Global convergence of the second order Ricker equation

Accepted Manuscript Global convergence of the second order Ricker equation J. Per´an, D. Franco PII: DOI: Reference:

S0893-9659(15)00088-9 http://dx.doi.org/10.1016/j.aml.2015.02.022 AML 4742

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Applied Mathematics Letters

Received date: 20 December 2014 Revised date: 27 February 2015 Accepted date: 27 February 2015 Please cite this article as: J. Per´an, D. Franco, Global convergence of the second order Ricker equation, Appl. Math. Lett. (2015), http://dx.doi.org/10.1016/j.aml.2015.02.022 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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1

Global convergence of the second order Ricker equation

2

J. Per´ana,1,∗, D. Francoa,1 a

3 4

5

Departamento de Matem´ atica Aplicada de la UNED. ETSI Industriales. C/ Juan del Rosal, 12. Madrid 28040. Spain.

Abstract We present a short analytic proof of a 1976 conjecture on the global dynamics of the equation xn = xn−1 ex−xn−2 , where x ∈ (0, 1) and x0 , x1 ∈ (0, +∞). The proof is based in considering the parameter x in the previous equation as a complex variable. This transforms the problem in studying the asymptotic behaviour of a sequence of analytic functions.

6

Keywords: Delayed Ricker Equation, Global attractor, Difference

7

equation, Montel’s Theorem, Population dynamics

8

1. Introduction. The difference-delayed Ricker equation,

9

xn = xn−1 ex−xn−2

,

x0 , x1 ∈ (0, +∞),

(1)

10

where x > 0, arises as the natural generalization of the celebrated Ricker

11

population model [7], xn = xn−1 ex−xn−1 , when there are explicit time lags in

12

the density dependent regulatory mechanisms of the population.

13

In 1976, Levin and May [3] conjectured that the sequence in (1) converges

14

to the equilibrium x, whenever x ∈ (0, 1). Recently, Bartha, Garab and

15

Krisztin provided a computer aided proof for this conjecture [1]. Indeed, they Corresponding author Email address: [email protected] (J. Per´ an) Preprint submitted to Applied Mathematics Letters. ∗

February 27, 2015

2 16

showed that the sequence in (1) converges for the critical parameter value

17

x = 1 as well. Interestingly, their proof is computer-aided for x ∈ (0.5, 1],

18

but it is fully analytic for x ∈ (0, 0.5]. Therefore, they gave an analytic

19

proof partially solving the conjecture. However, as they noted, there were

20

other previous analytic proofs partially solving the conjecture in a bigger

21

subinterval of (0, 1); e.g. [4, 8]. The best of these results was obtained by

22

Tkachenko and Trofimchuk in [8], proving it for x ∈ (0, 0.875); see [1] for

23

more details and other contributions related to this problem.

24

In this letter, we give an analytic proof of the conjecture. Our approach

25

is new: we treat the parameter x in (1) as a complex variable, studying the

26

asymptotic behaviour of the sequence of analytic functions thus obtained.

27

The proof is based on a combination of classical complex analysis theory and

28

the study of the asymptotic behaviour of a linear difference equation. The

29

organization of the paper is as follows. In Section 2 we prove some general

30

properties of the sequence in (1) which are needed for later sections. In

31

particular, we show that if the sequence converges, then it does it rapidly (in

32

the sense that the series of the absolute differences between xn and its limit

33

converges). In Section 3 we focus on the dynamics of a second order linear

34

difference equation, providing asymptotic bounds for its solutions (Theorem

35

2). This result is used in Section 4 to get a priori bounds for the derivatives

36

of the sequence of analytic functions commented above, but we consider

37

Theorem 2 of independent interest. Finally, also in Section 4 we analytically

38

prove the conjecture showing, in addition, that the convergence is uniform

39

on compact sets of (0, 1).

3 40

2. Preliminaries.

41

In the following result we provide a new short proof for the conjecture

42

in the case x ∈ (0, 1/2] and we prove that, if xn converges, then it does it

43

rapidly.

44

Theorem 1. Let x, x0 , x1 ∈ (0, +∞) and define xn = xn−1 ex−xn−2 for n ≥ 2.

45

Then, 0 < xn ≤ e2x−1 for n ≥ 2 and lim inf n→∞ xn ≤ x ≤ lim supn→∞ xn .

46

Moreover,

47

a) if x ∈ (0, 1/2], then limn→∞ xn = x.

48

b) if x ∈ (0, 1) and limn→∞ xn = x, then

49

Proof. Notice that

P∞

n=0

|xn − x| < ∞.

0 < xn = xn−1 ex−xn−2 = xn−2 ex−xn−2 ex−xn−3 ≤ ex−1 ex = e2x−1 . 50

Let µ and ν be the upper and lower limits of xn respectively. We have

51

that ν ≤ x ≤ µ. Otherwise, xn would be an eventually decreasing sequence in

52

(x, ∞) or an eventually increasing sequence in (0, x), thus it would converge

53

to the equilibrium x and the conclusion would follow.

54

Assume now that x ∈ (0, 1/2], so xn ≤ 1 for n ≥ 2. Let xσ(n) be a subse-

55

quence of xn converging to µ. The sequence (xσ(n) , xσ(n)−1 , xσ(n)−2 , xσ(n)−3 ) is

56

bounded in R4 , thus it has a subsequence converging to a point (µ, γ1 , γ2 , γ3 ) ∈

57

58

R4 , which satisfies µ = γ1 ex−γ2 (thus γ2 ≤ x) and µ = γ2 ex−γ2 ex−γ3 . As

h(t) = te−t is an increasing function in (−∞, 1], we have

µ = γ2 ex−γ2 ex−γ3 ≤ xex−x ex−ν ≤ µex−µ ex−ν ,

4 59

60

61

and therefore µ ≤ xex−ν and 2x − µ − ν ≥ 0. Analogously, by considering a

subsequence xρ(n) → ν, we have 2x−µ−ν ≤ 0. Now, µ ≤ xex−ν = xex−(2x−µ) ,

that is, µe−µ ≤ xe−x , which implies µ = x, ν = 2x − µ = x.

Finally, assume x ∈ (0, 1) and limn→∞ xn = x. It can be easily seen that Pn−2 P xn = x1 e k=0 (x−xk ) , so ∞ k=0 (x − xk ) = log(x/x1 ). To verify that the series converges absolutely, we consider the following sets

M1 := {n ∈ N : xn ≤ min{xn−1 , xn+1 }} , M2 := {n ∈ N : xn ≥ max{xn−1 , xn+1 }} , and M := M1 ∪ M2 . If M1 or M2 are finite sets, then the sequence x − xk is eventually monotonic and then the series converges absolutely. In other case, let σ : N → M be the increasing bijection. Notice that σ(n) ∈ M1 ⇔ σ(n + 1) ∈ M2 . Moreover, if σ(n) ∈ M1 , then · · · ≥ xσ(n)−2 ≥ x ≥ xσ(n)−1 ≥ xσ(n) ≤ xσ(n)+1 ≤ · · · ≤ xσ(n+1)−2 ≤ x ≤ xσ(n+1)−1 ≤ xσ(n+1) ≥ xσ(n+1)+1 ≥ · · · ≥ xσ(n+2)−2 ≥ x ≥ xσ(n+2)−1 ≥ . . . and xσ(n) = xσ(n)−2 ex−xσ(n)−2 ex−xσ(n)−3 ≥ xex−xσ(n)−3 ≥ xex−xσ(n−1) . 62

Analogously, xσ(n) ≤ xex−xσ(n−1) when σ(n) ∈ M2 .

Now consider  > 0 and j0 > 1 such that xe < 1 and that xσ(j) − x < 

5 for all j ≥ j0 − 1. Then,

X ∞ σ(j+1)−2 ∞ X X log(xσ(j+1) ) − log(xσ(j) ) |xn − x| = (x − x) = n j=j0 j=j0 n=σ(j)−1 n=σ(j0 )−1 ∞ X

∞ ∞ X X xσ(j) − xσ(j−1) ≤ (log x + x − xσ(j) ) − (log x + x − xσ(j−1) ) ≤ j=j0

j=j0

≤ xσ(j0 ) − xσ(j0 −1) +

j=j0 +1 ∞ X

= xσ(j0 ) − xσ(j0 −1) +

≤ ··· ≤

∞ X k=0

∞ X

xex−ηj xσ(j−1) − xσ(j−2)

j=j0 +1 ∞ X

≤ xσ(j0 ) − xσ(j0 −1) +

(xe )k xσ(j0 ) − xσ(j0 −1) =

x−x xe σ(j−1) − xex−xσ(j−2)

j=j0 +1

xe xσ(j−1) − xσ(j−2)

1 xσ(j0 ) − xσ(j0 −1) < ∞ .  1 − xe

63

Here, of course, ηj is a certain point between xσ(j−1) and xσ(j−2) .

64

3. Asymptotic behaviour of a linear equation.

65

In the next section, we will reduce the problem on the convergence of the

66

nonlinear equation (1) to the study of the asymptotics of the second order

67

linear difference equation yn = yn−1 − an−2 yn−2 + bn−2 , P∞

y0 , y1 ∈ C,

(2)

68

where bn ∈ C and an ∈ R satisfies

69

Therefore, our following result describes the asymptotic behaviour of (2).

70

71

n=0

|an − a| < ∞ with a ∈ (1/4, 1).

A fundamental step in the proof of this result consists in factorizing (2) an by using that there exists β0 ∈ C such that the sequence βn+1 = 1 − is βn

6 72

convergent. The existence of β0 with such a property can be derived from

73

the existence of convergent tail sequences for the continued fraction K(−an /1) = −

74

a1 . a2 1− a3 1− . 1 − ..

Indeed, let sn be the map sn (w) = −an /(1 + w) and write λ :=

√ 1−i 4a−1 . 2

By

77

Theorem 1.1 in [5], the sequence cn := s1 ◦ · · · ◦ sn (−λ) converges to a limit an β0 ∈ C and then, the sequence defined by βn+1 = 1 − does converge to βn λ. We will refer to the sequence βn as the convergent tail for an and to the

78

sequence

75

76

79

80

81

82

83

84

  √ n a+1 , for all m ≥ η ηn = min η ∈ N : |βm |, |1 − βm+1 | < n+1 √ as the tail index sequence for an . Notice that |λ| = |1 − λ| = a.

(3)

We stress that it is possible to define convergent tails and tail index √ sequences when an converges to a ∈ (0, 1/4], just replacing a in (3) with √ 1+ 1−4a . 2

This situation is simpler than the elliptic case covered by Theorem P 1.1 in [5], in the sense that the condition ∞ n=0 |an − a| < ∞ is unnecessary.

See Theorem 4.13 in [6] for a ∈ (0, 1/4) (known in the continued fractions

85

literature as loxodromic case) and Theorem 4.17 in [6] for a = 1/4 (known

86

as parabolic case).

87

Theorem 2. Let an ∈ R, bn ∈ C, for all n, a ∈ (1/4, 1), y0 , y1 ∈ C, and let

88

yn ∈ C be the sequence defined by

yn = yn−1 − an−2 yn−2 + bn−2 .

(4)

7 89

If

P∞

n=0

|an − a| < ∞, then,  max |yηm | + |yηm +1 | , supn≥ηm |bn | |yn | ≤ , 2  √ a+1 1 − mm+1

(5)

90

for all m > 0 and n ≥ ηm , where ηm is the tail index sequence for am .

91

Proof. Consider a fixed natural number m through the proof.

92

93

94

be the convergent tail for an and αn := 1 − βn+1 . Write ebn := bn for n ≥ ηm ,

ebηm −1 := yηm +1 − βηm yηm , ebηm −2 := yηm . For n > ηm , the second order difference equation (4) can be factorized as zn − αn−1 zn−1 = ebn−1 ,

95

zηm = ebηm −1 ,

yn − βn−1 yn−1 = zn−1 ,

96

(just replace zn from (7) into (6)).

97

98

Qq

Equation (6) is solved by zn =

j=p

100

n−1 X

k=ηm −1

Analogously, yn = so it follows that sup |yn | ≤

n≥ηm

101

Pn−1

|ebk |

Pn−1



n > ηm

n > ηm

e Qn−1

k=ηm −1 bk

αj := 1 whenever p > q), and so |zn | ≤

99

Let βn ∈ C

j=k+1

(6) (7)

αj (where we write

n−k−1 supk≥ηm −1 |ebk | sup |αj | ≤ . 1 − supk≥ηm |αk | j≥ηm

k=ηm −1 zk

Qn−1

j=k+1



βj (where we write zηm −1 := ebηm −2 ),

supn≥ηm −2 |e bn | 1−supn≥ηm |αn |



supn≥ηm −1 |zn | supn≥ηm −2 |ebn | ≤ ≤ 2 . √ 1 − supn≥ηm |βn | 1 − supn≥ηm |βn | m a+1 1 − m+1

Finally, observe that ebηm −1 = |yηm +1 − βηm yηm | ≤ |yηm | + |yηm +1 |

8 102

103

104

4. Main result. Let x0 , x1 ∈ (0, ∞). These points remain fixed in all what follows. Define

a sequence of analytic functions fn : C → C by

fn (z) = fn−1 (z) exp (z − fn−2 (z)) , 105

where f0 (z) = x0 , f1 (z) = x1 are constant functions. Notice that fn (R) ⊂ R.

106

Differentiating the equation above we obtain

0 0 fn0 = fn−1 /fn−1 fn + fn 1 − fn−2 ,

107

and writing gn = fn0 /fn we get, for each z ∈ C, the non-autonomous linear

108

difference equation gn = gn−1 − fn−2 gn−2 + 1

109

g0 = g1 = 0 .

(8)

Differentiating repeatedly, gnk)

110

;

=

k) gn−1

−

k) fn−2 gn−2

−

k   X k j=1

j

j)

k−j)

fn−2 gn−2

;

k)

k)

g0 = g1 = 0 .

(9)

Next, Theorem 2 is applied to equations (8) and (9).

111

Lemma 1. Let x ∈ (1/4, 1) be such that limn→∞ fn (x) = x and let ηn be the

112

tail index sequence for fn (x). Then, k k) fn (x) ≤  (Cm (x)) k! 2k √ x+1 1 − mm+1

113

and

k+1 k) gn (x) ≤  (Cm (x))  k! √ 2k+2 x+1 1 − mm+1

(10)

for m > 0, k ≥ 0 and n ≥ ηm , where Cm (x) := max{1, |gηm +1 (x)|+|gηm (x)|}.

9 Proof. We proceed by induction on k, while holding m fixed and considering  √ 2 x+1 n ≥ ηm throughout the proof. For k = 0, one has |fn (x)| ≤ mm+1 <1

and by applying Theorem 2 to equation (8), we see that |gn (x)| ≤

max{1, |gηm +1 (x)| + |gηm (x)|} . 2  √ x+1 1 − mm+1

Now, assume (10), for n ≥ ηm , as induction hypothesis. We have, k   X k+1) k j) k−j) fn (x) = (fn (x)gn (x))k) ≤ fn (x) gn (x) j j=0

≤

114

k X j=0

k! 

j Cm

√ m x+1 m+1

1−

2j 

k−j+1 k+1 Cm Cm (k + 1)! ≤ 2k+2 ,  2(k−j)+2 √ √ m x+1 m x+1 1 − m+1 1 − m+1

s instead of (Cm (x))s for simplicity. where we write Cm

k)

Next, we apply Theorem 2 to equation (9) with yn = gn (x), an = fn (x),
j) P k−j) bn = − kj=1 kj fn (x) gn (x), to obtain |bn | ≤

k   X k j=1

j

|fnj) (x)| |gnk−j) (x)| ≤ 

k+2 (k + 1)! Cm 2k+2 , √ m x+1 1 − m+1

C k+2 (k + 1)! k+2 gηm (x) + |gηm +1 (x)| ≤ Cm ≤ Cm ≤  m √ 2k+2 , x+1 1 − mm+1 

k+1) gn (x) ≤ 

k+2 Cm (k+1)!  √ m x+1 2k+2 1− m+1

1−

√ m x+1 m+1

2 = 

k+2 Cm (k + 1)! 2(k+1)+2 . √ m x+1 1 − m+1

115

116

117

Let B = {b ∈ (0, 1) : limn→∞ fn (t) = t for all t ∈ (0, b]}. From Theorem 1, we know that 1/2 ∈ B, so B 6= ∅ and sup B ≥ 1/2.

10 118

Lemma 2. If x ∈ (1/4, 1)∩B, then limn→∞ fn (ω) = ω uniformly on compact

119

subsets of the disk Ω(x) :=

(

√ 2) x (1 − x) ω ∈ C : |ω − x| < . 2

(11)

120

Proof. Consider again the tail index sequence ηn for fn (x) and fix m ∈ N.

121

By applying Lemma 1, we get k k) fn (x) ≤  (Cm (x)) k! 2k √ x+1 1 − mm+1

122

123

for all k ≥ 0 and n ≥ ηm . Taylor expansion shows that the family {fn }n≥ηm is uniformly bounded in the disk ( Ωm (x) :=

ω ∈ C : |ω − x| <



√

x+1 1− mm+1

Cm (x)

2

m m+1

)

,

124

thus it is a normal family (Montel’s Theorem, [2]), that is, it contains a

125

subsequence converging uniformly on compact sets to a holomorphic function.

126

As fn (ω) → ω pointwise in B (which has accumulation points in Ωm (x)), the

127

Identity Theorem for analytic functions assures that fn (ω) → ω uniformly

128

on compact sets of Ωm (x).

129

130

131

132

As a consequence, gn (ω) =

fn0 (ω) fn (ω)

1 uniformly on compact sets of Ωm (x) ω √ m x+1 and Cm (x) are as close as we want m+1

→

and then we can choose m ∈ N so as √ to x and 2/x, respectively. Therefore, fn (ω) → ω uniformly on compact sets of Ω(x).

133

Once we have proved that (0, 1) ⊂ B, the uniform convergence on compact

134

subsets of (1/4, 1) will be a consequence of Lemma 2, while that result for

135

(0, 1/4] will follow from an obvious modification of Theorem 2 and Lemmas

11

0.25

0.50

0.75

1.00

Figure 1: Region formed by disks Ω(x) for x ∈ [1/4, 1), used to extend B from (0, 1/2] to (0, 1) and thus proving the conjecture.

√

√

√ 1+ 1−4a 2

or

√ 1+ 1−4x ). 2

136

1 and 2 (just replace

137

absolute summability will follow from Theorem 1 b).

138

139

140

141

a or

x with

Analogously, the

Theorem 3. For all x ∈ (0, 1) one has limn→∞ fn (x) = x. The convergence P is uniform on compact subsets of (0, 1) and ∞ n=0 |fn (x) − x| < ∞.

Proof. Suppose that b := sup B < 1, let h : (0, 1) → (0, 1) be the increasing √ 2 x(1− x) function h(x) = x + and b∗ = h−1 (b). As h(1/4) = 13/48 < 1/2, 3 b+b∗ 2

142

we have 1/4 < b∗ < b, and then, c =

143

c ∈ B ∩ (1/4, 1), one has h(c) ∈ Ω(c) ∩ (0, 1) ⊂ B, a contradiction, because

144

b = sup B. Therefore, (0, 1) ⊂ B.

145

< b = h(b∗ ) < h(c). Since

Since we have not covered the critical parameter value x = 1 considered

146

in [1], it remains open to find an analytic proof for such a case.

147

Acknowledgements

148

We are indebted to Prof. Fernando Bombal for helpful discussions. Both

149

authors were supported by the Spanish Ministry of Science and Innovation

150

and FEDER, grant MTM2013-43404-P, and by the E.T.S. de Ingenieros In-

151

dustriales (UNED), grant 2014-MAT09.

12 152

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153

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154

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