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

Salma M. Farris, Karam N. Abdul-Kareem

Department of Mathematics, College of Computer and Mathematical Sciences, University of Mosul, Mosul, Iraq.

DOI: 10.4236/oalib.1107300                

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 W^u(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.

Keywords

Local Unstable Set, Unstable Set, Homoclinic Point, Homoclinic Orbit, Non-Degenerate, Homoclinic Tangency, Homoclinic Bifurcation

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 [1] (like, period-doubling and saddle-node of bifurcation [2] ).

According to [3] [4] [5] 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: [6]

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

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

2.2. Definition 2: [7]

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. $\left|f\left(x\right)-f\left(P\right)\right|>\left|x-P\right|,\forall x\in I$ where $x\ne P$.

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

2.3. Definition 3: [8]

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: [9]

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 ${\lim }_{i\to -\infty }{q}_i=P$.

2.5. Definition 5: [10]

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: [11]

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: [3]

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 $\phi =\hat{\phi }$ if:

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

2) For $\phi =\hat{\phi }$, $f_{\hat{\phi }}$ has a point of homoclinic tangency $x_{cr}$ associated to p.

3) For $\phi >\hat{\phi }$ ( $\phi <\hat{\phi }$ ), 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 [12].

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 $w_{loc}^u\left(P_1\right)=\left(\frac{1}{2\left|a\right|},\infty \right)$.

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 $w^u\left(P_1\right)=\left(\frac{1}{\left|a\right|}-P_1,\infty \right)$.

d) Remark: [13]

The local unstable set of the fixed point P₂ is $w_{loc}^u\left(P_2\right)=\left(-\infty ,\frac{-1}{2\left|a\right|}\right)$, and the unstable set of the fixed point P₂ is $w^u\left(P_2\right)=\left(-\infty ,\frac{-1}{\left|a\right|}-P_2\right)$. 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), $h_{a,b}^{-1}\left(P_1\right)=\pm P_1$.

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

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

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

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

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

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

By Lemma (3-b) $h_{a,b}^{-1}\left(P_1\right)=\pm P_1$, then

$h_{a,\frac{-2}{a}}^{-1}\left(\frac{2}{a}\right)=\pm \frac{2}{a}$. 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) $h_{a,b}^{-1}\left(P_1\right)=\pm P_1$, 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

$-P_1<b$, 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) $h_{a,b}^{-1}\left(P_1\right)=\pm P_1$.

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

$h_{1,-1}^{-1}\left(x\right)=\pm \sqrt{x+1}$.

The backward orbit of $P_1=\frac{1+\sqrt{5}}{2}$

$h_{1,-1}^{-1}\left(\frac{1+\sqrt{5}}{2}\right)=\pm \frac{1+\sqrt{5}}{2}$, 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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