Families of 6-Cycles of Third Order
Antonio Linero Bas and Daniel Nieves Roldán
Abstract The present paper deals with the existence of 6-cycles of a certain type of difference equations of third order. In concrete, it focuses on equations of the form x n +3 = x i g ( x j )h ( x k ), where i , j , k ∈{n , n + 1, n + 2} are pairwise distinct and g , h : (0, ∞) → (0, ∞) are continuous. The main result of the paper assures that the unique 6-cycle displaying such form is given by the potential one x n +3 =
(
)2
Keywords p -cycle · Functional equations · Equilibrium points · Homeomorphism · Monotonicity
1 Introduction
In general, for a set X ,amap f : Ω ⊂ X k → X defined on some subset of a finite Cartesian product of X , and x 1 ,..., x k elements of X (so-called initial conditions), we say that
(1)
is an autonomous difference equation of order k . Notice that, once we have introduced the initial conditions, the recurrence constructed by f gives a unique solution ( x n )∞ n =1 if that map f is well defined for any element ( x n +k 1 ,..., x n +1 , x n ). When we obtain a constant solution ( x , x ,... ) we say that x is an equilibrium point of the difference equation. Notice that x satisfies the equation x = f ( x ,..., x ).
Recall that ( x n )∞ n =1 is a periodic solution if x n +m = x n for all n ≥ 1 and some positive integer m . The smallest of such values is called the period of ( x n )∞ n =1 .If, additionally, all the solutions are periodic and p is the least common multiple of their
A. Linero Bas · D. Nieves Roldán (B) Departamento de Matemáticas, Universidad de Murcia (Spain), Murcia, Spain e-mail: daniel.nieves@um.es
A. Linero Bas e-mail: lineroba@um.es
© The Author(s), under exclusive license to Springer Nature Switzerland AG 2024 S. Olaru et al. (eds.), Difference Equations, Discrete Dynamical Systems and Applications, Springer Proceedings in Mathematics & Statistics 444, https:// doi.org/ 10.1007/ 978- 3- 031- 51049- 6_1
periods, we say that the difference equation is a p -cycle or is globally periodic of period p .
Maybe, the most popular p -cycle is the Lyness’ cycle x n +1 = 1+ x n x n 1 ,a 5-cycle. It was already known, in an implicit way, by Gauss when working in the spherical geometry of the pentagramma mirificum, a spherical pentagram formed by five successively orthogonal great-circle arcs. Its construction and the relation with the 5-cycle can be consulted in [9].
Let us comment that this 5-cycle receives the name of Lyness because R.C. Lyness accounted for it in a series of papers dealing with the existence of cycles, see [14–16].
Related to the study of global periodicity, different approaches are employed. For instance, techniques of discrete dynamical systems [3, 6, 7]; the resolution of functional equations [2, 17]; and even direct arguments of real analysis [18]. Moreover, we must highlight the problems of existence of families of p -cycles and the classification of p -cycles by conjugation. The main families of p -cycles that can be found in the literature are rational cycles [8, 10]; and potential cycles [5].
For the classification by conjugation let us recall that, given a metric space X ,we can associate to the difference Eq. (1), the following map F : X k → X k given by F ( x 1 ,..., x k ) =
(2)
So, if X 1 , X 2 are two metric spaces and g1 : X 1 → X 1 , g2 : X 2 → X 2 ,are two continuous maps, we way that ( X 1 , g1 ) and ( X 2 , g2 ) are topologically conjugate if there exists a homeomorphism h : X 1 → X 2 such that h ◦ g1 = g2 ◦ h . So, if Eq. (1) is a p -cycle of order k , and we consider an arbitrary homeomorphism α defined from (0, ∞) into itself, then F ( x 1 , x 2 ,..., x k ) = ( x 2 ,..., x k , f ( x k ,..., x 1 )) is topologically conjugate to
G ( x 1 , x 2 ,..., x k ) = ( x 2 ,..., x k ,α 1 ( f (α( x k ),...,α( x 1 ))))
Therefore if x n +k = f ( x n +k 1 ,..., x n ) is a p -cycle, then x n +k = α 1 ( f (α( x n +k 1 ),...,α( x k ))) is also a p -cycle. For instance, all the p -cycles of second order x n +2 = f ( x n +1 , x n ), with p ≥ 3, are topologically conjugate to rotations of the plane [3].
In [13], we proved that x n +3 = x n ( x n +2 x n +1 )2 is the unique potential 6-cycle of the form x n +3 = x i f ( x j , x k ), with i , j , k ∈{n , n + 1, n + 2} pairwise distinct and f : (0, ∞)2 → (0, ∞) continuous. This means that the cycle must have the form x n +3 = x i x α j j x αk k , where α j and αk are real numbers. Here, we go further and prove that, in fact, is the unique 6-cycle of third order exhibiting the form x n +3 = x i g ( x j )h ( x k ), (3)
with i , j , k ∈{n , n + 1, n + 2}, pairwise distinct, and g , h : (0, ∞) → (0, ∞) continuous.
The main goal of this paper is the proof of the following result, which will follow directly from below Theorems 2, 3 and 4
Theorem 1 Let g , h : (0, ∞) → (0, ∞) be continuous maps. Then, the unique 6cycle of the form
x n +3 = x i g ( x j )h ( x k ),
where i , j , k ∈{n , n + 1, n + 2} are pairwise distinct, is given by x n +3 = x n ( x n +2 x n +1 )2 .
It is worth mentioning that this gives a new proof for the location of 6-cycles different from that presented in [4, Theorem 3], where the proof was based on an incorrect result. In concrete,
Lemma 1 [1, Lemma 3.2.] Suppose that every solution of Eq. (1) is periodic of prime period p .If S : D p → D is continuous and symmetric, where D is a non-degenerate interval of real numbers, then the function
I S
),
where ( yn )∞ n =−k +1 is a solution of Eq. (1) with y0 = x 1 ,..., y k +1 = x k , is an invariant of Eq. (1) such that I S ( x k ,..., x 2 , x 1 ) = I S ( x 1 , x 2 ,..., x k )
The problem with such statement resides in the fact that the equality
is not consistent with the form which I S is defined. For instance, let us consider a 3-cycle of order k = 2, x n +2 = f ( x n +1 , x n ), and take the symmetric function S ( x 1 , x 2 , x 3 ) = x 1 + x 2 + x 3 . Notice that we can choose x 1 , x 2 so that x 3 = f ( x 2 , x 1 ) /= f ( x 1 , x 2 ) =˜ x 3 , since f is not symmetric. Hence, I S ( x 1 , x 2 ) = S ( x 1 , x 2 , x 3 ) = x 1 + x 2 +˜ x 3 /= x 1 + x 2 + x 3 = S ( x 1 , x 2 , x 3 ) = I S ( x 2 , x 1 ).
The paper is organized as follows. In Sect. 2 we present some general considerations concerning the set of equilibrium points of Eq. (3) and the value of g (1) and h (1). These considerations will apply for each of the possible combinations of indices i , j , k . In Sect. 3 we discuss the case i = n + 2, j = n + 1, k = n and we show that there are no 6-cycles, see Theorem 2. The same occurs for the combination i = n + 1, j = n + 2, k = n , as we will see in Theorem 3 of Sect. 4. Nevertheless, in Sect. 5 we will find in Theorem 4 the unique 6-cycle of the form enunciated in
Theorem 1. In each of the sections, the used tools will be the analysis of the sets of equilibrium points and the solution of appropriate functional equations. Finally, in Sect. 6 we state some open problems and forthcoming lines of research related to the topic.
2 General Considerations
In this section we are going to present two general considerations about the set of equilibrium points of Eq. (3) which will be useful in the development of each one of the possible combinations for such equation. Unless otherwise stated, we will assume that Eq. (3)isa 6-cycle. Observe that x is an equilibrium point of this equation if and only if x = x g ( x )h ( x ), i.e. g ( x )h ( x ) = 1. This allows us to describe its set of equilibrium points by the closed set
F ={ x > 0 : g ( x )h ( x ) = 1}.
We prove here the non-emptiness character of F . In the proof, Id |(0,∞) represents the identity map from (0, ∞) into itself.
Lemma 2 F =∅
Proof Notice that belonging to F is equivalent to being a fixed point of the discrete dynamical system φ associate to the difference equation, namely Φ : (0, ∞)3 → (0, ∞)3 ,Φ( x 1 , x 2 , x 3 ) = ( x 2 , x 3 , x i g ( x j )h ( x k )).
Since the difference equation is a p -cycle, p = 6, then Φ 6 ≡ Id |(0,∞) ,so Φ is a periodic homeomorphism. In that case, being Φ defined in a space homeomorphic to R3 , it is well-known (see [11], where the reader will find a brief explanation on the fact that any periodic homeomorphism on Rn has fixed points if n ≤ 4) that Φ possesses necessarily a fixed point, which means that the initial difference equation has an equilibrium point, and F =∅ ◻
Remark 1 The emptiness character of F in the previous lemma is crucial for our study. In general, a periodic homeomorphism, F : Rk → Rk , does not have necessarily fixed points (for instance, for an example in R7 , consult [11]).
Nevertheless, all the known p -cycles in the literature present equilibrium points or, equivalently, its associate discrete dynamical systems have fixed points. It is still an open problem to determine whether a p -cycle has necessarily fixed points. For a scope of the problem, see [12].
Concerning the value of the period p of the cycle, if the order k of the difference equation x n +k = f ( x n +k 1 ,..., x n +1 , x n ) is k = 1, necessarily p = 1 or p = 2 (the identity or an involution in the real line, for instance, x n +1 = x n or x n +1 = 1 x n for (0, ∞)); when k ≥ 2, the period p can reach the value of any positive integer,
to see it, take for instance a linear homogeneous difference equation of order k whose characteristic polynomial have p th-roots of the unity as eigenvalues, and select them by conjugated pairs, adding if necessary 1 or 1 (for instance, for the 7th-roots λ1 = cos ( 2π 7 ) + i · sin ( 2π 7 ), λ2 = λ1 , we form the difference equation x n +2 = 2cos ( 2π 7 ) x n +1 x n ,a 7-cycle). ◻
Remark 2 Observe that, without loss of generality, we can assume that 1 ∈ F and g (1) = h (1) = 1. Indeed, since F =∅,let z 0 ∈ F and, firstly, write Eq. (3)as x n +3 = x i ( g ( x j ) g (
).
In this case, we have g ( z 0 ) = 1 = h ( z 0 ) since g ( z 0 )h ( z 0 ) = 1. Secondly, apply the transformation x = z 0 y to obtain z 0 yn +3 = z 0 yi g ( z 0 y j ) h ( z 0 yk ) , yn +3 = yi ( g ( z 0 y j ) g ( z 0 ) ) (g ( z 0 )h ( z 0 yk )) =: yi γ( y j )η( yk ).
It is easily seen that this new equation continues to be a 6-cycle, and γ(1) = η(1) = 1 Thus, unless otherwise stated, in the next sections we will assume that g (1) = h (1) = 1, 1 ∈ F ◻
3 The Case xn+3 = xn+2 g ( xn+1 ) h ( xn )
In order to prove that the difference equation
cannot be a 6-cycle, firstly we present some general necessary properties for the equation to be a 6-cycle, related with the fact that h must be a homeomorphism and with the set of equilibrium points F associate to the equation. Secondly, we will divide our study in several steps, depending on the cardinality of F (≥ 2 or 1) and the monotonic character of h . The complete analysis will be collected in Theorem 2
Lemma 3 Suppose that Eq. (4)isa 6-cycle. Then h is a homeomorphism.
Proof First, we prove that h is surjective. Given a value y > 0, take the initial conditions x 1 = 1, x 2 = 1, x 3 = y g (1). By the global periodicity, we obtain a periodic sequence of period a divisor of 6. In particular, y g (1) = x 3 = x 9 = x 8 g ( x 7 )h ( x 6 ) = x 2 g ( x 1 )h ( x 6 ) = g (1)h ( x 6 ). Then, h ( x 6 ) = y which proves the surjectivity of h
To prove the injectivity, suppose that h ( x ) = h ( y ) for some values x , y > 0 Consider, on the one hand, the initial conditions x 1 = x , x 2 = x 3 = 1, and on the
other hand, y1 = y , y2 = y3 = 1.If h ( x ) = h ( y ) it is easily seen that x n = yn for all n ≥ 2. In particular, x = x 1 = x 7 = y7 = y1 = y ,so x = y . ◻
Remark 3 Notice that Lemma 3 implies that h is strictly monotonic.
In the development of our discussion, the following properties on the set of equilibrium points F will be essential for the case in which Card (F ) ≥ 2.
Lemma 4 Assume that Eq. (4)isa 6-cycle. Let u , x ∈ F , with u /= x , u < x . Then:
(a) x = u g (u )h ( x g ( x )h (u ))
(b) u g (u )h ( x ) ∈ F . In fact, if we put x 0 = x , then the sequence ( x k )k defined recursively as x k +1 = u g (u )h ( x k ), k ≥ 0, is contained in F . Moreover, x 1 / ∈{ x , u }.
(c) x g ( x )h (u ) ∈ F . In fact, if we put u 0 = u , then the sequence (u k )k defined recursively as u k +1 = x g ( x )h (u k ), k ≥ 0, is contained in F . Moreover, u 1 / ∈{ x , u }.
(d) It holds u x n = g ( x n ) g ( x n +1 ) = h ( x n +1 ) h ( x n ) for all n ≥ 0.
(e) If h is increasing, x 1 , u 1 ∈ (u , x ).
(f) If h is decreasing, then x 1 < u < x 2 < x and u < u 1 < x < u 2 , where x 1 , x 2 , u 1 , u 2 are defined in Part (b) and Part (c).
(g) If h is decreasing, then inf F = 0 and sup F =∞. In fact, F = (0, ∞).
Proof (a) Set z 1 = u , z 2 = u , z 3 = x to obtain (recall that g ( z )h ( z ) = 1 if z ∈ { x , u }): z 4 = x , z 5 = x g ( x )h (u ), z 6 = x g ( x )h (u ). From the global periodicity,
x = z 3 = z 9 = z 8 g ( z 7 ) h ( z 6 ) = z 2
( z 1 ) h ( z 6 ) = u g (u ) h ( x g ( x )h (u )).
(b)Take z 1 = x , z 2 = x , z 3 = u .Then z 4 = u , z 5 = u g (u )h ( x ), z 6 = u g (u )h ( x ). Again, by global periodicity we find x = z 2 = z 8 = z 7 g ( z 6 )h ( z 5 ) = z 1 g ( z 6 )h ( z 5 ) = x g (u g (u )h ( x ))h (u g (u )h ( x ))
Then 1 = g (u g (u )h ( x ))h (u g (u )h ( x )), which implies u g (u )h ( x ) ∈ F .
If we repeat the process with the points x 1 := u g (u )h ( x ) and u , we get an element of F , namely x 2 := u g (u )h ( x 1 ). By induction it is a simple matter to see that x k +1 defined recursively as x k +1 = u g (u )h ( x k ), k ≥ 0, is a point in F .
We proceed to prove that x 1 / ∈{ x , u }. Suppose that u g (u )h ( x ) = x 1 = x . Returning to the above sequence generated by z 1 = x , z 2 = x , z 3 = u , we would find u = z 3 = z 9 = z 8 g ( z 7 )h ( z 6 ) = x g ( x )h (u g (u )h ( x )) = x g ( x )h ( x ) = x , so u = x ,a contradiction.
Suppose that u g (u )h ( x ) = x 1 = u . Now, the initial conditions z j , j = 1, 2, 3, would generate the sequence ( z j ) j ≥1 = ( x , x , u , u , u , u ,...), consequently u = x by global periodicity, a new contradiction.
(c) Its proof is completely analogous to Part (b). Now it suffices to consider initial conditions u , u , x
(d) Let x 0 = x and x n +1 = u g (u )h ( x n ) for all n ≥ 0. We have seen in (a) that, from the initial terms x , x , u , we deduce that u = x g ( x )h ( x 1 ). Thus, x 1 = u g (u )h ( x ) =
( x g ( x )h ( x 1 ))g (u )h ( x ) = xh ( x 1 )g (u )(g ( x )h ( x )) = xh ( x 1 )g (u ) and uh ( x ) = xh ( x 1 ). Therefore, u x = h ( x 1 ) h ( x ) Since x j ∈ F , j = 0, 1, then h ( x j ) = 1 g ( x j ) , j = 0, 1, and we obtain
u x 0 = h ( x 1 ) h ( x 0 ) = g ( x 0 ) g ( x 1 )
Now, if we do the same reasoning with the initial conditions x n , x n , u , n ≥ 1, analogously
u x n = h ( x n +1 ) h ( x n ) = g ( x n ) g ( x n +1 ) .
(e) Suppose that h is increasing. Since u < x ,wehave x 1 = u g (u )h ( x ) ≥ u g (u )h (u ) = u . Since x 1 /= u , we find x 1 > u . To see now that x 1 < x , assume on the contrary that u < x < x 1 = u g (u )h ( x ) (notice that x 1 /= x ). In this case, by Part (a), interchanging the roles of u and x , and using that h is increasing with u < x < x 1 , we obtain u = x g ( x )h (u g (u )h ( x )) = x g ( x )h ( x 1 ) ≥ x g ( x )h ( x ) = x , so u > x (recall u /= x ), a contradiction. Therefore, u < x 1 < x .
The proof of u < u 1 < x is completely analogous and we omit it.
(f) Suppose that h is decreasing. Firstly, we claim that x 2 /= u . Indeed, otherwise, u = x 2 = u g (u )h ( x 1 ) would imply g (u )h ( x 1 ) = 1, and from Part (d) we would find u x = h ( x 1 ) h ( x ) = 1 g (u ) h ( x ) , that is, u g (u )h ( x ) = x . Being u g (u )h ( x ) = x 1 , we would obtain x 1 = x = u , a contradiction.
Next, since u < x ,the value x 1 satisfies x 1 = u g (u )h ( x ) ≤ u g (u )h (u ) = u , hence x 1 < u < x (we use that x 1 /= u ). Next, take into account that x 2 = u g (u )h ( x 1 ), with x 1 < u < x to deduce that x 2 ≥ u g (u )h (u ) = u (recall that x 2 /= u ). Moreover, by force x 2 < x , because otherwise, if x ≤ x 2 , from Part (d) and the monotonicity of h we would obtain 1 < u x 1 = h ( x 2 ) h ( x 1 ) ≤ h ( x 1 ) h ( x 1 ) = 1, a contradiction.
(g) Firstly, let us see that inf F = 0 and sup F =∞. Arguing by contradiction, if inf F = a > 0, due to the continuity of the functions g and h , a = min F ∈ F . So, we could apply the same reasoning as in part (f) with u = a < x and we would get values of F that are smaller than a , arriving to a contradiction. A similar reasoning applies to see that sup F =∞. Therefore, for any two values in F ,by ( f ) with suitable values u , x , there exists another value of F between them. Thus, by density we obtain F = (0, ∞).
3.1 The Case Card (F ) ≥ 2
To prove that this case does not provide 6-cycles of the form of Eq. (4), we distinguish two cases, according to the monotonicity of h .
The case Card (F ) ≥ 2, h increasing: In the increasing case, we will use the general properties of F established in Lemma 4 to derive the non-existence of 6cycles.
Proposition 1 Consider Eq. (4) and assume that h is an increasing homeomorphism with Card (F ) ≥ 2. Then, the difference equation cannot be a 6-cycle.
Proof Let x , y be two different points in F , with x < y . By Lemma 4-(b)-(c)-(e), respectively, we know that x g ( x )h ( y ) and y g ( y )h ( x ) belong to F , they are different from x , y , and they are in the interior of the subinterval having endpoints x , y .
Since h is increasing and x < y g ( y )h ( x )< y , from the equality established in Lemma 4-(a), y = x g ( x ) h ( y g ( y )h ( x )), we deduce that y = x g ( x ) h ( y g ( y )h ( x )) < x g ( x ) h ( y ), therefore y < x g ( x )h ( y ), which contradicts our previous observation on the location of the value x g ( x )h ( y ) in the open interval ( x , y ) ◻
The case Card (F ) ≥ 2, h decreasing: When h is decreasing, we will present a more elaborated proof lying in the study of certain functional equation. Firstly, by Lemma 4-(g), F = (0, ∞). This means that Eq. (4) can be formulated as x n +3 = x n +2 g ( x n +1 ) g ( x n ) or x n +3 = x n +2 h ( x n ) h ( x n +1 ) . We are going to prove that there are no 6-cycles displaying this form. To see it, we argue by contradiction, by supposing that the difference equation is a 6-cycle. Recall that we are supposing that 1 ∈ F , with g (1) = h (1) = 1
Lemma 5 The functional equation
x = ϕ ( x ϕ(
(5) has no solutions in the family of continuous maps defined from (0, ∞) into itself.
Proof Let ϕ : (0, ∞) → (0, ∞) be a solution of (5). If ϕ satisfies the functional equation, obviously Im (ϕ) = (0, ∞). In particular, 1 ∈ Im (ϕ). Next, we claim that ϕ(1) = 1,so ϕ( x ) /= x if x /= 1. To see it, being ϕ 1 ({1}) =∅, suppose that ϕ( z ) = 1 for some z > 0. Then, z = ϕ ( z ϕ( z ) ) = ϕ ( z 1 ) = ϕ( z ) = 1, which implies that z = 1 This ends the claim.
According to the above properties, we can distinguish three possibilities for the solution of (5) (the cases ϕ( x )< 1 for all x /= 1 and ϕ( x )> 1 for all x /= 1 are rejected due to the fact that Im (ϕ) = (0, ∞)):
(i) x <ϕ( x )< 1 (see Fig. 1). Here, x ϕ( x ) < 1 if x < 1 and, according to the case, with ˜ x = x ϕ( x ) , we deduce x = ϕ ( x ϕ( x ) ) > x ϕ( x ) . Therefore, ϕ( x )> 1 for all x < 1, a contradiction.
(ii) ϕ( x )< x < 1 (see Fig. 1). Now, necessarily ϕ( z )> z if z > 1, otherwise we would have, from the initial claim, that ϕ(w)<w for all w /= 1 and then w = ϕ ( w ϕ(w) ) < w ϕ(w) , so ϕ(w)< 1 for all w /= 1, a contradiction, since Im(ϕ) = (0, ∞). Then, if z > 1 we have z ϕ( z ) < 1 and, by hypothesis, z = ϕ ( z ϕ( z ) ) < 1, a contradiction.
(iii) ϕ( x )> 1 for all x < 1 (see Fig. 1). In that situation, x ϕ( x ) < 1 for all x < 1, and consequently, x = ϕ ( x ϕ( x ) ) > 1, a new contradiction. ◻
Fig. 1 The solution ϕ , Cases (i)–(ii) ϕ( x )< 1 and Case (iii) ϕ( x )> 1 if x < 1
Corollary 1 The functional equation x = ϕ ( x ϕ( x ) ) , x > 0, (6) has no solutions in the family H of homeomorphisms ϕ from (0, ∞) into itself.
Proposition 2 If Card (F ) ≥ 2 and h is a decreasing homeomorphism, then Eq. (4) is not a 6-cycle.
Proof Recall that F = (0, ∞), g ( x ) = 1 h ( x ) for all x > 0 and g (1) = h (1) = 1.
Consider the initial conditions x 1 = 1, x 2 = 1, x 3 = x , with x > 0 arbitrarily taken. Then, x 4 = x 3 g ( x 2 )h ( x 1 ) =
, x 5 = x 4 g ( x 3 )h ( x 2 ) = x g ( x ) = x
) and x 6 = x 5 g ( x 4 )h ( x 3 ) = x h ( x ) g ( x )h ( x ) = x h ( x ) . From the global periodicity and the fact that F = (0, ∞), we obtain x = x 3 = x 9 = x 8 g ( x 7 )h ( x 6 ) = x 2 g ( x 1 )h ( x 6 ) = h ( x h ( x ) ) , that is, x = h ( x h ( x ) ) forall x > 0 (7)
Finally, Corollary 1 gives the non-existence of solutions for this equation, therefore Eq. (4) cannot provide a 6-cycle. ◻
Remark 4 We emphasize that the last result also works if h is increasing and F = (0, ∞). Nevertheless, we have only payed attention to the decreasing case, because in the increasing case we would need to demonstrate previously that the existence of two different points in F imply that really F = (0, ∞). Another reason to separate the increasing and decreasing cases into two propositions, Proposition 1 and Proposition 2, is to present two perspectives to attack the problem, the second one related with the resolution of functional equations. Maybe, the decreasing case could be studied with the same strategy applied to the increasing case, following the ideas in Lemma 4.
3.2 The Case Card (F ) = 1
Suppose that F is a singleton, say F ={ z 0 }. Hence, g ( z 0 )h ( z 0 ) = 1. Recall that, without loss of generality, we can assume that z 0 = 1 and g (1) = h (1) = 1,aswe stated at the end of Sect. 2
Lemma 6 Under the above conditions, if Eq. (4)isa 6-cycle, then for all x > 0 it holds:
(a) h 1 ( x ) = x (g ( x ))2 h ( x ).
(b) h 1 ( x ) = 1 g ( x g ( x ))h ( x )
(c) 1 = g ( 1 g ( x g ( x ))h ( x ) ) h ( x g ( x )).
Proof Since h is a homeomorphism by Lemma 3, given an arbitrary value x > 0,we take the initial conditions x 1 = h 1 ( x ), x 2 = 1, x 3 = 1 and apply Eq. (4) to obtain x 4 = x , x 5 = x , x 6 = x g ( x ). Then by the global periodicity, h 1 ( x ) = x 1 = x 7 = x 6 g ( x 5 )h ( x 4 ) = x (g ( x ))2 h ( x ), 1 = x 2 = x 8 = x 7 g ( x 6 )h ( x 5 ) = x 1 g ( x 6 )h ( x 5 ) = h 1 ( x )g ( x g ( x ))h ( x ), 1 = x 3 = x 9 = x 8 g ( x 7 )h ( x 6 ) = x 2 g ( x 1 )h ( x 6 ) = g (h 1 ( x )) h ( x g ( x )),
which are precisely the three formulas of the statement. ◻
Lemma 7 Let Eq. (4)bea 6-cycle. Suppose that F ={1}. Then:
(a) x ∈ Fix(h ) if and only if (g ( x ))2 = 1 x
(b) If x ∈ Fix(h ), then √ x ∈ Fix(h ). Moreover, x = 1 is the unique fixed point of h
Proof (a) This follows as a direct consequence of Lemma 6-(a). (b) Let x ∈ Fix(h ),so h ( x ) = x = h 1 ( x ). By Lemma 6-(b), we deduce that x = 1 g ( x g ( x )) x , or x 2 g ( x g ( x )) = 1 (8)
By Part (a), g ( x ) = / 1 x . Replacing this equality into (8) yields x 2 g ( x / 1 x ) = 1, that is, g (√ x ) = 1 x 2 . Next, we use Lemma 6-(c) to obtain 1 = g ( 1 g ( x g ( x ))h ( x ) ) h ( x g ( x )) = g ( 1 g (√ x ) h ( x ) ) h (√ x ) = g ( 1 1 x 2 x ) h (√ x ) = g ( x ) h (√ x ) = / 1 x h (√ x ) .
Therefore, h (√ x ) = √ x and √ x ∈ Fix(h ).
Finally, suppose that x ∈ Fix(h ) and prove that x = 1. By Part (a), g ( x ) = 1 √ x , and also √ x ∈ Fix(h ). In particular g (√ x ) = 1 √√ x . Furthermore, we have seen that g (√ x ) = 1 x 2 . We then deduce that 1 √√ x = 1 x 2 , or simplifying, x = x 8 . The unique positive real solution of this equation is x = 1. ◻
Lemma 8 Let Eq. (4)bea 6-cycle. Suppose that F ={1}. Then, for all x > 0,
and
Set
h ( x ). By global periodicity, 1 = x
hence, we conclude that h ( x ) =
. On the other hand, 1 = x 2 = x
), and since h is a homeomorphism, x g ( x ) = h 1 ( 1 g ( x (g ( x ))2 h ( x )) ). Finally, by Lemma 6-(a), we obtain x g ( x ) = h 1 ( 1 g (h 1 ( x )) )
From now on, since 1 is the unique fixed point of h , we distinguish three cases, namely:
(a) h is increasing, with h ( x )> x if 0 < x < 1 (b) h is increasing, with h ( x )< x if 0 < x < 1 (c) h is decreasing.
In the above cases, we will use that Card (F ) = 1, with F ={1}, therefore for the rest of values x /= 1 it must be either g ( x )h ( x )< 1 for all x ∈ (0, 1) or the reverse inequality g ( x )h ( x )> 1 for all x ∈ (0, 1).
Case (a): h increasing, h ( x )> x if 0 < x < 1 (see Fig. 2). Suppose that Eq. (4) is a 6-cycle.
Fig. 2 Case h increasing, h ( x )> x if 0 < x < 1
(a.1) Let h ( x )g ( x )> 1, for all x ∈ (0, 1). According to Lemma 6-(a), h 1 ( x ) = x (g ( x ))2 h ( x ), and the symmetry of h 1 and h with respect to the diagonal y = x , we have
> h 1 ( x ) = x (g (
))2 h (
) = x g ( x )(g ( x )h ( x )) > x g ( x ),
that is, g ( x )< 1 if 0 < x < 1. This implies that h ( x )> 1 for all x ∈ (0, 1), a contradiction. So, we reject this case.
(a.2) Let h ( x )g ( x )< 1, for all x ∈ (0, 1). Reasoning as in Case (a.1), we find h 1 ( x ) = x (g ( x ))2 h ( x )< x g ( x ). On the other hand, being h ( x )> x in (0, 1) we also deduce that h 1 ( x ) = x (g ( x ))2 h ( x )> x 2 (g ( x ))2 From the two inequalities for h 1 ( x ) we conclude that x g ( x )< 1 in (0, 1). Since, in addition, x g ( x ) = h 1 ( 1 g (h 1 ( x )) ) (see Eq. (10)), we deduce that h 1 ( 1 g (h 1 ( x )) ) < 1 for all x ∈ (0, 1), which implies that 1 g (h 1 ( x )) < 1 or g (h 1 ( x ))> 1 in (0, 1). Being h 1 |(0,1) a homeomorphism from (0, 1) into itself, to obtain g (w)> 1 for all w ∈ (0, 1). Thus, x < x g ( x )< 1 in (0, 1).
By Lemma 6-(b), and our hypothesis h 1 ( x )< 1,if 0 < x < 1 we have 1 g ( x g ( x ))h ( x ) < 1, that is 1 < g ( x g ( x ))h ( x ). Finally, from the monotonicity of h applied to x < x g ( x )< 1 we obtain 1 < g ( x g ( x ))h ( x )< g ( x g ( x ))h ( x g ( x )) < 1 due to our assumption on the product h ( z )g ( z )< 1 in (0, 1). We derive a contradiction, thus we must also reject Case (a.2).
We summarize our above study in this result:
Proposition 3 In Eq. (4), suppose that h is increasing, with h ( x )> x if 0 < x < 1, and that the set of equilibrium points reduces to F ={1}. Then, Eq. (4) cannot be a 6-cycle.
Case (b): h increasing, h ( x )< x if 0 < x < 1. In this part, we suppose that Eq. (4) is a 6-cycle, with F ={1}, h increasing and h ( x )< x < h 1 ( x )< 1 if 0 < x < 1 (interchange the roles of h and h 1 in Fig. 2). Again, since F ∩ (0, 1) =∅, we have to discuss two cases, namely, (b.1) h ( x )g ( x )> 1 for all x ∈ (0, 1) or (b.2) h ( x )g ( x )< 1 for all x ∈ (0, 1).
(b.1) Let g ( x )h ( x )> 1 for all x ∈ (0, 1). Since h ( x )< 1, we deduce that g ( x )> 1 in (0, 1). Also, by Lemma 6-(a) and the monotonicity of h , x > h ( x ) = h 1 ( x )
(g ( x ))2 >
(g ( x ))2 = 1 (g ( x ))2 ,
hence x (g ( x ))2 > 1. At the same time, we also find x g ( x )> h ( x )g ( x )> 1, and consequently by Eq. (10), x g ( x ) = h 1 ( 1 g (h 1 ( x )) ) > 1,
that implies g (h 1 ( x )) < 1 for all 0 < x < 1. Since h 1 is an increasing homeomorphism with h (1) = 1, putting w = h 1 ( x ) we conclude that g (w)< 1 for all w ∈ (0, 1), contrary to our initial hypothesis. Therefore, Case (b.1) does not generate 6-cycles of the form of (4).
(b.2) Suppose that g ( x )h ( x )< 1 for all x ∈ (0, 1). Since x < h 1 ( x ) = x (g ( x ))2 h ( x ) = x g ( x )(g ( x )h ( x )) < x g ( x ) (we have applied Lemma 6-(a)), we obtain that g ( x )> 1 in (0, 1)
If, additionally, x g ( x )> 1 for some x < 1, then from Eq. (10) we get that h 1 ( 1 g (h 1 ( x )) ) > 1. Hence we obtain g (h 1 ( x ))< 1. Setting w = h 1 ( x ), with w< 1, we deduce that g (w)< 1, a contradiction with the fact that g (w)> 1 in (0, 1). On the other hand, if x g ( x )< 1 for some x < 1, then we have x < x g ( x )< 1 and from Lemma 6-(b), i.e. 1 > h 1 ( x ) = 1 g ( x g ( x ))h ( x ) , we deduce that g ( x g ( x )) h ( x )> 1. Then apply that h is increasing to arrive to 1 < g ( x g ( x ))h ( x )< g ( x g ( x ))h ( x g ( x )) < 1, a contradiction. We conclude that x g ( x ) = 1 for all x ∈ [0, 1]. Then, Lemma 6-(b) implies h 1 ( x ) = 1 h ( x ) , and therefore, h ( x )h 1 ( x ) = 1 for all x ∈[0, 1], which is impossible since h ( x )< 1 and h 1 ( x )< 1 for all x ∈[0, 1]. To sum up, Case (b.2) does not produce 6-cycles of the form (4). Our discussion can be collected in this result:
Proposition 4 In Eq. (4), suppose that h is increasing, with h ( x )< x if 0 < x < 1, and that the set of equilibrium points reduces to F ={1}. Then, Eq. (4) cannot be a 6-cycle.
Case (c): h decreasing. Next, we suppose that Eq. (4)isa 6-cycle, with F ={1}, h decreasing (see Fig. 3). Since F ∩ (0, 1) =∅, it is necessary to distinguish two cases, (c.1) h ( x )g ( x )< 1 for all x ∈ (0, 1) or (c.2) h ( x )g ( x )> 1 for all x ∈ (0, 1).Asa preliminary, notice that if h is decreasing, by the symmetry of the inverse map h 1 with respect to the diagonal, we know that h 1 is also decreasing with h 1 ( x )> 1 if x < 1.
Fig. 3 Case h decreasing, h ( x )> 1 if 0 < x < 1 and h ( x )< 1 if x > 1
(c.1) Here, by force, g ( x )< 1 for all x ∈ (0, 1). Then x g ( x )< x < 1 in (0, 1). We claim that g ( z )< 1 for all z /= 1. It only remains to see that g ( z )< 1 for all z > 1.
By Eq. (10) and the fact that x g ( x )< 1 in (0, 1),wehave 1 > x g ( x ) = h 1 ( 1 g (h 1 ( x )) ) for all 0 < x < 1. Then, being h decreasing, 1 g (h 1 ( x )) > 1 or g (h 1 ( x ))< 1 for all x ∈ (0, 1). Since h 1 |(0,1) is a homeomorphism from (0, 1) to (1, ∞), setting w = h 1 ( x ) we have g (w)< 1 for all w> 1. This ends the claim. Next, we distinguish the following cases:
– h ( z )g ( z )> 1 in (1, ∞). Then, according to the above claim, necessarily h ( z )> 1 for all z > 1, a contradiction (see Fig. 3).
– h ( z )g ( z )< 1 in (1, ∞). Use Lemma 6-(b) to any z > 1 to obtain 1 > h 1 ( z ) = 1
g ( z g ( z ))h ( z ) , so g ( z g ( z ))h ( z )> 1 As h ( z )< 1, we deduce that g ( z g ( z )) > 1 for each z > 1, which is impossible because g (u ) ≤ 1 for all u ∈ (0, ∞).
Therefore, Case (c.1) derives in a contradiction and we reject this case. (c.2) Let h ( x )g ( x )> 1 for all 0 < x < 1. Notice that if u g (u ) = 1 for some u < 1, then from Lemma 6-(b) we obtain h 1 (u ) = 1 g (u g (u ))h (u ) = 1 h (u ) , therefore h (u ) · h 1 (u ) = 1, which is impossible. As a consequence, either x g ( x )> 1 for all 0 < x < 1 or x g ( x )< 1 for all 0 < x < 1.
(★) Suppose that x g ( x )> 1 for all 0 < x < 1. By force, g ( x )> 1 in (0, 1).
On the other hand, from Eq. (10), 1 < x g ( x ) = h 1 ( 1 g (h 1 ( x )) ) , it follows that 1 g (h 1 ( x )) < 1,so g (h 1 ( x )) > 1 for all x < 1. Thus, being the restriction h 1 |(0,1) a homeomorphism from (0, 1) onto (1, ∞), it implies that g (w)> 1 for all w> 1 and consequently g (u ) ≥ 1 for all u > 0. Finally, use Lemma 6-(b), h 1 ( x ) = 1 g ( x g ( x ))h ( x ) , to an arbitrary point x < 1 to deduce that 1 = h 1 ( x ) · h ( x ) · g ( x g ( x )) > 1, a contradiction.
(
★★) Suppose that x g ( x )< 1 for all 0 < x < 1. Firstly, we claim that g (w)< 1 for all w> 1. To see it, use Eq. (10) to an arbitrary point x < 1, x g ( x ) =
h 1 ( 1
g (h 1 ( x )) ) < 1, and apply that h 1 is decreasing with h 1 (1) = 1 to deduce that 1 g (h 1 ( x )) > 1 or g (h 1 ( x )) < 1. Since x < 1 was arbitrarily taken and h 1 is a homeomorphism from (0, 1) onto (1, ∞) we obtain that g (w)< 1 for all w> 1, as we claimed.
Additionally, g ( x ) /= 1 for all x < 1, since otherwise by Lemma 6-(b), for some x /= 1, we would find
h 1 ( x ) = 1 g ( x g ( x )) h ( x ) = 1 g ( x )h ( x ) = 1 h ( x ) ,
therefore h ( x )h 1 ( x ) = 1, with x < 1, a contradiction. Thus, we can divide our study in two additional subcases, namely, g ( x )< 1 for all x < 1,or g ( x )> 1 for all x < 1.
(∗)If g ( x )< 1 for all x < 1, then, according to the initial claim of this case, we have g (u ) ≤ 1 for all u > 0. Hence, by Lemma 6-(b) and the decreasing character of h , h 1 ,for w> 1 we have 1 = h 1 (w) · g (w g (w))h (w)< 1, a contradiction.
(∗∗) Suppose that g ( x )> 1 for all x < 1. Moreover, g (w)< 1 for all w> 1, and recall that x g ( x )< 1 and h ( x )g ( x )> 1 for all x < 1. Moreover h (w)g (w)< 1 if w> 1, since h (w)< 1 and g (w)< 1 if w> 1. We finally distinguish two possibilities:
• If w g (w)> 1 for some w> 1, then g (w g (w)) < 1 and Lemma 6-(b) yields h 1 (w) = 1 g (w g (w)) h (w) > 1 h (w) > 1,
hence h 1 (w)> 1, which is impossible for values w> 1.
• If w g (w) ≤ 1 for some w> 1,byEq. (10), we have 1 ≥ w g (w) = h 1 ( 1 g (h 1 (w)) ) , therefore 1 g (h 1 (w)) ≥ 1 and consequently g (h 1 (w)) ≤ 1, which is impossible because h 1 (w)< 1 and we had supposed that g ( x )> 1 in (0, 1).This concludes the discussion of (∗∗).
So, Case (c.2) does not provide any 6-cycle of type (4) and we can gather the discussion of Cases (c.1) and (c.2) in the following result:
Proposition 5 In Eq. (4), suppose that h is decreasing and that the set of equilibrium points reduces to F ={1}. Then, Eq. (4) cannot be a 6-cycle.
As a result of putting together Propositions 1 and 2, for the case in which Card (F ) ≥ 2, and Propositions 3, 4 and 5, for the case Card (F ) = 1, we obtain the main result of Sect. 3.
Theorem 2 There are no 6-cycles displaying the form
n +3 = x n +2 g ( x n +1 ) h ( x n ),
with g , h : (0, ∞) −→ (0, ∞) continuous.
4 The Case xn+3 = xn+1 g ( xn+2 ) h ( xn )
In this section, we consider the third order difference equation
with g , h : (0, ∞) → (0, ∞) continuous. We are going to prove that there are no 6-cycles of this form, see Theorem 3.
Our first observation is to recall that, according to that established at the end of Sect. 2, we can assume without loss of generality that 1 ∈ F , with g (1) = h (1) = 1.
As a first necessary condition to obtain a 6-cycle, we prove that h must be bijective.
Lemma 9 Suppose that Eq. (11)isa 6-cycle, with 1 ∈ F , g (1) = h (1) = 1. Then, h is a homeomorphism.
Proof The proof is completely analogous to the one developed in Lemma 3.Now, it suffices to take again the initial conditions x 1 = 1, x 2 = 1, x 3 = y to prove that h is surjective, and if h ( x ) = h ( y ), to use the pair of initial conditions x 1 = x , x 2 = x 3 = 1 and y1 = y , y2 = y3 = 1 to deduce that x = y ◻
Lemma 10 Suppose that Eq. (11)isa 6-cycle, with 1 ∈ F , g (1) = h (1) = 1.If g (u ) = 1 then u = 1.
Proof Set x 1 = 1, x 2 = 1 and x 3 = u . Then, by Eq. (11), x 4 = 1, x 5 = u , x 6 = h (u ), and 1 = x 2 = x 8 = x 6 g ( x 7 )h ( x 5 ) = x 6 g ( x 1 )h ( x 5 ) = h (u )h (u ). Therefore, h (u ) = 1 and since h is a homeomorphism with h (1) = 1, we conclude that u = 1.
Lemma 11 Suppose that Eq. (11)isa 6-cycle, with 1 ∈ F , g (1) = h (1) = 1. Then h 1 ( x ) = 1 h ( x g (g ( x ))) for all x > 0
Proof Let x > 0 be arbitrary. Take x 1 = h 1 ( x ), x 2 = 1, x 3 = 1 and apply the recurrence to obtain x 4 = x , x 5 = g ( x ), x 6 = x g (g ( x )). By global periodicity, 1 = x 3 = x 9 = x 7 g ( x 8 )h ( x 6 ) = x 1 g ( x 2 )h ( x 6 ) = h 1 ( x )h ( x g (g ( x ))).
Therefore, h 1 ( x ) = 1 h ( x g (g ( x )))
According to the last result, we have h 1 ( x ) = 1 h (ψ( x )) , forall x > 0, (12) where
Notice that from (12) and the fact that h , h 1 are homeomorphisms, we can also write
Lemma 12 Suppose that Eq. (11)isa 6-cycle, with 1 ∈ F , g (1) = h (1) = 1. Then, ψ : (0, ∞) → (0, ∞) is a decreasing homeomorphism.
Proof Since h 1 is a homeomorphism, it is immediate to deduce that ψ is a homeomorphism from (0, ∞) into itself. If h 1 is decreasing (respectively, increasing), then 1 h 1 is increasing (decreasing), and consequently h 1 ( 1 h 1 ) is decreasing, both in the cases h 1 increasing and decreasing.
We conclude Sect. 4 with its main result: Theorem 3 There are no 6-cycles displaying the form x n +3 = x n +1 g ( x n +2 ) h ( x n ),
with g , h : (0, ∞) −→ (0, ∞) continuous.
Proof Suppose that Eq. (11)isa 6-cycle. Then, by Lemma 12 and g (g ( x )) = ψ( x ) x , we know that the map g (g ( x )) is a decreasing homeomorphism from (0, ∞) into itself. In this case, it is a simple matter to show that g : (0, ∞) → (0, ∞) is also a homeomorphism, either increasing or decreasing. In both cases, however, we find that g ◦ g is increasing, a contradiction.
5 The Case xn+3 = xn h ( xn+1 )g ( xn+2 )
Contrarily to the study developed in Sects. 3 and 4, in the remaining case, namely the difference equation
x n +3 = x n h ( x n +1 ) g ( x n +2 ), (15) where g , h : (0, ∞) → (0, ∞) are continuous maps, we are able to discover the potential 6-cycle
x n +3 = x n ( x n +2 x n +1 )2 . (16)
Then the question arises whether there exist or not another 6-cycles displaying the form of (15), and different from the potential one above mentioned. We are going to prove that the answer is negative, the unique 6-cycle of type (15) is precisely the potential one given by Eq. (16).
Recall that by Remark 2, since F =∅, we can assume that 1 ∈ F jointly with g (1) = h (1) = 1. The first result exposes a sufficient condition on the maps g , h in order that Eq. (15)bea 6-cycle.
Proposition 6 Let (15)bea 6-cycle. Suppose that g ( x )h ( x ) = 1 for all x > 0. Then, the 6-cycle must be x n +3 = x n ( x n +2 x n +1 )2 .
Proof Observe that the condition g ( x )h ( x ) = 1 for all x > 0 is equivalent to state that F = (0, ∞).Let x > 0 be arbitrary. If we take initial conditions x 1 = 1, x 2 = x , x 3 = x and proceed to apply the difference equation, we obtain x 4 = 1, x 5 = xh ( x ), x 6 = x g ( xh ( x )) By the 6-periodicity of the solutions, x = x 3 = x 9 = x 6 h ( x 7 )g ( x 8 ) = x 6 h ( x 1 )g ( x 2 ) = x g ( xh ( x ))g ( x ).
Hence, g ( x g ( x ) ) = 1 g ( x ) , forall x > 0 (17)
On the other hand, we generate a new solution from the initial conditions x 1 = x , x 2 = 1, x 3 = 1.Now, x 4 = x , x 5 = g ( x ), x 6 = h ( x ) g (g ( x )), and 1 = x 3 = x 9 = x 6 h ( x 7 )g ( x 8 ) = x 6 h ( x 1 )g ( x 2 ) = h ( x ) g (g ( x ))h ( x ), so since h ( x ) = 1 g ( x ) ,we deduce that [g ( x )]2 = g (g ( x )), forall x > 0 (18)
In [4] it was proved that the unique solutions to the system of functional Eqs. (17) and (18)are g ( x ) = 1 or g ( x ) = x 2 for all x > 0. The first one provides the 3cycle x n +3 = x n , whereas the second one generates the potential 6-cycle x n +3 = x n ( x n +2 x n +1 )2 , which ends the proof. ◻
In the following, we use the notation ϕ (k ) , k ≥ 2, to denote the iterate ϕ ◦ k times ◦ ϕ. We continue the study to prove the result in the general case, that is, without assuming g ( x )h ( x ) = 1 for all x > 0.
Lemma 13 Let Eq. (15)bea 6-cycle. Assume that h (1) = g (1) = 1. Then, for all x > 0 it holds:
h ( x ) = / 1 g (g ( x )) , (19) 1 = / 1 g (3) ( x ) g (√g (2) ( x )) , (20) [g ( x )]2 = /g (2) (√g (2) ( x )), (21)
(3) ( x ) = /
Proof From the proof of Proposition 6, where we have considered the initial conditions x 1 = x , x 2 = x 3 = 1, we know that 1 = h ( x )g (g ( x ))h ( x ). So, h ( x ) = / 1 g (g ( x )) , which proves (19).
Also, x = x 1 = x 7 = x 4 h ( x 5 )g ( x 6 ) = xh (g ( x )) g (h ( x )g (g ( x ))).Hence h (g ( x ))g (h ( x )g (g ( x ))) = 1, and using (19) twice, the first one with g ( x ) instead of x , we obtain / 1 g (g (g ( x ))) · g (√g (2) ( x )) = 1, or1 = / 1 g (3) ( x ) · g (√g (2) ( x )) .
We have proved (20).
Again by the global periodicity, 1 = x 2 = x 8 = x 5 h ( x 6 )g ( x 7 ) = x 5 h ( x 6 )g ( x 1 ) = g ( x )h (h ( x )g (g ( x ))) g ( x ), and in this case we deduce that 1 = [g ( x )]2 h (h ( x )g (g ( x ))) and if we use (19) twice, in fact we obtain (21): 1 = [g ( x )]2 h (/ 1 g (g ( x )) · g (g ( x ))) = [g ( x )]2 h (√g (2) ( x )) = [g ( x )]2 ┌ | | √ 1 g (2) (√g (2) ( x ))
Now, by (20), we have g (3) ( x ) = [g (√g (2) ( x ))]2 , and using (21) with √g (2) ( x ) instead of x , we get Eq. (22):
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“Let us not forget that it is not the want of generous sentiment, but of sufficient information, that prevents the American people from being united in action against the aggressive policy of the slave power. Were these simple questions submitted to-day to the people of the United States: Are you in favor of the extension of slavery? Are you in favor of such extension by the aid or connivance of the federal government? And could they be permitted to record their votes in response, without embarrassment, without constraint of any kind, nineteentwentieths of the people of the free States, and perhaps more than half of the people of the slave States, would return a decided negative to both.
“Let us have faith in the people. Let us believe, that at heart they are hostile to the extension of slavery, desirous that the territories of the Union be consecrated to free labor and free institutions; and that they require only enlightenment as to the most effectual means of securing this end, to convert their cherished sentiment into a fixed principle of action.
“The times are pregnant with warning. That a disunion party exists in the South, no longer admits of a doubt. It accepts the election of Mr. Buchanan as affording time and means to consolidate its strength and mature its plans, which comprehend not only the enslavement of Kansas, and the recognition of slavery in all territory of the United States, but the conversion of the lower half of California into a slave State, the organization of a new slavery territory in the Gadsden purchase, the future annexation of Nicaragua and subjugation of Central America, and the acquisition of Cuba; and, as the free States are not expected to submit to all this, ultimate dismemberment of the Union, and the formation of a great slaveholding confederacy, with foreign alliances with Brazil and Russia. It may assume at first a moderate tone, to prevent the sudden alienation of its Northern allies; it may delay the development of its plot, as it did under the Pierce administration; but the repeal of the Missouri compromise came at last, and so will come upon the country inevitably the final acts of the dark conspiracy. When that hour shall come, then will the honest Democrats of the free States be driven into our ranks, and the men of the slave States who prefer the republic of Washington, Adams and Jefferson a republic of law, order and liberty to an oligarchy of slaveholders and slavery propagandists, governed by Wise, Atchison, Soulé, and Walker, founded in fraud and violence and seeking aggrandizement by the spoliation of nations, will bid God speed to the labors of the Republican party to preserve liberty and the Union, one and inseparable, perpetual and all powerful.
“Washington, D. C., Nov. 27, 1856.”
The Lincoln and Douglas Debate.
The Senatorial term of Douglas was drawing near to its close, when in July, 1858, he left Washington to enter upon the canvass for reelection. The Republican State Convention of Illinois had in the month previous met at Springfield, and nominated Abraham Lincoln as a candidate for United States Senator, this with a view to pledge all Republican members of the Legislature to vote for him—a practice since gone into disuse in most of the States, because of the rivalries which it engenders and the aggravation of the dangers of defeat sure to follow in the selection of a candidate in advance. “First get your goose, then cook it,” inelegantly describes the basic principles of improved political tactics. But the Republicans, particularly of the western part of Illinois, had a double purpose in the selection of Lincoln. He was not as radical as they, but he well represented the growing Republican sentiment, and he best of all men could cope with Douglas on the stump in a canvass which they desired should attract the attention of the Nation, and give shape to the sentiment of the North on all questions pertaining to slavery. The doctrine of “popular sovereignty” was not acceptable to the Republicans, the recent repeal of the Missouri compromise having led them, or the more radical portion of them, to despise all compromise measures.
The plan of the Illinois Republicans, if indeed it was a well-settled plan, accomplished even more than was anticipated, though it did not result in immediate success. It gave to the debate which followed between Lincoln and Douglas a world-wide celebrity, and did more to educate and train the anti-slavery sentiment, taken in connection with the ever-growing excitement in Kansas, than anything that could have happened.
Lincoln’s speech before the convention which nominated him, gave the first clear expression to the idea that there was an “irrepressible conflict” between freedom and slavery. Wm. H. Seward
on October 25th following, at Rochester, N. Y., expressed the same idea in these words:
“It is an irrepressible conflict between opposing and enduring forces, and it means that the United States will sooner or later become either an entire slaveholding Nation, or an entirely free labor Nation.”
Lincoln’s words at Springfield, in July, 1858, were:
“If we could first know where we are, and whither we are tending, we could better judge what to do, and how to do it. We are now far into the fifth year, since a policy was initiated with the avowed object, and confident promise of putting an end to the slavery agitation. Under the operation of that policy, that agitation has not only not ceased, but has constantly augmented. In my opinion it will not cease, until a crisis shall have been reached and passed. ‘A house divided against itself cannot stand.’ I believe this government cannot endure permanently half slave and half free. I do not expect the Union to be dissolved—I do not expect the house to fall—but I do expect it will cease to be divided. It will become all one thing, or all the other. Either the opponents of slavery will arrest the further spread of it, and place it where the public mind shall rest in the belief that it is in the course of ultimate extinction; or its advocates will push it forward, till it shall become alike lawful in all the States, old as well as new—North as well as South.”
Douglas arrived in Chicago on the 9th of July, and was warmly received by enthusiastic friends. His doctrine of “popular sovereignty” had all the attractions of novelty and apparent fairness. For months it divided many Republicans, and at one time the New York Tribune showed indications of endorsing the position of Douglas—a fact probably traceable to the attitude of jealousy and hostility manifested toward him by the Buchanan administration. Neither of the great debaters were to be wholly free in the coming contest. Douglas was undermined by Buchanan, who feared him as a rival, and by the more bitter friends of slavery, who could not see that the new doctrine was safely in their interest; but these things were dwarfed in the State conflict, and those who shared such feelings had to make at least a show of friendship until they saw the result. Lincoln was at first handicapped by the doubts of that class of
Republicans who thought “popular sovereignty” not bad Republican doctrine.
On the arrival of Douglas he replied to Lincoln’s Springfield speech; on the 16th he spoke at Bloomington, and on the 17th, in the afternoon, at Springfield. Lincoln had heard all three speeches, and replied to the last on the night of the day of its delivery. He next addressed to Douglas the following challenge to debate:
C , July 24th, 1858.
H . S. A. D : My Dear Sir: Will it be agreeable to you to make an arrangement to divide time, and address the same audience, during the present canvass? etc. Mr. Judd is authorized to receive your answer, and if agreeable to you, to enter into terms of such agreement, etc.
Your obedient servant, A. L .
Douglas promptly accepted the challenge, and it was arranged that there should be seven joint debates, each alternately opening and closing, the opening speech to occupy one hour, the reply one hour and a half, and the closing half an hour. They spoke at Ottawa, August 21st; Freeport, August 27th; Jonesboro’, September 15th; Charleston, September 18th; Galesburg, October 7th; Quincy, October 13th; and Alton, October 15th. We give in Book III of this volume their closing speeches in full.
Great crowds attended, and some of the more enterprising daily journals gave phonographic reports of the speeches. The enthusiasm of the North soon ran in Lincoln’s favor, though Douglas had hosts of friends; but then the growing and the aggressive party was the Republican, and even the novelty of a new and attractive doctrine like that of “popular sovereignty” could not long divert their attention. The prize suspended in view of the combatants was the United States Senatorship, and to close political observers this was plainly within the grasp of Douglas by reason of an apportionment which would give his party a majority in the Legislature, even though the popular majority should be twenty thousand against him—a system of apportionment, by the way, not confined to Illinois alone, or not peculiar to it in the work of any of the great parties at any period when party lines were drawn.
Buchanan closely watched the fight, and it was charged and is still believed by the friends of the “Little Giant,” that the administration secretly employed its patronage and power to defeat him. Certain it is that a few prominent Democrats deserted the standard of Douglas, and that some of them were rewarded. In the heat of the battle, however, Douglas’ friends were careless of the views of the administration. He was a greater leader than Buchanan, and in Illinois at least he overshadowed the administration. He lacked neither money nor friends. Special trains of cars, banners, cannon, bands, processions, were all supplied with lavish hands. The democracy of Illinois, nor yet of any other State, ever did so well before or since, and if the administration had been with him this enthusiasm might have spread to all other States and given his doctrine a larger and more glorious life. Only the border States of the South, however, saw opportunity and glory in it, while the officeholders in other sections stood off and awaited results.
Lincoln’s position was different. He, doubtless, early realized that his chances for election were remote indeed, with the apportionment as it was, and he sought to impress the nation with the truth of his convictions, and this without other display than the force of their statement and publication. Always a modest man, he was never more so than in this great battle. He declared that he did not care for the local result, and in the light of what transpired, the position was wisely taken. Douglas was apparently just as earnest, though more ambitious; for he declared in the vehemence of the advocacy of his doctrine, that “he did not care whether slavery was voted up or voted down.” Douglas had more to lose than Lincoln—a place which his high abilities had honored in the United States Senate, and which intriguing enemies in his own party made him doubly anxious to hold. Beaten, and he was out of the field for the Presidency, with his enthroned rival a candidate for re-election. Successful, and that rival must leave the field, with himself in direct command of a great majority of the party. This view must have then been presented, but the rapid rise in public feeling made it in part incorrect. The calculation of Douglas that he could at one and the same time retain the good will of all his political friends in Illinois and those of the South failed him, though he did at the time, and until his death, better represent the majority of his party in the whole country than any other leader.
At the election which followed the debate, the popular choice in the State as a whole was for Lincoln by 126,084 to 121,940 for Douglas; but the apportionment of 1850 gave to Douglas a plain majority of the Senators and Representatives.
At the Freeport meeting, August 27th, there were sharp questions and answers between the debaters. They were brought on by Lincoln, who, after alluding to some questions propounded to him at Ottawa, said:
“I now propose that I will answer any of the interrogatories, upon condition that he will answer questions from me not exceeding the same number, to which I give him an opportunity to respond. The judge remains silent; I now say that I will answer his interrogatories, whether he answer mine or not, and that after I have done so I shall propound mine to him.
“I have supposed myself, since the organization of the Republican party at Bloomington in May, 1856, bound as a party man by the platforms of the party, there, and since. If, in any interrogatories which I shall answer, I go beyond the scope of what is within these platforms, it will be perceived that no one is responsible but myself.
“Having said thus much, I will take up the judge’s interrogatories as I find them printed in the Chicago Times, and answer them seriatim. In order that there may be no mistake about it, I have copied the interrogatories in writing, and also my answers to them. The first one of these interrogatories is in these words:
Question 1.—I desire to know whether Lincoln to-day stands, as he did in 1854, in favor of the unconditional repeal of the Fugitive Slave Law?
Answer.—I do not now, nor ever did, stand in favor of the unconditional repeal of the Fugitive Slave Law.
Q. 2.—I desire him to answer whether he stands pledged to-day, as he did in 1854, against the admission of any more slave States into the Union, even if the people want them?
A.—I do not now, nor ever did, stand pledged against the admission of any more slave States into the Union.
Q. 3—I want to know, whether he stands pledged against the admission of a new State into the Union, with such a Constitution as the people of the State may see fit to make?
A.—I do not stand pledged against the admission of a new State into the Union, with such a Constitution as the people of the State may see fit to make.
Q. 4.—I want to know whether he stands to-day pledged to the abolition of slavery in the District of Columbia?
A.—I do not stand to-day pledged to the abolition of slavery in the District of Columbia.
Q. 5.—I desire him to answer whether he stands pledged to the prohibition of the slave trade between the different States?
A.—I do not stand pledged to prohibition of the slave trade between the different States.
Q. 6.—I desire to know whether he stands pledged to prohibit slavery in all the Territories of the United States, North as well as South of the Missouri Compromise line?
A.—I am impliedly, if not expressly, pledged to a belief in the RIGHT and DUTY of Congress to prohibit slavery in all of the United States’ Territories.
Q. 7.—I desire him to answer, whether he is opposed to the acquisition of any new territory, unless slavery is first prohibited therein?
A.—I am not generally opposed to honest acquisition of territory; and in any given case, I would or would not oppose such acquisition, according as I might think such acquisition would or would not aggravate the slavery question among ourselves.
“Now, my friends, it will be perceived upon an examination of these questions and answers, that so far, I have only answered that I was not pledged to this, that, or the other.
The judge has not framed his interrogatories to ask me anything more than this and I have answered in strict accordance with the interrogatories, and have answered truly, that I am not pledged at all upon any of the points to which I have answered. But I am not disposed to hang upon the exact form of his interrogatories. I am rather disposed to take up, at least some of these questions, and state what I really think upon them.
“The fourth one is in regard to the abolition of slavery in the District of Columbia. In relation to that, I have my mind very
distinctly made up. I should be very glad to see slavery abolished in the District of Columbia. I believe that Congress possesses the constitutional power to abolish it. Yet, as a member of Congress, I should not, with my present views, be in favor of endeavoring to abolish slavery in the District of Columbia, unless it should be upon these conditions: F , That the abolition should be gradual; S , That it should be on a vote of a majority of qualified voters in the District; and T , That compensation should be made to unwilling owners. With these three conditions, I confess I would be exceedingly glad to see Congress abolish slavery in the District of Columbia, and in the language of Henry Clay, ‘sweep from our Capital that foul blot upon our nation.’
I now proceed to propound to the judge the interrogatories, so far as I have framed them. I will bring forward a new instalment when I get them ready. I will bring now only four. The first one is:—
1. If the people of Kansas shall, by means entirely unobjectionable in all other respects, adopt a State Constitution and ask admission into the Union under it before they have the requisite number of inhabitants, according to the English bill—some ninety-three thousand—will he vote to admit them?
2. Can the people of the United States Territory, in any lawful way, against the wish of any citizen of the United States, exclude slavery from its limits prior to the formation of a State Constitution?
3. If the Supreme Court of the United States shall decide that States cannot exclude slavery from their limits, are you in favor of acquiescing in, adopting and following such decision as a rule of political action?
4. Are you in favor of acquiring additional territory in disregard of how much acquisition may affect the nation on the slavery question?
To these questions Mr. Douglas said: “In reference to Kansas, it is my opinion that, as she has population enough to constitute a slave State, she has people enough for a free State. I hold it to be a sacred rule of universal application, to require a Territory to contain the requisite population for a member of Congress, before it is admitted as a State into the Union.
2. “It matters not what way the Supreme Court may hereafter decide, as to the abstract question whether slavery may or may not
go into a Territory under the Constitution, the people have the lawful means to introduce it, or exclude it as they please, for the reason that slavery cannot exist a day, or an hour, anywhere, unless it is supported by local police regulations. These police regulations can only be established by the local legislature, and if the people are opposed to slavery, they will elect representatives to that body, who will, by unfriendly legislation, effectually prevent the introduction of it into their midst. If, on the contrary, they are for it, their legislation will favor its extension. Hence, no matter what the decision of the Supreme Court may be on that abstract question, still the right of the people to make a slave Territory or a free Territory is perfect and complete under the Nebraska bill.
“3. The third question which Mr. Lincoln presented is, if the Supreme Court of the United States shall decide that a State of this Union cannot exclude slavery from its own limits, will I submit to it? I am amazed that Mr. Lincoln should ask such a question.
“He casts an imputation upon the Supreme Court of the United States by supposing that they would violate the constitution of the United States. I tell him that such a thing is not possible. It would be an act of moral treason that no man on the bench could ever descend to. Mr. Lincoln, himself, would never, in his partisan feelings, so far forget what was right as to be guilty of such an act.
“4. With our natural increase, growing with a rapidity unknown in any other part of the globe, with the tide of emigration that is fleeing from despotism in the old world, to seek refuge in our own, there is a constant torrent pouring into this country that requires more land, more territory upon which to settle, and just as fast as our interests and our destiny require an additional territory in the North, in the South, or on the Island of the Ocean, I am for it, and when we require it, will leave the people, according to the Nebraska bill, free to do as they please on the subject of slavery, and every other question.”
The bitterness of the feelings aroused by the canvass and boldness of Douglas, can both be well shown by a brief abstract from his speech at Freeport. He had persisted in calling the Republicans “Black Republicans,” although the crowd, the great majority of which was there against him, insisted that he should say “White Republican.” In response to these oft repeated demands, he said:—
“Now, there are a great many Black Republicans of you who do not know this thing was done. (“White, white,” and great clamor). I wish to remind you that while Mr. Lincoln was speaking, there was not a Democrat vulgar and blackguard enough to interrupt him. But I now that the shoe is pinching you. I am clinching Lincoln now, and you are scared to death for the result. I have seen this thing before. I have seen men make appointments for discussions and the moment their man has been heard, try to interrupt and prevent a fair hearing of the other side. I have seen your mobs before and defy your wrath. (Tremendous applause.)
“My friends, do not cheer, for I need my whole time.
“I have been put to severe tests. I have stood by my principles in fair weather and in foul, in the sunshine and in the rain. I have defended the great principle of self-government here among you when Northern sentiment ran in a torrent against me, and I have defended that same great principle when Southern sentiment came down like an avalanche upon me. I was not afraid of any test they put to me. I knew I was right—I knew my principles were sound—I knew that the people would see in the end that I had done right, and I knew that the God of Heaven would smile upon me if I was faithful in the performance of my duty.”
As an illustration of the earnestness of Lincoln’s position we need only quote two paragraphs from his speech at Alton:—
“Is slavery wrong? That is the real issue. That is the issue that will continue in this country when these poor tongues of Judge Douglas and myself shall be silent. It is the eternal struggle between these two principles—right and wrong—throughout the world. They are two principles that have stood face to face from the beginning of time; and will ever continue to struggle. The one is the common right of humanity, and the other the divine right of Kings. It is the same principle in whatever shape it develops itself. It is the same spirit that says, ‘you work and toil, and earn bread, and I’ll eat it.’ No matter in what shape it comes, whether from the mouth of a King who seeks to bestride the people of his own nation and life by the fruit of their labor, or from one race of men as an apology for enslaving another race, it is the same tyrannical principle.”
And again:—
“On this subject of treating it as a wrong, and limiting its spread, let me say a word. Has anything ever threatened the existence of this Union save and except this very institution of slavery? What is it that we hold most dear among us? Our own liberty and prosperity. What has ever threatened our liberty and prosperity save and except this institution of slavery? If this is true, how do you propose to improve the condition of things? by enlarging slavery?—by spreading it out and making it bigger? You may have a wen or cancer upon your person and not be able to cut it out, lest you bleed to death; but surely it is no way to cure it, to engraft it and spread it over your whole body. That is no proper way of treating what you regard a wrong. You see this peaceful way of dealing with it as a wrong— restricting the spread of it, and not allowing it to go into new countries where it has not already existed. That is the peaceful way, the old-fashioned way, the way in which the fathers themselves set us the example.”
The administration of Pierce had left that of Buchanan a dangerous legacy. He found the pro-slavery party in Congress temporarily triumphant, it is true, and supported by the action of Congress in rejecting the Topeka constitution and recognizing the territorial government, but he found that that decision was not acceptable either to the majority of the people in the country or to a rapidly rising anti-slavery sentiment in the North. Yet he saw but one course to pursue, and that was to sustain the territorial government, which had issued the call for the Lecompton convention. He was supported in this view by the action of the Supreme Court, which had decided that slavery existed in Kansas under the constitution of the United States, and that the people therein could only relieve themselves of it by the election of delegates who would prohibit it in the constitution to be framed by the Lecompton convention. The Free State men refused to recognize the call, made little, if any, preparation for the election, yet on the last day a number of them voted for State officials and a member of Congress under the Lecompton constitution. This had the effect of suspending hostilities between the parties, yet peace was actually maintained only by the intervention of U. S. troops, under the command of Col. Sumner, who afterwards won distinction in the war of the rebellion. The Free State people stood firmly by their Topeka constitution, and refused to vote on questions affecting delegates to the Lecompton
convention. They had no confidence in Governor Walker, the appointee of President Buchanan, and his proclamations passed unheeded. They recognized their own Governor Robinson, who in a message dated December 7th, 1857, explained and defended their position in these words:
“The convention which framed the constitution at Topeka originated with the people of Kansas territory. They have adopted and ratified the same twice by a direct vote, and also indirectly through two elections of State officers and members of the State Legislature. Yet it has pleased the administration to regard the whole proceeding as revolutionary.”
The Lecompton convention, proclaimed by Governor Walker to be lawfully constituted, met for the second time, Sept. 4th, 1857, and proceeded to frame a constitution, and adjourned finally Nov. 7th. A large majority of the delegates, as in the first, were of course proslavery, because of the refusal of the anti-slavery men to participate in the election. It refused to submit the whole constitution to the people, it is said, in opposition to the desire of President Buchanan, and part of his Cabinet. It submitted only the question of whether or not slavery should exist in the new State, and this they were required to do under the Kansas-Nebraska act, if indeed they were not required to submit it all. Yet such was the hostility of the pro-slavery men to submission, that it was only by three majority the proposition to submit the main question was adopted—a confession in advance that the result was not likely to favor their side of the controversy. But six weeks’ time was also allowed for preparation, the election being ordered for Dec. 21st, 1857. Still another advantage was taken in the printing of the ballots, as ordered by the convention. The method prescribed was to endorse the ballots, “Constitution with Slavery,” and “Constitution with no Slavery,” thus compelling the voter, however adverse his views, as to other parts of the Constitution, to vote for it as a whole. As a consequence, (at least this was given as one of the reasons) the Free State men as a rule refused to participate in the election, and the result as returned was 6,143 votes in favor of slavery, and 589 against it. The constitution was announced as adopted, an election was ordered on the first Monday of January, 1858, for State officers, members of the Legislature, and a member of Congress. The opponents of the Lecompton constitution
did not now refrain from voting, partly because of their desire to secure the representative in Congress, but mainly to secure an opportunity, as advised by their State officers, to vote down the Lecompton constitution. Both parties warmly contested the result, but the Free State men won, and with their general victory secured a large majority in the Legislature.
The ballots of the Free State men were now headed with the words “Against the Lecompton Constitution,” and they returned 10,226 votes against it, to 134 for it with slavery, and 24 for it against slavery. This return was certified by J. W. Denver, “Secretary and Acting Governor,” and its validity was endorsed by Douglas in his report from the Senate Territorial Committee. It was in better accord with his idea of popular sovereignty, as it showed almost twice as large a vote as that cast under the Lecompton plan, the fairness of the return not being disputed, while that of the month previous was disputed.
But their previous refusal to vote on the Lecompton constitution gave their opponents an advantage in position strangely at variance with the wishes of a majority of the people. The President of that convention, J. Calhoun, forwarded the document to the President with an official request that it be submitted to Congress. This was done in a message dated 2d February, 1858, and the President recommended the admission of Kansas under it.
This message occasioned a violent debate in Congress, which continued for three months. It was replete with sectional abuse and bitterness, and nearly all the members of both Houses participated. It finally closed with the passage of the “Act for the admission of the State of Kansas into the Union,” passed May 4th, 1858. This Act had been reported by a committee of conference of both Houses, and was passed in the Senate by 31 to 22, and in the House by 112 to 103. There was a strict party vote in the Senate with the exception of Mr. Douglas, C. E. Stuart of Michigan, and D. C. Broderick of California, who voted with the Republican minority. In the House several antiLecompton democrats voted with the Republican minority. These were Messrs. Adrian of New Jersey; Chapman of Pennsylvania; Clark of New York; Cockerill of Ohio; Davis of Indiana; Harris of Illinois; Haskin of New York; Hickman of Pennsylvania; McKibben of California; Marshall of Illinois; Morgan of New York; Morris, Shaw,
and Smith of Illinois. The Americans who voted with the Republicans were Crittenden of Kentucky; Davis of Maryland; Marshall of Kentucky; Ricaud of Maryland; Underwood of Kentucky. A number of those previously classed as Anti-Lecompton Democrats voted against their colleagues of the same faction, and consequently against the bill. These were Messrs. Cockerill, Gwesheck, Hall, Lawrence, Pendleton and Cox of Ohio; English and Foley of Indiana; and Jones of Pennsylvania. The Americans who voted against the bill were Kennedy of Maryland; Anderson of Missouri; Eustis of Louisiana; Gilmer of North Carolina; Hill of Georgia; Maynard, Ready and Zollicoffer of Tennessee; and Trippe of Georgia.
