Stability for an Klein–Gordon equation type with a boundary dissipation of fractional derivative type

Abstract

We consider an Klein–Gordon relativistic equation with a boundary dissipation of fractional derivative type. We study of stability of the system using semigroups theory and classical theorems over asymptotic behavior.

Keywords

Semigroup theory Klein–Gordon equation polynomial stability

1. Introduction

The classical works of Klein (1927) and Gordon (1926) derived a relativistic equation for a charged particle in an electromagnetic field. This equation is of conservative dispersive type and has played an important role in the study of elementary particles. In general, the Klein–Gordon equation is used to describe dispersive wave phenomena. In a series of papers [5,6], Mainardi has considered the Klein–Gordon equation with dissipation. These works are relevant for provide interesting examples of normal dispersion (usually met in the absence of dissipation) and anomalous dispersion (always present on viscoelasticity waves). In this work, we consider the following problem $\begin{array}{l} (1.1) & \{\begin{matrix} \frac{\partial^{2} u}{\partial t^{2}} + 2 γ \frac{\partial u}{\partial t} + β^{2} u = c^{2} \frac{\partial^{2} u}{\partial x^{2}}, x \in [0, 1], t > 0, \\ u (x, 0) = u_{0} (x), u_{t} (x, 0) = u_{1} (x) . \end{matrix} \end{array}$ This equation can be thought as a simple prototype of two regimes of dispersion that depend on the relative weight of their two independent parameters γ and β. For $γ = 0$ , the equation is the linear version of the Klein–Gordon equation, and when $β = 0$ . the equation gives the telegraph equation. Moreover, if $0 ⩽ γ < β$ the occurrence of normal dispersion is observed and, the case $0 ⩽ β < γ$ the anomalous dispersion is obtained, in fact, this case turns out to be similar to viscoelastic waves in view of the common anomalous dispersion in their full range of frequencies. If $γ = β$ , this equation corresponds to the distortion-less wave propagation, where there is attenuation with no dispersion. For the special case $β = 0$ , it reproduces the Maxwell model of viscoelasticity.

In [8], Mbodje and Montseny shown that the fractional derivative $\partial_{t}^{α}$ forces the system to become dissipative and the solution approaches the equilibrium state. Hence, when dissipation is applied on the boundary, we can consider them as controllers helping to reduce vibrations.

In this article we are interested in studying the stabilization in the case $γ = 0$ in (1.1). In order to achieve this, we have eliminated the dissipation given in the domain and placed the dissipation on the border. It is importante to note that dissipation imposed is of a fractional derivative type and thus we focused on the following system: $\begin{array}{rcl} (1.2) & \{\begin{matrix} \frac{\partial^{2} u}{\partial t^{2}} + β^{2} u = c^{2} \frac{\partial^{2} u}{\partial x^{2}}, x \in [0, 1], t ⩾ 0, \\ u (x, 0) = u_{0} (x), u_{t} (x, 0) = u_{1} (x), \\ u (0, t) = 0, for t ⩾ 0, \\ c^{2} u_{x} (1, t) = - \partial_{t}^{α, η} u (1, t), for t ⩾ 0, \end{matrix} \end{array}$ where $u = u (x, t)$ is a real-value function, $0 < α < 1$ denote the order of the exponential fractional derivative, and $η ⩾ 0$ the weight in the exponential factor of this derivative (see (2.4)).

In recent decades, the equations that involving real order derivatives have assumed an important role in modeling the anomalous dynamics of many processes related to complex systems in the most diverse areas of sciences and engineering. The interest in the specific topic of fractional calculus surged only at the end of the last century, however, the study of fractional differential equations as a separate topic arose some 40 years ago. Questions about the existence of solutions to Cauchy type problems involving the Caputo fractional derivatives and integral. Several articles published during the last decade have developed engineering applications in viscoelastic damping and structural mechanics (see [7] and references therein). A systematic and rigorous study of some problems of this kind involving fractional differential equations can be found in recent literature [4]. Although numerous theoretical applications of fractional calculation operators have been found during their long history, many applied researchers have tried to model processes using this theory but it has not been easy. This may be due, in part, to the fact that many of the useful properties of the entera derivative are not transferred analogously to fractional case, such as, for example, a clear geometric or physical meaning, product rules, chain rules, and so on.

In this work, we have obtained the polynomial stability of a thermoelastic mixture with a boundary dissipation of fractional derivative type. In order to obtain the partial result we used a similar approach that Mbodje and Montseny [8], which is concerned on the fact that the input-output relationship in a certain diffusion equation realizes the fractional derivative operator.

This paper is organized as follows. In Section 2 briefly outlines the notation and also we reformulate the model (1.2) into an augmented system, coupling the system (1.2) with a suitable diffusion equation. In Section 3, we establish the well-posedness of the system (2.9). In Section 4 we show the lack of exponential stability. In Section 5 we obtain the polynomial stability of the corresponding semigroup.

Finally, throughout this paper, C is a generic constant, not necessarily the same at each occasion (it will change from line to line) and depends on the indicated quantities.

2. Augmented model

This section is concerned with the reformulation of the model (1.2) with boundary conditions (1.2)_{3, 4} into an augmented system. Method based on the fact the input-output relationship in a certain diffusion equation realizes the fractional derivative operator. We review some basic properties of fractional integrals and derivatives, which we will use later in the analysis of our problem. Fractional calculus, in allowing integrals and derivatives of any positive order, can be considered a branch of mathematical physics which deals with integro-differential equations, where integrals are of convolution type and exhibit weakly singular kernels of power law type. There are multiplies ways in which fractional derivatives and integrals can be defined: These are not all equivalent to each other, but each of them has its own advantages and disadvantages. Let $0 < α < 1$ . We define the Caputo fractional integral of order $α > 0$ as $\begin{array}{rcl} (2.1) & J^{α} f (t) = \frac{1}{Γ (α)} \int_{0}^{t} {(t - τ)}^{α - 1} f (τ) d τ, \end{array}$ where Γ is the well-known gamma function and $f \in L^{1} (I)$ , and $t > 0$ .

Moreover, the Caputo fractional derivative operator of order $α > 0$ is defined by $\begin{matrix} (2.2) & D_{t}^{α} f (t) = J^{1 - α} D_{t} f (t) : = \frac{1}{Γ (1 - α)} \int_{0}^{t} {(t - τ)}^{- α} \frac{d f}{d τ} (τ) d τ, \end{matrix}$ with $f \in W^{1, 1} (I)$ , and $t > 0$ . Note that Caputo definition of fractional derivative does possess a very simple but interesting interpretation: if the function $f (t)$ represents the strain history within a viscoelastic material whose relaxation function is ${[Γ (1 - α) t^{α}]}^{- 1}$ then the material will experience at any time t a total stress given the expression $D^{α} f (t)$ . Also, it easy to show that $D^{α}$ is a left inverse of $J^{α}$ , but in general it is not a right inverse. More precisely, we have $\begin{array}{rcl} D_{t}^{α} J^{α} f = f, J^{α} D_{t}^{α} f (t) = f (t) - f (0) . \end{array}$ For the proof of above equalities and more properties of fractional calculus see [10].

In this work, we consider slightly different version of (2.1) and (2.2). In [3], Choi and MacCamy establish the following definition of fractional integro-differential operators with weight exponential. Let $0 < α < 1$ , $η ⩾ 0$ , the exponential fractional integral of order α is defined by $\begin{matrix} (2.3) & J^{α, η} f (t) = \frac{1}{Γ (α)} \int_{0}^{t} e^{- η (t - τ)} {(t - τ)}^{α - 1} f (τ) d τ, \end{matrix}$ with $f \in L^{1} (I)$ and $t > 0$ . The exponential fractional derivative operator of order α is defined by $\begin{matrix} (2.4) & \partial_{t}^{α, η} f (t) = \frac{1}{Γ (1 - α)} \int_{0}^{t} e^{- η (t - τ)} {(t - τ)}^{- α} \frac{d f}{d τ} (τ) d τ, \end{matrix}$ with $f \in W^{1, 1} (I)$ and $t > 0$ . Note that $\partial_{t}^{α, η} f (t) = J^{1 - α, η} f^{'} (t)$ .

Very little attention has been given to this type of non-local fractional feedback. In this article, we work with fractional derivatives that involve an integrable kernel. However, in future works, we are interested in considering feedback that involve singular and non-integrable kernels, this leads to substantial mathematical difficulties because the methods developed for terms of convolution with integrable kernel are no longer valid.

Lemma 2.1.
Let u be a solution of the system ( 1.2 ). Then, the energy functional $E_{1} (t)$ is given by $\begin{array}{rcl} (2.5) & E_{1} (t) = \frac{1}{2} \int_{0}^{1} [u_{t}^{2} + c^{2} u_{x}^{2} + β^{2} u^{2}] d x, \end{array}$ and satisfies $\begin{array}{rcl} (2.6) & \frac{d}{d t} E_{1} (t) = - \partial_{t}^{α, η} u (1, t) u_{t} (1, t) . \end{array}$
Proof.
In fact, multiplying (1.2)₁ by $u_{t}$ , integrating by part in x over [0,1] and performing straightforward calculations we obtain (2.5), deriving with respect to t we have $\begin{array}{rcl} (2.7) & \frac{d}{d t} E_{1} (t) = c^{2} u_{x} (1, t) u_{t} (1, t) . \end{array}$ Replacing (1.2)₄ the result follows. □

Now, we given a reformulation of the model (1.2) with boundary conditions (1.2)_{3, 4} into an augmented system. For this, we introduce the equation $\begin{array}{l} \partial_{t} ϕ (ξ, t) + (ξ^{2} + η) ϕ (ξ, t) - U (t) μ (ξ) = 0, ξ \in R, η ⩾ 0, t > 0, \\ ϕ (ξ, 0) = 0, \end{array}$ where $μ (ξ) = | ξ |^{(2 α - 1) / 2}$ . Multiplying the above equation by $e^{- (ξ^{2} + η) t}$ and integrating we get $\begin{matrix} ϕ (ξ, t) = \int_{0}^{t} μ (ξ) U (t) e^{- (ξ^{2} + η) (t - τ)} d τ \end{matrix}$ and $\begin{matrix} \int_{R} μ (ξ) ϕ (ξ, t) d ξ = \int_{R} \int_{0}^{t} μ {(ξ)}^{2} U (t) e^{- (ξ^{2} + η) (t - τ)} d τ d ξ, \end{matrix}$ using Fubini theorem and recalling the definition of the Gamma function we get that $\begin{matrix} (2.8) & \partial_{t}^{α, η} U = \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ . \end{matrix}$ For more details of this deduction see [7]. The next technical lemma will be very useful for our purposes.
Lemma 2.2.
Let $D = {λ \in C : Re λ + η > 0} \cup {λ \in C : Im λ \neq 0}$ . If $λ \in D$ then $\begin{array}{rcl} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + | λ |} d ξ = \frac{π}{sin (α π)} {(η + | λ |)}^{α - 1} . \end{array}$

For the proof see [7].

With the representation (2.8) for the fractional derivative, the system (1.2) can be written as $\begin{array}{rcl} (2.9) & \{\begin{matrix} u_{t t} - c^{2} u_{x x} + β^{2} u = 0 \\ \partial_{t} ϕ (ξ, t) + (ξ^{2} + η) ϕ (ξ, t) - u_{t} (1, t) μ (ξ) = 0, \\ u (x, 0) = u_{0} (x), u_{t} (x, 0) = u_{1} (x) \\ u (0, t) = 0 \\ c^{2} u_{x} (1, t) = - \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ . \end{matrix} \end{array}$
Remark 2.3.
From (2.8), the boundary conditions in (1.2) and (2.9) note that $c^{2} u_{x} (1, t) = - \partial_{t}^{α, η} u (1, t)$ .

The energy associated with to previous system is given by $\begin{matrix} (2.10) & E (t) = \frac{1}{2} [‖ u_{t} ‖_{L^{2} (0, 1)}^{2} + c^{2} ‖ u_{x} ‖_{L^{2} (0, 1)}^{2} + β^{2} ‖ u ‖_{L^{2} (0, 1)}^{2} + \frac{sin (α π)}{π} ‖ ϕ ‖_{L^{2} (R)}^{2}] \end{matrix}$ and satisfies $\begin{array}{l} (2.11) & \frac{d}{d t} E (t) = - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) ϕ^{2} (ξ, t) d ξ ⩽ 0 . \end{array}$
3. Setting of the semigroup

In this section we use results of the semigroup theory of linear operators to obtain an existence theorem of the system (2.9). We will use the following standard $L^{2} (0, 1)$ space, we are the scalar product and the norm are denoted by $\begin{matrix} {⟨ φ, ψ ⟩}_{L^{2} (0, 1)} = \int_{0}^{1} φ \overline{ψ} d x, ‖ ψ ‖_{L^{2} (0, 1)}^{2} = \int_{0}^{1} | ψ |^{2} d x . \end{matrix}$ In a similar way, let $L^{2} (R)$ be the Hilbert space of all measurable square integrable functions on the real line with the inner product $\begin{array}{rcl} {⟨ φ, ψ ⟩}_{L^{2} (R)} = \int_{R} φ \overline{ψ} d ξ, φ, ψ \in L^{2} (R) . \end{array}$ We define $H (0, 1) = {ψ \in H^{1} (0, 1) : ψ (0) = 0}$ . Then $\begin{array}{rcl} (3.1) & H = H (0, 1) \times L^{2} (0, 1) \times L^{2} (R), \end{array}$ equipped with the inner product given by $\begin{array}{rcl} (3.2) & {⟨ U, \tilde{U} ⟩}_{H} = c^{2} \int_{0}^{1} u_{x} {\overline{\tilde{u}}}_{x} d x + β^{2} \int_{0}^{1} u \overline{\tilde{u}} d x + \int_{0}^{1} v \overline{\tilde{v}} d x + \frac{sin (α π)}{π} \int_{R} ϕ \overline{\tilde{ϕ}} d x, \end{array}$ where $U = (u, v, ϕ)$ and $\tilde{U} = (\tilde{u}, \tilde{v}, \tilde{ϕ})$ . The norm is given by $\begin{array}{rcl} ‖ U ‖_{H}^{2} = c^{2} ‖ u_{x} ‖_{L^{2} (0, 1)}^{2} + β^{2} ‖ u ‖_{L^{2} (0, 1)}^{2} + ‖ v ‖_{L^{2} (0, 1)}^{2} + \frac{sin (α π)}{π} ‖ ϕ ‖_{L^{2} (R)}^{2} . \end{array}$ Now, we wish to transform the initial boundary value problem (2.9) to an abstract problem in the Hilbert space $H$ . We introduce the functions $u_{t} = v$ and rewrite the system (2.9) as the following initial value problem $\begin{array}{rcl} (3.3) & \frac{d}{d t} U (t) = A U (t), U (0) = U_{0}, \forall t > 0, \end{array}$ where $U = (u, v, ϕ)$ and $U_{0} = (u_{0}, u_{1}, ϕ_{0})$ , and the operator $A : D (A) \subset H \to H$ is given by $\begin{matrix} (3.4) & A (\begin{matrix} u \\ v \\ ϕ \end{matrix}) = (\begin{matrix} v \\ c^{2} u_{x x} - β^{2} u \\ - (ξ^{2} + η) ϕ + v (1, t) μ (ξ) \end{matrix}) . \end{matrix}$ with domain $\begin{array}{l} D (A) = & {U = (u, v, ϕ) \in H; u \in H^{2} (0, 1) \cap H, v \in H (0, 1), (ξ^{2} + η) ϕ + v (1, t) μ (ξ) \in L^{2} (R), \\ | ξ | ϕ \in L^{2} (R), c^{2} u_{x} (1, t) + \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ = 0} . \end{array}$

Remark 3.1.
For the linear operator $A$ we define the resolvent set by $\begin{array}{rcl} ϱ (A) & = & {λ \in C : (λ I - A) is invertible} \\ = & {λ \in C : {(λ I - A)}^{- 1} is a bounded linear operator}, \end{array}$ and the spectrum of $A$ is $σ (A) = C ∖ ϱ (A)$ .

Note that an operator may be injective, even bounded below, but not invertible. The unilateral shift on $l^{2} (N)$ is such an example. This shift operator is an isometry, therefore bounded below by 1. But it is not invertible as it is not surjective. The set of λ for which $λ I - A$ is injective but does not have dense range is known as the residual spectrum or compression spectrum of $A$ and is denoted by $σ_{r} (A)$ . $\begin{array}{rcl} σ_{r} (A) & = & {λ \in C . λ I - A has an inverse (bounded or unbounded) \\ with domain not dense in H} . \end{array}$
Proposition 3.2.
The operator $A$ is the infinitesimal generator of a contraction semigroup ${S_{A} (t)}_{t ⩾ 0}$ .
Proof.
We will show that $A$ is a dissipative operator and that the operator $λ I - A$ is surjective for any $λ > 0$ . Then our conclusion follows using the well known the Lumer–Phillips theorem. We observe that if $U = (u, v, ϕ) \in D (A)$ , then using (2.11) and (3.3), we get $\begin{array}{l} (3.5) & Re {⟨ A U, U ⟩}_{H} = - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) {| ϕ (ξ, t) |}^{2} d ξ ⩽ 0 . \end{array}$ In fact, $\begin{array}{rcl} {⟨ A U, U ⟩}_{H} & = & c^{2} \int_{0}^{1} v_{x} {\overline{u}}_{x} d x + β^{2} \int_{0}^{1} v \overline{u} d x + \int_{0}^{1} [c^{2} u_{x x} - β^{2} u] \overline{v} d x \\ + \frac{sin (α π)}{π} \int_{R} [- (ξ^{2} + η) ϕ + v (1, t) μ (ξ)] \overline{ϕ} d ξ \\ = & c^{2} \int_{0}^{1} v_{x} {\overline{u}}_{x} d x + β^{2} \int_{0}^{1} (v \overline{u} - \overline{v \overline{u}}) d x + c^{2} \int_{0}^{1} u_{x x} \overline{v} d x \\ - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ + v (1, t) \frac{sin (α π)}{π} \int_{R} μ (ξ) \overline{ϕ} d ξ \\ = & c^{2} \int_{0}^{1} v_{x} {\overline{u}}_{x} d x + β^{2} \int_{0}^{1} (v \overline{u} - \overline{v \overline{u}}) d x + c^{2} \int_{0}^{1} u_{x x} \overline{v} d x \\ - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ - c^{2} v (1, t) {\overline{u}}_{x} (1, t) . \end{array}$ Integrating by parts we have $\begin{array}{rcl} {⟨ A U, U ⟩}_{H} & = & c^{2} \int_{0}^{1} v_{x} {\overline{u}}_{x} d x + β^{2} \int_{0}^{1} (v \overline{u} - \overline{v \overline{u}}) d x + c^{2} u_{x} (1, t) \overline{v} (1, t) - c^{2} \int_{0}^{1} {\overline{v}}_{x} u_{x} d x \\ - 2 α \int_{0}^{1} | v |^{2} d x - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ - c^{2} v (1, t) {\overline{u}}_{x} (1, t) \\ = & c^{2} \int_{0}^{1} (v_{x} {\overline{u}}_{x} - \overline{v_{x} {\overline{u}}_{x}}) d x + β^{2} \int_{0}^{1} (v \overline{u} - \overline{v \overline{u}}) d x \\ - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ - c^{2} ({\overline{u}}_{x} (1, t) v (1, t) - \overline{{\overline{u}}_{x} (1, t) v (1, t)}) \\ = & 2 i c^{2} Im \int_{0}^{1} v_{x} {\overline{u}}_{x} d x + 2 i β^{2} Im \int_{0}^{1} v \overline{u} d x \\ - \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ - 2 i c^{2} Im ({\overline{u}}_{x} (1, t) v (1, t)) . \end{array}$ Taking the real part, it follows that $A$ is a dissipative operator and (3.5) follows.

Next, we will prove that the operator $λ I - A$ is surjective for $λ > 0$ . Let $F = (f_{1}, f_{2}, f_{3}) \in H$ and $U = (u, v, ϕ) \in D (A)$ such that $λ I - A U = F$ , that is, $\begin{array}{rcl} (3.6) & \{\begin{matrix} λ u - v = f_{1} in H (0, 1) \\ λ v - c^{2} u_{x x} + β^{2} u = f_{2} in L^{2} (0, 1) \\ λ ϕ + (ξ^{2} + η) ϕ - v (1, t) μ (ξ) = f_{3} in L^{2} (R) . \end{matrix} \end{array}$ From (3.6)₃ we have $\begin{array}{rcl} (3.7) & ϕ (ξ, t) = \frac{f_{3} (ξ) + v (1, t) μ (ξ)}{ξ^{2} + η + λ}, \end{array}$ and by (3.6)₁ we obtain $\begin{array}{rcl} (3.8) & v = λ u - f_{1} \in H (0, 1) . \end{array}$ Replacing (3.8) into (3.6)₂ yields $\begin{array}{rcl} (3.9) & (λ^{2} + β^{2}) u - c^{2} u_{x x} = f_{2} + λ f_{1} . \end{array}$ To solve (3.9) is equivalent to finding $u \in H^{2} (0, 1) \cap H$ such that $\begin{array}{rcl} (3.10) & \int_{0}^{1} [(λ^{2} + β^{2}) u - c^{2} u_{x x}] ζ d x = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x \end{array}$ for all $ζ \in H$ . Then $\begin{array}{rcl} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x - c^{2} \int_{0}^{1} u_{x x} ζ d x = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x \end{array}$ Integrating by parts we have $\begin{array}{rcl} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x - c^{2} u_{x} (1, t) ζ (1, t) + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x . \end{array}$ Using (2.9)₅ it follows that $\begin{array}{l} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x + [\frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ) d ξ] ζ (1, t) + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x \\ = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x . \end{array}$ From (3.7) we have $\begin{array}{l} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x + [\frac{sin (α π)}{π} \int_{R} μ (ξ) \frac{f_{3} (ξ) + v (1, t) μ (ξ)}{ξ^{2} + η + λ} d ξ] ζ (1, t) \\ + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x . \end{array}$ Then $\begin{array}{l} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x + [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + λ} d ξ] ζ (1, t) \\ + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + λ} d ξ] v (1, t) ζ (1, t) + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x \\ (3.11) & = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x . \end{array}$ Replacing (3.8) into (3.11) we obtain $\begin{array}{l} \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x + [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + λ} d ξ] ζ (1, t) \\ + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + λ} d ξ] λ u (1, t) ζ (1, t) + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x \\ = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + λ} d ξ] f (1) ζ (1, t) . \end{array}$ This equality is equivalent at the problem $\begin{array}{rcl} (3.12) & a (u, ζ) = L (ζ), \end{array}$ where the bilinear form continuous and coercive $a : H (0, 1) \times H (0, 1) \to R$ and the continuous linear form $L : H (0, 1) \to R$ are defined by $\begin{array}{rcl} a (u, ζ) & = & \int_{0}^{1} (λ^{2} + β^{2}) u ζ d x + [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + λ} d ξ] ζ (1, t) . \\ + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + λ} d ξ] λ u (1, t) ζ (1, t) + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x \end{array}$ and $\begin{array}{rcl} L (ζ) = \int_{0}^{1} [f_{2} + λ f_{1}] ζ d x + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + λ} d ξ] f (1) ζ (1, t) . \end{array}$ It follows by the Lax–Milgram theorem that for all $ζ \in H (0, 1)$ the problem (3.12) admits a unique solution $u \in H (0, 1)$ . Using elliptic regularity, it follows from (3.12) that $u \in H^{2} (0, 1)$ . therefore, the operator $λ I - A$ is surjective for any $λ > 0$ . □

Now, using the Hille–Yosida theorem, we have the following results. Theorem 3.3.

If $U_{0} \in H$ , then the system ( 3.3 ) has a unique mild solution $\begin{array}{rcl} U \in C (R_{0}^{+}, H) . \end{array}$

If $U_{0} \in D (A)$ , then the system ( 3.3 ) has a unique strong solution $\begin{array}{rcl} U \in C (R_{0}^{+}, D (A)) \cap C^{1} (R_{0}^{+}, H) . \end{array}$

4. The lack of exponential stability

In this section, we show the lack of exponential stability. For this, the following results are fundamental.

Lemma 4.1 ([7]).

Let $i = \sqrt{- 1}$ , and suppose $ω \in R$ . For any fixed $η > 0$ , the integral $\begin{array}{rcl} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i ω} d ξ \to 0, as | ω | \to \infty . \end{array}$

Theorem 4.2 ([9]).

Let ${S (t)}_{t ⩾ 0}$ be a $C_{0}$ -semigroup of contractions on Hilbert space $H$ with generator $A$ . The following statements are equivalent:

The semigroup ${S (t)}_{t ⩾ 0}$ is exponentially stable: $\begin{array}{rcl} \exists M > 0, \exists a > 0, ‖ S (t) ‖ ⩽ M e^{- a t}, \end{array}$

The set $i R \subset ϱ (A)$ and $\begin{array}{rcl} sup {‖ {(i ω I - A)}^{- 1} ‖, ω \in R} < \infty . \end{array}$

Equipped with the foregoing theorem, we going to prove our claim.

Theorem 4.3.
The semigroup ${S (t)}_{t ⩾ 0}$ generated by the operator $A$ is not exponentially stable.
Proof.
Case $η = 0$ . We will show that $i ω \notin ϱ (A)$ .

In fact, we consider $(sin (x), 0, 0) \in H$ , then denoting by $(u, v, ϕ)$ the image of $(sin (x), 0, 0)$ by $A^{- 1}$ , we can see that $ϕ (ξ) = | ξ |^{(2 α - 5) / 2} sin (1)$ . But $ϕ \notin L^{2} (R)$ for $α \in (0, 1)$ . This way $(u, v, ϕ) \notin D (A)$ .

We are going to conclude the proof by contradiction. Suppose that $i ω \in ϱ (A)$ , this implies that $i ω I - A$ is invertible and ${(i ω I - A)}^{- 1} V \in D (A)$ for all $V \in H$ . From the equality $\begin{matrix} {(i ω I - A)}^{- 1} (i ω I - A) = I \end{matrix}$ we have $\begin{matrix} i ω {(i ω I - A)}^{- 1} A^{- 1} W - {(i ω I - A)}^{- 1} W = A^{- 1} W \end{matrix}$ for all $W \in H$ . If $W = (sin (x), 0, 0)$ then the left side of the equality belongs to $D (A)$ and the right side does not belong to $D (A)$ which it is a contradiction. This implies that $i ω$ is not in the resolvent set of $A$ .

Case $η \neq 0$ . Using the Theorem 4.2, it suffices to prove that $‖ {(i ω I - A)}^{- 1} ‖$ is unbounded for $ω \in R$ .

In fact, let $ω \in R$ , $F = (f_{1}, f_{2}, f_{3}) \in H$ , and $U = (u, v, ϕ) \in D (A)$ be such that: $\begin{array}{rcl} (4.1) & \{\begin{matrix} i ω u - v = f_{1} \\ i ω v - c^{2} u_{x x} + β^{2} u = f_{2} \\ (ξ^{2} + η + i ω) ϕ (ξ) - v (1, t) μ (ξ) = f_{3} \\ u (0, t) = 0 \\ c^{2} u_{x} (1, t) + \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ = 0 \\ | ξ | ϕ \in L^{2} (R) . \end{matrix} \end{array}$ Replacing (4.1)₁ into (4.1)₂ and performing straightforward calculations we have $\begin{array}{rcl} u_{x x} + \frac{(ω^{2} - β^{2})}{c^{2}} u = - \frac{(f_{2} (x) + i ω f_{1} (x))}{c^{2}} . \end{array}$ Writing $m^{2} = \frac{ω^{2} - β^{2}}{c^{2}} > 0$ for a good choice of β. Let $F (x) = - \frac{(f_{2} (x) + i ω f_{1} (x))}{c^{2}}$ , we have the equality $\begin{array}{rcl} (4.2) & u_{x x} + m^{2} u = F (x) . \end{array}$ The solution to the homogeneous problem associated with (4.2) is given by $\begin{array}{rcl} u_{h} (x) = c_{1} cos (m x) + c_{2} sin (m x) . \end{array}$ Using (4.1)₄ it follows that $\begin{array}{rcl} (4.3) & u_{h} (x) = c_{2} sin (m x) . \end{array}$ Since the wronskian is $W (cos (m x), sin (m x)) = m$ , we have that the particular solution of equation (4.2) is $\begin{array}{rcl} (4.4) & u_{p} (x) = c_{1} (x) cos (m x) + c_{2} (x) sin (m x), \end{array}$ where $\begin{array}{rcl} (4.5) & \{\begin{matrix} c_{1} (x) = - \frac{1}{m} \int_{0}^{x} F (σ) sin (m σ) d σ \\ c_{2} (x) = \frac{1}{m} \int_{0}^{x} F (σ) cos (m σ) d σ . \end{matrix} \end{array}$ Hence $\begin{array}{rcl} u_{p} (x) & = & \frac{1}{m} \int_{0}^{x} F (σ) sin [m (x - σ)] d σ \\ (4.6) & = & - \frac{1}{m c^{2}} \int_{0}^{x} (f_{2} (σ) + i ω f_{1} (σ)) sin [m (x - σ)] d σ . \end{array}$ From (4.3) and (4.6) we obtain that the general solution of (4.2) is given by $\begin{array}{rcl} (4.7) & u (x, t) = c_{2} sin (m x) - \frac{1}{m c^{2}} \int_{0}^{x} (f_{2} (σ) + i ω f_{1} (σ)) sin [m (x - σ)] d σ, \end{array}$ Moreover, $\begin{array}{rcl} (4.8) & v (x, t) = i ω c_{2} sin (m x) - f_{1} (x) - \frac{i ω}{m c^{2}} \int_{0}^{x} (f_{2} (σ) + i ω f_{1} (σ)) sin [m (x - σ)] d σ, \end{array}$ and $\begin{array}{rcl} (4.9) & v (1, t) = i ω c_{2} sin (m) - f_{1} (1) - \frac{i ω}{m c^{2}} \int_{0}^{1} (f_{2} (σ) + i ω f_{1} (σ)) sin [m (1 - σ)] d σ . \end{array}$ From (4.1)₃ we have $ϕ (ξ) = \frac{μ (ξ)}{ξ^{2} + η + i ω} v (1, t) + \frac{f_{3}}{ξ^{2} + η + i ω}$ , hence $\begin{array}{rcl} (4.10) & \int_{R} μ (ξ) ϕ (ξ) d ξ & = & \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i ω} d ξ v (1, t) + \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + i ω} d ξ . \end{array}$ Replacing (4.7), (4.9) and (4.10) into (4.1)₃ we obtain $\begin{array}{rcl} (4.11) & (c^{2} m cos (m) + i ω \frac{sin (α π)}{π} (\int_{R} \frac{μ^{2} (ξ) d ξ}{ξ^{2} + η + i ω}) sin (m)) c_{2} = G (m), \end{array}$ where $\begin{array}{rcl} G (m) & = & \frac{i ω}{m c^{2}} \frac{sin (α π)}{π} (\int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i ω} d ξ) \int_{0}^{1} (f_{2} (σ) + i ω f_{1} (σ)) sin [m (1 - σ)] d σ \\ + \frac{sin (α π)}{π} f_{1} (1) (\int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i ω} d ξ) - \frac{sin (α π)}{π} (\int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + i ω} d ξ) \\ (4.12) & + \frac{1}{m} \int_{0}^{1} (f_{2} (σ) + i ω f_{1} (σ)) cos [m (1 - σ)] d σ . \end{array}$ Let be the sequence $F_{n} = (f_{1 n}, f_{2 n}, f_{3 n}) \in H$ defines by $\begin{array}{rcl} (4.13) & f_{1 n} (x) = \frac{1}{ω_{n}} sin (m_{n} x), f_{2 n} (x) = cos (m_{n} x), f_{3 n} = 0, \end{array}$ with $m_{n} = \frac{2 n + 1}{2} π$ , $n = 1, 2, \dots$ . Remembering that $m^{2} = \frac{ω^{2} - β^{2}}{c^{2}}$ and solving ω we have $ω_{n}^{2} = \frac{{(2 n + 1)}^{2}}{4} π^{2} c^{2} + β^{2}$ , hence $ω_{n} \to \infty$ if $n \to \infty$ . $\begin{array}{rcl} ‖ F_{n} ‖_{H}^{2} & = & {‖ (f_{1 n}, f_{2 n}, f_{3 n}) ‖}_{H}^{2} \\ = & c^{2} ‖ f_{{1 n}_{x}} ‖_{L^{2} (0, 1)}^{2} + β^{2} ‖ f_{1 n} ‖_{L^{2} (0, 1)}^{2} + ‖ f_{2 n} ‖_{L^{2} (0, 1)}^{2} + \frac{sin (α π)}{π} ‖ f_{3 n} ‖_{L^{2} (R)}^{2} \\ = & c^{2} \int_{0}^{1} | f_{{1 n}_{x}} |^{2} d x + β^{2} \int_{0}^{1} | f_{1 n} |^{2} d x + \int_{0}^{1} | f_{2 n} |^{2} d x \\ = & [\frac{m_{n}^{2}}{ω_{n}^{2}} c^{2} + 1] \int_{0}^{1} {cos}^{2} (m_{n} x) d x + \frac{β^{2}}{ω_{n}^{2}} \int_{0}^{1} {sin}^{2} (m_{n} x) d x \\ (4.14) & = & \frac{1}{2} [\frac{m_{n}^{2}}{ω_{n}^{2}} c^{2} + 1] + \frac{β^{2}}{2 ω_{n}^{2}} ⟶ \frac{c^{2} + 1}{2} as n ⟶ \infty . \end{array}$ Now, replacing (4.13) into (4.6) and deriving, we can to obtain $\begin{array}{rcl} - c^{2} u_{p_{x}} (x, t) & = & sin (m_{n} x) \int_{0}^{x} e^{i m_{n} σ} sin (m_{n} σ) d σ + cos (m_{n} x) \int_{0}^{x} e^{i m_{n} σ} cos (m_{n} σ) d σ \\ = & sin (m_{n} x) [\int_{0}^{x} cos (m_{n} σ) sin (m_{n} σ) d σ + i \int_{0}^{x} \frac{1 - cos (2 m_{n} σ)}{2} d σ] \\ + cos (m_{n} x) [\int_{0}^{x} \frac{1 + cos (2 m_{n} σ)}{2} d σ + i \int_{0}^{x} sin (m_{n} σ) cos (m_{n} σ) d σ] \\ = & sin (m_{n} x) {[- \frac{{cos}^{2} (m_{n} σ)}{m_{n}} + i \frac{σ}{2} - i \frac{sin (2 m_{n} σ)}{4 m_{n}}]}_{0}^{x} \\ + cos (m_{n} x) {[\frac{σ}{2} + \frac{sin (2 m_{n} σ)}{4 m_{n}} + i \frac{{sin}^{2} (m_{n} σ)}{m_{n}}]}_{0}^{x} \\ = & sin (m_{n} x) [\frac{{sin}^{2} (m_{n} x)}{m_{n}} + i \frac{x}{2} - i \frac{sin (2 m_{n} x)}{4 m_{n}}] \\ + cos (m_{n} x) [\frac{x}{2} + \frac{sin (2 m_{n} x)}{4 m_{n}} + i \frac{{sin}^{2} (m_{n} x)}{m_{n}}] \\ = & e^{i m_{n} x} \frac{x}{2} + i \frac{1 - cos (2 m_{n} x)}{2 m_{n}} e^{- i m_{n} x} + \frac{sin (2 m_{n} x)}{4 m_{n}} e^{- i m_{n} x} . \end{array}$

Hence $\begin{matrix} u_{x} (x, t) = m_{n} c_{2} cos (m_{n} x) - \frac{e^{- i m_{n} x}}{c^{2}} [\frac{x}{2} e^{i 2 m_{n} x} + \frac{sin (2 m_{n} x)}{4 m_{n}} + i \frac{1 - cos (2 m_{n} x)}{2 m_{n}}] . \end{matrix}$ Thus, $\begin{array}{rcl} (4.15) & | u_{x} (x, t) | ⩾ | | m_{n} c_{2} cos (m_{n} x) | - \frac{1}{c^{2}} | \frac{x}{2} e^{i 2 m_{n} x} + \frac{sin (2 m_{n} x)}{4 m_{n}} + i \frac{1 - cos (2 m_{n} x)}{2 m_{n}} | | . \end{array}$ On the other hand, for all $P, Q ⩾ 0$ and $0 < ε < 1$ , we have $\begin{array}{rcl} (4.16) & 2 P Q ⩽ 2 \sqrt{ε} P \frac{1}{\sqrt{ε}} Q ⩽ ε P^{2} + \frac{1}{ε} Q^{2} ⟺ - 2 P Q ⩾ - ε P^{2} - \frac{1}{ε} Q^{2}, \end{array}$ Then, into (4.15) $\begin{matrix} (4.17) & {| u_{x} (x, t) |}^{2} ⩾ (1 - ε) | c_{2} |^{2} m_{n}^{2} {cos}^{2} (m_{n} x) - (\frac{1}{ε} - 1) \frac{1}{c^{4}} H (n), \end{matrix}$ where $\begin{array}{rcl} H (n) & = & {| \frac{x}{2} e^{i 2 ω_{n} x} + \frac{sin (2 m_{n} x)}{4 m_{n}} + i \frac{1 - cos (2 m_{n} x)}{2 m_{n}} |}^{2} \\ = & {(\frac{x}{2} cos (2 m_{n} x) + \frac{sin (2 m_{n} x)}{4 m_{n}})}^{2} + {(\frac{x}{2} sin (2 m_{n} x) + \frac{1 - cos (2 m_{n} x)}{2 m_{n}})}^{2} \\ = & \frac{x^{2}}{4} {cos}^{2} (2 m_{n} x) + x \frac{cos (2 m_{n} x) sin (2 m_{n} x)}{4 m_{n}} + \frac{{sin}^{2} (2 m_{n} x)}{16 m_{n}^{2}} \\ + \frac{x^{2}}{4} {sin}^{2} (2 m_{n} x) + x \frac{sin (2 m_{n} x)}{2 m_{n}} - x \frac{sin (2 m_{n} x) cos (2 m_{n} x)}{2 m_{n}} \\ + \frac{1}{4 m_{n}^{2}} - \frac{cos (2 m_{n} x)}{2 m_{n}^{2}} + \frac{{cos}^{2} (2 m_{n} x)}{4 m_{n}^{2}} \\ = & \frac{x^{2}}{4} + \frac{1}{4 m_{n}^{2}} - x \frac{sin (2 m_{n} x) cos (2 m_{n} x)}{4 m_{n}} + \frac{{sin}^{2} (2 m_{n} x)}{16 m_{n}^{2}} \\ + x \frac{sin (2 m_{n} x)}{2 m_{n}} - \frac{cos (2 m_{n} x)}{2 m_{n}^{2}} + \frac{{cos}^{2} (2 m_{n} x)}{4 m_{n}^{2}} \\ = & \frac{x^{2}}{4} + \frac{5}{16 m_{n}^{2}} - x \frac{sin (2 m_{n} x) cos (2 m_{n} x)}{4 m_{n}} + x \frac{sin (2 m_{n} x)}{2 m_{n}} - \frac{cos (2 m_{n} x)}{2 m_{n}^{2}} + 3 \frac{{cos}^{2} (2 m_{n} x)}{16 m_{n}^{2}} \end{array}$ integrating $H (n)$ respect to x on $[0, 1]$ and remembering that $m_{n} = \frac{2 n + 1}{2} π$ , $n = 1, 2, \dots$ , we obtain $\begin{array}{rcl} \int_{0}^{1} H (n) d x & = & {[\frac{x^{3}}{12} + \frac{5 x}{16 m_{n}^{2}} + \frac{cos (4 m_{n} x)}{32 m_{n}} - \frac{sin (4 m_{n} x)}{128 m_{n}^{2}} + \frac{sin (2 m_{n} x)}{8 m_{n}^{3}} - x \frac{cos (2 m_{n} x)}{4 m_{n}^{2}}]}_{0}^{1} \\ + {[- \frac{sin (2 m_{n} x)}{4 m_{n}^{3}} + \frac{3 x}{32 m_{n}^{2}} + \frac{3 sin (4 m_{n} x)}{128 m_{n}^{3}}]}_{0}^{1} \\ = & \frac{1}{12} + \frac{5}{16 m_{n}^{2}} + \frac{cos (4 m_{n})}{32 m_{n}} - \frac{1}{32 m_{n}} - \frac{sin (4 m_{n})}{128 m_{n}^{2}} + \frac{sin (2 m_{n})}{8 m_{n}^{3}} - \frac{cos (2 m_{n})}{4 m_{n}^{2}} \\ - \frac{sin (2 m_{n})}{4 m_{n}^{3}} + \frac{3}{32 m_{n}^{2}} + \frac{3 sin (4 m_{n})}{128 m_{n}^{3}} \\ = & \frac{1}{12} + \frac{5}{16 m_{n}^{2}} + \frac{1}{4 m_{n}^{2}} + \frac{3}{32 m_{n}^{2}} = \frac{1}{12} + \frac{21}{32 m_{n}^{2}} \end{array}$ and the following conditions holds: $\begin{array}{rcl} (4.18) & 0 ⩽ \frac{1}{2 c^{4}} (\frac{1}{ε} - 1) \int_{0}^{1} H (n) d x ⩽ \frac{1}{2 c^{4}} (\frac{1}{ε} - 1) [\frac{1}{2} + \frac{1}{m_{n}}] ⩽ \frac{3}{4 c^{4}} (\frac{1}{ε} - 1) . \end{array}$ Moreover $\begin{array}{rcl} {‖ {(i ω_{n} I - A)}^{- 1} F_{n} ‖}_{H}^{2} ⩾ ‖ u_{n} ‖_{H (0, 1)}^{2} = \frac{1}{2} \int_{0}^{1} | u_{x} |^{2} d x, \end{array}$ which, together with (4.17), implies that $\begin{array}{l} {‖ {(i ω_{n} I - A)}^{- 1} F_{n} ‖}_{H}^{2} \\ (4.19) & ⩾ \frac{1}{2} (1 - ε) | c_{2} |^{2} m_{n}^{2} (\frac{1}{2} + \frac{1}{4 m_{n}} sin (2 m_{n})) - \frac{1}{2 c^{4}} (\frac{1}{ε} - 1) \int_{0}^{1} H (n) d x . \end{array}$

Replacing (4.13) in (4.12) and (4.11) and using (4.16) $\begin{array}{rcl} | c_{2} |^{2} m_{n}^{2} & ⩾ & (1 - \frac{1}{ε}) {| - \frac{i}{2} e^{i 2 m_{n}} + \frac{{sin}^{2} (m_{n})}{m_{n}} + i \frac{sin (2 m_{n})}{4 m_{n}} + \frac{m_{n} sin (m_{n})}{ω_{n}} |}^{2} \\ + | 1 - ε | {| \frac{- \frac{1}{2} + \frac{sin (2 m_{n})}{4 m_{n}} + i \frac{1 - cos (2 m_{n})}{2 m_{n}}}{\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i ω_{n}} d ξ} |}^{2}, \end{array}$ which combined with Lemma 4.1 we obtain that $| c_{2} |^{2} m_{n}^{2} \to \infty$ as $n \to \infty$ . This fact together with (4.19), (4.18) and (4.14), we reach to $\begin{array}{rcl} lim_{n \to \infty} \frac{‖ {(i ω_{n} I - A)}^{- 1} F_{n} ‖_{H}}{‖ F_{n} ‖_{H}} = + \infty, \end{array}$ the result follow of Theorem 4.2. □

5. Asymptotic behavior

In this section, we show the asymptotic behavior of the solution of problem. First, we prove that the semigroup ${S_{A} (t)}_{t ⩾ 0}$ is asymptotically stable, for this we use Theorem 5.1 due to Arend–Batty [1]. Next, we show that ${S_{A} (t)}_{t ⩾ 0}$ is polynomially stable using the Theorem 5.7 of Borichev–Tomilov, see [2].

To achieve our first proof we use the following result.

Theorem 5.1 ([1]).

Let $A$ be the generator of a uniformly bounded C ₀ -semigroup ${S (t)}_{t ⩾ 0}$ on a Hilbert space $H$ . If $σ (A) \cap i R$ is at most a countable set and $σ_{r} (A) \cap i R = \emptyset$ then the semigroup ${S (t)}_{t ⩾ 0}$ is asymptotically stable, that is, $\begin{array}{rcl} {‖ S (t) z ‖}_{H} \to 0 as t \to \infty, for any z \in H . \end{array}$

In order to obtain our result, we will test the following lemmas.

Lemma 5.2.
We have $σ (A) \cap i R = \emptyset$ .
Proof.
By contradiction. We suppose that there $λ \in R$ , $λ \neq 0$ and $U \neq 0$ , such that $A U = i λ U$ , that is, $(i λ - A) U = 0$ . Then $\begin{array}{rcl} (5.1) & \{\begin{matrix} i λ u - v = 0 \\ i λ v - c^{2} u_{x x} + β^{2} u = 0 \\ i λ ϕ + (ξ^{2} + η) ϕ - v (1, t) μ (ξ) = 0 . \end{matrix} \end{array}$ From (3.5) we have $ϕ (ξ, t) = 0$ , and by (5.1)₃ we obtain $v (1, t) = 0$ . Hence, from (5.1)₁ and (2.9)₅ it follows that $u (1, t) = 0$ and $u_{x} (1, t) = 0$ . On the other hand, replacing (5.1)₁ into (5.1)₂ we obtain $\begin{array}{l} (5.2) & - λ^{2} u - c^{2} u_{x x} + β^{2} u = 0, \end{array}$ We can rewrite (5.2) as the initial value problem $\begin{array}{rcl} (5.3) & \{\begin{matrix} - c^{2} u_{x x} - (λ^{2} - β^{2}) u = 0, \\ u (1, t) = 0, u_{x} (1, t) = 0 . \end{matrix} \end{array}$ In the Theorem 5.7, the condition (2) holds if we show that any point $σ (A) \cap {i R}$ is at most a countable set.

In fact, we will prove that the operator $i λ I - A$ is surjective for $λ \neq 0$ . For this purpose, let $F = (f_{1}, f_{2}, f_{3}) \in H$ , we seek $U = (u, v, ϕ) \in D (A)$ such that $i λ I - A U = F$ , that is, $\begin{array}{rcl} (5.4) & \{\begin{matrix} i λ u - v = f_{1} \\ i λ v - c^{2} u_{x x} + β^{2} u = f_{2} \\ i λ ϕ + (ξ^{2} + η) ϕ - v (1, t) μ (ξ) = f_{3}, \end{matrix} \end{array}$ with the initial condition $\begin{array}{rcl} (5.5) & c^{2} u_{x} (1, t) = - \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ . \end{array}$ Suppose that we have v, from (5.4)₁ it follows that $v (1, t) = i λ u (1, t) - f_{1} (1)$ . Thus, $v \in H$ . On the other hand, replacing (5.4)₁ into (5.4)₂ we get $\begin{array}{l} - λ^{2} u - c^{2} u_{x x} + β^{2} u = f_{2} + i λ f_{1}, \end{array}$ this equation is equivalent to find $u \in H^{2} (0, 1) \cap H$ such that $\begin{matrix} (5.6) & \begin{matrix} \int_{0}^{1} [- λ^{2} u - c^{2} u_{x x} + β^{2} u] ζ d x = \int_{0}^{1} [f_{2} + i λ f_{1}] ζ d x, \end{matrix} \end{matrix}$ for all $ζ \in H (0, 1)$ . We estimate (5.6), $\begin{array}{rcl} \int_{0}^{1} - λ^{2} u ζ d x + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x + β^{2} \int_{0}^{1} u ζ d x - c^{2} u_{x} (1, t) ζ (1, t) = \int_{0}^{1} [f_{2} + i λ f_{1}] ζ d x . \end{array}$ Clearing ϕ in (5.4)₃ and using (2.9)₃, we get $\begin{array}{l} \int_{0}^{1} - λ^{2} u ζ d x + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x + β^{2} \int_{0}^{1} u ζ d x \\ + [\frac{sin (α π)}{π} \int_{R} μ (ξ) \frac{f_{3} (ξ) + v (1, t) μ (ξ)}{ξ^{2} + η + i λ} d ξ] ζ (1, t) = \int_{0}^{1} [f_{2} + i λ f_{1}] ζ d x . \end{array}$ From (5.4)₁, we have $\begin{array}{l} (β^{2} - λ^{2}) \int_{0}^{1} u ζ d x + c^{2} \int_{0}^{1} u_{x} ζ_{x} d x + i [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i λ} d ξ] λ u (1, t) ζ (1, t) \\ = \int_{0}^{1} [f_{2} + i λ f_{1}] ζ d x - [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η + i λ} d ξ] ζ (1, t) \\ (5.7) & + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i λ} d ξ] f_{1} (1) ζ (1, t) \end{array}$

Rewrite this equality as $\begin{array}{rcl} (5.8) & - {⟨ L_{λ} U, V ⟩}_{H (0, 1)} + {⟨ U, V ⟩}_{H (0, 1)} = Φ (V), \end{array}$ with $\begin{array}{rcl} {⟨ L_{λ} U, V ⟩}_{H (0, 1)} = (λ^{2} - β^{2}) \int_{0}^{1} u ζ d x - i [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i λ} d ξ] λ u (1, t) ζ (1, t) \end{array}$ and ${⟨ U, V ⟩}_{H (0, 1)} = c^{2} \int_{0}^{1} u_{x} ζ_{x} d x$ .

Using that $L^{2} (0, 1) \overset{c}{↪} H^{- 1} (0, 1)$ and $H (0, 1) \overset{c}{↪} L^{2} (0, 1)$ it follows that the operator $L_{λ}$ is compact from $L^{2} (0, 1)$ into $L^{2} (0, 1)$ . This way, by Fredholm alternative, proving the existence of U solution of (5.8) reduces to show that 1 is not a eigenvalue of $L_{λ}$ . In fact, if 1 is an eigenvalue, then there exists $U \neq 0$ , such that $\begin{array}{rcl} (5.9) & {⟨ L_{λ} U, V ⟩}_{H (0, 1)} = {⟨ U, V ⟩}_{H (0, 1)}, \forall V \in H (0, 1) . \end{array}$ In particular, for $U = V$ we obtain $\begin{array}{rcl} (λ^{2} - β^{2}) \int_{0}^{1} | u |^{2} d x + i [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η + i λ} d ξ] λ {| u (1, t) |}^{2} = c^{2} \int_{0}^{1} | u_{x} |^{2} d x, \end{array}$ and, of this $u (1, t) = 0$ .

Moreover, from (5.9), integrating by parts and performing straightforward calculations we can to deduce that $u_{x} (1, t) = 0$ , and from (5.6) we have $\begin{matrix} λ^{2} u + c^{2} u_{x x} - β^{2} u = 0 \end{matrix}$ We can rewrite the above as the initial value problem $\begin{array}{rcl} (5.10) & \{\begin{matrix} c^{2} u_{x x} = - λ^{2} u + β^{2} u = - (λ^{2} - β^{2}) u, \\ u (1, t) = 0, u_{x} (1, t) = 0 . \end{matrix} \end{array}$ Using the Picard theorem (ordinary differential equations), (5.10) has a unique solution $u = 0$ . It follows from (5.1) that $v = 0$ . Therefore, $U = (u, v, ϕ) = 0$ . □
Lemma 5.3.
If $η \neq 0$ then $0 \in ϱ (A)$ .
Proof.
From (3.4) we have that $U = (u, v, ϕ) \in ker (A)$ if and only if $A U = 0$ , that is, $\begin{array}{rcl} (5.11) & \{\begin{matrix} v = 0 \\ c^{2} u_{x x} - β^{2} u = 0 \\ (ξ^{2} + η) ϕ - v (1, t) μ (ξ) = 0 . \end{matrix} \end{array}$ From (5.11) we have $v = ϕ = 0$ . Multiplying (5.11)₂ by u, integrating over $x \in (0, 1)$ and using the definition of $H (0, 1)$ it follows that $\begin{array}{rcl} (5.12) & - c^{2} \int_{0}^{1} u_{x}^{2} d x - β^{2} \int_{0}^{1} u^{2} d x + u (1, t) u_{x} (1, t) = 0 . \end{array}$ Moreover, using that $ϕ = 0$ into (2.9)₅ we obtain $u_{x} (1, t) = 0$ . This way, (5.12) is given by $\begin{array}{rcl} \int_{0}^{1} u_{x}^{2} d x + β^{2} \int_{0}^{1} u^{2} d x = 0 . \end{array}$ Hence $u \equiv 0$ and $(u, v, ϕ) = 0$ and $A$ is injective.

Given $F = (f_{1}, f_{2}, f_{3}) \in H$ , we must show that there exists a unique $U = (u, v, ϕ)$ in $D (A)$ , such that $A U = F$ , namely, $\begin{array}{rcl} (5.13) & \{\begin{matrix} - v = f_{1} \\ - c^{2} u_{x x} + β^{2} u = f_{2} \\ (ξ^{2} + η) ϕ - v (1, t) μ (ξ) = f_{3}, \end{matrix} \end{array}$ with the initial condition $\begin{array}{rcl} (5.14) & c^{2} u_{x} (1, t) = - \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ . \end{array}$ From (5.7) it follows that $\begin{array}{rcl} c^{2} \int_{0}^{1} u_{x} ζ_{x} d x + β^{2} \int_{0}^{1} u ζ d x & = & \int_{0}^{1} f_{2} ζ d x - [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η} d ξ] ζ (1, t) \\ (5.15) & + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η} d ξ] f_{1} (1) ζ (1, t) . \end{array}$ The equation (5.15) is equivalent to the problem $\begin{array}{rcl} (5.16) & a_{η} (u, ζ) = L_{η} (ζ), \end{array}$ where the bilinear form continuous and coercive $a_{η} : H (0, 1) \times H (0, 1) \to R$ and, the continuous linear form $L_{η} : H (0, 1) \to R$ are defined by $\begin{array}{rcl} a_{η} (u, ζ) = c^{2} \int_{0}^{1} u_{x} ζ_{x} d x + β^{2} \int_{0}^{1} u ζ d x \end{array}$ and $\begin{array}{rcl} L_{η} (ζ) & = & \int_{0}^{1} f_{2} ζ d x - [\frac{sin (α π)}{π} \int_{R} \frac{μ (ξ) f_{3} (ξ)}{ξ^{2} + η} d ξ] ζ (1, t) \\ + [\frac{sin (α π)}{π} \int_{R} \frac{μ^{2} (ξ)}{ξ^{2} + η} d ξ] f_{1} (1) ζ (1, t) . \end{array}$ Applying the Lax–Milgran theorem, we have that for all $ζ \in H (0, 1)$ the problem (5.16) admits an unique solution $u \in H (0, 1)$ . Using elliptic regularity, it follows from (5.15) that $u \in H^{2} (0, 1)$ . Therefore, the operator $A$ is surjective. □
Lemma 5.4.
Let $A$ be defined by ( 3.4 ), then $\begin{matrix} (5.17) & A^{} (\begin{matrix} u \\ v \\ ϕ \end{matrix}) = (\begin{matrix} - v \\ c^{2} u_{x x} - β^{2} u \\ - (ξ^{2} + η) ϕ - v (1, t) μ (ξ) \end{matrix}), \end{matrix}$ with the domain* $\begin{array}{l} D (A^{}) = & {(u, v, ϕ) \in H : u \in H^{2} (0, 1) \cap H, v \in H (0, 1), (ξ^{2} + η) ϕ + v (1, t) μ (ξ) \in L^{2} (R) \\ c^{2} u_{x} (1, t) + \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ = 0, | ξ | ϕ \in L^{2} (R)} . \end{array}$
Proof.
It is not difficult to show that $⟨ A U, U ⟩ = ⟨ U, A^{} U ⟩$ . □
Lemma 5.5.
We have that $σ_{r} (A) = \emptyset$ .
Proof.
Since $λ \in σ_{r} (A)$ , $\overline{λ} \in σ_{p} (A^{})$ the proof will be successful if we can show that $σ_{r} (A) = σ_{p} (A^{})$ . This is because we have considering that the eigenvalues of $A$ are symmetric on the real axis. In fact, we will consider the eigenvalue problem $A^{} U = λ U$ for $λ \in C$ and $0 \neq (u, v, ϕ) \in D (A^{})$ , that is, from (5.17) $\begin{array}{rcl} (5.18) & \{\begin{matrix} λ u + v = 0 \\ λ v - [c^{2} u_{x x} - β^{2} u] = 0 \\ λ ϕ + (ξ^{2} + η) ϕ + v (1, t) μ (ξ) = 0 . \end{matrix} \end{array}$ Replacing (5.18)₁ into (5.18)₂ we obtain $\begin{array}{rcl} (5.19) & - λ^{2} u - [c^{2} u_{x x} - β^{2} u] = 0 \end{array}$ with the following boundary condition $\begin{array}{rcl} (5.20) & \{\begin{matrix} u (0, t) = 0 \\ c^{2} u_{x} (1, t) = - λ {(λ + η)}^{α - 1} u (1, t) . \end{matrix} \end{array}$ However, the system (5.19)–(5.20) is exactly the eigenvalue problem of $A$ . Thus, $A^{}$ has the same eigenvalues with $A$ . The lemma follows. □

With the previous lemmas and using the Theorem 5.1 we have proved that.
Theorem 5.6.
The* $C_{0}$ -semigroup of contraction $S_{A} {(t)}_{t ⩾ 0}$ is asymptotically stable, that is, $\begin{array}{rcl} {‖ S_{A} (t) U_{0} ‖}_{H} \to 0 as t \to \infty, for U_{0} \in H . \end{array}$

To achieve our second proof we use the following result.
Theorem 5.7 ([2]).

Let $S (t) = e^{A t}$ be a C ₀ -semigroup of contractions on Hilbert space $H$ . If $\begin{array}{rcl} i R \subseteq ϱ (A) and sup_{| β | ⩾ 1} \frac{1}{β^{ℓ}} {‖ {(i β I - A)}^{- 1} ‖}_{L (H)} < M \end{array}$ for some ℓ, then there exist c such that $\begin{array}{rcl} {‖ e^{A t} U_{0} ‖}^{2} ⩽ \frac{c}{t^{\frac{2}{ℓ}}} ‖ U_{0} ‖_{D (A)}^{2} . \end{array}$

Now, we will prove the second main theorem of this section.

Theorem 5.8.

If $η \neq 0$ then the semigroup ${S_{A} (t)}_{t ⩾ 0}$ is polynomially stable and $\begin{array}{rcl} (5.21) & {‖ S_{A} (t) U_{0} ‖}_{H} ⩽ \frac{1}{t^{1 / 2 (1 - α)}} ‖ U_{0} ‖_{D (A)} . \end{array}$

Proof.

We will study the resolvent equation $(i λ I - A) U = F$ , $λ \in R$ . That is, $\begin{array}{rcl} (5.22) & \{\begin{matrix} i λ u - v = f_{1} \\ i λ v - c^{2} u_{x x} + β^{2} u = f_{2} \\ (ξ^{2} + η + i λ) ϕ - v (1, t) μ (ξ) = f_{3}, \end{matrix} \end{array}$ where $F = (f_{1}, f_{2}, f_{3})$ . Taking the inner product in $H$ with U and using (3.5) we have $\begin{array}{rcl} | Re {⟨ A U, U ⟩}_{H} | ⩽ ‖ U ‖_{H} ‖ F ‖_{H}, \end{array}$ that is, $\begin{array}{rcl} (5.23) & \frac{sin (α π)}{π} \int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ ⩽ ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ Moreover, from (5.22)₁ we have $| | λ | | u (1, t) | - | f_{1} (1) | | ⩽ | v (1, t) |$ , then $\begin{array}{rcl} (5.24) & | λ |^{2} {| u (1, t) |}^{2} ⩽ C {| f_{1} (1) |}^{2} + C {| v (1, t) |}^{2} . \end{array}$ On the other hand, from (2.9)₅ and using the Cauchy-Schwartz inequality we have $\begin{array}{rcl} c^{4} {| u_{x} (1, t) |}^{2} & = & {| \frac{sin (α π)}{π} \int_{R} μ (ξ) ϕ (ξ, t) d ξ |}^{2} \\ ⩽ & {| \frac{sin (α π)}{π} |}^{2} {| \int_{R} {(ξ^{2} + η)}^{- 1 / 2} μ (ξ) {(ξ^{2} + η)}^{1 / 2} ϕ (ξ, t) d ξ |}^{2} \\ ⩽ & (\int_{R} {(ξ^{2} + η)}^{- 1} {| μ (ξ) |}^{2} d ξ) {| \frac{sin (α π)}{π} |}^{2} \int_{R} (ξ^{2} + η) {| ϕ (ξ, t) |}^{2} d ξ \\ ⩽ & C ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ Hence $\begin{matrix} (5.25) & {| u_{x} (1, t) |}^{2} ⩽ C ‖ U ‖_{H} ‖ F ‖_{H} . \end{matrix}$ From (5.22)₃ we obtain $\begin{array}{l} v (1, t) μ (ξ) = (ξ^{2} + η + i λ) ϕ - f_{3} (ξ) . \end{array}$ Multiplying this equality by ${(ξ^{2} + η + i λ)}^{- 1} μ (ξ)$ we have $\begin{array}{rcl} {(ξ^{2} + η + i λ)}^{- 1} μ^{2} (ξ) v (1, t) & = & μ (ξ) ϕ - {(ξ^{2} + η + i λ)}^{- 1} μ (ξ) f_{3} (ξ) . \end{array}$ Applying absolute values and integrating over $ξ \in R$ we have $\begin{array}{l} (\int_{R} {| (ξ^{2} + η + i λ) |}^{- 1} {| μ (ξ) |}^{2} d ξ) | v (1, t) | \\ ⩽ \int_{R} | μ (ξ) | | ϕ | d ξ + \int_{R} {| (ξ^{2} + η + i λ) |}^{- 1} | μ (ξ) | | f_{3} (ξ) | d ξ \\ ⩽ \int_{R} {(ξ^{2} + η)}^{- 1 / 2} | μ (ξ) | {(ξ^{2} + η)}^{1 / 2} | ϕ | d ξ + \int_{R} {| (ξ^{2} + η + i λ) |}^{- 1} | μ (ξ) | | f_{3} (ξ) | d ξ . \end{array}$ Applying the Cauchy-Schwartz inequality and straightforward estimates it follows that $\begin{array}{rcl} (\int_{R} \frac{| μ (ξ) |^{2} d ξ}{(ξ^{2} + η + | λ |)}) | v (1, t) | & ⩽ & {(\int_{R} \frac{| μ (ξ) |^{2}}{ξ^{2} + η} d ξ)}^{1 / 2} {(\int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ)}^{1 / 2} \\ + {(\int_{R} \frac{| μ (ξ) |^{2} d ξ}{{(ξ^{2} + η + | λ |)}^{2}})}^{1 / 2} {(\int_{R} | f_{3} |^{2} d ξ)}^{1 / 2} . \end{array}$ Applying power squared on both sides of the inequality and using $2 a b ⩽ a^{2} + b^{2}$ we obtain $\begin{array}{rcl} {(\int_{R} \frac{| μ (ξ) |^{2} d ξ}{(ξ^{2} + η + | λ |)})}^{2} {| v (1, t) |}^{2} & ⩽ & 2 (\int_{R} \frac{| μ (ξ) |^{2} d ξ}{(ξ^{2} + η)}) (\int_{R} (ξ^{2} + η) | ϕ |^{2} d ξ) \\ (5.26) & + 2 (\int_{R} \frac{| μ (ξ) |^{2} d ξ}{{(ξ^{2} + η + | λ |)}^{2}}) (\int_{R} {| f_{3} (ξ) |}^{2} d ξ) . \end{array}$ Hence, using (5.23) $\begin{array}{l} {(\int_{R} \frac{| μ (ξ) |^{2}}{| λ | + ξ^{2} + η} d ξ)}^{2} {| v (1, t) |}^{2} \\ (5.27) & ⩽ C (\int_{R} \frac{| μ (ξ) |^{2}}{(ξ^{2} + η)} d ξ) ‖ U ‖_{H} ‖ F ‖_{H} + C (\int_{R} \frac{| μ (ξ) |^{2}}{{(| λ | + ξ^{2} + η)}^{2}} d ξ) ‖ F ‖_{H}^{2} . \end{array}$ From Lemma 2.2, it follows that $\begin{array}{l} (5.28) & {| v (1, t) |}^{2} ⩽ C | λ |^{2 - 2 α} ‖ U ‖_{H} ‖ F ‖_{H} + C ‖ F ‖_{H}^{2} . \end{array}$

To conclude the proof of the theorem we need the following lemma:

Lemma 5.9.

Let $q \in H^{2} (0, 1)$ . Then we have $\begin{array}{l} \int_{0}^{1} [2 q | v |^{2} + 2 c^{2} q | u_{x} |^{2} - c^{2} q_{x x} | u |^{2} + 2 β^{2} q | u |^{2}] d x \\ (5.29) & = 2 c^{2} q Re (\overline{u} u_{x}) - q_{x} | u |^{2} /_{0}^{1} + R, \end{array}$ where $\begin{array}{rcl} (5.30) & R = 2 \int_{0}^{1} q Re (\overline{u} f_{2}) d x - 2 \int_{0}^{1} q Re (v {\overline{f}}_{1}) d x . \end{array}$

Remark 5.10.

For R, we have $\begin{array}{rcl} (5.31) & | R | ⩽ C ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$

Proof.

Proof of Lemma Multiplying (5.22)₂ by $q \overline{u}$ and integrating over $(0, 1)$ we have $\begin{array}{l} - \int_{0}^{1} q (\overline{i λ u}) v d x - c^{2} \int_{0}^{1} q u_{x x} \overline{u} d x + β^{2} \int_{0}^{1} q | u |^{2} d x = \int_{0}^{1} q \overline{u} f_{2} d x . \end{array}$ From (5.22)₁ we have $- (\overline{i λ u}) = \overline{v} + {\overline{f}}_{1}$ . Then $\begin{array}{l} - \int_{0}^{1} q (\overline{v} + {\overline{f}}_{1}) v d x - c^{2} \int_{0}^{1} q u_{x x} \overline{u} d x + β^{2} \int_{0}^{1} q | u |^{2} d x = \int_{0}^{1} q \overline{u} f_{2} d x . \end{array}$ Thus $\begin{array}{l} \int_{0}^{1} q | v |^{2} d x - c^{2} \int_{0}^{1} q u_{x x} \overline{u} d x + β^{2} \int_{0}^{1} q | u |^{2} d x = \int_{0}^{1} q \overline{u} f_{2} d x - \int_{0}^{1} q v {\overline{f}}_{1} d x . \end{array}$ Then integrating by parts $\begin{array}{l} \int_{0}^{1} q | v |^{2} d x + c^{2} \int_{0}^{1} q | u_{x} |^{2} d x + c^{2} \int_{0}^{1} q_{x} \overline{u} u_{x} d x + β^{2} \int_{0}^{1} q | u |^{2} d x \\ = \int_{0}^{1} q \overline{u} f_{2} d x - \int_{0}^{1} q v {\overline{f}}_{1} d x + c^{2} q \overline{u} u_{x} /_{0}^{1} . \end{array}$ Taking the real part it follows that $\begin{array}{l} \int_{0}^{1} q | v |^{2} d x + c^{2} \int_{0}^{1} q | u_{x} |^{2} d x + c^{2} \int_{0}^{1} q_{x} Re (\overline{u} u_{x}) d x + β^{2} \int_{0}^{1} q | u |^{2} d x \\ (5.32) & = \int_{0}^{1} q Re (\overline{u} f_{2}) d x - \int_{0}^{1} q Re (v {\overline{f}}_{1}) d x + c^{2} q Re (\overline{u} u_{x}) /_{0}^{1} \end{array}$ On the other hand, $Re (f {\overline{f}}_{x}) = \frac{1}{2} \frac{d}{d x} | f |^{2}$ . Hence into (5.32) we have $\begin{array}{l} \int_{0}^{1} q | v |^{2} d x + c^{2} \int_{0}^{1} q | u_{x} |^{2} d x + \frac{c^{2}}{2} \int_{0}^{1} q_{x} \frac{d}{d x} | u |^{2} d x + β^{2} \int_{0}^{1} q | u |^{2} d x \\ = \int_{0}^{1} q Re (\overline{u} f_{2}) d x - \int_{0}^{1} q Re (v {\overline{f}}_{1}) d x + c^{2} q Re (\overline{u} u_{x}) /_{0}^{1} \end{array}$ Then $\begin{array}{l} \int_{0}^{1} q | v |^{2} d x + c^{2} \int_{0}^{1} q | u_{x} |^{2} d x - \frac{c^{2}}{2} \int_{0}^{1} q_{x x} | u |^{2} d x + β^{2} \int_{0}^{1} q | u |^{2} d x \\ = \int_{0}^{1} q Re (\overline{u} f_{2}) d x - \int_{0}^{1} q Re (v {\overline{f}}_{1}) d x + c^{2} q Re (\overline{u} u_{x}) - \frac{c^{2}}{2} q_{x} | u |^{2} /_{0}^{1} . \end{array}$ It follows that $\begin{array}{l} (5.33) & \int_{0}^{1} [2 q | v |^{2} + 2 c^{2} q | u_{x} |^{2} - c^{2} q_{x x} | u |^{2} + 2 β^{2} q | u |^{2}] d x . \\ = 2 \int_{0}^{1} q Re (\overline{u} f_{2}) d x - 2 \int_{0}^{1} q Re (v {\overline{f}}_{1}) d x + 2 c^{2} q Re (\overline{u} u_{x}) - c^{2} q_{x} | u |^{2} /_{0}^{1} \end{array}$ and, the lemma follows. □

Now, we return to the proof of the Theorem 5.8. Taking $q (x) = 1$ in Lemma 5.9 we have $\begin{array}{l} (5.34) & \int_{0}^{1} [2 | v |^{2} + 2 c^{2} | u_{x} |^{2} + 2 β^{2} | u |^{2}] d x = 2 c^{2} Re [\overline{u} (1, t) u_{x} (1, t)] + R . \end{array}$ Using the Young inequality and (5.25) we get $\begin{array}{rcl} \int_{0}^{1} [| v |^{2} + c^{2} | u_{x} |^{2} + β^{2} | u |^{2}] d x & ⩽ & c^{2} [{| u (1, t) |}^{2} + {| u_{x} (1, t) |}^{2}] + C ‖ U ‖_{H} ‖ F ‖_{H} \\ ⩽ & c^{2} {| u (1, t) |}^{2} + c^{2} C ‖ U ‖_{H} ‖ F ‖_{H} + C ‖ U ‖_{H} ‖ F ‖_{H} \\ (5.35) & ⩽ & c^{2} {| u (1, t) |}^{2} + C ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ Moreover, from (5.23) $\begin{array}{rcl} (5.36) & \frac{sin (α π)}{π} \int_{R} | ϕ |^{2} d ξ ⩽ C ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ Hence, adding (5.35) and (5.36) it follows that $\begin{array}{rcl} (5.37) & ‖ U ‖_{H}^{2} ⩽ c^{2} {| u (1, t) |}^{2} + C ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ From (5.22)₁ for $λ \neq 0$ , we have $| u (1, t) |^{2} = \frac{| v (1, t) |^{2} + 2 | v (1, t) | | f_{1} | + | f_{1} |^{2}}{| λ |^{2}}$ . Then for $λ \neq 0$ , $\begin{array}{rcl} ‖ U ‖_{H}^{2} & ⩽ & C | λ |^{2 - 2 α} ‖ U ‖_{H} ‖ F ‖_{H} + C ‖ U ‖_{H} ‖ F ‖_{H} + C ‖ F ‖_{H}^{2} + \frac{C}{| λ |^{2}} ‖ U ‖_{H}^{2} \\ (5.38) & + \frac{C}{| λ |^{2}} ‖ F ‖_{H}^{2} + \frac{C}{| λ |^{2}} ‖ U ‖_{H} ‖ F ‖_{H} . \end{array}$ If $| λ | > 1$ we get $‖ U ‖_{H}^{2} ⩽ | λ |^{4 (1 - α)} ‖ F ‖_{H}^{2} ⟺ ‖ U ‖_{H} ⩽ | λ |^{2 (1 - α)} ‖ F ‖_{H}$ . It follows that $\begin{array}{rcl} \frac{1}{| λ |^{2 (1 - α)}} {‖ {(i λ I - A)}^{- 1} ‖}_{L (H)} ⩽ C, \forall λ \in R, \end{array}$ for a positive constant C. The conclusion follows applying the Theorem 5.7. □

Footnotes

Acknowledgements

The first author was partially supported by CNPq-Brazil: grants 158706/2014-5, 164793/2015-1 and 402689/2012-7. The authors were partially financed by project Fondecyt 1191137.

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