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( 61 .)


Proof: Assume for the moment that we were given a solution x(z), holo-
morphic in D. Then the coordinates of the vector function x(z) all can be
1.1 Simply Connected Regions 3

expanded into power series about z0 , with a radius of convergence at least
ρ. Combining these series into a vector, we can expand x(z) into a vector
power series 1

xn (z ’ z0 )n , |z ’ z0 | < ρ,
x(z) = (1.2)

where xn ∈ C ν are the coe¬cient vectors. Likewise, we can expand the
coe¬cient matrix A(z) into a matrix power series

An (z ’ z0 )n , |z ’ z0 | < ρ,
A(z) = (1.3)

with coe¬cient matrices An ∈ C ν—ν . Inserting these expansions into the
system (1.1) and comparing coe¬cients leads to the identities
n ≥ 0.
(n + 1) xn+1 = An’m xm , (1.4)

Hence, given x0 , we can recursively compute xn for n ≥ 1 from (1.4), which
proves the uniqueness of the solution. To show existence, it remains to
check whether the formal power series solution of our initial value problem,
resulting from (1.4), converges for |z ’ z0 | < ρ. To do this, note that
convergence of (1.3) implies An ¤ c K n for every constant K > 1/ρ and
su¬ciently large c > 0, depending on K. Hence, equation (1.4) implies (n+
1) xn+1 ¤ c m=0 K n’m xm , n ≥ 0. De¬ning a majorizing sequence
(cn ) by c0 = x0 , resp. (n + 1) cn+1 = c m=0 K n’m cm , n ≥ 0, we
conclude by induction that xn ¤ cn , for n ≥ 0. The power series f (z) =
∞ n
0 cn z can be easily checked to formally satisfy the linear ODE y =
c (1’Kz)’1 y. This equation has the solution y(z) = c0 (1’Kz)’c/K , which
is holomorphic in the disc |z| < 1/K. Expanding this function into its power
series about the origin, inserting into the ODE and comparing coe¬cients,
one checks that the coe¬cients satisfy the same recursion relation as the
cn , hence are, in fact, equal to the numbers cn . This proves that the radius
of convergence of (1.2) is at least 1/K, and since K was arbitrary except
for K > 1/ρ, the proof is completed.2 2
Using the above local result together with the monodromy theorem on
p. 225 in the Appendix, it is now easy to show the following global version
of the same result:

1 Observe that whenever we write |z ’ z0 | < ρ, or a similar condition on z, we wish to
state that the corresponding formula holds, and here in particular the series converges,
for such z.
2 An alternative proof for convergence of f (z) is as follows: Show (n + 1) c
n+1 =
n n’m c
m = (c + Kn) cn , hence the quotient test implies convergence of f (z)
c m=0 K
for |z| < 1/K. While this argument is simpler, it depends on the structure of the recursion
relation for (cn ) and fails in more general cases; see, e.g., the proof of Lemma 2 (p. 28).
4 1. Basic Properties of Solutions

Theorem 1 Let a system (1.1), with A(z) holomorphic in a simply con-
nected region G ‚ C , be given. Then for every z0 ∈ G and every x0 ∈ C ν ,
there exists a unique vector-valued function x(z), holomorphic in G, such
x (z) = A(z) x(z), z ∈ G, x(z0 ) = x0 . (1.5)

Proof: Given any path γ in G originating at z0 , we may cover the path with
¬nitely many circles in G, such that when proceeding along γ, each circle
contains the midpoint of the next one. Applying Lemma 1 successively to
each circle, one can show that the unique local solution of the initial value
problem can be holomorphically continued along the path γ. Since any two
paths in a simply connected region are always homotopic, the monodromy
theorem mentioned above completes the proof.

Exercises: In the following exercises, let G be a simply connected region,
and A(z) a matrix-valued function, holomorphic in G.
1. Show that the set of all solutions of (1.1) is a vector space over C .
For its dimension, see Theorem 2 (p. 6).
2. Give a di¬erent proof of Lemma 1, analogous to that of Picard-
Lindel¨f™s theorem in the real variable case.
3. Check that the proof of the previous exercise, with minor modi¬ca-
tions, may be used to prove the same result with the disc D replaced
by the largest subregion of G, which is star-shaped with respect to
z0 .
4. Use Riemann™s mapping theorem [1] and Lemma 1 to obtain another
proof of Theorem 1.
5. Consider the following νth-order linear ODE:

y (ν) ’ a1 (z) y (ν’1) ’ . . . ’ aν (z) y = 0, (1.6)

where ak (z) are (scalar) holomorphic functions in G. Introducing the
® 
0 1 ... 0 0
0 0
0 ... 0
 
 . ,
. . .
A(z) =  .
. . .
. . . .

°0 1»
0 ... 0
aν (z) aν’1 (z) . . . a2 (z) a1 (z)

conclude from Theorem 1 that all solutions of (1.6) are holomorphic
in G. A matrix of the above form will be called a companion matrix
corresponding to the row vector a(z) = (aν (z), . . . , a1 (z)).
1.2 Fundamental Solutions 5

6. Consider the following second-order ODE, usually called Legendre™s
[(1 ’ z 2 ) x ] + µ x = 0, µ ∈ C. (1.7)

(a) Write (1.7) in the form (1.6) and determine where the coe¬cients
ak (z) are holomorphic resp. singular.

(b) Insert x(z) = 0 xn z n into (1.7), compare coe¬cients, and ¬nd
the resulting recursion for the xn . Without explicitly ¬nding the
coe¬cients, ¬nd the radius of convergence of the power series.
(c) For µ = m (m + 1), m ∈ N0 , show that (1.7) has a solution
that is a polynomial of degree m. These polynomials are called
Legendre™s polynomials.
(d) Verify that the values µ = m (m + 1) are the only ones for which
a nontrivial polynomial solution can exist.

7. For a system (1.1), choose an arbitrary row vector t0 (z) that is holo-
morphic in G, and de¬ne inductively

z ∈ G, k ≥ 0.
tk+1 (z) = tk (z) + tk (z) A(z), (1.8)

Let T (z) be the matrix with rows t0 (z), . . . , tν’1 (z); hence T (z) is
holomorphic in G.

(a) Show that one can choose t0 (z) so that det T (z) does not vanish
identically on G. If this is so, the vector t0 (z) will be called a
cyclic vector for A(z). More precisely, for arbitrary z0 ∈ G and
su¬ciently small ρ > 0, show the existence of a cyclic vector for
which det T (z) = 0 on D(z0 , ρ).
(b) For T (z) as above, de¬ne b(z) by tν (z) = b(z) T (z) on D, and let
B(z) be the companion matrix corresponding to b(z). Conclude
for x(z) = T (z) x(z) that x(z) solves (1.1) (on G) if and only if
x (z) = B(z) x(z), z ∈ D. Compare this to the previous exercise.
˜ ˜

1.2 Fundamental Solutions
As is common in the real theory of linear systems of ODE, we say that a
ν — ν matrix-valued function X(z) is a fundamental solution of (1.1), if all
columns are solutions of (1.1), so that in particular X(z) is holomorphic
in G, and if in addition the determinant of X(z) is nonzero; note that
according to the following proposition det X(z0 ) = 0 for some z0 ∈ G
already implies det X(z) = 0 for every z ∈ G:
6 1. Basic Properties of Solutions

Proposition 1 (Wronski™s Identity) Consider a holomorphic matrix-
valued function X(z) satisfying X (z) = A(z) X(z) for z ∈ G, and let
w(z) = det X(z) and a(z) = trace A(z). Then
z ∈ G,
w(z) = w(z0 ) exp a(u) du ,

for arbitrary z0 ∈ G. Hence in particular, either w(z) ≡ 0 or w(z) = 0 for
every z ∈ G.

Proof: The de¬nition of determinants shows w (z) = k=1 wk (z), where
wk (z) is the determinant of the matrix obtained from X(z) by di¬erenti-
ating its kth row and leaving the others. If rk (z) denotes the kth row of
X(z), then X (z) = A(z) X(z) implies rk (z) = j=1 akj (z) rj (z). Using
this and observing that the determinant of matrices with two equal rows
vanishes, we obtain wk (z) = akk (z) w(z), hence

w (z) = a(z) w(z).

Solving this di¬erential equation then completes the proof.
Existence of fundamental solutions is clear: For ¬xed z0 ∈ G, take the
unique solution of the initial value problem (1.5) with x0 = ek , the kth
unit vector, for 1 ¤ k ¤ ν. Combining these vectors as columns of a matrix
X(z), we see that det X(z0 ) = 1; hence X(z) is a fundamental solution.
The signi¬cance of fundamental solutions is that, as for the real case, every
solution of (1.1) is a linear combination of the columns of a fundamental

Theorem 2 Suppose that X(z) is a fundamental solution of (1.1), and
let x(z) be the unique solution of the initial value problem (1.5). For c =
X ’1 (z0 ) x0 , we then have

z ∈ G.
x(z) = X(z) c,

Thus, the C -vector space of all solutions of (1.1) is of dimension ν, and
the columns of X(z) are a basis.

Proof: First observe that X(z) c, for any c ∈ C ν , is a solution of (1.1).
De¬ning c as in the theorem then leads to a solution satisfying the initial
value condition, so the proof is completed.
In principle, the computation of the power series expansion of a funda-
mental solution of (1.1) presents no new problem: For z0 ∈ G and x0 = ek ,
1 ¤ k ¤ ν, compute the power series expansion of the solution of (1.5) as
in the proof of Lemma 1, thus obtaining a power series representation of a
1.2 Fundamental Solutions 7

Xn (z ’ z0 )n , with
fundamental solution near z0 of the form X(z) = 0
X0 = I, and
n ≥ 0.
(n + 1) Xn+1 = An’m Xm , (1.9)
Re-expanding this power series, one then can holomorphically continue the
fundamental solution into all of G. However, in practically all cases, it will
not be possible to compute all coe¬cients Xn , and even if we succeeded,
the process of holomorphic continuation would be extremely tedious, if not
impossible. So on one hand, the recursion equations (1.9) contain all the
information about the global behavior of the corresponding fundamental so-
lution, but to extract such information explicitly must be considered highly
non-trivial. Much of what follows will be about other ways of represent-
ing fundamental solutions, which then allow us to learn more about their
behavior, e.g, near a boundary point of the region G.

Exercises: Throughout the following exercises, let a system (1.1) on a
simply connected region G be arbitrarily given.
1. Let X(z) be a fundamental solution of (1.1). Show that X(z) is an-
other fundamental solution of (1.1) if and only if X(z) = X(z) C for
some constant invertible ν — ν matrix C.
2. For A(z) = A z k , with k ∈ N0 and A ∈ C ν—ν (and G = C ), show
that X(z) = exp[A z k+1 /(k + 1)] is a fundamental solution of (1.1).
3. For B(z) commuting with A(z) and B (z) = A(z), z ∈ G, show that
X(z) = exp[B(z)], z ∈ G, is a fundamental solution of (1.1).
4. For ν = 2 and a11 (z) = a22 (z) ≡ 0, a21 (z) ≡ 1, a12 (z) = z, show
that no B(z) exists so that B(z) commutes with A(z) and B (z) =
A(z), z ∈ G, for whatever region G.
5. Let X(z) be a holomorphic matrix, with det X(z) = 0 for every z ∈ G.
Find A(z) such that X(z) is a fundamental solution of (1.1).
6. Show that X(z) is a fundamental solution of (1.1) if and only if
[X ’1 (z)]T is one for x = ’AT (z) x.
˜ ˜
7. Let x1 (z), . . . , xµ (z), µ < ν, be solutions of (1.1). Show that the rank
of the matrix X(z) = [x1 (z), . . . , xµ (z)] is constant, for z ∈ G.
8. Let x1 (z), . . . , xµ (z), µ < ν, be linearly independent solutions of (1.1).
Show existence of holomorphic vector functions xµ+1 (z), . . . , xν (z),
z ∈ G, so that for some, possibly small, subregion G ‚ G we have
det[x1 (z), . . . , xν (z)] = 0 on G. For T (z) = [x1 (z), . . . xν (z)], set x =
T (z) x and conclude that x(z) satis¬es (1.1) if and only if x solves
˜ ˜
˜ ˜
z ∈ G,
x = A(z) x,


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