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. 51
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2πi
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

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( 61 .)



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