For h a , b ( x ) ∈ H , h a , b − 2 ( P 1 ) = ∓ − P 1 − b a .

Proof:

By lemma (4.1), h a , b − 1 ( P 1 ) = ∓ P 1 . Then h a , b − 2 ( P 1 ) = h a , b − 1 ( ± P 1 ) = ∓ ± P 1 − b a . For h a , b − 2 ( P 1 ) = h a , b − 1 ( + P 1 ) = ∓ + P 1 − b a = ± P 1 , so we have the same state, so is omitted. For h a , b − 2 ( P 1 ) = h a , b − 1 ( − P 1 ) = ∓ − P 1 − b a .

Now if – P 1 ≥ b , in remark (5.1.1), then put h a , b − 2 ( P 1 ) = ∓ − P 1 − b a = ∓ q 1 , j where j ∈ N . For h a , b − 1 ( q 1 , j ) = h a , b − 3 ( P 1 ) = ∓ ± q 1 , j − b a = ∓ q 2 , j , where ± q 1 , j > b . Similarly for h a , b − 1 ( ± q 2 , j ) = h a , b − 4 ( P 1 ) = ∓ ± q 2 , j − b ą = ∓ q 3 , j , where ± q 2 , j > b .

And so on (see Figure 3). Thus we have sequences of preimage points of h a , b ( x ) , q i , j ∈ ℝ . If ∓ q i , j ∈ w l o c u ( P 1 ) , for some i , j ∈ N , then these points are the homoclinic points of the repelling fixed point P 1 .

In following propositions and examples we study the homoclinic points of the fixed point P 1 of h a , b ( x ) ∈ H for various values of a and b.

For h a , b ( x ) ∈ H . If b = 0 then the fixed point P 1 has no homoclinic points.

Proof:

It is clear that h a , 0 ( x ) = a x 2 and P 1 = 1 a . By proposition (3.1), w l o c u ( P 1 ) = ( 1 2 a , ∞ ) . Now the first preimage of h a , 0 ( x ) is h a , 0 − 1 ( x ) = ∓ x a where x > 0 . For P 1 = 1 a and by lemma (4.1), we have h a , 0 − 1 ( 1 a ) = ∓ 1 a 2 = ∓ 1 a = ∓ P 1 . But + P 1 = 1 a is a fixed point, and − P 1 = − 1 a ∉ w l o c u ( P 1 ) = ( 1 2 a , ∞ ) . Moreover, by proposition (5.1.1), we have

h a , 0 − 2 ( P 1 ) = ∓ − P 1 a = ± − 1 a 2 ∉ ℝ , since 1 a 2 > 0 . There for h a , 0 − n ( P 1 ) is undefined in ℝ for n ≥ 2 . Thus h a , 0 ( x ) = a x 2 has no homoclinic point to the fixed point P 1 .

For h 1 , 0 ( x ) = x 2 has no homoclinic points to the fixed point P 1 = 1 .

Solution:

By proposition (3.1), w l o c u ( P 1 ) = ( 1 2 , ∞ ) . The first preimage of h 1 , 0 ( x ) is

h 1 , 0 − 1 ( x ) = ∓ x where x > 0 . For h 1 , 0 − 1 ( 1 ) = ∓ 1 = ± 1 = ± P 1 . But P 1 = 1 is a fixed point of h 1 , 0 ( x ) , and − 1 ∉ w l o c u ( P 1 ) . Now for h 1 , 0 − 1 ( − 1 ) = h 1 , 0 − 2 ( 1 ) = ∓ − 1 ∉ ℝ . So h 1 , 0 ( x ) = x 2 has no homoclinic points to the fixed point P 1 .

For h a , b ( x ) ∈ H , the first preimage of the fixed point P 1 cannot be a homoclinic point to the fixed point P 1 .

Proof:

The first preimage of h a , b ( x ) is h a , b − 1 ( x ) = ∓ x − b a where x > b . For P 1 = 1 + 1 − 4 a b 2 a and by lemma (4.1) we have h a , b − 1 ( P 1 ) = ∓ P 1 − b a = ∓ P 1 where P 1 > b . But + P 1 is a fixed point, by remark (2.2) P 1 > 1 2 a , we have − P 1 < − 1 2 a and w l o c u ( P 1 ) = ( 1 2 a , ∞ ) .

So − P 1 ∉ w l o c u ( P 1 ) = ( 1 2 a , ∞ ) . Thus the fixed point P 1 has no homoclinic points at the first preimage.

h 1 , − 2 ( x ) = x 2 − 2 has no homoclinic points to the fixed point P 1 in the first preimage.

Solution:

The fixed point P 1 = 2 , and the first preimage of h 1 , − 2 ( x ) is h 1 , − 2 − 1 ( x ) = ∓ x + 2 . From proposition (3.1), w l o c u ( P 1 ) = ( 1 2 , ∞ ) . For P 1 = 2 then h 1 , − 2 − 1 ( 2 ) = ∓ 2 + 2 = ± 2 = ± P 1 . But P 1 = 2 is a fixed point of h 1 , − 2 ( x ) and − 2 ∉ w l o c u ( P 1 ) . So h 1 , − 2 ( x ) = x 2 − 2 has no homoclinic points to the fixed point P 1 in the first preimage.

Following remarkes assert that some inverse images of P 1 , h a , b − n ( P 1 ) are belong to the local unstable set of P 1 for b ≤ − 2 a .

If b < − ( 5 + 2 5 ) 4 a of h a , b ∈ H , then the second preimage of the fixed point P 1 belongs to the local unstable set of P 1 .

Proof:

Let b < − 5 − 2 5 4 a , it is clearly that b < − 5 + 2 5 4 a . There for a b + 5 + 2 5 4 < 0 and a b + 5 − 2 5 4 < 0 . So ( a b + 5 + 2 5 4 ) ( a b + 5 − 2 5 4 ) > 0 . Then 4 a 2 b 2 + 10 a b + 5 4 > 0 , thus 1 − 4 a b < 4 a 2 b 2 + 6 a b + 9 4 , which implies 1 − 4 a b < ( 2 a b + 3 2 ) 2 . Since ( 2 a b + 3 2 ) 2 = ( − ( 2 a b + 3 2 ) ) 2 , there for 1 − 4 a b < ( − 2 a b − 3 2 ) 2 , then 1 − 4 a b < − 2 a b − 3 2 . Thus 1 + 1 − 4 a b 2 a < − b − 1 4 a , i.e. P 1 < − b − 1 4 a . There for − P 1 − b a > 1 2 a . By remark (5.1.1) h a , b − 2 ( P 1 ) = ∓ − P 1 − b a , So h a , b − 2 ( P 1 ) > 1 2 a . Thus h a , b − 2 ( P 1 ) ∈ w l o c u ( P 1 ) . (See proposition (3.1)).

If − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a of h a , b ∈ H , then the third preimage of the fixed point P 1 belongs to the local unstable set of P 1 .

The proof is same of the above remark with some more complicated details.

For the family H = { h a , b ( x ) = a x 2 + b } , there exist homoclinic points to the fixed point P 1 whenever b ≤ − 2 a .

Proof:

According to a proposition (5.1.4), we begin with the second preimage of the fixed point P 1 . By remark (5.1.1), we have

h a , b − 2 ( P 1 ) = ± − P 1 − b a . Suppose that ± − P 1 − b a ∈ ℝ , i.e. − P 1 ≥ b , then P 1 ≤ − b . Since P 1 = 1 + 1 − 4 a b 2 a , where b < 1 4 a . Then 1 + 1 − 4 a b 2 a ≤ − b , which implies

1 − 4 a b ≤ − 2 a b − 1 .

Let c = − 2 a b − 1 . Since b < 1 4 a , then 1 − 4 a b > 0 , which implies 0 < 1 − 4 a b ≤ c , i.e. c ∈ ( 0 , ∞ ) ………(*).

Then we have three cases for b

Case 1: If b > 0 , then a b > 0 , thus − 2 a b < 0 , there for − 2 a b − 1 < − 1 , which implies c < − 1 which is a contradiction with c ∈ ( 0 , ∞ ) in (*).

Case 2: If b = 0 , then by proposition (5.1.2) there is no homoclinic points to the fixed point P 1 .

Case 3: If b < 0 , then a b < 0 . Thus − 2 a b > 0 , there for − 2 a b − 1 > − 1 , which implies c > − 1 , i.e. c ∈ ( − 1 , ∞ ) . Since c ∈ ( 0 , ∞ ) in (*), then c ∈ ( − 1 , ∞ ) ∩ ( 0 , ∞ ) = ( 0 , ∞ ) . There for 1 − 4 a b ≤ − 2 a b − 1 , implies that 1 − 4 a b ≤ 4 a 2 b 2 + 4 a b + 1 , thus 4 a b ( a b + 2 ) ≥ 0 ………(**).

Since a > 0 and b < 0 , then 4 a b < 0 , so by (**), a b + 2 ≤ 0 , thus a b ≤ − 2 . It follows that b ≤ − 2 a . By remark (2.1), h ′ a , b ( P 1 ) > 1 .

Now:

1) For b < − ( 5 + 2 5 ) 4 a , let h a , b − 2 ( P 1 ) = q 1 , 1 , by remark (5.1.6),

q 1 , 1 ∈ w l o c u ( P 1 ) and it is clear that h a , b 2 ( q 1 , 1 ) = P 1 (see Figure 3).

2) For − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a , let h a , b − 3 ( P 1 ) = q 2 , 1 , by remark (5.1.7), q 2 , 1 ∈ w l o c u ( P 1 ) and it is clear that h a , b 3 ( q 2 , 1 ) = P 1 (see Figure 3).

There for q 1 , 2 for − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a and q 1 , 1 for b < − ( 5 + 2 5 ) 4 a are the first homoclinic points for the repelling fixed point P 1 .

If b > − 2 a , then h a , b ( x ) = a x 2 + b has no homoclinic points to the fixed point P 1 .

Proof:

According to a proposition (5.1.4), we begin with the second preimage of the fixed point P 1 . By remark (5.1.1), we have

h a , b − 2 ( P 1 ) = ± − P 1 − b a . Since P 1 is a repelling fixed point for b < 1 4 a . So, with our assumption, we have − 2 a < b < 1 4 a . Now we divide the proof into three cases:

Case 1: If − 2 a < b < 0 , then − 2 < a b < 0 . Thus 0 < a b + 2 < 2 , and − 8 < 4 a b < 0 . So 4 a b ( a b + 2 ) < 0 , which implies 4 a 2 b 2 + 4 a b + 1 < 1 − 4 a b . Thus

( 2 a b + 1 ) 2 < 1 − 4 a b ……… (*).

Since ( 2 a b + 1 ) 2 = ( − ( 2 a b + 1 ) ) 2 and − 2 a < b < 0 , then 1 − 4 a b > 0 . Thus (*) will being − 2 a b − 1 < 1 − 4 a b , which implies, − b < 1 + 1 − 4 a b 2 a , i.e. − b < P 1 , which implies − P 1 − b a < 0 . So

h a , b − 2 ( P 1 ) = ± − P 1 − b a ∉ ℝ . There for h a , b − n ( P 1 ) ∉ ℝ , ∀ n ∈ N . Then h a , b ( x ) has no homoclinic points to the fixed point P 1 for − 2 a < b < 0 .

Case 2: If b = 0 , then by proposition (5.1.2) there is no homoclinic point to the fixed point P 1 .

Case 3: If 0 < b < 1 4 a , then 0 < a b < 1 4 . Thus 2 < a b + 2 < 9 4 , and 0 < 4 a b < 1 . So 4 a b ( a b + 2 ) > 0 , which implies 4 a 2 b 2 + 4 a b + 1 > 1 − 4 a b , thus ( 2 a b + 1 ) 2 > 1 − 4 a b ……… (*), Since

( 2 a b + 1 ) 2 = ( − ( 2 a b + 1 ) ) 2 and 0 < b < 1 4 a , then 1 − 4 a b > 0 . Thus (*) will being − 2 a b − 1 > 1 − 4 a b , which implies, − b > 1 + 1 − 4 a b 2 a , i.e. − b > P 1 , which implies − P 1 − b a > 0 . So h a , b − 2 ( P 1 ) = ± − P 1 − b a ∈ ℝ . But by theorem (5.1.8), if ± − P 1 − b a ∈ ℝ , then b ≤ − 2 a which is a contradiction with 0 < b < 1 4 a . There for h a , b ( x ) has no homoclinic points to the fixed point P 1 for 0 < b < 1 4 a .

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

For h 2 , − 3 ( x ) = 2 x 2 − 3 , has a homoclinic point to the fixed point P 1 at (0.86602540378).

Solution:

The fixed point P 1 = 3 2 = 1.5 and the first preimage of h 2 , − 3 ( x ) is h 2 , − 3 − 1 ( x ) = ∓ x + 3 2 . From proposition (3.1), w l o c u ( P 1 ) = ( 1 4 , ∞ ) . Clearly h ′ 1 , − 2 ( 1.5 ) > 1 . For P 1 = 1.5 , then h 2 , − 3 − 1 ( 1.5 ) = ∓ 1.5 + 3 2 = ± 1.5 = ± P 1 . But P 1 = 1.5 is a fixed point of h 2 , − 3 ( x ) , and − 1.5 ∉ w l o c u ( P 1 ) . Now for

h 2 , − 3 − 1 ( − 1.5 ) = h 2 , − 3 − 2 ( 1.5 ) = ∓ − 1.5 + 3 2 = ± 0.86602540378 ,

where

0.86602540378 ∈ w l o c u ( P 1 ) and − 0.86602540378 ∉ w l o c u ( P 1 ) . Moreover h 2 , − 3 ( 0.86602540378 ) = − 1.5 and h 1 , − 2 2 ( 0.86602540378 ) = 1.5 So (0.86602540378) is a homoclinic point to the fixed point P 1 .

Here we consider 0.86602540378 (the first) homoclinic point for the fixed point (1.5). In fact, there are many points belong to the local unstable set of 1.5 (i.e. homoclinic points to P 1 = 1.5 ). In fact

h 2 , − 3 − 1 ( 0.86602540378 ) = h 2 , − 3 − 3 ( 1.5 ) = ∓ 0.86602540378 + 3 2 = ∓ 1.390328271 .

Now, 1.390328271 ∈ w l o c u ( P 1 ) , and − 1.390328271 ∉ w l o c u ( P 1 ) . For

h 2 , − 3 − 1 ( − 0.86602540378 ) = h 2 , − 3 − 3 ( 1.5 ) = ∓ − 0.86602540378 + 3 2 = ∓ 1.032950772 ,

and

1.032950772 ∈ w l o c u ( P 1 ) , and − 1.032950772 ∉ w l o c u ( P 1 ) . If we continue with this way, we get a set { 0.86602540378 , 1.390328271 , 1.032950772 , ⋯ } . Every point in this set belongs to w l o c u ( P 1 ) . Each point of this set is a homoclinic point to the fixed point P 1 = 1.5 . See Figure 4.

For h 1 , − 2 ( x ) = x 2 − 2 , 2 is a homoclinic point to the fixed point P 1 .

Solution:

It is clear that P 1 = 2 , and the first preimage of h 1 , − 2 ( x ) is h 1 , − 2 − 1 ( x ) = ∓ x + 2 .

From proposition (3.1), w l o c u ( P 1 ) = ( 1 2 , ∞ ) . Clearly h ′ 1 , − 2 ( 2 ) > 1 . For P 1 = 2 , then h 1 , − 2 − 1 ( 2 ) = ∓ 2 + 2 = ± 2 = ± P 1 . But P 1 = 2 is a fixed point of h 1 , − 2 ( x ) , and − 2 ∉ w l o c u ( P 1 ) . Now for h 1 , − 2 − 1 ( − 2 ) = h 1 , − 2 − 2 ( 2 ) = ∓ − 2 + 2 = 0 ∉ w l o c u ( P 1 ) . Now for h 1 , − 2 − 1 ( 0 ) = h 1 , − 2 − 3 ( 2 ) = ∓ 0 + 2 = ∓ 2 .

2 ∈ w l o c u ( P 1 ) and − 2 ∉ w l o c u ( P 1 ) . Moreover h 1 , − 2 ( 2 ) = 0 , h 1 , − 2 2 ( 2 ) = − 2 and h 1 , − 2 3 ( 2 ) = 2 . So 2 is a homoclinc point to the fixed point P 1 = 2 .

In fact, any preimage point contained in w l o c u ( P 1 ) is a homoclinic point for P 1 . See Figure 5.

h 1 , − 1 ( x ) = x 2 − 1 , has no homoclinic points to the fixed point P 1 .

Solution:

It is clear that P 1 = 1 + 5 2 = 1.618033989 , and the first preimage of h 1 , 0 ( x ) is h 1 , − 1 − 1 ( x ) = ∓ x + 1 .

From proposition (3.1), w l o c u ( P 1 ) = ( 1 2 , ∞ ) . For P 1 = 1.618033989 , then h 1 , − 1 − 1 ( 1.618033989 ) = ∓ 1.618033989 + 1 = ± 1.618033989 = ± P 1 . But

+ P 1 = 1.618033989 is a fixed point of h 1 , 0 ( x ) , and − 1.618033989 ∉ w l o c u ( P 1 ) . Now for h 1 , − 1 − 1 ( − 1.618033989 ) = h 1 , − 1 − 2 ( 1.618033989 ) = ∓ − 1.618033989 + 1 = ∓ − 0.618033989 ∉ ℝ . So h 1 , − 1 ( x ) = x 2 − 1 has no homoclinic points to the fixed point P 1

For b > − 2 a , we proved in theorem (5.1.9), h a , b ( x ) has no homoclinic points, so h a , b ( x ) has no homoclinic orbits for the repelling fixed point P 1 .

To study the homoclinic orbits of h a , b ( x ) for b ≤ − 2 a , we introduce the following theorems and lemmas.

For h a , b ( x ) ∈ H with b ≤ − 2 a , if q < c where c ∈ ℝ is a constant and q is a homoclinic point to P 1 , then P 2 + 1 a < c .

Proof:

Let b ≤ − 2 a , then 1 − 4 a b ≥ 3 , which implies P 2 = 1 − 1 − 4 a b 2 a ≤ − 1 a , thus P 2 + 1 a ≤ 0 .

Now since q ∈ w l o c u ( P 1 ) , then q > 1 2 a > 0 (see proposition (3.1)). So it is clearly P 2 + 1 a < q , thus P 2 + 1 a < c .

( P 1 + P 2 + 1 a ) c − P 1 ( P 2 + 1 a ) = 2 c − b − P 1 a , where c ∈ ℝ is a constant.

Proof:

Since P 1 = 1 + 1 − 4 a b 2 a and P 2 = 1 − 1 − 4 a b 2 a .

( P 1 + P 2 + 1 a ) c − P 1 ( P 2 + 1 a ) = ( 1 + 1 − 4 a b 2 a + 1 − 1 − 4 a b 2 a + 1 a ) c − ( 1 + 1 − 4 a b 2 a ) ( 1 − 1 − 4 a b 2 a + 1 a ) = ( 2 2 a + 1 a ) c − ( 1 − ( 1 − 4 a b ) 4 a 2 ) − ( 1 + 1 − 4 a b 2 a a ) = ( 2 a ) c − ( 4 a b 4 a 2 ) − ( P 1 a ) = ( 2 a ) c − ( b a ) − ( P 1 a ) = 2 c − b − P 1 a

Let b ≤ − 2 a for h a , b ∈ H . If P 2 ≤ x ≤ P 1 , then 〈 h a , b − n ( x ) 〉 is an increasing sequence.

Proof:

The first preimage of h a , b ( x ) is h a , b − 1 ( x ) = x − b a .

Claim: h a , b − 1 ( x ) is increasing for P 2 ≤ x ≤ P 1 . To show this, let h a , b − 1 ( x ) ≥ x ,

that is x − b a ≥ x , thus a x 2 − x + b ≤ 0 , there for ( x − P 1 ) ( x − P 2 ) ≤ 0 . Then

either x ≤ P 1 and x ≥ P 2 , or x ≥ P 1 and x ≤ P 2 (omitted because there is no intersection). Hence h a , b − 1 ( x ) is increasing whenever P 2 ≤ x ≤ P 1 . Now h a , b − 1 ( h a , b − 1 ( x ) ) ≥ h a , b − 1 ( x ) . Which implies h a , b − 2 ( x ) ≥ h a , b − 1 ( x ) , ( h a , b − 1 is increasing). Thus, we have h a , b − ( n + 1 ) ( x ) ≥ h a , b − n ( x ) , for any n ∈ N .

Let b ≤ − 2 a and h a , b ( x ) ∈ H . If x ≥ P 1 , then 〈 h a , b − n ( x ) 〉 is a decreasing sequence.

The proof is the same as the above theorem.

For h a , b ( x ) ∈ H with b ≤ − 2 a , if P 2 ≤ x ≤ P 1 then the upper bound of the increasing sequence of preimages of x, 〈 h a , b − n ( x ) 〉 is P 1 .

Proof:

It is clear that h a , b − 1 ( x ) = ∓ x − b a . We will prove h a , b − n ( x ) ≤ P 1 by induction

Since x ≤ P 1 , then x − b a ≤ P 1 − b a . Since x ≥ P 2 and P 2 > b (because h a , b − 1 ( P 2 ) = ∓ P 2 − b a = ∓ P 2 so h a , b − 1 ( P 2 ) is undefined if P 2 < b ), then x > b . There for x − b a ≤ P 1 − b a . So, by lemma (4.1) then h a , b − 1 ( x ) ≤ P 1 ……….(*).

Now since h a , b ( h a , b − 2 ( x ) ) = h a , b − 1 ( x ) , i.e. a ( h a , b − 2 ( x ) ) 2 + b = h a , b − 1 ( x ) . Then by (*) we have a ( h a , b − 2 ( x ) ) 2 + b ≤ P 1 , thus h a , b − 2 ( x ) ≤ P 1 − b a . Thus by lemma (4.1), then h a , b − 2 ( x ) ≤ P 1 .

Now assume that h a , b − ( n − 1 ) ( x ) ≤ P 1 is true and we have to show h a , b − n ( x ) ≤ P 1 .

Since a ( h a , b − n ( x ) ) 2 + b = h a , b − ( n − 1 ) ( x ) . Then, with our assumption h a , b − ( n − 1 ) ( x ) ≤ P 1 , we get a ( h a , b − n ( x ) ) 2 + b ≤ P 1 , thus h a , b − n ( x ) ≤ P 1 − b a . So by lemma (4.1), h a , b − n ( x ) ≤ P 1 .

For h a , b ( x ) ∈ H with b ≤ − 2 a , if x ≥ P 1 then the lower bound of the decreasing sequence 〈 h a , b − n ( x ) 〉 is P 1 .

The proof is the same as the above theorem.

If b ≤ − 2 a , h a , b ( x ) ∈ H and, P 2 ≤ x ≤ P 1 then the supremum of the increasing sequence 〈 h a , b − n ( x ) 〉 is P 1 . sup { h a , b − n ( x ) : n ≥ 1 } = P 1 .

Proof:

The first preimage of h a , b ( x ) is h a , b − 1 ( x ) = x − b a .

If sup { h a , b − n ( x ) } ≠ P 1 . Let c < P 1 and sup { h a , b − n ( x ) } = c . There for h a , b − n ( x ) → c . By lemma (5.2.2) P 2 + 1 a < c , there for P 2 + 1 a < c < P 1 , which implies P 1 − c > 0 .

Now let ∈   = P 1 − c . Then ∃ k ∈ N such that | h a , b − n ( x ) − c | < P 1 − c , ∀ n > k , thus | ( a ( h a , b − ( n + 1 ) ( x ) ) 2 + b ) − c | < P 1 − c , which implies

c − P 1 < ( a ( h a , b − ( n + 1 ) ( x ) ) 2 + b ) − c < P 1 − c , thus

2 c − b − P 1 a < h a , b − ( n + 1 ) ( x ) < P 1 − b a . By lemma (4.1) and lemma (5.2.3), ( P 1 + P 2 + 1 a ) c − P 1 ( P 2 + 1 a ) < h a , b − ( n + 1 ) ( x ) < P 1 ……(*).

But c > P 2 + 1 a and c < P 1 , which implies ( c − ( P 2 + 1 a ) ) ( c − P 1 ) < 0 , thus c 2 − ( P 1 + P 2 + 1 a ) c + P 1 ( P 2 + 1 a ) < 0 , hence c 2 < ( P 1 + P 2 + 1 a ) c − P 1 ( P 2 + 1 a ) , there for c < ( P 1 + P 2 + 1 a ) c − P 1 ( P 2 + 1 a ) . Thus by (*), c < h a , b − ( n + 1 ) ( x ) < P 1 which is a contradiction with sup { h a , b − n ( x ) } = c . Thus sup { h a , b − n ( x ) } = P 1 .

For b ≤ − 2 a of the functions h a , b ( x ) ∈ H and x ≥ P 1 then the infimum of a decreasing sequence 〈 h a , b − n ( x ) 〉 is P 1 . inf { h a , b − n ( x ) : n ≥ 1 } = P 1 .

The proof is the same as the above theorem.

Finally, we introduce the main theorem in this section.

For h a , b ( x ) = a x 2 + b with b ≤ − 2 a and x ∈ ℝ , the homoclinic orbist of the (first) homoclinic points q 2 , 1 and q 1 , 1 are O ( q 2 , 1 ) = { P 1 , − P 1 , q 1 , 1 , q 2 , 1 , ⋯ , P 1 } for − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a and O ( q 1 , 1 ) = { P 1 , − P 1 , q 1 , 1 , ⋯ , P 1 } for b < − ( 5 + 2 5 ) 4 a .

Proof:

Since P 1 = 1 + 1 − 4 a b 2 a . Then

1) By Remark (2.1), for b < 1 4 a , h ′ a , b ( P 1 ) > 1 .

2) By theorem (5.1.8), h a , b 3 ( q 2 , 1 ) = P 1 for − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a and h a , b 2 ( q 1 , 1 ) = P 1 for b < − ( 5 + 2 5 ) 4 a (see Figure 3). So

h a , b n ( q 2 , 1 ) = P 1 for n ≥ 3 , h a , b n ( q 1 , 1 ) = P 1 for n ≥ 2 .

3) By theorems (5.2.4) and (5.2.8), (resp. (5.2.5) and (5.2.9)), 〈 h a , b − n ( q j , 1 ) 〉 where

j = 1 , 2 is an increasing with supremum (resp. decreasing with infimum) P 1 . Thus, h a , b − n ( q j , 1 ) → P 1 . There for, 1, 2 and 3 show that q 2 , 1 , q 1 , 1 are the homoclinic points for P₁ with the homoclinic orbits

O ( q 2 , 1 ) = { P 1 , − P 1 , q 1 , 1 , q 2 , 1 , ⋯ , P 1 } for − ( 5 + 2 5 ) 4 a ≤ b ≤ − 2 a and

O ( q 1 , 1 ) = { P 1 , − P 1 , q 1 , 1 , ⋯ , P 1 } for b < − ( 5 + 2 5 ) 4 a . See Figure 3.

Following examples explain the cases for b = − 2 a and b < − 2 a respectively.

For h 1 , − 2 ( x ) = x 2 − 2 , a homoclinic orbit of a homoclinic point 2 is

O ( 2 ) = { 2 , − 2 , 0 , 2 , ⋯ , 2 } .

Solution:

The forward orbit of 2 , h 1 , − 2 ( 2 ) = 0 , h 1 , − 2 ( 0 ) = − 2 , h 1 , − 2 ( − 2 ) = 2 , thus h 1 , − 2 3 ( 2 ) = 2 . So h 1 , − 2 n ( 2 ) = 2 for n ≥ 3 .

The backward orbit of 2 is h 1 , − 2 − n ( 2 ) = { 2 , 2 + 2 , 2 + 2 + 2 , ⋯ } .

To prove h 1 , − 2 − n ( 2 ) → 2 . Consider the sequence 〈 h 1 , − 2 − n ( 2 ) 〉 = 〈 a n 〉 .

〈 a n 〉 is an increasing sequence: It is easily shown that for − 1 ≤ x ≤ 2 the function x + 2 is an increasing function. Hence ⋯ ≥ h 1 , − 2 − 2 ( x ) ≥ h 1 , − 2 − 1 ( x ) ≥ x , thus 〈 a n 〉 is an increasing sequence.

Moreover for − 1 ≤ x ≤ 2 , by theorem (5.2.6) h 1 , − 2 − n ( x ) ≤ 2 , ∀ n (i.e. 2 is an upper bound for 〈 a n 〉 ). Thus to show that 〈 h 1 , − 2 − n ( x ) 〉 converges to 2, it is enough prove that 2 = sup { 〈 a n 〉 } . If 2 ≠ sup { 〈 a n 〉 } , let c < 2 and c = sup { 〈 a n 〉 } . By lemma (5.2.2) then 0 < c < 2 , since 〈 a n 〉 is an increasing sequence and c = sup { a n : n ∈ N } , then a n → c . Now, since 0 < c < 2 , then 2 − c > 0 . Let ∈   = 2 − c .

Then ∃ k ∈ N such that | a n − c | < 2 − c , ∀ n > k .

Since the iteration of this sequence is a n = a n + 1 2 − 2 , thus | ( a n + 1 2 − 2 ) − c | < 2 − c , which implies 2 c < a n + 1 2 < 4 , then 2 c < a n + 1 < 2 ………(*).

But c > 0 and c < 2 which implies c ( c − 2 ) < 0 , therefor c < 2 c . Thus by (*), c < a n + 1 < 2 which is a contradiction with c = sup { A 〈 a n 〉 } . Thus 2 = sup { A 〈 a n 〉 } , and a n → 2 . So O ( 2 ) = { 2 , − 2 , 0 , 2 , ⋯ , 2 } is a homoclinic orbit of a homoclinic point 2 for h 1 , − 2 ( x ) .

For h 1 , − 6 ( x ) = x 2 − 6 , a homoclinic orbit of a homoclinic point 3 is

O ( 3 ) = { 3 , − 3 , 3 , ⋯ , 3 } .

Solution:

The forward orbit of 3 , h 1 , − 6 ( 3 ) = − 3 , h 1 , − 6 ( − 3 ) = h 1 , − 6 2 ( 3 ) = 3 . So h 1 , − 6 n ( 3 ) = 3 for n ≥ 2 .

The backward orbit of 3 is h 1 , − 6 − n ( 3 ) = { 3 , 3 + 6 , 3 + 6 + 6 , ⋯ } .

To prove h 1 , − 6 − n ( 3 ) → 3 . Consider the sequence 〈 h 1 , − 6 − n ( 3 ) 〉 = 〈 a n 〉 .

〈 a n 〉 is an increasing sequence: It is easily shown that for − 2 ≤ x ≤ 3 the function x + 6 is an increasing function. Hence ⋯ ≥ h 1 , − 6 − 2 ( x ) ≥ h 1 , − 6 − 1 ( x ) ≥ x , thus 〈 a n 〉 is an increasing sequence.

Moreover, for − 2 ≤ x ≤ 3 , by theorem (5.2.6) h 1 , − 6 − n ( x ) ≤ 3 , ∀ n (i.e. 3 is an upper bound for 〈 a n 〉 ). Thus to show that 〈 h 1 , − 6 − n ( x ) 〉 converges to 3, it is enough prove that 3 = sup { 〈 a n 〉 } . If 3 ≠ sup { 〈 a n 〉 } , let c < 3 and c = sup { 〈 a n 〉 } . By lemma (5.2.2) then − 1 < c < 3 , since 〈 a n 〉 is an increasing sequence and c = sup { a n : n ∈ N } , then a n → c . Now, since − 1 < c < 3 , then 3 − c > 0 . Let ∈   = 3 − c .

Then ∃ k ∈ N such that | a n − c | < 3 − c , ∀ n > k .

Since the iteration of this sequence is a n = a n + 1 2 − 6 , thus | ( a n + 1 2 − 6 ) − c | < 3 − c , which implies 2 c + 3 < a n + 1 2 < 9 , then 2 c + 3 < a n + 1 < 3 ………(*).

But c > − 1 and c < 3 which implies ( c + 1 ) ( c − 3 ) < 0 , there for c < 2 c + 3 . Thus by (*), c < a n + 1 < 3 which is a contradiction with c = sup { 〈 a n 〉 } . Thus 3 = sup { 〈 a n 〉 } , and a n → 3 . So O ( 3 ) = { 3 , − 3 , 3 , ⋯ , 3 } is a homoclinic orbit of a homoclinic point 3 for h 1 , − 6 ( x ) .