Energy decay for solutions of the wave equation with general memory boundary conditions
aa r X i v : . [ m a t h . O C ] F e b Energy decay for solutions of the wave equation with generalmemory boundary conditions.
Pierre CornilleauEcole Centrale de LyonInstitut Camille Jordan, UMR CNRS 520836 avenue Guy de Collongue69134 Ecully Cedex [email protected] NicaiseUniversit´e de Valenciennes et du Hainaut Cambr´esisLAMAV, FR CNRS 2956,Institut des Sciences et Techniques de ValenciennesF-59313 - Valenciennes Cedex 9 [email protected] 27, 2018
Abstract
We consider the wave equation in a smooth domain subject to Dirichlet boundary conditions onone part of the boundary and dissipative boundary conditions of memory-delay type on the remainderpart of the boundary, where a general borelian measure is involved. Under quite weak assumptions onthis measure, using the multiplier method and a standard integral inequality we show the exponentialstability of the system. Some examples of measures satisfying our hypotheses are given, recoveringand extending some of the results from the literature.
Introduction
We consider the wave equation subject to Dirichlet boundary conditions on one part of the boundaryand dissipative boundary conditions of memory-delay type on the remainder part of the boundary. Moreprecisely, let Ω be a bounded open connected set of R n ( n ≥
2) such that, in the sense of Neˇcas ([8]), itsboundary ∂ Ω is of class C . Throughout the paper, I denotes the n × n identity matrix, while A s denotesthe symmetric part of a matrix A . Let m be a C vector field on ¯Ω such thatinf ¯Ω div( m ) > sup ¯Ω (div( m ) − λ m ) (1)where λ m ( x ) is the smallest eigenvalue function of the real symmetric matrix ∇ m ( x ) s . Remark 1
The set of all C vector fields on ¯Ω such that (1) holds is an open cone. If m is in this set,we denote c ( m ) = 12 (cid:18) inf ¯Ω div( m ) − sup ¯Ω (div( m ) − λ m ) (cid:19) . Example 1 • An affine example is given by m ( x ) = ( A + A )( x − x ) , where A is a definite positive matrix, A a skew-symmetric matrix and x any point in R n .1 A non linear example is m ( x ) = ( dI + A )( x − x ) + F ( x )where d > A is a skew-symmetric matrix, x any point in R n and F is a C vector field on ¯Ωsuch that sup x ∈ ¯Ω k ( ∇ F ( x )) s k < dn ( k · k stands for the usual 2-norm of matrices).We define a partition of ∂ Ω in the following way. Denoting by ν ( x ) the normal unit vector pointingoutward of Ω at a point x ∈ ∂ Ω , we consider a partition ( ∂ Ω N , ∂ Ω D ) of the boundary such that themeasure of ∂ Ω D is positive and that ∂ Ω N ⊂ { x ∈ ∂ Ω , m ( x ) · ν ( x ) > } , ∂ Ω D ⊂ { x ∈ ∂ Ω , m ( x ) · ν ( x ) } . (2)Furthermore, we assume ∂ Ω D ∩ ∂ Ω N = ∅ or m · n ≤ ∂ Ω D ∩ ∂ Ω N (3)where n stands for the normal unit vector pointing outward of ∂ Ω N when considering ∂ Ω N as a sub-manifold of ∂ Ω.On this domain, we consider the following delayed wave problem:( S ) u ′′ − ∆ u = 0 u = 0 ∂ ν u + m · ν (cid:16) µ u ′ ( t ) + R t u ′ ( t − s ) dµ ( s ) (cid:17) = 0 u (0) = u u ′ (0) = u in R ∗ + × Ω , on R ∗ + × ∂ Ω D , on R ∗ + × ∂ Ω N , in Ω , in Ω , where u ′ (resp. u ′′ ) is the first (resp. second) time-derivative of u , ∂ ν u = ∇ u · ν is the normal outwardderivative of u on ∂ Ω. Moreover µ is some positive constant and µ is a borelian measure on R + .The above problem covers the case of a problem with memory type as studied for instance in [1, 3, 5, 9],when the measure µ is given by dµ ( s ) = k ( s ) ds, (4)where ds stands for the Lebesgue measure and k is non negative kernel. But it also covers the case of aproblem with a delay as studied for instance in [10, 11, 12], when the measure µ is given by µ = µ δ τ , (5)where µ is a non negative constant and τ > dµ ( s ) = k ( s ) χ [ τ ,τ ] ( s ) ds, (6)where 0 ≤ τ < τ , χ [ τ ,τ ] is the characteristic equation of the interval [ τ , τ ] and k is a non negativefunction in L ∞ ([ τ , τ ]).A closer look at the decay results obtained in these references shows that there are different ways toquantify the energy of ( S ). More precisely for the measure of the form (4), the exponential or polynomialdecay of an appropriated energy is proved in [1, 3, 5, 9], by combining the multiplier method (or differentialgeometry arguments) with the use of suitable Lyapounov functionals (or integral inequalities) under theassumptions that the kernel k is sufficiently smooth and has a certain decay at infinity. On the otherhand for a measure of delay type like (5) or (6), the exponential stability of the system was proved in[10, 11, 12] by proving an observability estimate obtained by assuming that the term R t u ′ ( t − s ) dµ ( s ) issufficiently small with respect to µ u ′ ( t ). Consequently, our goal is here to obtain some uniform decayresults in the general context described above with a similar assumptions than in [10, 11, 12]. Moreprecisely, we will show in this paper that if there exists α > µ tot := Z + ∞ e αs d | µ | ( s ) < µ (7)2here | µ | is the absolute value of the measure µ , then the above problem ( S ) is exponentially stable.The paper is organized as follows: in the first two sections, we explain how to define an energy usingsome basic measure theory. Using well-known results, we obtain the existence of energy solutions. Inthis setting we present and prove our stabilization result in the third section. Examples of measures µ satisfying our hypotheses are given in the end of the paper, where we show that we recover and extendsome of the results from the references cited above.Finally in the whole paper we use the notation A . B for the estimate A ≤ CB with some constant C that only depends on Ω, m or µ . In this section we show that the assumption (7) implies the existence of some borelian finite measure λ such that λ ( R + ) < µ , | µ | ≤ λ (8)(in the sense that, for every measurable set B , | µ | ( B ) ≤ λ ( B )) andfor all measurable set B , Z B λ ([ s, + ∞ )) ds ≤ α − λ ( B ) (9)Indeed we show the following equivalence: Proposition 1
Let µ be a borelian positive measure on R + and µ some positive constant. The followingproperties are equivalent: • ∃ α > such that Z + ∞ e αs dµ ( s ) < µ . • There exists a borelian measure λ on R + such that λ ( R + ) < µ , µ ≤ λ and, for some constant β > ,for all measurable set B , Z B λ ([ s, + ∞ )) ds ≤ β − λ ( B ) . Proof
We introduce the application T from the set of positive borelian measures into itself as follows:if µ is some positive borelian measure, we define a positive borelian measure T ( µ ) by T ( µ )( B ) = Z B µ ([ s, + ∞ )) ds, if B is any measurable set.( ⇐ ) If λ fulfills the second property, then it immediately follows that ∀ n ∈ N , β n T n ( µ ) ≤ λ, where as usual T n is the composition T ◦ T · · · ◦ T n -times. A summation consequently gives, for any r ∈ (0 , ∞ X n =0 ( rβ ) n T n ( µ ) ≤ ∞ X n =0 r n λ = (1 − r ) − λ. Using Fubini theorem, we can now compute T n ( µ )( R + ) = Z + ∞ Z + ∞ s n +1 · · · Z + ∞ s (cid:18)Z + ∞ s dµ ( s ) (cid:19) ds · · · ds n ! ds n +1 = Z + ∞ (cid:18)Z s · · · (cid:18)Z s n ds n +1 (cid:19) · · · ds (cid:19) dµ ( s )= Z + ∞ s n n ! dµ ( s ) 3o that, using monotone convergence theorem, one can obtain Z + ∞ e rβs dµ ( s ) ≤ (1 − r ) − λ ( R + )and our proof ends using that (1 − r ) − λ ( R + ) < µ for sufficiently small r .( ⇒ ) For any measurable set B , we define λ ( B ) = ∞ X n =0 α n T n ( µ )( B ) . It is clear that λ is a borelian measure such that µ ≤ λ . Moreover, if B is a measurable set, one has,thanks to monotone convergence theorem T ( λ )( B ) = Z B λ ([ s, + ∞ )) ds = ∞ X n =0 α n Z B T n ( µ )([ s, + ∞ )) ds ≤ α − ∞ X n =0 α n +1 T n +1 ( µ )( B ) , that is, T ( λ ) ≤ α − λ .Finally, another use of monotone convergence theorem gives λ ( R + ) = Z + ∞ e αs dµ ( s ) < µ . (cid:3) Remark 2 • If µ satisfies our first property, one can choose β = α in our second property. • If µ is supported in (0 , τ ], it is straightforward to see that, for some small enough constant c , dλ ( s ) = dµ ( s ) + c χ [0 ,τ ] ( s ) ds fulfills (7). This observation allows us to recover the choices of energy in [10, 11].In the sequel, we can thus consider the measure λ obtained by the application of Proposition 1 to | µ | . Defining H D (Ω) := { u ∈ H (Ω); u = 0 on ∂ Ω D } and H (Ω) := { u ∈ H (Ω); u = 0 on ∂ Ω } , we here present an application of Theorem 4.4 of Propst and Pr¨uss paper (see [13]) in the framework ofhypothesis (3). Theorem 1
Suppose u ∈ H D (Ω) , u ∈ L (Ω) . Then ( S ) admits a unique solution u ∈ C ( R + , H (Ω)) ∩C ( R + , L (Ω)) in the weak sense of Propst and Pr¨uss. Moreover, if u ∈ H (Ω) ∩ H D (Ω) , u ∈ H (Ω) ,then u ∈ C ( R + , H (Ω)) ∩ C ( R + , L (Ω)) and the additional results hold ∀ t ≥ , ∆ u ( t ) ∈ L (Ω) ∂ ν u ( t ) | ∂ Ω N ∈ H / ( ∂ Ω N ) . roof The proof is the one proposed in [13], Theorem 4.4 except that, for smoother data, we can notuse elliptic result in the general context of (3) to get more regularity. (cid:3)
Inspired by [10, 11], we now define the energy of the solution of (S) at any positive time t by the followingformula: E ( t ) = 12 Z Ω ( u ′ ( t, x )) + |∇ u ( t, x ) | dx + 12 Z ∂ Ω N m · ν Z t (cid:18)Z s ( u ′ ( t − r, x )) dr (cid:19) dλ ( s ) dσ + 12 Z ∂ Ω N m · ν Z ∞ t (cid:18)Z s ( u ′ ( s − r, x )) dr (cid:19) dλ ( s ) dσ. Remark 3 • In the definition of energy, the measure λ can be replaced by any positive borelianmeasure ν such that ν ≤ λ such as, for instance, | µ | . In fact, we will see later that conditions (8) and (9) are only here toensure that the corresponding energy E λ is non increasing, but the decay of another energy E ν isimplied by the decay of E λ . • If µ is compactly supported in [0 , τ ], for times greater than τ , one can recover the energies from[10, 11] by choosing the measure λ supported in [0 , τ ] given by Remark 2. Indeed, the last term inthe energy is null for t > τ , and the second term is reduced to12 Z ∂ Ω N m · ν Z τ (cid:18)Z s ( u ′ ( t − r, x )) dr (cid:19) dλ ( s ) dσ. We now identify our energy space.
Proposition 2 If u ∈ H (Ω) ∩ H D (Ω) , u ∈ H (Ω) , then u ′ ∈ L ∞ ( R + , H (Ω)) . Consequently, for suchinitial conditions, the energy E ( t ) is well defined for any t > and it uniformly depends continuously onthe initial data. Proof
Let us first pick some solution of ( S ) with u ∈ H (Ω) ∩ H D (Ω) , u ∈ H D (Ω). We define thestandard energy as E ( t ) = 12 Z Ω ( u ′ ( t, x )) + |∇ u ( t, x ) | dx. As in [6], it is classical that E (0) − E ( T ) = − Z T Z ∂ Ω N ∂ ν uu ′ dσdt. Using the form of our boundary condition and Young inequality, one gets, for any ǫ > E (0) − E ( T ) = Z T Z ∂ Ω N ( m · ν ) (cid:18) µ u ′ ( t ) + u ′ ( t ) Z t u ′ ( t − s ) dµ ( s ) (cid:19) dσdt ≥ Z T Z ∂ Ω N ( m · ν ) (cid:16) µ − ǫ (cid:17) u ′ ( t ) − ǫ (cid:18)Z t u ′ ( t − s ) dµ ( s ) (cid:19) ! dσ Using that µ ≤ | µ | and Cauchy-Schwarz inequality consequently give us E (0) − E ( T ) ≥ Z ∂ Ω N ( m · ν ) (cid:16) µ − ǫ (cid:17) Z T u ′ ( t ) dt − µ tot ǫ Z T Z t ( u ′ ( t − s )) d | µ | ( s ) ! dσ. Z T Z t ( u ′ ( t − s )) d | µ | ( s ) dt = Z T Z Ts u ′ ( t − s ) dt ! d | µ | ( s )= Z T Z T − s u ′ ( t ) dt ! d | µ | ( s )= Z T Z T − t d | µ | ( s ) ! u ′ ( t ) dt so that, using | µ | ([0 , T − t ]) ≤ µ , E (0) − E ( T ) ≥ Z ∂ Ω N ( m · ν ) (cid:18) µ − ǫ − µ ǫ (cid:19) Z T u ′ ( t ) dt ! dσ. The choice of ǫ = µ tot finally gives us that E ( T ) is bounded. Using the density of H (Ω) ∩ H D (Ω) × H D (Ω)in H D (Ω) × L (Ω), we get the boundedness of E for solutions with initial data u ∈ H D (Ω) , u ∈ L (Ω).In particular, if u ∈ H D (Ω) , u ∈ L (Ω), we obtain that u ∈ L ∞ ( R + , H (Ω)).Let now u be a solution of ( S ) with u ∈ H (Ω) ∩ H D (Ω), u ∈ H (Ω). Using Theorem 1, one candefine the limit in L (Ω) u of u ′′ ( t ) as t → u ′ issolution of ( S ) with initial data u ∈ H D (Ω) and u ∈ L (Ω).Indeed, one can see that Fubini’s theorem gives Z t u ′ ( t − s ) dµ ( s ) = Z t (cid:18)Z s u ′′ ( s − r ) dµ ( r ) (cid:19) ds, provided u = 0 on ∂ Ω N , so that ddt (cid:18)Z t u ′ ( t − s ) dµ ( s ) (cid:19) = Z t u ′′ ( t − s ) dµ ( s ) . Using the proof above, one concludes that u ′ ∈ L ∞ ( R + , H (Ω)) which, thanks to a classical trace result,give that u ′ ∈ L ∞ ( R + , L ( ∂ Ω)).The first three terms of the energy E ( t ) are consequently defined for any time t >
0. We only needto take a look at the last one to achieve our result. Using Fubini theorem again, one has Z + ∞ t (cid:18)Z s ( u ′ ( t − r )) dr (cid:19) dλ ( s ) = Z + ∞ u ′ ( r ) Z + ∞ max( r,t ) dλ ( s ) ! dr so that, using (9), Z ∂ Ω N m · ν Z ∞ t (cid:18)Z s ( u ′ ( s − r, x )) dr (cid:19) dλ ( s ) dσ ≤ k m k ∞ k u ′ k L ∞ ( L ( ∂ Ω)) Z + ∞ λ ([ r, + ∞ )) dr ≤ α − k m k ∞ λ ( R + ) k u ′ k L ∞ ( L ( ∂ Ω)) . (cid:3) In this second approach, we assume that µ is supported in [0 , τ ] and that ∂ Ω D ∩ ∂ Ω N = ∅ . We heresimply follow the result obtained by Nicaise-Pignotti ([11]).First, observe that, for t > τ , ( S ) is reduced to u ′′ − ∆ u = 0 u = 0 ∂ ν u + m · ν (cid:0) µ u ′ ( t ) + R τ u ′ ( t − s ) dµ ( s ) (cid:1) = 0 u (0) = u u ′ (0) = u in ( τ, + ∞ ) × Ω , on ( τ, + ∞ ) × ∂ Ω D , on ( τ, + ∞ ) × ∂ Ω N , in Ω , in Ω .
6e define X τ = L ( ∂ Ω N × (0 , × (0 , τ ) , dσdρsdµ ( s ))) and Y τ = L ( ∂ Ω N × (0 , τ ); H (0 , , dσsdµ ( s )) . One can use the same strategy as in the proof of Theorem 2.1 in [11] to get
Theorem 2 • If u ( τ ) ∈ H D (Ω) , u ′ ( τ ) ∈ L (Ω) and u ′ ( τ − ρs, x ) ∈ X τ , ( S ) has a unique solution u ∈ C ([ τ, + ∞ ) , H D (Ω)) ∩ C ([ τ, + ∞ ) , L (Ω)) . Moreover, if u ( τ ) ∈ H (Ω) ∩ H D (Ω) , u ′ ( τ ) ∈ H (Ω) and u ′ ( τ − ρs, x ) ∈ Y τ , then (cid:26) u ∈ C ([ τ, + ∞ ) , H D (Ω)) ∩ C ([ τ, + ∞ ) , H (Ω)); t su ′′ ( t − ρs, x ) ∈ C ([ τ, + ∞ ) , X τ ) . • If ( u nτ ( x ) , v nτ ( x ) , g n ( s, ρ, x )) → ( u ( τ, x ) , u ′ ( τ, x ) , u ′ ( τ − ρs, x )) in H D (Ω) × L (Ω) × X τ , then thesolution u n of u ′′ − ∆ u = 0 u = 0 ∂ ν u + m · ν (cid:0) µ u ′ ( t ) + R τ u ′ ( t − s ) dµ ( s ) (cid:1) = 0 u ( τ ) = u nτ u ′ ( τ ) = u nτ u ′ ( x, τ − ρs ) = g n ( x, s, ρ ) in ( τ, + ∞ ) × Ω , on ( τ, + ∞ ) × ∂ Ω D , on ( τ, + ∞ ) × ∂ Ω N , in Ω , in Ω , in Ω N × (0 , τ ) × (0 , is such that E ( u n ) converges uniformly with respect to time towards E ( u ) . Proof
We define z ( x, ρ, s, t ) = u ′ ( t − ρs, x ) for x ∈ ∂ Ω N , t > τ , s ∈ (0 , τ ), ρ ∈ (0 , S ) is then equivalent to u ′′ − ∆ u = 0 sz t ( x, ρ, s, t ) + z ρ ( x, ρ, s, t ) = 0 u = 0 ∂ ν u + m · ν ( µ u ′ ( t ) + R τ u ′ ( t − s ) dµ ( s )) = 0 u ( τ ) = u ( τ ) u ′ ( τ ) = u ( τ ) z ( x, , t, s ) = u ′ ( t, x ) z ( x, ρ, τ, s ) = f ( x, ρ, s ) in ( τ, + ∞ ) × Ω , in ∂ Ω N × (0 , × (0 , τ ) × ( τ, + ∞ ) , on ( τ, + ∞ ) × ∂ Ω D , on ( τ, + ∞ ) × ∂ Ω N , in Ω , in Ω , on ∂ Ω N × ( τ, + ∞ ) × (0 , τ ) , on ∂ Ω N × (0 , × (0 , τ ) , where f ( x, ρ, s ) = u ′ ( τ − ρs, x ).Consequently, ( S ) can be rewritten as (cid:26) U ′ = A UU ( τ ) = ( u ( τ ) , u ′ ( τ ) , f ) T where the operator is defined by A uvz = v ∆ u − s − z ρ with domain D ( A ) = { ( u, v, z ) T ∈ H D (Ω) × L (Ω) × Y τ ; ∆ u ∈ L (Ω) ,∂ ν u ( x ) = − ( m · ν ) (cid:18) µ v ( t ) + Z τ z ( x, , s ) dµ ( s ) (cid:19) on ∂ Ω N , v ( x ) = z ( x, , s ) on ∂ Ω N × (0 , τ ) } . A is a maximal monotone operator on the Hilbert space H := H D (Ω) × L (Ω) × X τ endowed with the product topology. It consequently generates a contractionsemigroup on H . Moreover, if ( u ( τ, x ) , u ′ ( τ, x ) , u ′ ( τ − ρs, x )) ∈ D ( A ), one gets that (cid:26) u ∈ C ([ τ, + ∞ ) , H D (Ω)) ∩ C ([ τ, + ∞ ) , H (Ω)); t su ′′ ( t − ρs, x ) ∈ C ([ τ, + ∞ ) , X τ ) . This ends the proof. (cid:3)
We can consequently deduce another way to obtain solutions:
Corollary 1
Suppose that u ∈ H (Ω) ∩ H D (Ω) , u ∈ H (Ω) , then ( S ) has a unique solution u ∈C ([ τ, + ∞ ) , H D (Ω)) ∩ C ([ τ, + ∞ ) , L (Ω)) . Proof
Thanks to Theorem 1, one only needs to check that if u ∈ C ([0 , τ ] , H D (Ω)) then u ′ ( x, τ − ρs ) ∈ X τ ;and this is straightforward using Fubini theorem. (cid:3) We begin with a classical elementary result due to Komornik [6]:
Lemma 1
Let E : [0 , + ∞ [ → R + be a non-decreasing function that fulfils: ∀ t ≥ , Z ∞ t E ( s ) ds ≤ T E ( t ) , for some T > . Then, one has: ∀ t > T, E ( t ) ≤ E (0) exp (cid:18) − tT (cid:19) . We will now show the following stabilization result:
Theorem 3
Assume (1)-(7). Then, if u ∈ H (Ω) ∩ H D (Ω) , u ∈ H (Ω) , there exists T > such thatthe energy E ( t ) of the solution u of ( S ) satisfies: ∀ t > T, E ( t ) ≤ E (0) exp (cid:18) − tT (cid:19) . Proof
Our goal is to perform the multiplier method and to deal with the delay terms to show that onecan apply Lemma 1 to the energy.
Lemma 2
There exists
C > , such that, for any solution u of ( S ) and any S ≤ T , E ( S ) − E ( T ) > C Z TS Z ∂ Ω N ( m · ν ) (cid:18) ( u ′ ( t )) + Z t ( u ′ ( t − s )) dλ ( s ) (cid:19) dσdt. In particular, the energy is a non-increasing function of time.
Proof
We start from the classical result that E ( S ) − E ( T ) = − Z TS Z ∂ Ω ∂ ν uu ′ dσdt. As above, one gets, for any ǫ > E ( S ) − E ( T ) ≥ Z TS Z ∂ Ω N ( m · ν ) (cid:16) µ − ǫ (cid:17) u ′ ( t ) − µ tot ǫ Z T Z t ( u ′ ( t − s )) d | µ | ( s ) ! dσdt
8e will now split E − E in two terms:[ E − E ] TS = − (cid:18)Z ∂ Ω N ( m · ν )[ f ( t, x ) − g ( t, x )] TS dσ (cid:19) , where f ( t, x ) = Z t (cid:18)Z s ( u ′ ( t − r )) dr (cid:19) dλ ( s ) ,g ( t, x ) = Z t (cid:18)Z s u ′ ( r ) dr (cid:19) dλ ( s ) . A change of variable allows us to get f ( t, x ) = Z t Z t u ′ ( r ) drdλ ( s ) − Z t Z t − s u ′ ( r ) drdλ ( s ) . An application of Fubini theorem consequently gives us f ( t, x ) − g ( t, x ) = Z t u ′ ( r ) λ ([0 , r ]) dr − Z t Z t − s u ′ ( r ) drdλ ( s )and, as above, one can use Fubini theorem to deduce that Z TS Z t ( u ′ ( t − s )) dλ ( s ) dt = (cid:20)Z t Z t − s u ′ ( r ) drdλ ( s ) (cid:21) TS . One now uses λ ([0 , r ]) λ ( R + ) to conclude that[ E − E ] TS ≥ Z TS Z ∂ Ω N m · ν (cid:18)Z t ( u ′ ( t − s )) dλ ( s ) − λ ( R + ) u ′ ( t ) (cid:19) dσdt. Summing up and using that | µ | ≤ λ , we have obtained that E ( S ) − E ( T ) > Z TS Z ∂ Ω N ( m · ν ) (cid:18)(cid:18) µ − λ ( R + ) + ǫ (cid:19) u ′ ( t ) + 12 (cid:16) − µ tot ǫ (cid:17) Z t ( u ′ ( t − s )) dλ ( s ) (cid:19) dσdt. We finally chose ǫ = µ which gives us our result since λ ( R + ) < µ . (cid:3) In the multiplier method, one may use Rellich’s relation, especially in the context of singularities. Inour framework (3), the following Rellich inequality (see the proof of Theorem 4 in [4] or Proposition 4 in[2]) is useful
Proposition 3
For any u ∈ H (Ω) such that ∆ u ∈ L (Ω) , u | ∂ Ω D ∈ H ( ∂ Ω D ) and ∂ ν u | ∂ Ω N ∈ H ( ∂ Ω N ) . Then it satisfies ∂ ν u ( m. ∇ u ) − ( m · ν ) |∇ u | ∈ L ( ∂ Ω) and we have the following inequality Z Ω △ u ( m. ∇ u ) dx ≤ R Ω (div( m ) I − ∇ m ) s )( ∇ u, ∇ u ) dx + Z ∂ Ω (2 ∂ ν u ( m. ∇ u ) − ( m · ν ) |∇ u | ) dσ. With this result, we can prove the following multiplier estimate:
Lemma 3
Let
M u = 2 m. ∇ u + a u , where a := (inf ¯Ω div( m )) + sup ¯Ω (div( m ) − λ m ) (cid:1) . Then underthe assumptions of Theorem 3, the following inequality holds true: c ( m ) Z TS Z Ω ( u ′ ) + |∇ u | dxdt ≤ − [ Z Ω u ′ M u ] TS + Z TS Z ∂ Ω N M u∂ ν u + ( m · ν )(( u ′ ) − |∇ u | ) dσdt. roof Firstly, we consider M = 2 m · ∇ u + au where a will be fixed later. Using the fact that u is aregular solution of ( S ) and noting that u ′′ M u = ( u ′ M u ) ′ − u ′ M u ′ , an integration by parts gives:0 = Z TS Z Ω ( u ′′ − △ u ) M udxdt = (cid:20)Z Ω u ′ M udx (cid:21) TS − Z TS Z Ω ( u ′ M u ′ + △ uM u ) dxdt. Now, thanks to Proposition 3, we have : Z Ω △ uM udx ≤ a Z Ω △ uudx + Z Ω (div( m ) I − ∇ m ) s )( ∇ u, ∇ u ) dx + Z ∂ Ω (2 ∂ ν u ( m. ∇ u ) − ( m.ν ) |∇ u | ) dσ. Consequently, Green-Riemann formula leads to: Z Ω △ uM udx = Z Ω ((div( m ) − a ) I − ∇ m ) s )( ∇ u, ∇ u ) dx + Z ∂ Ω ( ∂ ν uM u − ( m · ν ) |∇ u | ) dσ. Using the fact that ∇ u = ∂ ν uν on ∂ Ω D and m · ν ∂ Ω D , we have then: Z Ω △ uM udx ≤ Z Ω ((div( m ) − a ) I − ∇ m ) s )( ∇ u, ∇ u ) dx + Z ∂ Ω N ( ∂ ν uM u − ( m · ν ) |∇ u | ) dσ. On the other hand, another use of Green formula gives us: Z Ω u ′ M u ′ dx = Z Ω ( a − div( m ))( u ′ ) dx + Z ∂ Ω N ( m · ν ) | u ′ | dσ. Consequently Z TS Z Ω (div( m ) − a )( u ′ ) + (( a − div( m )) I + 2( ∇ m ) s )( ∇ u, ∇ u ) dxdt ≤ − (cid:20)Z Ω u ′ M udx (cid:21) TS + Z TS Z ∂ Ω N ∂ ν uM u + ( m · ν )(( u ′ ) − |∇ u | ) dσdt. Our goal is now to find a such that div( m ) − a and ( a − div( m )) I + 2( ∇ m ) s are uniformly minorized onΩ. One has to find a such that, uniformly on Ω, (cid:26) div( m ) − a ≥ c λ m + ( a − div( m )) ≥ c (10)for some positive constant c . The latter condition is then equivalent to find a which fulfillsinf ¯Ω div( m ) > a > sup ¯Ω (div( m ) − λ m ) , and its existence is now guaranteed by (1). Moreover, it is straightforward to see that the greatest valueof c such that (10) holds is c ( m ) = 12 (cid:18) inf ¯Ω div( m ) − sup ¯Ω (div( m ) − λ m ) (cid:19) and is obtained for a = a . This ends the proof. (cid:3) Consequently, the following result holds
Lemma 4
For every τ ≤ S < T < ∞ , the following inequality holds true: Z TS Z Ω ( u ′ ) + |∇ u | dxdt . E ( S ) . roof We start from Lemma 3.First of all, Young and Poincar´e inequalities give | Z Ω u ′ M udx | . E ( t ) , so that − (cid:20)Z Ω u ′ M udx (cid:21) TS . E ( S ) + E ( T ) CE ( S ) . Now, from the boundary condition, one has
M u∂ ν u + ( m · ν )(( u ′ ) − |∇ u | ) = ( m · ν ) (cid:18)(cid:18) µ u ′ + Z t u ′ ( t − s ) dµ ( s ) (cid:19) M u + ( u ′ ) − |∇ u | (cid:19) . Using the definition of
M u and Young inequality, we get for any ǫ > M u∂ ν u +( m · ν )(( u ′ ) −|∇ u | ) ( m · ν ) (cid:18) k m k ∞ + µ a ǫ (cid:19) ( u ′ ) + a ǫ (cid:18)Z t u ′ ( t − s ) dµ ( s ) (cid:19) + ǫu ! . Another use of Poincar´e inequality consequently allow us to choose ǫ > ǫ Z ∂ Ω N ( m · ν ) u dσ c ( m )2 Z Ω |∇ u | dx. Cauchy-Schwarz inequality consequently leads to c ( m )2 Z TS Z Ω ( u ′ ) + |∇ u | dxdt . E ( S ) + Z TS Z ∂ Ω N ( m · ν ) (cid:18) u ′ ( t ) + Z t ( u ′ ( t − s )) d | µ | ( s ) (cid:19) dσdt and, since | µ | ≤ λ , Lemma 2 gives us the desired result: c ( m ) Z TS Z Ω ( u ′ ) + |∇ u | dxdt . E ( S ) . (cid:3) To conclude we need to absorb the two last integral terms for which we use the following result.
Lemma 5 • For any solution u and any S < T , Z TS Z ∂ Ω N m · ν Z t (cid:18)Z s ( u ′ ( t − r, x )) dr (cid:19) dλ ( s ) dσdt . Z TS Z ∂ Ω N m · ν Z t ( u ′ ( t − s, x )) dλ ( s ) dσdt. • For any solution u and any S < T , Z TS Z ∂ Ω N m · ν Z + ∞ t (cid:18)Z s ( u ′ ( s − r, x )) dr (cid:19) dλ ( s ) dσdt . Z TS Z ∂ Ω N m · ν Z t ( u ′ ( t − s, x )) dλ ( s ) dσdt + Z + ∞ S Z ∂ Ω N m · ν u ′ dσdt. Proof • We start from the left hand side term. We fix x ∈ ∂ Ω N , t ∈ [ S, T ] and we use Fubini theorem toestimate integrals with respect to time: Z t (cid:18)Z s ( u ′ ( t − r, x )) dr (cid:19) dλ ( s ) = Z t ( u ′ ( t − r, x )) λ ([ r, t ]) dr ≤ Z t ( u ′ ( t − r, x )) λ ([ r, + ∞ )) dr ≤ α − Z t ( u ′ ( t − r, x )) dλ ( r )which gives the required result after an integration with respect to t and x .11 As above, fixing x ∈ ∂ Ω N , we obtain Z TS Z + ∞ t (cid:18)Z s ( u ′ ( s − r, x ) dr (cid:19) dλ ( s ) dt = Z TS Z + ∞ ( u ′ ( r, x )) λ ([max( r, t ) , + ∞ ]) drdt = Z TS Z t ( u ′ ( t − r, x )) drλ ([ t, + ∞ )) dt + Z TS (cid:18)Z + ∞ t ( u ′ ( r, x )) λ ([ r, + ∞ )) dr (cid:19) dt. Since for all r ≤ t , λ ([ t, + ∞ ) ≤ λ ([ r, + ∞ )), we first have Z TS Z t ( u ′ ( t − r, x )) drλ ([ t, + ∞ )) dt ≤ Z TS Z t ( u ′ ( t − r, x )) λ ([ r, + ∞ )) drdt ≤ α − Z TS Z t ( u ′ ( t − r, x )) dλ ( r ) dt. On the other hand, Fubini theorem gives us Z TS (cid:18)Z + ∞ t ( u ′ ( r, x )) λ ([ r, + ∞ )) dr (cid:19) dt = Z + ∞ S u ′ ( r ) λ ([ r, + ∞ ))(min( T, r ) − S ) dr. We now note that rλ ([ r, + ∞ )) ≤ Z + ∞ r sdλ ( s ) ≤ Z + ∞ sdλ ( s )and Z + ∞ sdλ ( s ) = Z + ∞ λ ([ t, + ∞ )) dt ≤ α − λ ( R + ) . We consequently obtain Z TS (cid:18)Z + ∞ t ( u ′ ( r, x )) λ ([ r, + ∞ )) dr (cid:19) dt . Z + ∞ S u ′ , which give the required result after an integration over ∂ Ω N . (cid:3) Up to now, we have proven that Z TS E ( t ) dt . E ( S ) + Z TS Z ∂ Ω N m · ν Z t ( u ′ ( t − s, x )) dλ ( s ) dσdt + Z + ∞ S Z ∂ Ω N m · ν u ′ dσdt. Lemma 2 allows us to conclude since it gives Z + ∞ S Z ∂ Ω N m · ν u ′ dσdt . E ( S )and Z TS Z ∂ Ω N m · ν Z t ( u ′ ( t − s, x )) dλ ( s ) dσdt . E ( S ) . (cid:3) Remark 4
In the case of some compactly supported measure µ , one can also obtain exponential decayresult for the following problem u ′′ − ∆ u = 0 u = 0 ∂ ν u + µ u ′ ( t ) + R t u ′ ( t − s ) dµ ( s ) = 0 u (0) = u u ′ (0) = u in R ∗ + × R ∗ + , on R ∗ + × ∂ Ω D , on R ∗ + × ∂ Ω N , in Ω , in Ω ,
12s it was done in [11] using the work of Lasiecka-Triggiani-Yao [7] and since the system is time invariantfor t ≫ µ provided thatinf ∂ Ω N m · ν > . We start with two general results and then particularize them to recover results from the literature.
Example 2 If µ is some borelian measure such that | µ | ( R + ) < µ and Z + ∞ e βs d | µ | ( s ) < + ∞ for some β >
0, then µ fulfils the assumption (7) for an appropriate α . Indeed for any 0 ≤ α ≤ β , theexpression Z + ∞ e αs d | µ | ( s )is finite and by the dominated convergence Theorem of Lebesgue we have Z + ∞ e αs d | µ | ( s ) → | µ | ( R + ) as α → . Consequently by the assumption | µ | ( R + ) < µ , we get (7) for α small enough. Example 3
One can choose µ = ∞ X i =1 µ i δ τ i , where ( τ i ) ∞ i =1 , ( µ i ) ∞ i =1 are some families such that τ i > ∞ X i =1 | µ i | e ατ i < µ for some α > Example 4
If we choose dµ ( s ) = k ( s ) ds where k is a kernel satisfying Z + ∞ | k ( s ) | ds < µ and Z + ∞ | k ( s ) | e βs ds < ∞ for some β >
0. Then as a consequence of Example 2, we get an exponential decay rate for the system(S) under the (very weak) condition above, in particular we do not need any differentiability assumptionson k , nor uniform exponential decay of k at infinity as in [1, 3, 5, 9]. Example 5
Choose dµ ( s ) = k ( s ) χ [ τ ,τ ] ( s ) ds, where k is an integrable function in [ τ , τ ] such that Z τ τ | k ( s ) | ds < µ , then we get an exponential decay for the system (S) as a consequence of Example 2 because the secondassumption trivially holds. In that case we extend the results of [11] to a larger class of kernels k , forinstance in the class of bounded variations functions.13 xample 6 Take µ ( s ) = µ δ τ ( s ) , where µ is a constant and τ > | µ | < µ , then we recover the decay results from [10, 12]. References [1]
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