POLONICI MATHEMATICI LXXIV (2000)
On 1-regular ordinary differential operators
by Grzegorz Lysik (Warszawa)
To the memory of Bogdan Ziemian , my great friend and adviser
Abstract. Solutions to singular linear ordinary differential equations with analytic coefficients are found in the form of Laplace type integrals.
Introduction. Let
(1) P
x, d
dx
= X
ni=0
a
i(x) d
idx
ibe a linear differential operator of order n ∈ N with coefficients a
i(x) = P
∞j=0
a
ijx
jconvergent for |x| < r, i = 0, . . . , n−1, and a
n(x) = x
mwith some m ∈ N
0. Let κ
Pbe the Katz invariant for P , i.e. the smallest κ ∈ R such that there are no points of N
Pbelow the line {(i, j) ∈ N
0×Z : j = κ(i−n)+m−n}
where
N
P= {(i, j) ∈ N
0× Z : a
ii+j6= 0}
is the Newton diagram for P . If κ
P≤ 0 then zero is a regular or regular singular point for P , and the well known Fuchs theorem states that the fundamental system of solutions of P u = 0 consists of convergent series of Taylor type, whose coefficients can be easily determined (cf. [CL]). On the other hand, in the case κ
P> 0, zero is an irregular singular point for P , and there exist power series solutions to P u = 0 but they need not be convergent.
During the last several years a special method called multisummability was worked out to deal with divergent solutions of differential equations. By this method, starting from a formal power series solution, one constructs a holomorphic solution in a sector in C g \ {0} having the formal one as its
2000 Mathematics Subject Classification: 34A20, 34A30 44A15.
Key words and phrases: singular differential equations, Laplace integrals, Mellin transformation.
[201]
asymptotic expansion (cf. [B], [E], [M]). Unfortunately, the method cannot be applied directly to the study of the Cauchy problem. We shall describe how the Cauchy problem can be treated by a method based on the Mellin transformation. We shall concentrate on the study of the Cauchy problem for the homogeneous equation P u = 0 where P is a 1-regular operator , i.e.
an operator with κ
P≤ 1. Observe that any operator P with κ
P> 0 can be reduced to a 1-regular operator e P by the change of variable e x = x
κP. The coefficients of e P are analytic functions in the variable e x
1/κP, but this should not cause any essential difficulties.
Our method of treatment of the Cauchy problem for P u = 0 with the Cauchy data at a non-singular point 0 < t < r can be described as follows.
Firstly, we note that any 1-regular operator P given by (1) with a
n(x) = x
2ncan be written in the form
(2) P
x, d
dx
= Q
x
2d
dx
+
n−1
X
i=0
g
i(x)
x
2d
dx
i,
where Q is a polynomial of degree n and g
i, i = 0, . . . , n − 1, are functions analytic in the disc B(r) and vanishing at zero. Next, after the change of variable s(x) = exp{1/t − 1/x} the original Cauchy problem is transformed into the one for the equation R(s, sd/ds)w = 0 with the Cauchy data at 1, where
R
s, s d
ds
= Q
s d
ds
+
n−1
X
i=0
eg
i(s)
s d
ds
iand eg
i(s) = g
i((1/t − log s)
−1), i = 0, . . . , n − 1. The operator R has a regular singular point at zero and its coefficients eg
i, i = 0, . . . , n − 1, are generalized analytic functions, i.e. they can be represented in the form eg
i(s) =
T∞
0
ψ
i(α)s
αe
−α/tdα with some entire functions ψ
i, i = 0, . . . , n − 1, of exponential growth. Now, applying the Mellin transformation we obtain a convolution equation for the function G(z) :=
Tt
0
w(s)s
−z−1ds, Q(z)G(z) +
∞
\
0
A(α, z)G(z − α)e
−α/tdα = Φ(z), where A(α, z) = P
n−1i=0
ψ
i(α)(z − α)
iand Φ is a polynomial determined by the Cauchy data. We solve the convolution equation by the method of successive approximations. Its solution G is a holomorphic function on C \ S
mµ=1
(̺
µ+ R
+), where ̺
1, . . . , ̺
mare the roots of Q. Furthermore, assuming that arg(̺
ν−̺
µ) 6= 0 for any 1 ≤ ν < µ ≤ m, the jump of G across the half-line ̺
µ+ R
+is a Laplace ultradistribution S
µ, µ = 1, . . . , m, on the half-line. Finally, the solution to the Cauchy problem for Rw = 0 is given by w(s) = P
mµ=1
S
µ[s
·] and putting u(x) = w(s(x)) we get the solution to
the original Cauchy problem. A closer examination of the ultradistributions S
µallows representing the solution u in the form of Laplace integrals. This type of representation can be viewed as parallel to the one obtained by the multisummability method. The author believes that it can give a new insight into the Stokes phenomenon.
0. Notation. The open disc with centre at z
0∈ C and radius r > 0 is denoted by B(z
0; r) or simply by B(r) if z
0= 0.
By e B(r) (resp. e C ) we denote the universal covering space of the punc- tured disc B(r) \ {0} (resp. C \ {0}). A point z ∈ e B(r) is written as z = |z|e
i arg zwith 0 < |z| < r and arg z ∈ R.
For θ ∈ R we set l
θ= (0, e
i∞θ) = {z ∈ C \ {0} : arg z = θ}. If θ = 0 then l
θ= R
+.
By a left (resp. right) tubular neighbourhood of a ray ̺ + l
θ, ̺ ∈ C, θ ∈ R, we mean a set {z : dist(z, ̺ + l
θ) < b, θ < arg(z − ̺) < θ + π/2} with some b > 0 (resp. {z : dist(z, ̺ + l
θ) < b, θ − π/2 < arg(z − ̺) < θ}).
For ̺ ∈ C and θ
−< θ
+with θ
+− θ
−< 2π we set
S(̺; (θ
−, θ
+)) = {z ∈ C \ {̺} : θ
−< arg(z − ̺) < θ
+}.
If θ
+− θ
−≥ 2π the set S(̺; (θ
−, θ
+)) is interpreted as a subset of C g \ {̺}.
For θ ∈ R and ω ∈ R we set
Ω
ωθ= {z ∈ e C : cos θ log |z| − sin θ arg z < ω}.
For z ∈ C we put hzi = 1 + |z|.
1. Generalized analytic functions, the Laplace and Mellin trans- formations. To fit our purposes we slightly modify the theory of generalized analytic functions given in [Z], and the definitions of the Laplace and Mellin transformations. We do not give the proofs of the stated facts since the proofs follow the ones given in [Z], [ L2] and [ L3].
Fix ̺
1, . . . , ̺
m∈ C and θ ∈ R. Set Γ
θ= S
mµ=1
(̺
µ+ l
θ). For a ∈ Re
−iθand ω ∈ Re
−iθdefine
L
a(Γ
θ) = {ϕ ∈ C
∞(Γ
θ) : kϕk
a,h= sup
0≤α≤h
sup
y∈Γθ
|e
−ayD
αϕ(y)| < ∞ for any h ∈ N}, L
(ω)(Γ
θ) = lim −→
a<θω
L
a(Γ
θ),
where a <
θω means that ae
iθ< ωe
iθ. The dual space L
′(ω)(Γ
θ) of L
(ω)(Γ
θ) is
called the space of Laplace distributions on Γ
θ. Replacing the norms kϕk
a,hby
kϕk
(Ma,hp)= sup
α∈N0
sup
y∈Γθ
|e
−ayh
αD
αϕ(y)|
M
α,
where (M
p)
∞p=0is a sequence of positive numbers satisfying conditions (M.1), (M.2) and (M.3) of [K], we obtain the space of Laplace ultradistributions L
(M(ω)p)′(Γ
θ) (see also [ L2]).
Observe that the function Γ
θ∋ y 7→ exp
z(y) := e
yzbelongs to L
∗(ω)(Γ
θ) where ∗ = ∅ or (M
p) if and only if Re(e
iθz) < ωe
iθ. Thus, we can define the Laplace transform of S ∈ L
∗′(ω)(Γ
θ) by
LS(z) = S[exp
z] for Re(e
iθz) < ωe
iθ.
Note that LS(−1/x) is defined in the disc (2ωe
iθ)
−1B(e
iθ; 1) if ωe
iθ< 0, in the half-plane Re(xe
−iθ) > 0 if ω = 0, and outside (2ωe
iθ)
−1B(−e
iθ; 1) if 0 < ωe
iθ.
Analogously, the function Γ
θ∋ y 7→ ϕ
s(y) := s
ybelongs to L
∗(ω)(Γ
θ) iff s ∈ Ω
ωeθ iθ:= {s ∈ e C : cos θ log |s| − sin θ arg s < ωe
iθ}. So, we can define the Taylor transform of S ∈ L
∗′(ω)(Γ
θ) by
T S(s) = S[ϕ
s] for s ∈ Ω
ωeθ iθ.
We call the image of L
∗′(ω)(Γ
θ) under the Taylor transformation the space of generalized analytic functions determined by L
∗′(ω)(Γ
θ) and denote it by GAF(L
∗′(ω)(Γ
θ)). If (̺
ν+ l
θ) ∩ (̺
µ+ l
θ) = ∅ for 1 ≤ ν < µ ≤ m, we have a natural decomposition
(3) GAF(L
∗′(ω)(Γ
θ)) = M
m µ=1GAF(L
∗′(ω)(̺
µ+ l
θ)) = M
m µ=1s
̺µ· GAF(L
∗′(ω)(l
θ)).
The space GAF(L
∗′(ω)(l
θ)) can be characterized (cf. [ L2], Th. 6) as the set of w ∈ O(Ω
ωeθ iθ) such that for any a <
θω one can find k < ∞ such that
|w(s)| ≤
C(1 + |log s|)
kfor s ∈ Ω
θaeiθif ∗ = ∅, C exp{M (k|log s|)} for s ∈ Ω
θaeiθif ∗ = (M
p), where M is the associated function of the sequence (M
p) defined by
M (̺) = sup
p∈N0
log ̺
pM
0M
pfor ̺ > 0.
Fix t ∈ Ω
ωeθ iθand ̺ ∈ C. We define the Mellin transform of w ∈ GAF(L
∗′(ω)(̺ + l
θ)) by
(4) M
θtw(z) =
\
γtθ
w(s)s
−z−1ds,
where γ
tθ= {s ∈ e C : s = t exp{−e
−iθr}, 0 ≤ r < ∞} with the orien- tation reverse to that induced by the above parametrization. Then M
θtw is holomorphic on {Re((z − ̺)e
−iθ) < 0}. Since the integration curve γ
tθin (4) can be replaced by γ
tθ′for any |θ − θ
′| ≤ π/2 we conclude that M
θtw ∈ O(C \ (̺ + l
θ)). Furthermore (cf. [ L3], Th. 4, [Z], Th. 10.1), there exists C < ∞ such that for 0 < dist(z, ̺ + l
θ) ≤ 1,
(5) |M
θtw(z)| ≤
C|t
̺−z|(dist(z, ̺ + l
θ))
−Cif ∗ = ∅, C|t
̺−z| exp
M
∗C
dist(z, ̺ + l
θ)
if ∗ = (M
p), where M
∗is the growth function of the sequence (M
p) given by
(6) M
∗(̺) = sup
p∈N0
log ̺
pp!M
0M
pfor ̺ > 0.
Moreover, the boundary value of M
θtw, S = b(M
θtw), belongs to L
∗′(ω)(̺+l
θ) and w = (2πi)
−1T S.
Conversely (cf. [ L3], Th. 5), if G ∈ O(C \ (̺ + l
θ)) satisfies (5) (with G in place of M
θtw) for 0 < dist(z, ̺ + l
θ) ≤ 1 and |G(z)| ≤ C|t
̺−z|/hzi for dist(z, ̺ + l
θ) ≥ 1 then G = M
θtw with a unique w given by w = (2πi)
−1T b(G).
Analogously, using the decomposition (3), we define the Mellin transform of w ∈ GAF(L
∗′(ω)(Γ
θ)), which is a holomorphic function on C \ Γ
θand satisfies appropriate estimates.
The Mellin transformation has the following operational property, which makes it useful in the study of the Cauchy problem.
If w ∈ GAF(L
∗′(ω)(Γ
θ)) and t ∈ Ω
ωeθ iθthen for i ∈ N
0, (7) M
θts d ds
iw
(z) = z
iM
θtw(z) + W
i(z) for z ∈ C \ Γ
θ,
where W
iis a polynomial of degree ≤ i − 1 depending on w(t), . . . , w
(i−1)(t).
2. The main result. Let P be a differential operator (1) with coeffi- cients analytic in B(r), r > 0. Assume that P is 1-regular and a
n(x) = x
2n. Then for i = 0, . . . , n − 1, a
i(x) = P
∞j=2i
a
ijx
jfor |x| < r. Furthermore, it follows by Lemma 1.3 of Chapter 4 of [T] that P can be written in the form (2), where Q(z) = z
n+ P
n−1i=0
a
i2iz
iand g
i(x) = P
∞j=1
g
jix
jfor
|x| < r, i = 0, . . . , n − 1. Fix 0 < t < r and consider the Cauchy problem (8)
P u = 0,
u(t) = u
0, . . . , u
(n−1)(t) = u
n−1.
It is well known that the solution u of (8) is unique and it extends holomor- phically to a function on e B(r). Our aim is to represent the u in the form of Laplace type integrals. To formulate the main result denote by ̺
1, . . . , ̺
mthe roots of Q with multiplicities k
1, . . . , k
m, respectively. Define the set Θ
sof singular directions by
Θ
s= {θ ∈ R : θ mod(2π) = arg(̺
ν− ̺
µ) for some 1 ≤ ν 6= µ ≤ m}
(Θ
s= ∅ if m = 1). Choose θ 6∈ Θ
ssuch that t ∈ (r/2)B(e
iθ; 1) and denote by θ
−(resp. θ
+) the greatest (resp. smallest) singular direction less (resp.
greater) than θ (θ
±= ±∞ if Θ
s= ∅).
Main Theorem. Let P be a 1-regular operator (2). Fix 0 < t < r and retain the preceding notations. Then the unique solution u of the Cauchy problem (8) is given by
(9) u(x) =
X
m µ=1LS
µ(1/t − 1/x) for x ∈ (r/2)B(e
iθ; 1),
with a unique S
µ∈ L
∗′(ω)(̺
µ+ l
θ) (µ = 1, . . . , m), where ω = (cos(θ/t) − 1/r)e
−iθand
(10) ∗ =
∅ if k
µ= 1,
p!(p/log p)
p/(kµ−1)if k
µ> 1.
Furthermore, S
µ, µ = 1, . . . , m, restricted to ̺
µ+l
θextends holomorphically to a function Ψ
µ∈ O(S(̺
µ; (θ
−, θ
+))) such that for any r
′< r and θ
−< e θ
−< e θ
+< θ
+,
(11) |Ψ
µ(̺
µ+ γ)e
γ/t| ≤
C|γ|
−Cexp{|γ|/r
′} if k
µ= 1, C exp
C
|γ|
kµ−1log C
|γ| + |γ|
r
′if k
µ> 1, for e θ
−≤ arg γ ≤ e θ
+with some C < ∞.
Thus, for any θ
−< θ
′< θ
+, u can be written in the form (12) u(x) =
X
m µ=1e
−̺µ(1/t−1/x)reg
\
lθ′
Ψ
µ(̺
µ+ γ)e
γ/t−γ/xdγ
for x ∈ (r/2)B(e
iθ′; 1), where the regularization of the integral is distributional if k
µ= 1 and ultra- distributional of class p!(p/log p)
p/(kµ−1)if k
µ> 1.
Remark . We conjecture that Ψ
µis a multivalued holomorphic function
on C with the set of branching points {̺
1, . . . , ̺
µ}.
3. Auxiliary lemmas. In the proof of the main theorem we shall use the following lemmas.
Lemma 1. For ν ∈ N put I
ν(γ, z) =
\
Tν(γ)
dα
hz − α
1i . . . hz − α
1− . . . − α
νi for γ ∈ R
+, z ∈ C with T
ν(γ) = {α ∈ (R
+)
ν: α
1+ . . . + α
ν≤ γ}. Then
|I
ν(γ, z)| ≤ 2
νν! log
ν(1 + |γ|) for γ ∈ R
+, z ∈ C.
P r o o f. We can consider only the case z = x ∈ R
+. Let 0 < γ ≤ x. By induction we show that
I
ν(γ, x) = 1 ν! log
ν1 + x 1 + x − γ
,
which is bounded by
ν!1log
ν(1 + γ). In fact, I
1(γ, x) = log
1+x−γ1+xand for ν ≥ 2 we derive
I
ν(γ, x) =
γ
\
0
1 1 + x − α
1I
ν−1(γ − α
1, x − α
1) dα
1= 1
(ν − 1)!
γ
\
0
log
ν−1 1+x−α1+x−γ11 + x − α
1dα
1= 1 ν! log
ν1 + x 1 + x − γ
. Now let 0 < x ≤ γ. We observe that T
ν(γ) = S
νk=0
T
kν(γ) with T
kν(γ) = {α ∈ R
ν+: α
1≤ x, . . . , α
1+ . . . + α
ν−k≤ x, x ≤ α
1+ . . . + α
ν−k+1, α
1+ . . . + α
ν≤ γ}. Now for k ∈ {0, 1, . . . , ν} we compute
\
Tkν(γ)
1 1 + x − α
1. . . 1
1 + x − α
1− . . . − α
ν−k× 1
1 + α
1+ . . . + α
ν−k+1− x . . . 1
1 + α
1+ . . . + α
ν− x dα
=
\
Tν−k(x)
1 1 + x − α
1. . . 1
1 + x − α
1− . . . − α
ν−kdα
×
\
Tk(γ−x)
1 1 + β
1. . . 1
1 + β
1+ . . . + β
kdβ
= 1
(ν − k)! log
ν−k(1 + x) · 1
k! log
k(1 + γ − x).
So
I
ν(γ, x) = X
ν k=01
(ν − k)!k! log
ν−k(1 + x) · log
k(1 + γ − x)
= 1
ν! (log(1 + x) + log(1 + γ − x))
ν, which is bounded by
2ν!νlog
ν(1 + γ).
Lemma 2. Let |Ψ (γ)| ≤ Ce
|γ|/rfor γ ∈ e
iθ+ l
θwith r > 0. Then the integral
w
θ(s) =
\
eiθ+lθ
Ψ (γ)s
γe
−γdγ
converges on the set of s ∈ e C such that s/e ∈ Ω
−1/rθand u
θ(x) = w
θ(exp{1 − 1/x}) is defined in the disc (r/2)B(e
iθ; 1).
P r o o f. Indeed
w
θ(s) =
∞\
1
Ψ (te
iθ)(s/e)
teiθe
iθdt and the integral converges if
1/r + Re(e
iθlog(s/e)) = 1/r + cos θ log(|s|/e) − sin θ arg s < 0.
To prove the second statement observe that for s = exp{1 − 1/x} we have log(|s|/e) = − Re(1/x) = − Re x/|x|
2and arg s = − Im(1/x) = Im x/|x|
2. So, if s/e ∈ Ω
θ−1/rthen x satisfies (cos θ Re x + sin θ Im x)|x|
−2= Re(e
−iθx)|x|
−2> r
−1and hence x ∈ (r/2)B(e
iθ; 1).
Lemma 3. Let s > 0, M
0= M
1= 1 and M
p= p!(p/log p)
psfor p ∈ N, p ≥ 2. Then M
∗(̺) ∼ ̺
1/slog ̺ as ̺ → ∞.
P r o o f. By (8) we have, for ̺ > 1, M
∗(̺) = max(log ̺, sup
p∈N, p≥2
p(log ̺ + s log log p − s log p)).
To compute the supremum define
g(̺, x) = x(log ̺ + s log log x − s log x) for x > 1, ̺ > 0.
Since g
x′(̺, x) = log ̺ + s log log x − s log x + s/log x − s, for ̺ ≥ e
sthere exists a unique x(̺) ≥ e such that g
′x(̺, x(̺)) = 0. Put g(̺) = g(̺, x(̺)) for
̺ ≥ e
s. Then g(̺) = sx(̺)(1−1/log x(̺)) and so g(̺) ∼ x(̺) as ̺ → ∞. Put f (x) =
ex log x exp
− 1 log x
sand h(̺) = ̺
1/slog ̺
1/s.
Then for ̺ > e
s, e
−sf (h(̺)) ≤ ̺ ≤ 2
s−1f (h(2̺)). Since x(̺) = f
−1(̺) this
implies h(̺) ∼ x(̺) as ̺ → ∞. Finally, in a standard way (see [ L1]) we show
that M
∗(̺) ∼ g(̺) as ̺ → ∞.
4. Proof of the main theorem. Let P be given by (2), where Q is a polynomial of degree n and g
i(x) = P
∞j=1
g
jix
j, i = 0, . . . , n−1, are functions analytic in the disc B(r), r > 0. Consider the Cauchy problem (8). Putting, if necessary, x
′= x/t we can assume that t = 1 and r > 1. Observe that by the change of independent variable s(x) = exp{1 − 1/x}, (8) is transformed into
(13)
Q
s d
ds
w +
n−1
X
i=0
eg
i(s)
s d
ds
iw = 0, w(1) = w
0, . . . , w
(n−1)(1) = w
n−1,
where eg
i(s) = g
i((− log(s/e))
−1), i = 0, . . . , n−1, w(s) = u((− log(s/e))
−1), w
0= u
0, w
1= u
1, w
2= u
1+ u
2and so on.
Since lim
s→0eg
i(s) = 0 for i = 0, . . . , n − 1, we have obtained an equa- tion with a regular singular point at zero, but with coefficients which are generalized analytic functions of the form (cf. [Z], Th. 14.1)
(14) eg
i(s) =
\
lθ
ψ
i(α)s
αe
−αdα for s/e ∈ Ω
−1/rθ, θ ∈ R, where ψ
i(α) = P
∞j=1
g
jiα
j−1/(j − 1)! for α ∈ C is the Borel transform of g
i, which is an entire function satisfying |ψ
i(α)| ≤ C
r′exp{|α|/r
′} for any r
′< r (i = 0, . . . , n − 1).
The equations of this type were studied by Bogdan Ziemian in [Z]. Under suitable conditions he proved the existence of generalized analytic solutions with positive radii of convergence. However his theorem ([Z], Theorem 16.2) cannot be applied here without additional assumptions on the functions g
iand does not guarantee that the radius of convergence of a solution is greater than 1.
We shall solve (13) by applying the Mellin transformation. Fix a non- singular direction θ 6∈ Θ
ssuch that cos θ > 1/r (this assumption ensures that 1 ∈ Ω
−1/rθ). Observe that by (14) and (7),
M
θ1eg
i(s)
s d
ds
iw
(z)
=
\
lθ
ψ
i(α)((z − α)
iM
θ1w(z − α) + W
i(z − α))e
−αdα
=
\
lθ
ψ
i(α)(z − α)
iM
θ1w(z − α)e
−αdα + f W
i(z),
where f W
iis a polynomial of degree ≤ i − 1, i = 0, . . . , n − 1. Thus applying
the Mellin transformation to (13) we get the convolution equation
(15) Q(z)G
θ(z) +
\
lθ
A
0(α, z)G
θ(z − α)e
−αdα = Φ(z), where
G
θ(z) = M
θ1w(z), A
0(α, z) =
n−1
X
i=0
ψ
i(α)(z − α)
iand Φ is a polynomial of degree ≤ n − 1 depending on w
0, . . . , w
n−1.
We solve (15) by the approximation scheme G
0θ(z) = Φ(z)
Q(z) , G
ν+1θ(z) = 1
Q(z)
n Φ(z) −
\
lθ
A
0(α, z)G
νθ(z − α)e
−αdα o
, ν ∈ N.
Put e G
ν+1θ= G
ν+1θ− G
νθfor ν ∈ N
0. Then we find G e
ν+1θ(z) = (−1)
ν+1Q(z)
\
lθ
A
νθ(γ, z) Φ(z − γ) Q(z − γ) e
−γdγ where for γ ∈ l
θ, ν ∈ N,
A
νθ(γ, z) =
\
α1∈lθ
|α1|≤|γ|
A
0(α
1, z)A
ν−1(γ − α
1, z − α
1) Q(z − α
1) dα
1=
\
Tθν(γ)
A
0(α
1, z)
Q(z − α
1) . . . A
0(α
ν, z − α
1− . . . − α
ν−1) Q(z − α
1− . . . − α
ν)
× A
0(γ − α
1− . . . − α
ν, z − α
1− . . . − α
ν) dα with T
θν(γ) = {α ∈ (l
θ)
ν: |α
1+ . . . + α
ν| ≤ |γ|}, γ ∈ l
θ.
Assume that dist(z, S
mµ=1
(̺
µ+ l
θ)) ≥ b with some b > 0. Then we can find C
bsuch that hzi
n≤ C
b|Q(z)|. Since |A
0(α, z)| ≤ Ce
|α|/r′hzi
n−1we derive
|A
0(α
1, z)|
|Q(z)| ≤ CC
be
|α1|/r′hzi , |A
0(α
2, z − α
1)|
|Q(z − α
1)| ≤ CC
be
|α2|/r′hz − α
1i , . . . ,
|A
0(α
ν, z − α
1− . . . − α
ν−1)|
|Q(z − α
1− . . . − α
ν−1)| ≤ CC
be
|αν|/r′hz − α
1− . . . − α
ν−1i ,
|A
0(γ − α
1− . . . − α
ν, z − α
1− . . . − α
ν)|
|Q(z − α
1− . . . − α
ν)| ≤ CC
be
|γ−α1−...−αν|/r′hz − α
1− . . . − α
νi . So by Lemma 1,
|A
νθ(γ, z)|
|Q(z)| ≤ (CC
b)
ν+1hzi e
|γ|/r′2
νν! log
ν(1 + |γ|).
Thus
(16) G
θ(z) = Φ(z) Q(z) +
\
lθ
A
θ(γ, z)
Q(z) · Φ(z − γ)
Q(z − γ) e
−γdγ, where A
θ(γ, z) = P
∞ν=0
(−1)
ν+1A
νθ(γ, z) satisfies, with K = 2CC
b,
(17) |A
θ(γ, z)|
|Q(z)| ≤ CC
bhzi e
|γ|/r′(1 + |γ|)
Kfor γ ∈ l
θ, dist
z, [
m µ=1(̺
µ+ l
θ)
≥ b.
Finally, since |Φ(z)|hzi ≤ C
b|Q(z)| for dist(z, {̺
1, . . . , ̺
m}) ≥ b we get, with some C < ∞,
(18) |G
θ(z)| ≤ C
hzi for dist z,
[
m µ=1(̺
µ+ l
θ)
≥ b.
Now assume that z is close to ̺
µwith a fixed µ ∈ {1, . . . , m}. To shorten notation put k = k
µ. Assume d ≤ |z−̺
µ| ≤ b, |arg(z−̺
µ−θ)| ≥ β with some β > 0 and 0 < d < b ≤ 1 with dist(̺
µ+ l
θ, S
ν6=µ
(̺
ν+ l
θ)) ≥ 2b. Since for α ∈ l
θwe have hz −αi
n−k|z −̺
µ−α|
k≤ C|Q(z −α)|, hz −αi
k−1≤ C
1hαi
k−1and (d + |α|)
k≤ C
2|z − ̺
µ− α|
kwe get, with a constant C independent of d,
|A
0(α
1, z)|
|Q(z)| ≤ Ce
|α1|/r′d
k,
|A
0(α
2, z − α
1)|
|Q(z − α
1)| ≤ Ce
|α2|/r′hz − α
1i
k−1|z − ̺
µ− α
1|
k≤ Ce
|α2|/r′C
1C
2hα
1i
k−1(d + |α
1|)
k, . . . ,
|A
0(α
ν, z − α
1− . . . − α
ν−1)|
|Q(z − α
1− . . . − α
ν−1)| ≤ Ce
|αν|/r′C
1C
2hα
1+ . . . + α
ν−1i
k−1(d + |α
1+ . . . + α
ν−1|)
k,
|A
0(γ − α
1− . . . − α
ν, z − α
1− . . . − α
ν)|
|Q(z − α
1− . . . − α
ν)|
≤ Ce
|γ−α1−...−αν|/r′C
1C
2hα
1+ . . . + α
νi
k−1(d + |α
1+ . . . + α
ν|)
k. So for γ ∈ l
θ,
|A
νθ(γ, z)|
|Q(z)| ≤ C
d
k(CC
1C
2)
νe
|γ|/r′×
\
Tθν(γ)
hα
1i
k−1. . . hα
1+ . . . + α
νi
k−1(d + |α
1|)
k. . . (d + |α
1+ . . . + α
ν|)
kdα
≤ C
d
ke
|γ|/r′1 ν!
L
d
k−1log d + |γ|
d
ν,
where L = CC
1C
2(for k > 1 we use hαi
k−1d
k−1≤ (d + |α|)
k−1). Thus
|A
θ(γ, z)|
|Q(z)| ≤ C d
ke
|γ|/r′d + |γ|
d
L/dk−1. Finally, since
\
lθ
1 + |γ|
d
L/dk−1e
|γ|/r′e
−|γ||dγ| ≤ C
C d
L/dk−1Γ
L
d
k−1+ 1
(here C = 2r
′/(r
′− 1) and Γ is the Euler function), we obtain, with C independent of d,
(19) |G
θ(z)| ≤
Cd
−L−1if k = 1,
C d
kexp
L
d
k−1log CL d
kif k > 1 for d ≤ |z − ̺
µ| ≤ b, |arg(z − ̺
µ− θ)| ≥ β.
Now, observe that θ can be changed within the interval (θ
−, θ
+), where θ
−(resp. θ
+) is the greatest (resp. smallest) singular direction less (resp.
greater) than θ. Also β can be chosen arbitrarily small positive. Thus, re- striction of G
θto a small left (resp. right) tubular neighbourhood of ̺
µ+l
θextends to a holomorphic function defined on θ
−< arg(z − ̺
µ) ≤ 0 (resp.
0 ≤ arg(z − ̺
1) < θ
+). The extension of G
θobtained this way also satis- fies (19) for d ≤ |z − ̺
µ| ≤ b and (18) for |z − ̺
µ| ≥ b.
Thus, by Lemma 3 and the results of Section 1, we get 1
2πi b(G
θ) = X
m µ=1S
θµ,
where S
θµ∈ L
∗′(0)(̺
µ+ l
θ) with ∗ given by (10). So, the solution w of (13) is given by w(s) = P
mµ=1
T S
θµ(s) for s ∈ Ω
0θ, and u(x) = w(e
1−1/x) is defined only for x ∈
12B(e
iθ; 1). However, the estimate (17) gives an additional information about S
θµ, µ = 1, . . . , m. Namely, changing θ within (θ
−, θ
+), we note that the restriction of S
θµto an open ray ̺
µ+ l
θis analytic, and extends holomorphically to a function Ψ
µdefined in a sector S(̺
µ; (θ
−, θ
+)).
To estimate Ψ
µput (with k = k
µ)
F
µ(α, z) = A(α, z)Φ(z − α)(z − ̺
µ− α)
kQ(z)Q(z − α)
for α ∈ l
θ′, z 6∈
[
m ν=1(̺
ν+ l
θ′), θ
−< θ
′< θ
+,
where A(α, z) = A
θ′(α, z) for (α, z) as above, θ
−< θ
′< θ
+. Since
b(Φ/Q)|
̺µ+lθ= 0, (16) implies that for γ ∈ S(0; (θ
−, θ
+)), Ψ
µ(̺
µ+ γ) = 1
2πi (G
θe−(̺
µ+ γ) − G
θe+(̺
µ+ γ))
= 1 2πi
\
lθ−e −lθ+e
F
µ(α, ̺
µ+ γ)
(γ − α)
ke
−αdα where θ
−< e θ
−< arg γ < e θ
+< θ
+. So for γ with θ
−< arg γ < θ
+,
Ψ
µ(̺
µ+ γ) = e Ψ
µ(γ)e
−γ, where Ψ e
µ(γ) =
kµ−1
X
l=0
C
l∂
l∂α
lF
µ(α, ̺
µ+ γ)
α=γ
with some constants C
l, l = 0, . . . , k
µ− 1 (µ = 1, . . . , m). Observe that (17), (19), and the Cauchy formula imply that for any θ
−< e θ
−< e θ
+< θ
+and r
′< r, | e Ψ
µ(γ)| can be estimated by the right hand side of (11) for θ e
−≤ arg γ ≤ e θ
+with some C < ∞. Since the above holds for any r
′< r, we conclude that e
γS
θµ∈ L
∗′(−1/re−iθ)(̺
µ+ l
θ) and so S
θµ∈ L
∗′(ω)(̺
µ+ l
θ) with ω = (cos θ − 1/r)e
−iθ. Now, Lemma 2 implies that w(s) is defined for s ∈ e C with s/e ∈ Ω
−1/rθFinally, u(x) = w(e
1−1/x) = P
mµ=1
L(e
γS
θµ)(−1/x) is defined for x ∈ (r/2)B(e
iθ; 1), and a direct computation shows that u can be written in the form (12) (with t = 1).
5. An example. Let us solve the Cauchy problem for the Euler equa- tion x
2u
′= u − x, u(1) = u
0. Putting s(x) = exp{1 − 1/x} and w(s) = u(1/(− log(s/e))) we get sw
′− w = 1/log(s/e), w(1) = u
0. Applying the Mellin transformation (4) with 0 < |θ| < π/2 and t = 1 we obtain the equation for G
θ= M
θ1w,
(z − 1)G
θ(z) = −u
0+
\
lθ
e
−αz − α dα.
Its solution is given by
G
θ(z) = −u
0z − 1 + 1 z − 1
\
lθ
e
−αz − α dα.
Now, we compute the boundary value S = (2πi)
−1b(G
θ):
S = (u
0+ A)δ
(1)+
\
lθ
log(α − 1) d
dα (e
−αδ
(α)) dα with A = Γ
′(1) − P
∞j=1 1 j!j
/e. Thus, the solution w = T S is given by w(s) = (u
0+ A)s + log(s/e)
\
lθ
log(α − 1)s
αe
−αdα for s/e ∈ Ω
θ0.
Finally, u(x) = w(e
1−1/x) is given by u(x) = (u
0+ A)e
1−1/x− 1
x
\
lθ