On the decay rate for the wave equation with viscoelastic boundary damping
aa r X i v : . [ m a t h . A P ] S e p ON THE DECAY RATE FOR THE WAVE EQUATION WITHVISCOELASTIC BOUNDARY DAMPING
REINHARD STAHN
Abstract.
We consider the wave equation with a boundary condition of mem-ory type. Under natural conditions on the acoustic impedance ˆ k of the bound-ary one can define a corresponding semigroup of contractions [9]. With the helpof Tauberian theorems we establish energy decay rates via resolvent estimateson the generator −A of the semigroup. We reduce the problem of estimatingthe resolvent of −A to the problem of estimating the resolvent of the corre-sponding stationary problem. Under not too strict additional assumptions onˆ k we establish an upper bound on the resolvent. For the wave equation onthe interval or the disk or for certain acoustic impedances making 0 a spectralpoint of A we prove our estimates to be sharp. Introduction
Let Ω ⊂ R d be a bounded domain with Lipschitz boundary and k : R → [0 , ∞ )be an integrable function, depending on the time-variable only and vanishing on( −∞ , ( U tt ( t, x ) − ∆ U ( t, x ) = 0 ( t ∈ R , x ∈ Ω) ,∂ n U ( t, x ) + k ∗ U t ( t, x ) = 0 ( t ∈ R , x ∈ ∂ Ω) . The function U is called the velocity potential . One can derive the acoustic pressure p ( t, x ) = U t ( t, x ) and fluid velocity v ( t, x ) = −∇ U ( t, x ) from U . The second formulagives the velocity potential its name. The convolution is defined by the usualformula k ∗ U t ( t, x ) = R ∞ k ( r ) U t ( t − r, x ) dr . Here n is the outward normal vector of ∂ Ω, which exists almost everywhere for Lipschitz domains. Furthermore ∂ n denotesthe normal derivative on the boundary.We assume that k ∈ L (0 , ∞ ) is a completely monotonic function . That is, thereexists a positive Radon measure ν on [0 , ∞ ) such that k ( t ) = R [0 , ∞ ) e − τt dν ( τ ). Wenote here that the integrability assumption on k is easily checked to be equivalentto(2) ν ( { } ) = 0 and Z ∞ τ − dν ( τ ) < ∞ . Let e τ ( t ) = e − τt [0 , ∞ ) ( t ) and ψ ( t, τ, x ) = e τ ∗ U t ( t, x ) ( t ∈ R , τ ≥ , x ∈ ∂ Ω) . MSC2010: Primary 35B40, 35L05. Secondary 35P20, 47D06.Keywords and phrases: wave equation, viscoelastic, energy, resolvent estimates, singularity atzero, memory, C -semigroups. We use the convention to identify functions defined on the interval [0 , ∞ ) with functionsdefined on R but zero to the left of t = 0. Informally it is not difficult to see that (1) for t > p = U t and v = −∇ U at time t = 0 and R ∞ k ( t ) U t ( − t ) dt at theboundary (the “essential” data from the past) is equivalent to(3) p t ( t, x ) + div v ( t, x ) = 0 ( t > , x ∈ Ω) ,v t ( t, x ) + ∇ p ( t, x ) = 0 ( t > , x ∈ Ω) , [ ψ t + τ ψ − p ]( t, τ, x ) = 0 ( t > , τ > , x ∈ ∂ Ω) , (cid:2) − v · n + R ∞ ψ ( τ ) dν ( τ ) (cid:3) ( t, x ) = 0 ( t > , x ∈ ∂ Ω) , and the information of the initial state x = ( p , v , ψ ) of the system at time t = 0.It is important to observe that p and v cannot fully describe the system’s stateat t = 0 since there are memory effects at the boundary. The missing data fromthe past is stored in the auxiliary function ψ .Let us define the energy of the system to be the sum of potential, kinetic andboundary energy: E ( x ) = Z Ω | p ( x ) | + | v ( x ) | dx + Z ∞ Z ∂ Ω | ψ ( τ, x ) | dS ( x ) dν ( τ ) . Furthermore we introduce the homogeneous first order energy by E hom ( x ) = Z Ω |∇ p | + | div v | dx + Z ∞ Z ∂ Ω | τ ψ − p | dSdν ( τ ) . The first order energy is defined by E = E + E hom . Let us define the (zerothorder) energy space, and the first order energy space by H = H = L (Ω) × ∇ H (Ω) × L ν ((0 , ∞ ) τ ; L ( ∂ Ω)) , (4) H = { x ∈ H : E ( x ) < ∞ and (cid:20) − v · n | ∂ Ω + Z ∞ ψ ( τ ) dν ( τ ) (cid:21) = 0 } . (5)Here ∇ H (Ω) is the space of vector fields v ∈ ( L (Ω)) d for which there exists afunction (potential) U ∈ H (Ω) such that v = −∇ U . We note that the spaceof gradient fields ∇ H (Ω) is a closed subspace of ( L (Ω)) d since Ω satisfies thePoincar´e inequality . To make the boundary condition, appearing in the definitionof H , meaningful we use that the trace operator Γ : H (Ω) → H / ( ∂ Ω) , u u | ∂ Ω is continuous and has a continuous right inverse. Therefore we see that v · n | ∂ Ω is welldefined as an element of H − / ( ∂ Ω) = ( H / ( ∂ Ω)) ∗ for vector fields v ∈ ( L (Ω)) d with div v ∈ L (Ω) by the relation(6) h v · n, Γ u i H − × H ( ∂ Ω) = Z Ω div vu + Z Ω v · ∇ u for all u ∈ H (Ω). Also note that E ( x ) < ∞ implies ψ ∈ L ν since ψ ( τ ) = ψ ( τ )1+ τ + Γ p τ + τψ ( τ ) − Γ p τ and ( τ τ ) ∈ L ν ∩ L ν (0 , ∞ ) by (2). The quadraticforms E and E turn H and H into Hilbert spaces respectively.An initial state x is called classical if its first order energy is finite and theboundary condition is satisfied (i.e. x ∈ H ). We say that x ∈ C ([0 , ∞ ); H ) ∩ C ([0 , ∞ ); H ) is a (classical) solution of (3) if it satisfies the first two lines in thesense of distributions and the last two lines in the trace sense, i.e. with v · n Poincar´e inequality: If Ω is a bounded Lipschitz domain then there exists a
C > p ∈ H (Ω) with R Ω p = 0 we have R Ω | p | ≤ C R Ω |∇ p | . Here and in the following we abbreviate L pν ((0 , ∞ ) τ ; L ( ∂ Ω)) simply by L pν for p ∈ { , } . ECAY VIA VISCOELASTIC BOUNDARY DAMPING 3 defined by (6) and p replaced by Γ p . From Theorem 1 below plus basics from thetheory of C -semigroups it follows that the initial value problem corresponding to(3) is well-posed in the sense that for all classical initial data x ∈ H there is aunique solution x with x (0) = x and the mapping H j ∋ x x ∈ C ([0 , ∞ ); H j )is continuous for j ∈ { , } . For a solution x with x = x (0) we also write e.g. E ( t, x ) instead of E ( x ( t )). Note that E hom ( x ( t )) = E ( ˙ x ( t )) - this justifies theadjective “homogeneous” for the quadratic form E hom .Our aim is to find the optimal decay rate of the energy, uniformly with respectto classical initial states. This means that we want to find the smallest possibledecreasing function N : [0 , ∞ ) → [0 , ∞ ) such that E ( t, x ) ≤ N ( t ) E ( x )for all x ∈ H . Because of Theorem 29, 30 and 31 this is essentially equivalentto estimate the resolvent of the wave equation’s generator A (defined in Section 2below) along the imaginary axis near infinity and near zero.Our two main results are Theorem 4 and 9. The Sections 3 and 4 are devoted tothe proofs. We illustrate the application of our main results to energy decay by sev-eral examples in Section 5. Our first main result (Theorem 4) implies in particularthat the task of estimating the resolvent of the complicated 3 × A is equivalent to estimate the resolvent of the corresponding (and much simpler)stationary operator. Our second main result (Theorem 9) thus determines an upperresolvent estimate of A at infinity. Unfortunately we need additional assumptions on the acoustic impedance (see (21)). However in our separate treatment of theΩ = (0 , The semigroup approach
We reformulate (3) as an abstract Cauchy problem in a Hilbert space:(7) ( ˙ x ( t ) + A x ( t ) = 0 , x (0) = x ∈ H . Following the approach of [9] we define the energy/state space H as in (4) and write x = ( p, v, ψ ) for its elements (the states). Again let Γ : H (Ω) → H / ( ∂ Ω) , u u | ∂ Ω be the trace operator on Ω. By abuse of notation let τ denote the multiplicationoperator on L ν (0 , ∞ ) mapping ψ ( τ ) to τ ψ ( τ ). We define the wave operator by A = ∇ − Γ 0 τ with D ( A ) = H . Note that E ( x ) = k x k D ( A ) = k x k H + kA x k H for all x ∈ D ( A ). REINHARD STAHN
Theorem 1 ([9]) . The Cauchy problem (7) is well posed. More precisely −A isthe generator of a C -semigroup of contractions in H . Taking formal Laplace transform of the wave equation (1) yields(8) ( z u ( x ) − ∆ u ( x ) = f ( x ∈ Ω) ,∂ n u ( x ) + z ˆ k ( z ) u ( x ) = g ( x ∈ ∂ Ω) . Here z is a complex number and formally u = ˆ U ( z ) = R ∞ e − zt U ( t ) dt , f = zU (0) + U t (0) and g = ˆ k ( z ) U (0) | ∂ Ω . A way to give (8) a precise meaning is via the methodof forms. Thus for z ∈ C \ ( −∞ ,
0) let us define the bounded sesquilinear form a z : H × H (Ω) → C by a z ( p, u ) = z Z Ω pu + Z Ω ∇ p · ∇ u + z ˆ k ( z ) Z ∂ Ω Γ p Γ udS. If we replace the right-hand side f, g by F ∈ H (Ω) ∗ (dual space of H ), given by h F, η i = R Ω f η + R ∂ Ω g Γ ηdS , then a functional analytic realization of (8) is given by(9) ∀ η ∈ H (Ω) : a z ( u, η ) = h F, η i ( H ) ∗ ,H (Ω) . For all z ∈ C \ ( −∞ ,
0) for which (9) has for all F ∈ H (Ω) ∗ a unique solution u ∈ H (Ω) we define the stationary resolvent operator R ( z ) : H (Ω) ∗ → H (Ω) , F u . Theorem 2 ([9]) . The spectrum of the wave operator satisfies σ ( −A ) \ ( −∞ ,
0] = { z ∈ C \ ( −∞ ,
0] : R ( z ) does not exist. }⊆ { z ∈ C : ℜ z < } . Furthermore all spectral points in C \ ( −∞ , are eigenvalues. Following the proof of the preceding theorem given by [9] one sees that for s ∈ C \ i [0 , ∞ )(10) ( is + A )( p, v, ψ ) = ( q, w, ϕ ) ∈ H is equivalent to ∀ u ∈ H (Ω) : a is ( p, u ) = h F, u i ( H ) ∗ ,H (Ω) (11) and v = w + ∇ pis , ψ ( τ ) = Γ p + ϕ ( τ ) is + τ , where h F, u i = is Z Ω qu − Z Ω w · ∇ u − is Z ∂ Ω (cid:20)Z ∞ ϕ ( τ ) is + τ dν ( τ ) (cid:21) Γ u dS =: h F , u i + h F , u i + h F , u i . (12)Observe that the adjoint operator of R ( z ) is given by R ( z ) ∗ = R ( z ) for all z ∈ C \ ( −∞ ,
0) for which R ( z ) is defined. Finally mention: Theorem 3 ([9]) . The wave operator A is injective. In the next section we characterize all kernels k for which A is invertible. ECAY VIA VISCOELASTIC BOUNDARY DAMPING 5 A correspondence between ( is + A ) − and R ( is )In this section we prove our first main result. Theorem 4.
The following holds: (i)
Let M : (0 , ∞ ) → [1 , ∞ ) be an increasing function. Then (cid:2) ∃ s > ∀ | s | ≥ s : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) ≤ CM ( | s | ) (cid:3) ⇔ h ∃ s > ∀ | s | ≥ s : k R ( is ) k L → L ≤ C | s | − M ( | s | ) i . (ii) ∃ s > ∀ | s | ≤ s : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) ≤ C | s | − . (iii) A is invertible iff ( τ τ − ) ∈ L ∞ ν , i.e. ∃ ε > ν | (0 ,ε ) = 0 . If A is not invertible we deduce from Theorem 3 that A can not be surjective inthis case. In Section 3.4 we characterize the range of A .3.1. Singularity at ∞ . In this subsection we prove Theorem 4 (i). Therefore letus first define the auxiliary spaces X θ by the real interpolation method: X θ = L (Ω) resp. H (Ω) if θ = 0 resp. 1 , ( L (Ω) , H (Ω)) θ, if θ ∈ (0 , , ( X θ ) ∗ if θ ∈ [ − , . For θ ∈ (0 ,
1) the space X θ coincides with the Besov space B θ, (Ω).Let us explain why we use the Besov spaces X θ instead of the Bessel potentialspaces H θ (Ω). The reason is that while the trace operator Γ : H θ (Ω) → H θ − / ( ∂ Ω)is continuous for θ ∈ (1 / ,
1] this is no longer true for θ = 1 / H = L ). On the other hand Γ : X / → L ( ∂ Ω) is indeed continuous (seeProposition 26 in the appendix). A corollary of this fact is that for some
C > ∀ u ∈ H (Ω) : k Γ u k L ( ∂ Ω) ≤ C k u k L (Ω) k u k H (Ω) . Actually, by Lemma 28, the preceding trace inequality is equivalent to the continuityof the trace operator Γ : X / → L ( ∂ Ω).Let us prove the following extrapolation result.
Proposition 5.
Let M : (1 , ∞ ) → [1 , ∞ ) be an increasing function. If (14) k R ( is ) k X − a → X b = O ( | s | a + b − M ( | s | )) as | s | → ∞ is true for a = b = 0 , then it is also true for all a, b ∈ [0 , .Proof. Throughout the proof we may assume | s | to be sufficiently large. Assumethat (14) is true for a = b = 0. Let f ∈ L (Ω) and p = R ( is ) f , i.e. ∀ u ∈ H (Ω) : a is ( p, u ) = Z Ω f u. Because of (13) and the uniform boundedness of ˆ k ( is ) there are constants c, C > ℜ a is ( p, p ) ≥ c k p k H − Cs k p k L . This helps us to estimate c k p k H ≤ ℜ a is ( p, p ) + Cs k p k L ≤ k f k L k p k L + Cs k p k L ≤ s − k f k L + Cs k p k L ≤ CM ( | s | ) k f k L . REINHARD STAHN
In other words, (14) is true for a = 0 , b = 1. By duality (recall R ( z ) ∗ = R ( z )) itis also true for a = − , b = 0. Almost the same calculation as above but now withthe help of (14) for the now known case a = − , b = 0 shows that (14) is also truefor a = − , b = 1.It remains to interpolate. First interpolate between the parameters ( a = 0 , b = 1)and ( a = 1 , b = 1) to get (14) for a ∈ [0 , , b = 1. Then interpolate between theparameters ( a = 0 , b = 0) and ( a = 1 , b = 0) to get (14) for a ∈ [0 , , b = 0. Onelast interpolation gives us the desired result. (cid:3) Let us proceed with the proof of Theorem 4 part (i). The implication “ ⇒ ”follows immediately from the equivalence of (10) and (11) with w, ϕ = 0. Thereforewe have to show k x k H ≤ CM ( | s | ) k y k H , for all large | s | and for all x = ( p, v, ψ ) ∈ D ( A ) , y = ( q, w, ϕ ) ∈ H satisfying (10) where C does not depend on s and y .Let F j for j ∈ { , , } be defined by (12) and let p j satisfy ∀ u ∈ H (Ω) : a is ( p j , u ) = h F j , u i ( H ) ∗ ,H (Ω) . Case j = 1 . It is clear that k F k L = | s | k q k L . By Proposition 5 we have k p k X b = O ( | s | b M ( | s | )) k q k L for all b ∈ [0 , Case j = 2 . It is clear that k F k X − ≤ k w k L .By Proposition 5 we have k p k X b = O ( | s | b M ( | s | )) k w k L for all b ∈ [0 , Case j = 3 . By the continuity of the trace Γ : X / → L ( ∂ Ω), H¨older’s inequality and(2) we have k F k X − ≤ C | s | (cid:13)(cid:13)(cid:13)(cid:13)Z ∞ ϕ ( τ ) is + τ dν ( τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) ≤ C | s | k ϕ k L ν . Again by Proposition 5 this yields k p k X b = O ( | s | b M ( | s | )) k ϕ k L ν for all b ∈ [0 , k p k X b = O ( | s | b M ( | s | )) k y k H for all b ∈ [0 , k v k L ≤ C | s | − ( k w k L + k p k H ) ≤ CM ( | s | ) k y k H and k ψ k L ν ≤ | s | − k ϕ k L ν + k Γ p k L Z ∞ | is + τ | dν ( τ ) ! ≤ | s | − k ϕ k L ν + C | s | − k p k X ≤ CM ( | s | ) k y k H . This concludes the proof of Theorem 4 part (i).3.2.
Singularity at . Now we prove Theorem 4 (ii). For s = 0 we equip theSobolev space H (Ω) with the equivalent norm k u k H s := k u k L + (cid:13)(cid:13) s − ∇ u (cid:13)(cid:13) L . Inwhat follows we are interested in the asymptotics s → s = 0. As in thepreceding subsection we introduce some auxiliary spaces by the real interpolation ECAY VIA VISCOELASTIC BOUNDARY DAMPING 7 method X θs = L (Ω) resp. H s (Ω) if θ = 0 resp. 1 , ( L (Ω) , H s (Ω)) θ, if θ ∈ (0 , , ( X θs ) ∗ if θ ∈ [ − , . We prove an analog of Proposition 5 - but without the unknown function M . Proposition 6.
Let a, b ∈ [0 , and θ + = max { a + b − , } , then (15) k R ( is ) k X − as → X bs = O ( | s | − − θ + ) as s → . Before we can prove this proposition we show
Lemma 7.
There is a constant C (Ω) solely depending on the dimension and volumeof Ω such that for all u ∈ H (Ω) Z Ω |∇ u | + Z ∂ Ω | u | dS ≥ C (Ω) Z Ω | u | . Proof.
For the dimension d = 1 this is an easy exercise for the reader. For d ≥ v ∈ W , (Ω): Z Ω |∇ v | + Z ∂ Ω | v | dS ≥ d √ π Γ(1 + d ) d (cid:18)Z Ω | v | dd − (cid:19) d − d . The right-hand side can easily be estimated from below by a constant times the L (Ω)-norm of v since Ω is bounded. The conclusion now follows by plugging in v = u . (cid:3) Proof of Proposition 6.
Because of (13) and the continuity of R ∋ s ˆ k ( is ) at zerowe have for all u ∈ H (Ω) a is ( u, u ) = Z Ω |∇ u | + is ˆ k (0) Z ∂ Ω | u | dS + o (1) k∇ u k L + O ( s ) k u k L . Thus for sufficiently small | s | we deduce from Lemma 7 and the fact ˆ k (0) > p ∈ H (Ω) of the stationary wave equation (11) with F = f ∈ L (Ω)the following estimate holds: | s | k p k L ≤ C | a is ( p, p ) | = C |h f, p i|≤ C | s | − k f k L + | s | k p k L . This shows (15) in the case a = b = 0.Let us define the semi-linear functional G s ( u ) = − s Z Ω u + i ˆ k ( is ) Z ∂ Ω udS for u ∈ H (Ω). Observe that G s (1) → i ˆ k (0) | ∂ Ω | 6 = 0 as s tends to 0. It is easyto see from Poincar´e’s inequality (recall that Ω has Lipschitz boundary) that theexpression k∇ u k L + | G s ( u ) | defines a norm on H (Ω) which is equivalent to theusual one - uniformly for small | s | . In particular p
7→ k∇ p k L is an equivalent normon the kernel of G s . REINHARD STAHN
Remember that p is the solution of (11) for F = f ∈ L (Ω). We decompose p = p + p G with p G = G s ( p ) = const. ∈ L (Ω) and G s ( p ) = 0. Then a is ( p, p ) = a is ( p , p ) = (1 + O ( | s | )) Z Ω |∇ p | . This implies k∇ p k L ≤ C | a is ( p, p ) | ≤ C |h f, p i| ≤ C k f k L k∇ p k . This in combination with (15) for a = b = 0 implies k p k H s ≤ C | s | − k f k L whichis (15) for the parameters a = 0 , b = 1. By duality (recall R ( z ) ∗ = R ( z )) equation(15) is also true for a = 1 , b = 0. A similar calculation as above with f replaced by F ∈ H (Ω) ∗ and (15) for a = 1 , b = 0 shows (15) for a = 1 , b = 1.What remains to do is some interpolation. It is important to interpolate in theright order. First, one has to show k R ( is ) k X s → X b s , k R ( is ) k X a s → X s = O ( | s | − )for a , b ∈ [0 , a = 0 , b = 0) and( a = 0 , b = 1) for the first estimate and between ( a = 0 , b = 0) and ( a = 1 , b = 0)for the second estimate. Choosing a and b appropriately, the preceding estimatesimply (15) in the case a + b ≤
1. Interpolation between the preceding case and a = 1 , b = 1 yields the remaining part of the proposition. (cid:3) Let us proceed with the proof of Theorem 4 part (ii) in a similar fashion asfor part (i). We have to show k x k H ≤ C | s | − k y k H for all small | s | and for all x = ( p, v, ψ ) ∈ D ( A ) , y = ( q, w, ϕ ) ∈ H satisfying (10) where C does not depend on s and y . Let F j for j ∈ { , , } be defined by (12) and let p j satisfy ∀ u ∈ H (Ω) : a is ( p j , u ) = h F j , u i ( H ) ∗ ,H (Ω) Case j = 1 . It is clear that k F k L = | s | k q k L . By Proposition 6 we have k p k X bs = O (1) k q k L for all b ∈ [0 , Case j = 2 . It is clear that k F k X − s ≤ | s | k w k L . ByProposition 6 we have k p k X bs = O ( | s | − b ) k w k L for all b ∈ [0 , Case j = 3 . Bythe continuity of the trace Γ : X / → L ( ∂ Ω) and by H¨older’s inequality we havefor all | s | ≤ k F k X − s ≤ k F k X − ≤ C | s | (cid:13)(cid:13)(cid:13)(cid:13)Z ∞ ϕ ( τ ) is + τ dν ( τ ) (cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) ≤ C | s | k ϕ k L ν . By Proposition 6 this yields k p k X bs = O ( | s | − − ( b − ) + ) k ϕ k L ν for all b ∈ [0 , k p k X bs = O ( | s | − − ( b − ) + ) k y k H for all b ∈ [0 , k v k L ≤ C | s | − ( k w k L + k∇ p k L ) ≤ C | s | − k w k L + C k p k H s ≤ C | s | − k y k H ECAY VIA VISCOELASTIC BOUNDARY DAMPING 9 and because of k p k X ≤ k p k X s for | s | ≤ k ψ k L ν ≤ | s | − k ϕ k L ν + k Γ p k L Z ∞ | is + τ | dν ( τ ) ! ≤ | s | − k ϕ k L ν + C | s | − k p k X s ≤ C | s | − k y k H . This concludes the proof of Theorem 4 part (ii).3.3.
Spectrum at . Let us prove part (iii) of Theorem 4.“ ⇒ ”. Let us first assume that y = ( q, w, ϕ ) ∈ H and x = ( p, v, ψ ) ∈ D ( A )satisfies (10) for s = 0. There is a function u ∈ H (Ω) such that w = ∇ u . We mayassume R Ω u = 0 to make u unique. Then (10) for s = 0 is(16) div v ( x ) = q ( x ) ( x ∈ Ω) , ∇ p ( x ) = w ( x ) = ∇ u ( x ) ( x ∈ Ω) ,τ ψ ( τ, x ) − p ( x ) = ϕ ( τ, x ) ( τ > , x ∈ ∂ Ω) , − v · n ( x ) + R ∞ ψ ( τ, x ) dν ( τ ) = 0 ( x ∈ ∂ Ω) . From the second line we see that necessarily p = u + α for some complex number α . We have(17) ψ = ϕ + Γ u + ατ ∈ ( L ν ∩ L ν )(0 , ∞ ; L ( ∂ Ω)) . The L ν -inclusion follows by the definition of D ( A ) as explained in the paragraphfollowing (5). Let us now specialize to the situation q, w = 0 and k ϕ k L ν ≤
1. Then u = 0. By the existence of A − there must be a uniform bound | α | ≤ C wherethe constant does not depend on ϕ . Because of this, (17) and R ∞ τ − dν ( τ ) < ∞ we deduce a bound (cid:13)(cid:13) τ − ϕ (cid:13)(cid:13) L ν = k ψ k L ν + C ≤ C where C does not depend on ϕ . Since this is true for all ϕ ∈ L ν (0 , ∞ ; L ( ∂ Ω)) we deduce that the function(0 , ∞ ) ∋ τ τ − is in L ν (0 , ∞ ). If we use this in the L ν -inclusion in (17) we seethat (cid:13)(cid:13) τ − ϕ (cid:13)(cid:13) L ν = k ψ k L ν + C ≤ C where C does not depend on ϕ . Thus τ − is an L ν -multiplier and thus it must be bounded with respect to the measure ν .“ ⇐ ”. Assume now that ν | (0 ,ε ) = 0 for some ε >
0. Given y = ( q, w, ϕ ) ∈ H weshow that there is a unique solution x = ( p, v, ψ ) ∈ D ( A ) of (16). From the secondline of (16) we see that necessarily p = u + α for some complex number α and u asin the first part of the proof. The definition of H forces the necessity of the ansatz v = −∇ U for some function U ∈ H (Ω) with R Ω U = 0 for uniqueness purposes. Itremains to uniquely determine α and U since then ψ is uniquely given by (17). Let h = − R ∞ ψdν ∈ L ( ∂ Ω). Then the first and the last line of (16) are equivalent to ( − ∆ U ( x ) = q ( x ) ( x ∈ Ω) ,∂ n U ( x ) = h ( x ) ( x ∈ ∂ Ω) . By the Poincar´e inequality this equation has a solution U - which is unique underthe constraint R Ω U = 0 - if and only if0 = Z Ω q + Z ∂ Ω hdS = Z Ω q − Z ∂ Ω (cid:18) ˆ k (0)Γ u + Z ∞ ε ϕ ( τ ) τ dν ( τ ) (cid:19) dS − α | ∂ Ω | ˆ k (0) . (18)In the second equality we also used (17). Since ˆ k (0) = 0 this determines α and thusalso U uniquely. This completes the proof.3.4. The range of A . In the case that A is not invertible (i.e. ( τ τ − ) / ∈ L ∞ ν ) in spite of Theorem 31 it is important to know the image R ( A ) of A . Tocharacterize the range we have to distinguish two cases: (i) ( τ τ − ) ∈ L ν and(ii) ( τ τ − ) / ∈ L ν . In case (ii) for a given ϕ ∈ L ν (0 , ∞ ; L ( ∂ Ω)) there might existno p ∈ H (Ω) such that (cid:18) τ ϕ ( τ ) + Γ pτ (cid:19) ∈ L ν (0 , ∞ ; L ( ∂ Ω)) . In the case that p exists, its boundary value Γ p is uniquely determined and thefunction ( τ ϕ ( τ ) /τ ) is integrable with respect to ν . Therefore we can define thecomplex number(19) m ϕ,p = Z ∂ Ω Z ∞ ϕ ( τ ) + Γ pτ dν ( τ ) dS. Equipped with this notation we can now formulate:
Theorem 8.
Assume that A is not invertible (i.e. ( τ τ − ) / ∈ L ∞ ν ). (i) If ( τ τ − ) ∈ L ν , then R ( A ) = ( ( q, w, ϕ ) ∈ H ; Z ∞ (cid:13)(cid:13)(cid:13)(cid:13) ϕ ( τ ) τ (cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) dν ( τ ) < ∞ ) . (ii) If ( τ τ − ) / ∈ L ν , then R ( A ) = (cid:26) ( q, w, ϕ ) ∈ H ; ∃ p ∈ H (Ω) : w = ∇ p, Z Ω q = m ϕ,p and Z ∞ (cid:13)(cid:13)(cid:13)(cid:13) ϕ ( τ ) + Γ pτ (cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) dν ( τ ) < ∞ ) where m ϕ,p is given by (19). If ( q, w, ϕ ) is in the image of A then p is unique. Infact it is the first component of the pre-image of ( q, w, ϕ ) .Proof. Let y = ( q, w, ϕ ) ∈ H . Clearly y ∈ R ( A ) if and only if we can find x =( p, v, ψ ) ∈ H such that A x = y . Let u ∈ H (Ω) be such that ∇ u = w and R Ω u = 0. As in the proof of Theorem 4(iii) we see that necessarily p = u + α forsome complex number α and(20) ϕ + Γ pτ = ψ ∈ L ν (0 , ∞ ; L ( ∂ Ω)) . Let us assume that case (i) is valid. Then the so defined ψ is in L ν if and onlyif ( τ ϕ ( τ ) /τ ) is square integrable with respect to ν . Now one can proceed as in ECAY VIA VISCOELASTIC BOUNDARY DAMPING 11 the “ ⇐ ”-part of the proof of Theorem 4(iii) to find the unique p and v such that A x = y .Let us now assume that case (ii) is valid. By (20) it is clear that the existenceof p as in the definition of R ( A ) is necessary. From the fact that ( τ τ − ) is notsquare integrable we see that Γ p is uniquely defined. Now we can again proceedas in the “ ⇐ ”-part of the proof of Theorem 4(iii) to find the unique p and v suchthat A x = y . The condition R Ω q = m ϕ,p on y comes from (18), where we have toreplace ˆ k (0)Γ u + R ∞ ε ϕ ( τ ) τ dν ( τ ) + α ˆ k (0) by R ∞ ϕ ( τ )+Γ pτ dν ( τ ) in our situation. (cid:3) An upper estimate for (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) if s → ∞ We are seeking for an increasing function M : [1 , ∞ ) → [1 , ∞ ) such that for someconstant C > (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) ≤ CM ( | s | ) ( | s | ≥ . In this section we want to show that the function M ( s ) = 1 / ℜ ˆ k ( is ) is an upperbound (up to a constant) for the norm of ( is + A ) − when | s | is large and if someadditional assumptions on the acoustic impedance ˆ k and the domain are satisfied.More precisely we assume that the acoustic impedance satisfies (cid:12)(cid:12)(cid:12) ˆ k (cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12) ˆ k (cid:12)(cid:12)(cid:12) ( ℜ ˆ k ) ( is ) = o (cid:18) L ( s ) (cid:19) as s → ∞ , (21) where L ( s ) = s α (1 + log( s )) for s ≥ . The real number α ∈ [0 ,
1) is a domain dependent constant which will be definedbelow. Note that for α ≥ u j ) be the sequence of normalized eigenfunctions of the Neumann Laplacianwith respect to the corresponding (non-negative) frequencies ( λ j ). That is(22) λ j u j ( x ) + ∆ u ( x ) = 0 ( x ∈ Ω) ,∂ n u j ( x ) = 0 ( x ∈ ∂ Ω) , k u j k L (Ω) = 1 . The eigenfrequencies are counted with multiplicity and we may order them so that0 ≤ λ ≤ λ ≤ . . . . We call a function p ∈ L (Ω) a spectral cluster of width δ > {| λ j − λ i | ; a j , a i = 0 } ≤ δ where p = P a j u j is the expan-sion of p into eigenfunctions. We define the (mean) frequency λ ( p ) ≥ p by λ ( p ) = P | ( a j / k p k L ) | λ j . We assume that the domain has the property thatfor sufficiently small δ > c, C > p of width δ the following estimate is true c k p k L (Ω) ≤ Z ∂ Ω | Γ p | dS ≤ Cλ ( p ) α k p k L (Ω) . (23)We call the left inequality the lower estimate and the right inequality the upperestimate . Note that the upper estimate is trivially satisfied for α = 1 by applyingthe trace inequality from Lemma 27. It is indeed reasonable to assume that thisestimate holds for some α strictly smaller than 1. For example if the boundary of ∂ Ω is of class C ∞ then both estimates hold with α = 2 /
3. See [3] for this result.For Ω being an interval one can choose α = 0 and for a square α = 1 / Theorem 9.
Assume that (21) is satisfied, where α ∈ [0 , is such that (23) holdsfor all spectral cluster p of sufficiently small width δ > . Then there is a constant C > such that k R ( s ) k L → L ≤ Cs ℜ ˆ k ( is ) for all s ≥ . Compare this result to Theorem 4 to obtain that the norm of (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) isbounded by C ℜ ˆ k ( is ) under the constraints of the preceding theorem.4.1. Some auxiliary definitions.
We fix a δ > δ . For p, q ∈ H (Ω) we define the Neumann form by a Nz ( p, q ) = z Z Ω pq + Z Ω ∇ p · ∇ q. We cover [0 , ∞ ) by disjoint intervals I k = [ kλ, ( k + 1) λ ) for k = 0 , , , . . . such that(i) λ ∈ [2 δ, δ ],(ii) ∃ k c ∈ N : I k c ⊃ ( s − δ, s + δ ).The covering depends on s ≥ p ∈ L (Ω)in terms of spectral clusters in the following way: p = ∞ X k =0 c k p k where p k = X λ j ∈ I k a j u j , k p k k L (Ω) = 1 . Let s k ( p ) ∈ I k be such that s k ( p ) = Z Ω |∇ p k | . Let p − ) = P k> ( < ) k c c k p k and p = p − + p . Let p c = c k c p k c . Obviously p = p + p c . Define p + = ( p + p c if a Nis ( p c ) ≥ ,p else , and let p − be given by p = p + + p − . Finally let ˜ p = p + − p − .4.2. Some auxiliary lemmas.
For the remaining part of Section 4 we use thenotation introduced in Subsection 4.1 and we assume that | s | ≥ Lemma 10.
For all p ∈ H (Ω) we have a Nis ( p, ˜ p ) ≥ | s | δ (cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) .Proof. a Nis ( p, ˜ p ) ≥ a Nis ( p , (˜ p ) ) = X k = k c (cid:12)(cid:12) s − s k (cid:12)(cid:12) | c k | ≥ sδ X k = k c | c k | = sδ (cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) . (cid:3) A little bit more involved is the proof of the next lemma.
ECAY VIA VISCOELASTIC BOUNDARY DAMPING 13
Lemma 11.
There is a constant
C > (depending on δ and α ) such that for all p ∈ H (Ω) Z ∂ Ω (cid:12)(cid:12) Γ p (cid:12)(cid:12) dS ≤ C | s | α (1 + log( | s | )) a Nis ( p, ˜ p ) | s | Proof.
Since a Nis ( p, ˜ p ) ≥ a Nis ( p , (˜ p ) ) we may assume that p c = 0. Because of Z ∂ Ω | Γ p | dS ≤ Z ∂ Ω | Γ p − | + | Γ p + | dS and a Nis ( p, ˜ p ) = a Nis ( p + ) − a Nis ( p − )we may assume without loss of generality that either p = p + or p = p − . We showthe proof in detail for the case p = p + . The case p = p − is analogous and thereforewe omit it. Z ∂ Ω | Γ p + | dS = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X k>k c c k Γ p k (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) ≤ X k>k c | c k | k Γ p k k L ( ∂ Ω) ! ≤ Cδ α X k>k c | c k | k α ! ≤ Cδ α X k>k c | c k | ( s k − s ) !| {z } a Nis ( p + ) X k>k c k α s k − s !| {z } =: J In the first line we used the continuity of the trace operator Γ : H (Ω) → L ( ∂ Ω).From the second to the third line we used the upper estimate (23) together with s k ∈ I k = λ [ k, k + 1) with λ ∈ [2 δ, δ ]. It remains to estimate J . It is a well known trickto estimate sums of positive and decreasing summands by corresponding integrals. J = X k>k c k α s k − s ≤ X k>k c k α λ k − s ≤ ( k c + 1) α λ ( k c + 1) − s + Z ∞ k c +1 x α λ x − s dx =: J + J . It is not difficult to see that J can be estimated by a constant times δ − − α s α − .For J we substitute y = λx/s and use that λ ( k c + 1) ≥ δ . This yields J ≤ Cδ − − α s α − Z ∞ δs y α y − dy ≤ Cδ − − α s α − Z δs y − dy + Z ∞ y − α dy ! ≤ Cδ − − α s α − (log( sδ ) + 1) . This concludes the proof. (cid:3)
Proof of Theorem 9.
Let p ∈ H (Ω) and | s | ≥
1. We have to verify thatsup {| a is ( p, u ) | ; u ∈ H (Ω) , k u k L (Ω) ≤ } ≥ c | s | ℜ ˆ k ( is ) k p k L (Ω) is true for some constant c > p and s . In the following we assumethat a Nis ( p c ) ≥
0. This implies that p + = ( p ) + + p c and p − = ( p ) − . The case a Nis ( p c ) < Z ∂ Ω | Γ p + | + | Γ p − | dS = Z ∂ Ω (cid:12)(cid:12) Γ p (cid:12)(cid:12) + (cid:12)(cid:12) Γ p − (cid:12)(cid:12) + | Γ p c | + 2 ℜ (Γ p Γ p c ) dS ≤ Z ∂ Ω (cid:12)(cid:12) Γ p (cid:12)(cid:12) + (cid:12)(cid:12) Γ p − (cid:12)(cid:12) + 2 | Γ p c | dS ≤ CL ( s ) a Nis ( p, ˜ p ) | s | + 2 Z ∂ Ω | Γ p c | dS. (24)Let us define L ( s ) = (cid:12)(cid:12)(cid:12) ˆ k ( is ) (cid:12)(cid:12)(cid:12) ℜ ˆ k ( is ) L ( s ) ≥ L ( s ) . Our assumption (21) on k is equivalent to | ˆ k | ( is ) = o (1 /L ( s )) as | s | → ∞ . Nowwe come to the final part of the proof which consists of distinguishing two cases.Essentially the first case means that p is roughly the same as p and the secondcase means that p is roughly the same as p c . We fix a constant ε ∈ (0 ,
1) to bechosen later. The choice of ε does not depend on s . Case 1: L ( s ) a Nis ( p, ˜ p ) ≥ ε | s | R ∂ Ω | Γ p c | dS . We first show that in this case theNeumann form dominates the form a is for | s | big enough in the following sense: (cid:12)(cid:12) a is ( p, ˜ p ) − a Nis ( p, ˜ p ) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) s ˆ k ( is ) Z ∂ Ω (Γ p + + Γ p − )(Γ p + − Γ p − ) dS (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12) s ˆ k ( is ) (cid:12)(cid:12)(cid:12) Z ∂ Ω | Γ p + | + | Γ p − | dS ≤ C (cid:12)(cid:12)(cid:12) s ˆ k ( is ) (cid:12)(cid:12)(cid:12) ε − L ( s ) a Nis ( p, ˜ p ) | s |≤ a Nis ( p, ˜ p ) . From the second to the third line we used the assumption of case 1 and (24). By(21) the last line is valid for all s ≥ s , where s is sufficiently large depending onhow small ε is. Therefore we have | a is ( p, ˜ p ) | ≥ ( 14 + 14 ) a Nis ( p, ˜ p ) ≥ sδ (cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) + ε | s | L ( s ) Z ∂ Ω | Γ p c | dS ≥ cε | s | L ( s ) (cid:16)(cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) + k p c k L (Ω) (cid:17) ≥ cε | s | ℜ ˆ k ( is ) k p k L (Ω) . ECAY VIA VISCOELASTIC BOUNDARY DAMPING 15
From the second to the third line we used the lower estimate (23) and in the laststep we used our assumptions on the acoustic impedance (21). The theorem isproved for case 1.
Case 2: L ( s ) a Nis ( p, ˜ p ) < ε | s | R ∂ Ω | Γ p c | dS . By Lemma 10 and lim | s |→∞ L ( s ) = ∞ this yields Z ∂ Ω | Γ p c | dS ≥ (cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) (25)for all | s | ≥ s with an s > ε . We show now that in case 2the form a is is dominated by the contribution from the boundary. By Lemma 11we have (cid:12)(cid:12)(cid:12)(cid:12)Z ∂ Ω p p c dS (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) CL ( s ) a Nis ( p, ˜ p ) | s | (cid:19) (cid:18)Z ∂ Ω | p c | dS (cid:19) ≤ C √ ε (cid:18) L ( s ) L ( s ) (cid:19) Z ∂ Ω | p c | dS ≤ ℜ ˆ k ( is )2 (cid:12)(cid:12)(cid:12) ˆ k ( is ) (cid:12)(cid:12)(cid:12) Z ∂ Ω | p c | dS. In the last step we choose ε so small that C √ ε ≤ /
2. Finally from this, (25)and the lower estimate (23) we deduce that ℑ a is ( p, p c ) ≥ | s | ℜ ˆ k ( is ) Z ∂ Ω | p c | dS ≥ c | s | ℜ ˆ k ( is )( k p c k L (Ω) + (cid:13)(cid:13) p (cid:13)(cid:13) L (Ω) )= c | s | ℜ ˆ k ( is ) k p k L (Ω) which yields the claimed result. 5. Examples
To illustrate our main results, Theorem 4 and Theorem 9, we want to considerspecial standard kernels k = k β,ε (with ε > < β <
1) introduced below.These standard kernels have the property that ℜ ˆ k ( is ) ≈ | ˆ k ( is ) | ≈ | s | β − for large | s | . This makes it easy to check whether (21) is satisfied or not. We take a closerlook at Ω being a square or a disk. In the case of the disk we show the optimality ofthe resolvent estimate, that is we show that (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) is not only bounded fromabove by a constant times 1 / ℜ ˆ k ( is ) but also from below. The standard kernels aredesigned in such a way that A is invertible (i.e. ( τ τ − ) ∈ L ν ; see Theorem 4).We have assumed this for the simplicity of exposition. However, in Subsection 5.5we briefly show that our results yield (optimal) decay rates also in the presence ofa singularity at zero.The case Ω = (0 ,
1) is treated separately in Section 6.5.1.
Properties of the standard kernels.
For ε > < β < k β,ε ( t ) = e − εt t − (1 − β ) for t > . To keep the notation short we fix ε and β now and write k instead of k β,ε throughoutthis section. Obviously k ∈ L (0 , ∞ ) and for all n ∈ N we have ( − n d n k/dt n ( t ) >
0. The last property is a characterization of completely monotonic functions. Thusthe kernel k is admissible in the sense that the semigroup from Section 2 is defined.Let Γ denote the Gamma function. Taking Laplace transform yields for z > − ε ˆ k ( z ) = Z ∞ e − ( ε + z ) t t − (1 − β ) dt = 1( ε + z ) β Z ∞ s − (1 − β ) e − s ds = Γ( β )( ε + z ) β . By analyticity the equality between the left end and the right end of this chain ofequations extends to C \ ( −∞ , − ε ].For s ∈ R , let ϕ ( s ) ∈ ( − π , π ) be the argument of ε − is . Note that ϕ ( s ) → ∓ π as s → ±∞ . Then we haveˆ k ( is ) = Γ( β ) (cid:12)(cid:12)(cid:12)(cid:12) ε − isε + s (cid:12)(cid:12)(cid:12)(cid:12) β (cos( βϕ ( s )) + i sin( βϕ ( s ))) . In particular ℜ ˆ k ( is ) ≈ (cid:12)(cid:12)(cid:12) ℑ ˆ k ( is ) (cid:12)(cid:12)(cid:12) ≈ | s | β for | s | ≥ . Here by ≈ we mean that the left-hand side is up to a constant, which does notdepend on s , an upper bound for the right-hand side and vice versa. The first ≈ -relation implies that the condition (21) is equivalent to the simpler estimate ℜ ˆ k ( is ) = o (1 /L ( s )) as | s | tends to infinity. More precisely we have(26) (21) ⇔ β > α. It is well known that for z > β ∈ (0 , z − β = sin( πβ ) π Z ∞ τ + z dττ β . Thus ˆ k ( z ) = sin( βπ ) π Γ( β ) Z ∞ ε τ + z dτ ( τ − ε ) β . In the notation of Section 1 this means dν ( τ ) = sin( βπ ) π Γ( β ) · [ ε, ∞ ) ( τ − ε ) β dτ. By Theorem 4 (iii) we see that A is invertible.5.2. Smooth domains.
Let us suppose that Ω has a C ∞ boundary and let k = k β,ε for some ε > < β <
1. By [3] we know that (23) is satisfied for α = 2 / β >
23 = ⇒ ∀ s ∈ R : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) ≤ C (1 + | s | ) β . By Theorem 30 this implies
Proposition 12.
Let ∂ Ω be of class C ∞ and k = k β,ε . If β > / then, for all t > and x ∈ H , E ( t, x ) ≤ Ct − β E ( x ) . ECAY VIA VISCOELASTIC BOUNDARY DAMPING 17
The disk.
Let Ω = D be the unit disk in R . The smallest possible choice of α in (23) is indeed 2 /
3. The simple proof is based on a
Rellich-type identity , see forinstance [3, page 5]. So the circle already realizes the “worsed case scenario” withrespect to the upper bounds for Neumann eigenfunctions. Thus in Proposition 12we cannot replace the condition β > / A . Lemma 13.
Let
Ω = D and k = k β,ε . Then there exists a sequence ( z n ) in thespectrum of −A such that ( ℑ z n ) is positive and increasing and such that there existsa constant C > such that < −ℜ z n ≤ C ( ℑ z n ) β holds for all n . As a corollary we have ∀ s > | σ |≤ s (cid:13)(cid:13) ( iσ + A ) − (cid:13)(cid:13) ≥ C (1 + s ) β . By Theorem 30 and the remark after Theorem 29 this implies
Proposition 14.
Let
Ω = D and k = k β,ε . If β > / then we have for all t ≥ that ct − β ≤ sup E ( x ) ≤ E ( t, x ) ≤ Ct − β . If β is arbitrary the left inequality remains valid.Proof of Lemma 13. Except for the rate of convergence of ( z n ) towards the imagi-nary axis the content of our lemma is included in [10, Theorem 5.2]. Therefore weonly sketch the existence of a sequence ( z n ) with imaginary part tending to infinityand real part tending to zero.First recall that an eigenvalue is a complex number z n such that (9) with F = 0and z = z n has a non-zero solution u . After a transformation to polar coordinates,by a separation of variables argument one can show that the existence of u isequivalent to the existence of a non-zero solution v of v ′′ ( r ) + r v ′ ( r ) − ( l r + z ) v ( r ) = 0 (0 < r < ,v ′ (1) + z ˆ k ( z ) v (1) = 0 ,v (0+) is finite , for some l ∈ N . The first and the third line forces that v ( r ) is proportional to J l ( izr ), where J l is the l -th order Bessel function of the first kind (see e.g. [1,Chapter 9]). Therefore the second line implies(28) J ′ l ( iz ) J l ( iz ) = i ˆ k ( z ) . We have seen that a complex number z n / ∈ ( −∞ ,
0] is an eigenvalue of the waveoperator if and only if it is a zero of (28) for some l . Let us fix l now. Following theapproach of [10] one can prove the existence of a sequence of zeros ( z n ) = ( is n − ξ n )with s n = nπ + (2 l + 1) π/ ℜ ξ n > ξ n tending to zero, by a Rouch´e argument. It remains to prove that ξ n = O (( ℑ z n ) − (1 − β ) ). By [1, Formula 9.2.1] the follow-ing asymptotic formula holds if z tends to infinity while ℜ z stays bounded (and l is fixed):(29) J l ( iz ) = r πz cos (cid:18) iz − (2 l + 1)4 π (cid:19) + O ( | z | − ) . A naive way to get the corresponding asympotic formula for J ′ and J ′′ wouldbe to take derivatives of the cosine. In fact this yields the correct leading term.The error term is again O ( | z | − ) in both cases. For the first derivative this is[1, Formula 9.2.11]. The formula for the second derivative then follows from theordinary differential equation satisfied by J l .Thus by a Taylor expansion of (28) we get:0 + iξ n + O ( | ξ n | + n − ) = i ˆ k ( is n ) − iξ n ˆ k ′ ( is n ) + O ( | ξ n | + n − ) . This implies ξ n = (1 + o (1))ˆ k ( is n )(30) = (1 + o (1)) Γ( β ) s βn (cos( βϕ ( s n )) + i sin( βϕ ( s n ))) . Here ϕ ( s ) is the argument of ε − is (see Section 5.1). (cid:3) Note that in the undamped case k = 0 we have z n = s n + O ( s − n ) by [1, Formula9.5.12] for the eigenvalues z n . Here again s n = nπ + (2 l + 1) π/ l is fixed. Thus(30) implies that z n = z n − (1 + o (1))ˆ k ( is n ).5.4. The square.
Let Ω = Q = (0 , π ) be a square. In terms of upper boundsfor boundary values of spectral clusters the square behaves slightly better than thedisk. It seems to be reasonable to believe that this is due to the fact that the squarehas no whispering gallery modes . Lemma 15.
Let
Ω = Q , k = k β,ε and δ > . If δ is sufficiently small then for each L ( Q ) -normalized spectral cluster p of width δ of the Neumann-Laplace operator c ≤ Z ∂ Ω | Γ p | dS ≤ Cs ( p ) . The constants c, C > do not depend on p . Furthermore the exponent α ( Q ) = 1 / is optimal, i.e. one cannot replace it by a smaller one. The optimality assertion of Lemma 15 may be somewhat surprising. If p was re-stricted to be a (pure) eigenfunction of the Neumann-Laplace operator the optimalexponent would be α = 0. This is a direct consequence of the explicit formula avail-able for the eigenfunctions. However, it will be clear from the proof why spectralclusters behave differently.As in the preceding examples the lemma implies Proposition 16.
Let
Ω = Q , k = k β,ε . If β > / then, for all t > and x ∈ H , E ( t, x ) ≤ Ct − β E ( x ) . Proof of Lemma 15.
The explicit form of the normalized Neumann eigenfunctions u m,n and its eigenfrequencies λ m,n ≥ u m,n ( x, y ) = 2 cos( mx ) cos( ny ) , λ m,n = m + n . ECAY VIA VISCOELASTIC BOUNDARY DAMPING 19
Let p = P m,n a n,m u n,m be a normalized spectral cluster of width δ . We choose s ≥ m, n ) with a m,n = 0 is included in I which isgiven by I = { ( m, n ) ∈ N ; s ≤ m + n ≤ ( s + δ ) } ,I = { ( m, n ) ∈ I ; m ≥ n } . Without loss of generality we may assume that P I | a m,n | ≥ /
2. We first provethe lower bound: Z ∂ Ω | Γ p | dS = X n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X m a m,n Γ u m,n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) ≥ π X I | a m,n | ≥ π k p k L (Ω) . In the first line we use the orthogonality relation for the cosine functions withrespect to the y variable. In the second line we use k u m,n k L ( ∂ Ω) = 4 √ π and thefact that the partial sum over m in the preceding step includes only one member if δ is small and if the index set is restriced to I .Let N n be the number of non-zero summands with respect to the inner sum inline one. It is not difficult to see that N n ≤ C √ s for a constant independent of n and s . Therefore we have Z ∂ Ω | Γ p | dS = X n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)X m a m,n Γ u m,n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) = X n N n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) N n X m a m,n Γ u m,n (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( ∂ Ω) ≤ C X m,n N n | a m,n | ≤ Cs k p k L (Ω) . It remains to prove optimality of the exponent α = 1 /
2. For n ∈ N we consider aspecial spectral cluster p of the form p = 2 N − X m =0 a m cos( mx ) cos( n y )where N = N ( n ) = ⌈ ε √ n ⌉ . If ε > n large enough we see that p is a spectral clusterof width δ . If we set a m = 1 / √ N we see that the L (Ω)-norm of p is 1 and Z { }× (0 , | Γ p | dS = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) N X m =1 a m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = N ( n ) ≥ ε √ n . This finishes the proof since s ( p ) ∈ [ n , n + δ ]. (cid:3) Decay in the presence of a singularity at zero.
So far in this sectionwe have excluded the case when A has a singularity at zero. The purpose of thissubsection is to show that getting decay rates in this case is not more difficult thanin the case where there is no singularity at zero. As in the previous subsectionwe simplify our presentation by considering a special family (ˆ k ′ α,β ) α,β of acousticimpedances given by the measures dν ′ α,β = τ α dτ | (0 , + ( τ − − β dτ | (1 , ∞ ) ( α ∈ (0 , ∞ ) , β ∈ (0 , . Obviously ( τ τ − ) is integrable with respect to ν ′ α,β (thus k ′ α,β is integrable)but it is not bounded with respect to that measure. Observe that α > τ τ − ) is square integrable with respect to ν ′ . In the following we assumefor simplicity that α >
1. The reason is that by Theorem 8 the range of A has asimpler representation in this case. Lemma 17.
Let α ∈ (1 , ∞ ) , β ∈ (0 , . Then ( τ τ − ) is integrable, squareintegrable but unbounded with respect to ν ′ α,β . Moreover ˆ k ′ α,β ( z ) = π sin( βπ ) (1 + z ) − β + O ( | z | − ) as z tends to infinity avoiding R − .Proof. We only have to prove the last statement. We calculateˆ k ′ ( z ) = Z τ α z + τ dτ + Z ∞ z + τ dτ ( τ − β =: I + II.
It is easy to see that the modulus of I is bounded by ( | z | − − for all z with | z | >
1. With regard to II we see that the well known identity z − β = sin( βπ ) π Z ∞ z + τ dττ β finishes the proof. (cid:3) Proposition 18.
Let α ∈ (0 , ∞ ) , β ∈ (2 / , and k = k ′ α,β . Let ∂ Ω be a C ∞ -manifold. Then (cid:13)(cid:13) ( is − A ) − (cid:13)(cid:13) = O ( | s | β ) as | s | > tends to infinity.Proof. This is an immediate consequence of Lemma 17 together with Theorem 4(i)and Theorem 9. (cid:3)
We are now in the position to prove an optimal decay estimate.
Proposition 19.
Let α ∈ (1 , ∞ ) , β ∈ (2 / , and k = k α,β . Let ∂ Ω be a C ∞ -manifold. Then E ( t, x ) ≤ C t (cid:20) E ( x ) + Z ∞ k ψ ( τ ) k L ( ∂ Ω) dν ( τ ) τ (cid:21) holds for all t ≥ and for all x = ( p , v , ψ ) ∈ H for which the right-hand side isfinite. The constant C > does not depend on x or t . Moreover this estimate issharp in the sense that it would be invalid if one replaces C/ (1 + t ) by o (1 / (1 + t )) as t tends to infinity. ECAY VIA VISCOELASTIC BOUNDARY DAMPING 21
Proof.
Proposition 18, Theorem 4(ii) together with [6, Theorem 8.4] yield (cid:13)(cid:13) e t A x (cid:13)(cid:13) ≤ C t k x k D ( A ) ∩ R ( A ) for all x ∈ D ( A ) ∩ R ( A ) . We know that the norm of D ( A ) is (equivalent to) the square root of the first orderenergy E . By Theorem 8 the norm on R ( A ) is given by k x k R ( A ) = E ( x ) + Z ∞ k ψ ( τ ) k L ( ∂ Ω) dν ( τ ) τ . This gives the desired estimate. The sharpness of this estimate follows from [6,Theorem 6.9 and the remarks in Chapter 8]. (cid:3) Optimal decay rates for the 1D case
Throughout this section Ω = (0 ,
1) and k is a completely monotonic, integrablefunction. We aim to show that in the 1D setting the conclusion of Theorem 9remains true without any further hypothesis - like (21) - on the acoustic impedance.Even more can be done - we prove that the upper estimate is optimal. Moreprecisely we prove Theorem 20.
Let
Ω = (0 , . Then there are constants c, C > such that for all s > we have c ℜ ˆ k ( is ) ≤ sup ≤| σ |≤ s (cid:13)(cid:13) ( iσ + A ) − (cid:13)(cid:13) ≤ C ℜ ˆ k ( is ) . We prove the lower bound by investigating the spectrum of −A which is close tothe imaginary axis (Subsection 6.1). Furthermore we give a more or less concreteformula for the stationary resolvent operator R ( is ) which allows to prove the upperbound (Subsection 6.2). Subsection 6.3 contains implications of Theorem 20 forthe decay rates of the energy of the wave equation.6.1. The spectrum.
The spectrum of −A satisfies a characteristic equation whichis implicitly contained in [10]. For convenience of the reader we give a completeproof. Proposition 21.
A number z ∈ C \ ( −∞ , is in the spectrum of −A , and hencean eigenvalue, if and only if it satisfies (31) (cid:18) ˆ k ( z ) − i tan (cid:18) iz (cid:19)(cid:19) · (cid:18) ˆ k ( z ) + i cot (cid:18) iz (cid:19)(cid:19) = 0 Proof.
By Theorem 2 together with the equivalence between (10) and (11) we seethat z is a spectral point if and only if there is a non-zero function p solving z p ( x ) − p ′′ ( x ) = 0 ( x ∈ (0 , , − p ′ (0) + z ˆ k ( z ) p (0) = 0 ,p ′ (1) + z ˆ k ( z ) p (1) = 0 . Up to a scalar factor the first two lines are equivalent to the following ansatz p ( x ) = cos( izx ) − i ˆ k ( z ) sin( izx ) . Plugging this into the third line yields that z is an eigenvalue if and only if(32) (cid:16) ˆ k ( z ) + 2 i ˆ k ( z ) cot( iz ) + 1 (cid:17) z sin( iz ) = 0 . Note that the zeros of the sine function do not lead to an eigenvalue since thecotangent function has a singularity at the same point. Actually we already knowfrom the situation of general domains that an eigenvalue which is neither zero nora negative number must have negative real-part. Thus we may simplify (32) bydividing by z sin( iz ). The claim now follows from the formula cot( ζ ) − tan( ζ ) =2 cot(2 ζ ) which is valid for all complex numbers ζ . (cid:3) Let
H, R >
0. The reader may consider H and R as large numbers. We areinterested in the part of the spectrum of −A contained in the strip U RH = { z ∈ C ; |ℑ z | > R and 0 < −ℜ z < H } . Proposition 22.
Let
H > . Then for R > large enough there exists a naturalnumber n > such that the part of the spectrum of −A which is contained in U RH is given by a doubly infinite sequence ( z n ) ∞ n = ± n with z − n = z n for all n and ℑ z n = πn − h (2 + O ( | ˆ k | )) ℑ ˆ k i ( iπn ) , ℜ z n = − h (2 + O ( | ˆ k | )) ℜ ˆ k i ( iπn ) . As a consequence the lower bound in Theorem 20 is proved.Note that the two asymptotic formulas given by the proposition imply z n =(2 + o (1))ˆ k ( inπ ) for n tending to plus or minus infinity. This formula can be provedby the same Taylor expansion argument as in the proof of Lemma 13. See also theremark after the proof of the mentioned lemma. But this is not enough in orderto prove the lower bound in Theorem 20 since it might happen that the real partof ˆ k ( is ) tends much faster to zero then its imaginary part! This explains the moreelaborate Taylor expansion technique in the proof below. Proof of Proposition 22.
We are searching for the solutions z ∈ U RH of the charac-teristic equation (31). For simplicity we only consider the solutions of z ∈ U RH and F ( z ) := ˆ k ( z ) − i tan (cid:18) iz (cid:19) = 0 . We apply a Rouch´e argument to show that the zeros of this equation are close tothe zeros is n = 2 nπi of the tangens-type function on the right-hand side. Let ( ε n )be a null-sequence of positive real numbers, smaller than H , to be fixed later. Let B n be the open ball of radius ε n around the center is n . For r > V RH ( r ) = { z ∈ C ; R < ℑ z < R + r and − H < ℜ z < H } . Take K ( r ) to be the boundary of the set V RH ( r ) \ ( S n B n ). Since ˆ k ( z ) tends to zeroas z tends to infinity with bounded real part we can choose R so large and ( ε n )so slowly decreasing such that | ˆ k ( z ) | < | i tan( iz/ | for z ∈ K ( r ). Thus Rouch´e’stheorem for meromorphic functions says that for F and for ( z i tan( iz/ V RH ( r ) the number of zeros minus the number of poles (counted withmultiplicity) is the same for all r >
0. The poles of F are actually the same as forfor the tangens type function. Thus it is proved that for large enough n ∈ N thezeros of F from U RH for R = (2 n − π are simple and contained in the balls B n for | n | ≥ n . Note that we used that we already know that zeros of the characteristicequation must have negative real part. ECAY VIA VISCOELASTIC BOUNDARY DAMPING 23
We have verified that all zeros z n of F | U RH are given by the following ansatz: z n = is n − ξ n with ℜ ξ n > ξ n = o (1) . In the remaining part of the proof we want to simplify the notation by droppingthe indices from z, s and ξ . We also write ˆ k instead of ˆ k ( z ). It is not difficult toverify that F ( z ) = 0 is equivalent to e z = 1 − ˆ k k = (1 − i ℑ ˆ k ) − ℜ ˆ k (1 + i ℑ ˆ k ) + ℜ ˆ k . Let α = arg(1 + i ℑ ˆ k ) be the argument of 1 + i ℑ ˆ k and L = (1 + ( ℑ ˆ k ) ) / . Thenarg(1 ± ˆ k ) = ± α (1 + O ( ℜ ˆ k )) = ± (1 + O ( | ˆ k | )) ℑ ˆ k, thus ℑ ξ = 2(1 + O ( | ˆ k | )) ℑ ˆ k. This yields the first asymptotic formula claimed in the proposition. The secondasympotic formula is a direct consequence of e −ℜ ξ = L − (1 + O ( | α | )) ℜ ˆ kL + (1 + O ( | α | )) ℜ ˆ k = 1 − L (1 + O ( |ℑ ˆ k | )) ℜ ˆ k + O (( ℜ ˆ k ) ) . (cid:3) Upper resolvent estimate.
We prove the upper estimate stated in Theorem20. By Theorem 4 part (i) it suffices to show
Proposition 23.
For all | s | ≥ we have k R ( is ) k L → L ≤ C ( | s | ℜ ˆ k ( is )) − .Proof. For some f ∈ L (0 ,
1) let p be the solution of − s p ( x ) − p ′′ ( x ) = f ( x ∈ (0 , , − p ′ (0) + is ˆ k ( is ) p (0) = 0 ,p ′ (1) + is ˆ k ( is ) p (1) = 0 . (34)Let us define two auxiliary functions p f and p by p f ( x ) = − s Z x sin( s ( x − y )) f ( y ) dy and p ( x ) = cos( sx ) + i ˆ k ( is ) sin( sx ) . It is easy to see that p = ap + p f with a ∈ C is the only possible ansatz whichsatisfies the first two lines in (34). The parameter a is uniquely defined by thecondition from the third line. A short calculation yields that this condition isequivalent to as · (cid:16) ˆ k ( is ) + i tan (cid:16) s (cid:17)(cid:17) (cid:16) ˆ k ( is ) − i cot (cid:16) s (cid:17)(cid:17)| {z } =: D ( s ) · sin( s ) = − p ′ f (1) − is ˆ k ( is ) p f (1) . Note that the singularities of D cancel the zeros of the sine function. Thus we havean explicit formula for a in terms of f . Further note that the absolute values of sp f (1) and p ′ f (1) can be estimated from above by a constant times k f k L . Thus | a | ≤ C | s | · | D ( s ) sin( s ) | · k f k L (0 , . By the presence of the tangent and contangent type function the factor D ( s ) sin( s )can only be small in a neighbourhood of s = 2 nπ or s = (2 n + 1) π . But in this case the real part of ˆ k prevents D from getting too small. We thus have an estimate | D ( s ) sin( s ) | ≥ c ℜ ˆ k ( is ) for | s | ≥ | a | .Since the L -norm of p can be estimated from above by a constant the proof isfinished. (cid:3) Decay rates.
An immediate consequence of Theorem 20, 4(iii), 29 and theremark after Theorem 29 is
Theorem 24.
Assume that ν | (0 ,ε ) = 0 for some ε > . Then there are constants c, C > and t > such that for all t ≥ t cM − ( t/c ) ≤ sup E ( x ) ≤ E ( t, x ) ≤ CM − log ( t/C ) where the increasing function M : (0 , ∞ ) → (0 , ∞ ) is given by M ( s ) = ( ℜ ˆ k ( is )) − . We made the assumption ν | (0 ,ε ) = 0 (i.e. A is invertible) only to simplify theformulation of the theorem. A recipe how to adapt the formulation in case of anon-invertible A is given in Subsection 5.5.7. Further research
For a complete treatment of resolvent estimates for wave equations like (1) itwould be desirable to answer at least the following two questions:
Question 1.
Is the upper bound on (cid:13)(cid:13) ( is − A ) − (cid:13)(cid:13) , given by Theorem 9, optimal? Question 2.
Can one discard the additional assumption (21) on ˆ k withoutchanging the conclusion of Theorem 9?A strategy to positively answer question 1 is to show that there exists a sequenceof eigenvalues of −A which tend to infinity and approach the imaginary axis suf-ficiently fast. We have seen that this strategy works at least for Ω = (0 ,
1) andΩ = D (see Section 6 and Subsection 5.3). For the disk we restricted to kernels k = k β,ε . However, with the more elaborate Taylor argument which proved Propo-sition 22 one can discard this restriction from Lemma 13. We believe that there isa general argument for any bounded Lipschitz domain Ω yielding the existence ofsuch a sequence of eigenvalues.By our investigations in Section 6 we already have a positive answer for question2 in the 1D setting. Moreover, if Ω = D is the disk we already know from thespectrum that an increasing function M with M ( s ) = o (( ℜ ˆ k ( is )) − ) can never bean upper bound for (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) for all large | s | . We think that the answer toquestion 2 is either “yes”, or if “no” then the upper bound solely depends on ℜ ˆ k and the infimum of all α making the upper estimate in (23) true for all spectralcluster p .Concerning the application of resolvent estimates to energy decay there is also athird question. Let us assume for a moment that the answers to questions 1 and 2were positive. Then Theorem 24 was true for any Ω. In general it is not possible toreplace M log by M in Theorem 29. However, does our particular situation allow fora smaller upper bound? In our opinion the most elegant result would be a positiveanswer to Question 3.
Is Theorem 24 true for all bounded Lipschitz domains Ω - evenwith M log replaced by M ? ECAY VIA VISCOELASTIC BOUNDARY DAMPING 25
Appendix A. Besov spaces and the trace operator
In this article we work with fractional Sobolev spaces, Besov spaces and thetrace operator acting on them. Note also that we work with the space H s ( ∂ Ω)which is not only a fractional Sobolev space but also is a function space on a closedsubset of R d which has non-empty interior. In this appendix we aim at providingsome results from the literature about Sobolev/Besov spaces and their relation tointerpolation spaces which is necessary to follow the arguments from our article.Of exceptional importance for the proof of Theorem 4 (i) and (ii) is the validityof the borderline trace theorem - Proposition 26. This borderline case seems to bewell-known to the experts - also for Lipschitz domains - but unfortunately we werenot able to find it in the literature except in [18, Theorem 18.6]. But the proofgiven there is not in our spirit - the Besov spaces are not defined as interpolationspaces there. Therefore we give a simple direct proof via the characterization ofBesov spaces as interpolation spaces which is true if Ω has the so called extensionproperty.A.1. Fractional Sobolev- and Besov spaces.
Let Ω ⊂ R d be a bounded Lip-schitz domain. Here by Lipschitz we mean that locally near any boundary pointand in an appropriate coordinate system one can describe Ω as the set of pointswhich are above the graph of some Lipschitz continuous function from R d − into R . Let 1 ≤ p ≤ ∞ . We assume the reader to be familiar with the usual Sobolevspace W ,p (Ω) which consists of all functions u ∈ L p (Ω) for which all distributionalderivatives ∂ j u are in L p (Ω). There are different methods of defining Besov spaces.For our purposes it is most convenient to define the Besov spaces for 0 < s < ≤ q ≤ ∞ as real interpolation spaces:(35) B s,pq (Ω) = ( L p (Ω) , W ,p (Ω)) s,q . Another approach is to define B s,pq ( R d ) for example via interpolation and then todefine the Besov space on Ω as restrictions to Ω of Besov function on R d . In generalthese approaches are not equivalent but if Ω satisfies the extension property theyare equivalent [17, Chapter 34]. In our setting (0 < s <
1) we say that Ω satisfiesthe extension property if there is a linear and continuous operator Ext : W ,p (Ω) → W ,p ( R d ) such that (Ext u ) | Ω = u for each u from W ,p (Ω). The extension propertyis fulfilled if Ω is bounded and has a Lipschitz boundary. In the following we alwaysassume that this extension property is fulfilled - otherwise some statements frombelow are not valid.The Sobolev-Slobodeckij spaces are defined as special Besov spaces W s,p (Ω) = B s,pp (Ω). It is common to write H s instead of W s, in the Hilbert space setting.For 0 ≤ s ≤ fractional Sobolev spaces (also known as Bessel potential spaces ) H s,p (Ω) via Fourier methods for the specialcase Ω = R d and via restriction for the general case. These spaces form a scale ofcomplex interpolation spaces. In general the fractional Sobolev spaces differ fromthe Sobolev-Slobodeckij spaces but coincide in the case p = 2 (see [2, Chapter 7.67].Note that in Adam’s and Fournier’s book the letter W stands for the fractionalSobolev spaces. We also have H ,p (Ω) = W ,p (Ω) for 1 < p < ∞ - which isCalder´on’s Theorem (see [11, page 7]). We mention that for all 0 < s ≤ s < q, q ∈ [1 , ∞ ] with the restriction q ≤ q if s = s : B s,pq (Ω) ֒ → B s ,pq (Ω) . This is a direct consequence of a general result about the real interpolation method(see e.g. [17, Lemma 22.2]).It is possible to define the Besov space B s,pq ( A ) on a general class of closed subsets A of R d - the so called d -sets . For Ω having a Lipschitz boundary its boundary ∂ Ωis such a set, since it is a d − H s ( A ) = B s, ( A ) in theHilbert space setting.A.2. Traces for functions with /p or more derivatives. Throughout thissubsection Ω ⊆ R d is a bounded domain with Lipschitz boundary and we let 1
Let /p < s < . Then the trace operator Γ : C (Ω) → C ( ∂ Ω) , u u | ∂ Ω extends continuously to an operator Γ : B s,pq (Ω) → B s − p ,pq ( ∂ Ω) . Furthermore Γ has a continuous right inverse: Ext : B s − p ,pq ( ∂ Ω) → B s,pq (Ω) , Γ ◦ Ext = id B s − p ,pq ( ∂ Ω) . The theorem remains valid for s = 1 , q = p if one replaces B s,pq (Ω) by W ,p (Ω) . Unfortunately this theorem is false for any 1 ≤ q ≤ ∞ in the borderline case s = 1 /p if one replaces the target space of Γ by L p ( ∂ Ω). But for our purposes it issufficient that a weakened version remains valid.
Proposition 26.
The trace operator
Γ : B p ,p (Ω) → L p ( ∂ Ω) is continuous. Actually Γ from this proposition is indeed surjective (but we do not need thisproperty in our article) and a more general version is proved in [18, Section 18.6].However there is no linear extension operator from L p ( ∂ Ω) back to the Besov space(See [18] and references therein).We indicate a simple direct proof of Proposition 26. It is based on two lemmaswhich have very simple proofs on their own. The first one is
Lemma 27.
For every C ∞ function u with compact support in R d it is true that k Γ u k L p ( ∂ Ω) ≤ C k u k − p L p (Ω) k u k p W ,p (Ω) . The straightforward proof can be found in [17, Lemma 13.1]. For a differentproof in the case p = 2 we refer to [14]. The second ingredient to the proof ofProposition 26 is [17, Lemma 25.3] which we recall here for convenience of thereader. ECAY VIA VISCOELASTIC BOUNDARY DAMPING 27
Lemma 28.
Let ( E , E ) be an interpolation couple, F a Banach space and let < θ < . Then a linear mapping L : E ∩ E → F extends to a continuousoperator L : ( E , E ) θ, → F if and only if there exists a C > such that for all u ∈ E ∩ E we have k Lu k E ∩ E ≤ C k u k − θE k u k θE .Proof of Proposition 26. Apply the if-part of Lemma 28 to E = L p (Ω), E = W ,p (Ω), F = L p ( ∂ Ω), L = Γ and θ = s . Use Lemma 27 to verify the converse. (cid:3) Appendix B. Semiuniform decay of bounded semigroups
We briefly recall three important results connecting resolvent estimates of gen-erators to the decay rate of their corresponding semigroups. In addition to theliterature mentioned below we recommend the reader to consult [6] for a generaloverview and finer results.Let X be a Banach space and B ( X ) the algebra of bounded operators acting on X . Throughout this section we assume that −A is a the generator of a bounded C -semigroup T : [0 , ∞ ) → B ( X ). By D ( A ) , R ( A ) we denote domain and range of A and by σ ( A ) its spectrum.B.1. Singularity at infinity.
The phrase “Singularity at infinity” refers to thesituation when the resolvent of ( is + A ) − of A has no poles on the imaginary axisbut is allowed to blow up in operator norm if s tends to infinity. The followingtheorem is due to Batty and Duyckaerts [4] but we also refer to [8] for a differentproof and to [5] for a generalization. Theorem 29 ([4]) . Assume that σ ( A ) ∩ i R = ∅ and that there exist constants C, s > and an increasing function M : [ s , ∞ ) → [1 , ∞ ) such that ∀ | s | ≥ s : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) X → X ≤ CM ( C | s | ) . (36) Then there exist constants
C, t > such that ∀ t ≥ t , x ∈ D ( A ) : k T ( t ) x k X ≤ CM − log ( tC ) k x k D ( A ) . Here M log ( s ) = M ( s )(log(1 + M ( s )) + log(1 + s )) . It is comparatively easy to see that a semiuniform decay rate for T as in theconclusion of the Batty-Duyckaerts theorem implies that A has no spectrum on theimaginary axis. Furthermore if the semigroup decays at least like the decreasingfunction m : [0 , ∞ ) → (0 , ∞ ) then the resolvent can not grow faster than thefunction M at infinity, where M ( s ) = 1 + m − r (1 / (2 | s | + 1)) (see [4, Proposition1.3]). Here m − r denotes the right inverse of a decreasing function.Note that for M ( s ) = | s | γ with γ >
0, this theorem tells us that the decay rate isestimated from above by (log( t ) /t ) /γ . One may wonder if the logarithmic term isnecessary. In general it is, as was shown in [7], but in the same article one can findthe following nice characterization of polynomial decay rates in the Hilbert spacesetting: Theorem 30 ([7]) . Let X be a Hilbert space and γ > . Assume that σ ( A ) ∩ i R = ∅ and that there exist constants C, s > such that ∀ | s | ≥ s : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) X → X ≤ C | s | γ . Then there exist constants
C, t > such that ∀ t ≥ t , x ∈ D ( A ) : k T ( t ) x k X ≤ Ct − γ k x k D ( A ) . B.2.
Singularity at zero and infinity. If A is our wave operator, by Theorem4 (iii), it may happen that 0 is a spectral point. Therefore it is convenient to havethe following generalization of Theorem 29 at hand: Theorem 31 ([12]) . Assume that σ ( A ) ∩ i R = { } . Assume that in additionto the condition (36) there exist C > , < s < and a decreasing function m : (0 , s ) → [1 , ∞ ) such that ∀ | s | ≤ s : (cid:13)(cid:13) ( is + A ) − (cid:13)(cid:13) X → X ≤ Cm ( C | s | ) . Then there exist constants
C, t > such that for all t ≥ t and x ∈ D ( A ) ∩ R ( A ) we have k T ( t ) x k X ≤ C " M − log ( tC ) + m − log ( tC ) + 1 t k x k D ( A ) ∩ R ( A ) . Here M log is defined as in Theorem 29 and m log ( s ) = m ( s )(log(1 + m ( s )) − log( s )) . Concerning the relevance of the 1 /t -term we refer the reader to [6, Section 8].In [6, Theorem 8.4] it was shown that in case of polynomial resolvent bounds on aHilbert space one can get rid of the logarithmic loss. Acknowledgments.
I am most grateful to Ralph Chill and Eva Fa˘sangov´a forhelpful discussions during my work on the topic of the article and for reading andcorrecting the first version of this paper. I am also grateful to Otared Kavian foruncovering a gap in the proof of Theorem 4 related to my ignorance about theborderline case of the trace theorem. Finally I want to thank Lars Perlich forhelpful comments.
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