Escher J., Guidotti P., Hieber M. et al. (editors) Parabolic Problems: The Herbert Amann Festschrift

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488 T. Nau and J. Saal

Deﬁnition 3.8. Let F be a Banach space, G ⊂ R

, m ∈ N,anda

∈L(F ). The

L(F )-valued homogeneous polynomial

a(ξ):=

|α|=2m

(ξ ∈ R

)

is called parameter-elliptic, if there exists an angle φ ∈ [0,π) such that the spectrum

σ(a(ξ)) of a(ξ)inL(F )satisﬁes

σ(a(ξ)) ⊂ Σ

(ξ ∈ R

, |ξ| =1). (3.5)

Then

ϕ := inf{φ : (3.5) holds}

is called angle of ellipticity of a.

A diﬀerential operator A(x, D):=

|α|≤2m

(x)D

with coeﬃcients a

: G →L(F)

is called parameter-elliptic in G with angle of ellipticity ϕ, if the principal part of

its symbol

(x, ξ):=

|α|=2m

(x)ξ

is parameter-elliptic with this angle of ellipticity for all x ∈ G.

Deﬁnition 3.9. Let F be a Banach space and let G ⊂ R

be a C

-domain. Let

: G →L(F )andb

j,β

: ∂G →L(F ). Set B

(x, D):=

|β|=m

β,j

(x)D

and

let A

(x, D):=

|α|=2m

(x)D

be parameter-elliptic in G of angle of ellipticity

ϕ ∈ [0,π). For each x

∈ ∂G we write the boundary value problem in local

coordinates about x

. The boundary value problem (3.4) is said to satisfy the

Lopatinskii-Shapiro condition,ifforeachφ>ϕthe ODE on R

(λ + A

,ξ



))v(x

)=0,x

> 0,

,ξ



)v(0) = h

,j=1,...,m,

v(x

) → 0,x

→∞,

has a unique solution v ∈ C((0, ∞),F)foreach(h

,...,h

)

∈ F

and each

λ ∈ Σ

π−φ

and ξ



∈ R

n−1

with |ξ



| + |λ| =0.

We refer to [22] for an introduction to the Lopatinskii-Shapiro condition for

scalar-valued boundary value problems and to [11] for an extensive treatment of

the F -valued case. Parameter-ellipticity of a boundary value problem now reads

as follows.

Deﬁnition 3.10. The boundary value problem (A, B) given through (3.4) is called

parameter-elliptic in G of angle ϕ ∈ [0,π), if A(·,D) is parameter-elliptic in G of

angle ϕ ∈ [0,π) and if the Lopatinskii-Shapiro condition holds. To indicate that ϕ

is the angle of ellipticity of the boundary value problem (A, B) we use the subscript

notation ϕ

(A,B)

R-sectoriality of Cylindrical Boundary Value Problems 489

4. Proof of the main result

We denote by

D(A

):={u ∈ W

2m,p

(V,F); B

2,j

(·,D)u =0(j =1,...,m)}

u := A

(·,D)u, u ∈ D(A

the L

(V,F)-realization of the induced boundary value problem

λu + A

,D)u = f in V,

2,j

,D)u =0 on∂V (j =1,...,m),

(4.1)

on the cross-section V of Ω. As the original boundary value problem (2.1) is as-

sumed to be parameter-elliptic with ellipticity angle ϕ

(A,B)

∈ [0,π/2), it is easy

to see that the same is valid for the boundary value problem (4.1) and that the

corresponding angle ϕ

)

is no larger than ϕ

(A,B)

. By employing ﬁnite open

coverings of V , in [11] the following result is proved.

Proposition 4.1. Let V ⊂ R

be a standard domain of class C

. Given the as-

sumptions (2.2) on the coeﬃcients a

and b

,foreachφ>ϕ

)

there exists a

= δ

(φ) ≥ 0 such that A

+ δ

∈RS(L

(V,F)) with φ

+δ

≤ φ.Moreover,we

have

R({λ

1−

|γ|

(λ + A

+ δ

)

−1

; λ ∈ Σ

π−φ

, 0 ≤|γ|≤2m}) < ∞. (4.2)

Remark 4.2. In [11] just the case k ≥ 2 is treated, whereas the case k = 1 is well

known.

From the deﬁnition it is clear that the coeﬃcients of the cylindrical parts

and A

of A only depend on x

or x

, respectively. For the sake of simplicity

we therefore drop the special indications for x, if no confusion seems likely. To be

precise we write

,D)=A

(x, D)=

|α|≤2m

(x)D

for i =1, 2, where

(x)=



0 ,α

=0,

),α

=0,

(x)=



0 ,α

=0,

),α

=0.

Further we set E := L

(Ω,F)andX := L

n−k

,E)

∼

(Ω,F). Given an

operator T : D(T ) ⊂ E → E, its canonical extension is deﬁned by

T ):=L

n−k

,D(T ))

(

Tu)(x):=T (u(x)),u∈ D(

T ),x∈ R

n−k

490 T. Nau and J. Saal

4.1. Constant coeﬃcients a

In the ﬁrst step we consider the model problem for the cylindrical boundary value

problem (2.1), i.e., we assume A

(x, D)onR

n−k

to be given as homogeneous

diﬀerential operator

(D):=

|α|=2m

with constant coeﬃcients a

∈ C.LetA

denote its realization in X with domain

D(A

):=W

2m,p

n−k

,E). We set

(·,D):=A

(D)+A

(·,D)

and

:= A

,D(A

):=D(A

) ∩D(

Note that no further restrictions on A

(x, D) have to be assumed.

Since it will always be clear from the context what we mean, from now on

we do not distinguish between

and A

. In other words, from this point on we

drop again the tilde notation and just write A

for simplicity.

Let φ>ϕ

,B)

, λ ∈ Σ

π−φ

and u ∈S(R

n−k

,D(A

)) ⊂ D(A

). Applying

E-valued Fourier transform F to f := (λ + A

+ δ

)u gives us

(λ + a

(·)+A

+ δ

)Fu = Ff.

Hence we formally have

u = F

−1

Ff,

where m

is given by the operator-valued symbol

(ξ):=(λ + a

(ξ)+A

+ δ

)

−1

,ξ∈ R

n−k

Note that m

∈ C

∞

n−k

, L(E)) is well deﬁned if

−(λ + a

(ξ)) ∈ ρ(A

+ δ

)(ξ ∈ R

n−k

In view of ϕ

)

≤ ϕ

,B)

and Proposition 4.1 this is obviously satisﬁed in

case that λ + a

(ξ) ∈ Σ

π−φ

. This, however, follows directly from the parame-

ter-ellipticity of A

(D), which is obtained as an immediate consequence of the

parameter-ellipticity of (A

,B), and since the ellipticity angle ϕ

of A

fulﬁlls

≤ ϕ

,B)

<φ.

In order to obtain

(λ + A

+ δ

)

−1

= F

−1

Ff ∈L(X),

the idea is to apply the operator-valued multiplier result of Proposition 3.7 to m

For this purpose, we next establish suitable representation formulas for derivatives

of m

Lemma 4.3. Let φ>ϕ

,B)

. Given α ∈{0, 1}

n−k

,let

⎧

⎨

⎩

W =(ω

,...,ω

) ∈ ({0, 1}

n−k

)

; r ≤ n − k, ω

=0,

j=1

= α

⎫

⎬

⎭

R-sectoriality of Cylindrical Boundary Value Problems 491

denote the set of all additive decompositions of α into r = r

many positive

multiindices. Then, with C

:= (−1)

r!,theformula

(ξ)=(λ + a

(ξ)+A

+ δ

)

−1

W∈Z

⎛

⎝

j=1

(ξ)

⎞

⎠

(λ + a

(ξ)+A

+ δ

)

−r

holds for all λ ∈ Σ

π−φ

and ξ ∈ R

n−k

Proof. Let |α| = 1. Then there exists i ∈{1,...,n− k} such that α = e

.Inthis

case Z

= {(α)} and we get immediately

(ξ)=−ξ

)(ξ)(λ + a

(ξ)+A

+ δ)

−2

=(λ + a

(ξ)+A

+ δ

)

−1

(−1)ξ

)(ξ)(λ + a

(ξ)+A

+ δ)

−1

Now assume the statement to be true for α ∈{0, 1}

n−k

with |α| <n−k.

Then for l ∈{1,...,n− k} such that α

=0weobtain

(ξ)

= ξ

W∈Z

⎛

⎝

j=1

(ξ)

⎞

⎠

(λ + a

(ξ)+A

+ δ

)

−(1+r

)

= ξ

W∈Z

⎡

⎣

i=1

)(ξ)

⎛

⎝

j=i

(ξ)

⎞

⎠

(λ + a

(ξ)+A

+ δ

)

−(1+r

)

⎛

⎝

(ξ)

⎞

⎠

(−(1 + r

))(D

)(ξ)(λ + a

(ξ)+A

+ δ

)

−(2+r

)

⎤

⎦

=(λ + a

(ξ)+A

+ δ

)

−1

W∈Z

α+e

⎛

⎝

j=1

(ξ)

⎞

⎠

(λ + a

(ξ)+A

+ δ

)

−r

. 

Inthesequelwedenoteby(β,γ) ∈ N

n−k

× N

a multiindex such that β is

the part corresponding to the variables x

∈ R

n−k

and γ corresponding to the

variables x

∈ V . In order to obtain the general estimate (2.3) for the full operator

A, we also have to consider the more involved symbols

(ξ):=λ

1−

|β|+|γ|

(ξ)=λ

1−

|β|+|γ|

(λ + a

(ξ)+A

+ δ

)

−1

for λ ∈ Σ

π−φ

, ξ ∈ R

n−k

,and|β| + |γ|≤2m.

492 T. Nau and J. Saal

Lemma 4.4. Let φ>ϕ

,B)

.Forα ∈{0, 1}

n−k

we have

(ξ)=λ

1−

|β|+|γ|

(λ + a

(ξ)+A

+ δ

)

−1



≤α



,β

W∈Z

α−α



j=1



)(ξ)(λ + a

(ξ)+A

+ δ

)

−1



for all λ ∈ Σ

π−φ

, ξ ∈ R

n−k

,and(β, γ) ∈ N

n−k

×N

such that |β|+ |γ|≤2m,and

with certain constants C



,β

∈ Z.

Proof. We ﬁrst show

(ξ)=λ

1−

|β|+|γ|



≤α

(

i;α



=0

)ξ

α−α



α−α



(λ + a

(ξ)+A

+ δ

)

−1

(4.3)

Let α = e

for some i ∈{1,...,n− k}.Then

(ξ)

= λ

1−

|β|+|γ|



(λ + a

(ξ)+A

+ δ

)

−1

+ ξ

(λ + a

(ξ)+A

+ δ

)

−1



already proves (4.3) for the case |α| = 1. Assume the statement to be true for

α ∈{0, 1}

n−k

with |α| <n− k.Forl ∈{1,...,n− k} such that α

=0wehave

(ξ)

= λ

1−

|β|+|γ|



≤α

(

i;α



=0

)ξ

α−α



α−α



(λ + a

(ξ)+A

+ δ

)

−1

= λ

1−

|β|+|γ|

⎛

⎝



≤α

(

i;α



=0

)β

α−α



α−α



(λ + a

(ξ)+A

+ δ

)

−1



≤α

(

i;α



=0

)ξ

α+e

−α



α+e

−α



(λ + a

(ξ)+A

+ δ

)

−1

⎞

⎠

= λ

1−

|β|+|γ|

⎛

⎝



≤α+e

;α



(

i;α



=0

)ξ

α+e

−α



α+e

−α



(λ + a

(ξ)+A

+ δ

)

−1



≤α+e

;α



(

i;α



=0

)ξ

α+e

−α



α+e

−α



(λ + a

(ξ)+A

+ δ

)

−1

⎞

⎠

= λ

1−

|β|+|γ|



≤α+e

(

i;α



=0

)ξ

α−α



α−α



(λ + a

(ξ)+A

+ δ

)

−1

R-sectoriality of Cylindrical Boundary Value Problems 493

This proves (4.3). Setting C



,β

i;α



=0

and applying Lemma 4.3 now yields

(ξ)=λ

1−

|β|+|γ|



≤α



,β

W∈Z

α−α



⎛

⎝

j=1

(ξ)

⎞

⎠

(λ + a

(ξ)+A

+ δ

)

−(r+1)

= λ

1−

|β|+|γ|

(λ + a

(ξ)+A

+ δ

)

−1



≤α



,β

W∈Z

α−α



j=1



)(ξ)(λ + a

(ξ)+A

+ δ

)

−1



This proves the assertion. 

With the above formulas at hand we can prove R-sectoriality for the model

problem.

Proposition 4.5. For each φ>ϕ

,B)

we have A

+ δ

∈RS(X) with δ

= δ

(φ)

as in Proposition 4.1. Moreover, φ

+δ

≤ φ and it holds that

R({λ

1−

|β|+|γ|

(λ + A

+ δ

)

−1

; λ ∈ Σ

π−φ

, 0 ≤|β|+ |γ|≤2m}) < ∞. (4.4)

Furthermore, the domain of A

is given as

D(A

)=L

n−k

,D(A

)) ∩

j=1

j,p

n−k

2m−j,p

(V,F)).

Proof. Let φ>ϕ

,B)

. We show that m

fulﬁlls the assumptions of the multiplier

result Proposition 3.7, i.e., that

R({ξ

(ξ); ξ ∈ R

n−k

,λ∈ Σ

π−φ

,α∈{0, 1}

n−k

}) < ∞.

As R-boundedness by virtue of Lemma 3.5 is preserved under summation and

composition of R-bounded operator families, it suﬃces by Lemma 4.4 to prove

that

R({λ

1−

|β|+|γ|

(λ + a

(ξ)+A

+ δ

)

−1

;

ξ ∈ R

n−k

,λ∈ Σ

π−φ

, 0 ≤|β| + |γ|≤2m}) < ∞

and that

R({ξ

)(ξ)(λ + a

(ξ)+A

+ δ

)

−1

;

ξ ∈ R

n−k

,λ∈ Σ

π−φ

,α∈{0, 1}

n−k

}) < ∞.

494 T. Nau and J. Saal

Thanks to (4.2) this follows by the contraction principle of Kahane if we can show

that both

(λ, ξ):=

1−

|β|+|γ|

(λ + a

(ξ))

1−

|γ|

and κ

(λ, ξ):=

(ξ)

λ + a

(ξ)

are uniformly bounded in (λ, ξ) ∈ Σ

π−φ

×R

n−k

.Toseethis,weﬁrstobservethat

λ, sξ)=κ

(λ, ξ)(s>0),

hence that κ

is quasi-homogeneous of degree zero. We set

K := {(λ, ξ) ∈ Σ

π−φ

× R

n−k

: |λ| + |ξ|

=1}. (4.5)

By the ellipticity condition, we obtain a

(ξ) ∈ Σ

for all ξ ∈ R

n−k

\{0}.Since

<φ, it therefore easily follows that

λ + a

(ξ) =0 onK.

Consequently, κ

is a continuous function on the compact set K and we obtain

|κ

(λ, ξ)|≤M ((λ, ξ) ∈ K).

By the quasi-homogeneity of κ

this implies

|κ

λ, sξ)|≤M ((λ, ξ) ∈ K, s > 0).

We have

λ| + |sξ|

= s

(|λ| + |ξ|

Thus, if we set s =(|λ| + |ξ|

)

−1/2m

we deduce

λ, sξ) ∈ K ((λ, ξ) ∈ Σ

π−φ

× R

n−k

)

and therefore that

|κ

(λ, ξ)| = |κ

λ, sξ)|≤M ((λ, ξ) ∈ Σ

π−φ

× R

n−k

The uniform boundedness of κ

canbeprovedinexactlythesameway.Byapplying

Proposition 3.7, relation (4.4) follows.

In particular, we have

(λ + A

+ δ

)

−1

= F

−1

Ff ∈L(X)

and

D(A

) ⊂

j=1

j,p

n−k

2m−j,p

(V,F)).

Furthermore, we can represent the resolvent applied to f ∈S(R

n−k

,E)asa

Bochner integral via

(λ + A

+ δ

)

−1

f(x

(2π)

(n−k)/2



n−k

·ξ

(λ + a

(ξ)+A

+ δ

)

−1

Ff(ξ)dξ.

Since taking the trace acts as a bounded operator on E,itcommuteswiththe

integral sign. This yields

2,j

(λ + A

+ δ

)

−1

f =0 (f ∈S(R

n−k

,E)).

R-sectoriality of Cylindrical Boundary Value Problems 495

Employing a density argument we conclude that

D(A

)=L

n−k

,D(A

)) ∩

j=1

j,p

n−k

2m−j,p

(V,F)).

Assuming that (A

+ δ

)u =0foru ∈ D(A

)nextimpliesthat

(ξ)+A

+ δ

)Fu(ξ)=0 (ξ ∈ R

n−k

Since A

+ δ

is sectorial and a

parameter-elliptic this yields Fu = 0, hence

u = 0. By permanence properties for sectorial operators, i.e., in this case for

+ δ

, we obtain that the same is true for the dual operator of A

+ δ

.This

implies that A

+ δ

is injective and has dense range. Hence we have proved that

+ δ

∈RS(X). 

4.2. Slightly varying coeﬃcients a

By a perturbation argument in this paragraph we generalize the R-sectoriality

for constant coeﬃcients to the case of slightly varying coeﬃcients of A

.Tothis

end, we will employ the following perturbation result which is based on a standard

Neumann series argument.

Lemma 4.6. Let R be a linear operator in X such that D(A

) ⊂ D(R) and let δ

be given as in Proposition 4.5. Assume that there are η>0 and δ>δ

such that

Rx

≤ η(A

+ δ)x

(x ∈ D(A)).

Then A

+ R + δ ∈RS(X), φ

+R+δ

≤ φ

+δ

, and for every φ>ϕ

,B)

have

R({λ



(λ+A

+R+ δ)

−1

; λ ∈ Σ

π−φ

, 0 ≤ +|β|+|γ|≤2m}) < ∞, (4.6)

whenever η<R({(A

+ δ)(λ + A

+ δ)

−1

})

−1

Proof. As

R(λ + A

+ δ)

−1



L(X)

≤ η(A

+ δ)(λ + A

+ δ)

−1



L(X)

≤ ηR({(A

+ δ)(λ + A

+ δ)

−1

})

< 1

by assumption, we see that

λ + A

+ R + δ =



1+R(λ + A

+ δ)

−1



(λ + A

+ δ)

is invertible. This implies



(λ + A

+ R + δ)

−1

= λ



(λ + A

+ δ)

−1

∞

j=0

(−R(λ + A

+ δ)

−1

)

By assumption we have δ

:= δ −δ

> 0. The fact that

|λ + δ

|≥c

(λ ∈ Σ

π−φ

)

496 T. Nau and J. Saal

for some c

> 0 yields the existence of a M

> 0 such that

|λ

/2m

|(λ + δ

)

1−(|β|+|γ|)/2m

≤ M

(λ ∈ Σ

π−φ

Thanks to the contraction principle of Kahane and Proposition 4.5 we deduce

R({λ



(λ + A

+ δ)

−1

})

≤ CR({(λ + δ

)

1−

|β|+|γ|

((λ + δ

)+A

+ δ

)

−1

}) ≤ C.

Lemma 3.5(a) then yields

R({λ



(λ + A

+ δ)

−1

(−R(λ + A

+ δ)

−1

)

})

≤R({λ



(λ + A

+ δ)

−1

})R({(R(λ + A

+ δ)

−1

)

})

≤ Cη

R({(A

+ δ)(λ + A

+ δ)

−1

)})

≤ Cν

(j ∈ N

)

with ν := ηR({(A

+ δ)(λ + A

+ δ)

−1

}) < 1. Employing again Lemma 3.5(a), in

particular the fact that the R-bound is preserved when taking the closure in the

strong operator topology, the assertion follows. 

Corollary 4.7. Let R(x

,D):=

|α

|=2m

(α

,0)

be given such that the

condition

|α

|=2m

r



∞

<ηis satisﬁed. Set

(x, D):=A

,D)+R(x

,D),x∈ Ω, (4.7)

and denote its X-realization by A

deﬁned on D(A

)=D(A

). Then there

exists a δ>0 such that A

+ δ ∈RS(X) with φ

+δ

≤ φ

+δ

provided that η

is suﬃciently small. In this case for φ>ϕ

,B)

we have

R({λ



(λ + A

+ δ)

−1

; λ ∈ Σ

π−φ

, 0 ≤ l + |β| + |γ|≤2m}) < ∞. (4.8)

Proof. By Proposition 4.5, in particular by relation (4.4), there exists a C>0

such that

D

(α

,0)

+ δ)

−1



L(X)

≤ C (α

∈ N

n−k

, |α

| =2m)

for each δ>δ

.Foraﬁxedδ>δ

this implies

Ru

≤

|α

|=2m

r



∞

D

(α

,0)

+ δ)

−1

+ δ)u

≤ Cη(A

+ δ)u

(u ∈ D(A

)).

Thus, if we assume that η<1/CR({(A

+δ)(λ+A

+δ)

−1

}), the assertion follows

from Lemma 4.6. 

R-sectoriality of Cylindrical Boundary Value Problems 497

4.3. Variable coeﬃcients a

In the next lemma we establish estimates that will turn out to be crucial for the

localization procedure.

Lemma 4.8. Let 1 <p<∞, (β

, 0) ∈ N

n−k

×N

, |(β

, 0)| = ν<2m,andr

≥ p

such that 2m−ν>

n−k

.Letb ∈ [L

∞

+ L

](R

n−k

), A

be the operator as deﬁned

in (4.7), and assume that φ>ϕ

(A,B)

(a) For every ε>0 there exists C(ε) > 0 such that

bD

(β

,0)

u

≤ εu

p,2m

+ C(ε)u

(u ∈ W

2m,p

n−k

,E)).

(b) For every ε>0 there exists a δ = δ(ε) > 0 such that

R({bD

(β

,0)

(λ + A

+ δ)

−1

; λ ∈ Σ

π−φ

}) ≤ ε.

Proof. (a) Let ε>0 be arbitrary. For simplicity we set β =(β

, 0). For b ∈

∞

n−k

)weobtainbyH¨older’s inequality and vector-valued complex interpola-

tion (see, e.g., [5]) that

bD

u

≤b

∞

u

p,ν

≤ Cb

∞

u

p,2m

u

1−

(u ∈ W

2m,p

n−k

,E)).

With the help of Young’s inequality we then can achieve that

bD

u

≤ εu

p,2m

+ C(ε)u

(u ∈ W

2m,p

n−k

,E)).

Now let b ∈ L

n−k

), r :=

,and



=1.ThenH¨older’s inequality and the

vector-valued version of the Gagliardo-Nirenberg inequality (see [19]) imply

bD

u

≤ Cb

D

u



≤ Cb

u

p,2m

u

1−τ

where τ =

n−k

(2m−ν)

∈ (0, 1) by our assumption on r

. Again an application of

Young’s inequality yields

bD

u

≤ εu

p,2m

+ C(ε)u

(u ∈ W

2m,p

n−k

,E)).

(b) Let (ε

)

j∈N

be a family of independent symmetric {−1, 1}-valued random

variables on a probability space ([0, 1], M,P), λ

∈ Σ

π−φ

,andf

∈ X.Forb ∈

∞

n−k

), δ

> 0, and arbitrary t ∈ [0, 1] we have

j=1

(t)bD

(λ

+ δ

+ A

+ δ)

−1

≤b

∞

j=1

(t)D

(λ

+ δ

+ A

+ δ)

−1

Note that there is a c

> 0 such that

|λ + δ

|≥c

(λ ∈ Σ

π−φ

,δ

> 0).