Proof. Step 1. Let
. For any constant skew-symmetric matrix
A, we construct a generalized solution
of the Dirichlet–Robin problem for the system (
1) in
with the boundary conditions
satisfying the conditions
, and
Such a solution may be constructed by the variational method, minimizing the functional
in the class of admissible functions
,
has compact support in
}. The boundedness of the Dirichlet follows from Korn’s inequality ([
7]; §3, inequality (43)). Condition (
6) is a consequence of Hardy’s inequality [
7].
Let be linearly independent constant skew-symmetric –matrices. We consider the solutions .
Step 2. Now in the same way, for any constant vector
,
we construct a generalized solution
of the Dirichlet–Robin problem for the system (
1) with the boundary conditions
and with conditions
,
Such a solution may be constructed by a variational method, minimizing the corresponding functional in the class of functions
has compact support in
. The boundedness of the Dirichlet follows from Korn’s inequality ([
7]; §3, inequality (43)). Condition (
7) follows from Hardy’s inequality ([
7]; §3, inequality (27)).
Step 3. The solutions
are linearly independent. Indeed, if
where
and
are constants, then
, since
has a finite Dirichlet integral
. Therefore
. Hence
where
B is a constant vector. Since the
satisfy (
7), we have
. The vectors
are linearly independent, and therefore
.
Thus, we have proved that the homogenous Dirichlet–Robin problem has at least linearly independent generalized solutions.
Step 4. Let us now prove that any generalized solution
u of the homogenous Dirichlet–Robin problem with the condition
is a linear combination of the constructed solutions. According to Korn’s inequality ([
7]; §3, inequality (43)), there is skew-symmetric matrix
A such that
Let
. For the function
we have
, since
. Hardy’s inequality implies that
where
B is a constant vector.
Let
. We set
, where
. It is easy to see that
Let us show that
. We substitute in the integral identity (
4) for
w the vector-valued function
, where
,
for
for
, we get
where
.
We claim that the right-hand side of (
8) approaches zero as
. Indeed, the Cauchy–Schwartz inequality yields that
where
Since
it follows that
and
as
. Hence,
Using the integral identity
we find that if
w is a solution of the homogeneous problem (
1), (
2), then
. The set of points where
is a linear manifold of dimension less than
, since the rank of the matrix
is
if
. Consequently,
. This conclusion follows from the fact that
and hence
on a subset of
of positive measure. This means that
. The theorem is proved for
.
Let now
. For a nontrivial constant skew-symmetric matrix
A, we construct a generalized solution
of the Dirichlet–Robin problem for the system (
1) in
with the boundary conditions (
5), minimizing the corresponding functional
in the class of functions
has a compact support in
}. This solution satisfies
and
. By Hardy’s inequality [
6] we get
where
is such that
.
We prove further that any generalized solution
u of the homogeneous Dirichlet–Robin problem (
1), (
2) has the form
, where
,
A is a skew-symmetric matrix, and
.
We observe that
, since
. By Korn’s inequality ([
7]; §3, inequality (43)), there is a skew-symmetric matrix
such that
We set
, where
, if
. It is easy to see that
, since (
10) and
by construction. Therefore, for
w, by Hardy’s inequality, the inequality of the form (
9) holds.
Let us prove that
. Substituting in the integral identity (
4) for
w the function
, where
,
for
for
. Further, as above, we obtain that
. This concludes the proof. □
Proof. We first consider the case , . Obviously, for .
We claim that
if
. Indeed, let
. By Lemma 2 we have equality (
3):
where
is a constant skew-symmetric matrix,
B is a constant vector,
It is easy to verify that and for . Hence , that is, . Therefore .
Thus, and . Using Theorem 1 we have .
Consider now the case when and .
Let
, where
. By Korn’s inequality ([
7]; §3, inequality (43)), there is a constant skew-symmetric matrix
A such that
where the constant
C is independent of
u.
For the function
we have
and
. Moreover,
v is a solution of (
1) in
. Hence, by Lemma 1, it has the form (
3):
where
is a polynomial, and
Let us prove that . First, establish the inequality . We have , and it is easy to verify that for . Hence by the triangle inequality..
Let
, where
. Then we have
in the interior of some cone
K. Hence,
The last integral converges if and only if . Hence and, therefore, and , where B is a constant vector. Thus, .
On the other hand, . Hence we have , where A is a constant skew-symmetric matrix and B is a constant vector. It is easy to verify that and . Hence , that is, . We obtain the embedding . In addition, it is obvious that for .
Thus, and By Theorem 1 we obtain . Hence .
Step 1. Now let
. For a non-trivial constant skew-symmetric matrix
A, we construct a generalized solution
of the mixed Dirichlet–Robin problem for the system (
1) in
with boundary conditions
by minimizing the functional
in the class of admissible functions
has compact support in
}. The resulting solution satisfies
and
. By Hardy’s inequality [
6] we obtain
where
and
.
Step 2. In this same way we obtain generalized solutions of the mixed Dirichlet–Robin for the system (
1) in
with the boundary conditions
and with the properties
, and
The solutions
and
are linearly independent. Indeed, if
for some constants
and
, then
because
has a finite Dirichlet integral
and
Thus, , where .
Since
and
, it follows that
. Hence
, and applying the elasticity operator to this equation, we obtain
where
I is the
unit matrix,
is the Dirac function. Hence it follows that
. It is easy to verify that
and
for
.
Hence the Dirichlet–Robin problem (
1), (
2) has at least three linearly independent solutions satisfying
.
Step 3. We claim that each generalized solution
u of the Dirichlet–Robin problem (
1), (
2) with condition
is a linear combination of the solutions constructed above. By Korn’s inequality ([
7]; §3, inequality (43)), there is a constant skew-symmetric matrix
such that
where the constant
C is independent of
u. For the function
we have
and
. Since
is a solution of the system (
1) in
, it follows by Lemma 1 that
where
is a polynomial and
In a similar way to the above we can show that
and
,
is a constant vector. Thus,
where
Let
and
. We set
Obviously,
w is a solution of the system (
1) in
,
and
It is easy to see that
and
Thus,
is a solution of the following problem
:
Let us prove that the solution
of problem
is unique, that is,
,
. To this end, we write the integral identity (
4) for the vector-valued function
, where
,
for
for
, we get
where
.
We claim that the right-hand side of (
11) approaches zero as
. Indeed, the Cauchy–Schwartz inequality yields that
where
Since
it follows that
and
as
. Hence
Using the integral identity
we find that if
w is a solution to the homogeneous problem (
1), (
2), then
. The set of all
x such that
is a linear manifold whose dimension is less than
, since the rank of the matrix
is
if
. Therefore,
. The relation
implies that
on a set of positive measure on
, and therefore,
The theorem is proved. □
Proof. Step 1. Assume that
. For each constant vector
:
we construct a generalized solution
of the Dirichlet–Robin problem for the system (
1) with the boundary conditions
and with the additional conditions
, and
Such a solution is constructed by the variational method. We minimize the corresponding functional over the class of admissible functions
has compact support in
. The boundedness of the Dirichlet integral follows from Korn’s inequality ([
7]; §3, inequality (43)). Condition (
13) follows from Hardy’s inequality ([
7]; §3, inequality (27)). By Lemma 2 the solution
takes the form
where
is a polynomial,
, with
A being a constant skew-symmetric matrix and
B being a constant vector, and
We claim that
. Assume that
. Then in the interior of a certain cone
K we have
and
This contradiction shows that
. Thus,
For a proof we consider a ball
with centre at the origin suck that
. Let
,
. Then
There exists a sequence of domains such that and .
We claim that the integrals in the right- and left-hand sides of (
17) converge. Indeed, the Cauchy–Schwartz inequality yields that
because
and
.
In fact, by the Cauchy–Schwartz inequality and the estimates
we obtain
as
for
. The constants
c and
C are independent of
R.
Letting
R in (
17) tend to infinity, we obtain the required Equation (
16):
and bearing in mind that
, we get
where
.
We claim that the constant
is non-zero in (
15). Indeed, if
, then
and
. Taking the scalar product of the system (
1) and 1 and integrating over
, we obtain
Hence,
and by (
18) we obtain
Using the integral identity
we get
, since
. By [
5], it follows that
, where
A is a constant skew-symmetric matrix and
B is a constant vector.
On the other hand, Formula (
15) with
and
yields that
Hence,
that is,
, which is possible only if
and
, that is,
.
However, and . This contradiction shows that if , then .
Step 2. Let be an arbitrary vector in . We consider the solution such that , , and , where .
We can associate with each vector in the corresponding vector in , thus obtaining a transformation such that , where . It is easy to verify that the transformation S is linear and non-degenerate.
Let
be a basis in
. For arbitrary linearly independent vectors
there exists a unique linear transformation (matrix)
S such that
. Then
Step 3. Consider now the elasticity system (
1) in
with boundary conditions
where
A is a constant skew-symmetric matrix. For every such matrix
A we construct a generalized solution of the system (
1) with the boundary conditions (
20) and the properties
,
Such a solution can be constructed using the variational method and minimizing the corresponding functional over the set of admissible functions
has compact support in
}. The Dirichlet integral is bounded by Korn’s inequality ([
7]; §3, inequality (43)). Condition (
21) follows from Hardy’s inequality [
7]. By Lemma 2 we have
As before, we can show that
. Hence,
Step 4. Consider now the difference
where
and
and
are defined by (
15) and (
23) respectively. Obviously,
v is a solution of (
1) in
and
.
We claim that
for
. Since
it follows by (
15) and (
23) that
where
.
It is easy to verify that and Hence Note also that . Therefore .
We now claim that
. For let
, that is,
where
and
as
. Then we obtain
On the other hand, as . This contradiction shows that .
Let us prove that if
is a basis in the space of skew-symmetric matrices, then
are linearly independent solutions, i.e., from the equality
follows that
. Indeed, assume that
where
, that is, let
Then we set
, so that
To prove that
, we put
, where
. Then
because
and
On the other hand,
and the integral on the right-hand side is finite if and only if
, that is,
and
, so that,
.
Thus, the mixed Dirichlet–Robin problem (
1), (
2) has at least
linearly independent generalized solutions.
Step 5. Let us show that each generalized solution
of the homogeneous problem (
1), (
2) such that
is a linear combination of the solutions
, that is,
By Lemma 2, the solution of the system (
1) in
has a representation (
3). Let us prove that there are
, such that the following equation holds for all
:
Since
is a basis in the space of skew-symmetric matrices, there are
, such that
. We put
where
. Obviously,
is a solution of (
1) in
,
, and
for
.
Step 6. Consider now the difference By construction, and . It follows by the triangle inequality that for .
We claim that
in
. Indeed, let
. Then
where
that is,
and
.
It is easy to see that
and
. Thus, we obtain problem
:
By construction, we have problem
:
We shall now prove the uniqueness of a solution of problem
. Let
and
be solutions such that
Then the function
satisfies
We claim that
in
. Indeed, consider the integral identity (
4) for
and put
, where
for
for
,
. We get
where
.
In the same way as in (
11) (Theorem 2, case n = 2), we can show that the right-hand side of (
25) tends to zero as
. Hence,
Using the integral identity
we find that if
is a solution to the homogeneous problem (
1), (
2), then
. The set of all
x such that
is a linear manifold whose dimension is less than
, since the rank of the matrix
is
if
. Therefore,
. The relation
implies that
on a set of positive measure on
, and therefore,
Thus, the solution to problem
is unique.
First of all we show that if
in (
15), then
By the properties of the fundamental solution of the elasticity system [
38], if
, where
is a homogeneous function of order zero, then
is a homogeneous function of order
, that is,
where
is a homogeneous function of order zero. We fix a point
such that
. By continuity,
in a neighborhood
of
. We consider a cone
K with vertex at the origin such that
. Then
for
Applying the triangle inequality to the Formula (
15) of the type
, where
we obtain
It is easy to verify that
. Hence
for
, and we obtain the problem
:
For the function
we have
This contradiction shows that
, that is,
. Hence
is a solution of the following problem
:
By the unique solubility of the problem , we have in . Hence , and since , it follows that in . This proves the theorem for .
The proof in the case
is carried out in a similar way. For a non-trivial constant skew-symmetric matrix
A, we construct a generalized solution
of the mixed Dirichlet–Robin problem for the system (
1) in
with the boundary conditions (
20) by minimizing the corresponding functional over the class of admissible functions
,
has a compact support in
}. This solution satisfies
and
. By Hardy’s inequality [
6] we obtain
where
and
.
Let us prove that each generalized solution
u of the problem (
1), (
2) satisfying the condition
has the following form:
where
A is a constant skew-symmetric matrix. By Lemma 2, the solution of the system (
1) has the form (
22) with
, where
A is a constant skew-symmetric matrix and
B is a constant vector.
We claim that
. For assuming that
, we can write (
22) in the following form:
By construction,
that is,
for each
R. It is easy to see that
Hence,
for any
Since
is a fundamental solution of (
1),
, where
and
are
–matrices whose entries are homogeneous functions of order zero (see [
39]), and so
. It follows that
By (
22) and the triangle inequality, we have
Hence for each . This contradiction shows that and . Hence,
We now claim that in (
22) the constant
. Assume that
. Then by the triangle inequality we obtain
On the other hand,
. Hence in a certain cone
K we have the inequality
Consequently,
his contradiction shows that
. Thus
It is easy to verify that
for
. Hence
and
, that is, the problem (
1), (
2), with condition
has at least one non-zero solution, so that
.
On the other hand, for , and therefore, . By Theorem 1, . Thus, we have for . The theorem is proved. □
Proof. Consider the case
. Let
. We shall prove the theorem by contradiction. Assume that
. Then there is a
u such that
and
. Since
, we have
. Hence by Theorem 3 we obtain
where
(see (
19)) and
is defined by Formula (
23). Substituting (
15) and (
23) in (
27), we obtain
where
. Since
, it follows that
. Hence,
where
We claim that
in (
28). Indeed, we assume that
. Taking the scalar product of (
1) with
u and integrating over
, we obtain
It is easy to verify that
and
for
. Next, using the Cauchy–Schwartz inequality we obtain
There exists a sequence of domains such that , .
In equality (
29) we pass to the limit as
. By the Cauchy–Schwartz inequality,
because
and
Thus,
On the other hand,
and
. Hence,
Using the integral identity
we find that if
u is a solution to the homogeneous problem (
1), (
2), then
. The set of all
x such that
is a linear manifold whose dimension is less than
, since the rank of the matrix
is
if
. Therefore,
. The relation
implies that
on a set of positive measure on
, and therefore,
This is a contradiction, since
and
. Thus,
in (
28).
By assumption,
. It is easy to verify that
and
. Now, by the triangle inequality we obtain
By the properties of the fundamental solution of the system (
1) (see [
38]) we have
, where
is a homogeneous function of order zero. Hence
is a homogeneous function of order
, that is,
where
is a homogeneous function of order zero. We fix a point
such that
. By continuity,
in a neighborhood
of
. We consider a cone
K with vertex at the origin such that
. Then
This contradiction shows that . This completes the proof for .
Consider now the case
. It sufficient to show that
for
. Assume that
, that is, there exists
u such that
and
. Since
, it follows that
. Hence by Theorem 3 we obtain
and for
Lemma 2 yields a representation (
22), that is,
where
,
A is a constant skew-symmetric matrix and
B is a constant vector. Substituting the expansion of
in the representation of
, we obtain
We prove that by contradiction. Indeed, assume that . Then we have .
On the other hand,
, where
and
are
– matrices whose entries are homogeneous functions of order zero (see [
39]). Hence,
in some cone
K, and therefore
for
. This contradiction shows that
. Thus,
Taking the scalar product of (
1) and
u and integrating over
, we obtain
In view of the boundary conditions
and
, we have
Since
and
, it follows that
Passing to the limit as
in equality (
30), we obtain
Using the obtained integral identity, we conclude that if
u is a solution to the homogeneous problem (
1), (
2), then
, where
is a constant skew-symmetric matrix,
is a constant vector. Hence,
where
as
. Thus, we have
, which is possible only if
and
. In view of boundary conditions (
2),
and
, so that
for
and
. The theorem is proved. □
Proof. Assume that
. To prove the theorem, we need to determine the number of linearly independent polynomial solutions of a system (
1), the degree of which does not exceed the fixed number.
Let
be a polynomial solution of the system (
1) of degree
r. Then the degree of the polynomial
does not exceed
r, and
P can be represented in the following form:
where
is a homogeneous polynomial of degree
s, satisfying the system (
1) (see [
38]).
The space of polynomials in
of degree at most
r has dimension
(see [
41]). Hence the dimension of the space of vector-valued polynomials in
of degree at most
r is equal to
Polynomials of this kind solving the elasticity system form a space of dimension
because each equation of the elasticity system is equivalent to the vanishing of some polynomial of degree
.
We denote by
the number of linearly independent polynomial solutions of (
1) whose degree is at most
r, and let
be the number of linearly independent homogeneous polynomials of degree
r that are solutions of (
1). Using representation
we obtain
where
We now prove the following statements are true:
(i) The Dirichlet–Robin problem (
1), (
2) with the condition
has
linearly independent solutions for
;
(ii) Each system of solutions is linearly dependent.
(i) Let
be a basis in the space of polynomial solutions of (
1) whose degrees do not exceed
r. Since
, it follows that
for
. For each
we consider the solution
of the system (
1) such that
and
Such a solution we can construct by the variational method, minimizing the corresponding functional over the class of admissible functions has compact support in .
Consider next the difference: . We have in , , and .
Let us prove that
, are linearly independent. Indeed, if
that is,
Hence,
and
By Lemma 1, the solution
V of the system (
1) in
has the following form:
where
is a polynomial, and
It is easy to verify that
By the triangle inequality,
We claim that
. Indeed, assume that
. Then in the interior of a certain cone
K we have
. Hence,
This integral converges only when . Therefore, .
Thus,
, where
as
, that is,
as
. Hence,
and by the estimates (
31) we obtain
Since
is a basis in the space of polynomial solutions of (
1) whose degrees do not exceed
r, it follows that
. Hence the problem has at least
linearly independent solutions.
(ii) Let us prove that each solution
u of the system (
1) with boundary conditions
,
and
can be represented as a linear combination of the solutions
. By Lemma 1, every solution of the system (
1) in
may be written as
where
is a polynomial of degree
,
Since , it follows that and, therefore, . Hence .
We claim that
is a solution of the system (
1). Indeed,
Since
is a polynomial and
as
, it follows that
, that
is a polynomial solution of the system (
1). Hence it is represented as a linear combination of the functions
:
We claim that
We set
By our construction of the solutions, after elementary transformations we obtain
Let us prove that
. Indeed,
is a solution, that is,
in
,
, and
. By the construction of the solutions
we have
Moreover, it is easy to verify that
Since is a unique solution of problem in Theorem 3, it follows that . This proves the theorem for .
The proof for is similar. We claim that
(i) The Dirichlet–Robin problem (
1), (
2) along with the condition
has
linearly independent solutions for
;
(ii) Each system of solutions is linearly dependent.
(i) Let
be a basis in the space of polynomial solutions of (
1) whose degrees do not exceed
r. The vector-valued functions
and
are linearly independent polynomial solutions of (
1). Therefore we can assume without loss of generality that
. The condition
shows that
and, therefore,
for
.
For each
we consider a solution
of the system (
1) such that
and, by Hardy’s inequality [
6], we have
where
such that
Such a solution may be constructed by the variational method, by minimizing the corresponding functional over the class of admissible functions. In the same way, we can construct solutions of (
1) with boundary conditions
, where
and
such that
and
. By Hardy’s inequality [
6] we obtain
where
and
Let
where
and
. Then
in
,
, and
We claim that
are linearly independent. Indeed, if
that is,
Hence
,
, and
By Lemma 1, the solution
V of the system (
1) in
has a representation
where
is a polynomial and
.
We prove that
by contradiction. Indeed, assume that
, where
. Then in the interior of a certain cone
K we have
. In addition,
. Hence,
for
This yields the inequalities
The resulting integral diverges for
. Hence
and
. Thus,
We now claim that
. Indeed, if
, then
because
for
By Hardy’s inequality [
6],
This contradiction shows that
. Hence,
and, therefore,
Let
, then by our choice of the function
and
, we obtain
Since
and
it follows that
The left-hand side of (
33) is a polynomial while its right-hand side has logarithmic growth as
, therefore they both vanish, and
Since form a basis in the space of polynomial solutions, it follows that , and in view of the linear independence of the vectors and , we obtain
Hence the
, are linearly independent solutions, that is, the Dirichlet–Robin problem (
1), (
2) supplemented with the condition
has at least,
linearly independent solutions.
(ii) Let us prove that each solution
u of (
1) with the boundary conditions (
2) and
for
may be represented as a linear combination of the sulutions
.
By Lemma 1, a sloution
u of (
1) in
has the form
where
is a polynomial of degree
and
.
Since , we have and, therefore, . Hence .
We claim that
is a solution of (
1). Indeed, we have
where
as
Since
is a polynomial and
as
, it follows that
, that is,
is a polynomial solution of (
1). Hence it is a finite linear combination of the functions
:
Let
To show that
, we put
After elementary transformations we obtain
where
We claim that
. Indeed, we have
in
,
,
. By the construction of the solutions
we get
It is easy to verify that
and
Hence it follows by the triangle inequality that
and
Using the unique solubility of problem
, we now see from Theorem 3 that
. Hence,
Thus the problem (
1), (
2) has
linearly independent solutions. The proof is complete. □