Skip to main content

differential equations - Plotting separatrices for nonlinear system


Consider the system:


x′=(1−x−y)xy′=(4−7x−3y)y



The system has a saddle point at (1/4,3/4). How can I plot the separatrices on the phase portrait having domain {(x,y): 0≤x≤1, 0≤y≤2}?


Here is my attempt. Start with a vector plot.


Clear[x, y];
Clear[Derivative];
f[x_, y_] = (1 - x - y) x;
g[x_, y_] = (4 - 7 x - 3 y) y;
vp = VectorPlot[{f[x, y], g[x, y]}, {x, 0, 1}, {y, 0, 2},
   VectorScale -> {0.045, 0.5, None},
   VectorStyle -> {GrayLevel[0.8]},
   VectorPoints -> 16,

   Axes -> True, AxesLabel -> {x, y}];

Get the equilibrium points.


eqpts = Solve[{f[x, y] == 0, g[x, y] == 0}, {x, y}]

There is a saddle point at (1/4,3/4). Set an eps value.


eps=1/10000;

Define a function.


sol[{x0_, y0_, tmin_, tmax_}] := 

 NDSolveValue[{x'[t] == f[x[t], y[t]], y'[t] == g[x[t], y[t]], 
   x[0] == x0, y[0] == y0}, {x[t], y[t]}, {t, tmin, tmax}]

Spend time (brute force) adjusting arguments to sol function to plot the separatrices.


sep1 = ParametricPlot[
   Evaluate@sol[{1/4 + eps, 3/4, 0, 40}], {t, 0, 40},
   PlotStyle -> {Red, Thick}];
sep2 = ParametricPlot[
   Evaluate@sol[{1/4 - eps, 3/4, 0, 40}], {t, 0, 40},
   PlotStyle -> {Red, Thick}];

sep3 = ParametricPlot[
   Evaluate@sol[{1/4 + eps, 3/4 + eps, -2.8, 0}], {t, -2.8, 0},
   PlotStyle -> {Red, Thick}];
sep4 = ParametricPlot[
   Evaluate@sol[{1/4 - eps, 3/4 - eps, -40, 0}], {t, -40, 0},
   PlotStyle -> {Red, Thick}];

Show everything together.


Show[vp, Graphics[{Black, PointSize[Large], 
   Point[{x, y}] /. eqpts}], sep1, sep2, sep3, sep4]


Result:


enter image description here


Which worked. Just wondering if there would be a simpler approach, one that beginning students can easily understand.


P.S. All code pasted below for an easy copy and paste.


Clear[x, y];
Clear[Derivative];
f[x_, y_] = (1 - x - y) x;
g[x_, y_] = (4 - 7 x - 3 y) y;
vp = VectorPlot[{f[x, y], g[x, y]}, {x, 0, 1}, {y, 0, 2},

   VectorScale -> {0.045, 0.5, None},
   VectorStyle -> {GrayLevel[0.8]},
   VectorPoints -> 16,
   Axes -> True, AxesLabel -> {x, y}];
eqpts = Solve[{f[x, y] == 0, g[x, y] == 0}, {x, y}];
eps = 1/10000;
sol[{x0_, y0_, tmin_, tmax_}] := 
  NDSolveValue[{x'[t] == f[x[t], y[t]], y'[t] == g[x[t], y[t]], 
    x[0] == x0, y[0] == y0}, {x[t], y[t]}, {t, tmin, tmax}];
sep1 = ParametricPlot[

   Evaluate@sol[{1/4 + eps, 3/4, 0, 40}], {t, 0, 40},
   PlotStyle -> {Red, Thick}];
sep2 = ParametricPlot[
   Evaluate@sol[{1/4 - eps, 3/4, 0, 40}], {t, 0, 40},
   PlotStyle -> {Red, Thick}];
sep3 = ParametricPlot[
   Evaluate@sol[{1/4 + eps, 3/4 + eps, -2.8, 0}], {t, -2.8, 0},
   PlotStyle -> {Red, Thick}];
sep4 = ParametricPlot[
   Evaluate@sol[{1/4 - eps, 3/4 - eps, -40, 0}], {t, -40, 0},

   PlotStyle -> {Red, Thick}];
Show[vp, Graphics[{Black, PointSize[Large], 
   Point[{x, y}] /. eqpts}], sep1, sep2, sep3, sep4]

Answer



We can solve (approximately) for the initial conditions of solutions that approach an equilbrium by comparing the displacement vector from the equilibrium with the vector field of the ODE. Such trajectory is characterized by the condition that these two vectors become parallel as the solution nears the equilibrium. I used a similar idea before, which is buried in this answer.


sys = {(1 - x - y) x, (4 - 7 x - 3 y) y};
vars = {x, y};
equilibria = Solve[sys == {0, 0}, vars, Reals]
(*  {{x -> 0, y -> 4/3}, {x -> 1/4, y -> 3/4}, {x -> 1, y -> 0}, {x -> 0, y -> 0}}  *)


saddles = Pick[equilibria, Sign@Det@D[sys, {vars}] /. equilibria, -1]
(*  {{x -> 1/4, y -> 3/4}}  *)

Here is a function to get initial conditions for the separatrices.


sepICS[p0_, eps_] := With[{p1 = p0 + eps * Norm[p0] {Cos[t], Sin[t]}},
   p1 /. NSolve[Det[{p1 - p0, sys /. Thread[vars -> p1]}] == 0 && 0 <= t < 2 Pi]
   ];

We get parametrizations for the separatrices, stopping the integration when the solution gets close to the saddle and when it leaves the plot domain. One problem with approaching a saddle point is that the initial condition, as well as the subsequent integration, is approximate. If the solution is pushed too far, it will ricochet off along another separatrix (approximately).


separatrices = Flatten[

   Module[{eps = 10^-7,         (* tunable distance from equilibrium *)
     X0, dX},
    With[{Xa = 0, Xb = 1, Yc = 0, Yd = 2,  (* plot domain boundaries *)
        p0 = vars /. #,         (* equilibrium *)
        X = Through[vars[t]]},  (* variables at t *)
       X0 = X /. t -> 0;        (* initial values *)
       dX = D[X, t];            (* derivatives *)
       With[{X1 = X[[1]], X2 = X[[2]]},
        First@NDSolve[{
             dX == (sys /. v : Alternatives @@ vars :> v[t]), 

             X0 == #,
             (*stop when close to saddle*)
             WhenEvent[Norm[X - p0] < 0.5 eps * Norm[p0], "StopIntegration"],
             (*stop when solution leaves plot domain*)
             WhenEvent[Abs[X1 - (Xa + Xb)/2] > (Xb - Xa)/2, "StopIntegration"], 
             WhenEvent[Abs[X2 - (Yc + Yd)/2] > (Yd - Yc)/2, "StopIntegration"]},
            vars, {t, -100, 100}] & /@ sepICS[p0, eps]
        ]] & /@ saddles
    ],
 1];



sepPlots = ParametricPlot @@@ ({{x[t], y[t]}, 
       Hold[Flatten][{t, x["Domain"]}]} /. separatrices // ReleaseHold);

Show[
 background,
 vp,
 sepPlots,
 PlotRange -> All, Frame -> True, Axes -> False, AspectRatio -> 1]


Mathematica graphics


The background is some attempt at mimicking ubpqdn's. One can use the lines from the plots of the separatrices to construct poiygon's to illustrate the regions created by them. It is a bit awkward to add the corner points.


sepLines = First@Cases[#, _Line, Infinity] & /@ sepPlots;
background = Graphics[
  Riffle[
   Lighter[#, 0.6] & /@ {Red, Purple, Yellow, Green},
   MapAt[Reverse, 
     Partition[sepLines, 2, 1, 1], {{2}, {4}}] /. {Line[p1_], 
      Line[p2_]} :> Polygon[

      Join[p1, p2,
       Nearest[Tuples[{{0, 1}, {0, 2}}], {p2[[-1, 1]], p1[[1, 2]]}, {1, 0.01}],
       Nearest[Tuples[{{0, 1}, {0, 2}}], {p1[[1, 1]], p2[[-1, 2]]}, {1, 0.01}]]]
   ],
  PlotRange -> All, Frame -> True, Axes -> False, AspectRatio -> 1
  ]


Addendum: Notes on the code.


1. v : x | y :> v[t]: The : is short for Pattern; | is short for Alternatives. So v : x | y defines the pattern symbol v to represent x or y. The whole v : x | y :> v[t], means replace x or y by x[t] or y[t] respectively. The rules {x -> x[t], y -> y[t]} are equivalent.



2. ParametricPlot @@@ ...: The main problem here are the domains of the solutions. The variable separatrices contains a list of solutions of the form


{{x -> x1ifn, y -> y1ifn}, {x -> x2ifn, y -> y2ifn},...}

where x1ifn, y1ifn etc. are interpolating functions. Each pair x1ifn, y1ifn has the same domain, but another pair x2ifn, y2ifn will have a different domain. So what is an easy way to plot all of the solutions? If ifn is an InterpolatingFunction, then ifn["Domain"] returns a list of domains for each input; in this case, it will have the form {{tmin, tmax}}. Flatten[{t, x["Domain"]}] will have the form {t, tmin, tmax} as needed for ParametricPlot. The problem is that the x in x["Domain"] has to be replaced by an InterpolatingFunction and evaluated before Flatten is evaluated. Hence the Hold[Flatten], to prevent flattening until after the /. separatrices has been executed; the ReleaseHold then lets the domains be evaluated and flattened. Since separatrices is a list of solutions (each of which is a list of Rules), the replacement yields a list of the form:


{{{x1ifn[t], y1ifn[t]}, Hold[Flatten][{t, {{tmin1, tmax1}}}]},
 {{x2ifn[t], y2ifn[t]}, Hold[Flatten][{t, {{tmin2, tmax2}}}]},
 ...}

After ReleaseHold, these elements will be ready to have ParametricPlot applied to them with @@@. This replaces the {} around each element with ParametricPlot[]:


{ParametricPlot[{x1ifn[t], y1ifn[t]}, {t, tmin1, tmax1}],

 ParametricPlot[{x2ifn[t], y2ifn[t]}, {t, tmin2, tmax2}],
 ...}

These automatically evaluate to the plots of each separatrix.


Comments

Post a Comment

Popular posts from this blog

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

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

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more