z0 e

X ’1 (u) b(u) du,

c=

˜

z0

and compare this to (1.11). A su¬cient condition is that the left-hand

side of (1.12) vanishes only for c = 0.

Section 1.5:

3. Verify that it su¬ces to treat the case µ = 2. Then, verify that the

diagonal blocks of the monodromy factor for X(z) are equal to C1 ,

C2 , and let the unknown block below the diagonal be denoted by B.

Next, show Y21 (u exp[2πi]) = Y21 (u) C1 and use this to conclude

z

’1

2πi ˜

X21 (ze ) = X22 (z) C2 (C21 + C21 ) + X22 (u) Y21 (u) du C1 ,

z0

˜

with C21 given by a de¬nite integral. From this identity, conclude

˜

B = C2 (C21 + C21 ) ’ C21 C1 .

4. Use Exercise 8 on p. 7 and Theorem 4.

Section 1.6:

4. For z0 = (ρ + µ) exp[i„ ], ± ¤ „ ¤ β, observe X(z) = X(z0 ) +

z

A(w) X(w) dw. Setting n(x) = X(x1/r ei„ ) , conclude n(x) ¤

z0

x

c + a x0 n(t) dt, for a su¬ciently large c > 0. Then, show by induc-

tion

m

[a(x ’ x0 )]k (x ’ t)m

x

n(x) ¤ c + am+1 n(t) dt

k! m!

x0

k=0

and let m ’ ∞. Finally, discuss the dependence of c upon z0 .

5. Consider X(z) = exp[A z r /r], with A not nilpotent.

6. Consider X(z) = exp[N z r /r], with N nilpotent, but not the zero

matrix. For ν = 1, observe that every solution x(z) can be represented

in the form x(z) = c exp[b(z)], with b (z) = A(z).

8. Consider G = C and X(z) = f (z) I, with an entire function having

in¬nitely many zeros, e.g., f (z) = sin z. Try to generalize this to

arbitrary regions G.

Solutions 247

Section 2.1:

1. (a) Conclude that B(z) = A(z) PN (z) ’ zPN (z) ’ PN (z) A0 is holo-

morphic in D(0, ρ) and vanishes at the origin of order N + 1.

z

(b) For N su¬ciently large, let S0 (z) = 0 B(u) uA0 ’I du z ’A0 , and

˜

z

Sk (z) = 0 A(u) Sk’1 (u) uA0 ’I du z ’A0 , k ≥ 1. Show that each

˜ ˜

˜

Sk (z) is holomorphic in D(0, ρ) and vanishes at the origin of

order N + 1.

(c) For N as above and ρ < ρ, show the existence of a ∈ R, c, K > 0,

˜

with K and a independent of N , such that (with arg z ∈ [0, 2π])

Sk (z) z A0 ¤ c (K/(N ’ a))k |z|N ’a , k ≥ 0, |z| ¤ ρ. Use this

˜ ˜

and the Maximum Principle to show, for su¬ciently large N ,

∞˜

˜

uniform convergence of S(z) = 0 Sk (z), |z| ¤ ρ, and conclude

˜

˜

that S(z) is holomorphic in D(0, ρ) and vanishes at the origin

of N + 1st order.

˜ ˜

(d) Show that X(z) = S(z) z A0 is a solution of (2.4).

3. For existence, use the previous exercise and the form of z J . The re-

n’1

cursion formula is [(n + µ)I ’ A0 ] sn = m=0 An’m sm , n ≥ 1.

4. Follows from the previous exercise.

5. Compute z A0 .

Section 2.2:

1. Let »k denote the two distinct eigenvalues of A, and de¬ne β =

µ1 ’ µ2 , » = »2 ’ »1 , ± = (a ’ »1 ) β/». Then one Floquet solution

is given by x1 (z) = e»1 z z µ1 F (±; β; »z), x2 (z) = c e»1 z z µ1 +1 (1 +

β)’1 F (1 + ±; 2 + β; »z). The second one is given similarly.

2. For »1 , β as above and ω 2 = 4(»1 ’ a)β, resp. µ = (µ1 + µ2 + 1)/2,

one solution is given by x1 (z) = e»1 z z µ “(β) (ω/2)1’β Jβ’1 (ωz 1/2 ),

x2 (z) = e»1 z z µ c “(1 + β) (ω/2)’β’1 Jβ+1 (ωz 1/2 ).

4. Show that the transformation y(z) = ez y (’z) carries the con¬‚uent

˜

hypergeometric equation into one of the same kind, but with ± re-

placed by β ’ ±, and use that the equation has exactly one solution

in form of a power series with constant term equal to 1.

5. Use the Beta Integral (B.11).

7. For z ∈ C , show

d

F (±1 , . . . , ±m ; β1 , . . . , βm ; z) =

dz

248 Solutions

m

±j /βj F (±1 + 1, . . . , ±m + 1; β1 + 1, . . . , βm + 1; z).

j=1

(δ + ±1 ) F (±1 , . . . , ±m ; β1 , . . . , βm ; z) =

±1 F (±1 + 1, ±2 . . . , ±m ; β1 , . . . , βm ; z).

(δ + β1 ) F (±1 , . . . , ±m ; β1 + 1, β2 , . . . , βm ; z) =

β1 F (±1 , . . . , ±m ; β1 , . . . , βm ; z).

From these identities, the di¬erential equation follows easily.

Section 2.3:

2. Use analogous arguments as in Exercise 7 on p. 25.

4. Expand (1 ’ zt)’± , for |z| < 1, into its power series, justify termwise

integration, and use the Beta Integral (B.11).

5. Use Stirling™s formula (B.12) to show convergence of the series; then,

restrict to Re γ > Re β > 0 ≥ Re ± and use (2.10), and ¬nally, use

holomorphy of both sides of (2.11) in all variables ±, β, γ to remove

this restriction.

6. Show that the right-hand side solves the hypergeometic equation.

Then show that the equation has only one power series solution with

constant term equal to 1.

Section 2.4:

1. Observe that a fundamental solution is X(z) = z K z M .

2. Note that in (2.18) we have to set A0 = B0 = B, A1 = A, and the

remaining An equal to 0. For n = 1, use T0 = I to see that the

equation becomes solvable if we take

00

B1 = ,

c0

and then

a b/2

T1 =

0d

follows, if we select the undetermined entry in the (2, 1) position equal

to zero. For n ≥ 2, the recursion equations following from (2.18) are

(n) (n’1) (n’1) (n’1)

’ c t12

n t11 = a t11 + b t21 ,

(n) (n’1) (n’1) (n’1)

(n ’ 1) t21 ’ c t22

= c t11 + d t21 ,

(n) (n’1) (n’1)

(n + 1) t12 = a t12 + b t22 ,

(n) (n’1) (n’1)

n t22 = c t12 + d t22 .

Solutions 249

Solving the system z x = B(z) x, one ¬nds as fundamental solution

˜ ˜

1 0

˜ = zK zM ,

X(z) =

zc log z z

with

0 0

K = diag [0, 1], M= .

c 0

Section 2.5:

(k) (k)

ν ν

3. Set p0 (x) = [x]ν ’ n ≥ 1.

k=1 b0 [x]ν’k , pn (x) = k=1 bn [x]ν’k ,

Then

n’1

n ≥ 1.

p0 (µ + n) yn = pn’m (µ + m) ym ,

m=0

4. For part (c), di¬erentiate the inhomogeneous ODE of (a) with respect

to w.

Section 3.1:

1. Estimate ty by one.

2. Use the previous exercise.

3. Find the recursion relation for the coe¬cients of the inverse, and

estimate similarly to the proof of the Splitting Lemma in Section 3.2.

4. Use Proposition 5 (p. 40), and split T1 (z) = P (z ’1 )(I + O(z ’µ )),

ˆ

with a matrix polynomial P and su¬ciently large µ ∈ N.

6. Use Stirling™s formula (p. 229).

Section 3.2:

(12)

1. Tn = b (1 + a)n’1 , n ≥ 1. Hence, the rate of growth roughly is as

n!, except when the sequence terminates.

4. Use Theorem 8 (p. 44) and the previous exercise.

250 Solutions

Section 3.3:

1. Let c1 , . . . , cν denote the elements in the ¬rst row of C, and de-

note the characteristic polynomial by p(», c1 , . . . , cν ). Expansion with

respect to the last column shows p(», c1 , . . . , cν ) = (’1)ν+1 cν ’

»p(», c1 , . . . , cν’1 ), from which follows

ν

p(», c1 , . . . , cν ) = (’1) [» ’

ν ν

cj »ν’j ].

j=1

2. The only case for A0 = 0 but nilpotent is

0 0

A0 = .

1 0

If the (1, 2)-element of A1 is nonzero, use T (z) = diag [1, z 1/2 ]; oth-

erwise use T (z) = diag [1, z].

3. For any T (z) as in (3.12), make suitable assumptions on the coef-

¬cients A1 , A2 , . . . so that this shearing transformation produces a

system whose leading term has several eigenvalues.

4. From the results of this section, show existence of a q-meromorphic

transformation for which the transformed system, after the change of

variable z = wq , has Poincar´ rank strictly smaller than qr, and then

e

use Exercise 4 on p. 15.

Section 3.4:

1. Set B(z) = z r n=0 An z ’n . Assuming that T (z) commutes with

r ˆ

ˆ ˆ ˆ

B(z), the problem reduces to solving z T (z) = C(z) T (z), C(z) =

∞

A(z) ’ B(z) = n=1 Cn z ’n . This leads to a simple recursion equa-

ˆ

tion for Tn .

2. The pole order is r.

Section 3.5:

An z ’n .

r’1

1. Use the exponential shift x = exp[q(z)], z q (z) = z r n=0

2. In case (a), we have q = 1. Theoretically, an HLFFS is obtained in