Variational-Potential-Methods--Igor-Chudinovich--Christian-Constanda

The Initial-Boundary Value Problems

All problem statements in this chapter are formal; rigorous versions will be presented after the introduction of the necessary function spaces.

This article examines initial-boundary value problems for time-dependent homogeneous equations with homogeneous initial data In Chapter 9, we demonstrate how to simplify the general case into a homogeneous framework.

By anelastic plate we understand an elastic body that occupies a region

S¯×[−h 0 /2, h 0 /2] in R 3 , where S is a domain in R 2 bounded by a simple closed curve∂S and 0< h 0 = const diamS is called thethickness.

Throughout the book we use the following notation and conventions.

Unless otherwise speciﬁed, Greek and Latin subscripts and superscripts in all formulas take the values 1,2 and 1,2,3, respectively, and summation over repeated indices is adopted.

The standard inner product inR 3 is (a, b) =a i b i

A generic point inR 2 referred to a Cartesian system of coordinates in the middle planex 3 = 0 of the plate is written asx= (x 1 , x 2 ).

Partial derivatives are denoted by∂ α =∂/∂x α and∂ t =∂/∂ t

A superscript T denotes matrix transposition A superscript ∗ denotes conjugation and transposition of a complex matrix.

The columns of a matrixM are denoted byM (i)

Matrix-valued functions and scalar functions are collectively known as functions When a space of scalar functions is denoted as Y, a matrix-valued function g is said to belong to Y if every entry of g is an element of Y.

A three-component vector q = (q 1 , q 2 , q 3 ) T may be written alternatively as q= (¯q T , q 3 ) T , where ¯q= (q 1 , q 2 ) T

S + is the ﬁnite domain enclosed by∂S, and S − =R 2 \(S + ∪∂S).

The boundary ∂S is a C 2 -curve with a uniquely deﬁned outward (with respect to S + ) normaln= (n 1 , n 2 ) T

Ifϕis a smooth function deﬁned in S + (S − ), thenϕ + (ϕ − ) denotes the limiting value (if it exists) of ϕas its argument tends to∂S from within S +

If the function ϕ is not smooth but has a trace on the boundary ∂S, it is represented as γ + ϕ(γ − ϕ) This notation is consistently applied to functions defined in G + (G −) and their limiting values or traces on the boundary Γ, ensuring clarity and avoiding ambiguity.

The operators of restriction fromR 2 (orS + ∪S − ) toS ± , or fromR 2 ×(0,∞) (orG + ∪G − ) toG ± , are denoted by π ±

Operators of extension from∂S to S ± , or from Γ toG ± , are denoted by l ± , respectively.

∆ is the Laplacian andδ ij is the Kronecker delta.

L and L −1 are, respectively, the Laplace transformation with respect to t, and its inverse The Laplace transform of a functionu(x, t) is denoted by ˆ u(x, p), where pis the transformation parameter.

Other notation will be introduced as the need arises.

Suppose that the material is homogeneous and isotropic, of densityρand Lam´e constantsλandà, which satisfy the inequalities [9] λ+à >0, à >0, ρ >0 (1.1)

The behavior of a plate as a three-dimensional elastic body, under specified initial and boundary conditions, is described by three key groups of equations These include the kinematic formulas, which express the deformation tensor components as ε ij = 1/2 (∂ i v j + ∂ j v i ); the stress-strain relations, represented by generalized Hooke’s law as t ij = λ ε kk δ ij + 2μ ε ij; and the equations of motion that govern the system's dynamics.

In addition, t i =t ij n j are the components of the stress vector on ∂S.

The model of bending of plates with transverse shear deformation that we intend to study here postulates a displacement ﬁeld of the form v α (x, x 3 , t) =x 3 u α (X), v 3 (x, x 3 , t) =u 3 (X) (1.5)

This assumption is valid only for plates whose ratio of thickness to diameter falls within a certain range (see the Preface).

The expressions and geometry of the plate indicate a method for simplifying equations (1.2)–(1.4) through a recognized technique that utilizes the averaging operators I α and J α, where α= 0,1.

N 3α =J 0 t 3α , q α =J 1 f α +I 1 t α3 , q 3 =J 0 f 3 +I 0 t 33 , h 2 =h 2 0 /12, system (1.4) yields the plate equations of motion

Also, from (1.2), (1.3), and (1.5), we obtain the plate constitutive relations

Finally, substituting (1.7) into (1.6) leads to the alternative equations of motion

A is the matrix diﬀerential operator with entries [9]

It is easily veriﬁed that, under conditions (1.1),Ais a strongly elliptic operator and satisﬁes G˚arding’s inequality [18].

The quantities N αβ and N α3 represent the average bending and twisting moments through the plate's thickness at the mid-plane (x 3 = 0), along with the transverse shear forces Meanwhile, q α and q 3 are the resultant combinations of body moments and forces, as well as the moments and forces applied on the plate's faces at x 3 = ±h 0 /2.

N 3 =à(∂ α u 3 +u α )n α , which can be written as

N i = (T u) i , where T is the matrix boundary operator with entries

The vector \( T \) represents the averaged moments and shear force acting on the lateral boundary \( \partial S \times (-\frac{h_0}{2}, \frac{h_0}{2}) \) The displacement vector \( u \) uniquely characterizes this boundary condition based on the assumption outlined in (1.5).

In Chapters 2–8, we deal almost exclusively with the homogeneous equation (1.8), that is,

To (1.9) we adjoin appropriate boundary conditions and homogeneous initial conditions The functions occurring on the right-hand side in all the boundary conditions below are prescribed.

Each problem we analyze is denoted by a symbolic name beginning with "D," signifying its dynamic nature, followed by initials that reflect the specific problem type or boundary condition For instance, the classical interior and exterior problems, represented as DD ± with Dirichlet boundary conditions, involve identifying functions u that belong to the class C 2 (G ± ).

In the interior and exterior initial boundary-value problems (DN ± ) with Neumann boundary conditions, we seek solutionsu∈C 2 (G ± )∩C 1 ( ¯G ± ) of

Consider two open arcs∂S 1 and∂S 2 of∂Ssuch that mes(∂S α )>0,

The interior and exterior initial-value problems (DM ± ) with mixed boundary conditions consist in ﬁndingu∈C 2 (G ± )∩C 1 ( ¯G ± ) satisfying

In the interior and exterior initial-boundary value problems (DC ± 1 ) with combined boundary conditions of the ﬁrst kind, we look for u ∈ C 2 (G ± )∩

If the boundary conditions are of the second kind, then the solutionu∈

In regions S ± ×[−h 0 /2, h 0 /2], two distinct elastic materials characterized by Lamé constants λ ±, μ ± and densities ρ ± are present The initial-boundary value problem (DT) with transmission boundary conditions aims to determine a pair of functions u ± that belong to the space C²(G ±) and C¹(¯G ±).

B ± (∂ t 2 u ± )(X) + (A ± u ± )(X) = 0, X ∈G ± , u ± (x,0+) = (∂ t u ± )(x,0+) = 0, x∈S ± , u + + (X)−u − − (X) =f(X), X∈Γ, (T + u + ) + (X)−(T − u − ) − (X) =g(X), X ∈Γ, where A ± ,B ± , and T ± have the obvious meaning.

Consider an open arc∂S 0 of∂Sthat models a crack, and let

We denote a function \( u \) belonging to the class \( C^k(\bar{G}) \) for \( k = 0, 1, 2, \ldots \) if the restrictions \( u^+ \) and \( u^- \) on the regions \( G^+ \) and \( G^- \) are both of class \( C^k(\bar{G}^+) \) and \( C^k(\bar{G}^-) \), respectively Additionally, the limiting values of \( u^+ \) and its derivatives up to order \( k \) on the boundary \( \Gamma_1 \) must match those of \( u^- \), although discrepancies may exist on \( \Gamma_0 \) In the context of the initial-boundary value problem (DKD) with Dirichlet boundary conditions, we aim to find \( u \) that lies in the intersection of \( C^2(G) \) and \( C^1(\bar{G}) \).

The problem (DKN) with Neumann boundary conditions consists in ﬁnd- ingu∈C 2 (G)∩C 1 ( ¯G) such that

LetKbe a (3×3)-matrix of the form

In the study of a plate on an elastic foundation governed by Dirichlet boundary conditions, we consider the initial-boundary value problems denoted as (DD ± K), where k is greater than zero and the (2×2) matrix ¯K = h²(k αβ) is confirmed to be positive definite Our focus is on finding the solution u within the specified parameters.

The corresponding problems (DN ± K ) with Neumann boundary conditions consist in ﬁnding functionsu∈C 2 (G ± )∩C 1 ( ¯G ± ) satisfying

Throughout what follows, we work frequently with the Laplace transforms of vector-valued functions u(X) =u(x, t),t∈R, which vanish fort κ}, κ >0, and derive estimates that show how the solutions depend on the complex parameter p.

(iii) Using Parseval’s equality, we then return to the spaces of originals and prove the existence of weak solutions to the given initial-boundary value problems.

A similar scheme will also be used to study the solvability of the associated time-dependent boundary integral equations.

A Matrix of Fundamental Solutions

To develop single-layer and double-layer potentials in dynamic scenarios, it is essential to utilize a matrix of fundamental solutions D(X) = D(x, t) that is zero for t < 0 Notably, only the Laplace transform ˆD(x, p) of this matrix is necessary, as demonstrated in [15] This approach allows for the design of numerical methods that rely exclusively on estimates of ˆD, eliminating the need for direct access to the matrix D itself.

Consequently, we seek a (3×3)-matrix ˆD(x, p) deﬁned forp∈C0, which has polynomial growth asp→ ∞ and satisﬁes

Bp 2 D(x, p) + (Aˆ D)(x, p) =ˆ δ(x)I, (1.10) where δ is the Dirac distribution and I is the identity (3×3)-matrix To ﬁnd ˆD(x, p), we apply the (generalized) Fourier transformation with respect to x∈R 2 in (1.10) and arrive at

A(ξ) ˜D (ξ, p) =I, or Θ(ξ, p) ˜D(ξ, p) =I, (1.11) where ˜D(ξ, p) is the Fourier transform of ˆD(x, p) and Θ(ξ, p) is the (3×3)- matrix of elements Θ αβ (ξ, p) =h 2 (λ+à)ξ α ξ β +δ αβ (ρh 2 p 2 +à+h 2 à|ξ| 2 ), Θ 33 (ξ, p) =ρp 2 +à|ξ| 2 , Θ α3 (ξ, p) =−Θ 3α (ξ, p) =iàξ α

From this it follows that det Θ(ξ, p) is invariant with respect to rotations in

R 2 ; that is, det Θ(ξ, p) depends only on|ξ| Then we takeξ 1 =|ξ|andξ 2 = 0 in (1.12) and ﬁnd that det Θ(ξ, p) =R(|ξ| 2 , p) ρp 2 h 2 +à+h 2 (λ+ 2à)|ξ| 2 0 −ià|ξ|

Let|ξ| 2 =s We denote the roots of the equationR(s, p) = 0 bys i =−χ 2 i and choose χ i so that Reχ i ≥0 It is easy to check that χ 2 1,2 =p

1.1 Lemma (i)The equationR(s, p) = 0does not have a triple root for any p∈C 0

(ii)The equationR(s, p) = 0has a double root if and only if p= 2à h(λ+à) −1 ρ −1 (λ+ 2à) 1/2

Proof (i) Suppose thatχ 2 1 =χ 2 2 =χ 2 3 Then the explicit expressions of the roots show that χ 2 1 =χ 2 2 implies that ρ 2 p 2 h 2 (λ+à) 2 −4ρà 2 (λ+ 2à) = 0; hence, p= 2à h(λ+à) −1 ρ −1 (λ+ 2à) 1/2

Now the equalityχ 2 1 =χ 2 3 yieldsλ+à= 0, which contradicts (1.1).

(ii) Ifχ 2 1 =χ 2 3 or χ 2 2 =χ 2 3 , then, as a straightforward calculation shows, we obtain p 2 =−à(ρh 2 ) −1 , which is impossible, since Rep >0.

Initially, we observe that χ i = 0 If Reχ i equals 0 for any index i, it follows that χ 2 i is less than 0, indicating that ˜s=|ξ˜| 2 =−χ 2 i serves as a positive root of the equation R(s, p) = 0 We define ξ˜ as (|ξ˜|, 0) and represent a nonzero solution of the linear algebraic equations system Θ( ˜ξ, p)g = 0 as g = (g 1, g 2, g 3) T By multiplying this equation by g ∗, we derive a significant relationship.

−2à|ξ˜|Re(g ∗ 3 ig 1 ) = 0; consequently, p 2 ∈Randp 2 >0 Since

2à|ξ˜|Re(g ∗ 3 ig 1 )≥ −à(|g 1 | 2 +|ξ˜| 2 |g 3 | 2 ), it follows that ρp 2 h 2 (|g 1 | 2 +|g 2 | 2 ) +ρp 2 |g 3 | 2

This contradiction completes the proof.

Hence, the elements of the matrix of fundamental solutions ˆD(x, p) are

Dˆ α3 (x, p) =−Dˆ 3α (x, p) =à(h 2 à∆−ρp 2 h 2 −à)∂ α Ψ(x, p), where Ψ(x, p) is the inverse Fourier transform of ˜Ψ(ξ, p); that is, Ψ(x, p) = (4π 2 ) −1

According to Lemma 1.1, there are two scenarios to consider In the first scenario, the roots of the equation R(s, p) = 0 are simple A direct calculation reveals that in this case, Ψ(x, p) can be expressed as Ψ(x, p) = (2πh^4) * (λ + 2à)^{-1} * c * K_0(χ_i |x|), where K_0 denotes the modified Bessel function of order zero, and c is defined as c = (χ_2^1 - χ_2^2)(χ_2^1 - χ_2^3)^{-1}.

In the second case, the equationR(s, p) = 0 has a double root: χ 2 1 =χ 2 2 =χ 2 3

, (1.16) where K 1 is the modiﬁed Bessel function of order one and ˜ c 1 =−˜c 3 =−(χ 2 3 −χ 2 1 ) −2 , ˜ c 2 = (χ 2 3 −χ 2 1 ) −1 (1.17)

1.2 Lemma For anyp∈C0 , the functionΨ(x, p)can be represented in the neighborhood of x= 0 in the form Ψ(x, p) =−

+O(|x| 6 ln|x|) + Ψ 0 (x, p), (1.18) whereΨ 0 (x, p)is an inﬁnitely diﬀerentiable function; in addition,Ψ(x, p)→0 exponentially as |x| → ∞.

Proof In the case of simple roots, from (1.14) and the asymptotic behavior of the modiﬁed Bessel functionK 0 (z) asz→0 [1] it follows that Ψ(x, p) =−

+O(|x| 6 ln|x|) + Ψ 0 (x, p). Using (1.15), it is easy to verify that c 1 +c 2 +c 3 = 0, c i χ 2 i = 0, c i χ 4 i = 1; therefore, (1.18) holds.

In the case of a double root, (1.16) implies that Ψ(x, p) =−

+O(|x| 6 ln|x|) + Ψ 0 (x, p), and we immediately regain (1.18) from (1.17).

The last assertion follows from the fact that Reχ i >0 and the asymptotic behavior ofK n (z),n= 0,1, asz→ ∞in such a way that Rez≥κ >0.

1.3 Corollary For anyp∈C0 , the elements of the matrix of fundamental solutions D(x, p)ˆ can be represented in the neighborhood of x= 0 in the form

(1.19) where Dˆ 0,ij (x, p) are inﬁnitely diﬀerentiable functions.

The proof of this assertion follows from (1.13) and (1.18).

1.4 Remark Representation (1.19) shows that for anyp∈C 0 , the asymptotic behavior of ˆD(x, p) in the neighborhood of x= 0 coincides with that of the matrix of fundamental solutionsD(x) for the equilibrium equation [10].

Time-dependent Plate Potentials

Let α and β be three-component vector-valued functions with compact support in ¯Γ, defined on the boundary of S and R We begin by introducing single-layer and double-layer potentials using Laplace transforms Specifically, we define the single-layer potential \( V_p \hat{\alpha} \) with density \( \hat{\alpha} \), where \( \hat{\alpha}(x, p) = L\alpha(x, t) \) and \( \hat{\beta}(x, p) = L\beta(x, t) \) for every \( p \in C_0 \).

D(xˆ −y, p) ˆα(y, p)ds y , x∈R 2 , and a double-layer potentialW p βˆ of density ˆβ by

T y is the moment-force boundary operator acting with respect to the pointy, ande j is thejth coordinate unit vector inR 3

According to Remark 1.4, for any fixed p in C0, the boundary properties of both potentials align with those of the corresponding single-layer and double-layer potentials in the equilibrium scenario Here, we outline the most significant of these properties.

(i) The single-layer potentialV p αˆ satisﬁes the equation

(ii)V p αˆ is continuous inR 2 ; in particular, the direct value (V p α)ˆ 0 on∂S of the corresponding weakly singular integral is given by

(iii) There hold the jump formulas

(T V p α)ˆ ± (x, p) =± 1 2 α(x, p) + (T Vˆ p α)ˆ 0 (x, p), x∈∂S, where (T V p α)ˆ 0 is the direct value of the corresponding singular integral on

(iv) The double-layer potentialW p βˆ satisﬁes (1.20) and can be extended by continuity from S ± to ¯S ± , respectively These extensions are of class

(v) There hold the jump formulas

(W p β)ˆ ± (x, p) =∓ 1 2 βˆ(x, p) + (W p β)ˆ 0 (x, p), x∈∂S, where (W p β)ˆ 0 is the direct value of the corresponding singular integral on∂S.

(vi) There holds the equality

This equality enables us to introduce the notation

Since ˆD(x, p) has polynomial growth with respect top∈C κ ,κ >0, we can deﬁne the single-layer and double-layer time-dependent (retarded) potentials

The inverse Laplace transforms of V p αˆ and W p β are denoted as V α and W β, respectively Notably, the inverse transform of the product of two transforms results in the convolution of their corresponding original functions.

P(x−y, t−τ)β(y, τ)ds y dτ, where P(x−y, t) is the inverse Laplace transform of ˆP(x−y, p); that is,

The retarded potentials, derived from the properties of V p αˆ and W p βˆ, exhibit specific characteristics when applied to smooth densities α and β that vanish for t < 0 and possess compact support concerning the time variable Notably, both V α and W β fulfill the governing equation, highlighting their fundamental role in this context.

(ii)V αandW β satisfy the initial conditions u(x,0+) = (∂ t u)(x,0+) = 0, x∈S + ∪S −

(iii) ForX∈Γ, there hold the jump formulas

In this article, we introduce Sobolev-type spaces and analyze the properties of boundary operators generated by potentials within these spaces We demonstrate that the established jump formulas are applicable to broader classes of densities, α and β, provided that the limiting values of the potentials are interpreted as traces on the boundary Γ.

Nonstationary Boundary Integral Equations

We aim to find comprehensive integral representations for the solutions to all initial-boundary value problems outlined in §1.1 This article specifically examines these representations for the Dirichlet-Dirichlet (DD ±) and Dirichlet-Neumann (DN ±) problems.

To represent the solutions of the operator (DD ±) as u(X) = (V α)(X) for X ∈ G ±, we introduce an unknown density α While V α meets the equation of motion and initial conditions, the function u must adhere solely to the Dirichlet boundary conditions Consequently, α is required to be a solution of the nonstationary boundary integral equation.

The kernel D(X −Y) of this equation has a retarded time argument For every ﬁxed value of this argument, the kernel is weakly singular with respect to the space variables.

If we now seek the solutions of (DD ± ) in the form u(X) = (W β)(X), X ∈G ± , (1.23) then, according to the properties ofW β, the unknown densityβ must satisfy the corresponding boundary integral equation

(W β) ± (X) =f(X), X∈Γ, (1.24) whose kernel again has a retarded time argument but is singular with respect to the space variables.

In the initial-boundary value problems (DN ± ), the representations (1.21) and (1.23) yield, respectively, the nonstationary boundary integral equations

The equation (N β)(X) = g(X), where X belongs to the set Γ, represents the inverse Laplace transform of N p β The kernel in this context shares a similar structure with that of a preceding equation, while the kernel in the latter equation is hypersingular concerning the spatial variables Notably, both kernels incorporate a retarded time argument.

In contrast to the equilibrium problems outlined in reference [10], the representations (1.21) and (1.23) do not include additive rigid displacements This observation applies to the integral representations of solutions for all initial-boundary value problems under consideration.

This article discusses the application of the Laplace transformation to nonstationary boundary integral equations, leading to new equations dependent on the transformation parameter p A key challenge is to establish estimates for the solutions, which reveal their dependence on p Once these estimates are established, it becomes possible to demonstrate the unique solvability of the time-dependent boundary equations.

Problems with Dirichlet Boundary Conditions

Function Spaces

In this article, we maintain consistent notation for norms and inner products across scalar and vector-valued function spaces, following the convention established in §1.1 It is important to note that all discussed function spaces are complex.

We start by introducing spaces of functions that depend on a complex parameter p; their properties, studied in [2], are listed below without proof.

H m (R 2 ), H m (S), and H m (∂S) are the standard Sobolev spaces whose elements are deﬁned on R 2 ,S, and the boundary∂S ofS, respectively.

H m,p (R 2 ) is the space that coincides as a set withH m (R 2 ) but is equipped with the norm u m,p R 2

, where ˜uis the (distributional) Fourier transform of the three-component distribution u ∈ S (R 2 ) (see the Appendix) Clearly, for any ﬁxed p ∈ C, the norms onH m,p (R 2 ) andH m (R 2 ) are equivalent.

H˚ m,p (S) is the subspace of allu∈H m,p (R 2 ) such that suppu⊂S.¯

H m,p (S) is the space of the restrictions to S of all v ∈ H m,p (R 2 ) The norm ofu∈H m,p (S) is deﬁned by u m,p;S = inf v∈H m,p (R 2 ): v| S =u v m,p

The inner products inL 2 (R 2 ), L 2 (S), andL 2 (∂S) are denoted by (ã,ã) 0 , (ã,ã) 0;S , and (ã,ã) 0;∂S , respectively.

H −m,p (R 2 ) is the dual ofH m,p (R 2 ) with respect to the duality generated by (ã,ã) 0

H 1/2,p (∂S) is the space of the traces on∂S of all the elements ofH 1,p (S).

It coincides as a set with H 1/2 (∂S) but is equipped with the norm f 1/2,p;∂S = inf u∈H 1,p (S):γu=f u 1,p;S

Hereγis the trace operator, which mapsH 1,p (S) continuously toH 1/2,p (∂S).

The uniform continuity of the trace operator γ with respect to p ∈ C is established, satisfying the inequality γu 1/2,p;∂S ≤ cu 1,p;S, where c is independent of p Additionally, the trace operators for the interior and exterior domains, S ±, are represented as γ ±.

H −1/2,p (∂S) is the dual ofH 1/2,p (∂S) with respect to the duality generated by (ã,ã) 0;∂S

Next,l + andl − are extension operators that, in the context of our function spaces, map H 1/2,p (∂S) to H 1,p (S + ) and H 1,p (S − ) continuously and uniformly with respect top∈C.

H m,k,κ (S) is the space of all ˆu(x, p),x∈S,p∈C κ , such that the mapping

U(p) = ˆu(ã, p) is holomorphic fromC κ toH m (S) and uˆ 2 m,k,κ;S = sup σ>κ

The norm on H m,k,κ (S) is defined by Formula (2.1), indicating that U(p) is an element of H m,p (S) for any p in C κ To emphasize that ˆ u(x, p) belongs to H m,p (S), we use this notation, while U(p) is used when considering it as a mapping from C κ to H m (S).

H ±1/2,k,κ (∂S) are introduced similarly; that is, these spaces consist of all fˆ(x, p),x∈∂S,p∈C κ , such that the corresponding mappingF(p) = ˆf(ã, p) is holomorphic fromC κ toH ±1/2 (∂S) and fˆ 2 ±1/2,k,κ;∂S = sup σ>κ

Once again, the above equality deﬁnes the norms on these spaces and, as above, we interpret ˆf(x, p) as an element ofH ±1/2,p (∂S) andF(p) as a mapping fromC κ to H ±1/2 (∂S).

H m,k,κ (G) and H ±1/2,k,κ (Γ) consist, respectively, of the inverse Laplace transformsuandf of the elements ˆuand ˆf ofH m,k,κ (S) andH ±1/2,k,κ (∂S); these spaces are equipped with the norms u m,k,κ;G =uˆ m,k,κ;S , f ±1/2,k,κ;Γ =fˆ ±1/2,k,κ;∂S

By the Paley–Wiener theorem and Parseval’s equality [12], for a nonnegative integer kthe spacesH 1,k,κ (G) consist of all three-component distributionsu deﬁned on S×Rthat vanish fort 0, problem(D + p ) has a unique weak solutionu∈H 1,p (S + )and u 1,p;S + ≤c|p|(q −1,p;S + +f 1/2,p;∂S ) (2.8)

Proof First we consider (D + p ) with homogeneous boundary conditions, which consists in ﬁnding u 0 ∈H˚ 1,p (S + ) such that p 2 (B 1/2 u 0 , B 1/2 v) 0;S ++a + (u 0 , v) = (q, v) 0;S + ∀v∈H˚ 1,p (S + ) (2.9)

By reiterating the proof of Lemma 2.3 from reference [7], we can establish that the form a + (u, v) is coercive on the space [ ˚H 1,p (S + )] 2 Additionally, since this form is continuous within this space, it follows that for any q in H −1,p (S + ), the variational equation a + (u 0 , v) = (q, v) 0;S + for all v in H˚ 1,p (S + ) has a unique solution u 0 in H˚ 1,p (S + ) This solution satisfies the estimate u 0 1 ≤ cq −1;S +.

The bounded antilinear functional \( a^+ (u_0, v) \) is defined on \( \dot{H}^{1,p}(S^+) \) for any \( u_0 \in H^{1,p}(S^+) \) and can be expressed in the form \( (2.10) \) with \( q \in H^{-1,p}(S^+) \) The operator \( A \) is defined by \( a^+ (u_0, v) \), which maps \( u_0 \in \dot{H}^{1,p}(S^+) \) to \( q \in H^{-1,p}(S^+) \) Consequently, the relationship is given by \( a^+ (u_0, v) = (Au_0, v)_{0;S^+} \) for all \( u_0, v \in \dot{H}^1(S^+) \).

Ais a homeomorphism from ˚H 1 (S + ) toH −1 (S + ) Additionally, from (2.6) it follows that Ais self-adjoint in the sense that

This is easily veriﬁed, since for anyu 0 , v∈H˚ 1,p (S + ),

Equation (2.9) can now be written in the form p 2 Bu 0 +Au 0 =q (2.11)

By applying the operator A −1 to both sides of equation (2.11), we derive the equivalent equation p 2 A −1 Bu 0 + u 0 = A −1 q (2.12) within the Banach space ˚H 1,p (S + ) We define B 0 as the restriction of A −1 B from H −1 (S + ) to ˚H 1 (S + ) and assert that B 0 is compact on ˚H 1 (S + ).

{u n } ∞ n=1 be a weakly convergent sequence in ˚H 1 (S + ) SinceS + is a bounded domain, Rellich’s lemma implies that the set {u n } ∞ n=1 is strongly compact in

L 2 (S + ) Therefore, there is a subsequence{u n j } ∞ j=1 that converges strongly inL 2 (S + ), consequently, also inH −1 (S + ) BecauseA −1 Bis continuous from

H −1 (S + ) to ˚H 1 (S + ), the sequence{B 0 u n j } ∞ j=1 is strongly convergent in the space ˚H 1 (S + ), which proves thatB 0 is a compact operator.

In the context of the Fredholm Alternative, the solvability of equation (2.12) ensures that it has a unique solution \( u_0 \in H^1(S^+) \) if and only if the homogeneous equation \( p^2 A^{-1} Bu_0 + u_0 = 0 \) (equation 2.13) admits only the trivial solution This homogeneous equation can be equivalently expressed as \( p^2 Bu_0 + Au_0 = 0 \), or in the form \( p^2 (B^{1/2} u_0, B^{1/2} v)_{0;S} + (u_0, v) = 0 \) for all \( v \in H^1(S^+) \) (equation 2.14).

Taking v = u 0 in (2.14), writing p = σ+iτ, and separating the real and imaginary parts, we arrive at

Ifτ= 0, then from (2.16) it follows thatB 1/2 u 0 = 0; hence,u 0 = 0 Ifτ= 0, then the equalityu 0 = 0 follows from (2.15) So (2.11) is uniquely solvable in

We now establish estimate (2.8) Taking v =u 0 in (2.9) and separating the real and imaginary parts, we obtain

Multiplying (2.18) by σ −1 τ and adding the new equality to (2.17), we ﬁnd that

Because p∈ C¯ κ , it follows that σ ≥ κ Taking into account the inequality a + (u 0 , u 0 )≥cu 0 2 1 , we obtain u 0 2 1,p;S + ≤c|p||(q, u 0 ) 0;S + |, from which u 0 1,p;S + ≤c|p|q −1,p;S +

We now return to the full problem (2.7) Letw=l + f ∈H 1 (S + ) We recall that since the extension operatorl + is continuous (uniformly with respect to p∈C), we have w 1,p;S + ≤cf 1/2,p;∂S (2.19)

Representing uin the formu=w+u 0 , we see thatu 0 ∈H˚ 1,p (S + ) satisﬁes the equation p 2 (B 1/2 u 0 , B 1/2 v) +a + (u 0 , v)

We claim that the form p 2 (B 1/2 w, B 1/2 v) 0;S ++a + (w, v) deﬁnes a bounded antilinear (conjugate linear) functional on ˚H 1,p (S + ); true,

Consequently, there is ˜q∈H −1,p (S + ) such that p 2 (B 1/2 w, B 1/2 v) 0;S ++a + (w, v) = (˜q, v) 0;S + ∀v∈H˚ 1,p (S + ) and q˜ −1,p;S + ≤cw 1,p;S + ≤cf 1/2,p;∂S

As we already know, the latter problem is uniquely solvable and its solution satisﬁes the estimate u 0 1,p ≤c|p|(q −1,p;S + +q˜ −1,p;S + )

Combining (2.21) and (2.19), we arrive at (2.8).

Ifu 1 andu 2 are two solutions of (2.7), thenu 0 =u 1 −u 2 ∈H˚ 1,p (S + ) is a solution of (2.14); hence,u 0 = 0, and the theorem is proved.

Due to the invalidity of Rellich’s lemma for unbounded domains, we must adjust our approach rather than replicate the previous proof for the case of S −.

2.2 Theorem For any f ∈H 1/2,p (∂S)and q∈H −1,p (S − ), p∈C¯ κ ,κ >0, problem(D − p )has a unique solutionu∈H 1,p (S − )and u 1,p;S − ≤c|p|(q −1,p;S − +f 1/2,p;∂S ).

Proof Again, ﬁrst we assume thatf = 0 In this case we seeku 0 ∈H 1,p (S − ) such that p 2 (B 1/2 u 0 , B 1/2 v) 0;S − +a − (u 0 , v) = (q, v) 0;S − ∀v∈H˚ 1,p (S − ) (2.22)

To prove the unique solvability of (2.22), we consider an auxiliary variational problem that consists in ﬁndingu 0 ∈H˚ 1 (S − ) such that

1 2 κ 2 (B 1/2 u 0 , B 1/2 v) 0;S − +a − (u 0 , v) = (q, v) 0;S − ∀v∈H˚ 1 (S − ), (2.23) where q∈H −1 (S − ) is prescribed Repeating the proof of Lemma 2.2 in [7], we ﬁnd that there is a constant c >0 such that a − (u, u) +u 2 0;S − ≥cu 2 1;S − ∀u∈H 1 (S − ) (2.24) From (2.24) it follows that the form a −,κ (u, v) = 1 2 κ 2 (B 1/2 u, B 1/2 v) 0;S − +a − (u, v), which is continuous onH˚ 1 (S − ) 2

The Lax–Milgram lemma indicates that equation (2.23) has a unique solution \( u_0 \in H^1_0(S^-) \) for any \( q \in H^{-1}(S^-) \) Additionally, for any \( u_0 \in H^1_0(S^-) \), the form \( a_{-,\kappa}(u_0, v) \) produces a bounded antilinear functional on \( H^1_0(S^-) \), allowing it to be expressed in the form of equation (2.23) This leads to the definition of a corresponding operator.

(A κ u 0 , v) 0;S − =a −,κ (u 0 , v) ∀v∈H˚ 1 (S − ), which is a homeomorphism from ˚H 1 (S − ) to H −1 (S − ) Equation (2.23) can be rewritten asA κ u 0 =q In turn, (2.22) can be written in the form

ApplyingA −1 κ on both sides in (2.25), we arrive at the equivalent equation u 0 + p 2 − 1 2 κ 2

A −1 κ Bu 0 =A −1 κ q (2.26) SettingB 1/2 u 0 =u b , we again rewrite (2.26) in the equivalent form u b + p 2 − 1 2 κ 2

Equation (2.27) is solvable in ˚H 1 (S − ) If u b ∈H˚ 1 (S − ) is its solution, then u b is, at the same time, the solution of (2.27) in L 2 (S − ) Conversely, let u b ∈L 2 (S − ) be a solution of (2.27) Since

B 1/2 A −1 κ B 1/2 u b ∈H˚ 1 (S − ), B 1/2 A −1 κ q∈H˚ 1 (S − ), it follows that u b ∈H˚ 1 (S − ) This means that problem (2.27) in ˚H 1 (S − ) is equivalent to itself inL 2 (S − ) We now study the properties of the restriction

Letq, ψ∈L 2 (S − ) be arbitrary, and let u 0 =A −1 κ B 1/2 q, v=A −1 κ B 1/2 ψ.

From the deﬁnition ofA κ it follows that a −,κ (u 0 , v) = (B 1/2 q, v) 0;S − , a −,κ (v, u 0 ) = (B 1/2 ψ, u 0 ) 0;S − ; hence,

In turn, (2.28) can be rewritten as

By (2.29),B κ is a symmetric operator onL 2 (S − ); therefore, it is self-adjoint (as a symmetric operator deﬁned on the whole of a Hilbert space [3]). Once again, letq∈L 2 (S − ) andu 0 =A −1 κ B 1/2 q; then

The expression (u 0 ,A κ u 0 ) 0;S − =a −,κ (u 0 , u 0 )≥0 indicates that B κ is nonnegative This implies that the spectrum of a self-adjoint nonnegative operator is confined to the half-line [0,∞) in the complex plane, establishing that every point in C\[0,∞) is a regular point for B κ It is important to note that this holds true for any p∈C κ.

Equation (2.27) is uniquely solvable for any \( q \in H^{-1}(S^{-}) \), leading to the conclusion that equation (2.22) has a unique solution \( u_0 \in H^{\circ 1,p}(S^{-}) \) for any \( q \in H^{-1,p}(S^{-}) \) To finalize the proof, we replicate the concluding steps of Theorem 2.1, substituting the extension operator \( l^{+} \) with \( l^{-} \).

Solvability of the Time-dependent Problems

We start with the variational version of problems (DD ± ) for the nonhomogeneous equation of motion The classical formulation of (DD + ) asks for a functionu∈C 2 (G + )∩C 1 ( ¯G + ) such that

Multiplying the ﬁrst equation (2.30) byv ∗ , wherev ∈C 0 ∞ ( ¯G + ) is such that v + = 0, integrating the new equality over S + with respect to x and over

[0,∞) with respect tot, and taking into account the initial data foruand the boundary value ofv, we arrive at

If \( u \in C^2(G^+) \cap C^1(\overline{G^+}) \) satisfies equation (2.31) for any \( v \in C_0^\infty(\overline{G^+}) \) such that \( v^+ = 0 \), \( u(x,0+) = 0 \) for \( x \in S^+ \), and \( u^+ = f \), then by integrating by parts in (2.31), we conclude that \( u \) is the solution to equation (2.30) This indicates that the variational problem (DD+) aims to find \( u \in H^{1,0,\kappa}(G^+) \) that meets these conditions.

(q, v) 0;S + dt ∀v∈C 0 ∞ ( ¯G + ), v + = 0, (2.32) γ + u=f, whereq andf are prescribed Similarly, the variational problem (DD − ) consists in ﬁndingu∈H 1,0,κ (G − ) that satisﬁes

2.3 Theorem For anyq∈H −1,1,κ (G + )andf∈H 1/2,1,κ (Γ),κ >0, problems (2.32)and (2.33) have unique solutionsu∈H 1,0,κ (G ± ) If q∈H −1,k,κ (G ± ) andf ∈H 1/2,k,κ (Γ),k∈R, thenu∈H 1,k−1,κ (G ± ) and u 1,k−1,κ;G ± ≤c(q −1,k,κ;G ± +f 1/2,k,κ;Γ ) (2.34)

Proof We prove the assertion for (DD + ); the case of (DD − ) is treated similarly.

Let \( \hat{u} \in H^{1,p}(S^+) \) denote the weak solution to the equation \( p^2 B u(x, p) + (A \hat{u})(x, p) = \hat{q}(x, p) \) for \( x \in S^+ \), alongside the boundary condition \( \gamma + u(x, p) = \hat{f}(x, p) \) for \( x \in \partial S \) This solution is derived using the Laplace transformation in the context of a non-homogeneous equation of motion, with its existence established in the preceding section For convenience, we redefine \( \hat{u}(\tilde{a}, p) \) as \( U(p) \), \( \hat{q}(\tilde{a}, p) \) as \( Q(p) \), and \( \hat{f}(\tilde{a}, p) \) as \( F(p) \), treating \( U \), \( Q \), and \( F \) as functions mapping from \( C^\kappa \) to \( H^{1}(S^+) \), \( H^{-1}(S^+) \), and \( H^{0}(S^+) \) respectively.

We claim that the inverse Laplace transformuof ˆubelongs to the space

H 1,k−1,κ (G + ) if q ∈ H −1,k,κ (G + ) and f ∈ H 1/2,k,κ (Γ) To show this, ﬁrst we verify that U is holomorphic from C κ to H 1 (S + ) Let p 0 ∈ C κ , and let

K R (p 0 ) be a circle with center atp 0 and radiusR (to be speciﬁed later), and such that ¯K R (p 0 )⊂C κ We recall that the solutionU(p 0 ) of the problem p 2 0 BU(p 0 ) + (AU)(p 0 ) =Q(p 0 ), γ + U(p 0 ) =F(p 0 ) satisﬁes the estimates

Rewriting (2.35) in the form p 2 0 BU(p) + (AU)(p) =Q(p)−(p 2 −p 2 0 )BU(p), γ + U(p) =F(p),

SinceU(p) −1;S + ≤ U(p) 1;S + , it follows that for psatisfying c|p 2 −p 2 0 | ≤ 1 2 (2.37) we have

We chooseR >0 so that ¯K R (p 0 )⊂C κ and (2.37) holds forp∈K R (p 0 ) Since

Q and F are holomorphic from C κ to H −1 (S + ) andH 1/2 (∂S), respectively, they are bounded in these spaces forp∈K¯ R (p 0 ) Estimate (2.38) shows that

U is also bounded inH 1 (S + ) forp∈K¯ R (p 0 ) By (2.36), p 2 0 B

AsU(p) is bounded inH 1 (S + )—hence, also inH −1 (S + )—forp∈K R (p 0 ), it follows that p→p lim0 U(p)−U(p 0 ) 1;S + = 0, which means that U is continuous from C κ to H 1 (S + ) at p 0 Finally, let

V ∈H 1 (S + ) be the solution of the problem p 2 0 BV −AV =Q (p 0 )−2p 0 U(p 0 ), γ + V =F (p 0 ); then the function

SinceU is continuous atp 0 , p→p lim0 W(p) 1;S + = 0, which means that U (p 0 ) exists andU (p 0 ) = V The arbitrariness of p 0 in

C κ implies that the mappingU is holomorphic fromC κ to H 1 (S + ).

To complete the proof of the theorem, we need to check thatuis the only solution of (2.32) We recall [11] that any two functionsf 1 (t) and f 2 (t) such that

The equation (2.40), represented as F₁(κ₁ + iτ)F₂(κ₂ + iτ)dτ, involves the Laplace transforms of f₁(t) and f₂(t) We consider a function v ∈ C₀∞(G⁺) such that γ + v = 0, and denote v(x, 0) as v₀(x), where v₀ belongs to H˚₁(S⁺) The Laplace transform is expressed as v(ã, p) = V(p) and v₀(ã) = V₀, with p defined as σ + iτ and p* as -σ + iτ By selecting σ > κ and fixing it, we set κ₁ = σ and κ₂ = -σ in (2.40) This results in the Laplace transform of ∂ₜv at the point p* being p*V(p*) - V₀.

SinceU(p) is a weak solution of (2.4), it follows that a +

We claim that the right-hand side in (2.43) vanishes First, we remark that qˆ 2 −1;S + ≤(1 +|p| 2 )qˆ 2 −1,p;S + ∀qˆ∈H −1 (S + ) (2.44)

For if ˆq∈H −1 (R 2 ) and ˜qis its Fourier transform, then qˆ 2 −1

Inequality (2.44) follows from (2.45) and the deﬁnition of the norms on the spacesH −1 (S + ) andH −1,p (S + ) By (2.44),

0 e −2σt |ϕ(t)| 2 dt 0 It is obvious that u∈H 1 ((0, T);L 2 (S + ))∩L 2 ((0, T); ˚H 1 (S + )); that is, U(t) = u(ã, t), regarded as a vector-valued function from (0, T) to

The same function, regarded as a mapping from (0, T) to ˚H 1 (S + ), belongs to

From (2.47) and (2.48) we see that H 1 ((0, T);L 2 (S + ))∩L 2 ((0, T); ˚H 1 (S + )) can be equipped with the norm u 2 1;G +

We now construct the functionZ(t) =z(ã, t), where z(x, t) ⎧⎨

Clearly, the restriction ofZto (0, T) belongs toH 1 ((0, T);L 2 (S + ))∩L 2 ((0, T);

H˚ 1 (S + )) and that z can be approximated with any accuracy in the norm ã 1;G +

T by means of elementsv ∈C 0 ∞ ( ¯G + ) such that γ + v = 0 Hence, we may setv=zin (2.46) and obtain

0, t > T, (2.50) and rewrite (2.46) in the form

Since U, regarded as a mapping from (0, T) to L 2 (S + ), belongs toH 1 (0, T), it is absolutely continuous on [0, T]; hence,

From (2.49) and (2.50), it follows thatZ, as a mapping from (0, T) to ˚H 1 (S + ), belongs toH 1 (0, T); hence,Z is absolutely continuous on [0, T] and

Formulas (2.51)–(2.53) now imply that a + (Z(0), Z(0)) +B 1/2 U(T) 2 0;S + = 0; therefore,u(T) =u(ã, T) = 0 for anyT >0, which completes the proof.

Problems with Neumann Boundary Conditions

The Poincar´ e–Steklov Operators

In this section we study the properties of the dynamic analogs of the Poincar´e– Steklov operators introduced in [7] We begin by considering these operators in Sobolev spaces with a parameter.

Letf ∈H 1/2,p (∂S),p∈C0, and letu∈H 1,p (S ± ) be the (unique) solutions of the problems p 2 (B 1/2 u, B 1/2 v) 0;S ± +a ± (u, v) = 0 ∀v∈H˚ 1,p (S ± ), γ ± u=f.

Also, let ϕ be an arbitrary element ofH 1/2,p (∂S), and let w∈H 1,p (S ± ) be such thatγ ± w=ϕ For everyp∈C0, we deﬁne a pair of Poincar´e–Steklov operators T p ± onH 1/2,p (∂S) by means of the equality

It is obvious that (3.2) deﬁnes T p ± correctly For if w 1 , w 2 ∈ H 1,p (S ± ) are such that γ ± w 1 =γ ± w 2 =ϕ, thenv=w 1 −w 2 ∈H˚ 1,p (S ± ) and, by (3.1), p 2 (B 1/2 u, B 1/2 w 1 ) 0;S ± +a ± (u, w 1 )

3.1 Lemma For anyp∈C 0 , the operators T p ± are homeomorphisms from

Proof Let f, ϕ ∈ H 1/2,p (∂S), let w = l ± ϕ, and let u ∈ H 1,p (S ± ) be the solutions of (3.1) inS ± By (3.2),

T p ± f −1/2,p;∂S ≤cu 1,p;S ± (3.5) Takingϕ=f andw=uin (3.1), we obtain p 2 B 1/2 u 2 0;S ± +a ± (u, u) =±(T p ± f, f) 0;∂S , where p=σ+iτ Next, we separate the real and imaginary parts above to

Multiplying (3.7) by σ −1 τ and adding the result to (3.6), we arrive at the equalities

(3.8) Since, as shown in [7], a ± (u, u) +u 2 0;S ± ≥cu 2 1;S ± , we easily deduce from (3.8) that u 2 1,p;S ± ≤ ±cσ −3 Re p(¯T p ± f, f) 0;∂S

; (3.9) hence, forp∈C¯ κ , u 2 1,p;S ± ≤c|p|T p ± f −1/2,p;∂S f 1/2,p;∂S , (3.10) and the trace theorem implies that u 1,p;S ± ≤c|p|T p ± f −1/2,p;∂S (3.11) From (3.5) and (3.10) it follows that

This proves the continuity ofT p ± fromH 1/2,p (∂S) toH −1/2,p (∂S), and (3.3).

We demonstrate the existence of the inverse operators (T p ± ) −1, which are continuous from H −1/2,p;∂S to H 1/2,p (∂S), as supported by the trace theorem If the ranges of T p ± are not dense in H −1/2,p (∂S), there exists a nonzero function ϕ ∈ H 1/2,p (∂S) such that (T p ± f, ϕ) 0;∂S = 0 for all f ∈ H 1/2,p (∂S) Setting f = ϕ leads to (T p ± ϕ, ϕ) 0;∂S = 0 According to (3.8), if z is the solution to the related boundary value problem with ϕ, then z must equal 0, resulting in a contradiction since ϕ cannot be zero This contradiction confirms the density of the ranges of T p ± in H −1/2,p (∂S).

At this stage, we can deﬁne operators ˆT ± and ( ˆT ± ) −1 on the elements

F(p) = ˆf(ã, p) and G(p) = ˆg(ã, p) ofH 1/2,k,κ (∂S) andH −1/2,k,κ (∂S),k∈R, respectively, by setting

Finally, we return to the spaces of originals and deﬁne the Poincar´e– Steklov operators T ± and (T ± ) −1 on the elements f ∈ H 1/2,k,κ (Γ) and g∈H −1/2,k,κ (Γ),k∈R, by means of the equalities

Theorem 3.2 states that for any κ > 0 and k ∈ R, the operators T ± are continuous and injective mappings from the space H 1/2,k,κ (Γ) to H −1/2,k−1,κ (Γ), with their ranges being dense in H −1/2,k−1,κ (Γ) Furthermore, the inverses of these operators, denoted as (T ± ) −1, are extended by continuity from the ranges of T ± to H −1/2,k,κ (Γ) These inverses are also continuous and injective from H −1/2,k,κ (Γ) to H 1/2,k−1,κ (Γ) for any k ∈ R, and their ranges are dense in the respective spaces.

First, we show that if F(p) = ˆf(ã, p) is holomorphic from C κ to H 1/2 (∂S), then Φ(p) = (T p ± fˆ)(ã, p) is holomorphic from C κ to the space H −1/2 (∂S).

In the proof of Theorem 2.3 we established that ifF(p) is holomorphic from

C κ to H 1/2 (∂S), thenU(p) = ˆu(ã, p), where ˆu(x, p) is the solution of either of the problems (3.1) with boundary data ˆf(x, p), is holomorphic from C κ to H 1 (S ± ) We now take any ϕ ∈ H 1/2 (∂S) and construct w = l ± ϕ Let p 0 ∈C κ By (3.2), Φ(p)−Φ(p 0 ), ϕ

LetK R (p 0 ) be a circle with the center atp 0 and radiusRsuch that ¯K R (p 0 )⊂

C κ Since uˆ 1,p;S ± ≤cuˆ 1;S ± , p∈K¯ R (p 0 ), where c= const>0 is independent ofp, it follows that p→p lim0 Φ(p)−Φ(p 0 ) p−p 0 , ϕ

This shows that Φ(p) is weakly holomorphic fromC κ toH −1/2 (∂S) According to Dunford’s theorem [18], any weakly holomorphic mapping is holomorphic in the strong sense, so our assertion is proved.

Next, by (3.3) withp=σ+iτ and the deﬁnition ofT ± ,

We now go over to the case of (T ± ) −1 For any p ∈ C 0 , the Neumann boundary value problems

The variational formulation of the problem involves finding \( \hat{u} \in H^{1,p}(S^\pm) \) such that the equation \( p^2(B_{1/2} \hat{u}, B_{1/2} v)_{S^\pm} + a^\pm(\hat{u}, v) = \pm(\hat{g}, \gamma^\pm v)_{ \partial S} \) holds for all \( v \in H^{1,p}(S^\pm) \), where \( \hat{g} \in H^{-1/2,p}(\partial S) \) is given By defining \( \hat{f} = (T_p^\pm)^{-1} \hat{g} \in H^{1/2,p}(\partial S) \) and recognizing that \( \hat{g} = T_p^\pm \hat{f} \), it can be concluded that \( \hat{u} \) are the unique solutions to the variational problems presented Furthermore, it follows from the inequality that \( \|\hat{u}\|_{1,p;S^\pm} \leq c |\hat{p}| \|\hat{g}\|_{-1/2,p;\partial S} \).

Considering the established inequality and revisiting the proof of Theorem 2.3, we deduce that if G(p) = ˆg(ã, p) is holomorphic from C κ to H −1/2 (∂S), then the solutions U(p) = ˆu(ã, p) of equation (3.12) exhibit holomorphic properties from C κ to H 1 (S ± ) Additionally, it is important to note that the trace operators γ ± maintain continuity from H 1 (S ± ).

(p) =γ ± U(p) are holomorphic fromC κ toH 1/2 (∂S) Also, from (3.4) it follows that (T ± ) −1 are continuous from H −1/2,k,κ (Γ) toH 1/2,k−1,κ (Γ) for anyk∈R.

To complete the proof, we only need to show that the ranges of T ± and (T ± ) −1 are dense in the corresponding spaces Let H ± k be the ranges of the operators

We take any g ∈H −1/2,k+1,κ (Γ) and set f = (T ± ) −1 g∈ H 1/2,k,κ (Γ) Since g =T ± (T ± ) −1 g, we have H −1/2,k+1,κ (Γ)⊂ H ± k It is obvious that for any k ∈ R, the space H −1/2,k+1,κ (Γ) is dense in H −1/2,k−1,κ (Γ); hence, H ± k are dense in H −1/2,k−1,κ (Γ) The case of (T ± ) −1 is treated similarly.

Solvability of the Problems

Consider problems (DN ± ), whose classical formulation was given in §1.1. Their variational versions consist in ﬁnding u ∈ H 1,0,κ (G ± ) such that, for g prescribed on Γ,

3.3 Theorem For anyg∈H −1/2,1,κ (Γ),κ >0, problems(3.13)have unique solutions u∈ H 1,0,κ (G ± ) If g ∈ H −1/2,k,κ (Γ), k ∈R, then these solutions belong toH 1,k−1,κ (G ± ), respectively, and u 1,k−1,κ;G ± ≤cg −1/2,k,κ;Γ (3.14)

The solutions of the problems (3.12), denoted as ˆu(x, p) for p ∈ C¯ κ, have been established to be holomorphic from C κ to H 1 (S ± ) This relationship, along with (3.11), confirms that u = L −1 U belongs to H 1,k−1,κ (G ± ), validating the condition in (3.14) The verification that u satisfies (3.13) follows a similar approach to that used for the corresponding case of (DD ± ) Therefore, it remains essential to confirm the uniqueness of the solution to (3.13).

Letu 1 andu 2 be two solutions of (3.13) Thenu=u 1 −u 2 ∈H 1,0,κ (G ± ) satisﬁes

Repeating the proof of Theorem 2.3 with H 1 (S ± ) instead of ˚H 1 (S ± ), we conclude thatu= 0, as required.

The more general problems (DN ± ) for the nonhomogeneous equation of motion can be reduced to the above by means of an appropriate substitution(see Chapter 9).

Boundary Integral Equations for Problems with Dirichlet and Neumann Boundary

Time-dependent Potentials

We recall that the single-layer (retarded) potential V αof densityα deﬁned on∂S×Rwas introduced in§1.3 as

Ifα(X) = 0 fort 0, u 1,p ≤c|p|(q −1,p +g −1/2,p;∂S ) (4.9)

Proof First, we remark that (g, γv) 0;∂S deﬁnes a bounded antilinear (conjugate linear) functional onH 1,p (R 2 ) since, by the trace theorem,

≤cg −1/2,p;∂S v 1,p , so there is ˜q∈H −1,p (R 2 ) such thatq˜ −1,p ≤cg −1/2,p;∂S and

WritingQ=q+ ˜q∈H −1,p (R 2 ), we bring (4.8) to the form p 2 (B 1/2 u, B 1/2 v) 0 +a(u, v) = (Q, v) 0 ∀v∈H 1,p (R 2 ) (4.10)

We consider a fixed κ > 0 and revisit the proof of Theorem 2.2 We introduce the bilinear form aκ(u, v) = 1/2 κ²(B¹/²u, B¹/²v)₀ + a(u, v), which is continuous, symmetric, and coercive on the space [H¹,p(ℝ²)]² This form leads to the definition of a self-adjoint nonnegative operator Aκ.

Then we rewrite (4.10) in the form

A κ u+ (p 2 − 1 2 κ 2 )Bu=Q and prove the unique solvability of this equation inH 1,p (R 2 ), just as we did in Theorem 2.2.

Takingv =u in (4.10) and separating the real and imaginary parts, we obtain

We now deﬁne a pair of operators ˆV 0 and ˆV 0 −1 on the elements ˜α(p) ˆ α(ã, p) and F(p) = ˆf(ã, p), p ∈ C κ , of H −1/2,k,κ (∂S) and H 1/2,k,κ (∂S), respectively, by

We also deﬁne the single-layer potential ˆVα˜ by

Returning to the spaces of originals, we deﬁne operatorsV 0 andV 0 −1 and the single-layer potentialV αby setting

4.3 Theorem For any κ > 0 and k ∈ R, the operator V 0 is continuous and injective from H −1/2,k,κ (Γ) to H 1/2,k−1,κ (Γ), and its range is dense in

H 1/2,k−1,κ (Γ) The inverse V 0 −1 , extended by continuity from the range of

V 0 , is continuous and injective from H 1/2,k,κ (Γ) to H −1/2,k−1,κ (Γ) for any k ∈ R, and its range is dense in H −1/2,k−1,κ (Γ) In addition, for any α∈

Proof Let ˜α(p)∈H −1/2,k,κ (∂S),p∈C κ Since V p αˆ is the solution of (4.8) with q= 0, g= (T p + − T p − )V p,0 αˆ= ˆα, from (4.9) it follows that π + V p αˆ 1,p;S + +π − V p αˆ 1,p;S − ≤c|p|αˆ −1/2,p;∂S

We claim that π ± V p α(ˆ ã, p) are holomorphic mappings from C κ to H 1 (S ± ), respectively To show this, we choose any p 0 ∈C κ , consider a circle K R (p 0 ) with the center atp 0 and radiusRsuch that ¯K R (p 0 )⊂C κ , and rewrite (4.8) as p 2 0 (B 1/2 V p α, Bˆ 1/2 v) 0 +a(V p α, v)ˆ

Considering the implications of equation (4.9) and revisiting the proof of Theorem 3.2, we establish that the function V p α(ˆ ã, p) is a bounded, continuous, and holomorphic mapping from C κ to H 1 (R 2) As a result, the mappings π ± V p α(ˆ ã, p) are also holomorphic from C κ to H 1 (S ±).

V p,0 αˆ=γ + π + V p αˆ =γ − π − V p αˆ is holomorphic fromC κ toH 1/2 (∂S) Since

=α 2 −1/2,k,κ;Γ , p=σ+iτ, the mappingV 0 :H −1/2,k,κ (Γ)→H 1/2,k−1,κ (Γ) is continuous for anyk∈R If

V 0 αˆ = 0, then ( ˆV 0 α)(p) = 0, ˜˜ α(p) = ˆα(ã, p), which means that (V p,0 α)(ˆ ã, p) = 0 for every p∈C κ ; hence, ˆα= 0.

The continuity ofV 0 −1 follows from the equality

The relationship V 0 −1 = T + − T − is established, leading to Theorem 3.2, which asserts that the ranges of V 0 and V 0 −1 are dense in their respective spaces This conclusion is supported by the density of H ±1/2,k+1,κ (Γ) within H ±1/2,k−1,κ (Γ) Additionally, the inequality (4.11) is derived from (4.3), thereby confirming the assertion.

We now go over to the properties of the (retarded) double-layer potential

W β with density β deﬁned on Γ This potential was introduced in §1.3 for smooth densities as

∂S β(y, τ),(T y D (j) )(y−x, t−τ) e j ds y dτ, and its Laplace transform is

4.4 Lemma For any p ∈ C0 and βˆ ∈ C 2 (∂S), the double-layer potential

W p βˆ can be written in the form

In this proof, we fix a point \( x \) in the set \( S^+ \) and analyze the function \( \hat{D}(j)(y-x, p) \) for \( y \) in the boundary set \( \bar{S}^- \) We consider functions \( f \) and \( \phi \) belonging to the Sobolev space \( H^{1/2}(\partial S) \), and denote \( u \) and \( v \) as the solutions to the exterior problem defined by boundary conditions \( f \) and \( \phi \), respectively This exploration is grounded in the definition of \( T^- \).

Since the right-hand sides above are equal, we deduce that

Takingf(y) = ˆβ(y, p) andϕ(y) = ˆD (j) (y−x, p) in this equality, we ﬁnd that

Multiplying both sides above bye j and summing overjfrom 1 to 3, we obtain the ﬁrst equality (4.12) The second one is derived analogously.

From the properties of the double-layer potential with a smooth density it follows that π ± W p βˆ∈H 1,p (S ± ) Consequently, we can deﬁne the operators

W p ± of the limiting values on∂S ofW p βˆwith ˆβ ∈C 2 (∂S) by

The double-layer potential \( W_p \hat{\beta} \) and the limiting value operators \( W_p^\pm \) can be continuously extended from \( C^2(\partial S) \) to \( H^{1/2,p}(\partial S) \) for any \( p \in C_0 \) These extended operators \( W_p^\pm \) serve as homeomorphisms between the spaces \( H^{1/2,p}(\partial S) \) and \( H^{1/2,p}(\partial S) \) Furthermore, the extended functions \( \pi^\pm W_p \hat{\beta} \) maintain continuity from \( H^{1/2,p}(\partial S) \) to \( H^{1,p}(S^\pm) \).

This assertion and inequalities (4.13)–(4.15) follow from Lemmas 3.1 and 4.1.

We now deﬁne pairs of operators ˆW ± and ( ˆW ± ) −1 on the elements ˜β(p) βˆ(ã, p) andF(p) = ˆf(ã, p),p∈C κ , ofH 1/2,k,κ (∂S) by

(ã, p), and the double-layer potential ˆWβ˜by

Returning to the spaces of originals, we deﬁne operatorsW ± and (W ± ) −1 and the double-layer potential W βby setting

Theorem 4.6 establishes that for any κ > 0 and k ∈ R, the operators W ± are continuous and injective mappings from H 1/2,k,κ (Γ) to H 1/2,k−2,κ (Γ) Their inverses, denoted as (W ± ) −1, are also continuous and injective when extended by continuity to H 1/2,k,κ (Γ) from the ranges of W ± Furthermore, these inverses have dense ranges in H 1/2,k−2,κ (Γ) Additionally, for any β within H 1/2,k,κ (Γ), the inequality π + W β 1,k−2,κ;G + + π − W β 1,k−2,κ;G − is bounded above by cβ 1/2,k,κ;Γ.

Proof The assertion follows from the equalities

It is now easy to see that (4.16) and (4.17) follow from (4.18) and (4.19), respectively.

We write N p =T p + W p + =T p − W p − and use the above procedure to deﬁne the operators ˆN ,Nˆ −1 andN, N −1

4.8 Theorem For any κ > 0 and k ∈ R, the operator N is continuous and injective from H 1/2,k,κ (Γ) to H −1/2,k−3,κ (Γ), and its range is dense in

H −1/2,k−3,κ (Γ) Its inverseN −1 , extended by continuity from the range ofN to H −1/2,k,κ (Γ), is continuous and injective from the space H −1/2,k,κ (Γ) to

H 1/2,k−1,κ (Γ)for anyk∈R, and its range is dense inH 1/2,k−1,κ (Γ).

Proof The assertion concerning N follows from Theorems 3.2 and 4.6 Al- ternatively, it can be established from the estimate

N p βˆ −1/2,p;∂S ≤c|p| 3 βˆ 1/2,p;∂S , which is obtained from (3.3) and (4.14).

Let ˆβ ∈H 1/2,p (∂S) andp∈C¯ κ By (4.18), (4.19), and the trace theorem, βˆ 2 1/2,p;∂S =W p − βˆ−W p + βˆ 2 1/2,p;∂S

This inequality and Theorems 3.2 and 4.6 complete the proof.

Nonstationary Boundary Integral Equations

Problems (DD ± ) consist in solving

B(∂ t 2 u)(X) + (Au)(X) = 0, X ∈G ± , u(x,0+) = (∂ t u)(x,0+) = 0, x∈S ± , u ± (X) =f(X), X ∈Γ, where f is prescribed on Γ We seek their solutions in the form u(X) = (V α)(X), X ∈G + or X∈G − , (4.20) where α is an unknown density deﬁned on∂S×R, which is zero for t 0, k ∈ R, systems (4.25) and (4.26) have unique solutions α∈H −1/2,k−2,κ (Γ) andβ ∈H 1/2,k−1,κ (Γ), in which case the functions u deﬁned by (4.20)or (4.22)belong to the space

H 1,k−1,κ (G ± ) Ifk≥1, then these functions are the weak solutions of problems (DN ± ), respectively.

The proof of this assertion is a repeat of that of Theorem 4.9, use being made of Theorems 3.2, 4.3, and 4.8.

The Direct Method

This article introduces the dynamic analog of the third Green’s formula, also referred to as the Somigliana representation formula in elasticity theory We consider \( u \in H^{1,k,\kappa}(G^+) \) with \( k \in \mathbb{R} \) and \( \kappa > 0 \) as the weak solution to the problem (DD+).

(4.27) where f ∈H 1/2,k,κ (Γ) Then for any p∈C κ , ˆu(x, p) =Lu(x, t) is the weak solution of the problem

(γ + u)(x, p) = ˆˆ f(x, p), x∈∂S, (4.28) where ˆf ∈H 1/2,p (∂S) We now consider the function w=π + V p T p + fˆ−π + W p f ,ˆ which belongs toH 1,p (S + ) Given that γ + w=V p,0 T p + fˆ−W p + fˆ=V p,0 (T p + − T p − ) ˆf = ˆf , w is also a weak solution of (4.28) Since (4.28) has a unique solution in

Returning to the spaces of originals, we obtain u(X) = (π + VT + f)(X)−(π + W f)(X), X ∈G + (4.29)

From this it follows that f =V 0 T + f −W + f Setting α=T + f and taking into account thatf+W + f =W − f, we arrive at

This nonstationary boundary integral equation for the densityαhas a direct physical meaning, sinceαis the boundary moment-force ﬁeld Solving (4.30), we obtain the solution of (4.27) as u(X) = (π + V α)(X)−(π + W f)(X), X∈G +

In the case of the exterior problem (DD − ), it is not diﬃcult to show that representation (4.29) takes the form u(X) =−(π − VT − f)(X) + (π − W f)(X), X ∈G − , (4.31) which leads to the system of boundary integral equations

4.11 Theorem For anyf ∈H 1/2,k,κ (Γ),κ >0,k∈R, systems(4.30)and

(4.32)have unique solutionsα∈H −1/2,k−1,κ (Γ), in which case u=±(π ± V α−π − W f)∈H 1,k−1,κ (G ± ).

If k ≥1, then these functions u are the weak solutions of problems (DD ± ), respectively.

The assertion is proved by using the representations α = T ± f and the properties of the boundary operators deﬁned in§4.1.

Going over to (DN ± ), we remark that, since here the boundary moment- force vector T ± γ ± u = g is known, we write the representations (4.29) and (4.31) as u(X) =±

, X ∈G ± , (4.33) where β = γ ± uis the unknown displacement ﬁeld on the boundary Since γ ± u=±(V 0 g−W ± β), we arrive at the systems of boundary integral equations

4.12 Theorem For any g ∈ H −1/2,k,κ (Γ), κ > 0, k ∈ R, systems (4.34) have unique solutionsβ ∈H 1/2,k−1,κ (Γ), in which case the functionsudeﬁned by (4.33)belong to H 1,k−1,κ (G ± ) Ifk≥1, then these functions are the weak solutions of problems (DN ± ), respectively.

The assertion follows from the representationsβ= (T ± ) −1 gand the properties of the boundary operators discussed above.

Inﬁnite Plate with a Finite Inclusion

In this study, we consider the middle plane domains S + and S − of plates characterized by distinct Lamé constants (λ +, à + for S + and λ −, à − for S −), varying densities (ρ + for S + and ρ − for S −), and different thickness parameters (h + for S + and h − for S −) The corresponding diagonal matrices are denoted as B + and B − Additionally, we define A ±, a ±, and T ± as the matrix differential operators, internal energy bilinear forms, and boundary moment-force operators associated with the respective domains S ±.

Ifu∈H 1,k,κ (R 2 ×(0,∞)), then we make the notation u ± =π ± u∈H 1,k,κ (G ± ), u={u + , u − }.

Similarly, for the Laplace transform ˆu∈H 1,p (R 2 ) ofuwe write ˆ u={uˆ + ,uˆ − }, uˆ ± =π ± uˆ∈H 1,p (S ± ).

In§1.1 we stated that the classical transmission (contact) problem (DT) consists in ﬁnding u + ∈C 2 (G + )∩C 1 ( ¯G + ), u − ∈C 2 (G − )∩C 1 ( ¯G − ) satisfying

Letv ∈ C 0 ∞ (R 2 ×[0,∞)), and let v ± =π ± v Multiplying the equations of motion for the two plates by v + ∗ and v ∗ − , integrating over G + and G − , respectively, and adding the resulting equalities, we arrive at the variational equation

(g, v 0 ) 0;∂S dt ∀v∈C 0 ∞ (R 2 ×[0,∞)), (5.2) where v 0 =v + + =v − − Equality (5.2) suggests that the variational version of problem (DT) in the more general case of the nonhomogeneous equation of motion should consist in ﬁndingu={u + , u − },u ± ∈H 1,0,κ (G ± ), such that ∞

(q, v) 0 + (g, v 0 ) 0;∂S dt ∀v∈C 0 ∞ (R 2 ×[0,∞)), (5.3) γ + u + −γ − u − =f, where q ∈H −1,k,κ (R 2 ×(0,∞)), f ∈H 1/2,k,κ (Γ), and g ∈H −1/2,k,κ (Γ) are prescribed.

5.1 Theorem For any q ∈ H −1,1,κ (R 2 ×(0,∞)), f ∈ H 1/2,1,κ (Γ), and g ∈H −1/2,1,κ (Γ),κ > 0, problem (5.3)has a unique solution u={u + , u − }, where u ± ∈H 1,0,κ (G ± ) If q ∈H −1,k,κ (R 2 ×(0,∞)), f ∈ H 1/2,k,κ (Γ), and g∈H −1/2,k,κ (Γ),k∈R, thenu ± ∈H 1,k−1,κ (G ± )and u + 1,k−1,κ;G + +u − 1,k−1,κ;G −

Proof We begin by rewriting (5.3) in terms of Laplace transforms Then, with the usual notation ˆ u=Lu, qˆ=Lq, fˆ=Lf, ˆg=Lg,

(DT) turns into the problem of finding û∈H 1,p (R 2 ), where ˆ u(x, p) ={uˆ + (x, p),uˆ − (x, p)}, û ± ∈H 1,p (S ± ), p∈C κ , such that a + (û + , v + ) +a − (û − , v − )

Here, v ± =π ± v andγvis the trace of von∂S.

First, we consider the case ˆf = 0 Then (5.5) reduces to the problem a + (ˆu + , v + ) +a − (ˆu − , v − )

Since ˆg∈H −1/2,p (∂S), the form (ˆg, γv) 0;∂S deﬁnes a bounded antilinear (conjugate linear) functional onH 1,p (R 2 ); hence, it can be expressed as

WritingQ= ˆq+ ˜q, we arrive at the problem a + (ˆu + , v + ) +a − (ˆu − , v − )

= (Q, v) 0 ∀v∈H 1,p (R 2 ) (5.7) Next, we introduce a bilinear forma κ (u, v) on [H 1,p (R 2 )] 2 by a κ (u, v) = 1 2 κ 2

It is obvious thata κ (u, v) is continuous and coercive on [H 1,p (R 2 )] 2 ; therefore, it generates a self-adjoint nonnegative operator A κ through the equality

A κ is a homeomorphism fromH 1,p (R 2 ) toH −1,p (R 2 ) We can now write (5.7) in the form

A κ uˆ+ p 2 − 1 2 κ 2 B˜uˆ=Q, (5.8) where ˜Buˆ= (B + uˆ + , B − uˆ − )∈H −1,p (R 2 ) In turn, (5.8) can be written as ˆ u+ p 2 − 1 2 κ 2

The equation B κ u b = ˜B 1/2 A −1 κ Q demonstrates that, when analyzed within the context of H 1,p (R 2), it is equivalent to its representation in L 2 (R 2) Additionally, it is established that B κ is self-adjoint and nonnegative in L 2 (R 2), ensuring the unique solvability of the equation As a result, both (5.7) and (5.6) possess a unique solution ˆu in H 1,p (R 2) for any given Q in H −1,p (R 2).

Taking v = ˆu in (5.7) and separating the real and imaginary parts, we obtain uˆ 2 1,p ≤c|p||(Q,u)ˆ 0 |; therefore, uˆ 1,p ≤c|p|Q −1,p ≤c|p|(qˆ −1,p + ˆg −1/2,p;∂S ) (5.10)

In the general case we construct w + =l + fˆ∈H 1,p (S + ), w={w + ,0}, then seek a solution of (5.5) of the form ˆu =w+ ˜u Clearly, ˜u∈ H 1,p (R 2 ) satisﬁes a + (˜u + , v + ) +a − (˜u − , v − )

Since the extension operator l + is continuous (uniformly with respect to the parameter p∈C) from H 1/2,p (∂S) toH 1,p (S + ), we have l + fˆ 1,p;S + ≤cfˆ 1/2,p;∂S , which leads to

Therefore, (5.11) has a solution ˜u∈H 1,p (R 2 ) and, by (5.10), u˜ 1,p ≤c|p|(qˆ −1,p +fˆ 1/2,p;∂S +gˆ −1/2,p;∂S ).

This implies that (5.5) also has a solution ˆu, which satisﬁes uˆ + 1,p;S + +uˆ − 1,p;S −

We now use (5.12) to show successively that ˆU ={Uˆ + ,Uˆ − }, where ˆU ± (p) ˆ u ± (ã, p), is bounded, continuous, and holomorphic from C κ to H 1 (S + )ì

H 1 (S − ) This and (5.12) prove that problem (DT) has a solution, which satisﬁes estimate (5.4).

To show that this solution is unique, suppose thatu∈H 1,0,κ (R 2 ×(0,∞)) satisﬁes

Repeating the relevant part of the proof of Theorem 2.3 withL 2 (S + ), ˚H 1 (S + ), and a + (u, v) replaced byL 2 (R 2 ), H 1 (R 2 ), and a + (u + , v + ) +a − (u − , v − ), we arrive at the desired conclusion.

We have confirmed the existence of a unique weak solution to our problem and now focus on expressing this solution through various combinations of time-dependent (retarded) single-layer and double-layer potentials Additionally, we demonstrate the unique solvability of the associated systems of nonstationary boundary integral equations.

Let V ± α ± and W ± β ± be the retarded single-layer and double-layer potentials constructed for the plates corresponding to S ± , respectively The boundary operators generated by these potentials are denoted by V +,0 , V −,0 ,

W + + , W − − , andN + , N − The Poincar´e–Steklov operators forS ± are denoted by T ± Similar notation is used for their Laplace transform counterparts, with the only diﬀerence that the latter carry a subscriptp.

First we represent the (weak) solutionu={u + , u − }of (5.1) in the form u + (X) = (V + α + )(X), X ∈G + , u − (X) = (V − α − )(X), X ∈G − , (5.13) where α ± are unknown densities deﬁned on Γ This yields the system of nonstationary boundary equations

(5.15) where β ± are unknown densities deﬁned on Γ Representation (5.15) yields the system of boundary integral equations

In the third case we seek the solution as u + (X) = (V + α + )(X), X ∈G + , u − (X) = (W − β − )(X), X ∈G − ,

(5.17) which leads to the system

Finally, seeking the solution in the form u + (X) = (W + β + )(X), X ∈G + , u − (X) = (V − α − )(X), X ∈G − ,

(5.19) we arrive at the system

5.2 Theorem For any f ∈ H 1/2,k,κ (Γ) and g ∈ H −1/2,k,κ (Γ), κ > 0, systems(5.14), (5.16), (5.18), and(5.20) have unique solutions α ± ∈H −1/2,k−2,κ (Γ), β ± ∈H 1/2,k−2,κ (Γ).

In each case, the corresponding functionudeﬁned by(5.13), (5.15), (5.17), or (5.19)belongs toH 1,k−1,κ (G + )×H 1,k−1,κ (G − ) Ifk≥1, thenuis the(weak) solution of problem(DT).

Proof We rewrite all the systems of boundary equations in terms of theirLaplace transforms For p ∈ C κ , systems (5.14), (5.16), (5.18), and (5.20) take, respectively, the form

Next, we consider the solution ˆu(x, p) = {uˆ + (x, p),uˆ − (x, p)} of problem (5.5) for anyp∈C κ In§5.1 it was shown that uˆ + 1,p;S + +uˆ − 1,p;S − ≤c|p| fˆ 1/2,p;∂S + ˆg −1/2,p;∂S

We write ˆf ± =γ ± uˆ ± By the trace theorem and (5.25), fˆ + 1/2,p;∂S +fˆ − 1/2,p;∂S

We also write ˆg + =T p,+ + fˆ + and ˆg − =T p,− − fˆ − By (5.25) and (3.5), gˆ + −1/2,p;∂S + ˆg − −1/2,p;∂S

We now consider system (5.21) Let ˆα ± = V p,± −1 fˆ ± Then (5.26) shows that αˆ + −1/2,p;∂S +αˆ − −1/2,p;∂S

Clearly, ˆu + = V p,+ αˆ + and ˆu − = V p,− αˆ − (hence, also ˆα + and ˆα − ) satisfy(5.21) The assertion concerning (5.14) follows from (5.28) and Theorem 5.1.

In the case of (5.22), we take ˆβ + =N p,+ −1 gˆ + and ˆβ − =N p,− −1 gˆ − From the properties of the operators (N p,± ) −1 and (5.27) it follows that βˆ + 1/2,p;∂S +βˆ − 1/2,p;∂S

The desired statement now follows from (5.29) and Theorem 5.1.

For systems (5.23) and (5.24) we set ˆ α + =V p,+,0 −1 fˆ + , βˆ − =N p,− −1 ˆg − , βˆ + =N p,+ −1 gˆ + , αˆ − =V p,−,0 −1 fˆ − , respectively The proof is completed by repeating the above arguments, with the obvious changes.

Multiply Connected Finite Plate

Consider a plate with Lam´e constants λ + , à + and density ρ + , and characterized by the thickness parameterh + and a middle plane domainS + whose boundary ∂S consists of two simple closedC 2 -curves∂S 1 and∂S 2 such that

The boundary ∂S 1 is completely contained within the finite area defined by ∂S 2 The outward-facing unit normals on both boundaries, ∂S 1 and ∂S 2, are oriented away from the region S+ The interior region defined by ∂S 1 is referred to as S i −, while the area outside of ∂S 2 is designated as S e −.

Fork∈R,κ >0, andp∈C0, we deﬁne the spaces

Iff ={f 1 , f 2 }andϕ={ϕ 1 , ϕ 2 }are elements of

, then theirL 2 (∂S)- inner product is deﬁned as

The trace operator γ + = {γ 1 + , γ 2 + } continuously maps functions from the Sobolev space H 1,p (S + ) to H 1/2,p (∂S), allowing us to express the trace of a function u ∈ H 1,p (S + ) as γ + u = {γ 1 + u, γ 2 + u} Following the notation established in §1.1, the trace on the boundary Γ is also denoted by γ + u The operator γ + is continuous for any k ∈ R, mapping from H 1,k,κ (G + ) to H 1/2,k,κ (Γ) Additionally, we define extension operators l ν + for ν = 1,2, which ensure that for any f ν ∈ H 1/2,p (∂S ν ), the extended function l + ν f ν belongs to H 1,p (S + ), with the condition γ 3−ν + l + ν f ν = 0 for ν = 1,2.

If f ={f 1 , f 2 } ∈ H 1/2,p (∂S), then we set l + f = l 1 + f 1 +l 2 + f 2 It is obvious thatγ + l + f =f and thatl + is continuous (uniformly with respect top) from

We now formulate three initial-boundary value problems for the plate with middle section S + The classical problem (DMCD) consists in ﬁnding u∈

B + (∂ t 2 u)(X) + (A + u)(X) = 0, X∈G + , u(x,0+) = (∂ t u)(x,0+) = 0, x∈S + , u + (X) =f 1 (X), X ∈Γ 1 , u + (X) =f 2 (X), X ∈Γ 2 , where f 1 and f 2 are prescribed Consequently, in the variational version of (DMCD) we seeku∈H 1,0,κ (G + ) satisfying

∀v∈C 0 ∞ ( ¯G + ), v + = 0, γ + u=f, where a + andB + have the obvious meaning.

The classical problem (DMCN) consists in ﬁndingu∈C 2 (G + )∩C 1 ( ¯G + ) such that

(T + u) + (X) =g 2 (X), X ∈Γ 2 , where g 1 andg 2 are prescribed In its corresponding variational version, we seeku∈H 1,0,κ (G + ) satisfying

The classical problem (DMCM) consists in ﬁndingu∈C 2 (G + )∩C 1 ( ¯G + ) such that

(T + u) + (X) =g 1 (X), X ∈Γ 1 , u + (X) =f 2 (X), X ∈Γ 2 , whereg 1 andf 2 are prescribed In the variational version of this problem, we seeku∈H 1,0,κ (G + ) satisfying

5.3 Theorem For any f ν ∈H 1/2,1,κ (Γ ν )andg ν ∈H −1/2,1,κ (Γ ν ),ν = 1,2, κ > 0, problems (DMCD), (DMCN), and (DMCM) have unique solutions u ∈H 1,0,κ (G + ) If f ν ∈ H 1/2,k,κ (Γ ν ) and g ν ∈H −1/2,k,κ (Γ ν ), k ∈ R, then each of these solutions usatisﬁes the corresponding estimate u 1,k−1,κ;G + ≤c f 1 1/2,k,κ;Γ 1 +f 2 1/2,k,κ;Γ 2

Proof Going over to Laplace transforms in problems (DMCD), (DMCN), and (DMCM), we arrive at new problems for everyp∈C κ Thus, in (MCD p ), we seek ˆu∈H 1,p (S + ) such that p 2 (B + 1/2 u, Bˆ 1/2 + v) 0;S ++a + (ˆu, v) = 0 ∀v∈H˚ 1,p (S + ), γ + uˆ= ˆf

In (MCN p ), we seek ˆu∈H 1,p (S + ) such that p 2 (B 1/2 + u, Bˆ 1/2 + v) 0;S ++a + (ˆu, v)

= (ˆg, γ + v) 0;∂S ∀v∈H 1,p (S + ), (5.31) where γ + v={γ 1 + v, γ 2 + v} is the trace ofv on∂S.

Finally, in (MCM p ), we seek ˆu∈H 1,p (S + ) such that p 2 (B 1/2 + u, Bˆ 1/2 + v) 0;S ++a + (ˆu, v)

= (ˆg 1 , γ 1 + v) 0;∂S 1 ∀v∈H˚ 1,p (S + , ∂S 1 ), (5.32) γ + 2 uˆ= ˆf 2 , where ˚H 1,p (S + , ∂S 1 ) is the subspace of alluinH 1,p (S + ) such thatγ 2 + u= 0.

We represent the solution of (MCD p ) in the form ˆu = w+l + fˆ It is obvious that w∈H˚ 1,p (S + ) satisﬁes p 2 (B 1/2 + w, B + 1/2 v) 0;S ++a + (w, v)

= (Q 1 , v) 0;S + ∀v∈H˚ 1,p (S + ), (5.33) where Q 1 ∈H −1,p (S + ) is deﬁned forv∈H˚ 1,p (S + ) by

The unique solvability of (5.33) and the estimates w 1,p ≤c|p|Q 1 −1,p;S + ≤c|p|fˆ 1/2,p;∂S are established by the method used in the proof of Theorem 2.1 Therefore, (5.30) is uniquely solvable in H 1,p (S − ) and ˆu 1,p;S + ≤c|p|fˆ 1/2,p;∂S (5.34)

In the case of equation (5.31) we remark that for ˆg ∈ H −1/2,p (∂S), the form (ˆg, γ + v) 0;∂S deﬁnes a bounded antilinear (conjugate linear) functional on the space H 1,p (S + ), so it can be written as

≤cgˆ −1/2,p;∂S v 1,p;S + , it follows that Q 2 ∈H˚ −1,p (S + ) satisﬁes

We then write (5.31) in the form a +,κ (ˆu, v) + (p 2 − 1 2 κ 2 )(B + 1/2 u, Bˆ + 1/2 v) 0;S +

= (Q 2 , v) 0;S + ∀v∈H 1,p (S + ), (5.35) where the bilinear form a +,κ is deﬁned on

It is obvious that a +,κ is continuous and coercive on

; hence, it generates an operator A κ that is a homeomorphism from H 1,p (S + ) to

H˚ −1,p (S + ), by means of the equality

We now rewrite (5.35) in the form

Applying the Fredholm Alternative, we prove the unique solvability of (5.36)— hence, of (5.31)—in H 1,p (S + ) for any ˆg ∈ H −1/2,p (∂S), p∈C κ Separating the real and imaginary parts in (5.35), we obtain uˆ 1,p;S + ≤c|p|Q 2 −1,p ≤c|p|gˆ −1/2,p;∂S (5.37)

The solution of (5.32) is sought in the form ˆu = w+l + 2 fˆ 2 , where w ∈

H˚ 1,p (S + , ∂S 1 ) Let ˚H −1,p (S + , ∂S 2 ) be the dual of ˚H 1,p (S + , ∂S 1 ) with respect to the duality generated by the inner product in L 2 (S + ) It is obvious that

H˚ −1,p (S + , ∂S 2 ) is the subspace ofH −1,p (S + ∪S¯ i − ) that consists of allqwith suppq⊂S¯ + We denote the norm ofq∈H˚ −1,p (S + , ∂S 2 ) byq −1,p;S + ∪ S ¯ − i

We see that (ˆg 1 , γ + 1 v) 0;∂S 1 deﬁnes a bounded antilinear (conjugate linear) functional on ˚H 1,p (S + , ∂S 1 ); hence, it can be expressed as

∀v∈H˚ 1,p (S + , ∂S 1 ) (5.38) Since p 2 (B + 1/2 l + 2 fˆ 2 , B + 1/2 v) 0;S ++a + (l + 2 fˆ 2 , v) = (q, v) 0;S + , where q∈H −1,p (S + , ∂S 2 ) and q −1,p;S + ∪ S ¯ − i ≤cfˆ 2 1/2,p;∂S 2 , we make the notation ˜Q 3 −q=Q 3 ∈H˚ −1,p (S + , ∂S 2 ) and see that

Taking into account the continuity and coerciveness of the form a + (u, v) on [ ˚H 1,p (S + , ∂S 1 )] 2 , we adapt the proof of Theorem 2.1 to our situation and conclude that (5.39) has a unique solution w∈H˚ 1,p (S + , ∂S 1 ) and that w 1,p;S + ∪ S ¯ i − ≤c|p| fˆ 2 1/2,p;∂S 2 +gˆ 1 −1/2,p;∂S 1 ).

Consequently, (5.32) has a unique solution ˆu∈H 1,p (S + ) and uˆ 1,p;S + ≤c|p| fˆ 2 1/2,p;∂S 2+gˆ 1 −1/2,p;∂S 1

To arrive at the desired assertion, we use (5.34), (5.37), and (5.40), and repeat the arguments in the proof of Theorem 2.3 with the obvious changes.

In this article, we develop an algebra of boundary operators specifically for multiply connected plates, utilizing Sobolev spaces with a parameter This approach allows for straightforward formulation of the related assertions in spaces of originals We denote this algebra using the notation a −,i (u, v).

E(u, v)dx, where the bilinear formE is for the plate with the Lam´e coeﬃcients λ + and à +

Let \( f = \{f_1, f_2\} \) and \( \phi = \{\phi_1, \phi_2\} \) be elements of \( H^{1/2,p}(\partial S) \) The solution \( u^+ \in H^{1,p}(S^+) \) satisfies the boundary condition defined by \( f \) Additionally, \( u(i)^- \in H^{1,p}(S_i^-) \) and \( u(e)^- \in H^{1,p}(S_e^-) \) meet specific conditions involving bilinear forms and trace operators \( \gamma_1^- \) and \( \gamma_2^- \) that map to \( H^{1/2,p}(\partial S_1) \) and \( H^{1/2,p}(\partial S_2) \), respectively Furthermore, the elements \( v(i)^- \) and \( v(e)^- \) in \( H^{1,p}(S_i^-) \) and \( H^{1,p}(S_e^-) \) are defined such that their traces correspond to \( \phi_1 \) and \( \phi_2 \) The Poincaré–Steklov operators \( T_p^\pm \) are introduced for further analysis.

5.4 Lemma For anyp∈C0 , the operators T p ± are homeomorphisms from

Proof The assertion concerning (5.42) and (5.43) for T − has already been proved in Lemma 3.1 Takingϕ=f andv + =u + in (5.41), we obtain p 2 B + 1/2 u + 2 0;S ++a + (u + , u + ) = (T p + f, f) 0;∂S (5.44) Separating the real and imaginary parts in (5.44), we arrive at

=σ −1 Re{p(¯T p + f, f) 0;∂S }, p=σ+iτ, and then, ﬁnally, at u + 2 1,p;S + ≤c|p||(T p + f, f) 0;∂S | ∀p∈C¯ κ

The rest of the argument is a verbatim repeat of the proof of Lemma 3.1.

We deﬁne the single-layer potential V p αˆ of density ˆα ∈ H −1/2,p (∂S), ˆ α={αˆ 1 ,αˆ 2 }, by setting

D(xˆ −y, p) ˆα ν (y, p)ds y , ν= 1,2 (not summed), and deﬁne a boundary operator V p,0 by

The second equality above can also be written as

5.5 Lemma For any p∈C0 , the operator V p,0 is a homeomorphism from

Proof We remark that, as follows from the properties of single-layer potentials, for any ˆα∈H −1/2,p (∂S) we have the jump formula

The argument now continues as in the proof of Lemma 4.1.

Letπ − i andπ e − be the operators of restriction fromR 2 (orR 2 \∂S) toS − i and S e − , respectively We deﬁne the double-layer potential W p βˆ of density βˆ∈H 1/2,p (∂S), ˆβ ={βˆ 1 ,βˆ 2 }, by

Pˆ(x−y, p) ˆβ ν (y, p)ds y , ν= 1,2 (not summed), and the boundary operators W p ± of its limiting values on∂Sby

W p ± βˆ=γ ± (W p βˆ), where, for a function udeﬁned onR 2 (or onR 2 \∂S), γ + u={γ 1 + π + u, γ 2 + π + u}, γ − u={γ 1 − π i − u, γ 2 − π e − u}.

5.6 Lemma For any p∈C 0 , the operatorsW p ± are homeomorphisms from

W p ± βˆ 1/2,p;∂S ≤c|p| 2 βˆ 1/2,p;∂S , βˆ 1/2,p;∂S ≤c|p| 2 W p ± βˆ 1/2,p;∂S Proof The assertion follows from the equalities

(V p T p − β)(x, p),ˆ x∈S + , (V p T p + β)(x, p),ˆ x∈S − i ∪S e − , which are proved as in Lemmas 4.4, 5.4, and 5.5.

We now deﬁne an operatorN p by setting

5.7 Lemma For every p∈C0 , the operator N p is a homeomorphism from

(5.46)This assertion is proved in exactly the same way as Theorem 4.8.

We deﬁne a pair of boundary operatorsχ 1 andχ 2 by χ 1 f ={g 1 , f 2 } ∈H −1/2,p (∂S 1 )×H 1/2,p (∂S 2 ), χ 2 g={g 1 , f 2 } ∈H −1/2,p (∂S 1 )×H 1/2,p (∂S 2 ).

5.8 Lemma For any p ∈ C 0 , the operator χ 1 is a homeomorphism from

H 1/2,p (∂S)toH −1/2,p (∂S 1 )×H 1/2,p (∂S 2 ), whileχ 2 is a homeomorphism from

Proof The continuity of χ ν , ν = 1,2, follows from Lemma 5.4 Estimates (5.47) and (5.49) follow from (5.42) and (5.43), respectively Let {g 1 , f 2 } ∈

H −1/2,p (∂S 1 )×H 1/2,p (∂S 2 ), and letu∈H 1,p (S + ) be the (unique) solution of (5.32) with boundary data {g 1 , f 2 } It is obvious thatf ={γ 1 + u, f 2 }satisﬁes χ 1 f = {g 1 , f 2 } Since (5.48) follows from (5.40), the statement for χ 1 is proved.

The function \( g = T_p + f \) satisfies \( \chi^2 g = \{ g_1, f_2 \} \), indicating that \( \chi^2 \) is a homeomorphism from \( H^{-1/2,p}(\partial S) \) to \( H^{-1/2,p}(\partial S_1) \times H^{1/2,p}(\partial S_2) \) To finalize the proof, it is essential to demonstrate that equation (5.50) is valid Let \( u \) represent the solution of equation (5.32) with boundary data \( \{ g_1, f_2 \} \), where \( f = \gamma_1 + u \) and \( g = T_p + f \) This definition of the operators \( T_p + \) is crucial for establishing the relationship between the functions involved.

|(T p + f, ϕ) 0;∂S | ≤cu 1,p;S + ϕ 1/2,p;∂S ; hence, g −1/2,p;∂S ≤cu 1,p;S + , and (5.50) follows from (5.40).

We introduce the time-dependent (or retarded) single-layer and double- layer potentialsV αandW βand their corresponding boundary operatorsV 0 ,

W ± ,T ± , andN in the spaces of originals in the usual way, and represent the solutionuof problem (DMCD) in the form u(X) = (V α)(X), X ∈G + , (5.51) or u(X) = (W β)(X), X ∈G + (5.52)

Representations (5.51) and (5.52) yield, respectively, the systems of nonstationary boundary integral equations

5.9 Theorem For any f ∈H 1/2,k,κ (Γ),κ > 0, k∈R, systems (5.53) and

(5.54) have unique solutions α∈H −1/2,k−1,κ (Γ) and β ∈ H 1/2,k−2,κ (Γ), respectively, in which caseudeﬁned by(5.51)or(5.52)belongs toH 1,k−1,κ (G + ).

If k≥1, thenuis the weak solution of problem(DMCD).

This assertion is proved by repeating the arguments in Theorem 4.9 and using Lemmas 5.5 and 5.6.

If we seek the solutionuof (DMCN) in the form (5.51) or (5.52), then we arrive at the systems of boundary equations

5.10 Theorem For any g ∈ H −1/2,k,κ (Γ), κ > 0, k ∈ R, systems (5.55) and (5.56) have unique solutions α∈H −1/2,k−2,κ (Γ) and β ∈H 1/2,k−1κ (Γ), in which caseudeﬁned by(5.51)or(5.52)belongs toH 1,k−1,κ (G + ) Ifk≥1, then uis the weak solution of problem(DMCN).

This assertion is proved just like Theorem 4.10, use being made of Lemmas 5.4, 5.5, and 5.7.

Finally, we seek the solution uof (DMCM) in the form (5.51) or (5.52) and arrive at the systems of boundary equations

5.11 Theorem For any {g 1 , f 2 } ∈ H −1/2,k,κ (Γ)×H 1/2,k,κ (Γ), κ > 0, k ∈ R, systems (5.57) and (5.58) have unique solutions α∈ H −1/2,k−2,κ (Γ) andβ ∈H 1/2,k−2,κ (Γ), in which case udeﬁned by(5.51) or(5.52) belongs to

H 1,k−1,κ (G + ) Ifk≥1, thenuis the weak solution of problem(DMCM).

Proof In terms of Laplace transforms, (5.57) and (5.58) become

By Lemmas 5.5, 5.7, and 5.8, systems (5.59) and (5.60) have unique solutions ˆ α=V p,0 −1 χ −1 1 {gˆ 1 ,fˆ 2 } ∈H −1/2,p (∂S) (5.61) and βˆ=N p −1 χ −1 2 {ˆg 1 ,fˆ 2 } ∈H 1/2,p (∂S) (5.62)

In turn, by (5.61), (5.62), (5.45), (5.46), (5.48), and (5.50), αˆ −1/2,p;∂S ≤c|p| 2 {ˆg 1 −1/2,p;∂S +fˆ 2 1/2,p;∂S }, (5.63) βˆ 1/2,p;∂S ≤c|p| 2 {gˆ 1 −1/2,p;∂S +fˆ 2 1/2,p;∂S } (5.64)The rest of the proof now follows from (5.63), (5.64), and Theorem 5.3.

Finite Plate with an Inclusion

Keeping the notation used in §5.2, below we consider an initial-boundary value problem for a piecewise homogeneous ﬁnite plate occupying the domain

The equation S = ¯S i − ∪S + simplifies to S − for notation purposes We define the plates with middle sections S + and S −, characterized by Lamé constants λ +, à + and λ −, à −, as well as densities ρ +, ρ − and thickness parameters h +, h − If u is a function defined on S, it can be expressed as u = {u +, u −}, where u ± represent the restrictions of u from S to S ± This notation is similarly applied when u is defined within the domain G.

Let (DTD) be the initial-boundary value problem with transmission conditions across ∂S 1 and Dirichlet condition on∂S 2 which consists in ﬁnding u + ∈C 2 (G + )∩C 1 ( ¯G + ), u − ∈C 2 (G − )∩C 1 ( ¯G − ) such that

Proceeding as in §5.1, we readily see that in the variational version of (DTD) we seek u={u + , u − },u ± ∈H 1,0,κ (G ± ), satisfying

(5.65) γ + 1 u + −γ 1 − u − =f 1 , γ + 2 u + =f 2 , where the trace operatorsγ 1 ± andγ + 2 have the obvious meaning,v 0 = (v + + ) 1 v − − , andf ={f 1 , f 2 } ∈H 1/2,k,κ (Γ) andg 1 ∈H −1/2,k,κ (Γ 1 ) are prescribed.

5.12 Theorem For any f ∈ H 1/2,1,κ (Γ) and g 1 ∈ H −1/2,1,κ (Γ 1 ), κ >0, problem (5.65)has a unique solution u={u + , u − }, where u ± ∈H 1,0,κ (G ± ).

The assertion is proved exactly as Theorem 5.1, withH 1,p (R 2 ) replaced by ˚H 1,p (S) and (5.12) written for this case, where it takes the form uˆ + 1,p;S + +uˆ − 1,p;S − ≤c|p|{fˆ 1/2,p;∂S +gˆ 1 −1/2,p;∂S 1 } (5.66)

We now represent the solution of (5.65) in terms of single-layer potentials. LetV − α − be the time-dependent (retarded) single-layer potential deﬁned by (V − α − )(X)

The equation D − (x−y, t−τ)α − (y, τ)ds y dτ, where X∈G, incorporates an unknown density α − defined on Γ 1 Additionally, D − represents the matrix of fundamental solutions related to the motion equation for the plate corresponding to S − Furthermore, we examine the single-layer potential V + α + defined within this context.

D + (x−y, t−τ)α + (y, τ)ds y dτ, X∈G + , where α + ={α +,1 , α +,2 }, α +,1 andα +,2 are unknown densities deﬁned on Γ 1 and Γ 2 , respectively, and

D + represents the matrix of fundamental solutions for the plate with a mid-plane section S + The characteristics of V + α + were examined in §5.2 The boundary operators produced by these potentials are referred to as V +,0 and V −,0.

Analogous notationW + + and W − − is used for the double-layer potentials

W + + β + ={(W + + β + ) 1 ,(W + + β + ) 2 } are the boundary operators generated by the limiting values of W + β + and

In our analysis, we denote the Poincaré–Steklov operators as T++ and T−−, while the moment-force boundary value operators derived from double-layer potentials are represented by N+ and N− Additionally, in the context of Laplace transform spaces, all these symbols are further specified with a subscript p.

We seek the solutionu={u + , u − }of (5.65) in the form u + (X) = (V + α + )(X), X ∈G + , u − (X) = (V − α − )(X), X ∈G −

Representation (5.67) yields the system of boundary integral equations

5.13 Theorem For any f ∈ H 1/2,k,κ (Γ) and g 1 ∈ H −1/2,k,κ (Γ 1 ), where κ >0 andk∈R, system(5.68) has a unique solution α + ∈H −1/2,k−2,κ (Γ), α − ∈H −1/2,k−2,κ (Γ 1 ), in which caseudeﬁned by (5.67)belongs to the space

H 1,k−1,κ (G + )×H 1,k−1,κ (G − ) Ifk≥1, thenuis the weak solution of problem

Proof In terms of Laplace transforms, (5.68) takes the form

Next, (5.65) turns into the problem of finding û = {uˆ + ,uˆ − }, where û ± ∈

We recall that the solution ˆuof (5.70) satisﬁes (5.66) We writeγ + 1 uˆ + = ˆf + , γ 1 − uˆ − = ˆf − and solve the systems of equations

From the properties of V p,+,0 and V p,−,0 it follows that (5.71) has a unique solution{αˆ + ,αˆ − } for anyp∈C κ , and that αˆ + −1/2,p;∂S ≤c|p|{fˆ + ,fˆ 2 } 1/2,p;∂S , αˆ − −1/2,p;∂S 1 ≤c|p|fˆ − 1/2,p;∂S 1

The trace theorem and (5.66) imply that αˆ + −1/2,p;∂S ≤c|p|uˆ + 1,p;S +

By noting that {αˆ + ,αˆ − } meets the criteria of equation (5.69) and considering equations (5.72) and (5.73), we finalize the proof using the standard methodology employed in similar assertions found in previous theorems.

If we now seek the solution of (5.65) in the form u + (X) = (W + β + )(X), X ∈G + , u − (X) = (W − β − )(X), X ∈G − ,

(5.74) then we arrive at the system of boundary equations

5.14 Theorem For any f ∈ H 1/2,k,κ (Γ) and g 1 ∈ H −1/2,k,κ (Γ 1 ), where κ > 0 and k ∈ R, system (5.75) has a unique solution β + ∈ H 1/2,k−2,κ (Γ), β − ∈H 1/2,k−2,κ (Γ 1 ), in which case u deﬁned by (5.74) belongs to the space

H 1,k−1,κ (G + )×H 1,k−1,κ (G − ) Ifk≥1, thenuis the weak solution of problem

Proof In terms of Laplace transforms, (5.75) becomes

Once more, let ˆube the solution of (5.70), and let ˆ g + =T p,+ + {fˆ + ,fˆ 2 }, ˆg − =T p,− − fˆ − , where ˆf + =γ 1 + uˆ + and ˆf − =γ 1 − uˆ − It is obvious that {βˆ + ,βˆ − }, where βˆ ± =N p,± −1 ˆg ± , (5.77) is a solution of (5.76) The properties ofN p,± −1 and (5.77) imply that for any p∈C κ , βˆ + 1/2,p;∂S ≤c|p|ˆg + −1/2,p;∂S , βˆ − 1/2,p;∂S 1 ≤c|p|ˆg − −1/2,p;∂S 1

Since ˆg + −1/2,p;∂S ≤cuˆ + 1,p;S + , ˆg − −1/2,p;∂S 1 ≤cuˆ − 1,p;S − , from (5.66) it follows that βˆ + 1/2,p;∂S +βˆ − 1/2,p;∂S 1 ≤c|p| 2 (fˆ 1/2,p;∂S + ˆg 1 −1/2,p;∂S 1 ).The rest of the proof is now completed in the usual way.

Formulation and Solvability of the Problems

Let∂S 0 be an open arc of a simple closedC 2 -curve∂S in R 2 , and let

We denote byπ i ,i= 0,1, the operators of restriction from∂S to∂S i As usual, S ± are the interior and exterior domains bounded by∂S.

The operators of restriction from S (or R 2 ) to S ± , or from G (or R 2 ×

Ifuis a function deﬁned onS, then we write u ± =π ± u, u= (u + , u − ).

Let γ i ± = π i γ ± , i = 0,1, be the operators of trace on ∂S i (or Γ i ) for functions deﬁned onS ± (orG ± ).

H 1,p (S), p∈C, is the space of allu(x, p) = (u + (x, p), u − (x, p)) such that u ± ∈H 1,p (S ± ) andγ + 1 u + =γ 1 − u − This space is equipped with the norm u 1,p;S =u + 1,p;S + +u − 1,p;S −

H˚ −1,p (S) is the dual of H 1,p (S) with respect to the duality generated by the inner product (ã,ã) 0;S in L 2 (S).

H˚ 1,p (S) is the subspace ofH 1,p (R 2 ) that consists of allusuch thatγ 0 + u + γ 0 − u − = 0.

H −1,p (S) is the dual of ˚H 1,p (S) with respect to the duality generated by (ã,ã) 0;S

H˚ 1/2,p (∂S 0 ) is the subspace of all functions f ∈ H 1/2,p (∂S) such that suppf ∈∂S 0

H −1/2,p (∂S 0 ) is the dual of ˚H 1/2,p (∂S 0 ) with respect to the duality generated by the inner product (ã,ã) 0;∂S 0 inL 2 (∂S 0 ).

H 1/2,p (∂S 0 ) is the space of the restrictions ϕ from ∂S to ∂S 0 of all f ∈

H 1/2,p (∂S), equipped with the norm ϕ 1/2,p;∂S 0 = inf f∈H 1/2,p (∂S): π 0 f=ϕ f 1/2,p;∂S

H˚ −1/2,p (∂S 0 ) is the dual ofH 1/2,p (∂S 0 ) with respect to the duality generated by (ã,ã) 0;∂S 0

The spaces H ±1,k,κ (G), H ±1/2,k,κ (Γ 0 ), and ˚H ±1/2,k,κ (Γ 0 ) are defined with k ∈ R and κ > 0, following standard conventions It is important to note that the trace operators γ 0 ± provide a continuous mapping from H 1,k,κ (G ± ) to H 1/2,k,κ (Γ 0 ) Additionally, for a function u = (u + , u − ) in H 1,k,κ (G), the expression γ 0 + u − γ 0 − u is contained within H˚ 1/2,k,κ (Γ 0 ), where γ 0 ± u denotes the application of the trace operators to u ±.

In what follows, we denote bya S (u, v) the energy bilinear form constructed for the plate with middle plane domainS.

In the variational version of the initial-boundary value problem (DKD), we seeku∈H 1,0,κ (G) satisfying

∀v∈C 0 ∞ (S×[0,∞)), (6.1) γ 0 + u=f + , γ 0 − u=f − , where f + andf − are prescribed on Γ 0

6.1 Theorem For any givenf + , f − ∈H 1/2,1,κ (Γ 0 ),κ >0, such thatδf f + −f − ∈H˚ 1/2,1,κ (Γ 0 ), problem (6.1) has a unique solution u∈H 1,0,κ (G).

Proof Going over to Laplace transforms in (6.1), we arrive at a problem that consists in ﬁnding u∈H 1,p (S) such that p 2 (B 1/2 u, B 1/2 v) 0;S +a S (u, v) = 0 ∀v∈H˚ 1,p (S), γ 0 + u=f + , γ 0 − u=f −

In this study, we investigate an auxiliary problem aimed at finding \( w \in H^{\circ}_{1,p}(S) \) that satisfies the equation \( p_2(B_{1/2}w, B_{1/2}v)_{0;S} + a_S(w,v) = (q,v)_{0;S} \) for all \( v \in H^{\circ}_{1,p}(S) \), where \( q \in H^{-1,p}(S) \) The unique solvability of this equation is established using standard arguments Additionally, Theorem 2.1 demonstrates that the form \( a_S(w,v) \) is symmetric, continuous, and coercive on \( [H^{\circ}_{1,p}(S)]^2 \), which leads to the conclusion that it defines a self-adjoint operator \( A \) on \( H^{\circ}_{1,p}(S) \).

(Au, v) 0;S =a S (u, v) ∀u, v∈H˚ 1,p (S), and thatAis a homeomorphism from ˚H 1,p (S) to H −1,p (S) We now rewrite (6.3) as

The equation Aw + 2Bw = q, or ws + p^2Bs = B^(1/2)A^(-1)q, where ws = B^(1/2)w and B = B^(1/2)A^(-1)B^(1/2), is verified to be equivalent in both H^(1,p)(S) and L^2(S) Additionally, B is established as a self-adjoint nonnegative operator on L^2(S) This confirms that both equations (6.4) and (6.3) possess a unique solution w in H^(1,p)(S) for any q in H^(-1,p)(S) By substituting v = w into (6.3) and separating the real and imaginary components, we derive the estimate w_(1,p) ≤ c|p|q_(−1,p;S).

Letl 0 be an extension operator from∂S 0 to∂S, which mapsH 1/2,p (∂S 0 ) continuously toH 1/2,p (∂S); that is, l 0 f + 1/2,p;∂S ≤cf + 1/2,p;∂S 0 ∀f + ∈H 1/2,p (∂S 0 ).

Let F + =l 0 f + , and letF − be an extension off − from ∂S 0 to ∂S such that π 1 F + =π 1 F − We take u 0 = (l + F + , l − F − )∈H 1,p (S) and seek a solutionu of (6.2) in the form u=u 0 +w Clearly,w∈H˚ 1,p (S) satisﬁes p 2 (B 1/2 w, B 1/2 v) 0;S +a S (w, v)

≤c f + 1/2,p;∂S 0 +δf 1/2,p;∂S v 1,p , we can write the right-hand side in (6.6) as (q, v) 0;S , whereq∈H −1,p (S) and q −1,p;S ≤c f + 1/2,p;∂S 0 +δf 1/2,p;∂S

Therefore, (6.2) has a unique solutionu∈H 1,p (S) and, as follows from (6.5), u 1,p;S ≤c|p| f + 1/2,p;∂S 0 +δf 1/2,p;∂S

Using (6.7), we complete the proof in the standard way.

Let C₀^∞(¯G) denote the space of functions with compact support in ¯G that are members of both C₀^∞(¯G⁺) and C₀^∞(¯G⁻) These functions must satisfy the condition that their limiting values, as well as the limiting values of all their derivatives, coincide on the boundary Γ₁ when approached from the interior of G⁺ and G⁻.

In the variational version of problem (DKN), we seeku∈H 1,0,κ (G) that satisﬁes

6.2 Theorem For any giveng + , g − ∈H −1/2,1,κ (Γ 0 ),κ >0, such thatδg g + −g − ∈H˚ −1/2,1,κ (Γ 0 ), problem(6.8)has a unique solutionu∈H 1,0,κ (G) If g + , g − ∈H −1/2,k,κ (Γ 0 )andδg ∈H˚ −1/2,k,κ (Γ 0 ),k∈R, then u∈H 1,k−1,κ (G) and u 1,k−1,κ;G ≤c δg −1/2,k,κ;Γ +g − −1/2,k,κ;Γ 0

Proof In terms of Laplace transforms, (6.8) becomes p 2 (B 1/2 u, B 1/2 v) 0;S +a S (u, v)

= (g + , γ 0 + v) 0;∂S 0 −(g − , γ − 0 v) 0;∂S 0 ∀v∈H 1,p (S), (6.9) or, which is the same, p 2 (B 1/2 u, B 1/2 v) 0;S +a S (u, v)

= (δg, γ 0 + v) 0;∂S 0 + (g − , γ 0 + v−γ 0 − v) 0;∂S 0 ∀v∈H 1,p (S). Since for anyv∈H 1,p (S), γ 0 + v−γ 0 − v∈H˚ 1/2,p (∂S 0 ), it follows that

≤c δg −1/2,p;∂S +g − −1/2,p;∂S 0 v 1,p;S , so we can write (6.9) as p 2 (B 1/2 u, B 1/2 v) 0;S +a S (u, v) = (Q, v) 0;S ∀v∈H 1,p (S), (6.10) where Q∈H˚ −1,p (S) and

To prove that (6.10) is uniquely solvable, we consider the bilinear form a κ,S (u, v) = 1 2 κ 2 (B 1/2 u, B 1/2 v) 0;S +a S (u, v).

The symmetric, continuous, and coercive nature of the form on [H 1,p (S)] 2 leads to the definition of a self-adjoint operator A κ, which acts as a homeomorphism from H 1,p (S) to ˚H −1,p (S) Utilizing standard arguments, we establish that equation (6.9) has a unique solution in H 1,p (S), and it follows that u 1,p;S is bounded by c|p| δg −1/2,p;∂S + g − −1/2,p;∂S 0.

Taking (6.11) into account, we complete the proof by following the usual procedure.

Tiêu đề	Variational and Potential Methods for a Class of Linear Hyperbolic Evolutionary Processes
Tác giả	Igor Chudinovich, Christian Constanda
Trường học	The University of Guanajuato
Chuyên ngành	Mathematics
Thể loại	monograph
Năm xuất bản	2005
Thành phố	Salamanca

Định dạng
Số trang	152
Dung lượng	1,8 MB