0387986995Basic TheoryC

it is easy to show that n'(x) satisfies integral equation (V.2.12). It is also evident that i7(x) is analytic for IxI < 6. Thus, the proof of Theorem V-2-5 is completed. 0 Now, finally, by using the argument given in Observations V-2-2 and V-2-3, we

obtain the following theorem.

Theorem V-2-7. Every formal solution E xC[[zjJ" of system (V.2.1) is conver- gent, i.e., E xC{x}".

Remark V-2-8. In general, (V.2.1) may not have any formal solutions. However,

Theorem V-2-7 states that if (V.2.1) has formal solutions, then every formal solution is convergent.

Remark V-2-9. If yi(x) = xA f (x) E XAC[[xfj" is a formal solution of a linear system X!LY = A(x)y, the formal power series f is a formal solution of xd _

[A(x) - A!, }i. Therefore, f E C{x}" by virtue of Theorem V-2-7. This proves convergence of the formal solution constructed in Theorem V-1-2.

V-3. The S-N decomposition of a matrix of infinite order

In §V-4, we shall define the S-N decomposition of a linear differential operator.

As a linear differential operator will be represented by a lower block-triangular matrix of infinite order, we derive, in this section, the S-N decomposition of such

N", x Nm matrix, while Bm is an nm x N,"_ 1 matrix. Also, let Amm = Smm +Nmm

3. THE S-N DECOMPOSITION OF A MATRIX OF INFINITE ORDER 119

and Am = Sm + Aim be the S-N decompositions of Amm and Am, respectively, where Smm is an nm x nm diagonalizable matrix, Sm is an Nm x Nm diagonalizable matrix, Nmm is an nm x nm nilpotent matrix, Aim is an N. x Nn nilpotent matrix,

SmmNmm = NmmSmm, and SmAm = NmSm. The following lemma shows how the two matrices Sm and Aim look.

Lemma V-3-1. The matrices Sm and Aim have the following forms:

where Cm and fm are am x N,_1 matrices.

Proof

Consider the case m > 2. Since the matrices Sm and Aim are polynomials in Am with constant coefficients, it follows that

where 13m-1 and Dm-1 are Nm-1 x Nm_1 matrices, Cm and fm are nm x Nm_1 matrices, and µm and vm are nm x nm matrices. Furthermore, Dm_1 and Vm are nilpotent. Also, Bm-1Dm-1 = Dm-1Bm-i and tlmvm = 1.m µm. Hence, it suffices to show that Bm-1 and µm are diagonalizable.

Note that since Sm is diagonalizable, Sm has Nm linearly independent eigen

vectors. An eigenvector of Sm has one of two forms k and [?.] , where p is an

P J

eigenvector of Bm_1i whereas q is an eigenvector of µm. Therefore, if we count those independent eigenvectors, it can be shown that Bm-1 has N,, -I linearly independent eigenvectors, while Pm has nm linearly independent eigenvectors. Note that Nm = Nm_ 1 + nm. This completes the proof of Lemma V-3-1. 0

Lemma V-3-1 implies that the matrices Sm and Nn have the following forms:

S1 = SI1,

Fm 1 Fm2 ... ... Nmm

where C,k and Fjk are n3 x nk matrices.

Set

We call (V.3.1) the S-N decomposition of the matrix A.

V-4. The S-N decomposition of a differential operator

Consider a differential operator

t>o

by the matrix

A =

L

matrix, the matrices S and N have the forms

respectively, where the a. and v, are n x n matrices and S2o = So +No is the S-N decomposition of the matrix S2o.

differential operator Lo[y1 = x + a(x)y, while N represents multiplication by v(x) (i.e., the operator: y" -+ v(x)y-). Since A = S + N and SN = NS, it follows

that

We call (V.4.4) the S-N decomposition of operator (V.4.1). We shall show in th next section that Lo[yj is diagonalizable and v(x)n = O.

V-5. A normal form of a differential operator

00

where f2(x) = Ex1fle and Sgt E WC). In §V-4, we derived the S-N decomposi-

t=o

tion

(cf. (V.4.4)). Notice that N = So + No is the S-N decomposition of S2o and that the operator Go[ul and multiplication by v(x) are represented respectively by matrices (V.4.2) and (V.4.3).

Since So is diagonalizable, there exist an invertible n x n matrix P0 and a diagonal matrix Ao such that

SoPo = PoAo, NO = diag[al, A2, ... , . \n)-

Note that A, (j = 1, 2,... , n) are eigenvalues of So. Hence, A. (j = 1, 2,... , n) are also eigenvalues of flo. For every positive integer m, Sm is diagonalizable. Hence,

m is sufficiently large, we can further determine matrices Pt for t > m + 1 by the equations

h=1

Equation (V.5.4) can be solved with respect to Pt, since the linear operator Pt -+

(tl + So)P1 - PtAo is invertible if m is sufficiently large. In this way, we can find

A

P1

entries of P(x)-1 are also formal power series in x and

124 V. SINGULARITIES OF THE FIRST KIND

(5) the transformation

(V.5.9) P(x)u

changes the differential operator (V.5.1) to another differential operator

and

r=

Ynl Inn

with constants "ljk such that

Furthermore, the matrix r as nilpotent.

Remark V-5-2. It is easily verified that the matrix P(x) is a formal solution of the system

xd) = P(x)(Ao + vo(x)) - Sl(x)P(x).

Since the entries of vo(x) are polynomials in x, the power series P(x) is convergent if f2(x) is convergent (cf. Theorem V-2-7). Therefore, in such a case, v(x) is convergent and, hence, o(x) is convergent.

Observation V-5-3. Choose integers l1, ... , to so that A.,+f j = Ak +fk if Aj -Ak is an integer. Then, vo(x) = x4rx-L, where L = diag(e1, f2, ... , en]. If we set

7t{V1 = x'! 1C(xf and ho{v") = x-LICo(x' i , it follows that

dv'

?{o(vl = x'L fXLTf + zLLv + Aoz' 7} = xdx + (Ao + L)v

Hence,

Note that

Ao + L = diag(A1 + P1, A2 + e2, ... , An + enj

and

It was already shower in §V-4 that r is nilpotent. Thus, the following theorem is

proved.

This equation is equivalent to the system

To begin with, let us remark that the S-N decomposition of the matrix 11 is given by S2o = So + No, where

0

2

Now, we calculate in the following steps:

Step 1. Fix a non-negative integer m so that m > 2 - (_a) = a.

Step 2. Find three 2(m + 1) x 2(m + 1) matrices A,,,, Sm, and Nm (cf. §§V-4 and

V-5).

Po

Step 4. Find two 2 x 2 matrices Mm(x) = No + >2xtvt and Qm(x) _ >xtPt

(cf. §§V-4 and V-5).

Step 5. We must obtain Qm(x)-1Mm(x)Qm(x) = vo(x) +O(xm+i), where

where a(a) is a real constant depending on a (cf. Theorem V-5-1).

After these calculations have been completed, we come to the following conclu-

sion.

+00

Conclusion V-5-9. There exists a unique 2 x 2 matrix P(x) = >xtPe such that

1=o

(1)the matrices Pl for t = 0,1, ... , m are given in Step 8,

(2)the power series P(x) converges for every x,

(8)the transformation y = P(x)u" changes (V.5.17) to

We illustrate the scheme given above in the case when a = 0. First we fix m = 2. Then,

that S2P2 = 0. Note that, in this case, Ao = 0.