8 1. Basic Properties of Solutions

for A(z) = T ’1 (z) [A(z) T (z) ’ T (z)]. Show that the ¬rst µ columns

˜

˜

of A(z) vanish identically. Compare this to Exercise 4 on p. 14.

1.3 Systems in General Regions

We now consider a system (1.1) in a general region G. Given a fundamen-

tal solution X(z), de¬ned near some point z0 ∈ G, we can holomorphically

continue X(z) along any path γ in G beginning at z0 and ending, say, at

z1 . Clearly, this process of holomorphic continuation produces a solution

of (1.1) near the point z1 . Since the path can be split into ¬nitely many

pieces, such that each of them is contained in a simply connected subregion

of G to which the results of the previous section apply, det X(z) cannot

vanish. Thus, X(z) remains fundamental during holomorphic continuation.

According to the monodromy theorem, for a di¬erent path from z0 to z1

the resulting fundamental solution near z1 will be the same provided the

two paths are homotopic. In particular, if γ is a Jordan curve whose inte-

rior region belongs to G, so that γ does not wind around exterior points

of G, then holomorphic continuation of X(z) along γ reproduces the same

fundamental solution that we started with. However, if the interior of γ

contains points from the complement of G, then simple examples in the

exercises below show that in general we shall obtain a di¬erent one. Hence

Theorem 1 (p. 4) fails for multiply connected G, since holomorphic con-

tinuation may not lead to a fundamental solution that is holomorphic in

G, but rather on a Riemann surface associated with G. We shall not go

into details about this, but will be content with the following result for a

punctured disc R(z0 , ρ) = {z : 0 < |z ’ z0 | < ρ}, or slightly more general,

an arbitrary ring R = {z : ρ1 < |z ’ z0 | < ρ}, 0 ¤ ρ1 < ρ:

Proposition 2 Let a system (1.1), with G = R as above, be given. Let

X(z) denote an arbitrary fundamental solution of (1.1) in a disc D =

D(z1 , ρ) ‚ R, ρ > 0. Then there exists a matrix M ∈ C ν—ν such that, for

˜ ˜

a ¬xed but arbitrary choice of the branch of (z ’ z0 )M = exp[M log(z ’ z0 )]

in D, the matrix

S(z) = X(z) (z ’ z0 )’M

is single-valued in R.

Proof: Continuation of X(z) along the circle |z ’ z0 | = |z1 ’ z0 | in the

˜

positive sense will produce a fundamental solution, say, X(z), of (1.1) in D.

According to Exercise 1 on p. 7, there exists an invertible matrix C ∈ C ν—ν

˜

so that X(z) = X(z) C for z ∈ D. Choose M so that C = exp[2πi M ], e.g.,

2πi M = log M . Then, continuation of (z ’z0 )M along the same circle leads

to exp[M (log(z ’ z0 ) + 2πi)] = (z ’ z0 )M C, which completes the proof. 2

1.3 Systems in General Regions 9

While the matrix C occurring in the above proof is uniquely de¬ned by

the fundamental solution X(z), the matrix M is not! We call any such M

a monodromy matrix for X(z). The unique matrix C is sometimes called

the monodromy factor for X(z). Observe that M can be any matrix with

exp[2πi M ] = C. So, in general, 2πi M may have eigenvalues di¬ering by

nonzero integers, and then 2πi M is not a branch for the matrix function

log C.

For G = R, it is convenient to think of solutions of (1.1) as de¬ned on the

Riemann surface of the natural logarithm of z ’ z0 , as described on p. 226

in the Appendix. This surface can best be visualized as a spiraling staircase

with in¬nitely many levels in both directions. For simplicity, take z0 = 0,

then traversing a circle about the origin in the positive, i.e., counterclock-

wise, direction, will not take us back to the same point, as it would in the

complex plane, but to the one on the next level, directly above the point

where we started. Thus, while complex numbers zk = r ei•k , r > 0, are the

same once their arguments •k di¬er by integer multiples of 2π, the corre-

sponding points on the Riemann surface are di¬erent. So strictly speaking,

instead of complex numbers z = r e2πi• , we deal with pairs (r, •). On this

surface, the matrix z M = exp[M log z] becomes a single-valued holomorphic

function by interpreting log z = log r + i•.

The above proposition shows that, once we have a monodromy matrix

M , we completely understand the branching behavior of the corresponding

fundamental solution X(z). It pays to work out the general form of z M for

M = J in Jordan form, in order to understand the various cases that can

occur for the branching behavior of X(z).

The computation of monodromy matrices and/or their eigenvalues is

a major task in many applications. In principle, it should be possible to

¬nd them by ¬rst computing a fundamental solution X(z) by means of

the recursions (1.9), and then iteratively re-expanding the resulting power

series to obtain the analytic continuation. In reality there is little hope of

e¬ectively doing this. So it will be useful to obtain other representations for

fundamental solutions providing more direct ways for ¬nding monodromy

matrices. For singularities of the ¬rst kind, which are discussed in the next

chapter, this can always be done, while for other cases this problem will

prove much more complicated.

Exercises: Throughout these exercises, let M ∈ C ν—ν .

1. Verify that X(z) = z M = exp[M log z] is a fundamental solution of

x = z ’1 M x near, e.g., z0 = 1, if we select any branch for the mul-

tivalued function log z, e.g., its principal value, which is real-valued

along the positive real axis.

2. Verify that X(z) = z M in general cannot be holomorphically contin-

ued (as a single-valued holomorphic function) into all of R(0, ∞).

10 1. Basic Properties of Solutions

3. Verify that M is a monodromy matrix for X(z) = z M .

4. Let Mk ∈ C ν—ν , 1 ¤ k ¤ µ, be such that they all commute with one

another, let A(z) = k=1 (z ’ zk )’1 Mk , with all distinct zk ∈ C ,

µ

and G = C \ {z1 , . . . , zµ }. For each k, 1 ¤ k ¤ µ, and ρ su¬ciently

small, show the existence of a fundamental solution of (1.1) in R(zk , ρ)

having monodromy matrix Mk .

5. For M as above and any matrix-valued S(z), holomorphic and single-

valued with det S(z) = 0 in z ∈ R(0, ρ), for some ρ > 0, ¬nd A(z) so

that X(z) = S(z) z M is a fundamental solution of (1.1).

6. For G = R(0, ρ), ρ > 0, show that monodromy factors for di¬erent

fundamental solutions of (1.1) are similar matrices. Show that the

eigenvalues of corresponding monodromy matrices always are con-

gruent modulo one in the following sense: If M1 , M2 are monodromy

matrices for fundamental solutions X1 (z), X2 (z) of (1.1), then for ev-

ery eigenvalue µ of M1 there exists k ∈ Z so that k+µ is an eigenvalue

of M2 .

7. Under the assumptions of the previous exercise, let M1 be a mon-

odromy matrix for some fundamental solution. Show that one can

choose a monodromy matrix M2 for another fundamental solution

so that both are similar. Verify that, for a given fundamental solu-

tion, one can always choose a unique monodromy matrix that has

eigenvalues with real parts in the half-open interval [0, 1).

8. Under the assumptions of the previous exercises, show the existence

of at least one solution vector of the form x(z) = s(z) z µ , with µ ∈ C

and s(z) a single-valued vector function in G.

9. Consider the scalar ODE (1.6) for ak (z) holomorphic in G = R(0, ρ),

ρ > 0. Show that (1.6) has at least one solution of the form y(z) =

s(z) z µ , with µ ∈ C and a scalar single-valued function s(z), z ∈ G.

1.4 Inhomogeneous Systems

We now return to a simply connected region G, but will consider an inho-

mogeneous system

x = A(z) x + b(z), z ∈ G, (1.10)

where b(z) is a vector-valued holomorphic function on G. We then refer to

(1.1) as the corresponding homogeneous system. As in the real variable case,

we can solve (1.10) as soon as a fundamental solution of the corresponding

homogeneous system (1.1) is known:

1.4 Inhomogeneous Systems 11

Theorem 3 (Variation of Constants Formula) For a simply con-

nected region G, and A(z), b(z) holomorphic in G, all solutions of (1.10)

are holomorphic in G and given by the formula

z

X ’1 (u) b(u) du , z ∈ G,

x(z) = X(z) c + (1.11)

z0

where z0 ∈ C and c ∈ C ν can be chosen arbitrarily.

Proof: It is easily checked that (1.11) represents solutions of (1.10). Con-

versely, if x0 (z) is any solution of (1.10), then for c = X ’1 (z0 ) x0 (z0 ) the

solution x(z) given by (1.11) satis¬es the same initial value condition at z0

as x0 (z). Their di¬erence satis¬es the corresponding homogeneous system,

2

hence is identically zero, owing to Theorem 1 (p. 4).

The somewhat strange name for (1.11) results from the following obser-

vation: For constant c ∈ C ν , the vector X(z) c solves the homogeneous

system (1.1), so we try an “Ansatz” for the inhomogeneous one by re-

placing c by a vector-valued function c(z). Di¬erentiation of X(z) c(z) and

insertion into (1.10) then leads to (1.11).

While (1.11) represents all solutions of (1.10), it requires that we know

a fundamental solution of (1.1), and this usually is not the case. In the

exercises below, we shall obtain at least local representations, in the form

of convergent power series, of solutions of (1.10) without knowing a funda-

mental solution of (1.1).

Exercises: If nothing else is said, let G be a simply connected region in

C and consider an inhomogeneous system (1.10).

1. Expanding A(z) and b(z) into power series about a point z0 ∈ G, ¬nd

the recursion formula for the power series coe¬cients of solutions.

2. In the case of a constant matrix A(z) ≡ A, ¬nd a necessary and

su¬cient condition on A, so that for every vector polynomial b(z) a

solution of (1.10) exists that is also a polynomial of the same degree

as b(z).

3. For

0 0

B(z) = ,

b(z) A(z)

show that x(z) solves (1.10) if and only if x(z) = [1, x(z)]T solves the

˜

homogeneous system x = B(z) x (of dimension ν + 1). Compare this

˜ ˜

to the next section on reduced systems.

4. For G = R(0, ρ), ρ > 0, let X(z) be a fundamental solution of (1.1)

with monodromy matrix M . Show that (1.10) has a single-valued

12 1. Basic Properties of Solutions

solution x(z), z ∈ G, if and only if we can choose a constant vector c

such that for some z0 ∈ G

z0 e2πi

’2πiM

X ’1 (u)b(u) du,

’ I)c =

(e (1.12)

z0

integrating, say, along a circle centered at the origin. Show that a

su¬cient condition for this to be true is that no nontrivial solution

of the homogeneous system (1.1) is single-valued in G.

1.5 Reduced Systems

In this section we shall be concerned with a system (1.1) whose coe¬cient

matrix is triangularly blocked. While corresponding results hold true for

upper triangularly blocked matrices, we choose A(z) to have the following

lower triangular block structure:

®

A11 (z) 0 ... 0

A21 (z) A22 (z) ... 0

A(z) = , (1.13)

. . .

..

° »

. . .

.

. . .

Aµ1 (z) Aµ2 (z) . . . Aµµ (z)

with µ ≥ 2, blocks Ajk (z) that are holomorphic in a (common) simply

connected region G, and such that the diagonal blocks are all square ma-

trices of arbitrary sizes. Such systems will be called reduced. If the diagonal

blocks of (1.13) are of type νk — νk , we sometimes say that (1.1) is reduced

of type (ν1 , . . . , νµ ).

Along with the “large” system (1.1), it is natural to consider the smaller

systems

1 ¤ k ¤ µ.

xk = Akk (z) xk , (1.14)

We show that the computation of a fundamental solution of (1.1) is, aside

from ¬nitely many integrations, equivalent to computing fundamental so-

lutions for (1.14), for every such k:

Theorem 4 Given a matrix A(z) as in (1.13), the system (1.1) has a

fundamental solution of the form

®

X11 (z) 0 ... 0

X21 (z) X22 (z)

... 0

X(z) = ,

. . .

..

° »

. . .

.

. . .

Xµ1 (z) Xµ2 (z) ... Xµµ (z)

1.5 Reduced Systems 13

with Xkk (z) being fundamental solutions of (1.14), and the o¬-diagonal

blocks Xjk (z), for 1 ¤ k < j ¤ µ, recursively given by

z

’1

z ∈ G,

Xjk (z) = Xjj (z) Cjk + Xjj (u) Yjk (u) du , (1.15)

z0