10
$\begingroup$

I would like to graphically show Bendixson’s criterion.

The Bendixson Criterion:

If $f_1$ and $f_2$ are continuous in a region $R$ which is simply-connected (i.e., without holes), and

$$\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}\ne0$$ at any point of $R$,

then the system

$$x_1' = f_1(x_1, x_2)$$

$$x_2' = f_2(x_1, x_2)$$

has no closed trajectories inside $R$.

Basically you can use this theorem to proof that there is no limit cycle within a system:

$$x' = f(x,y) $$

($x$ being a state vector and $f(x,y)$ the dynamic equation vector)

I wanted to visualize graphically that, if we HAVE a limit cycle $\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}$ will be equal to zero at some points.

Hence I draw a limit cycle (simple circle in the $x_1-x_2$ plane). Since $x'=f(x_1,x_2) \rightarrow f(x_1,x_2)$ will be tangent to the circle at all points, if the circle is a limit cycle (the trajectory can not escape).

enter image description here

Clear[t, x, y, z, P];
x[t_] = -Sin[t];
y[t_] = Cos[t];

P[t_] = {x[t], y[t]};
V[t_] = {x'[t], y'[t]};
curveplot = 
  ParametricPlot[P[t], {t, 0, 2*Pi}, PlotStyle -> Thickness[0.01]];

ar = Table[{P[t], P[t] + V[t]}, {t, 0, 2*Pi, Pi/4}];
Show[curveplot, 
 Graphics[{Arrow[ar], Red, AbsolutePointSize@10, Point@ar[[All, 1]]}],
  PlotRange -> All, AxesLabel -> {"x1", "x2"}, Ticks -> None]

But how could I visualize: $\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}$ ?? Any suggestions ?

Based on a comment from Rahul: $\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}$ is simply the divergence.

Hence I do the follwing:

Angle with x1 and x2:

f1[x1_, x2_] = -Sin[ArcTan[x2/x1]];
f2[x1_, x2_] = Cos[ArcTan[x2/x1]];

a = StreamPlot[{f1[x1, x2], f2[x1, x2]}, {x1, -1, 1}, {x2, -1, 1}];

Show[curveplot, a, 
 Graphics[{Arrow[ar], Red, AbsolutePointSize@10, Point@ar[[All, 1]]}],
  PlotRange -> All, AxesLabel -> {"x1", "x2"}, Ticks -> None]

enter image description here

EDIT: Thanks to J.M, I changed the ArcTan function to:

-Sin[ArcTan[x1, x2]]

and

Cos[ArcTan[x1, x2]]

(putting a comma between x1 and x2, gives the angle for this coordinate)

the output no looks better:

enter image description here

Now the question is:...what could you interpret ? ...and is there a better way to visualize the divergence ?

Any help is highly appreciated ! :)

$\endgroup$
  • 2
    $\begingroup$ $\partial f_1/\partial x_1+\partial f_2/\partial x_2$ is nothing but the divergence of the vector field $f$. You could draw the vector field itself and allow the viewer to interpret how much it is spreading out or contracting. $\endgroup$ – Rahul Jan 24 '17 at 19:10
  • $\begingroup$ @Rahul nice suggestion, how do I draw the divergence onto the same plot ? $\endgroup$ – james Jan 24 '17 at 20:31
  • $\begingroup$ @Rahul have a look at my question. Added your suggestion... but there is probably a mistake. $\endgroup$ – james Jan 24 '17 at 20:50
  • $\begingroup$ @C.E. Would you mind to explain a bit more, I am still confused ... sorry. $\endgroup$ – james Jan 24 '17 at 21:12
  • $\begingroup$ @C.E and do you think the result is correct as visualized ? $\endgroup$ – james Jan 24 '17 at 21:18
5
+50
$\begingroup$

Perhaps just exploring the linear case is sufficient. $\{x'(t),y'(t)\}=\vec{F}(x,y)=\{a x + b y, cx+dy\}$. Closed trajectories occurs when $ad-bc>0$ and $a+d=0=\nabla\cdot \vec{F}$:

f[a_, b_, c_, d_, x_, y_] := {{a, b}, {c, d}}.{x, y}
Manipulate[Column[{
   Grid[{{"a+d= ",
      a + d}, {"ad-bc= ",
      a d - b c}}],

   Show[
    StreamPlot[f[a, b, c, d, x, y], {x, -2, 2}, {y, -2, 2}, 
     Epilog -> {Red, PointSize[0.02], Point[{0, 0}], Green, 
       Point[{1, 1}]}, ImageSize -> 400],
    ParametricPlot[
      Evaluate[{x[t], y[t]} /. 
        First@DSolve[{{x'[t], y'[t]} == 
            f[a, b, c, d, x[t], y[t]], {x[0], y[0]} == {1, 1}}, {x[t],
            y[t]}, t]], {t, 0, 10}, PlotStyle -> {Red, Thick}] /. 
     Line[q__] :> Arrow[q]
    ]
   }, Alignment -> Center],
 {a, {-1/4, 1/4}}, {b, -1, 1, Appearance -> "Labeled"}, {c, -1, 1, 
  Appearance -> "Labeled"}, {d, {-1/4, 1/4}}]

enter image description here

$\endgroup$
  • $\begingroup$ Thanks a lot ! Looks great, but what would be your interpretation ? When circles are formed, a limit cycle exists ? $\endgroup$ – james Jan 27 '17 at 7:46
  • $\begingroup$ @totyped sorry for the delay in replying. I put this as a way of visually illustrating which also includes case where div F=0 and there are no closed trajectories as well as the contrapositive to the statement of the theorem. I am certain there will be better answers. The linear case is just the simplest case (I thought of), :) $\endgroup$ – ubpdqn Jan 28 '17 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.