Maximal regularity for the damped wave equation

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Maximal regularity for the damped wave equation

Mahdi Achache

To cite this version:

Mahdi Achache. Maximal regularity for the damped wave equation. 2018. �hal-01719547�

Maximal regularity for the damped wave equation

Mahdi Achache

Abstract

We consider the problem of maximal regularity for non-autonomous Cauchy problems

¨

u(t) + B (t) ˙ u(t) + A (t)u(t) = f (t) (t ∈ [0, τ ]), u(0) = u

, u(0) = ˙ u

. Here, the time dependent operator A (t) is bounded from V to V

and B (t) is associated with a family of sesquilinear forms with domain V . We prove maximal L

-regularity results and other regularity properties for the solutions of the previous problems under minimal regularity assumptions on the operators.

keywords: Damped wave equation, maximal regularity, non-autonomous evolution equations.

Mathematics Subject Classification (2010): 35K90, 35K45, 47D06.

1 Introduction

Let ( H , ( · , · ), k · k ) be a separable Hilbert space over R or C . We consider an- other separable Hilbert space V which is densely and continuously embedded into H . We denote by V

the (anti-) dual space of V so that

V ֒ →

H ֒ →

V

. Hence there exists a constant C > 0 such that

k u k ≤ C k u k

(u ∈ V ), where k · k

denotes the norm of V . Similarly,

k ψ k

≤ C k ψ k (ψ ∈ H ).

We denote by h , i the duality V

- V and note that h ψ, v i = (ψ, v) if ψ, v ∈ H . In this paper we consider maximal regularity for second order Cauchy prob- lems. We focus on the damped wave equation. We consider a family of sesquilinear forms

b : [0, τ ] × V × V → C , such that

∗

Univ. Bordeaux, Institut de Mathématiques (IMB). CNRS UMR 5251. 351, Cours

de la Libération 33405 Talence, France. Mahdi.Achache@math.u-bordeaux.fr

• [H1]: D( b (t)) = V (constant form domain),

• [H2]: | b (t, u, v) | ≤ M k u k

k v k

( V -uniform boundedness),

• [H3]: Re b (t, u, u)+ν k u k

≥ δ k u k

( ∀ u ∈ V ) for some δ > 0 and some ν ∈ R (uniform quasi-coercivity).

We denote by B(t), B (t) the usual associated operators with b (t)(as oper- ators on H and V

). Let A (t) ∈ L ( V , V

) such that there exists a function h : [0, τ ] → [0, ∞ ) satisfies ^R

t

h(t)

< ∞ and

kA (t) k

≤ h(t), t ∈ [0, τ ].

We denote by A(t) the part of A (t) in H , with domain D(A(t)) := { u ∈ V : A (t)u ∈ H}

A(t)u := A (t)u.

Given a function f defined on [0, τ ] with values either in H or in V

and consider the second order evolution equation



 

 

u

(t) + B (t)u

(t) + A (t)u(t) = f (t) u(0) = u

, u(0) = ˙ u

.

(1.1) This is an abstract damped non-autonomous wave equation and our aim is to prove well-posedness and the maximal L

-regularity for p ∈ (1, ∞ ) in V

and in H .

Definition 1.1. Let X = H or V

. We said that Problem (1.1) has the maximal L

-regularity in X, if for all f ∈ L

(0, τ ; X) and all (u

, u

) in the trace space (see Sections 2 and 3 for more details) there exists a unique u ∈ W

(0, τ ; X) ∩ W

(0, τ ; V ) wich satisfies (1.1) in the L

-sense.

The maximal regularity in V

in the case where p = 2 was first considered by Lions [27] (p. 151). He assumes that A (t) is associated with a sesquilinear forms a (t) wich satisfies the same properties as b (t) and by assuming an ad- ditional regularity assumption on the forms t → a (t, u, v) and t → b (t, u, v) for every fixed u, v ∈ V . Dautray-Lions [17] (p.667) proved the maximal L

-regularity in V

without the regularity assumption as before by taking f ∈ L

(0, τ ; H ) and considering mainly symmetric forms. Recently, Batty, Chill, Strivastava [14] proved the maximal L

-regularity for general forms B (.) and A (.) for the case u

= u

= 0 and h ∈ L

(0, τ ) by reducing the problem to a first order non-autonomous Cauchy problem. Dier-Ouhabaz [20] proved the maximal L

-regularity in V

for u

∈ V , u

∈ H and A (t) is also associated with a V -bounded quasi-coercive non-autonomous form a (t).

We improve the result in [14], by proving the maximal L

-regularity in V

for u

and u

are not necessarily 0 and t → t

h(t) ∈ L

(0, τ ). Our proof is based on the result of the first order problem as in [14], but the main difference being that we use a fixed point argument.

More interesting is the question of second order maximal regularity in H , i.e. whether the solution u of (1.1) is in H

(0, τ ; H ) provided that f ∈ L

(0, τ ; H ). A first answer of this question was giving by Batty, Chill, Strivas- tava [14] in the particular case B (.) = k A (.) for some constant k and that A (.) has the maximal regularity in H . By using the form method, Dier and Ouhabaz [20], proved the maximal L

-regularity in H without the rather strong assumption B (.) = k A (.), but A (t) is also associated with V -bounded quasi-coercive form a (t) and t → a (t, u, v), b (t, u, v) are symmetric and Lip- schitz continuous for all u, v ∈ V . We extend the results in [20] in three directions. The first one is to consider general forms which may not be symmetric. The second direction is to deal with maximal L

-regularity, for all p ∈ (1, ∞ ). The third direction, which is our main motivation, is to as- sume less regularity on the operators A (t), B (t) with respect to t. Our main results can be summarized as follows (see Theorems 3.7 and 3.9 for more general and precise statements). For p ∈ (1, ∞ ) we assume the following

- | b (t, u, v) − b (s, u, v) | ≤ w( | t − s | ) k u k

k v k

, for all u, v ∈ V . - kA (t) − A (s) k

≤ w( | t − s | ).

Such that

Z

w(t)

t

dt < ∞ , (1.2)

- For p 6 = 2 or p = 2 with D(B(0)

) ֒ → V , we assume Z

0

w(t)

t

dt < ∞ . (1.3)

- In the case p = 2 but D(B(0)

) 6 ֒ → V , we assume Z

0

w(t)

t

dt < ∞ , (1.4)

for some ǫ > 0.

Here w : [0, τ ] → [0, ∞ ) is a non-decreasing function. We prove that for f ∈ L

(0, τ ; H ) and one of the following conditions holds

1- for p ≥ 2, u

∈ ( V , D(A(0)))

p,p

and u

∈ ( H , D(B(0)))

p,p

, 2- for p < 2, u

∈ V and u

∈ ( H , D(B (0)))

p,p

,

then (1.1) has the maximal L

-regularity in H . Assume in addition that D(B(t)

) = V for all t ∈ [0, τ ] and w(t) ≤ Ct

, for some ǫ > 0, then for all f ∈ L

(0, τ ; H ) and u

, u

∈ V , we prove that the solution u ∈ H

(0, τ ; H ) ∩ C

(0, τ ; V ).

By induction, we observe that our approach allows to consider Cauchy prob- lem of order N for any N ≥ 3.

Finally, using similar ideas as in [22] and [6], we give counterexamples where the maximal regularity may fails.

Notation. We denote by C, C

or c... all inessential positive constants.

Their values may change from line to line.

2 Maximal regularity for the damped wave equa- tion in V ^′

In this section we prove the maximal regularity in V

for the Problem (1.1), we start by recalling a well-known result on maximal regularity for the first order non autonomous problem.

Following [7], we introduce the following definition

Definition 2.1. Let ( a (t))

be a family of V -bounded, sesquilinear forms.

A function t → a (t) is called relatively continuous if for each t ∈ [0, τ ] and all ǫ > 0 there exists α > 0, β ≥ 0 such that for all u, v ∈ V , s ∈ [0, τ ], | t − s | ≤ α implies that

| a (t, u, v) − a (s, u, v) | ≤ (ǫ k u k

+ β k u k

) k v k

.

Theorem 2.2. Let p ∈ (1, ∞ ). We assume one of the following conditions - for p = 2, t → b (t) is measurable .

- for p 6 = 2, t → b (t) is piecewise relatively continuous.

Then for all u

∈ ( V

, V )

p,p

and f ∈ L

(0, τ ; V

) there exists a unique solution v ∈ M R

( V , V

) = W

(0, τ ; V

) ∩ L

(0, τ ; V ) of the problem



 

 

v

(t) + B (t)v(t) = f (t) v(0) = u

.

(2.1)

In addition, there exists a positive constant C such that k v k

≤ C[ k u

k

p ,p

+ k f k

].

Here, k v k

= k v k

+ k v k

.

Proof. For the case p = 2 the result is due to Lions [27]. Now, since A (s)+ ν is a generator of an analytic semigroup in V

for all s ∈ [0, τ ] we then apply [7] (Theorem 2.7) to get the desired result for p 6 = 2.

From [3] (Theorem III 4.10.2) we have the following lemma

Lemma 2.3. Let E

, E

be two Banach spaces such that E

⊆ E

. Then W

(0, τ ; E

) ∩ L

(0, τ, E

) ֒ → C([0, τ ]; (E

, E

)

p,p

).

Let us consider the space

M R

( V , V , V

) = W

(0, τ ; V

) ∩ W

(0, τ ; V ), with norm

k u k

= k u

k

+ k u

k

.

Let v ∈ M R

( V , V , V

) be a solution of (2.1). For u

∈ V , we set w(t) = u

+ ^R

v(s)ds. Then w

(t) = v(t) and



 

 

w

(t) + B (t)w

(t) = f (t) w(0) = u

, w

(0) = u

.

(2.2) Moreover, we have the following estimate

k w k

≤ C[ k u

k

+ k u

k

+ k f k

]. (2.3) We note also that the solution of the Problem (2.2) is unique. Indeed, we suppose there are two solutions v

, v

, then v = v

− v

is a solution of the problem



 

 

v

(t) + B (t)v

(t) = 0 v(0) = 0, v

(0) = 0.

(2.4) Therefore, for t > 0 we get

2Re ^Z

0

h v

(s), v

(s) i ds + 2Re ^Z

0

hB (s)v

(s), v

(s) i ds = 0.

Then by using [17] (Theorem 1, p. 473) we obtain k v

(t) k

+ 2δ ^Z

0

k v

(s) k

ds ≤ ν Z

0

k v

(s) k

ds.

By Gronwall’s lemma we get v(t) = 0 for all t ∈ [0, τ ] and so v

(t) = v

(t).

Using Lemma 2.3 and the Sobolev embedding we have M R

( V , V , V

) ֒ → C

([0, τ ]; ( V

, V )

p,p

) ∩ C

([0, τ ]; V ). (2.5)

We define the associated trace space to M R

( V , V , V

) by T R

( V , V

) = { (u(0), u

(0)) : u ∈ M R

( V , V , V

) } , endowed with norm

k (u(0), u

(0)) k

= inf {k v k

: v(0) = u(0), v(0) = ˙ ˙ u(0) } . Proposition 2.4. For all p ∈ (1, ∞ ) we have

T R

( V , V

) = V × ( V

, V )

p,p

with equivalent norms.

Proof. The first injection T R

( V , V

) ֒ → V × ( V

, V )

p,p

is obtained by (2.5). For the second injection " ← ֓" let us take u

∈ ( V

, V )

p,p

. Then by [28] (Corollary 1.14) there exists w ∈ M R

( V , V

) such that w(0) = u

. We set u(t) = u

+ ^R

w(s)ds, where u

∈ V . Thus, u ∈ M R

( V , V , V

) and

k u k

≤ C

[ k u

k

+ k w k

].

We note that the trace space associated to M R

( V , V

) is isomorphic to the real interpolation space ( V

, V )

p,p

(see [29], Chapter 1). Then inf {k u k

: u(0) = u

, u

(0) = u

}

≤ C

[ k u

k

+ inf {k w k

: w(0) = u

} ]

≤ C[ k u

k

+ k u

k

].

Thus, V × ( V

, V )

p,p

֒ → T R

( V , V

).

Remark 2.5. From the previous proposition and (2.5), we can deduce that the operator

M R

( V , V , V

) → C([0, τ ]; T R

( V , V

)) u → (u, u

)

is well defined and bounded.

Our main result on maximal L

-regularity in V

is the following

Theorem 2.6. Let p ∈ (1, ∞ ). We assume one of the following conditions - for p = 2, t → b (t) is measurable.

- for p 6 = 2, t → b (t) is piecewise relatively continuous.

Let A (t) ∈ L ( V , V

) for all t ∈ [0, τ ] such that kA (t) k

≤ h(t) and R

0

t

h(t)

< ∞ . Then for all f ∈ L

(0, τ ; V

) and (u

, u

) ∈ T R

( V , V

), there exists a unique solution u ∈ M R

( V , V , V

) of the problem



 

 

u

(t) + B (t)u

(t) + A (t)u(t) = f (t) u(0) = u

, u

(0) = u

.

(2.6) Moreover, there exists positive constant C independent of u

, u

and f such that the following estimate holds

k u k

≤ C[ k (u

, u

) k

+ k f k

]. (2.7) As mentioned in the introduction, this theorem was proved by Batty, Chill and Srivastava [14] but they consider only the case u

= u

= 0 and suppose t → kA (t) k

∈ L

(0, τ ).

Proof. We introduce the subspace

M R

( V , V , V

) = { u ∈ M R

( V , V , V

) : u(0) = 0, u

(0) = 0 } .

We equip this subspace with the norm u → k u

k

+ k u

k

. To prove the existence and the uniqueness of the solution we use the contraction fixed point theorem and the existence of a solution in M R

( V , V , V

) for the Problem (2.2). Indeed, let z ∈ M R

( V , V , V

) and u ∈ M R

( V , V , V

) be the solution of the problem



 

 

u

(t) + B (t)u

(t) = f(t) − A (t)z(t) u(0) = 0, u

(0) = 0

(2.8) for a given f ∈ L

(0, τ ; V

). Let us consider the operator L : z → u. Then L : M R

( V , V , V

) → M R

( V , V , V

). Now, let z

, z

∈ M R

( V , V , V

) and u

= Lz

, u

= Lz

. We set u = u

− u

, w = z

− z

. Hence, u is the solution of the problem



 

 

u

(t) + B (t)u

(t) = −A (t)w(t) u(0) = 0, u

(0) = 0.

(2.9) Therefore, by (2.3) we get

k Lz

− Lz

k

0(V,V,V^′)

≤ C kA (.)w k

≤ C Z

0

h(t)

k w(t) k

dt

≤ C Z

0

h(t)

t

dt k w k

C^1−1p([0,τ];V)

≤ C Z

0

h(t)

t

dt k z

− z

k

0(V,V,V^′)

.

We choose τ small enough such that C ^R

h(t)

t

dt < 1. Thus, L is a contraction operator. So by the contraction fixed point theorem, there exists a unique solution of the problem



 

 

u

(t) + B (t)u

(t) + A (t)u(t) = f (t) u(0) = 0, u(0) = 0 ˙

(2.10) for all f ∈ L

(0, τ ; V

) and τ > 0 small enough and we get from the estimate (2.3)

k u k

≤ C k f k

. Now, let u

, u

∈ V × ( V

, V )

p,p

. Since by Proposition 2.4 T R

( V , V

) = V × ( V

, V )

p,p

, then there exists v ∈ M R

( V , V , V

) (with minimal norm) such that v(0) = u

and v

(0) = u

. We set z = u + v. Thus, z belongs to M R

( V , V , V

) and satisfies



 

 

z

(t) + B (t)z

(t) + A (t)z(t) = g(t) z(0) = u

, z

(0) = u

,

(2.11) with g = f + v

(t) + B (t)v

(t) + A (t)v(t) ∈ L

(0, τ ; V

). We get

k z k

≤ C

[ k g k

+ k (u

, u

) k

].

This completes the proof when τ is small enough. We note that by Remark 2.5 (z, z

) ∈ C([0, τ ]; T R

( V , V

)). For arbitrary τ > 0, we split [0, τ ] into a finite number of subintervals with small sizes and proceed exactly as in the previous proof. Finally, we stick the solutions and we get the desired result.

3 Maximal regularity for the damped wave equa- tion in H

Let A (t) and B (t) be as before and we assume that kA (t) k

≤ M, for all t ∈ [0, τ ]. Let us define the spaces

M R(p, H ) = { u ∈ W

(0, τ ; H ) ∩ W

(0, τ ; V ) s.t B (.)u

+ A (.)u ∈ L

(0, τ ; H ) } .

T r(p, H ) = { (u(0), u(0)) s.t : ˙ u ∈ M R(p, H ) } , endowed with norms respectively

k u k

= k u

k

+ k u k

+ kB (.)u

(.) + A (.)u(.) k

.

k (u(0), u

(0)) k

= inf {k v k

s.t :

v ∈ M R(p, H ), v(0) = u(0), v

(0) = u

(0) } . The maximal L

-regularity in H for the Problem (2.6) consists of proving ex- istence and uniqueness of a solution u ∈ M R(p, H ) provided f ∈ L

(0, τ ; H ) and (u

, u

) ∈ T r(p, H ).

3.1 Preparatory lemmas

Proposition 3.1. The maximal regularity on H of the problem



 

 

v

(t) + B (t)v

(t) + A (t)v(t) = f(t) v(0) = u

, v

(0) = u

+ γu

(3.1)

is equivalent to the maximal regularity of the problem



 

 

u

(t) + ( B (t) + 2γ )u

(t) + ( A (t) + γ B (t) + γ

)u(t) = e

f (t) u(0) = u

, u

(0) = u

(3.2) for all γ ∈ C .

Proof. Let v be the solution of (3.1) and γ ∈ C . We set u(t) = e

v(t).

By a simple computation we obtain that u satisfies (3.2). In addition, f ∈ L

(0, τ ; H ) if and only if t → e

f (t) ∈ L

(0, τ ; H ) and it is clear that v ∈ W

(0, τ ; H ) ∩ W

(0, τ ; V ) if and only if u ∈ W

(0, τ ; H ) ∩ W

(0, τ ; V ).

We deduce that we may replace B (t) by B (t) + γ. Therefore, by adding a large constant c from the previous proposition we may suppose without loss of generality that [H3] holds with ν = 0.

We note that for γ > 0 is big enough (γ > max {

, ^p | ν |} ) and for t ∈ [0, τ ], we have that C (t) = A (t)+γ B (t)+ γ

is associated with V -bounded coercive form c (t). In fact, let u ∈ V . We get

Re c (t, u, u) = Re hA (t)u, u i + γRe b (t, u, u) + γ

k u k

≥ − M k u k

+ γδ k u k

+ (γ

− ν) k u k

≥ (γδ − M ) k u k

.

We denote by S

the open sector S

= { z ∈ C

: | arg(z) | < θ } with vertex 0.

Lemma 3.2. For any t ∈ [0, τ ], the operators − B (t) and −B (t) generate

strongly continuous analytic semigroups of angle γ =

− arctan(

) on H

and V

,respectively. In addition, there exist constants C and C

, independent

of t, such that

1- k e

k

≤ 1 and k e

k

≤ C for all z ∈ S

.

2- k B(t)e

k

≤

and kB (t)e

k

≤

for all s ∈ R . 3- k e

k

≤

.

4- k (z − B(t))

k

≤ √

|z|

and k (z − B (t))

k

≤ √

|z|

for all z / ∈ S

with fixed θ > γ.

5- k (z − B (t))

k

≤

(1+|z|)^1+β2

for all β ∈ [0, 1], z / ∈ S

and p ∈ (1, ∞ ).

6- All the previous estimates hold for B (t)+α with constants independent of α for α > 0.

Proof. For assertions 1, 2, 3 and 4, 6 we refer to [24] (Proposition 2.1). For 5 we use the interpolation.

For p ∈ (1, ∞ ) and f ∈ L

(0, τ ; H ), we define the operator L by L(f )(t) = B(t) ^Z

0

e

f(s)ds, with t ∈ [0, τ ].

Proposition 3.3. Let p ∈ (1, ∞ ). We suppose that kB (t) − B (s) k

≤ w(t − s), where w : [0, τ ] → [0, ∞ ) is a non-decreasing function such that

Z

w(t)

t dt < ∞ . Then the operator L is bounded on L

(0, τ ; H ).

Proposition 3.3 follows from [24] (Lemmas 2.5 and 2.6).

Let p ∈ (1, ∞ ). We introduce the following assumptions

- for p 6 = 2 : t → b (t) is relatively continuous and for p = 2 : t → b (t) is measurable.

- | b (t, u, v) − b (s, u, v) | ≤ w( | t − s | ) k u k

k v k

for all u, v ∈ V . - kA (t) − A (s) k

≤ w( | t − s | ),

where w : [0, τ ] → [0, ∞ ) is a non-decreasing function satisfying Z

0

w(t)

t

dt < ∞ . (3.3)

We assume also that

- for p 6 = 2 (or p = 2 with D(B(0)

) ֒ → V ) Z

0

w(t)

t

dt < ∞ . (3.4)

- In the case where p = 2, but D(B(0)

) 6 ֒ → V Z

0

w(t)

t

dt < ∞ , (3.5)

for arbitrary small ǫ > 0.

Let γ > 0 be big enough such that C (t) = A (t) + γ B (t) + γ

is associated with a V -bounded coercive forms c (t). We denote by C(t) the part of C (t) on H .

It is clear that | c (t, u, v) − c (s, u, v) | ≤ (1 + γ )w( | t − s | ) k u k

k v k

for all u, v ∈ V .

We set X

= V for all p ∈ (1, 2[ and X

= ( V , D(C(0)))

p,p

for p ≥ 2.

In the following lemma B = B(0).

Lemma 3.4. We have

(1)- V ֒ → D(B

) if and only if D(B

) ֒ → V . (2)- If B = B

, we get D(B

) = D(B

) = V and

√ δ k u k

≤ k B

u k ≤ √

M k u k

. (3)- D(B

) = [ H , V ]

for all α <

.

(4)- D(B

) ֒ → V for all α <

.

Proof. Let u ∈ D(B

). If V ֒ → D(B

) we get k u k

≤ 1

δ Re (B

u, B

u)

≤ 1

δ k B

u kk B

u k

≤ C k u k

k B

u k . Then by the density of D(B

) on D(B

) we obtain

k u k

≤ C k B

u k for all u ∈ D(B

). Then D(B

) ֒ → V .

Now, we assume that D(B

) ֒ → V . It follows that B

∈ L ( H , V ). Let x ∈ H and we write B

x = A

A

x. Then we get

kB

x k

≤ kB

k

k B

x k

≤ M k B

k

k x k .

The boundedness implies B

∈ L ( H , V

) and by duality we have B

∈ L ( V , H ). Then V ⊆ D(B

) and we get for all x ∈ V

k x k

D(B¹²)

= k x k

+ k B

x k

≤ (C

+ k B

k

) k x k

. Thus, V ֒ → D(B

). This shows (1).

We assume that B = B

. By the density of D(B) in V and D(B

), we get for all u ∈ V

δ k u k

≤ Re b (0, u, u)

= k B

u k

≤ M k u k

. This shows (2).

For (3), we refer to [26] (Theorem 3.1).

Let α <

and u ∈ D(B ). We have k u k

≤ 1

δ k B

u kk B

u k

≤ 1

δ k B

u kk u k

≤ C(α)

δ k B

u kk u k

k u k

,

where C(α) > 0 depending on α. Thus, for all u ∈ D(B

) we get k u k

≤ C

C(α)

δ k B

u k . This shows (4).

Lemma 3.5. Let u

∈ ( H , D(B(0)))

p,p

and u

∈ X

, then the operators R

u

(t) = B(t)e

u

R

u

(t) = e

C (t)u

are bounded from ( H , D(B(0)))

p,p

and X

into L

(0, τ ; H ), respectively.

Remark 3.6. We note that the operator R

is already studied in [1] (Theo-

rem 2.2). Here we assume less regularity on the operators B (t) with respect

to t compared with [1] (Theorem 2.2).

Proof. First, we note that in case p < 2 we have ( H , D(B(0)))

p,p

= ( H , V )

p),p

(see [23] (p. 5)). Then by Lemma 3.2 k (λ − B (0))

k

p ,p;V)

≤ C

| λ |

. In case p > 2, the embedding ( H , D(B (0)))

p,p

֒ → V holds. In fact, we use the inclusion properties of the real interpolation spaces ([28], Proposition 1.1.4) to obtain

( H , D(B(0)))

p,p

= ( H , D(B(0)))

2+[¹₂−¹_p],p

֒ → ( H , D(B(0)))

2+[¹₂−¹p]−ǫ,2

= D(B (0)

),

with ǫ < [

−

]. Since by Proposition 3.4 D(B(0)

) ֒ → V , it follows that

k (λ − B (0))

k

p ,p;V)

≤ C

| λ | . Now, if D(B(0)

) ֒ → V we get

k (λ − B (0))

u

k

≤ C

k B(0)

k

L(H,D(B(0)¹²))

k (λ − B (0))

B(0)

u

k

≤ C 1

| λ | k u

k

.

For the other case (D(B (0)

) 6 ֒ → V ), since D(B (0)

) ֒ → V for all ǫ > 0 (see Proposition 3.4) then

k (λ − B (0))

u

k

≤ C

k B(0)

k

L(H,D(B(0)^1+ǫ2 ))

k B(0)

(λ − B (0))

B(0)

u

k

≤ C 1

| λ |

k u

k

. We write

R

u

(t) = (B(t)e

− B(0)e

)u

+ B(0)e

u

.

Choose a contour Γ in the positive half-plane we write by the holomorphic functional calculus

B(t)e

− B(0)e

= 1 2πi

Z

Γ

λe

(λ − B(t))

( B (t) −B (0))(λ − B (0))

dλ.

Therefore

k (B(t)e

− B(0)e

)u

k

≤ C Z

0

| λ | e

k (λ − B (0))

k

× k (λ − B (t))

k

d | λ |kB (t) − B (0) k

× k u

k

. Then

- for p 6 = 2 or p = 2 with D(B(0)

) ֒ → V , we have k (B(t)e

− B(0)e

)u

k ≤ C w(t)

t

k u

k

.

- for p = 2 and D(B(0)

) 6 ֒ → V , we get

k (B(t)e

− B(0)e

)u

k ≤ C w(t)

t

k u

k

.

Using the definition of the interpolation space of the domain for a sectorial operator (see [28]), we obtain

Z

0

k B (0)e

u

k

dt ≤ k u

k

. Then

- for p 6 = 2 (or p = 2 with D(B(0)

) ֒ → V ), we have k R

u

k

≤ C(( ^Z

0

w(t)

t

dt)

+ 1) k u

k

< ∞ .

- if p = 2 and D(B(0)

) 6 ֒ → V , we have k R

u

k

≤ C ( ^Z

0

w(t)

t

dt)

+ 1

!

k u

k

.

For R

with p < 2, we get

k e

C (t)u

k ≤ k e

k

kC (t)u

k

≤ C

√ t k u

k

.

Therefore k R

u

k

≤ C k u

k

. Now for p ≥ 2, we write

R

u

(t) = B (t)e

B (t)

C (t)u

= B (t)e

B (t)

( C (t) − C (0))u

+ B(t)e

( B (t)

− B (0)

) C (0)u

+ B(t)e

B (0)

C (0)u

= I

(t) + I

(t) + I

(t).

For i = 1 or i = 2, we have the following estimate k I

(t) k ≤ Cw(t) k e

k

k u

k

≤ C

w(t)

√ t k u

k

. Then

k I

k

≤ C

( ^Z

0

w(t)

t

dt)

k u

k

. We note that

C (0) :

( D(C(0)) → H V → V

. Then

D( C (0) |

) = ( V , D(C(0)))

, where C (0) |

is the part of C (0) on ( V

, H )

. Thus,

B (0)

C (0) ∈ L (( V , D(C(0)))

p,p

; ( V , D(B(0))

).

Observing that for t ∈ [0, τ ], I

(t) = R

B (0)

C (0)u

(t). Using the first part of the proposition, we obtain

k I

k

≤ C k u

k

. This shows that t → R

u

(t) ∈ L

(0, τ ; H ).

3.2 The Main Result

Our main result in this chapter is the following

Theorem 3.7. Let f ∈ L

(0, τ ; H ) and u

∈ ( H , D(B(0)))

p,p

, u

∈ X

, with p ∈ (1, ∞ ). Then the problem



 

 

v

(t) + B (t)v

(t) + A (t)v(t) = f(t) v(0) = u

, v

(0) = u

+ γu

(3.6)

has the maximal L

-regularity in H . In addition, there exists a positive con- stant C such that

k v k

≤ C k u

k

+ k u

k

+ k f k

. Proof. Let f ∈ L

(0, τ ; H ) and (u

, u

) ∈ ( V × ( V

, V )

p,p

). By Theorem 2.6 there exists a unique solution v ∈ W

(0, τ ; V

) ∩ W

(0, τ ; V ) of the Problem (3.6). Then by Proposition 3.1, there exists a unqiue solution u ∈ W

(0, τ ; V

) ∩ W

(0, τ ; V ) of the Problem (3.2).

To simplify the writing, we replace B (t) + γ by B (t).

Fix 0 ≤ t ≤ τ. We get from the equation in (3.2) B (t) ^Z

0

e

u

(s)ds + B (t) ^Z

0

e

B (s)u

(s)ds + B (t) ^Z

0

e

C (s)u(s)ds = B (t) ^Z

0

e

g(s)ds.

Here g(s) = e

f (s) for all s ∈ [0, τ ]. Hence,

B (t) ^Z

0

e

u

(s)ds + B (t) ^Z

0

e

B (s)u

(s)ds + B (t) ^Z

0

e

( C (s) − C (t))u(s)ds + B (t) ^Z

0

e

C (t)u(s)ds

= B (t) ^Z

0

e

g(s)ds. (3.7)

By performing an integration by parts, we have B (t) ^Z

0

e

u

(s)ds = B (t)u

(t) − B(t)e

u

(0)

− B (t) ^Z

0

e

B (t)u

(s)ds (3.8) and

B (t) ^Z

0

e

C (t)u(s)ds = C (t)u(t) − e

C (t)u(0)

− Z

0

e

C (t)u

(s)ds. (3.9)

Combining (3.8), (3.9) and (3.7) we obtain B (t)u

(t) − B(t)e

u

− B (t) ^Z

0

e

( B (t) − B (s))u

(s)ds + B (t) ^Z

0

e

( C (s) − C (t))u(s)ds + C (t)u(t) − e

C (t)u

−

Z

e

C (t)u

(s)ds

= B(t) ^Z

0

e

g(s)ds. (3.10)

Therefore

B (t)u

(t) + C (t)u(t) = B(t)e

u

+ e

C (t)u

+ B (t) ^Z

0

e

( B (t) − B (s)) B (s)

( B (s)u

(s) + C (s)u(s))ds

− B (t) ^Z

0

e

( B (t) − B (s)) B (s)

C (s)u(s)ds + ^Z

0

e

C (t)u

(s)ds + B (t) ^Z

0

e

( C (t) − C (s))u(s)ds + B(t) ^Z

0

e

g(s)ds.

This allows us to write

(I − Q)( B (.)u

+ C (.)u)(t) = + B (t)e

u

+ e

C (t)u

− B (t) ^Z

0

e

( B (t) − B (s)) B (s)

C (s)u(s)ds + B (t) ^Z

0

e

( C (t) − C (s))u(s)ds + ^Z

0

e

C (s)u

(s)ds + B (t) ^Z

0

e

g(s)ds, (3.11)

where

(Qh)(t) := B (t) ^Z

0

e

( B (t) − B (s))B(s)

h(s)ds.

Then, if I − Q is invertible on L

(0, τ ; H ) we obtain B (t)u

(t) + C (t)u(t)

= (I − Q)

[R

u

+ R

u

+ W

(u) + L(f ) + L

(u) + W

(u)](t), (3.12)