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Linear constant coefficient ordinary differential systems
Ian Tice
Department of Mathematical Sciences
Carnegie Mellon University
March 29, 2022
Contents
0 Overview 2
1 Elliptic ordinary differential systems in arbitrary dimension 4
2 Scalar differential equations with general initial conditions 14
3 General finite dimensional ordinary differential systems 25
4 General systems on real intervals and with real coefficients and data 40
5 Systems on the half line with decay conditions at infinity 44
A Some reminders from complex analysis 57
0 Overview
As the title suggests, these notes concern systems of linear ordinary differential equations with constant, complex coefficients. The reader possessing some familiarity with such problems will likely immediately wonder how such a study can occupy a document of this length. Indeed, one could attempt to summarize the standard theory in a single sentence: rewrite the problem as a first order system, multiply by the appropriate matrix exponential integrating factor, and integrate. The only real subtlety comes in computing the matrix exponential, but this problem is readily dispatched with the Jordan normal form. QED, right? Wrong! There are three hidden assumptions in this pithy description. The first is that the problem can naturally be rewritten as a first order system, making it amenable to the above attack. The second assumption is that we are only interested in specifying the most basic initial conditions, namely the values of the unknown function and its derivatives up to order one less than the order of the system. The third is that the system is finite dimensional, so the Jordan normal form is available for use. The purpose of these notes is to study what happens when we negate these standard assumptions and consider systems that do not naturally rewrite in first order form, systems with other choices of initial conditions, and systems taking values in infinite dimensional complex Banach spaces. When we begin flipping these switches, the above simple picture falls apart pretty quickly and leaves behind some rather tricky issues. Why should we care about flipping these switches? In brief: partial differential equations. All of these variants arise naturally when we attempt to use the tools of ordinary differential equations and systems to study partial differential equations and systems. In fact, the author first encountered a discussion of such general ordinary differential systems with general initial conditions (and more!) while reading the seminal PDE paper Estimates near the boundary for solutions of elliptic partial differential equations satisfying general boundary conditions II by Agmon, Douglis, and Nirenberg [1], in which this theory plays an essential role. In the ADN paper this material is developed quite rapidly, and these notes began as the author’s attempt to fill in some of the details and more deeply understand the material. The only references the author could find addressing general systems were the rather old books of Ince [3] and Poole [4], but while they were certainly helpful they did not contain all of the material needed to process ADN. Hence the existence of these notes, which one could think of as a primer on the ODE analysis needed to understand ADN, though there is more here than strictly needed for that purpose. The reader interested in going further will need a few key tools to make headway. First, it’s a good idea to have some basic experience with ordinary differential equations on R; any undergrad- uate course should suffice. Second, a good grasp of linear algebra over C is required. Third, the basics of complex analysis will be routinely used: holomorphic functions, path integrals, the Cauchy- Goursat theorem, and the residue theorem will play starring roles. In dealing with systems, we will need to work with holomorphic functions f : C → X where X is a complex Banach space. When X is finite dimensional, this is easy, as we can just think of each component being holomorphic, and we define path integrals of such maps in the obvious way, component-wise. However, when X is infinite dimensional some care is needed to work out the properties of holomorphic functions, path integrals, and the versions of Cauchy-Goursat and the residue theorem. The reader who doesn’t know or care to learn this material can simply replace all appearances of X with CN^ in Section 1, which is the only place where the infinite dimensional setting is considered. The reader who doesn’t know but cares to learn this material is directed either to the author’s complex analysis crash course [5] or else to the fantastic book by Dieudonn´e [2], which does all that is needed and more. The reader who knows this material already is given a gold star. Fourth, an ε worth of Lebesgue integration
differentiable on all of C (and hence smooth and analytic by complex analysis). We will write
H(C; X) = {f : C → X | f is holomorphic}. (0.1)
We will use the terminology of [2, 5] when describing complex path integrals. In particular, we refer to “nice” (roughly speaking, almost continuously differentiable) paths as roads and closed paths as loops.
- Given f ∈ H(C; X) we write its zero set as
Z(f ) = {z ∈ C | f (z) = 0}. (0.2)
- We will only explicitly talk about meromorphic functions with values in C. These can be thought of as functions f : C\P (f ) → C that are holomorphic, with each point of P (f ) isolated and consisting of an isolated singularity that is at worst a pole of finite order. The set P (f ) is the polar set.
- Given z 0 ∈ C and R > 0 we write ∂B(z 0 , R) to mean both the set {w ∈ C | |z 0 − w| = R} and the simple counter-clockwise loop parameterized by γ : [0, 1] → C defined via γ(t) = z 0 + Re^2 πit. The latter will always appear in path integrals: ∫
∂B(z 0 ,R)
f (z)dz. (0.3)
- We write L(X, Y ) for the bounded linear maps between complex Banach spaces X and Y , and we write L(X) = L(X, X).
- We follow the common ODE convention of writing the unknown as x : C → X. To highlight the connection with ODE on R we usually write the variable as τ ∈ C rather than t ∈ R. In this way we can think of τ as a sort of complex time variable.
1 Elliptic ordinary differential systems in arbitrary dimen-
sion
We begin our survey of constant coefficient ordinary differential systems by studying the nicest case, in which the system is elliptic. In this case most of the theory works just as well in infinite dimensional complex Banach spaces as it does in C, so we present the Banach framework for the sake of generality. We begin with a definition.
Definition 1.1. Let X and Y be complex Banach spaces and consider a polynomial p : C → L(X, Y ) with deg(p) = n ∈ N of the form p(z) =
∑n k=0 Akz
k, where Ak ∈ L(X, Y ) for 0 ≤ k ≤ n.
- We define the constant coefficient differential operator
p(D) =
∑^ n
k=
AkDk, (1.1)
where D^0 = I is the identity. More precisely, for f : C → X holomorphic, p(D)f : C → Y is defined via
p(D)f =
∑^ n
k=
AkDkf =
∑^ n
k=
Akf (k). (1.2)
The operators A 0 ,... , An are called the coefficients of p(D).
- This induces a linear map p(D) : H(C; X) → H(C; Y ) that we call a linear differential operator of order n = deg(p). We define
ker(p(D)) = {x ∈ H(C; X) | p(D)x = 0} (1.3)
for the kernel of p(D), which is also called the space of homogenous solutions to p(D)x = 0 (here homogenous refers to the fact that the right side of the equation is 0 ).
- We say that p(D) is elliptic if An ∈ L(X, Y ) is invertible.
Some remarks are in order.
Remark 1.2. If p(D) is elliptic, then the invertibility of An ∈ L(X, Y ) requires that X and Y are isomorphic.
Remark 1.3. If X = Y = C, then every constant coefficient differential operator of order n ≥ 0 is elliptic by the definition of the degree of a polynomial.
Remark 1.4. When dim(X) ≥ 2 (possibly infinite) an equation of the form p(D)x = f for a given f ∈ H(C; Y ) is called an ordinary differential system. The term system is used to contrast with the case when dim(X) = 1, in which case the word system is typically replaced with equation.
Our focus for the moment will be elliptic differential operators. If p(D) is elliptic of order 0, then p(D) = A 0 ∈ L(X, Y ) is an isomorphism, so there is nothing to study: the unique solution to p(D)x = f ∈ H(C; Y ) is x = A− 0 1 f ∈ H(C; X). As such, we will restrict our attention to elliptic operators of order n ≥ 1. Our goal is to find conditions to complement the equation p(D)x = f that lead to unique solvability. The most important feature of an elliptic differential operator is found in the following lemma, which establishes an equivalence between elliptic differential operators of arbitrary order and first order elliptic operators. In essence, in the elliptic case it suffices to only consider first order systems.
Lemma 1.5. Let X and Y be complex Banach spaces and let p(D) : H(C; X) → H(C; Y ) be an elliptic differential operator of order n ≥ 1 , written p(D) =
∑n k=0 AkD
k. Define A ∈ L(Xn) in
block-form via
A =
0 I 0 · · · 0
0 0 · · · I 0
0 0 · · · 0 I
−A− n 1 A 0 −A− n 1 A 1 · · · · · · −A− n 1 An− 1
Let f : C → Y be holomorphic and define the holomorphic map F : C → Xn^ via
F = (0,... , 0 , A− n 1 f ). (1.5)
Let ξ 0 ,... , ξn− 1 ∈ X. Then the following are equivalent.
which in turn implies that p(D)x = f. Moreover, the equation Ξ(0) = (ξ 0 ,... , ξn− 1 ) implies that Dkx(0) = ξk for 0 ≤ k ≤ n − 1. It remains only to prove the uniqueness assertion of the first item. If ˜x : C → X is a holo- morphic function such that p(D)˜x = f and Dk^ x˜(0) = ξk, then we may argue as above to pro- duce Ξ = (˜˜ x, ˜x′,... , x˜(n−1)) : C → Xn, a holomorphic function satisfying Ξ˜′^ = AΞ +˜ F and Ξ(0) = (˜ ξ 0 ,... , ξn− 1 ). By the uniqueness assertion of the second item we then have that Ξ = Ξ and˜ hence that x = ˜x.
Remark 1.6. The conditions Dkx(0) = ξk are called initial conditions. This terminology is not completely obvious when we allow any τ ∈ C. Its origin lies in applications in which τ is restricted to [0, ∞) ⊂ R, and is thought of as a time variable parameterizing some evolving process. In this setting the meaning of initial conditions is clear as the time 0 is the initial time in the process.
The lemma establishes the equivalence of two seemingly different problems. We now show that the first order problem is solvable, which allows us to solve both problems.
Theorem 1.7. Let X and Y be complex Banach spaces and p(D) =
∑n k=0 AkD
k (^) : H(C; X) →
H(C; Y ) be an elliptic differential operator of order n ≥ 1. Then the following hold.
- Let A ∈ L(Xn) be determined by the coefficients of p(D) as in Lemma 1.5. For every holo- morphic map F : C → Xn^ and (ξ 0 ,... , ξn− 1 ) ∈ Xn^ there exists a unique holomorphic function Ξ : C → Xn^ satisfying (^) { Ξ′^ = AΞ + F Ξ(0) = (ξ 0 ,... , ξn− 1 ).
Moreover, Ξ is given by the formula
Ξ(τ ) = exp(τ A)(ξ 0 ,... , ξn− 1 ) +
λτ
exp((τ − z)A)F (z)dz, (1.16)
where for any τ ∈ C the road λτ : [0, 1] → C is given by λτ (t) = tτ.
- For every holomorphic function f : C → Y and ξ 0 ,... , ξn− 1 ∈ X there exists a unique holomorphic function x : C → X satisfying { p(D)x = f Dkx(0) = ξk for 0 ≤ k ≤ n − 1.
- The map F : H(C; Xn) → H(C; Xn) × Xn^ given by
F(Ξ) = (Ξ′^ − AΞ, Ξ(0)) (1.18)
is a linear isomorphism.
- The map Φ : H(C; X) → H(C; Y ) × Xn^ given by
Φ(x) = (p(D)x, x(0), x′(0),... , x(n−1)(0)) (1.19)
is a linear isomorphism.
Proof. The third item and fourth items are simply linear algebraic restatements of the first and second items, respectively, so it suffices to prove the first and second. In turn, Lemma 1.5 shows that it suffices to prove the first item. We will thus only prove the first item. The uniqueness of such a Ξ follows from the same argument used to prove the first item implies the second in Lemma 1.5, so we may further reduce to proving the existence of a solution, and for this we will show that Ξ : C → Xn^ defined by (1.16) is holomorphic and satisfies (1.15). Define Ψ : C → Xn^ via Ψ(τ ) =
λτ exp(−zA)F^ (z)dz. We compute
Ψ(τ ) =
0
τ exp(−tτ A)F (tτ )dt, (1.20)
and hence Ψ is holomorphic with
Ψ′(τ ) =
0
[exp(−tτ A)F (tτ ) − τ tA exp(−tτ A)F (tτ ) + τ t exp(−tτ A)F ′(tτ )] dt. (1.21)
Integrating by parts shows that
∫ (^1)
0
τ t exp(−tτ A)F ′(tτ )dt =
0
t exp(−tτ A)
d dt
[F (tτ )]dt
= t exp(−tτ A)F (tτ )|t t=1=0 −
0
[exp(−tτ A) − τ tA exp(−tτ A)]F (tτ )dt
= exp(−τ A)F (τ ) −
0
[exp(−tτ A) − τ tA exp(−tτ A)]F (tτ )dt. (1.22)
Hence, Ψ′(τ ) = exp(−τ A)F (τ ), (1.23)
and we conclude that Ξ is holomorphic and satisfies
Ξ′(τ ) = A exp(τ A)(ξ 0 ,... , ξn− 1 ) + A exp(τ A)
λτ
exp(−zA)F (z)dz + exp(τ A)Ψ′(τ )
= AΞ(τ ) + F (τ ). (1.24)
Moreover,
Ξ(0) = (ξ 0 ,... , ξn− 1 ) +
λ 0
exp(−zA)F (z)dz = (ξ 0 ,... , ξn− 1 ), (1.25)
and existence is proved.
A differential operator p(D) : H(C; X) → H(C; Y ) can be lifted to be viewed as an operator p(D) : H(C; L(X)) → H(C; L(X, Y )), which leads to some very useful theoretical tools called propagators. We define these now.
Definition 1.8.∑ Let X and Y be complex Banach spaces and fix a differential operator p(D) = n k=0 AkD
k (^) with A k ∈ L(X, Y^ ).
- p(D) induces a linear differential operator p(D) : H(C; L(X)) → H(C; L(X, Y )) via
p(D)L =
∑^ n
k=
AkDkL. (1.26)
- Σ(τ ) ∈ L(Xn) for each τ ∈ C, and in block-form we have that
Σ(τ ) =
L 0 (τ ) · · · Ln− 1 (τ ) L′ 0 (τ ) · · · L′ n− 1 (τ ) .. .
L
(n−1) 0 (τ^ )^ · · ·^ L
(n−1) n− 1 (τ^ )
where Lk : C → L(X) is the holomorphic map from Definition 1.8. In particular, the map Σ : C → L(Xn) is holomorphic.
- Σ(0) = I, and for every τ, ω ∈ C we have that Σ(ω)Σ(τ ) = Σ(ω + τ ). In particular, Σ(τ ) ∈ G(L(Xn)) for each τ ∈ C, where G(L(Xn)) ⊂ L(Xn) denotes the set of invertible elements of the Banach algebra L(Xn), and Σ : C → G(L(Xn)) is a group homomorphism.
- We have that Σ(τ ) = exp(τ A), where A ∈ L(Xn) is as in Lemma 1.5.
Proof. The first item follows directly from Theorem 1.9. To prove the second it suffices to prove the third since τ A and ωA commute, and hence
exp(τ A) exp(ωA) = exp((τ + ω)A). (1.33)
However, we will give a direct proof of the second item as it is more enlightening. Let ω ∈ C and let Σ(τ ) = (y 0 ,... , yn− 1 ). Then Σ(ω)(y 0 ,... , yn− 1 ) = (y(ω), y′(ω),... , y(n−1)(ω)), where y = S(y 0 ,... , yn− 1 ). In particular, this means that y : C → X satisfies
{ p(D)y = 0 Dky(0) = yk = Dkx(τ ),
where x = S(ξ 0 ,... , ξn− 1 ). Define z : C → X via z = x(· + τ ). Then p(D)z = 0 and Dkz(0) = Dkx(τ ), and so by uniqueness we have that z = y, i.e. y = x(· + τ ). Hence
Σ(ω)Σ(τ )(ξ 0 ,... , ξn− 1 ) = Σ(ω)(y 0 ,... , yn− 1 ) = (y(ω), y′(ω),... , y(n−1)(ω)) = (x(ω + τ ), x′(ω + τ ),... , x(n−1)(ω + τ )) = Σ(ω + τ )(ξ 0 ,... , ξn− 1 ). (1.35)
This holds for arbitrary (ξ 0 ,... , ξn− 1 ) ∈ Xn, so we conclude that
Σ(ω)Σ(τ ) = Σ(ω + τ ) for all ω, τ ∈ C. (1.36)
This proves the second item. To prove the third item we simply note that, in the language of Lemma 1.5 the map Σ(τ ) is given by Σ(τ )(ξ 0 , · · · , ξn− 1 ) = Ξ(τ ), where Ξ′^ = AΞ and Ξ(0) = (ξ 0 ,... , ξn− 1 ). Theorem 1.7 then shows that Ξ(τ ) = exp(τ A)(ξ 0 ,... , ξn− 1 ), and the third item is proved.
The propagators from Definition 1.8 now give us the ability to write down explicit formulas for the solutions to elliptic differential systems.
Theorem 1.11. Let X and Y be complex Banach spaces and p(D) : H(C; X) → H(C; Y ) be an elliptic differential operator of order n ≥ 1. Let x : C → X and f : C → Y be holomorphic and ξ 0 ,... , ξn− 1 ∈ X. Then the following are equivalent.
- x satisfies (^) { p(D)x = f Dkx(0) = ξk for 0 ≤ k ≤ n − 1.
- x is given by
x(τ ) = S(ξ 0 ,... , ξn− 1 )(τ ) +
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz
n∑− 1
k=
Lk(τ )ξk +
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz
where λτ : [0, 1] → C is the road given by λτ (t) = tτ , and Lk and S are as in Definition 1.8.
Proof. Let Ξ : C → Xn^ be the holomorphic map given by
Ξ(τ ) = exp(τ A)(ξ 0 ,... , ξn− 1 ) +
λτ
exp((τ − z)A)F (z)dz, (1.39)
where F = (0,... , 0 , A− n 1 f ) and A is determined by the coefficients of p(D) as in Lemma 1.5. Theorem 1.7 shows that Ξ satisfies { Ξ′^ = AΞ + F Ξ(0) = (ξ 0 ,... , ξn− 1 ).
Theorem 1.10 then shows that
Ξ(τ ) = Σ(τ )(ξ 0 ,... , ξn− 1 ) +
λτ
Σ(t − z)(0,... , 0 , A− n 1 f )dz, (1.41)
and
Ξ 1 (τ ) =
n∑− 1
k=
Lk(τ )ξk +
λτ
Ln− 1 ((τ − z))A− n 1 f (z)dz. (1.42)
The result then follows immediately from these identities, Lemma 1.5, and the definition of S : Xn^ → ker(p(D)).
Next we turn our attention to certain linear algebraic questions related to ker(p(D)). Our first result establishes some equivalent conditions to check for linearly independent and spanning sets.
Theorem 1.12. Let X and Y be complex Banach spaces and p(D) : H(C; X) → H(C; Y ) be an elliptic differential operator of order n ≥ 1. The following hold.
- Let x, y ∈ ker(p(D)). Then the following are equivalent.
(a) x = y in ker(p(D)). (b) For each τ ∈ C we have that
(x(τ ), x′(τ ),... , x(n−1)(τ )) = (y(τ ), y′(τ ),... , y(n−1)(τ )). (1.43)
On the other hand, (y 0 ,... , yn− 1 ) ∈ Xn^ if and only if
(y 0 ,... , yn− 1 ) = (x(τ ), x′(τ ),... , x(n−1)(τ )) for x = SΣ(−τ )(y 0 ,... , yn− 1 ), (1.48)
so the previous chain of equivalences proves the third item.
An obvious byproduct of this result is that if p(D) has order n ≥ 1, then ker(p(D)) is infinite dimensional when X is and is finite dimensional when X is. In the finite dimensional case it remains to compute the exact dimension of the space.
Theorem 1.13. Let X and Y be finite dimensional complex Banach spaces of dimension N ≥ 1 , and let p(D) : H(C; X) → H(C; Y ) be an elliptic differential operator of order n ≥ 1. The following hold.
- ker(p(D)) is finite dimensional, and dim ker(p(D)) = nN = n dim(X).
- Let ∅ 6 = B ⊆ ker(p(D)). Then the following are equivalent.
(a) B is a basis of ker(p(D)). (b) For each τ ∈ C the set {(x(τ ), x′(τ ),... , x(n−1)(τ ))}x∈B is a basis of Xn. (c) There exists τ ∈ C such that the set {(x(τ ), x′(τ ),... , x(n−1)(τ ))}x∈B is a basis of Xn.
Proof. The first item follows from Theorem 1.9, which shows that ker(p(D)) and Xn^ are isomorphic. The second item follows by combining the second and third items of Theorem 1.12.
The following theorem shows how we can use the propagators {Lk}n k−=0^1 ⊂ L(X) to produce certain nice bases of ker(p(D)) and, conversely, how we can recover these operators from these nice bases of ker(p(D)).
Theorem 1.14. Let X and Y be finite dimensional complex Banach spaces of dimension N ≥ 1 and let p(D) : H(C; X) → H(C; Y ) be an elliptic differential operator of order n ≥ 1. For 0 ≤ k ≤ n − 1 let Bk = {bk, 1 ,... , bk,N } ⊂ X be a basis, and suppose that E = {xkj | 0 ≤ k ≤ n − 1 and 1 ≤ j ≤ N } ⊆ ker(p(D)). Then the following are equivalent.
- For 0 ≤ k ≤ n − 1 and 1 ≤ j ≤ N the function xkj ∈ ker(p(D)) is given by xkj (τ ) = Lk(τ )bk,j.
- E is a basis of ker(p(D)) and for 0 ≤ `, k ≤ n − 1 and 1 ≤ j ≤ N we have that
Dxkj (0) = δkbk,j. (1.49)
In either case, for each 0 ≤ k ≤ n − 1 and τ ∈ C we have that
Lk(τ ) =
∑^ N
j=
xkj (τ )b∗ k,j , (1.50)
where {b∗ k, 1 ,... , b∗ k,N } ⊂ X∗^ is the dual basis associated to Bk.
Proof. The fact that the first item implies the second follows directly from the second item of Theorem 1.13, together with the facts that DLk(0) = δkI, and Bk is a basis of X. Suppose, then, that the second item holds. For 0 ≤ k ≤ n − 1 define RK : C → L(X) via
Rk(τ ) =
∑^ N
j=
xkj (τ )b∗ k,j. (1.51)
i.e. if y =
∑N
j=1 αj^ bk,j^ (and hence^ αj^ =^ b
∗ k,j (y)^ ∈^ C), then
Rk(τ ) =
∑^ N
j=
αj xkj (τ ). (1.52)
We then compute
p(D)Rk(τ ) =
∑^ N
j=
p(D)xkj (τ )b∗ k,j = 0. (1.53)
On the other hand, for any y ∈ X we have that
D`Rk(0)y =
∑^ N
j=
b∗ k,j (y)Dxkj (0) = δk
∑N
j=
b∗ k,j (y)bk,j = δk`y, (1.54)
and hence DRk(0) = δkI. However, Lk : C → L(X) is the unique solution to
{ p(D)Lk = 0 DLk(0) = δkI,
so we deduce that Rk = Lk. Then
Lk(τ )bk,j =
∑^ N
m=
xkm(τ )b∗ k,m(bk,j ) = xkj (τ ) (1.56)
and we conclude that the first item holds.
Remark 1.15. The most common use of the first item of Theorem 1.14 occurs when all of the Bk are the same. In practice, though, using the different bases Bk might be convenient in the second item when the basis {xkj } of ker(p(D)) is found through some ad hoc means.
2 Scalar differential equations with general initial condi-
tions
We now turn our attention to the special case X = C, in which case we call the equation p(D)x = f a scalar ODE. Our goals are two-fold. First, we aim to derive some representation formulas for solutions that are more useful than those found in Theorem 1.11. While the formula from the theorem is useful from a theoretical perspective, it is impractical to work with in most situations. The issue is that for the formula to be useful, we first need to know the propagators, but these
The above representation formula was derived under the assumption that p had n distinct roots, but the resulting formula is agnostic to this fact, which suggests we might try to use it more generally. Before doing this, we we need to make a key observation. The formula produces a solution xh ∈ ker(p(D)) for each h ∈ H(C; C), but ker(p(D)) is of dimension n, while H(C; C) is infinite dimensional. Thus, it’s wild overkill to use generic functions h ∈ H(C; C) in the representation formula. Basic linear algebra suggests that we could reduce to using only h belonging to some subspace of H(C; C) of dimension n, and an obvious choice of such a space is the set of complex polynomials of degree at most n − 1. This will be our strategy. To proceed we first need a couple technical tools. The first examines how this representation formula behaves in a more general context.
Proposition 2.1. Let f : C → C ∪ {∞} be meromorphic such that P (f ) ⊂ B(0, R). Then the function x : C → C defined by
x(τ ) =
2 πi
∂B(0,R)
eτ z^ f (z)dz (2.7)
is holomorphic, and for each k ∈ N we have that
Dkx(τ ) =
2 πi
∂B(0,R)
zkeτ z^ f (z)dz. (2.8)
Moreover, for each z ∈ P (f ) there exists a polynomial pz : C → C such that deg(pz ) ≤ ord(f, z) − 1 , and x(τ ) =
z∈P (f )
eτ z^ pz (τ ) for all τ ∈ C. (2.9)
Proof. If P (f ) = ∅, then Cauchy-Goursat implies that x = 0 and the result follows trivially. Assume then that P (f ) 6 = ∅. Since P (f ) is bounded, it must be finite, so we can write P (f ) = {z 1 ,... , zn} for the n distinct poles of f. Write nk = ord(f, zk) for 1 ≤ k ≤ n. The residue theorem and Proposition A.1 then imply that
x(τ ) =
∑^ n
k=
Res(eτ^ ·f, zk) =
∑^ n
k=
(nk − 1)!
lim z→zk
d dz
)nk − 1 ((z − zk)nk^ eτ z^ f (z)). (2.10)
Define fk : C → C via fk(z) = (z − zk)nk^ f (z), which is holomorphic by the definition of the order of a pole. From the Leibniz rule, we compute
Dnk^ −^1 (eτ z^ fk(z)) =
n∑k − 1
j=
(nk − 1)! j!(nk − 1 − j)!
τ j^ eτ z^ Dnk^ −j−^1 fk(z), (2.11)
and hence
x(τ ) =
∑^ n
k=
eτ zk
n∑k − 1
j=
cjkτ j^ (2.12)
for some constants cjk ∈ C for 1 ≤ k ≤ n and 0 ≤ j ≤ nk − 1. Hence x is a linear combination of exponentials multiplied by polynomials and is thus holomorphic. In particular, (2.9) is proved. It remains only to prove (2.8). According to Cauchy-Goursat, we have that
x(τ ) =
2 πi
∂B(0,R)
eτ z^ f (z)dz =
0
R exp(τ Re^2 πit)f (Re^2 πit)dt, (2.13)
and hence
Dkx(τ ) =
0
R(Re^2 πit)k^ exp(τ Re^2 πit)f (Re^2 πit)dt =
2 πi
∂B(0,R)
zkeτ z^ f (z)dz, (2.14)
which is (2.8).
The second technical tool associates to a polynomial p : C → C of degree n a collection of polynomials q 0 ,... , qn− 1 with some properties that will be extremely useful in working with the linear algebra associated to our representation formula.
Proposition 2.2.∑ Suppose that p : C → C is a polynomial of degree n ≥ 1 given by p(z) = n m=0 amz
n−m. For 0 ≤ j ≤ n − 1 define the polynomials qj : C → C via
qj (z) =
n ∑− 1 −j
m=
amzn−j−^1 −m. (2.15)
Let 0 < R be such that Z(p) ⊆ B(0, R). Then for each 0 ≤ j, k ≤ n − 1 we have that
1 2 πi
∂B(0,R)
zkqj (z) p(z)
dz = δjk (2.16)
Proof. First note that the degree of the polynomial z 7 → zkqj (z) is n − 1 − j + k. If k < j, then n − 1 − j + k ≤ n − 2, and so Proposition A.2 implies that
1 2 πi
∂B(0,R)
zkqj (z) p(z)
dz = 0. (2.17)
On the other hand, if j ≤ k, then
zkqj (z) = zk(a 0 zn−j−^1 + · · · + an−j− 1 ) = zk−j−^1 (a 0 zn^ + · · · + an−j− 1 zj+1) = zk−j−^1 (p(z) − (an + an− 1 z + · · · + an−j zj^ )) =: zk−j−^1 p(z) − z−^1 rj,k(z), (2.18)
where deg(rj,k) ≤ k ≤ n − 1 ≤ deg(zp(z)) − 2, so Proposition A.2 again implies
2 πi
∂B(0,R)
zkqj (z) p(z)
dz =
2 πi
∂B(0,R)
zk−j−^1 dz +
2 πi
∂B(0,R)
rj,k(z) zp(z)
dz
2 πi
∂B(0,R)
zk−j−^1 dz = δjk. (2.19)
This suggests some notation.
Definition 2.3.∑ Suppose that p : C → C is a polynomial of degree n ≥ 1 given by p(z) = n m=0 amz
n−m. The polynomials q 0 ,... , qn− 1 : C → C given by Proposition 2.2 are called the
polynomials associated to p.
The third technical tool defines a useful linear map from H(C; C) to C∞(C^2 ; C).
(b) x is given by
x(τ ) =
n∑− 1
k=
2 πi
∂B(0,R)
eτ z^
ξkqk(z) p(z)
dz +
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz
2 πi
∂B(0,R)
eτ z
n∑− 1
k=
ξkqk(z) + T f (τ, z)qn− 1 (z)
dz p(z)
where λτ : [0, 1] → C is the road given by λτ (t) = tτ and T f is as in Lemma 2.4.
Proof. We begin with the proof of the first item. Define Rk : C → C via
Rk(τ ) =
2 πi
∂B(0,R)
eτ z^
qk(z) p(z)
dz. (2.27)
Proposition 2.1 shows that Rk is holomorphic and
p(D)Rk(τ ) =
2 πi
∂B(0,R)
p(z)eτ z^
qk(z) p(z)
dz =
2 πi
∂B(0,R)
eτ z^ qk(z)dz = 0, (2.28)
where the last equality follows from Cauchy-Goursat. On the other hand, the properties of the associated polynomials {qk}n k−=0^1 imply that
Dj^ Rk(0) =
2 πi
∂B(0,R)
zj^
qk(z) p(z)
dz = δjk, (2.29)
and so Rk = Lk by uniqueness. The first item and Theorem 1.11 then imply the equivalence of (2.25) and the first identity in (2.26). It remains only to prove that ∫
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz =
2 πi
∂B(0,R)
T f (τ, z)
qn− 1 (z) p(z)
dz. (2.30)
To this end we first use the first item to write
Ln− 1 (z) =
2 πi
∂B(0,R)
ezw^
qn− 1 (w) p(w)
dw =
0
R 22 πiθ^ exp(zRe^2 πiθ)
qn− 1 (Re^2 πiθ) p(Re^2 πiθ)
dθ. (2.31)
Then ∫
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz =
0
τ Ln− 1 ((1 − t)τ )A− n 1 f (tτ )dt
0
τ A− n 1 f (tτ )
0
Re^2 πiθ^ exp((1 − t)τ Re^2 πiθ)
qn− 1 (Re^2 πiθ) p(Re^2 πiθ)
dθ
dt. (2.32)
All of the functions being integrated are smooth functions valued in C. We may then expand into real and imaginary parts and apply Fubini’s theorem to compute
∫
λτ
Ln− 1 (τ − z)A− n 1 f (z)dz =
0
Re^2 πiθ^
qn− 1 (Re^2 πiθ) p(Re^2 πiθ)
0
τ exp((1 − t)τ Re^2 πiθ)A− n 1 f (tτ )dt
dθ
2 πi
∂B(0,R)
qn− 1 (z) p(z)
A− n^1
λτ
e(τ^ −w)z^ f (w)dw
dz =
2 πi
∂B(0,R)
qn− 1 (z) p(z)
T f (τ, z)dz. (2.33)
This is (2.30).
The benefit of our new representation formula is that the solution is solely expressed in terms of the polynomial p and the data ξ 0 ,... , ξn− 1 and f. We don’t need to compute the propagators first. The following example illustrates one way this presents an advantage.
Example 2.6. Fix n ≥ 1 and let Ω be a metric space. Suppose that a 0 ,... , an : Ω → C are continuous and that an(ω) 6 = 0 for all ω ∈ Ω. Define π : C × Ω → C via
π(z, ω) =
∑^ n
j=
aj (ω)zj^ , (2.34)
which in particular means, thanks to our assumption about an, that π(·, ω) : C → C is a polynomial of degree n for every ω ∈ Ω. Fix ξ 0 ,... , ξn− 1 ∈ C and f ∈ H(C; C). We define x : C × Ω → C via the condition that x(·, ω) ∈ H(C; C) is the unique solution to
{ π(D, ω)x(τ, ω) = f (τ ) Dkx(0, ω) = ξk for 0 ≤ k ≤ n − 1
for each ω ∈ Ω, where D acts only in the first variable. A natural question arises: how does x change as we change ω? We can get some very useful information from our representation, but first we need a key observation. The associated polynomials will now also depend on ω since the coefficients aj do. We write qj (z, ω) to emphasize this. The formula for these shows that
qj (z, ω) =
n− ∑ 1 −j
m=
an−m(ω)zn−j−^1 −m. (2.36)
Fix ω 0 ∈ Ω and let R 0 be such that Z(π(·, ω 0 )) ⊂ B(0, R 0 ). Since the roots of π(·, ω) vary continuously with ω (see Theorem A.3), we can pick ε > 0 such that Z(π(·, ω)) ⊂ B(0, R 0 ) for all ω ∈ BΩ(ω 0 , ε). For such ω we can plug into our representation formula to see that
x(τ, ω) =
2 πi
∂B(0,R 0 )
eτ z
∑n−^1
k=
ξk
qk(z, ω) π(z, ω)
dz for all τ ∈ C. (2.37)
Using this and the dominated convergence theorem, we deduce that
lim (τ,ω)→(τ 0 ,ω 0 )
x(τ, ω) = x(τ 0 , ω 0 ) (2.38)
for every τ 0 ∈ C and ω 0 ∈ Ω. Hence, x ∈ C^0 (C × Ω; C). Suppose now that we have the extra information that Ω is an open subset of a normed vector space and that each ak is Cm^ for some m ≥ 1. Then this argument can be readily modified to deduce that for 1 ≤ j ≤ m,
∂ 2 jx(τ, ω 0 ) =
2 πi
∂B(0,R 0 )
eτ z
∑n−^1
k=
ξk ∂ 2 j
qk(z, ω) π(z, ω)
ω=ω 0
dz, (2.39)
which is well-defined since the quotient rule shows the poles of the term in parentheses are a subset of the zeros of π(·, ω 0 ). In turn, we can also prove that x ∈ Cm(C × Ω; C).
