Skip to main content

differential equations - Putting NDSolve into ParametricPlot


I am having issues using Manipulate to plot the (numeric) solution to an ODE for different parameter values.


I have a code that has several stages, which seem to all work when I do them one after another. This code solves a system of ODEs (for particular parameter values), then does a parametric plot of the solution. The problem is I need to put the lines together to wrap them around with Manipulate (for I can do this code again easily for different parameter values), and this is causing me a lot of pain.


My initial code is:


(*this is my ODE*)

unforced[x0_, p0_, α_:α, δ_:δ] :=

{x'[t] == p[t],
p'[t] == -α x[t] - δ p[t] + α (x[t])^3,
x[0] == x0,
p[0] == p0
}

(*Choose some parameter values*)

α = -1, δ = 1


(*Solve my ODE*)

s = NDSolve[unforced[x0 = 1, p0 = 1], {x, p}, {t, 20}]

(*Plot It*)

ParametricPlot[Evaluate[{x[t], p[t]} /. s], {t, 0, 20}]

But now I want to be able to Manipulate the parameter values α and δ. So I start putting the lines of code together... and problems happen.


ParametricPlot[Evaluate[{x[t], p[t]}/.

NDSolve[{x'[t] == p[t],
p'[t] == -α x[t] - δ p[t] + α (x[t])^3,
x[0] == 1, p[0] == 0}, {x, p}, {t, 20}]], {t, 0, 20}]

This plots an empty graph. This confuses me because it seems like all I did was substitute into my previous code. Because this doesn't work, I can't put a Manipulate around this. If it worked then I would have tried:


Manipulate[ParametricPlot[Evaluate[{x[t], p[t]} /.
NDSolve[{x'[t] == p[t],
p'[t] == -α x[t] - δ p[t] + α (x[t])^3,
x[0] == 1, p[0] == 0}, {x, p}, {t, 20}]], {t, 0, 20}],
{{α, -1, "α"}, -2, 0}, {{δ, 0, "δ"}, 0, 2}]


How do I get around this problem?




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....