Phase Portraits of Random Planar Homogeneous Vector Fields

Abstract

In this paper, we study the probability of occurrence of phase portraits in the set of random planar homogeneous polynomial vector fields, of degree n. In particular, for $n=1,2,3,$ we give the complete solution of the problem; that is, we either give the exact value of each probability of occurrence or we estimate it by using the Monte Carlo method. Remarkably is that all but two of these phase portraits are characterized by the index at the origin and by the number of invariant straight lines through this point.

Appendix

In this appendix we give conditions on the coefficients of polynomial equations of degree 3 or 4 to know its number real zeros zeros and for degree 3 also their signs.

Lemma 15

Set $p(x)=x^3+ax^2+bx+c,$ $c_2=ab-9c,$ $d_2=a^2-3b$ and $d_3=-27c^2+18abc-4a^3c+a^2b^2-4b^3.$ Assume that $a\cdot b\cdot c \cdot c_2\cdot d_2\cdot d_3\ne 0.$ Then the following holds:

• (i) p has a unique real root of multiplicity one, if and only if $d_3<0.$ This root is negative (resp. positive) if $c>0$ (resp. $c<0$).

• (ii) p has three negative real roots if and only if $d_3>0,\,c_2>0,\,b>0,c>0.$

• (iii) p has two negative roots and one positive one if and only if either, $d_3>0,\,c<0,c_2>0$ or $d_3>0,\,c<0,\,c_2<0,b<0.$

• (iv) p has one negative root and two positive roots if and only if either, $d_3>0,\,c>0,c_2<0$ or $d_3>0,\,c>0,\,c_2>0,b<0.$

• (v) p has three positive roots if and only if $d_3>0,\,c_2<0,\,b>0,c<0.$

Proof

The proof is based in Sturm’s method which asserts that if $\big (p_0=p,p_1\ldots ,p_n\big )$ is a Sturm’s sequence of p in [a, b] with $p(a)\cdot p(b)\ne 0,$ then the number of real zeros of p in (a, b) is $V(a)-V(b)$ where V(x) is the number of changes of sign in the ordered sequence $\big (p_0(x),p_1(x),\ldots ,p_n(x)\big ),$ where the zeroes are disregarded. For any polynomial without multiple roots such a sequence always exists, see [27]. For p,  without multiple roots and $d_2\ne 0,$ one Sturm sequence is $p_0=p,$ $p_1=p'$ and

$$\begin{aligned} p_2(x) = \frac{1}{9}\left( 2d_2x+c_2\right) ,\quad p_3(x) = \frac{9}{4\,d_2^2}\,d_3. \end{aligned}$$

The quantity $d_3$ is the classical discriminant of p and it is known that $d_3<0$ if and only if p has a real root and two complex ones whereas $d_3>0$ if and only if p has three real roots (see [11] for instance). We are going to consider the following two tables depending on the sign of $d_3$ where $b,c,d_2,c_2$ stands for their respective signs.

	$-\infty$	0	$\infty$		$-\infty$	0	$\infty$
$d_3<0$				$d_3>0$
$p_0$	−	c	$+$	$p_0$	−	c	$+$
$p_1$	$+$	b	$+$	$p_1$	$+$	b	$+$
$p_2$	$-d_2$	$c_2$	$d_2$	$p_2$	$-d_2$	$c_2$	$d_2$
$p_3$	−	−	−	$p_3$	$+$	$+$	$+$

We separate the proof in two cases, depending on the sign of $d_3.$

Case 1 Assume that $d_3<0.$ We observe that $V(-\infty )=2$ and $V(+\infty )=1.$ Then

• The polynomial p has a negative real root if and only if $V(0)=1.$ It is easy to see that it happens when $c>0$ and one of the three following conditions hold:

  $$\begin{aligned} c_2>0,b>0;\quad c_2<0,b>0;\quad c_2<0,b<0. \end{aligned}$$

  We observe that $d_3<0,\,c_2>0,\,b<0,\,c>0$ is not compatible because then $V(0)-V(+\infty )=3$ but $V(-\infty )-V(+\infty )=1.$ Summarizing, p has a negative real root and two more complex ones if and only if $d_3<0,\,c>0.$

• The polynomial p has a positive real root if and only if $V(0)=2.$ It is easy to see that it happens when $c<0$ and one of the three following conditions hold:

  $$\begin{aligned} c_2>0,b>0;\quad c_2>0,b<0;\quad c_2<0,b>0. \end{aligned}$$

  As before conditions $d_3<0,c_2<0,b<0,c<0$ are incompatible and then p has a positive real root and two more complex ones if and only if $d_3<0,\,c<0$ as announced.

Case 2 Assume that $d_3>0$ and consider the above right-hand side table of signs. We see that in that case $d_2$ must be positive. Otherwise $V(-\infty )=1,\,V(\infty )=2$ which is not possible (we can also argue that since $d_3>0$ implies that p has three real roots, its derivative has two real roots and hence its discriminant which is equal to $4\,d_2$ has to be positive).

It is straightforward to see that items (ii), (iii), (iv), (v) are equivalent to $V(0)=0,V(0)=1,V(0)=2,V(0)=3$ respectively and that these number of changes of sign are satisfied exactly when the conditions on $b,c,c_2$ are the ones stated in the lemma. $\square$

Lemma 16

Let $p(x)=x^4+ax^3+bx^2+cx+d.$ Assume that $c\cdot d\cdot c_2\cdot c_3\cdot d_2\cdot d_3\cdot d_4\ne 0.$ Then the following holds:

• (a) p has two simple real and two complex roots if and only if $d_4<0.$

• (b) p has four different real roots if and only if $d_4>0,d_2>0$ and $d_3>0.$

• (c) p has no real root if and only if $d_4>0$ and either, $d_2<0$ or $d_3<0.$

Proof

A Sturm sequence of p,  is $p_0=p,$ $p_1=p',$

$$\begin{aligned} p_2(x) = \frac{1}{16}\left( d_2x^2+2(ab-2c)x+c_2\right) ,\,\, p_3(x) = \frac{16(2d_3x+c_3)}{d_2^2},\,\, p_4(x) = \frac{d_2^2d_4}{64d_3^2}. \end{aligned}$$

where

$$\begin{aligned} d_2= & {} 3\,{a}^{2}-8\,b,\\ c_2= & {} ac-16 d,\\ d_3= & {} -3\,{a}^{3}c+{a}^{2}{b}^{2}-6\,{a}^{2}d+14\,abc-4\,{b}^{3}+16\,bd-18\, {c}^{2} ,\\ c_3= & {} -9\,{a}^{3}d+{a}^{2}bc+32\,abd+3\,a{c}^{2}-4\,{b}^{2}c-48\,cd,\\ d_4= & {} -27\,{a}^{4}{d}^{2}+18\,{a}^{3}bcd -4\,{a}^{3}{c}^{3}-4\, {a}^{2}{b}^{3}d+{a}^{2}{b}^{2}{c}^{2}\\&+144\,{a}^{2}b{d}^{2}-6\,{a}^{2}{c}^{2}d-80\,a{ b}^{2}cd+\\&+18\,ab{c}^{3}+16\,{b}^{4}d-4\,{b}^{3}{c}^{2}-192\,ac{d}^{2}\\&- 128\,{b}^{2}{d}^{2}+144\,b{c}^{2}d-27\,{c}^{4}+256\,{d}^{3}. \end{aligned}$$

We note that $d_4$ is the discriminant of p and since $d_4\ne 0$ all its roots are simple.

It is known, see [11] for instance, that $d_4<0$ if and only if p has two simple real and two complex roots. And that $d_4>0$ if and only if p has either, four different real roots or no real root. It is very easy to distinguish between these last two possibilities using the Sturm method. Consider the corresponding table when $d_4>0,$ where again each value stands for its sign.

	$-\infty$	0	$\infty$
$p_0$	$+$	d	$+$
$p_1$	−	c	$+$
$p_2$	$d_2$	$c_2$	$d_2$
$p_3$	$-d_3$	$c_3$	$d_3$
$p_4$	$+$	$+$	$+$

If p has four real roots then $V(-\infty )-V(\infty )$ must be four and this only can happen if $V(-\infty )=4$ and $V(\infty )=0$ what immediately says that $d_2$ and $d_3$ must be positive. If $d_4>0$ and $d_2<0$ or $d_3<0$ then $V(-\infty )-V(+\infty )=2-2=0$ and the result follows.  $\square$