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
Vlp,β(G)
= R
G
X
|α|≤l
r
p(β−l+|α|)|D
αu|
pdx
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
Wlp,β(G)
=
R
G
X
|α|≤l
r
pβ|D
αu|
pdx
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]
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(Γ
±) and B
p,βs−1/p(Γ
±) 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
(Γ
±) 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
±kwith constant coefficients on Γ
+and Γ
−, respectively. In the spaces V
p,βsthe 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
±ksatisfy 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
(Γ
±) then u ∈ W
p,β+ss+2m(D).
The corresponding result for the spaces V
p,βshas 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
nwhere 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 = (−
12ω
0,
12ω
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 = ±
12ω
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
Vsp,β(D)
= R
D
X
|α|≤s
r
p(β−s+|α|)|D
αu|
pdx
1/p(D
α= D
xα= D
xα11. . . D
αxnn, D
xj= (1/i)∂
xj= (1/i)
∂x∂j
, r = |y|).
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
Vsp,β(D)
=
R
D
X
|α|≤l
|y|
p(β−s+|α|)|(D
αu)(x)|
pdx
+ R
D
R
D
X
|α|=l
| |y|
β(D
αu)(y, z) − |y
0|
β(D
αu)(y
0, z
0)|
pdx 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
αu)(x)|
pdx
+ X
|α|=l
R R
D D
|x−x0|<|y|/2
|y|
pβ|(D
αu)(y, z) − (D
αu)(y
0, z
0)|
pdx 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−s0for |x − x
0| <
12|y| . In the case l < s < s
0= l + 1 we can apply the equation
(D
αu)(x) − (D
αu)(x
0) = −
1
R
0
d
dt (D
αu)(x + t(x
0− x)) dt
= (x − x
0)
1
R
0
(∇D
αu)(x + t(x
0− x)) dt to obtain
R R
D D
|x−x0|<|y|/2
|y|
pβ|(D
αu)(x) − (D
αu)(x
0)|
pdx dx
0|x − x
0|
n+p(s−l)≤
1
R
0
R R
D D
|x−x0|<|y|/2
|y|
pβ|x − x
0|
−n+p(l+1−s)|(∇D
αu)(x + t(x − x
0))|
pdx dx
0dt
≤ R
D
|y|
pβ0|(∇D
αu)(x)|
pdx for |α| = l, i.e. kuk
Vsp,β(D)
≤ ckuk
Vs0 p,β0(D).
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
rjζ
ν(r)| < c
j2
−νjwith constants c
jindependent of r and ν. If we set r = |y| = (y
12+ y
22)
1/2we 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=−∞
kζ
νuk
pVs p,β(D) 1/p.
Let B
◦s−1/pp,β(Γ
±) (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
B◦s−1/pp,β (Γ±)
= inf{kvk
Vsp,β(D)
: v ∈ V
p,βs(D), v|
Γ±= u} . Lemma 2. The norm (1.5) is equivalent to the norm
kuk = X
∞ν=−∞
kζ
νuk
pB◦s−1/pp,β (Γ±)
1/p.
P r o o f. Let v ∈ V
p,βs(D) be an extension of u ∈ B
◦s−1/pp,β(Γ
±) such that kvk
Vsp,β(D)
≤ 2kuk
B◦s−1/pp,β (Γ±)
. Since ζ
νv is an extension of ζ
νu we get
X
∞ν=−∞
kζ
νuk
pB◦s−1/pp,β (Γ±)
1/p≤ X
∞ν=−∞
kζ
νvk
pVs p,β(D) 1/p≤ ckvk
Vsp,β(D)
≤ 2ckuk
B◦s−1/pp,β (Γ±)
. Furthermore, there exist extensions v
νof ζ
νu such that kv
νk
Vsp,β(D)
≤ 2kζ
νuk
B◦s−1/pp,β (Γ±)
(ν = . . . , −1, 0, +1, . . .). Since w
ν= (ζ
ν−1+ ζ
ν+ ζ
ν+1)v
ν= ζ
νu on Γ
±and w = P
∞ν=−∞
w
ν= u on Γ
±we obtain kuk
pB◦s−1/pp,β (Γ±)
≤ kwk
pVsp,β(D)
≤ c
∞
X
ν=−∞
kζ
νwk
pVsp,β(D)
= c
∞
X
ν=−∞
ζ
ν∞
X
k=−∞
w
kp Vp,βs (D)
= c
∞
X
ν=−∞
ζ
νν+2
X
k=ν−2
(ζ
k−1+ ζ
k+ ζ
k+1)v
kp Vp,βs (D)
≤ c
0∞
X
ν=−∞
kv
νk
pVsp,β(D)
≤ c
02
p∞
X
ν=−∞
kζ
νuk
pB◦s−1/pp,β (Γ±)
.
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)|
pdξ
+ R R
R+×Rn−2R+×Rn−2
|ξ−ξ0|<ξ1/2
ξ
pβ1k
X
j=0
(−1)
jk j
u jξ + (k − j)ξ
0k
p
dξ dξ
0|ξ − ξ
0|
n−2+ps 1/p.
P r o o f. Let u∈B
◦s−1/pp,β(Γ
±). Define u
ν(ξ) = ζ
ν(2
νξ)u(2
νξ) (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
1ku
νk
pB◦s−1/pp,β (Γ±)
≤ R
R+×Rn−2
|u
ν(ξ)|
pdξ
+ R R
R+×Rn−2 R+×Rn−2
|ξ−ξ0|<1/2
k
X
j=0
(−1)
jk j
u
νjξ + (k − 1)ξ
0k
pdξ dξ
0|ξ − ξ
0|
n−2+ps≤ c
2ku
νk
pB◦s−1/pp,β (Γ±)
where the constants c
1, c
2are independent of u and ν. It can be easily verified that ku
νk
B◦s−1/pp,β (Γ±)
= 2
ν(s−β−n/p)kζ
νuk
B◦s−1/pp,β (Γ±)
. Hence, c
12
ν(ps−pβ−n)kζ
νuk
pB◦s−1/pp,β (Γ±)
≤ 2
−ν(n−1)R
R+×Rn−2
|ζ
ν(ξ)u(ξ)|
pdξ
+ 2
−ν(n−ps)R R
R+×Rn−2 R+×Rn−2
|ξ−ξ0|<ξ1/2
k
X
j=0
(−1)
jk j
× ζ
νjξ + (k − j)ξ
0k
u jξ + (k − j)ξ
0k
p
dξ dξ
0|ξ − ξ
0|
n−2+ps≤ c
22
ν(ps−pβ−n)kζ
νuk
pB◦s−1/pp,β (Γ±)
.
Applying Lemma 2 we obtain the assertion of the theorem.
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|)+11
|(D
αu)(ξ)|
pdξ
+ X
|α|=l
R R
R+×Rn−2 R+×Rn−2
|ξ−ξ0|<ξ1/2
ξ
1pβ|(D
αu)(ξ) − (D
αu)(ξ
0)|
pdξ dξ
0|ξ − ξ
0|
n−1+pσ 1/por to the norm kuk =
R
R+×Rn−2
ξ
1p(β−s)+1|u(ξ)|
pdξ
+ X
|α|=l
R R
R+×Rn−2 R+×Rn−2
|ξ−ξ0|<ξ1/2
ξ
1pβ|(D
αu)(ξ) − (D
αu)(ξ
0)|
pdξ 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)|
pdx
+ R R
D D
|x−x0|<|y|/2
|y|
pβ|(D
αu)(y, z) − (D
αu)(y
0, z
0)|
p|x − x
0|
n+pσdx 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
Wsp,β(D)
= X
|α|≤s
R
D
r
pβ|D
αu|
pdx
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
Wsp,β(D)
=
X
|α|≤l
R
D
r
pβ|D
αu|
pdx
+ X
|α|=l
R R
D D
|x−x0|<|y|/2
r
pβ|(D
αu)(x) − (D
αu)(x
0)|
p|x − x
0|
n+pσdx dx
0 1/p.
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
Wsp,β(R+×Rn−2)
=
R
R+×Rn−2
r
pβX
|α|≤s
|(D
αu)(r, z)|
pdr dz
1/pif s is an integer, and (1.10) kuk
Wsp,β(R+×Rn−2)
=
R
R+×Rn−2
r
pβX
|α|≤l
|(D
αu)(r, z)|
pdr dz
+ X
|α|=l
R R
R+×Rn−2 R+×Rn−2
|(r−r0,z−z0)|<r/2
r
pβ|(D
αu)(r, z) − (D
αu)(r
0, z
0)|
p|(r, z) − (r
0, z
0)|
n−1+pσdr dz dr
0dz
0 1/pif 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
0ω0/2
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
Wsp,β+1/p(R+×Rn−2)
≤ ckuk
Wsp,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
jϕ(t) dt = δ
0,j,
supp ψ ⊂ (−1/4, 1/4),
∞
R
−∞
t
jψ(t) dt = δ
0,j(j = 0, 1, . . . , k). Here δ
0,jdenotes the Kronecker symbol.
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
jrD
γz(Kg)(r, z)|
pdr dz ≤ ckgk
pWsp,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)|
pdr dz ≤ ckgk
pWsp,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
jrD
γz(Kg)(r, z) is a linear combination of terms of the form
T = R
R
R
R+
(D
αg)(tr, z + τ r)t
µϕ(t)τ
νψ(τ ) dt dτ
= r
−2R
R
R
R+
(D
αg)(t, τ ) t r
µϕ t r
τ − z r
νψ 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
−2t r
µϕ t r
τ − z r
νψ τ − z r
dt dτ = R
R
R
R+
t
µϕ(t)τ
νψ(τ ) dt dτ is a constant, the function D
jrD
γzKg can be written as a finite sum of expressions of the form
T
0= cr
−2+hsi−j−|γ|R
R
R
R+
((D
αg)(t, τ ) − (D
αg)(r, z))
× t r
µ1ϕ
(µ2)t r
τ − z r
ν1ψ
(ν2)τ − z r
dt dτ
= cr
hsi−j−|γ|R
R
R
R+
((D
αg)(tr, z + τ r) − (D
αg)(r, z))
× t
µ1ϕ
(µ2)(t)τ
ν1ψ
(ν2)(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
rjD
γz(Kg)(r, z)|
pdr dz
≤ c X
|α|=s
R
R
R
R+
r
p(β−σ)1/4
R
−1/4 5/4
R
3/4
|(D
αg)(tr, z + τ r) − (D
αg)(r, z)|
pdt dτ dr dz
= c X
|α|=s
R
R
R
R+
r
p(β−σ)−2 z+r/4R
z−r/4 5r/4
R
3r/4
|(D
αg)(r
0, z
0) − (D
αg)(r, z)|
pdr
0dz
0drdz
≤ c X
|α|=s
R
R
R
R+
r
pβR R
R R+
|(r0−r,z0−z)|<r/2
|(D
αg)(r
0, z
0) − (D
αg)(r, z)|
p(|r − r
0|
2+ |z − z
0|
2)
1+pσ/2dr
0dz
0drdz
≤ ckgk
pWsp,β(R+×Rn−2)
.
Analogously, the inequality (1.14) can be proved. Furthermore, it can be shown that r
βD
zαKg ∈ 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)|
pdr ≤
p
|β − p + 1|
p ∞R
0
r
β|f
0(r)|
pdr , 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
αu (|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
αu ∈ W
p,β0(D) and by Lemma 5 we obtain D
αu ∈ W
p,β0(D) for
|α| ≤ l.
(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
αu ∈ W
p,β0(D) for |α| ≤ l. Using the fact that β − β
0= σ − σ
0we get
R R
D D
|x−x0|<|y|/2
|y|
pβ|(D
αu)(x) − (D
αu)(x
0)|
pdx dx
0|x − x
0|
n+pσ≤ c R R
D D
|x−x0|<|y|/2
|y|
pβ0|(D
αu)(x) − (D
αu)(x
0)|
pdx dx
0|x − x
0|
n+pσ0for |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
αu)(x) − (D
αu)(x
0) = (x − x
0)
1
R
0
(∇D
αu)(x + t(x
0− x)) dt yields
R R
D D
|x−x0|<|y|/2
|y|
pβ|(D
αu)(x) − (D
αu)(x
0)|
pdx dx
0|x − x
0|
n+pσ≤ c R
D
|y|
pβ0X
|α0|=l+1
|(D
α0u)(x)|
pdx 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
pLp(M )
+ R
M
R
M
|∆
kζu(z)|
pdz dζ
|ζ|
n−2+pκ 1/pwhere ∆
kζu(z) = P
kν=0
(−1)
ν kνu(z + νζ).
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
non M , i.e. f
n(z) = u
n(0, 0, z) = u
◦n(0, z) (the function u
◦nis defined by (1.11)). Since the functions u
◦ncan 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)|
pdr dz < ∞ .
Consequently, u
◦n(0, z) − (Ku
◦n)(0, z), i.e. f
nis the trace of Ku
◦non 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
nk
Bs−β−2/pp (M )
≤ ckKu
◦− Ku
◦nk
Wp,β−s+hsi+1hsi+1 (D)
≤ c
0ku − u
nk
Wsp,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∂
yj2u (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∂
zαu −
hsi−i−j−|α|
X
µ+ν=0
(∂
zαKu
◦i+µ, j+ν) y
1µy
ν2µ!ν!
p
dx
≤ ckuk
pWs p,β(D).
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
i1y
2ji!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∂
zαu −
hsi
X
i+j=0
(Ku
◦ij) y
1iy
j2i!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
1iy
2ji!j! ∈ V
p,βs(D) .
Now we investigate the question whether the v
ijare (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)|
pdr 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)|
pdr dz < ∞ .
(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|
pdr dz
≤ c R
R+×Rn−2
r
p(β−s+1)+1∂
∂r K(f − g)
p
dr dz ≤ ckf − gk
pWsp,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|
pdr dz < ∞ ,
i.e. f |
M= g|
M. Conversely, if f |
M= g|
Mthen (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 |
pdr dz
≤ ckf k
pWp,β−hsi+1/ps−hsi (R+×Rn−2)
≤ ckf k
pWsp,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
iβ−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
iKf ∈ V
p,β+1/ps−1(R
+× R
n−2) , i.e. ∂f /∂z
iβ−s+1,p
∼ 0.
2) Let ∂f /∂z
1β−s+1,p
∼ 0. If β − s < −1 − 2/p this means
∂z∂1
(f |
M) =
∂z∂f1
|
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)|
pdr dz ≤ c R
R+×Rn−2
(r
−1|f (r, z)|
p+r
−2|f (r, z)−f (0, z)|
p) dr dz
≤ c R
R+×Rn−2
r
−1z1
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
1p
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
1iy
j2∈ V
p,βs(D) if and only if v
ijβ−s+i+j,p
∼ u
◦ijfor i + j ≤ hsi.
P r o o f. 1) Let v
ijβ−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
1iy
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
µy1D
yν2hsi
X
i+j=0
(K(u
◦ij− v
ij)) y
i1y
2ji!j!
p
dx < ∞ .
Since D
γyK(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
1iy
2ji!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
◦µνβ−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
i1y
j2i!j!
p
dx < ∞ .
For µ + ν = hsi − 1 this yields u
◦µνβ−s+µ+ν,p
∼ v
µν. Analogously, by induction on µ + ν we show that u
◦µνβ−s+µ+ν,p
∼ v
νµfor µ + ν ≤ hsi − 2.
1.5. Connection between the spaces of traces. We denote by B
p,βs−1/p(Γ
±) (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(Γ±)= inf{kvk
Wp,βs (D): v ∈ W
p,βs(D), v|
Γ±= u} .
By Corollary 1, B
p,βs−1/p(Γ
±) ⊂ B
◦s−1/pp,β(Γ
±) for β > s − 2/p. Let u ∈ B
p,βs−1/p(Γ
+) 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
i1y
2ji!j! + v
0Γ+
=
hsi
X
k=0
(Kv
k) r
kk! + u
0where
v
k=
k
X
i=0
k i
cos ω
02
isin ω
02
k−iv
i,k−i∈ W
p,β+1/ps−k(R
+× R
n−2)
and u
0∈ B
◦s−1/pp,β(Γ
+).
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
kk! ∈ V
p,β−s+1/p0(R
+× R
n−2) iff v
kβ−s+k,p
∼ 0 .
P r o o f. 1) If v
kβ−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
hsik=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
hsik=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)+1D
rµhsi
X
k=0
(Kv
k) r
kk!
p
dr dz < ∞
for µ = 0, 1, . . . , hsi. Analogously to the proof of Theorem 5, this implies that v
kβ−s+k,p
∼ 0.
As a consequence of Lemma 8 we obtain the following theorem.
Theorem 6. Let u ∈ B
p,βs−1/p(Γ
+). 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
kk! ∈ B
◦s−1/pp,β(Γ
+) .
The functions v
kin (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
kk! ∈ B
◦s−1/pp,β(Γ
+) iff w
kβ−s+k,p
∼ v
k.
In particular , v
kβ−s+k,p
∼ ∂
rku for k ≤ hs−1/pi and v
kβ−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
kβ−s+k,p
∼ ∂
rku for k ≤ hs−1/pi. Differentiating (1.22) we get
∂
rνu −
hsi
X
k=0
(Kv
k) r
kk!
∈ B
◦s−ν−1/pp,β(Γ
+) . Since ∂
rjKv
k∈ V
p,β+1/ps−k−j(Γ
+) for j ≥ 1 (see Lemma 4) this implies
(1.23) ∂
rνu −
hsi−ν
X
k=0
(Kv
k+ν) r
kk! ∈ B
◦s−ν−1/pp,β(Γ
+) .
It can be easily verified that K(r
kv)∈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(∂
rνu) − KKv
ν∈ V
p,β+ν+1/ps(R
+× R
n−2), i.e. ∂
νru
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
kk! ∈ 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
kk! ∈ V
p,βs−1/p(R
+× R
n−2) = B
◦s−1/pp,β(R
+× R
n−2)
where u
k= ∂
rku ∈ 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
1kk!
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
(Γ
±) 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
(Γ
±)) 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;µνγD
µy1D
yν2D
zγu = 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
(Γ
±)) belongs to B
◦s−m± k−1/p
p,β
(Γ
±) 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
iy
ji!j! + u
0where u
◦ij∈ W
p,β+1/ps−i−j(R
+×R
n−2), u
0∈ V
p,βs(D), D
µrD
νzKu
◦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+νb
±k;µνγj
X
σ=0
j σ
cos ω
02
σ± sin ω
02
j−σ×D
zγu
◦µ+σ,ν+j−σβ−s+m±k+j,p