Evgeny V Semenko

Tatiana I Semenko

Two linear conjugation problems, arising in the theory of the ow over an obstacle

The tho linear conjugation problems on the real axis are considered. These problems additionally include the symmetry condition and the asymptotic condition at the innity and arise in the hydrodynamics, in some models of the ow over an obstacle. The coecients and right hand sides of the conjugation conditions (and, naturally, the solution) depend on the complex parameter (ω in the text). The specic feature of these problems is that the right hand sides of conjugation conditions include the values of the sought analytic functions in given points. Additionally the right hand side in the rst problem includes the unknown function, that should be determined together with the solution.

The solutions of the problems are constructed and some their properties are discussed, in particular the asymptotics of the solutions, when the parameter ω tends to zero, is found.

), that G(-χ, ω) = G -1 (χ, ω), G 0 (-χ, ω) = G -1 0 (χ, ω),

Introduction, problems statement

This article is organized as follows. The section 1 contains the statement of two problems (problem 1 and 2 respectively). In the key section 2, we construct the solution of problem 1 for complex parameters ω, Im ω > 0. The following section 3 is devoted to the solution at the real value of the parameter ω, when the coecient in the linear conjugation condition degenerates. In the short section 4, we construct the solution of problem 2. In the section 5, we nd the asymptotics of the solutions at ω → 0. At last, in the appendix (section 6), all necessary subsidiary technical results are collected.

In the whole text, ν = const ∈ R, ν > 1, and we use on default the notations ξ, ω for the complex numbers (Im ω 0) and χ, µ, s for the real numbers.

Engineering Mathematical Department, Novosibirsk State Technical University, Rus- sia 1 1.1

The problem 1 At each xed ω, Im ω 0, we consider the linear conjugation (Riemann) problem

(ν(ω -χ) 2 cosh(χ) -χ sinh(χ))V + (χ, ω) = -(ν(ω + χ) 2 cosh(χ) -χ sinh(χ))V -(χ, ω) -χ sinh(χ)V + (ω, ω)- sinh(χ)(ω 2 -χ 2 ) χ 1 2π 2 + h(ω) + cosh(χ)(ω 2 -χ 2 )h(ω), χ ∈ R, (1)
with the symmetry condition

V + (ξ, ω) = V -(-ξ, ω), ξ ∈ C, (2)
and the asymptotic condition at the innity

V ± (ξ, ω) = - h(ω) 2ν + O 1 ξ , ξ → ∞, ( 3
)
where the unknowns V ± (χ, ω) are the boundary values of the functions V ± (ξ, ω), analytic with respect to ξ at ±Im ξ > 0 respectively. The right hand side of the equation ( 1) contains the value of unknown V + (ω, ω) and the function h(ω), which is unknown too and should be determined together with the solution V ± (ξ, ω).

The equation (1) we write in a form

V + (χ, ω) = -G(χ, ω)V -(χ, ω) + g(χ, ω), χ ∈ R, (4)
where

G(χ, ω) = ν(χ + ω) 2 -χ tanh(χ) ν(χ -ω) 2 -χ tanh(χ)
, or G(χ, ω) = F (-χ, ω) F (χ, ω) ,

(5) here

F (ξ, ω) = ν cosh(ξ)(ξ -ω) 2 -ξ sinh(ξ) (6)
is analytic in a whole plane (entire) function; g(χ, ω) = -g 1 (χ, ω) 1 2π 2 + h(ω) + g 2 (χ, ω)h(ω) -g 3 (χ, ω)V + (ω, ω), (7) where the functions

g 1 (ξ, ω) = sinh(ξ)(ω 2 -ξ 2 ) ξF (ξ, ω) , g 2 (ξ, ω) = cosh(ξ)(ω 2 -ξ 2 ) F (ξ, ω) , g 3 (ξ, ω) = ξ sinh(ξ) F (ξ, ω) (8)
are meromorphic in a whole plane and their poles are the zeroes of F (ξ, ω).

1.2

The problem 2

Like in the previous subsection, we consider the linear conjugation problem for unknown analytic functions V ± 0 (ξ, ω):

V + 0 (χ, ω) + V - 0 (χ, ω) -V + 0 (ω, ω) = 1 2π 2 + h(ω) (cosh(χ) -1)(ω 2 -χ 2 ) χ 2 cosh(χ) + V + (χ, ω) + V -(χ, ω) -V + (ω, ω) cosh(χ) , ( 9
)
with the symmetry condition

V + 0 (ξ, ω) = V - 0 (-ξ, ω), ( 10
)
and the asymptotic condition at the innity

V ± 0 (ξ, ω) = V 02 (ω) + O 1 ξ , ξ → ∞, (11)
where the function V 02 (ω) is unknown and should be determined together with the solution V ± 0 (ξ, ω). Here V ± (ξ, ω) and h(ω) is the solution of the problem 1.

The solution of the problem 1

In this section, we assume on default Im ω > 0 . To construct the solution of linear conjugation problem, we use absolutely classical way, see e.g. [1]. As usual, we introduce the notation for the Cauchy integrals

S ± (f (s) | ξ) = 1 2πi R f (s) ds s -ξ , ± Im ξ > 0, S(f (s) | χ) = 1 2πi R f (s) ds s -χ , χ ∈ R,
and use the well known Plemelj formulas (see e.g. [1, 2]):

S ± (f (s) | χ) = lim ξ→χ∈R S ± (f (s) | ξ) = S(f (s) | χ) ± f (s) 2 . ( 12
)
In particular,

S ± (1 | ξ) = 1 2πi R ds s -ξ = ± 1 2 , ± Im ξ > 0; S(1 | χ) = 0, χ ∈ R. ( 13
)
We note more two evident properties of the Cauchy integrals for the even and odd functions:

φ(-s) = φ(s) =⇒ S + (φ | ξ) = -S -(φ | -ξ), S(φ | 0) = 0, S ± (φ | 0) = ± φ(0) 2 ;
(14)

φ(-s) = -φ(s) =⇒ S + (φ | ξ) = S -(φ | -ξ), S ± (φ | 0) = S(φ | 0). ( 15
)
2.1

The solution of the homogeneous problem

Here we consider the homogeneous problem

X + (χ, ω) = G(χ, ω)X -(χ, ω), χ ∈ R, (16)
where G(χ, ω) has a form (5), and X ± (ξ, ω) are unknown functions, analytic with respect to ξ at ± Im ξ > 0 respectively. Lemma 1

1. F (χ, ω) ̸ = 0 and so G(χ, ω) is smooth, bounded and doesn't vanish, and ind G(χ, ω)

χ∈R = 1 2π ∆ arg G(χ, ω) χ∈R = -2.
2. In class of functions with no more than power-law growth with respect to ξ at the innity, there exists only (up to an ω-depended factor C(ω)) analytic at ±Im ξ > 0 and non vanishing solution of problem (16). This solution X ± (ξ, ω) is analytic with respect to ω in the upper half plane.

3. X + (-ξ, ω) = X -(ξ, ω) and the solution X ± (i.e. the factor C(ω)) may be chosen so that

X ± (ξ, ω) = ξ 2 ± 2ωξ + O(1), ξ → ∞. ( 17
)
So, the additional condition

lim ξ→∞ X ± (ξ, ω) ξ 2 = 1 (18)
determines the non vanishing solution of the homogeneous problem (16) unambiguously.

The function

Y (ξ, ω) = F (ξ, ω)X + (ξ, ω), Im ξ 0, F (-ξ, ω)X -(ξ, ω), Im ξ 0, (19)
is even and entire with respect to ξ and

1 Y (χ, ω) = 1 νχ 4 cosh(χ) 1 + O 1 χ , χ → ∞, χ ∈ R. ( 20
)
Proof. The function (6) F (χ, ω) we represent in a form

F (χ, ω) = ν cosh(χ) χ - χ tanh(χ) ν -ω χ + χ tanh(χ) ν -ω . ( 21
)
Since χ± χ tanh(χ)/ν ∈ R, then at Im ω ̸ = 0 we have F (χ, ω) ̸ = 0. Besides, it follows from (5), ( 21), that

G(χ, ω) =     χ - χ tanh(χ) ν + ω χ - χ tanh(χ) ν -ω     •     χ + χ tanh(χ) ν + ω χ + χ tanh(χ) ν -ω     . ( 22
)
Let x(χ) = χ -χ tanh(χ)/ν. The function x(χ) is real and continuous and x(±∞) = ±∞, and so

ind χ - χ tanh(χ) ν + ω χ - χ tanh(χ) ν -ω χ∈R = ind x + ω x -ω x∈R = -1,
since the function (z + ω)/(z -ω), z = x + iy is analytic in a whole plane, including z = ∞ and has one zero z = -ω in the lower half plane and one pole z = ω in the upper half plane. Similarly

ind χ + χ tanh(χ) ν + ω χ + χ tanh(χ) ν -ω χ∈R = -1,
and it follows from (22), that the index of G(χ, ω) as a whole is equal to -2.

The rst lemma statement proved. Now we construct the solution X ± (ξ, ω). We choose arbitrary in the upper half plane two points ξ 1 and ξ 2 (they may be coincident and may be depended on ω) and let

G 0 (χ, ω) = G(χ, ω) (χ -ξ 1 )(χ -ξ 2 ) (χ + ξ 1 )(χ + ξ 2 ) . ( 23
)
Evidently, the index of G 0 (χ, ω) is equal to zero, i.e. one may choose the one-valued and continuous branch of ln G 0 (χ, ω). Let

X ± 0 (ξ, ω) = exp S ± (ln G 0 (s, ω) | ξ) , X ± (ξ, ω) = X ± 0 (ξ, ω)(ξ ± ξ 1 )(ξ ± ξ 2 ). (24)
According to the Plemelj formulas ( 12) and ( 23), (24), we obviously have

X + 0 (χ, ω) X - 0 (χ, ω) = G 0 (χ, ω) and X + (χ, ω) X -(χ, ω) = G(χ, ω),
i.e. X ± (ξ, ω) is the solution of the homogeneous problem (16). The functions

X ± 0 (ξ, ω)
, and together with them the functions X ± (ξ, ω), are analytic with respect to ξ and don't vanish at ±Im ξ > 0.

The function F (ξ, ω) and together with it the coecient G(χ, ω) are evidently analytic with respect to ω. So, if the chosen points ξ 1 , ξ 2 are analytic with respect to ω, then the functions (24) X ± 0 (ξ, ω) and X ± (ξ, ω) are obviously analytic with respect to ω.

Let's prove the uniqueness of X ± (ξ). If X ± 1 (ξ, ω) is the other non vanishing solution of the problem (16), i.e.

X + 1 (χ, ω) X - 1 (χ, ω) = X + (χ, ω) X -(χ, ω) ,
and so the function ln G 0 (χ, ω) is odd with respect to χ, then according to (15

) S ± (ln G 0 | ξ) is even: S + (ln G 0 | -ξ) = S -(ln G 0 | ξ),
and so the functions (24) X ± 0 (ξ, ω), X ± (ξ, ω) are symmetric:

X + 0 (-ξ, ω) = X - 0 (ξ, ω), X + (-ξ, ω) = X -(ξ, ω). ( 25
)
The representation (5) evidently follows that

G(χ, ω) = 1 + G 1 (ω) χ + O 1 χ 2 , χ → ∞,
consequently according to the (23) we have the same representation for G 0 :

G 0 (χ, ω) = 1 + G 0 1 (ω) χ + O 1 χ 2 , χ → ∞.
Whence one may choose the branch of ln G 0 (χ) in such a way, that

ln G 0 (χ) = G 0 1 (ω) χ + O 1 χ 2 , χ → ∞.
So according to the formula (24) and lemma 2 (Appendix, subsection 6.1), we obtain

ln X ± 0 (ξ, ω) = X ± 01 (ω) ξ + O 1 ξ 2 , ξ → ∞, hence X ± 0 (ξ, ω) = 1 + X ± 01 (ω) ξ + O 1 ξ 2 , ξ → ∞,
and so

X ± (ξ, ω) = (ξ ± ξ 1 )(ξ ± ξ 2 )X ± 0 (ξ, ω) = ξ 2 + A ± 1 (ω)ξ + O(1), ξ → ∞, ( 26
) and the symmetry condition X + (-ξ, ω) = X -(ξ, ω) follows that

A ± 1 (ω) = ±A 1 (ω).
Let's nd A 1 . Up to the branch choice, we have at

χ → ∞ ln G(χ, ω) = -4πiθ(χ) + 4ω χ + O 1 χ 2 , ln (χ -ξ 1 )(χ -ξ 2 ) (χ + ξ 1 )(χ + ξ 2 ) = 4πiθ(χ) - 2(ξ 1 + ξ 2 ) χ + O 1 χ 2 ,
where θ(χ) is a Heaviside function, whence

ln G 0 (χ) = 4ω -2(ξ 1 + ξ 2 ) χ + O 1 χ 2 ,
and then according to the the formula (24) and lemma 2 we obtain

ln X ± 0 (ξ, ω) = ± 2ω -(ξ 1 + ξ 2 ) ξ + O 1 ξ 2 , ξ → ∞,
i.e.

X ± 0 (ξ) = 1 ± 2ω -(ξ 1 + ξ 2 ) ξ + O 1 ξ 2 , ξ → ∞,
and since

(ξ ± ξ 1 )(ξ ± ξ 2 ) = ξ 2 ± (ξ 1 + ξ 2 )ξ + O(1), ξ → ∞, then X ± (ξ) = (ξ ± ξ 1 )(ξ ± ξ 2 )X ± 0 (ξ) = ξ 2 ± 2ωξ + O(1), ξ → ∞, hence A ± 1 = ±A 1 = ±2ω.
Thus the third lemma statement follows from (25), (26). Now we consider the last lemma statement. It follows from the equalities

G(χ, ω) = F (-χ, ω)/F (χ, ω) = X + (χ, ω)/X -(χ, ω), that F (χ, ω)X + (χ, ω) = F (-χ, ω)X -(χ, ω), χ ∈ R, i.e. the function (19) Y (ξ, ω) is entire with respect to ξ. The equality Y (-ξ, ω) = Y (ξ, ω) obviously follows from (25). At last, from the representation F (ξ, ω) (6) it follows that F (χ, ω) = νχ 2 cosh(χ) 1 + O 1 χ , χ → ∞, χ ∈ R,
together with the condition (17) it implies the condition (20). So, the lemma proved completely. Remark. We emphasize, at the proving of the lemma, we directly construct the solution X ± (ξ, ω) of the homogeneous problem (16), satisfying to the condition (18). From the uniqueness of this solution (the third statement of the lemma) it follows that this solution doesn't depend on the choice of the points ξ j , j = 1, 2.

2.2

The solution of the inhomogeneous problem

We construct the solution of the inhomogeneous problem (4) by the absolutely classical way: since the coecient index ind G = -2 < 0 (the rst statement of the lemma 1), then the only bounded solution of problem (4) must have a form

V ± (ξ, ω) = ±X ± (ξ, ω)S ± g(s, ω) X + (s, ω) | ξ , ( 27
)
and the solution (27) is bounded if and only if the right hand side g(χ, ω) satises two solvability conditions, see e.g. [1].

So, we have to investigate the function (27). According to the representations (7), (8) of g(χ, ω), it has a form

V ± (ξ, ω) = ∓X ± (ξ, ω)S ± (g 01 (s, ω) | ξ) 1 2π 2 + h(ω) ± X ± (ξ, ω)S ± (g 02 (s, ω) | ξ)h(ω) ∓ X ± (ξ, ω)S ± (g 03 (s, ω) | ξ)V + (ω, ω), ( 28
)
where

g 01 (χ, ω) = g 1 (χ, ω) X + (χ, ω) = sinh(χ)(ω 2 -χ 2 ) χY (χ, ω) , g 02 (χ, ω) = g 2 (χ, ω) X + (χ, ω) = cosh(χ)(ω 2 -χ 2 ) Y (χ, ω) , g 03 (χ, ω) = g 3 (χ, ω) X + (χ, ω) = χ sinh(χ) Y (χ, ω) , ( 29
)
and Y (ξ, ω) have a form (19). Since Y (χ, ω) is even with respect to χ (the last statement of the lemma 1), then the functions (29) g 0j (χ, ω), j = 1, 3 are even too, and so according to the Cauchy integral property (14), the functions S(g 0j | ξ) are odd:

S + (g 0j | ξ) = -S -(g 0j | -ξ), j = 1, 3, therefore the solution (28) is symmet- ric: V + (ξ, ω) = V -(-ξ, ω), i.e. the condition (2) fullled.
To construct the solution of the problem 1, it remains to fulll the asymptotic condition (3) and to nd the functions h(ω) and yet unknown V + (ω, ω).

The asymptotics of 1/Y (χ, ω) (20) evidently follows that the functions (29) g 0j have a representation

g 01 (χ, ω) = O 1 χ 3 , g 02 = - 1 νχ 2 +O 1 χ 3 , g 03 = O 1 χ 3 , χ → ∞.
Using the lemma 2 (Appendix, subsection 6.1), we obtain

S ± (g 0j (s, ω) | ξ) = - J j (ω) ξ - K j (ω) ξ 2 ± R j 2ξ 2 + O 1 ξ 3 , ξ → ∞, ( 30
)
where

J j (ω) = 1 2πi R g 0j (s, ω) ds, j = 1, 3;
(31)

K j (ω) = 1 2πi
R sg 0j (s, ω) ds, j = 1, 3;

R j = 0, j = 1, 3, -1/ν, j = 2.
(32) But the functions g 0j (s, ω) are even with respect to s, whence follows K j (ω) ≡ 0, j = 1, 3. Taking in mind the asymptotics X ± (ξ, ω) (17), the formula (30) with K j (ω) ≡ 0, j = 1, 3 and R j of the form (32), we nally obtain the asymptotics of the solution (28):

V ± (ξ, ω) = ±(ξ ± 2ω)J(ω) - h(ω) 2ν + O 1 ξ , ξ → ∞,
where

J(ω) = J 1 (ω) 1 2π 2 + h(ω) -h(ω)J 2 (ω) + J 3 (ω)V + (ω, ω),
and J j (ω) have a form (31). So to satisfy the condition (3), we have to fulll the equality J(ω) = 0 or

J 1 (ω) 1 2π 2 + h(ω) -h(ω)J 2 (ω) + J 3 (ω)V + (ω, ω) = 0. ( 33
)
To analyze the equation (33), we introduce the notations

P j (ω) = 1 2πi R s j cosh(s) Y (s, ω) ds, Q j (ω) = 1 2πi R s j sinh(s) Y (s, ω) ds. ( 34
)
The analysis and calculation of these integrals we adduce in the appendix, subsection 6.3. In the terms (34), we have (see (29))

J 1 (ω) = ω 2 Q -1 (ω) -Q 1 (ω), J 2 (ω) = ω 2 P 0 (ω) -P 2 (ω), J 3 (ω) = Q 1 (ω),
and since Q 1 = νP 2 + νω 2 P 0 (Appendix, subsection 6.3, formula (145)), then the equation (33) takes a form

ω 2 Q -1 (ω) 1 2π 2 + h(ω) -P 0 (ω)h(ω) + νP 0 (ω)V + (ω, ω) = = Q 1 (ω) 1 2π 2 + h(ω) -P 2 (ω)h(ω) -νP 2 (ω)V + (ω, ω). ( 35
)
On the other hand, we have from the formula (28) the equation

V + (ω, ω) = -X + (ω, ω)S + (g 01 (s, ω) | ω) 1 2π 2 + h(ω) + X + (ω, ω)S + (g 02 (s, ω) | ω)h(ω) -X + (ω, ω)S + (g 03 (s, ω) | ω)V + (ω, ω). ( 36
)
But for each even function φ(s) we evidently have

S + (φ(s)(ω 2 -s 2 ) | ω) = 1 2πi R φ(s)(ω 2 -s 2 ) s -ω ds = - 1 2πi R sφ(s) ds - ω 2πi R φ(s) ds = - ω 2πi R φ(s) ds, ( 37
)
and from the representations (29) in accordance with (37) and in terms of (34) it follows

S + (g 01 (s, ω) | ω) = -ωQ -1 (ω), S + (g 02 (s, ω) | ω) = -ωP 0 (ω). (38)
For S + (g 03 | ω) we use the formula (146) (Appendix, subsection 6.3), and thus the equation (36) takes a form

Q -1 (ω) 1 2π 2 + h(ω) -P 0 (ω)h(ω) + νP 0 (ω)V + (ω, ω) = 0. (39)
Together with the equation (35), we obtain the system

       Q -1 (ω0 νP 0 (ω) - 1 ν h(ω) + V + (ω, ω) = - Q -1 (ω) 2π 2 νP 0 (ω) Q 1 (ω) νP 2 (ω) - 1 ν h(ω) -V + (ω, ω) = - Q 1 (ω) 2π 2 νP 2 (ω) (40)
The solution of this system is

h(ω) = - 1 2π 2 ∆ h (ω) ∆(ω) , V + (ω, ω) = - 1 2π 2 ∆ V (ω) ∆(ω) , (41)
where

∆(ω) = Q -1 νP 0 + Q 1 νP 2 - 2 ν , ∆ h = Q -1 νP 0 + Q 1 νP 2 = ∆(ω) + 2 ν , ∆ V = 1 ν Q 1 νP 2 - Q -1 νP 0 . ( 42
)
In the appendix, subsection 6.4, we investigate the functions ∆(ω), ∆ h (ω), ∆ V (ω) in (42), in particular we establish for them other representations, show the solvability of the system (40), i.e. ∆(ω) ̸ = 0, for each complex ω ̸ = 0, Im ω 0, and nd their asymptotics at ω → 0. Thus, the solution of the inhomogeneous problem 1 at Im ω > 0 is constructed completely.

To formulate the nal result, we should return to the representation (21) of F (χ, ω). We introduce at χ ∈ R the functions

µ ± (χ) = χ ± ψ(χ) √ ν , ψ(χ) = sign(χ) χ tan(χ). ( 43
)
The functions µ ± (χ) are evidently odd, smooth, monotonically increase (at ν > 1) and µ ± (χ)

̸ = 0, µ + (χ) ̸ = µ -(χ), sign(µ ± (χ)) = sign(χ) if χ ̸ = 0.
Therefore the inverse functions χ ± (µ) = µ -1 ± exist, are smooth and monotonically increase too, and

χ ± (µ) ̸ = 0, χ + (µ) ̸ = χ -(µ), sign(χ ± (µ)) = sign(µ) if µ ̸ = 0.
The full description of the properties of the functions µ ± (χ), χ ± (µ) see in Appendix, subsection 6.2.

Just in terms of χ ± (s) we formulate the nal result. Let's introduce the functions

h 0 (s) = ln s 2 χ + (s)χ -(s) , h 1 (s) = 1 2
s 2 χ 2 + (s) + s 2 χ 2 -(s) , s ∈ R; H j (ω) = ω πi R h j (s) s 2 -ω 2 ds, Im ω > 0. (44)
Using the representations (162), (165) (Appendix, subsection 6.4) for the functions ∆(ω), ∆ h (ω) and ∆ V (ω) from (42), we nally formulate the solution of the problem 1 in the following theorem.

Theorem 1 The solution of the problem 1 has a form

h(ω) = H 1 (ω) -e H 0 (ω) H 1 (ω) -e H 0 (ω) -2/ν , V ± (ξ, ω) = ±X ± (ξ, ω)S ± (g 00 (s, ω) | ξ),
where H j (ω), j = 0, 1 have a form (44), X ± (ξ, ω) is a solution of homogeneous problem, constructed in the lemma 1 (see formula (24)),

g 00 (s, ω) = -g 01 (s, ω) 1 2π 2 + h(ω) + g 02 (s, ω)h(ω)- g 03 (s, ω) 2 -H 1 (ω) -e H 0 (ω) ν(H 1 (ω) -e H 0 (ω) ) -2 , g 01 (s, ω) = sinh(s)(ω 2 -s 2 ) sY (s, ω) , g 02 (s, ω) = cosh(s)(ω 2 -s 2 ) Y (s, ω) , g 03 (s, ω) = s sinh(s) Y (s, ω) , Y (s, ω) = F (s, ω)X + (s, ω), F (s, ω) = ν(s -ω) 2 cosh(s) -s sinh(s).
3 The solution at real ω ̸ = 0

Here we essentially use the introduced in the previous subsection functions χ ± (µ). It follows from the representation (21): at real ω = µ the function F (ξ, ω) has two real roots ξ = χ ± (µ). So, the coecient G(χ, ω) = F (-χ, ω)/F (χ, ω) degenerates (vanishes or approaches innity) at real ω and so the solution V ± (ξ, ω) may become singular. Here we investigate the limit value

V ± (ξ, ω), when ω → µ ∈ R, µ ̸ = 0.
The real roots ξ = χ ± (µ) are evidently the values of the complex roots ξ ± (ω) for the real ω = µ: ξ ± (µ) = χ ± (µ). The functions ξ ± (ω) are dened and analytic with respect to ω at least when ω is situated near the real axis:

Im ω ≈ 0. Since the index of G(χ, ω) = F (-χ, ω)/F (χ, ω) is equal to -2,
then the roots ξ ± (ω) are situated in the upper half plane, when Im ω > 0. We use exactly the points

ξ 1 = ξ 1 (ω) = ξ -(ω), ξ 2 = ξ 2 (ω) = ξ + (ω)
for the construction of the solution X ± (ξ, ω) of the homogeneous problem (16) (we remind, this solution doesn't depend on the choice of ξ 1,2 ). Then the function

F 0 (ξ, ω) = F (ξ, ω) (ξ -ξ 1 (ω))(ξ -ξ 2 (ω)) (45)
is entire (analytic in a whole plane) with respect to ξ and has a continuous limit, when ω → µ ∈ R, i.e. F 0 (ξ, ω) → F 0 (ξ, µ) and F 0 (χ, µ) ̸ = 0, χ ∈ R. Hence the function (23) G 0 (χ, ω) = F 0 (-χ, ω)/F 0 (χ, ω) has a continuous limit too: G 0 (χ, ω) → G 0 (χ, µ) ̸ = 0, and so for the function X ± 0 (ξ, ω) in (24) we obtain

X ± 0 (ξ, ω) ---→ ω→µ exp S ± (ln G 0 (s, µ) | ξ) = X ± 0 (ξ, µ) ̸ = 0.
We introduce the function

Y 0 (ξ, ω) = F 0 (ξ, ω)X + 0 (ξ, ω), Im ξ 0, F 0 (-ξ, ω)X - 0 (ξ, ω), Im ξ 0. (46)
Absolutely similar to the proof of the lemma 1, we conclude, the function

Y 0 (ξ, ω) is entire with respect to ξ, is even (Y 0 (-ξ, ω) = Y 0 (ξ, ω))
and the previous considerations imply, that

Y 0 (ξ, ω) ---→ ω→µ Y 0 (ξ, µ) ̸ = 0. ( 47
)
In terms of the introduced function Y 0 (ξ, ω), we rewrite the solution (28) (see also (24)) in a form 2, where, according to the (29),

V ± (ξ, ω) = (ξ ± ξ 1 )(ξ ± ξ 2 )X ± 0 (ξ, ω) 2πi R (s 2 + 1) 2 g 0 (s, ω) ds (s -ξ)(s 2 -ξ 2 1 )(s 2 -ξ 2 2 ) , ( 48
)
ξ j = ξ j (ω), j = 1,
g 0 (χ, ω) = - sinh(χ)(ω 2 -χ 2 ) χ(χ 2 + 1) 2 Y 0 (χ, ω) 1 2π 2 + h(ω) + cosh(χ)(ω 2 -χ 2 ) (χ 2 + 1) 2 Y 0 (χ, ω) h(ω) - χ sinh(χ) (χ 2 + 1) 2 Y 0 (χ, ω) V + (ω, ω). ( 49
)
Since the function Y 0 (χ, ω) is even with respect to χ, then the function g 0 (χ, ω) is even with respect to χ too.

In the Appendix, subsection 6.4, it is shown: the functions h(ω), V + (ω, ω) are continuous at Im ω 0, ω ̸ = 0, i.e. have a continuous limit at ω → µ ∈ R:

h(ω) ---→ ω→µ h(µ), V + (ω, ω) ---→ ω→µ V + (µ, µ). ( 50
)
The representation (49) and the statements (47) and (50) evidently imply

g 0 (χ, ω) ---→ ω→µ g 0 (χ, µ) = - sinh(χ)(µ 2 -χ 2 ) χ(χ 2 + 1) 2 Y 0 (χ, µ) 1 2π 2 + h(µ) + cosh(χ)(µ 2 -χ 2 ) (χ 2 + 1) 2 Y 0 (χ, µ) h(µ) - χ sinh(χ) (χ 2 + 1) 2 Y 0 (χ, µ) V + (µ, µ). ( 51
)
In order to go to the limit in the representation (48), we use the expansions into a sum of simple fractions:

1 (ξ ∓ ξ 1 )(ξ ∓ ξ 2 ) = ±1 ξ 1 -ξ 2 1 ξ ∓ ξ 1 - 1 ξ ∓ ξ 2 , ( 52
) (s 2 + 1) 2 (s -ξ)(s 2 -ξ 2 1 )(s 2 -ξ 2 2 ) = (ξ 2 + 1) 2 (ξ 2 -ξ 2 1 )(ξ 2 -ξ 2 2 ) • 1 s -ξ + + (ξ 2 1 + 1) 2 2ξ 1 (ξ 2 1 -ξ 2 2 ) 1 (ξ + ξ 1 )(s + ξ 1 ) - 1 (ξ -ξ 1 )(s -ξ 1 ) + + (ξ 2 2 + 1) 2 2ξ 2 (ξ 2 2 -ξ 2 1 ) 1 (ξ + ξ 2 )(s + ξ 2 ) - 1 (ξ -ξ 2 )(s -ξ 2 ) . ( 53
)
Using the formula (53) for the solution (48), we obtain

V ± (ξ, ω) = X ± 0 (ξ, ω)(ξ 2 + 1) 2 (ξ ∓ ξ 1 )(ξ ∓ ξ 2 ) S ± (g 0 (s, ω) | ξ)+ (ξ ± ξ 1 )(ξ ± ξ 2 )X ± 0 (ξ, ω)(ξ 2 1 + 1) 2 2ξ 1 (ξ 2 1 -ξ 2 2 ) S -(g 0 (s, ω) | -ξ 1 ) ξ + ξ 1 - S + (g 0 (s, ω) | ξ 1 ) ξ -ξ 1 + (ξ ± ξ 1 )(ξ ± ξ 2 )X ± 0 (ξ, ω)(ξ 2 2 + 1) 2 2ξ 2 (ξ 2 2 -ξ 2 1 ) S -(g 0 (s, ω) | -ξ 2 ) ξ + ξ 2 - S + (g 0 (s, ω) | ξ 2 ) ξ -ξ 2 .
But the function g 0 (s, ω) is even with respect to s, and so S + (g 0 | ξ j ) = -S -(g 0 | -ξ j ), j = 1, 2 and the previous formula takes a form

V ± (ξ, ω) = X ± 0 (ξ, ω)(ξ 2 + 1) 2 (ξ ∓ ξ 1 )(ξ ∓ ξ 2 ) S ± (g 0 (s, ω) | ξ)∓ ξ(ξ ± ξ 2 )X ± 0 (ξ, ω)(ξ 2 1 + 1) 2 ξ 1 (ξ ∓ ξ 1 )(ξ 2 1 -ξ 2 2 ) S ± (g 0 (s, ω) | ±ξ 1 )∓ ξ(ξ ± ξ 1 )X ± 0 (ξ, ω)(ξ 2 2 + 1) 2 ξ 2 (ξ ∓ ξ 2 )(ξ 2 2 -ξ 2 1 ) S ± (g 0 (s, ω) | ±ξ 2 ). ( 54
)
Using the formula (52) for the rst summand in (54), we convert the expression (54) into the form

V ± (ξ, ω) = ±X ± 0 (ξ, ω) (ξ 1 -ξ 2 )(ξ ∓ ξ 1 ) × (ξ 2 + 1) 2 S ± (g 0 (s, ω) | ξ) - ξ(ξ ± ξ 2 )(ξ 2 1 + 1) 2 ξ 1 (ξ 1 + ξ 2 ) S ± (g 0 (s, ω) | ±ξ 1 ) ± X ± 0 (ξ, ω) (ξ 2 -ξ 1 )(ξ ∓ ξ 2 ) × ( 55
) (ξ 2 + 1) 2 S ± (g 0 (s, ω) | ξ) - ξ(ξ ± ξ 1 )(ξ 2 2 + 1) 2 ξ 2 (ξ 1 + ξ 2 ) S ± (g 0 (s, ω) | ±ξ 2 ) .
In (55), we go to the limit at

ω → µ ̸ = 0, µ ∈ R and ξ → χ ∈ R. Then ξ 1,2 (ω) → χ 1,2 = χ 1,2 (µ), χ 1 (µ) = χ -(µ), χ 2 (µ) = χ + (µ)
, and according to the Plemelj formulas (12)

S ± (g 0 (s, ω) | ξ) → S(g 0 (s, µ) | χ) ± g 0 (χ, µ) 2 , S ± (g 0 (s, ω) | ±ξ j ) → S(g 0 (s, µ) | ±χ j ) ± g 0 (±χ j , µ) 2 , j = 1, 2,
and so we obtain

V ± (χ, µ) = Φ ± 1 (χ, µ) + Φ ± 2 (χ, µ) + Φ ± 3 (χ, µ) + Φ ± 4 (χ, µ), ( 56
)
where [START_REF] Carrier | Functions of a complex variable[END_REF] .

Φ ± 1 (χ, µ) = ±X ± 0 (χ, µ)(χ 2 + 1) 2 (χ 1 -χ 2 )(χ ∓ χ 1 ) × S(g 0 (s, µ) | χ) - χ(χ ± χ 2 )(χ 2 1 + 1) 2 χ 1 (χ 1 + χ 2 )(χ 2 + 1) 2 S(g 0 (s, µ) | ±χ 1 ) , Φ ± 2 (χ, µ) = X ± 0 (χ, µ)(χ 2 + 1) 2 2(χ 1 -χ 2 )(χ ∓ χ 1 ) g 0 (χ, µ) - χ(χ ± χ 2 )(χ 2 1 + 1) 2 g 0 (±χ 1 , µ) χ 1 (χ 1 + χ 2 )(χ 2 + 1) 2 , Φ ± 3 (χ, µ) = ±X ± 0 (χ, µ)(χ 2 + 1) 2 (χ 2 -χ 1 )(χ ∓ χ 2 ) × S(g 0 (s, µ) | χ) - χ(χ ± χ 1 )(χ 2 2 + 1) 2 χ 2 (χ 1 + χ 2 )(χ 2 + 1) 2 S(g 0 (s, µ) | ±χ 2 ) , Φ ± 4 (χ, µ) = X ± 0 (χ, µ)(χ 2 + 1) 2 2(χ 2 -χ 1 )(χ ∓ χ 2 ) g 0 (χ, µ) - χ(χ ± χ 1 )(χ 2 2 + 1) 2 g 0 (±χ 2 , µ) χ 2 (χ 1 + χ 2 )(χ 2 + 1)
The functions Φ ± j (χ, µ), j = 1, 4 are continuous (and even smooth) for every real (χ, µ). Indeed, the only discontinuity points can be ±χ 1 for Φ ± j , j = 1, 2, and ±χ 2 for Φ ± j , j = 3, 4. But obviously

Φ ± 2 (±χ 1 , µ) = X ± 0 (±χ 1 , µ)(χ 2 1 + 1) 2 2(χ 1 -χ 2 ) (g 0 ) ′ χ (±χ 1 , µ) -(φ ± 1 ) ′ (±χ 1 )g 0 (±χ 1 , µ) ,
where

φ ± 1 (χ) = χ(χ ± χ 2 )(χ 2 1 + 1) 2 χ 1 (χ 1 + χ 2 )(χ 2 + 1) 2 .
Absolutely similarly with the using of the formula S(1 | χ) = 0, see (13), we obtain

Φ ± 1 (±χ 1 , µ) = ±X ± 0 (±χ 1 , µ)(χ 2 1 + 1) 2 (χ 1 -χ 2 ) ×   1 2πi R g 0 (s, µ) -g 0 (±χ 1 , µ) (s ∓ χ 1 ) 2 ds -(φ ± 1 ) ′ (±χ 1 )S(g 0 (s, µ) | ±χ 1 )   .
By the same way

Φ ± 3 (±χ 2 , µ) = ±X ± 0 (±χ 2 , µ)(χ 2 2 + 1) 2 (χ 2 -χ 1 ) ×   1 2πi R g 0 (s, µ) -g 0 (±χ 2 , µ) (s ∓ χ 2 ) 2 ds -(φ ± 2 ) ′ (±χ 2 )S(g 0 (s, µ) | ±χ 2 )   , Φ ± 4 (±χ 2 , µ) = X ± 0 (±χ 2 , µ)(χ 2 2 + 1) 2 2(χ 2 -χ 1 ) (g 0 ) ′ χ (±χ 2 , µ) -(φ ± 2 ) ′ (±χ 2 )g 0 (±χ 2 , µ) ,
where

φ ± 2 (χ) = χ(χ ± χ 1 )(χ 2 2 + 1) 2 χ 2 (χ 1 + χ 2 )(χ 2 + 1) 2 .
Thus, we established the continuity of V ± (χ, µ) for every real χ and real µ ̸ = 0. So, the solution V ± (ξ, ω) of the problem (1), ( 2), (3) is regular (continuous and even smooth) for every real ω, except the points, where χ 1 = χ 2 or χ 1 = 0 or χ 2 = 0, or χ 1 + χ 2 = 0, in any case it is an only point

ω = µ = 0.

The solution of the problem 2

We proceed to the problem (9), (10), (11). The conjugation condition (9) has a form

V + 0 (χ, ω) + V - 0 (χ, ω) = V + 0 (ω, ω) - 1 2π 2 + h(ω) + g 0 (χ, ω), ( 57
)
where

g 0 (χ, ω) = 1 2π 2 + h(ω) ω 2 (cosh(χ) -1) + χ 2 χ 2 cosh(χ) + V + (χ, ω) + V -(χ, ω) -V + (ω, ω) cosh(χ) , (58)
and V ± (χ, ω), h(ω) is a solution of the problem 1. In accordance with the conditions (2), (3), the function g 0 (χ, ω) (58) is even with respect to χ and

g 0 (χ, ω) = O 1 χ 2 , χ → ∞. ( 59
)
The bounded solution of the problem (57) has evidently form (see e.g. [1])

V ± 0 (ξ, ω) = ±C(ω) + V + 0 (ω, ω) 2 - 1 2 1 2π 2 + h(ω) ± S ± (g 0 | ξ), (60)
and since the function g 0 is even, then

S + (g 0 | -ξ) = -S -(g 0 | ξ).
Then the symmetry condition (10) gives C(ω) = 0, i.e. the solution is unique. At that, by virtue of the condition (59) and lemma 2 (Appendix, subsection 6.1), we obtain

V ± 0 (ξ, ω) = V + 0 (ω, ω) 2 - 1 2 1 2π 2 + h(ω) + O 1 ξ , ξ → ∞,
i.e. we've nd the unknown V 02 (ω) in the asymptotic condition (11):

V 02 (ω) = V + 0 (ω, ω) 2 - 1 2 1 2π 2 + h(ω) .
It remains to determine V + 0 (ω, ω).

We have from the representation (60), where C(ω) = 0, that

V + 0 (ω, ω) = V + 0 (ω, ω) 2 - 1 2 1 2π 2 + h(ω) + S + (g 0 | ω),
i.e.

V + 0 (ω, ω) = 2S + (g 0 | ω) - 1 2π 2 + h(ω) .
Theorem 2 The solution of the problem 2 has a form

V ± 0 (ξ, ω) = S + (g 0 (s, ω) | ω) - 1 2π 2 + h(ω) ± S ± (g 0 (s, ω) | ξ), V 02 (ω) = S + (g 0 (s, ω) | ω) - 1 2π 2 + h(ω) ,
where

g 0 (s, ω) = 1 2π 2 + h(ω) ω 2 (cosh(s) -1) + s 2 s 2 cosh(s) + V + (s, ω) + V -(s, ω) -V + (ω, ω) cosh(s) ,
and h(ω), V ± (ξ, ω) is a solution of the problem 1.

Obviously, the solution of problem 2 is continues with respect to ξ at ±Im ξ 0, continues with respect to ω at Im ω 0, ω ̸ = 0 and analytic at Im ω > 0.

5 The asymptotics of the solutions at ω → 0

5.1

The asymptotics of the solution of problem 1

We start from the asymptotics of the functions h(ω) and V + (ω, ω). The representation (41) and the asymptotic formulas (166) (Appendix, subsection (6.4)) imply

h(ω) = iA 0 ω + A 1 + O(ω), V + (ω, ω) = -A 2 + O(ω), ω → 0, (61)
where

A 0 = 1 2πνA , A 1 = ν + 3 24ν(ν -1) 2 A 2 - ν + 1 2π 2 (ν -1) , A 2 = 1 2π 2 ν , A = A(ν) = R cosh(s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s)
ds.

(

) 62
Now we go to the asymptotics of the solution V ± (ξ, ω). Let U ± (ξ, ω) = ωV ± (ξ, ω), then (see ( 1), ( 2), (3)) the function U ± is a solution of the linear conjugation problem

(ν(ω -χ) 2 cosh(χ) -χ sinh(χ))U + (χ, ω) = -(ν(ω + χ) 2 cosh(χ) -χ sinh(χ))U -(χ, ω)- χ sinh(χ)ωV + (ω, ω) - sinh(χ)(ω 2 -χ 2 ) χ ω 2π 2 + ωh(ω) + cosh(χ)(ω 2 -χ 2 )ωh(ω). (63)
Whence, based on the asymptotics (61), we conclude: at ω → 0 the function

U ± have a form U ± (ξ, ω) = U ± 0 (ξ) + U ± 1 (ξ)ω + O(ω 2 )
, where the functions U ± j (ξ), j = 0, 1 satisfy the conjugation conditions

U + 0 (χ) = -U - 0 (χ) -iA 0 cosh(χ) -sinh(χ)/χ ν cosh(χ) -sinh(χ)/χ , ( 64
) (ν cosh(χ) -sinh(χ)/χ)U + 1 (χ) = -(ν cosh(χ) -sinh(χ)/χ)U - 1 (χ)+ cosh(χ) 2ν U + 0 (χ) -U - 0 (χ) χ -A 1 + sinh(χ) χ 1 2π 2 + A 1 + A 2 . ( 65
)
At that the symmetry condition (2) evidently implies, that the functions

U ± j (ξ) are symmetric too: U + j (ξ) = U - j (-ξ), j = 0, 1.
The bounded solution of the problem (64) has a form

U ± 0 (ξ) = ±C ∓ iA 0 S ± (W 0 (s) | ξ), W 0 (s) = cosh(χ) -sinh(χ)/χ ν cosh(χ) -sinh(χ)/χ ,
where C = const. But the function W 0 (s) is even, therefore

S + (W 0 | -ξ) = -S -(W 0 | ξ) and the symmetry condition U + 0 (ξ) = U - 0 (-ξ)
implies C = 0. Thus, we have the representation

U ± 0 (ξ) = ∓iA 0 S ± (W 0 (s) | ξ), W 0 (χ) = cosh(χ) -sinh(χ)/χ ν cosh(χ) -sinh(χ)/χ , ( 66
)
where A 0 has a form (62). Now let's go to the problem (65). We note, the function W 0 (s)/s (see (66)) is smooth and odd, in particular S(W 0 (s) | 0) = 0, and therefore, taking in mind the Plemelj formulas (12), we have

U + 0 (χ) -U - 0 (χ) χ = - 2iA 0 χ (S(W 0 | χ) -S(W 0 | 0)) = -2iA 0 S W 0 (s) s | χ .
Thus the equation (65) takes a form

U + 1 (χ) = -U - 1 (χ) + W 1 (χ), W 1 (χ) = 1 ν cosh(χ) -sinh(χ)/χ × cosh(χ) -4iνA 0 S W 0 (s) s | χ -A 1 + 1 2π 2 + A 1 + A 2 sinh(χ) χ ,
and absolutely similar to the previous considerations, we obtain the representation

U ± 1 (ξ) = ±S ± (W 1 | ξ).
We summarize all results in the next theorem.

Theorem 3 The solution of the problem 1 has the following asymptotics at ω → 0:

h(ω) = iA 0 ω + A 1 + O(ω), V ± (ξ, ω) = U ± 0 (ξ) ω + U ± 1 (ξ) + O(ω),
where

U ± 0 (ξ) = ∓iA 0 S ± (W 0 (s) | ξ), U ± 1 (ξ) = ±S ± (W 1 (s) | ξ), W 0 (χ) = cosh(χ) -sinh(χ)/χ ν cosh(χ) -sinh(χ)/χ , W 1 (χ) = 1 ν cosh(χ) -sinh(χ)/χ × cosh(χ) -4iνA 0 S W 0 (s) s | χ -A 1 + 1 2π 2 + A 1 + A 2 sinh(χ) χ , A 0 = 1 2πνA , A 1 = ν + 3 24ν(ν -1) 2 A - ν + 1 2π 2 (ν -1) , A 2 = 1 2π 2 ν , A = A(ν) = R cosh(s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s) ds. 20 5.2
The asymptotics of the solution of problem 2

In this subsection, we use on default the notations of the theorem 3. By virtue of the theorem 3 and formula (61), the function g 0 (s, ω) (58) has the following asymptotics at ω → 0:

g 0 (s, ω) = iA 0 + U + 0 (s) + U - 0 (s) ω cosh(s) + 1/(2π 2 ) + A 1 + U + 1 (s) + U - 1 (s) + A 2 cosh(s) + O(ω),
and since

iA 0 + U + 0 (s) + U - 0 (s) = iA 0 -iA 0 W 0 (s) = iA 0 (ν -1) cosh(s) ν cosh(s) -sinh(s)/s , U + 1 (s) + U - 1 (s) = W 1 (s),
then we obtain

g 0 (s, ω) = q -1 (s) ω + q 0 (s) + O(ω), (67)
where

q -1 (s) = iA 0 (ν -1) ν cosh(s) -sinh(s)/s , q 0 (s) = 1/(2π 2 ) + A 1 + W 1 (s) + A 2 cosh(s) . ( 68
)
Now we should consider the asymptotics S + (g 0 | ω), see theorem 2. Absolutely similar to the proof of the lemma 3 (Appendix, subsection 6.1), we easy obtain

S + (g(s) | ω) = S(g(s) | 0) + g(0) 2 + ω 2πi R g(s) -g(0) s 2 ds + O(ω 2 ). ( 69
)
But the functions q j (s) are obviously even, i.e. S(q j | 0) = 0, j = -1, 0, whence the formulas (67) and (69) imply

S + (g 0 (s, ω) | ω) = q -1 (0) 2ω + q 0 (0) 2 -A 3 + O(ω), A 3 = - 1 2πi R q -1 (s) -q -1 (0) s 2 ds. ( 70
)
We evidently have from the formula (68), that q -1 (0) = iA 0 . Next,

A 3 = - A 0 2π R ν(1 -cosh(s)) + sinh(s)/s -1 s 2 (ν cosh(s) -sinh(s)/s) ds = - A 0 2π   R (ν -1)(1 -cosh(s)) s 2 (ν cosh(s) -sinh(s)/s) ds + R sinh(s)/s -cosh(s) s 2 (ν cosh(s) -sinh(s)/s) ds   = - 1 4π 2 νA   -A -(ν -1) R cosh(s) -1 s 2 (ν cosh(s) -sinh(s)/s) ds   ,
and nally

A 3 = 1 4π 2 ν + ν -1 4π 2 νA R cosh(s) -1 s 2 (ν cosh(s) -sinh(s)/s)
ds.

(71)

Then, using the representation W 1 (s) in the theorem 3, and taking in mind, that

A(ν) = 2πiS(W 0 (s)/s | 0), we have W 1 (0) = -1/(2π 2 ν) = -A 2 (see theorem 3
), therefore by virtue of the (68) q 0 (0) = 1/(2π 2 ) + A 1 , and so, according to the formula (70),

S + (g 0 (s, ω | ω) = iA 0 2ω + A 1 2 + 1 4π 2 -A 3 + O(ω), ( 72
)
where A 3 = A 3 (ν) have a form (71). Substituting the asymptotics of h(ω) (61), g 0 (s, ω) (67), (68), and S + (g 0 | ω) (72) in the formulas for the solution of problem 2, given in the theorem 2, we obtain the asymptotics of the mentioned solution.

Theorem 4 At ω → 0

V ± 0 (ξ, ω) = V ± 00 (ξ) ω + V ± 10 (ξ) + O(ω), V 02 (ω) = - iA 0 2ω -A 4 + O(ω),
where

V ± 00 (ξ) = - iA 0 2 ± S ± (q -1 (s) | ξ), V ± 10 (ξ) = - A 1 2 - 1 4π 2 -A 3 ± S ± (q 0 (s) | ξ), q -1 (s) = iA 0 (ν -1) ν cosh(s) -sinh(s)/s , q 0 (s) = 1/(2π 2 ) + A 1 + W 1 (s) + A 2 cosh(s) , A 4 = A 1 2 + 1 4π 2 + A 3 ,
and the constants A 0 , A 1 , A 3 have a form (62) and (71) respectively.

6 Appendix

The asymptotics of Cauchy integrals

Lemma 2 Let the function φ(χ) be smooth at χ ∈ R and

φ(χ) = φ 1 χ + φ 2 χ 2 + O 1 χ 3 , χ → ∞, φ j = const, j = 1, 2.
(73) Then the Cauchy integral S ± (φ | ξ) has an asymptotics

S ± (φ | ξ) = φ ± 1 ξ + φ ± 2 ξ 2 + O 1 ξ 3 , ξ → ∞, ( 74
)
where

φ ± 1 = ± φ 1 2 - 1 2πi R φ(s) ds, φ ± 2 = ± φ 2 2 - 1 2πi R (sφ(s) -φ 1 )ds. ( 75
)
Proof. We mean ξ → ∞ on default. The condition (73) obviously implies the representation

φ(χ) = φ 1 χ + φ 2 χ 2 + φ 3 (χ) χ 3 , |χ| > 1, (76)
where the function φ 3 (χ) is bounded. Let's denote S ± (φ | ξ) = Φ ± (ξ) for short. We have

Φ ± (ξ) = 1 2πi 1 -1 φ(s) ds s -ξ + 1 2πi |s|>1 φ(s) ds s -ξ = Φ ± 1 (ξ) + Φ ± 2 (ξ).
At |s| < 1, we use the decomposition

1 s -ξ = - 1 ξ - s ξ 2 + O 1 ξ 3 , then Φ ± 1 (ξ) = - 1 2πiξ 1 -1 φ(s) ds - 1 2πiξ 2 1 -1 sφ(s) ds + O 1 ξ 3 . ( 77
)
At |s| > 1, we use the representation

1 2πi |s|>1 ds s -ξ = 1 2πi R ds s -ξ - 1 2πi 1 -1 ds s -ξ = = S ± (1 | ξ) - 1 2πi ln ξ -1 ξ + 1 , (78)
and, having in mind the formula (13) and the asymptotic decomposition

ln ξ -1 ξ + 1 = - 2 ξ + O 1 ξ 3 , ξ → ∞,
we rewrite (78) it in a form

1 2πi |s|>1 ds s -ξ = sign(Im ξ) 2 + 1 2πiξ 1 -1 ds + O 1 ξ 3 . ( 79
)
Using the expansion into a sum of simple fractions

1 s j (s -ξ) = 1 ξ j (s -ξ) - j k=1 1 ξ k s j-k+1 , j 1,
and the formula (79), we obtain

1 2πi |s|>1 ds s(s -ξ) = sign(Im ξ) 2ξ - 1 2πiξ 2 1 -1 ds - 1 2πiξ |s|>1 ds s + O 1 ξ 3 ; 1 2πi |s|>1 ds s 2 (s -ξ) = sign(Im ξ) 2ξ 2 - 1 2πiξ 2 |s|>1 ds s - 1 2πiξ |s|>1 ds s 2 + O 1 ξ 3 ; 1 2πi |s|>1 φ 3 (s) ds s 3 (s -ξ) = - 1 2πiξ |s|>1 φ 3 (s) ds s 3 - 1 2πiξ 2 |s|>1 φ 3 (s) ds s 2 + O 1 ξ 3 ;
and in total according to the representation (76)

Φ ± 2 (ξ) = 1 ξ    φ 1 sign(Im ξ) 2 - 1 2πi |s|>1 φ 1 s + φ 2 s 2 + φ 3 (s) s 3 ds    + 1 ξ 2    1 2πi 1 -1 φ 1 ds + φ 2 sign(Im ξ) 2 - 1 2πi |s|>1 φ 2 s + φ 3 (s) s 2 ds    + O 1 ξ 3 .
Together with the formula (77), we obtain

Φ ± (ξ) = Φ ± 1 (ξ) + Φ ± 2 (ξ) = sign(Im ξ) 2 φ 1 ξ + φ 2 ξ 2 - 1 2πiξ R φ(s) ds - 1 2πiξ 2 R (sφ(s) -φ 1 )ds + O 1 ξ 3 . ( 80
)
The lemma statement (the formulas (74), (75)) obviously follows from the formula (80).

Remark. Absolutely similar if φ(χ) = φ 1 /χ+O(1/χ 2 ) at χ → ∞, where

φ 1 = const, then S ± (φ | ξ) = φ ± 1 /ξ + O(1/ξ 2 ), ξ → ∞ where φ ± 1 has a form (75).
Now we proceed to the specic Cauchy integrals, which we use to investigate the function, depending on one variable ω only, and accordingly we use the notations

S + 0 (φ(s) | ω) = ω 2πi R φ(s) ds s 2 -ω 2 , Im ω > 0; ( 81
) S 0 (φ(s) | µ) = µ 2πi R φ(s) ds s 2 -µ 2 , µ ∈ R. ( 82
)
We obviously have the representations

S + 0 (φ | ω) = S + (φ | ω) -S -(φ | -ω), Im ω > 0; S 0 (φ | µ) = S(φ | µ) -S(φ | -µ), µ ∈ R, (83)
and, in particular (see ( 13))

S + 0 (1 | ω) = 1, Im ω > 0; S 0 (1 | µ) = 0, µ ∈ R. ( 84
)
If the function φ(s) is even, then according to the Plemelj formulas (12)

S + 0 (φ | µ) = S 0 (φ | µ) + φ(µ), µ ∈ R. ( 85
)
Lemma 3 Let the function φ(s) be real, even and smooth and

φ(s) = A + B |s| + O 1 s 2 , s → ∞, A, B = const.
Then:

1. At µ ∈ R Re S + 0 (φ | µ) = φ(µ), Im S + 0 (φ | µ) = -iS 0 (φ | µ).

At

ω → 0 S + 0 (φ | ω) = φ(0) + ω πi R φ(s) -φ(0) s 2 ds + φ ′′ (0)ω 2 2 + O(ω 3 ). 3. At ω → ∞ S + 0 (φ | ω) = A + o(1).
Proof. The rst lemma statement evidently follows from (85). We proceed to the second statement. According to the (84), we have

S + 0 (φ | ω) = φ(0) + S + 0 (φ(s) -φ(0) | ω) = φ(0) + ω πi R s 2 φ 1 (s) ds s 2 -ω 2 , ( 86
)
where φ 1 (s) = (φ(s) -φ(0))/s 2 . But since φ(s) is even and smooth, and so φ ′ (0) = 0, then φ 1 (s) is even and smooth too, therefore

lim ω→0 1 πi R s 2 φ 1 (s) ds s 2 -ω 2 = 1 πi R φ 1 (s) ds.
Whence

S + 0 (φ | ω) = φ(0) + O(ω),
and so

S + 0 (φ 1 | ω) = φ 1 (0) + O(ω) = φ ′′ (0) 2 + O(ω).
Then it follows from (86), that

S + 0 (φ | ω) = φ(0) + ω πi R (s 2 -ω 2 + ω 2 )φ 1 (s) ds s 2 -ω 2 = φ(0) + ω πi R φ 1 (s) ds + ω 2 S + 0 (φ 1 (s) | ω) = φ(0) + ω πi R φ(s) -φ(0) s 2 ds + ω 2 φ ′′ (0) 2 + O(ω 3 ).
We've proved the second lemma statement. Let now ω → ∞. First (see (84)),

S + 0 (φ | ω) = A + S + 0 (φ(s) -A | ω). ( 87
)
We represent

S + 0 (φ(s) -A | ω) = S + 0 (φ 0 (s) | ω) + S + 0 (φ 1 (s) | ω), ( 88
)
where

φ 0 (s) = φ(s) -A, |s| < 1, 0, |s| > 1, φ 1 (s) = 0, |s| < 1, φ(s) -A, |s| > 1.
Obviously,

S + 0 (φ 0 | ω) = o(1), ω → ∞.
(89) In turn, at |s| > 1 we have the lemma condition

φ 1 (s) = B |s| + φ 2 (s) s 2 , ( 90
)
the function φ 2 is bounded and we can assume, that φ 2 (s) ≡ 0 at |s| < 1.

As it was shown at the proof of the previous lemma,

S ± φ 2 (s) s 2 | ω = - 1 2πiω |s|>1 φ 2 (s) ds s 2 + O 1 ω 2 = o(1),
whence, according to the formula (83),

S + 0 φ 2 (s) s 2 | ω = S + φ 2 (s) s 2 | ω -S -φ 2 (s) s 2 | -ω = o(1). ( 91
)
We now turn to the integral from 1/|s|. Similar to the previous considerations,

ω πi |s|>1 ds |s|(s 2 -ω 2 ) = 1 2πi |s|>1 ds |s|(s -ω) - 1 2πi |s|>1 ds |s|(s + ω) = S + 1 (ω)-S - 1 (-ω).
We directly compute

S ± 1 (ω) = - ln(1 -ω 2 ) 2πiω ,
and so, independently on the choice of a branch of the logarithm,

ω πi |s|>1 ds |s|(s 2 -ω 2 ) = o(1), ω → ∞. ( 92
)
Combining the formulas (87), ( 88), ( 89), ( 90), ( 91) and (92), we obtain the last lemma statement. The lemma is proved completely.

6.2

The functions χ ± (µ).

These functions are the inverse functions for the µ ± (χ) of the form (43), i.e.

µ ± (χ) χ = 1 ± tan(χ) νχ . ( 93
)
We list several obvious properties of these functions, which follow from formulas (43) or (93):

• the functions µ ± (χ) and, respectively, χ ± (µ) are odd, smooth, monotonically increase, µ ± (0) = χ ± (0) = 0;

• µ ± (χ)/χ and, respectively, µ/χ ± (µ) are even and smooth;

• µ 1,2 (χ)/χ > 0 and, respectively, µ/χ 1,2 (µ) > 0;

• at s ∈ (0, +∞) the functions µ -(s)/s and s/χ -(s) monotonically increase from the value 1 -1/ √ ν ∈ (0, 1) at s = 0 till the value 1 at s = +∞;

• at s ∈ (0, +∞) the functions µ + (s)/s and s/χ + (s) monotonically decrease from the value 1 + 1/

√

ν > 1 at s = 0 till the value 1 at s = +∞.

In the next lemma, we establish the asymptotics of the functions χ ± (s) at s → 0 and s → ∞. We use the notation

β = 1/ √ ν. Lemma 4 1. At s → 0 s χ ± (s) = 1 ∓ β ± βs 2 6(1 ∓ β) 2 + O(s 4 ), ( 94
) ln s χ ± (s) = ln(1 ∓ β) ± βs 2 6(1 ∓ β) 3 + O(s 4 ), ( 95
) s χ ± (s) 2 = (1 ∓ β) 2 ± 2βs 2 6(1 ∓ β) + O(s 4 ). ( 96
) 2. At s → ∞ s χ ± (s) = 1 ± A 1 β |s| + A 2 β 2 |s| ± A 3 β 3 |s| |s| + O 1 s 2 , ( 97
) ln s χ ± (s) = ± A 4 β |s| + A 5 β 2 |s| ± A 6 β 3 |s| |s| + O 1 s 2 , ( 98
)
s χ ± (s) 2 = 1 ± A 7 β |s| + A 8 β 2 |s| ± A 9 β 3 |s| |s| + O 1 s 2 , ( 99
)
here A j = const, j = 1, 9.

Remark. One may compute the values of constants A j , but, as we see below, there is no need.

Proof. By the denition, the functions χ ± (s) are the solutions of the equations

χ ± (s) = s ± βψ(χ ± (s)), β = 1 √ ν , ψ(s) = sign(s) s tanh(s). ( 100
)
Let s → 0. It's easy to see, the function ψ(s) is odd and

ψ(0) = ψ ′′ (0) = ψ (4) (0) = 0, ψ ′ (0) = 1, ψ ′′′ (0) = -1. ( 101
)
Since the functions χ ± (s) are odd, then χ ′′ ± (0) = χ (4)

± (0) = 0.
The equation (100) and the formula (101) implies

χ ′ ± (s) = 1 ± βψ ′ (χ ± )χ ′ ± =⇒ χ ′ ± (0) = 1 1 ∓ βψ ′ (0) = 1 1 ∓ β ,
and similarly

χ ′′′ ± = ±βψ ′′′ (χ ± )(χ ′ ± ) 3 ± 3βψ ′′ (χ ± )χ ′ ± χ ′′ ± ± βψ ′ (χ ± )χ ′′′ ± =⇒ χ ′′′ ± (0) = ±βψ ′′′ (0)(χ ′ ± (0)) 3 ± 3βψ ′′ (0)χ ′ ± (0)χ ′′ ± (0) 1 ∓ βψ ′ (0) = ∓ β (1 ∓ β) 4 . Hence χ ± (s) s = χ ′ ± (0) + χ ′′′ ± (0)s 2 6 + O(s 4 ) = 1 1 ∓ β ∓ βs 2 6(1 ∓ β) 4 + O(s 4 ),
and

s χ ± (s) = 1 ∓ β ± βs 2 6(1 ∓ β) 2 + O(s 4 ),
i.e. we proved the formula (94). The formulas (95) and (96) obviously follows from (94). The rst lemma statement proved. Now let s → ∞. Then tanh(s) = sign(s) + O(e -|s| ), whence s tanh(s) = |s| + O(e -|s| ), and so

µ ± (χ) = χ ± sign(χ) √ ν |χ| + O(e -|χ| ), or s = χ ± (s) ± sign(s) √ ν |χ ± (s)| + O(e -|s| ).

29

This equation evidently implies the statement (97). The statements (98) and (99) obviously follows from (97). The lemma proved completely. Now we return to the function h j (s), H j (ω), j = 0, 1 of the form (44), i.e.

h 0 (s) = ln s 2 χ + (s)χ -(s) , H 0 (ω) = S + 0 (h 0 (s) | ω), ( 102
)
h 1 (s) = 1 2
s 2 χ 2 + (s) + s 2 χ 2 -(s) , H 1 (ω) = S + 0 (h 1 (s) | ω), ( 103
)
where the Cauchy integral S + 0 we've introduced in the previous subsection, the formula (81).

Theorem 5

1. The functions h j (s) are even and smooth and at real ω = µ H j (-µ) = H j (µ), j = 0, 1.

At s

→ 0 h 0 (s) = ln ν -1 ν + ν(3ν + 1)s 2 6(ν -1) 3 + O(s 4 ), h 1 (s) = 1 + 1 ν + s 2 3(ν -1) + O(s 4 ). (104) 3
. At s → ∞ h 0 (s) = A 0 |s| + O 1 s 2 , h 1 (s) = 1 + A 1 |s| + O 1 s 2 , ( 105
) A 0 , A 1 = const. 4. At ω → 0 H 0 (ω) = ln ν -1 ν -iωB ν ν -1 + ν(3ν + 1)ω 2 6(ν -1) 3 + O(ω 3 ), e H 0 (ω) = ν -1 ν -iωB -ω 2 νB 2 ν -1 - 3ν + 1 6(ν -1) 2 + O(ω 3 ), H 1 (ω) = 1 + 1 ν + ω 2 3(ν -1) + O(ω 3 ), (106)
where

B = 2 π R cosh(s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s)
ds.

(107)

5. At ω → ∞ H 0 (ω) = o(1), exp(H 0 (ω)) = 1 + o(1), H 1 (ω) = 1 + o(1). ( 108
)
Proof. The rst theorem statement evidently follows from the denitions h j (s), j = 0, 1 (102), (103), the simplest properties of χ ± (µ) (see the beginning of this subsection) and the rst statement of the lemma 3.

We go to the second statement. The denition h 0 (s) (102) and the formula (95) from the previous lemma immediately imply

h 0 (s) = ln(1 -β 2 ) + β 2 (3 + β 2 )s 2 3(1 -β 2 ) 3 + O(s 4 ), s → 0,
and taking into account β = 1/ √ ν, we obtain the formula (104) for h 0 (s). Absolutely similar we obtain from (96) the second part of (104).

We proceed to the H j (ω), j = 0, 1. It follows from the lemma 3 (previous subsection), that

H j (ω) = h j (0) + ω πi R h j (s) -h j (0) s 2 ds + h ′′ j (0)ω 2 2 + O(ω 3 ), ω → 0, (109)
H 0 (ω) = o(1), H 1 (ω) = 1 + o(1), ω → ∞.
But (104) and (105) imply

h 0 (0) = ln ν -1 ν , h ′′ 0 (0) = ν(3ν + 1) 3(ν -1) 3 , h 1 (0) = 1 + 1 ν , h ′′ 1 (0) = 2 3(ν -1)
,

and so to prove (106), ( 107) and (108) we only need to nd the integrals

B j = - 1 π R h j (s) -h j (0) s 2 ds, j = 0, 1. (110)
Let's start by the B 1 calculating. According to the h 1 (s) denition (103) and h 1 (0) = 1 + 1/ν = 1 + β 2 , we represent

B 1 = - 1 2π R 1 χ 2 + (s) + 1 χ 2 -(s) - (1 -β) 2 + (1 + β) 2 s 2 ds = J + +J -, (111)
where

J ± = - 1 2π R 1 χ 2 ± (s) - (1 ∓ β) 2 s 2 ds.
We make the change of variable s = µ ± (p) in the integrals J ± respectively and obtain

J ± = - 1 2π R µ ′ ± (p) p 2 - (1 ∓ β) 2 µ ′ ± (p) µ 2 ± (p) dp,
and the representation (111) takes a form

B 1 = - 1 2π R b(p) dp, b(p) = µ ′ + (p) + µ ′ -(p) p 2 - (1 -β) 2 µ ′ + (p) µ 2 + (p) - (1 + β) 2 µ ′ -(p) µ 2 -(p) . ( 112
)
But (see ( 43)

) µ + (p) + µ -(p) = 2p, i.e. µ ′ + (p) + µ ′ -(p) = 2.
Using that, and taking in mind, that the integrand b(p) in ( 112) is even, we calculate the integral B 1 in (112) in the sense of principal value:

B 1 = - 1 π lim ε→0   ∞ ε 2 dp p 2 - ∞ ε (1 -β) 2 µ ′ + (p) µ 2 + (p) dp - ∞ ε (1 + β) 2 µ ′ -(p) µ 2 -(p) dp   = - 1 π lim ε→0 - 2 p ∞ ε + (1 -β) 2 µ + (p) ∞ ε + (1 + β) 2 µ -(p) ∞ ε ,
or

B 1 = - 1 π lim ε→0 2 ε - (1 -β) 2 µ + (ε) - (1 + β) 2 µ -(ε) . ( 113
)
It is easy follows from (43), that µ ± (s) = s(1 ∓ β) + O(s 3 ), s → 0 and therefore

1 µ ± (ε) = 1 ε(1 ∓ β) + O(ε), ε → 0. (114)
Substituting the expressions (114) in (113), we nally obtain

B 1 = - 1 π lim ε→0 2 ε - 1 -β ε - 1 + β ε + O(ε) = 0.
Thus we actually proved the formula (106) for H 1 (ω). Now we proceed to the B 0 calculating. We at once integrate its denition (110) by part and, using the denition of h 0 (s) (102), obtain

B 0 = - 1 π R 2 s - χ ′ + (s) χ + (s) - χ ′ -(s) χ -(s) ds s = J + + J -, (115)
where

J ± = - 1 π R 1 s - χ ′ ± (s) χ ± (s) ds s .
We make again the change of variable s = µ ± (p) in the integrals J ± respectively and the formula (115) takes a form

B 0 = 1 π   R 1 p - µ ′ + (p) µ + (p) dp µ + (p) + R 1 p - µ ′ -(p) µ -(p) dp µ -(p)   . ( 116
)
Calculating the integral (116) in the sense of the principal value, and using the obvious formulas (see (43))

µ + (p) + µ -(p) = 2p, µ + (p)µ -(p) = p 2 - p tanh(p) ν ,
and the formula (114), we obtain

B 0 = 1 π lim ε→0    |s|>ε µ + (s) + µ -(s) sµ + (s)µ -(s) ds - |s|>ε µ ′ + (s) µ 2 + (s) ds - |s|>ε µ ′ -(s) µ 2 -(s) ds    = 1 π lim ε→0    |s|>ε 2ν cosh(s) s 2 (ν cosh(s) -sinh(s)/s) ds - 2 µ + (ε) - 2 µ -(ε)    = 1 π lim ε→0    |s|>ε 2ν cosh(s) s 2 (ν cosh(s) -sinh(s)/s) ds - 4ν (ν -1)ε + O(ε)    = 1 π lim ε→0    |s|>ε 2ν cosh(s) s 2 (ν cosh(s) -sinh(s)/s) ds - |s|>ε 2ν ds (ν -1)s 2    = - 2ν π(ν -1) lim ε→0 |s|>ε cosh(s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s) ds,
and nally

B 0 = - 2ν π(ν -1) R cosh(s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s) ds. ( 117
)
Substituting the expressions for B 0 (117) and (110) in the expression for H 0 (ω) (109), we obviously obtain the representation (106), (107) for H 0 (ω) at ω → 0. The theorem proved completely.

6.3

The analysis and calculation of some integrals

Here we analyze some important for the problems 1 and 2 functions and express them in terms of the introduced above functions H 0 (ω) and H 1 (ω).

We start with the values ln X + (0, ω) and (ln X + ) ′′ ξξ (0, ω) for the introduced in the subsection 2.1 solution of the homogeneous problem X + (ξ, ω). We take the representation (24) and choose ξ 1 = ξ 2 = ω (we remind, the solution X + (ξ, ω) doesn't depend on the choice of points ξ 1,2 , located in the upper half plane). Then the representation (23), (24) gives

ln X + (0, ω) = 2 ln ω + 1 2πi R φ(s, ω) s ds, (ln X + ) ′′ ξξ (0, ω) = - 2 ω 2 + 1 πi R φ(s, ω) s 3 ds, (118)
where (see (23), (5))

φ(s, ω) = ln F (-s, ω)(s -ω) 2 F (s, ω)(s + ω) 2 . ( 119
)
The integrals converge because

φ(s, ω) = O 1 s , s → ∞ and φ(0, ω) = φ ′ s (0, ω) = φ ′′ ss (0, ω) = 0.
We integrate the integrals in (118) by part:

ln X + (0, ω) = 2 ln ω - 1 2πi R φ ′ s (s, ω) ln |s| ds, (ln X + ) ′′ ξξ (0, ω) = - 2 ω 2 + 1 2πi R φ ′ s (s, ω) ds s 2 . ( 120
)
We present (see ( 21), (43))

F (s, ω) = ν cosh(s)(µ + (s) -ω)(µ -(s) -ω),
then, according to the (119)

φ ′ s (s, ω) = -2ω µ ′ + (s) µ 2 + (s) -ω 2 + µ ′ -(s) µ 2 -(s) -ω 2 - 2 s 2 -ω 2 , ( 121
)
and we obtain the representation for ln X + (0, ω):

ln X + (0, ω) = 2 ln ω + ω πi ×   R µ ′ + (s) ln |s| ds µ 2 + (s) -ω 2 + R µ ′ -(s) ln |s| ds µ 2 -(s) -ω 2 - R 2 ln |s| ds s 2 -ω 2   = 2 ln ω + ω πi   R ln |χ + (s)| ds s 2 -ω 2 + R ln |χ -(s)| ds s 2 -ω 2 - R 2 ln |s| ds s 2 -ω 2   ,
and nally, according to the denition S + 0 (81) and the formula (102), we have the representation

ln X + (0, ω) = 2 ln ω -H 0 (ω).
(122) Now we consider (ln X + ) ′′ ξξ (0, ω): according to the formulas (120), (121), we present

(ln X + ) ′′ ξξ (0, ω) = - 2 ω 2 - ω πi R µ ′ + (s) µ 2 + (s) -ω 2 + µ ′ -(s) µ 2 -(s) -ω 2 + 2 ω 2 ds s 2 + ω πi R 2 ω 2 + 2 s 2 -ω 2 ds s 2 ,
and since (see ( 43)

) µ + (s) + µ -(s) = 2s, i.e. µ ′ + (s) + µ ′ -(s) = 2,
then, according to the denition S + 0 (81), we can write

(ln X + ) ′′ ξξ (0, ω) = - 2 ω 2 - ω πi R µ ′ 1 (s) µ 2 + (s) -ω 2 + µ ′ + (s) ω 2 ds s 2 - ω πi R µ ′ -(s) µ 2 -(s) -ω 2 + µ ′ -(s) ω 2 ds s 2 + 2 πiω R ds s 2 -ω 2 = - 2 ω 2 - 1 ω 2 S + 0 s 2 χ 2 + (s) | ω - 1 ω 2 S + 0 s 2 χ 2 -(s) | ω + 2 ω 2 S + 0 (1 | ω),
and nally, using the formulas (84) and (103), we obtain the representation

(ln X + ) ′′ ξξ (0, ω) = - 2H 1 (ω) ω 2 .
(123)

Now we proceed to the analysis and computation of the introduced in the subsection 2.2 (formula (34)) integrals

P j (ω) = 1 2πi R s j cosh(s) Y (s, ω) ds, Q j (ω) = 1 2πi R s j sinh(s) Y (s, ω) ds, ( 124
)
where on default Im ω > 0. As we know (the lemma 1), Y (s, ω) ̸ = 0 and the only singular points of integrals (124) are s = ∞ and s = 0 at j < 0. We consider the integrals (124) in the sense of the principal value, i.e.

R • • • ds = lim ε→0 R→∞ ε<|s|<R • • • ds.
The asymptotics of Y (χ, ω) (20) shows, the integrals P j (ω) converge at j = -1, 3 and Q j (ω) converge at j = -2, 2. We dene

P j (ω) = 1 2πi R s j cosh(s) Y (s, ω) - 1 Y (0, ω) ds, j = -3, -2. ( 125
)
Since the function φ(s) = cosh(s)/Y (s, ω) is even, then φ(s)-φ(0) = As 2 +O(s 4 ), s → 0, A = const, i.e. the integrals (125) converge in the sense of the principal value too, see e.g. [1, 2]. We note, the integral from the odd function is equal to zero, and so

P -3 (ω) = P -1 (ω) = P 1 (ω) = P 3 (ω) = Q -2 (ω) = Q 0 (ω) = Q 2 (ω) = 0. (126)
Before calculating the integrals, we establish several auxiliary statements. We introduce the curves L m , L ε and the region Π m :

L m = { ξ = πm + iβ, β ∈ [0, πm] }∪ { ξ = -πm + iβ, β ∈ [0, πm] }∪ { ξ = α + iπm, |α| πm }, m ∈ N;
(127)

Π m = { ξ | Re ξ ∈ (-πm, πm), Im ξ ∈ (0, πm) }, m ∈ N;
(128) and similarly tanh(ξ) = -1 + O(e -2πm ) at ξ = -πm + iβ, as a whole we obtain the lemma statement.

L ε = { ξ | |ξ| = ε, Im ξ > 0 }, ε > 0; (129
Lemma 6 Let the function φ(ξ) is dened in the upper half plane and

φ(ξ) ξ 4 cosh(ξ) ξ∈Lm = α ξ + O 1 ξ 2 , m → ∞. ( 130
)
Then

lim m→∞ 1 2πi Lm φ(ξ) dξ Y (ξ, ω) = α 2ν . ( 131
)
Proof. It follows from the lemma 5, that at ξ ∈ L m and m → ∞

F (ξ, ω) νξ 2 cosh(ξ) = 1 - ω ξ 2 - tanh(ξ) ξ = 1 + O 1 ξ ,
and therefore

νξ 2 cosh(ξ) F (ξ, ω) = 1 + O 1 ξ , ξ ∈ L m , m → ∞. ( 132
)
Using the representation (132) and the asymptotic of X ± (ξ, ω) (17) (see lemma 1), we obtain

1 2πi Lm φ(ξ) dξ Y (ξ, ω) = 1 2πi Lm φ(ξ) ξ 4 cosh(ξ) • ξ 2 cosh(ξ) F (ξ, ω) • ξ 2 X + (ξ, ω) dξ = 1 2πi Lm α ξ + O 1 ξ 2 1 ν + O 1 ξ 1 + O 1 ξ dξ,
and nally

1 2πi Lm φ(ξ) dξ Y (ξ, ω) = α 2πiν Lm dξ ξ + Lm O 1 ξ 2 dξ = I 1 + I 2 . ( 133
)
For the rst summand, we may deform the curve L m into a semicircle and so calculate I 1 :

I 1 = α 2πiν |ξ|=πm Im ξ>0 dξ ξ = α 2ν . ( 134
)
Passing to the second summand, we note that the length of L m is L(L m ) = 4πm, and at ξ ∈ L m we have |ξ| πm, whence follows

|I 2 | = Lm O 1 ξ 2 dξ const • L(L m ) |ξ| 2 = o(1), m → ∞. ( 135
)
The lemma statement evidently follows from (133), ( 134) and (135).

Corollary. The lemma condition (130) is satised, and so the lemma statement is valid, for the functions ξ j cosh(ξ), j = -3, 3 with α = 1 at j = 3 and α = 0 at j < 3, and, according to the lemma 5, for the functions ξ j sinh(ξ), j = -2, 2 with α = 0.

We calculate the integrals P j , Q j with the help of residues. Let ξ n = ξ n (ω), n = 1, ∞ be the zeroes of Y (ξ, ω) (i.e. the zeroes of F (ξ, ω), see the formulas (19), ( 6)) in the upper half plane and

ψ n = Res 1 Y (ξ n , ω) . ( 136
)
Lemma 7 Let the function φ(ξ) be analytic in the upper half plane and

φ(ξ) ξ 5 cosh(ξ) ξ∈Lm = α ξ + O 1 ξ 2 , ( 137
)
where L m have a form (127). Then

1 2πi R φ(s) ds sY (s, ω) = φ(0) 2Y (0, ω) - α 2ν + n φ(ξ n ) ξ n ψ n , ( 138
)
here ψ n are the residues (136). On the other hand, ε<|s|<m φ(s) ds sY (s, ω)

+ Lm φ(ξ) dξ ξY (ξ, ω) - Lε φ(ξ) dξ ξY (ξ, ω) = 2πi ξn∈Πm φ(ξ n ) ξ n ψ n ,
where the region Π m and curve L ε have a form (128) and (129) respectively. Hence,

1 2πi R φ(s) ds sY (s, ω) = lim ε→0 m→∞   ξn∈Πm φ(ξ n ) ξ n ψ n - 1 2πi Lm φ(ξ) dξ ξY (ξ, ω) + 1 2πi Lε φ(ξ) dξ ξY (ξ, ω)   . ( 139
)
But for any function f (ξ), analytic in a neighborhood of zero (see e.g. [1, 2])

lim ε→0 1 2πi Lε f (ξ) ξ dξ = lim ε→0   1 2πi Lε f (ξ) -f (0) ξ dξ + f (0) 2πi Lε dξ ξ   = f (0) 2 ,
and so

lim ε→0 1 2πi Lε φ(ξ) dξ ξY (ξ, ω) = φ(0) 2Y (0, ω) . ( 140
)
The lemma statement (138) obviously follows from (139), (140) and lemma 6 (formula (131)). (142)

Q -1 (ω) = νP 0 (ω) -2 - 1 ν + ω 2 (1 -Z(ω)) 2 , ( 143
)
where

Z(ω)) = Y ′′ ξξ (0, ω) Y (0, ω) ; (144) Q 1 (ω) = ν(P 2 (ω) + ω 2 P 0 (ω)); (145)
S + s sinh(s) Y (s, ω) | ω = - 1 X + (ω, ω) -νωP 0 (ω). (146)
Proof. We apply the lemma 7 sequentially to the integrals P j (ω), j = -3, 3 and Q j (ω), j = -2, 2 of the form (124) and (125). As follows from the corollary after the lemma 6, the lemma 6, and hence the lemma 7, is true for all of them.

For convenience, we introduce the notation

I j (ω) = n ξ j n cosh(ξ n )ψ n , ( 147
)
where ψ n are the residues of 1/Y (ξ, ω), see (136).

We start from P -3 (ω) of the form (125). Since the function

f (ξ) = cosh(ξ)/Y (ξ, ω) is even, then f (ξ) = f (0) + f ′′ (0)ξ 2 /2 + O(ξ 4 ), ξ → 0,
and so (see the proof of the previous lemma)

lim ε→0 1 2πi Lε f (ξ) -f (0) ξ 3 dξ = f ′′ (0) 4 .
So, having in mind the formula (126), the statement of lemma 7 for P -3 takes a form

0 = P -3 (ω) = I -3 (ω) + 1 4 cosh(ξ) Y (ξ, ω) ′′ ξξ ξ=0
, and so

I -3 (ω) = 1 4 Y ′′ ξξ (0, ω) Y 2 (0, ω) - 1 Y (0, ω) = Z(ω) -1 4Y (0, ω) , ( 148
)
where Z(ω) has a form (144). Absolutely similarly

P -2 (ω) = I -2 (ω). (149)
Now we apply the lemma 7 for P -1 (ω) of a form (124). Here φ(ξ) = cosh(ξ) and, again taking into account the formula (126), we obtain

0 = P -1 (ω) = I -1 (ω) + 1 2Y (0, ω) ,
and so together with the formula (148) we have

I -1 (ω) = - 1 2Y (0, ω) , I -3 (ω) = I -1 (ω)(1 -Z(ω)) 2 .
(150)

Next in the same way we obtain

P 0 (ω) = I 0 (ω), (151) 0
= P 1 (ω) = I 1 (ω), (152)
P 2 (ω) = I 2 (ω), (153)
and at last

0 = P 3 (ω) = I 3 (ω) - 1 2ν ,
and so

I 3 (ω) = 1 2ν . ( 154
)
Now we proceed to the integrals Q j (ω), j = -2, 2 of the form (124). According to the form of F (ξ, ω) (6), the equation

0 = F (ξ n , ω) = ν(ξ n -ω) 2 cosh(ξ n ) -ξ n sinh(ξ n ) implies sinh(ξ n ) = ν cosh(ξ n )(ξ n -ω) 2 ξ n . ( 155
)
Using the formula (126), lemma 7, formulas (155) and (148), we obtain

0 = Q -2 = n sinh(ξ n ) ξ 2 n ψ n + 1 2Y (0, ω) = ν n cosh(ξ n )(ξ n -ω) 2 ξ 3 n ψ n -I -1 = ν(I -1 -2ωI -2 + ω 2 I -3 ) -I -1 ,
whence (see (153), (150))

P -2 = I -2 = ν -1 2νω I -1 + ω 2 I -3 = I -1 ν -1 2νω + 1 -Z(ω) 2 . ( 156
)
Absolutely similarly

Q -1 = ν n cosh(ξ n )(ξ n -ω) 2 ξ 2 n ψ n = ν(I 0 -2ωI -1 + ω 2 I -2 ); (157)
then using ( 126) and ( 155)

0 = Q 0 = n sinh(ξ n )ψ n = ν(I 1 -2ωI 0 + ω 2 I -1 ),
and so according to the formulas (150), (151), (152)

P 0 = I 0 = ω 2 I -1 = - ω 4Y (0, ω) , (158)
i.e. we prove the formula (141). At the same time, the formula (157), taking into account the formulas (158) and ( 156), takes a form (143). Now we proceed to Q 1 . Using (155), ( 151), ( 152) and (153), we have

Q 1 = ν(I 2 -2ωI 1 + ω 2 I 0 ) = ν(P 2 + ω 2 P 0 ),
i.e. we obtain the formula (145). We go to Q 2 . Using once again (126) and (155), we have

0 = Q 2 = ν(I 3 -2ωI 2 + ω 2 I 1 ),
and so according to the formulas (152), ( 153), ( 154)

P 2 = I 2 = I 3 2ω = 1 4νω ,
i.e. we obtain (142). It remains to prove the formula (146). Absolutely similar to the proof of the lemma 7, and taking into account the equality (see ( 19), (6)) Y (ω, ω) = F (ω, ω)X + (ω, ω) = -ω sinh(ω)X + (ω, ω), we have i.e. the theorem proved completely. Now we express the results of the theorem 6 in terms of the functions H j (ω), j = 0, 1 of the form (44) or (102), (103). According to the (19), ( 6) and (122), we have Y (0, ω) = F (0, ω)X + (0, ω) = νω 4 e -H 0 (ω) .

(159)

We go to the function Z(ω) of the form (144). Since Y (ξ, ω) is even with respect to ξ (see the lemma 1), i.e. Y ′ ξ (0, ω) = 0, then, taking in mind the formulas ( 19) and (6) again, we obtain and nally, according to the representation (123),

Z(ω) = 1 - 2(ν + 1) νω 2 - 2H 1 (ω) ω 2 .
(160) Thus, we can supplement the statements of the theorem 6 with the following ones: (161) Indeed, both formulas evidently follow from the statements of the theorem 6 and the formulas (159) and (160).

6.4

On the functions ∆(ω), ∆ h (ω), ∆ V (ω)

In the subsection 2.2, we obtained the system of equations (40) for the unknowns h(ω) and V + (ω, ω) and, respectively, the expressions for them (41) and (42). In this subsection, we establish other representations for the functions ∆(ω), ∆ h (ω), ∆ V (ω) in (42), show the solvability of the system (40) at ω ̸ = 0, i.e. the condition ∆(ω) ̸ = 0, and nd the asymptotics ∆(ω), ∆ h (ω), ∆ V (ω) at ω → 0.

First, the formulas (42) and (161) give the representation

∆(ω) = - 2 ν + H 1 (ω) -e H 0 (ω) . ( 162
)
Theorem 8 ∆(ω) ̸ = 0 at Im ω 0, ω ̸ = 0.

Proof. It follows from the theorem 5, formulas (106), ( 107) and (108), that Additionally the rst statement of the theorem 5 implies ∆(-µ) = ∆(µ), µ ∈ R.

To prove this theorem, we consider the index (the increment of the argument) of ∆(ω):

ind ∆ = 1 2π ∆ arg ∆(ω) L along the curve L = { ω | |ω| = R, Im ω > } ∪ [-R, -ε] ∪ { ω | |ω| = ε, Im ω > 0 } ∪ [ε, R],
traversed counterclockwise and at R >> 1, ε << 1. If ∆(µ) ̸ = 0 at µ ∈ [ε, R], then (see (163)) ∆(ω) ̸ = 0 at ω ∈ L and ind ∆ is dened correctly, and at that if ind ∆ = 0, then ∆(ω) ̸ = 0 at |ω| ∈ [ε, R], Im ω 0, what means ∆(ω) ̸ = 0 in a whole upper half plane, excluding the point ω = 0. In turn, to prove the last statement we consider the dependance of ∆(ω) on the parameter ν > 1, i.e. on β = 1/ √ ν ∈ (0, 1). Namely, if ∆(µ) ̸ = 0 at µ ∈ [ε, R], then ind ∆ doesn't depend on β, and so if ind ∆ = 0 at least for only one β ∈ (0, 1), then Ind ∆ = 0 for each β ∈ (0, 1), and so ∆(ω) ̸ = 0 at ω ̸ = 0.

First we consider the dependance of χ ± (s) on β at β → 0. For χ ± (s) we have the equation (100) χ ± (s) = s ± βψ(χ ± (s)), where ψ(s) = sign(s) s tanh(s), and so

χ ± (s) β=0 = s; ∂χ ± ∂β = ±ψ(χ ± ) ± βψ ′ (χ ± ) ∂χ ± ∂β =⇒ ∂χ ± ∂β β=0 = ±ψ(χ ± ) 1 ∓ βψ ′ (χ ± ) β=0 = ±ψ(

  ) we go around the curves L m and L ε counterclockwise.Lemma 5 tanh(ξ) = O(1) at ξ ∈ L m , m → ∞. Proof. If ξ = α + πim, α ∈ R, then tanh(ξ) = tanh(α) = O(1), in its turn at ξ = πm + iβ we havetanh(ξ) = 1 -e -2πm e -2iβ1 + e -2πm e -2iβ = 1 + O(e -2πm ),

S

  + s sinh(s) Y (s, ω) | ω = 1 2πi R s sinh(s) (s -ω)Y (s, ω) ds = ω sinh(ω) Y (ω, ω) + n ξ n sinh(ξ n ) ξ n -ω ψ n = -1 X + (ω, ω) + n ν(ξ n -ω) 2 cosh(ξ n ) ξ n -ω ψ n = -1 X + (ω, ω) + ν 2πi R (s -ω) cosh(s) Y (s, ω) ds,and since the function cosh(s)/Y (s, ω) is even with respect to s, i.e.R s cosh(s) Y (s, ω) ds = 0, then S + χ sinh(χ) Y (χ, ω) | ω = -1 X + (ω, ω)-νωP 0 (ω),

Z 2 =

 2 (ω) = (ln Y ) ′′ ξξ (0, ω) + Y ′ ξ (0, ω) Y (0, ω) (ln F ) ′′ ξξ (0, ω) + (ln X + ) ′′ ξξ (0, ω) = 1 -2(ν + 1) νω 2 + (ln X + ) ′′ ξξ (0, ω),

Theorem 7 Q

 7 -1 (ω) νP 0 (ω) = -1 + H 1 (ω), Q 1 (ω) νP 2 (ω = 1 -e H 0 (ω) .

∆

  (ω) = iωB + O(ω 2 ), ω → 0; ∆(ω) = -2 ν + o(1), ω → ∞, (163) what in particular means ∆(ω) ̸ = 0 at |ω| = ε << 1 or |ω| = R >> 1.

2

 2 (s) + O(β 3 ), then according to the (102), (103) and (162)H 0 (ω) = -β 2 S + 0 1 cosh 2 (s) | ω + O(β 3 ), e H 0 (ω) = 1 -β 2 S | ω + O(β 3 ), ∆(ω) = -2β 2 + β 2 S + 0 2 tanh(µ) µ | ω + O(β 3 ),

  s);

	Then we successively obtain
			χ ± (s) s	= 1 ± β	ψ(s) s	+ β 2 ψ(s)ψ ′ (s) s	+ O(β 3 ),
	s χ ± (s)	= 1 ∓ β	ψ(s) s	+ β 2 ψ 2 (s) s 2 -	ψ(s)ψ ′ (s) s	+ O(β 3 ),
	ln	s χ ± (s)	= ∓β	ψ(s) s	+	β 2 2	ψ 2 (s) s 2 -	2ψ(s)ψ ′ (s) s	+ O(β 3 ),
	s χ (s)	2	= 1 ∓ β	2ψ(s) s	+ β 2 3ψ 2 (s) s 2 -	2ψ(s)ψ ′ (s) s	+ O(β 3 ),
										-	β 2 cosh 2 (s)	+ O(β 3 ),
		h 1 (s) = 1 + β 2 3ψ 2 (s) s 2 -	2ψ(s)ψ ′ (s) s	+ O(β 3 ) =
					1 + β 2 2ψ 2 (s) s 2 -	ψ 2 (s) s
	absolutely similarly					∂ 2 χ ± ∂β 2 β=0	= 2ψ(s)ψ ′ (s);
	and in total
		χ ± = χ ±	β=0	+β	∂χ ± ∂β β=0	+	β 2 2	∂ 2 χ ± ∂β 2 β=0	+O(β 3 ) =
										44

s ± βψ(s) + β 2 ψ(s)ψ ′ (s) + O(β 3 ). so, using the denitions (102), (103) and ψ(s) = sign(s) s tanh(s), we have

h 0 (s) = β 2 ψ 2 (s) s 2 -2ψ(s)ψ ′ (s) s + O(β 3 ) = -β 2 ψ 2 (s) s ′ + O(β 3 ) =

and nally, with the help of the (85), we obtain the boundary values of ∆(ω) at ω = µ ∈ R:

First, the formula (164

But then, taking into account the asymptotics (163), we have Re ∆(ω) < 0 at ω ∈ L, which automatically follows ind ∆ = 0. Thus, ind ∆ = 0, when β ≈ 0. On the other hand, similarly to the previous considerations one can show, Re ∆(µ) < 0 at all β ∈ (0, 1), and so

what nally implies the theorem statement. The theorem proved.

Remark. Since ∆(ω) ̸ = 0, then the unknowns h(ω) and V + (ω, ω) of a form (41) are analytic at Im ω > 0 and continues at Im ω 0, ω ̸ = 0. Now we express the functions ∆ h (ω), ∆ V (ω) of the form (42) through the functions H j (ω), j = 0, 1. The formulas (42) and ( 161) have already given the representation for ∆(ω) (162) and also give the representations

At last, we establish the asymptotics of ∆(ω), ∆ h (ω), ∆ V (ω) at ω → 0. The statements of the theorem 5, formulas (106) and (107), evidently imply

where

s) -sinh(s)/s s 2 (ν cosh(s) -sinh(s)/s) ds.


S F A Batista

Ludovic Leclercq
email: ludovic.leclercq@entpe.fr

Jean Krug
email: jean.krug@entpe.fr

Nikolas Geroliminis
email: nikolas.geroliminis@epfl.ch

Trip length estimation for the macroscopic traffic simulation: scaling microscopic into macroscopic networks

Keywords: Macroscopic Trip Lengths, Accumulation-based Macroscopic Simulation, Macroscopic Fundamental Diagram, Congestion, MFD Dynamics

. However, no study focus on the definition of the macroscopic trip lengths based on the microscopic network.

In this paper, we discuss three methods to calculate the macroscopic trip lengths based on the aggregation of microscopic trips, within each reservoir considering: (i) no further information;

(ii) the next reservoir to be traveled; and (iii) the related macro-path (i.e., the list of reservoirs that are crossed from the Origin to the Destination). Based on a static analysis of the network, we show that these methods: yield different average trip lengths at the reservoir level; and give similar average trip lengths at the macroscopic-path level, where the aggregation effect is on the standard deviation. We analyze the dependence of the trip lengths on the Origin-Destination matrix. We discuss a procedure to estimate the macroscopic trip lengths based on a new Origin-Destination matrix. We show that the estimated trip lengths show a good agreement with the ones calculated based on a new calculation of the microscopic trips, for the new Origin-Destination matrix. We analyze the impact of the macroscopic trip lengths on the traffic states, that are simulated using an accumulation-based simulator. We show that the trip lengths have a strong impact on the traffic states. And, the trip lengths are congestion-dependent when calculated at the reservoir level.

INTRODUCTION

The first ideas of macroscopic traffic simulation based on the decomposition of a city into reservoirs were introduced by Godfrey [START_REF] Godfrey | The mechanism of a road network[END_REF]. But, it really started attracting more attention from the scientific community after the seminal works of Daganzo [START_REF] Daganzo | Urban gridlock: Macroscopic modeling and mitigation approaches[END_REF] and Geroliminis and Daganzo [START_REF] Geroliminis | Existence of urban-scale macroscopic fundamental diagrams: Some experimental findings[END_REF].

Each reservoir is defined by a well-defined relation between mean flow and accumulation, the Macroscopic Fundamental Diagram (MFD). Based on Yokohama (Japan) traffic data, Geroliminis and Daganzo [START_REF] Geroliminis | Existence of urban-scale macroscopic fundamental diagrams: Some experimental findings[END_REF] provided ground truth evidence of the MFD existence. This has been confirmed by further studies: e.g., Buisson and Ladier, Ambühl and Menendez, Derrman et al., Lodera et al. [START_REF] Buisson | Exploring the Impact of Homogeneity of Traffic Measurements on the Existence of Macroscopic Fundamental Diagrams[END_REF][START_REF] Ambühl | Data fusion algorithm for macroscopic fundamental diagram estimation[END_REF][START_REF] Derrman | Towards Estimating Urban Macroscopic Fundamental Diagrams From Mobile Phone Signaling Data: A Simulation Study[END_REF][START_REF] Lodera | Data fusion algorithm for macroscopic fundamental diagram estimation[END_REF]. The applications of the MFD in simulation has mostly been for now for testing different control algorithms: Haddad and Geroliminis [START_REF] Haddad | On the stability of traffic perimeter control in two-region urban cities[END_REF]; Ekbatani et al. [START_REF] Ekbatani | Urban congestion gating control based on reduced operational network fundamental diagrams[END_REF]; Ramezani et al. [START_REF] Ramezani | Dynamics of heterogeneity in urban networks: aggregated traffic modeling and hierarchical control[END_REF]; and Haddad [START_REF] Haddad | Optimal perimeter control synthesis for two urban regions with aggregate boundary queue dynamics[END_REF], for some examples.

The mathematical formulation of the MFD was introduced by Daganzo (1), for a single reservoir. The traffic dynamics is governed by a state equation that relates the vehicle's accumulation (n(t)) with the balance between the inflow (f(t)) and outflow (g(t)). Depending on the assumption made on g(t), one can distinguish two models in the literature: the accumulation-based model (Daganzo,Geroliminis and Daganzo,[START_REF] Daganzo | Urban gridlock: Macroscopic modeling and mitigation approaches[END_REF][START_REF] Geroliminis | Existence of urban-scale macroscopic fundamental diagrams: Some experimental findings[END_REF]; and the trip-based model (Arnott,[START_REF] Arnott | A bathtub model of downtown traffic congestion[END_REF]. One main difference between both models lies in the definition of the trip lengths for a macroscopic path (macro-path). A macro-path is defined as a sequence of reservoirs in a multi-region network level approach.

The first implementation of the accumulation-based model assumes a constant mean trip length ( L) for all vehicles traveling inside the same reservoir (Daganzo,[START_REF] Daganzo | Urban gridlock: Macroscopic modeling and mitigation approaches[END_REF]. This assumption might not be realistic in cases where the mean trip lengths are highly influenced by origins and destinations (Leclercq et al.,[START_REF] Leclercq | Macroscopic traffic dynamics with heterogeneous route patterns[END_REF]. While, empirical results in Yokohama showed that this might be a reasonable assumption for a range of cases. Yildirimoglu and Geroliminis (3) proposes different average trip lengths for the different macro-paths crossing the same reservoir. The same work also proposes to split the conservation equation into different classes of vehicles that share a common travel distance. Ramezani et al. [START_REF] Ramezani | Dynamics of heterogeneity in urban networks: aggregated traffic modeling and hierarchical control[END_REF] proposes a more refined approach, where the trip lengths are dynamically calculated. The trip lengths are calculated for all vehicles traveling on reservoir i and going to a common adjacent reservoir j, depending on the exchange flows and accumulation. An accumulation-based model considers that the network outflow depends on the average trip length (without specifying in details which type of average is taken).

A trip-based model (Arnott, Fosgerau, Lamotte and Geroliminis, Leclercq et al., Mariotte et al.,[START_REF] Arnott | A bathtub model of downtown traffic congestion[END_REF][START_REF] Fosgerau | Congestion in the bathtub[END_REF][START_REF] Lamotte | The morning commute in urban areas: Insights from theory and simulation[END_REF][START_REF] Leclercq | Dynamic macroscopic simulation of on-street parking search: A trip-based approach[END_REF][START_REF] Mariotte | A trip-based multi-reservoir MFD simulation framework[END_REF] emphasizes that the outflow depends on the rate at which vehicles complete a generating trip length in the beginning of the trip and keeps track of the distance traveled for each vehicle. Recent work (Lamotte and Geroliminis, Mariotte et al., [START_REF] Lamotte | The morning commute in urban areas: Insights from theory and simulation[END_REF][START_REF] Mariotte | A trip-based multi-reservoir MFD simulation framework[END_REF] showed that hysteresis loops might be generated in the outflow MFD between the onset and offset of congestion, even if the production MFD has no hysteresis.

Both accumulation-and trip-based approaches thus require a proper estimation of trip lengths within reservoirs. On the other hand, a definition of the trip lengths at the macro-path level is required for the first traffic assignment iteration. Thus, dedicated methods able to scale up trips from the real (microscopic) network to the reservoirs (macroscopic) network should be designed and validated. These methods should be able to easily calculate trip lengths at the reservoir and at the macro-path levels.

In this paper, we propose to investigate three methods to calculate macroscopic trip lengths.

The idea is to use the microscopic network to generate individual trips and then aggregate them TRB 2018 Annual Meeting Original paper submittal within the reservoirs considering: (i) no further information; (ii) the next reservoir to be traveled (i.e., the adjacent destination reservoir); and (iii) the related macro-path. These methods allow to obtain a trip length distribution at the reservoir level. The calculation of the trip lengths at the macro-path level is directly obtained by averaging the means of the trip length distribution over the sequence of reservoirs crossed. We note that these methods can be directly applied to both the accumulation-and trip-based models. We test these aggregation methods on the 6 th Lyon district network (France), that is divided into eight reservoirs. We analyze the differences in the distributions of trip lengths obtained through the three methods. We then discuss an approach to estimate the macroscopic trip lengths, considering the Origin-Destination (OD 1 ) matrix between reservoirs without requiring the re-sampling of micro-trips. Finally, we test the three calculation methods by performing an accumulation-based dynamic simulations, considering the Wardrop equilibrium on the macro-paths. Finally, we show that the calculation methods highly influences the simulation results and the network equilibrium. We show that trip lengths are sensible to congestion at the reservoir level. 1 Capital letters refer to macroscopic OD's, while lower case letters refer to microscopic od's.

TRB 2018 Annual Meeting Original paper submittal

THEORETICAL BACKGROUND ON THE ACCUMULATION-BASED MACROSCOPIC

SIMULATION

The traffic dynamics inside a single reservoir is governed by the evolution of the accumulation (n(t)):

dn dt = f (t) -g(t) (1)
where f(t) is the inflow function and g(t) is the outflow function.

Daganzo [START_REF] Daganzo | Urban gridlock: Macroscopic modeling and mitigation approaches[END_REF] proposes that the outflow function (g(t)) depends on n(t); and is proportional to the average trip length L:

g(t) ≈ P(n(t)) L ( 2
)
where P(n(t)) is the travel production that depends on the accumulation n(t).

Daganzo [START_REF] Daganzo | Urban gridlock: Macroscopic modeling and mitigation approaches[END_REF] proposes to consider an average trip length L for all vehicles traveling on the same reservoir. Some authors also consider an average trip length at the reservoir level and independent of the OD, for: a macroscopic simulator set-up (Daganzo, Ramezani et

L i (t) = P(n i (t)) q i (t) (3)
L i j (t) = n i j (t) n i (t) P(n i (t)) q i j (t) (4)
where Eq. 3 is valid for internal trips (or internal trip endings) to reservoir i; and Eq. 4 is valid for the trips going to the adjacent reservoirs. L i (t) is the internal trip length for reservoir i. q i (t) is the internal flow on reservoir i. n i j (t) is the accumulation that goes from reservoir i to the adjacent reservoir j. q i j (t) is the transfer flow between reservoirs i and j.

TRB 2018 Annual Meeting Original paper submittal

TRIP LENGTH DISTRIBUTIONS: METHODOLOGICAL FRAMEWORK

In this section, we describe the proposed methods to calculate the trip lengths. We start from the microscopic network and perform a uniform sampling of N trips od pairs. For each od pair, we calculate effective micro-trips using the Dijkstra algorithm. Micro-trips represent shortest-paths between the od pairs. Let Γ be the set of the calculated micro-trips. We describe now three methods to aggregate micro-trips and calculate macroscopic trip lengths. We present the mathematical formulation of these methods at: (i) the reservoir level that is crucial for the macroscopic simulation; and (ii) the macro-path level that is important for the first assignment iteration.

At the reservoir level, these methods are defined as:

1. Method 1: no information. All micro-trips that travel on the i-th reservoir are aggregated to calculate the average trip length Li of reservoir i:

Li = ∑ k l k i ∑ k δ ik , ∀k ∈ Γ (5)
where l k i is the portion of micro-trip k that belongs to reservoir i; and δ ik is a dummy variable that equals 1 if micro-trip k travels on reservoir i.

The set of trip lengths of reservoir i is:

L i = {l k i }, ∀k ∈ Γ (6)
2. Method 2: next destination adjacent reservoir. All micro-trips traveling on the i-th reservoir and going to the j-th adjacent reservoir are aggregated to define the average trip length Li j to go from reservoir i to j:

Li j = ∑ k δ i jk l k i ∑ k δ i jk , ∀k ∈ Γ ∧ ∀ j ∈ Λ (7)
where δ i jk is a dummy variable that equals 1 if micro-trip k travels on reservoir i and goes to reservoir j; and Λ is the set of adjacent reservoirs to i.

The set of trip lengths of reservoir i that goes to j is:

L i j = {δ i jk l k i }, ∀k ∈ Γ (8)
3. Method 3: related macro-path p. All micro-trips that travel on reservoir i and both belong to macro-path p are aggregated to define the average trip length of p on reservoir i ( Lp i ):

Lp i = ∑ k δ p ik l k i ∑ k δ p ik , ∀k ∈ Γ (9)
where δ p ik is a dummy variable that equals 1 if micro-trip k travels on reservoir i and both define macro-path p.

TRB 2018 Annual Meeting Original paper submittal

The set of trip lengths for macro-path p on reservoir i is:

L p i = {δ p ik l k i }, ∀k ∈ Γ (10)
Consider that macro-path p is defined as the following sequence of reservoirs:

p = (p 1 , . . . p m , . . . , p R ), ∀m = 1, . . . , R (11)
where R is the number of reservoirs that define p.

The average trip lengths at the macro-path level (L p ) are directly calculated from the previous definitions at the reservoir level:

1. Method 1:

Lp = p R ∑ m=p 1 Lm ( 12
)
2. Method 2:

Lp = p R ∑ m=p 1 p R-1 ∑ l=p m+1 L ml + Lp R p R ( 13
)
3. Method 3:

Lp = p R ∑ m=p 1 Lp m ( 14
)
At the macro-path level, the variances of L p are calculated through the convolution of the variances of the set of trip lengths of the R reservoirs that define p. This means that we assume independent trip length distributions at the reservoir level.

Note that, the set of the trip lengths previously defined at the reservoir level strongly depend on the algorithm used to calculate the micro-trips. For now, we focus on shortest-paths in distance over the microscopic network. More refined approaches like Frejinger et al. [START_REF] Frejinger | Sampling of alternatives for route choice modeling[END_REF] and Flotterod and Bierlaire (21) will be investigated in future research.

There are also other factors that may play an important role. Leclercq et al. [START_REF] Leclercq | Macroscopic traffic dynamics with heterogeneous route patterns[END_REF] shows that (i) traffic conditions and (ii) od matrix influence the trip lengths. The (iii) value of N trips and (iv)

the definition of a macro-path are other important factors. Points (iii) and (iv) will be subject to future research, while points (i) and (ii) will be investigated in this study.

TRB 2018 Annual Meeting Original paper submittal

STATIC ANALYSIS OF THE MACROSCOPIC TRIP LENGTHS Network definition

To test the proposed methods to calculate the trip length distributions, we consider the Lyon 6 th district network shown in Fig. 1. This network has 757 links, 431 nodes and arbitrarily divided into 8 reservoirs.

x The MFD functions (Fig. 1) are determined using a microscopic simulator called Symuvia (Leclercq,[START_REF] Leclercq | Hybrid approaches to the solutions of the "Lighthill-Whitham-Richards" model[END_REF]. We consider a larger network that incorporates the 3 rd and 6 th districts of Lyon and the neighbor city of Villeurbarnne. This network covers a surface of 40 square kilometers. We consider two types of demand: a static demand that consists of ∼ 800.000 vehicles and is estimated using the model of Cabrera Delgado and Bonnel [START_REF] Cabrera Delgado | Level of aggregation of zoning and temporal transferability of the gravity distribution model: The case of Lyon[END_REF]; and a dynamic demand that is obtained by TRB 2018 Annual Meeting Original paper submittal combining daily demand and loop detectors measurements. We consider MFD sensors placed on the larger network and the data is aggregated every 10 minutes. We assume a bi-parabolic shape to fit the simulated data.

We sample N trips = 10000 od pairs and calculate the micro-trips. We exclude the microtrips that cross more than one time the same reservoir and we are left with ∼ 9200 micro-trips, that define 205 macro-paths. We hereafter refer to this set of macro-paths as Ψ and the corresponding set of micro-trips as Ξ. The aggregation of the od pairs into the OD matrix is listed in Table 1.

We refer to this OD matrix as M. In this static analysis, we also have interest to investigate the impact of the OD matrix on the trip lengths. For this, we consider a second arbitrary sampling of N trips = 10000 od pairs, where there are more od pairs samples on the origin reservoir

Analysis of the trip length distributions

We first consider the aggregation of the micro-trips by reservoir. In Fig. 2, we show the average trip length for the aggregation methods 1 and 2. For method 3, we show the histogram of trip lengths of all macro-paths that cross a reservoir. As observed in Fig. 2, it is clear that the destination and the specific macro-paths matter as the trip lengths are different than the mean of method 1, when we segment the micro-trips at the reservoir level. In fact, as shown by Leclercq et al. [START_REF] Leclercq | Macroscopic traffic dynamics with heterogeneous route patterns[END_REF], the hypothesis of method 1 is clearly a strong restriction for MFD simulation. This will be investigated in the next section.

We now investigate the impact of the three methods at the macro-path level. For this, we consider all possible OD pairs and the most sampled macro-path for each OD (i.e., the macro-paths that have the largest number of aggregated micro-trips, at the macro-path level). We show that the impact of the different aggregation methods is on the macroscopic trip length standard deviations and not on the averages. The averages are approximately similar for the three methods and the different OD's. But there are some exceptions, such as for the following OD's: 2-8; 3-8; 5-8;

8-1; 3-1; 4-8; 7-8; or 8-4. As is it shown in Fig. 3, method 1 shows the largest standard deviation of the distributions, comparing the three methods. And, as expected the standard deviation is reduced from method 1 to 2 and to 3. Method 1 has the largest standard deviation because we are: aggregating a larger heterogeneity of micro-trips; and, assuming that the distributions of trip lengths are independent for different reservoirs, which is clearly not the case.

Impact of the OD matrix on the trip lengths

To analyze the impact of the OD matrix on the trip lengths, we calculate the associated micro-trips to M * . We investigate this impact on methods 1 and 2. The micro-trips are aggregated at the reservoir level and the average values that are calculated are listed in Table 2, for methods 1 and 2. As observed, method 1 is impacted, as shown for the cases of reservoirs 1, 2, 6 and 8. And, for method 2, the impact is observed for the cases of reservoirs: 1 to 2; 2 to 4; 2 to 5; 3 to 5; 3 to 7; 4 to 2; 4 to 6; 6 to 4; 6 to 8; and 7 to 6. Indeed, for both methods, the differences are of the order of 10 to 50 meters. But, the study network is also small. And, these differences will indeed be more significant in larger networks.

We discuss now a procedure to estimate the trip lengths based on the original calculation of micro-trips and M * . One solution is to re-calculate the micro-trips for M * , but this is time consuming. However for comparison, we do so for M * .

The trip lengths for method 1, are estimated as:

Lk = 1 N * k ∑ i ∑ j α * i j Lk i j , ∀(i, j) ∈ Λ ∧ i ̸ = j ( 15
)
where Lk is the estimated trip length for reservoir k; N * k is the total number of micro-trips of M * that cross reservoir k; α * i j is the number of micro-trips of M * that are crossing reservoir k and coming from reservoir i and going to j; Lk i j is the average trip length calculated from the aggregation of all micro-trips of M that cross reservoir k. Such calculations are simple and allow to update trip lengths to dynamic variations of the OD matrix.

We aim to show based on Eq. 15 that: Lk ≈ Lk . We estimate average trip length for all reservoirs based on M and M * . The results are listed in For method 2, the trip lengths are estimated as:

Lk = 1 N * k R ∑ i α * id Lk id , ∀i ∈ Λ ∧ i ̸ = d ( 16
)
where α * id is the number of micro-trips crossing reservoir k and going to the destination adjacent reservoir d, calculated based on the new sampling; R is the number of adjacent reservoirs to k; and Lk id is the average trip length calculated through method 2, considering M.

To test the estimation of average trip lengths for method 2, using Eq. 16, we consider: the OD matrices M and M * ; and the micro-trips that are sampled based on M * . We aggregate these micro-trips according to method 2, to estimate the average trip lengths. The results are listed in Table 2. We also check that Lk = Lk , by setting M = M * . The estimated average trip lengths ( L)

show a good agreement with the ones ( L) calculated directly from the aggregation of the micro-trips through method 2.

TRB 2018 Annual Meeting Original paper submittal

Simulation settings

For the accumulation-based macroscopic simulation, we consider the network and MFD functions shown in Fig. 1. A total simulation period of T = 10000 seconds is considered, discretized into intervals of δt = 10 seconds.

We consider three OD pairs: 1 to 6; 8 to 4; and 3 to 4. For each OD pair, we consider a maximum of three macro-paths. A macro-path is given by the aggregation of the micro-trips by the set of crossed reservoirs. We stress out, that a complete investigation of the macro-path definition based on the aggregation of micro-trips will be subject to a future research. For the three OD pairs, we have seven macro-paths. These macro-paths and corresponding average trip lengths calculated at the reservoir level using each method, are listed in Table 3. We consider the following demand levels. For OD 1-6, we consider the following demand levels: 0. The traffic flow assignment is performed using the Wardrop principle. That is, we assume that users are perfectly rational and utility minimizers. To solve the Stochastic User Equilibrium fixed point problem, we use the Method of Successive Averages (MSA). The convergence criterium of the MSA is defined by the root mean square error (RMSE):

RMSE = ∑ O ∑ D ∑ k (Q j+1 k -Q j k ) 2 N ≤ tol (17)
where

Q j+1 k
and Q j k are the new and old flows of macro-path k; N is the number of macro-paths for each OD; and tol is the convergence criterium. We set tol = 0.01.

Does trip length definition affect the simulated traffic states?

We investigate the influence of the three methods used to calculate the trip lengths, on the traffic states. The evolution of the accumulation and mean speed inside each reservoir is shown in Fig. 4.

We observe a clear impact on the congestion pattern and duration. For methods 1 and 2, the congestion lasts between 2000 and 5000 seconds of the simulation period. Whilst, for method 3, it lasts until 8000 seconds. Indeed, different trip lengths at the reservoir level means completely different traffic states. Method 3 has a larger heterogeneity of trip lengths at the reservoir level than methods 1 and 2 (see Table 3). And, the succession of trip lengths for the same macro-path is different for the three methods. Consider, for example, the macro-path 3-2-4. For this macro-path, the trip length for reservoir 2 is: 684 meters, according to method 3; and 363 meters according to method 1. A larger trip length means a potential bottleneck (i.e., a reduction on the inflow capacity for this reservoir and specific macro-path) of this macro-path, considering method 3.

Then, the congestion patterns are different for methods 1 and 3. One can also observe the evolution of the accumulation inside reservoir 2, for methods 1 and 3. The peak of the accumulation, is much larger for method 3, when compared to method 1. This is also explained by the larger heterogeneity of average trip lengths given by method 3 (see Table 3). The impact of considering a larger heterogeneity of trip lengths at the reservoir level is also discussed by 3 : Average trip lengths (m) at the reservoir level, for the seven macro-paths considered for the accumulation-based simulation. Top: Method 1. The number of micro-trips that define each macro-path is also listed. Middle: Method 2. Bottom: Method 3.

Geroliminis [START_REF] Yildirimoglu | Approximating dynamic equilibrium conditions with macroscopic fundamental diagrams[END_REF]. For now, the trip lengths were calculated based on shortest-paths in distance on the microscopic network, without considering travel times and thus traffic states. Hereby, we investigate the dependence of the trip lengths on the traffic states. For this purpose, we calculate time-dependent micro-trips, based on the mean speed given by the MFD dynamics. We consider the evolution of the mean speed, for Method 3 (Fig. 4), between 2000 and 8000 seconds. Based on this, we estimate an average mean speed for each reservoir. Considering the latter and the link lengths of the microscopic network, we calculate time-dependent micro-trips based on the same microscopic od pairs of the micro-trips listed in Ξ. We recover the microscopic lengths of these micro-trips and aggregate them according to the three methods the reservoir level. We then compare these results against the static trip lengths, calculated based on Ξ. For the comparison at the reservoir level, we define a parameter α that defines the absolute difference between the dynamic (L d ) and static (L s ) average trip lengths, as:

R 1 R 2 R 3 R 4 R 5 R 6 R 7
α = L d -L s ( 18
)
The static (L s ) and dynamic (L d ) average trip lengths and corresponding standard deviations are listed on Table 2, for methods 1 and 2. The absolute differences defined by α for the three methods, at the reservoir level, is shown in Fig. 5. As a reference for method 3, we consider the set of macro-paths Ψ. The histograms represent the relative differences of trip lengths, for the same macro-path, defined at the reservoir level.

As observed in Table 2, the average trip lengths at the reservoir level are affected by ∼ 100 meters for method 1 and 2. There are a few exceptions. Consider as an example, the case of method 2 for micro-trips going from reservoir 1 to 3. At the static level, there are only 13 microtrips that are aggregated, while for the dynamic one there are 635 micro-trips. In fact, performing the static sampling, micro-trips that departure from nodes close to the border between reservoirs 1 and 3 are more probable to cross to reservoir 3 rather than 2. However, when reservoir 2 is congested, the link costs increase and consequently micro-trips that begin in nodes that are more far away from the border between reservoirs 1 and 3, will more probably cross it. On the other hand, the static micro-trips are not sensible to the reservoir definition. While in the dynamic case, we are considering the mean speed calculated through the MFD simulation and this induces also a dependence of the macroscopic network. Thus, the number of micro-trips that are aggregated to calculate the dynamic trip lengths is different. Method 3 is the more sensible to the number of aggregated micro-trips (Fig. 5). For this consider, as example, macro-path 1-3. The set of corresponding average trip lengths (in meters) are: 47 (reservoir 1) and 434 (reservoir 3) meters with 6 micro-trips associated, for the static case; and, 535 (reservoir 1) and 471 (reservoir 3) meters with 116 micro-trips associated, for the dynamic case. The latter also highlights the reasons previously mentioned. And, for this particular case, the difference between the static and dynamic cases are also due to the special geometry of the macroscopic network. Note that, there are only two connecting nodes between reservoirs 1 and 3 (Fig. 1).

TRB 2018 Annual Meeting Original paper submittal

CONCLUSIONS

In this paper, we discuss three methods to calculate macroscopic trip lengths at both the reservoir and macro-path levels. These methods are based on the aggregation of micro-trips. We show that the different methods give different average trip lengths at the reservoir level. And, this has a significant impact on the macroscopic simulation. A trip-based simulation considering variable versus average trip lengths inside the same reservoir will be presented in the extended version of the paper, to put in evidence the importance of more refined definitions of trip lengths for the macroscopic simulation. Different trip lengths at the reservoir level yield completely different traffic dynamics. At the macro-path level, the impact of the different methods is on the standard deviation and not on the average of the trip lengths. We show that the trip lengths are congestiondependent at the reservoir level. Method 3 is the more sensible to congestion, at the reservoir level.

We also show the dependence of the trip lengths on the OD matrix. And, we discuss a procedure to quickly estimate the trip lengths based on a new definition of the OD matrix, without the need to recalculate the micro-trips. This is very useful when OD matrix is changing with time. We show that this procedure gives coherent average trip lengths for methods 1 and 2, compared to when we recalculate the micro-trips.

As future research directions, we plan to investigate the dependence of the trip lengths on:

the number of micro-trips N trips ; and on the algorithm used to calculate the micro-trips. A groundtruth validation of these methods is also foreseen. We also plan to further tackle the question of the macro-path definition, based on the micro-trips aggregation.

TRB 2018 Annual Meeting Original paper submittal

8 FIGURE 3 :

 83 FIGURE 3 : Average and standard deviations for the most sampled macro-paths for each OD of the network. The results are calculated at the macro-path level. Each subplot correspond to an origin reservoir. The vertical dashed lines separate the different destination reservoirs.

1 3 time

 3 Despite at the macro-path level, the three methods yield similar average trip lengths, the 2 succession of trip lengths for the macro-paths calculated by the three methods have a clear impact 3 on the traffic states inside each reservoir.[START_REF] Ramezani | Dynamics of heterogeneity in urban networks: aggregated traffic modeling and hierarchical control[END_REF] In order to highlight which method is more realistic, a comparison with a trip-based simula-5 TRB 2018 Annual MeetingOriginal paper submittal tion, where vehicles follow the same trip sequence with the estimated micro-trips, will be discussed 1 in the extended version of this paper. This aims to put in evidence the importance of variable trip 2 lengths versus average trip lengths assumption, for different trips inside the same reservoir.

3 FIGURE 4 :

 34 FIGURE 4 : Evolution of n(t) and v(t) during the simulation period, considering the three methods to calculate the trip lengths.

,

FIGURE 5 :

 5 FIGURE 5 : Absolute difference α (km) at the reservoir level, between the trip lengths calculated based on the time-dependent micro-trips and the micro-trips listed on Ξ.

Table 1 .

 1 1 and destination reservoirs 4, 6 and 8. This means, we are considering more micro-trips that go from the north to the south of the network (i.e., from reservoir 1 to 4, 6 and 8). For each of the previous OD, we sample 600 od pairs, out of the 10000. The remaining od pairs are uniformly sampled on the microscopic network. We refer to this second OD matrix as M * and the results are listed in We consider M * to investigate the influence of the OD matrix on the trip lengths; and to introduce a procedure to calibrate the trip lengths based on the OD matrices information, without the need to recalculate the micro-trips.

				Destination reservoir
		1	2	3	4	5	6	7	8
		1 171 204 125 141 96 147 120 163
	Origin Reservoir	2 210 294 202 189 149 214 190 270 3 196 245 121 141 87 103 115 174 4 146 214 113 110 87 128 111 170 5 87 160 66 88 67 109 92 126 6 134 154 109 79 82 80 129 147 7 124 201 120 88 89 72 106 139
		8 211 227 161 156 85 118 181 187
				Destination reservoir
		1	2	3	4	5	6	7	8
		1 171 204 125 141 96 147 120 163
	Origin Reservoir	2 145 250 85 589 125 589 133 572 3 199 276 175 207 110 249 174 333 4 144 181 72 110 77 108 87 148 5 86 102 49 77 48 83 76 119 6 105 136 78 66 56 52 100 116 7 141 122 79 77 67 48 93 149
		8 181 202 140 113 71 130 123 135

TABLE 1 :

 1 Number of trips between each macroscopic origin and destination reservoirs, for the 6

th Lyon district network. Top: Matrix M. Bottom: Matrix M * . TRB 2018 Annual Meeting Original paper submittal

Table 2

 2

		40		Reservoir 1			40		Reservoir 2
	Counts	20 30				M 3 M 1 M 2 : 1 to 2 M 2 : 1 to 3 M 2 : 1 to 1	Counts	20 30			M 3 M 1 M 2 : 2 to 1 M 2 : 2 to 3 M 2 : 2 to 4 M 2 : 2 to 5 M 2 : 2 to 2
		10						10
		0						0
		0	200	400	600	800 1000		0	200	400	600	800 1000
				L [m]					L [m]
		40		Reservoir 3			40		Reservoir 4
	Counts	20 30				M 3 M 1 M 2 : 3 to 1 M 2 : 3 to 2 M 2 : 3 to 5 M 2 : 3 to 7 M 2 : 3 to 3	Counts	30 20			M 3 M 1 M 2 : 4 to 2 M 2 : 4 to 5 M 2 : 4 to 6 M 2 : 4 to 4
		10						10
		0						0
		0	200	400	600	800 1000		0	200	400	600	800 1000
				L [m]					L [m]
		40		Reservoir 5			40		Reservoir 6
	Counts	20 30				M 3 M 1 M 2 : 5 to 2 M 2 : 5 to 4 M 2 : 5 to 6 M 2 : 5 to 7 M 2 : 5 to 5	Counts	20 30			M 3 M 1 M 2 : 6 to 4 M 2 : 6 to 5 M 2 : 6 to 7 M 2 : 6 to 8 M 2 : 6 to 6
		10						10
		0						0
		0	200	400	600	800 1000		0	200	400	600	800 1000
				L [m]					L [m]
		40		Reservoir 7			40		Reservoir 8
	Counts	20 30				M 3 M 1 M 2 : 7 to 3 M 2 : 7 to 5 M 2 : 7 to 6 M 2 : 7 to 8 M 2 : 7 to 7	Counts	30 20			M 3 M 1 M 2 : 8 to 6 M 2 : 8 to 7 M 2 : 8 to 8
		10						10
		0						0
		0	200	400	600	800 1000		0	200	400	600	800 1000
				L [m]					L [m]

. For a first test, we check that Lk = Lk , by setting M = M * . The second test is based on M * . As a reference for comparison, TRB 2018 Annual Meeting Original paper submittal

FIGURE 2 : Average trip lengths for the three methods at the reservoir level.

we sample micro-trips based on M * and calculated the average trip lengths using method 1. The 1 results are listed on the fourth column of Table 2. As it can be observed, the estimated average trip 2 TRB 2018 Annual Meeting Original paper submittal lengths ( L) show a very good agreement with the average trip length ( L) calculated directly from the micro-trips aggregation. The largest absolute difference is 10 meters for reservoir 4.

TABLE 2 :

 2

	TRB 2018 Annual Meeting

Average ( L) and estimated ( L) (m) trip lengths (at the reservoir level), for each reservoir and considering the two trip matrices: M; and M * . Average ( L) and standard deviation (σ L ) of the trip lengths calculated based on the time-dependent micro-trips are also listed. Top: Method 1. Bottom: Method 2. TRB 2018 Annual Meeting Original paper submittal TRIP LENGTH DISTRIBUTIONS AND THE MACROSCOPIC SIMULATION

  3 (veh/s) for δt ∈ [0, 1000]; 1.1 (veh/s) for δt ∈ [1000, 3000]; and 0.2 (veh/s) for δt ∈ [3000, 10000]. For OF 8-4, we consider the following demand levels: 0.2 (veh/s) for δt ∈ [0, 2000]; 0.9 (veh/s) for δt ∈ [2000, 4000]; and 0.3 (veh/s) for δt ∈ [4000, 10000]. For OD 3-4, we consider the following demand levels: 0.2 (veh/s)

δt ∈ [0, 3000]; 0.7 (veh/s) for δt ∈ [3000, 5000]; and 0.1 (veh/s) for δt ∈ [5000, 10000].

ACKNOWLEDGEMENTS

This project is supported by the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation programme (grant agreement No 646592 -MAGnUM project). S. F. A. Batista also acknowledges funding support by the region Auvergne-Rhône-Alpes (ARC7 Research Program) and from the University of Lyon supported by the PALSE mobility program. TRB 2018 Annual Meeting Original paper submittal