Asymptotic Domain Decomposition Method for Approximation the Spectrum of the Diffusion Operator in a Domain Containing Thin Tubes

Abstract

The spectral problem for the diffusion operator is considered in a domain containing thin tubes. A new version of the method of partial asymptotic decomposition of the domain is introduced to reduce the dimension inside the tubes. It truncates the tubes at some small distance from the ends of the tubes and replaces the tubes with segments. At the interface of the three-dimensional and one-dimensional subdomains, special junction conditions are set: the pointwise continuity of the flux and the continuity of the average over a cross-section of the eigenfunctions. The existence of the discrete spectrum is proved for this partially reduced problem of the hybrid dimension. The conditions of the closeness of two spectra, i.e., of the diffusion operator in the full-dimensional domain and the partially reduced one, are obtained.

1. Introduction

The method of asymptotic partial decomposition of the domain (MAPDD) was introduced for partial derivative equations set in thin tube structures in [1] (cf. also [2,3,4]). The thin tube structures (also called thin rod structures) are some unions of thin cylinders. In this case, the method gives an important gain in computational resources reducing the dimension to one everywhere except for small two- or three-dimensional parts of the domain. This method was applied to the spectral problems in [5], where the asymptotic of the spectrum of the Laplacian in two joint thin rectangles with Neumann’s boundary conditions on the lateral boundary were considered. The MAPDD was justified for such a structure. The asymptotic behavior of the spectrum of the Laplacian with Dirichlet’s or Neumann’s boundary conditions in thin domains was considered in a vast body of literature (see [6,7,8,9,10,11,12,13], and the references therein). Furthermore, let us mention some works on the eigenvalues for the Laplacian for several special cases of domains containing thin tubes, described with the help of a small parameter (cf. [14,15,16,17] among others); these spectral problems were treated using asymptotic analysis. As a matter of fact, Refs. [6,7,8,13] address spectral problems in thin planar domains, while Refs. [10,11,12] deal with three-dimensional domains in very different structures from those considered here. This topic is of great interest due to its multiple applications in scattering theory, wave-guides, etc.

In the present paper, we introduce a new version of the MAPDD with special interface conditions between the full-dimensional and one-dimensional parts, that is, the pointwise continuity of the flux and the continuity of the average over a cross-section of the eigenfunctions. These conditions were used for MAPDD approximation of the heat equation in [18]. In particular, some numerical methods also use junction conditions with jumps of the unknown function (see, e.g., [19,20]). Here, we deal with a spectral problem for the Neumann diffusion operator.

The results obtained in former works on the asymptotic analysis of spectral problems in thin domains mostly use the presence in the model of a small parameter, that is, the ratio of the characteristic sizes in the transverse and longitudinal (or in-plane) directions. The justification for these results was provided via the proof of theorems concerning the convergence of the spectrum of the original problem to that of the spectrum of the reduced problem as the small parameter tends to zero. In the present paper, we use another approach related to the method of justification for the MAPDD in [21], but we introduce different junction conditions. There is no explicit small parameter in the description of the domain, but, implicitly, it is introduced via the assumption that the first positive eigenvalues of the Neumann diffusion operator on the cross-sections of the tubes are sufficiently large. The closeness of the spectra of the original and reduced models is proved under the condition that the eigenvalues of both the original and the reduced models are smaller than these first eigenvalues on the cross-sections of the tubes. In addition, here, we do not assume that the domain is thin everywhere. It may have some thick parts connected by thin tubes. The boundary is assumed to be Lipschitz, and we do not require the regularity of the coefficient in the diffusion operator out of the tubes. The proof of the aforementioned closeness of the spectra also relies on the distance from the ends of the tubes to the one-dimensional domain in the reduced model, a distance that needs to be adjusted to achieve the desired accuracy for the approximation between the spectra.

The main results are as follows. The spectral problem for the diffusion operator is considered in a domain containing thin tubes. The reduced model is obtained from the original one by the truncation of the three-dimensional tubes at some small distance from the ends of the tubes replacing the truncated parts of the tubes by the segments. At the interface of the three-dimensional and one-dimensional subdomains, special junction conditions are set: the pointwise continuity of the flux and continuity of the average of the eigenfunctions over a cross-section. The existence of the discrete spectrum is proved for this partially reduced problem of the hybrid dimension. After prescribing an accurate precision $\epsilon$, to obtain it, the conditions of the closeness of two spectra (i.e., of the diffusion operator in the full-dimensional domain and the partially reduced one) are obtained.

The structure of the paper is as follows: Section 2 contains some preliminary results, the definition of the domain, and the setting of the spectral problem. Moreover, the required properties on the smoothness of the eigenfunctions are obtained (cf. Section 2.3). In Section 3, the reduced approximate spectral problem is formulated. Because of the hybrid dimension, notations, weak formulation, and properties of spaces become more complicated. The required smoothness results for the eigenfunctions are also proved (cf. Section 3.3). In Section 4, we state the main results on the closeness of the spectra of the original and reduced problems (cf. Theorems 2 and 3). As a consequence, we claim the closeness of the spectra in the case where the tubes are cylinders (cf. Theorems 4 and 5). Section 5 and Section 6 contain the proofs of Theorems 2 and 3, respectively. Finally, some concluding remarks are outlined in Section 7.

2. Preliminary Results and Setting of the Problem

In this section, we introduce some notations and preliminary results that will be used throughout the whole paper. Section 2.2 contains the setting of the spectral problem under consideration, and, in Section 2.3, we provide certain smoothness properties for the eigenfunctions.

2.1. Preliminaries

By a tube, we mean a set $\mathcal{T}\subset {\mathbb{R}}^3$ of the form $\mathcal{T}=\Omega \times (0,\ell )$ in the local coordinates $(y,z)=(y_1,y_2,z)$, where the base $\Omega$ of the tube is a bounded domain in ${\mathbb{R}}^2$ with a Lipschitz boundary $\partial \Omega$. The set $\partial \mathcal{T}=\partial \Omega \times (0,\ell )$ is called the lateral surface of the tube $\mathcal{T}$, and ℓ stands for its length. Define

$$\begin{array}{cc}\hfill & D_1\phi ={\displaystyle \frac{\partial \phi }{\partial y_1}},D_2\phi ={\displaystyle \frac{\partial \phi }{\partial y_2}},D_z\phi ={\displaystyle \frac{\partial \phi }{\partial z}},{\nabla }_y\phi =(D_1\phi ,D_2\phi ),{\Delta }_y\phi =D_1^2\phi +D_2^2\phi .\hfill \end{array}$$

Let ${\mu }_1(\Omega )$ be the first positive eigenvalue of the following problem

$$\begin{array}{cc}& -{\Delta }_y\phi =\mu \phi ,y\in \Omega ,\\ & {\nabla }_y\phi ·\nu =0,y\in \partial \Omega ,\end{array}$$

where $\nu$ is the outward normal to $\partial \Omega$.

Recall that for a disk $\Omega$ with a diameter D, ${\mu }_1(\Omega )=4p^2/D^2$ where $p=1.8142\dots$ is the smallest positive root of the derivative $J_1^{\prime }$ of the Bessel function $J_1$. In the general case, we have [22]

$${\mu }_1(\Omega )\le {\mu }_1\left({\Omega }^{*}\right)={\displaystyle \frac{\pi p^2}{\mathrm{meas}(\Omega )}},$$

(1)

where ${\Omega }^{*}$ is a disk of the same measure as $\Omega$. Inequality (1) becomes an equality only if $\Omega$ is the disk.

In the case where $\Omega$ is a convex domain of diameter D, we have the lower estimate [23]:

$${\displaystyle \frac{{\pi }^2}{2D^2}}\le {\mu }_1(\Omega ).$$

Note that $\rho (\Omega )=1/\sqrt{{\mu }_1(\Omega )}$ is an optimal constant in the Poincaré inequality

$${\parallel \phi \parallel }_{L^2(\Omega )}\le \rho (\Omega ){\parallel {\nabla }_y\phi \parallel }_{L^2(\Omega )}\forall \phi \in {\tilde{H}}^1(\Omega ).$$

Here, ${\tilde{H}}^1(\Omega )=\{\phi \in H^1(\Omega )|\underset{\Omega }{\int }\phi dx=0\}$ is a Hilbert space equipped with the inner product and norm, respectively

$${(\phi ,\psi )}_{{\tilde{H}}^1(\Omega )}={({\nabla }_y\phi ,{\nabla }_y\psi )}_{L^2(\Omega )}{,\parallel \phi \parallel }_{{\tilde{H}}^1(\Omega )}={\parallel {\nabla }_y\phi \parallel }_{L^2(\Omega )}.$$

Obviously, ${\mu }_1(ϵ\Omega )={ϵ}^{-2}{\mu }_1(\Omega )$ for the contracted domain, and $ϵ\Omega =\left\{ϵy\right|y\in \Omega \}$ for all $0<ϵ$.

We see that for a wide class of domains, the value $\rho (\Omega )$ characterizes the size and geometry of $\Omega$. So, we refer to a thin tube $\mathcal{T}$ when $\rho (\Omega )\ll \ell$. For an $\Omega$ that is a disk with a radius of $ϵ$, the tube is thin if $ϵ\ll \ell$; in this case, $\rho (\Omega )=ϵ/p$.

2.2. Spectral Problem

Let G be a bounded domain in ${\mathbb{R}}^3$ with a Lipschitz boundary $\partial G$. Consider the following spectral problem: find a couple $(\lambda ,u)\in \mathbb{R}\times H^1\left(G\right)$ such that ${\parallel u\parallel }_{L^2\left(G\right)}=1$, satisfying the equation

$$\begin{array}{cc}\hfill -\mathrm{div}(K\nabla u)=\lambda u,& x\in G\hfill \end{array}$$

and the boundary condition

$$\begin{array}{cc}\hfill K\nabla u·n=0,& x\in \partial G.\hfill \end{array}$$

Here, $K\in L^{\infty }\left(G\right)$, $\underset{x\in G}{\mathrm{ess}\inf }K\left(x\right)>0$, n is the outward normal to $\partial G$.

The weak formulation is given by the following identity:

$$\begin{array}{cc}& {(K\nabla u,\nabla \phi )}_{L^2\left(G\right)}=\lambda {(u,\phi )}_{L^2\left(G\right)}\forall \phi \in H^1\left(G\right).\end{array}$$

(2)

It is well known that this problem has a countable set $0={\lambda }_0<{\lambda }_1\le {\lambda }_2\le \dots$ of eigenvalues and that the corresponding set ${\left\{{\phi }_k\right\}}_{k=0}^{\infty }$ of eigenfunctions can be chosen to form an orthogonal base in $H^1\left(G\right)$ and an orthonormal base in $L^2\left(G\right)$.

Assume that G contains a set of disjoint thin tubes ${\mathcal{T}}_j$, $1\le j\le N$ such that their lateral surfaces $\partial {\mathcal{T}}_j$ belong to the boundary $\partial G$. Let each tube ${\mathcal{T}}_j$ in the local coordinate system $(y,z)$ have the form ${\mathcal{T}}_j={\Omega }_j\times (0,{\ell }_j)$ and let $\partial {\mathcal{T}}_j=\partial {\Omega }_j\times (0,{\ell }_j)$. Consider the subtube ${\mathcal{T}}_{j,h_j}={\Omega }_j\times (h_j,{\ell }_j-h_j)$, where $h_j\in (0,{\ell }_j/2)$, and denote ${\mathcal{T}}_h=\underset{j=1}{\stackrel{N}{\cup }}{\mathcal{T}}_{j,h_j}$, $G_h=G\setminus {\overline{\mathcal{T}}}_h$.

Define the average of the function $u\in L^2\left({\Omega }_j\right)$ over the cross-section of the tube ${\mathcal{T}}_j$

$${\left[u\right]}_j={\displaystyle \frac{1}{\mathrm{meas}\left({\Omega }_j\right)}}\underset{{\Omega }_j}{\int }u\left(y\right)dy.$$

Assume that K satisfies the following condition:

$$\begin{array}{cc}\hfill & K\left(x\right)=K_j=\mathrm{const}\mathrm{in}{\mathcal{T}}_j,1\le j\le N.\hfill \end{array}$$

2.3. Some Properties of Eigenfunctions

Let u be an eigenfunction corresponding to the eigenvalue $\lambda$. Denote by $u_j$ the restriction of the function u to ${\mathcal{T}}_j$, $1\le j\le N$. Let $r_j=u_j-{\left[u_j\right]}_j$. We will consider $u_j$ and $r_j$ as functions on $(0,{\ell }_j)$ with values in $H^1\left({\Omega }_j\right)$. Note that $u_j,r_j\in L^2(0,{\ell }_j;H^1\left({\Omega }_j\right))$, $D_z{u}_j,D_z{r}_j\in L^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$. We set ${\tilde{\lambda }}_j=\lambda /K_j$.

Lemma 1.

The function ${\left[u_j\right]}_j$ belongs to the space $H^2(0,{\ell }_j)$, and

$$\begin{array}{c}\hfill -D_z^2{\left[u_j\right]}_j={\tilde{\lambda }}_j{\left[u_j\right]}_j.\end{array}$$

(3)

Proof. 

It is clear that ${\left[u_j\right]}_j\in H^1(0,{\ell }_j)$ and $D_z{\left[u_j\right]}_j={\left[D_z{u}_j\right]}_j$.

Let $\eta \in C_0^{\infty }(0,{\ell }_j)$. Substituting into (2) the function $\phi$, equal to $K_j^{-1}\eta \left(z\right)$ on ${\mathcal{T}}_j$ and equal to zero on $G\setminus {\mathcal{T}}_j$, we obtain the identity

$$\begin{array}{c}\hfill \underset{0}{\overset{{\ell }_j}{\int }}{D}_z{\left[u_j\right]}_j\left(z\right){\eta }^{\prime }\left(z\right)dz=\underset{0}{\overset{{\ell }_j}{\int }}{\tilde{\lambda }}_j{\left[u_j\right]}_j\left(z\right)\eta \left(z\right)dz\forall \eta \in C_0^{\infty }(0,{\ell }_j),\end{array}$$

(4)

which implies the existence of the weak derivative $D_z^2{\left[u_j\right]}_j$ by definition and equality (3).

The lemma is proved. □

Lemma 2.

The functions $u_j$ and $r_j$ have the derivatives $D_z^2{u}_j$ and $D_z^2{r}_j$, which belong to the space $L_{loc}^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$.

Proof. 

We introduce the finite-difference analogues of the derivative $D_z\Phi$

$${\partial }^h\Phi \left(z\right)={\displaystyle \frac{\Phi (z+h)-\Phi \left(z\right)}{h}},{\overline{\partial }}^h\Phi \left(z\right)={\displaystyle \frac{\Phi \left(z\right)-\Phi (z-h)}{h}},$$

where $h>0$.

Assume that $(a,b)\subset (0,{\ell }_j)$, and let $\zeta \in C_0^{\infty }(0,{\ell }_j)$ be a cut-off function, equal to one on $(a,b)$. By substituting into identity (2) the test function $\phi$ equal to $-K_j^{-1}{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{u}_j\right)$ on ${\mathcal{T}}_j$ and zero on $G\setminus {\mathcal{T}}_j$, we obtain

$$\begin{array}{cc}\hfill & -\underset{0}{\overset{{\ell }_j}{\int }}{\left(D_z{u}_j,{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{D}_z{u}_j\right)\right)}_{L^2\left({\Omega }_j\right)}dz-\underset{0}{\overset{{\ell }_j}{\int }}{\left(D_z{u}_j,{\overline{\partial }}^h\left(2\zeta {\zeta }^{\prime }{\partial }^h{u}_j\right)\right)}_{L^2\left({\Omega }_j\right)}dz\hfill \\ \hfill & -\underset{0}{\overset{{\ell }_j}{\int }}{\left({\nabla }_y{u}_j,{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{\nabla }_y{u}_j\right)\right)}_{L^2\left({\Omega }_j\right)}dz=-{\tilde{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{\left(u_j,{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{u}_j\right)\right)}_{L^2\left({\Omega }_j\right)}dz.\hfill \end{array}$$

Here, h is small enough.

Using the finite-difference analogue of the formula for integration by parts, we obtain

$$\begin{array}{cc}\hfill & \underset{0}{\overset{{\ell }_j}{\int }}\parallel \zeta {\partial }^h{D}_z{u}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+\underset{0}{\overset{{\ell }_j}{\int }}{\parallel \zeta {\partial }^h{\nabla }_y{u}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz\hfill \\ \hfill & =-\underset{0}{\overset{{\ell }_j}{\int }}{(\zeta {\partial }^h{D}_z{u}_j,2{\zeta }^{\prime }{\partial }^h{u}_j)}_{L^2\left({\Omega }_j\right)}dz+{\tilde{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{\parallel \zeta {\partial }^h{u}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\underset{0}{\overset{{\ell }_j}{\int }}\parallel \zeta {\partial }^h{D}_z{u}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+{\displaystyle \frac{1}{2}}\underset{0}{\overset{{\ell }_j}{\int }}\parallel 2{\zeta }^{\prime }{\partial }^h{u}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+{\tilde{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{\parallel \zeta {\partial }^h{u}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz.\hfill \end{array}$$

From here,

$$\begin{array}{cc}\hfill & \underset{0}{\overset{{\ell }_j}{\int }}\parallel \zeta {\partial }^h{D}_z{u}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz\le c_{\zeta }{\parallel D_z{u}_j\parallel }_{L^2(0,{\ell }_j;L^2\left({\Omega }_j\right))}^2.\hfill \end{array}$$

Thus, we have the following estimate uniform in h

$$\begin{array}{cc}\hfill & {\parallel {\partial }^h{D}_z{u}_j\parallel }_{L^2(a,b;L^2\left({\Omega }_j\right))}^2\le c_{\zeta }{\parallel D_z{u}_j\parallel }_{L^2(0,{\ell }_j;L^2\left({\Omega }_j\right))}^2,\hfill \end{array}$$

where $c_{\zeta }$ is a constant depending on $\zeta$.

It implies the existence of the derivative $D_z^2{u}_j\in L^2(a,b;L^2\left({\Omega }_j\right))$. Thus, $D_z^2{u}_j\in L_{loc}^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$. As a consequence, there exists a derivative $D_z^2{r}_j\in L_{loc}^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$.

The lemma is proved. □

Lemma 3.

For almost all $z\in (0,{\ell }_j)$ the following identities hold

$$\begin{array}{cc}\hfill & {(D_z^2{u}_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}={({\nabla }_y{u}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\tilde{\lambda }}_j{(u_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\forall \phi \in H^1\left({\Omega }_j\right).\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill & {(D_z^2{r}_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}={({\nabla }_y{r}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\tilde{\lambda }}_j{(r_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\forall \phi \in H^1\left({\Omega }_j\right).\hfill \end{array}$$

(6)

Proof. 

Let $\eta \in C_0^{\infty }(0,{\ell }_j)$ and $\phi \in H^1\left({\Omega }_j\right)$. Since $D_z^2{u}_j\in L_{loc}^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$, then the substitution into (2) of the test function equal to $K_j^{-1}\eta \left(z\right)\phi \left(y\right)$ on ${\mathcal{T}}_j$ and zero on $G\setminus {\mathcal{T}}_j$ and an integration by parts gives

$$\begin{array}{cc}\hfill & -\underset{0}{\overset{{\ell }_j}{\int }}{(D_z^2{u}_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\eta \left(z\right)dz+\underset{0}{\overset{{\ell }_j}{\int }}{(\nabla u_j\left(z\right),\nabla \phi )}_{L^2\left({\Omega }_j\right)}\eta \left(z\right)dz={\tilde{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{(u\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\eta \left(z\right)dz.\hfill \end{array}$$

This implies identity (5). As a consequence (cf. (4)), identity (6) holds.

The lemma is proved. □

3. Approximate Spectral Problem

This section is devoted to the setting of the problem referred to by us as the approximate spectral problem of hybrid dimensions. Section 3.1. contains the notations and preliminary results convenient for the formulation of the problem. In Section 3.2, we formulate the spectral problem along with its variational formulation in the suitable Hilbert spaces and show the discreteness of the spectrum. Moreover, we prove the required smoothness results for the eigenfunctions which, due to the junctions conditions (cf. (15) and (16)), can be weaker than those obtained for the eigenfunctions of the original problem in Section 2.3, but which somehow justify the junction conditions (cf. Corollaries 2 and 3).

3.1. Spaces ${\mathcal{L}}_h^2\left(G\right)$, ${\tilde{\mathcal{L}}}_h^2\left(G\right)$, ${\mathcal{H}}_h^1\left(G\right)$, and ${\tilde{\mathcal{H}}}_h^1\left(G\right)$

Henceforth, we will use the notation $u_j$ for the restriction of a function $u\in L^2\left(G\right)$ to ${\mathcal{T}}_j$, $1\le j\le N$.

Let ${\mathcal{L}}_h^2\left(G\right)$ be the close subspace of $L^2\left(G\right)$ of functions u such that $u_j(y,z)={\left[u_j\right]}_j\left(z\right)$ for $x=(y,z)\in {\mathcal{T}}_{j,h_j}$, $1\le j\le N$. Introduce also the space

$${\tilde{\mathcal{L}}}_h^2\left(G\right)=\{u\in {\mathcal{L}}_h^2\left(G\right)|\underset{G}{\int }udx=0\}.$$

As noted in the introduction, the approximate spectral problem deals with the domain G with truncated tube-like parts. The following lemma pertains to the topological structure of the set $G_h=G\setminus \underset{j=1}{\stackrel{N}{\cup }}\overline{{\mathcal{T}}_{j,h_j}}$.

Lemma 4.

The set $G_h$ has the following structure:

$$\begin{array}{c}\hfill G_h=\underset{k=1}{\stackrel{m}{\cup }}{G}_{h,k},\end{array}$$

(7)

where $m\le N+1$ and $G_{h,k}$ are domains with Lipschitz boundaries such that $\overline{G_{h,k}}\cap \overline{G_{h,\ell }}=\varnothing$ for $k\ne \ell$.

Proof. 

Let $\tilde{G}$ be an open subset of the set G. We will write that points $x,y\in \tilde{G}$ are connected in $\tilde{G}$ if there exists a continuous curve $L\subset \tilde{G}$ starting at x and ending at y or $y=x$.

We set $G_h^0=G$ and $G_h^j=G_h^{j-1}\setminus \overline{{\mathcal{T}}_{j,h_j}}$, $1\le j\le N$. Note that the boundary $\partial G_h^j$ of each of the sets $G_h^j$ satisfies the Lipschitz condition because $\partial G_h^j=\partial G_h^{j-1}\setminus (\partial {\Omega }_j\times (0,{\ell }_j))\cup ({\Omega }_j\times \{0,{\ell }_j\})$.

Let us show that for all $1\le j\le N$ the following equality holds:

$$\begin{array}{c}\hfill G_h^j=\underset{k=1}{\stackrel{m_j}{\cup }}{G}_{h,k},\end{array}$$

(8)

where $m_j\le j+1$ and $G_{h,k}$ are domains with Lipschitz boundaries such that $\overline{G_{h,k}}\cap {\overline{G}}_{h,\ell }=\varnothing$ for $k\ne \ell$.

Introduce the sets ${\mathcal{T}}_{j,h_j}^{\ell }$ and ${\mathcal{T}}_{j,h_j}^r$, $1\le j\le N$, which in the local coordinate system associated with ${\mathcal{T}}_j$ have the following form:

$${\mathcal{T}}_{j,h_j}^{\ell }={\Omega }_j\times (0,h_j),{\mathcal{T}}_{j,h_j}^r={\Omega }_j\times ({\ell }_j-h_j,{\ell }_j).$$

Step 1. Let us show that equality (8) is valid for $j=1$. If the set $G_h^1=G\setminus \overline{{\mathcal{T}}_{1,h_1}}$ is connected, then (8) is true with $m_1=1$ and $G_{h,1}=G_h^1$.

Assume that $G_h^1$ is not connected. Fix points $x^{\ell }\in {\mathcal{T}}_{1,h_1}^{\ell }$ and $x^r\in {\mathcal{T}}_{1,h_1}^r$. Denote by $G_h^{1,\ell }$ the set of points $y\in G_h^1$, connected with $x^{\ell }$ in $G_h^1$. Similarly, denote by $G_h^{1,r}$ the set of points $y\in G_h^1$, connected with $x^r$ in $G_h^1$. It is clear that $G_h^{1,\ell }$ and $G_h^{1,r}$ are disjoint domains and ${\mathcal{T}}_{1,h_1}^{\ell }\subset G_h^{1,\ell }$, ${\mathcal{T}}_{1,h_1}^r\subset G_h^{1,r}$.

Let us prove that

$$\begin{array}{c}\hfill G_h^1=G_h^{1,\ell }\cup G_h^{1,r}\end{array}$$

(9)

Assume that $y\in G_h^1\setminus G_h^{1,\ell }$. Since set G is connected, there is a curve $L=\{x=\zeta (t),$$0\le t\le 1\}\subset G$, where the function $\zeta :[0,1]\to G$ is continuous, and $\zeta \left(0\right)=y$, $\zeta \left(1\right)=x^{\ell }$.

Since $y\notin G_h^{1,\ell }$, then $L\cap \overline{{\mathcal{T}}_{1,h_1}}\ne \varnothing$, and there exists $t_{*}\in (0,1)$ such that $\zeta \left(t_{*}\right)\in {\overline{\mathcal{T}}}_{1,h_1}$ and $\zeta \left(t\right)\in G_h^1$ for $0\le t<t_{*}$. Moreover, there exists $t\in (0,t_{*})$ such that $\zeta \left(t\right)\in {\mathcal{T}}_{1,h_1}^{\ell }\cup {\mathcal{T}}_{1,h_1}^r$. If $\zeta \left(t\right)\in {\mathcal{T}}_{1,h_1}^{\ell }$ then points y and $x^{\ell }$ are connected in $G_h^1$, which contradicts the assumption $y\notin G_h^{1,\ell }$. Hence, $\zeta \left(t\right)\in {\mathcal{T}}_{1,h_1}^r$, and so $y\in G_h^{1,r}$. Thus, equality (9) is true, i.e., (8) is true with $m_1=2$ and $G_{h,1}=G_h^{1,\ell }$, $G_{h,2}=G_h^{1,r}$.

Note also that $\partial G_h^1=\partial G_h^{1,\ell }\cup \partial G_h^{1,r}$, where $\partial G_h^{1,\ell }\cap \partial G_h^{1,r}=\varnothing$. Since $\partial G_h^1$ satisfies the Lipschitz condition, then $\partial G_h^{1,\ell }$ and $\partial G_h^{1,r}$ satisfy the Lipschitz condition also.

Step n. Assume that (8) is true for $j=n-1$. Since the set ${\mathcal{T}}_{j,h_j}\subset G_h^{j-1}$ is connected, then ${\mathcal{T}}_{j,h_j}\subset G_{h,k}$ for some $1\le k\le m_{n-1}$. If ${\tilde{G}}_{h,k}=G_{h,k}\setminus \overline{{\mathcal{T}}_{j,h_j}}$ is connected then

$$\begin{array}{c}\hfill G_h^n=\underset{1\le i\le m_{n-1},i\ne k}{\cup }{G}_{h,i}\cup {\tilde{G}}_{h,k}.\end{array}$$

If the set ${\tilde{G}}_{h,k}$ is not connected, then repeating the argument at step 1 (with $G_h^1$ replaced by ${\tilde{G}}_{h,k}$ and ${\mathcal{T}}_{1,h_1}^{\ell }$, ${\mathcal{T}}_{1,h_1}^r$ replaced by ${\mathcal{T}}_{j,h_j}^{\ell }$, ${\mathcal{T}}_{j,h_j}^r$), we see that ${\tilde{G}}_{h,k}$ is represented as the union of two non-intersecting regions ${\tilde{G}}_{h,k}^{\ell }$ and ${\tilde{G}}_{h,k}^r$. In this case

$$\begin{array}{c}\hfill G_h^n=\underset{1\le i\le m_{n-1},i\ne k}{\cup }{G}_{h,i}\cup {\tilde{G}}_{h,k}^{\ell }\cup {\tilde{G}}_{h,k}^r.\end{array}$$

Thus, representation (8) also holds for $j=n$.

Having done N steps, we claim equality (7).

The lemma is proved. □

Introduce the space ${\mathcal{H}}_h^1\left(G\right)$ consisting of functions $u\in {\mathcal{L}}_h^2\left(G\right)$ such that $u\in H^1\left(G_{h,k}\right)$ for all $1\le k\le m$, $u_j\in H^1\left({\mathcal{T}}_{j,h}\right)$ for all $1\le j\le N$, and the following junction conditions

$${\left[u\right]}_j{|}_{z=h_j-0}=u_j{|}_{z=h_j+0},u_j{{|}_{z={\ell }_j-h_j-0}={\left[u\right]}_j|}_{z={\ell }_j-h_j+0}$$

(10)

hold.

Naturally, $u_j={\left[u_j\right]}_j\in H^1(0,{\ell }_j)$ and $\nabla u_j\left(x\right)=(0,0,D_z{u}_j\left(z\right))$ for $x=(y,z)\in {\mathcal{T}}_{j,h}$, $1\le j\le N$.

Note that ${\mathcal{H}}_h^1\left(G\right)$ is a Hilbert space with the inner product and the norm

$$\begin{array}{cc}\hfill & {(u,v)}_{{\mathcal{H}}_h^1\left(G\right)}={(K\nabla u,\nabla v)}_{L^2\left(G\right)}+{(u,v)}_{L^2\left(G\right)},\hfill \\ \hfill & {\parallel u\parallel }_{{\mathcal{H}}_h^1\left(G\right)}={\left(\parallel \sqrt{K}{\nabla u\parallel }_{L^2\left(G\right)}^2+{\parallel u\parallel }_{L^2\left(G\right)}^2\right)}^{1/2}.\hfill \end{array}$$

Lemma 5.

The embedding of ${\mathcal{H}}_h^1\left(G\right)$ into ${\mathcal{L}}_h^2\left(G\right)$ is compact.

Proof. 

Assume that ${\left\{u_n\right\}}_{n=1}^{\infty }\subset {\mathcal{H}}_h^1\left(G\right)$ is a sequence that converges weakly in ${\mathcal{H}}_h^1\left(G\right)$. As a result, it weakly converges in $H^1\left(G_{h,k}\right)$ for all $1\le k\le m$ and in $H^1\left({\mathcal{T}}_{j,h}\right)$ for all $1\le j\le N$. So, ${\left\{u_n\right\}}_{n=1}^{\infty }$ converges strongly in $L^2\left(G_{h,k}\right)$ for all $1\le k\le m$ and in $L^2\left({\mathcal{T}}_{j,h}\right)$ for all $1\le j\le N$. Thus, it converges strongly in ${\mathcal{L}}_h^2\left(G\right)$.

The lemma is proved. □

Lemma 6.

Suppose that $u\in {\mathcal{H}}_h^1\left(G\right)$ and $\nabla u=0$. Then, $u=\mathrm{const}$.

Proof. 

It follows from $\nabla u=0$ that $u\left(x\right)=c_{h,k}=\mathrm{const}$ in $G_{h,k}$ for all $1\le k\le m$ and $u\left(x\right)=c_j=\mathrm{const}$ in ${\mathcal{T}}_{j,h_j}$ for all $1\le j\le m$. Hence, due to the connectedness of G and condition (10), it implies that $u\left(x\right)=\mathrm{const}$ in G.

The lemma is proved. □

The following analogue of the Poincaré inequality holds in ${\mathcal{H}}_h^1\left(G\right)$ (see analogous inequalities in some spaces of discontinuous functions in [19,24]).

Lemma 7.

There exists a constant $C>0$ such that

$$\begin{array}{c}\hfill {\parallel u\parallel }_{{\mathcal{H}}_h^1\left(G\right)}^2\le C\left[{(K\nabla u,\nabla u)}_{L^2\left(G\right)}+{(\underset{G}{\int }udx)}^2\right]\forall u\in {\mathcal{H}}_h^1\left(G\right).\end{array}$$

(11)

Proof. 

Assume the opposite. Then, there is a sequence ${\left\{u_n\right\}}_{n=1}^{\infty }\subset {\mathcal{H}}_h^1\left(G\right)$ such that

$$\begin{array}{c}\hfill {\parallel u_n\parallel }_{{\mathcal{H}}_h^1\left(G\right)}^2>n\left[{(K\nabla u_n,\nabla u_n)}_{L^2\left(G\right)}+{(\underset{G}{\int }{u}_{n}dx)}^2\right]\forall n\ge 1.\end{array}$$

Putting $v_n={\displaystyle \frac{1}{\parallel u_n{\parallel }_{{\mathcal{H}}_h^1\left(G\right)}}}{u}_n$, we have ${\parallel v_n\parallel }_{{\mathcal{H}}_h^1\left(G\right)}=1$ and

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{n}}>{(K\nabla v_n,\nabla v_n)}_{L^2\left(G\right)}+{(\underset{G}{\int }{v}_{n}dx)}^2\forall n\ge 1.\end{array}$$

(12)

Using the compactness of embedding of ${\mathcal{H}}_h^1\left(G\right)$ into ${\mathcal{L}}_h^2\left(G\right)$, we choose a subsequence ${\left\{v_{n_k}\right\}}_{k=1}^{\infty }$ such that $v_{n_k}\to v$ weakly in ${\mathcal{H}}_h^1\left(G\right)$ and strongly in ${\mathcal{L}}_h^2\left(G\right)$.

It follows from (12) that $\parallel \nabla v_{n_k}{\parallel }_{L^2\left(G\right)}\to 0$ and $\underset{G}{\int }{v}_{n_k}dx\to 0$. So, $\nabla v=0$, $v=c=\mathrm{const}$ and $\underset{G}{\int }vdx=c·\mathrm{meas}G=0$. Thus, $v=0$ and $v_{n_k}\to 0$ in ${\mathcal{H}}_h^1\left(G\right)$ which contradicts the equality ${\parallel v_{n_k}\parallel }_{{\mathcal{H}}_h^1\left(G\right)}=1$.

The lemma is proved. □

Let us introduce in ${\mathcal{H}}_h^1\left(G\right)$ the closed subspace

$${\tilde{\mathcal{H}}}_h^1\left(G\right)={\mathcal{H}}_h^1\left(G\right)\cap {\tilde{\mathcal{L}}}_h^2\left(G\right)=\{u\in {\mathcal{H}}_h^1\left(G\right)|\underset{G}{\int }udx=0\}.$$

It follows from (11) that ${\tilde{\mathcal{H}}}_h^1\left(G\right)$ is a Hilbert space with the following inner product and the norm:

$${(u,v)}_{{\tilde{\mathcal{H}}}_h^1\left(G\right)}={(K\nabla u,\nabla v)}_{L^2\left(G\right)}{,\parallel u\parallel }_{{\tilde{\mathcal{H}}}_h^1\left(G\right)}={\parallel \sqrt{K}\nabla u\parallel }_{L^2\left(G\right)}.$$

Moreover, it follows from Lemma 5 that the embedding of ${\tilde{\mathcal{H}}}_h^1\left(G\right)$ into ${\tilde{\mathcal{L}}}_h^2\left(G\right)$ is compact.

3.2. Setting of the Approximate Spectral Problem

Consider the following spectral problem: find a couple $(\widehat{\lambda },U)\in ℝ\times {\mathcal{H}}_h^1\left(G\right)$, such that ${\parallel U\parallel }_{L^2\left(G\right)}=1$, satisfying the equation

$$-\mathrm{div}(K\Delta U)=\widehat{\lambda }U,x\in G_h\cup {\mathcal{T}}_h,$$

(13)

the boundary condition

$$K\nabla U·n=0,x\in \partial G$$

(14)

and the following junction conditions:

$$\begin{array}{cc}\hfill & \left[U_j\right]{|}_{z=h_j-0}=U_j{|}_{z=h_j+0},U_j{{|}_{z={\ell }_j-h_j-0}=\left[U_j\right]|}_{z={\ell }_j-h_j+0},1\le j\le N,\hfill \end{array}$$

(15)

$$\begin{array}{cc}\hfill & D_z{U}_j{|}_{z=h_j-0}=D_z{U}_j{|}_{z=h_j+0},D_z{U}_j{{|}_{z={\ell }_j-h_j-0}=D_z{U}_j|}_{z={\ell }_j-h_j+0},1\le j\le N.\hfill \end{array}$$

(16)

where $U_j$ is the restriction of U on ${\mathcal{T}}_j$.

The weak formulation of this problem is given by the following identity:

$$\begin{array}{cc}\hfill & {(K\nabla U,\nabla \phi )}_{L^2\left(G\right)}=\widehat{\lambda }{(U,\phi )}_{L^2\left(G\right)}\forall \phi \in {\mathcal{H}}_h^1\left(G\right).\hfill \end{array}$$

(17)

Note that conditions (15) are satisfied since $U\in {\mathcal{H}}_h^1\left(G\right)$. We will show also that (17) implies a validity of junction conditions (16) (see Corollaries 2 and 3).

It is clear that each eigenvalue $\widehat{\lambda }$ is non-negative, the minimal eigenvalue is ${\widehat{\lambda }}_0=0$, and the corresponding eigenfunction ${\widehat{U}}_0$ is a constant.

Since the embedding of ${\tilde{\mathcal{H}}}_h^1\left(G\right)$ into ${\tilde{\mathcal{L}}}_h^2\left(G\right)$ is dense and compact, we are in a classical abstract framework of bilinear, continuous, coercive forms on a couple of Hilbert spaces ${\tilde{\mathcal{H}}}_h^1\left(G\right)$ and ${\tilde{\mathcal{L}}}_h^2\left(G\right)$, cf., for instance, Theorem 5.5 of Chapter I in [25]. So, the following theorem is true.

Theorem 1.

There exists a system ${\left\{{\widehat{\phi }}_k\right\}}_{k=1}^{\infty }$ of eigenfunctions to problem (17) corresponding to the eigenvalues $0<{\widehat{\lambda }}_1\le {\widehat{\lambda }}_2\le \cdots \le {\widehat{\lambda }}_k\le \dots$ forming an orthonormal base in ${\tilde{\mathcal{H}}}_h^1\left(G\right)$ and an orthogonal base in ${\tilde{\mathcal{L}}}_h^2\left(G\right)$.

Corollary 1.

There is a system ${\left\{{\widehat{\phi }}_k\right\}}_{k=0}^{\infty }$ of eigenfunctions to problem (17) corresponding to the eigenvalues $0={\widehat{\lambda }}_0<{\widehat{\lambda }}_1\le {\widehat{\lambda }}_2\le \dots {\widehat{\lambda }}_k\le \dots$, forming an orthonormal base in ${\mathcal{L}}_h^2\left(G\right)$ and an orthogonal base in ${\mathcal{H}}_h^1\left(G\right)$.

3.3. Some Properties of Eigenfunctions

Henceforth, in this section, $\widehat{\lambda }$ is an eigenvalue and $U\in {\mathcal{H}}_h^1\left(G\right)$ is corresponding eigenfunction such that ${\parallel U\parallel }_{L^2\left(G\right)}=1$. Moreover, ${\overline{\lambda }}_j=\widehat{\lambda }/K_j$, $1\le j\le N$.

Lemma 8.

The function U satisfies the identity

$$\begin{array}{cc}\hfill & {(K\nabla U,\nabla \phi )}_{L^2\left(G\right)}=\widehat{\lambda }{(U,\phi )}_{L^2\left(G\right)}\forall \phi \in H^1\left(G\right).\hfill \end{array}$$

(18)

Proof. 

Let $\phi \in H^1\left(G\right)$. Note that the function

$$\begin{array}{c}\hfill \tilde{\phi }\left(x\right)=\left\{\begin{array}{c}\phi \left(x\right),x\in G_h,\hfill \\ {\phi ]}_j\left(x\right),x\in {\mathcal{T}}_{j,h_j},1\le j\le N\hfill \end{array}\right.\end{array}$$

belongs to ${\mathcal{H}}_h^1\left(G\right)$ and

$$\begin{array}{cc}\hfill & {(K\nabla U,\nabla \phi )}_{L^2\left(G\right)}-\widehat{\lambda }{(U,\phi )}_{L^2\left(G\right)}={(K\nabla U,\nabla \tilde{\phi })}_{L^2\left(G\right)}-\widehat{\lambda }{(U,\tilde{\phi })}_{L^2\left(G\right)}\hfill \end{array}$$

So, (18) holds.

The lemma is proved. □

Remember that $U_j$ is the restriction of U on ${\mathcal{T}}_j$, $1\le j\le m$. We will consider $U_j$ as a function on $(0,{\ell }_j)$ with values in $H^1\left({\Omega }_j\right)$. We put $W_j\left(z\right)=D_z{U}_j\left(z\right)$ for $z\in (0,h_j)\cup ({\ell }_j-h_j,{\ell }_j)$ and $W_j\left(z\right)=D_z{\left[U_j\right]}_j\left(z\right)$ for $z\in (h_j,{\ell }_j-h_j)$. It is clear that $U_j\in L^2(0,{\ell }_j;H^1\left({\Omega }_j\right))$ and $W_j\in L^2(0,{\ell }_j;L^2\left({\Omega }_j\right))$. Moreover, ${\left[U_j\right]}_j\in L^2(0,{\ell }_j)$ and $D_z{\left[U_j\right]}_j\in L^2(0,{\ell }_j)$.

Lemma 9.

The function $W_j$ has a derivative $D_z{W}_j\in L^2(0,{\ell }_j;{\left(H^1\left({\Omega }_j\right)\right)}^{\prime })$.

Proof. 

It follows from (17) that

$$\begin{array}{cc}\hfill & \underset{0}{\overset{{\ell }_j}{\int }}{(W_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}{\eta }^{\prime }\left(z\right)dz+\underset{0}{\overset{{\ell }_j}{\int }}\left[{({\nabla }_y{U}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\overline{\lambda }}_j{(U_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\right]\eta \left(z\right)dz=0\hfill \end{array}$$

(19)

for all $\phi \in H^1\left({\Omega }_j\right)$ and $\eta \in C_0^{\infty }(0,{\ell }_j)$. Remember that ${\overline{\lambda }}_j=\widehat{\lambda }/K_j$

Therefore, the function $W_j$ has a derivative $D_z{W}_j:(0,{\ell }_j)\to {\left(H^1\left({\Omega }_j\right)\right)}^{\prime }$ such that

$$\begin{array}{c}\hfill {⟨D_z{W}_j\left(z\right),\phi ⟩}_{{\left(H^1\left({\Omega }_j\right)\right)}^{\prime }\times H^1\left({\Omega }_j\right)}={({\nabla }_y{U}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\overline{\lambda }}_j{(U_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\forall \phi \in H^1\left({\Omega }_j\right).\end{array}$$

for almost all $z\in (0,{\ell }_j)$.

It follows from the estimate

$$\begin{array}{c}\left|{⟨D_z{W}_j\left(z\right),\phi ⟩}_{{\left(H^1\left({\Omega }_j\right)\right)}^{\prime }\times H^1\left({\Omega }_j\right)}\right|\le \parallel {\nabla }_y{U}_j{\left(z\right)\parallel }_{L^2\left({\Omega }_j\right)}\parallel {\nabla }_y{\phi \parallel }_{L^2\left({\Omega }_j\right)}+{\overline{\lambda }}_j\parallel U_j{\left(z\right)\parallel }_{L^2\left({\Omega }_j\right)}{\parallel \phi \parallel }_{L^2\left({\Omega }_j\right)}\\ \le max\{1,{\overline{\lambda }}_j\}\parallel U_j\left(z\right){\parallel }_{H^1\left({\Omega }_j\right)}{\parallel \phi \parallel }_{H^1\left({\Omega }_j\right)}\end{array}$$

that

$$\begin{array}{c}\hfill {\parallel D_z{W}_j\parallel }_{L^2(0,{\ell }_j;{\left(H^1\left({\Omega }_j\right)\right)}^{\prime })}\le max\{1,{\overline{\lambda }}_j\}{\parallel U_j\parallel }_{L^2(0,{\ell }_j;H^1\left({\Omega }_j\right))}.\end{array}$$

Thus, $D_z{W}_j\in L^2(0,{\ell }_j;{\left(H^1\left({\Omega }_j\right)\right)}^{\prime })$.

The lemma is proved. □

Corollary 2.

$W_j\in C([0,{\ell }_j];{\left(H^1\left({\Omega }_j\right)\right)}^{\prime })$, and so, the eigenfunction U satisfies the junction conditions (16) in ${\left(H^1\left({\Omega }_j\right)\right)}^{\prime }$.

Lemma 10.

The function ${\left[U_j\right]}_j$ belongs to the space $H^2(0,{\ell }_j)$, and

$$-D_z^2{\left[U_j\right]}_j={\overline{\lambda }}_j{\left[U_j\right]}_j.$$

Proof. 

The substitution of $\phi =K_j^{-1}\eta$ with $\eta \in C_0^{\infty }(0,{\ell }_j)$ into (17) gives

$$\begin{array}{c}\hfill \underset{0}{\overset{{\ell }_j}{\int }}{D}_z{\left[U_j\right]}_j\left(z\right){\eta }^{\prime }\left(z\right)dz={\overline{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{\left[U_j\right]}_j\left(z\right)\eta \left(z\right)dz=0\forall \eta \in C_0^{\infty }(0,{\ell }_j).\end{array}$$

(20)

So, $D_z^2{\left[U_j\right]}_j=-{\overline{\lambda }}_j{\left[U\right]}_j$.

The lemma is proved. □

We put $R_j=U_j-{\left[U_j\right]}_j$, $1\le j\le N$ and will consider $R_j$ as a function on $(0,{\ell }_j)$ with values in $H^1\left({\Omega }_j\right)$. It is clear that $R_j=0$, $D_z{R}_j=0$ on $(h_j,{\ell }_j-h_j)$ and $D_z{R}_j=W_j-D_z{\left[U_j\right]}_j$ on $(0,h_j)\cup ({\ell }_j-h_j,{\ell }_j)$.

It follows from Lemmas 9 and 10 that $D_z{R}_j\in C([0,{\ell }_j];{\left(H^1\left({\Omega }_j\right)\right)}^{\prime })$. So $D_z{R}_j\left(h_j\right)=0$ and $D_z{R}_j({\ell }_j-h_j)=0$ in ${\left(H^1\left({\Omega }_j\right)\right)}^{\prime }$.

Lemma 11.

For each $0<a<h_j$ the function $R_j$ has derivatives $D_z^2{R}_j\in L^2(a,h_j;L^2\left({\Omega }_j\right))$ and $D_z^2{R}_j\in L^2({\ell }_j-h_j,{\ell }_j-a;L^2\left({\Omega }_j\right))$.

Proof. 

We put ${\tilde{R}}_j\left(z\right)=R_j\left(z\right)$ for $0<z\le h_j$ and ${\tilde{R}}_j\left(z\right)=R_j(2h_j-z)$ for $h_j\le z<2h_j$. Note that ${\tilde{R}}_j\in C(0,2h_j;L^2\left({\Omega }_j\right))$ and $D_z{\tilde{R}}_j\in L^2(0,2h_j;L^2\left({\Omega }_j\right))$.

Let $\phi \in L^2(0,2h_j;H^1\left({\Omega }_j\right))$ be an arbitrary function identically equal to zero in some neighborhoods of the points $z=0$ and $z=2h_j$ and such that $D_z\phi \in L^2(0,2h_j;L^2\left({\Omega }_j\right))$.

As $R_j\left(z\right)=0$ for $z\in (h_j,{\ell }_j-h_j)$ then (17) and (20) imply that

$$\begin{array}{cc}\hfill & \underset{0}{\overset{h_j}{\int }}{(D_z{R}_j,D_z\phi )}_{L^2\left({\Omega }_j\right)}dz+\underset{0}{\overset{h_j}{\int }}{({\nabla }_y{R}_j,{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}dz={\overline{\lambda }}_j\underset{0}{\overset{h_j}{\int }}{(R_j,\phi )}_{L^2\left({\Omega }_j\right)}dz.\hfill \end{array}$$

Moreover,

$$\begin{array}{cc}\hfill & -\underset{h_j}{\overset{2h_j}{\int }}{(D_z{\tilde{R}}_j,D_z\tilde{\phi })}_{L^2\left({\Omega }_j\right)}dz+\underset{h_j}{\overset{2h_j}{\int }}{({\nabla }_y{\tilde{R}}_j,{\nabla }_y\tilde{\phi })}_{L^2\left({\Omega }_j\right)}dz={\overline{\lambda }}_j\underset{h_j}{\overset{2h_j}{\int }}{({\tilde{R}}_j,\tilde{\phi })}_{L^2\left({\Omega }_j\right)}dz,\hfill \end{array}$$

where $\tilde{\phi }\left(z\right)=\phi (2h_j-z)$. Consequently,

$$\begin{array}{cc}\hfill & \underset{0}{\overset{2h_j}{\int }}{(D_z{\tilde{R}}_j,D_z\phi )}_{L^2\left({\Omega }_j\right)}dz+\underset{0}{\overset{2h_j}{\int }}{({\nabla }_y{\tilde{R}}_j,\nabla \phi )}_{L^2\left({\Omega }_j\right)}dz={\overline{\lambda }}_j\underset{0}{\overset{2h_j}{\int }}{({\tilde{R}}_j,\phi )}_{L^2\left({\Omega }_j\right)}dz.\hfill \end{array}$$

(21)

Assume that $0<a<h_j<b<2h_j$ and $\zeta \in C_0^{\infty }(0,2h_j)$ is a cutoff function, equal to one on $(a,b)$. Substituting into (21) the test function $\phi =-{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{\tilde{R}}_j\right)$, where h is small enough, we have

$$\begin{array}{cc}\hfill & -\underset{0}{\overset{2h_j}{\int }}{(D_z{\tilde{R}}_j,{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{D}_z{\tilde{R}}_j\right))}_{L^2\left({\Omega }_j\right)}dz-\underset{0}{\overset{2h_j}{\int }}{(D_z{\tilde{R}}_j,{\overline{\partial }}^h\left(2\zeta {\zeta }^{\prime }{\partial }^h{\tilde{R}}_j\right))}_{L^2\left({\Omega }_j\right)}dz\hfill \\ \hfill & -\underset{0}{\overset{2h_j}{\int }}{({\nabla }_y{\tilde{R}}_j,{\nabla }_y{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{\tilde{R}}_j\right))}_{L^2\left({\Omega }_j\right)}dz=-{\overline{\lambda }}_j\underset{0}{\overset{2h_j}{\int }}{({\tilde{R}}_j,{\overline{\partial }}^h\left({\zeta }^2{\partial }^h{\tilde{R}}_j\right))}_{L^2\left({\Omega }_j\right)}dz.\hfill \end{array}$$

Using the difference analogue of the formula for integration by parts, we obtain

$$\begin{array}{cc}\hfill & \underset{0}{\overset{2h_j}{\int }}\parallel \zeta {\partial }^h{D}_z{\tilde{R}}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+\underset{0}{\overset{2h_j}{\int }}{\parallel \zeta {\partial }^h{\nabla }_y{\tilde{R}}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz\hfill \\ \hfill & =-\underset{0}{\overset{2h_j}{\int }}{(\zeta {\partial }^h{D}_z{\tilde{R}}_j,2{\zeta }^{\prime }{\partial }^h{\tilde{R}}_j)}_{L^2\left({\Omega }_j\right)}dz+{\tilde{\lambda }}_j\underset{0}{\overset{2h_j}{\int }}{\parallel \zeta {\partial }^h{\tilde{R}}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\underset{0}{\overset{2h_j}{\int }}\parallel \zeta {\partial }^h{D}_z{\tilde{R}}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+{\displaystyle \frac{1}{2}}\underset{0}{\overset{2h_j}{\int }}\parallel 2{\zeta }^{\prime }{\partial }^h{\tilde{R}}_j{\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz+{\overline{\lambda }}_j\underset{0}{\overset{2h_j}{\int }}{\parallel \zeta {\partial }^h{\tilde{R}}_j\parallel }_{L^2\left({\Omega }_j\right)}^{2}dz.\hfill \end{array}$$

From here, we have the following estimate uniform in h

$$\begin{array}{cc}\hfill & {\parallel {\partial }^h{D}_z{\tilde{R}}_j\parallel }_{L^2(a,b;L^2\left({\Omega }_j\right))}^2\le c_{\zeta }\parallel D_z{\tilde{R}}_j{\parallel }_{L^2(0,2h_j;L^2\left({\Omega }_j\right))}^2=2c_{\zeta }{\parallel D_z{R}_j\parallel }_{L^2(0,h_j;L^2\left({\Omega }_j\right))}^2.\hfill \end{array}$$

This implies the existence of the derivative $D_z^2{\tilde{R}}_j\in L^2(a,b;L^2\left({\Omega }_j\right))$. Thus, $D_z^2{R}_j\in L^2(a,h_j;L^2\left({\Omega }_j\right))$.

The existence of the derivative $D_z^2{R}_j\in L^2({\ell }_j-h_j,{\ell }_j-a;L^2\left({\Omega }_j\right))$ is proved in a similar way.

The lemma is proved. □

Corollary 3.

The following properties hold:

(1) 

$D_z{R}_j\in C((0,h_j];L^2\left({\Omega }_j\right))$ and $D_z{R}_j(h_j-0)=0$ in $L^2\left({\Omega }_j\right)$.

(2) 

$D_z{R}_j\in C([{\ell }_j-h_j,{\ell }_j);L^2\left({\Omega }_j\right))$ and $D_z{R}_j({\ell }_j-h_j+0)=0$ in $L^2\left({\Omega }_j\right)$.

(3) 

The eigenfunction U satisfies the junction conditions in (16) in $L^2\left({\Omega }_j\right)$.

Proof. 

To prove properties (1) and (2), it should be taken into account that, due to Corollary 2, $D_z{R}_j(h_j-0)=0$, $D_z{R}_j({\ell }_j-h_j+0)=0$ in ${\left(H^1\left({\Omega }_j\right)\right)}^{\prime }$, and $H^1\left({\Omega }_j\right)$ is dense in $L^2\left({\Omega }_j\right)$.

Property (3) holds because $D_z{U}_j=D_z{\left[U_j\right]}_j+D_z{R}_j$, where $D_z{R}_j\left(z\right)=0$ for $z\in (h_j,{\ell }_j-h_j)$ and $D_z{\left[U_j\right]}_j\in C[0,{\ell }_j]$.

The corollary is proved. □

Lemma 12.

For almost all $z\in (0,h_j)\cup ({\ell }_j-h_j,{\ell }_j)$, the following identities hold

$$\begin{array}{cc}\hfill & {(D_z^2{U}_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}={({\nabla }_y{U}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\overline{\lambda }}_j{(U_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\forall \phi \in H^1\left({\Omega }_j\right).\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & {(D_z^2{R}_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}={({\nabla }_y{R}_j\left(z\right),{\nabla }_y\phi )}_{L^2\left({\Omega }_j\right)}-{\overline{\lambda }}_j{(R_j\left(z\right),\phi )}_{L^2\left({\Omega }_j\right)}\forall \phi \in H^1\left({\Omega }_j\right).\hfill \end{array}$$

(23)

The proof is similar to the proof of Lemma 3.

4. Formulation of the Main Results

Let ${\left\{{\lambda }_k\right\}}_{k=0}^{\infty }$ be the set of eigenvalues to problem (2), and ${\left\{{\widehat{\lambda }}_k\right\}}_{k=0}^{\infty }$ be the set of eigenvalues to problem (17).

Let $\epsilon$ be the desired accuracy, $0<\epsilon <1$ and ${\epsilon }_1={\displaystyle \frac{\epsilon }{1+\epsilon }}$.

Let $0={\mu }_{0,j}<{\mu }_{1,j}\le {\mu }_{2,j}\le \cdots$ be the set of eigenvalues to the Neumann problem

$$\begin{array}{cc}\hfill & -{\Delta }_y\psi =\mu \psi ,y\in {\Omega }_j,\hfill \\ \hfill & {\nabla }_y\psi ·\nu =0,y\in \partial {\Omega }_j,\hfill \end{array}$$

and ${\left\{{\psi }_{k,j}\right\}}_{k=0}^{\infty }$ be the set of corresponding eigenfunctions, forming an orthogonal base in $H^1\left({\Omega }_j\right)$ and an orthonormal base in $L^2\left({\Omega }_j\right)$.

Theorem 2.

Let λ be the eigenvalue of problem (2) and ${\tilde{\lambda }}_j=\lambda /K_j<{\mu }_{1,j}$ for all $1\le j\le N$. Let the numbers $h_j$ be such that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{\nu }_{1,j}}}max\left\{ln{\displaystyle \frac{2}{{\epsilon }_1^2}},ln{\displaystyle \frac{{96}^2{K}_j^2{\nu }_{1,j}^4}{{\epsilon }_1^2}},ln{\displaystyle \frac{{24}^2{K}_j\lambda {\nu }_{1,j}^2}{{\epsilon }_1^2}}\right\}\le h_j<{\ell }_j/2,1\le j\le N,\end{array}$$

(24)

where ${\nu }_{1,j}=\sqrt{{\mu }_{1,j}-{\tilde{\lambda }}_j}$. Then, the following estimate holds:

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\widehat{\lambda }}_k-\lambda |\le \epsilon .\end{array}$$

(25)

Theore 3.

Let $\widehat{\lambda }$ be the eigenvalue of problem (17) and ${\overline{\lambda }}_j=\widehat{\lambda }/K_j<{\mu }_{1,j}$ for all $1\le j\le N$. Let the numbers $h_j$ be such that

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{{\widehat{\nu }}_{1,j}}}max\left\{ln{\displaystyle \frac{4}{{\epsilon }_1^2}},ln{\displaystyle \frac{{96}^2{K}_j^2{\widehat{\nu }}_{1,j}^4}{{\epsilon }_1^2}},ln{\displaystyle \frac{{12}^2{K}_j\widehat{\lambda }{\widehat{\nu }}_{1,j}^2}{{\epsilon }_1^2}}\right\}\le h_j<{\ell }_j/2,1\le j\le N,\end{array}$$

(26)

where ${\widehat{\nu }}_{1,j}=\sqrt{{\mu }_{1,j}-{\overline{\lambda }}_j}$. Then, the following estimate holds:

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\lambda }_k-\widehat{\lambda }|\le \epsilon .\end{array}$$

(27)

Consider a special case where, for each of the tubes ${\mathcal{T}}_j$, its cross section is a disk with a radius of ${ϵ}_j$. Remember that, in this case, ${\mu }_{1,j}=p^2/{ϵ}_j^2$, where $p=1.8142\dots$ is the smallest positive root of the derivative $J_1^{\prime }$ of the Bessel function $J_1$. So, Theorems 2 and 3 can be reformulated as follows.

Theorem 4.

Let λ be the eigenvalue of problem (2) and $\lambda <K_j{p}^2/{ϵ}_j^2$ for all $1\le j\le N$. Let the numbers $h_j$ be such that

$$\begin{array}{c}\hfill {\displaystyle \frac{\sqrt{K_j}}{\sqrt{{\rho }_j}}}max\left\{ln{\displaystyle \frac{2}{{\epsilon }_1^2}},ln{\displaystyle \frac{{96}^2{\rho }_j^2}{{\epsilon }_1^2}},ln{\displaystyle \frac{{24}^2\lambda {\rho }_j}{{\epsilon }_1^2}}\right\}\le h_j<{\ell }_j/2,1\le j\le N,\end{array}$$

where ${\rho }_j=K_j{p}^2/{ϵ}_j^2-\lambda$. Then, the following estimate holds:

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\widehat{\lambda }}_k-\lambda |\le \epsilon .\end{array}$$

Theorem 5.

Let $\widehat{\lambda }$ be the eigenvalue of problem (17) and $\widehat{\lambda }<K_j{p}^2/{ϵ}_j^2$ for all $1\le j\le N$. Let the numbers $h_j$ be such that

$$\begin{array}{c}\hfill {\displaystyle \frac{\sqrt{K_j}}{{\widehat{\rho }}_j}}max\left\{ln{\displaystyle \frac{4}{{\epsilon }_1^2}},ln{\displaystyle \frac{{96}^2{\widehat{\rho }}_j^2}{{\epsilon }_1^2}},ln{\displaystyle \frac{{12}^2\widehat{\lambda }{\widehat{\rho }}_j}{{\epsilon }_1^2}}\right\}\le h_j<{\ell }_j/2,1\le j\le N,\end{array}$$

where ${\widehat{\rho }}_j=\sqrt{K_j{p}^2/{ϵ}_j^2-\widehat{\lambda }}$. Then, the following estimate holds:

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\lambda }_k-\widehat{\lambda }|\le \epsilon .\end{array}$$

5. Proof of Theorem 2

Let $u\in H^1\left(G\right)$ be the eigenfunction of problem (2) corresponding to the eigenvalue $\lambda$ such that ${\parallel u\parallel }_{L^2\left(G\right)}=1$. Recall that according to the hypothesis of the theorem, ${\tilde{\lambda }}_j=\lambda /K_j<{\mu }_{1,j}$ for all $1\le j\le N$ and ${\nu }_{k,j}=\sqrt{{\mu }_{k,j}-{\tilde{\lambda }}_j}$ for $k\ge 1$.

Henceforth, we will use the following notations:

$$\begin{array}{cc}\hfill & {\mathcal{T}}_{j,h_j/2,h_j}={\Omega }_j\times [(h_j/2,h_j)\cup ({\ell }_j-h_j,{\ell }_j-h_j/2)],\hfill \\ \hfill & {\mathcal{T}}_{j,h_j/4,h_j}={\Omega }_j\times [(h_j/4,h_j)\cup ({\ell }_j-h_j,{\ell }_j-h_j/4)].\hfill \end{array}$$

We introduce the function $\widehat{u}\in {\mathcal{H}}_h^1\left(G\right)$ by the formula

$$\begin{array}{cc}\hfill & \widehat{u}\left(x\right)=\left\{\begin{array}{c}u\left(x\right),x\in G_{h/2}=G\setminus \underset{j=1}{\stackrel{N}{\cup }}\overline{{\mathcal{T}}_{j,h_j/2}},\hfill \\ u_j(y,z)-{\zeta }_j\left(z\right)r_j(y,z),x=(y,z)\in {\mathcal{T}}_{j,h_j/2},1\le j\le N,\hfill \end{array}\right.\hfill \end{array}$$

where $r_j(y,z)=u_j(y,z)-{\left[u_j\right]}_j\left(z\right),(y,z)\in {\mathcal{T}}_{j,h_j/2},1\le j\le N$ and

$$\begin{array}{c}\hfill {\zeta }_j\left(z\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{h_j}}(z-3h_j/4)-{\displaystyle \frac{16}{h_j^3}}{(z-3h_j/4)}^3,z\in [h_j/2,h_j],\hfill \\ {\displaystyle \frac{1}{2}}-{\displaystyle \frac{3}{h_j}}(z-{\ell }_j+3h_j/4)+{\displaystyle \frac{16}{h_j^3}}{(z-{\ell }_j+3h_j/4)}^3,z\in [{\ell }_j-h_j,{\ell }_j-h_j/2],\hfill \\ 1,z\in [h_j,{\ell }_j-h_j].\hfill \end{array}\right.\end{array}$$

(28)

Note that ${\zeta }_j\in H^2(h_j/2,{\ell }_j-h_j/2)$. Moreover,

$$\begin{array}{cc}\hfill & {\zeta }_j(h_j/2)=0,{\zeta }_j^{\prime }(h_j/2)=0,{\zeta }_j({\ell }_j-h_j/2)=0,{\zeta }_j^{\prime }({\ell }_j-h_j/2)=0,\hfill \\ \hfill & 0\le {\zeta }_j\le 1,|{\zeta }_j^{\prime }|\le {\displaystyle \frac{3}{h_j}},\left|{\zeta }_j^{\prime \prime }\right|\le {\displaystyle \frac{24}{h_j^2}}.\hfill \end{array}$$

(29)

Let us expand $r_j\left(z\right)$ and $D_z{r}_j\left(z\right)$ for almost all $z\in (0,{\ell }_j)$ into a Fourier series converging in $L^2\left({\Omega }_j\right)$

$$\begin{array}{cc}\hfill & r_j\left(z\right)=\sum _{k=1}^{\infty }{c}_{k,j}\left(z\right){\psi }_{k,j},D_z{r}_j(z,y)=\sum _{k=1}^{\infty }{c}_{k,j}^{\prime }\left(z\right){\psi }_{k,j},\hfill \end{array}$$

where $c_{k,j}\left(z\right)={(r_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}$, $c_{k,j}^{\prime }\left(z\right)={(D_z{r}_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}$. It is clear that $c_{0,j}\left(z\right)=0$, $c_{0,j}^{\prime }\left(z\right)=0$.

It follows from (2) and (20) that

$$\begin{array}{cc}\hfill & \underset{0}{\overset{{\ell }_j}{\int }}{(D_z{r}_j\left(z\right),{\eta }^{\prime }\left(z\right){\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz+\underset{0}{\overset{{\ell }_j}{\int }}{({\nabla }_y{r}_j\left(z\right),\eta \left(z\right){\nabla }_y{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz\hfill \\ \hfill & ={\tilde{\lambda }}_j\underset{0}{\overset{{\ell }_j}{\int }}{(r_j\left(z\right),\eta \left(z\right){\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz\forall \eta \in C_0^{\infty }(0,{\ell }_j).\hfill \end{array}$$

Taking into account that

$${({\nabla }_y{r}_j\left(z\right),{\nabla }_y{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}={\mu }_{k,j}{(r_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)},$$

we have

$$\begin{array}{cc}\hfill & \underset{0}{\overset{{\ell }_j}{\int }}{c}_{k,j}^{\prime }\left(z\right){\eta }^{\prime }\left(z\right)dz+\underset{0}{\overset{{\ell }_j}{\int }}{\nu }_{k,j}^2{c}_{k,j}\left(z\right)\eta \left(z\right)dz=0\forall \eta \in C_0^{\infty }(0,{\ell }_j).\hfill \end{array}$$

Thus, for all $k\ge 1$,

$$\begin{array}{c}\hfill c_{k,j}^{\prime \prime }\left(z\right)={\nu }_{k,j}^2{c}_{k,j}\left(z\right),z\in (0,{\ell }_j).\end{array}$$

(30)

Lemma 13.

Let $h_j<{\ell }_j/2$, $1\le j\le N$. Then, for all $k\ge 1$, the following estimates hold

$$\begin{array}{cc}\hfill & {\parallel c_{k,j}\parallel }_{L^2(h_j/2,h_j)}^2+\parallel c_{k,j}{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2,\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & {\parallel c_{k,j}\parallel }_{L^2(h_j/2,{\ell }_j-h_j/2)}^2\le 2e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2,\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & {\parallel c_{k,j}^{\prime }\parallel }_{L^2(h_j/2,h_j)}^2+\parallel c_{k,j}^{\prime }{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}^{\prime }\parallel }_{L^2(0,{\ell }_j)}^2,.\hfill \end{array}$$

(33)

Proof. 

Since the coefficients $c_{k,j}$ satisfy Equation (30), then for all $0<\xi <h_j/2<z<h_j$ the following formulas hold:

$$\begin{array}{cc}\hfill & c_{k,j}\left(z\right)\hfill \\ \hfill & ={\displaystyle \frac{[e^{-{\nu }_{k,j}(z-\xi )}-e^{-{\nu }_{k,j}(2{\ell }_j-3\xi -z)}]c_{k,j}\left(\xi \right)+[e^{-{\nu }_{k,j}({\ell }_j-\xi -z)}-e^{-{\nu }_{k,j}({\ell }_j-3\xi +z)}]c_{k,j}({\ell }_j-\xi )}{1-e^{-2{\nu }_{k,j}({\ell }_j-2\xi )}}}\hfill \\ \hfill & ={\displaystyle \frac{e^{-{\nu }_{k,j}(z-\xi )}[1-e^{-2{\nu }_{k,j}({\ell }_j-\xi -z)}]c_{k,j}\left(\xi \right)+e^{-{\nu }_{k,j}({\ell }_j-\xi -z)}[1-e^{-2{\nu }_{k,j}(z-\xi )}]c_{k,j}({\ell }_j-\xi )}{1-e^{-2{\nu }_{k,j}({\ell }_j-2\xi )}}},\hfill \\ \hfill & c_{k,j}({\ell }_j-z)\hfill \\ \hfill & ={\displaystyle \frac{[e^{-{\nu }_{k,j}({\ell }_j-z-\xi )}-e^{-{\nu }_{k,j}({\ell }_j-3\xi +z)}]c_{k,j}\left(\xi \right)+[e^{-{\nu }_{k,j}(z-\xi )}-e^{-{\nu }_{k,j}(2{\ell }_j-3\xi -z)}]c_{k,j}({\ell }_j-\xi )}{1-e^{-2{\nu }_{k,j}({\ell }_j-2\xi )}}}\hfill \\ \hfill & ={\displaystyle \frac{e^{-{\nu }_{k,j}({\ell }_j-z-\xi )}[1-e^{-2{\nu }_{k,j}(z-\xi )}]c_{k,j}\left(\xi \right)+e^{-{\nu }_{k,j}(z-\xi )}[1-e^{-2{\nu }_{k,j}({\ell }_j-z-\xi )}]c_{k,j}({\ell }_j-\xi )}{1-e^{-2{\nu }_{k,j}({\ell }_j-2\xi )}}},\hfill \end{array}$$

From here,

$$\begin{array}{cc}\hfill & c_{k,j}^2\left(z\right)\le 2\left[e^{-2{\nu }_{k,j}(z-\xi )}{c}_{k,j}^2\left(\xi \right)+e^{-2{\nu }_{k,j}({\ell }_j-\xi -z)}{c}_k^2({\ell }_j-\xi )\right],\hfill \\ \hfill & c_{k,j}^2({\ell }_j-z)\le 2\left[e^{-2{\nu }_{k,j}({\ell }_j-\xi -z)}{c}_{k,j}^2\left(\xi \right)+e^{-2{\nu }_{k,j}(z-\xi )}{c}_{k,j}^3({\ell }_j-\xi )\right].\hfill \end{array}$$

Summing up these estimates, we have

$$\begin{array}{cc}\hfill c_{k,j}^2\left(z\right)+c_{k,j}^2({\ell }_j-z)& \le 2\left[e^{-2{\nu }_{k,j}(z-\xi )}+e^{-2{\nu }_{k,j}({\ell }_j-\xi -z)}\left]\right[c_{k,j}^2\left(\xi \right)+c_{k,j}^2({\ell }_j-\xi )\right]\hfill \\ \hfill & \le 2\left[e^{-2{\nu }_{1,j}(z-\xi )}+e^{-2{\nu }_{1,j}({\ell }_j-\xi -z)}\left]\right[c_{k,j}^2\left(\xi \right)+c_{k,j}^2({\ell }_j-\xi )\right].\hfill \end{array}$$

From here,

$$\begin{array}{cc}\hfill & e^{-2{\nu }_{1,j}\xi }\left[c_{k,j}^2\left(z\right)+c_{k,j}^2(\ell -z)\right]\le 2\left[e^{-2{\nu }_{1,j}z}+e^{-2{\nu }_{1,j}({\ell }_j-z)}\right]\left[c_{k,j}^2\left(\xi \right)+c_{k,j}^2(\ell -\xi )\right].\hfill \end{array}$$

(34)

Integrating this inequality over $\xi \in (0,h_j/2)$ and over $z\in (h_j/2,h_j)$, we arrive at the estimate

$$\begin{array}{cc}\hfill & {\parallel c_{k,j}\parallel }_{L^2(h_j/2,h_j)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\hfill \\ \hfill & \le 2\left[e^{-{\nu }_{1,j}{h}_j}+e^{-{\nu }_{1,j}(2{\ell }_j-3h_j)}\right]\left[\parallel c_{k,j}{\parallel }_{L^2(0,h_j/2)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2\right]\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2.\hfill \end{array}$$

Integrating inequality (34) over $\xi \in (0,h_j/2)$ and over $z\in (h_j/2,{\ell }_j/2)$, we have

$$\begin{array}{cc}\hfill & (1-e^{-{\nu }_{1,j}{h}_j}){\parallel c_{k,j}\parallel }_{L^2(h_j/2,{\ell }_j-h_j/2)}^2\hfill \\ \hfill & \le 2(e^{-{\nu }_{1,j}{h}_j}-e^{-2{\nu }_{1,j}({\ell }_j-h_j/2)})\left[\parallel c_{k,j}{\parallel }_{L^2(0,h_j/2)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2\right]\hfill \\ \hfill & \le 2e^{-{\nu }_{1,j}{h}_j}\left[\parallel c_{k,j}{\parallel }_{L^2(0,h_j/2)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2\right].\hfill \end{array}$$

(35)

From here,

$$\begin{array}{cc}\hfill \parallel c_{k,j}{\parallel }_{L^2(h_j/2,{\ell }_j-h_j/2)}^2& \le 2e^{-{\nu }_{1,j}{h}_j}\left[\parallel c_{k,j}{\parallel }_{L^2(0,h_j/2)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2\right]\hfill \\ \hfill & +e^{-{\nu }_{1,j}{h}_j}\parallel c_{k,j}{\parallel }_{L^2(h_j/2,{\ell }_j-h_j/2)}^2\le 2e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2.\hfill \end{array}$$

It follows from (30) that the coefficients $c_{k,j}^{\prime }$ satisfy the equation ${\left(c_{k,j}^{\prime }\right)}^{\prime \prime }={\nu }_{k,j}^2{c}_{k,j}^{\prime }$. Therefore, estimate (33) is proved in exactly the same way as estimate (31).

The lemma is proved. □

Corollary 4.

Let $h_j<{\ell }_j/2$, $1\le j\le N$. Then, the following estimates hold:

$$\begin{array}{cc}\hfill & {\parallel r_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel u_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2,\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill & {\parallel D_z{r}_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel D_z{u}_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2.\hfill \end{array}$$

(37)

Proof. 

Indeed, it follows from (31) that

$$\begin{array}{cc}\hfill & {\parallel r_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2=\sum _{k=1}^{\infty }\left[\parallel c_{k,j}{\parallel }_{L^2(h_j/2,h_j)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\right]\hfill \\ \hfill & \le 4e^{-{\nu }_{1,j}{h}_j}\sum _{k=1}^{\infty }\parallel c_{k,j}{\parallel }_{L^2(0,{\ell }_j)}^2=4e^{-{\nu }_{1,j}{h}_j}\parallel u_j-{\left[u_j\right]}_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\le 4e^{-{\nu }_{1,j}{h}_j}{\parallel u_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2.\hfill \end{array}$$

Similarly, estimate (37) follows from (33).

The corollary is proved. □

Lemma 14.

Let ${\displaystyle \frac{1}{{\nu }_{1,j}}}ln{\displaystyle \frac{2}{{\epsilon }_1^2}}\le h_j\le {\ell }_j/2$, $1\le j\le N$. Then, the following estimate holds:

$$\begin{array}{cc}\hfill & {\parallel \widehat{u}-u\parallel }_{L^2\left(G\right)}\le {\epsilon }_1.\hfill \end{array}$$

(38)

Proof. 

Using estimate (32) and noting that $2e^{-{\nu }_{1,j}{h}_j}\le {\epsilon }_1^2,$ we have

$$\begin{array}{cc}\hfill & {\parallel \widehat{u}-u\parallel }_{L^2\left(G\right)}^2\le \sum _{j=1}^N{\parallel r_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2}\right)}^2\hfill \\ \hfill & =\sum _{j=1}^N\sum _{k=1}^{\infty }\parallel c_{k,j}{\parallel }_{L^2(h_j/2,{\ell }_j-h_j/2)}^2\le \sum _{j=1}^N\sum _{k=1}^{\infty }2e^{-{\nu }_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2\hfill \\ \hfill & \le \sum _{j=1}^N\parallel u_j-{\left[u_j\right]}_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2{\epsilon }_1^2\le \sum _{j=1}^N\parallel u_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2{\epsilon }_1^2\le {\parallel u\parallel }_{L^2\left(G\right)}^2{\epsilon }_1^2={\epsilon }_1^2.\hfill \end{array}$$

The lemma is proved. □

Lemma 15.

The function $\widehat{u}$ satisfies the identity

$$\begin{array}{c}\hfill {(K\nabla \widehat{u},\nabla \phi )}_{L^2\left(G\right)}-\lambda {(\widehat{u},\phi )}_{L^2\left(G\right)}={(\Psi ,\phi )}_{L^2\left(G\right)}\forall \phi \in {\mathcal{H}}_h^1\left(G\right),\end{array}$$

(39)

where

$$\Psi \left(x\right)=\left\{\begin{array}{c}0,x\in G\setminus \underset{j=1}{\stackrel{N}{\cup }}{\mathcal{T}}_{j,h_j/2,h_j},\hfill \\ {\Psi }_j\left(x\right)=K_j[{\zeta }_j^{\prime \prime }\left(z\right)r_j(y,z)+2{\zeta }_j^{\prime }\left(z\right)D_z{r}_j(y,z)],x=(y,z)\in {\mathcal{T}}_{j,h_j/2,h_j},1\le j\le N.\hfill \end{array}\right.$$

Proof. 

Let $\phi \in {\mathcal{H}}_h^1\left(G\right)$, and let functions ${\eta }_j\in C^{\infty }[0,{\ell }_j]$, where $1\le j\le N$, be such that ${\eta }_j\left(z\right)=0$ for $z\in [0,h_j/4]\cup [{\ell }_j-h_j/4,{\ell }_j]$, and ${\eta }_j\left(z\right)=1$ for $z\in [h_j/2,{\ell }_j-h_j/2]$.

Introduce functions

$$\begin{array}{cc}\hfill & {\psi }_0\left(x\right)=\left\{\begin{array}{c}\phi \left(x\right),x\in G\setminus \underset{j=1}{\stackrel{N}{\cup }}{\mathcal{T}}_{j,h_j/4},\hfill \\ (1-{\eta }_j\left(z\right))\phi (y,z),x=(y,z)\in {\mathcal{T}}_{j,h_j/4},1\le j\le N.\hfill \end{array}\right.\hfill \\ \hfill & {\psi }_j\left(x\right)=\left\{\begin{array}{c}0,x\in G\setminus {\mathcal{T}}_{j,h_j/4},\hfill \\ {\eta }_j\left(z\right)\phi (y,z),x=(y,z)\in {\mathcal{T}}_{j,h_j/4}.\hfill \end{array}\right.\hfill \end{array}$$

As $\varphi =\sum _{j=0}^N{\psi }_j$ then

$$\begin{array}{cc}\hfill {(K\nabla \widehat{u},\nabla \phi )}_{L^2\left(G\right)}-\lambda {(\widehat{u},\phi )}_{L^2\left(G\right)}& ={(K\nabla \widehat{u},\nabla {\psi }_0)}_{L^2\left(G\right)}-\lambda {(\widehat{u},{\psi }_0)}_{L^2\left(G\right)}\hfill \\ \hfill & +\sum _{j=1}^N{K}_j\left[{(\nabla \widehat{u},\nabla {\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}-{\tilde{\lambda }}_j{(\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}\right]\hfill \end{array}$$

Note that ${\psi }_0\in H^1\left(G\right)$, ${\psi }_0=0$ in $G\setminus G_{h/2}$ and $\widehat{u}=u$ in $G_{h/2}$. So,

$$\begin{array}{cc}\hfill & {(K\nabla \widehat{u},\nabla {\psi }_0)}_{L^2\left(G\right)}-\lambda {(\widehat{u},{\psi }_0)}_{L^2\left(G\right)}={(K\nabla u,\nabla {\psi }_0)}_{L^2\left(G\right)}-\lambda {(u,{\psi }_0)}_{L^2\left(G\right)}=0.\hfill \end{array}$$

Note also that

$$\begin{array}{cc}\hfill & {(\nabla \widehat{u},\nabla {\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}-{\tilde{\lambda }}_j{(\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}\hfill \\ \hfill & ={(D_z\widehat{u},D_z{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}+{({\nabla }_y\widehat{u},{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}-{\tilde{\lambda }}_j{(\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}\hfill \\ \hfill & +{(D_z{\left[u_j\right]}_j,D_z{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j}\right)}-{\tilde{\lambda }}_j{({\left[u_j\right]}_j,{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j}\right)}.\hfill \end{array}$$

Integrating by parts over z and taking into account that ${\psi }_j(h_j/4)=0$, ${\psi }_j({\ell }_j-h_j/4)=0$, and $-D_z^2{\left[u_j\right]}_j={\tilde{\lambda }}_j{\left[u_j\right]}_j$ in ${\mathcal{T}}_{j,h_j}$, we have

$$\begin{array}{cc}\hfill & {(\nabla \widehat{u},\nabla {\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}-{\tilde{\lambda }}_j{(\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}\hfill \\ \hfill & =-{(D_z^2\widehat{u}+{\tilde{\lambda }}_j\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}+{({\nabla }_y\widehat{u},{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}\hfill \\ \hfill & +{(D_z\widehat{u},{\psi }_j)}_{L^2\left({\Omega }_j\right)}{{|}_{z=h_j-0}+{(D_z{\left[u_j\right]}_j,{\psi }_j)}_{L^2\left({\Omega }_j\right)}|}_{z={\ell }_j-h_j-0}\hfill \\ \hfill & -{(D_z{\left[u_j\right]}_j,{\psi }_j)}_{L^2\left({\Omega }_j\right)}{{|}_{z=h_j+0}-{(D_z\widehat{u},{\psi }_j)}_{L^2\left({\Omega }_j\right)}|}_{z={\ell }_j-h_j-0}.\hfill \end{array}$$

Noting that $D_z\widehat{u}(h_j-0)=D_z{\left[u_j\right]}_j(h_j-0)$ does not depend on y and ${\left[{\psi }_j\right]}_j(h_j-0)={\psi }_j(h_j+0)$, we have

$$\begin{array}{cc}\hfill & {(D_z\widehat{u},{\psi }_j)}_{L^2\left({\Omega }_j\right)}{{|}_{z=h_j-0}-{(D_z{\left[u_j\right]}_j,{\psi }_j)}_{L^2\left({\Omega }_j\right)}|}_{z=h_j+0}\hfill \\ \hfill & ={(D_z{\left[u_j\right]}_j,{\left[{\psi }_j\right]}_j)}_{L^2\left({\Omega }_j\right)}{{|}_{z=h_j-0}-{(D_z{\left[u_j\right]}_j,{\psi }_j)}_{L^2\left({\Omega }_j\right)}|}_{z=h_j+0}=0.\hfill \end{array}$$

Likewise

$${(D_z\widehat{u},{\psi }_j)}_{L^2\left({\Omega }_j\right)}{{|}_{z={\ell }_l-h_j-0}-{(D_z{\left[u\right]}_j,{\psi }_j)}_{L^2\left({\Omega }_j\right)}|}_{{\ell }_j-h_j+0}=0.$$

Thus, using (5) we obtain

$$\begin{array}{cc}\hfill & {(\nabla \widehat{u},\nabla {\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}-{\tilde{\lambda }}_j{(\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4}\right)}\hfill \\ \hfill & =-{(D_z^2\widehat{u}+{\tilde{\lambda }}_j\widehat{u},{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}+{({\nabla }_y\widehat{u},{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}\hfill \\ \hfill & =-{(D_z^{2}u+{\tilde{\lambda }}_{j}u,{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}+{({\nabla }_{y}u,{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/4,h_j}\right)}\hfill \\ \hfill & +{(D_z^2\left({\zeta }_j{r}_j\right)+{\tilde{\lambda }}_j\left({\zeta }_j{r}_j\right),{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}-{({\nabla }_y\left({\zeta }_j{r}_j\right),{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}\hfill \\ \hfill & ={({\zeta }_j^{\prime \prime }{r}_j+2{\zeta }_j^{\prime }{D}_z{r}_j,{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}+\hfill \\ \hfill & +{(D_z^2{r}_j+{\tilde{\lambda }}_j{r}_j,{\zeta }_j{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}-{({\nabla }_y{r}_j,{\zeta }_j{\nabla }_y{\psi }_j)}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}.\hfill \end{array}$$

By virtue of identity (6), the last two terms vanish. Noting that ${\psi }_j=\phi$ on ${\mathcal{T}}_{j,h_j/2,h_j}$, we obtain

$$\begin{array}{c}\hfill {(K\nabla \widehat{u},\nabla \phi )}_{L^2\left(G\right)}-\lambda {(\widehat{u},\phi )}_{L^2\left(G\right)}=\sum _{j=1}^N{K}_j{({\zeta }_j^{\prime \prime }{r}_j+2{\zeta }_j^{\prime }{D}_z{r}_j,{\psi }_j)}_{L^2\left(\right)}={(\Psi ,\phi )}_{L^2\left(G\right)}.\end{array}$$

The lemma is proved. □

Lemma 16.

Assume that the numbers $h_j$ satisfy the conditions in (24). Then, the following inequalities hold:

$$\begin{array}{c}\hfill 2K_j^2{\displaystyle \frac{{48}^2}{h_j^4}}{e}^{-{\nu }_{1,j}{h}_j}\le {\displaystyle \frac{1}{2}}{\epsilon }_1^2,2K_j{\displaystyle \frac{{12}^2}{h_j^2}}{e}^{-{\nu }_{1,j}{h}_j}\le {\displaystyle \frac{1}{2\lambda }}{\epsilon }_1^2.\end{array}$$

(40)

Proof. 

Let us set $t={\nu }_{1,j}{h}_j$ and transform the inequalities in (40) to the form

$$\begin{array}{cc}\hfill & t+4lnt\ge {\alpha }_j,t+2lnt\ge {\beta }_j\hfill \end{array}$$

(41)

where ${\alpha }_j=ln{\displaystyle \frac{{96}^2{K}_j^2{\nu }_{1,j}^4}{{\epsilon }_1^2}}$, ${\beta }_j=ln{\displaystyle \frac{{24}^2{K}_j\lambda {\nu }_{1,j}^2}{{\epsilon }_1^2}}$.

The function $g\left(t\right)=t+4lnt$ is increasing. If ${\alpha }_j\le 1$, then $g\left(t\right)\ge g\left(1\right)=1\ge {\alpha }_j$ for $t\ge 1$. If ${\alpha }_j\ge 1$, then $g\left(t\right)\ge g\left({\alpha }_j\right)\ge {\alpha }_j$ for $t\ge {\alpha }_j$. Thus, the first inequality in (41) holds for $h_j\ge {\displaystyle \frac{1}{{\nu }_{1,j}}}max\left\{{\alpha }_j,1\right\}$.

Similarly, the second inequality (41) holds for $h_j\ge {\displaystyle \frac{1}{{\nu }_{1,j}}}max\left\{{\beta }_j,1\right\}$.

It should be noted that (24) implies that $h_j{\nu }_{1,j}\ge max\left\{{\alpha }_j,{\beta }_j,1\right\}$. Furthermore, note that ${\nu }_{1,j}{h}_j>ln(2/{\epsilon }_1^2)$ already implies that ${\nu }_{1,j}{h}_j>1$.

The lemma is proved. □

Lemma 17.

Assume that the numbers $h_j$ satisfy the conditions in (24). Then, the following estimate is true:

$$\begin{array}{c}\hfill {\parallel \Psi \parallel }_{L^2\left(G\right)}\le {\epsilon }_1.\end{array}$$

(42)

Proof. 

Using estimates (29), (36), (37), we have

$$\begin{array}{cc}\hfill & {\parallel \Psi \parallel }_{L^2\left(G\right)}^2=\sum _{j=1}^N{\parallel {\Psi }_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le 2\sum _{j=1}^N{K}_j^2\left[\parallel {\zeta }_j^{\prime \prime }{r}_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2+{\parallel 2{\zeta }_j^{\prime }{D}_z{r}_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\right]\hfill \\ \hfill & \le 2\sum _{j=1}^N{K}_j^2\left[{\displaystyle \frac{{24}^2}{h_j^4}}\parallel r_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2+{\displaystyle \frac{6^2}{h_j^2}}{\parallel D_z{r}_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\right]\hfill \\ \hfill & \le 2\sum _{j=1}^N{K}_j^2{e}^{-{\nu }_{1,j}{h}_j}\left[{\displaystyle \frac{{48}^2}{h_j^4}}\parallel u_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2+{\displaystyle \frac{{12}^2}{h_j^2}}{\parallel \nabla u_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\right].\hfill \end{array}$$

Due to the estimates in (40), we have

$$\begin{array}{cc}\hfill & {\parallel \Psi \parallel }_{L^2\left(G\right)}^2\le \sum _{j=1}^N\left[\parallel u_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2+{\displaystyle \frac{1}{\lambda }}{\parallel \sqrt{K_j}\nabla u_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\right]{\displaystyle \frac{{\epsilon }_1^2}{2}}\le \left[{\parallel u\parallel }_{L^2\left(G\right)}+{\displaystyle \frac{1}{\lambda }}{\parallel \sqrt{K}\nabla u\parallel }_{L^2\left(G\right)}^2\right]{\displaystyle \frac{{\epsilon }_1^2}{2}}.\hfill \end{array}$$

Taking into account that ${\displaystyle \frac{1}{\lambda }}\parallel \sqrt{K}{\nabla u\parallel }_{L^2\left(G\right)}^2={\parallel u\parallel }_{L^2\left(G\right)}^2=1$, we arrive at inequality (42).

The lemma is proved. □

Let us now show that estimate (25) holds under the conditions of Theorem 2.

Let ${\left\{{\widehat{\lambda }}_k\right\}}_{k=0}^{\infty }$ be the set of eigenvalues to problem (17) and ${\left\{{\widehat{\phi }}_k\right\}}_{k=0}^{\infty }$ be the set of corresponding eigenfunctions, forming an orthonormal base in ${\mathcal{L}}_h^2\left(G\right)$.

From (39) and equality

$${(K\nabla \widehat{u},\nabla {\widehat{\phi }}_k)}_{L^2\left(G\right)}={\widehat{\lambda }}_k{(\widehat{u},{\widehat{\phi }}_k)}_{L^2\left(G\right)},k\ge 0$$

it follows that

$$\begin{array}{cc}\hfill & ({\widehat{\lambda }}_k-\lambda ){(\widehat{u},{\widehat{\phi }}_k)}_{L^2\left(G\right)}={(\Psi ,{\widehat{\phi }}_k)}_{L^2\left(G\right)},k\ge 0.\hfill \end{array}$$

So,

$$\underset{k\ge 0}{min}{({\widehat{\lambda }}_k-\lambda )}^2\parallel \widehat{u}{\parallel }_{L^2\left(G\right)}^2\le \sum _{k=0}^{\infty }{({\widehat{\lambda }}_k-\lambda )}^2{(\widehat{u},{\widehat{\phi }}_k)}_{L^2\left(G\right)}^2\le \sum _{k=0}^{\infty }{(\Psi ,{\widehat{\phi }}_k)}_{L^2\left(G\right)}^2\le {\parallel \Psi \parallel }_{L^2\left(G\right)}^2.$$

Hence,

$$\underset{k\ge 0}{min}|{\widehat{\lambda }}_k-{\lambda \left|\right(\parallel u\parallel }_{L^2\left(G\right)}-\parallel \widehat{u}{-u\parallel }_{L^2\left(G\right)})\le \underset{k\ge 0}{min}|{\widehat{\lambda }}_k-\lambda |\parallel \widehat{u}{\parallel }_{L^2\left(G\right)}\le {\parallel \Psi \parallel }_{L^2\left(G\right)}.$$

As ${\parallel u\parallel }_{L^2\left(G\right)}=1$, $\parallel \widehat{u}{-u\parallel }_{L^2\left(G\right)}\le {\epsilon }_1$ and ${\parallel \Psi \parallel }_{L^2\left(G\right)}\le {\epsilon }_1$, we obtain the estimate

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\widehat{\lambda }}_k-\lambda |(1-{\epsilon }_1)\le {\epsilon }_1,\end{array}$$

which is equivalent to estimate (25).

Theorem 2 is proved.

6. Proof of Theorem 3

Let $U\in {\mathcal{H}}_h^1\left(G\right)$ be the eigenfunction of problem (17) corresponding to the eigenvalue $\widehat{\lambda }$ and such that ${\parallel U\parallel }_{L^2\left(G\right)}=1$. Recall that according to the hypothesis of the theorem, ${\overline{\lambda }}_j=\widehat{\lambda }/K_j<{\mu }_{1,j}$ for all $1\le j\le N$ and ${\widehat{\nu }}_{k,j}=\sqrt{{\mu }_{k,j}-{\overline{\lambda }}_j}$ for $k\ge 1$.

We introduce the function $\widehat{U}\in H^1\left(G\right)$ by the formula

$$\begin{array}{cc}\hfill & \widehat{U}\left(x\right)=\left\{\begin{array}{c}U\left(x\right),x\in G\setminus \underset{j=1}{\stackrel{N}{\cup }}{\mathcal{T}}_{j,h_j/2,h_j}\hfill \\ U_j(y,z)-{\zeta }_j\left(z\right)R_j(y,z),x=(y,z)\in {\mathcal{T}}_{j,h_j/2,h_j},1\le j\le N,\hfill \end{array}\right.\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill & R_j(y,z)=U_j(y,z)-{\left[U_j\right]}_j\left(z\right),(y,z)\in {\mathcal{T}}_{j,h_j/2,h_j},1\le j\le N,\hfill \end{array}$$

and the function ${\zeta }_j$ is defined by Formula (28).

Let us expand $R_j\left(z\right)$ and $D_z{R}_j\left(z\right)$ for almost all $z\in (0,h_j)\cup ({\ell }_j-h_j,{\ell }_j)$ into a Fourier series converging in $L^2\left({\Omega }_j\right)$:

$$\begin{array}{cc}\hfill & R_j\left(z\right)=\sum _{k=1}^{\infty }{c}_{k,j}\left(z\right){\psi }_{k,j},D_z{R}_j\left(z\right)=\sum _{k=1}^{\infty }{c}_{k,j}^{\prime }\left(z\right){\psi }_{k,j}.\hfill \end{array}$$

where $c_{k,j}\left(z\right)={(R_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}$, $c_{k,j}^{\prime }\left(z\right)={(D_z{R}_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}$.

It follows from (17) that

$$\begin{array}{cc}\hfill & \underset{0}{\overset{h_j}{\int }}{(D_z{R}_j\left(z\right),{\eta }^{\prime }\left(z\right){\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz+\underset{0}{\overset{h_j}{\int }}{({\nabla }_y{R}_j\left(z\right),\eta \left(z\right){\nabla }_y{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz\hfill \\ \hfill & ={\overline{\lambda }}_j\underset{0}{\overset{h_j}{\int }}{(R_j\left(z\right),\eta \left(z\right){\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}dz\forall \eta \in C_0^{\infty }(0,h_j).\hfill \end{array}$$

Taking into account that

$${({\nabla }_y{R}_j\left(z\right),{\nabla }_y{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)}={\mu }_{k,j}{(R_j\left(z\right),{\psi }_{k,j})}_{L^2\left({\Omega }_j\right)},$$

for almost all $z\in (0,h_j)\cup ({\ell }_j-h_j,{\ell }_j)$, we have

$$\begin{array}{cc}\hfill & \underset{0}{\overset{h_j}{\int }}{c}_{k,j}^{\prime }\left(z\right){\eta }^{\prime }\left(z\right)dz+\underset{0}{\overset{h_j}{\int }}{\widehat{\nu }}_{k,j}^2{c}_{k,j}\left(z\right)\eta \left(z\right)dz=0\forall \eta \in C_0^{\infty }(0,h_j)\cup C_0^{\infty }({\ell }_j-h_j,{\ell }_j).\hfill \end{array}$$

Thus,

$$\begin{array}{c}\hfill c_{k,j}^{″}\left(z\right)={\widehat{\nu }}_{k,j}^2{c}_{k,j}\left(z\right),z\in (0,h_j)\cup ({\ell }_j-h_j,{\ell }_j).\end{array}$$

(43)

Remember that $D_z{R}_j\left(h_j\right)=0$, $D_z{R}_j({\ell }_j-h_j)=0$ (cf. Corollary 3.5). Hence,

$$c_{k,j}^{\prime }\left(h_j\right)=0,c_{k,j}^{\prime }({\ell }_j-h_j)=0\forall k\ge 1.$$

Lemma 18.

Let $h_j\le {\ell }_j/2$, $1\le j\le N$. Then, for all $k\ge 1$, the following estimates hold:

$$\begin{array}{cc}\hfill & \parallel c_{k,j}{\parallel }_{L^2(h_j/2,h_j)}^2+\parallel c_{k,j}{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,{\ell }_j)}^2,\hfill \end{array}$$

(44)

$$\begin{array}{cc}\hfill & \parallel c_{k,j}^{\prime }{\parallel }_{L^2(h_j/2,h_j)}^2+\parallel c_{k,j}^{\prime }{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_{k,j}^{\prime }\parallel }_{L^2(0,{\ell }_j)}^2.\hfill \end{array}$$

(45)

Proof. 

Let $0<\xi <h_j/2<z<h_j$. As the coefficients $c_{k,j}$ satisfy equation (43) and the condition $c_{k,j}^{\prime }\left(h_j\right)=0$, we have

$$\begin{array}{cc}\hfill & c_{k,j}\left(z\right)=c_{k,j}\left(\xi \right)e^{-{\widehat{\nu }}_{k,j}(z-\xi )}{\displaystyle \frac{1+e^{-2{\widehat{\nu }}_{k,j}(h_j-z)}}{1+e^{-2{\widehat{\nu }}_{k,j}(h_j-\xi )}}}.\hfill \end{array}$$

From here,

$$\begin{array}{cc}\hfill & c_{k,j}^2\left(z\right)\le 4c_{k,j}^2\left(\xi \right)e^{-2{\widehat{\nu }}_{k,j}(z-\xi )}\le 4c_{k,j}^2\left(\xi \right)e^{-2{\widehat{\nu }}_{1,j}(z-\xi )}.\hfill \end{array}$$

So,

$$\begin{array}{cc}\hfill & c_{k,j}^2\left(z\right)e^{-2{\widehat{\nu }}_{1,j}\xi }\le 4c_{k,j}^2\left(\xi \right)e^{-2{\widehat{\nu }}_{1,j}z}.\hfill \end{array}$$

Integrating this inequality over $\xi \in (0,h_j/2)$ and $z\in (h_j/2,h_j)$, we come to an inequality

$$\begin{array}{cc}\hfill & \parallel c_{k,j}{\parallel }_{L^2(h_j/2,h_j)}^2\le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_{k,j}\parallel }_{L^2(0,h_j/2)}^2.\hfill \end{array}$$

(46)

The following inequality is proved in a similar way

$$\begin{array}{cc}\hfill & \parallel c_{k,j}{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_k\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2.\hfill \end{array}$$

(47)

By adding inequalities (46) and (47) and coarsening the result, we arrive at inequality (44).

From (43), it follows that the coefficients $c_{k,j}^{\prime }$ satisfy the equation ${\left(c_k^{\prime }\right)}^{\prime \prime }={\widehat{\nu }}_{k,j}^2{c}_k^{\prime }$. Taking into account that $c_{k,j}^{\prime }\left(h_j\right)=0$, we have for all $0<\xi <h_j/2<z<h_j$

$$\begin{array}{c}\hfill c_{k,j}^{\prime }\left(z\right)=c_{k,j}^{\prime }\left(\xi \right)e^{-{\widehat{\nu }}_{k,j}(z-\xi )}{\displaystyle \frac{1-e^{-2{\widehat{\nu }}_{k,j}(h_j-z)}}{1-e^{-2{\widehat{\nu }}_{k,j}(h_j-\xi )}}}.\end{array}$$

From here,

$$\begin{array}{c}\hfill |c_k^{\prime }{\left(z\right)|}^2\le |c_k^{\prime }{\left(\xi \right)|}^2{e}^{-2{\widehat{\nu }}_{k,j}(z-\xi )}\le {\left|c_k^{\prime }\left(\xi \right)\right|}^2{e}^{-2{\widehat{\nu }}_{1,j}(z-\xi )}.\end{array}$$

Hence,

$$\begin{array}{c}\hfill e^{-2{\widehat{\nu }}_{1,j}\xi }|c_{k,j}^{\prime }{\left(z\right)|}^2\le {|c_{k,j}^{\prime }\left(\xi \right)|}^2{e}^{-2{\widehat{\nu }}_{1,j}z}.\end{array}$$

Integrating this inequality over $\xi \in (0,h_j/2)$ and $z\in (h_j/2,h_j)$, we come to an inequality

$$\begin{array}{c}\hfill \parallel c_{k,j}^{\prime }{\parallel }_{L^2(h_j/2,h_j)}^2\le e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_{k,j}^{\prime }\parallel }_{L^2(0,h_j/2)}^2.\end{array}$$

(48)

In the same way, we prove the inequality

$$\begin{array}{c}\hfill \parallel c_{k,j}^{\prime }{\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\le e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel c_{k,j}^{\prime }\parallel }_{L^2({\ell }_j-h_j/2,{\ell }_j)}^2.\end{array}$$

(49)

By adding inequalities (48) and (49) and coarsening the result, we arrive at (45).

The lemma is proved. □

Corollary 5.

Let $h_j<{\ell }_j/2$, $1\le j\le N$. Then, the following estimates hold:

$$\begin{array}{cc}\hfill & \parallel R_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel U_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2,\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill & \parallel D_z{R}_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel D_z{U}_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2.\hfill \end{array}$$

(51)

Proof. 

Indeed,

$$\begin{array}{cc}\hfill & \parallel R_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2=\sum _{k=1}^{\infty }\left[\parallel c_{k,j}{\parallel }_{L^2(h_j/2,h_j)}^2+{\parallel c_{k,j}\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\right]\hfill \\ \hfill & \le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}\sum _{k=1}^{\infty }\parallel c_{k,j}{\parallel }_{L^2(0,{\ell }_j)}^2=4e^{-{\widehat{\nu }}_{1,j}{h}_j}\parallel U_j-{\left[U_j\right]}_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\le 4e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel U_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2,\hfill \\ \hfill & \parallel D_z{R}_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2=\sum _{k=1}^{\infty }\left[\parallel c_{k,j}^{\prime }{\parallel }_{L^2(h_j/2,h_j)}^2+{\parallel c_{k,j}^{\prime }\parallel }_{L^2({\ell }_j-h_j,{\ell }_j-h_j/2)}^2\right]\hfill \\ \hfill & \le e^{-{\widehat{\nu }}_{1,j}{h}_j}\sum _{k=1}^{\infty }\parallel c_{k,j}^{\prime }{\parallel }_{L^2(0,{\ell }_j)}^2=e^{-{\widehat{\nu }}_{1,j}{h}_j}\parallel D_z{U}_j-{\left[D_z{U}_j\right]}_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\le e^{-{\widehat{\nu }}_{1,j}{h}_j}{\parallel D_z{U}_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2.\hfill \end{array}$$

The lemma is proved. □

Lemma 19.

Let ${\displaystyle \frac{1}{{\widehat{\nu }}_{1,j}}}ln{\displaystyle \frac{4}{{\epsilon }_1^2}}\le h_j<{\ell }_j/2$ for all $1\le j\le N$. Then, the following estimate holds:

$$\begin{array}{c}\hfill \parallel \widehat{U}{-U\parallel }_{L^2\left(G\right)}\le {\epsilon }_1.\end{array}$$

(52)

Proof. 

Using estimate (50) and taking into account that $4e^{-{\widehat{\nu }}_{1,j}{h}_j}\le {\epsilon }_1^2$, we have

$$\begin{array}{cc}\hfill & \parallel \widehat{U}{-U\parallel }_{L^2\left(G\right)}^2=\sum _{j=1}^N\parallel {\zeta }_j{R}_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\le \sum _{j=1}^N{\parallel R_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\hfill \\ \hfill & \le \sum _{j=1}^{N}4e^{-{\widehat{\nu }}_{1,j}{h}_j}\parallel U_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\le \sum _{j=1}^N\parallel U_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2{\epsilon }_1^2\le {\parallel U\parallel }_{L^2\left(G\right)}^2{\epsilon }_1^2={\epsilon }_1^2.\hfill \end{array}$$

The lemma is proved. □

Lemma 20.

The following identity is true:

$$\begin{array}{c}\hfill {(K\nabla \widehat{U},\nabla \phi )}_{L^2\left(G\right)}-\widehat{\lambda }{(\widehat{U},\phi )}_{L^2\left(G\right)}={(\Phi ,\phi )}_{L^2\left(G\right)}\forall \phi \in H^1\left(G\right),\end{array}$$

(53)

where

$$\begin{array}{c}\hfill \Phi \left(x\right)=\left\{\begin{array}{c}0,x\in G\setminus \underset{j=1}{\stackrel{N}{\cup }}{\mathcal{T}}_{j,h_j/2,h_j},\hfill \\ {\Phi }_j\left(x\right)=K_j[{\zeta }_j^{\prime \prime }\left(z\right)R_j(y,z)+2{\zeta }_j^{\prime }\left(z\right)D_z{R}_j(y,z)],x=(y,z)\in {\mathcal{T}}_{j,h_j/2,h_j},1\le j\le N.\hfill \end{array}\right.\end{array}$$

Proof. 

From (18), it follows that

$$\begin{array}{cc}\hfill & {(K\nabla \widehat{U},\nabla \phi )}_{L^2\left(G\right)}-\widehat{\lambda }{(\widehat{U},\phi )}_{L^2\left(G\right)}={(K\nabla (\widehat{U}-U),\nabla \phi )}_{L^2\left(G\right)}-\widehat{\lambda }{(\widehat{U}-U,\phi )}_{L^2\left(G\right)}\hfill \\ \hfill & =-\sum _{j=1}^N{K}_j\left[{(D_z\left({\zeta }_j{R}_j\right),D_z\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}+{({\zeta }_j{\nabla }_y{R}_j,{\nabla }_y\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}-{\overline{\lambda }}_j{({\zeta }_j{R}_j,\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}\right].\hfill \end{array}$$

Using the fact that $D_z\left({\zeta }_j{R}_j\right)\left(z\right)=0$ in $L^2\left({\Omega }_j\right)$ for $z=h_j/2,h_j,{\ell }_j-h_j,{\ell }_j-h_j/2$, we perform the transformation

$$\begin{array}{cc}\hfill & {(D_z\left({\zeta }_j{R}_j\right),D_z\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}=-{(D_z^2\left({\zeta }_j{R}_j\right),\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}\hfill \\ \hfill & =-{({\zeta }_j{D}_z^2{R}_j,\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}-{({\zeta }_j^{\prime \prime }{R}_j+2{\zeta }_j^{\prime }{D}_z{R}_j,\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}.\hfill \end{array}$$

As a result, we obtain

$$\begin{array}{cc}\hfill & {(K\nabla \widehat{U},\nabla \phi )}_{L^2\left(G\right)}-\widehat{\lambda }{(\widehat{U},\phi )}_{L^2\left(G\right)}={(\Phi ,\phi )}_{L^2\left(G\right)}\hfill \\ \hfill & +\sum _{j=1}^N{K}_j[{(D_z^2{R}_j+{\overline{\lambda }}_j{R}_j,{\zeta }_j\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}-{({\nabla }_y{R}_j,{\zeta }_j{\nabla }_y\phi )}_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}].\hfill \end{array}$$

Using identity (23), we arrive at (53).

The lemma is proved. □

Lemma 21.

Assume that the numbers $h_j$ satisfy the conditions in (26). Then, the following inequalities hold:

$$\begin{array}{cc}\hfill & 2K_j^2{\displaystyle \frac{{48}^2}{h_j^4}}{e}^{-{\widehat{\nu }}_{1,j}{h}_j}\le {\displaystyle \frac{{\epsilon }_1^2}{2}},2K_j{\displaystyle \frac{6^2}{h_j^2}}{e}^{-{\widehat{\nu }}_{1,j}{h}_j}\le {\displaystyle \frac{{\epsilon }_1^2}{2\widehat{\lambda }}}.\hfill \end{array}$$

(54)

The proof of this lemma repeats the proof of lemma 16.

Lemma 22.

Assume that the numbers $h_j$ satisfy the conditions in (26). Then, the following estimate holds:

$$\begin{array}{c}\hfill {\parallel \Phi \parallel }_{L^2\left(G\right)}\le {\epsilon }_1.\end{array}$$

(55)

Proof. 

Using estimates (29), (50), and (51), we have

$$\begin{array}{cc}\hfill & {\parallel \Phi \parallel }_{L^2\left(G\right)}^2=\sum _{j=1}^N{\parallel {\Phi }_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\hfill \\ \hfill & \le 2\sum _{j=1}^N{K}_j^2\left[\parallel {\zeta }^{\prime \prime }{R}_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2+{\parallel 2{\zeta }^{\prime }{D}_z{R}_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\right]\hfill \\ \hfill & \le 2\sum _{j=1}^N{K}_j^2\left[{\displaystyle \frac{{24}^2}{h_j^4}}\parallel R_j{\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}+{\displaystyle \frac{6^2}{h_j^2}}{\parallel D_z{R}_j\parallel }_{L^2\left({\mathcal{T}}_{j,h_j/2,h_j}\right)}^2\right]\hfill \\ \hfill & \le 2\sum _{j=1}^N{K}_j^2{e}^{-{\widehat{\nu }}_{1,j}{h}_j}\left[{\displaystyle \frac{{48}^2}{h_j^4}}\parallel U_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2+{\displaystyle \frac{6^2}{h_j^2}}{\parallel D_z{U}_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\right].\hfill \end{array}$$

Using estimates (54), we have

$$\begin{array}{cc}\hfill & {\parallel \Phi \parallel }_{L^2\left(G\right)}^2\le \sum _{j=1}^N\left[\parallel U_j{\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2+{\displaystyle \frac{1}{\widehat{\lambda }}}{\parallel \sqrt{K_j}{D}_z{U}_j\parallel }_{L^2\left({\mathcal{T}}_j\right)}^2\right]{\displaystyle \frac{{\epsilon }_1^2}{2}}\le \left[{\parallel U\parallel }_{L^2\left(G\right)}^2+{\displaystyle \frac{1}{\widehat{\lambda }}}{\parallel \sqrt{K}\nabla U\parallel }_{L^2\left(G\right)}^2\right]{\displaystyle \frac{{\epsilon }_1^2}{2}}.\hfill \end{array}$$

Taking into account that ${\displaystyle \frac{1}{\widehat{\lambda }}}\parallel \sqrt{K}{\nabla U\parallel }_{L^2\left(G\right)}^2={\parallel U\parallel }_{L^2\left(G\right)}^2=1$, we arrive at inequality (55).

The lemma is proved. □

Let us now establish the validity of Theorem 3.

Remember that $\widehat{U}\in H^1\left(G\right)$. So

$$\begin{array}{c}\hfill {(K\nabla \widehat{U},\nabla {\phi }_k)}_{L^2\left(G\right)}={\lambda }_k{(\widehat{U},{\phi }_k)}_{L^2\left(G\right)},k\ge 0\end{array}$$

(56)

It follows from (53) and (56) that

$$\begin{array}{c}\hfill ({\lambda }_k-\widehat{\lambda }){(\widehat{U},{\phi }_k)}_{L^2\left(G\right)}={(\Phi ,{\phi }_k)}_{L^2\left(G\right)},k\ge 0.\end{array}$$

So,

$$\underset{k\ge 0}{min}{({\lambda }_k-\widehat{\lambda })}^2\parallel \widehat{U}{\parallel }_{L^2\left(G\right)}^2\le \sum _{k=0}^{\infty }{({\lambda }_k-\widehat{\lambda })}^2{(\widehat{U},{\phi }_k)}_{L^2\left(G\right)}^2\le \sum _{k=0}^{\infty }{(\Phi ,{\phi }_k)}_{L^2\left(G\right)}^2={\parallel \Phi \parallel }_{L^2\left(G\right)}^2.$$

Hence,

$$\underset{k\ge 0}{min}|{\lambda }_k-\widehat{\lambda }{\left|\right(\parallel U\parallel }_{L^2\left(G\right)}-\parallel \widehat{U}{-U\parallel }_{L^2\left(G\right)})\le \underset{k\ge 0}{min}|{\lambda }_k-\widehat{\lambda }|\parallel \widehat{U}{\parallel }_{L^2\left(G\right)}\le {\parallel \Phi \parallel }_{L^2\left(G\right)}.$$

As ${\parallel U\parallel }_{L^2\left(G\right)}=1$, $\parallel \widehat{U}{-U\parallel }_{L^2\left(G\right)}\le {\epsilon }_1$ and ${\parallel \Phi \parallel }_{L^2\left(G\right)}\le {\epsilon }_1$, we obtain the estimate

$$\begin{array}{c}\hfill \underset{k\ge 0}{min}|{\lambda }_k-\widehat{\lambda }|(1-{\epsilon }_1)\le {\epsilon }_1,\end{array}$$

which is equivalent to estimate (27).

Theorem 3 is proved.

7. Conclusions

A new method reducing computational resources is introduced to find a set of first eigenvalues of the Neumann diffusion operator in a three-dimensional domain containing thin tubes. The method consists of the truncation of the tubes at some small distance from the ends of the tubes. The truncated parts are replaced by one-dimensional segments, and special junction conditions are stated on the interfaces of the three-dimensional parts and one-dimensional segments: pointwise continuity of the fluxes and continuity of an average of the eigenfunction. The method is justified by the theorems pointing out at what distance we can truncate the tubes while keeping the given accuracy for the approximations of eigenvalues. In the case in which the domain consists of thin tubes only, the method significantly reduces the computational time: if the ratio of the thicknesses of tubes to their lengths is $1/m$, the time is reduced m times. This acceleration brings a significant gain in time and allows the computations of the eigenvalues in domains of complex geometry. Using the technique developed here, the approach to the eigenvalues of the original and reduced problems preserving their multiplicities (up to a prescribed accuracy), and the approaches of the corresponding eigenfunctions will be addressed in a forthcoming paper by the authors. This extension shell likely involve the construction of sets of almost orthogonal eigenfunctions of each problem (cf. [26,27] for an abstract framework and [28] for the technique in a singularly perturbed spectral problem). The analysis of the method will be extended to the comparison of the eigenfunction of the original and partially decomposed problems.