OPTIMAL SEMIGROUP REGULARITY FOR VELOCITY COUPLED ELASTIC SYSTEMS: A DEGENERATE FRACTIONAL DAMPING CASE

. In this note, we consider an abstract system of two damped elastic systems. The damping involves the average velocity and a fractional power of the principal operator, with power θ in [0 , 1]. The damping matrix is degenerate, which makes the regularity analysis more delicate. First, using a combination of the frequency domain method and multipliers technique, we prove the following regularity for the underlying semigroup:

Later on, Liu and Liu proposed a method for proving analyticity or Gevrey class regularity for semigroups associated with elastic systems [21]; their method combines resolvent estimates using a contradiction argument and mulptipliers technique.Other closely related works include e.g.[1,10,[15][16][17][18][19]22]; those works discuss either regularity or stability issues, which are the main themes of the present contribution.
In a recent work [2], the authors consider a system of two damped abstract elastic systems in a Hilbert space, (see System (2.1) below).The damping involves a fractional power θ of the principal operator, and the average velocity, so that it is degenerate (see below, just after System (2.1), a brief explanation of why we use the term degenerate.)The power θ lies in the interval [−1 , 1].They prove the following regularity and stability results for the associated semigroup: -The semigroup is not analytic for every θ in (0, 1]. -The semigroup is eventually differentiable for each θ in (0, 1).
-The semigroup is polynomially stable of order O(t 1 2θ ), as t goes to infinity for each θ in [−1, 0).-Further, the resolvent estimate for the Gevrey class regularity in the case where θ in (0, 1/4] is optimal, and so is the polynomial stability for each θ in [−1, 0).
All of those results are established using resolvent estimates and multipliers technique, combined with semigroup stability characterization results e.g.[4,14,27], or regularity results [26,28].One notes that, unlike the case of a single equation where analyticity holds for every θ in [1/2, 1], in [2] the authors did not get analyticity for any value of θ.Though they were able to prove Gevrey class regularity for θ in (0, 1/2), one notices that they provide two different Gevrey classes depending on whether θ lies in (0, 1/4] or in (1/4, 1/2]; the Gevrey class for θ lying in (0, 1/4] is in agreement with the case of a single similarly damped elastic system.However, the Gevrey class corresponding to θ in (1/4, 1/2] is off by (1 − 4θ)/6θ.A natural question is then: why this discrepancy?Is this due to the degeneracy of the damping operator matrix, or to some technical difficulties?Besides, is the differentiablity of the semigroup the best regularity that one should expect for θ in (1/2, 1)?
The purpose of this note is to carefully examine those questions.The plan of the rest of our work is as follows: in Section 2, we discuss new regularity results.Section 3 deals with the eigenvalue asymptotics and the optimality of our regularity results.Details about eigenvalues computation and asymptotics are provided in Section 4. In Section 5, we present some examples of application.

New regularity results
Let H be a complex Hilbert space with the inner product • , • and the induced norm | • |.We consider the following abstract system of coupled equations: where A is a self-adjoint, positive definite (unbounded) operator on a complex Hilbert space H, a, b, γ > 0, a = b, and θ ∈ [0, 1].
Notice that the matrix defining the damping is given by and it is a singular matrix; the associated quadratic form is degenerate, hence the terminology "degenerate fractional damping".
2 ), and assume that V ⊂ H ⊂ V with compact injections, where V denotes the topological dual of V .We also assume that for any r, s in R with r < s, the embedding D(A s ) ⊂ D(A r ) is compact.
Introduce the Hilbert space H = V × H × V × H, over the field C of complex numbers, equipped with the norm Throughout this note, we shall assume: Introduce the operator with domain , where τ = max{θ, 1 2 }.One easily checks that for every (2.4) so that the operator A θ is dissipative.Further, the operator A θ is densely defined.It is straight-forward to show that A θ is bijective from D(A θ ) to H. Therefore, the Lumer-Phillips Theorem shows that the operator A θ generates a strongly continuous semigroup of contractions (S θ (t)) t≥0 on the Hilbert space H.One also checks that where ρ(A θ ) denotes the resolvent set of A θ .This shows that the semigroup (S θ (t)) t≥0 is strongly stable on the Hilbert space H, thanks to the strong stability criterion of e.g.Arendt and Batty [3].
Remark 2.1.We find it useful to give some helpful information about how the inclusion (2.5) is established.First, notice that for each θ ∈ [0, 1] we have D(A θ ) contained in ((D(A) + D(A 3 2 −θ )) × V ) 2 ; notice that when θ lies is in [0, 1/2], the sum D(A) + D(A 3 2 −θ ) reduces to D(A), while for θ in (1/2, 1], this sum simplifies to D(A 3 2 −θ ).Thus, one readily deduces that for θ in [0, 1), A θ has a compact resolvent.Therefore the spectrum of A θ is discrete.Proving (2.5) reduces to showing that for any λ in R, one has Ker(iλI − A θ ) = {0 H }, which is fairly easy.When θ = 1, the operator A θ no longer has a compact resolvent; in this case, to prove (2.5), not only do we have to show that Ker(iλI − A θ ) = {0 H } for any λ in R, which is easy, but we also have to show that for any λ in R, the operator iλI − A θ is surjective; we rely on the Fredholm alternative and the closed graph theorem to show the latter as follows: Set θ = 1.Let λ in R. Let F = (f, g, h, ) in H, we shall find U = (u, v, w, z) in D(A 1 ) such that (iλI − A 1 )U = F, which reduces to (2.6) Now, set L(u, w) = (aAu + iλγA(u + w), bAw + iλγA(u + w)).
One readily checks that L : Notice that the system (2.6) may be rewritten The Fredholm alternative, e.g.Theorem VI.6, p. 92 of [5] shows that solving (2.7) amounts to proving that the equation (u, w) − λ 2 L −1 (u, w) = (0, 0) has the unique solution (u, w) = (0, 0), or equivalently that (0,0) is the unique solution of the equation L(u, w) − λ 2 (u, w) = (0, 0).Taking the duality product between V × V and V × V on both sides of the latter equation with (u, w), we derive Taking the imaginary part in (2.8), one deduces u = −w.Therefore, we have on the one hand and on the other hand −λ 2 w + bAw = 0, which is equivalent to The combination of (2.9) and (2.10) yields (a − b)Au = 0, from which it readily follows u = 0, since a = b.Therefore w = 0. Thus, for θ = 1, combining the fact that iλI − A 1 is bijective and the closed graph theorem, we derive that (2.5) holds.
By interpolation, where, here and in the sequel, C denotes a generic positive constant independent of n.
Next, we take inner product of (2.19) with λ −1+2θ The inner product term in the above equality is o(1) due to (2.20) and (2.21).Therefore, we obtain Furthermore, equation (2.22) can now be rewritten as where we have replaced Combining the last equality with (2.12), (2.20) and (2.23), we arrive at To estimate the term A 1 2 w n , we take the inner product of (2.19) with λ −1+2θ We replace the term λ −1 n A (2.28) Finally, we take the inner product of (2.17 , we see that On the other hand, by (2.12) and 1 2 < θ, we also have .

By interpolation,
= O(1). (2.30) Next, we take the inner product of (2.19) with The inner product term in the above equality is o(1) due to (2.20) and (2.30).Therefore, we obtain Furthermore, (2.31) can now be rewritten as where we have replaced (2.18).The inner product of (2.17) with Combining the last equality with (2.20) and (2.32), we arrive at Finally, we take the inner product of (2.17 Remark 2.5.In Theorem 1.1 of [2], the auhors prove that for every θ ∈ (1/2, 1], and every r in (2(1 − θ), 1], one has : That equality shows that our resolvent estimate leading to the claimed Gevrey regularity class for θ in (1/2, 1) is optimal.Similarly, in Theorem 1.2 of [2] they prove: for every θ ∈ (0, 1/2) and every r ∈ (2θ, 1], one has: which shows the optimality of our resolvent estimate for θ in (0, 1/2) as well.These optimality results prove that the analyticity of the semigroup occurs for the only value θ = 1/2.Thus, in the case of interacting elastic systems with degenerate damping as in this contribution, analyticity of the underlying semigroup occurs only when the damping is structural; the degeneracy prevents analyticity for above structural damping mechanisms.
A careful reader may have noticed that in Remark 2.5, we do not claim the optimality of the Gevrey class; this is due to the fact that we are using a sufficient condition for Gevrey regularity, namely, Theorem 2.3.In the next section, we shall discuss the eigenvalue problem, state asymptotic expressions for the eigenvalues, and use those to show that our analyticity or Gevrey class regularity results are indeed optimal.We shall also prove the nondifferentiability of the semigroup for the values θ = 0 and θ = 1.

Let
Consequently, the eigenvalue system reads where a, b, and γ are positive constants, and θ ∈ [−1, 1].From the first three equations, Therefore, using the fourth equation, and denoting {µ n } the sequence of eigenvalues of the operator A, we find where , and e n = abµ 2 n .
From now on, we assume This assumption leads us to the simplified characteristic equation: Our main results for this section read Here, ψ n,1 , ψ n,2 , ψ n,3 , and ψ n,4 are two pairs of complex conjugate numbers. (3.5) (i) For every θ in (0, 1), there exists a sequence (λ n ) of eigenvalues of A θ such that for every ε > 0, one has (ii) For θ = 0 or θ = 1, there exists a sequence (λ n ) of eigenvalues of A θ such that for some real number .
The proof of Theorem 3.1 is very technical and will be dealt with in the next section.The proof of Theorem 3.2 follows at once from Theorem 3.1.
In particular, thanks to Theorem 3.2 and the following two results, one derives the optimality of our regularity results, as well as the nondifferentiability of the semigroup for θ = 0 and θ = 1.(i) Suppose that the spectrum σ(A) of A contains a sequence (λ n ) such that for some in R. Then the semigroup T is not differentiable.
(ii) Let δ ≥ 1. Suppose that there exists a sequence (λ n ) in σ(A) such that Then the semigroup T is not of Gevrey class δ.
Remark 3.5.In Theorem 2.4, we prove regularity results for System (2.1).As noted above, the combination of Theorem 3.2, Lemma 3.3 and Lemma 3.4 show that those regularity results are optimal.Accounting for the stability results established in [2], we can summarize regularity and stability results for System (2.1) in the following table (see next page).

Point spectrum analysis
In this section, our goal is to analyze the point spectrum by solving the quartic characteristic equation (3.4).To this end, first, we describe how to compute the analytic solution to a general quartic equation.Then, we describe how to solve (3.4).

Solutions to a general quartic equation
We begin by describing how to solve a general quartic equation: analytically.Our procedure follows that described in [6].To this end, we define According to [6], we have the following cubic resolvent associated with (4.1).
To solve (4.1), we first need to solve (4.3).For this purpose, we define h = 3s − m 2 , g = 2m 3 − 9ms + 27l, P = h/3, Q = g/27. (4.4) We further define the discriminant as: We also need to define: where, for any ξ = |ξ|e iθ , we consider ξ + + Φ With (4.7) and m from (4.2), the solutions to the cubic resolvent can then be expressed via the Cardano's formula [24]: + , and − , such that k 0 k + k − = r, we can denote the four roots to (4.1) according to [6] as: 4.2.Compute the roots of the characteristic equation (3.4)In this section, we describe how to solve (3.4) based on the analytic solutions to a general quartic equation described in Section 4.1.To this end, for the sequence of eigenvalues of the operator A, {µ n }, recall that where β n , c n , d n , and e n are directly associated with β, c, d, and e in (4.1), representing the coefficients of the quartic characteristic equation associated with µ n .In particular, we use the subscript n to stress the dependency of these coefficients on the sequence {µ n } and we shall use the same convention subsequently for the quantities that are associated with {µ n }.So, from (4.2) and (4.4), we have that that are associated with {µ n }.As a result, we also have ∆ n from (4.5), j n from (4.3), Φ n,± from (4.6), Λ n and Ω n from (4.7), j n,0 and j n,± from (4.8), and k n,0 , k n,± , λ n,1 , λ n,2 , λ n,3 , and λ n,4 from (4.9) that are associated with {µ n }.
With these definitions, we introduce Lemma 4.1 and Lemma 4.2, which are concerned with the expression of Λ n and Ω n . .12) where κ n,1 , κ n,2 , and x are quantities related to − Qn 2 and In (4.14), 3x is the power of the leading term shared by − Qn 2 and . We further assume that κ n,1 ± κ n,2 = 0. We omit the discussion when κ n,1 ± κ n,2 = 0 as it is not involved in our problem.
-Finally, when , without loss of generality, we can write: where we further assume that κ n,1 ± κ n,2 = 0. Then While Lemma 4.1 provides a characterization of the leading term of Λ n , later on, we will show that using just the leading term of Λ n does not suffice to estimate the leading terms of λ n,0 and λ n,± as desired.This motivates us to further characterize the lower order terms of Λ n , yielding Lemma 4.2.Lemma 4.2.Let Λ n = Z n + ω n , where Z n is the leading term of Λ n as n → ∞, and ω n is the lower order term.i.e. ω n = o(Z n ).Then Proof.First, we define By Cardano's formula, x n,0 , x n,± are solutions to x 3 n + P n x n + Q n = 0. We further define where Λ n = Z n + ω n , Z n is the leading term of Λ n , and ω n = o(Z n ).Clearly, Therefore, y n,0 and y n,± are the three solutions for: On the other hand, using (4.16), by Vieta's formula with respect to y n,0 , and y n,± , which indicates that (4.15) is true.
. We describe how the leading terms of quantity is determined.As an illustrative example, we show how to determine the leading term of ∆ n when θ ∈ [−1, 1 2 ).Using the formula for ∆ n given in (4.18), there are four exponential terms in total with µ n as the base.Therefore, it suffices to identify the largest exponent among the four exponential terms by solving the following optimization problem: {2θ + 5, 4θ + 4, 6θ + 3, 6} .
Our next goal is to identify the solutions to the characteristic equation (3.4) across all the regions of θ defined as aforementioned.These solutions have been summarized in Theorem 3.1.
The rest of this section is dedicated to prove Theorem 3.1.To this end, in Section 4.2.1, first, we will compute a detailed expression for the roots of the cubic resolvent j n,0 and j n,± through the use of Cardano's formula, Lemma 4.1, and Lemma 4.2.Upon knowing j n 's, we can compute λ n 's, through the use of (4.9) in Section 4.2.2.

Compute the roots of the cubic resolvent
In this section, we would like to compute the roots of the cubic resolvent (4.3).For this purpose, first, we compute the leading terms of Λ n and Ω n using Lemma 4.1.Then, we compute the lower order terms of Λ n .Upon having sufficient knowledge about the expression of Λ n and Ω n , we will compute the roots to the cubic resolvent via the Cardano's formula (4.8).
By Lemma 4.1, to compute the leading terms of Λ n and Ω n , we need to know about the leading terms P n , Q n , and √ ∆ n .We would also need to know the relative order of √ ∆ n and − Qn 2 .These quantities are given in Table 2.
With the quantities presented in Table 2, we can compute the leading terms of Λ n and Ω n .The results are summarized in Table 3.Note that in Table 3, For the convenience of applying the Cardano's formula, we also report m n in Table 3.Now that we know Λ n and Ω n , we would like to apply Cardano's formula in (4.8) to compute j n 's.We discuss the computation case by case according to the value of θ.We first discuss when θ ∈ [−1, 1  2 ).In this case, using (4.8) and Table 3, we notice that j n,0 = o(µ n ).This motivates us to use Lemma 4.2 to acquire lower order terms for Λ n .From (4.15), Table 2, and We then discuss the case when θ = 1 2 .By (4.8) and Table 3, we have that Finally, we discuss the case when θ ∈ ( 1 2 , 1].From (4.8) and Table 3, it is obvious that We need to compute lower order terms of j n 's.To this end, consider z n = j n − 4µ 2θ n .Then, z n is the solution to the cubic equation Examining the expression of z n 's in a similar fashion to that of j n 's, we have that As a result, We have finished the discussion of all three cases.To sum up, The roots to the cubic resolvent are

Compute the roots of the characteristic equation given the roots to the cubic resolvent
Now that we know the roots of the cubic resolvent as given in (4.20), we are ready to proceed to compute the roots of the characteristic equation (3.4).To this end, first, we compute k n 's.Then, we compute the λ n 's via (4.9).
From (4.20), the definition of k n 's, and using the fact that r n = −8µ 3θ n , we have that: With (4.9) and (4.21), the roots to the characteristic equation are given as in Theorem 3.1.

Some examples of application
In this section, we provide a few concrete examples of application of our regularity results.To this end, let Ω be a bounded domain in R N with smooth enough boundary Γ.These examples and the abstract system investigated in this contribution are inspired by Haraux' work [12] on the simultaneous stabilization of wave equations with different speeds of propagation.The work of Haraux followed from discussions with J.L. Lions about simultaneous control of uncoupled elastic systems [13].

(
iii) The semigroup T is differentiable if and only if for any positive real number b, there are real constants a and C with C > 0 depending on b such thatρ(A) ⊇ Σ b = {λ ∈ C; Reλ > a − b log |Imλ|},and||(λ − A) −1 || ≤ C|Imλ|, ∀λ ∈ Σ b with Reλ ≤ 0. Lemma 3.4.[19, Corollary 2.2] Let T = (T (t)) t≥0 be a strongly continuous and bounded semigroup on a Hilbert space X.Let A denote the infinitesimal generator A of the semigroup T .

Table 2 .
P n , Q n , and