1.3. Examples
1.3.3. Integro-differential equations
In this subsection we study integro-differential equations. The results presented here are based on [Tro15c]. We focus on equations of the form
(∂0,ϱ(1 + k∗) + A) u = f, (1.12)
as well as
(∂0,ϱ(1 − k∗)−1+ A) u = f, (1.13)
where in both cases A : D(A) ⊆ H → H is an m-accretive operator on some Hilbert space H. The kernel k is a suitable operator-valued function in the space L1,ϱ(R≥0; L(H)) defined as follows.
Definition. Let ϱ ∈ R. We call a function k : R≥0 → L(H) admissible, if k is weakly measurable (i.e., for each x, y ∈ H the mapping t ↦→ ⟨k(t)x|y⟩H is measurable) and t ↦→ ∥k(t)∥
is measurable1. We define
L1,ϱ(R≥0; L(H)) :=
1We note that the measurability of t ↦→ ∥k(t)∥ follows from the weak measurability of k, if H is separable.
as well as
L1,ϱ(R≥0; L(H)) := L1,ϱ(R≥0; L(H))
⧸∼=,
where the relation ∼= is the usual equality almost everywhere. We equip L1,ϱ(R; L(H)) with the usual norm defined by
|k|L1,ϱ :=
which proves that the functional is indeed bounded. The linearity is trivial. Hence, by the Riesz representation theorem there exists a unique element (k ∗ (χIx)) (t) ∈ H with
⟨(k ∗ (χIx)) (t)|y⟩H =
= sup
where Sim(R; H) denotes the space of simple functions with values in H, is well-defined. More-over, k∗ extends to a bounded linear operator on Hϱ(R; H) for each ϱ ≥ ϱ0 with ∥k ∗ ∥L(Hϱ)≤
|k|L1,ϱ. In particular ∥k ∗ ∥L(Hϱ)→ 0 as ϱ → ∞.
Proof. By Lemma 1.3.4 we have defined k∗ for functions φ = χIx where I ⊆ R is a bounded interval and x ∈ H. Thus, for φ ∈ Sim(R; H), k ∗ φ is a continuous function, in particular, it is measurable. Moreover, choosing pairwise disjoint intervals, we derive from (1.14) that
| (k ∗ φ) (t)|H ≤
which shows the first assertion. The second assertion follows from ∥k ∗ ∥L(Hϱ) ≤ |k|L1,ϱ and
|k|L1,ϱ → 0 for ϱ → ∞, by monotone convergence.
In case of a separable Hilbert space H, we have the usual integral expression for the function k ∗ f as the next lemma shows.
Lemma 1.3.6. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; L(H)), H separable. Then for f ∈ Hϱ(R; H)
Proof. We first prove that
R≥0 ∋ t ↦→ k(t)f (t) ∈ H
is measurable. Indeed, first we note that for all x, y ∈ H we have that t ↦→ ⟨x|k(t)∗y⟩H = ⟨k(t)x|y⟩H
is measurable, i.e. t ↦→ k(t)∗y is weakly measurable for each y ∈ H. By the Theorem of Pettis (see Theorem B.7), we infer that t ↦→ k(t)∗y is measurable for each y ∈ H. Hence,
t ↦→ ⟨k(t)f (t)|y⟩H = ⟨f (t)|k(t)∗y⟩H
is measurable and thus, again by Theorem B.7 we derive the measurability of t ↦→ k(t)f (t).
In particular, the function
by Young’s inequality. Consider now the function g : R → H defined by
g(t) :=
{∫∞
0 k(s)f (t − s) ds, if ∫∞
0 |k(s)f (t − s)|H ds < ∞,
0, otherwise.
Then g is measurable. Indeed, for y ∈ H we have that
∫
is measurable, we obtain the measurability of
t ↦→ ⟨g(t)|y⟩H
by Fubini’s Theorem. Again by Theorem B.7, the measurability of g follows and by the estimate shown above, we have that
∫ Summarizing we have shown that
˜k∗ : f ↦→
is a well-defined and bounded operator on Hϱ(R; H). Let now I ⊆ R be a bounded interval and x ∈ H. Then
i.e. k∗ and ˜k∗ coincide on the dense set Sim(R; H), which yields the assertion.
Definition. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; L(H)). Then we define ˆk(z) ∈ L(H) for z ∈
Remark 1.3.7. We note that ˆk(z) is well-defined by the Riesz representation theorem. More-over, the mapping CRe>ϱ0 ∋ z ↦→ ˆk(z) ∈ L(H) is bounded by √1
2π|k|L1,ϱ0 and analytic accord-ing to [HP57, Theorem 3.10.1].
Lemma 1.3.8. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; L(H)). Then, k∗ = √
2πˆk(∂0,ϱ) for each ϱ > ϱ0.
Proof. Since ˆk is bounded on CRe>ϱ0, the operator ˆk(∂0,ϱ) is bounded for each ϱ > ϱ0 as well.
Moreover, by Lemma 1.3.5 the operator k∗ is bounded on Hϱ(R; H), too. Thus, it suffices to show k ∗ φ = √
2πˆk(∂0,ϱ)φ for φ = χIy for some bounded interval I ⊆ R, y ∈ H and ϱ > ϱ0. So, let ϱ > ϱ0 and φ = χIy for some interval I ⊆ R, y ∈ H. Using that k ∗ φ ∈ L1,ϱ(R; H) by
(1.14), we can compute
⟨x|√
2πˆk(i t + ϱ) (Lϱφ) (t)⟩H =
∫
R
e−(i t+ϱ)s⟨x|k(s) (Lϱφ) (t)⟩H ds
= 1
√2π
∫
R
∫
R
e−(i t+ϱ)(s+r)⟨x|k(s)φ(r)⟩H dr ds
= 1
√2π
∫
R
∫
R
e−(i t+ϱ)r⟨x|k(s)y⟩HχI(r − s) dr ds
= 1
√2π
∫
R
e−(i t+ϱ)r⟨x| (k ∗ φ) (r)⟩H dr
= ⟨x|Lϱ(k ∗ φ)(t)⟩H for each x ∈ H, t ∈ R. That shows the assertion.
The latter lemma gives that (1.12) and (1.13) are indeed evolutionary problems with M (z) = 1 +√
2πˆk(z) and M (z) = (
1 +√ 2πˆk(z)
)
, respectively. We now address the well-posedness of the problems (1.12) and (1.13). For doing so, we formulate the following conditions.
Condition 1.3.9. Let ϱ0∈ R and k ∈ L1,ϱ0(R≥0; L(H)). We say that k satisfies the condition (a),(b) and (c), respectively, if
(a) For almost every t ∈ R, the operator k(t) is selfadjoint.
(b) For almost every t, s ∈ R we have k(t)k(s) = k(s)k(t).
(c) There exists ϱ1 ≥ ϱ0 and d ≤ 0 such that
t Im⟨ˆk(i t + ϱ1)x|x⟩H ≥ d|x|2H for each t ∈ R, x ∈ H.
We first show, that a kernel k satisfying Condition 1.3.9 (a) and (c), also satisfies a similar inequality like in (c) for all ϱ > ϱ1. The precise statement is as follows.
Lemma 1.3.10. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; L(H)) satisfying Condition 1.3.9 (a) and (c).
Then for each t ∈ R, ϱ ≥ ϱ1 (where ϱ1 is chosen according to Condition 1.3.9 (c)) and x ∈ H one has
t Im⟨ˆk(i t + ϱ)x|x⟩H ≥ 4d|x|2H.
Proof. The proof is based on the proof presented in [CS03, Lemma 3.4] in case of scalar-valued kernels. For x ∈ H we define
f (t) := ⟨k(t)x|x⟩H (t ∈ R≥0)
and get f ∈ L1,ϱ0(R≥0; R) by the selfadjointness of k(t). Moreover, we get harmonic function, we employ the Poisson formula for the half-plane (see e.g. [SS03, p.149]) and get
With this result at hand, we are able to prove the well-posedness of the integro-differential equations. In fact, we show that the material laws satisfy (1.7) and thus, the well-posedness follows from Proposition 1.2.18. We start with (1.12).
Proposition 1.3.11. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; H). Moreover, assume that k satisfies Condition 1.3.9 (a) and (c). Then, the material law M defined by M (z) := 1 +√
2πˆk(z) satisfies (1.7) on CRe>ϱ for some ϱ ≥ max{0, ϱ0}.
Proof. For x ∈ H and t ∈ R, ϱ ≥ max{0, ϱ1} where ϱ1 is chosen according to Condition 1.3.9
(c) we estimate
where we have used Lemma 1.3.10. Thus, choosing ϱ large enough, the assertion follows, since
|k|L1,ϱ → 0 as ϱ → ∞ by monotone convergence.
To deal with (1.13), we additionally need to impose Condition 1.3.9 (b). We start with the following observation.
Lemma 1.3.12. Let ϱ0 ∈ R and k ∈ L1,ϱ0(R≥0; H) such that |k|L1,ϱ0 < 1 and assume that k satisfies Condition 1.3.9 (a) and (b). Let ϱ > ϱ0and t ∈ R. Then the operator |1−√
2πˆk(i t+ϱ)|
is boundedly invertible and for each x ∈ H we have Re⟨(i t + ϱ)(1 − is boundedly invertible due to the Neumann series. Moreover, by Condition 1.3.9 (a) we have that ˆk(i t + ϱ)∗ = ˆk(− i t + ϱ) and thus, we have that (1 −√
2πˆk(i t + ϱ))∗ is boundedly invertible, too. The latter gives that |1 −√
2πˆk(i t + ϱ)| is boundedly invertible. Moreover, Condition 1.3.9 (b) yields that ˆk(i t + ϱ) is normal and so is 1 −√
and D(i t + ϱ) pairwise commute. Thus, we have (1 −√
2πˆk(z) is boundedly invertible for each z ∈ CRe>ϱ. If k satisfies Condition 1.3.9 (a)
-(c), then there is some ϱ ≥ max{0, ϱ0}, such that M (z) := (1 −√
2πˆk(z))−1 satisfies (1.7) on CRe>ϱ.
Proof. First we choose ϱ2 ≥ ϱ1, where ϱ1 is chosen according to Condition 1.3.9 (c), such that
|k|L
We conclude this section with some classical examples of kernels, satisfying Condition 1.3.9 (a)-(c).
Example 1.3.14. Let k : R≥0 → R measurable such that ∫∞
0 |k(t)| e−ϱ0t dt < ∞ for some ϱ0 ∈ R. Then clearly, k satisfies Condition 1.3.9 (a) and (b).
(a) Assume that k is absolutely continuous with ∫∞
0 |k′(t)| e−ϱ1t dt < ∞ for some ϱ1 ∈ R.
and thus, for x ∈ H we can estimate
(b) In [Prü09] the kernel is assumed to be non-negative and non-increasing. In fact, this also yields that k satisfies Condition 1.3.9 (c) with d = 0. Indeed, we can even generalize this fact to kernels k ∈ L1,ϱ0(R≥0; L(H)) satisfying Condition 1.3.9(a) such that
⟨k(t)x|x⟩H ≥ 0
which yields
t Im⟨ˆk(i t + ϱ)x|x⟩H ≥ 0
for t > 0. For t = 0 the inequality holds trivially and for t < 0 we use the fact that ˆk(i t + ϱ) = (ˆk(− i t + ϱ))∗ by Condition 1.3.9(a) and hence,
t Im⟨ˆk(i t + ϱ)x|x⟩H = t Im⟨x|ˆk(− i t + ϱ)x⟩H = −t Im⟨ˆk(− i t + ϱ)x|x⟩H ≥ 0.