1. Introduction
The engineering practice demonstrates a big variety of composite structures having thin inclusions. In particular, thin inclusions may cross an external boundary of the elastic body at zero angle. It is well known that, in general cases, Korn’s inequalities are not valid in domains with non-smooth boundaries. This implies difficulties with a solution existence of such problems. This paper focuses on the equilibrium problem for a 2D elastic body with a thin rigid exfoliated inclusion crossing the external boundary of the body at zero angle. The exfoliation implies a presence of the crack between the inclusion and the elastic body. Moreover, the zero angle provides a non-smooth boundary in the mathematical formulation of the problem. Consequently, a proof of the solution existence requires additional arguments. We impose inequality type boundary conditions at the crack faces. In addition, these boundary conditions depend on a positive damage parameter characterizing a cohesion between the crack faces. Over the last decades, many papers have been published concerning problems for elastic bodies with exfoliated thin inclusions and inequality type boundary conditions. We refer the reader to [
1,
2,
3,
4,
5,
6,
7,
8,
9]. Boundary value problems with inequality type boundary conditions for cohesive cracks can be found in [
10,
11,
12,
13]. The results for problems with a zero angle between the inclusion and the external boundaries are presented in [
14], where a fictitious domain method was used to prove a solution existence. As for derivation of suitable models for thin inclusions in elastic structures, see [
15,
16,
17,
18,
19] and the references therein. There are a number of applied papers related to thin inclusions in elastic bodies [
20,
21,
22,
23,
24,
25,
26], as well as books with general approaches to the description of nonhomogeneous bodies [
27,
28].
The model considered in this paper belongs to the class of problems with a free boundary. The results obtained may be useful for other models with free boundaries, see for example [
29], as well as for models of functionally graded materials [
30,
31].
From the practical point of view, the obtained results can be used for the modeling and analysis of different elastic and nonelastic structures having non-smooth boundaries.
This paper is structured as follows: In
Section 2, we prove a solution existence of the problem. Variational and differential formulations of the problem are discussed. Passages to limits, as the damage parameter tends to infinity and to zero, are investigated in
Section 3 and
Section 4. Mixed boundary conditions on the external boundary are considered in
Section 5. A case of the inclusion partially extending beyond the elastic body is analyzed in
Section 6. A solution existence of the optimal control problem with a suitable cost functional is proved.
2. Problem Formulation
Bounded domain in
with a smooth boundary
is denoted by
, and
is a smooth curve such that
. The domain
fits to an elastic body, and
corresponds to a thin rigid inclusion. We assume that the angle between
and
at the point
is equal to zero for the undeformed state of the elastic body, see
Figure 1. Denote
and introduce Sobolev space
Assume that
is a given elasticity tensor with the usual properties of symmetry and positive definiteness,
. Despite the cusp of the domain
, Korn’s inequality is fulfilled in the space
Indeed, consider a fictitious domain
with a smooth boundary
as depicted in
Figure 2 assuming that angles between
and
at the points
are nonzero. Extended domain
has the smooth cut
Let
Since
on
, this function can be extended by zero to
Denote by
the extended function defined in
. It is clear that Korn’s inequality holds for all such extended functions
. Hence, it is valid for all
since
is zero in
i.e., there exists a constant
such that
Here
is defined according to Hooke’s law
To simplify the formulae, we write
instead of
.
Denote by
the space of infinitesimal rigid rotations,
Let
be a given external force acting on the elastic body. Introduce a set of admissible displacements
where
,
are traces of the function
h on the crack faces
, respectively. The signs ± fit to positive and negative crack faces with respect to the outward unit normal vector
to
An equilibrium problem for the body
and the inclusion
is formulated as follows: unknown functions are the displacement field
the stress tensor
defined in
, as well as
satisfying the following equations and boundary conditions
Here
is a damage parameter characterizing a cohesion between the crack faces;
Relations (
2) are the equilibrium equation and constitutive law ( Hooke’s law). The first inequality in (
4) provides a mutual non-penetration between the crack faces
. The identity (
6) provides a zero moment acting on the inclusion
The contact set between the crack faces is unknown a priori, and the model (
2)–(
6) corresponds to a free boundary approach.
The problem (
2)–(
6) has a unique solution. Indeed, consider the energy functional
Then the minimization problem
has a solution. To prove this statement, it suffices to use Korn’s inequality (
1) and note that the set
S is weakly closed. Indeed, we have
Hence, the set
S is closed and consequently weakly closed. The solution of the problem (
7) satisfies the variational inequality
The following statement takes place.
Theorem 1. For smooth solutions, problem formulations (2)–(6) and (8) and (9) are equivalent. Proof. To simplify notations, we omit the symbol
We first check that the equilibrium equation follows from (
8) and (
9), see (
2). To this end, we have to substitute in (
9) test functions of the form
Next, we choose test functions
on
and substitute in (
9). This implies
and consequently, taking into account the equilibrium equation, we obtain
Since
is arbitrary on
, from (
10) we derive the third relation of (
4). The relation (
10) can be written in the form
which gives
providing, by the third relation of (
4), the identity (
6). □
Now we can choose test functions in (
9) in the form
on
see
Figure 3. It provides the following relation
Consequently, integrating by parts, we have
In view of the choice of
v and the last relation of (
4), from (
13) the second relation of (
4) follows.
It remains to check the condition (
5). To this end, suppose that at a given point
we have
. Take test functions in (
9) of the following form:
is a small parameter and
Q is a small neighborhood of the point
,
is a quite smooth function, see
Figure 3. We obtain
consequently, by the third relation of (
4),
and thus
i.e.,
On the other hand, assuming that
we easily derive
and (
5) easily follows. Thus, from (
8) and (
9) we have derived all equations and boundary conditions (
2)–(
6).
Now we prove the converse statement. Let (
2)–(
6) be fulfilled. We take
and multiply the first equation from (
2) by
Integrating over
, one obtains
and consequently,
To derive the variational inequality (
9) from (
14), it suffices to prove that
But the inequality (
15) easily follows from boundary conditions (
4)–(
5). Thus, the equivalence of (
8) and (
9) and (
2)–(
6) is completely proved.
In addition to (
2)–(
6), we can provide one more differential formulation of the problem (
8) and (
9). It reads as follows (again omitting the symbol
): it is necessary to find a displacement field
a stress tensor
defined in
and a thin inclusion displacement
defined on
such that
Let us check that formulations (
16)–(
20) and (
8) and (
9) are equivalent for smooth solutions. Assume that (
8) and (
9) take place. This provides the equilibrium equation, see (
2). From (
8) and (
9) we easily derive
Making the integration by parts in (
21), we obtain (
19). By (
21), the variational inequality (
9) can be rewritten as
thus, integrating by parts, the inequality (
20) is obtained. Consequently, all relations (
16)–(
20) are derived from (
8) and (
9).
Conversely, let (
16)–(
20) be fulfilled. We take
and multiply the first equation from (
16) by
. Integrating over
we obtain
Thus,
We see that for obtaining the variational inequality (
9) from (
22), it is enough to prove the inequality
But the inequality (
23) easily follows from (
19)–(
20). Thus, the equivalence of (
8) and (
9) and (
16)–(
20) is proved for smooth solutions.
4. Passage to Limit in (8) and (9) as
This section concerns a passage to limit as
in the model (
8) and (
9). We will prove that the limit model corresponds to the case without exfoliation of the rigid inclusion from the surrounding elastic body, i.e., to the case without a crack between
and the elastic body.
Since the relation (
24) takes place, we have the uniform in
estimate
Moreover, (
24), (
33) imply
Introduce a set of admissible displacements for the limit problem
Note that we have
on
for
By (
33), (
34), we can assume that as
In particular, we have
on
Take in (
9) test functions of the form
and pass to the limit as
taking into account (
35)–(
37). It gives
Hence, the following assertion has been proved.
Theorem 4. Solutions of the problems (8) and (9) converge in the sense (35)–(37) to the solution of (38) as Notice that the problem (
38) corresponds to minimization of the energy functional over the set
The minimization problem (
39) has a unique solution satisfying (
38).
To conclude this section, we provide a differential formulation of the problem (
38): find a displacement field
a stress tensor
defined in
, and
such that
We see that the limit problem (
40)–(
42) fits to the rigid inclusion without exfoliation from the surrounding elastic body.
Problems (
38) and (
40)–(
42) are equivalent for smooth solutions. The proof of this statement is simpler compared with that of Theorem 1.
5. Mixed Boundary Conditions on External Boundary
Along with the problem (
2)–(
6), it is possible to consider mixed boundary conditions on
Denote by
n a unit normal vector to
Like in the previous sections, the angle between
and
at the point
is zero. Suppose that
is a smooth part of
as depicted in
Figure 4, meas
. Problem formulation is as follows: unknown functions are the displacement field
the stress tensor
defined in
, as well as
satisfying the following equations and boundary conditions
We are planning to solve the problem (
43)–(
47) by minimizing a suitable functional. To this end, preliminary arguments are needed. Introduce the space
By adding a fictitious domain
consider an extended domain
with a smooth cut
and Lipschitz external boundary
, see
Figure 2. Taking
we can extend this function by zero to
. The extended function is equal to zero on
and belongs to the space
where
Consequently, Korn’s inequality holds for such extended functions. This implies that Korn’s inequality is valid in the space
To proceed, we introduce a set of admissible displacements
The set
is closed and convex; hence, it is weakly closed. The above arguments imply that the problem
has (a unique) solution satisfying the following variational inequality
The following statement holds.
Theorem 5. For smooth solutions, problem formulations (43)–(47) and (48)–(49) are equivalent. The arguments are omitted since they are similar to those of Theorem 1.
6. Optimal Control of the Inclusion Shape
We devote this section to the case of the rigid inclusion crossing the boundary
and partially located outside of the elastic body. Denote by
,
internal and external parts of the rigid inclusion
,
, see
Figure 5. The part
of the inclusion
is assumed to be a Lipschitz curve and
is a continuous one. The angle between
and
at the point
is zero. Tip points of the inclusion
are
and
Moreover,
. We suppose that the inclusion shape located between the points
and
may change. The aim of this section is to analyze an optimal control problem assuming that a control function is a shape of the inclusion located between
,
, and a cost functional is a displacement of the inclusion at the tip point
To be more precise, consider a bounded and weakly closed set
assuming that the inclusion shape located between the points
and
is described as a graph of the function
In this case, the curve
corresponding to
is denoted by
and
is denoted by
An equilibrium problem for the body
with a given function
and a fixed damage parameter
is formulated as follows: unknown functions are the displacement field
the stress tensor
defined in
, as well as
satisfying the following equations and boundary conditions
The problem (
50)–(
54) can be written in the variational form. To this end, introduce a set of admissible displacements
Then the problem (
50)–(
54) is equivalent to the following variational inequality
Moreover, we know that the problem (
55) and (
56) corresponds to minimization of the energy functional over the set
According to
Section 2, the set
is weakly closed, and the energy functional is coercive on
for any fixed
Hence, the problem (
57) indeed has a solution
satisfying (
55) and (
56).
Note that a displacement of the
points is defined according to the function
from (
51). If
then
In particular, we can find a displacement of the inclusion
at the tip point
, i.e.,
Thus, for any
a value of a cost functional
can be defined. An optimal control problem to be analyzed below is formulated as follows
We see that the inclusion shape is defined by the function
Solving optimal control problem (
58), we find the best inclusion shape that minimizes the cost functional. The following statement takes place.
Theorem 6. There exists a solution of the optimal control problem (58). Proof. Consider a minimizing sequence
By boundedness of
and the imbedding theorem, we can assume that as
Introduce the notation
where
. For any
a solution of the problem
can be found. We can provide a transformation of the independent variables
where
A smooth function
with compact support in
is chosen in such a way that
in a neighborhood of the union
where all functions
are extended by zero outside of
. Denote
Next, we can use the arguments of paper [
32] where a perturbation of the inclusion shapes is analyzed, and obtain a priori estimates from (
60) and (
61) being uniform in
By (
63), we assume that as
and moreover, it is proved
Making a transformation of the independent variables (
62) in (
60) and (
61) and passing to the limit as
by (
64), we obtain
which means
It is turned out (see [
32]) that if
then as
for all
, and consequently, for all
where
. Then, by (
67),
Thus, the limit function
from (
59) is a solution of the optimal control problem (
58). Theorem 6 is proved. □