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Integral representations for solutions of some BVPs for the Lame' system in
multiply connected domains
Boundary Value Problems 2011, 2011:53 doi:10.1186/1687-2770-2011-53
Alberto Cialdea ()
Vita Leonessa ()
Angelica Malaspina ()
ISSN 1687-2770
Article type Research
Submission date 21 May 2011
Acceptance date 12 December 2011
Publication date 12 December 2011
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Integral representations for solutions of some BVPs for the
Lam´e system in multiply connected domains
Alberto Cialdea
∗1
, Vita Leonessa
1
and Angelica Malaspina
1
1
Department of Mathematics and Computer Science, University of Basilicata,

and the theory of differential forms (see, e.g., [11, 12]) was introduced in [13] for the n-dimensional Laplace
equation and later extended to the three-dimensional elasticity in [14]. This method can be considered as
an extension of the one given by Muskhelishvili [15] in the complex plane. The double layer potential ansatz
for the traction problem can be treated in a similar way, as shown in [16].
In the present paper we are going to consider these two last approaches in a multiply connected bounded
domain of R
n
(n ≥ 2). Similar results for Laplace equation have been recently obtained in [17]. We
remark that we do not require the use of pseudo-differential operators nor the use of hypersingular integrals,
differently from other methods (see, e.g., [18, Chapter 4] for the study of the Neumann problem for Laplace
equation by means of a double layer potential).
After giving some notations and definitions in Section 2, we prove some preliminary results in Section 3.
They concern the study of the first derivatives of a double layer potential. This leads to the construction of
a reducing operator, which will be useful in the study of the integral system of the first kind arising in the
Dirichlet problem.
Section 4 is devoted to the case n = 2, where there exist some exceptional boundaries in which we need
to add a constant vector to the simple layer potential. In particular, after giving an explicit example of such
boundaries, we prove that in a multiply connected domain the boundary is exceptional if, and only if, the
external boundary is exceptional.
In Section 5 we find the solution of the Dirichlet problem in a multiply connected domain by means of a
simple layer potential. We show how to reduce the problem to an equivalent Fredholm equation (see Remark
5.5).
Section 6 is devoted to the traction problem. It turns out that the solution of this problem does exist in
the form of a double layer potential if, and only if, the given forces are balanced on each connected component
of the boundary. While in a simply connected domain the solution of the traction problem can be always
represented by means of a double layer potential (provided that, of course, the given forces are balanced on
the boundary), this is not true in a multiply connected domain. Therefore the presence or absence of “holes”
makes a difference.
We mention that lately we have applied the same method to the study of the Stokes system [19]. Moreover
2

and Ω
j
∩ Ω
k
= ∅, j, k = 1, . . . , m, j = k. For brevity, we shall
call such a domain an (m + 1)-connected domain. We denote by ν the outwards unit normal on Σ = ∂Ω.
Let E be the partial differential operator
Eu = ∆u + k∇divu,
where u = (u
1
, . . . , u
n
) is a vector-valued function and k > (n − 2)/n is a real constant. A fundamental
solution of the operator −E is given by Kelvin’s matrix whose entries are
Γ
ij
(x, y) =







1
2π

−
k + 2
2(k + 1)

+
k
2(k + 1)
(x
i
− y
i
)(x
j
− y
j
)
|x − y|
n

, if n ≥ 3,
(1)
i, j = 1, . . . , n, ω
n
being the hypersurface measure of the unit sphere in R
n
.
As usual, we denote by E(u, v) the bilinear form defined as
E(u, v) = 2σ
ih
(u) ε
ih
(v) = 2σ
ih
(v) ε

jj
(u) .
Let us consider the boundary operator L
ξ
whose components are
L
ξ
i
u = (k − ξ)(div u) ν
i
+ ν
j
∂
j
u
i
+ ξν
j
∂
i
u
j
, i = 1, . . . , n, (2)
ξ being a real parameter. We remark that the operator L
1
is just the stress operator 2σ
ih
ν
h
, which we shall

p
is integrable over Σ (j = 1, . . . , n). If h is any non-negative integer, L
p
h
(Σ) is
the vector space of all differential forms of degree h defined on Σ such that their components are integrable
functions belonging to L
p
(Σ) in a coordinate system of class C
1
and consequently in every coordinate
system of class C
1
. The space [L
p
h
(Σ)]
n
is constituted by the vectors (v
1
, . . . , v
n
) such that v
j
is a differential
form of L
p
h
(Σ) (j = 1, . . . , n). [W
1,p

being the adjoint of S (for more details
see, e.g., [9, 10]).
We end this section by defining the spaces in which we look for the solutions of the BVPs we are going
to consider.
Definition 2.1. The vector-valued function u belongs to S
p
if, and only if, there exists ϕ ∈ [L
p
(Σ)]
n
such
that u can be represented by a simple layer potential
u(x) =

Σ
Γ(x, y) ϕ(y) dσ
y
, x ∈ Ω. (3)
Definition 2.2. The vector-valued function w belongs to D
p
if, and only if, there exists ψ ∈ [W
1,p
(Σ)]
n
such that w can be represented by a double layer potential
w(x) =

Σ
[L
y

ω
n

2 + (1 − ξ)k
2(1 + k)
δ
ij
+
nk(ξ + 1)
2(k + 1)
(y
i
− x
i
)(y
j
− x
j
)
|y − x|
2

(y
p
− x
p
)ν
p
(y)
|y − x|

) , (6)
while for ξ = k/(2 + k) the kernels L
ξ
i,y
[Γ
j
(x, y)] have a strong singularity on Σ.
Let us denote by w
ξ
the double layer potential
w
ξ
j
(x) =

Σ
u
i
(y)L
ξ
i,y
[Γ
j
(x, y)] dσ
y
, j = 1, . . . , n. (7)
It is known that the first derivatives of a harmonic double layer potential with density ϕ belonging to
W
1,p
(Σ) can be written by means of the formula proved in [13, p. 187]

|x − y|
2−n
, if n ≥ 3
and s
h
(x, y) is the double h-form introduced by Hodge in [22]
s
h
(x, y) =

j
1
< <j
h
s(x, y)dx
j
1
. . . dx
j
h
dy
j
1
. . . dy
j
h
.
Since, for a scalar function f and for a fixed h, we have ∗df ∧ dx
h
= (−1)

, x ∈ Ω. (10)
The following lemma can be considered as an extension of formula (9) to elasticity. Here du denotes the
vector (du
1
, . . . , du
n
) and ψ = (ψ
1
, . . . , ψ
n
) is an element of [L
p
1
(Σ)]
n
.
Lemma 3.1. Let w
ξ
be the double layer potential (7) with density u ∈ [W
1,p
(Σ)]
n
. Then
∂
s
w
ξ
j
(x) = K
ξ

j
3
. . . dy
j
n
, (12)
K
ξ
hj
(x, y) =
1
ω
n
k(ξ + 1)
2(k + 1)
(y
l
− x
l
)(y
j
− x
j
)
|y − x|
n
+
k − (2 + k)ξ
2(k + 1)
δ

δ
ij
x
h
ν
h
|x|
n
− n
x
i
x
j
x
h
ν
h
|x|
n+2
,
we find in Ω
w
ξ
j
(x) = −
1
ω
n

Σ

)
|y − x|
n

+
k − (2 + k)ξ
2(k + 1)
M
ij
y

|y − x|
2−n
2 − n

dσ
y
=
−

Σ
u
j
(y)
∂s(x, y)
∂ν
y
dσ
y
+

y
[s(x, y)]

dσ
y
.
An integration by parts on Σ leads to
w
ξ
j
(x) = −

Σ
u
j
(y)
∂s(x, y)
∂ν
y
dσ
y
−

Σ
M
hi
[u
i
(y)]


dσ
y
−

Σ
M
hi
[u
i
(y)] K
ξ
hj
(x, y) dσ
y
.
6
Therefore, by recalling (9),
∂
s
w
ξ
j
(x) = Θ
s
(du
j
)(x) −

Σ
M

.
This identity is established by observing that on Σ we have
1
(n − 2)!
δ
123 n
hij
3
j
n
df ∧ dx
j
3
. . . dx
j
n
=
1
(n − 2)!
δ
123 n
hij
3
j
n
∂
j
2
fdx
j

2
ν
j
1
∂
j
2
f dσ = (ν
h
∂
i
f − ν
i
∂
h
f)dσ .
Then we can rewrite (14) as
∂
s
w
ξ
j
(x) = Θ
s
(du
j
)(x) −
1
(n − 2)!
δ

Σ
f(y)∂
x
s
(y
p
− x
p
)(y
j
− x
j
)
|y − x|
n
dσ
y
=
(15)
ω
n
2
(δ
pj
− 2ν
j
(η)ν
p
(η)) ν
s

−n
. Since h ∈ C
∞
(R
n
\ {0}) is even and homogeneous of degree 2 − n, due to
the results proved in [23], we have
lim
x→η

Σ
f(y)∂
x
s
(y
p
− x
p
)(y
j
− x
j
)
|y − x|
n
dσ
y
= −ν
s
(η)γ

F(h)(x) =

R
n
h(y) e
−2π i x·y
dy
(see also [24] and note that in [23, 24] ν is the inner normal). On the other hand
F(h
pj
)(x) =
1
2 − n
F(x
p
∂
j
(|x|
2−n
)) = −
1
(2 − n) 2πi
∂
p
F(∂
j
(|x|
2−n
)) = −
1

−2
) =
π
n/2−2
(n − 2) Γ(n/2 − 1)
(δ
pj
|x|
−2
− 2x
j
x
p
|x|
−4
).
Finally, keeping in mind that ω
n
= n π
n/2
/Γ(n/2 + 1) and Γ(n/2 + 1) = n(n − 2) Γ(n/2 − 1)/4, we obtain
γ
pj
(η) = −2
π
n/2
(n − 2) Γ(n/2 − 1)
(δ
pj
− 2ν

x→η
Θ
s
(ψ)(x) = −
1
2
ψ
s
(η) + Θ
s
(ψ)(η), (18)
where Θ
s
is given by (10) and the limit has to be understood as an internal angular boundary value.
Proof. First we note that the assumption (17) is not restrictive, because, given the 1-form ψ on Σ, there
exist scalar functions ψ
h
defined on Σ such that ψ = ψ
h
dx
h
and (17) holds (see [26, p. 41]). We have
Θ
s
(ψ)(x) =

j
1
< <j
n−2

j
1
< <j
n−2
δ
1 2 n
kj
1
j
n−2
h
δ
12 n
ij
1
j
n−2
s

Σ
∂
x
i
[s(x, y)]ν
k
(y)ψ
h
(y) dσ
y
= δ

(η) + Θ
s
(ψ)(η)
a.e. on Σ. From (17) it follows that δ
is
kh
ν
i
ν
k
ψ
h
= ν
i
ν
i
ψ
s
− ν
i
ν
s
ψ
i
= ψ
s
and (18) is proved.
Lemma 3.4. Let ψ ∈ L
p
1

=
−

k − ξ
2(k + 1)
ν
j
(η)ψ
i
(η) +
ξ
2
ν
i
(η)ψ
j
(η)

ν
s
(η)+
1
(n − 2)!
δ
123 n
lij
3
j
n


x
s
K
ξ
lj
(x, y) ∧ ψ(y) ∧ dy
j
3
. . . dy
j
n
=
1
(n − 2)!
δ
123 n
lij
3
j
n
δ
123 n
rhj
3
j
n

Σ
∂
x

lim
x→η
1
(n − 2)!
δ
123 n
lij
3
j
n

Σ
∂
x
s
K
ξ
lj
(x, y) ∧ ψ(y) ∧ dy
j
3
. . . dy
j
n
=
δ
li
rh

k(ξ + 1)

j
n

Σ
∂
x
s
K
ξ
lj
(η, y) ∧ ψ(y) ∧ dy
j
3
. . . dy
j
n
.
On the other hand
δ
li
rh

k(ξ + 1)
4(k + 1)
(δ
lj
− 2ν
j
ν
l

l
ν
s

ν
r
ψ
h
=

ξ
2
δ
lj
ν
s
−
k(ξ + 1)
2(k + 1)
ν
j
ν
l
ν
s

(ν
l
ψ
i

1
(Σ)]
n
. Then, for almost every η ∈ Σ,
lim
x→η
[(k − ξ)K
ξ
jj
(ψ)(x)ν
i
(η) + ν
j
(η)K
ξ
ij
(ψ)(x) + ξν
j
(η)K
ξ
ji
(ψ)(x)] =
(k − ξ)K
ξ
jj
(ψ)(η)ν
i
(η) + ν
j
(η)K

js
(ψ)(x) = −
1
2
ψ
js
(η) +

k − ξ
2(k + 1)
ν
j
(η)ψ
hh
(η) +
ξ
2
ν
h
(η) ψ
hj
(η)

ν
s
(η) + K
ξ
js
(ψ)(η).
9

j
(η)K
ξ
ji
(ψ)(η),
where
Φ(ψ) = (k − ξ)

−
1
2
ψ
jj
+

k − ξ
2(k + 1)
ν
j
ψ
hh
+
ξ
2
ν
h
ψ
hj

ν


+ ξ ν
j

−
1
2
ψ
ji
+

k − ξ
2(k + 1)
ν
j
ψ
hh
+
ξ
2
ν
h
ψ
hj

ν
i

.
Conditions (21) lead to

be the double layer potential (7) with density u ∈ [W
1,p
(Σ)]
n
. Then
L
ξ
+,i
(w
ξ
) = L
ξ
−,i
(w
ξ
) = (k − ξ)K
ξ
jj
(du)ν
i
+ ν
j
K
ξ
ij
(du) + ξν
j
K
ξ
ji

be the following singular integral operator
Rϕ(x) =

Σ
d
x
[Γ(x, y)] ϕ(y) dσ
y
. (23)
10
Let us define R
ξ
: [L
p
1
(Σ)]
n
→ [L
p
(Σ)]
n
to be the singular integral operator
R
ξ
i
(ψ)(x) = (k − ξ)K
ξ
jj
(ψ)(x)ν
i

x
[Γ(x, y)] ϕ(y) dσ
y
. (26)
Proof. Let u be the simple layer potential with density ϕ ∈ [L
p
(Σ)]
n
. In view of Lemma 3.7, we have a.e.
on Σ
R
ξ
i
(Rϕ) = (k − ξ)K
ξ
jj
(du)ν
i
+ ν
j
K
ξ
ij
(du) + ξν
j
K
ξ
ji
(du) = L
ξ

[u(y)] Γ
ij
(x, y) dσ
y
and then, on account of (26),
L
ξ
w
ξ
= −
1
2
L
ξ
u + T
ξ
(L
ξ
u) = −
1
2

1
2
ϕ + T
ξ
ϕ

+ T
ξ

, x ∈ Σ, (27)
and denote by T
∗
its adjoint.
In this subsection we determine the dimension of the following eigenspaces
V
±
=

ϕ ∈ [L
p
(Σ)]
n
: ∓
1
2
ϕ + T
∗
ϕ = 0

; (28)
11
W
±
=

ϕ ∈ [L
p
(Σ)]
n

Σ
j
: h = 1, . . . , n(n + 1)/2, j = 1, . . . , m} ,
where {v
h
: h = 1 . . . , n(n + 1)/2} is an orthonormal basis of the space R and χ
Σ
j
is the characteristic
function of Σ
j
.
Proof. We define the vector-valued functions α
j
, j = 1, . . . , m as α
j
(x) = (a + Bx)χ
Σ
j
(x), x ∈ Σ. For a
fixed j = 1, . . . , m, the function α
j
(x) belongs to V
+
; indeed
−
1
2
(a+Bx)χ
Σ

−
1
2
(a + Bx) +
1
2
(a + Bx) = 0, x ∈ Σ
j
,
because of

Σ
[L
y
Γ(x, y)]

α
j
(y) dσ
y
=





α
j
(x) x ∈ Ω
j

hj
= 0, we have
n(n+1)/2

h=1
c
hj
v
h
(x) = 0, x ∈ Σ
j
, j = 1, . . . , m.
Then, by applying a classical uniqueness theorem to the domain Ω
j
,
n(n+1)/2

h=1
c
hj
v
h
(x) = 0, x ∈ Ω
j
, j = 1, . . . , m,
12
from which it easily follows that
c
hj
= 0, h = 1, . . . , n(n + 1)/2, j = 1, . . . , m.

−→ (R
n
× S
n
)
m
ϕ −→ (a
1
, B
1
, . . . , a
m
,B
m
).
If τ (ϕ) = 0, from a classical uniqueness theorem, we have that ϕ ≡ 0 in R
n
. Thus, τ is an injective map and
dim W
−
≤ n(n + 1)m/2. The assertion follows from (30).
Lemma 3.12. The spaces V
−
and W
+
have dimension n(n + 1)/2. Moreover V
−
is constituted by the
restrictions to Σ of the rigid displacements.
Proof. Let α ∈ R. If x ∈ Σ, we have




−α(x) x ∈ Ω,
−α(x)/2 x ∈ Σ,
0 x /∈ Ω.
This shows that the restriction to Σ of α belongs to V
−
and then dim V
−
≥ dim R = n(n + 1)/2. On the
other hand, suppose φ ∈ W
+
and let u be the simple layer potential with density φ. Since Eu = 0 in Ω and
L
+
u = 0 on Σ, u = a + Bx in Ω. Let σ be the linear map
σ : W
+
−→ R
n
× S
n
φ −→ (a, B).
If n ≥ 3, we have that σ(φ) = 0 implies u ≡ 0 in R
n
and then φ ≡ 0 on Σ, in view of classical uniqueness
theorems.
If n = 2, define W
0

≤ 3. In any case, dim W
+
≤ n(n + 1)/2 and the result follows from (30).
13
4 The bidimensional case
The case n = 2 requires some additional considerations. It is well-known that there are some domains in
which no every harmonic function can be represented by means of a harmonic simple layer potential. For
instance, on the unit disk we have

|y|=1
log |x − y| ds
y
= 0, |x| < 1.
Similar domains occur also in elasticity. In order to give explicitly such an example, let us prove the
following lemma.
Lemma 4.1. Let Σ
R
be the circle of radius R centered at the origin. We have

Σ
R
|x − y|
2
log |x − y| ds
y
= 2πR(R
2
log R + (1 + log R)|x|
2
), |x| < R. (32)

R
. Moreover
∆u(x) = 4

Σ
R
(1 + log |x − y|) ds
y
and then also ∆u is constant on Σ
R
. Since ∆u is harmonic in Ω
R
and continuous on
Ω
R
, it is constant in
Ω
R
and then
∆u(x) = ∆u(0) = 4

Σ
R
(1 + log |y|) ds
y
= 8πR(1 + log R), x ∈ Ω
R
.
The function u(x) − 2πR(1 + log R)|x|
2

ij
(x, y) ds
y
= δ
ij
R
4(k + 1)
(k − 2(k + 2) log R), |x| < R. (33)
14
Proof. Since
∂
11

Σ
R
|x − y|
2
log |x − y| ds
y
= 2

Σ
R
log |x − y| ds
y
+ 2

Σ
R
(x

2
− y
2
)
2
|x − y|
2
ds
y
= πR, |x| < R.
From (32) we have also
∂
12

Σ
R
|x − y|
2
log |x − y| ds
y
= 2

Σ
R
(x
1
− y
1
)(x
2

If there exists some constant vector which cannot be represented in the simply connected domain Ω by
a simple layer potential, we say that the boundary of Ω is exceptional. We have proved that
Lemma 4.3. The circle Σ
R
with R = exp[k/(2(k + 2))] is exceptional for the operator ∆ + k∇div.
Due to the results in [28], one can scale the domain in such a way that its boundary is not exceptional.
Here we show that also in some (m + 1)-connected domains one cannot represent any constant vectors
by a simple layer potential and that this happens if, and only if, the exterior boundary Σ
0
(considered as
the boundary of the simply connected domain Ω
0
) is exceptional.
We note that, if any constant vector c can be represented by a simple layer potential, then any sufficiently
smooth solution of the system Eu = 0 can be represented by a simple layer potential as well (see Section 5
below).
We first prove a property of the singular integral system

Σ
ϕ
j
(y)
∂
∂s
x
Γ
ij
(x, y) ds
y
= 0, x ∈ Σ, i = 1, 2. (34)

− x
i
)(y
j
− x
j
)
|x − y|
2

ds
y
and, since
∂
∂s
x
(x
i
− y
i
)(x
j
− y
j
)
|x − y|
2
= ˙x
i
x

i
∂
∂x
j
log |x − y| + ˙x
j
∂
∂x
i
log |x − y| − 2
(x
i
− y
i
)(x
j
− y
j
)
|x − y|
2
∂
∂s
x
log |x − y| =
2

˙x
i
˙x

j
− y
j
)
|x − y|
2
= O(|y − x|
h−1
).
We have proved that
4
∂
∂s
x
Γ
ij
(x, y) = −
1
2π
k + 2
2(k + 1)
δ
ij
∂
∂s
x
log |x − y| + O(|y − x|
h−1
)
and then the system (34) is of regular type (see [15, 29]). From the general theory we know that such a

∞
(R
2
)]
2
. This can be siproved by the same method in [13, pp. 189–190]. Therefore ψ
has to be constant on each curve Σ
j
(j = 0, . . . , m), i.e. ψ is a linear combination of e
i
χ
Σ
j
(i = 1, 2,
j = 0, 1, . . . , m).
Theorem 4.5. Let Ω ⊂ R
2
be an (m + 1)-connected domain. The following conditions are equivalent:
16
I. there exists a H¨older continuous vector function ϕ ≡ 0 such that

Σ
Γ(x, y) ϕ(y) ds
y
= 0, x ∈ Σ; (36)
II. there exists a constant vector which cannot be represented in Ω by a simple layer potential (i.e., there
exists c ∈ R
2
such that c /∈ S
p

where
C =








α
1,0
. . . α
2m+2,0
. . . . . . . . .
α
1,m
. . . α
2m+2,m
β
1,0
. . . β
2m+2,0
. . . . . . . . .
β
1,m
. . . β
2m+2,m



1
and ψ
2
such that

Σ
Γ(x, y) ψ
j
(y) ds
y
= e
j
, x ∈ Ω, j = 1, 2.
Arguing as before, we find ψ
j
= 0 on Σ
k
, k = 1, . . . , m, j = 1, 2, and then

Σ
0
Γ(x, y) ψ
j
(y) ds
y
= e
j
, x ∈ Ω
0
, j = 1, 2.

+ µ
2
ψ
2
= 0 on Σ
0
. (38)
This implies

Σ
0
Γ(x, y) (λϕ(y) + µ
1
ψ
1
(y) + µ
2
ψ
2
(y)) ds
y
= 0, x ∈ Ω
0
,
i.e. µ
1
e
1
+ µ
2

0
,
0 y ∈ Σ \ Σ
0
,
we can write

Σ
Γ(x, y) ϕ(y) ds
y
= c, x ∈ Ω,
and this contradicts II.
III ⇒ IV. Let us suppose det C = 0. For any c = (α, β) ∈ R
2
there exists λ = (λ
1
, . . . , λ
2m+2
) solution
of the system
2m+2

j=1
λ
j
α
jk
= α,
2m+2


2m+2

j=1
λ
j
ϕ
j
= 0 on Σ
k
for k = 1, . . . , m. Then Σ
0
is not exceptional.
IV ⇒ I. From (37) it follows that there exists an eigensolution λ = (λ
1
, . . . , λ
2m+2
) of the homogeneous
system
2m+2

j=1
λ
j
c
jk
= 0, k = 0, . . . , m.
Set
ϕ(x) =
2m+2


.
5 The Dirichlet problem
The purpose of this section is to represent the solution of the Dirichlet problem in an (m + 1)-connected
domain by means of a simple layer potential. Precisely we give an existence and uniqueness theorem for the
problem



u ∈ S
p
,
E u = 0 in Ω,
u = f on Σ,
(40)
where f ∈ [W
1,p
(Σ)]
n
.
We establish some preliminary results.
Theorem 5.1. Given ω ∈ [L
p
1
(Σ)]
n
, there exists a solution of the singular integral system

Σ
d
x

R
∗
j
ψ(x) =

Σ
ψ
i
(y) ∧ d
y
[Γ
ij
(x, y)], x ∈ Σ.
Thanks to Corollary 3.10, the integral system (41) admits a solution ϕ ∈ [L
p
(Σ)]
n
if, and only if,

Σ
ψ
i
∧ ω
i
= 0 (43)
for any ψ = (ψ
1
, . . . , ψ
n
) ∈ [L

(23).
Proof. Consider the following singular integral system:

Σ
d
x
[Γ(x, y)] ϕ(y) dσ
y
= df(x), x ∈ Σ, (45)
in which the unknown is ϕ ∈ [L
p
(Σ)]
n
and the datum is df ∈ [L
p
1
(Σ)]
n
. In view of Theorem 5.1, there exists
a solution ϕ of system (45) because conditions (42) are satisfied.
In the next result we consider the eigenspace F of the Fredholm integral system
−
1
2
ψ(x) +

Σ
L
k/(k+2)
x

h=1
n

i=1
(c
i
h
− c
i
0
)

Σ
Γ(x, y) Ψ
h,i
(y) dσ
y
+ c
0
, x ∈ Ω, (47)
where Ψ
h,i
∈ F (h = 1, . . . , m, i = 1, . . . , n) satisfy the following conditions

Σ
Γ(x, y) Ψ
h,i
(y) dσ
y
= δ

component of R
n
\ Ω. Then V
j
= 0 in R
n
\ Ω
0
5
and V
j
(x) = a
k
j
in Ω
k
(k = 1, . . . , m). For every k = 1, . . . , m,
consider the n × nm matrix D
k
defined as follows
D
k
=






a

. . . D
m
)

has a not vanishing determinant. Indeed, if det D = 0, the linear
system Dλ = 0 admits an eigensolution λ = (λ
1
, . . . , λ
nm
) ∈ R
nm
. Hence the potential
W (x) =
nm

j=1
λ
j
V
j
(x)
vanishes not only on R
n
\ Ω
0
, but also on Ω
k
(k = 1, . . . , m). Since this implies W = 0 on Σ, we find W = 0
in Ω, thanks to the classical uniqueness theorem for the Dirichlet problem. Accordingly, W = 0 all over R
n

e
i
, k = 1, . . . , m.
Setting
V
h,i
(x) =
nm

j=1
λ
h
i,j
V
j
(x), x ∈ Ω,
we get E
V
h,i
= 0, V
h,i
|
Σ
0
= 0 and
V
h,i
|
Σ
k

h,i
(x) + c
0
.
The potential v belongs to S
p
, thanks to the isomorphism σ introduced in the proof of Lemma 3.12 (for
n = 2 see Definition 4.6). Moreover
v(x)|
Σ
k
=
m

h=1
n

i=1
(c
i
h
− c
i
0
)δ
hk
e
i
+ c
0

follows that the condition u = 0 on Σ implies that
−
1
4
ϕ +

T
k/(k+2)

2
ϕ = 0, (48)
where T
k/(k+2)
is the compact operator given by (26). By bootstrap techniques, (48) implies that ϕ is a
H¨older function on Σ. Then u belongs to [C
1,λ
(Ω) ∩ C
2
(Ω)]
n
and we get that

Ω
E(u, u) dx = 0,
from which
E(u, u) = 0 in Ω. (49)
The solution of (49) is u(x) = a + Bx, where a ∈ R
n
and B ∈ S
n

df.
Therefore, even if we do not have an equivalent reduction in the usual sense, such Fredholm system is
equivalent to the Dirichlet problem (40).
6 The traction problem
The aim of this section is to study the possibility of representing the solution of the traction problem by
means of a double layer potential. As we shall see, in an (m + 1)-connected domain this is possible if, and
only if, the given forces are balanced on each connected component Σ
j
of the boundary.
More precisely, we consider the problem



w ∈ D
p
,
Eu = 0 in Ω,
Lw = f on Σ,
(50)
22
where f ∈ [L
p
(Σ)]
n
is such that

Σ
f(x) (a + Bx) dσ
x
= 0, a ∈ R

n
(0 < h < λ) such that ψ
k
→ ψ in [W
1,2
(Σ)]
n
.
Setting
w
k
(x) =

Σ
[L
y
Γ(x, y)]

ψ
k
(y) dσ
y
,
we have that w
k
∈ [C
1,h
(Ω)]
n
, Ew

On the other hand we have that K
sj
(dψ
k
) → K
sj
(dψ) in L
2
(Ω). By applying formula (11), we see that
∇w
k
→ ∇w in [L
2
(Ω)]
n
. Moreover, since K
sj
(dψ
k
) → K
sj
(dψ) also in L
2
(Σ), (22) shows that Lw
k
→ Lw
in [L
2
(Σ)]
n

2
φ = f, (56)
23
where T is given by (27).
Proof. Assume that conditions (54) hold. If u is the double layer potential with density ψ ∈ [W
1,p
(Σ)]
n
, in
view of (22) the boundary condition Lu = f turns into the equation
R

(dψ) = f, (57)
where R

is given by (24) with ξ = 1.
On account of Theorem 5.4, if n = 2 and Σ
0
is exceptional, any ψ ∈ [W
1,p
(Σ)]
2
can be written as

Σ
Γ(x, y)φ(y) dσ
y
+ c,
with φ ∈ [L
p

−
1
2
φ + Tφ = γ. (59)
From Lemma 3.11 the dimension of the kernel N (−I/2 + T
∗
) = V
+
is n(n + 1)m/2 and {v
h
χ
Σ
j
: j =
1, . . . , m, h = 1, . . . , n(n + 1)/2} is a basis of it. The equation (59) has a solution if, and only if,

Σ
j
γv
h
dσ = 0, j = 1, . . . , m, h = 1, . . . , n(n + 1)/2. (60)
Since γ is solution of (58), conditions (60) are fulfilled. Indeed, picking j = 1, . . . , m and h = 1, . . . , n(n+1)/2,
by integrating (58) on Σ
j
we find (see (31))

Σ
j
fv
h

dσ +

Σ
γ(y) dσ
y

Σ
j
[L
x
Γ(x, y)]

v
h
(x) dσ
x
=

Σ
j
γv
h
dσ .
24