An alternative method to estimate the splitting of separatrices for the discretized pendulum equation

Abstract

We consider the discretization $q (t + ε) + q (t - ε) - 2 q (t) = ε^{2} sin (q (t)),$ $ε > 0$ a small parameter, of the pendulum equation $q^{″} = sin (q)$ ; in system form, we have the discretization $q (t + ε) - q (t) = ε p (t + ε), p (t + ε) - p (t) = ε sin (q (t))$ of the system $q^{'} = p, p^{'} = sin (q) .$

The latter system of ordinary differential equations has two saddle points at $A = (0, 0)$ , $B = (2 π, 0)$ and near both, there exist stable and unstable manifolds. It also admits a heteroclinic orbit connecting the stationary points B and A parametrized by $q_{0} (t) = 4 arctan (e^{- t})$ and which contains the stable manifold of this system at A as well as its unstable manifold at B. We prove that the stable manifold of the point A and the unstable manifold of the point B do not coincide for the discretization. More precisely, we show that the vertical distance between these two manifolds is exponentially small but not zero and in particular we give an asymptotic estimate of this distance. For this purpose we use a method adapted from the article of Schäfke and Volkmer [J. Reine Angew. Math. 425 (1992), 9–60] using formal series and accurate estimates of the coefficients. Our result is a variant of the results of Gelfreich [Comm. Math. Phys. 201 (1999), 155–216], Lazutkin et al. [Physica D 40 (1989), 235–248] for the pendulum problem and our method of proof, however, is quite different. This method will be useful for other problems of this type.

Keywords

difference equation manifolds linear operator formal solution Gevrey asymptotic quasi-solution

1. Introduction

We consider the following difference equation $q (t + ε) + q (t - ε) - 2 q (t) = ε^{2} sin (q (t)),$ (1.1) where ε is a small parameter. This second-order equation is a discretization of the pendulum equation $q^{″} = sin (q)$ . It is equivalent to the following system of first-order difference equations $\{\begin{matrix} q (t + ε) = q (t) + ε p (t + ε), \\ p (t + ε) = p (t) + ε sin (q (t)), \end{matrix}$ (1.2) which can be considered as a discretization of the system $\{\begin{matrix} q^{'} = p, \\ p^{'} = sin (q) . \end{matrix}$ (1.3)

The latter system has centers at $((2 n + 1) π; 0)$ and saddle points at $(2 n π; 0)$ , $n \in Z$ . We focus on the saddle points $A = (0, 0)$ , $B = (2 π, 0)$ and their associated stable and unstable manifolds. For the discretized Eq. (1.2) and sufficiently small $ε > 0$ , these manifolds still exist.

The system (1.3) has $(q_{0} (t), q_{0}^{'} (t))$ , where $q_{0} (t) = 4 arctan (e^{- t})$ , as a heteroclinic orbit connecting the stationary points B and A; it is a parametrization of the curve $p = - 2 sin (q / 2)$ and contains the stable manifold of (1.3) at the point A as well as its unstable manifold at B. This curve, together with $p = 2 sin (q / 2)$ , separates regions with periodic orbits from regions with non-periodic orbits and is therefore often called a separatrix. Our purpose is to study the behavior of this separatrix under discretization of the equation – it turns out that there is no longer a heteroclinic orbit for system (1.2) and the stable manifold at A and the unstable manifold at B no longer coincide. This already follows from a general theorem due to Ushiki [10] which states that a system like (1.2) cannot have analytic invariant curves connecting two stationary points. Our aim is a more quantitative result: we want to estimate the distance between the stable manifold $W_{s, ε}^{-}$ of (1.2) at A and the unstable manifold $W_{u, ε}^{+}$ of (1.2) at B as a function of the parameter ε.

Lazutkin et al. [5], Gelfreich [2] (see also Lazutkin [3,4]) had given an asymptotic estimate of the splitting angle between the manifolds. In [6] Sauzin had studied the parametric resurgence of the problem, whereas in our work, we use an asymptotic approximation of the coefficients of the formal solution following the method of [7]. Starting from a heteroclinic solution of the differential equation, they study the behavior of analytic solutions of the difference equation in the neighbourhood of its singularities $t = \pm \frac{π}{2} i$ .

We show that the distance between these two manifolds is exponentially small but not zero and we give an asymptotic estimate of this distance. Our result is a variant of the results of Gelfreich [2], Lazutkin et al. [5] for the pendulum problem; our method of proof, however, is quite different.

The basic idea used in this paper is to read the exponentially small informations in the asymptotic series. The advantage is that we use only information provided by the formal solutions and nothing else. We use a method adapted from the article of Schäfke and Volkmer [7] using a formal power series solution and accurate estimates of the coefficients. It turns out that the adaptation of this method for the pendulum equation is more difficult than in the case of the logistic equation that we treated in Sellama [8].

For any positive $t_{0}$ , it is known that for sufficiently small $ε_{0} > 0$ and all $t \in] - \infty, t_{0}]$ there is exactly one point $w_{s, ε}^{-} (t) = (q_{0} (t), {\tilde{p}}_{s, ε}^{-} (t))$ on the stable manifold having first coordinate $q_{0} (t)$ . We will show the following result.

Theorem 1.1.

There exists a constant $α \neq 0$ , such that for any positive $t_{0}$ ${dist}_{v} (w_{s, ε}^{-} (t), W_{u, ε}^{+}) = \frac{4 π α}{ε^{2}} cosh (t) sin (\frac{2 π t}{ε}) e^{- \frac{π^{2}}{ε}} + O (\frac{1}{ε} e^{- \frac{π^{2}}{ε}}) as ε ↘ 0,$ uniformly for $- t_{0} < t < t_{0}$ , where ${dist}_{v} (P, W_{u, ε}^{+})$ denotes the vertical distance of a point P from the unstable manifolds $W_{u, ε}^{+}$ .

Remark.

This result can be improved to obtain more terms with the same techniques. It corresponds to the result of Lazutkin et al. [5], Gelfreich [2] as the angle between the manifolds at an intersection point is asymptotically equivalent to $\frac{1}{q_{0}^{'} (t)} \frac{d}{d t} {dist}_{v} (w_{s, ε}^{-} (t), W_{u, ε}^{+}),$ but we do not want to give any detail here.

Our proof uses the following steps. First, we construct a formal solution for the difference Eq. (1.1) in the form of a power series in $d = 2 arsinh (ε / 2)$ , whose coefficients are polynomials in $u = tanh (d t / ε)$ . This is done in Section 2; the introduction of d is necessary because polynomials are desired as coefficients. Then, we give asymptotic approximations of these coefficients using appropriate norms on spaces of polynomials. To that purpose we introduce operators on polynomials series. In Section 6 we use the truncated Laplace transform to construct a function which satisfies (1.1) except for an exponentially small error. The next and last step is to give an asymptotic estimate for the distance of some point of the stable manifold from the unstable manifold. A calculation shows that $α \approx 89.0334$ and therefore $4 π α \approx 1118.8267$ (see Remark 5.4); the corresponding constants of Lazutkin have already been calculated with high precision (see Lazutkin et al. [5]). Suris [9] had shown that $α \neq 0$ . The formal series of Section 2 are the same as the formal series of Gelfreich [2], except for a trivial change in notation.

2. Formal solutions

The purpose of this section is to find a convenient formal solution for Eq. (1.1). First, we need some preparations. We put $\begin{array}{rcl} u : = tanh (\frac{d}{ε} t), \\ q_{0 d} (t) : = 4 arctan (exp (- \frac{d}{ε} t)), \\ q_{d} (t) = \sqrt{1 - u^{2}} A_{d} (u) + q_{0 d} (t), A_{d} (u) = \sum_{n = 1}^{\infty} A_{n} (u) d^{n} \end{array}$ for a formal solution of (1.1), where $d = ε + \sum_{n = 3}^{\infty} d_{n} ε^{n}$ is a formal powers series in ε to be determined.

Remark.
The linearization of Eq. (1.1) at the point A gives the following equation $Z (t + ε) + Z (t - ε) - 2 Z (t) = ε^{2} Z (t) .$ The parameter d is such that $Z (t) = e^{- d t}$ is a solution of this equation. It can be written as $d = log (λ)$ , where λ is the multiplier of the fixed point.

By Taylor expansion, we obtain $q_{0 d} (t + ε) + q_{0 d} (t - ε) - 2 q_{0 d} (t) = 2 \sum_{n = 1}^{+ \infty} \frac{1}{(2 n)!} q_{0 d}^{(2 n)} (t) ε^{2 n},$ (2.1) where $\frac{2}{(2 n)!} q_{0 d}^{(2 n)} (t) ε^{2 n} / d^{2 n}$ is an odd polynomial $I_{2 n - 1} (u)$ multiplied by $\sqrt{1 - u^{2}}$ ; we find $I_{2 n - 1} (1) = 4 / (2 n)!$ .

Using $cos (q_{0 d}) = 2 u^{2} - 1$ , $sin (q_{0 d}) = 2 u \sqrt{1 - u^{2}}$ , we can express our Eq. (1.1) in the form $A_{d} (T^{+}) \sqrt{\frac{1 - {(T^{+})}^{2}}{1 - u^{2}}} + A_{d} (T^{-}) \sqrt{\frac{1 - {(T^{-})}^{2}}{1 - u^{2}}} - 2 A_{d} (u) = f (ε, u, A_{d} (u))$ (2.2) or equivalently $\frac{A_{d} (T^{+})}{cosh (d) + u sinh (d)} + \frac{A_{d} (T^{-})}{cosh (d) - u sinh (d)} - 2 A_{d} (u) = f (ε, u, A_{d} (u)),$ (2.3) where $\begin{array}{rcl} f (ε, u, A_{d} (u)) = ε^{2} (2 u cos (A_{d} (u) \sqrt{1 - u^{2}}) + \frac{2 u^{2} - 1}{\sqrt{1 - u^{2}}} sin (A_{d} (u) \sqrt{1 - u^{2}})) \\ - \sum_{n = 1}^{+ \infty} I_{2 n - 1} (u) d^{2 n}, \\ T^{+} = T^{+} (d, u) = \frac{u + tanh (d)}{1 + u tanh (d)} = tanh (\frac{d}{ε} (t + ε)), \\ T^{-} = T^{-} (d, u) = \frac{u - tanh (d)}{1 - u tanh (d)} = tanh (\frac{d}{ε} (t - ε)) . \end{array}$ As $u \to 1$ , the expressions $T^{+}$ and $T^{-}$ reduce to 1, the denominators in (2.3) simplify to $e^{\pm d}$ and hence Eq. (2.3) reduces to $(e^{- d} + e^{d} - 2) A_{d} (1) = ε^{2} (2 + A_{d} (1)) - 4 (cosh (d) - 1) .$ Therefore, $(2 cosh (d) - 2 - ε^{2}) (2 + A_{d} (1)) = 0 .$

Because $A_{d} (1) = \sum_{n = 1}^{\infty} A_{n} (1) d^{n} = O (d) \neq - 2$ , this is equivalent to $2 cosh (d) - 2 - ε^{2} = 0$ and hence we necessarily have $ε = 2 sinh (d / 2)$ if we want a formal solution such that the coefficients have limits as $u \to 1$ .
Theorem 2.1 (On the formal solution).

If $ε = 2 sinh (d / 2)$ , then Eq.( 2.2 ) has a unique formal solution of the form $A_{d} (u) = \sum_{n = 1}^{+ \infty} A_{2 n - 1} (u) d^{2 n},$ (2.4) where $A_{2 n - 1} (u)$ are odd polynomials of degree $⩽ 2 n - 1$ .

Remark.

A similar formal solution was found using another method in [9].

Proof of Theorem 2.1.

We will use the induction principle to show that there exist unique odd polynomials $A_{1}, A_{3}, A_{5}, \dots, A_{2 n - 1}$ such that $Z_{n} (d, u) = \sum_{k = 1}^{n} A_{2 k - 1} (u) d^{2 k}$ (2.5) satisfy $R_{n} (d, u) = O (d^{2 n + 4}),$ (2.6) where $\begin{array}{rclr} R_{n} (d, u) & = & Z_{n, d} (T^{+}) \sqrt{\frac{1 - {(T^{+})}^{2}}{1 - u^{2}}} + Z_{n, d} (T^{-}) \sqrt{\frac{1 - {(T^{-})}^{2}}{1 - u^{2}}} \\ - 2 Z_{n, d} (u) - f (ε, u, Z_{n, d} (u)) . & (2.7) \end{array}$ For $n = 1$ , a short calculation shows that we must have $A_{1} (u) = - \frac{1}{4} u$ and hence $Z_{1, d} (u) = - \frac{1}{4} u d^{2}$ . We obtain $R_{1} (d, u) = (\frac{- 91}{48} u^{5} + \frac{137}{48} u^{3} - \frac{23}{24}) d^{6} + O (d^{8}) .$ Suppose now that there exist $A_{1}, A_{3}, A_{5}, \dots, A_{2 n - 1}$ such that $Z_{n} (d, u) = \sum_{k = 1}^{n} A_{2 k - 1} (u) d^{2 k}$ (2.8) satisfies (2.6), (2.7). We show that there is a unique polynomial $A_{2 n + 1} (u)$ such that $Z_{n + 1} (d, u) = Z_{n} (d, u) + A_{2 n + 1} (u) d^{2 n + 2}$ (2.9) satisfies (2.6). We put $R_{n} (d, u) = R_{2 n + 3} (u) d^{2 n + 4} + O (d^{2 n + 6}),$ (2.10) where $R_{2 n + 3} (u)$ is odd and $deg (R_{2 n + 3} (u)) ⩽ 2 n + 3$ .

We substitute $Z_{n + 1} (d, u)$ in Eq. (2.7). Using Taylor expansion, (2.9), (2.10) and $ε = 2 sinh (d / 2)$ , we obtain $\begin{array}{rcl} Z_{n + 1, d} (T^{+}) \sqrt{\frac{1 - {(T^{+})}^{2}}{1 - u^{2}}} - Z_{n + 1, d} (T^{-}) \sqrt{\frac{1 - {(T^{-})}^{2}}{1 - u^{2}}} - 2 Z_{n + 1, d} (u) - f (ε, u, Z_{n + 1, d}) \\ = [(u^{4} - 2 u^{2} + 1) A_{2 n + 1}^{″} (u) + (4 u^{3} - 4 u) A_{2 n + 1}^{'} (u) + R_{2 n + 3} (u)] d^{2 n + 4} + O (d^{2 n + 6}) . \end{array}$ We notice that (2.10) is satisfied if and only if ${[{(1 - u^{2})}^{2} A_{2 n + 1}^{'} (u)]}^{'} + R_{2 n + 3} (u) = 0 .$ (2.11)

This differential equation has a unique solution vanishing at $u = 0$ without singularity at $u = 1$ , namely $A_{2 n + 1} (u) = - \int_{0}^{u} \frac{\int_{1}^{t} R_{2 n + 3} (s) d s}{{(1 - t^{2})}^{2}} d t .$ (2.12) We now show that this solution is an odd polynomial of u. It is clear that $\int_{1}^{t} R_{2 n + 3} (s) d s$ vanishes for $t = 1$ and as $R_{2 n + 3} (s)$ is odd, it also vanishes for $t = - 1$ . It suffices to show that $R_{2 n + 3} (s)$ also vanishes at $s = \pm 1$ . Indeed, taking the limit of (2.7) as $u \to 1$ as we did for (2.3) and using $lim_{u \to 1} f (ε, u, Z (d, u)) = ε^{2} Z (d, 1)$ we obtain $R_{2 n + 3} (1) d^{2 n + 4} = (e^{d} + e^{- d} - 2 - ε^{2}) Z (d, 1) + O (d^{2 n + 6}) .$ By our choice of $ε = 2 sinh (d / 2)$ , we obtain $R_{2 n + 3} (1) d^{2 n + 4} = O (d^{2 n + 6})$ . Consequently $R_{2 n + 3} (1) = 0$ . As $R_{2 n + 3} (u)$ is odd, we also have $R_{2 n + 3} (- 1) = - R_{2 n + 3} (1) = 0$ . This proves that $A_{2 n + 1} (u)$ is an odd polynomial of degree at most $2 n + 1$ . □

The first polynomials

A_{2 n - 1} (u)

with

n > 0

can be calculated using Maple (see Table 1).

Table 1

The first coefficients of the formal solution

n	1	2	3
$A_{2 n - 1} (u)$	$- \frac{1}{4} u$	$(\frac{91}{864} u^{3} - \frac{47}{576} u)$	$(- \frac{319}{2880} u^{5} + \frac{185}{1152} u^{3} - \frac{3703}{69, 120} u)$

Now, we introduce the operators $C_{2}$ , $C$ , $S_{2}$ , $S$ defined by $\begin{matrix} C (Z) (d, u) = \frac{1}{2} (Z (d, T^{+ \frac{1}{2}}) + Z (d, T^{- \frac{1}{2}})), \\ S (Z) (d, u) = \frac{1}{2} (Z (d, T^{+ \frac{1}{2}}) - Z (d, T^{- \frac{1}{2}})), \\ C_{1} (Z) (d, u) = \frac{1}{2} (Z (d, T^{+}) + Z (d, T^{-})), \\ S_{1} (Z) (d, u) = \frac{1}{2} (Z (d, T^{+}) - Z (d, T^{-})), \end{matrix}$ (2.13) where $T^{+ \frac{1}{2}} = T^{+} (\frac{d}{2}, u)$ , $T^{- \frac{1}{2}} = T^{-} (\frac{d}{2}, u)$ and $Z (d, u)$ is a formal power series in d whose coefficients are polynomials. We can easily prove that $\begin{matrix} C_{1} = 2 S^{2} + Id, \\ S_{1} = 2 S C \end{matrix}$ (2.14) and $\begin{matrix} C_{1} (Q \cdot G) = C_{1} (Q) C_{1} (G) + S_{1} (Q) S_{1} (G), \\ S_{1} (Q \cdot G) = C_{1} (Q) S_{1} (G) + S_{1} (Q) C_{1} (G), \\ C (Q \cdot G) = C (Q) C (G) + S (Q) S (G), \\ S (Q \cdot G) = C (Q) S (G) + S (Q) C (G), \\ SC (Q) = CS (Q) \end{matrix}$ (2.15) if Q, G are formal power series in d whose coefficients are polynomials of u.

3. Norms for polynomials and basis

In this section we recall some definitions and results of [7]. Using a certain sequence of polynomials, we define convenient norms on spaces of polynomials which satisfy some useful proprieties. We denote by

$P$ the set of all polynomial whose coefficients are complex,

$P_{n}$ the space of all polynomials of degree less than or equal to n.

Proposition 3.1 ([7]).

We define the sequence of polynomials $τ_{n} (u)$ by $τ_{0} (u) = 1, τ_{1} (u) = u, τ_{n + 1} (u) = \frac{1}{n} D τ_{n} (u) for n ⩾ 1,$ where the operator D is defined by $D : = (1 - u^{2}) \frac{\partial}{\partial u} .$ Then we have

$T^{+} (d, u) = \sum_{n = 0}^{\infty} τ_{n + 1} (u) d^{n}$ ,

$τ_{n} (u)$ has exactly degree n and hence $τ_{0} (u), \dots, τ_{n} (u)$ form a basis of $P_{n}$ ,

$τ_{n} (tanh (z)) = \frac{1}{(n - 1)!} {(\frac{d}{d z})}^{n - 1} (tanh (z))$ .

Definition 3.2.
Let $p \in P_{n}$ . As $τ_{0} (u), \dots, τ_{n} (u)$ form a basis of $P_{n}$ , we can write $p \in P_{n}$ as $p = \sum_{k = 0}^{n} a_{k} τ_{k} (u) .$ Then we define the norms $∥ p ∥_{n} = \sum_{i = 0}^{n} | a_{i} | {(\frac{π}{2})}^{n - i} .$ (3.1)
Theorem 3.3 ([7]).

Let n, m be positive integers and $p \in P_{n}$ , $q \in P_{m}$ . The norms ( 3.1 ) have the following properties:

$∥ D p ∥_{n + 1} ⩽ n ∥ p ∥_{n}$ .

If the constant term of p in the basis ${τ_{0}, τ_{1}, \dots, τ_{n}}$ is zero, we have $∥ p ∥_{n} ⩽ ∥ D p ∥_{n + 1} .$

There exists a constant $M_{1}$ such that, for all n, m, $∥ p q ∥_{n + m} ⩽ M_{1} ∥ p ∥_{n} ∥ q ∥_{m}$ .

There is a constant $M_{2}$ such that for all $n > 1$ , $| p (u) | ⩽ M_{2} {(\frac{2}{π})}^{n} ∥ p ∥_{n}$ ( $- 1 ⩽ u ⩽ 1$ ) .

There is a constant $M_{3}$ such that for all $n > 1$ with $p (1) = p (- 1) = 0$ ${∥ \frac{p}{τ_{2}} ∥}_{n - 2} ⩽ M_{3} ∥ p ∥_{n} .$

4. Operators

In this section we will use some definitions of Schäfke and Volkmer [7] and adapt their results on operators on polynomial series to our context. Let $Q : = {Q (d, u) = \sum_{n = 0}^{\infty} Q_{n} (u) d^{n}, where Q_{n} (u) \in P_{n} for all n \in N} .$ By abuse of notation, let $∥ Q ∥_{n} = ∥ Q_{n} ∥_{n}$ for a polynomial series $Q (d, u) = \sum_{n = 0}^{\infty} Q_{n} (u) d^{n} .$

Definition 4.1.
Let $f (z) = \sum_{i = 0}^{\infty} f_{i} z^{i}$ be a formal power series of z whose coefficients are complex. We define the linear operator $f (dD)$ on $Q$ by $f (dD) Q (d, u) = \sum_{n = 0}^{\infty} (\sum_{i = 0}^{n} f_{i} D^{i} Q_{n - i} (u)) d^{n} .$ (4.1)

By the above definition and (1) of Proposition 3.1 we can show that $Q (d, T^{+} (θ d, u)) = (exp (θ dD) Q) (d, u) for Q \in Q and all θ \in C .$ Thus with (2.14) and (1) of Proposition 3.1 we obtain $\begin{matrix} C (Q) = cosh (\frac{d}{2} D) Q, S (Q) = sinh (\frac{d}{2} D) Q, \\ C_{1} (Q) = cosh (dD) Q, S_{1} (Q) = sinh (dD) Q \end{matrix}$ (4.2) for polynomial series Q in $Q$ .
Remark.
According to the definition of norms in (3.1), we have $If Q \in Q, then d Q \in Q and ∥ d Q ∥_{n} = \frac{π}{2} ∥ Q ∥_{n - 1} for all n ⩾ 1 .$ (4.3)
Theorem 4.2 ([7]).

Let $f (z)$ be formal power series having a radius of convergence greater than $2 π$ and let k be a positive integer. There is a constant K such that: If Q is a polynomial series having the following property $∥ Q ∥_{n} ⩽ \{\begin{matrix} 0 & for n < k, \\ M (n - k)! {(2 π)}^{- n} & for n ⩾ k, \end{matrix}$ where M is independent of n and $Q \in Q$ then the polynomial series $f (dD) Q$ satisfies ${∥ f (dD) Q ∥}_{n} ⩽ \{\begin{matrix} 0 & for n < k, \\ M K (n - k)! {(2 π)}^{- n} & for n ⩾ k . \end{matrix}$

Proof.
The proof of this theorem is completely analogous to that of [7]. □

Now we define on $Q$ the following operator $J = \frac{S}{dD} .$ (4.4) Because $S Q = sinh (dD / 2) Q$ , the notation $\frac{S}{dD}$ simply means $F (dD)$ with $F (z) = \frac{1}{z} sinh (\frac{z}{2})$ .
Lemma 4.3.
For each integer k there exist a positive constant K such that: If Q is a polynomial series with odd $Q_{n}$ of degree at most n, $∥ Q ∥_{n} = 0$ for $n < k$ in case of positive k and $∥ dD Q ∥_{n} ⩽ M (n - k)! {(2 π)}^{- n} for n ⩾ max (0, k),$ where M is independent of n, then the polynomial series $J^{- 1} (Q)$ satisfies ${∥ J^{- 1} (Q) ∥}_{n} ⩽ M K {(2 π)}^{- n} \{\begin{matrix} (n - k + 1)! & for k ⩽ 1, \\ (n - 1)! log (n) & for k = 2, \\ (n - 1)! & for k ⩾ 3 . \end{matrix}$
Proof.
We can see easily that $J^{- 1} = π {\tilde{C}}^{- 1} + g (dD)$ , where $\tilde{C} = cosh (\frac{1}{4} dD)$ and $g (z)$ is analytic for $| z | < 4 π$ , and use the proof of [7]. □

We have $S = dD J = J dD$ , but using this relation for the inversion of $S$ would give an insufficient result. Using the formula $1 = \frac{2}{z} sinh (\frac{z}{2}) + F (z) z, where F (z) = z^{- 2} (z - 2 sinh (\frac{z}{2}))$ is an entire function, we obtain the relation $Q = 2 J Q + F (dD) dD Q$ (4.5) for polynomial series $Q \in Q$ . This will be essential in the proof of the following theorem.
Theorem 4.4.
For each integer k there exists a positive constant K such that: If Q is a polynomial series with odd $Q_{n}$ of degree at most n, $∥ Q ∥_{n} = 0$ for $n < k$ in case of positive k, and ${∥ S (Q) ∥}_{n} ⩽ M (n - k)! {(2 π)}^{- n} for n ⩾ max (0, k),$ where M is independent of n, then the polynomial series Q satisfies $∥ Q ∥_{n} ⩽ M K {(2 π)}^{- n} \{\begin{matrix} (n - k + 1)! & for k ⩽ 1, \\ (n - 1)! log (n) & for k = 2, \\ (n - 1)! & for k ⩾ 3 . \end{matrix}$
Proof.
By the preceding theorem, we have the wanted inequalities for $dD Q = J^{- 1} S Q$ in the place of Q. Here we used again $∥ dD Z ∥_{n} ⩽ (n - 1) ∥ Z ∥_{n - 1}$ for any polynomial series $Z \in Q$ . Using Theorem 4.2 implies the same for $F (dD) dD Q$ with the entire function F of (4.5). As $∥ Z ∥_{n} ⩽ ∥ dD Z ∥_{n + 1}$ by Theorem 3.3, we find the wanted inequalities (and even something better in the cases $k ⩾ 2$ ) also for $J Q$ because $dD J = S$ . Thus formula (4.5) yields the result. □

In order to obtain an asymptotic approximation for the coefficients of the formal solution, we will need to reverse some operators. This is not possible for the operators $S$ and $dD$ on the set $Q$ , but we can define a subset $Q^{}$ of $Q$ on which these operators have a right inverses.

If we define $Q^{} : = {Q (d, u) = \sum_{n = 1}^{\infty} P_{n} (u) d^{n}, where P_{n} (u) \in P_{n}^{} for all n ⩾ 1},$ where $P_{n}^{}$ is the subspace of $P_{n}$ defined by $P_{n}^{} : = {\sum_{i = 0}^{n} α_{i} τ_{i} \in P_{n} | α_{0} = 0} .$ Then, the restrictions of the operators $dD$ , $S$ on $Q^{}$ , denoted here by the same symbols $\begin{array}{rcl} dD : Q^{} \to (1 - u^{2}) d^{2} Q, \\ S : Q^{} \to (1 - u^{2}) d^{2} Q, \end{array}$ are bijective. We denote by $T$ the inverse of the restriction of $S$ to $Q^{}$ , and we have $T S = Id on Q^{} .$ Theorem 4.5 ([7]).

Let $α : = \frac{4}{π} \sum_{n = 1}^{\infty} α_{n},$ where $α_{n} = O (n^{- k})$ as $n \to \infty$ with some integer $k ⩾ 2$ .

If we consider a polynomial series $Q_{α} (d, u) = \sum_{n = 2}^{\infty} α_{n} (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} (u) d^{n},$ then the coefficients ${T (Q_{α})}_{n}$ of $T (Q_{α})$ satisfy ${∥ {T (Q_{α})}_{n} - α (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} ∥}_{n} = O ((n - k)! {(2 π)}^{- n})$ as $n \to \infty$ for n.

Proof.
The proof of this theorem is completely analogous to that of [7]. □
Theorem 4.6 ([7]).

Let k, l, p, q be integers with $p ⩾ k$ and $q ⩾ l$ . Define m as the minimum of $k + q$ and $l + p$ . Then there is a constant K with the following property:

If P and Q are polynomial series such that $∥ P ∥_{n} = 0$ for $n < p$ , $∥ Q ∥_{n} = 0$ for $n < q$ and $\begin{matrix} ∥ P ∥_{n} ⩽ M_{1} (n - k)! {(2 π)}^{- n} for n ⩾ p, \\ ∥ Q ∥_{n} ⩽ M_{2} (n - l)! {(2 π)}^{- n} for n ⩾ q \end{matrix}$ then $∥ P Q ∥_{n} ⩽ K M_{1} M_{2} (n - m)! {(2 π)}^{- n} for n ⩾ p + q .$

Remark 4.7.
Observe that the results of this section can also be applied, if the constants M are replaced by any increasing sequence ${(M_{n})}_{n \in N}$ . In Theorems 4.2 and 4.6 the first n terms of the resulting polynomial series only depend on the first n terms of the given series, so the “M” in the result simply has to be replaced by “ $M_{n}$ ”. In Lemma 4.3 and Theorem 4.4, the first n terms of the result depend of the first $n + 1$ given terms, so “M” in the result has to be replaced by “ $M_{n + 1}$ ”.

5. Asymptotic approximation of the coefficients of the formal solution

In this section we will estimate the coefficients of the formal solution obtained previously (Section 2). The idea is to write Eq. (2.2) essentially in the form $V (d, u) S (Q_{1} S (Q_{2} A)) (d, u) = g (d, u, A (d, u)),$ (5.1) where V, $Q_{1}$ and $Q_{2}$ are known polynomials of d and u, and g is a certain function of d, u and A involving the operators $S$ , $C$ and $J$ multiplied by sufficiently high powers of d.

Thanks to this equation, we will estimate the coefficients of the formal solution using the results of the previous section. We show that the coefficients of this formal solution is Gevrey-1, more precisely $∥ A ∥_{n} = O (n! {(2 π)}^{- n})$ .

5.1. Rewriting of Eq. (2.2)

Consider the decomposition $A (d, u) = U (d, u) + F (d, u),$ (5.2) where U is the initial part of A calculated before $U (d, u) = - \frac{1}{4} u d^{2} + (\frac{91}{864} u^{3} - \frac{47}{576} u) d^{4} + (- \frac{319}{2880} u^{5} + \frac{185}{1152} u^{3} - \frac{3703}{69, 120} u) d^{6} .$ We insert this into (2.3), with (2.14) and (2.15), and obtain $2 cosh (d) \cdot C_{1} (F) - 2 u sinh (d) \cdot S_{1} (F) = W_{0} \cdot F + f_{1} (d, u, F),$ (5.3) where $\begin{array}{rcl} W_{0} = ({cosh}^{2} (d) - u^{2} {sinh}^{2} (d)) [2 + (2 u^{2} - 1) ε^{2} cos (U \cdot \sqrt{1 - u^{2}}) \\ - 2 u ε^{2} sin (U \cdot \sqrt{1 - u^{2}}) \cdot \sqrt{1 - u^{2}}], \\ f_{1} (d, u, F (d, u)) = ε^{2} ({cosh}^{2} (d) - u^{2} {sinh}^{2} (d)) [(\frac{2 u^{2} - 1}{\sqrt{1 - u^{2}}} sin (U \cdot \sqrt{1 - u^{2}}) \\ + 2 u cos (U \cdot \sqrt{1 - u^{2}})) cos (F (d, u) \sqrt{1 - u^{2}}) \\ + (\frac{2 u^{2} - 1}{\sqrt{1 - u^{2}}} cos (U \cdot \sqrt{1 - u^{2}}) - 2 u sin (U \cdot \sqrt{1 - u^{2}})) \\ \times (sin (F (d, u) \sqrt{1 - u^{2}}) - F (d, u) \sqrt{1 - u^{2}})] \\ - ({cosh}^{2} (d) - u^{2} {sinh}^{2} (d)) \sum_{n = 1}^{+ \infty} I_{2 n - 1} (u) d^{2 n} - 2 cosh (d) C_{1} (U) \\ - 2 u sinh (d) S_{1} (U) - 2 ({cosh}^{2} (d) - u^{2} {sinh}^{2} (d)) \cdot U . \end{array}$ Observe that $f_{1}$ has the form $\begin{matrix} f_{1} (d, u, F (d, u)) = & y_{0} (d, u) + y_{1} (d, u) \sum_{n = 1}^{\infty} \frac{1}{(2 n)!} {(1 - u^{2})}^{n} F {(d, u)}^{2 n} \\ + y_{2} (d, u) \sum_{n = 1}^{\infty} \frac{1}{(2 n + 1)!} {(1 - u^{2})}^{n} F {(d, u)}^{2 n + 1}, \end{matrix}$ (5.4) where $y_{n} (d, u)$ , $n = 1, 2, 3$ are convergent polynomial series.

Now, we let $\begin{matrix} F (d, u) = Q (d, u) \cdot G (d, u), \\ J (d, u) = Q_{1} (d, u) \cdot S (G), \end{matrix}$ (5.5) where G is a formal power series whose first term contains $d^{8}$ and $\begin{array}{rclr} \begin{matrix} Q (d, u) = 1 + \frac{1}{4} (1 - u^{2}) d^{2} + (\frac{91}{432} u^{4} - \frac{13}{48} u^{2} + \frac{13}{216}) d^{4} \\ + (- \frac{319}{960} u^{6} + \frac{1079}{1728} u^{4} - \frac{937}{2880} u^{2} + \frac{287}{8640}) d^{6}, \\ Q_{1} (d, u) = (u^{2} - 1) d^{2} + \frac{1}{4} (1 - u^{4}) d^{4} - \frac{5}{48} (1 - u^{2}) (\frac{4}{9} u^{4} + u^{2} + 1) d^{6} \\ + (1 - u^{2}) (- \frac{367}{2160} u^{6} + \frac{185}{432} u^{4} - \frac{997}{4320} u^{2}) d^{8} . \end{matrix} & (5.6) \end{array}$ The choice of $Q_{1} (d, u)$ and $Q (d, u)$ depends in a precise way of the form of Eq. (5.1) and has been determined using Maple.

Using (5.5), (5.6) and (2.15), we can rewrite Eq. (5.3) in the form $\begin{array}{rcl} W_{0} Q G + f_{1} (d, u, F (d, u)) & = & [2 cosh (d) C_{1} (Q) - 2 u sinh (d) S_{1} (Q)] C_{1} (G) \\ + [2 cosh (d) S_{1} (Q) - 2 u sinh (d) C_{1} (Q)] S_{1} (G) . \end{array}$ Using (2.14), we obtain $V \cdot S^{2} (G) + W \cdot S C (G) = W_{1} G + f_{2} (d, u, F (d, u)),$ (5.7) where $f_{2} (d, u, F (d, u)) = \frac{1}{4} f_{1} (d, u, F (d, u))$ , $\begin{matrix} V (d, u) = cosh (d) C_{1} (Q) - u sinh (d) S_{1} (Q), \\ W (d, u) = cosh (d) S_{1} (Q) - u sinh (d) C_{1} (Q), \\ W_{1} (d, u) = \frac{1}{4} (- 2 cosh (d) C_{1} (Q) + 2 u sinh (d) S_{1} (Q) + W_{0} Q) . \end{matrix}$ (5.8) The calculation of the first terms of the series $W_{1}$ by Maple shows that the convergent polynomial series $W_{1} (d, u)$ begins with a term containing $d^{10}$ .

Using (5.5) and (2.15), we find $S (J) = S (Q_{1} S (G)) = C (Q_{1}) S^{2} (G) + S (Q_{1}) S C (G) .$ (5.9) Using $\begin{array}{rcl} V_{1} (d, u) & = & 1 + (1 - u^{2}) d^{2} + (- \frac{71}{432} u^{4} - \frac{1}{12} u^{2} + \frac{107}{432}) d^{4} \\ + (\frac{1351}{2160} u^{6} - \frac{193}{144} u^{4} + \frac{49}{60} u^{2} - \frac{11}{108}) d^{6}, \end{array}$ we obtain $\begin{array}{rcl} V_{1} \cdot C (Q_{1}) = Q_{1} V + W_{2}, \\ V_{1} \cdot S (Q_{1}) = Q_{1} W + W_{3}, \end{array}$ where $W_{2} (d, u)$ and $W_{3} (d, u)$ are convergent polynomial series beginning with $d^{10}$ . With (5.7) and (5.9), this implies $\begin{array}{rclr} V_{1} \cdot S (Q_{1} S (G)) & = & W_{2} \cdot S^{2} (G) + W_{3} S C (G) + Q_{1} W_{1} G \\ + Q_{1} f_{2} (d, u, F (d, u)) . & (5.10) \end{array}$ This allows us to prove the following theorem.

Theorem 5.1.

We have $\begin{array}{rclr} G (d, u) & = & (\frac{α}{d^{2}} + (β + \frac{α}{3})) \frac{u H_{0} (d, u)}{τ_{2} (u)} - (β d + \frac{α}{d}) H_{2} (d, u) \\ + δ d H_{1} (d, u) + S (d, u), & (5.11) \end{array}$ where α, β, δ are constants and the polynomial series $H_{0}$ , $H_{1}$ , $H_{2}$ , S are defined by $\begin{array}{rclr} H_{0} (d, u) : = \sum_{\begin{array}{c} n = 10 \\ n even \end{array}}^{\infty} (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) d^{n}, \\ H_{1} (d, u) : = \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} (n - 1)! {(\frac{i}{2 π})}^{n + 1} τ_{n} (u) d^{n}, & (5.12) \\ H_{2} (d, u) : = \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} n! {(\frac{i}{2 π})}^{n + 1} τ_{n} (u) d^{n} \end{array}$ and $S (d, u)$ is a polynomial series satisfying $∥ S ∥_{n} = O ((n - 3)! {(2 π)}^{n}) .$

To prove this theorem we need to estimate the coefficients of the polynomial series $S (Q_{1} S (G))$ . This will be the subject of the following paragraph. Remark.

Observe that the series F, G are odd in u, even in d and beginning with $d^{8}$ . The series J is even in u, odd in d and begins with $d^{11}$ . In the series F, G, A, J, the degree of the polynomial that is the coefficient of $d^{n}$ is at most $n - 1$ . Thus the results of Section 4 can still be applied and $d^{- 1} F, d^{- 1} G, d^{- 1} A, d^{- 1} J \in Q$ .

5.2. Upper bounds for the coefficients of

S (Q_{1} S (G))

In this subsection, we will use Eq. (5.10), together with the definitions of $V_{1}$ and $Q_{1}$ , J and G, $W_{i}$ , $i = 1, 2, 3$ , to prove the following lemma.

Lemma 5.2.
${∥ \frac{1}{d} S (Q_{1} S (G)) ∥}_{n} = O ((n - 8)! {(2 π)}^{- n}) as n \to \infty .$
Proof.
We set $e_{n} : = \frac{{(2 π)}^{n} ∥ S (J) ∥_{n}}{(n - 8)!} for n ⩾ 12 .$ (5.13) We must show that $e_{n} = O (n^{- 1})$ . In the sequel, we will use the following convention: if $a_{n}$ , $n = 0, 1, \dots$ is any sequence of positive real numbers, then $a_{n}^{+} : = max (a_{0}, a_{1}, \dots, a_{n}) for all n ⩾ 0 .$ We have ${∥ S (J) ∥}_{n} ⩽ e_{n} (n - 8)! {(2 π)}^{- n} for n ⩾ 12 .$ (5.14) Using Theorem 4.4 and Remark 4.7, we obtain $∥ J ∥_{n} ⩽ K_{1} e_{n + 1}^{+} (n - 1)! {(2 π)}^{- n} for n ⩾ 11,$ (5.15) where $K_{1}$ denotes the constant associated with the operator $S$ in Theorem 4.4; it is independent of the present context. In this proof, $K_{i}$ , $i = 1, \dots, 9$ , will always denote constants independent of n and the sequence $e_{n}$ .

Using (5.5), we obtain ${∥ Q_{1} S (G) ∥}_{n} ⩽ K_{1} e_{n + 1}^{+} (n - 1)! {(2 π)}^{- n} for n ⩾ 11 .$ (5.16) We use (5) of Theorem 3.3 and Theorem 4.6. Since $\frac{Q_{1} (d, u)}{τ_{2} (u) d^{2}}$ is a convergent power series beginning with 1, there is a constant $K_{2}$ such that ${∥ S (G) ∥}_{n} ⩽ K_{2} e_{n + 3}^{+} (n + 1)! {(2 π)}^{- n} for n ⩾ 9 .$ (5.17) Using again Theorem 4.4 (and Remark 4.7) and the fact that $F = G / Q$ where Q is given in (5.6), we obtain $\begin{matrix} ∥ G ∥_{n} ⩽ K_{3} e_{n + 4}^{+} (n + 2)! {(2 π)}^{- n} for n ⩾ 8, \\ ∥ F ∥_{n} ⩽ K_{3} e_{n + 4}^{+} (n + 2)! {(2 π)}^{- n} for n ⩾ 8, \end{matrix}$ (5.18) where $K_{3}$ is a constant independent of n.

This, together with Theorem 4.6, implies that there are constants $K_{4}$ , L such that for all $k ⩾ 2$ ${∥ F^{k} ∥}_{n} ⩽ K_{4} L_{1}^{k} f_{n}^{(k)} (n - 5)! {(2 π)}^{- n} for n ⩾ 8 k,$ (5.19) where $\begin{matrix} f_{n}^{(2)} = \sum_{i = 8}^{n - 8} e_{i + 4}^{+} e_{n - i + 4}^{+} \frac{(i + 2)! (n - i + 2)!}{(n - 5)!} for n ⩾ 16, \\ f_{n}^{(k + 1)} = \sum_{i = 8}^{n - 8 k} e_{i + 4}^{+} f_{n - i}^{(k)} \frac{(i + 2)! (n - i - 5)!}{(n - 5)!} for n ⩾ 8 (k + 1), \end{matrix}$ with $f_{n}^{(k)} : = 0$ for $n < 8 k$ .

Using Theorems 4.2 and 4.6 and $W_{i} = O (d^{10})$ , $i = 1, 2, 3$ , we obtain $\begin{array}{rclr} {∥ W_{2} \cdot S^{2} (G) ∥}_{n} ⩽ K_{5} e_{n - 6}^{+} (n - 9)! {(2 π)}^{- n} for n ⩾ 20, & (5.20) \\ {∥ W_{3} S C (G) ∥}_{n} ⩽ K_{6} e_{n - 6}^{+} (n - 9)! {(2 π)}^{- n} for n ⩾ 19, & (5.21) \\ ∥ Q_{1} W_{1} G ∥_{n} ⩽ K_{7} e_{n - 8}^{+} (n - 10)! {(2 π)}^{- n} for n ⩾ 20, & (5.22) \\ {∥ Q_{1} f_{2} (d, u, F_{d} (u)) ∥}_{n} ⩽ K_{8} (1 + \sum_{k ⩾ 2} \frac{L_{2}^{k}}{k!} f_{n - 4}^{(k) +}) (n - 9)! {(2 π)}^{- n} for n ⩾ 12 . & (5.23) \end{array}$

Now, let us take Eq. (5.10) ${∥ V_{1} \cdot S (J) ∥}_{n} ⩽ {∥ W_{2} \cdot S^{2} (G) ∥}_{n} + {∥ W_{3} S C (G) ∥}_{n} + ∥ Q_{1} W_{1} G ∥_{n} + {∥ Q_{1} f_{2} (d, u, F_{d} (u)) ∥}_{n} .$ Using (5.20), (5.21), (5.22) and (5.23), we obtain ${∥ V_{1} \cdot S (J) ∥}_{n} ⩽ K_{9} (1 + e_{n - 6}^{+} + \sum_{k ⩾ 2} \frac{L_{2}^{k}}{k!} f_{n - 4}^{(k) +}) (n - 9)! {(2 π)}^{- n} .$ (5.24) Since $V_{1}$ is a convergent polynomial series beginning with 1, we also have ${∥ S (J) ∥}_{n} ⩽ K_{10} (1 + e_{n - 6}^{+} + \sum_{k ⩾ 2} \frac{L_{2}^{k}}{k!} f_{n - 4}^{(k) +}) (n - 9)! {(2 π)}^{- n} .$ (5.25)

Using (5.13), we obtain $e_{n}^{+} ⩽ \frac{K}{n} (1 + e_{n - 6}^{+} + \sum_{k ⩾ 2} \frac{L_{2}^{k}}{k!} f_{n - 4}^{(k) +}) for n ⩾ 12 .$ (5.26) This completes the proof of the theorem. □
Lemma 5.3.
Under the condition ( 5.26 ), we have $e_{n} = O (n^{- 1})$ as $n \to \infty$ .
Proof.
Let $K_{1} ⩾ 10! e_{12}^{+}$ be an arbitrary number. We assume that $e_{n} ⩽ \frac{K_{1} (n + p - 1)!}{(n - 2)! (p + 11)!} for 12 ⩽ n ⩽ N - 4$ (5.27) with some $p ⩾ - 1$ , $N ⩾ 16$ . This gives for $16 ⩽ n ⩽ N$ $(n - 5)! f_{n}^{(2)} ⩽ K_{1}^{2} \sum_{i = 8}^{n - 8} \frac{(i + p + 3)! (n + p - i + 3)!}{{((p + 11)!)}^{2}} .$ The first and last term of the above sum are the largest, so we can easily estimate $\sum_{i = 8}^{n - 8} (i + p + 3)! (n + p - i + 3)! ⩽ (p + 11)! (n + p - 4)! .$ We obtain $f_{n}^{(2)} ⩽ K_{1}^{2} \frac{(n + p - 4)!}{(p + 11)! (n - 5)!} for 16 ⩽ n ⩽ N .$ In a similar way, we prove by induction that $f_{n}^{(k)} ⩽ K_{1}^{k} \frac{(n + p - 4)!}{(p + 11)! (n - 5)!} ⩽ K_{1}^{k} \frac{(n + p - 1)!}{(p + 11)! (n - 2)!} for 8 k ⩽ n ⩽ N .$ Using the assumption of the lemma, we obtain $e_{n} ⩽ \frac{K}{n} e^{(1 + L_{2}) K_{1}} \frac{(n + p - 1)!}{(p + 11)! (n - 2)!} for 16 ⩽ n ⩽ N .$

Now we choose $N_{0} ⩾ 16$ so large that $\frac{K exp ((1 + L_{2}) K_{1})}{N_{0}} ⩽ K_{1}$ and then p so large that (5.27) holds for $N = N_{0}$ . In a first step, our considerations imply by induction over N that (5.27) holds for all N and hence $e_{n} = O (\frac{(n + p - 1)!}{(n - 2)!}) as n \to \infty$ for this possibly large value of p.

As $K_{1}$ is arbitrary in (5.27), we have also shown for any $p ⩾ - 1$ that $e_{n} = O (\frac{(n + p - 1)!}{(n - 2)!}) as n \to \infty$ implies that $e_{n} = O (\frac{(n + p - 2)!}{(n - 2)!}) as n \to \infty .$ Consequently the last assertion is proved for $p = - 1$ and we have shown $e_{n} = O (n^{- 1}) as n \to \infty .$ Finally we have proved that ${∥ S (J) ∥}_{n} = O ((n - 9)! {(2 π)}^{- n}) as n \to \infty$ and hence that ${∥ \frac{1}{d} S (J) ∥}_{n} = O ((n - 8)! {(2 π)}^{- n}) as n \to \infty$ which completes the proof of the lemma. □

5.3. Proof of Theorem 5.1

Let $E : = \frac{1}{d} S (J) = S (d^{- 1} J)$ . The polynomial series E is odd in d and its coefficients are odd in u. We partition it $\begin{array}{rclr} E_{n} (u) & = & α_{n} (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} (u) + β_{n - 2} (n - 3)! {(\frac{i}{2 π})}^{n - 3} τ_{n - 2} (u) \\ + γ_{n - 4} (n - 5)! {(\frac{i}{2 π})}^{n - 5} τ_{n - 4} (u) + {\overline{E}}_{n - 6} (u) & (5.28) \end{array}$ for odd $n ⩾ 11$ , where $α_{n}$ , $β_{n}$ and $γ_{n}$ are real numbers and also ${\overline{E}}_{n}$ have at most degree n for all n. For the whole series E this is equivalent to $S (d^{- 1} J) = E = E_{1} + d^{2} E_{2} + d^{4} E_{3} + d^{6} \overline{E},$ (5.29) where $\begin{array}{rcl} E_{1} = \sum_{n = 11}^{+ \infty} α_{n} (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} (u) d^{n}, \\ E_{2} = \sum_{n = 9}^{+ \infty} β_{n} (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} (u) d^{n}, \\ E_{3} = \sum_{n = 7}^{+ \infty} γ_{n} (n - 1)! {(\frac{i}{2 π})}^{n - 1} τ_{n} (u) d^{n}, \\ \overline{E} = \sum_{n = 5}^{+ \infty} {\overline{E}}_{n} (u) d^{n} . \end{array}$ Lemma 5.2 implies that $α_{n} = O (n^{- 7}), β_{n} = O (n^{- 5}), γ_{n} = O (n^{- 3}) and ∥ {\overline{E}}_{n} ∥_{n} = O ((n - 2)! {(2 π)}^{- n}) .$ Applying $T$ to (5.29) we obtain $\frac{1}{d} J = T (E_{1}) + d^{2} T (E_{2}) + d^{4} T (E_{3}) + d^{6} T (\overline{E}) .$ (5.30) To the first three summands we apply Theorem 4.5. Thus we obtain $\begin{array}{rcl} {∥ {T (E_{1})}_{n} - α (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) ∥}_{n} = O ((n - 7)! {(2 π)}^{- n}), \\ {∥ {T (E_{2})}_{n} - β (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) ∥}_{n} = O ((n - 5)! {(2 π)}^{- n}), \\ {∥ {T (E_{3})}_{n} - γ (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) ∥}_{n} = O ((n - 3)! {(2 π)}^{- n}), \end{array}$ where $α = \frac{4}{π} \sum_{\begin{array}{c} n = 11 \\ n odd \end{array}}^{\infty} α_{n}, β = \frac{4}{π} \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} β_{n}, γ = \frac{4}{π} \sum_{\begin{array}{c} n = 7 \\ n odd \end{array}}^{\infty} γ_{n} .$ To the last part of (5.30) we apply Theorem 4.4 and obtain ${∥ {T (\overline{E})}_{n} ∥}_{n} = O ((n - 1)! {(2 π)}^{- n} log (n)),$ thus altogether $\begin{array}{rcl} {∥ {(\frac{1}{d} J)}_{n} - α (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} - β (n - 3)! {(\frac{i}{2 π})}^{n - 2} τ_{n - 2} - γ (n - 5)! {(\frac{i}{2 π})}^{n - 4} τ_{n - 4} ∥}_{n} \\ = O ((n - 7)! {(2 π)}^{- n} log (n)) . \end{array}$ Using (4.3) we obtain $\begin{array}{rcl} {∥ {J}_{n} - α (n - 2)! {(\frac{i}{2 π})}^{n - 1} τ_{n - 1} - β (n - 4)! {(\frac{i}{2 π})}^{n - 3} τ_{n - 3} - γ (n - 6)! {(\frac{i}{2 π})}^{n - 5} τ_{n - 5} ∥}_{n} \\ = O ((n - 8)! {(2 π)}^{- n} log (n)) . \end{array}$

Remark 5.4.

The asymptotic of $J_{n}$ gives a good approximation of α; it suffices to calculate, using a formal calculation software (for example, PARI), the first $40$ terms of $A (d, u)$ by the recursion of Section 2 and to evaluate the highest coefficients of $J_{n}$ to get the approximation $α \approx 89.0334$ .

Next we observe that $J = Q_{1} S (G)$ , where $Q_{1}$ is given in (5.6). Using part (5) of Theorem 3.3, we obtain $\begin{array}{rclr} ∥ {S (G)}_{n} + α n! {(\frac{i}{2 π})}^{n + 1} \frac{τ_{n + 1}}{τ_{2}} + (β - α p_{2} (u)) (n - 2)! {(\frac{i}{2 π})}^{n - 1} \frac{τ_{n - 1}}{τ_{2}} \\ {+ (γ - β p_{2} (u) - α p_{4} (u)) (n - 4)! {(\frac{i}{2 π})}^{n - 3} \frac{τ_{n - 3}}{τ_{2}} ∥}_{n} \\ = O ((n - 6)! {(2 π)}^{- n} log (n)), & (5.31) \end{array}$ where $\begin{array}{rcl} p_{2} (u) = - \frac{1}{4} - \frac{u^{2}}{4} = - \frac{1}{2} + \frac{τ_{2}}{4}, \\ p_{4} (u) = \frac{1}{24} - \frac{u^{2}}{48} - \frac{7 u^{4}}{432} = \frac{1}{216} + \frac{55}{1296} τ_{2} + \frac{7}{432} τ_{4} . \end{array}$

Remark.

Observe that the approximation (5.31) of the coefficients ${S (G)}_{n}$ is polynomial. Indeed, the polynomials $τ_{n} (u)$ , $n ⩾ 2$ are divisible by $τ_{2} (u)$ .

In order to find an asymptotic estimation for the coefficients of the formal solution, we need to apply the inverse of operator $S$ . To this purpose, we show the following lemma.

Lemma 5.5.

Let $H_{0}$ , $H_{1}$ , $H_{2}$ be the polynomial series in ( 5.12 ) and define the operator $C_{\frac{1}{4}} = cosh (\frac{dD}{4})$ . Then

the polynomial series $C_{\frac{1}{4}} (H_{0})$ , $C_{\frac{1}{4}} (H_{1})$ are convergent,

the polynomial series $S (H_{0})$ , $S (H_{1})$ are convergent,

$C (H_{0}) = - H_{0} + μ_{1} (d, u)$ , where $μ_{1} (d, u)$ is a convergent series,

$S (H_{2}) = \frac{1}{2} H_{0} + μ_{2} (d, u)$ , where $H_{2} = d \frac{\partial}{\partial d} H_{1}$ and $μ_{2} (d, u)$ is a convergent series,

$S (\frac{H_{0}}{τ_{2} (u)}) = - u sinh (d) \frac{H_{0}}{τ_{2} (u)} + μ_{3} (d, u)$ , where $μ_{3} (d, u)$ is a convergent series,

$S (\frac{u H_{0}}{τ_{2} (u)}) = - \frac{1}{2} (u^{2} + 1) sinh (d) \frac{H_{0}}{τ_{2} (u)} + μ_{4} (d, u)$ , where $μ_{4} (d, u)$ is a convergent series,

$S (\frac{u H_{0}}{sinh (d) τ_{2} (u)} - H_{2}) = - \frac{H_{0}}{τ_{2} (u)} + μ_{5} (d, u)$ , where $μ_{5} (d, u)$ is a convergent series.

Proof.

(1) We have $C_{\frac{1}{4}} (H_{0}) = \sum_{\begin{array}{c} m = 0 \\ m even \end{array}}^{\infty} \frac{1}{4^{m} m!} d^{m} D^{m} (\sum_{\begin{array}{c} n = 10 \\ n even \end{array}}^{\infty} (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) d^{n}) .$ Using the definition of the operator D in Proposition 3.1, we obtain $\begin{array}{rclr} C_{\frac{1}{4}} (H_{0}) & = & \sum_{\begin{array}{c} m ⩾ 0 \\ n ⩾ 10 even \end{array}} \frac{1}{4^{m} m!} (n + m - 1)! {(\frac{i}{2 π})}^{n} τ_{n + m} (u) d^{n + m} \\ = & \sum_{\begin{array}{c} k = 10 \\ k even \end{array}}^{\infty} γ_{k} (k - 1)! {(\frac{i}{2 π})}^{k} τ_{k} d^{k}, & (5.32) \end{array}$ where $γ_{k} : = \sum_{\begin{array}{c} m = 0 \\ m even \end{array}}^{k - 10} \frac{1}{4^{m} m!} {(\frac{2 π}{i})}^{m} .$ Hence $γ_{k} = \sum_{\begin{array}{c} m = 0 \\ m even \end{array}}^{\infty} \frac{1}{m!} {(\frac{π}{2 i})}^{m} - \sum_{\begin{array}{c} m = k - 8 \\ m even \end{array}}^{\infty} \frac{1}{m!} {(\frac{π}{2 i})}^{m} .$ Using $\sum_{\begin{array}{c} m = 0 \\ m even \end{array}}^{\infty} \frac{1}{m!} {(\frac{π}{2 i})}^{m} = \sum_{l = 0}^{\infty} \frac{{(- 1)}^{l}}{(2 l)!} {(\frac{π}{2})}^{2 l} = cos (\frac{π}{2}) = 0,$ we find $γ_{k} = - \sum_{\begin{array}{c} m = k - 8 \\ m even \end{array}}^{\infty} \frac{1}{m!} {(\frac{π}{2 i})}^{m}$ which implies $\begin{array}{rcl} | γ_{k} | & ⩽ & \frac{1}{(k - 8)!} {(\frac{π}{2})}^{k - 8} \sum_{\begin{array}{c} m = 0 \\ m even \end{array}}^{\infty} \frac{1}{m!} {(\frac{π}{2})}^{m} \\ ⩽ & \frac{cosh (π / 2)}{(k - 8)!} {(\frac{π}{2})}^{k - 8} . \end{array}$

This with (5.32) imply that $C_{\frac{1}{4}} (H_{0}) = μ (d, u)$ is convergent. For $C_{\frac{1}{4}} (H_{1})$ , we use the same method.

(2) As $S = 2 S_{\frac{1}{4}} C_{\frac{1}{4}}$ , where $S_{\frac{1}{4}} = sinh (\frac{dD}{4})$ , we obtain using (1), $S (H_{0}) = 2 S_{\frac{1}{4}} (C_{\frac{1}{4}} (H_{0})) = 2 S_{\frac{1}{4}} (μ) .$ This implies that $S (H_{0})$ is convergent. For $S (H_{1})$ , we use the same method.

(3) We have $C (H_{0}) = (2 C_{\frac{1}{4}}^{2} - Id) H_{0} = - H_{0} + 2 C_{\frac{1}{4}}^{2} (H_{0})$ . This with (1) imply $C (H_{0}) = - H_{0} + μ_{1} (d, u),$ where $μ_{1} (d, u)$ is a convergent series.

(4) We differentiate the equation $S (H_{1}) = C_{1}$ with respect to d. As $z \frac{d}{d z} (sinh (z / 2)) = \frac{z}{2} cosh (z / 2),$ we obtain $S (H_{2}) + \frac{1}{2} C (dD H_{1}) = d \frac{\partial C_{1}}{\partial d} .$ Using (3) of this lemma, we obtain $S (H_{2}) = \frac{1}{2} dD H_{1} + μ_{2} (d, u) = \frac{1}{2} H_{0} + μ_{2} (d, u),$ where $μ_{2} (d, u)$ is a convergent series.

(5) Using (2.15) and (2), (3) of this lemma we obtain $\begin{array}{rcl} S (H_{0}) = S (τ_{2} (u) \frac{H_{0}}{τ_{2} (u)}) = S (τ_{2} (u)) C (\frac{H_{0}}{τ_{2}}) + C (τ_{2} (u)) S (\frac{H_{0}}{τ_{2}}) \\ = μ_{3} (d, u), \\ C (H_{0}) = C (τ_{2} (u) \frac{H_{0}}{τ_{2} (u)}) = C (τ_{2} (u)) C (\frac{H_{0}}{τ_{2}}) + S (τ_{2} (u)) S (\frac{H_{0}}{τ_{2}}) \\ = - H_{0} . \end{array}$ This implies $\begin{array}{rcl} S (\frac{H_{0}}{τ_{2}}) & = & \frac{S (τ_{2} (u))}{C {(τ_{2} (u))}^{2} - S {(τ_{2} (u))}^{2}} H_{0} + μ_{4} (d, u) \\ = & - u sinh (d) \frac{H_{0}}{τ_{2} (u)} + μ_{4} (d, u), \end{array}$ where $μ_{3} (d, u)$ , $μ_{4} (d, u)$ are convergent series.

(6) The proof of (6) is similar to that of (5).

(7) Using (4) and (6), we obtain $\begin{array}{rcl} S (\frac{u H_{0}}{sinh (d) τ_{2} (u)} - H_{2}) & = & (- \frac{1}{2} (u^{2} + 1) - \frac{1}{2} τ_{2} (u)) \frac{H_{0}}{τ_{2} (u)} + μ_{5} (d, u) \\ = & - \frac{H_{0}}{τ_{2} (u)} + μ_{5} (d, u), \end{array}$ where $μ_{5} (d, u)$ is a convergent series. This completes the proof of the lemma. □

Using the definition of $H_{0}$ in (5.12) and (5.31), we can rewrite $\frac{1}{d} S (G)$ in the form $\frac{1}{d} S (G) = - α \frac{H_{0}}{d^{2} τ_{2} (u)} - (β + \frac{1}{2} α) \frac{H_{0}}{τ_{2} (u)} + \frac{1}{4} α H_{0} + \overline{X},$ (5.33) where $∥ \overline{X} ∥_{n} = O ((n - 3)! {(2 π)}^{- n}) .$ Using (2) and (5) of the previous lemma and applying also the inverse of the operator $S$ in (5.33), we obtain $\frac{1}{d} G = \frac{1}{d^{2}} (α + (β + \frac{1}{2} α) d^{2}) [\frac{1}{sinh (d)} \frac{u H_{0}}{τ_{2} (u)} - H_{2}] + \frac{1}{2} α H_{2} + δ H_{1} + {\overline{X}}_{1},$ (5.34) where δ is a constant and $\begin{array}{rclr} H_{2} (d, u) = d \frac{\partial H_{1}}{\partial d} (d, u) = \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} n! {(\frac{i}{2 π})}^{n + 1} τ_{n} (u) d^{n}, & (5.35) \\ ∥ {\overline{X}}_{1} ∥_{n} = O ((n - 2)! {(2 π)}^{- n}), & (5.36) \\ δ = \frac{4}{π} \sum_{n = 1}^{\infty} δ_{n}, & (5.37) \end{array}$ with $δ_{n} = O (n^{- 2})$ . In the expression of $\frac{1}{d} G$ , the term $δ H_{1}$ comes from the fact that the series X can be written $X (d, u) = D_{1} (d, u) + d^{2} D_{2} (d, u),$ where $\begin{array}{rcl} D_{1} (d, u) = \sum_{n = 8}^{\infty} δ_{n} (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) d^{n}, \\ ∥ D_{2} ∥_{n} = O ((n - 1)! {(2 π)}^{- n}) . \end{array}$ If we apply Theorem 4.5 to the series $D_{1} (d, u)$ and Theorem 4.4 to $D_{1} (d, u)$ , the term $δ H_{1}$ appears in the expression of $\frac{1}{d} G$ .

Since $\frac{1}{sinh (d)} = d^{- 1} - \frac{d}{6} + O (d^{3})$ , we obtain $\begin{array}{rclr} G (d, u) & = & (\frac{α}{d^{2}} + (β + \frac{α}{3})) \frac{u H_{0} (d, u)}{τ_{2} (u)} - (β d + \frac{α}{d}) H_{2} (d, u) \\ + δ d H_{1} (d, u) + S (d, u), & (5.38) \end{array}$ where $∥ S ∥_{n} = O ((n - 3)! {(2 π)}^{n})$ .

Observe that in $d^{- 2} \frac{u H_{0}}{τ_{2} (u)}$ , $d^{- 1} H_{2}$ the degree of the coefficients of $d^{n}$ exceeds n. This is due to the fact that the expressions $u \frac{τ_{n + 2}}{τ_{2}} - τ_{n + 1}$ etc., which are of degree $n - 1$ , were split.

It is not necessary (but would not be difficult) to write down asymptotic approximations for the coefficients of F, because Eqs (5.5) and (5.3) can be used. This completes the proof of Theorem 5.1. □

6. Functions and quasi-solutions

So far, we have shown that Eq. (2.2) has a formal solution and we have found an asymptotic approximation of the coefficients of the formal solution. We will use this to construct a quasi-solution, i.e. a function that satisfies Eq. (2.2) except for some exponentially small error. To that purpose, we define the functions $H_{n} (u) : = (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u)$ (6.1) and $\begin{array}{rclr} h_{0} (t, u) : = \sum_{\begin{array}{c} n = 10 \\ n even \end{array}}^{\infty} H_{n} (u) \frac{t^{n - 1}}{(n - 1)!}, & (6.2) \\ h_{1} (t, u) : = \frac{i}{2 π} \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} H_{n} (u) \frac{t^{n - 1}}{(n - 1)!}, & (6.3) \\ h_{2} (t, u) : = \frac{i}{2 π} \sum_{\begin{array}{c} n = 9 \\ n odd \end{array}}^{\infty} n H_{n} (u) \frac{t^{n - 1}}{(n - 1)!} . & (6.4) \end{array}$ This means that $\begin{matrix} h_{0} (t, u) = \frac{i}{2 π} \sum_{\begin{array}{c} n = 1 \\ n odd \end{array}}^{\infty} {(\frac{i}{2 π})}^{n} τ_{n + 1} (u) t^{n} - (H_{2} t + H_{4} \frac{t^{3}}{3!} + H_{6} \frac{t^{5}}{5!} + H_{8} \frac{t^{7}}{7!}), \\ h_{1} (t, u) = {(\frac{i}{2 π})}^{2} \sum_{\begin{array}{c} n = 0 \\ n even \end{array}}^{\infty} {(\frac{i}{2 π})}^{n} τ_{n + 1} (u) t^{n} \\ - \frac{i}{2 π} (H_{1} + H_{3} \frac{t^{2}}{2!} + H_{5} \frac{t^{4}}{4!} + H_{7} \frac{t^{6}}{6!}), \\ h_{2} (t, u) = h_{1} (t, u) + t \frac{\partial h_{1}}{\partial t} (t, u) . \end{matrix}$ (6.5) Using part (4) of Proposition 3.1, we obtain $\frac{i}{2 π} \sum_{\begin{array}{c} n = 1 \\ n odd \end{array}}^{\infty} {(\frac{i}{2 π})}^{n} τ_{n + 1} (u) t^{n} = \frac{i}{2 π} \sum_{\begin{array}{c} n = 1 \\ n odd \end{array}}^{\infty} \frac{1}{n!} {(\frac{i t}{2 π})}^{n} \frac{d^{n}}{d^{n} ξ} (tanh (ξ)),$ where $ξ = ξ (u) = artanh (u)$ .

Seeing this as the difference of two Taylor expansions, we can write $\begin{array}{rclr} h_{0} (t, u) & = & \frac{i}{4 π} [tanh (ξ + \frac{i t}{2 π}) - tanh (ξ - \frac{i t}{2 π})] \\ - (H_{2} t + H_{4} \frac{t^{3}}{3!} + H_{6} \frac{t^{5}}{5!} + H_{8} \frac{t^{7}}{7!}) . & (6.6) \end{array}$ Similarly, $\begin{array}{rcl} h_{1} (t, u) & = & - \frac{1}{4 π^{2}} \sum_{\begin{array}{c} n = 0 \\ n even \end{array}}^{\infty} \frac{1}{n!} {(\frac{i t}{2 π})}^{n} \frac{d^{n}}{d^{n} ξ} (tanh (ξ)) \\ - \frac{i}{2 π} (H_{1} + H_{3} \frac{t^{2}}{2!} + H_{5} \frac{t^{4}}{4!} + H_{7} \frac{t^{6}}{6!}), \end{array}$ or equivalently $\begin{array}{rclr} h_{1} (t, u) & = & \frac{- 1}{8 π^{2}} [tanh (ξ + \frac{i t}{2 π}) + tanh (ξ - \frac{i t}{2 π})] \\ - \frac{i}{2 π} (H_{1} + H_{3} \frac{t^{2}}{2!} + H_{5} \frac{t^{4}}{4!} + H_{7} \frac{t^{6}}{6!}) . & (6.7) \end{array}$ For fixed real u the functions $h_{k} (\cdot, u)$ , $k = 0, 1, 2$ , are analytic in $| t | < ρ$ , where $ρ = π^{2}$ , because the singularities of tanh are $i (\frac{π}{2} + n π)$ , n integer. The functional equations for the trigonometric and hyperbolic functions imply that $\begin{array}{rclr} h_{0} (t, u) = - \frac{1}{2 π} \frac{(1 - u^{2}) sin (t / (2 π)) cos (t / (2 π))}{cos {(t / (2 π))}^{2} + u^{2} sin {(t / (2 π))}^{2}} \\ + \frac{τ_{2}}{4 π^{2}} t - \frac{τ_{4}}{16 π^{4}} t^{3} + \frac{τ_{6}}{64 π^{6}} t^{5} - \frac{τ_{8}}{256 π^{8}} t^{7}, \\ h_{1} (t, u) = - \frac{1}{4 π^{2}} \frac{(u - u^{3}) sin {(t / (2 π))}^{2}}{cos {(t / (2 π))}^{2} + u^{2} sin {(t / (2 π))}^{2}} & (6.8) \\ + \frac{τ_{1}}{4 π^{2}} - \frac{τ_{3}}{16 π^{4}} t^{2} + \frac{τ_{5}}{64 π^{6}} t^{4} - \frac{τ_{7}}{256 π^{8}} t^{6}, \\ h_{2} (t, u) = h_{1} (t, u) + t \frac{\partial h_{1}}{\partial t} (t, u) . \end{array}$ In the subsequent definition, we consider real values of u, $0 < u ⩽ 1$ , here $h_{k}$ , $k = 0, 1, 2$ , are also analytic with respect to t on the positive real axis.

We define the functions $H_{k} (d, u)$ , $k = 0, \dots, 2$ , by $\begin{array}{rclr} H_{0} (d, u) : = \int_{0}^{+ \infty} e^{- \frac{t}{d}} h_{0} (t, u) d t for 0 < u ⩽ 1, \\ H_{1} (d, u) : = \int_{0}^{+ \infty} e^{- \frac{t}{d}} h_{1} (t, u) d t for 0 < u ⩽ 1, & (6.9) \\ H_{2} (d, u) : = \int_{0}^{+ \infty} e^{- \frac{t}{d}} h_{2} (t, u) d t for 0 < u ⩽ 1 . \end{array}$ We have $H_{2} (d, u) = d \frac{\partial H_{1}}{\partial d} (d, u) .$

Indeed $\begin{array}{rcl} d \frac{\partial H_{1}}{\partial d} (d, u) & = & \int_{0}^{+ \infty} (\frac{1}{d} e^{- \frac{t}{d}}) t \cdot h_{1} (t, u) d t \\ = & - \int_{0}^{+ \infty} \frac{\partial}{\partial t} (e^{- \frac{t}{d}}) \cdot t \cdot h_{1} (t, u) d t \\ = & \int_{0}^{+ \infty} e^{- \frac{t}{d}} (h_{1} (t, u) + t \cdot \frac{\partial}{\partial t} h_{1} (t, u)) d t \\ = & \int_{0}^{+ \infty} e^{- \frac{t}{d}} h_{2} (t, u) d t . \end{array}$

The functions $H_{k} (d, \cdot)$ are real analytic; they can be continued analytically to the interval $- 1 < u ⩽ 1$ in the following way. Choose some positive number M and let $Γ_{1}$ the path consisting of the segment from 0 to $M i$ and of the ray $t \mapsto t + M i$ , $t ⩾ 0$ . Let $Γ_{2}$ be the symmetric path that could also be obtained using $- M$ instead of M. Recalling (6.6), we can also define $\begin{array}{rclr} H_{0} (d, u) & : = & \frac{i}{4 π} [\int_{Γ_{2}} e^{- \frac{t}{d}} tanh (ξ + \frac{i t}{2 π}) d t - \int_{Γ_{1}} e^{- \frac{t}{d}} tanh (ξ - \frac{i t}{2 π}) d t] \\ + μ_{0} (d, u), & (6.10) \end{array}$ where $μ_{0} (d, u) : = \frac{1}{4 π^{2}} τ_{2} (u) d^{2} - \frac{3}{8 π^{4}} τ_{4} (u) d^{4} + \frac{15}{8 π^{6}} τ_{6} (u) d^{6} - \frac{315}{16 π^{8}} τ_{8} (u) d^{8}$ for $- tanh (\frac{2}{π} M) < u ⩽ 1$ . As M is arbitrary, this defines the analytic continuation of $H_{0} (d, \cdot)$ for $- 1 < u ⩽ 1$ . Similarly, the real analytic continuations of $H_{k}$ , $k = 1, 2$ , are defined.

In the sequel, we use the operator $C$ , $S$ also for functions.

Lemma 6.1.

Consider the functions $H_{k} (d, u)$ , $k = 0, 1, 2$ , defined in ( 6.9 ). Then, for $- 1 < u ⩽ 1$ :

For $k = 0, 1$ , $H_{k} (d, T^{\pm \frac{1}{2}}) = - H_{k} (d, u) + μ_{k}^{\pm} (d, u),$ (6.11) where $T^{+ \frac{1}{2}}$ , $T^{- \frac{1}{2}}$ are defined in ( 2.13 ) and the functions $μ_{k}^{\pm} (d, u)$ , $k = 0, 1$ , are analytic, beginning with $d^{10}$ , respectively $d^{9}$ .

For $k = 0, 1$ , $S (H_{k}) = μ_{k} (d, u)$ , where the functions $μ_{k} (d, u)$ , $k = 0, 1$ , are analytic, beginning with $d^{11}$ , respectively $d^{10}$ .

For $k = 0, 1$ , $C (H_{k}) = - H_{k} (d, u) + λ_{k} (d, u)$ , where the functions $λ_{k} (d, u)$ , $k = 0, 1$ , are analytic, beginning with $d^{10}$ , respectively $d^{9}$ .

$S (H_{2}) = \frac{1}{2} H_{0} (d, u) + μ_{2} (d, u)$ , where $μ_{2} (d, u)$ is a analytic function, beginning with $d^{10}$ .

$S (\frac{u H_{0}}{τ_{2} (u)}) = - \frac{1}{2} (u^{2} + 1) sinh (d) \frac{H_{0}}{τ_{2} (u)} + μ_{4} (d, u)$ , where the function $μ_{4} (d, u)$ is analytic, beginning with $d^{11}$ .

Proof.

(1) For $k = 0$ we replace u by $T^{+ \frac{1}{2}}$ in (6.6). Using (6.10) and $ξ (T^{+ \frac{1}{2}}) = ξ (u) + \frac{1}{2} d$ , we obtain for $0 < u ⩽ 1$ $H_{0} (d, T^{+ \frac{1}{2}}) = \int_{0}^{+ \infty} e^{- \frac{t}{d}} h_{0} (t, T^{+ \frac{1}{2}}) d t = \frac{i}{4 π} I^{+} + μ (d, T^{+ \frac{1}{2}}),$ (6.12) where $I^{+} = \int_{0}^{+ \infty} e^{- \frac{t}{d}} tanh (ξ + \frac{d}{2} + \frac{i t}{2 π}) d t - \int_{0}^{+ \infty} e^{- \frac{t}{d}} tanh (ξ - \frac{d}{2} + \frac{i t}{2 π}) d t .$ If we substitute $t + π i d$ in the first part, $t - π i d$ in the second part we obtain $I^{+} = - \int_{- π i d}^{+ \infty - π i d} e^{- \frac{t}{d}} tanh (ξ + \frac{i t}{2 π}) d t + \int_{π i d}^{+ \infty + π i d} e^{- \frac{t}{d}} tanh (ξ - \frac{i t}{π}) d t .$ Now, we apply Cauchy’s theorem $\begin{array}{rcl} I^{+} & = & - \int_{0}^{+ \infty} e^{- \frac{t}{d}} (tanh (ξ + \frac{i t}{2 π}) - tanh (ξ - \frac{i t}{2 π})) d t \\ + \int_{0}^{- π i d} e^{- \frac{t}{d}} tanh (ξ + \frac{i t}{2 π}) d t - \int_{0}^{π i d} e^{- \frac{t}{d}} tanh (ξ - \frac{i t}{2 π}) d t . \end{array}$ Substituting $t = - i d s$ in the second part, $t = i d s$ in the third part, we obtain $\begin{array}{rcl} I^{+} & = & \int_{0}^{+ \infty} e^{- \frac{t}{d}} (tanh (ξ + \frac{i t}{2 π}) - tanh (ξ - \frac{i t}{2 π})) d t \\ - 2 i d \int_{0}^{π} cos (s) tanh (ξ + d \frac{s}{2 π}) d s . \end{array}$ With (6.12) this implies for $0 < u ⩽ 1$ $H_{0} (d, T^{+}) = - H_{0} (d, u) + μ_{0}^{+} (d, u),$ (6.13) where $μ_{0}^{+} (d, u) = \frac{d}{2 π} \int_{0}^{π} cos (s) tanh (ξ + d \frac{s}{2 π}) d s + μ (d, T^{+ \frac{1}{2}}) .$ By real analytic continuation, this formula is valid for $- 1 < u ⩽ 1$ . We use the same method for $H_{0} (d, T^{- \frac{1}{2}})$ , $H_{1} (d, T^{\pm \frac{1}{2}})$ and obtain for $- 1 < u ⩽ 1$ $\begin{array}{rclr} H_{0} (d, T^{- \frac{1}{2}}) = - H_{0} (d, u) + μ_{0}^{-} (d, u), \\ H_{1} (d, T^{+ \frac{1}{2}}) = - H_{1} (d, u) + μ_{1}^{+} (d, u), & (6.14) \\ H_{1} (d, T^{- \frac{1}{2}}) = - H_{1} (d, u) + μ_{1}^{-} (d, u), \end{array}$ where $\begin{array}{rcl} μ_{0}^{-} (d, u) = \frac{d}{2 π} \int_{0}^{π} cos (s) tanh (ξ - \frac{d s}{2 π}) d s + μ_{0} (d, T^{- \frac{1}{2}}), \\ μ_{1}^{+} (d, u) = - \frac{d}{4 π^{2}} \int_{0}^{π} sin (s) tanh (ξ + \frac{d s}{2 π}) d s + μ_{1} (d, T^{+ \frac{1}{2}}), \\ μ_{1}^{-} (d, u) = - \frac{d}{4 π^{2}} \int_{0}^{π} sin (s) tanh (ξ - \frac{d s}{2 π}) d s + μ_{1} (d, T^{- \frac{1}{2}}), \\ μ_{1} (d, u) = \frac{1}{4 π^{2}} τ_{1} (u) d - \frac{1}{8 π^{4}} τ_{3} (u) d^{3} + \frac{3}{8 π^{6}} τ_{5} (u) d^{5} - \frac{45}{16 π^{8}} τ_{7} (u) d^{7} . \end{array}$

(2) Using the definition of the operator $S$ in (2.13) and (1) of this lemma, the result is immediate.

(3) The proof of (3) is similar to that of (2).

(4) For $k = 1$ , we differentiate (6.11) with respect to d. Using the fact that $H_{2} (d, u) = d \frac{\partial H_{1}}{\partial d} (d, u)$ and $\frac{\partial T^{\pm \frac{1}{2}}}{\partial d} = \pm \frac{1}{2} (1 - {(T^{\pm \frac{1}{2}})}^{2}),$ we obtain $H_{2} (d, u) \pm \frac{d}{2} (1 - {(T^{\pm \frac{1}{2}})}^{2}) \frac{\partial H_{1}}{\partial u} (d, T^{\pm \frac{1}{2}}) = - H_{2} (d, u) + d μ_{1}^{' \pm} (d, u) .$ This implies $\begin{array}{rcl} S (H_{2}) & = & - \frac{d}{2} C ((1 - u^{2}) \frac{\partial H_{1}}{\partial u}) + d (μ_{1}^{' +} (d, u) - μ_{1}^{' -} (d, u)) \\ = & \frac{d}{2} (1 - u^{2}) \frac{\partial H_{1}}{\partial u} + μ_{2} (d, u) \\ = & \frac{1}{2} H_{0} (d, u) + μ_{2} (d, u), \end{array}$ where $μ_{2} (d, u)$ is an analytic function beginning with $d^{10}$ .

(5) Using (2.15) and (2), (3) of previous lemma, we obtain $\begin{array}{rcl} S (\frac{u}{τ_{2}} H_{0}) & = & S (\frac{u}{τ_{2}}) C (H_{0}) + C (\frac{u}{τ_{2}}) S (H_{0}) \\ = & - \frac{1}{2} (u^{2} + 1) sinh (d) \frac{H_{0}}{τ_{2} (u)} + μ_{4} (d, u), \end{array}$ where the function $μ_{4} (d, u)$ is analytic, beginning with $d^{11}$ . This completes the proof of Lemma 6.1. □

In the sequel we consider $u_{0} \in] - 1, 0]$ .

Proposition 6.2.

We have

Uniformly for $u_{0} ⩽ u ⩽ 1$ , $\begin{array}{rclr} H_{0} (d, u) \sim \sum_{\begin{array}{c} n = 2 \\ n even \end{array}}^{\infty} (n - 1)! {(\frac{i}{2 π})}^{n} τ_{n} (u) d^{n} as d ↘ 0, \\ H_{1} (d, u) \sim \sum_{\begin{array}{c} n = 3 \\ n odd \end{array}}^{\infty} (n - 1)! {(\frac{i}{2 π})}^{n + 1} τ_{n} (u) d^{n} as d ↘ 0, & (6.15) \\ H_{2} (d, u) \sim \sum_{\begin{array}{c} n = 3 \\ n odd \end{array}}^{\infty} n! {(\frac{i}{2 π})}^{n + 1} τ_{n} (u) d^{n} as d ↘ 0 . \end{array}$

For $i = 1, 2, 3$ , $| \frac{\partial H_{i}}{\partial u} (d, u) | ⩽ K d$ for $u_{0} < u ⩽ 1$ ( $d > 0$ ) .

The proof of this proposition is similar to that of [8].

With the aim of applying the results of [7], we consider $S_{n - 2} (u) = R_{n} (u)$ , where $S_{n} (u)$ is the remainder term in (5.38). Then $R_{n} (u)$ , n is a sequence of polynomials of degree at most n and $∥ R_{n} ∥_{n} = O ((n - 5)! {(2 π)}^{- 1} log (n)) .$

Lemma 6.3 ([7]).

If we define $\begin{array}{rcl} r (t, u) : = \sum_{n = 10}^{\infty} R_{n} (u) \frac{t^{n - 1}}{(n - 1)!} (t \in C, | t | ⩽ π^{2}, u_{0} ⩽ u ⩽ 1), \\ r (t, u) : = r (π^{2}, u) + (t - π^{2}) \frac{\partial r}{\partial t} (π^{2}, u) (t > π^{2}, u_{0} ⩽ u ⩽ 1), \\ R (d, u) : = \int_{0}^{\infty} e^{- \frac{t}{d}} r (t, u) d t \end{array}$ then

r is continuously differentiable function on the set B of all $(t, u)$ such that u satisfies $u_{0} ⩽ u ⩽ 1$ and t is a complex number and satisfies $| t | ⩽ π^{2}$ or $t > π^{2}$ . The restriction of r to $u_{0} ⩽ u ⩽ 1$ , $| t | ⩽ π^{2}$ is twice continuously differentiable, for fixed $u_{0} ⩽ u ⩽ 1$ the function $r (t, u)$ is analytic in $| t | < π^{2}$ ,

$R (d, u)$ is continuous, partially differentiable with respect to u, has continuous partial derivative and $R (d, u) \sim \sum_{n = 10}^{\infty} R_{n} (u) d^{n} as d ↘ 0,$ (6.16)

$| R (d, u) | ⩽ K d^{3}$ , $| \frac{\partial R}{\partial u} (d, u) | ⩽ K d^{3}$ for $u_{0} ⩽ u ⩽ 1$ ( $d > 0$ ) .

The importance of our definition of $R$ lies in a certain compatibility with insertion of the functions $T^{+}$ , $T^{-}$ for u. First let $\begin{array}{rcl} \sum_{n = 10}^{\infty} R_{n}^{+} (u) d^{n} = \sum_{n = 11}^{\infty} R_{n} (T^{+}) d^{n}, \\ \sum_{n = 10}^{\infty} R_{n}^{-} (u) d^{n} = \sum_{n = 11}^{\infty} R_{n} (T^{-}) d^{n} . \end{array}$ We obtain new sequences $R_{n}^{+} (u)$ , $R_{n}^{-} (u)$ of polynomials of degree at most n. This follows from the relation $p (T^{+} (d, u)) = \sum_{k = 0}^{\infty} \frac{1}{k!} D^{k} p (u) d^{k} .$ (6.17) Theorem 4.2 implies $\begin{array}{rcl} {∥ R_{n}^{+} (u) ∥}_{n} = O ((n - 5)! {(2 π)}^{- n} log (n)), \\ {∥ R_{n}^{-} (u) ∥}_{n} = O ((n - 5)! {(2 π)}^{- n} log (n)) . \end{array}$ Therefore we can use the previous lemma for $R_{n}^{+} (u)$ , $R_{n}^{-} (u)$ and obtain functions $R^{+} (d, u)$ , $R^{-} (d, u)$ .

Theorem 6.4.

There is a positive constant K independent of d, u such that $\begin{array}{rcl} | R^{+} (d, u) - R (d, T^{+}) | ⩽ K d^{3} e^{- \frac{π^{2}}{d}} for d > 0, u_{0} < u ⩽ 1, \\ | R^{-} (d, u) - R (d, T^{-}) | ⩽ K d^{3} e^{- \frac{π^{2}}{d}} for d > 0, u_{0} < u ⩽ 1 . \end{array}$

Proof.

The proof is exactly the one of [7]. □

Definition 6.5.

Let $D (d, u)$ be a function defined for $0 < d < d_{0}$ and $u_{0} < u < 1$ . We say that $D (d, u)$ has property G if $D (d, u) = \int_{0}^{\infty} e^{- \frac{t}{d}} q (t, u) d t (0 < d < d_{0}, u_{0} < u < 1)$ is the Laplace transform of some function $q (t, u)$ that has the following properties:

$q (t, u)$ is defined if $u_{0} < u < 1$ and either t is complex and $| t | < π^{2}$ or t is real and $t ⩾ 0$ ,

$q (t, u)$ is analytic in $| t | < π^{2}$ for $u_{0} < u < 1$ ,

$q (t, u)$ restricted to $0 ⩽ t < π^{2}$ or $t ⩾ π^{2}$ is continuous and the ${lim}_{t \to π^{2}} q (t, u)$ exists for every $u_{0} < u < 1$ ,

there is a positive constant K such that $| q (d, u) | ⩽ K e^{K t} for t ⩾ 0 (0 < d < d_{0}, u_{0} < u < 1) .$

Lemma 6.6.

For $u_{0} < u ⩽ 1$ , we have

If $H_{i} (d, u)$ , $i = 0, 1, 2$ are the functions of ( 6.9 ) then $d^{2} H_{k} (d, u) = (1 - u^{2}) {\tilde{H}}_{k} (d, u) + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}), k = 0, 1$ and $d^{3} H_{2} (d, u) = (1 - u^{2}) {\tilde{H}}_{2} (d, u) + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}),$ where ${\tilde{H}}_{i} (d, u)$ , $i = 1, 2, 3$ , have property G.

If $D_{1}$ , $D_{2}$ have property G and their first terms in the Taylor development at $d = 0$ , begin with $d^{k}$ , $k ⩾ 2$ , then $D_{1} (d, u) D_{2} (d, u) = d^{k} D (d, u) + O (d^{k} e^{- \frac{π^{2}}{d}}),$ where $D (d, u)$ has property G.

Any function $D (d, u)$ analytic in a neighborhood of $d = 0$ has property G if $D (0, u) = 0$ for all u.

If $R (d, u)$ is defined by Lemma 6.3 then $\frac{1}{d^{2}} R (d, u)$ has property G.

If $D_{1}$ , $D_{2}$ have property G then so do $D_{1} + D_{2}$ , $D_{1} - D_{2}$ and $D_{1} \cdot D_{2}$ .

If $D (d, u)$ has property G then $| D (d, u) | ⩽ K d (0 < d < \frac{1}{K})$ (6.18) with some constant $K > 0$ independent of u.

Proof.

(1) Consider first the case $i = 0$ . We have two subcases:

If $u > 0$ , we have $d^{2} H_{0} = (1 - u^{2}) \int_{0}^{\infty} e^{- \frac{t}{d}} g_{2} (t, u) d t,$ (6.19) where $g_{2} (t, u) = \frac{1}{(1 - u^{2})} \int_{0}^{t} \int_{0}^{τ} h_{0} (s, u) d s d τ .$ $g_{2} (t, u)$ has a logarithmic singularity at $t_{k} (s) = (2 k + 1) π^{2} \pm d \frac{2 π s}{ε} i$ for $k ⩾ 0$ , $s > 0$ . It is analytic in $| t | < π^{2}$ and ${lim}_{t \to π^{2}} g_{2} (t, u)$ exists.

If we put ${\tilde{H}}_{0} (d, u) = \int_{0}^{\infty} e^{- \frac{t}{d}} {\tilde{g}}_{2} (t, u) d t,$ where ${\tilde{g}}_{2} (t, u) = \{\begin{matrix} g_{2} (t, u) & if t ⩽ π^{2}, \\ g_{2} (π^{2}, u) & if t ⩾ π^{2} \end{matrix}$ then ${\tilde{H}}_{0} (d, u)$ has property G and $d^{2} H_{0} (d, u) = (1 - u^{2}) {\tilde{H}}_{0} (d, u) + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}) .$

For $- u_{0} < u < 1$ , where $0 < u_{0} < 1$ , we have $\begin{array}{rcl} H_{0} (d, u) & = & \int_{0}^{\infty e^{i φ}} e^{- \frac{t}{d}} h_{0} (t, u) d t + 2 π i \sum_{k ⩾ 0} Res (e^{- \frac{t}{d}} h_{0} (t, u), t_{k} (s)) \\ = & \int_{0}^{\infty e^{i φ}} e^{- \frac{t}{d}} h_{0} (t, u) d t + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}), \end{array}$ where $\frac{π}{2} < φ < \frac{π}{4}$ . For $0 < u < u_{0}$ , this formula coincides with the formula $\int_{0}^{\infty} e^{- \frac{t}{d}} h_{0} (t, u) d t$ and extends it by real analytic continuation for $- u_{0} < u < 0$ .

This implies $d^{2} H_{0} (d, u) = (1 - u^{2}) \int_{0}^{\infty e^{i φ}} e^{- \frac{t}{d}} g_{2} (t, u) d t + O ((1 - u^{2}) d^{2} e^{- \frac{π^{2}}{d}})$ we obtain $d^{2} H_{0} (d, u) - (1 - u^{2}) {\tilde{H}}_{0} (d, u) = (1 - u^{2}) \int_{Γ} e^{- \frac{t}{d}} g_{2} (t, u) d Γ + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}),$ where Γ is the path following the real line from infinity to $(π^{2}, 0)$ , then along the vertical line from $(π^{2}, 0)$ to $(π^{2}, π^{2} tan (φ))$ and finally along the line $y = tan (φ) x$ from $(π^{2}, π^{2} tan (φ))$ to infinity.

Because $g_{2} (\cdot, u)$ is bounded on Γ, we have $| d^{2} H_{0} (d, u) - (1 - u^{2}) {\tilde{H}}_{0} (d, u) | ⩽ K (1 - u^{2}) e^{- \frac{π^{2}}{d}},$ where K is a positive constant. Finally $d^{2} H_{0} (d, u) = (1 - u^{2}) {\tilde{H}}_{0} (d, u) + O ((1 - u^{2}) e^{- \frac{π^{2}}{d}}) for 0 < u_{0} < u ⩽ 1 .$

For $i = 1$ we use the same method.

For $i = 2$ , we use the same method with $d^{3} H_{2} = (1 - u^{2}) \int_{0}^{\infty} e^{- \frac{t}{d}} g_{2} (t, u) d t,$ (6.20) where $g_{2} (t, u) = \frac{1}{(1 - u^{2})} \int_{0}^{t} \int_{0}^{σ} \int_{0}^{τ} h_{2} (s, u) d s d σ d τ .$

(2) We assume that $D_{1} (d, u)$ , $D_{2} (d, u)$ have property G and their first terms in the Taylor development at $d = 0$ begin with $d^{k}$ , $k ⩾ 2$ . Then $\begin{array}{rcl} D_{1} = \int_{0}^{\infty} e^{- \frac{t}{d}} f (t, u) d t, \\ D_{2} = \int_{0}^{\infty} e^{- \frac{t}{d}} g (t, u) d t, \end{array}$ where $f (t, u)$ , $g (t, u)$ are analytic in $| t | < π^{2}$ and $f (t, u) = O (t^{k - 1})$ , $g (t, u) = O (t^{k - 1})$ . Hence we have $D_{1} (d, u) D_{2} (d, u) = \int_{0}^{\infty} e^{- \frac{t}{d}} (f * g) (t, u) d t .$ (6.21) The function $h (t, u) = (f * g) (t, u)$ can be written $\begin{array}{rcl} h (t, u) & = & (f * g) (t, u) = \int_{0}^{t} f (s, u) g (t - s, u) d s \\ = & \int_{0}^{t} f (t - s, u) g (s, u) d s \\ = & \int_{t / 2}^{t} f (s, u) g (t - s, u) d s + \int_{t / 2}^{t} f (t - s, u) g (s, u) d s . \end{array}$ For $t < π^{2}$ , the function $h (t, u)$ is k times differentiable with respect to t and $\begin{array}{rcl} h^{'} (t, u) = f (t, u) g (0, u) + f (0, u) g (t, u) - f (\frac{t}{2}, u) g (\frac{t}{2}, u) \\ + \int_{t / 2}^{t} f (s, u) g^{'} (t - s, u) d s + \int_{t / 2}^{t} f^{'} (t - s, u) g (s, u) d s \\ = \int_{t / 2}^{t} f (s, u) g^{'} (t - s, u) d s + \int_{t / 2}^{t} f^{'} (t - s, u) g (s, u) d s \\ - f (\frac{t}{2}, u) g (\frac{t}{2}, u), \\ h^{(k)} (t, u) = \int_{t / 2}^{t} f (s, u) g^{(k - 1)} (t - s, u) d s + \int_{t / 2}^{t} f^{(k - 1)} (t - s, u) g (s, u) d s \\ - \sum_{n = 0}^{k - 1} f^{(n)} (\frac{t}{2}, u) g^{(k - 1 - n)} (\frac{t}{2}, u) . \end{array}$ Observe that $h^{(k)} (t, u)$ is continuous on $[0, 2 π^{2} [$ and analytic for $| t | < π^{2}$ . If we put $\tilde{h} (t, u) = \{\begin{matrix} h (t, u) & if t < π^{2}, \\ {(t - π^{2})}^{k} & if t ⩾ π^{2}, \end{matrix}$ then $\int_{0}^{\infty} e^{- \frac{t}{d}} {\tilde{h}}^{(k)} (t, u) d t$ has property G and $\begin{array}{rcl} (D_{1} \cdot D_{2}) (d, u) & = & \int_{0}^{\infty} e^{- \frac{t}{d}} h (t, u) d t \\ = & \int_{0}^{\infty} e^{- \frac{t}{d}} \tilde{h} (t, u) d t + O (d^{k} e^{- \frac{π^{2}}{d}}) \\ = & d^{k} \int_{0}^{\infty} e^{- \frac{t}{d}} {\tilde{h}}^{(k)} (t, u) d t + O (d^{k} e^{- \frac{π^{2}}{d}}) \\ = & d^{k} D (d, u) + O (d^{k} e^{- \frac{π^{2}}{d}}), \end{array}$ where $D (d, u)$ has property G.

For $0 ⩽ u ⩽ 1$ , the proof of (3), (4), (5) and (6) is exactly the one of [7]. This proof is valid for $u_{0} < u ⩽ 1$ .

Now we have a formal solution of Eq. (2.2) and an asymptotic estimate for its coefficients. With the results on the functions in the beginning of this section, we have enough information to give a precise function which satisfies Eq. (2.2) with an exponentially small error as $d \to 0$ .

In Theorem 2.1 we found that (2.2) has a uniquely determined formal power series solution $A (d, u) = U (d, u) + Q (d, u) G (d, u),$ (6.22) where U, Q are defined in (5.2), (5.6) and G is given by (5.38). This suggests that we put $\begin{array}{rclr} G (d, u) = (\frac{α}{d^{2}} + (β + \frac{α}{3})) \frac{u H_{0} (d, u)}{τ_{2} (u)} - (β d + \frac{α}{d}) H_{2} (d, u) \\ + δ d H_{1} (d, u) + d^{- 2} R (d, u), & (6.23) \\ A (d, u) = U (d, u) + Q (d, u) G (d, u) \end{array}$ for $d > 0$ , $u_{0} < u ⩽ 1$ , where $H_{i} (d, u)$ , $i = 0, \dots, 2$ are defined in (6.9) and $R (d, u)$ is the function corresponding to $R_{n}$ , $n = 10, \dots$ , according to Lemma 6.3. Using (1) of Proposition 6.2 and (2) of Lemma 6.3, we obtain $G (d, u) \sim G (d, u) as d ↘ 0 for every u_{0} < u ⩽ 1 .$ (6.24) Consequently $A (d, u) \sim A (d, u) as d ↘ 0 for every u_{0} < u ⩽ 1 . □$ (6.25)

Theorem 6.7.

The function $G (d, u)$ satisfies ( 5.10 ) except for an exponentially small error. More precisely, there exists $K > 0$ such that $| V_{1} \cdot S (Q_{1} S (G)) - W_{2} \cdot S^{2} (G) - W_{3} S C (G) - Q_{1} W_{1} G - Q_{1} f_{2} (d, u, Q G) | ⩽ K d^{3} (1 - u^{2}) e^{- \frac{π^{2}}{d}}$ uniformly for $u_{0} < u < 1$ , $0 < d < d_{0}$ .

In the proof of this theorem, the functions $D_{i} (d, u)$ , $i = 1, 2$ have property G. Proof of Theorem 6.7.

We set $F (d, u) = V_{1} \cdot S (Q_{1} S (G)) - W_{2} \cdot S^{2} (G) - W_{3} S C (G) - Q_{1} W_{1} G - Q_{1} f_{2} (d, u, Q G) .$ (6.26) Using (2), (4), (5) of Lemma 6.1, and (6.23), we obtain $S (G) = - (α_{1} sinh (d) \frac{1}{τ_{2}} - (\frac{α_{1}}{2} sinh (d) - \frac{β_{1}}{2})) H_{0} + (1 - u^{2}) d^{8} D_{1} (d, u) + d^{- 2} S (R),$ where $D_{1} (d, u)$ has property G and $\begin{array}{rclr} α_{1} : = \frac{α}{d^{2}} + (β + \frac{α}{3}), \\ β_{1} : = \frac{α}{d} + β d . & (6.27) \end{array}$ This with Lemma 6.1 imply $\begin{array}{rcl} V_{1} S (Q_{1} S (G)) & = & d^{5} D_{2} (d, u) H_{0} + (1 - u^{2}) d^{11} D_{3} (d, u) + d^{- 2} V_{1} S (Q_{1}) C S (R) \\ + d^{- 2} V_{1} C (Q_{1}) S^{2} (R), \end{array}$ where $D_{2} (d, u)$ , $D_{3} (d, u)$ have property G. Because $\begin{array}{rcl} C (Q_{1}) = (1 - u^{2}) d^{2} (1 + O (d^{2})), \\ S (Q_{1}) = (1 - u^{2}) d^{3} (u + O (d^{2})), \\ V_{1} = 1 + O (d^{2}), \end{array}$ it is sufficient to apply Theorem 6.4, (4) of Lemma 6.6 for $R$ and (1) of Lemma 6.6 for $H_{0}$ . Then we obtain $V_{1} S (Q_{1} S (G)) = d^{5} D_{2} (d, u) H_{0} + (1 - u^{2}) d^{2} D_{4} (d, u) + O ((1 - u^{2}) d^{3} e^{- \frac{π^{2}}{d}}),$ where $D_{4} (d, u)$ has property G. With (1) of Lemma 6.6, this implies $V_{1} S (Q_{1} S (G)) = (1 - u^{2}) d^{2} D_{5} (d, u) + O ((1 - u^{2}) d^{3} e^{- \frac{π^{2}}{d}}) .$ (6.28)

Using the same method for the terms $W_{2} \cdot S^{2} (G)$ , $W_{3} S C (G)$ and $Q_{1} W_{1} G$ , we obtain $\begin{array}{rclr} W_{2} \cdot S^{2} (G) = (1 - u^{2}) d^{10} D_{6} (d, u) + O ((1 - u^{2}) d^{10} e^{- \frac{π^{2}}{d}}), \\ W_{3} S C (G) = (1 - u^{2}) d^{9} D_{7} (d, u) + O ((1 - u^{2}) d^{10} e^{- \frac{π^{2}}{d}}), & (6.29) \\ Q_{1} W_{1} G = (1 - u^{2}) d^{8} D_{8} (d, u) + O ((1 - u^{2}) d^{10} e^{- \frac{π^{2}}{d}}) . \end{array}$ To study the term $Q_{1} f_{2} (d, u, Q G)$ , we first treat $G^{2}$ and $G^{3}$ , $\begin{array}{rclr} G {(d, u)}^{2} & = & α_{1}^{2} \frac{u^{2}}{τ_{2}^{2}} H_{0}^{2} + β_{1}^{2} H_{2}^{2} + δ^{2} d^{2} H_{1}^{2} + d^{- 4} R^{2} - 2 α_{1} β_{1} \frac{u}{τ_{2}} H_{0} H_{2} + 2 α_{1} δ \frac{u}{τ_{2}} H_{0} H_{1} \\ + 2 d^{- 2} α_{1} \frac{u}{τ_{2}} H_{0} R - 2 β_{1} δ d H_{1} H_{2} - 2 d^{- 2} β_{1} R H_{2} + 2 d^{- 1} δ R H_{1} . & (6.30) \end{array}$ Using (1), (2) and (4) of Lemma 6.6 we obtain $G {(d, u)}^{2} = d^{2} D_{9} (d, u) + O (d^{2} e^{- \frac{π^{2}}{d}}) .$ (6.31) With (2) of Lemma 6.6, this implies $G {(d, u)}^{3} = G (d, u) G {(d, u)}^{2} = d^{8} D_{10} (d, u) + O (d^{8} e^{- \frac{π^{2}}{d}}) .$ (6.32) We can rewrite $f_{2} (d, u, G) = y_{0} (d, u) + y_{1} (d, u) G^{2} f_{1, 1} (d, u, G^{2}) + y_{2} (d, u) d^{2} G^{3} f_{1, 2} (d, u, G^{2}),$ (6.33) where $f_{1, 1}$ , $f_{1, 2}$ and $y_{i}$ , $i = 0, 1, 2$ are analytic. □

Lemma 6.8.

For $i = 1, 2$ , $f_{1, i} (d, u, G^{2}) = \int_{0}^{\infty} e^{- \frac{t}{d}} f_{i} (t, u) d t + O (d^{2} e^{- \frac{π^{2}}{d}}) for d > 0, u_{0} < u ⩽ 1,$ (6.34) where $f_{i} (\cdot, u)$ is analytic in $| t | < π^{2}$ and continuous in $[0, π^{2} [$ and $[π^{2}, \infty [$ .

Proof.

Using (6.31), we can rewrite $G^{2} (d, u) = \int_{0}^{\infty} e^{- \frac{t}{d}} g (t, u) d t + O (d^{2} e^{- \frac{π^{2}}{d}}),$ where $g (\cdot, u)$ is analytic in $| t | < π^{2}$ , continuous in $[0, π^{2} [$ and $[π^{2}, \infty [$ . We obtain $f_{1, i} (d, u, G^{2}) = \sum_{n = 1}^{\infty} f_{n, i} (d, u) G^{2 n},$ where $f_{n, i} (d, u) = \int_{0}^{\infty} e^{- \frac{t}{d}} φ_{n, i} (t, u) d t$ and $φ_{n, i} (t, u)$ are entire functions.

Using the proof of Theorem 5.1 from [1], we find $f_{1, i} (d, u, G^{2}) = \int_{0}^{\infty} e^{- \frac{t}{d}} f_{i} (t, u) d t + O (d^{2} e^{- \frac{π^{2}}{d}}),$ (6.35) where the series $\sum_{n = 1}^{\infty} (φ_{n, i} * g^{* n}) (t, u), g^{* n} = g * \dots * g, n times,$ is uniformly convergent to a function $f_{i} (t, u)$ analytic in $| t | < π^{2}$ and continuous in $[0, π^{2} [$ and $[π^{2}, \infty [$ , which also satisfies $| f_{i} (t, u) | ⩽ K exp (K t) for t ⩾ 0, u_{0} < u ⩽ 1 .$ Then $f_{1, i} (d, u, G^{2}) = Y_{i} (d, u) + O (d^{2} e^{- \frac{π^{2}}{d}}) for d > 0, u_{0} < u ⩽ 1,$ where $Y_{i} (d, u)$ , $i = 1, 2$ have property G.

This lemma with (6.33) and (2) of Lemma 6.6 imply $Q_{1} f_{2} (d, u, G) = d^{6} (1 - u^{2}) D_{10} (d, u) + O (d^{6} (1 - u^{2}) e^{- \frac{π^{2}}{d}}) .$ (6.36) Combining (6.28), (6.29) and (6.36), we find $F (d, u) = d^{2} (1 - u^{2}) D (d, u) + K (d, u),$ where $D (d, u)$ has property G and $| K (d, u) | ⩽ K d^{3} (1 - u^{2}) d e^{\frac{- π^{2}}{d}} for 0 < d < d_{0}, u_{0} < u < 1 .$ (6.37)

Hence $F (d, u) = (1 - u^{2}) d^{2} \int_{0}^{\infty} e^{- \frac{t}{d}} q (t, u) d t + K (d, u),$ where $q (t, u)$ is analytic in $| t | < π^{2}$ , is continuous on $[0, π^{2} [$ and $] π^{2}, \infty [$ , has a limit as $t \to π^{2}$ for every $u_{0} < u < 1$ and satisfies $| q (t, u) | ⩽ K e^{K t},$ (6.38) with a constant K independent of u. If $q (t, u) = \sum_{n = 0}^{\infty} q_{n} (u) t^{n}$ is the power series of $q (t, u)$ near $t = 0$ , Watson’s lemma with (6.37) imply $F (d, u) \sim \sum_{n = 0}^{\infty} n! (1 - u^{2}) q_{n} (u) d^{n + 3} as d ↘ 0 for every u_{0} < u < 1 .$ On the other hand because of its definition $\begin{array}{rcl} F (d, u) & \sim & V_{1} \cdot S (Q_{1} S (G)) - W_{2} \cdot S^{2} (G) - W_{3} S C (G) \\ - Q_{1} W_{1} G - Q_{1} f_{2} (d, u, Q G) = 0 + 0 d + \dots, \end{array}$ since the formal series G satisfies (5.10). This means that all $q_{n} \equiv 0$ . Thus we obtain for $u_{0} < u < 1$ with (6.38) $\begin{array}{rcl} (1 - u^{2}) d^{2} \int_{0}^{\infty} e^{- \frac{t}{d}} | q (t, u) | d t & ⩽ & (1 - u^{2}) d^{2} \int_{π^{2}}^{\infty} e^{- \frac{t}{d}} K e^{K t} d t \\ ⩽ & K (1 - u^{2}) d^{3} e^{- \frac{π^{2}}{d}} (0 < d < d_{0}) \end{array}$ and thus $| F (d, u) | ⩽ K (1 - u^{2}) d^{3} e^{- \frac{π^{2}}{d}} (0 < d < d_{0}) .$ We have proved that $\begin{array}{rcl} | V_{1} \cdot S (Q_{1} S (G)) - W_{2} \cdot S^{2} (G) - W_{3} S C (G) - Q_{1} W_{1} G - Q_{1} f_{2} (d, u, Q G) | \\ ⩽ K (1 - u^{2}) d^{3} e^{- \frac{π^{2}}{d}} \end{array}$ for $0 < d < d_{0}$ , $u_{0} < u < 1$ , i.e. $Q (d, u)$ is a quasi-solution of (5.10) on this interval. This implies that the function $A (d, u)$ defined in (6.23) is a quasi-solution of (2.2). More precisely $\begin{array}{rclr} | \sqrt{\frac{1 - {(T^{+})}^{2}}{1 - u^{2}}} A (d, T^{+}) + \sqrt{\frac{1 - {(T^{-})}^{2}}{1 - u^{2}}} A (d, T^{-}) - 2 A (d, u) - f (ε, A (d, u)) | \\ ⩽ K d e^{- \frac{π^{2}}{d}} . □ & (6.39) \end{array}$

7. Distance between points of manifolds

Clearly, if $q_{ε} (t)$ is an exact solution of the difference Eq. (1.1), then $(q_{ε} (t), p_{ε} (t))$ , where $p_{ε} (t) = \frac{1}{ε} (q_{ε} (t) - q_{ε} (t - ε))$ , is an exact solution of system (1.2). In the Introduction, we have mentioned that the stable manifold $W_{s}^{-}$ of this system at $A = (0, 0)$ is parametrized by $t \to (q_{ε}^{-} (t), p_{ε}^{-} (t))$ and the unstable manifold $W_{u}^{+}$ of (1.2) at $B = (2 π, 0)$ is parametrized by $t \to (q_{ε}^{+} (t), p_{ε}^{+} (t))$ , where $(q_{ε}^{-} (t), p_{ε}^{-} (t))$ is an exact solution of (1.2) and $q_{ε}^{+} (t) = 2 π - q_{ε}^{-} (- t)$ , $p_{ε}^{+} (t) = p_{ε}^{-} (- t + ε)$ .

In the previous section, we have constructed a quasi-solution $A (d, u)$ for Eq. (2.2), i.e. it satisfies this equation with an exponentially small error. We denote by ${\tilde{W}}_{s}$ , ${\tilde{W}}_{u}$ the manifolds close to $W_{s}^{-}$ respectively $W_{u}^{+}$ parametrized by $t \mapsto (ξ_{-} (t), φ_{-} (t))$ respectively $t \mapsto (ξ_{+} (t), φ_{+} (t))$ , where $ξ_{-} (t) = \sqrt{1 - u {(t)}^{2}} A (d, u (t)) + q_{0 d} (t)$ and $ξ_{+} (t) = 2 π - ξ_{-} (- t)$ , $φ_{-} (t) = \frac{1}{ε} (ξ_{-} (t) - ξ_{-} (t - ε))$ , $φ_{+} (t) = \frac{1}{ε} (ξ_{+} (t) - ξ_{+} (t - ε))$ . Here and in the sequel, we often omit to indicate the dependence with respect to ε for the sake of simplicity of notation.

We will first show that the vertical distance between some point $(q_{1}, p_{1})$ of the stable manifold $W_{s}^{-}$ and the manifold ${\tilde{W}}_{s}$ is exponentially small. For this purpose, we consider the sequence $Z_{n} = (q_{n}, p_{n})$ on the stable manifold $W_{s}^{-}$ , defined by $Z_{n + 1} = (\binom{q_{n + 1}}{p_{n + 1}}) = ϕ (\binom{q_{n}}{p_{n}}) = (\binom{q_{n} + ε p_{n + 1}}{p_{n} + ε sin (q_{n})}) for n = 1, 2, 3, \dots .$ (7.1) There is a sequence $t_{n}$ such that $ξ_{-} (t_{n}) = q_{n} = \sqrt{1 - u {(t_{n})}^{2}} A (d, u (t_{n})) + q_{0 d} (t_{n}) .$ (7.2)

The vertical projection of the point $(q_{n}, p_{n})$ on the manifold ${\tilde{W}}_{s}$ is the point $(q_{n}, b_{n})$ , where $b_{n} = φ_{-} (t_{n}) = g (q_{n}) : = φ_{-} (ξ_{-}^{- 1} (q_{n}))$ for $n = 1, 2, 3, \dots$ . Let $Δ_{ε} (n)$ denote the vertical distance between the point $(q_{n}, p_{n})$ and the manifold ${\tilde{W}}_{s}$ . Then $Δ_{ε} (n) = p_{n} - b_{n} .$ (7.3) For the ϕ-image of the point $(q_{n}, b_{n})$ on ${\tilde{W}}_{s}$ , we find $\begin{array}{rclr} (\binom{{\tilde{q}}_{n + 1}}{{\tilde{b}}_{n + 1}}) & = & ϕ (\binom{q_{n}}{b_{n}}) = (\binom{q_{n} + ε {\tilde{b}}_{n + 1}}{b_{n} + ε sin (q_{n})}) \\ = & (\binom{ξ_{-} (t_{n}) + ε {\tilde{b}}_{n + 1}}{ε^{- 1} (ξ_{-} (t_{n}) - ξ_{-} (t_{n} - ε)) + ε sin (ξ_{-} (t_{n}))}) \\ = & (\binom{2 ξ_{-} (t_{n}) - ξ_{-} (t_{n} - ε) + ε^{2} sin (ξ_{-} (t_{n}))}{ε^{- 1} (ξ_{-} (t_{n}) - ξ_{-} (t_{n} - ε)) + ε sin (ξ_{-} (t_{n}))}) . & (7.4) \end{array}$ As the function $ξ_{-}$ satisfies Eq. (1.1) except for an exponentially small error because of (6.39) and (7.2), we have $\begin{matrix} {\tilde{q}}_{n + 1} = ξ_{-} (t_{n} + ε) + e_{n + 1} (ε), \\ {\tilde{b}}_{n + 1} = φ_{-} (t_{n} + ε) + \frac{1}{ε} e_{n + 1} (ε), \end{matrix}$ (7.5) where $| e_{n + 1} (ε) | ⩽ K ε exp (- \frac{π^{2}}{ε})$ with some K independent of ε and n. Since $g (ξ_{-} (t_{n} + ε)) = φ_{-} (t_{n} + ε)$ , we have $\begin{array}{rclr} {\tilde{b}}_{n + 1} & = & g (ξ_{-} (t_{n} + ε)) + \frac{1}{ε} e_{n + 1} (ε) \\ = & g ({\tilde{q}}_{n + 1} - e_{n + 1} (ε)) + \frac{1}{ε} e_{n + 1} (ε) . & (7.6) \end{array}$ On the other hand, using (7.3) and the definition of ϕ, we obtain $\begin{array}{rclr} (\binom{{\tilde{q}}_{n + 1}}{{\tilde{b}}_{n + 1}}) & = & ϕ (\binom{q_{n}}{p_{n} - Δ_{ε} (n)}) = (\binom{q_{n} + ε p_{n} + ε^{2} sin (q_{n}) - ε Δ_{ε} (n)}{p_{n} + ε sin (q_{n}) - Δ_{ε} (n)}) \\ = & (\binom{q_{n + 1} - ε Δ_{ε} (n)}{p_{n + 1} - Δ_{ε} (n)}) . & (7.7) \end{array}$ With (7.3) and (7.6) this implies $\begin{array}{rcl} Δ_{ε} (n + 1) & = & p_{n + 1} - b_{n + 1} = p_{n + 1} - g (q_{n + 1}) \\ = & Δ_{ε} (n) + {\tilde{b}}_{n + 1} - g ({\tilde{q}}_{n + 1} + ε Δ_{ε} (n)) \\ = & Δ_{ε} (n) + g ({\tilde{q}}_{n + 1} - e_{n + 1} (ε)) - g ({\tilde{q}}_{n + 1} + ε Δ_{ε} (n)) + \frac{1}{ε} e_{n + 1} (ε) . \end{array}$ Using Taylor expansion we obtain $Δ_{ε} (n + 1) = (1 - ε g^{'} (θ_{n + 1})) Δ_{ε} (n) + \frac{1}{ε} (1 - ε g^{'} (θ_{n + 1})) e_{n + 1} (ε),$ (7.8) where ${\tilde{q}}_{n + 1} - e_{n + 1} (ε) < θ_{n + 1} < {\tilde{q}}_{n + 1} + ε Δ_{ε} (n)$ .

Now g is analytic and ε-close to the curve $p = - 2 sin (q / 2)$ , hence $g^{'} (θ_{n + 1}) = - cos (\frac{θ_{n + 1}}{2}) + O (d) .$ Thus given any positive $μ < π$ , there is a positive constant c such that for all $q_{1} ⩽ μ$ , all n and sufficiently small d, $1 - ε g^{'} (θ_{n + 1}) ⩾ 1 + ε c .$ It is now convenient to write (7.8) in the form $Δ_{ε} (n) = {(1 - ε g^{'} (θ_{n + 1}))}^{- 1} Δ_{ε} (n + 1) - \frac{1}{ε} e_{n + 1} (ε) .$ As $Δ_{ε} (n) \to 0$ as $n \to \infty$ , this implies that there is a positive constant K such that $| Δ_{ε} (n) | ⩽ K e^{- \frac{π^{2}}{ε}} \sum_{k = 0}^{\infty} {(1 + ε c)}^{- k} .$ Consequently $Δ_{ε} (n) = O (\frac{1}{ε} exp (- \frac{π^{2}}{ε})) .$ (7.9) In particular ${dist}_{v} ((q_{1}, p_{1}), {\tilde{W}}_{s}) = O (\frac{1}{ε} exp (- \frac{π^{2}}{ε})),$ (7.10) where $(q_{1}, p_{1})$ is any point on the stable manifold $W_{s}^{-}$ , provided $q_{1} ⩽ μ < π$ ; here ${dist}_{v}$ denotes the vertical distance.

The estimate (7.10) can be extended to any $μ < 2 π$ and a starting point $(q_{1}, p_{1})$ with $q_{1} ⩽ μ$ in the following way. The relation (7.8) remains valid, except that now we just have the existence of some constant $c > 0$ such that $1 - ε g^{'} (θ_{n + 1}) ⩾ 1 - ε c$ for all n. As system (1.2) can be regarded as a one-step numerical method for the system (1.3) of differential equations and the starting point is at a distance $O (ε)$ of its solution $(q_{0} (t), q_{0}^{'} (t))$ , results on the convergence of one-step methods can be applied and yield that $q_{k} = q_{0} (t_{1} + (k - 1) ε) + O (ε)$ , where $q_{0} (t_{1}) = q_{1}$ , uniformly for integer k, $1 ⩽ k ⩽ L / ε$ , where L is any positive constant. We choose L such $q_{0} (t_{1} + L) ⩽ μ / 2 < π$ . Repeated application of (7.8) now gives $| Δ_{ε} (n) - Δ_{ε} (n + [L / ε]) | ⩽ K e^{- \frac{π^{2}}{ε}} \sum_{k = 0}^{[L / ε]} {(1 - ε c)}^{- k} ⩽ \frac{K e^{L c}}{c ε} e^{- \frac{π^{2}}{ε}} .$ To the quantity $Δ_{ε} (n + [L / ε])$ , inequality (7.9) can be applied, because $q_{n + [L / ε]} ⩽ μ / 2 < π$ . Thus (7.9) and hence also (7.10) remain valid also uniformly for $0 < q_{1} ⩽ μ$ , provided $μ < 2 π$ .

The analogous reasoning applies to the vertical distance of a point $({\tilde{q}}_{1}, {\tilde{p}}_{1})$ on the unstable manifold $W_{u}^{+}$ from the manifold ${\tilde{W}}_{u}$ and yields ${dist}_{v} (({\tilde{q}}_{1}, {\tilde{p}}_{1}), {\tilde{W}}_{u}) = O (\frac{1}{ε} exp (- \frac{π^{2}}{ε})) .$ (7.11) Another method to obtain (7.11) consists in using (7.10) and symmetry.

Now we will estimate the vertical distance between the two manifolds ${\tilde{W}}_{s}$ and ${\tilde{W}}_{u}$ . As the quasi-solution $A (d, u)$ is defined for $- 1 < u = : tanh (t) < 1$ , we can define $\begin{array}{rclr} \begin{matrix} A^{+} (d, u) = - A (d, - u), \\ ξ_{+} (t) = \sqrt{1 - u {(t)}^{2}} A^{+} (d, u (t)) + 2 π - q_{0 d} (- t) = 2 π - ξ_{-} (- t), \end{matrix} & (7.12) \\ D_{ε} (t) = ξ_{+} (t) - ξ_{-} (t) for - \frac{4}{3} < t < \frac{4}{3} . & (7.13) \end{array}$ Using (7.12) and the definition of $ξ_{-} (t)$ we find $\begin{array}{rclr} D_{ε} (t) & = & - \sqrt{1 - u {(t)}^{2}} (A (d, u (t)) + A (d, - u (t))) - q_{0 d} (t) - q_{0 d} (- t) + 2 π \\ = & - \sqrt{1 - u {(t)}^{2}} (A (d, u (t)) + A (d, - u (t))) . & (7.14) \end{array}$ With (5.5) and (6.23) this implies $\begin{array}{rcl} D_{ε} (t) & = & - \sqrt{1 - u {(t)}^{2}} Q (d, u) [α_{1} \frac{u}{τ_{2}} (H_{0} (d, u) - H_{0} (d, - u)) \\ - β_{1} (H_{2} (d, u) + H_{2} (d, - u)) + δ d (H_{1} (d, u) + H_{1} (d, - u))], \end{array}$ where $Q (d, u)$ is defined in (5.6), $α_{1}$ , $β_{1}$ are defined in (6.27) and $H_{i} (d, u)$ are defined in (6.9) and can be continued analytically to $- 1 < u ⩽ 1$ as in (6.10).

Using the fact that the functions $u h_{0} (s, u)$ , $h_{1} (s, u)$ , $h_{2} (s, u)$ in (6.9) are odd, we can apply the residue theorem and obtain for $- 1 < u < 1$ that $\begin{array}{rcl} H_{i} (d, u) + H_{i} (d, - u) & = & \sum_{Im (s_{k} (t)) < 0} 2 π i Res (e^{- \frac{s}{d}} h_{i} (s, u), s_{k} (t)) \\ - \sum_{Im (s_{k} (t)) > 0} 2 π i Res (e^{- \frac{s}{d}} h_{i} (s, u), s_{k} (t)), i = 0, 1, 2, \end{array}$ where $s_{k} (t) = π^{2} \pm \frac{2 d π t}{ε} i + 2 k π^{2}$ for $k ⩾ 0$ . We obtain $\begin{array}{rcl} Res (e^{- \frac{s}{d}} h_{0} (s, u), s_{k} (t)) = \frac{1}{2} exp (- \frac{(k + 1) π^{2}}{d}) exp (\mp \frac{2 π t i}{ε}), \\ Res (e^{- \frac{s}{d}} h_{1} (s, u), s_{k} (t)) = \pm \frac{i}{4 π} exp (- \frac{(k + 1) π^{2}}{d}) exp (\mp \frac{2 π t i}{ε}), \\ Res (e^{- \frac{s}{d}} h_{2} (s, u), s_{k} (t)) = \pm \frac{i}{4 ε d} (π ε i \mp 2 d t) exp (- \frac{(k + 1) π^{2}}{d}) exp (\mp \frac{2 π t i}{ε}) \end{array}$ and hence $D_{ε} (t) = \frac{1}{d^{2}} ϕ_{1} (t, ε) exp (- \frac{π^{2}}{d}) + O (e^{- \frac{π^{2}}{d}}),$ (7.15) where $\begin{array}{rclr} ϕ_{1} (t, ε) & = & 2 π α [sinh (\frac{d t}{ε}) + t / cosh (\frac{d t}{ε})] sin (\frac{2 π t}{ε}) \\ + [\frac{π α}{cosh ((d t) / ε)}] cos (\frac{2 π t}{ε}) . & (7.16) \end{array}$ As a consequence of (7.15) and (7.12), we obtain immediately that $ξ^{\pm} (0) = π + O (ε^{- 2} e^{- π^{2} / ε}) .$ (7.17)

Now, let us take a point $(ξ_{+} (t), φ_{+} (t))$ on the manifold ${\tilde{W}}_{u}$ . We suppose that the point $(ξ_{-} (t_{1}), φ_{-} (t_{1}))$ is its vertical projection on the manifold ${\tilde{W}}_{s}$ . We will evaluate the vertical distance between these two points ${dist}_{v} (t) = φ_{+} (t) - φ_{-} (t_{1}) = φ_{+} (t) - g (ξ_{+} (t)),$ (7.18) where $g (x) = φ_{-} (ξ_{-}^{- 1} (x))$ . Thus by (7.13) ${dist}_{v} (t) = φ_{+} (t) - g (ξ_{-} (t) + D_{ε} (t)) .$ (7.19) Using Taylor expansion, we find ${dist}_{v} (t) = φ_{+} (t) - g (ξ_{-} (t)) - D_{ε} (t) g^{'} (η (t)),$ (7.20) where $ξ_{-} (t) < η (t) < ξ_{-} (t) + D_{ε} (t)$ . Here $\begin{array}{rclr} η (t) = q_{0 d} (t) + O (d), hence & (7.21) \\ g^{'} (η (t)) = - cos (\frac{η (t)}{2}) + O (d) = - tanh (\frac{d t}{ε}) + O (d) . & (7.22) \end{array}$ As $g (ξ_{-} (t)) = φ_{-} (t)$ , this yields $\begin{array}{rclr} {dist}_{v} (t) & = & φ_{+} (t) - φ_{-} (t) - D_{ε} (t) g_{-}^{'} (η (t)) \\ = & \frac{1}{ε} (ξ_{+} (t) - ξ_{+} (t - ε)) - \frac{1}{ε} (ξ_{-} (t) - ξ_{-} (t - ε)) - D_{ε} (t) g_{-}^{'} (η (t)) \\ = & \frac{1}{ε} (D_{ε} (t) - D_{ε} (t - ε)) - D_{ε} (t) g^{'} (η (t)) . & (7.23) \end{array}$ Now formula (7.15) applies and we obtain $D_{ε} (t - ε) = D_{ε} (t) + \frac{1}{d} ϕ_{2} (t, ε) exp (- \frac{π^{2}}{d}) + O (e^{- \frac{π^{2}}{d}}),$ (7.24) where $\begin{array}{rclr} ϕ_{2} (t, ε) & = & - 2 π α [\frac{cosh {((d t) / ε)}^{2} + 1}{cosh ((d t) / ε)} - t \frac{tanh ((d t) / ε)}{cosh ((d t) / ε)}] sin (\frac{2 π t}{ε}) \\ + π α [\frac{tanh ((d t) / ε)}{cosh ((d t) / ε)}] cos (\frac{2 π t}{ε}) . & (7.25) \end{array}$ With (7.23) and (7.22) this implies ${dist}_{v} (t) = \frac{1}{d^{2}} [- ϕ_{2} (t, ε) + tanh (\frac{d t}{ε}) ϕ_{1} (t, ε)] e^{- \frac{π^{2}}{d}} + O (\frac{1}{d} e^{- \frac{π^{2}}{d}}) .$ (7.26) Consequently, for $- \frac{4}{3} ⩽ t ⩽ \frac{4}{3}$ ${dist}_{v} (t) = \frac{4 π α}{ε^{2}} cosh (t) sin (\frac{2 π t}{ε}) e^{- \frac{π^{2}}{ε}} + O (\frac{1}{ε} e^{- \frac{π^{2}}{ε}}) as ε ↘ 0 .$ (7.27)

Combining this result with (7.10) and (7.11), we finally obtain our main result Theorem 1.1, because $ξ_{\pm} (t) = q_{0} (t) + O (ε^{2})$ uniformly with respect to t on any finite interval.

References

[1]

Canalis-Durand,

J.P.

Ramis,

Schäfke and

Sibuya, Gevrey solutions of singularly perturbed differential equations, J. Reine Angew. Math. 518 (2000), 95–129.

[2]

V.G.

Gelfreich, A proof of the exponentially small transversality of the separatrices for the standard map, Comm. Math. Phys. 201 (1999), 155–216.

[3]

V.F.

Lazutkin, Splitting of separatrices for the Chirikov’s standard map, Preprint VINITI no. 6372/84, 1984 (in Russian).

[4]

V.F.

Lazutkin, Exponential splitting of separatrices and an analytical integral for the semistandard map, Preprint, Université Paris VII, 1991, cf. MR 94a:58108.

[5]

V.F.

Lazutkin,

I.G.

Schachmannski and

M.B.

Tabanov, Splitting of separatrices for standard and semistandard mappings, Physica D 40 (1989), 235–248.

[6]

Sauzin, Résurgence paramétrique et exponentielle petitesse de l’écart des séparatrices du pendule rapidement forcé, Annales de l’institut Fourier (Grenoble) 45 (1995), 453–511.

[7]

Schäfke and

Volkmer, Asymptotic analysis of the equichordal problem, J. Reine Angew. Math. 425 (1992), 9–60.

[8]

Sellama, On the distance between separatrices for the discretized logistic differential equation, J. Difference Equ. Appl. 16 (2010), 1057–1099.

[9]

Y.B.

Suris, On the complex separatrices of some standard-like maps, Nonlinearity 7 (1994), 1225–1236.

10.

[10]

Ushiki, Sur les liasons-cols des systèmes dynamiques analytiques, C. R. Acad. Sci. Paris Sér. A–B 291 (1980), A447–A449.

An alternative method to estimate the splitting of separatrices for the discretized pendulum equation

Abstract

Keywords

1. Introduction

Proposition 3.1 ([7]).

Definition 3.2. Let p ∈ P n . As τ 0 ( u ) , … , τ n ( u ) form a basis of P n , we can write p ∈ P n as p = ∑ k = 0 n a k τ k ( u ) . Then we define the norms ∥ p ∥ n = ∑ i = 0 n | a i | ( π 2 ) n − i . (3.1) Theorem 3.3 ([7]).

4. Operators

Proof. The proof of this theorem is completely analogous to that of [7]. □ Theorem 4.6 ([7]).

References

Definition 3.2.
Let $p \in P_{n}$ . As $τ_{0} (u), \dots, τ_{n} (u)$ form a basis of $P_{n}$ , we can write $p \in P_{n}$ as $p = \sum_{k = 0}^{n} a_{k} τ_{k} (u) .$ Then we define the norms $∥ p ∥_{n} = \sum_{i = 0}^{n} | a_{i} | {(\frac{π}{2})}^{n - i} .$ (3.1)
Theorem 3.3 ([7]).

Proof.
The proof of this theorem is completely analogous to that of [7]. □
Theorem 4.6 ([7]).