On the Differentiability of Weak Solutions of an Abstract Evolution Equation with a Scalar Type Spectral Operator

Proposition 3.1. Let n = 1,2, … and I be a subinterval of an interval [0, T)  (0 < T ≤ ∞). A weak solution y(·) of (1.1) on [0, T) is n times strongly differentiable on I if and only if

in which case,

Proof. “Only if” part.

Let n = 1,2, … and suppose that a weak solution y(·) of (1.1) on [0, T) is n times strongly differentiable on a subinterval I⊆[0, T).

Then for any g^* ∈ D(A^*),

Hence, by the closedness of the operator A (cf. [1]), This concludes proving the “only if” part for n = 1, in which case, as is to be noted, the subinterval I can shrink to a single point t ∈ [0, T).

Let n ≥ 2 and let the interval I be not a singleton. Then differentiating (3.4) at an arbitrary t ∈ I, we obtain

with the increments Δt being such that t + Δ ∈ I.

Since

by the closedness of A, we infer that Considering (3.4) and (3.7), Continuing inductively in this manner, we arrive at “If” part.

Let y(·) be a weak solution of (1.1) on an interval [0, T) (0 < T ≤ ∞) such that for an n = 1,2, … and a subinterval I⊆[0, T),

As was discussed (see Section 1), with some f ∈ ⋂_0≤t<TD(e^tA).

The fact that e^tAf ∈ D(Aⁿ), t ∈ I, implies by the properties of the o.c. and [2], Proposition 3.1, that, for any g^* ∈ X^*,

Given a k = 1, …, n and an arbitrary t ∈ I, let us choose a segment [a, b] ⊂ I (a < b) so that t is its midpoint if 0 < t < T or a = 0 if t = 0. For increments Δt such that a ≤ t + Δt ≤ b and any g^* ∈ X^*, we have Indeed, for any k = 1, …, n and an arbitrary λ ∈ σ(A),

Therefore, for any g^* ∈ X^*,

For Δ_m : = {λ ∈ ℂ∣|λ| ≤ m}, m = 1,2, …, let us fix an arbitrary k = 1, …, n and consider the sequence of vector functions Since f ∈ D(A^ke^tA), t ∈ I, by the properties of the o.c., Due to the boundedness of Δ_m, m = 1,2, …, by the properties of the o.c., AE_A(Δ_m), m = 1,2, …, is a bounded linear operator on X (∥AE_A(Δ_m)∥≤4Mm, m = 1,2, …, [7]). Hence, the vector functions are strongly continuous on I.

For an arbitrary segment [a, b]⊆I, we have

Hence, for any k = 1, …, n, the vector function A^ke^tAf, t ∈ I, is strongly continuous on I, being the uniform limit of the sequence of strongly continuous on I vector functions $$ on any segment [a, b]⊆I.

Let us fix an arbitrary a ∈ I and integrate (3.15) for k = 1 between a and an arbitrary t ∈ I. Considering the strong continuity of Ae^tAf, t ∈ I, we have

Whence, as follows from the Hahn-Banach Theorem, By the strong continuity of Ae^tAf, t ∈ I, Consequently, by (3.15), for n = 2, Whence, analogously, Continuing inductively in this manner, we infer that, for any k = 1, …, n,

Corollary 3.2. Let I be a subinterval of an interval [0, T)  (0 < T ≤ ∞). A weak solution y(·) of (1.1) on [0, T) is strongly infinite differentiable on I if and only if

in which case,

Theorem 4.1. Every weak solution of (1.1) on [0, ∞) is strongly infinite differentiable on [0, ∞) if and only if there is a b₊ > 0 such that the set $$ is bounded.

Proof. “If” part.

Let b₊ > 0 be such that the set $$ is bounded and y(·) a weak solution of (1.1) on [0, ∞). Then (see Section 1)

with some $$

For any n = 1,2, …, t ≥ 0 and an arbitrary g^* ∈ X^*,

Indeed, the former integral is finite due to the boundedness of the set $$, the finiteness of the measure v(f, g^*, ·) and the continuity of the integrated function on ℂ.

For the latter one, we have

Further, for any n = 1,2, … and t ≥ 0, Indeed, since, due to the boundedness of the set $$, the set $$ is void for all sufficiently large m′s, In addition to this

By the properties of the o.c. and [2], Proposition 3.1, (4.2) and (4.4) imply that

Then, by Corollary 3.2, y(·) is strongly infinite differentiable on [0, ∞).

“Only if” part.

We will prove this part by contrapositive.

Assume that, for any b₊ > 0, the set $$ is unbounded.

In particular, for any n = 1,2, …, unbounded is the set $$.

Hence, we can choose a sequence of points in the complex plane $$ as follows:

The latter implies in particular that the points λ_n are distinct (λ_i ≠ λ_j, i ≠ j).

Since, for any n = 1,2, …, the set

is open in ℂ, there exists an ɛ_n > 0 such that this set along with the point λ_n contains the open disk of radius ɛ_n centered at λ_n Then, for any λ ∈ Δ_n, n = 1,2, …, Further, since the points λ_n, n = 1,2, …, are distinct, we can regard the radii of the disks, ɛ_n, n = 1,2, …, to be small enough so that Whence, by the properties of the s.m., Observe also, that the subspaces E_A(Δ_n)X, n = 1,2, …, are nontrivial since Δ_n∩σ(A) ≠ ∅, with Δ_n being an open set.

By choosing a unit vector e_n ∈ E_A(Δ_n)X (∥e_n∥ = 1), n = 1,2, …, we obtain a vector sequence such that

(δ_ij is the Kronecker delta symbol).

As is readily verified, (4.14) implies that the vectors {e₁, e₂, …} are linearly independent.

Moreover, there is an ɛ > 0 such that

Indeed, the opposite implies the existence of a subsequence $$ such that Then, for any k = 1,2, …, by selecting a vector f_n(k) ∈ span ({e_j∣j = 1,2, …,   j ≠ n(k)}) such that ∥e_n(k) − f_n(k)∥<d_n(k) + 1/k, considering (4.14) and (2.6), we arrive at the contradiction: As follows from the Hahn-Banach Theorem, for any n = 1,2, …, there is an $$ such that Concerning the sequence of the real parts, $$, there are two possibilities: it is either bounded or unbounded above.

Suppose that the sequence $$ is bounded above, that is, there is such an ω > 0 that

Let By (4.14), For any t ≥ 0 and an arbitrary g^* ∈ X^*, Similarly, for any t ≥ 0, By [2], Proposition 3.1, (4.22) and (4.23) imply that $$

Therefore (see Section 1), y(t): = e^tAf, 0 ≤ t < ∞, is a weak solution of (1.1) on [0, ∞).

Let

(cf. (4.18)).

Considering (4.18) and (4.15), we have

Similarly to (4.22), Whence, by [2], Proposition 3.1, y(0) = f ∉ D(A).

Consequently, by Proposition 3.1, the weak solution y(t) = e^tAf, 0 ≤ t < ∞, of (1.1) on [0, ∞) is not strongly differentiable at 0.

Now, assume that the sequence $$ is unbounded above.

Therefore, there is a subsequence $$ such that

Then the vector is well defined.

By (4.14),

For any t ≥ 0 and an arbitrary g^* ∈ X^*, similar to (4.22), Indeed, for all sufficiently large k′s, n(k) − t ≥ 1 and by (4.27), Analogously, for an arbitrary t ≥ 0,

From (4.30) and (4.32), by [2], Proposition 3.1, we infer that f ∈ ⋂_0≤t<∞D(e^tA).

Therefore (see Section 1), y(t): = e^tAf, 0 ≤ t < ∞, is a weak solution of (1.1) on [0, ∞).

For any λ ∈ Δ_n(k), k = 1,2, …, by (4.11), (4.12), and (4.27),

Hence, for λ ∈ Δ_n(k), k = 1,2, …, Using this estimate, we obtain Whence, by [2], Proposition 3.1, y(0) = f ∉ D(A).

Therefore, by Proposition 3.1, the weak solution y(t) = e^tAf, 0 ≤ t < ∞, of (1.1) on [0, ∞) is not strongly differentiable at 0.

With every possibility concerning $$ considered, we infer that the opposite to the assumption that, for a certain b₊ > 0, the set $$ is bounded, allows to single out a weak solution of (1.1) on [0, ∞) that is not strongly differentiable at 0, much less strongly infinite differentiable on [0, ∞).

Thus, the “only if” part has been proved by contrapositive.

Theorem 4.2. Every weak solution of (1.1) on [0, ∞) is strongly infinite differentiable on (0, ∞) if and only if there is a b₊ > 0 such that, for any b₋ > 0, the set $$ is bounded.

Proof. “If” part.

Let b₊ > 0 be such that, for an arbitrary b₋ > 0, the set $$ is bounded.

Let y(·) be a weak solution of (1.1) on [0, ∞). Then (see Section 1)

with some f ∈ ⋂_0≤t<∞D(e^tA).

For any n = 1,2, …, t > 0, and an arbitrary g^* ∈ X^*,

The first integral in this sum is finite due the boundedness of the set $$, the finiteness of the measure v(f, g^*, ·), and the continuity of the integrated function on ℂ.

The finiteness of the second integral is proved in absolutely the same manner as for the corresponding integral in the proof of the “if” part of Theorem 4.1.

Finally for the third one, considering that b₋ > 0 is arbitrary and setting b₋ : = nt⁻¹, we have

Further, for any n = 1,2, … and t > 0, Indeed, since, due to the boundedness of the set $$, the set $$ is void for all sufficiently large m′s, Similarly to the “if” part of Theorem 4.1 and (4.38), we have By the properties of the o.c. and [2], Proposition 3.1, (4.37) and (4.39) imply that Whence, by Corollary 3.2, y(·) is strongly infinite differentiable on (0, ∞).

“Only if” part.

As well as in Theorem 4.1, we will prove this part by contrapositive.

Thus, we assume that, for any b₊ > 0, there is such a b₋ > 0 that the set $$ is unbounded.

Let us show that this assumption can even be strengthened. To wit: there is such a b₋ > 0 that, for any b₊ > 0, the set $$ is unbounded.

Indeed, there are two possibilities:

• (1)

  for a certain b₋ > 0, the set {λ ∈ σ(A)∣ − b₋ln  | Im λ | < Reλ ≤ 0} is unbounded;

• (2)

  for any b₋ > 0, the set {λ ∈ σ(A)∣ − b₋ln  | Im λ | < Reλ ≤ 0} is bounded.

In the first case, the set $$ is unbounded with the very b₋ > 0, for which the set {λ ∈ σ(A)∣ − b₋ln  | Im λ | < Reλ ≤ 0} is unbounded, and an arbitrary b₊ > 0.

In the second case, based on the premise we infer that, for any b₊ > 0, the set {λ ∈ σ(A)∣0 < Reλ < b₊ln  | Im λ|} is unbounded. Then so is the set $$ for any b₋ > 0 and b₊ > 0.

Thus, let us fix a b₋ > 0 such that the set $$ is unbounded for an arbitrary b₊ > 0.

In particular, for any n = 1,2, …, the set $$ is unbounded.

Hence, we can select a sequence of points in the complex plane $$ in the following manner:

The latter, in particular, implies that the points λ_n are distinct.

Since, for any n = 1,2, …, the set

is open in ℂ, there exists such an ɛ_n > 0 that this set, along with the point λ_n, contains the open disk of radius ɛ_n centered at λ_n Hence, for any λ ∈ Δ_n, n = 1,2, …, Since the points λ_n are distinct, we can regard the radii of the disks, ɛ_n, to be small enough so that

By the properties of the s.m., the latter implies

Observe that the subspaces E_A(Δ_n)X, n = 1,2, …, are nontrivial since Δ_n∩σ(A) ≠ ∅, Δ_n being an open set.

By choosing a unit vector e_n ∈ E_A(Δ_n)X (∥e_n∥ = 1), n = 1,2, …, we obtain a vector sequence such that

In the same manner as in the proof of Theorem 4.1, one can show that there is an ɛ > 0 such that As follows from the Hahn-Banach Theorem, for any n = 1,2, …, there is an $$ such that Concerning the sequence of the real parts, $$, there are two possibilities: it is either bounded or unbounded.

First, assume that the sequence $$ is bounded, that is, there is an ω > 0 such that

Let By (4.49), In absolutely the same fashion as it was done in the case of bounded above sequence $$ in the proof of the “only if” part of Theorem 4.1, it is shown that f ∈ ⋂_0≤t<∞D(e^tA).

Therefore (see Section 1), y(t): = e^tAf, 0 ≤ t < ∞, is a weak solution of (1.1) on [0, ∞).

Let

(cf. (4.51)).

Taking (4.50) and (4.51) into account, we have

Thus, Whence, by the properties of the o.c. and [2], Proposition 3.1, y(1) = e^Af ∉ D(A).

Consequently, by Proposition 3.1, the weak solution y(t) = e^tAf,   0 ≤ t < ∞, of (1.1) on [0, ∞) is not once strongly differentiable on (0, ∞).

Now, assume that the sequence $$ is unbounded. Therefore, there is a subsequence $$ such that

Suppose that Reλ_n(k) → ∞ as k → ∞.

Without restricting generality, we can regard that

Considering this case just like the analogous one in the proof of the “only if” part of Theorem 4.1, one can show that the vector belongs to ⋂_0≤t<∞D(e^tA) and, hence, (see Section 1) y(t): = e^tAf, 0 ≤ t < ∞, is a weak solution of (1.1) on [0, ∞).

For any λ ∈ Δ_n(k), k = 1,2, …, by (4.46), (4.47), and (4.59),

Hence, for λ ∈ Δ_n(k), k = 1,2, …, Using this estimate, we have Whence, by the properties of the o.c. and [2], Proposition 3.1, y(1) = e^Af ∉ D(A).

Therefore, by Proposition 3.1, the weak solution y(t) = e^tAf, 0 ≤ t < ∞, of (1.1) on [0, ∞) is not once strongly differentiable on (0, ∞).

Now, suppose that Reλ_n(k) → −∞ as k → ∞.

Without restricting generality, we can regard that

Let For any t ≥ 0 and an arbitrary g^* ∈ X^*, Analogously, for an arbitrary t ≥ 0, From (4.66) and (4.67), by [2], Proposition 3.1, we infer that f ∈ ⋂_0≤t<∞D(e^tA).

Therefore (see Section 1), y(t): = e^tAf, 0 ≤ t < ∞, is a weak solution of (1.1) on [0, ∞).

Let

(cf. (4.51)).

Taking into account (4.50) and (4.51), we have

For any λ ∈ Δ_n(k), k = 1,2, …, by (4.47) and (4.64), Hence, for λ ∈ Δ_n(k), k = 1,2, …, Using these estimates, we have Whence, by the properties of the o.c. and [2], Proposition 3.1, $$.

Therefore, by Proposition 3.1, the weak solution y(t) = e^tAf, 0 ≤ t < ∞, of (1.1) on [0, ∞) is not once strongly differentiable on (0, ∞).

With every possibility concerning $$ considered, we infer that the opposite to the assumption that there is a b₊ > 0 such that, for any b₋ > 0, the set $$ is bounded, allows to single out a weak solution of (1.1) on [0, ∞) that is not once strongly differentiable on (0, ∞), much less strongly infinite differentiable on (0, ∞).

Thus, the “only if” part has been proved by contrapositive.

Proposition 5.1. If every weak solution of (1.1) on [0, ∞) is strongly differentiable at 0, then all of them are strongly infinite differentiable on [0, ∞).

Proposition 5.2. If every weak solution of (1.1) on [0, ∞) is once strongly differentiable on (0, ∞), then all of them are strongly infinite differentiable on (0, ∞).