Homoclinic Bifurcation of a Quadratic Family of Real Functions with Two Parameters

Abstract

In this work the homoclinic bifurcation of the family H={h(a,b)(x)=ax2+b:a∈R/{0},b∈R} is studied. We proved that this family has a homoclinic tangency associated to x=0 of P1 for b=-2/a. Also we proved that Wu(P1) does not intersect the backward orbit of P1 for b>-2/a, but has intersection for b<-2/a with a>0. So H has this type of the bifurcation.

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Farris, S.M. and Abdul-Kareem, K.N. (2021) Homoclinic Bifurcation of a Quadratic Family of Real Functions with Two Parameters. Open Access Library Journal, 8, 1-11. doi: 10.4236/oalib.1107300. 1. Introduction

There are various definitions for the homoclinic bifurcation. In the sense of Devaney, the homoclinic bifurcation is a global type of bifurcations, that is, this type of bifurcation is a collection of local and simple types of bifurcations  (like, period-doubling and saddle-node of bifurcation  ).

According to    we have another definition of the homoclinic bifurcation via the notions of the unstable sets of a repelling periodic point (fixed point) and the intersection of this set with the backward orbits of this point.

The purpose of this work is to prove the family

$H=\left\{{h}_{a,b}\left(x\right)=a{x}^{2}+b:a\in ℝ/\left\{0\right\},b\in ℝ\right\}$ has homoclinic bifurcation at $b=\frac{-2}{a}$ following the second definition.

2. Definitions and Basic Concepts

2.1. Definition 1: 

A fixed point P is said to be expanding for a map f, if there exists a neighborhood $U\left(P\right)$ such that $|{f}^{\prime }\left(x\right)|>1$ for any $x\in U\left(P\right)$.

The neighborhood in the previous definition is exactly the local unstable set.

2.2. Definition 2: 

Let P be a repelling fixed point for a function $f:I\to I$ on a compact interval $I\subset R$, then there is an open interval about P on which f is one-to-one and satisfies the expansion property. $|f\left(x\right)-f\left(P\right)|>|x-P|,\forall x\in I$ where $x\ne P$.

The interval in the previous definition is exactly the unstable set of P.

2.3. Definition 3: 

Let P is fixed point and ${f}^{\prime }\left(P\right)>1$ for a map $f:ℝ\to ℝ$. A point q is called homoclinic point to P if $q\in {w}_{loc}^{u}\left(P\right)$ and there exists $n>0$ such that ${f}^{n}\left(q\right)=P$.

2.4. Definition 4: 

The union of the forward orbit of q with a suitable sequence of preimage of q is called the homoclinic orbit of P. That is $O\left(q\right)=\left\{P,\cdots ,{q}_{-2},{q}_{-1},q,{q}_{1},{q}_{2},\cdots ,{q}_{m}=P\right\}$ where ${q}_{i+1}=f\left({q}_{i}\right)$ for $i\le m-1$, ${q}_{m}=P$ and ${\mathrm{lim}}_{i\to -\infty }{q}_{i}=P$.

2.5. Definition 5: 

The critical x point is non-degenerate if ${f}^{″}\left(x\right)\ne 0$. The critical point x is degenerate if ${f}^{″}\left(x\right)=0$.

2.6. Definition 6: 

Let f be a smooth map on $I\subset R$, and let p be a hyperbolic fixed point for the map f. If ${W}^{u}\left(p\right)$ intersects the backward orbit of p at a nondegenerate critical point ${x}_{cr}$ of f, then ${x}_{cr}$ is called a point of homoclinic tangency associated to p.

2.7. Definition 7: 

Let ${f}_{\phi }$ be a smooth map on $I\subset R$, and let p be a hyperbolic fixed point for a map ${f}_{\phi }$. We say that ${f}_{\phi }$ has homoclinic bifurcation associated to p at $\phi =\stackrel{^}{\phi }$ if:

1) For $\phi <\stackrel{^}{\phi }$ ( $\phi >\stackrel{^}{\phi }$ ), ${W}^{u}\left(p\right)$ and the backward orbit of p has no intersect.

2) For $\phi =\stackrel{^}{\phi }$, ${f}_{\stackrel{^}{\phi }}$ has a point of homoclinic tangency ${x}_{cr}$ associated to p.

3) For $\phi >\stackrel{^}{\phi }$ ( $\phi <\stackrel{^}{\phi }$ ), the intersection of ${W}^{u}\left(p\right)$ with the backward orbit of p is nonempty.

3. Homoclinic Bifurcation of the Family $H=\left\{{h}_{a,b}\left(x\right)=a{x}^{2}+b:a\in ℝ/\left\{0\right\},b\in ℝ\right\}$

In this section, we show that the family H has a point of homoclinic tangency associated to P1 at $b=\frac{-2}{a}$, and H has a homoclinic bifurcation.

We need the following results proved in .

At the first, the fixed points of ${h}_{a,b}\left(x\right)$ are

${P}_{1}=\frac{1+\sqrt{1-4ab}}{2a},{P}_{2}=\frac{1-\sqrt{1-4ab}}{2a}$.

a) Proposition:

For ${h}_{a,b}\left(x\right)\in H$ with $a>0$ the local unstable set of the fixed point P1 is ${w}_{loc}^{u}\left({P}_{1}\right)=\left(\frac{1}{2|a|},\infty \right)$.

b) Lemma:

For ${h}_{a,b}\left(x\right)\in H$, ${h}_{a,b}^{-1}\left({P}_{1}\right)=\mp \sqrt{\frac{{P}_{1}-b}{a}}=\mp {P}_{1}$ where ${P}_{1}>b$ for $a>0.$

c) Theorem:

For ${h}_{a,b}\left(x\right)\in H$ with $a>0$, the unstable set of the fixed point P1 is ${w}^{u}\left({P}_{1}\right)=\left(\frac{1}{|a|}-{P}_{1},\infty \right)$.

d) Remark: 

The local unstable set of the fixed point P2 is ${w}_{loc}^{u}\left({P}_{2}\right)=\left(-\infty ,\frac{-1}{2|a|}\right)$, and the unstable set of the fixed point P2 is ${w}^{u}\left({P}_{2}\right)=\left(-\infty ,\frac{-1}{|a|}-{P}_{2}\right)$. In this work we will omit the result about P2 because ( ${{h}^{\prime }}_{a,b}\left({P}_{2}\right)<-1$, when $b<\frac{-3}{4a}$ for $a>0$ $b>\frac{-3}{4a}$ for $a<0$ ). Thus we will not care for the fixed point P2. (See definition (2.3)).

e) Remark:

For ${h}_{a,b}\left(x\right)\in H$, ${h}_{a,b}^{-2}\left({P}_{1}\right)=\mp \sqrt{\frac{-{P}_{1}-b}{a}}$.

f) Proposition:

For ${h}_{a,b}\in H$, if $b<\frac{-\left(5+2\sqrt{5}\right)}{4a}$ with $a>0$, then the second preimage of the fixed point P1 belongs to the local unstable set of P1.

g) Proposition:

For ${h}_{a,b}\in H$, if $\frac{-\left(5+2\sqrt{5}\right)}{4a}\le b\le \frac{-2}{a}$ with $a>0$, then the third preimage of the fixed point P1 belongs to the local unstable set of P1.

h) Theorem:

For the family $H=\left\{{h}_{a,b}\left(x\right)=a{x}^{2}+b:a>0\right\}$, there exist homoclinic points to the fixed point P1 whenever $b\le \frac{-2}{a}$. Moreover ${h}_{a,b}^{-2}\left({P}_{1}\right)={q}_{1,1},{h}_{a,b}^{-3}\left({P}_{1}\right)={q}_{2,1}$ are the first homoclinic points for $b<\frac{-\left(5+2\sqrt{5}\right)}{4a}$, $\frac{-\left(5+2\sqrt{5}\right)}{4a}\le b\le \frac{-2}{a}$ respictivelity.

i) Example:

For ${h}_{1,-6}\left(x\right)={x}^{2}-6$, a homoclinic orbit of a homoclinic point $\sqrt{3}$ is: $O\left(\sqrt{3}\right)=\left\{3,-3,\sqrt{3},\cdots ,3\right\}$.

The main result:

3.1. Lemma 1

For ${h}_{a,b}\left(x\right)=a{x}^{2}+b$ with $a\in ℝ/\left\{0\right\}$, the critical point of ${h}_{a,b}\left(x\right)$ is 0, and it is a non-degenerate critical point.

Proof:

Clearly that the critical point of ${h}_{a,b}\left(x\right)$ is zero.

Since $a\ne 0$, then

${{h}^{″}}_{a,b}\left(x\right)=2a\ne 0$.

So ${h}_{a,b}\left(x\right)$ has a non-degenerate critical point at $x=0$. ∎

3.2. Lemma 2

If $b=0$ of ${h}_{a,b}\left(x\right)\in H$ with $a\in ℝ/\left\{0\right\}$, then the backward orbit of the repelling fixed point P1 is undefined in $ℝ$.

Proof:

${h}_{a,0}\left(x\right)=a{x}^{2}$, clearly ${P}_{1}=\frac{1}{a}$.

Now the first preimage of ${h}_{a,0}\left(x\right)$ is

${h}_{a,0}^{-1}\left(x\right)=\mp \sqrt{\frac{x}{a}}$, where $x>0$ for $a>0$.

By Lemma (3-b), we have

${h}_{a,0}^{-1}\left(\frac{1}{a}\right)=\mp \sqrt{\frac{1}{{a}^{2}}}=\mp \frac{1}{a}=\mp {P}_{1}$.

But +P1 is a fixed point, and $-{P}_{1}=-\frac{1}{a}\notin {w}_{loc}^{u}\left({P}_{1}\right)=\left(\frac{1}{2a},\infty \right)$, see Proposition (3-a).

By Remark (3-e) we have

${h}_{a,0}^{-2}\left({P}_{1}\right)=\mp \sqrt{\frac{-{P}_{1}}{a}}=\mp \sqrt{\frac{-\frac{1}{ą}}{a}}=\mp \sqrt{\frac{-1}{{a}^{2}}}\notin ℝ$,

since $\frac{1}{{a}^{2}}>0$, $\forall a\in ℝ/\left\{0\right\}$.

Therefore ${h}_{a,0}^{-n}\left({P}_{1}\right)$ are undefined in $ℝ$ with $n\ge 2$.

Thus the backward orbit of the repelling fixed point P1 is undefined in 3.3. Theorem 1

For the family ${h}_{a,b}\left(x\right)=a{x}^{2}+b$, 0 belong to the backward orbit of P1 whenever $b=\frac{-2}{a}$ with $a\in ℝ/\left\{0\right\}$, and the backward orbit of P1 is:

${h}_{a,\frac{-2}{a}}^{-n}\left({P}_{1}=\frac{2}{a}\right)=\left\{\frac{2}{a},-\frac{2}{a},0,\frac{\sqrt{2}}{a},\cdots \right\}$.

Proof:

We test the values of n which makes ${h}_{a,b}^{-n}\left({P}_{1}\right)=0$.

By Lemma (3-b), ${h}_{a,b}^{-1}\left({P}_{1}\right)=±{P}_{1}$.

So ${h}_{a,b}^{-1}\left({P}_{1}\right)\ne 0$.

Now suppose that ${h}_{a,b}^{-2}\left({P}_{1}\right)=0$, by Remark (3-e) then

${h}_{a,b}^{-2}\left({P}_{1}\right)=\mp \sqrt{\frac{-{P}_{1}-b}{a}}=0$, thus

$\frac{-{P}_{1}-b}{a}=0$

$-{P}_{1}-b=0$

${P}_{1}=b$.

Since the fixed point ${P}_{1}=\frac{1+\sqrt{1-4ab}}{2a}$, therefore

$\frac{1+\sqrt{1-4ab}}{2a}=-b$,

then

$1+\sqrt{1-4ab}=-2ab$

$\sqrt{1-4ab}=-2ab-1$

$1-4ab=4{a}^{2}{b}^{2}+4ab+1$

$4{a}^{2}{b}^{2}+8ab=0$, which implies

$4ab\left(ab+2\right)=0$, then either

$b=0$, but by the above Lemma (3.2) the backward orbit of P1 is undefined, so we omit this case.

Or $ab+2=0$, thus

$b=\frac{-2}{a}$.

Now, ${P}_{1}=\frac{2}{a}$ and to find the backward orbit of P1, we consider

${h}_{a,\frac{-2}{a}}^{-1}\left(x\right)=±\frac{\sqrt{ax+2}}{a}$.

By Lemma (3-b) ${h}_{a,b}^{-1}\left({P}_{1}\right)=±{P}_{1}$, then

${h}_{a,\frac{-2}{a}}^{-1}\left(\frac{2}{a}\right)=±\frac{2}{a}$. But $+\frac{2}{a}$ is a fixed point, therefor

${h}_{a,\frac{-2}{a}}^{-1}\left(\frac{2}{a}\right)=-\frac{2}{a}$.

So

${h}_{a,\frac{-2}{a}}^{-2}\left(\frac{2}{a}\right)=\frac{\sqrt{a\left(\frac{-2}{a}\right)+2}}{a}=0$.

${h}_{a,\frac{-2}{a}}^{-3}\left(\frac{2}{a}\right)=\frac{\sqrt{a\left(0\right)+2}}{a}=\frac{\sqrt{2}}{a}$, and so on.

Therefore the backward orbit of ${P}_{1}=\frac{2}{a}$ is:

${h}_{a,\frac{-2}{a}}^{-n}\left({P}_{1}=\frac{2}{a}\right)=\left\{\frac{2}{a},-\frac{2}{a},0,\frac{\sqrt{2}}{a},\cdots \right\}$. ∎

3.4. Example

For ${h}_{1,-2}\left(x\right)={x}^{2}-2$, 0 belongs to the backward orbit of ${P}_{1}=2$ (Figure 1), and the backward orbit of P1 is ${h}_{1,-2}^{-n}\left(2\right)=\left\{2,-2,0,\sqrt{2},\cdots ,2\right\}$.

Figure 1. For ${h}_{1,-2}\left(x\right)={x}^{2}-2$, the backward orbit of ${P}_{1}=2$.

3.5. Theorem 2

If $b>\frac{-2}{a}$ for ${h}_{a,b}\left(x\right)\in H$ with $a>0$, then there is no intersection of the backward orbit with the unstable set of P1.

Proof:

The backward orbit of P1

By Lemma (3-b) ${h}_{a,b}^{-1}\left({P}_{1}\right)=±{P}_{1}$, since +P1 is a fixed point, then we consider

${h}_{a,b}^{-1}\left({P}_{1}\right)=-{P}_{1}$.

By Remark (3-e), ${h}_{a,b}^{-2}\left({P}_{1}\right)=\mp \sqrt{\frac{-{P}_{1}-b}{a}}$.

If $-{P}_{1}>b$, then by Theorem (3-h),

$b\le \frac{-2}{a}$ which is a contradiction with $b>\frac{-2}{a}$. Therefore

$-{P}_{1}, which implies

${h}_{a,b}^{-2}\left({P}_{1}\right)\notin ℝ$.

So ${h}_{a,b}^{-n}\left({P}_{1}\right)$ are undefined in $ℝ$ with $n\ge 2$.

Thus the backward orbit of P1 is undefined .

So the intersection of ${W}^{u}\left({P}_{1}\right)$ with the backward orbit of P1 is also undefined. ∎

3.6. Theorem 3

If $b=\frac{-2}{a}$ for ${h}_{a,b}\left(x\right)\in H$ with $a>0$, then ${h}_{a,\frac{-2}{a}}$ has a point of homoclinic tangency at 0 associated to P1.

Proof:

By Theorem (3.3), ${h}_{a,\frac{-2}{a}}^{-n}\left({P}_{1}=\frac{2}{a}\right)=\left\{\frac{2}{a},-\frac{2}{a},0,\frac{\sqrt{2}}{a},\cdots \right\}$.

By Theorem (3-c), ${W}^{u}\left({P}_{1}\right)=\left(\frac{1}{a}-{P}_{1},\infty \right)$, then

${W}^{u}\left({P}_{1}=\frac{2}{a}\right)=\left(\frac{1}{a}-\frac{2}{a},\infty \right)$, i.e.

${W}^{u}\left({P}_{1}=\frac{2}{a}\right)=\left(-\frac{1}{a},\infty \right)$. Now

${h}_{a,\frac{-2}{a}}^{-n}\left(\frac{2}{a}\right)$ intersects ${W}^{u}\left({P}_{1}=\frac{2}{a}\right)$ at 0.

By Lemma (3.1) 0 is a non-degenerate critical point. So ${h}_{a,\frac{-2}{a}}$ has a point of homoclinic tangency at 0 associated to P1. ∎

3.7. Theorem 4

If $b<\frac{-2}{a}$ for ${h}_{a,b}\left(x\right)\in H$ with $a>0$, then the backward orbit of P1 crosses the unstable set ${W}^{u}\left({P}_{1}\right)$.

Proof:

First consider the backward orbit of P1.

By Lemma (3-b) ${h}_{a,b}^{-1}\left({P}_{1}\right)=±{P}_{1}$.

But + P1 is a fixed point, therefore we consider

${h}_{a,b}^{-1}\left({P}_{1}\right)=-{P}_{1}$.

By Remark (3-e), ${h}_{a,b}^{-2}\left({P}_{1}\right)=\mp \sqrt{\frac{-{P}_{1}-b}{a}}$.

Since $b<\frac{-2}{a}$, then by Theorem (3-h)

${h}_{a,b}^{-2}\left({P}_{1}\right)\in ℝ$.

Let ${h}_{a,b}^{-2}\left({P}_{1}\right)={q}_{1,1}$, ${h}_{a,b}^{-3}\left({P}_{1}\right)={q}_{2,1}$.

By Proposition (3-f), if $b<\frac{-\left(5+2\sqrt{5}\right)}{4a}$, then ${q}_{1,1}\in {W}_{loc}^{u}\left({P}_{1}\right)$.

By Proposition (3-g), if $\frac{-\left(5+2\sqrt{5}\right)}{4a}\le b<\frac{-2}{a}$, then ${q}_{2,1}\in {W}_{loc}^{u}\left({P}_{1}\right)$.

Now since the local unstable set of the repelling fixed point contained in the unstable set of the repelling fixed point. Therefore

${h}_{a,b}^{-n}\left({P}_{1}\right)\cap {W}^{u}\left({P}_{1}\right)\ne \varnothing$. ∎

Following examples explain the cases for $b>\frac{-2}{a}$, $b=\frac{-2}{a}$ and $b<\frac{-2}{a}$ (with $a>0$ ) respectively.

3.8. Example 1

For ${h}_{1,-1}\left(x\right)={x}^{2}-1$, we have no intersection of the backward orbit of P1 with the unstable set of P1.

Solution:

Consider the fixed point of ${h}_{1,-1}\left(x\right)$ is ${P}_{1}=\frac{1+\sqrt{5}}{2}$, and

${h}_{1,-1}^{-1}\left(x\right)=±\sqrt{x+1}$.

The backward orbit of ${P}_{1}=\frac{1+\sqrt{5}}{2}$

${h}_{1,-1}^{-1}\left(\frac{1+\sqrt{5}}{2}\right)=±\frac{1+\sqrt{5}}{2}$, where $+\frac{1+\sqrt{5}}{2}$ is a fixed point, therefore we consider

${h}_{1,-1}^{-1}\left(\frac{1+\sqrt{5}}{2}\right)=-\frac{1+\sqrt{5}}{2}$. Now

${h}_{1,-1}^{-2}\left(\frac{1+\sqrt{5}}{2}\right)=\mp \sqrt{-\frac{1+\sqrt{5}}{2}+1}\notin ℝ$.

So ${h}_{1,-1}^{-n}\left(\frac{1+\sqrt{5}}{2}\right)$ are undefined in $ℝ$ with $n\ge 2$.

Thus the backward orbit of P1 is undefined.

So the intersection of ${W}^{u}\left(\frac{1+\sqrt{5}}{2}\right)$ with the backward orbit of P1 is also undefined. ∎

3.9. Example 2

For ${h}_{1,-2}\left(x\right)={x}^{2}-2$, then ${h}_{1,-2}$ has a point of tangency at 0 associated to P1.

Solution:

Consider the fixed point of ${h}_{1,-2}\left(x\right)$ is ${P}_{1}=2$.

By Example (3.4), The backward orbit of ${P}_{1}=2$ is

${h}_{1,-2}^{-n}\left(2\right)=\left\{2,-2,0,\sqrt{2},\cdots ,2\right\}$.

On the other hand, the unstable set of ${P}_{1}=2$ is

${W}^{u}\left(2\right)=\left(-1,\infty \right)$, (see Theorem (3-c)). Now

${h}_{1,-2}^{-n}\left(2\right)$ intersects ${W}^{u}\left(2\right)$ at 0.

By Lemma (3.1), 0 is a non-degenerate critical point. So ${h}_{1,-2}$ has a point of tangency at 0 associated to ${P}_{1}=2$. ∎

3.10. Example 3

For ${h}_{1,-6}\left(x\right)={x}^{2}-6$, the backward orbit of P1 crosses the unstable set ${W}^{u}\left({P}_{1}\right)$.

Solution:

First consider the fixed point ${P}_{1}=3$.

The backward orbit of 3 is:

${h}_{1,-6}^{-n}\left(3\right)=\left\{3,-3,\sqrt{3},\cdots ,3\right\}$ (see Example (3-i)), with

${h}_{1,-6}^{-1}\left(3\right)={h}_{1,-2}^{2}\left(\sqrt{3}\right)$, and ${h}_{1,-6}^{-2}\left(3\right)=\sqrt{3}$.

Since $\sqrt{3}$ is a homoclinic point of ${P}_{1}=3$, then

$\sqrt{3}\in {W}_{loc}^{u}\left(3\right)$.

Now since the local unstable set of the repelling fixed point ${P}_{1}=3$ contained in the unstable set of the repelling fixed point ${P}_{1}=3$. Therefore

${h}_{1,-6}^{-n}\left(3\right)\cap {W}^{u}\left(3\right)\ne \varnothing$. ∎

Note , the main theorem in the work :

3.11. Theorem 5

${h}_{a,b}\left(x\right)=a{x}^{2}+b,a>0$ has a homoclinic bifurcation associated to the repelling fixed point of ${h}_{a,b}$, ${P}_{1}=\frac{1+\sqrt{1-4ab}}{2a}$, at $b=\frac{-2}{a}$.

Proof:

1) For $b>\frac{-2}{a}$, by Theorem (3.5) the intersection of the backward orbit of P1 and the unstable set of P1 is undefined.

2) For $b=\frac{-2}{a}$, by Theorem (3.6) ${h}_{a,\frac{-2}{a}}$ has a point of homoclinic tangency associated to P1 at $x=0$.

3) For $b<\frac{-2}{a}$, by Theorem (3.7) the backward orbit of P1 crosses the unstable set of P1, ${W}^{u}\left({P}_{1}\right)$.

Therefore ${h}_{a,b}$ has a homoclinic bifurcation associated to P1 at $b=\frac{-2}{a}$. ∎

3.12. Example

${h}_{1,-2}\left(x\right)={x}^{2}-2$ has a homoclinic bifurcation associated to the repelling fixed point of ${h}_{1,-2}$, ${P}_{1}=2$, at $b=-2$.

${h}_{1,-2}\left(x\right)$ has a homoclinic bifurcation associated to the repelling fixed point of ${h}_{1,-2}$, ${P}_{1}=2$, at $b=-2$. See examples (3.8), (3.9), (3.10).

3.13. Remark

For $a<0$, we have same results which proved above for $a>0$. In fact, we can prove in similar ways, that: ${h}_{a,b}\left(x\right)=a{x}^{2}+b,a<0$ has a homoclinic bifurcation associated to the repelling fixed point of ${h}_{a,b}$, ${P}_{1}=\frac{1+\sqrt{1-4ab}}{2a}$, at $b=\frac{-2}{a}$.

4. Conclusion

We conclude that the family $H=\left\{{h}_{a,b}\left(x\right)=a{x}^{2}+b:a\in ℝ/\left\{0\right\},b\in ℝ\right\}$ has homoclinic tangency associated to P1 at the critical point $x=0$. Also for $b>\frac{-2}{a}$ we have no intersection between the backward orbit of P1 and the unstable set of P1, and the backward orbit of P1 crosses the unstable set of P1 for $b<\frac{-2}{a}$. So we have homoclinic bifurcation at $b=\frac{-2}{a}$.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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