• Nie Znaleziono Wyników

Introduction. In this paper some problems of the teory of weighted Sobolev–

N/A
N/A
Protected

Academic year: 2021

Share "Introduction. In this paper some problems of the teory of weighted Sobolev–"

Copied!
26
0
0

Pełen tekst

(1)

INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1992

ON TWO CLASSES OF

WEIGHTED SOBOLEV–SLOBODETSKI˘ I SPACES IN A DIHEDRAL ANGLE

J. R O S S M A N N

Department of Mathematics, University of Rostock Universit¨ atsplatz 1, D-O-2500 Rostock, Germany

Introduction. In this paper some problems of the teory of weighted Sobolev–

Slobodetski˘ı spaces are dealt with. Weighted Sobolev spaces play an important role in the study of elliptic boundary value problems in non-smooth domains (see e.g. [2]–[4], [6]). Here two classes of weighted spaces are of special interest: the spaces V

p,βl

(G) with the homogeneous norms

(0.1) kuk

Vl

p,β(G)

=  R

G

X

|α|≤l

r

p(β−l+|α|)

|D

α

u|

p

dx 

1/p

(r = r(x) denotes the distance of x to the set of singularities on the boundary) and the spaces W

p,βl

(G) with the inhomogeneous norms

(0.2) kuk

Wl

p,β(G)

=

 R

G

X

|α|≤l

r

|D

α

u|

p

dx



1/p

.

We restrict ourselves to the case that G = D is a dihedral angle. This is the model case for domains with smooth non-intersecting edges. For domains with conical points the weighted Sobolev spaces V

p,βl

(G) and W

p,βl

(G) were investigated e.g.

in [2], [5], [8].

The present paper consists of two sections. In Section 1 the main properties of the weighted Sobolev spaces V

p,βl

(D), W

p,βl

(D) (l an integer, l ≥ 0, β ∈ R, 1 < p < ∞) and of the corresponding weighted Sobolev–Slobodetski˘ı spaces V

p,βs

(D), W

p,βs

(D) (s real, s ≥ 0, β ∈ R, 1 < p < ∞) will be investigated. It will be shown e.g. that the following imbeddings are valid:

V

p,βs

(D) ⊂ V

p,βs00

(D) if s ≥ s

0

, s − β = s

0

− β

0

,

W

p,βs

(D) ⊂ W

p,βs0 0

(D) if s ≥ s

0

, s − β = s

0

− β

0

, β

0

> −2/p .

[399]

(2)

Furthermore, we give a connection between the spaces V

p,βs

(D) and W

p,βs

(D). It will be proved that every function u ∈ W

p,βs

(D) can be written as the sum of a “quasi-polynomial” and a function from V

p,βs

(D) (see Theorem 5). This is a generalization of Lemma 1.3 of [6] (see also [7]) where such a representation has been shown if s is an integer and β + 2/p is not an integer. Analogous results are obtained for the space B

s−1/pp,β

±

) and B

p,βs−1/p

±

) of the traces of functions from V

p,βs

(D) and W

p,βs

(D) on the sides Γ

+

and Γ

of D, respectively.

Section 2 is concerned with two applications of the given results. In Section 2.1 we investigate conditions on g

±k

∈ B

s−m

± k−1/p

p,β

±

) under which there exists u ∈ W

p,βs

(D) satisfying the boundary conditions

(0.3) B

k±

u = g

k±

on Γ

±

(k = 1, . . . , p

±

)

where {B

k+

} and {B

k

} are normal systems of homogeneous boundary operators of order m

±k

with constant coefficients on Γ

+

and Γ

, respectively. In the spaces V

p,βs

the existence of u is always ensured (see [4]). We will show that u ∈ W

p,βs

(D) satisfying (0.3) exists if and only if the boundary data g

±k

satisfy some compati- bility conditions on the edge M (see Theorem 7).

In Section 2.2 the following regularity assertion for solutions of elliptic bound- ary value problems will be proved: If u ∈ W

p,β2m

(D) is a solution of an elliptic boundary value problem Lu = f in D, B

k±

u = g

k±

on Γ

±

(k = 1, . . . , m) where f ∈ W

p,β+ss

(D), g

±k

∈ B

s+2m−m

± k−1/p

p,β+s

±

) then u ∈ W

p,β+ss+2m

(D).

The corresponding result for the spaces V

p,βs

has been proved in [4].

1. Weighted Sobolev–Slobodetski˘ ı spaces in a dihedral angle

1.1. The spaces V

p,βs

(D). Let D = K × R

n−2

= {x = (y, z) ∈ R

n

: y ∈ K, z = (z

1

, . . . , z

n−2

) ∈ R

n−2

} be a dihedral angle in R

n

where K is a plane wedge which has the following representation in polar coordinates r, ω:

K = {y = (y

1

, y

2

) ∈ R

2

: 0 < r < ∞, ω ∈ Ω = (−

12

ω

0

,

12

ω

0

)}

(0 < ω

0

≤ 2π). The boundary of D consists of the (n − 1)-dimensional half-planes Γ

±

= {x = (x, z) ∈ R

n

: 0 < r < ∞, ω = ±

12

ω

0

, z ∈ R

n−2

}

and of the edge M = {(0, 0)} × R

n−2

. In the sequel we will denote the coordinates of a point x = (x

1

, . . . , x

n

) by y

1

, y

2

, z

1

, . . . , z

n−2

, i.e. y

1

= x

1

, y

2

= x

2

, z

j

= x

j+2

(j = 1, . . . , n − 2).

If s is a non-negative integer and p, β are real numbers, 1 < p < ∞, then V

p,βs

(D) denotes the closure of C

0

(D \ M ) = {u ∈ C

(D) : supp u ⊂ D \ M , supp u compact} with respect to the norm

(1.1) kuk

Vs

p,β(D)

=  R

D

X

|α|≤s

r

p(β−s+|α|)

|D

α

u|

p

dx 

1/p

(D

α

= D

xα

= D

xα11

. . . D

αxnn

, D

xj

= (1/i)∂

xj

= (1/i)

∂x

j

, r = |y|).

(3)

For non-integer positive s = l + σ (l an integer, 0 < σ < 1) the space V

p,βs

(D) is defined as the closure of C

0

(D \ M ) with respect to the norm

(1.2) kuk

Vs

p,β(D)

=



R

D

X

|α|≤l

|y|

p(β−s+|α|)

|(D

α

u)(x)|

p

dx

+ R

D

R

D

X

|α|=l

| |y|

β

(D

α

u)(y, z) − |y

0

|

β

(D

α

u)(y

0

, z

0

)|

p

dx dx

0

|x − x

0

|

n+pσ



1/p

.

It can easily be shown that the norm (1.2) is equivalent to the norm (1.3) kuk =



R

D

X

|α|≤l

|y|

p(β−s+|α|)

|(D

α

u)(x)|

p

dx

+ X

|α|=l

R R

D D

|x−x0|<|y|/2

|y|

|(D

α

u)(y, z) − (D

α

u)(y

0

, z

0

)|

p

dx dx

0

|x − x

0

|

n+pσ



1/p

.

Theorem 1. If s

0

> s and β

0

− s

0

= β − s then V

p,βs00

(D) is continuously imbedded in V

p,βs

(D).

P r o o f. If l ≤ s < s

0

< l + 1 (l an integer) then the imbedding immediately follows from the inequality

|y|

β−β0

= |y|

s−s0

≤ 2

s−s0

|x − x

0

|

s−s0

for |x − x

0

| <

12

|y| . In the case l < s < s

0

= l + 1 we can apply the equation

(D

α

u)(x) − (D

α

u)(x

0

) = −

1

R

0

d

dt (D

α

u)(x + t(x

0

− x)) dt

= (x − x

0

)

1

R

0

(∇D

α

u)(x + t(x

0

− x)) dt to obtain

R R

D D

|x−x0|<|y|/2

|y|

|(D

α

u)(x) − (D

α

u)(x

0

)|

p

dx dx

0

|x − x

0

|

n+p(s−l)

1

R

0

R R

D D

|x−x0|<|y|/2

|y|

|x − x

0

|

−n+p(l+1−s)

|(∇D

α

u)(x + t(x − x

0

))|

p

dx dx

0

dt

≤ R

D

|y|

0

|(∇D

α

u)(x)|

p

dx for |α| = l, i.e. kuk

Vs

p,β(D)

≤ ckuk

Vs0 p,β0(D)

.

(4)

In the sequel let ζ

ν

(ν = . . . , −1, 0, +1, . . .) be smooth functions on R

+

with support in the interval [2

ν−1

, 2

ν+1

] such that

(1.4)

X

ν=−∞

ζ

ν

(r) = 1 and |D

rj

ζ

ν

(r)| < c

j

2

−νj

with constants c

j

independent of r and ν. If we set r = |y| = (y

12

+ y

22

)

1/2

we can interpret ζ

ν

= ζ

ν

(r) as functions on D. Analogously to Lemma 1.1 in [4] the following assertion can be proved.

Lemma 1. The norm k · k

Vp,βs (D)

is equivalent to the norm kuk =  X

ν=−∞

ν

uk

pVs p,β(D)



1/p

.

Let B

s−1/pp,β

±

) (s > 1/p) be the space of the traces of functions from V

p,βs

(D) on Γ

+

and Γ

, respectively, provided with the norm

(1.5) kuk

Bs−1/pp,β±)

= inf{kvk

Vs

p,β(D)

: v ∈ V

p,βs

(D), v|

Γ±

= u} . Lemma 2. The norm (1.5) is equivalent to the norm

kuk =  X

ν=−∞

ν

uk

p

Bs−1/pp,β±)



1/p

.

P r o o f. Let v ∈ V

p,βs

(D) be an extension of u ∈ B

s−1/pp,β

±

) such that kvk

Vs

p,β(D)

≤ 2kuk

Bs−1/pp,β±)

. Since ζ

ν

v is an extension of ζ

ν

u we get

 X

ν=−∞

ν

uk

p

Bs−1/pp,β±)



1/p

≤  X

ν=−∞

ν

vk

pVs p,β(D)



1/p

≤ ckvk

Vs

p,β(D)

≤ 2ckuk

Bs−1/pp,β±)

. Furthermore, there exist extensions v

ν

of ζ

ν

u such that kv

ν

k

Vs

p,β(D)

≤ 2kζ

ν

uk

Bs−1/pp,β±)

(ν = . . . , −1, 0, +1, . . .). Since w

ν

= (ζ

ν−1

+ ζ

ν

+ ζ

ν+1

)v

ν

= ζ

ν

u on Γ

±

and w = P

ν=−∞

w

ν

= u on Γ

±

we obtain kuk

p

Bs−1/pp,β±)

≤ kwk

pVs

p,β(D)

≤ c

X

ν=−∞

ν

wk

pVs

p,β(D)

= c

X

ν=−∞

ζ

ν

X

k=−∞

w

k

p Vp,βs (D)

= c

X

ν=−∞

ζ

ν

ν+2

X

k=ν−2

k−1

+ ζ

k

+ ζ

k+1

)v

k

p Vp,βs (D)

≤ c

0

X

ν=−∞

kv

ν

k

pVs

p,β(D)

≤ c

0

2

p

X

ν=−∞

ν

uk

p

Bs−1/pp,β±)

.

(5)

In the half-planes Γ

+

, Γ

we introduce the Cartesian coordinates ξ

1

= r =

|y| = dist(x, M ), ξ

2

= z

1

, . . . , ξ

n−1

= z

n−2

. Using Lemmas 1 and 2 we can give the following norm equivalent to (1.5).

Theorem 2. Let p, β and s (1 < p < ∞, s > 1/p) be real numbers and k an arbitrary integer such that k > s − 1/p. Then the norm (1.5) is equivalent to the norm

(1.6) kuk =



R

R+×Rn−2

ξ

1p(β−s)+1

|u(ξ)|

p

+ R R

R+×Rn−2R+×Rn−2

|ξ−ξ0|<ξ1/2

ξ

1

k

X

j=0

(−1)

j

k j



u  jξ + (k − j)ξ

0

k



p

dξ dξ

0

|ξ − ξ

0

|

n−2+ps



1/p

.

P r o o f. Let u∈B

s−1/pp,β

±

). Define u

ν

(ξ) = ζ

ν

(2

ν

ξ)u(2

ν

ξ) (ν = 0, ±1, ±2, . . .).

Since suppu

ν

⊂ {ξ ∈ R

+

× R

n−2

: 1/2 < ξ

1

< 2} the norm of u can be estimated by the usual Besov space norm (see [11]), i.e.

c

1

ku

ν

k

p

Bs−1/pp,β±)

≤ R

R+×Rn−2

|u

ν

(ξ)|

p

+ R R

R+×Rn−2 R+×Rn−2

|ξ−ξ0|<1/2

k

X

j=0

(−1)

j

k j

 u

ν

 jξ + (k − 1)ξ

0

k



p

dξ dξ

0

|ξ − ξ

0

|

n−2+ps

≤ c

2

ku

ν

k

p

Bs−1/pp,β±)

where the constants c

1

, c

2

are independent of u and ν. It can be easily verified that ku

ν

k

Bs−1/pp,β±)

= 2

ν(s−β−n/p)

ν

uk

Bs−1/pp,β±)

. Hence, c

1

2

ν(ps−pβ−n)

ν

uk

p

Bs−1/pp,β±)

≤ 2

−ν(n−1)

R

R+×Rn−2

ν

(ξ)u(ξ)|

p

+ 2

−ν(n−ps)

R R

R+×Rn−2 R+×Rn−2

|ξ−ξ0|<ξ1/2

k

X

j=0

(−1)

j

k j



× ζ

ν

 jξ + (k − j)ξ

0

k



u  jξ + (k − j)ξ

0

k



p

dξ dξ

0

|ξ − ξ

0

|

n−2+ps

≤ c

2

2

ν(ps−pβ−n)

ν

uk

p

Bs−1/pp,β±)

.

Applying Lemma 2 we obtain the assertion of the theorem.

(6)

R e m a r k 1. By means of other equivalent norms in the usual Besov space (see [11], Section 4.4) analogously to Theorem 2 equivalence of other weighted norms to the norm (1.5) can be proved. In particular, if s − 1/p is not an integer, s − 1/p = l + σ (l an integer, l ≥ 0, 0 < σ < 1) then (1.5) is equivalent to the norm

kuk

Vp,βs−1/p(R+×Rn−2)

=

 X

|α|≤l

R

R+×Rn−2

ξ

p(β−s+|α|)+1

1

|(D

α

u)(ξ)|

p

+ X

|α|=l

R R

R+×Rn−2 R+×Rn−2

|ξ−ξ0|<ξ1/2

ξ

1

|(D

α

u)(ξ) − (D

α

u)(ξ

0

)|

p

dξ dξ

0

|ξ − ξ

0

|

n−1+pσ



1/p

or to the norm kuk =



R

R+×Rn−2

ξ

1p(β−s)+1

|u(ξ)|

p

+ X

|α|=l

R R

R+×Rn−2 R+×Rn−2

|ξ−ξ0|<ξ1/2

ξ

1

|(D

α

u)(ξ) − (D

α

u)(ξ

0

)|

p

dξ dξ

0

|ξ − ξ

0

|

n−1+pσ



1/p

.

Analogously, the norm (1.2) is equivalent to the norm kuk =

 R

D

|y|

p(β−s)

|u(x)|

p

dx

+ R R

D D

|x−x0|<|y|/2

|y|

|(D

α

u)(y, z) − (D

α

u)(y

0

, z

0

)|

p

|x − x

0

|

n+pσ

dx dx

0



1/p

.

1.2. The spaces W

p,βs

(D). Let p, β, s be real numbers, 1 < p < ∞, β > −2/p, s ≥ 0. If s is an integer then we define the space W

p,βs

(D) as the closure of C

0

(D) with respect to the norm

(1.7) kuk

Ws

p,β(D)

=  X

|α|≤s

R

D

r

|D

α

u|

p

dx 

1/p

.

If s = l + σ (l an integer, 0 < σ < 1) then the space W

p,βs

(D) will be defined as the closure of C

0

(D) with respect to the norm

(1.8) kuk

Ws

p,β(D)

=

 X

|α|≤l

R

D

r

|D

α

u|

p

dx

+ X

|α|=l

R R

D D

|x−x0|<|y|/2

r

|(D

α

u)(x) − (D

α

u)(x

0

)|

p

|x − x

0

|

n+pσ

dx dx

0



1/p

.

(7)

Furthermore, we define W

p,βs

(R

+

× R

n−2

) (s ≥ 0, 1 < p < ∞, β > −1/p) as the closure of C

0

(R

+

× R

n−2

) with respect to the norm

(1.9) kuk

Ws

p,β(R+×Rn−2)

=

 R

R+×Rn−2

r

X

|α|≤s

|(D

α

u)(r, z)|

p

dr dz



1/p

if s is an integer, and (1.10) kuk

Ws

p,β(R+×Rn−2)

=



R

R+×Rn−2

r

X

|α|≤l

|(D

α

u)(r, z)|

p

dr dz

+ X

|α|=l

R R

R+×Rn−2 R+×Rn−2

|(r−r0,z−z0)|<r/2

r

|(D

α

u)(r, z) − (D

α

u)(r

0

, z

0

)|

p

|(r, z) − (r

0

, z

0

)|

n−1+pσ

dr dz dr

0

dz

0



1/p

if s = l + σ (l an integer, l ≥ 0, 0 < σ < 1).

Let u = u(y

1

, y

2

, z) be an arbitrary function on D. Then we denote by u

the function

(1.11) u

(r, z) = 1 ω

0

ω0/2

R

−ω0/2

u(r cos ω, r sin ω, z) dω .

Lemma 3. If u ∈ W

p,βs

(D) then u

∈ W

p,β+1/ps

(R

+

× R

n−2

) and ku

k

Ws

p,β+1/p(R+×Rn−2)

≤ ckuk

Ws

p,β(D)

.

If s is an integer then this assertion immediately follows from the definition of the spaces W

p,βs

(D), W

p,βs

(R

+

× R

n−2

). For s not an integer it is proved in [9].

We introduce the operator (Kg)(r, z) = χ(r) R

Rn−2

R

R+

g(tr, z + τ r)K(t, τ ) dt dτ

where χ is a smooth cut-off function on R

+

, equal to unity in [0, 1] and to zero in (2, ∞) and K(t, τ ) = ϕ(t)ψ(τ

1

) . . . ψ(τ

n−2

) is a product of smooth functions ϕ ∈ C

0

(R

+

), ψ ∈ C

0

(R) satisfying

(1.12)

supp ϕ ⊂ (3/4, 5/4),

R

0

t

j

ϕ(t) dt = δ

0,j

,

supp ψ ⊂ (−1/4, 1/4),

R

−∞

t

j

ψ(t) dt = δ

0,j

(j = 0, 1, . . . , k). Here δ

0,j

denotes the Kronecker symbol.

(8)

Lemma 4. If g ∈ W

p,βs

(R

+

×R

n−2

) then Kg ∈ T

ν=1

W

p,β+hsi−s+νhsi+ν

(R

+

×R

n−2

).

Furthermore, (1.13) R

Rn−2

R

R+

r

p(β−s+j+|γ|)

|D

jr

D

γz

(Kg)(r, z)|

p

dr dz ≤ ckgk

pWs

p,β(R+×Rn−2)

for j ≥ 1 or |γ| ≥ hsi + 1, and

(1.14) R

Rn−2

R

R+

r

p(β−s+hsi)+1

|D

γz

(Kg − g)|

p

dr dz ≤ ckgk

pWs

p,β(R+×Rn−2)

for |γ| = hsi. Here hsi denotes the largest integer less than s.

P r o o f. For simplicity we restrict ourselves to the case n = 3. For n > 3 the lemma can be proved analogously.

If j + |γ| = hsi and r < 1 then D

jr

D

γz

(Kg)(r, z) is a linear combination of terms of the form

T = R

R

R

R+

(D

α

g)(tr, z + τ r)t

µ

ϕ(t)τ

ν

ψ(τ ) dt dτ

= r

−2

R

R

R

R+

(D

α

g)(t, τ )  t r



µ

ϕ  t r

 τ − z r



ν

ψ  τ − z r

 dt dτ where |α| = j + |γ|, µ + ν = j.

Let j + |γ| ≥ hsi + 1, r < 1. Since

R

R

R

R+

r

−2

 t r



µ

ϕ  t r

 τ − z r



ν

ψ  τ − z r



dt dτ = R

R

R

R+

t

µ

ϕ(t)τ

ν

ψ(τ ) dt dτ is a constant, the function D

jr

D

γz

Kg can be written as a finite sum of expressions of the form

T

0

= cr

−2+hsi−j−|γ|

R

R

R

R+

((D

α

g)(t, τ ) − (D

α

g)(r, z))

×  t r



µ1

ϕ

2)

 t r

 τ − z r



ν1

ψ

2)

 τ − z r

 dt dτ

= cr

hsi−j−|γ|

R

R

R

R+

((D

α

g)(tr, z + τ r) − (D

α

g)(r, z))

× t

µ1

ϕ

2)

(t)τ

ν1

ψ

2)

(τ ) dt dτ where |α| = hsi. Consequently, for j + |γ| ≥ hsi + 1 we obtain

R

R 1

R

0

r

p(β−s+j+|γ|)

|D

rj

D

γz

(Kg)(r, z)|

p

dr dz

(9)

≤ c X

|α|=s

R

R

R

R+

r

p(β−σ)

1/4

R

−1/4 5/4

R

3/4

|(D

α

g)(tr, z + τ r) − (D

α

g)(r, z)|

p

dt dτ dr dz

= c X

|α|=s

R

R

R

R+

r

p(β−σ)−2



z+r/4

R

z−r/4 5r/4

R

3r/4

|(D

α

g)(r

0

, z

0

) − (D

α

g)(r, z)|

p

dr

0

dz

0

 drdz

≤ c X

|α|=s

R

R

R

R+

r



R R

R R+

|(r0−r,z0−z)|<r/2

|(D

α

g)(r

0

, z

0

) − (D

α

g)(r, z)|

p

(|r − r

0

|

2

+ |z − z

0

|

2

)

1+pσ/2

dr

0

dz

0

 drdz

≤ ckgk

pWs

p,β(R+×Rn−2)

.

Analogously, the inequality (1.14) can be proved. Furthermore, it can be shown that r

β

D

zα

Kg ∈ L

p

(R

+

× R

n−2

). Together with (1.13) and (1.14) this implies Kg ∈ T

ν=1

W

p,β+hsi−s+νhsi+ν

(R

+

× R

n−2

).

R e m a r k 2. If we interpret Kg as a function on D (i.e. we define v(y, z) = (Kg)(|y|, z) = (Kg)(r, z)) then

Kg ∈

\

ν=1

W

p,β+hsi−s+ν−1/phsi+ν

(D) for g ∈ W

p,βs

(R

+

× R

n−2

) . The following lemma is a consequence of the Hardy inequality

R

0

r

β−p

|f (r)|

p

dr ≤

 p

|β − p + 1|



p ∞

R

0

r

β

|f

0

(r)|

p

dr , which is satisfied if f (0) = 0, β < p − 1 or if f (∞) = 0, β > p − 1.

Lemma 5. Let u ∈ W

p,β0

(D) (β > −2/p) be such that ∇u ∈ W

p,β0 0

(D) where β

0

> 1 − 2/p. Then u ∈ V

p,β1 0

(D).

Now we can prove an imbedding analogous to Theorem 1 for the spaces W

p,βs

(D).

Theorem 3. Let s

0

≥ s, β

0

− s

0

= β − s and β > −2/p. Then W

p,βs0 0

(D) is continuously imbedded in W

p,βs

(D).

P r o o f. Without loss of generality we can assume that l ≤ s < s

0

≤ l + 1 where l is a non-negative integer. We consider the following cases: (a) s = l, s

0

= l + σ (0 < σ < 1), (b) s = l + σ, s

0

= l + σ

0

(0 < σ < σ

0

< 1), (c) s = l + σ, s

0

= l + 1 (0 < σ < 1).

(a) Let u ∈ W

p,βs0 0

(D) and v

α

= D

α

u (|α| = l). By Lemma 5 and (1.14) we have Kv

α

∈ W

p,β+1−σ1

(D) = W

p,β+11

(D) ⊂ W

p,β0

(D) and v

α

− Kv

α

∈ W

p,β0

(D).

Hence, v

α

= D

α

u ∈ W

p,β0

(D) and by Lemma 5 we obtain D

α

u ∈ W

p,β0

(D) for

|α| ≤ l.

(10)

(b) Analogously to (a) we obtain Kv

α

∈ W

p,β1 0+1−σ0

(D) = W

p,β+1−σ1

(D) for

|α| = l. Since Kv

α

= 0 for r = |y| > 2 this implies Kv

α

∈ W

p,β+11

(D) ⊂ W

p,β0

(D) for |α| = l. Furthermore, by (1.14), v

α

−Kv

α

∈ W

p,β−σ0

(D)∩W

p,β0 0

(D) ⊂ W

p,β0

(D) and analogously to (a) it follows that D

α

u ∈ W

p,β0

(D) for |α| ≤ l. Using the fact that β − β

0

= σ − σ

0

we get

R R

D D

|x−x0|<|y|/2

|y|

|(D

α

u)(x) − (D

α

u)(x

0

)|

p

dx dx

0

|x − x

0

|

n+pσ

≤ c R R

D D

|x−x0|<|y|/2

|y|

0

|(D

α

u)(x) − (D

α

u)(x

0

)|

p

dx dx

0

|x − x

0

|

n+pσ0

for |α| = l. Hence, u ∈ W

p,βs

(D).

(c) By Lemma 5 the space W

p,βl+10

(D) is imbedded in W

p,βl

(D). Furthermore, the equation

(D

α

u)(x) − (D

α

u)(x

0

) = (x − x

0

)

1

R

0

(∇D

α

u)(x + t(x

0

− x)) dt yields

R R

D D

|x−x0|<|y|/2

|y|

|(D

α

u)(x) − (D

α

u)(x

0

)|

p

dx dx

0

|x − x

0

|

n+pσ

≤ c R

D

|y|

0

X

0|=l+1

|(D

α0

u)(x)|

p

dx for |α| = l. This implies W

p,βl+10

(D) ⊂ W

p,βs

(D).

Corollary 1. If β > s − 2/p then

W

p,βs

(D) ⊂ W

p,β−s+hsihsi

(D) ⊂ W

p,β−s+hsi−1hsi−1

(D) ⊂ . . . ⊂ W

p,β−s0

(D) , i.e. W

p,βs

(D) ⊂ V

p,βs

(D).

R e m a r k 3. Analogously to Theorem 3 it can be proved that W

p,βs0 0

(R

+

× R

n−2

) is continuously imbedded in W

p,βs

(D) if s

0

≥ s, β

0

− s

0

= β − s, β > −1/p.

1.3. Traces of functions from W

p,βs

(D) on M. In [6] (Lemma 1.1) it has been proved that the trace of a function u ∈ W

p,βl

(D) belongs to the Besov space B

l−β−2/pp

(M ) for l an integer, l > β + 2/p > 0. Here the norm in B

pκ

(M ) is defined by

(1.15) kuk

Bκ p(M )

=

 kuk

pL

p(M )

+ R

M

R

M

|∆

kζ

u(z)|

p

dz dζ

|ζ|

n−2+pκ



1/p

where ∆

kζ

u(z) = P

k

ν=0

(−1)

ν kν

u(z + νζ).

(11)

Theorem 4. Let u ∈ W

p,βs

(D), s > β + 2/p > 0. Then the trace of u on M exists and belongs to the Besov space B

ps−β−2/p

(M ).

P r o o f. Let {u

n

} be a sequence of functions from C

0

(D) which converges to u in W

p,βs

(D). Furthermore, let {f

n

} be the sequence of the traces of u

n

on M , i.e. f

n

(z) = u

n

(0, 0, z) = u

n

(0, z) (the function u

n

is defined by (1.11)). Since the functions u

n

can be interpreted as functions from W

p,−1/p+1/21

(R

+

× R

n−2

) Lemma 4 yields

R

R+×Rn−2

r

−1−p/2

|u

n

(r, z) − (Ku

n

)(r, z)|

p

dr dz < ∞ .

Consequently, u

n

(0, z) − (Ku

n

)(0, z), i.e. f

n

is the trace of Ku

n

on M . We denote the trace of Ku

on M by f . Then by Lemma 1.1 of [6] and Lemmas 3 and 4 we get

kf − f

n

k

Bs−β−2/pp (M )

≤ ckKu

− Ku

n

k

Wp,β−s+hsi+1hsi+1 (D)

≤ c

0

ku − u

n

k

Ws

p,β(D)

. Hence, {f

n

} converges to f in B

ps−β−2/p

(M ).

Conversely, it can be proved that every f ∈ B

ps−β−2/p

(M ) (s > β + 2/p > 0) can be extended to a function v ∈ W

p,βs

(D). We define the extension operator K as follows:

(Kf )(r, z) = χ(r) R

Rn−2

g(z + τ r)ψ(τ

1

) . . . ψ(τ

n−2

) dτ

where ψ ∈ C

0

(R) satisfies (1.12). By Lemma 1.2 of [6] the operator K defines a continuous map from B

pκ

(M ) into W

p,s−κ−2/ps

(D) for κ > 0, κ not an integer, s an integer, s + h−κ − 2/pi > −2. From Theorem 3 it follows that this is also true for real s, s + h−κ − 2/pi > −2.

1.4. Connection between V

p,βs

(D) and W

p,βs

(D). In [6], [7] it has been proved that every u ∈ W

p,βs

(D) is the sum of a “quasi-polynomial” and a function from V

p,βs

(D) if s is a non-negative integer and β + 2/p is not an integer. We will give a similar connection between V

p,βs

(D) and W

p,βs

(D) without any restrictions on s and β.

Let u ∈ W

p,βs

(D) (β > −2/p). We denote the derivatives ∂

yi1

yj2

u (i + j ≤ hsi) by u

ij

. By using the properties of the operator K the following lemma can be easily proved (see [9]).

Lemma 6. If u∈W

p,βs

(D) then the following inequality holds for i+j+|α|≤hsi:

R

D

r

p(β−s+i+j+|α|)

iy1

yj2

zα

u −

hsi−i−j−|α|

X

µ+ν=0

(∂

zα

Ku

i+µ, j+ν

) y

1µ

y

ν2

µ!ν!

p

dx

≤ ckuk

pWs p,β(D)

.

(12)

Lemma 6 implies the following corollary.

Corollary 2. If u ∈ W

p,βs

(D) then u −

hsi

X

i+j=0

(Ku

ij

)(r, z) y

i1

y

2j

i!j! ∈ V

p,βs

(D) .

P r o o f. By Lemma 4 we have ∂

yµ1

yν2

zα

(Ku

ij

) ∈ V

p,β−s+i+j+µ+ν+|α|0

(D) if µ + ν ≥ 1 or |α| ≥ s − i − j. Therefore, from Lemma 6 it follows that

R

D

r

p(β−s+µ+ν+|α|)

yµ1

νy2

zα

 u −

hsi

X

i+j=0

(Ku

ij

) y

1i

y

j2

i!j!



p

dx < ∞ for µ + ν + |α| ≤ hsi, which yields the desired conclusion.

By Corollary 2 for every u ∈ W

p,βs

(D) there exist v

ij

∈ W

p,β+1/ps

(R

+

× R

n−2

) such that

u −

hsi

X

i+j=0

(Kv

ij

)(r, z) y

1i

y

2j

i!j! ∈ V

p,βs

(D) .

Now we investigate the question whether the v

ij

are (in some sense) uniquely determined by u. Similarly to [8] we introduce the following equivalence relation in W

p,β+1/ps

(R

+

× R

n−2

) (s > 0, β > −2/p):

(1.16) f

β−s,p

∼ g ⇔ R

R+×Rn−2

r

p(β−s)+1

|(Kf )(r, z) − (Kg)(r, z)|

p

dr dz < ∞ .

Another characterization is the following:

(1.17) f

β−s,p

∼ g ⇔ K(f − g) ∈

\

ν=0

V

p,β−s+ν+1/pν

(R

+

× R

n−2

) .

P r o o f. If f, g ∈ W

p,β+1/ps

(R

+

× R

n−2

) and f

β−s,p

∼ g then by Lemma 4 K(f − g) ∈

\

ν=hsi+1

W

p,β−s+ν+1/pν

(R

+

× R

n−2

) ∩ V

p,β−s+1/p0

(R

+

× R

n−2

) , and (1.17) follows from W

p,βl

(R

+

×R

n−2

)∩V

p,β−l0

(R

+

×R

n−2

) ⊂ V

p,βl

(R

+

×R

n−2

) (cf. Remark 1).

R e m a r k 4. Let f, g ∈ W

p,β+1/ps

(R

+

× R

n−2

).

(a) If β − s < −2/p then f

β−s,p

∼ g ⇔ f |

M

= g|

M

. (b) If β − s = −2/p then

f

β−s,p

∼ g ⇔ R

R+×Rn−2

r

p(β−s)+1

|f (r, z) − g(r, z)|

p

dr dz < ∞ .

(13)

(c) If β − s > −2/p then f

β−s,p

∼ g

β−s,p

∼ 0.

P r o o f. (a) If β − s < −2/p then the trace of f − g on M exists and coincides with that of K(f − g). The Hardy inequality and Lemma 4 imply

(1.18) R

R+×Rn−2

r

p(β−s)+1

|K(f − g) − (f − g)|

M

|

p

dr dz

≤ c R

R+×Rn−2

r

p(β−s+1)+1

∂r K(f − g)

p

dr dz ≤ ckf − gk

pWs

p,β+1/p

(R

+

× R

n−2

) . Hence, from (1.16) it follows that

R

R+×Rn−2

r

p(β−s)+1

|(f − g)|

M

|

p

dr dz < ∞ ,

i.e. f |

M

= g|

M

. Conversely, if f |

M

= g|

M

then (1.16) follows from (1.18).

(b) If β − s = −2/p then W

p,β+1/ps

(R

+

× R

n−2

) ⊂ W

p,β−hsi+1/ps−hsi

(R

+

× R

n−2

) (see Remark 3) and Lemma 4 yields

R

R+×Rn−2

r

p(β−s)+1

|Kf − f |

p

dr dz

≤ ckf k

p

Wp,β−hsi+1/ps−hsi (R+×Rn−2)

≤ ckf k

pWs

p,β+1/p(R+×Rn−2)

. This proves the assertion.

(c) If β − s > −2/p then W

p,β+1/ps

(R

+

× R

n−2

) is imbedded in V

p,β+1/ps

(R

+

× R

n−2

) (cf. Corollary 1, Remark 3). Consequently, (1.16) is satisfied for all f, g ∈ W

p,β+1/ps

(R

+

× R

n−2

).

Lemma 7. Let f ∈ W

p,β+1/ps

(R

+

× R

n−2

), β − s ≤ −1 − 2/p. Then f

β−s,p

∼ 0 iff ∂f /∂z

i

β−s+1,p

∼ 0 for one i ∈ {1, . . . , n − 2}.

P r o o f. 1) If f

β−s,p

∼ 0 then Kf ∈ V

p,β+1/ps

(R

+

× R

n−2

) and K ∂f

∂z

i

= ∂

∂z

i

Kf ∈ V

p,β+1/ps−1

(R

+

× R

n−2

) , i.e. ∂f /∂z

i

β−s+1,p

∼ 0.

2) Let ∂f /∂z

1

β−s+1,p

∼ 0. If β − s < −1 − 2/p this means

∂z

1

(f |

M

) =

∂z∂f

1

|

M

= 0 where f |

M

∈ B

ps−β−2/p

(M ). Hence, f |

M

= 0, i.e. f

β−s,p

∼ 0. If β − s = −1 − 2/p we first assume that f (r, z) = 0 for r > 1 and |z

1

| > 1. Using the Hardy inequality we get

R

R+×Rn−2

r

−1

|f (0, z)|

p

dr dz ≤ c R

R+×Rn−2

(r

−1

|f (r, z)|

p

+r

−2

|f (r, z)−f (0, z)|

p

) dr dz

(14)

≤ c R

R+×Rn−2

 r

−1

z1

R

−1

∂t f (r, t, z

2

, . . . , z

n−2

) dt

p

+

∂f (r, z)

∂r

p

 dr dz

≤ c

 R

R+×Rn−2

r

−1

∂f

∂z

1

p

dr dz +

∂f

∂r

p

Lp(R+×Rn−2)

 .

Hence, f |

M

= 0, i.e. f

β−s,p

∼ 0. If f has an arbitrary support then we show that (ϕf )|

M

= 0 for every cut-off function ϕ with compact support.

Theorem 5. Let u, v

ij

(i + j ≤ hsi) be arbitrary functions from W

p,βs

(D) and W

p,β+1/ps−i−j

(R

+

× R

n−2

), respectively. Then

(1.19) u −

hsi

X

i+j=0

1

i!j! (Kv

ij

)(r, z)y

1i

y

j2

∈ V

p,βs

(D) if and only if v

ij

β−s+i+j,p

∼ u

ij

for i + j ≤ hsi.

P r o o f. 1) Let v

ij

β−s+i+j,p

∼ u

ij

. Then K(v

ij

− u

ij

) ∈ V

p,β−s+hsi+i+j+1+1/phsi+1

(R

+

× R

n−2

) for i + j ≤ hsi. Interpreting K(v

ij

− u

ij

) as functions on D we get K(v

ij

− u

ij

) ∈ V

p,β−s+hsi+i+j+1hsi+1

(D) and K(v

ij

− u

ij

)y

1i

y

2j

∈ V

p,β−s+hsi+1hsi+1

(D) ⊂ V

p,βs

(D). Using Corollary 2 we obtain (1.19).

2) Assume that (1.19) is satisfied. Then by Corollary 2

(1.20) R

D

r

p(β−s+µ+ν)

D

µy1

D

yν2

hsi

X

i+j=0

(K(u

ij

− v

ij

)) y

i1

y

2j

i!j!

p

dx < ∞ .

Since D

γy

K(u

ij

− v

ij

) ∈ V

p,β−s+i+j+|γ|0

(D) for |γ| ≥ 1 (see Lemma 4), (1.20) yields

(1.21) R

D

r

p(β−s+µ+ν)

hsi−µ−ν

X

i+j=0

(K(u

i+µ,j+ν

− v

i+µ,j+ν

)) y

1i

y

2j

i!j!

p

dx < ∞ .

For µ + ν = hsi, (1.21) implies K(u

µν

− v

µν

) ∈ V

p,β−s+µ+ν0

(D), i.e. u

µν

β−s+µ+ν,p

∼ v

µν

. Then from (1.21) it follows that

R

D

r

p(β−s+µ+ν)

hsi−1−µ−ν

X

i+j=0

(K(u

i+µ,j+ν

− v

i+µ,j+ν

)) y

i1

y

j2

i!j!

p

dx < ∞ .

For µ + ν = hsi − 1 this yields u

µν

β−s+µ+ν,p

∼ v

µν

. Analogously, by induction on µ + ν we show that u

µν

β−s+µ+ν,p

∼ v

νµ

for µ + ν ≤ hsi − 2.

(15)

1.5. Connection between the spaces of traces. We denote by B

p,βs−1/p

±

) (s >

1/p, β > −2/p) the space of the traces of functions from W

p,βs

(D) on the sides Γ

+

and Γ

, respectively, provided with the norm

kuk

Bp,βs−1/p±)

= inf{kvk

Wp,βs (D)

: v ∈ W

p,βs

(D), v|

Γ±

= u} .

By Corollary 1, B

p,βs−1/p

±

) ⊂ B

s−1/pp,β

±

) for β > s − 2/p. Let u ∈ B

p,βs−1/p

+

) and let v ∈ W

p,βs

(D) be an extension of u. Then by Theorem 5, u has the repre- sentation

u = v|

Γ+

=

 X

i+j≤hsi

(Kv

ij

) y

i1

y

2j

i!j! + v

0



Γ+

=

hsi

X

k=0

(Kv

k

) r

k

k! + u

0

where

v

k

=

k

X

i=0

k i



cos ω

0

2



i

 sin ω

0

2



k−i

v

i,k−i

∈ W

p,β+1/ps−k

(R

+

× R

n−2

)

and u

0

∈ B

s−1/pp,β

+

).

Lemma 8. Let v

k

∈ W

p,β+1/ps−k

(R

+

× R

n−2

) (k = 0, 1, . . . , hsi). Then

hsi

X

k=0

(Kv

k

) r

k

k! ∈ V

p,β−s+1/p0

(R

+

× R

n−2

) iff v

k

β−s+k,p

∼ 0 .

P r o o f. 1) If v

k

β−s+k,p

∼ 0 then Kv

k

∈ V

p,β−s+k+1/p0

(R

+

×R

n−2

) and (Kv

k

)r

k

∈ V

p,β−s+1/p0

(R

+

× R

n−2

).

2) If P

hsi

k=0

(Kv

k

)r

k

/k! ∈ V

p,β−s+1/p0

(R

+

× R

n−2

) then from the properties of K (see Lemma 4) it follows that P

hsi

k=0

(Kv

k

)r

k

/k! ∈ V

p,β−s+1/ps

(R

+

× R

n−2

), i.e.

R

R+×Rn−2

r

p(β−s+µ)+1

D

rµ

hsi

X

k=0

(Kv

k

) r

k

k!

p

dr dz < ∞

for µ = 0, 1, . . . , hsi. Analogously to the proof of Theorem 5, this implies that v

k

β−s+k,p

∼ 0.

As a consequence of Lemma 8 we obtain the following theorem.

Theorem 6. Let u ∈ B

p,βs−1/p

+

). Then there exist v

k

∈ W

p,β+1/ps−k

(R

+

×R

n−2

) (k = 0, 1, . . . , hsi) such that

(1.22) u −

hsi

X

k=0

(Kv

k

)(r, z) r

k

k! ∈ B

s−1/pp,β

+

) .

(16)

The functions v

k

in (1.22) are uniquely determined in the following sense: if w

k

∈ W

p,β+1/ps−k

(R

+

× R

n−2

) (k = 0, 1, . . . , hsi) then

u −

hsi

X

k=0

(Kw

k

)(r, z) r

k

k! ∈ B

s−1/pp,β

+

) iff w

k

β−s+k,p

∼ v

k

.

In particular , v

k

β−s+k,p

∼ ∂

rk

u for k ≤ hs−1/pi and v

k

β−s+k,p

∼ 0 for k > [s−β −2/p]

(here [κ] denotes the largest integer less than or equal to κ, i.e. [κ] = −h−κi−1).

P r o o f. It remains to show that v

k

β−s+k,p

∼ ∂

rk

u for k ≤ hs−1/pi. Differentiating (1.22) we get

rν

 u −

hsi

X

k=0

(Kv

k

) r

k

k!



∈ B

s−ν−1/pp,β

+

) . Since ∂

rj

Kv

k

∈ V

p,β+1/ps−k−j

+

) for j ≥ 1 (see Lemma 4) this implies

(1.23) ∂

rν

u −

hsi−ν

X

k=0

(Kv

k+ν

) r

k

k! ∈ B

s−ν−1/pp,β

+

) .

It can be easily verified that K(r

k

v)∈V

p,β−k+1/ps

(R

+

× R

n−2

) for v∈W

p,β+1/ps

(R

+

× R

n−2

) and positive integers k. Then (1.23) yields that K(∂

rν

u) − KKv

ν

∈ V

p,β+ν+1/ps

(R

+

× R

n−2

), i.e. ∂

νr

u

β−s+ν,p

∼ Kv

νβ−s+ν,p

∼ v

ν

.

The following lemma gives a connection between the spaces B

p,βs−1/p

(R

+

× R

n−2

) and W

p,βs−1/p

(R

+

× R

n−2

).

Lemma 9. Let s − 1/p be non-integer , s > 1/p and β > −1/p. Then B

p,βs−1/p

(R

+

× R

n−2

) = W

p,βs−1/p

(R

+

× R

n−2

) .

P r o o f. 1) If u ∈ B

p,βs−1/p

(R

+

× R

n−2

) then there exist v

k

∈ W

p,β+1/ps−k

(R

+

× R

n−2

) such that

u −

hsi

X

k=0

(Kv

k

) r

k

k! ∈ B

s−1/pp,β

(R

+

× R

n−2

) = V

p,βs−1/p

(R

+

× R

n−2

) . Since (Kv

k

)r

k

∈ W

p,βs−1/p

(R

+

× R

n−2

) we get u ∈ W

p,βs−1/p

(R

+

× R

n−2

).

2) Let u ∈ W

p,βs−1/p

(R

+

× R

n−2

). Then analogously to Theorem 5 it can be shown that

v = u −

hs−1/pi

X

k=0

(Ku

k

) r

k

k! ∈ V

p,βs−1/p

(R

+

× R

n−2

) = B

s−1/pp,β

(R

+

× R

n−2

)

(17)

where u

k

= ∂

rk

u ∈ W

p,βs−k−1/p

(R

+

× R

n−2

). Without loss of generality we can assume that Γ

+

is the half-plane Γ

+

= {x = (y, z) : y

1

> 0, y

2

= 0, z ∈ R

n−2

} which can be identified with R

+

× R

n−2

. If v

0

∈ V

p,βs

(D) is an extension of v then

u

0

= v

0

+

hs−1/pi

X

k=0

(Ku

k

) y

1k

k!

is an extension of u which lies in W

p,βs

(D). Hence, u ∈ B

p,βs−1/p

(R

+

× R

n−2

).

2. Applications to boundary value problems

2.1. Compatibility conditions for boundary data on Γ

±

. For the spaces V

p,βs

(D) the following lemma has been proved (see [4], [9]).

Lemma 10. Let B

k±

(k = 1, . . . , p

±

) be homogeneous differential operators with constant coefficients of order m

±k

< s − 1/p. Assume that {B

k+

} and {B

k

} are normal on Γ

+

and Γ

, respectively. Then for all g

k±

∈ B

s−m

± k−1/p

p,β

±

) there exists u ∈ V

p,βs

(D) such that B

k±

u = g

k±

on Γ

±

for k = 1, . . . , p

±

.

We will investigate conditions on g

±k

(g

±k

∈ B

s−m

± k−1/p

p,β

±

)) under which there exists u ∈ W

p,βs

(D) satisfying the boundary conditions

(2.1) B

k±

u = X

µ+ν+|γ|=m±k

b

±k;µνγ

D

µy1

D

yν2

D

zγ

u = g

k±

on Γ

±

(k = 1, . . . , p

±

). Here we restrict ourselves to the case n = 3.

By Theorem 6, B

k±

u|

Γ±

− g

k±

(u ∈ W

p,βs

(D), g

k±

∈ B

s−m

± k−1/p

p,β

±

)) belongs to B

s−m

± k−1/p

p,β

±

) iff

(2.2) ∂

rj

(B

k±

u|

Γ±

− g

k±

)

β−s+m

± k+j,p

∼ 0

for j = 0, 1, . . . , [s − β − 2/p] − m

±k

. Using the representation u =

[s−β−2/p]

X

i+j=0

(Ku

ij

) y

i

y

j

i!j! + u

0

where u

ij

∈ W

p,β+1/ps−i−j

(R

+

×R

n−2

), u

0

∈ V

p,βs

(D), D

µr

D

νz

Ku

ij

∈ V

p,β+1/ps−i−j−µ−ν

(R

+

× R

n−2

) if µ ≥ 1 or ν ≥ s − β − 2/p − i − j (see Theorem 5) we get the following equivalences:

(2.3) X

µ+ν+γ=m±k

(−i)

µ+ν

b

±k;µνγ

j

X

σ=0

 j σ



cos ω

0

2



σ



± sin ω

0

2



j−σ

×D

zγ

u

µ+σ,ν+j−σ

β−s+m±k+j,p

∼ ∂

rj

g

k±

Cytaty

Powiązane dokumenty

[r]

D i b l´ık, On existence and asymptotic behaviour of solutions of singular Cauchy problem for certain system of ordinary differential equations, Fasc. H a l e, Theory of

In 1842 Dirichlet proved that for any real number ξ there exist infinitely many rational numbers p/q such that |ξ−p/q| &lt; q −2.. This problem has not been solved except in

The two new theorems in this paper provide upper bounds on the con- centration function of additive functions evaluated on shifted γ-twin prime, where γ is any positive even

It is known from [5] that any bounded pseudoconvex domain with Lip- schitz boundary is hyperconvex, that is, it admits a bounded continuous plurisubharmonic (psh) exhaustion

We show that a generalized upper and lower solution method is still valid, and develop a monotone iterative technique for finding minimal and maximal solutions.. In our situation,

Studying the sequential completeness of the space of germs of Banach- valued holomorphic functions at a points of a metric vector space some theorems on extension of holomorphic maps

The solution is constructed as the limit of the solutions of approxi- mating equations, for which the boundary is no more characteristic, so that the theory for