Skip to main content

plotting - 4-dimensional Lotka-Volterra plot


I was referring to the question I posted earlier in Mathematics forum, but then I came to see this example on Wikipedia. How did they plot the system in phase space?


enter image description here


The closet relevant topic I have done is the Lorenz Equation, where I used NDSolve and then ParametricPlot3D evaluated at the fix points.



Answer



Here's how to numerically solve the model.


nsp = 4;


{r[1], r[2], r[3], r[4]} = {1, 0.72, 1.53, 1.27};
{k[1], k[2], k[3], k[4]} = {1, 1, 1, 1};
amat = {{1, 1.09, 1.52, 0}, {0, 1, 0.44, 1.36}, {2.33, 0, 1, 0.47}, {1.21, 0.51, 0.35, 1}};
Do[a[i, j] = amat[[i, j]], {i, nsp}, {j, nsp}];

eqns = Table[x[i]'[t] ==
  r[i]*x[i][t]*(1 - Sum[a[i, j]*x[j][t]/k[i], {j, nsp}]), {i, nsp}];
ics = Table[x[i][0] == 0.1, {i, nsp}];
unks = Table[x[i], {i, nsp}];


tmax = 10000;
sol = NDSolve[{eqns, ics}, unks, {t, 0, tmax}][[1]];

ParametricPlot3D[Evaluate[{x[1][t], x[2][t], x[3][t]} /. sol], {t, 100, tmax},
  AxesLabel -> {"x1", "x2", "x3"}, PlotPoints -> 200]

Mathematica graphics


I don't know how to color the ParametricPlot3D according to x[4][t] -- that could be an interesting question if it hasn't already been answered on the site.


Update


Here's how to color according to x[4][t], using the slot #4 which represents t here:



ParametricPlot3D[Evaluate[{x[1][t], x[2][t], x[3][t]} /. sol], {t, 100, tmax}, 
  AxesLabel -> {"x1", "x2", "x3"}, PlotPoints -> 400, 
  ColorFunction -> (ColorData["SunsetColors", x[4][#4] /. sol] &), 
  ColorFunctionScaling -> False]

Mathematica graphics


Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

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}]
Post a Comment
Read more