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122 7. Cauchy-Heine Transform

one can therefore characterize the image of the operator Sk,d . This will be
done in detail in the context of multisummable series later on.

Exercises: Throughout the following exercises, let О» be any complex
number, let p(z) be a polynomial of degree r в‰Ґ 1 and highest coeп¬ѓcient
one, and let g(u) be holomorphic (and single-valued) for 0 в‰¤ |u| < ПЃ.

1. For j = 0, . . . , r, deп¬Ѓne
z
uв€’О» eв€’p(u) g(1/u) dur , 0 < |z| < ПЃ,
О» p(z)
fj (1/z) = z e
в€ћ(2jПЂ/r)

integrating from в€ћ along the line arg u = 2jПЂ/r to some arbitrarily
chosen point z0,j , |z0,j |в€’1 < ПЃ, and then to z. Show that each fj is
holomorphic for 0 < |z| < ПЃ, and fj (1/z) в€’ fjв€’1 (1/z) = cj z О» ep(z) ,
j = 1, . . . , r, with cj = Оіj uв€’О» eв€’p(u) g(1/u) du, where Оіj is a path
from в€ћ along arg u = 2jПЂ/r to z0,j , then to z0,jв€’1 , and back to в€ћ
along arg u = 2(j в€’ 1)ПЂ/r.

2. For z в€€ Sj = S(2jПЂ/r, 3ПЂ/r, ПЃ) and j = 0, . . . , r, show that fj (z) is
bounded at the origin.

3. Show
fjв€’1 (z) в€’ fj (z) в€€ A1/r,0 (Sjв€’1 в€© Sj , E ), 1 в‰¤ j в‰¤ r,
fr (z) = f0 (zeв€’2ПЂi ), z в€€ Sr .

4. Conclude fj в€€ A1/r (Sj , E ), 0 в‰¤ j в‰¤ r.
8
Solutions of Highest Level

In this chapter we are going to prove that the formal transformations oc-
curring in the Splitting Lemma (p. 42) and in Theorem 11 (p. 52) are, in
fact, r-summable in the sense of Chapter 6. Based upon this, we shall then
show that highest-level formal fundamental solutions are k-summable for
1/k = s equal to their Gevrey order. As an application of this result we
then prove the factorization of formal fundamental solutions according to
Ramis and the author, as described in Chapter 14. For every highest-level
formal fundamental solution (HLFFS for short), we then deп¬Ѓne normal so-
lutions of highest level that were п¬Ѓrst introduced by the author in , and
we shall investigate their properties in more detail, deп¬Ѓning corresponding
StokesвЂ™ directions, StokesвЂ™ multipliers, etc.
Since in Chapter 3 all power series have been in the variable 1/z, we
have to make some more or less obvious adjustments in the deп¬Ѓnition of
r-summability. In particular, we deп¬Ѓne sectorial regions at inп¬Ѓnity to be
such regions G that by the inversion z в†’ 1/z are mapped onto sectorial
Л†
regions in the previous sense, and from now on, we shall denote by f (z) a
formal power series in the variable z в€’1 . It is then natural to say that

вЂў a power series f (z) в€€ E [[z в€’1 ]] is k-summable in direction d if and
Л†
в€ћ
only if f (z в€’1 ) = 0 fn z n is k-summable in direction в€’d.
Л†

The formal Borel transform Bk f is deп¬Ѓned by applying Bk to f (z в€’1 ), and
Л†Л† Л†
Л†
its sum then is holomorphic in a small sector of bisecting direction в€’d, and
Л†
is of exponential growth at most k there. The k-sum of f (z) then is obtained
by summing f (z в€’1 ) followed by the change of variable z в†’ 1/z. Hence, if
Л†
124 8. Solutions of Highest Level

Л†
f (z) is the k-sum in direction d of a power series f (z) in inverse powers
of z, then f (z) is holomorphic in a sectorial region G(d, О±) (near inп¬Ѓnity)
of bisecting direction d and opening О± > ПЂ/k, and f (z) can be represented
by a Laplace integral (5.1), with z replaced by 1/z and the direction of
integration П„ close to в€’d. So here, turning the path of integration in the
positive sense leads to holomorphic continuation of f (z) in the negative
sense, and vice versa.
Because HLFFS in general are series in some root of z в€’1 , we need to
generalize the notion of k-summability to q-meromorphic transformations,
or more generally to arbitrary formal Laurent series in the variable z в€’1/q :

вЂў A power series in z в€’1/q will be called k-summable in direction d, if the
series obtained by the change of variable z = wq is qk-summable in
direction d/q. Compare Exercise 2 on p. 72 to see that this deп¬Ѓnition
gives good sense even if the series we start with accidentally happens
to not contain any roots.
в€ћ
вЂў A formal Laurent series n=в€’m fn z в€’n is called k-summable in di-
в€ћ
rection d if and only if its power series part f (z) = n=0 fn z в€’n is
Л†
summable in this sense, with the k-sum of the Laurent series equal
to the k-sum of the power series part plus the п¬Ѓnite principle part
в€’1 в€’n
n=в€’m fn z . Compare this to Exercise 5 on p. 72.

8.1 The Improved Splitting Lemma
In the construction of HLFFS we had to consider formal analytic transfor-
mations in two places: In the Splitting Lemma in Section 3.2, and later in
Section 3.4, п¬Ѓnding a transformation so that the new system is rational.
To proceed, we п¬Ѓrst reconsider the proof of the Splitting Lemma and show
that, beginning with a convergent system (3.1), the transformation as well
as the transformed system are r-summable, and we can even completely
describe the set of singular directions.
Lemma 11 (Improved Splitting Lemma) Consider a convergent sys-
tem (3.1) (p. 37) satisfying the assumptions of the Splitting Lemma on p. 42.
Л† Л†
Then the formal matrix power series T12 (z) and the block B22 (z) of the coef-
п¬Ѓcient matrix of the transformed system are r-summable in every direction
d except for the п¬Ѓnitely many singular ones of the form в€’rd = arg(Вµ1 в€’ Вµ2 )
(jj)
mod 2ПЂ, with Вµj an eigenvalue of A0 . The same statement, but with Вµ1 , Вµ2
Л† Л†
interchanged, holds for T21 (z) and B11 (z).

Proof: Analogously to the proof of Theorem 11 (p. 52), we deп¬Ѓne T (u) =
в€ћ в€ћ
nв€’r nв€’r
1 T12 u /О“(n/r), B(u) = 1 B22 u /О“(n/r), and (slightly abus-
в€ћ (jk) nв€’r
/О“(n/r). Then urв€’1 Ajk (u) are en-
ing notation) Ajk (u) = 1 An u
8.1 The Improved Splitting Lemma 125

tire functions of exponential growth at most r, owing to the convergence
of (3.1). Moreover, (3.6) and (3.7) (p. 42) are formally equivalent to the
integral equation
(22) (11)
в€’ (A0 + rur I) T (u)
T (u) A0 =
u
A11 ([ur в€’ tr ]1/r ) T (t) в€’ T (t) B([ur в€’ tr ]1/r ) dtr ,
A12 (u) + (8.1)
0

with u
A21 ([ur в€’ tr ]1/r ) T (t) dtr .
B(u) = A22 (u) + (8.2)
0

The Splitting Lemma ensures existence of T (u) and B(u), holomorphic and
single-valued near the origin and satisfying (8.1). We aim at proving that
both can be holomorphically continued into the largest star-shaped1 region
(22) (11)
G that does not contain any point u for which A0 and A0 + rur I have
an eigenvalue in common. Observe Exercise 3 on p. 214 to see that this
is the largest star-shaped set not containing any solution of the equation
rur = Вµ2 в€’ Вµ1 . Moreover, we shall show that in this region both matrices
have exponential growth at most r. This then implies r-summability of
Л† Л†
T12 (z) and B22 (z), with singular directions as stated, and one can argue
analogously for the other two blocks.
To do all this, we employ an iteration: Beginning with B(u; 0) в‰Ў 0,
T (u; 0) в‰Ў 0, we plug B(u; m) and T (u; m) into the right-hand sides of
(8.1), (8.2) and determine T (u; m + 1) from the left-hand side of (8.1),
resp. let B(u; m + 1) be equal to B(u) in (8.2). Both sequences so obtained
are holomorphic in G, except for a pole of order at most r в€’ 1 at the
origin. For the following estimates, we choose d so that the ray u = xeв€’id ,
x в‰Ґ 0, is in G, hence d is a non-singular direction. Then Ajk (u) в‰¤
в€ћ n nв€’r
/О“(n/r), with a в‰Ґ 0 independent of d. Inductively we show
1ax
в€ћ (m) nв€’r
estimates of the form T (u; m) в‰¤ /О“(n/r), B(u; m) в‰¤
1 tn x
в€ћ (m) nв€’r (0) (0)
1 bn x /О“(n/r): For m = 0 this is certainly correct with tn = bn =
0. Given this estimate for some m в‰Ґ 0, we use the recursion formulas and
the Beta Integral (p. 229) to show the same type of estimate for m + 1. In
particular, we may set
пЈј
(m+1) (m)
пЈґ
nв€’1
= an + 1 anв€’k tk
bn пЈЅ
n в‰Ґ 1,
(m) (m) пЈґ пЈѕ
(m+1) nв€’1 nв€’k
n
tn = c a + 1 (a + bnв€’k )tk

where c is a constant which arises when solving the left-hand side of (8.1)
for T (u) and then estimating the solution вЂ“ hence c is independent of n
and m, as well as independent of d provided we alter d only slightly so that

1 With respect to the origin.
126 8. Solutions of Highest Level

it keeps a positive distance from the singular directions. For every n, the
(m) (m)
numbers tn , bn can be seen to be monotonically increasing with respect
to m and become constant when m в‰Ґ n. Their limiting values tn , bn then
satisfy the same recursion equations, with the superscripts dropped. Setting
в€ћ в€ћ в€ћ
b(x) = 1 bn xn , t(x) = 1 tn xn , a(x) = 1 an xn = ax(1 в€’ ax)в€’1 , we
obtain formally b(x) = a(x)[1+t(x)], t(x) = c[a(x)+(a(x)+b(x))t(x)]. This
can be turned into a quadratic equation for t(x), having one solution t1 (x)
that is a holomorphic function near the origin, while the other one has a
pole there. The coeп¬ѓcients of t1 (x) satisfy the same recursion formulas as
tn , so are in fact equal to tn . This implies that both tn and bn cannot grow
faster than some constant to the power n. From this fact we conclude that
r
T (u; m) and B(u; m) can be estimated by c x1в€’r eKx with suitably large
c, K independent of m. Therefore, the proof will be completed provided
that we show convergence of T (u; m) and B(u; m) as m в†’ в€ћ. This can be
done by deriving estimates, similar to the ones above, for the diп¬Ђerences
T (u; m) в€’ T (u; m в€’ 1) and B(u; m) в€’ B(u; m в€’ 1) and turning the sequence
2
into a telescoping sum. For details, compare the exercises below.
It is important to note for later applications that the above lemma not
Л†
only ensures r-summability of the transformation T (z), but also allows ex-
plicit computation of the possible singular directions in terms of parameters
of the system (3.1) (p. 37). As we shall make clear in Chapter 9, it may
happen that some, or all, of these directions are nonsingular, depending on
whether some entries in StokesвЂ™ multipliers are zero. However, in generic
cases all these directions are singular!

Exercises: In the following exercises, make the same assumptions as in
the lemma above, and use the same notation as in its proof.
1. For m в‰Ґ 1 and u as in the proof of the above lemma, show
в€ћ
d(m) xnв€’r /О“(n/r),
T (u; m) в€’ T (u; m в€’ 1) в‰¤ n
n=m
в€ћ
d(m) xnв€’r /О“(n/r),
Лњ
B(u; m) в€’ B(u; m в€’ 1) в‰¤ n
n=m

(m) Лњ(m)
with constants dn , dn satisfying
nв€’1 nв€’1
(m) Лњ(m) (m)
d(m+1) d(m+1) =
Лњ
nв€’k
anв€’k dk ,
= b (dk + dk ),
n n
k=m k=m

for suitably large b > 0, and a as in the proof of the lemma.
(m)
в€ћ Лњ
2. Setting dm (x) = n=m dn xn , and analogously for dm (x), derive a
recursion for these functions вЂ“ in particular, conclude convergence of
the series for 0 < x < min(a, b).
8.2 More on Transformation to Rational Form 127

Лњ Лњ
3. Setting d(x) = m dm (x), d(x) = m dm (x), use the previous exer-
cise to show functional equations, and from these compute the func-
Лњ
tions in terms of d1 , d1 . In particular, show that both functions are
holomorphic near the origin. Use this to conclude existence of d so
(m) Лњ(m)
that dn , dn в‰¤ dn for every n, m.
4. Use the previous exercises to show that T (u; m) and B(u; m) converge
uniformly on every compact subset of the region G deп¬Ѓned in the
proof of the above lemma.

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