ISSN 1063-4541, Vestnik St. Petersburg University. Mathematics, 2008, Vol. 41, No. 4, pp. 324–335. © Allerton Press, Inc., 2008. Original Russian Text © A.A. Solov’ev, 2008, published in Vestnik Sankt-Peterburgskogo Universiteta. Seriya 1. Matematika, Mekhanika, Astronomiya, 2008, No. 4, pp. 46–58.
ON THE OCCASION OF THE 70th ANNIVERSARY OF BIRTHDAY OF VLADIMIR GILELEVICH MAZ’YA Dedicated to Vladimir Gilelevich Maz’ya
Lp-Boundedness of a Boundary Integral Operator on a Contour with a Peak A. A. Solov’ev Received May 18, 2008
Abstract—Results on the solvability of boundary integral equations on a plane contour with a peak obtained in collaboration with V.G. Maz’ya are developed. Earlier, it was proved that, on a contour Γ with an outward peak, the operator of the boundary equation of the Dirichlet boundary value problem maps the space p, β + 1 (Γ) continuously onto p, β(Γ). The norm of a function in p, β (Γ) is defined as ϕ
1
p, β ( Γ )
⎛ = ⎜ ( ∂/∂s )ϕ ( q ) p q ⎝Γ
∫
pβ
ds q +
∫ ϕ(q)
p
q
p(β – 1)
Γ
⎞ ds q⎟ ⎠
1/ p
,
−
provided that the peak is at the origin. In this case, the norms on the spaces +p, β ( Γ ) are defined by 1/ p
ϕ
−
+p, β ( Γ )
=
∫
p ϕ ( q+ ) − + ϕ ( q– ) q
p(β – µ)
ds q + ϕ
p p, β + 1
,
Γ ∪ { q < δ}
where q± are the intersection points of Γ with the circle {z: |z| = |q|} and δ > 0 is a fixed small number. On a contour with an inward peak, the operator of the boundary equation of the Dirichlet problem con+ tinuously maps p, β + 1 (Γ) onto p, β(Γ), where p, β(Γ) is the direct sum of p, β (Γ) and the space (Γ) of functions on Γ of the form p ( z ) =
∑
(k) k m t Rez k=0
with the parameter m = [µ – β – p–1].
The operator I – 2W of the boundary integral equation of plane elasticity theory, where W is the elastic double-layer potential, is considered. The main result is that the operator I – 2W continuously maps the – – space p, β + 1 × p, β + 1 (Γ) to the space p, β × p, β (Γ). On a contour with an inward peak, the obtained representation of the operator I – 2W and theorems on the boundedness of auxiliary integral operators imply that the images of vector-valued functions from – p, β + 1 × p, β + 1 (Γ) have components representable as sums of functions from the spaces p, β (Γ) and p, β(Γ). DOI: 10.3103/S1063454108040079
Our purpose in this article is to develop the results on solvability of boundary integral equations on a plane contour with a peak obtained in collaboration with V.G. Maz’ya. In short, in [2–4] three spaces were found, say X, Y, and Z, such that (a) on a contour with an outward peak, the boundary equation operator maps X continuously onto Y; (b) on a contour with an inward peak, the boundary equation operator maps X continuously onto Z. In this paper, we consider the operator I – 2W of the boundary integral equation of the plane elasticity theory, W being the elastic double-layer potential. Given a contour with an outward peak, we prove that operator I – 2W maps the space X × X of vector-valued functions continuously into the space Y × Y. Having obtained representation (1) of operator I –2W and employing the continuity of auxiliary integral operators substantiated in Theorem 6 below, we arrive at the conclusion that, on a contour with an inward peak, the components of the images of elements of X × X are representable as a sum of functions belonging the spaces Y and Z. We now proceed to explaining the results in detail. 324
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR
325
The boundary integral equation of the inner Dirichlet problem for the Lamé operator in a domain Ω bounded by the curve Γ takes the form ( – I + 2W )u = 2v. Here, I is the identity operator, u, v: Γ the double-layer potential
R2 are vector-valued functions, and Wu(z) is the direct value of
Wu ( z ) =
∫ { T ( ∂ , n )Γ ( z, q ) }*u ( q ) ds , q
q
q
Γ
where µ { T ( ∂ q, n q )Γ ( z, q ) }* = ---------------------------2π ( λ + 2µ ) ⎧⎛ ⎛ 2 ⎪ 2(λ + µ) ( x – u)( y – v ) ( x – u) × ⎨ ⎜ I + --------------------2- ⎜ ⎜ ⎜ 2 µ z – q ⎝ ( x – u)( y – v ) ⎪⎝ (y – v ) ⎩ ⎫ ⎞ ∂ 1 ⎪ ⎟ ------- log -------------- ⎬, z–q ⎪ ⎠ ∂s q ⎭
⎛ +⎜ 0 1 ⎝ –1 0
z = ( x, y ),
⎞⎞ ∂ 1 ⎟ ⎟ ------- log -------------⎟ ⎟ ∂n q z–q ⎠⎠
q = ( u, v ).
The Lp-boundedness of the double-layer potential W on piecewise smooth curves without cusp points has been investigated by many authors (see the survey article [1] for a comprehensive list of references and a historical account). We suppose that the curve Γ \{z = 0} is of the class C 2. The point z = 0 is called an outward (inward) peak if Ω (the complementary domain Ωc, respectively) is given near the peak by the inequalities κ–(x) < y < κ+(x), 0 < x < δ, where x–µ – 1κ±(x) ∈ C2[0, δ] and lim x
–µ–1
x → +0
κ±(x) = α± with µ > 0 and α+ > α–. We shall designate
by Γ± the arcs {(x, κ±(x)) : x ∈ [0, δ]}; points that lie on Γ+ and Γ– and have the same abscissa will be denoted by symbols q+ and q–. Let p, β (Γ) be the space of functions on Γ \{z = 0} with a finite norm 1
ϕ
1
p, β ( Γ )
⎛ p = ⎜ ( ∂/∂s )ϕ ( q ) q ⎝Γ
∫
pβ
ds q +
∫ ϕ(q)
p
q
p(β – 1)
Γ
⎞ ds q⎟ ⎠
1/ p
.
If |q|βϕ ∈ Lp(Γ), then ϕ is said to belong to p, β(Γ). Let the norm in this space be defined by the relation ϕ
p, β ( Γ )
=
β
q ϕ
L p(Γ) .
±
We now introduce the spaces p, β (Γ) of functions ϕ on Γ \{z = 0} with finite norms ϕ
±
p, β ( Γ )
⎛ = ⎜ ⎝Γ
∫
+
ϕ ( q+ ) ± ϕ ( q– ) q p
p(β – µ)
ds q + ϕ
⎞ p 1 ⎟ p, β + 1 ( Γ )
∪ Γ–
1/ p
⎠
.
In what follows, we will write p, β (Γ) in place of p, β(Γ). –
∑
Let (Γ) be the space of restrictions of real-valued functions of the kind p(z) = mk = 0 t Rez to Γ \{z = 0} (here, m = [µ – β – p–1]). The expression for the norm of the function p can be written in the form: m
p
(Γ)
=
∑t
(k)
.
k=0
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
k
k
326
SOLOV’EV (+)
Define the space p, β(Γ) as the direct sum of spaces p, β (Γ) and (Γ). We shall now express the kernel of the integral operator W in an appropriate form. To do this, we employ the relation ⎛ 2 ⎞ ⎞ 1 1 1⎛ -------2 ⎜ x xy ⎟ = --- I + --- ⎜ Re ( z/z ) Im ( z/z ) ⎟ . 2 2 ⎝ Im ( z/z ) – Re ( z/z ) ⎠ z ⎜⎝ xy y 2 ⎟⎠ Since µ 2(µ + λ) 1 ---------------------------- ⎛ I + --------------------- I⎞ = ------ I, ⎠ 2π ( λ + 2µ ) ⎝ 2µ 2π the matrix operator –I + 2W can be written as ⎛ λ+µ λ+µ ⎜ – 1 + --------------------------- R 3 2 + ---------------------------- I 3 2π ( λ + 2µ ) 2π ( λ + 2µ ) ⎜ ⎜ λ+µ λ+µ ⎜ – 2 + --------------------------- I 3 – 1 – ---------------------------- R 3 2π ( λ + 2µ ) 2π ( λ + 2µ ) ⎝
⎞ ⎟ ⎟. ⎟ ⎟ ⎠
(1)
Here, 1 ∂ 1 ( 1 σ ) ( z ) = σ ( z ) – --- σ ( q ) -------- ⎛ log --------------⎞ ds q , ⎝ π ∂n q z–q⎠
∫ Γ
( 2 σ ) ( z ) =
∂
- ⎛ log --------------⎞ ds ∫ σ ( q ) -----∂s ⎝ z–q⎠ 1
q
Γ
( R 3 σ ) ( z ) =
q
z = – σ' ( q ) ⎛ log --------------⎞ ds q , ⎝ z–q⎠
∫ Γ
z–q ∂
- -------- ⎛ log --------------⎞ ds , ∫ σ ( q )Re ---------z – q ∂n ⎝ z–q⎠ 1
q
q
Γ
and ( I 3 σ ) ( z ) =
z–q ∂
- -------- ⎛ log --------------⎞ ds . ∫ σ ( q )Im ---------z – q ∂n ⎝ z–q⎠ 1
q
q
Γ
The continuity of the first two integral operators 1 and 2 has been studied in the papers [2–4]; below, we formulate the results obtained there. Theorem 1. Suppose that 0 < β + p–1 < min{µ, 1}. The operator 1 maps the space p, β + 1 (Γ) continuously into the space p, β(Γ), provided that domain Ω has an outward peak, and into the space p, β(Γ), provided that Ω has an inward peak. 1
Theorem 2. Suppose that 0 < β + p–1 < min{µ, 1}. The operator 2 maps the space p, β + 1 (Γ) continuously into the space p, β(Γ), provided that domain Ω has an inward or outward peak. We shall always assume that 0 < β + p–1 < min{µ, 1}. Our primary purpose in this article is to establish the continuity of the operator 1
I – 2W : ( p, β + 1 × p, β + 1 ) ( Γ ) 1
1
( p , β × p, β ) ( Γ )
on a contour Γ with an outward peak. By Theorems 1 and 2, it suffices to consider operators R3 and I3. In Theorem 6, we prove the continuity of the operators R 3 : p, β + 1 ( Γ ) 1
p, β ( Γ ) and I 3 : p, β + 1 ( Γ ) 1
p, β ( Γ )
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR
327
for a domain Ω with an inward peak. Together with Theorems 1, 2, and representation (1), this fact implies 1 1 that, for u ∈ ( p, β + 1 × p, β + 1 )(Γ), each component of the vector function v = (–I + 2W)u is a sum of functions belonging to spaces p, β(Γ) and p, β(Γ). We shall now formulate some auxiliary statements, for the proof of which we refer the reader to the papers [2–4]. Consider an integral operator T of the kind Tf ( x ) =
∫ K ( x, y ) f ( y ) d y R
with kernel K(x, y) satisfying the inequality 1 1 K ( x, y ) ≤ c -------------- ------------------------------, x – y (1 + x – y J )
J ≥ 0.
Here and in what follows, symbol c denotes various positive constants. We introduce the space p, α(R) of functions on R with the norm ϕ
p, α ( R )
2 α/2
= (1 + x )
ϕ
L p(R) .
The following theorem can be proved in the same way as the theorem on boundedness of singular integral operators in the Lp-space with power weight [5]. Theorem 3. If an operator T : Lp(R) Lp(R), 1 < p < ∞, is bounded and – J < α + p–1 < J + 1, then T is continuous in the space p, α(R). We write ρ(u) = κ+(u) – κ–(u) and define the function h by the expression δ
dν
∫ ---------ρ( ν)
= τ,
τ ∈ ( 0, δ ).
h(τ)
The following Lemma contains all inequalities that we need. Lemma 1. Suppose that |1 – ξ/τ| < ε0, where τ, ξ ∈ (0, δ), and that ε0 is a sufficiently small positive number. Then ρ c 1 1 1 ζ (h(ξ ) )ρ ζ(h(τ)) ------------------------------------------------------------------ – -------------------------- ≤ -- ⎛ -------------------------- + ---⎞ 2 2 2 2 ⎝ τ ( ξ – τ ) + 1 ξ⎠ ( h ( ξ ) – h ( τ ) ) + (ρ ζ (h(ξ))) (ξ – τ) + 1
(2)
( h ( ξ ) – h ( τ ) ) + iρ ( h ( τ ) ) ( ξ – τ ) – i c ------------------------------------------------------------ – ------------------------ ≤ --. ( h ( ξ ) – h ( τ ) ) – iρ ( h ( τ ) ) ( ξ – τ ) + i τ
(3)
and
We now proceed to the proof of the results formulated above. Theorem 4. Suppose that Ω has an outward peak and that β + p–1 < min{µ, 1}. Then operator 3 maps
1 p, β + 1 (Γ)
continuously into the space p, β(Γ).
Proof. (i) Let ε > 0 be so small that |κ±(x) – κ −+ (u)| ≥ cuµ + 1 for all u satisfying the inequality |x – u| < εx. Denote by Γ ± (x), Γ ± (x), and Γ ± (x) the arcs on Γ± that are the inverse images of the segments [0, (1 – ε)x], [(1 – ε)x, (1 + ε)x] and [(1 + ε)x, δ] under the projection. We express σ as a sum of two functions σ0 and σ1 so that suppσ0 ⊂ Γ+ ∪ Γ– and suppσ1 ⊂ Γ \{|q| > δ/2}. The expression for (3σ)(z), z = x + iκ±(x) ∈ Γ±, may be written as l
c
r
∫
( 3 σ ) ( z ) = r
r
Γ+ ( x ) ∪ Γ– ( x )
z–q ∂ 1 σ 0 ( q ) ----------- -------- log ----- ds q z – q ∂n q q
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
328
SOLOV’EV
∫
+ r Γ+ ( x )
⎛ +⎜ ⎜ ⎝ Γ\ { q
∫
+
> δ/2 }
∫
+ r Γ+ ( x )
∪
r Γ– ( x )
∪
∫
r Γ– ( x )
z–q ∂ σ 0 ( q ) ----------- -------z – q ∂n q
∫
+
c Γ −+ ( x )
c
Γ± ( x )
N
∑ k=1
1 ⎛ z⎞ k --- Re --- ds q k ⎝ q⎠ (4)
⎞ –q ∂ 1 ⎟ σ ( q ) z---------- -------- log -------------- ds q + ⎟ 1 z – q ∂n q z–q l l Γ+ ( x ) ∪ Γ– ( x ) ⎠
∫
z–q ∂ ⎛ 1 σ 0 ( q ) ----------- -------- ⎜ log -------------- – z – q ∂n q ⎝ z–q
N
∑ k=1
1 ⎛ z ⎞ k⎞ --- Re --- ⎟ ds q = k ⎝ q⎠ ⎠
7
∑ I ( z ). k
k=1
First we evaluate the second term I2. To this end, we employ the equality ∂ z k k ∂ –k k ∂ –k Re -------- ⎛ ---⎞ = – Rez -------Imq + Imz -------Req ; ∂n q ⎝ q⎠ ∂s q ∂s q here, q = u + iκ±(u) ∈ Γ±. By means of the inequalities |Imzk | ≤ cxk + µ and |(∂/∂sq)q–k | ≤ cu–k – 1, we find that k+µ
x k ∂ –k -. Imz -------Req ≤ c ---------k+1 ∂s q u Therefore, applying the relation ⎛ k–r r – 1⎞ Rez = ( Rez ) – ⎜ Imz ( Rez ) ⎟ Imz ⎝r = 1 ⎠ k–1
k
∑
k
gives Rez – ( Rez ) ≤ cx k
k
k+µ
.
This yields the inequality k+µ
∂ z k x r r k ∂ –k - q ∈ Γ + ( x ) ∪ Γ – ( x ), Re -------- ⎛ ---⎞ – ( Rez ) -------Imq ≤ ---------k+1 ∂n q ⎝ q⎠ ∂s q u from which we get
∫ r
r
Γ+ ( x ) ∪ Γ– ( x )
= x
∫
k r
r
Γ+ ( x ) ∪ Γ– ( x )
z–q ∂ z k σ 0 ( q ) ----------- -------- Re ⎛ ---⎞ ds q z – q ∂n q ⎝ q⎠ (5)
z–q ∂ –k σ 0 ( q ) ----------- -------Imq ds q + ( Rσ ) ( z ) = ( J k σ 0 ) ( z ) + ( Rσ 0 ) ( z ). z – q ∂s q
The last term on the right can be estimated as δ
| ( Rσ 0 ) ( z )| ≤ cx
µ
∫
(1 – ε) x
σ0 ( τ ) --------------- dτ, τ
z ∈ Γ+ ∪ Γ– ;
this gives
∫
z
p(β – µ)
Rσ 0 ( z ) dσ z ≤ c σ' p
p p, β + 1 .
Γ+ ∪ Γ–
Let σ+ and σ– be the restrictions of σ0 to the arcs Γ+ and Γ–. Using relation (5) and the inequality VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR µ+1 z+ – q z– – q x µ + – z – ≤ ---------- ≤ cx , ------------- – ------------- ≤ 2 z-------------u z+ – q z– – q z– – q
329
(6)
we obtain δ
( Jσ 0 ) ( z + ) – ( Jσ 0 ) ( z – ) ≤ cx
µ
∫
(1 – ε) x
σ+ ( u ) + σ– ( u ) --------------------------------------- du, u
(7)
and, by Hardy’s inequality,
∫
z
p(β – µ)
I 2 ( z + ) – I 2 ( z – ) ds z ≤ c σ' p
p p, β + 1 .
(8)
Γ+ ∪ Γ–
Now we make use of the estimate ( ∂/∂n q ) log q ≤ cu
µ–1
as u
0
and inequality (6) to evaluate the difference I1(z+) – I1(z–):
∫
z
p(β – µ)
I 1 ( z + ) – I 1 ( z – ) ds z ≤ c σ' p
p p, β + 1 .
(9)
Γ+ ∪ Γ–
Since the function I3(z) is infinitely differentiable in a neighborhood of zero, it follows from Taylor’s formula that I3(z+) – I3(z–) = O(xµ + 1). Therefore, the difference I3(z+) – I3(z–) satisfies an estimate similar to (9). In what follows, we will always assume that z ∈ Γ+. To evaluate term I4, we introduce ρ(x) = κ+(x) – κ–(x). We will need the inequality ∂ ρ( x ) 1 2 – 1/2 -2 -------- log -------------- – ( 1 + κ '+ ( u ) ) ------------------------------------2 ∂n q z–q ( x – u) + ρ( x ) (10) 2µ + 1 x c µ–1 ⎛ ⎞ + ------------------------------------- ( q = u + iκ – ( x ) ∈ Γ – ( x ) ). ≤c x 2 2µ + 2⎠ ⎝ ( x – u) + x To prove (10), we note that κ – ( x ) – κ – ( u ) – κ '– ( u ) ( x – u ) ≤ cx
µ–1
( x – u) . 2
Therefore, ( u – x )κ '– ( u ) – ( κ – ( u ) – κ + ( x ) ) 1 ∂ 2 1/2 – -------- log -------------- ( 1 + ( κ '+ ( u ) ) = ------------------------------------------------------------------------2 2 z–q ∂n q ( x – u ) + ( κ–( u ) – κ+( x ) ) κ+( x ) – κ–( x ) µ–1 - + O ( x ). = ---------------------------------------------------------------2 2 ( x – u ) + ( κ–( u ) – κ+( x ) )
(11)
Since |κ–(x) – κ–(u)| ≤ cxµ|x – u|, we have 2µ + 1 κ+( x ) – κ–( x ) x ρ( x ) ⎛ -------------------------------------⎞ , ---------------------------------------------------------------- = ------------------------------------------2 + O 2 2 2 ⎝ ( x – u ) 2 + x 2µ + 2⎠ ( x – u ) + ( κ–( u ) – κ+( x ) ) ( x – u) + (ρ( x ))
and now (10) follows from (12) and (11). We write z = x + iκ+(x) and q = u + iκ–(u), q ∈ Γ – (x). Let z* denote x + iκ+(u). Since c
µ
x–u x z – q z* – q µ z – z* ≤ c -------------------------------1 ≤ cx , ----------- – -------------- ≤ 2 ------------µ + z – q z* – q z* – q x–u +x it follows from (12) that z – q z* – q ∂ x 1 µ–1 + -------------------------------------⎞ . ----------- – -------------- -------- log -------------- ≤ ⎛ x 2 2µ + 2⎠ ⎝ z – q z* – q ∂n q z–q ( x – u) + x 2µ + 1
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
(12)
330
SOLOV’EV
Employing this estimate gives δ
p
∫x
∫
p(β – µ)
c
0
Γ– ( x )
z – q z* – q ∂ 1 σ 0 ( q ) ⎛ ----------- – --------------⎞ -------- ⎛ log --------------⎞ ds q d x ≤ c σ' ⎝ z – q z* – q⎠ ∂n q ⎝ z–q⎠
p p, β + 1 .
Indeed, changing the variables τ = u–µ, ξ = x–µ and putting σ±(u) = σ(u + iκ±(u)), u ∈ [0, δ], we have δ
∫x
p(β – µ) ⎛
(1 + ε) x
p
2µ + 1 ⎞ σ – ( u )x ------------------------------------d u ⎜ ⎟ dx ≤ c 2 2µ + 2 ⎝ (1 – ε) x ( x – u ) + x ⎠
0
∫
+∞
∫ξ δ
p(1 – α) ⎛
p
– 1/µ µ σ – ( τ ) dτ⎞ -----------------------------2 -----⎟ dξ, ⎜ 2 ⎝R (ξ – τ) + µ τ ⎠
–µ
∫
where β + p–1 = µ(α – p–1). By Theorem 3, the right-hand side does not exceed δ
+∞
∫τ
c
δ
– pα
σ– ( τ
– 1/µ
p
) dτ ≤
–µ
∫ σ' ( u ) –
p
u
p(β + 1)
du.
0
Hence, it suffices to estimate the integral
∫ c Γ– ( x )
z* – q ∂ 1 σ ( q ) -------------- -------- log -------------- ds q . z* – q ∂n q z–q
(13)
The inequality ρ (ξ – τ) – i 1 ( h ( ξ ) – h ( τ ) ) + iρ ( h ( τ ) ) ζ ( h ( ξ ) )ρ ζ (h(τ)) - – ------------------------ -------------------------------------------------------------------------------------- -----------------------------------------------------------------( h ( ξ ) – h ( τ ) ) – iρ ( h ( τ ) ) ( h ( ξ ) – h ( τ ) ) 2 – ( ρ ( h ( ξ ) ) ) 2 ( ξ – τ ) + i ( ξ – τ ) 2 + 1 ζ c⎛ 1 1 - + ---⎞ ≤ -- -------------------------2 ⎝ τ ( ξ – τ ) + 1 ξ⎠
(14)
is easily justified with the help of (2) and (3). By changing the variables u = h(τ) and x = h(ξ) in (13), we find from (10) and (14) that
∫ c
Γ– ( x )
z* – q ∂ 1 σ 0 ( q ) -------------- -------- log -------------- ds q = z* – q ∂n q z–q
σ– ( h ( τ ) ) ( ξ – τ ) – i -------------------------- ------------------------ dτ + I ( ξ ). 2 (ξ – τ) + i ( ξ – τ ) + 1
∫
(15)
Here, the last term can be estimated as ξ
c I ( ξ ) ≤ -ξ
∫ –1
h (δ)
∞
∞
∫
∫
dτ dτ σ – ( h ( τ ) ) ----- + c σ – ( h ( τ ) ) ----2- + c τ τ ξ
–1
h (δ)
σ – ( h ( τ ) ) dτ -------------------------- -----. 2 (ξ – τ) + 1 τ
Now it follows from (15) that δ
(1 + ε) x
∫x
p(β – µ)
∞
≤
∫ –1
h (δ)
ξ
p(1 – α)
∫
(1 – ε) x
0
p
( x – u ) + iρ ( u ) ρ ( x ) du - dx σ – ( u ) ------------------------------------ ------------------------------------( x – u ) – iρ ( u ) ( x – u ) 2 + ρ ( x ) 2
(ξ – τ) – i
1
- --------------------------- dτ ∫ σ ( h ( τ ) ) ----------------------(ξ – τ) + i (ξ – τ) + 1 –
2
(16)
∞
p
dξ +
∫
ξ
p(1 – α)
I ( ξ ) dξ. p
h –1 ( δ )
By virtue of Theorem 3 and Hardy’s inequality, the last integral does not exceed ∞
∫ –1
h (δ)
δ
d p p(β + 1) p(1 – α) -----σ – ( h ( τ ) ) τ dτ = c σ '– ( u ) u du. dτ
∫ 0
Since VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR
331
∂ 1 τ–i 1 1 - = -----------------2 = – ----- ----------, ---------- ------------∂τ τ+i τ + i τ2 + 1 (τ + i) we have (ξ – τ) – i
1
dτ
d
- --------------------------- dτ = – ∫ -----σ ( h ( τ ) ) ------------------------ . ∫ σ ( h ( τ ) ) ----------------------(ξ – τ) + i (ξ – τ) + 1 (ξ – τ) + i dτ –
–
2
By the theorem on boundedness of a Hilbert operator in a weighted Lp-space, the first term on the right in (16) does not exceed δ
p d p p(β + 1) p(1 – α) -----σ – ( h ( τ ) ) τ dτ = c σ '– ( u ) u du. dτ
∫
∫
–1
0
h (δ)
On Γ + ∪ Γ + (x) ∪ Γ – (x), we have c
l
l
( ∂/∂n q ) log z – q ≤ cx
µ–1
.
Therefore, (1 + ε) x
I 5 ( z ) + I 6 ( z ) ≤ cx
∫
µ–1
( σ + ( u ) + σ – ( u ) )du.
0
An application of Hardy’s inequality shows that I5
p, β – µ
+ I6
p, β – µ
≤ c σ'
p, β + 1 .
It remains to treat I7. We use the expression ⎛ ∂ 1 -------- Re ⎜ log ---------------- – ∂n q ⎝ 1 – z/q
N
∑ k=1
( z/q ) ( z/q ) 1 ⎛ z ⎞ k⎞ --- --- ⎟ = – Re --------------------- cos ( n q, x ) + Im --------------------- cos ( n q, y ). k ⎝ q⎠ ⎠ q–z q–z N+1
N+1
The first term on the right does not exceed cxN + 1uµ – N – 2. We write the function Im ( ( z/q )
N+1
(q – z) ) –1
as z Re ⎛ ---⎞ ⎝ q⎠
N+1
⎛ 1 1 N+1 Im ----------- + ⎜ Im ----------+ Imz ( Rez ) N + 1 q–z ⎝ q
1 z N – k⎞ Re ----------( Rez ) ⎟ Re ----------- . N+1 q ⎠ q–z k=1 N
k
∑
Since |Im(q – z)–1| ≤ cuµ – 1 throughout Γ + (x) ∪ Γ – (x), we have r
z Re ⎛ ---⎞ ⎝ q⎠
N+1
r
1 N+1 µ–N–2 Im ----------- ≤ cx u . q–z
(17)
In addition, we have N
Imz
k
( Rez ) ∑ Re ----------q z
N+1
N–k
≤ cx
N+µ+1 –N–1
u
(18)
k=1
and 1 µ–N–1 Im ----------≤ cu . N+1 q It follows from estimates (17), (18), and (19) that
(19)
( z/q ) µ–N–2 N+1 r r Im --------------------- ≤ cu x on Γ + ( x ) ∪ Γ – ( x ). q–z N+1
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
332
SOLOV’EV
Finally, we have ⎛ ∂ q -------- Re ⎜ log ----------- – ∂n q ⎝ q – z
N
∑ k=1
1 ⎛ z ⎞ k⎞ N+1 µ–N–2 --- --- ⎟ ≤ cx u . k ⎝ q⎠ ⎠
Consequently, δ
I 7 ( z ) ≤ cx
N+1
∫ ( σ ( u ) + σ ( u ) )u +
–
µ–N–2
du.
(20)
x
Let us take N equal to [µ]. By Hardy’s inequality, the integral in (20) belongs to p, β – µ(0, δ), and the norm in p, β – µ(Γ+ ∪ Γ–) of the left-hand side of the inequality does not exceed c||σ'||p, β + 1(Γ). Therefore, we have
∫
( 3 σ0 ) ( z+ ) – ( 3 σ0 ) ( z– ) z p
p(β – µ)
ds z ≤ c σ'
(21)
p, β + 1 .
Γ+ ∪ Γ–
∂ 1 (ii) Now we prove that the operator ----- 3 : p, β + 1 (Γ) ∂s parts, we represent ∂/∂sz(3σ) as a sum of three operators: ∂ z–q
p, β + 1 (Γ) is bounded. On integrating by
∂
------- ⎛ -----------⎞ -------- log -------------- ds , ∫ σ ( q ) ∂s ⎝ z – q⎠ ∂n z–q
( 31 σ ) ( z ) =
1
q
z
Γ
q
z–q ∂ z ( 32 σ ) ( z ) = – σ' ( q ) ----------- -------- log -------------- ds q , z – q ∂n z z–q
∫ Γ
∂ z–q ∂ z ( 33 σ ) ( z ) = – σ ( q ) ------- ----------- -------- log -------------- ds q . ∂s q z – q ∂n z z–q
∫ Γ
The fact that the operator 32 :
p, β + 1 (Γ) is bounded is proved in [3]. We now show
1 p, β + 1 (Γ) 1 p, β + 1 (Γ)
that the operators 31 and 33 map operators 1* and 2* defined as
∫
( 1* σ 0 ) ( z ) =
continuously into p, β + 1 (Γ). It suffices to consider the
Γ+ ∪ Γ–
( 2* σ 0 ) ( z ) =
∫
Γ+ ∪ Γ–
∂ z–q ∂ 1 σ 0 ( q ) ------- ----------- -------- log -------------- , ∂s z z – q ∂n q z–q
∂ z–q ∂ 1 σ 0 ( q ) ------- ----------- -------- log -------------- ds q . ∂s q z – q ∂n z z–q
Let M(q, z) denote the expression ∂ z–q ∂ ∂ z–q ∂ 1 z ------- ----------- -------- log -------------- + ------- ----------- -------- log -------------- . ∂s z z – q ∂n q ∂s q z – q ∂n z z–q z–q We use the formula
∫
Γ– ∪ Γ+
∫
σ 0 ( q )M ( q, z ) ds q = c Γ– ( x )
∪
σ 0 ( q )M ( q, z ) ds q + J ( z ); c Γ+ ( x )
here, |J(z)| does not exceed VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR
c -----2 x
(1 – ε) x
∫
c ( σ + ( u ) + σ – ( u ) ) du + -x
0
333
δ
du ( σ + ( u ) + σ – ( u ) ) -----2- . u (1 + ε) x
∫
Employing Hardy’s inequality, we obtain J
p, β + 1 ( Γ + ∪ Γ – )
≤ c σ'
p, β + 1 ( Γ ) .
Suppose that z ∈ Γ+. If q ∈ Γ + , we have c
M ( q, z ) ≤ cx , –2
whence it follows that (1 + ε) x
∫
c σ 0 ( q )M ( q, z ) ds q ≤ -----2 σ – ( u ) du. x (1 – ε) x
∫
c Γ+ ( x )
(22)
Now the required estimate for the p, β + 1 -norm of the left-hand side of the last expression can be derived from Hardy’s inequality. That the formulas ∂ ρ( x ) 1 2 – 1/2 –1 -2 + O ( x ), -------- log -------------- = ( 1 + κ '+ ( x ) ) ------------------------------------2 ∂n q z–q ( x – u) + ρ( x )
(23)
∂ ρ(u) 1 2 – 1/2 –1 -2 + O ( x ), -------- log -------------- = ( 1 + κ '+ ( x ) ) ------------------------------------2 ∂n z z–q ( x – u) + ρ(u)
(24)
ρ( x ) ∂ z–q –1 -2 + O ( x ), ------- ----------- = 2i ------------------------------------2 ∂s q z – q ( x – u) + ρ( x )
(25)
ρ(u) ∂ z–q –1 -2 + O ( x ). ------- ----------- = 2i ------------------------------------2 ∂s z z – q ( x – u) + ρ(u)
(26)
are valid on the arc Γ – can be verified by direct computation. It follows from (23)–(26) that c
c ρ(u) c - + ----- . M ( q, z ) ≤ --- ------------------------------------u ( x – u )2 + ρ ( u )2 x2 In view of (22), it suffices to estimate the p, β + 1 -norm of the integral (1 + ε) x
J( x) =
∫
(1 – ε) x
ρ( x ) σ – ( u ) ------------------------------------- du. ------------u ( x – u )2 + ρ ( x )2
Changing the variables τ = u–µ and ξ = x–µ, we obtain ∞
δ
∫x
p(β + 1)
∫
J ( x ) dx ≤ c ξ p
0
δ
p(1 – α) ⎛
∞
p
– 1/µ ⎞ σ – ( τ ) dτ ⎞ - -----dξ⎟ ⎜ -------------------------2 –1 ⎟ ⎝ –∞ ( ξ – τ ) + 1 τ ⎠ ⎠
–µ
1/ p
∫
.
By Theorem 3, up to a constant, the integral on the right does not exceed ∞
∫ δ
δ
σ– ( τ
– 1/µ
p – ( pβ + µ + 1 )/µ
) τ
∫
p
dτ ≤ c σ '– ( u ) u
–µ
p(β + 1)
(27)
du.
0
This completes the proof of the boundedness of the operators 31 and 33. Hence, we arrive at the inequality ( ∂/∂s ) ( 3 σ 0 )
p, β + 1 ( Γ )
≤ c σ'
Now the boundedness of the operator 3 : p, β + 1 (Γ) 1
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
(28)
p, β + 1 ( Γ ) .
p, β(Γ) follows from (21) and (28). Vol. 41
No. 4
2008
334
SOLOV’EV
The following theorem, which is the main result of the paper, is a corollary to Theorems 1, 2, and 3. Theorem 5. Let Ω be a domain with an outward peak and let 0 < β + p–1 < min{µ, 1}. Then the operator I – 2W : ( p, β + 1 × p, β + 1 ) ( Γ ) 1
( p, β × p , β ) ( Γ )
1
(29)
is continuous. The next theorem considers the operators R3 and I3 on a contour with an inward peak. Theorem 6. Let Ω be a domain with an inward peak and let 0 < β + p–1 < min{µ, 1}. Then the operators R 3 : p, β + 1 ( Γ ) 1
p, β ( Γ )
I 3 : p, β + 1 ( Γ )
p, β ( Γ )
and 1
are continuous. Proof. Take ε > 0 and the arcs Γ ± (x), Γ ± (x), and Γ ± (x) as in Theorem 4. We express σ as a sum of two functions σ0 and σ1 so that supp σ0 ⊂ Γ+ ∪ Γ– and supp σ1 ⊂ Γ\{|q| > δ/2} and write (3σ)(z), z = x + iκ±(x) ∈ Γ± as follows: l
c
∫ r
r
Γ+ ( x ) ∪ Γ– ( x )
⎛ +⎜ + ⎜ ⎝ Γl+ ( x ) ∪ Γl– ( x )
∫
∫
+
Γ\ { q > δ/2 }
r
z–q ∂ 1 σ 0 ( q ) ----------- -------- log -------------- ds q z – q ∂n q z–q
∫ c
Γ± ( x )
⎞ –q ∂ 1 ⎟ σ ( q ) z---------- -------- log -------------- ds q + ⎟ 0 z – q ∂n q z–q c Γ −+ ( x ) ⎠
∫
5
z–q ∂ 1 σ 1 ( q ) ----------- -------- log -------------- ds q = z – q ∂n q z–q
∑ I ( z ), k
k=1
The terms Ik, k = 2, …, 5, can be estimated in the same way as in the preceding theorem. Hence we content ourselves with the first term I1. We use the equality 1 1 ∂ 1 -------- log -------------- = – Re ----------- cos ( n q, x ) + Im ----------- cos ( n q, y ). (30) q–z q–z ∂n q q–z Suppose that q ∈ Γ + (x) ∪ Γ – (x) and that z ∈ Γ ∩ {|q| < δ/2}. For the first term on the right in (30), we have r
r
1 µ –1 Re ----------- cos ( n q, x ) = O ( x u ). q–z To obtain an expression for the second term on the right in (30), we employ the relation m
Im ( q – z )
–1
= Im
m+1
k
∑
(31)
1 z z -------------------+ Im -----------Re m+1 k+1 q–z q q k=0
(32)
1 1 1 1 m+1 m+1 Im ----------- + Re -----------Imz Re ----------- . + Re -----------Rez m+1 m+1 q – z q –z q q For the first term in (32), we have m
∑
k
z Im ---------= k+1 q k=0
m
∑ x Imq k
–k–1
µ –1
+ O ( x u ).
k=0
The second term on the right in (32) does not exceed cxµu–1. Now, since Im ( q – z )
–1
≤ c(u
µ–1
µ –1
+ x u ),
for the third term in (32), we obtain Req
–m–1
Rez
m+1
Im ( q – z )
–1
µ –1
≤ cx u .
VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
Lp-BOUNDEDNESS OF A BOUNDARY INTEGRAL OPERATOR
335
Regarding the last term in (32), we get Req
–m–1
Imz
m+1
Re ( q – z )
µ –1
≤ cx u .
–1
Thus, if q ∈ Γ + (x) ∪ Γ – (x), we have r
r
1 ∂ 1 -------- log -------------- = Im ---------cos ( n q, y ) + I ( q, z ), k+1 ∂n q q–z q where |I(q, z)| ≤ cxµu–1. Writing the factor (z – q)( z – q )–1 in I1 in the form q⎛ z 1 --- 1 – 2iIm --- ----------------⎞ q⎝ q 1 – z/q⎠ and allowing for the representations q z–q µ µ ----------- = --- ( 1 + O ( x ) ) and q/q = 1 + O ( x ), q z–q we arrive at the relation z–q ∂ 1 ----------- -------- log -------------- = z – q ∂n q z–q
m–1
cos ( n , y ) + O ( x u ∑ x Im ---------q 1
k
µ –1
q
k+1
).
(33)
k=0
By Hardy’s inequality, the operators (1 + ε) x
τ( x )
x
k
∫
τ ( u )u
µ–k–1
du,
0 ≤ k ≤ m,
0 ∞
τ( x )
x
µ
∫
τ ( u )u du –1
(1 + ε) x
map the space p, β(0, ∞) continuously into the space p, β – µ(0, ∞). It follows that
∫ r
r
Γ+ ( x ) ∪ Γ– ( x )
∂ 1 σ 0 ( q ) -------- log -------------- ds q = ∂n q z–q
m
∑c
(k)
( σ 0 )x + ( R 1 σ 0 ) ( z ) , k
k=0
where (k)
c (σ) =
∫
σ 0 ( q )g k ( q ) ds q
Γ+ ∪ Γ–
are continuous linear functionals in
1 p, β + 1 (Γ),
R1 σ0
and then (R1σ0)(z) can be estimated as
p, β – µ ( Γ ∩ { q < δ/2 } )
≤ c σ'
p, β + 1 ( Γ ) .
This concludes the proof of the theorem. REFERENCES 1. V. G. Maz’ya, “Boundary Integral Equations,” in Analysis IV. Linear and Boundary Integral Equations, Encycl. Math. Sci. 27, 127–222 (VINITI, Moscow, 1988; Springer, New York, 1991). 2. V. Maz’ya and A. Soloviev, “Lp-Theory of a Boundary Integral Equation on a Cuspidal Contour,” Appl. Anal. 65, 289–305 (1997). 3. V. Maz’ya and A. Soloviev, “Lp-Theory of Boundary Integral Equations on a Contour with Outward Peak,” Integral Equations Oper. Theory 32, 75–100 (1998). 4. V. Maz’ya and A. Soloviev, “Lp-Theory of Boundary Integral Equations on a Contour with Inward Peak,” Z. Anal. Anwend. 17 (3), 641–673 (1998). 5. E. Stein, “Note on Singular Integrals,” Proc. Amer. Math. Soc. 8, 250–254 (1957). VESTNIK ST. PETERSBURG UNIVERSITY. MATHEMATICS
Vol. 41
No. 4
2008
