On Equilibrium Problem for T-Shape Elastic Structure

Abstract

This paper is concerned with an equilibrium problem for an elastic structure consisting of a plate and an elastic beam connected to each other at a given point. We consider two cases: In the first one, the elastic beam is connected to a rigid part of the elastic plate; in the second case, contact occurs between two elastic bodies. The elastic plate may contain a thin rigid delaminated inclusion. Neumann-type boundary conditions are considered at the external boundary of the plate. The existence of a solution to the considered problems is proven. A sufficient and necessary condition imposed onto the external forces for the solvability of the problems is found. Passages to the limit with respect to the rigidity parameter of the elastic beam are justified. For all problems, we analyze variational statements as well as differential ones.

1. Introduction

T-shape elastic and non-elastic structures are widely used in applications. The behavior of such structures depends on the geometry and material properties of the parts that it make up. The components of the considered structures may contain thin and volume inclusions, either elastic or rigid ones. Moreover, the junction conditions between the components have a big influence on the stresses emerging inside the structure. From the standpoint of mathematical modeling, various connections between the constitutive parts imply suitable boundary conditions imposed on the interfaces. Additional difficulties appear provided that the inclusions are delaminated from the surrounding deformable material. In recent decades, a lot of papers have been published related to the analysis of elastic bodies with thin and volume inclusions with delaminations. The presence of delamination means that there is an interfacial crack located between the inclusion and the surrounding body. In this case, suitable boundary conditions should be specified at the crack faces. We refer the reader to [1,2,3,4,5,6] and many other papers. In particular, in paper [5], one can find T-shape thin delaminated inclusions incorporated into elastic bodies, with the inequality-type boundary conditions imposed at the crack faces providing mutual non-penetration. In addition to the proof of the solution’s existence and an analysis of the solution’s dependence on the parameters, the junction conditions between the inclusions are found. Various junction problems are investigated in [7,8,9,10,11,12,13] and the literature therein. One more difficulty in these analyses is concerned with the Neumann-type boundary condition imposed at the external boundary. This leads to the need to find suitable conditions for the external forces to ensure the existence of a solution. In so doing, we have to present the Sobolev space as a sum of two orthogonal subspaces and find a unique solution in a suitable subspace. There are many results concerning equilibrium problems for elastic bodies with thin inclusions without delaminations [14,15,16,17,18,19,20,21,22,23]. As for general approaches to the analysis of composite structures, suitable results can be found in [24,25]. In the book [26], one can find results for contact problems between two inclined elastic plates, as well as for the contact between an elastic plate and an inclined elastic beam; see also papers [27,28,29,30,31] related to T-shape deformable structures.

In the present paper, we analyze a T-shape deformable structure consisting of a plate and a beam attached to the plate. The plate may contain volume and thin rigid inclusions. In fact, we can look at the considered T-shape structure with the junction between the bodies of different dimensions in mind. This paper is organized as follows. In Section 2 and Section 3, we analyze an equilibrium problem for a plate connected with an elastic beam provided that the Neumann-type boundary conditions are considered at the external boundary; the plate is assumed to contain a rigid volume inclusion. The existence of a solution is proven. An asymptotic analysis is provided as the rigidity parameter of the beam tends to infinity. Section 4 and Section 5 are concerned with a case without rigid inclusion. Sufficient and necessary conditions imposed on the external forces are found, proving the existence of a solution to the problem. To prove the solution’s existence in this case, we should present a suitable Sobolev space as a sum of two orthogonal subspaces. The passage to the limit is also analyzed as the rigidity parameter of the beam tends to infinity. In all cases, limit models are investigated. Section 6 takes the case of an elastic plate with a thin rigid delaminated inclusion.

2. Problem Statement

Let $\Omega \subset {\mathbb{R}}^2$ be a bounded domain with a smooth boundary $\Gamma$ and $\omega \subset \Omega$ a subdomain with a smooth boundary ${\Gamma }_0,$${\Gamma }_0\cap \Gamma =\varnothing$. Assume that $(0,0)\in \omega .$ The interval $(-1,0)$ of the axis $x_3$ is denoted by $\sigma .$ The axis $x_3$ is orthogonal to the plane $x_1{x}_2.$ The domain $\Omega \setminus \overline{\omega }$ fits to the mid-surface of an elastic plate, and $\omega$ is a rigid part of the plate; $\sigma$ corresponds to an elastic beam. Thus, the point $(0,0,0)$ is the joint point of the plate and the beam; see Figure 1. We denote using $n=(n_1,n_2),\nu =({\nu }_1,{\nu }_2)$ the unit external vectors to $\Gamma ,{\Gamma }_0,$ respectively.

Let $f\in L^2(\Omega ),g\in L^2\left(\sigma \right)$ be given as the external forces acting on the plate and the beam, respectively. We introduce the Sobolev space with an arbitrary constant c

$$\begin{array}{c}\hfill H^{2,\omega }(\Omega )=\{v\in H^2(\Omega )|v=c\mathrm{on}\omega ;c\in \mathbb{R}\}\end{array}$$

and the following space:

$$\begin{array}{c}\hfill W^{\omega }=\{(v,u)\in H^{2,\omega }(\Omega )\times H^1\left(\sigma \right)|v(0,0)=u\left(0\right),u(-1)=0\}.\end{array}$$

The norm in the space $W^{\omega }$ is defined as follows:

$$\begin{array}{c}\hfill {\parallel (v,u)\parallel }_{W^{\omega }}^2={\parallel v\parallel }_{H^2(\Omega )}^2+{\parallel u\parallel }_{H^1\left(\sigma \right)}^2.\end{array}$$

(1)

We first formulate the variational statement of the problem. Assume that $A=\left\{a_{ijkl}\right\}$ is a given elasticity tensor with the usual properties of symmetry and positive definiteness, $a_{ijkl}\in L_{loc}^{\infty }\left({\mathbb{R}}^2\right),$$i,j,k,l=1,2$. For a scalar function w, we put $\nabla \nabla w=\left\{w_{,ij}\right\},i,j=1,2.$ If $M=\left\{M_{ij}\right\},i,j=1,2,$ then $\nabla \nabla M=M_{ij,ij}.$ The summation convention over repeated indices is used; all functions with two lower indices are assumed to be symmetric in those indices.

Wentroduce notations for a bending moment $M^n$ and a transverse force $T^n=T^n\left(M\right)$

$$\begin{array}{c}\hfill M^n=-M_{ij}{n}_j{n}_i;T^n=-M_{ij,j}{n}_i-M_{ij,k}{\tau }_k{\tau }_j{n}_i,({\tau }_1,{\tau }_2)=(-n_2,n_1).\end{array}$$

(2)

We should take note of Green’s formula. For the smooth functions $w,M=\left\{M_{ij}\right\},i,j=1,2,$ defined in $\Omega ,$ we have [26]

$$\begin{array}{c}\hfill -\underset{\Omega }{\int }M·\nabla \nabla w=-\underset{\Omega }{\int }w\nabla \nabla M+\underset{\Gamma }{\int }{M}^n{w}_n-\underset{\Gamma }{\int }{T}^{n}w,\end{array}$$

(3)

where $w_n={\displaystyle \frac{\partial w}{\partial n}}$.

Consider the bilinear form

$$\begin{array}{c}\hfill D(w,\overline{w})=\underset{\Omega }{\int }{a}_{ijkl}{w}_{,kl}{\overline{w}}_{,ij}=-\underset{\Omega }{\int }{M}_{ij}{\overline{w}}_{,ij},\end{array}$$

where $M_{ij}=M_{ij}\left(w\right)=-a_{ijkl}{w}_{,kl},i,j=1,2,$ and introduce the energy functional $\Pi :W^{\omega }\to \mathbb{R},$

$$\begin{array}{c}\hfill \Pi (v,u)={\displaystyle \frac{1}{2}}D(v,v)-\underset{\Omega }{\int }fv+{\displaystyle \frac{1}{2}}\underset{\sigma }{\int }{u}_{,3}^2-\underset{\sigma }{\int }gu.\end{array}$$

The following variational problem would be analyzed:

$$\begin{array}{c}\hfill \mathrm{Find}(v,u)\in W^{\omega }\mathrm{such}\mathrm{that}\Pi (v,u)=\underset{W^{\omega }}{inf}\Pi .\end{array}$$

(4)

This problem has a solution satisfying the identity

$$\begin{array}{c}\hfill (v,u)\in W^{\omega },\end{array}$$

(5)

$$\begin{array}{c}\hfill D(v,\overline{v})-\underset{\Omega }{\int }f\overline{v}+\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}=0\forall (\overline{v},\overline{u})\in W^{\omega }.\end{array}$$

(6)

To prove the solvability of the problem (4), we have to check the coercivity of the functional $\Pi$ on the space $W^{\omega }$. In this case, the solvability of the problem (4) follows since the weak lower semicontinuity of the quadratic functional $\Pi$ is clear. To this end, we will prove the following statement.

Proposition 1. 

A constant $c>0$ exists such that

$$\begin{array}{c}\hfill D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2\ge c{\parallel (v,u)\parallel }_{W^{\omega }}^2\forall (v,u)\in W^{\omega }.\end{array}$$

(7)

Proof. 

Assume that the statement is not true. In this case, we can choose a sequence $(v^k,u^k)\in W^{\omega }$ such that

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_{W^{\omega }}^2=1,\end{array}$$

(8)

$$\begin{array}{c}\hfill D(v^k,v^k)+\underset{\sigma }{\int }{\left(u_{,3}^k\right)}^2\to 0,k\to \infty .\end{array}$$

(9)

Choosing a subsequence with the same notation, we can assume that as $k\to \infty ,$

$$\begin{array}{c}\hfill (v^k,u^k)\to (v,u)\mathrm{weakly} \mathrm{in}{W}^{\omega }\mathrm{and} \mathrm{strongly} \mathrm{in}{H}^1(\Omega )\times L^2\left(\sigma \right).\end{array}$$

(10)

According to (9), it follows that

$$\begin{array}{c}\hfill D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2=0.\end{array}$$

(11)

From (11), we conclude that $a_0,a_1,a_2,b\in \mathbb{R}$ exist such that

$$\begin{array}{c}\hfill v\left(x\right)=a_0+a_1{x}_1+a_2{x}_2,x=(x_1,x_2)\in \Omega ;u\left(x_3\right)=b,x_3\in \sigma .\end{array}$$

(12)

On the other hand, $v=\mathit{const}$ on $\omega$; hence, $v\left(x\right)=a_0.$ Since $(v,u)\in W^{\omega }$, we obtain $b=a_0.$ In addition to this, $u(-1)=0,$ and consequently,

$$\begin{array}{c}\hfill (v,u)=(0,0).\end{array}$$

(13)

We can use the norm in the space $W^{\omega }$ equivalent to (1). Then, according to (10), (11), we obtain as $k\to \infty$

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_{W^{\omega }}^2=\parallel v^k{\parallel }_{H^1(\Omega )}^2+{\parallel u^k\parallel }_{L^2\left(\sigma \right)}^2+D(v^k,v^k)+\underset{\sigma }{\int }{\left(u_{,3}^k\right)}^2\to \\ \hfill \to {\parallel v\parallel }_{H^1(\Omega )}^2+{\parallel u\parallel }_{L^2\left(\sigma \right)}^2+D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2={\parallel (v,u)\parallel }_{W^{\omega }}^2\end{array}$$

which implies

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_{W^{\omega }}\to {\parallel (v,u)\parallel }_{W^{\omega }}.\end{array}$$

Hence, from (8), it follows that

$$\begin{array}{c}\hfill {\parallel (v,u)\parallel }_{W^{\omega }}=1\end{array}$$

which contradicts (13). Proposition 1 is proven. □

Note that the solution to the problem (5)–(6) is unique. Indeed, assuming the existence of two solutions $(v^1,u^1),(v^2,u^2),$ from (5)–(6), we easily obtain the equality

$$\begin{array}{c}\hfill D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2=0\end{array}$$

with $v=v^1-v^2,u=u^1-u^2.$ Hence, according to (7), it gives $v=0,u=0.$

Now, we are able to write a differential statement of the problem (5)–(6): it is necessary to find the functions $v,u$ defined in $\Omega$ and $\sigma$, respectively, as well as a constant $a_0$ such that

$$\begin{array}{c}\hfill \nabla \nabla M+f=0\mathrm{in}\Omega \setminus \overline{\omega },\end{array}$$

(14)

$$\begin{array}{c}\hfill M+A\nabla \nabla v=0\mathrm{in}\Omega \setminus \overline{\omega },\end{array}$$

(15)

$$\begin{array}{c}\hfill -u_{,33}=g\mathrm{in}\sigma ,\end{array}$$

(16)

$$\begin{array}{c}\hfill M^n=T^n=0\mathrm{on}\Gamma ,\end{array}$$

(17)

$$\begin{array}{c}\hfill v=a_0\mathrm{on}\omega ;v=a_0,{\displaystyle \frac{\partial v}{\partial \nu }}=0\mathrm{on}{\Gamma }_0,\end{array}$$

(18)

$$\begin{array}{c}\hfill v(0,0)=u\left(0\right),u(-1)=0,\end{array}$$

(19)

$$\begin{array}{c}\hfill -\underset{\omega }{\int }f+\underset{{\Gamma }_0}{\int }{T}^{\nu }+u_{,3}\left(0\right)=0.\end{array}$$

(20)

Relations (14)–(15) are the equilibrium equation and the constitutive law for the elastic part of the plate; (16) is the equilibrium equation for the elastic beam. The boundary conditions (17) correspond to a free edge of the plate, i.e., the bending moment and the transverse force are equal to zero. The second group of boundary conditions (18) provides a glue condition at ${\Gamma }_0$ for v and ${\displaystyle \frac{\partial v}{\partial \nu }}$. As for condition (20), it provides a zero principal vector of the forces acting on the rigid inclusion $\omega .$ According to (19), the displacements of the plate and the beam coincide at the joint point; moreover, the displacement of the beam is zero at the point $x_3=-1.$

The following assertion takes place.

Theorem 1. 

For smooth solutions, problem statements (5)–(6) and (14)–(20) are equivalent.

Proof. 

Let (5)–(6) be fulfilled. We substitute into (6) test functions of the form $(\overline{v},\overline{u})=(\phi ,0),\phi \in C_0^{\infty }(\Omega \setminus \overline{\omega }),$$\phi \equiv 0$ in $\omega .$ This implies

$$\begin{array}{c}\hfill D(v,\phi )-\underset{\Omega }{\int }f\phi =0\forall \phi \in C_0^{\infty }(\Omega \setminus \overline{\omega }).\end{array}$$

This identity means that the equilibrium Equation (14) holds. Next, we substitute into (6) test functions of the form $(\overline{v},\overline{u})=(0,\psi ),$ where $\psi \in C_0^{\infty }\left(\sigma \right).$ In this case, the following identity follows:

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{u}_{,3}{\psi }_{,3}-\underset{\sigma }{\int }g\psi =0\forall \psi \in C_0^{\infty }\left(\sigma \right).\end{array}$$

Hence, Equation (16) is obtained.

Now, we integrate by the parts in (6) by using formula (3):

$$\begin{array}{c}\hfill -\underset{\Omega \setminus \overline{\omega }}{\int }\overline{v}\nabla \nabla M-\underset{\Gamma }{\int }{T}^n\overline{v}+\underset{\Gamma }{\int }{M}^n{\overline{v}}_n-\\ \hfill -\underset{{\Gamma }_0}{\int }{M}^{\nu }{\overline{v}}_{\nu }+\underset{{\Gamma }_0}{\int }{T}^{\nu }\overline{v}-\underset{\Omega }{\int }f\overline{v}-\underset{\sigma }{\int }{u}_{,33}\overline{u}-\underset{\sigma }{\int }g\overline{u}+\left(u_{,3}\overline{u}\right)\left(0\right)=0,\end{array}$$

(21)

where $M^{\nu },T^{\nu }$ are defined similarly to how they are defined in (2).

Taking into account the arbitrariness of $(\overline{v},\overline{u}),$ the identity (21) implies Equations (14) and (16), as well as

$$\begin{array}{c}\hfill a(-\underset{\omega }{\int }f+\underset{{\Gamma }_0}{\int }{T}^{\nu }+u_{,3}\left(0\right))=0,\end{array}$$

(22)

where $\overline{v}{|}_{\omega }=a\in \mathbb{R}.$ Since a is arbitrary, we obtain (20). The boundary conditions (17) can be derived from (6) by substituting the test functions $(\overline{v},0)\in W^{\omega }$ such that $\mathrm{supp}\overline{v}$ is located in a small neighborhood of the boundary $\Gamma .$

Hence, all the relations (14)–(20) are derived from (5)–(6).

Let us prove the converse. Assume that (14)–(20) hold. Then, for $(\overline{v},\overline{u})\in W^{\omega }$, it follows that

$$\begin{array}{c}\hfill \underset{\Omega \setminus \overline{\omega }}{\int }(-\nabla \nabla M-f)\overline{v}+\underset{\sigma }{\int }(-u_{,33}-g)\overline{u}=0.\end{array}$$

(23)

The integration by the parts in (23) implies that

$$\begin{array}{c}\hfill D(v,\overline{v})-\underset{\Omega }{\int }f\overline{v}+\underset{\omega }{\int }f\overline{v}-\underset{{\Gamma }_0}{\int }{T}^{\nu }\overline{v}+\\ \hfill +\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}-\left(u_{,3}\overline{u}\right)\left(0\right)=0.\end{array}$$

Through the relation $\overline{u}\left(0\right)=\overline{v}(0,0)$ and (20), we obtain (6). Theorem 1 is proven. □

3. The Rigidity Parameter Tends to Infinity in Problem (5)–(6)

In this section, we introduce a positive parameter characterizing the rigidity property of the elastic beam into the model (5)–(6) and the pass to the limit as $\alpha \to \infty .$ The energy functional $\Pi :W^{\omega }\to \mathbb{R}$ has the form

$$\begin{array}{c}\hfill {\Pi }^{\alpha }(v,u)={\displaystyle \frac{1}{2}}D(v,v)-\underset{\Omega }{\int }fv+{\displaystyle \frac{\alpha }{2}}\underset{\sigma }{\int }{u}_{,3}^2-\underset{\sigma }{\int }gu.\end{array}$$

For any fixed $\alpha >0,$ the minimization problem similar to (4) has a solution satisfying the identity

$$\begin{array}{c}\hfill (v^{\alpha },u^{\alpha })\in W^{\omega },\end{array}$$

(24)

$$\begin{array}{c}\hfill D(v^{\alpha },\overline{v})-\underset{\Omega }{\int }f\overline{v}+\alpha \underset{\sigma }{\int }{u}_{,3}^{\alpha }{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}=0\forall (\overline{v},\overline{u})\in W^{\omega }.\end{array}$$

(25)

Our aim is to pass to the limit in (24)–(25) as $\alpha \to \infty .$ At the first step, we obtain a priori estimates for the solutions of (24)–(25). From (24)–(25), it follows that

$$\begin{array}{c}\hfill D(v^{\alpha },v^{\alpha })-\underset{\Omega }{\int }f{v}^{\alpha }+\alpha \underset{\sigma }{\int }{\left(u_{,3}^{\alpha }\right)}^2-\underset{\sigma }{\int }g{u}^{\alpha }=0.\end{array}$$

(26)

According to Proposition 1, the relations (26) imply the estimate is uniform in $\alpha >{\alpha }_0>0$:

$$\begin{array}{c}\hfill \parallel (v^{\alpha },u^{\alpha }){\parallel }_{W^{\omega }}\le c.\end{array}$$

(27)

Choosing a subsequence, if necessary, we can assume that as $\alpha \to \infty ,$

$$\begin{array}{c}\hfill (v^{\alpha },u^{\alpha })\to (v,u)\mathrm{weakly} \mathrm{in}{W}^{\omega }.\end{array}$$

(28)

Moreover,

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{\left(u_{,3}^{\alpha }\right)}^2\le {\displaystyle \frac{c}{\alpha }}.\end{array}$$

(29)

The estimate (29) provides

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{u}_{,3}^2=0,\end{array}$$

and in view of the boundary condition $u(-1)=0,$ we obtain

$$\begin{array}{c}\hfill u\equiv 0\mathrm{on}\sigma .\end{array}$$

(30)

We introduce the set of admissible displacements for the limit problem

$$\begin{array}{c}\hfill V_{\omega }=\{v\in H^2(\Omega )|v\equiv 0\mathrm{on}\omega \}.\end{array}$$

Take any element $\overline{v}\in V_{\omega },\overline{u}\equiv 0\mathrm{on}\sigma .$ Then, $(\overline{v},\overline{u})\in W^{\omega }.$ Substitute $(\overline{v},\overline{u})$ into (25) as a test function and pass using (29), (30) to the limit as $\alpha \to \infty .$ This implies

$$\begin{array}{c}\hfill v\in V_{\omega },u\equiv 0\mathrm{on}\sigma ,\end{array}$$

(31)

$$\begin{array}{c}\hfill D(v,\overline{v})-\underset{\Omega }{\int }f\overline{v}=0\forall \overline{v}\in V_{\omega }.\end{array}$$

(32)

Thus, we have proven the following assertion.

Theorem 2. 

The solutions of the problem (24)–(25) converge according to (28), (30) with the solution of (31)–(32) as $\alpha \to \infty$.

The differential statement of the problem (31)–(32) is as follows. We have to find the functions $v,u$ defined in $\Omega ,\sigma ,$ respectively, such that $v\equiv 0$ in $\omega ,$ $u\equiv 0$ in $\sigma ,$

$$\begin{array}{c}\hfill \nabla \nabla M+f=0\mathrm{in}\Omega \setminus \overline{\omega },\end{array}$$

(33)

$$\begin{array}{c}\hfill M+A\nabla \nabla v=0\mathrm{in}\Omega \setminus \overline{\omega },\end{array}$$

(34)

$$\begin{array}{c}\hfill M^n=T^n=0\mathrm{on}\Gamma ,\end{array}$$

(35)

$$\begin{array}{c}\hfill v={\displaystyle \frac{\partial v}{\partial \nu }}=0\mathrm{on}{\Gamma }_0.\end{array}$$

(36)

It is easy to check that statements (31)–(32) and (33)–(36) are equivalent for smooth solutions.

4. A Case Without Rigid Inclusion

In this section, we consider a situation corresponding to a fully elastic plate. In this case, the rigid part of the plate is absent; see Figure 2. Due to the boundary conditions on $\Gamma$ corresponding to the free edge of the plate, a suitable boundary value problem would be non-coercive. In particular, for the problem to be solvable, we should impose a suitable condition onto the external forces.

Consider the space

$$\begin{array}{c}\hfill W=\{(v,u)\in H^2(\Omega )\times H^1\left(\sigma \right)|v(0,0)=u\left(0\right),u(-1)=0\}\end{array}$$

with the inner product

$$\begin{array}{c}\hfill \langle (v,u),(\overline{v},\overline{u})\rangle =\underset{\Omega }{\int }v\underset{\Omega }{\int }\overline{v}+\underset{\Omega }{\int }\nabla v\nabla \overline{v}+D(v,\overline{v})+\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}.\end{array}$$

(37)

This inner product induces the norm equivalent to the standard one. To check this, using the embedding theorem, we first note that for a small $\beta >0$, the following inequality holds:

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{2}}{(\underset{\Omega }{\int }v)}^2+{\displaystyle \frac{1}{2}}\underset{\Omega }{\int }{|\nabla v|}^2+{\displaystyle \frac{1}{2}}D(v,v)\ge \beta v^2(0,0).\end{array}$$

Then,

$$\begin{array}{c}\hfill \langle (v,u),(v,u)\rangle \ge {\displaystyle \frac{1}{2}}{(\underset{\Omega }{\int }v)}^2+{\displaystyle \frac{1}{2}}\underset{\Omega }{\int }{|\nabla v|}^2+{\displaystyle \frac{1}{2}}D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2+\beta v^2(0,0).\end{array}$$

On the other hand, $v^2(0,0)=u^2\left(0\right),$ and we can choose the following equivalent norm in the space $H^1\left(\sigma \right):$

$$\begin{array}{c}\hfill {\parallel u\parallel }_{H^1\left(\sigma \right)}^2=\underset{\sigma }{\int }{u}_{,3}^2+\beta u^2\left(0\right).\end{array}$$

Hence,

$$\begin{array}{c}\hfill \langle (v,u),(v,u)\rangle \ge {\displaystyle \frac{1}{2}}{\parallel v\parallel }_{H^2(\Omega )}^2+{\parallel u\parallel }_{H^1\left(\sigma \right)}^2,\end{array}$$

and the proof is complete.

Introduce the subspaces into $W:$

$$\begin{array}{c}\hfill S=\{(l,u)\in W|l\left(x\right)=a_1{x}_1+a_2{x}_2,x=(x_1,x_2)\in \Omega ;u\equiv 0\mathrm{on}\sigma ;a_1,a_2\in \mathbb{R}\},\\ \hfill S^{\perp }=\{(v,u)\in W|\underset{\Omega }{\int }v\underset{\Omega }{\int }{x}_i+\underset{\Omega }{\int }{v}_{,i}=0,i=1,2\}.\end{array}$$

The following statement applies.

Proposition 2. 

The space W can be presented as a direct sum of two orthogonal subspaces with respect to the inner product (37),

$$\begin{array}{c}\hfill W=S\oplus S^{\perp }.\end{array}$$

Proof. 

We take $(l,0)\in S,(v,u)\in W,l\left(x\right)=a_1{x}_1+a_2{x}_2.$ Then,

$$\begin{array}{c}\hfill \langle (l,0),(v,u)\rangle =a_i(\underset{\Omega }{\int }v\underset{\Omega }{\int }{x}_i+\underset{\Omega }{\int }{v}_{,i}).\end{array}$$

We see that a sufficient and necessary condition for the validity of the identity

$$\begin{array}{c}\hfill \langle (l,0),(v,u)\rangle =0\forall (l,0)\in S\end{array}$$

has the form $(v,u)\in S^{\perp }.$ Proposition 2 is proven.

Consider the energy functional ${\Pi }_0:W\to \mathbb{R},$

$$\begin{array}{c}\hfill {\Pi }_0(v,u)={\displaystyle \frac{1}{2}}D(v,v)-\underset{\Omega }{\int }fv+{\displaystyle \frac{1}{2}}\underset{\sigma }{\int }{u}_{,3}^2-\underset{\sigma }{\int }gu\end{array}$$

and the following problem:

$$\begin{array}{c}\hfill \mathrm{Find}(v,u)\in S^{\perp }\mathrm{such}\mathrm{that}{\Pi }_0(v,u)=\underset{S^{\perp }}{inf}{\Pi }_0.\end{array}$$

(38)

This problem has a solution satisfying the identity

$$\begin{array}{c}\hfill (v,u)\in S^{\perp },\end{array}$$

(39)

$$\begin{array}{c}\hfill D(v,\overline{v})-\underset{\Omega }{\int }f\overline{v}+\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}=0\forall (\overline{v},\overline{u})\in S^{\perp }.\end{array}$$

(40)

To prove this statement, it suffices to check the coercivity of the functional ${\Pi }_0$ on the space $S^{\perp }$ since the weak lower semicontinuity of ${\Pi }_0$ is clear. □

Proposition 3. 

A constant $c_1>0$ exists such that

$$\begin{array}{c}\hfill D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2\ge c_1{\parallel (v,u)\parallel }_W^2\forall (v,u)\in S^{\perp }.\end{array}$$

Proof. 

Assume that the statement is not true. In this case, we can choose a sequence $(v^k,u^k)\in S^{\perp }$ such that

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_W^2=1,\end{array}$$

(41)

$$\begin{array}{c}\hfill D(v^k,v^k)+\underset{\sigma }{\int }{\left(u_{,3}^k\right)}^2\to 0,k\to \infty .\end{array}$$

(42)

Choosing a subsequence, if necessary, we assume that as $k\to \infty ,$

$$\begin{array}{c}\hfill (v^k,u^k)\to (v,u)\mathrm{weakly} \mathrm{in}W\mathrm{and} \mathrm{strongly} \mathrm{in}{H}^1(\Omega )\times L^2\left(\sigma \right).\end{array}$$

(43)

Then, according to (42), we have

$$\begin{array}{c}\hfill D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2=0.\end{array}$$

(44)

From (44), we see that $a_0,a_1,a_2,b\in \mathbb{R}$ exist such that

$$\begin{array}{c}\hfill v\left(x\right)=a_0+a_1{x}_1+a_2{x}_2,x=(x_1,x_2)\in \Omega ;u\left(x_3\right)=b,x_3\in \sigma .\end{array}$$

Since $u(-1)=0$, we conclude that $b=0.$ On the other hand, $u\left(0\right)=v(0,0).$ Consequently, $v\left(x\right)=a_1{x}_1+a_2{x}_2,u\equiv 0$ on $\sigma$, i.e., $(v,u)\in S.$ Taking into account that $(v,u)\in S^{\perp }$, we conclude that

$$\begin{array}{c}\hfill (v,u)=(0,0).\end{array}$$

(45)

Next, using (43), (44), we obtain as $k\to \infty$

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_W^2=\parallel v^k{\parallel }_{H^1(\Omega )}^2+{\parallel u^k\parallel }_{L^2\left(\sigma \right)}^2+D(v^k,v^k)+\underset{\sigma }{\int }{\left(u_{,3}^k\right)}^2\to \\ \hfill \to {\parallel v\parallel }_{H^1(\Omega )}^2+{\parallel u\parallel }_{L^2\left(\sigma \right)}^2+D(v,v)+\underset{\sigma }{\int }{u}_{,3}^2={\parallel (v,u)\parallel }_W^2,\end{array}$$

i.e.,

$$\begin{array}{c}\hfill \parallel (v^k,u^k){\parallel }_W\to {\parallel (v,u)\parallel }_W.\end{array}$$

Then, according to (43), we conclude that

$$\begin{array}{c}\hfill (v^k,u^k)\to (v,u)\mathrm{strongly} \mathrm{in}W.\end{array}$$

In this case, from (41), it follows that

$$\begin{array}{c}\hfill {\parallel (v,u)\parallel }_W=1,\end{array}$$

and we obtain a contradiction to (45). Proposition 3 is proven.

Assume that the external forces satisfy the following condition:

$$\begin{array}{c}\hfill \underset{\Omega }{\int }fl=0\forall l\in L(\Omega ),\end{array}$$

(46)

where

$$\begin{array}{c}\hfill L(\Omega )=\left\{l\right|l\left(x\right)=a_1{x}_1+a_2{x}_2,x=(x_1,x_2)\in \Omega ;a_1,a_2\in \mathbb{R}\}.\end{array}$$

(47)

In this case, the relations (39)–(40) can be rewritten in the form

$$\begin{array}{c}\hfill (v,u)\in S^{\perp },\end{array}$$

(48)

$$\begin{array}{c}\hfill D(v,\overline{v}+l)-\underset{\Omega }{\int }f(\overline{v}+l)+\underset{\sigma }{\int }{u}_{,3}({\overline{u}}_{,3}+0)-\underset{\sigma }{\int }g(\overline{u}+0)=0\forall (\overline{v},\overline{u})\in S^{\perp },\forall (l,0)\in S.\end{array}$$

(49)

Taking into account Proposition 2, the problem (48)–(49) permits the following equivalent form:

$$\begin{array}{c}\hfill (v,u)\in S^{\perp },\end{array}$$

(50)

$$\begin{array}{c}\hfill D(v,\tilde{v})-\underset{\Omega }{\int }f\tilde{v}+\underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=0\forall (\tilde{v},\tilde{u})\in W.\end{array}$$

(51)

We therefore arrive at the following statement. □

Theorem 3. 

A unique solution to the problem (50)–(51) exists provided that condition (46) holds.

The uniqueness of the solution to the problem (50)–(51) is clear since for two possible solutions $(v^1,u^1),(v^2,u^2)$, we derive

$$\begin{array}{c}\hfill D(v^1-v^2,v^1-v^2)+\underset{\sigma }{\int }{(u_{,3}^1-u_{,3}^2)}^2=0.\end{array}$$

According to Proposition 3, this equality implies that $v^1=v^2,u^1=u^2.$

Denote using ${\delta }_0$ the Dirac delta-function, i.e.,

$$\begin{array}{c}\hfill {\delta }_0\left(\phi \right)=\phi (0,0)\forall \phi \in C_0^{\infty }(\Omega ).\end{array}$$

We can formulate the differential statement of the problem (50)–(51): find functions $v,u$ defined in $\Omega$ and $\sigma$, respectively, such that

$$\begin{array}{c}\hfill \nabla \nabla M+f=u_{,3}\left(0\right){\delta }_0\mathrm{in}\Omega ,\end{array}$$

(52)

$$\begin{array}{c}\hfill M+A\nabla \nabla v=0\mathrm{in}\Omega ,\end{array}$$

(53)

$$\begin{array}{c}\hfill -u_{,33}=g\mathrm{on}\sigma ,\end{array}$$

(54)

$$\begin{array}{c}\hfill M^n=T^n=0\mathrm{on}\Gamma ,\end{array}$$

(55)

$$\begin{array}{c}\hfill v(0,0)=u\left(0\right),u(-1)=0,\end{array}$$

(56)

$$\begin{array}{c}\hfill \underset{\Omega }{\int }v\underset{\Omega }{\int }{x}_i+\underset{\Omega }{\int }{v}_{,i}=0,i=1,2.\end{array}$$

(57)

Theorem 4. 

For smooth solutions, problem statements (50)–(51) and (52)–(57) are equivalent.

Proof. 

Let (50)–(51) be fulfilled. Substitute into (51) test functions of the form $(\tilde{v},\tilde{u})=(\phi ,0)\in W,\phi \in C_0^{\infty }\left({\Omega }_0\right),{\Omega }_0=\Omega \setminus \left\{(0,0)\right\}.$ This provides the identity

$$\begin{array}{c}\hfill D(v,\phi )-\underset{{\Omega }_0}{\int }f\phi =0\forall \phi \in C_0^{\infty }\left({\Omega }_0\right),\end{array}$$

i.e., we obtain the equation

$$\begin{array}{c}\hfill \nabla \nabla M+f=0\mathrm{in}{\Omega }_0.\end{array}$$

(58)

Next, we substitute into (51) test functions of the form $(0,\psi )\in W,\psi \in C_0^{\infty }\left(\sigma \right),$ which implies

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{u}_{,3}{\psi }_{,3}=\underset{\sigma }{\int }g\psi \forall \psi \in C_0^{\infty }\left(\sigma \right),\end{array}$$

i.e., the following equation is obtained:

$$\begin{array}{c}\hfill -u_{,33}=g\mathrm{on}\sigma .\end{array}$$

(59)

To prove the validity of the boundary conditions (55), we choose in (51) test functions of the form $(\tilde{v},0)\in W$ such that the support for $\tilde{v}$ is located in a small neighborhood of $\Gamma .$ This gives

$$\begin{array}{c}\hfill -\underset{\Omega }{\int }\nabla \nabla M·\tilde{v}-\underset{\Omega }{\int }f\tilde{v}+\underset{\Gamma }{\int }{M}^n{\tilde{v}}_n-\underset{\Gamma }{\int }{T}^n\tilde{v}=0.\end{array}$$

(60)

Accounting for (58) and the arbitrariness of $\tilde{v},{\tilde{v}}_n$ to $\Gamma$, the identity (60) implies the boundary conditions (55).

To conclude, we substitute into (51) test functions of the form $(\tilde{v},\tilde{u})=(\phi ,\psi )\in W,\phi \in C_0^{\infty }(\Omega ).$ According to (59) and the equality $\phi (0,0)=\psi \left(0\right),$ this substitution provides the identity

$$\begin{array}{c}\hfill D(v,\phi )-\underset{\Omega }{\int }f\phi +u_{,3}\left(0\right)\phi (0,0)=0\forall \phi \in C_0^{\infty }(\Omega ).\end{array}$$

This identity means that Equation (52) holds in the distributional sense. Thus, all relations (52)–(57) are obtained from (50)–(51).

Let us prove the converse. Assume that (52)–(57) hold. We multiply (52), (54) by $\tilde{v},\tilde{u},$ respectively, with $(\tilde{v},\tilde{u})\in W.$ The following two relations are obtained:

$$\begin{array}{c}\hfill \underset{\Omega }{\int }(\nabla \nabla M+f)\tilde{v}=u_{,3}\left(0\right)\tilde{v}(0,0),\end{array}$$

(61)

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=u_{,3}\left(0\right)\tilde{u}\left(0\right).\end{array}$$

(62)

Since $\tilde{v}(0,0)=\tilde{u}\left(0\right)$, relations (61)–(62) together with Green’s Formula (3) imply the identity

$$\begin{array}{c}\hfill D(v,\tilde{v})-\underset{\Omega }{\int }f\tilde{v}+\underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=0\forall (\tilde{v},\tilde{u})\in W,\end{array}$$

i.e., (51). Theorem 4 is proven. □

5. The Rigidity Parameter Tends to Infinity in Problem (50)–(51)

Similar to Section 3, we introduce a positive parameter $\alpha$ into the model (50)–(51) and pass to the limit as $\alpha \to \infty .$ Like in Section 3, this parameter describes the rigidity property of the beam $\sigma .$ The condition (46) is assumed to be fulfilled in this section.

The energy functional ${\Pi }_0^{\alpha }:W\to \mathbb{R}$ has the following form:

$$\begin{array}{c}\hfill {\Pi }_0^{\alpha }(v,u)={\displaystyle \frac{1}{2}}D(v,v)-\underset{\Omega }{\int }fv+{\displaystyle \frac{\alpha }{2}}\underset{\sigma }{\int }{u}_{,3}^2-\underset{\sigma }{\int }gu.\end{array}$$

We know that for any fixed $\alpha >0,$ the solution to the problem

$$\begin{array}{c}\hfill (v^{\alpha },u^{\alpha })\in S^{\perp },\end{array}$$

(63)

$$\begin{array}{c}\hfill D(v^{\alpha },\overline{v})-\underset{\Omega }{\int }f\overline{v}+\alpha \underset{\sigma }{\int }{u}_{,3}^{\alpha }{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}=0\forall (\overline{v},\overline{u})\in W\end{array}$$

(64)

exists and is unique. Relations (63)–(64) imply

$$\begin{array}{c}\hfill D(v^{\alpha },v^{\alpha })-\underset{\Omega }{\int }f{v}^{\alpha }+\alpha \underset{\sigma }{\int }{\left(u_{,3}^{\alpha }\right)}^2-\underset{\sigma }{\int }g{u}^{\alpha }=0.\end{array}$$

Consequently, according to Proposition 3, uniformly in $\alpha >{\alpha }_0>0,$ we obtain

$$\begin{array}{c}\hfill \parallel (v^{\alpha },u^{\alpha }){\parallel }_W\le c,\end{array}$$

(65)

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{\left(u_{,3}^{\alpha }\right)}^2\le {\displaystyle \frac{c}{\alpha }}.\end{array}$$

(66)

Assuming that as $\alpha \to \infty ,$ by (65), (66), for a suitable subsequence, it provides

$$\begin{array}{c}\hfill (v^{\alpha },u^{\alpha })\to (v,u)\mathrm{weakly} \mathrm{in}W,\end{array}$$

(67)

$$\begin{array}{c}\hfill u\equiv 0\mathrm{on}\sigma .\end{array}$$

(68)

Introduce the set of admissible functions for the limit problem

$$\begin{array}{c}\hfill V=\{v\in H^2(\Omega )|v(0,0)=0\}.\end{array}$$

Next, we take any element $\overline{v}\in V,\overline{u}\equiv 0\mathrm{on}\sigma .$ Then, $(\overline{v},\overline{u})\in W.$ This element $(\overline{v},\overline{u})$ can be substituted into (64) as a test function, and the passage to the limit can be fulfilled as $\alpha \to \infty .$ The limit relations are as follows:

$$\begin{array}{c}\hfill v\in V,u\equiv 0\mathrm{on}\sigma ,\end{array}$$

(69)

$$\begin{array}{c}\hfill D(v,\overline{v})-\underset{\Omega }{\int }f\overline{v}=0\forall \overline{v}\in V,\end{array}$$

(70)

$$\begin{array}{c}\hfill \underset{\Omega }{\int }v\underset{\Omega }{\int }{x}_i+\underset{\Omega }{\int }{v}_{,i}=0,i=1,2.\end{array}$$

(71)

Thus, the following assertion has been proven.

Theorem 5. 

The solutions to the problem (63)–(64) converge according to (67)–(68) with the solution to (69)–(71) as $\alpha \to \infty$ provided that the condition (46) holds.

The differential statement of the problem (69)–(71) is as follows: find functions $v,u$ defined in $\Omega ,\sigma ,$ respectively, such that $u\equiv 0$ in $\sigma$:

$$\begin{array}{c}\hfill \nabla \nabla M+f=0\mathrm{in}\Omega \setminus \left\{\right(0,0\left)\right\},\end{array}$$

(72)

$$\begin{array}{c}\hfill M+A\nabla \nabla v=0\mathrm{in}\Omega ,\end{array}$$

(73)

$$\begin{array}{c}\hfill M^n=T^n=0\mathrm{on}\Gamma ,\end{array}$$

(74)

$$\begin{array}{c}\hfill v(0,0)=0,\underset{\Omega }{\int }v\underset{\Omega }{\int }{x}_i+\underset{\Omega }{\int }{v}_{,i}=0,i=1,2.\end{array}$$

(75)

Theorem 6. 

For smooth solutions, problem statements (69)–(71) and (72)–(75) are equivalent.

We omit the proof of this assertion since it is easier compared to that for Theorem 1.

6. The Case of a Thin Delaminated Inclusion

In this section, we assume that the elastic plate contains a thin rigid inclusion. Let $\gamma$ be a smooth curve without self-intersections and $\nu =({\nu }_1,{\nu }_2)$ a unit normal vector to $\gamma ;$ $\overline{\gamma }\subset \Omega ,(0,0)\notin \overline{\gamma }$; see Figure 3. The form of the inclusion coincides with $\gamma$. Assume that the inclusion is delaminated from the surrounding elastic body at the positive face with respect to $\nu ,$ i.e, at ${\gamma }^{+}.$ Denote ${\Omega }_{\gamma }=\Omega \setminus \overline{\gamma }.$

Introduce the space

$$\begin{array}{c}\hfill H^{2,\gamma }\left({\Omega }_{\gamma }\right)=\{v\in H^2\left({\Omega }_{\gamma }\right){\left|v\right|}_{{\gamma }^{-}}\in L\left(\gamma \right)\},\end{array}$$

where

$$\begin{array}{c}\hfill L\left(\gamma \right)=\left\{l\right|l\left(x\right)=b_0+b_1{x}_1+b_2{x}_2;x=(x_1,x_2)\in \gamma ;b_i\in \mathbb{R},i=0,1,2\},\end{array}$$

In the definition of $L\left(\gamma \right)$, the constants $b_i$ are arbitrary.

Denote

$$\begin{array}{c}\hfill W^{\gamma }=\{(v,u)\in H^{2,\gamma }\left({\Omega }_{\gamma }\right)\times H^1\left(\sigma \right)|v(0,0)=u\left(0\right),u(-1)=0\}\end{array}$$

and consider the energy functional ${\Pi }_{\gamma }:W^{\gamma }\to \mathbb{R}$

$$\begin{array}{c}\hfill {\Pi }_{\gamma }(v,u)={\displaystyle \frac{1}{2}}{D}_{\gamma }(v,v)-\underset{{\Omega }_{\gamma }}{\int }fv+{\displaystyle \frac{1}{2}}\underset{\sigma }{\int }{u}_{,3}^2-\underset{\sigma }{\int }gu\end{array}$$

with

$$\begin{array}{c}\hfill D_{\gamma }(w,\overline{w})=\underset{{\Omega }_{\gamma }}{\int }{a}_{ijkl}{w}_{,kl}{\overline{w}}_{,ij}.\end{array}$$

Introduce the inner product into the space $W^{\gamma }:$

$$\begin{array}{c}\hfill \{(v,u),(\overline{v},\overline{u})\}=\underset{{\Omega }_{\gamma }}{\int }v\underset{{\Omega }_{\gamma }}{\int }\overline{v}+\underset{{\Omega }_{\gamma }}{\int }\nabla v\nabla \overline{v}+D_{\gamma }(v,\overline{v})+\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}.\end{array}$$

(76)

Like in Section 4, we can check that the inner product (76) induces the norm in the space $W^{\gamma }$ to be equivalent to the standard one.

Consider two subspaces in $W^{\gamma }:$

$$\begin{array}{c}\hfill S_{\gamma }=\{(l,u)\in W^{\gamma }|l\left(x\right)=a_1{x}_1+a_2{x}_2,a_i\in \mathbb{R},i=1,2;x=(x_1,x_2)\in {\Omega }_{\gamma }\\ \hfill u\equiv 0\mathrm{on}\sigma ;a_1,a_2\in \mathbb{R}\},\end{array}$$

$$\begin{array}{c}\hfill S_{\gamma }^{\perp }=\{(v,u)\in W^{\gamma }|\underset{{\Omega }_{\gamma }}{\int }v\underset{{\Omega }_{\gamma }}{\int }{x}_i+\underset{{\Omega }_{\gamma }}{\int }{v}_{,i}=0,i=1,2\}.\end{array}$$

We can prove the following assertion.

Proposition 4. 

The space $W^{\gamma }$ can be presented as a direct sum of two orthogonal subspaces with respect to the inner product (76):

$$\begin{array}{c}\hfill W^{\gamma }=S_{\gamma }\oplus S_{\gamma }^{\perp }.\end{array}$$

The proof of this proposition reminds us of that for Proposition 2, so we omit it.

Now, consider the following variational problem:

$$\begin{array}{c}\hfill \mathrm{Find}(v,u)\in S_{\gamma }^{\perp }\mathrm{such}\mathrm{that}{\Pi }_{\gamma }(v,u)=\underset{S_{\gamma }^{\perp }}{inf}{\Pi }_{\gamma }.\end{array}$$

(77)

The problem (77) has a solution satisfying the following identity:

$$\begin{array}{c}\hfill (v,u)\in S_{\gamma }^{\perp },\end{array}$$

(78)

$$\begin{array}{c}\hfill D_{\gamma }(v,\overline{v})-\underset{{\Omega }_{\gamma }}{\int }f\overline{v}+\underset{\sigma }{\int }{u}_{,3}{\overline{u}}_{,3}-\underset{\sigma }{\int }g\overline{u}=0\forall (\overline{v},\overline{u})\in S_{\gamma }^{\perp }.\end{array}$$

(79)

Indeed, the energy functional ${\Pi }_{\gamma }$ is coercive on $S_{\gamma }^{\perp }$ due to the following assertion.

Proposition 5. 

A constant $c_2>0$ exists such that

$$\begin{array}{c}\hfill D_{\gamma }(v,v)+\underset{\sigma }{\int }{u}_{,3}^2\ge c_1{\parallel (v,u)\parallel }_{W^{\gamma }}^2\forall (v,u)\in S_{\gamma }^{\perp }.\end{array}$$

We do not provide the proof of Proposition 5 since it follows the same lines of those for Proposition 3.

The weak lower semicontinuity of the functional ${\Pi }_{\gamma }$ is clear; hence, the problem (77) has a solution satisfying (78)–(79).

Now, assume that the following condition holds:

$$\begin{array}{c}\hfill \underset{{\Omega }_{\gamma }}{\int }fl=0\forall l\in L\left({\Omega }_{\gamma }\right),\end{array}$$

(80)

where the space $L\left({\Omega }_{\gamma }\right)$ is introduced similarly to in (47). In this case, from (78)–(79), we obtain

$$\begin{array}{c}\hfill D_{\gamma }(v,\overline{v}+l)-\underset{{\Omega }_{\gamma }}{\int }f(\overline{v}+l)+\underset{\sigma }{\int }{u}_{,3}({\overline{u}}_{,3}+0)-\underset{\sigma }{\int }g(\overline{u}+0)=0\\ \hfill \forall (\overline{v},\overline{u})\in S_{\gamma }^{\perp },\forall (l,0)\in S_{\gamma },\end{array}$$

and according to Proposition 4, it follows that

$$\begin{array}{c}\hfill (v,u)\in S_{\gamma }^{\perp },\end{array}$$

(81)

$$\begin{array}{c}\hfill D_{\gamma }(v,\tilde{v})-\underset{{\Omega }_{\gamma }}{\int }f\tilde{v}+\underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=0\forall (\tilde{v},\tilde{u})\in W^{\gamma }.\end{array}$$

(82)

Hence, we arrive at the following assertion.

Theorem 7. 

A unique solution to the problem (81)–(82) exists provided that the condition (80) applies.

Now, we are ready to write the differential statement for the problem (81)–(82): it is necessary to find functions $v,u$ defined in ${\Omega }_{\gamma }$ and $\sigma$, respectively, as well as a function $l_0\in S\left(\gamma \right)$ such that

$$\begin{array}{c}\hfill \nabla \nabla M+f=u_{,3}\left(0\right){\delta }_0\mathrm{in}{\Omega }_{\gamma },\end{array}$$

(83)

$$\begin{array}{c}\hfill M+A\nabla \nabla v=0\mathrm{in}{\Omega }_{\gamma },\end{array}$$

(84)

$$\begin{array}{c}\hfill -u_{,33}=g\mathrm{on}\sigma ,\end{array}$$

(85)

$$\begin{array}{c}\hfill M^n=T^n=0\mathrm{on}\Gamma ,\end{array}$$

(86)

$$\begin{array}{c}\hfill M^{\nu }=0,v=l_0\mathrm{on}{\gamma }^{-};M^{\nu }=T^{\nu }=0\mathrm{on}{\gamma }^{+},\end{array}$$

(87)

$$\begin{array}{c}\hfill v(0,0)=u\left(0\right),u(-1)=0,\end{array}$$

(88)

$$\begin{array}{c}\hfill \underset{{\gamma }^{-}}{\int }{T}^{\nu }w=0\forall w\in H^{2,\gamma }\left({\Omega }_{\gamma }\right),\end{array}$$

(89)

$$\begin{array}{c}\hfill \underset{{\Omega }_{\gamma }}{\int }v\underset{{\Omega }_{\gamma }}{\int }{x}_i+\underset{{\Omega }_{\gamma }}{\int }{v}_{,i}=0,i=1,2.\end{array}$$

(90)

The following assertion applies.

Theorem 8. 

For smooth solutions, problem statements (78)–(79) and (83)–(90) are equivalent.

Proof. 

Let (83)–(90) be fulfilled. Multiply (83), (90) by $\tilde{v},\tilde{u},$ respectively, and integrate over ${\Omega }_{\gamma },\sigma ,$ where $(\tilde{v},\tilde{u})\in W^{\gamma }.$ This provides the relations

$$\begin{array}{c}\hfill \underset{{\Omega }_{\gamma }}{\int }(\nabla \nabla M+f)\tilde{v}=u_{,3}\left(0\right)\tilde{v}(0,0),\end{array}$$

(91)

$$\begin{array}{c}\hfill \underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=u_{,3}\left(0\right)\tilde{u}\left(0\right).\end{array}$$

(92)

In (91), we can use the following Green’s formula valid for smooth functions $M=\left\{M_{ij}\right\},w$:

$$\begin{array}{c}\hfill -\underset{{\Omega }_{\gamma }}{\int }M·\nabla \nabla w=-\underset{{\Omega }_{\gamma }}{\int }w\nabla \nabla M-\underset{\gamma }{\int }\left[M^{\nu }{w}_{\nu }\right]+\underset{\gamma }{\int }\left[T^{\nu }w\right]+\\ \hfill +\underset{\Gamma }{\int }{M}^n{w}_n-\underset{\Gamma }{\int }{T}^{n}w,\end{array}$$

(93)

where $\left[p\right]=p^{+}-p^{-}$ is the jump of a function p on $\gamma .$ The signs ± correspond to the positive and negative directions of $\nu .$ This formula can be derived from (3) for the domain ${\Omega }_{\gamma }$ with the cut $\gamma .$ Taking into account that $\tilde{v}(0,0)=\tilde{u}\left(0\right)$ and boundary conditions (86)–(89), from (91)–(92), we obtain

$$\begin{array}{c}\hfill D_{\gamma }(v,\tilde{v})-\underset{{\Omega }_{\gamma }}{\int }f\tilde{v}+\underset{\sigma }{\int }{u}_{,3}{\tilde{u}}_{,3}-\underset{\sigma }{\int }g\tilde{u}=0\forall (\tilde{v},\tilde{u})\in W^{\gamma },\end{array}$$

which coincides with (82). □

Conversely, let us prove that (81)–(82) implies all relations (83)–(90). Assume that (81)–(82) holds. The proof of the validity of (83)–(86) can be obtained similarly to that for (52)–(55) in Theorem 4. Hence, it suffices to prove the first equality in (87) and the second group of relations (87), as well as (89). All of these relations follow from (82) using Green’s formula (93) and taking into account the arbitrariness of $\tilde{v},{\tilde{v}}_{\nu }$ to ${\gamma }^{+}$ and $\tilde{v}$ to ${\gamma }^{-}$. Theorem 8 is proven.

7. Conclusions

A rigorous analysis of the boundary value problems for a T-shape deformable structure is provided. The structure consists of a plate and a connected elastic beam. The plate may contain a volume rigid inclusion and a thin rigid delaminated inclusion. Neumann-type boundary conditions are considered at the external boundary of the plate, which imply the necessity of finding a suitable condition imposed onto the external forces acting on the structure for the problem to be solvable. The existence of solutions to the considered problems is established. To this end, the Sobolev space is presented as a sum of two orthogonal subspaces, and a unique solution is found in a suitable subspace. An asymptotic analysis of the solutions with respect to the rigidity parameter of the beam is performed, and statements of the limit problems are obtained. The technique proposed in this paper can be used for the analysis of various deformable T-shape structures with Neumann-type boundary conditions imposed at the external boundaries.