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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 in Mathematica (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]

$\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}$ is simply the divergence.

Hence I plot a streamplot in Mathematica:

enter image description here

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

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]

Now the question is:...what could you interpret ??

Any help is highly appreciated ! :)

1 Answers 1

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Note that if you have $$\dot{\mathbf x}=F(\mathbf x)$$ where $F(\mathbf x)=\begin{bmatrix}f_1(\mathbf x)&f_2(\mathbf x)\end{bmatrix}^T$ then the Bendixon criterion can be investigated. But in your example, $$x(t)=-\sin t\\y(t)=\cos t$$ simply implies: $$\dot{x}=-y\\ \dot{y}=x$$ which means $$f_1(\mathbf x)=-x_2,\quad f_2(\mathbf x)=x_1$$ and therefore $$\frac{\partial f_1}{\partial x_1}+\frac{\partial f_2}{\partial x_2}=0+0=0$$ As a side-note, these two relationships hold: $$\sin(\operatorname{atan2}(x_1,x_2))=\frac{x_1}{\sqrt{x_1^2+x_2^2}}\\ \cos(\operatorname{atan2}(x_1,x_2))=\frac{x_2}{\sqrt{x_1^2+x_2^2}}$$