Asymptotic Laplace transforms and evolution equations

Günter Lumer and Frank Neubrander

1 Introduction

The application of classical Laplace transform techniques to differential or integral equations requires one to impose restrictive growth assumptions on the functions involved. To avoid this flaw, the Argentinian mathematician J.C. Vignaux [22] extended the concept of asymptotic power series to Laplace transforms. In the first part of this paper we will give an introduction to Vignaux’s method of asymptotic Laplace transforms and prove the following representation results, among others. Consider analytic functions r : Ω → X, where Ω ⊂ ℂ and X is a Banach space. Then r is the Laplace transform

r(λ) = ∫_{0}^{∞} e^{−λt}u(t) dt

of a generalized function u : [0, ∞) → X if its domain Ω contains some right halfplane {λ ∈ ℂ : Reλ > ω} on which it grows less than exponentially.^{1} Moreover, the faster a function r grows at infinity, the less regular the generalized function u will be. If the domain Ω does not contain any right half-plane, then r cannot be represented as a Laplace transform. However, if Ω still contains a post-sectorial region^{2} on which r has less than exponential growth, then, for all 0 < T ≤ ∞, there exists a generalized function u : [0, T) → X such that r is an asymptotic Laplace transform of u. Moreover, more growth of an analytic function r at infinity, a larger interval [0, T), or a smaller post-sectorial region S lead to a decay in regularity of u.

1Precise definitions of generalized functions and their Laplace transforms will be given below.

2A region S is said to be post-sectorial if for all 0 < α < π/2 there exists a constant r_{α} > 0 such that the rays re^{iβ} are subsets of S for all r > r_{α} and all −α < β < α. The vector space of all analytic functions defined in some post-sectorial region S with values in X is denoted by A(P; X).

In the second part of the paper we show how asymptotic Laplace transform methods can be used to characterize the existence, uniqueness and regularity of classical or generalized solutions u : [0, T) → X (0 < T ≤ ∞) of evolution equations

u′(t) = Au(t), 0 ≤ t < T, u(0) = x,  (ACP)

where A is a closed linear operator with domain D(A) and range in X. We consider the general case of a closed linear operator; i.e., no a priori spectral assumptions will be made and we will allow nondense domains. The aim of the second section is twofold. First, for a given closed linear operator A we characterize the set of all initial values x for which (ACP) has a classical or generalized solution. Second, we give sufficient conditions on A which imply the existence of unique generalized solutions of (ACP) on [0, T) for all x ∈ X.

2 Asymptotic Laplace Transforms

The Laplace transform f̂(λ) = ∫_{0}^{∞} e^{−λt}f(t) dt of a function f ∈ L_{loc}^{1}([0, ∞); X) exists as an improper Bochner integral for some λ ∈ ℂ if and only if F (t) := ∫_{0}^{t} f(r) ds is of exponential type.^{3} In this case f̂ exists and is analytic in some right half-plane. The purpose of this section is to present an asymptotic version of the Laplace transform which has the additional benefit that any f ∈ L_{loc}^{1} becomes transformable while the basic operational properties of the Laplace transform remain valid.^{4}

For 0 < T < ∞, the T -asymptotic Laplace transform of f ∈ L_{loc}^{1}([0, ∞); X) is given by the set of analytic functions r defined on some post-sectorial region S with values in X which are asymptotically equal to the finite Laplace transform of f; i.e.,

{f}_{T} := {r ∈ A(P; X) : r(λ) − ∫_{0}^{T} e^{−λt}f(t) dt ≈_{T} 0},

3See, for example, G. Doetsch [15], Kap. 2.2. A function F is of exponential type if there exist positive constants M and ω such that ∥F (t)∥ ≤ Me^{ωt} for all t ≥ 0.

4The concept of asymptotic Laplace transforms has its roots in the theory of asymptotic power series of analytic functions; see, e.g., R. Remmert [21], 9.6. With a slightly different definition, asymptotic Laplace transforms were considered first by J.C. Vignaux [22] and further investigated by J.C Vignaux and M. Cotlar [23], W.A. Ditkin [14], L. Berg [7], [8], Yu. I. Lyubich [20], and M.

Deakin [13]. However, the method and the scope of its applicability remained largely unnoticed.

where a function a : (ω, ∞) → X satisfies a ≈_{T} 0 if ^{5}

lim sup_{λ→∞} \frac{1}{λ} ln ∥a(λ)∥ ≤ −T.

The asymptotic Laplace transform {f} of a function f ∈ L_{loc}^{1}([0, ∞); X) is defined as the intersection of the sets {f}_{T}; i.e.,

{f} := ∩_{T >0}{f}_{T} = {r ∈ A(P; X) : r(λ) − ∫_{0}^{T} e^{−λt}f(t) dt ≈_{T} 0 for all T > 0}.

To simplify notation let {f}_{∞} := {f}. It is easy to see that f → {f}_{T} is linear; i.e., {f}_{T} + {g}_{T} = {f + g}_{T} and {cf}_{T} = c{f}_{T} for 0 < T ≤ ∞, f, g ∈ L_{loc}^{1} and c ∈ ℂ.

Theorem 2.1. (Existence and Uniqueness)^{6} Let f, g ∈ L_{loc}^{1}([0, ∞); X) and 0 < T ≤ ∞. Then

(a) ∅ ≠ {f} ⊂ {f}_{T};

(b) if f̂ exists, then f̂ ∈ {f};

(c) {f}_{T} ∩ {g}_{T} ≠ ∅ if and only if f = g a.e. on [0, T].

Proof. (a) For f ∈ L_{loc}^{1} and 0 < T < ∞ define f_{T} := f ·χ_{[0,T]}. Then the finite Laplace transform f̂_{T} (λ) = ∫_{0}^{T} e^{−λt}f(t) dt = ∫_{0}^{∞} f_{T} (t) dt is an entire function and

{f}_{T} = f̂_{T} + {0}_{T} ≠ ∅,  (2.1)

where {0}_{T} := {a ∈ A(P; X) : a ≈_{T} 0}. Let 0 < T′ < T < ∞, F (t) := ∫_{0}^{t} f(s) ds (t ≥ 0) and h(λ) := ∫_{T′}^{T} e^{−λt}f(t) dt = e^{−λT} F (T) − e^{−λT′}F (T′) + λ ∫_{T′}^{T} e^{−λt}F (t) dt. Since F is bounded on [T′, T], it follows that h ≈_{T′} 0. Thus {f}_{T} = f̂_{T′} + h + {0}_{T} ⊂ f̂_{T′} + h + {0}_{T′} = f̂_{T′} + {0}_{T′} = {f}_{T′}. This shows that

{f} = ∩_{T >0}{f}_{T} ⊂ {f}_{T} ⊂ {f}_{T′} if 0 < T′ < T < ∞.

5If one requires a ≈_{T} 0 if and only if a ∈ C^{∞}((ω, ∞); X) and lim sup_{λ→∞} \frac{1}{λ} ln ∥a^{(n)}(λ)∥ ≤ −T for all n ∈ ℕ_{0}, then the asymtotic Laplace transform has the additional operational property {(−t)^{n}f}_{T} = {f}^{(n)}, while all other results presented here remain valid; see B. Bäumer [2]. However, the definition above suffices for the purposes of this paper and the proofs extend easily to the more general case. One should notice that ≈_{T} defines an equivalence relation on A(P; X). Furthermore, if r ∈ A(P; X) and q ∈ A(P; ℂ) with r ≈_{T} 0 and q ≈_{S} 0, then qr ≈_{T +S} 0.

6This theorem is essentially due to Vignaux [22] who extended the Ritt’s Theorem about asymtotic series to asymptotic integrals; see, e.g., R. Remmert [21], 9.6.4 and also J.C. Vignaux and M. Cotlar [23], L. Berg [7], and Yu. I. Lyubich [20].

To see that the asymptotic Laplace transform f → {f} is well defined (i.e., {f} ≠ ∅ for all f ∈ L_{loc}^{1}((0, ∞); X)), define

r(λ) := λ ∫_{0}^{∞} e^{−λt}(1 − e^{− \frac{d(λ)}{G(t)}} ) F (t) dt,  (2.2)

where G(t) := max{∥F (t)∥, 1}, and λ → d(λ) is an analytic function to be chosen below such that r ∈ {f}_{T} for all T > 0. Since |1 − e^{−z}| = |z ∫_{0}^{1} e^{−zt} dt| ≤ |z| for all

z ∈ ℂ with Re(z) ≥ 0, it follows that ∥e^{−λt}(1 − e^{− \frac{d(λ)}{G(t)}} ) F (t)∥ ≤ e^{−Re(λ)t}|d(λ)| for all t ≥ 0 and λ ∈ Ω := {λ ∈ ℂ : Re(λ) > 0 and Re(d(λ)) ≥ 0}. If we assume that Ω contains a post-sectorial region, then r ∈ A(P; X) and satisfies the growth estimate

∥r(λ)∥ ≤ \frac{1}{Re(λ)}|λd(λ)|.

Consider a(λ) = r(λ) − ∫_{0}^{T} e^{−λt}f(t) dt = r(λ) + λe^{−λT} F (T) − λ ∫_{0}^{T} e^{−λt}F (t)dt =

λ ∫_{T}^{∞} e^{−λt}(1−e^{− \frac{d(λ)}{G(t)}} )F (t) dt−λ ∫_{0}^{T} e^{−λt}e^{− \frac{d(λ)}{G(t)}} F (t) dt+λe^{−λT} F (T) = a_{1}(λ)−a_{2}(λ)+ a_{3}(λ). Then r ∈ {f} if and only if a ≈_{T} 0 for all T > 0. Clearly, a_{3}(λ) = λe^{−λT} F (T) satisfies a_{3} ≈_{T} 0 for all T > 0. Consider a_{2}(λ) = λ ∫_{0}^{T} e^{−λt}e^{− \frac{d(λ)}{G(t)}} F (t) dt. If λ ∈ Ω ∩ ℝ_{+}, then ∥a_{2}(λ)∥ ≤ MT λe^{−Re(d(λ))/M}, where M = sup_{0≤t≤T} ∥G(t)∥, and therefore lim sup_{λ→∞} \frac{1}{λ} ln ∥a_{2}(λ)∥ ≤ − lim inf_{λ→∞} \frac{1}{λ}Re(d(λ)). Thus, to ensure that a_{2} ≈_{T} 0 for all T > 0 we assume that \frac{1}{λ}Re(d(λ)) → ∞ as λ → ∞. Finally, consider a_{3}(λ) =

λ ∫_{T}^{∞} e^{−λt}(1 − e^{− \frac{d(λ)}{G(t)}} )F (t) dt. Then ∥a_{3}(λ)∥ ≤ |d(λ)|e^{−λT} for all λ ∈ Ω ∩ ℝ_{+}. This shows that a_{3} ≈_{T} 0 for all T > 0 if lim sup_{λ→∞} \frac{1}{λ} ln(|d(λ)|) ≤ 0.

The considerations above show that r as defined in (2.2) is an element of {f} if the damping function d is analytic on a post-sectorial subregion S of the open right half-plane, Re(d(λ)) ≥ 0 for all λ ∈ S, \frac{1}{λ}Re(d(λ)) → ∞ as λ → ∞, and lim sup_{λ→∞} \frac{1}{λ} ln(|d(λ)|) ≤ 0.

An example of a damping function with these properties is given by

d(λ) = λ ln(λ).

There, S = {λ ∈ ℂ : Re(λ) > 1} ∩ {λ = re^{iα} : r > e^{α tan(α)}, −π/2 < α < π/2}.

(b) Since f̂ exists if and only if there are constants M, ω > 0 such that ∥F (t)∥ ≤ Me^{ωt} for all t ≥ 0, there exists a constant C_{T} > 0 such that ∥ ∫_{T}^{∞} e^{−λt}f(t) dt∥ = ∥ − e^{−λT} F (T) + λ ∫_{T}^{∞} e^{−λt}F (t) dt∥ ≤ e^{−λT} ∥F (T)∥ + M \frac{λ}{λ−ω} e^{−(λ−ω)T} ≤ C_{T} e^{−λT} for

all λ > 2ω. This shows that f̂(λ) − ∫_{0}^{T} e^{−λt}f(t) dt = ∫_{T}^{∞} e^{−λt}f(t) dt ≈_{T} 0 for all T > 0. Thus,

{f}_{T} = f̂+ {0}_{T} for all T > 0, and {f} = f̂+ ∩_{T >0}{0}_{T} .  (2.3)

(c) Clearly, the approximate Laplace transform {f}_{T} of any f ∈ L_{loc}^{1} comprises a “large” set of analytic functions. Nevertheless, the sets are disjoint for different f. To see this let 0 < T ≤ ∞, and r ∈ {f}_{T} ∩ {g}_{T}. Then there exist a, b ≈_{T} 0 such that

r(λ) = ∫_{0}^{T} e^{−λt}f(t) dt + a(λ) = ∫_{0}^{T} e^{−λt}g(t) dt + b(λ)

for all λ > ω. Thus, ĥ_{T} (λ) = ∫_{0}^{T} e^{−λt}h(t) dt = c(λ), where h := f − g, c := b − a, and c ≈_{T} 0. Now statement (c) follows from the following proposition; see also B. Bäumer and F. Neubrander [5].

Proposition 2.2. Let T > 0, f ∈ L_{loc}^{1}([0, ∞); X) and assume f̂ exists. The following are equivalent.

(i) lim sup_{λ→∞} \frac{1}{λ} ln ∥f̂(λ)∥ ≤ −T.

(ii) f = 0 a.e. on [0, T].

Moreover, f does not identically vanish a.e. on any interval containing 0 if and only if lim sup_{λ→∞} \frac{1}{λ} ln ∥f̂(λ)∥ = 0.

Proof. The existence of f̂ implies the exponential boundedness of F (t) = ∫_{0}^{t} f(s) ds. Thus, G(t) := ∫_{0}^{t} F (s) ds is Lipschitz-continuous (∥G(t) − G(s)∥ ≤ M| ∫_{s}^{t} e^{ωr} dr| for some M, ω > 0 and all t, s ≥ 0) and f̂(λ) = λ^{2}Ĝ(λ) = λd̂G(λ) for Re(λ) > ω. Define r(λ) := \frac{1}{λ} f̂(λ) = ∫_{0}^{∞} e^{−λt} dG(t) = λ ∫_{0}^{∞} e^{−λt}G(t) dt for Re(λ) > ω. The Phragmén-Doetsch inversion formula, whose proof can be found in B. Bäumer and F. Neubrander [5], yields

G(t) = lim_{k→∞} ∑_{j=1}^{∞}(−1)^{j+1} \frac{1}{j!}e^{tkj}r(kj),  (2.4)

where the limit exists uniformly for t ∈ [0, ∞).

Suppose (i) holds. Then lim sup_{λ→∞} \frac{1}{λ} ln ∥r(λ)∥ ≤ −T. Thus, for 0 < ξ < T, there exists M, λ_{0} > 0 such that ∥r(λ)∥ ≤ Me^{−λξ} for all λ > λ_{0}. Let t ∈ [0, ξ). Then

∥ ∑_{j=1}^{∞}(−1)^{j+1} \frac{1}{j!}e^{tkj}r(kj)∥ ≤ M ∑_{j=1}^{∞} \frac{1}{j!}e^{(t−ξ)kj} = M (e^{e^{(t−ξ)k}} − 1) → 0  (2.5)

as k → ∞. Because r(λ) = ∫_{0}^{∞} e^{−λt} dG(t), it follows from (2.4) that G = 0 on [0, ξ) for all 0 < ξ < T. Thus, f = 0 on [0, T]. This proves that (i) implies (ii).

Suppose (ii) holds. Then r(λ) = λ ∫_{T}^{∞} e^{−λt}G(t) dt = e^{−λT} λ ∫_{0}^{∞} e^{−λt}G(t + T) dt. Since t → G(t+T) is exponentially bounded, it follows that ∥λ ∫_{0}^{∞} e^{−λt}G(t+T) dt∥ ≤ C for some C > 0 and therefore ∥r(λ)∥ ≤ Ce^{−λT} for all sufficiently large λ > 0. This proves that (ii) implies (i).

If f̂ exists, then it follows from the exponential boundedness of the antiderivative F and f̂(λ) = λ ∫_{0}^{∞} e^{−λt}F (t) dt that there exist M, λ_{0} > 0 such that ∥f̂(λ)∥ ≤ M for all λ > λ_{0}. Thus, if f̂ exists, then lim sup_{λ→∞} \frac{1}{λ} ln ∥f̂(λ)∥ ≤ 0. This finishes the proof of the proposition.

Example 2.3. The function f : t → e^{t^{2}} is clearly not Laplace transformable. As shown in the proof of Theorem 2.1, the function

r_{0}(λ) := λ ∫_{0}^{∞} e^{−λt}(1 − e^{− \frac{λ ln(λ)}{G(t)}} ) F (t) dt

is an asymptotic Laplace transform of f (i.e., r_{0} ∈ {f}), where F (t) = ∫_{0}^{t} e^{s^{2}} ds and G(t) := max{∥F (t)∥, 1}. Since r_{0} is not a classical function of analysis one might want to find more familar functions in {f}. We present two further methods of constructing elements of {f} which depend more on the special nature of the function f(t) = e^{t^{2}}. The first one is due to L. Berg [7]. Consider the entire function

r_{1}(λ) := ∫_{0}^{λ/2} e^{−λt+t^{2}} dt = \frac{λ}{2} ∫_{0}^{1} e^{− \frac{λ^{2}}{2} s+ \frac{λ^{2}}{4} s^{2}} ds.

Since the convex function t → e^{−λt+t^{2}} attains its minimum at t = \frac{λ}{2} it follows that r_{1}(λ) − ∫_{0}^{T} e^{−λt}e^{t^{2}} dt = ∫_{T}^{λ/2} e^{−λt+t^{2}} dt ≤ (\frac{λ}{2} − T)e^{−λT +T^{2}} for all sufficiently large λ > 0. Therefore, r_{1} ∈ {f}_{T} for all T > 0 and thus r_{1} ∈ {f}.

Since f is analytic, Cauchy’s theorem provides another possibility of finding an representative in {f}. Let Γ be the directed rectangle consisting of the paths Γ_{1} := {t ∈ [0, T]}, Γ_{2}(N) := {t = T + ir; r ∈ [0, N]}, Γ_{3}(N) := {t = r + iN; r ∈ [0, T]}, and Γ_{4}(N) := {t = ir; r ∈ [0, N]}. Since ∫_{Γ_{3}(N)} e^{−λt}e^{t^{2}} dt = ∫_{0}^{T} e^{−λ(r+iN)}e^{r^{2}−N^{2}+2irN} dr → 0 as N → ∞, it follows from Cauchy’s integral theorem that

∫_{0}^{T} e^{−λt}e^{t^{2}} dt = lim_{N→∞} (− ∫_{Γ_{2}(N)} + ∫_{Γ_{4}(N)}) e^{−λt}e^{t^{2}} dt.

Define ^{7}

r_{2}(λ) := lim_{N→∞} ∫_{Γ_{4}(N)} e^{−λt}e^{t^{2}} dt = i ∫_{0}^{∞} e^{−iλr}e^{−r^{2}} dr = \frac{i}{2}\sqrt{π}e^{− \frac{1}{4} λ^{2}}Erfc(\frac{iλ}{2}).

Then r_{2}(λ) − ∫_{0}^{T} e^{−λt}e^{t^{2}} dt = a(λ), where the remainder λ → a(λ) is given by a(λ) := lim_{N→∞} ∫_{Γ_{2}(N)} e^{−λt}e^{t^{2}} dt = i ∫_{0}^{∞} e^{−λ(T +ir)}e^{T^{2}−r^{2}+2iT r} dr. For λ > 0, the estimate

∥a(λ)∥ ≤ e^{−λT} e^{T^{2}} ∫_{0}^{∞} e^{−r^{2}} dr implies that r_{2} ∈ {f}. ♢ If f, g ∈ L_{loc}^{1} and one of them is scalar-valued, then

7The Laplace transform of the function t → e^{−t^{2}} is λ → \frac{1}{2} \sqrt{π}e^{− \frac{1}{4} λ^{2}}Erfc( \frac{λ}{2} ).

f ∗ g : t → ∫_{0}^{t} f(t − s)g(s) ds (t ≥ 0)

denotes the convolution of f and g. For all scalar, locally Lebesgue integrable functions functions g, h and locally Bochner integrable functions f the convolution f ∗ g is in L_{loc}^{1}, f ∗ g = g ∗ f, and (g ∗ h) ∗ f = g ∗ (h ∗ f).^{8} The usefulness of the Laplace transform in applications to differential and integral equations is due to the fact that it maps differentiation, integration, and more general, convolution onto multiplication. Next it will be shown that these crucial operational properties extend to the asymptotic Laplace transform as well.

Proposition 2.4. Let 0 < T ≤ ∞ and f ∈ L_{loc}^{1}([0, ∞); X). If f is absolutely continuous and f′ exists a.e., then

{f′}_{T} = λ{f}_{T} − f(0).

If g ∈ L_{loc}^{1}([0, ∞); ℂ) and T < ∞, then

{f ∗ g}_{T} = {f}_{T} · {g}_{T} + {0}_{T} .

In particular, {f}_{T} · {g}_{T} ⊂ {f ∗ g}_{T} for all 0 < T ≤ ∞.

Proof. Let 0 < T < ∞. If q ∈ {f′}_{T}, then there exist ω > 0 and a ≈_{T} 0 such that q(λ) = ∫_{0}^{T} e^{−λt}f′(t) dt+a(λ) = λ [∫_{0}^{T} e^{−λt}f(t) dt + \frac{1}{λ}e^{−λT} f(T) + \frac{1}{λ}a(λ)]−f(0) for all λ > ω. Thus q ∈ λ{f}_{T} − f(0). Reversely, if r ∈ λ{f}_{T} − f(0) then there exist ω > 0 and a ≈_{T} 0 such that r(λ) = λ [∫_{0}^{T} e^{−λt}f(t) dt + a(λ)] − f(0) = ∫_{0}^{T} e^{−λt}f′(t) dt −

e^{−λT} f(T) + λa(λ) for all λ > ω. Thus, r ∈ {f′}_{T}. This shows that {f′}_{T} = λ{f}_{T} − f(0) for all 0 < T < ∞ and thus also for T = ∞. To prove the other statement, let 0 < T < ∞, r(λ) = ∫_{0}^{T} e^{−λt}f(t) dt + a(λ), and q(λ) = ∫_{0}^{T} e^{−λs}g(s) ds + b(λ), where a, b ≈_{T} 0. Define w(λ) := a(λ) ∫_{0}^{T} e^{−λs}g(s) ds + b(λ) ∫_{0}^{T} e^{−λt}f(t) dt + a(λ)b(λ). Then w ≈_{T} 0 and

r(λ)q(λ) = ∫_{0}^{T} ∫_{0}^{T} e^{−λ(t+s)}f(t)g(s) dtds + w(λ)
= ∫_{0}^{T} ∫_{s}^{T +s} e^{−λt}f(t − s)g(s) dtds + w(λ)

= ∫_{0}^{T} ∫_{0}^{t} e^{−λt}f(t − s)g(s) dsdt + ∫_{T}^{2T} ∫_{t−T}^{T} e^{−λt}f(t − s)g(s) dsdt + w(λ)

= ∫_{0}^{T} e^{−λt}(f ∗ g)(t) dt + e^{−λT} ∫_{0}^{T} e^{−λt} ∫_{t}^{T} f(t + T − s)g(s) dsdt + w(λ).

8For a proof of the scalar case, see N. Dunford and J. Schwartz [16], VIII.1.24, 633-636.

Thus, {f}_{T} · {g}_{T} ⊂ {f ∗ g}_{T}. If r ∈ {f ∗ g}_{T}, then r(λ) = ∫_{0}^{T} e^{−λt}(f ∗ g)(t) dt + a(λ)

for some a ≈_{T} 0. Since ∫_{0}^{T} e^{−λt}(f ∗ g)(t) dt = (∫_{0}^{T} e^{−λt}f(t) dt) (∫_{0}^{T} e^{−λs}g(s) ds) −

e^{−λT} ∫_{0}^{T} e^{−λt} ∫_{t}^{T} f(t + T − s)g(s) dsdt it follows that r ∈ {f}_{T} {g}_{T} + {0}_{T}. This shows that {f ∗ g}_{T} ⊂ {f}_{T} {g}_{T} + {0}_{T} ⊂ {f ∗ g}_{T} + {0}_{T} = {f ∗ g}_{T}. Thus, {f ∗ g}_{T} = {f}_{T} · {g}_{T} + {0}_{T} and {f}_{T} · {g}_{T} ⊂ {f ∗ g}_{T} for all 0 < T < ∞. Therefore, {f}{g} ⊂ ∩_{T >0}{f}_{T} {g}_{T} ⊂ ∩_{T >0}{f ∗ g}_{T} = {f ∗ g}.

Example 2.5. Consider the ordinary differential equation v′+v = (2t+1)e^{t^{2}} (t ≥ 0), v(0) = 1. Suppose there exists an absolutely continuous solution on some interval [0, T] (let us pretend for the moment that we do not know this already). To determine v with the asymptotic Laplace transform, observe that {(2t + 1)e^{t^{2}}}_{T} = {v′ + v}_{T} = λ{v}_{T} − 1 + {v}_{T}. Therefore,

\{v\}_{T} = \frac{1}{λ + 1} + \frac{1}{λ + 1}\{(2t + 1)e^{t^{2}}\}_{T}
⊂ {e^{t}}_{T} + {e^{t}}_{T} {(2t + 1)e^{t^{2}}}_{T} ⊂ {e^{t} + e^{t} ∗ (2t + 1)e^{t^{2}}}_{T} = {e^{t^{2}}}_{T} .

It follows from Theorem 2.1 (c) that {v}_{T} = {e^{t^{2}}}_{T}. Thus, the unique absolutely continuous solution is given by v(t) = e^{t^{2}}. ♢

Remark 2.6. (Inversion) If f ∈ L_{loc}^{1}, then F := 1 ∗ f is bounded on [0, T], and for every y ∈ \{f\}_{T} one has that \frac{1}{λ}y ∈ \{1\}_{T} \{f\}_{T} ⊂ \{F \}_{T}. Thus, there exists a ≈_{T} 0 such that

\frac{1}{λ}y(λ) = ∫_{0}^{T} e^{−λt}F (t) dt + a(λ) = ∫_{0}^{∞} dG(t) + a(λ)

for all λ > 0, where G is piecewise defined by G(t) := 1 ∗ F (t) = ∫_{0}^{t}(t − s)f(s) ds for 0 ≤ t ≤ T and G(t) := 1 ∗ F (T) for all t ≥ T. By (2.5), the Phragmén-Doetsch inversion (2.4) is not affected by exponentially decaying perturbations. It follows that

G(t) = ∫_{0}^{t}(t − s)f(s) ds = lim_{k→∞} ∑_{j=1}^{∞}(−1)^{j+1} \frac{1}{j!}e^{tkj} \frac{y(kj)}{kj}  (2.6)

for all f ∈ L_{loc}^{1}, y ∈ {f}_{T}, and 0 ≤ t < T ≤ ∞. ^{9} ♢ Remark 2.7. (Multiplied Laplace Transforms) Let Ω be an open subset of ℂ and let m : Ω → ℂ be analytic. Then q : Ω → X is said to have an asymptotic, m-multipled Laplace representation, if there exist 0 < T ≤ ∞, and f ∈ C_{0}([0, ∞; X) such that

9A completely different inversion formula is due to Yu. I. Lyubich [20]. Let F : [0, ∞) → X be of bounded semivariation, F (0) = 0 and r(λ) = ∫_{0}^{∞} dF (t) + a(λ) (λ > ω_{0} > 0) for some a ∈ L_{loc}^{1}([ω_{0}, ∞), X) with a ≈_{T} 0. Let ω > ω_{0}. Then H(μ) := \frac{1}{2πi} ∫_{ω}^{∞} e^{μt} \frac{r(t)}{t} dt is welldefined for Re(μ) < 0, has an analytic continuation to the sliced half-plane ℂ \ [0, T) \ {μ : Re(μ) ≥ T }, and F (t) = lim_{ϵ→0} H(t + iϵ) − H(t − iϵ) for all t ∈ [0, T).

mq ∈ {f}_{T} .  (2.7)

The function q is said to have an m-multiplied Laplace representation, if there exists a Laplace transformable f ∈ C_{0}([0, ∞); X) (i.e., f is continuous and f(0) = 0) such that mq = f̂. To see how multiplied Laplace transforms of normalized continuous functions relate to Laplace transforms of generalized functions, let m = k̂ for some regularizing function k; i.e.,

k ∈ L_{loc}^{1}([0, ∞); ℂ) and lim sup_{λ→∞} \frac{1}{λ} ln |k̂(λ)| = 0.^{10}  (2.8)

Then, by Proposition 2.2 and the Titchmarsh-Foiaş Theorem in B. Bäumer, G. Lumer and F. Neubrander [3], the convolution operator K : f → k ∗ f is an injective operator on C([0, ∞); X) with dense range in the Frechét space C_{0}([0, ∞); X). Moreover, ∥f∥_{K,n} := sup_{t∈[0,n]} ∥Kf(t)∥ defines a family of seminorms on C([0, ∞); X) and K extends to an isomorphism between the Frechét completion C^{[k]}([0, ∞); X) of C([0, ∞); X) with respect to that family of seminorms and the Frechét space C_{0}([0, ∞); X).^{11}

A k-generalized function u ∈ C^{[k]} is Laplace transformable if the continuous function f := k ∗ u is Laplace transformable and the Laplace transform of u is defined as

û(λ) := \frac{f̂(λ)}{k̂(λ)}.

Similarily, the asymptotic Laplace transform is defined as \{u\}_{T} := \frac{1}{k̂(λ)}\{f\}_{T} (0 <

T ≤ ∞). Thus, if q has a m-multiplied Laplace representation mq ∈ {f}_{T} or mq = f̂ for some f ∈ C_{0}([0, ∞); X), where m = k̂ for some regularizing function k, then q has a Laplace representation q ∈ {u}_{T} or q = û for u = K^{−1}f ∈ C^{[k]}([0, ∞); X).

Next we give examples of classes of analytic functions which can be represented as Laplace transforms of generalized functions. Consider first analytic functions q : {Re(λ) > ω ≥ 0} → X with sup_{Re(λ)>ω} ∥λq(λ)∥ < ∞. Let b > 0 and

f(t) := lim_{n→∞} \frac{1}{2πi} ∫_{ϵ−in}^{ϵ+in} e^{λt} \frac{q(λ)}{λ^{b}} dλ.

10Typical examples of regularizing functions are k_{n}(t) = t^{n}/n! with k̂_{n}(λ) = 1/λ^{n+1} and k_{δ}(t) = \frac{1}{2πi} ∫_{ω+iℝ} e^{tλ−λ^{δ}} dλ (0 < δ < 1) with k̂_{δ}(λ) = e^{−λ^{δ}}. For the special case δ = 1/2, the regularizer k_{δ} is given by k_{1/2}(t) = \frac{1}{2\sqrt{π}} t^{−3/2}e^{− \frac{1}{4t}}. Also, all functions m(λ) = e^{−λ^{γ}} with γ ≥ 1 can not be represented as Laplace transforms of functions in k ∈ L_{loc}^{1}; see G. Doetsch [15], p.163, 229, 263.

11See alse B. Bäumer [2] and B. Bäumer, G. Lumer and F. Neubrander [3].

Then f ∈ C_{0}([0, ∞); X), sup_{t>0} ∥e^{−ωt} \frac{1}{t^{b}} f(t)∥ < ∞ and q(λ) = λ^{b}f̂(λ) for Re(λ) > ω;

see, e.g., B. Bäumer and F. Neubrander [5]. Thus q(λ) = û(λ) = \frac{f̂(λ)}{k̂(λ)}, where k(t) =

\frac{1}{Γ(b)}t^{b−1} and u = K^{−1}f ∈ C^{[k]} coincides with the b-th (distributional) derivative of f. More general, if q is an analytic function on some half-plane {Re(λ) > ω ≥ 0} with values in X which grows less than exponentially, i.e.,

sup_{Re(λ)>ω} ∥λk̂_{0}(λ)q(λ)∥ < ∞

for some regularizing function k_{0}, then it follows from the above mentioned representation theorem that there exists f ∈ C_{0}([0, ∞); X) such that \frac{1}{λ} k̂_{0}(λ)q(λ) = k̂(λ)q(λ) = f̂(λ) or q(λ) = û(λ), where k := 1 ∗ k_{0} and u ∈ C^{[k]}([0, ∞); X) is a generalized function such that f = k ∗ u.^{12} Thus, a faster growing q will have a less regular u such that q = û. ♢

Next we consider analytic functions q : Ω → X with sup_{λ∈Ω} ∥λq(λ)∥ < ∞, where the domain Ω contains a post-sectorial region

Ω_{Ψ} := {λ : Re(λ) ≥ β > 0, |Imλ| ≤ Ψ(Re(λ))},

where Ψ is a positive, strictly increasing C^{1}-function with Ψ(r) → ∞ as r → ∞ and sup_{r≥β} Ψ′(r)/r^{α}Ψ(r) < ∞ for some α ≥ 0; i.e., Ψ is dominated by some exponential function t → e^{cr^{γ}}. ^{13}

Theorem 2.8. Let Ω_{Ψ} ⊂ Ω, q : Ω → X analytic, and sup_{λ∈Ω} ∥λq(λ)∥ < ∞.

(a) If there exists c > 0 and an analytic function m : Ω_{Ψ} → ℂ such that

|m(λ)| ≤ e^{−cΨ^{−1}(|λ|)} for all λ ∈ Ω_{Ψ},

then there exists f ∈ C_{0}([0, c), X) such that mq ∈ {f}_{T} for all 0 < T < c.

(b) If there exists c > 0, d > 1, and an analytic function m : Ω_{Ψ} → ℂ such that

|m(λ)| ≤ e^{−c(Ψ^{−1}(|λ|))^{d}} for all λ ∈ Ω_{Ψ},

then there exists f ∈ C_{0}([0, ∞), X) such that mq ∈ {f}.

Proof. (a) Let Γ be the oriented boundary of the region Ω_{Ψ}; i.e., Γ = Γ_{−} ∪ Γ_{β} ∪ Γ_{+}, where Γ_{±} := {r ± iΨ(r); β ≤ r < ∞}, and Γ_{β} := {β + ir; −Ψ(β) ≤ r ≤ Ψ(r)}. Let

12If k_{i} are regularizing functions and k_{1} ∗ k_{2} = k_{3}, then C^{[k_{1}]} is continuously embedded in C^{[k_{3}]}.

13This growth condition on Ψ does not restrict generality; see Corollary 2.9 (a).

Asymptotic Laplace transforms and evolution equations 11

M, C_{0}, C_{1} > 0 be such that ∥λq(λ)∥ ≤ M and |\frac{1±iΨ′(r)}{r±iΨ(r)} | ≤ C_{0} + C_{1}r^{α} for all λ ∈ Ω_{Ψ} and all r ≥ β. Since Ψ^{−1} is increasing it follows that

Ψ^{−1}(|r ± iΨ(r)|) ≥ Ψ^{−1}(|Ψ(r)|) = r

for all r ≥ β. Thus, for λ = r ±iΨ(r) ∈ Γ_{±}, we have the estimate ∥e^{λt}m(λ)q(λ) dλ∥ ≤ Me^{rt}e^{−cΨ^{−1}(|r±iΨ(r)|)} |\frac{1±iΨ′(r)}{r±iΨ(r)} | dr ≤ M(C_{0} + C_{1}r^{α})e^{−r(c−t)}dr. It follows that

f(t) := \frac{1}{2πi} ∫_{Γ} e^{λt}m(λ)q(λ) dλ

is welldefined and continuous for 0 ≤ t < c. Next it will be shown that

∫_{0}^{T} e^{−λt}f(t) dt = m(λ)q(λ) − a_{T} (λ)

for all 0 < T < c and λ in the interior Ω_{Ψ}^{0} of Ω_{Ψ}, where

a_{T} (λ) := \frac{e^{−λT}}{2πi} ∫_{Γ} \frac{1}{μ − λ}e^{μT} m(μ)q(μ) dμ.

To see this, let Γ(n) := Γ ∩ {Re(λ) ≤ n} and Π_{n} := {n + ir; −Ψ(n) ≤ r ≤ Ψ(r)}. Then, for all λ ∈ Ω_{Ψ}^{0},

∫_{0}^{T} e^{−λt}f(t) dt = ∫_{0}^{T} e^{−λt} \frac{1}{2πi} ∫_{Γ} e^{μt}m(μ)q(μ) dμ dt

= \frac{1}{2πi} ∫_{Γ} \frac{1}{μ − λ} [e^{(μ−λ)T} − 1] m(μ)q(μ) dμ

= lim_{n→∞} \frac{1}{2πi} [∫_{Γ(n)} \frac{1}{μ − λ}e^{(μ−λ)T} m(μ)q(μ) dμ − ∫_{Γ(n)} \frac{1}{μ − λ}m(μ)q(μ) dμ]

= lim_{n→∞} \frac{1}{2πi}[ ∫_{Γ(n)} \frac{1}{μ − λ}e^{(μ−λ)T} m(μ)q(μ) dμ

+ (− ∫_{Γ(n)} + ∫_{Π_{n}} − ∫_{Π_{n}}) \frac{1}{μ − λ}m(μ)q(μ) dμ]

= m(λ)q(λ) + lim_{n→∞} \frac{1}{2πi}[ ∫_{Γ(n)} \frac{1}{μ − λ}e^{(μ−λ)T} m(μ)q(μ) dμ

− ∫_{Π_{n}} \frac{1}{μ − λ}m(μ)q(μ) dμ]

= m(λ)q(λ) + \frac{1}{2πi} ∫_{Γ} \frac{1}{μ − λ}e^{(μ−λ)T} m(μ)q(μ) dμ,

since, by the residue theorem,

\frac{1}{2πi} (− ∫_{Γ(n)} + ∫_{Π_{n}}) \frac{1}{μ − λ}m(μ)q(μ) dμ = m(λ)q(λ)

for n > Re(λ), and lim_{n→∞} \frac{1}{2πi} ∫_{Π_{n}} \frac{1}{μ−λ}m(μ)q(μ) dμ = 0. The last statement follows from the estimate

∥ ∫_{Π_{n}} \frac{1}{μ − λ}m(μ)q(μ) dμ∥ ≤ ∫_{−Ψ(n)}^{Ψ(n)} \frac{1}{|n + ir − λ|}e^{−cΨ^{−1}(|n+ir|)} \frac{M}{|n + ir|} dr

≤ Me^{−cΨ^{−1}(n)} ∫_{−∞}^{∞} \frac{1}{|n + ir − λ|^{2}} dr.

To show that a_{T} ≈_{T} 0, it suffices to prove the boundedness of ∫_{Γ} \frac{1}{μ−λ}e^{μT} m(μ)q(μ) dμ for λ → ∞. Clearly, ∫_{Γ_{β}} \frac{1}{μ−λ}e^{μT} m(μ)q(μ) dμ → 0 as λ → ∞. Moreover,

∥ ∫_{Γ_{±}} \frac{1}{μ − λ}e^{μT} m(μ)q(μ)dμ∥ ≤ ∫_{β}^{∞} \frac{M}{dist(λ, Γ_{±})}e^{rT} e^{−cΨ^{−1}(|r±iΨ(r)|)} |\frac{1 ± iΨ′(r)}{r ± Ψ(r)}| dr

≤ \frac{M}{dist(λ, Γ_{±})} ∫_{β}^{∞}(C_{0} + C_{1}r^{α})e^{−r(c−T)} dr.

This shows that a_{T} ≈_{T} 0. Thus, mq ∈ {f}_{T} for all 0 < T < c. Before proving the statement (b), an estimate will be given for ∥f(t)∥ for 0 ≤ t < c. Let t be fixed and γ ≥ β be arbitrary. Define Γ^{∗} = Γ_{−}^{∗} ∪ Π_{γ} ∪ Γ_{+}^{∗}, where Γ_{±}^{∗} := {r ± iΨ(r); γ ≤ r < ∞\}, and Π_{γ} := \{γ + Ψ(γ)e^{ir}; \frac{π}{2} ≤ r ≤ \frac{π}{2} \}. Then, by Cauchy’s theorem, f(t) = \frac{1}{2πi} ∫_{Γ^{∗}} e^{λt}m(λ)q(λ) dλ. Moreover,

∥\frac{1}{2πi} ∫_{Π_{γ}} e^{λt}m(λ)q(λ)dλ∥ ≤ \frac{M}{2π} ∫ \frac{π}{2} ; \frac{−π}{2} |e^{γt+tΨ(γ)e^{ir}}|e^{−cΨ^{−1}(|γ+Ψ(γ)e^{ir}|)} \frac{Ψ(γ)}{|γ + Ψ(γ)e^{ir}|}dr

≤ \frac{M}{2π} ∫_{−π/2}^{π/2} e^{γt+tΨ(γ) cos(r)}e^{−cΨ^{−1}(Ψ(γ))} dr

= \frac{M}{π} e^{−γ(c−t)} ∫_{0}^{π/2} e^{tΨ(γ) cos(r)} dr,

and, since ∥e^{λt}m(λ)q(λ) dλ∥ ≤ M(C_{0} + C_{1}r^{α})e^{−r(c−t)}dr along Γ_{±},

∥\frac{1}{2πi} ∫_{Γ_{±}^{∗}} e^{λt}m(λ)q(λ) dλ∥ ≤ \frac{M}{2π} ∫_{γ}^{∞}(C_{0} + C_{1}r^{α})e^{−r(c−t)} dr

= \frac{MC_{0}}{2π} e^{−γ(c−t)} \frac{1}{c − t} + \frac{MC_{1}}{2π} ∫_{γ}^{∞} r^{α}e^{−r(c−t)} dr.

Thus

∥f(t)∥ ≤ \frac{M}{π} e^{−γ(c−t)} (∫_{0}^{π/2} e^{tΨ(γ) cos(r)} dr + \frac{C_{0}}{c − t}) + \frac{MC_{1}}{π} ∫_{γ}^{∞} r^{α}e^{−r(c−t)} dr.

Now choose γ such that Ψ(γ) = \frac{Ψ(β)c}{t}; i.e., γ := Ψ^{−1}(\frac{Ψ(β)c}{t}). Then

∥f(t)∥ ≤ \frac{M}{π} e^{−Ψ^{−1}( \frac{Ψ(β)c}{t})(c−t)} (∫_{0}^{π/2} e^{cΨ(β) cos(r)} dr + \frac{C_{0}}{c − t})

+ \frac{MC_{1}}{π} ∫_{γ}^{∞} r^{α}e^{−r(c−t)} dr.

for 0 < t < c. In particular, f(0) = 0.

(b) Let the path Γ and the constants M, C_{0}, C_{1} be as above. Then, for λ = r ± iΨ(r) ∈ Γ_{±},

∥e^{λt}m(λ)q(λ) dλ∥ ≤ Me^{rt}e^{−c(Ψ^{−1}(|r±iΨ(r)|))^{d}} |\frac{1 ± iΨ′(r)}{r ± iΨ(r)}| dr
≤ M(C_{0} + C_{1}r^{α})e^{rt−cr^{d}}dr.

Thus, f(t) := \frac{1}{2πi} ∫_{Γ} e^{λt}m(λ)q(λ) dλ is welldefined and continuous for t ≥ 0. As in part (a) one shows that

∫_{0}^{T} e^{−λt}f(t) dt = m(λ)q(λ) + \frac{e^{−λT}}{2πi} ∫_{Γ} \frac{1}{μ − λ}e^{μT} m(μ)q(μ) dμ.

Since ∫_{Γ_{β}} \frac{1}{μ−λ}e^{μT} m(μ)q(μ) dμ → 0 as λ → ∞ and

∥ ∫_{Γ_{±}} \frac{1}{μ − λ}e^{μT} m(μ)q(μ)dμ∥ ≤ ∫_{β}^{∞} \frac{M}{dist(λ, Γ_{±})}e^{rT −c(Ψ^{−1}(|r±iΨ(r)|))^{d}} |\frac{1 ± iΨ′(r)}{r ± Ψ(r)}| dr

≤ \frac{M}{dist(λ, Γ_{±})} ∫_{β}^{∞}(C_{0} + C_{1}r^{α})e^{rT −cr^{d}} dr,

it follows that mq ∈ {f}_{T} for all T > 0. Thus, mq ∈ {f}. To estimate ∥f(t)∥ for 0 < t ≤ 1, let t be fixed, γ ≥ β be arbitrary, and define the path Γ^{∗} as above. Then, f(t) = \frac{1}{2πi} ∫_{Γ^{∗}} e^{λt}m(λ)q(λ) dλ,

∥ ∫_{Π_{γ}} e^{λt}m(λ)q(λ)dλ∥ ≤ M ∫− \frac{π}{2}^{\frac{π}{2}} |e^{γt+tΨ(γ)e^{ir}}|e^{−c(Ψ^{−1}(|γ+Ψ(γ)e^{ir}|))^{d}} \frac{Ψ(γ)}{|γ + Ψ(γ)e^{ir}|} dr

≤ M ∫_{−π/2}^{π/2} e^{γt+tΨ(γ) cos(r)}e^{−c(Ψ^{−1}(Ψ(γ)))^{d}} dr

= 2Me^{γt−cγ^{d}} ∫_{0}^{π/2} e^{tΨ(γ) cos(r)} dr,

and, since ∥e^{λt}m(λ)q(λ) dλ∥ ≤ M(C_{0} + C_{1}r^{α})e^{rt−cr^{d}}dr along Γ_{±},

∥\frac{1}{2πi} ∫_{Γ_{±}^{∗}} e^{λt}m(λ)q(λ) dλ∥ ≤ \frac{M}{2π} ∫_{γ}^{∞}(C_{0} + C_{1}r^{α})e^{rt−cr^{d}} dr.

Thus, for t ≥ 0, ∥f(t)∥ ≤ \frac{M}{π} e^{γt−cγ^{d}} ∫_{0}^{π/2} e^{tΨ(γ) cos(r)} dr + \frac{M}{π} ∫_{γ}^{∞}(C_{0} + C_{1}r^{α})e^{rt−cr^{d}} dr.

If 0 < t ≤ 1, one might choose γ such that Ψ(γ) = \frac{Ψ(β)}{t}; i.e., γ := Ψ^{−1}(\frac{Ψ(β)}{t}). Then, for this γ,

∥f(t)∥ ≤ \frac{M}{π} e^{γt−cγ^{d}} ∫_{0}^{π/2} e^{Ψ(β) cos(r)} dr + \frac{M}{π} ∫_{γ}^{∞}(C_{0} + C_{1}r^{α})e^{rt−cr^{d}} dr

for 0 < t < 1. In particular, f(0) = 0.

Corollary 2.9. Let Ω_{Ψ} ⊂ Ω, q : Ω → X analytic, and sup_{λ∈Ω} ∥λq(λ)∥ < ∞.

(a) Let Ψ(x) = e^{ax^{γ}} (a > 0, γ > 1). Then, for all b > 0, there exists f ∈ C_{0}([0, ∞); X) such that \frac{1}{λ^{b}} q(λ) ∈ \{f\}.

(b) Let Ψ(x) = e^{ax} (a > 0). Then

1. for all b > 0 there exists f ∈ C_{0}([0, ab); X) such that \frac{1}{λ^{b}} q(λ) ∈ \{f\}_{T} (0 < T < ab),

2. for all b > 0, 0 < δ < 1 there exists f ∈ C_{0}([0, ∞); X) such that e^{−bλ^{δ}}q(λ) ∈ {f}.

(c) Let Ψ(x) = x^{1/a} (0 < a < 1). Then

1. for all b > 0 there exists f ∈ C_{0}([0, b cos(aπ/2)); X) such that e^{−bλ^{a}}q(λ) ∈ {f}_{T} (0 < T < b cos(aπ/2)),

2. for all b > 0, a < γ < 1 there exists f ∈ C_{0}([0, ∞); X) such that e^{−bλ^{γ}} q(λ) ∈ {f}.

Proof. (a) Since Ψ^{−1}(x) = (\frac{1}{a} ln(x))^{1/γ}, it follows that e^{−ab(Ψ^{−1}(|λ|))^{γ}} = \frac{1}{|λ|^{b}}. Thus, if one takes m(λ) = \frac{1}{λ^{b}}, then (a) follows from Theorem 2.8 (b).

(b) Since Ψ^{−1}(x) = \frac{1}{a} ln(x), it follows that e^{−abΨ^{−1}(|λ|)} = \frac{1}{|λ|^{b}}. Define m(λ) := \frac{1}{λ^{b}}. Then 1. follows from Theorem 2.8 (a). To prove the second statement note that |e^{−z^{δ}}| ≤ e^{−|z|^{δ}} if Re(z) > 0 and 0 < δ < 1. Let m(λ) = e^{−bλ^{δ}}. Then there exists a constant c > 0 such that |m(λ)| ≤ e^{−b|λ|^{δ}} ≤ e^{−c( \frac{1}{a} ln(|λ|))^{2}} = e^{−c(Ψ^{−1}(|λ|))^{2}} for all λ ∈ Ω_{Ψ}. Now statement 2 follows from Theorem 2.8 (b).

(c) Let m(λ) = e^{−bλ^{a}}. The statement 1 follows from Theorem 2.8 (a) since |m(λ)| ≤ e^{−b cos(aπ/2)|λ|^{a}} = e^{−b cos(aπ/2)Ψ^{−1}(|λ|)} (Re(λ) > 0). If m(λ) = e^{−bλ^{γ}}, then statement 2 follows from Theorem 2.8 (b) since |m(λ)| ≤ e^{−b cos(γπ/2)|λ|^{γ}} = e^{−b cos(γπ/2)Ψ^{−1}(|λ|)^{γ/a}} (Re(λ) > 0).

Asymptotic Laplace transforms and evolution equations 15

Clearly, if q is an analytic function on some post-sector Ω_{Ψ} on which it grows less than exponentially, i.e., sup_{λ∈Ω_{Ψ}} ∥λk̂_{0}(λ)q(λ)∥ < ∞ for some regularizing function k_{0}, then the corollary above can be applied that to q̃(λ) := k_{0}(λ)q(λ). Thus Corollary 1.9 shows that more growth of an analytic function q ∈ A(P; X) at infinity, a larger interval [0, T), or a smaller post-sectorial region Ω_{Ψ} lead to a decay in regularity of the generalized function μ for which q ∈ {μ}_{T}.

3 The Abstract Cauchy Problem

This section is concerned with applications of asymptotic Laplace transform methods to abstract Cauchy problems

u′(t) = Au(t), t ∈ I, u(0) = x,  (ACP)

where A is a closed^{14} linear operator with domain D(A) and range in a Banach space X, and either I = [0, T] or I = [0, ∞). In this section, k will always be a regularizing function whose Laplace transform has no real zeros^{15}; i.e.,

k ∈ L_{loc}^{1}([0, ∞); ℂ), lim sup_{λ→∞} \frac{1}{λ} ln |k̂(λ)| = 0, and k̂(λ) ≠ 0 for all λ > ω.

Since A is closed and \frac{d}{dt}(k ∗ u)(t) = (k ∗ u′)(t) + k(t)u(0), the convolution of k with (ACP) yields (k∗u)′ = k∗u′+k(t)x = A(k∗u)+k(t)x or k∗u = A(1∗k∗u)+(1∗k)(t)x. Thus, a function u ∈ C^{[k]}(I; X) is called a k-generalized (mild) solution of (ACP) if v = k ∗ u ∈ C_{0}([0, ∞); X)^{16}solves

v(t) = A ∫_{0}^{t} v(s) ds + (1 ∗ k)(t)(x), t ∈ I.  (3.1)

Conversely, any v ∈ C_{0}(I; X) satisfying (3.1) is called a k-regularized solution of (ACP).^{17} One of the main tools used in this section are the following operational

14The closedness of A in X × X is assumed for convenience only; the transform methods extend naturally to linear operators whose graphs are closed in Z × X, where Z is a Banach space which contains D(A) and is continuously embedded in X, see B. Bäumer and F. Neubrander [4].

15Recall that if k is regularizing, then the convolution operator u → k∗u is an isomorphism between the generalized function space C^{[k]}(I; X) and C_{0}(I; X); see Proposition 2.2 and Remark 2.7. As seen below, the assumption k̂(λ) ≠ 0 leads to generation results for operators whose resolvent is analytic on some post-sectorial region Ω. Without this assumption one can obtain generation results for operators whose resolvents are meromorphic on Ω. However, the proofs rely on an extension of the Phragmén-Doetsch inversion (2.6) due to B. Bäumer [2] which is beyond the scope of this paper.

16If necessary, we continue v ∈ C_{0}([0, T]; X) onto [0, ∞) by defining v(t) := v(T) for t > T.

17The notion of k-generalized solutions was introduced by I. Cioranescu and G. Lumer [10][11]. It was shown by B. Bäumer [2] that if u ∈ C^{[k]}(I; X) is a k-generalized solution of (ACP), then there exists a sequence of continuous functions u_{n} ∈ C(I; [D(A)]) (where [D(A)] denotes the Banach space D(A) endowed with the graph norm) such that u_{n} → u and u_{n} − 1 ∗ Au_{n} − x → 0 in C^{[k]}(I; X). Because u coincides with the Cauchy sequence {u_{n}} (modulo the 0-sequences) in C^{[k]}, any k-generalized solution u of (ACP) can be regarded as a k-limit solution {u_{n}} of (ACP). In fact, the two notions are equivalent.

properties of the asymptotic Laplace transform. Combining them with those of the first section, it is easy to characterize the existence, uniqueness, and regularity of (generalized) solutions of (ACP) in terms of (local or global) spectral properties of the operator A.

Proposition 3.1. Let A be closed, 0 < T ≤ ∞, and v, w ∈ L_{loc}^{1}([0, ∞; X).

(a) If r ∈ {v}_{T}, and Ar ∈ {w}_{T}, then Av = w on [0, T].

(b) \{A(1 ∗ v)\}_{T} = \frac{A}{λ} v̂_{T} (λ) + \{0\}_{T}.

Proof. (a) If r ∈ {v}_{T}, r(λ) ∈ D(A) for all λ in a post-sectorial region S and Ar ∈ {w}_{T}, then the Phragmén-Doetsch inversion (2.6) yields that ∫_{0}^{t}(t − s)v(s) ds =

lim_{k→∞} ∑_{j=1}^{∞} \frac{(−1)^{j+1}}{j!} e^{tkj} \frac{r(kj)}{kj} and ∫_{0}^{t}(t−s)w(s) ds = lim_{k→∞} ∑_{j=1}^{∞} \frac{(−1)^{j+1}}{j!} e^{tkj}A\frac{r(kj)}{kj}

for all t ∈ [0, T). Because A is closed one obtains that ∫_{0}^{t}(t − s)v(s) ds ∈ D(A) and A ∫_{0}^{t}(t − s)v(s) ds = ∫_{0}^{t}(t − s)w(s) ds for all t ∈ [0, T). Again using the closedness of A, v(t) ∈ D(A) and Av(t) = w(t) for almost all t ∈ [0, T). (b) By the closedness of

A and (3.1) one obtains \{A(1 ∗ v)\}_{T} = { \atop [A(1 ∗ v)_{T}]} + \{0\}_{T} = ∫_{0}^{T} e^{−λt}A ∫_{0}^{t} v(s) ds

+ \{0\}_{T} = A ∫_{0}^{T} e^{−λt} ∫_{0}^{t} v(s) ds + \{0\}_{T} = e^{−λT} A ∫_{0}^{T} v(s) ds + \frac{A}{λ} ∫_{0}^{T} e^{−λt}v(t) dt + \{0\}_{T} = \frac{A}{λ} v̂_{T} + \{0\}_{T}

Proposition 3.2. Let A be closed, k ∈ L_{loc}^{1}([0, ∞); ℂ) be a regularizing function,^{18} x ∈ X, 0 < T < ∞, and v ∈ L_{loc}^{1}([0, ∞); X). Then the following are equivalent.

(a) v(t) = A ∫_{0}^{t} v(s) ds + (1 ∗ k)(t)x for almost all t ∈ [0, T].

(b) There exists a function r such that (λI − A)r(λ) − x ≈_{T} 0 and k̂r ∈ {v}_{T}.

Proof. (a) ⇒ (b): It follows from (2.1), (2.3), and the Propositions 2.4 and 3.1 that v̂_{T} (λ) + \{0\}_{T} = \{v\}_{T} = \{A(1 ∗ v) + (1 ∗ k)(t)x\}_{T} = \frac{A}{λ} v̂_{T} + \{0\}_{T} + \{x\}_{T} \{k\}_{T} = \frac{A}{λ} v̂_{T} + \{0\}_{T} + (x̂ + \{0\}_{T})(k̂ + \{0\}_{T}) = \frac{A}{λ} v̂_{T} + \frac{x}{λ} k̂(λ) + \{0\}_{T}. Thus

(λI − A)v̂_{T} − k̂(λ)x ≈_{T} 0.  (3.2)

(b) ⇒ (a): By Proposition 2.4, \frac{1}{λ} k̂(λ)r(λ) ∈ \frac{1}{λ}\{v\}_{T} ⊂ \{1∗v\}_{T}. Because k is a regularizer it follows that A(\frac{1}{λ} k̂(λ)r(λ))−k̂(λ)r(λ)+ \frac{1}{λ} k̂(λ)x ∈ \{0\}_{T} and thus A(\frac{1}{λ} k̂(λ)r(λ)) ∈ k̂(λ)r(λ) − \frac{1}{λ} k̂(λ)x + \{0\}_{T} ⊂ \{v\}_{T} + \{1 ∗ k\}_{T} x + \{0\}_{T} ⊂ \{v + (1 ∗ k)x\}_{T}. It follows from Proposition 3.1 (a) that (1 ∗ v)(t) ∈ D(A) and A(1 ∗ v)(t) = v(t) + (1 ∗ k)(t)x for almost all t ∈ [0, T].

18Here and in the following propostions one can also take k(t) = δ(t), k̂(λ) = 1.

The statement of Proposition 3.2 extends easily to the case T = ∞. The proof of the equivalence of the statements (a) and (c) below requires an obvious reformulation of Proposition 3.1 for Laplace transforms and is omitted; see also B. Bäumer and F. Neubrander [5].

Proposition 3.3. Let A be closed, k ∈ L_{loc}^{1}([0, ∞); ℂ) be a regularizing function, x ∈ X, and v ∈ L_{loc}^{1}([0, ∞); X). Then the following are equivalent.

(a) v(t) = A ∫_{0}^{t} v(s) ds + (1 ∗ k)(t)x for almost all t ≥ 0.

(b) There exists a function r such that (λI − A)r(λ) − x ≈_{T} 0 for all T > 0 and k̂r ∈ {v}.

If, in addition, v is Laplace transformable (i.e., 1 ∗ v is exponentially bounded), then (a) is equivalent to

(c) There exists a function r such that (λI − A)r(λ) − x = 0 and k̂r = v̂.

The Cauchy problem (ACP) has the uniqueness property on [0, T] if v ≡ 0 is the only continuous function [0, T] → X for which v(t) = A ∫_{0}^{t} v(s) for all t ∈ [0, T]. We say that (ACP) has the uniqueness property if it has the uniqueness property on [0, T] for all T > 0.

Theorem 3.4. Let A be closed and 0 < T < ∞. Then the following are equivalent.

(a) (ACP) has the uniqueness property on [0, T].

(b) If r is analytic and bounded on some half-plane {λ ∈ ℂ : Reλ > ω} then (λI − A)r(λ) ≈_{T} 0 implies that r ≈_{T} 0.

Proof. Suppose (a) holds, let r be analytic and bounded on some right half-plane and (λI − A)r(λ) ≈_{T} 0. Then, as seen in Remark 2.7, q(λ) = \frac{1}{λ}r(λ) is the Laplace transform of a continuous function v, (λI − A)q(λ) ≈_{T} 0 and q ∈ {v}. It follows from Proposition 3.3 that v(t) = A ∫_{0}^{t} v(s) ds and therefore v(t) = 0 for all t ∈ [0, T]. By (2.3), q ≈_{T} 0. Thus r ≈_{T} 0. Suppose (b) holds. Let v be a continuous function such that v(t) = A ∫_{0}^{t} v(s) ds for all t ∈ [0, T]. By (3.2), (λI − A)v̂_{T} (λ) ≈_{T} 0. Thus, v̂_{T} ≈_{T} 0. It follows from Proposition 2.2 that v(t) = 0 for all t ∈ [0, T].

Corollary 3.5. [Lyubich’s Uniqueness Theorem] If the resolvent R(λ, A) exists for λ > ω and lim sup_{λ→∞} \frac{1}{λ} ln ∥R(λ, A)∥ ≤ 0, then (ACP) has the uniqeness property.

Proof. If a(λ) := (λI − A)r(λ) ≈_{T} 0 (λ > ω), then r(λ) = R(λ, A)a(λ) ≈_{T} 0. Now the statement follows from Theorem 3.4.

The Cauchy problem (ACP) is said to be wellposed on [0, T] (or [0, ∞)) if there exists a regularizing function k (whose Laplace transform has at most finitely many real zeros) such that, for all x ∈ X, (ACP) has a unique, k-generalized (mild) solution u ∈ C^{[k]}([0, T]; X) (or u ∈ C^{[k]}([0, T]; X)).

Proposition 3.6. If (ACP) is wellposed, then the resolvent R(λ, A) exists for λ > ω and lim sup_{λ→∞} \frac{1}{λ} ln ∥R(λ, A)∥ ≤ 0.

Proof. Suppose there exists a regularizing function k such that, for all x ∈ X, there exists a unique, continuous function v : [0, ∞) → X such that v(t) = A ∫_{0}^{t} v(s) ds + (1 ∗ k)(t)x for all t ∈ [0, T]. For t ≥ 0 define linear operators S(t) by S(t)x = v(t) and S : X → C([0, T]; X) by x → S(·)x. Then S is closed and everywhere defined. Thus S is bounded; i.e., there exists M > 0 such that ∥Sx∥ = sup_{0≤t≤T} ∥S(t)x∥ ≤ M∥x∥ for all x ∈ X.^{19} It follows from Proposition 3.1 that Ŝ_{T} (λ)x ∈ D(A) for all x ∈ X and λ ∈ ℂ and ∥Ŝ_{T} (λ)x∥ ≤ MT ∥x∥ for x ∈ X and Reλ > 0. Moreover, (λI − A)Ŝ_{T} (λ)x = (λI − A) ∫_{0}^{T} e^{−λt}S(t)x dt = λŜ_{T} (λ)x − A[e^{−λT} ∫_{0}^{T} S(s)x ds + λ ∫_{0}^{T} e^{−λt} ∫_{0}^{t} S(s)x ds] = λŜ_{T} (λ)x−e^{−λT} S(T)x+e^{−λT} (1∗k)(T)x−λ ∫_{0}^{T} e^{−λt}(S(t)x− (1∗k)(t)x) dt = −e^{−λT} S(T)x+e^{−λT} (1∗k)(T)x+λ ∫_{0}^{T} e^{−λt}(1∗k)(t)x dt = −e^{−λT} S(T)x + ∫_{0}^{T} e^{−λt}k(t) dtx = −e^{−λT} S(T)x + k̂(λ)x − ∫_{T}^{∞} e^{−λt}k(t) dtx. Thus,

(λI − A)\frac{1}{k̂(λ)}Ŝ_{T} (λ)x = (I − B(λ))x,  (3.3)

where B(λ) = \frac{1}{k̂(λ)}[e^{−λT} S(T) + ∫_{T}^{∞} e^{−λt}k(t) dt]. Let λ_{0} ∈ ℂ such that k̂(λ_{0}) ≠ 0 and

(λ_{0}I −B(λ_{0}))^{−1} exists. Suppose there exists x_{0} ∈ X such that (λ_{0}I −A)x_{0} = 0. Then u(t) = e^{λ_{0}t}x_{0} solves (ACP) and S(t)x_{0} = v(t) = (k ∗ u)(t) = (k(t) ∗ e^{λ_{0}t})x_{0}. By (3.3), (I − B(λ_{0}))x_{0} = (λ_{0}I − A)\frac{1}{k̂(λ_{0})} Ŝ_{T} (λ_{0})x_{0} = \frac{1}{k̂(λ_{0})} Ŝ_{T} (λ_{0})(λ_{0}I − A)x_{0} = 0. Therefore,

x_{0} = 0. This shows that the set Ω := {λ ∈ ℂ : k̂(λ) ≠ 0 and (I − B(λ))^{−1} exists} is contained in the resolvent set ρ(A) and, for all λ ∈ Ω,

R(λ, A) = \frac{1}{k̂(λ)}Ŝ_{T} (λ)(I − B(λ))^{−1}.

By (2.3), lim sup_{λ→∞} \frac{1}{λ} ln ∥B(λ)∥ ≤ −T. Thus, there exists ω > 0 such that ∥B(λ)∥ ≤ 1/2 and k̂(λ) ≠ 0 for all λ > ω. Therefore, (ω, ∞) ⊂ ρ(A), ∥R(λ, A)∥ ≤ \frac{2MT}{|k̂(λ)|}, and

thus lim sup_{λ→∞} \frac{1}{λ} ln ∥R(λ, A)∥ ≤ 0.^{20}

19This shows that existence and uniqueness of k-generalized mild solutions of u′(t) = Au(t) (t ∈ [0, T]), u(0) = x, for all x ∈ X implies their continuous dependence on the initial data; i.e., ∥u∥_{[k]} = sup_{0≤t≤T} ∥(k ∗ u)(t)∥ = sup_{0≤t≤T} ∥v(t)∥ ≤ M∥x∥ for all x ∈ X.

20If 1/k̂(λ) is sufficiently well-behaved, then Ω contains a post-sectorial region Ω_{Ψ} because

Many authors have considered the abstract Cauchy problem (ACP) for operators whose resolvents exist on a post-sectorial region Ω_{Ψ} := {λ : Reλ ≥ β > 0, |Imλ| ≤ Ψ(Reλ)} and have less than exponential growth; i.e., there exists a regularizing function k_{0} as defined in (2.8))^{21} such that sup_{λ∈Ω_{Ψ}} ∥λk̂_{0}(λ)R(λ, A)∥ < ∞.^{22} The asymptotic Laplace transform methods developed in the first section, especially Theorem 2.8, provide a useful, additional tool for a fine, detailed analysis of the degree of wellposedness for such Cauchy problems. We demonstrate this by considering postsectorial regions Ω_{Ψ}, where either Ψ(x) = e^{ax^{γ}} (a > 0, γ > 1), Ψ(x) = e^{ax} (a > 0), or Ψ(x) = x^{1/a} (0 < a < 1). With the asymptotic representation results from Section 2 at hand (Theorem 2.8, Corollary 2.9) one obtains immediately the (well known) fact that in these cases the abstract Cauchy problem (ACP) is k-wellposed on [0, T] for all 0 < T ≤ ∞. Furthermore, we show that k = k_{0} ∗ k_{Ψ,T}, where the regularizing function k_{Ψ,T} is explicitely given in terms of Ψ and T. By specifying k in this manner, one gains a better understanding of how the regularity (or irregularity) of the generalized solutions is affected by the boundary of the resolvent set and the growth of the resolvent, and, how a larger time interval [0, T] leads to a decay of regularity of the solutions.

Theorem 3.7. Let A be a linear operator on a Banach space X such that Ω_{Ψ} ⊂ ρ(A) and sup_{λ∈Ω_{Ψ}} ∥λk̂_{0}(λ)R(λ, A)∥ < ∞ for some regularizing function k_{0}.

(a) Let Ψ(x) = e^{ax^{γ}} (a > 0, γ > 1). Then, for all b > 0, the abstract Cauchy problem (ACP) is (k_{b} ∗ k_{0})-wellposed on [0, ∞), where k_{b}(t) = t^{b−1}/Γ(b).

(b) Let Ψ(x) = e^{ax} (a > 0). Then,

1. for all b > 0 and 0 < T < ab, the abstract Cauchy problem (ACP) is (k_{b} ∗ k_{0})-wellposed on [0, T], where k_{b}(t) = t^{b−1}/Γ(b), and,

2. for all b > 0, 0 < δ < 1, the abstract Cauchy problem (ACP) is (k_{b,δ} ∗ k_{0})- wellposed on [0, ∞), where k̂_{b,δ}(λ) = e^{−bλ^{δ}}.

(c) Let Ψ(x) = x^{1/a} (0 < a < 1). Then,

1. for all b > 0 and 0 < T < b cos(aπ/2), the abstract Cauchy problem (ACP) is (k_{b} ∗ k_{0})-wellposed on [0, T], where k̂_{b}(λ) = e^{−bλ^{a}}, and,

2. for all b > 0, a < γ < 1 , the abstract Cauchy problem (ACP) is (k_{b,γ} ∗k_{0})- wellposed on [0, ∞), where k̂_{b,γ}(λ) = e^{−bλ^{γ}}.

∥B(λ)∥ ≤ 1/2 for all λ ∈ Ω_{Ψ}. Moreover, the estimate ∥k̂(λ)R(λ, A)∥ ≤ 2MT holds for all λ ∈ Ω_{Ψ}.

See, for example, I. Cioranescu and G. Lumer [10], [11], [12].

21As the proof below reveals one can also take k_{0}(t) = δ(t); i.e., k̂_{0}(λ) = 1.

22See, for example, W. Arendt, O. El-Mennaoui and V. Keyantuo [1], R. Beals [6], J. Chazarain [9], I. Cioranescu and G. Lumer [10], [11], [12], H. Emamirad [17], H. Komatsu [18], R. deLaubenfels [19], and the references therein.

Proof. It follows from lim sup_{λ→∞} \frac{1}{λ} ln |k̂(λ)| = 0 that lim sup_{λ→∞} \frac{1}{λ} ln ∥R(λ, A)∥ ≤ 0. Thus, by Corollary 3.5, (ACP) has the uniqueness property. Because (λI − A)R(λ, A)x−x = 0 for all x ∈ X and λ ∈ Ω_{Ψ}, the statements follow immediately from Corollary 2.9 (setting q(λ) = k̂_{0}(λ)R(λ, A)x) and the Propositions 3.2 and 3.3.

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