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Article

Elasticity Problem with a Cusp between Thin Inclusion and Boundary

Lavrentyev Institute of Hydrodynamics of SB RAS, Novosibirsk State University, Novosibirsk 630090, Russia
Axioms 2023, 12(12), 1081; https://doi.org/10.3390/axioms12121081
Submission received: 12 October 2023 / Revised: 7 November 2023 / Accepted: 21 November 2023 / Published: 27 November 2023
(This article belongs to the Special Issue Principles of Variational Methods in Mathematical Physics)

Abstract

:
This paper concerns an equilibrium problem for an an elastic body with a thin rigid inclusion crossing an external boundary of the body at zero angle. The inclusion is assumed to be exfoliated from the surrounding elastic material that provides an interfacial crack. To avoid nonphysical interpenetration of the opposite crack faces, we impose inequality type constraints. Moreover, boundary conditions at the crack faces depend on a positive parameter describing a cohesion. A solution existence of the problem with different conditions on the external boundary is proved. Passages to the limit are analyzed as the damage parameter tends to infinity and to zero. Finally, an optimal control problem with a suitable cost functional is investigated. In this case, a part of the rigid inclusion is located outside of the elastic body, and a control function is a shape of the inclusion.

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 R 2 with a smooth boundary Γ is denoted by Ω , and γ Ω is a smooth curve such that γ ¯ Γ = ( 0 , 0 ) , Ω γ = Ω γ ¯ . The domain Ω γ fits to an elastic body, and γ corresponds to a thin rigid inclusion. We assume that the angle between Γ and γ at the point ( 0 , 0 ) is equal to zero for the undeformed state of the elastic body, see Figure 1. Denote Γ 0 = Γ { ( 0 , 0 ) } and introduce Sobolev space
H Γ 0 1 ( Ω γ ) = { v H 1 ( Ω γ ) | v = 0 on Γ 0 } .
Assume that C = { c i j k l } is a given elasticity tensor with the usual properties of symmetry and positive definiteness, c i j k l L l o c ( R 2 ) , i , j , k , l = 1 , 2 . Despite the cusp of the domain Ω γ , Korn’s inequality is fulfilled in the space H Γ 0 1 ( Ω γ ) 2 . Indeed, consider a fictitious domain Ω 0 with a smooth boundary Ω 0 as depicted in Figure 2 assuming that angles between Ω 0 and Γ at the points x 1 , x 2 are nonzero. Extended domain Ω e = Ω γ Ω 0 γ 1 γ 2 has the smooth cut γ . Let v H Γ 0 1 ( Ω γ ) 2 . Since v = 0 on Γ 0 , this function can be extended by zero to Ω 0 . Denote by v ˜ the extended function defined in Ω e . It is clear that Korn’s inequality holds for all such extended functions v ˜ . Hence, it is valid for all v H Γ 0 1 ( Ω γ ) 2 , since v ˜ is zero in Ω 0 , i.e., there exists a constant c 0 > 0 such that
Ω γ σ ( v ) ε ( v ) c 0 v H Γ 0 1 ( Ω γ ) 2 2 v H Γ 0 1 ( Ω γ ) 2 .
Here σ ( v ) is defined according to Hooke’s law σ ( v ) = C ε ( v ) , ε ( v ) = { ε i j ( v ) } , ε i j ( v ) = 1 2 ( v i , j + v j , i ) , i , j = 1 , 2 . To simplify the formulae, we write σ ( v ) ε ( v ) instead of σ i j ( v ) ε i j ( v ) .
Denote by R 0 ( γ ) the space of infinitesimal rigid rotations,
R 0 ( γ ) = { ρ = ( ρ 1 , ρ 2 ) | ρ ( x ) = b ( x 2 , x 1 ) , b R , x = ( x 1 , x 2 ) γ } .
Let f L 2 ( Ω ) 2 be a given external force acting on the elastic body. Introduce a set of admissible displacements
S = { v H Γ 0 1 ( Ω γ ) 2 | [ v ν ] 0 on γ ; v | γ R 0 ( γ ) } ,
where v ν = v ν , [ h ] = h + h ; h ± are traces of the function h on the crack faces γ ± , respectively. The signs ± fit to positive and negative crack faces with respect to the outward unit normal vector ν to γ .
An equilibrium problem for the body Ω γ and the inclusion γ is formulated as follows: unknown functions are the displacement field u α = ( u 1 α , u 2 α ) , the stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω γ , as well as ρ 0 α R 0 ( γ ) , satisfying the following equations and boundary conditions
div σ = f , σ = C ε ( u α ) in Ω γ ,
u α = 0 on Γ 0 ; u α = ρ 0 α on γ ,
[ u ν α ] 0 , σ ν + + 1 α [ u ν α ] 0 , σ τ + + 1 α [ u τ α ] = 0 on γ ,
[ u ν α ] ( σ ν + + 1 α [ u ν α ] ) = 0 on γ ,
γ [ σ ν ] ρ = 0 ρ R 0 ( γ ) .
Here α > 0 is a damage parameter characterizing a cohesion between the crack faces; σ ν = σ i j ν j ν i ,   σ τ = σ i j ν j τ i , ν = ( ν 1 , ν 2 ) , τ = ( ν 2 , ν 1 ) .
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 γ ± . The identity (6) provides a zero moment acting on the inclusion γ . 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 G α : H Γ 0 1 ( Ω γ ) 2 R ,
G α ( v ) = 1 2 Ω γ σ ( v ) ε ( v ) Ω γ f v + 1 2 α γ [ v ] 2 .
Then the minimization problem
inf v S G α ( v )
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 [ v ν ] H 1 / 2 ( γ ) . Hence, the set S is closed and consequently weakly closed. The solution of the problem (7) satisfies the variational inequality
u α S ,
Ω γ σ ( u α ) ε ( u ¯ u α ) Ω γ f ( u ¯ u α ) + + 1 α γ [ u α ] [ u ¯ u α ] 0 u ¯ S .
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 α . 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 u ¯ = u ± v , v C 0 ( Ω γ ) 2 . Next, we choose test functions u ¯ = u ± v , [ v ν ] = 0 on γ ,   v S , and substitute in (9). This implies
Ω γ σ ( u ) ε ( v ) Ω γ f v + 1 α γ [ u ] [ v ] = 0 ,
and consequently, taking into account the equilibrium equation, we obtain
γ [ σ ν ] v ν γ [ σ τ v τ ] + 1 α γ [ u τ ] [ v τ ] = 0 .
Since v τ + is arbitrary on γ , from (10) we derive the third relation of (4). The relation (10) can be written in the form
γ [ σ ν ] v ν γ [ σ τ ] v τ + γ [ σ τ ] v τ + γ ( σ τ 1 α [ u τ ] ) v τ = 0
which gives
γ [ σ ν ] v + γ v τ ( σ τ + 1 α [ u τ ] ) = 0
providing, by the third relation of (4), the identity (6). □
Now we can choose test functions in (9) in the form u ¯ = u + v , v H Γ 1 ( Ω γ ) 2 , v ν + 0 on γ , supp v Q ¯ , see Figure 3. It provides the following relation
Ω γ σ ( u ) ε ( v ) Ω γ f v + 1 α γ [ u ] v + 0 .
Consequently, integrating by parts, we have
γ ( σ ν + v ν + + σ τ + v τ + ) 1 α γ [ u ν ] v ν + 1 α γ [ u τ ] v τ + 0 .
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 x 0 γ we have [ u ν ( x 0 ) ] > 0 . Take test functions in (9) of the following form: u ¯ = u ± β ψ ,   supp ψ Q ¯ , β is a small parameter and Q is a small neighborhood of the point x 0 , ψ is a quite smooth function, see Figure 3. We obtain
Ω γ σ ( u ) ε ( ψ ) Ω γ f ψ + 1 α γ [ u ] ψ + = 0 ,
consequently, by the third relation of (4),
γ σ ν + ψ ν + 1 α γ [ u ν ] ψ ν + = 0 ,
and thus σ ν + ( x 0 ) 1 α [ u ν ( x 0 ) ] = 0 , i.e., [ u ν ( x 0 ) ] ( σ ν + ( x 0 ) 1 α [ u ν ( x 0 ) ] ) = 0 . On the other hand, assuming that σ ν + ( x 0 ) + 1 α [ u ν ( x 0 ) ] < 0 , we easily derive [ u ν ( x 0 ) ] = 0 , 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 u ¯ S and multiply the first equation from (2) by u ¯ u . Integrating over Ω γ , one obtains
Ω γ ( div σ + f ) ( u ¯ u ) = 0 ,
and consequently,
Ω γ σ ( u ) ε ( u ¯ u ) Ω γ f ( u ¯ u ) ± 1 α γ [ u ] [ u ¯ u ] + + γ [ σ ν ( u ¯ u ) ] = 0 .
To derive the variational inequality (9) from (14), it suffices to prove that
γ [ σ ν ( u ¯ u ) ] + 1 α γ [ u ] [ u ¯ u ] 0 .
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 α ): it is necessary to find a displacement field u = ( u 1 , u 2 ) , a stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω γ , and a thin inclusion displacement ρ 0 R 0 ( γ ) defined on γ such that
div σ = f , σ = C ε ( u ) in Ω γ ,
u = 0 on Γ 0 ,
[ u ν ] 0 , u = ρ 0 on γ ,
γ [ σ ν · u ] + 1 α γ [ u ] 2 = 0 ,
γ [ σ ν · u ¯ ] + 1 α γ [ u ] [ u ¯ ] 0 u ¯ S .
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
Ω γ σ ( u ) ε ( u ) Ω γ f u + 1 α γ [ u ] 2 = 0 .
Making the integration by parts in (21), we obtain (19). By (21), the variational inequality (9) can be rewritten as
Ω γ σ ( u ) ε ( u ¯ ) Ω γ f u ¯ + 1 α γ [ u ] [ u ¯ ] 0 u ¯ S ,
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 u ¯ S and multiply the first equation from (16) by u ¯ u . Integrating over Ω γ , we obtain
Ω γ ( div σ + f ) ( u ¯ u ) = 0 .
Thus,
γ [ σ ν ( u ¯ u ) ] + Ω γ σ ( u ) ε ( u ¯ u ) Ω γ f ( u ¯ u ) ± 1 α γ [ u ] [ u ¯ u ] = 0 .
We see that for obtaining the variational inequality (9) from (22), it is enough to prove the inequality
γ [ σ ν ( u ¯ u ) ] 1 α γ [ u ] [ u ¯ u ] 0 .
But the inequality (23) easily follows from (19)–(20). Thus, the equivalence of (8) and (9) and (16)–(20) is proved for smooth solutions.

3. Passage to Limit in (8) and (9) as α

In this section, we justify a passage to limit in the model (8) and (9) as α . It turns out that the limit model fits to zero friction at the positive crack face γ + .
From (9), it follows
Ω γ σ ( u α ) ε ( u α ) Ω γ f u α + 1 α γ [ u α ] 2 = 0 .
By Korn’s inequality valid in the space H Γ 0 1 ( Ω γ ) 2 , the relation (24) implies a uniform in α estimate
u α H Γ 0 1 ( Ω γ ) 2 c .
Choosing a subsequence, if necessary, we assume that as α
u α u weakly   in H Γ 0 1 ( Ω γ ) 2 ,
[ u α ] [ u ] weakly   in L 2 ( γ ) 2 ,
u | γ = ρ R 0 ( γ ) .
By (25)–(27), a passage to the limit as α can be fulfilled in (8) and (9) which implies
u S , Ω γ σ ( u ) ε ( u ¯ u ) Ω γ f ( u ¯ u ) 0 u ¯ S .
Thus, we arrive at the following assertion.
Theorem 2.
Solutions of the problems (8) and (9) converge in the sense (25)–(27) to the solution of (28) as α .
The problem (28) admits a differential formulation. It is necessary to find a displacement field u = ( u 1 , u 2 ) , a stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω γ , and ρ R 0 ( γ ) such that
div σ = f , σ = C ε ( u ) in Ω γ ,
u = 0 on Γ 0 ; u = ρ on γ ,
[ u ν ] 0 , σ ν + 0 , σ τ + = 0 , [ u ν ] σ ν + = 0 on γ ,
γ [ σ ν ] ρ = 0 ρ R 0 ( γ ) .
We can prove the following assertion.
Theorem 3.
For smooth solutions, problem formulations (28) and (29)–(32) are equivalent.
We do not prove this theorem since the arguments used are simpler as compared to those of Theorem 1.
The model (29)–(32) is characterized by zero friction at the positive crack face γ + .

4. Passage to Limit in (8) and (9) as α 0

This section concerns a passage to limit as α 0 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 γ and the elastic body.
Since the relation (24) takes place, we have the uniform in α estimate
u α H Γ 0 1 ( Ω γ ) 2 c .
Moreover, (24), (33) imply
γ [ u α ] 2 c α .
Introduce a set of admissible displacements for the limit problem
S 0 = { v H Γ 0 1 ( Ω γ ) 2 | v | γ R 0 ( γ ) } .
Note that we have [ v ] = 0 on γ for v S 0 . By (33), (34), we can assume that as α 0 ,
u α u weakly   in H Γ 0 1 ( Ω γ ) 2 ,
[ u α ] [ u ] weakly   in L 2 ( γ ) 2 ,
u | γ = ρ 0 R 0 ( γ ) .
In particular, we have [ u ] = 0 on γ .
Take in (9) test functions of the form u ¯ = u α ± v , v S 0 , and pass to the limit as α 0 taking into account (35)–(37). It gives
u S 0 , Ω γ σ ( u ) ε ( v ) Ω γ f v = 0 v S 0 .
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 α 0 .
Notice that the problem (38) corresponds to minimization of the energy functional over the set S 0 ,
inf v S 0 { 1 2 Ω γ σ ( v ) ε ( v ) Ω γ f v } .
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 u = ( u 1 , u 2 ) , a stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω γ , and ρ 0 R 0 ( γ ) such that
div σ = f , σ = C ε ( u ) in Ω γ ,
u = 0 on Γ 0 ; u = ρ 0 on γ ,
γ [ σ ν ] ρ = 0 ρ R 0 ( γ ) .
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 Γ 0

Along with the problem (2)–(6), it is possible to consider mixed boundary conditions on Γ 0 . Denote by n a unit normal vector to Γ . Like in the previous sections, the angle between Γ and γ at the point ( 0 , 0 ) is zero. Suppose that γ 1 is a smooth part of Γ as depicted in Figure 4, meas Γ γ 1 > 0 . Problem formulation is as follows: unknown functions are the displacement field u α = ( u 1 α , u 2 α ) , the stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω γ , as well as ρ α R 0 ( γ ) , satisfying the following equations and boundary conditions
div σ = f , σ = C ε ( u α ) in Ω γ ,
u α = 0 on Γ γ 1 ; σ n = 0 on γ 1 ; u α = ρ α on γ ,
[ u ν α ] 0 , σ ν + + 1 α [ u ν α ] 0 , σ τ + + 1 α [ u τ α ] = 0 on γ ,
[ u ν α ] ( σ ν + + 1 α [ u ν α ] ) = 0 on γ ,
γ [ σ ν ] ρ = 0 ρ R 0 ( γ ) .
We are planning to solve the problem (43)–(47) by minimizing a suitable functional. To this end, preliminary arguments are needed. Introduce the space
H Γ γ 1 1 ( Ω γ ) = { v H 1 ( Ω γ ) | v = 0 on Γ γ 1 } .
By adding a fictitious domain Ω 0 , consider an extended domain Ω e = Ω γ Ω 0 γ 2 with a smooth cut γ γ 1 { ( 0 , 0 ) } and Lipschitz external boundary Γ e , see Figure 2. Taking v H Γ γ 1 1 ( Ω γ ) 2 , we can extend this function by zero to Ω 0 . The extended function is equal to zero on Γ e and belongs to the space H Γ e 1 ( Ω e ) 2 , where
H Γ e 1 ( Ω e ) = { v H 1 ( Ω e ) | v = 0 on Γ e } .
Consequently, Korn’s inequality holds for such extended functions. This implies that Korn’s inequality is valid in the space H Γ γ 1 1 ( Ω γ ) 2 .
To proceed, we introduce a set of admissible displacements
S N = { v H Γ γ 1 1 ( Ω γ ) 2 | [ v ν ] 0 on γ ; v | γ R 0 ( γ ) } .
The set S N is closed and convex; hence, it is weakly closed. The above arguments imply that the problem
inf v S N { 1 2 Ω γ σ ( v ) ε ( v ) Ω γ f v + 1 2 α γ [ v ] 2 }
has (a unique) solution satisfying the following variational inequality
u α S N ,
Ω γ σ ( u α ) ε ( v u α ) Ω γ f ( v u α ) + + 1 α γ [ u α ] [ v u α ] 0 v S N .
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 Γ and partially located outside of the elastic body. Denote by γ , γ e internal and external parts of the rigid inclusion γ 0 = γ γ e { ( 0 , 0 ) } , Γ γ 0 = ( 0 , 0 ) , see Figure 5. The part γ of the inclusion γ 0 is assumed to be a Lipschitz curve and γ e is a continuous one. The angle between γ 0 and Γ at the point ( 0 , 0 ) is zero. Tip points of the inclusion γ 0 are x 0 = ( 1 , 1 ) and z 0 = ( 1 , 1 ) . Moreover, y 0 γ , y 0 = ( 1 / 2 , 1 ) . We suppose that the inclusion shape located between the points x 0 and y 0 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 x 0 , y 0 , and a cost functional is a displacement of the inclusion at the tip point z 0 . To be more precise, consider a bounded and weakly closed set Ξ H 0 2 ( 1 / 2 , 1 ) assuming that the inclusion shape located between the points x 0 and y 0 is described as a graph of the function
x 2 = ξ ( x 1 ) 1 , x 1 ( 1 / 2 , 1 ) , ξ Ξ .
In this case, the curve γ corresponding to ξ is denoted by γ ξ , and Ω γ ξ is denoted by Ω ξ . An equilibrium problem for the body Ω ξ with a given function ξ and a fixed damage parameter α is formulated as follows: unknown functions are the displacement field u ξ = ( u 1 ξ , u 2 ξ ) , the stress tensor σ = { σ i j } , i , j = 1 , 2 , defined in Ω ξ , as well as ρ 0 ξ R 0 ( γ ξ ) , satisfying the following equations and boundary conditions
div σ = f , σ = C ε ( u ξ ) in Ω ξ ,
u ξ = 0 on Γ 0 ; u ξ = ρ 0 ξ on γ ξ ,
[ u ν ξ ] 0 , σ ν + + 1 α [ u ν ξ ] 0 , σ τ + + 1 α [ u τ ξ ] = 0 on γ ξ ,
[ u ν ξ ] ( σ ν + + 1 α [ u ν ξ ] ) = 0 on γ ξ ,
γ ξ [ σ ν ] ρ = 0 ρ R 0 ( γ ξ ) .
The problem (50)–(54) can be written in the variational form. To this end, introduce a set of admissible displacements
S ξ = { v H Γ 0 1 ( Ω ξ ) 2 | [ v ν ] 0 on γ ξ ; v | γ ξ R 0 ( γ ξ ) } .
Then the problem (50)–(54) is equivalent to the following variational inequality
u ξ S ξ ,
Ω ξ σ ( u ξ ) ε ( v u ξ ) Ω ξ f ( v u ξ ) + + 1 α γ ξ [ u ξ ] [ v u ξ ] 0 v S ξ .
Moreover, we know that the problem (55) and (56) corresponds to minimization of the energy functional over the set S ξ ,
inf v S ξ { 1 2 Ω ξ σ ( v ) ε ( v ) Ω ξ f v + 1 2 α γ ξ [ v ] 2 } .
According to Section 2, the set S ξ is weakly closed, and the energy functional is coercive on S ξ for any fixed ξ Ξ . Hence, the problem (57) indeed has a solution u ξ satisfying (55) and (56).
Note that a displacement of the γ e points is defined according to the function ρ 0 ξ from (51). If
ρ 0 ξ ( x ) = ( c 1 , c 2 ) + b ( x 2 , x 1 ) , x = ( x 1 , x 2 ) γ ξ ,
then
ρ 0 ξ ( x ) = ( c 1 , c 2 ) + b ( x 2 , x 1 ) , x = ( x 1 , x 2 ) γ e .
In particular, we can find a displacement of the inclusion γ 0 at the tip point z 0 , i.e., ρ 0 ξ ( 1 , 1 ) = ( ρ 01 ξ ( 1 , 1 ) , ρ 02 ξ ( 1 , 1 ) ) . Thus, for any ξ Ξ a value of a cost functional
J ( ξ ) = | ρ 02 ξ ( 1 , 1 ) |
can be defined. An optimal control problem to be analyzed below is formulated as follows
inf ξ ¯ Ξ J ( ξ ¯ ) .
We see that the inclusion shape is defined by the function ξ . 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 ξ m Ξ . By boundedness of Ξ and the imbedding theorem, we can assume that as m ,
ξ m ξ weakly   in H 0 2 ( 1 / 2 , 1 ) , ξ Ξ , ξ m ξ strongly   in C 1 [ 1 / 2 , 1 ] .
Introduce the notation
S m = { v H Γ 0 1 ( Ω m ) 2 | [ v ν ] 0 o n γ ξ m ;
v | γ ξ m R 0 ( γ ξ m ) } ,
where Ω m = Ω γ ¯ ξ m . For any m , a solution of the problem
u m S m ,
Ω m σ ( u m ) ε ( v u m ) Ω m f ( v u m ) + + 1 α γ ξ m [ u m ] [ v u m ] 0 v S m
can be found. We can provide a transformation of the independent variables
x 1 = y 1 x 2 = y 2 φ ( y ) ( ξ m ( y 1 ) ξ ( y 1 ) ) ,
where x = ( x 1 , x 2 ) Ω ξ , y = ( y 1 , y 2 ) Ω m . A smooth function φ with compact support in Ω ξ is chosen in such a way that φ = 1 in a neighborhood of the union
{ ( y 1 , y 2 ) | y 2 = ξ m ( y 1 ) 1 , y 1 [ 1 / 2 , 1 ] } ,
where all functions ξ , ξ m are extended by zero outside of [ 1 / 2 , 1 ] . Denote
u m ( x ) = u m ( y ) , x Ω ξ , y Ω m .
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 m ,
u m H Γ 0 1 ( Ω ξ ) 2 c .
By (63), we assume that as m ,
u m u weakly   in H Γ 0 1 ( Ω ξ ) 2 ,
and moreover, it is proved
u m u strongly in H Γ 0 1 ( Ω ξ ) 2 .
Making a transformation of the independent variables (62) in (60) and (61) and passing to the limit as m , by (64), we obtain
u S ξ ,
Ω ξ σ ( u ) ε ( v u ) Ω ξ f ( v u ) + + 1 α γ ξ [ u ] [ v u ] 0 v S ξ
which means u = u ξ . It is turned out (see [32]) that if
ρ m = u m o n γ ξ m ,
then as m ,
ρ m ( x ) ρ 0 ξ ( x )
for all x γ , and consequently, for all x γ e , where ρ 0 ξ = ( ρ 01 ξ , ρ 02 ξ ) = u ξ | γ ξ . Then, by (67),
inf ξ ¯ Ξ J ( ξ ¯ ) = lim inf m J ( ξ m ) = lim inf m | ρ m 2 ( 1 , 1 ) | = | ρ 02 ξ ( 1 , 1 ) | = = J ( ξ ) inf ξ ¯ Ξ J ( ξ ¯ ) .
Thus, the limit function ξ from (59) is a solution of the optimal control problem (58). Theorem 6 is proved. □

7. Conclusions

This paper provides a rigorous mathematical analysis of the problem for an elastic body with a thin rigid inclusion crossing the external boundary of the body at zero angle. It is supposed that the inclusion is exfoliated from the surrounding body which implies a presence of the interfacial crack. Inequality type constraints are imposed at the crack faces proving a mutual non-penetration between them. In addition to this factor, boundary conditions depend on a positive damage parameter describing the adhesion between the opposite crack faces. These boundary conditions imply that the problem considered refers to the problem with an unknown set of a contact. The solution existence of the problem is proved, and asymptotic analysis is fulfilled with respect to the damage parameter assuming that this parameter tends to infinity and to zero. Therefore, in the frame of a high–level mathematical model, we prove a correctness of the boundary value problem and analyze the limit models. Moreover, the existence of a solution to the optimal control problem is proved which allows us to find the optimal shape of the rigid inclusion. The approaches used in this paper can be useful for the analysis of various complex elastic and inelastic structures formulated in terms of free boundary problems, and in particular, for those having zero angles of the boundary.

Funding

This work is supported by the Mathematical Center in Akademgorodok under agreement No. 075-15-2022-282 with the Ministry of Science and Higher Education of the Russian Federation.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. Geometry of the problem.
Figure 1. Geometry of the problem.
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Figure 2. Fictitious domain Ω 0 .
Figure 2. Fictitious domain Ω 0 .
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Figure 3. Neighborhood Q.
Figure 3. Neighborhood Q.
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Figure 4. Partition of the external boundary.
Figure 4. Partition of the external boundary.
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Figure 5. Inclusion γ 0 = γ γ e { ( 0 , 0 ) } .
Figure 5. Inclusion γ 0 = γ γ e { ( 0 , 0 ) } .
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Khludnev, A. Elasticity Problem with a Cusp between Thin Inclusion and Boundary. Axioms 2023, 12, 1081. https://doi.org/10.3390/axioms12121081

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Khludnev A. Elasticity Problem with a Cusp between Thin Inclusion and Boundary. Axioms. 2023; 12(12):1081. https://doi.org/10.3390/axioms12121081

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Khludnev, Alexander. 2023. "Elasticity Problem with a Cusp between Thin Inclusion and Boundary" Axioms 12, no. 12: 1081. https://doi.org/10.3390/axioms12121081

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