∞ ˜

convergence of 1 tn z n , hence tn ¤ K n . Thus we obtain an estimate of

(12) (21)

the desired type for Tn . In the same fashion one can estimate Tn ,

hence T (z) is of Gevrey order s. Moreover, from (3.6) we conclude, using

Exercise 2 on p. 41, that both diagonal blocks of the transformed system

2

are of Gevrey order s.

Applying the above Lemma repeatedly and using Exercise 2 on p. 41, we

obtain an analogous result when A0 has more than two blocks:

44 3. Highest-Level Formal Solutions

Theorem 8 Let (3.3) be a formal system of Gevrey order s ≥ 0, and as-

(11) (µµ)

sume that A0 = diag [A0 , . . . A0 ], such that the kth diagonal block has

exactly one eigenvalue »k , and let »1 , . . . , »µ be distinct. Then there exists

a unique formal analytic transformation of Gevrey order s = max{s, 1/r},

˜

with diagonal blocks all equal to I and the o¬-diagonal ones having zero con-

stant terms, such that the transformed formal system is diagonally blocked,

with each of the diagonal blocks being formal of Gevrey order s. ˜

Remark 2: It follows from the form of the formal analytic transformation

that the diagonal blocks of the transformed system have leading terms with

exactly one eigenvalue. Assuming that this was already the case for (3.3),

we can use the scalar exponential shift x = exp[»z r /r] x to produce a

˜

system with nilpotent matrix A0 . Such systems will be investigated in the

3

following section.

Remark 3: Suppose that A0 has ν distinct eigenvalues; this situation usu-

ally is referred to as the distinct eigenvalue case. Then the above theorem

shows existence of a formal analytic transformation for which the trans-

formed system is diagonal, i.e., decoupled into ν one-dimensional systems.

These can easily be solved, and in this way one can compute a formal

3

fundamental solution of (3.3). For details, compare Exercise 4.

Exercises:

1. For r = 1, ν = 2 and

ab

, An = 0, n ≥ 2,

A0 = diag [0, 1], A1 =

00

(12)

compute Tn given by (3.8) and ¬nd its rate of growth, in depen-

dence upon a and b, as n ’ ∞.

2. Assume that a formal system (3.3) is given, whose coe¬cients An all

commute with one another (this is satis¬ed for ν = 1, or whenever all

An are diagonal). Show that B(z) = Ar log z + n=r An z r’n /(r ’ n)

commutes with A(z). In case the series in (3.3) converges for |z| > ρ,

conclude that (3.3) is elementary in the sense of Section 1.6.

3. Under the assumptions of the previous exercise, show that (3.3), at

least formally, i.e., disregarding the question of convergence of the

series involved, has the fundamental solution X(z) = T (z) z Λ eQ(z) ,

ˆ ˆ

r’1

with Q(z) = n=0 An z r’n /(r ’ n), Λ = Ar , and

∞

Tn z ’n = exp [

ˆ An z r’n /(r ’ n)].

T (z) = I +

1 n>r

3.3 Nilpotent Leading Term 45

4. In case the leading term A0 has all distinct eigenvalues, show that

(3.3) has a formal fundamental solution of the form

X(z) = T (z) z Λ eQ(z) ,

ˆ ˆ

ˆ

with a formal analytic transformation T (z), being of Gevrey order

s = max{s, 1/r} in case (3.3) is formal of Gevrey order s, a constant

˜

diagonal matrix Λ, and a diagonal matrix Q(z) with polynomials

along the diagonal satisfying Q(0) = 0. Explicitly describe an algo-

rithm for computing the matrices Λ and Q(z) and any ¬nite number

ˆ

of coe¬cients of T (z).

3.3 Nilpotent Leading Term

In this section, we are concerned with formal systems (3.3) whose leading

term only has one eigenvalue, so that the Splitting Lemma does not apply.

A scalar exponential shift then reduces such a system to one with nilpotent

leading term, and such a transformation commutes with every other one we

are using here. Hence it is without loss of generality when, for notational

convenience, we restrict our discussion to systems (3.3) having a nilpotent

leading term A0 = 0. We may also assume A0 in Jordan canonical form

A0 = diag [N1 , . . . , Nµ ], (3.9)

with nilpotent Jordan blocks of weakly decreasing sizes s1 ≥ . . . ≥ sµ “ if

this were not so, we could apply a constant transformation. In what follows,

ˆ ˆ

we shall block A(z) = [Ajk (z)] according to the block structure of the ma-

trix A0 , and similarly for transformations resp. corresponding transformed

equations. Our goal is to simplify the given system, in a sense to be made

clear later, using ¬nitely many transformations that are either terminating

analytic ones, or shearing transformations, or scalar exponential shifts. Ow-

ing to the nature of the transformations we see that in case the system we

begin with is convergent, then the one we obtain in the end converges, too.

Hence, divergent transformations or systems will only occur when applying

the Splitting Lemma!

As a ¬rst step we shall try to ¬nd terminating analytic transformations,

resp. unrami¬ed shearings, that leave the Poincar´ rank r of (3.3) un-

e

changed, but possibly increase the matrix rank of A0 or, in general, leave

the ranks of Ak ¬xed, for 1 ¤ k ¤ j ’ 1, and increase the rank of Aj . To

0 0

have a way of expressing this, we shall use the following terminology:

Order Relation for Nilpotent Matrices

˜ ˜

Given any two nilpotent matrices N, N ∈ C ν—ν , we say that N

˜

is superior to N , if for some j ≥ 1 we have rank N = rank N k

k

˜

for 1 ¤ k ¤ j ’ 1, and rank N < rank N .

j j

46 3. Highest-Level Formal Solutions

Note that the ranks of the powers determine the structure of a nilpotent

Jordan matrix up to a permutation of its blocks, so if we restrict ourselves

to nilpotent Jordan matrices whose Jordan blocks have decreasing sizes,

then the above relation becomes a total order, and it is easily seen that the

nilpotent matrix with rank equal to ν ’ 1 is the maximal element for this

order relation.

Classically, the case of nilpotent leading matrices has been treated by

¬nding transformations reducing the rank of A0 , or if possible, even making

A0 = 0, hence lowering the Poincar´ rank. Here we do the opposite, which

e

at ¬rst glance appears unreasonable. However, we shall see that this may be

considered as an example of what sometimes is called “worst case analysis,”

since in case of a maximal leading term we shall ¬nd that the system has

a very particular structure.

Finding the transformations that will produce a superior leading term

requires two steps, the ¬rst one being the following analogue to the Splitting

Lemma; see, e.g., Wasow [281] for an equivalent version.

Lemma 4 Let a formal system (3.3) with leading term as in (3.9) be given.

Then for every n0 ∈ N there exists an analytic transformation of the form

T (z) = I + [Tjk (z)], blocked according to the block structure of A0 , with

n0 (jk) ’n

matrices Tjk (z) = 1 Tn z , such that the coe¬cient matrix of the

ˆ ˆ ˆ

transformed system has the form B(z) = z r A0 + [Bjk (z)], with Bjk (z) =

∞ (jk) (jk)

z r 1 Bn z ’n , and so that for 1 ¤ n ¤ n0 the coe¬cients Bn

• have all zero rows except for the ¬rst one in case 1 ¤ j ¤ k ¤ µ,

• have all zero columns except for the last one in case 1 ¤ k < j ¤ µ.

The transformation T (z) is unique if we require for 1 ¤ n ¤ n0 that

(jk)

• all Tn have vanishing last row in case 1 ¤ j ¤ k ¤ µ,

(jk)

• all Tn have vanishing ¬rst column in case 1 ¤ k < j ¤ µ.

Proof: Insertion into (3.5) implies the following identities for the coe¬-

ˆ ˆ

cients of the blocks of A(z), B(z), T (z):

(jk) (jk) (jk) (jk)

Nj Tn ’ Tn Nk = Bn + Rn , n ≥ 1, 1 ¤ j, k ¤ µ, (3.10)

(jk)

where Rn only involves blocks of Tm , Bm with m < n. For n ¤ n0 ,

(jk)

Lemma 25 (p. 213) implies existence of a unique matrix Bn with nonzero

entries only in the ¬rst row, resp. last column, such that (3.10) has a so-

(jk)

lution Tn . This solution is unique if we require its last row, resp. ¬rst

column, to vanish. Which case applies depends on the shape of the block,

i.e., upon its position on or above, resp. below, the block-diagonal. For

(jk) (jk)

n ≥ n0 + 1, we have Tn = 0; hence take Bn so that (3.10) holds, thus

2

completing the proof.

3.3 Nilpotent Leading Term 47

Remark 4: We say that (3.3) (with A0 as above and µ ≥ 1) is normalized

(jk)

up to z ’n0 , if all coe¬cients An , for 1 ¤ n ¤ n0 , have nonzero entries

only in ¬rst row resp. last column (in case j ¤ k resp. j > k). If this is so for

some n0 , we brie¬‚y call (3.3) normalized. If (3.3) is normalized up to z ’n0 ,

(jk)

and in addition all An with j = k vanish completely, for 1 ¤ n ¤ n0 ,

then we say that (3.3) is reduced up to z ’n0 .

For later use, observe that if (3.3) is already normalized up to z ’˜ 0 , then

n

normalization up to z ’n0 , for n0 > n0 , can be done by a transformation

˜

T (z) with coe¬cients Tn = 0 for 1 ¤ n ¤ n0 ; hence the corresponding

˜

3

coe¬cients An remain unchanged.

For normalized systems (3.3) with µ > 1 we apply shearing transforma-

tions to produce another system with superior leading term:

Proposition 6 For some n0 ∈ N, let (3.3) be a formal system with lead-

ing term as in (3.9), normalized up to z ’n0 , and assume µ ≥ 2. Assume

(kµ)

the existence of n1 , 1 ¤ n1 ¤ n0 , so that Akµ (z) = z r n≥n1 An z ’n ,

ˆ

(kµ)

1 ¤ k ¤ µ ’ 1, and not all An1 vanish. Then the unrami¬ed shearing

transformation T (z) = diag [Is1 , . . . , Isµ’1 , z n1 Isµ ] produces a system with

a superior leading term.

(kµ)

Proof: Abbreviate Ck = An1 , 1 ¤ k ¤ µ ’ 1. Then, check that the

shearing transformation produces a transformed system with leading term

®

®

N1 0 . . . 0 C1

ck,1 . . . ck,sµ

0 N2 . . . C2

0

0

. , C = 0 ...

.

. .

..

B0 = . . . . .

. .

.

. . . . k

°. .»

. .

°0 0 . . . Nµ’1 Cµ’1 »

0 ... 0

0 0 ... 0 Nµ

≥ 1, we ¬nd

By assumption, Ck = 0 for at least one k. For

®

()

N1 0 ... 0 C1

()

0 C2

N2 ... 0

. ,

B0 = . .

. .

..

. . . .

.

. . .

()

°0 Cµ’1 »

0 . . . Nµ’1

0 0 ... 0 Nµ

() ( ’1)

with Ck = Nk’1 Ck + Ck Nµ , ≥ 2. This shows that the th row of

()

Ck equals the ¬rst one of Ck , while the following ones vanish. From this

we learn that those rows of B0 with a one in a subdiagonal coincide with

the corresponding rows of A0 . Therefore, we have rank A0 ¤ rank B0 , and

()

equality holds if and only if the rows of Ck , for every k = 1, . . . , µ ’ 1,

48 3. Highest-Level Formal Solutions

are linear combinations of the rows of Nµ . Now, take = sµ , the size of

the last, i.e., the smallest, Jordan block of A0 . Then we have Nµ = 0, but

()

Ck = 0 for at least one k, hence inequality holds for the ranks, thus B0 is

2

superior to A0 .

Suppose that we are given a formal system with a leading term as in (3.9),

normalized up to z ’n0 . Proposition 6 then implies that either Akµ (z) =

ˆ