A fixed point P is said to be expanding for a map f, if there exists a neighborhood U ( P ) such that | f ′ ( x ) | > 1 for any x ∈ U ( P ) .

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

Let P be a repelling fixed point for a function f : I → I on a compact interval I ⊂ R , then there is an open interval about P on which f is one-to-one and satisfies the expansion property. | f ( x ) − f ( P ) | > | x − P | , ∀ x ∈ I where x ≠ P .

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

Let P is fixed point and f ′ ( P ) > 1 for a map f : ℝ → ℝ . A point q is called homoclinic point to P if q ∈ w l o c u ( P ) and there exists n > 0 such that f n ( q ) = P .

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 ( q ) = { P , ⋯ , q − 2 , q − 1 , q , q 1 , q 2 , ⋯ , q m = P } where q i + 1 = f ( q i ) for i ≤ m − 1 , q m = P and lim i → − ∞ q i = P .

The critical x point is non-degenerate if f ″ ( x ) ≠ 0 . The critical point x is degenerate if f ″ ( x ) = 0 .

Let f be a smooth map on I ⊂ R , and let p be a hyperbolic fixed point for the map f. If W u ( p ) intersects the backward orbit of p at a nondegenerate critical point x c r of f, then x c r is called a point of homoclinic tangency associated to p.

Let f φ be a smooth map on I ⊂ R , and let p be a hyperbolic fixed point for a map f φ . We say that f φ has homoclinic bifurcation associated to p at φ = φ ^ if:

1) For φ < φ ^ ( φ > φ ^ ), W u ( p ) and the backward orbit of p has no intersect.

2) For φ = φ ^ , f φ ^ has a point of homoclinic tangency x c r associated to p.

3) For φ > φ ^ ( φ < φ ^ ), the intersection of W u ( p ) with the backward orbit of p is nonempty.

For h a , b ( x ) = a x 2 + b with a ∈ ℝ / { 0 } , the critical point of h a , b ( x ) is 0, and it is a non-degenerate critical point.

Proof:

Clearly that the critical point of h a , b ( x ) is zero.

Since a ≠ 0 , then

h ″ a , b ( x ) = 2 a ≠ 0 .

So h a , b ( x ) has a non-degenerate critical point at x = 0 . ∎

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

Proof:

h a , 0 ( x ) = a x 2 , clearly P 1 = 1 a .

Now the first preimage of h a , 0 ( x ) is

h a , 0 − 1 ( x ) = ∓ x a , where x > 0 for a > 0 .

By Lemma (3-b), we have

h a , 0 − 1 ( 1 a ) = ∓ 1 a 2 = ∓ 1 a = ∓ P 1 .

But +P₁ is a fixed point, and − P 1 = − 1 a ∉ w l o c u ( P 1 ) = ( 1 2 a , ∞ ) , see Proposition (3-a).

By Remark (3-e) we have

h a , 0 − 2 ( P 1 ) = ∓ − P 1 a = ∓ − 1 ą a = ∓ − 1 a 2 ∉ ℝ ,

since 1 a 2 > 0 , ∀ a ∈ ℝ / { 0 } .

Therefore h a , 0 − n ( P 1 ) are undefined in ℝ with n ≥ 2 .

Thus the backward orbit of the repelling fixed point P₁ is undefined in ∎

For the family h a , b ( x ) = a x 2 + b , 0 belong to the backward orbit of P₁ whenever b = − 2 a with a ∈ ℝ / { 0 } , and the backward orbit of P₁ is:

h a , − 2 a − n ( P 1 = 2 a ) = { 2 a , − 2 a , 0 , 2 a , ⋯ } .

Proof:

We test the values of n which makes h a , b − n ( P 1 ) = 0 .

By Lemma (3-b), h a , b − 1 ( P 1 ) = ± P 1 .

So h a , b − 1 ( P 1 ) ≠ 0 .

Now suppose that h a , b − 2 ( P 1 ) = 0 , by Remark (3-e) then

h a , b − 2 ( P 1 ) = ∓ − P 1 − b a = 0 , thus

− P 1 − b a = 0

− P 1 − b = 0

P 1 = b .

Since the fixed point P 1 = 1 + 1 − 4 a b 2 a , therefore

1 + 1 − 4 a b 2 a = − b ,

then

1 + 1 − 4 a b = − 2 a b

1 − 4 a b = − 2 a b − 1

1 − 4 a b = 4 a 2 b 2 + 4 a b + 1

4 a 2 b 2 + 8 a b = 0 , which implies

4 a b ( a b + 2 ) = 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 a b + 2 = 0 , thus

b = − 2 a .

Now, P 1 = 2 a and to find the backward orbit of P₁, we consider

h a , − 2 a − 1 ( x ) = ± a x + 2 a .

By Lemma (3-b) h a , b − 1 ( P 1 ) = ± P 1 , then

h a , − 2 a − 1 ( 2 a ) = ± 2 a . But + 2 a is a fixed point, therefor

h a , − 2 a − 1 ( 2 a ) = − 2 a .

So

h a , − 2 a − 2 ( 2 a ) = a ( − 2 a ) + 2 a = 0 .

h a , − 2 a − 3 ( 2 a ) = a ( 0 ) + 2 a = 2 a , and so on.

Therefore the backward orbit of P 1 = 2 a is:

h a , − 2 a − n ( P 1 = 2 a ) = { 2 a , − 2 a , 0 , 2 a , ⋯ } . ∎

For h 1 , − 2 ( x ) = 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 ( 2 ) = { 2 , − 2 , 0 , 2 , ⋯ , 2 } .

If b > − 2 a for h a , b ( x ) ∈ 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 ( P 1 ) = ± P 1 , since +P₁ is a fixed point, then we consider

h a , b − 1 ( P 1 ) = − P 1 .

By Remark (3-e), h a , b − 2 ( P 1 ) = ∓ − P 1 − b a .

If − P 1 > b , then by Theorem (3-h),

b ≤ − 2 a which is a contradiction with b > − 2 a . Therefore

− P 1 < b , which implies

h a , b − 2 ( P 1 ) ∉ ℝ .

So h a , b − n ( P 1 ) are undefined in ℝ with n ≥ 2 .

Thus the backward orbit of P₁ is undefined .

So the intersection of W u ( P 1 ) with the backward orbit of P₁ is also undefined. ∎

If b = − 2 a for h a , b ( x ) ∈ H with a > 0 , then h a , − 2 a has a point of homoclinic tangency at 0 associated to P₁.

Proof:

By Theorem (3.3), h a , − 2 a − n ( P 1 = 2 a ) = { 2 a , − 2 a , 0 , 2 a , ⋯ } .

By Theorem (3-c), W u ( P 1 ) = ( 1 a − P 1 , ∞ ) , then

W u ( P 1 = 2 a ) = ( 1 a − 2 a , ∞ ) , i.e.

W u ( P 1 = 2 a ) = ( − 1 a , ∞ ) . Now

h a , − 2 a − n ( 2 a ) intersects W u ( P 1 = 2 a ) at 0.

By Lemma (3.1) 0 is a non-degenerate critical point. So h a , − 2 a has a point of homoclinic tangency at 0 associated to P₁. ∎

If b < − 2 a for h a , b ( x ) ∈ H with a > 0 , then the backward orbit of P₁ crosses the unstable set W u ( P 1 ) .

Proof:

First consider the backward orbit of P₁.

By Lemma (3-b) h a , b − 1 ( P 1 ) = ± P 1 .

But + P₁ is a fixed point, therefore we consider

h a , b − 1 ( P 1 ) = − P 1 .

By Remark (3-e), h a , b − 2 ( P 1 ) = ∓ − P 1 − b a .

Since b < − 2 a , then by Theorem (3-h)

h a , b − 2 ( P 1 ) ∈ ℝ .

Let h a , b − 2 ( P 1 ) = q 1 , 1 , h a , b − 3 ( P 1 ) = q 2 , 1 .

By Proposition (3-f), if b < − ( 5 + 2 5 ) 4 a , then q 1 , 1 ∈ W l o c u ( P 1 ) .

By Proposition (3-g), if − ( 5 + 2 5 ) 4 a ≤ b < − 2 a , then q 2 , 1 ∈ W l o c u ( P 1 ) .

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 ( P 1 ) ∩ W u ( P 1 ) ≠ ∅ . ∎

Following examples explain the cases for b > − 2 a , b = − 2 a and b < − 2 a (with a > 0 ) respectively.

For h 1 , − 1 ( x ) = 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 ( x ) is P 1 = 1 + 5 2 , and

h 1 , − 1 − 1 ( x ) = ± x + 1 .

The backward orbit of P 1 = 1 + 5 2

h 1 , − 1 − 1 ( 1 + 5 2 ) = ± 1 + 5 2 , where + 1 + 5 2 is a fixed point, therefore we consider

h 1 , − 1 − 1 ( 1 + 5 2 ) = − 1 + 5 2 . Now

h 1 , − 1 − 2 ( 1 + 5 2 ) = ∓ − 1 + 5 2 + 1 ∉ ℝ .

So h 1 , − 1 − n ( 1 + 5 2 ) are undefined in ℝ with n ≥ 2 .

Thus the backward orbit of P₁ is undefined.

So the intersection of W u ( 1 + 5 2 ) with the backward orbit of P₁ is also undefined. ∎

For h 1 , − 2 ( x ) = 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 ( x ) is P 1 = 2 .

By Example (3.4), The backward orbit of P 1 = 2 is

h 1 , − 2 − n ( 2 ) = { 2 , − 2 , 0 , 2 , ⋯ , 2 } .

On the other hand, the unstable set of P 1 = 2 is

W u ( 2 ) = ( − 1 , ∞ ) , (see Theorem (3-c)). Now

h 1 , − 2 − n ( 2 ) intersects W u ( 2 ) 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 . ∎

For h 1 , − 6 ( x ) = x 2 − 6 , the backward orbit of P₁ crosses the unstable set W u ( P 1 ) .

Solution:

First consider the fixed point P 1 = 3 .

The backward orbit of 3 is:

h 1 , − 6 − n ( 3 ) = { 3 , − 3 , 3 , ⋯ , 3 } (see Example (3-i)), with

h 1 , − 6 − 1 ( 3 ) = h 1 , − 2 2 ( 3 ) , and h 1 , − 6 − 2 ( 3 ) = 3 .

Since 3 is a homoclinic point of P 1 = 3 , then

3 ∈ W l o c u ( 3 ) .

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 ( 3 ) ∩ W u ( 3 ) ≠ ∅ . ∎

Note , the main theorem in the work :

h a , b ( x ) = a x 2 + b , a > 0 has a homoclinic bifurcation associated to the repelling fixed point of h a , b , P 1 = 1 + 1 − 4 a b 2 a , at b = − 2 a .

Proof:

1) For b > − 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 = − 2 a , by Theorem (3.6) h a , − 2 a has a point of homoclinic tangency associated to P₁ at x = 0 .

3) For b < − 2 a , by Theorem (3.7) the backward orbit of P₁ crosses the unstable set of P₁, W u ( P 1 ) .

Therefore h a , b has a homoclinic bifurcation associated to P₁ at b = − 2 a . ∎

h 1 , − 2 ( x ) = 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 ( x ) 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).

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 ( x ) = a x 2 + b , a < 0 has a homoclinic bifurcation associated to the repelling fixed point of h a , b , P 1 = 1 + 1 − 4 a b 2 a , at b = − 2 a .