'[t] == y[t], y'[t] == x[t] + x[t]^2/4,
x[0] == p[[1]], y[0] == p[[2]]}, {x, y}, {t, 0, T}];
psol = ParametricPlot[Evaluate[{x[t], y[t]} /. sol], {t, 0, T},
PlotRange -> {{-6, 0}, {-3, 3}}, PlotStyle -> Red ];
Show[{region, psol}], {{p, {-2, 0}}, Locator}, {{T, 5}, 0, 12, 0.1}]

I suspect that Dynamic needs to be introduced here, but I don't know how to implement it successfully.



Answer







  • How can I constrain the Locator to stay within the region defined by RegionPlot?



    You can check if Locator's coordinates fulfill the condition defining your region. It can be done with the second argument of Dynamic if you introduce Locator explicitly. Take a look at line with Locator[Dynamic[p, With[{...





  • A strange behavior occurs whenever the left mouse button is pressed: the right end of the region is truncated. Why?




    The body of a Manipulate is effectively wrapped with Dynamic. Each time you move something, it will be evaluated. During dynamic evaluation $PerformanceGoal is set to "Speed" unless you change it. It results in less sampling points and cut corners.


    You can change it to "Quality" but here there is no point in evaluating region each time anyway. It is independent from T and p so let's do it outside, once for good.




Edit:





  • Your answer is just what I wanted though I need to create a CDF demo. Unfortunately I obtained an error message upon creating a CDF from your answer.




    That's because region definition is forgotten as soon as the Kernel is quit, in contrast to Manipulates and DynamicModules variables generated "on fly" like sol and psol here.


    You can use SaveDefinitions->True or inject the region with With.





  • I also tried to change the appearance of the locator to a disk but was challenged there as well.



    For custom Locators it is better to set Appearance->None and display whatever you want in its coordinates.





With[{
region = RegionPlot[y^2 < x^2*(1 + x/6), {x, -6, 0}, {y, -2.5, 2.5}]
},

Manipulate[
sol = NDSolve[
{x'[t] == y[t], y'[t] == x[t] + x[t]^2/4,
x[0] == p[[1]], y[0] == p[[2]]},
{x, y}, {t, 0, T}
];

psol = ParametricPlot[Evaluate[{x[t], y[t]} /. sol], {t, 0, T},
PlotRange -> {{-6, 0}, {-3, 3}}, PlotStyle -> Red
];

Show[{
region, psol,
Graphics[{
Dynamic @ Disk[p, .1],
Locator[Dynamic[p,
With[{x = #[[1]], y = #[[2]]}, If[y^2 < x^2*(1 + x/6), p = #]] &],

Appearance -> None
]
}]
},
AspectRatio -> Automatic
],
{{p, {-2, 0}}, None}, {{T, 5}, 0, 12, 0.1}
]]

enter image description here



Comments

Popular posts from this blog

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 - 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 - 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.