ISS-based robustness to various neglected damping mechanisms for the 1-D wave PDE

Abstract

This paper is devoted to the study of the robustness properties of the 1-D wave equation for an elastic vibrating string under four different damping mechanisms that are usually neglected in the study of the wave equation: (i) friction with the surrounding medium of the string (or viscous damping), (ii) thermoelastic phenomena (or thermal damping), (iii) internal friction of the string (or Kelvin-Voigt damping), and (iv) friction at the free end of the string (the so-called passive damper). The passive damper is also the simplest boundary feedback law that guarantees exponential stability for the string. We study robustness with respect to distributed inputs and boundary disturbances in the context of Input-to-State Stability (ISS). By constructing appropriate ISS Lyapunov functionals, we prove the ISS property expressed in various spatial norms.

Appendix A Well-posedness of 1-D wave models

We next present the well-posedness analysis for all studied models. However, it should be noted that such an analysis provides sufficient and not necessary conditions for the existence/uniqueness of a solution with specific regularity properties. On the other hand, Theorem 1, Theorem 2 and Theorem 3 assume less regularity properties for the external inputs. For example, the well-posedness analysis given below for Model (D) requires that $f\in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$ is a sufficient condition for the existence/uniqueness of a solution for Model (D), while Theorem 3 simply requires that $f\in C^{0} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$. This is important because one cannot exclude the possibility that a solution for Model (D) with the regularity properties that are specified in Theorem 3 exists for less regular inputs (see the discussion on pages 105-110 in [31]).

The results that follow use a similar notation (for example, the state is U, the state space is X, the linear unbounded operator is A, etc.). However, the reader should not be tempted, by an overlapping notation for different quantities, to compare the different quantities in the analysis of each one of the models.

(1) Well-posedness analysis of Model (B).

Model (B) consists of equations (2), (5), (6). Applying the transformation

$$\begin{aligned} w(t,x)=u_{t} (t,x)\quad v(t,x)=u(t,x)-d(t)x, \ for \ t\ge 0,x\in [0,1] \end{aligned}$$

(A1)

and assuming that $d\in C^{1} \left( {\mathbb {R}} _{+} \right)$, we obtain the following first-order in time model

$$\begin{aligned} U_{t}+AU=F \end{aligned}$$

(A2)

where

$$\begin{aligned}{} & {} U=(v,w) \end{aligned}$$

(A3)

$$\begin{aligned}{} & {} F(t,x)=\left( -{\dot{d}}(t)x,f(t,x)\right) , \ for \ t\ge 0,x\in [0,1] \end{aligned}$$

(A4)

and $A:D(A)\rightarrow X$ is the linear unbounded operator defined by

$$\begin{aligned} AU=\left( -w,-c^{2} v''+\mu w\right) , for \ U=(v,w)\in D(A) \end{aligned}$$

(A5)

with X being the Hilbert space

$$\begin{aligned} X=\left\{ \, v\in H^{1} (0,1)\,:\, v(0)=0\, \right\} \times L^{2} (0,1) \end{aligned}$$

(A6)

with scalar product

$$\begin{aligned} \left( U,{\bar{U}}\right)= & {} c^{2} \int _{0}^{1}v'(x){\bar{v}}'(x)dx +\int _{0}^{1}v(x){\bar{v}}(x)dx +\int _{0}^{1}w(x){\bar{w}}(x)dx, \nonumber \\{} & {} \qquad for \ U=(v,w)\in X, \ {\bar{U}}=({\bar{v}},{\bar{w}})\in X \end{aligned}$$

(A7)

and $D(A)\subset X$ being the following linear subspace of X

$$\begin{aligned} D(A)=\left\{ \, (v,w)\in H^{2} (0,1)\times H^{1} (0,1)\,:\, v(0)=w(0)=v'(1)+aw(1)=0\, \right\} \nonumber \\ \end{aligned}$$

(A8)

Let $I:X\rightarrow X$ be the identity operator. We next show that $A+I$ is a maximal monotone operator (see [5]). Indeed, we have for all $U=(v,w)\in D(A)$

$$\begin{aligned} \left( (A+I)U,U\right)= & {} -c^{2} \int _{0}^{1}w'(x)v'(x)dx -\int _{0}^{1}w(x)v(x)dx -c^{2} \int _{0}^{1}v''(x)w(x)dx \\{} & {} {+(\mu +1)\left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} } \\= & {} -c^{2} w(1)v'(1)-\int _{0}^{1}w(x)v(x)dx +(\mu +1)\left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} \\\ge & {} ac^{2} w^{2} (1)-\left\| w\right\| _{2} \left\| v\right\| _{2} +(\mu +1)\left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} \\\ge & {} ac^{2} w^{2} (1)+\left( \mu +\frac{1}{2} \right) \left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\frac{1}{2} \left\| v\right\| _{2}^{2} \ge 0 \end{aligned}$$

In the above we have used (A5), (A7), (A8), integration by parts, the Cauchy-Schwarz inequality and the fact that $\left\| w\right\| _{2} \left\| v\right\| _{2} \le \frac{1}{2} \left\| w\right\| _{2}^{2} +\frac{1}{2} \left\| v\right\| _{2}^{2}$. Moreover, for every ${\bar{F}}=\left( {\bar{F}}_{1},{\bar{F}}_{2} \right) \in X$ the equation $(A+2I)U={\bar{F}}\in X$ has a unique solution $U=(v,w)\in D(A)$. Indeed, the equation $(A+2I)U={\bar{F}}\in X$ gives $w=2v-{\bar{F}}_{1}$ and $v(x)={\tilde{v}}(x)+\frac{a{\bar{F}}_{1} (1)}{2a+1} x$ for $x\in [0,1]$, where ${\tilde{v}}\in H^{2} (0,1)$ is the unique solution of the boundary value problem

$$\begin{aligned}{} & {} {-c^{2} {\tilde{v}}''(x)+2(\mu +2){\tilde{v}}(x)=(\mu +2){\bar{F}}_{1} (x)+{\bar{F}}_{2} (x)-2(\mu +2)\frac{a{\bar{F}}_{1} (1)}{2a+1} x, }\nonumber \\{} & {} \quad {\ for \ x\in (0,1) \ a.e., \ with \ {\tilde{v}}(0)=0 \ and \ {\tilde{v}}'(1)=-2a{\tilde{v}}(1) } \end{aligned}$$

(A9)

The fact that the boundary-value problem (A9) has a unique solution follows from the methodology described on pages 221-229 in [5] and the Lax-Milgram Theorem (Corollary 5.8 on page 140 in [5]).

Thus $A+I$ is a maximal monotone operator and consequently (using the Hille-Yosida Theorem) A is the generator of a continuous semigroup of contractions $S(t):X\rightarrow X$. It follows from Theorem 7.10 on page 198 in [5] that for every $F\in C^{1} \left( {\mathbb {R}} _{+};X\right)$ and for every $U_{0} \in D(A)$ there exists a unique solution $U\in C^{1} \left( {\mathbb {R}} _{+};X\right) \cap C^{0} \left( {\mathbb {R}} _{+};D(A)\right)$ of the initial-value problem (A.2) with initial condition $U(0)=U_{0}$.

Going back to the original state variables (and using (A1)), we conclude that for every $d\in C^{2} \left( {\mathbb {R}} _{+} \right)$, $f\in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$ and for every $u_{0} \in H^{2} (0,1)$, $w_{0} \in H^{1} (0,1)$ with $u_{0} (0)=w_{0} (0)=0$, $u'_{0} (1)=-aw_{0} (1)+d(0)$ there exists a unique solution $u\in C^{1} \left( {\mathbb {R}} _{+};H^{1} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$ with $u_{t} \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{1} (0,1)\right)$ of the initial-boundary value problem (2), (5), (6) with $u[0]=u_{0}$, $u_{t} [0]=w_{0}$ that additionally satisfies $u_{t} (t,0)=0$ for all $t\ge 0$.

(2) Well-posedness analysis of Model (C).

Model (C) consists of equations (2), (6), (7), (8), (9). Applying the transformation (A1) and assuming that $d\in C^{1} \left( {\mathbb {R}} _{+} \right)$, we obtain the first-order in time model (A2), where

$$\begin{aligned}{} & {} U=(v,w,\theta ) \end{aligned}$$

(A10)

$$\begin{aligned}{} & {} F(t,x)=\left( -{\dot{d}}(t)x,f(t,x),0\right) , \ for \ t\ge 0,x\in [0,1] \end{aligned}$$

(A11)

and $A:D(A)\rightarrow X$ is the linear unbounded operator defined by

$$\begin{aligned} AU=\left( -w,-c^{2} v''+\mu w+b\theta ',-k\theta ''+\lambda w'\right) , \ for \ U=(v,w,\theta )\in D(A)\nonumber \\ \end{aligned}$$

(A12)

with X being the Hilbert space

$$\begin{aligned} X=\left\{ \, v\in H^{1} (0,1)\,:\, v(0)=0\, \right\} \times L^{2} (0,1)\times L^{2} (0,1) \end{aligned}$$

(A13)

with scalar product

$$\begin{aligned} \left( U,{\bar{U}}\right)= & {} c^{2} \int _{0}^{1}v'(x){\bar{v}}'(x)dx +\int _{0}^{1}v(x){\bar{v}}(x)dx \nonumber \\{} & {} {+\int _{0}^{1}w(x){\bar{w}}(x)dx +\frac{b}{\lambda } \int _{0}^{1}\theta (x){\bar{\theta }}(x)dx,} \nonumber \\{} & {} {\qquad \ for \ U=(v,w,\theta )\in X, \ {\bar{U}}=({\bar{v}},{\bar{w}},{\bar{\theta }})\in X } \end{aligned}$$

(A14)

and $D(A)\subset X$ being the following linear subspace of X

$$\begin{aligned} D(A)= {\left\{ \, (v,w,\theta )\in H^{2} (0,1)\times H^{1} (0,1)\times H^{2} (0,1)\,:\, \begin{array}{c} {v(0)=w(0)=\theta (0)=0} \\ {v'(1)+aw(1)=\theta (1)=0} \end{array}\, \right\} }\nonumber \\ \end{aligned}$$

(A15)

Let $I:X\rightarrow X$ be the identity operator. We next show that $A+I$ is a maximal monotone operator (see [5]). Indeed, we have for all $U=(v,w,\theta )\in D(A)$

$$\begin{aligned} \left( (A+I)U,U\right)= & {} -c^{2} \int _{0}^{1}w'(x)v'(x)dx -\int _{0}^{1}w(x)v(x)dx -c^{2} \int _{0}^{1}v''(x)w(x)dx \\{} & {} {+(\mu +1)\left\| w\right\| _{2}^{2} +b\int _{0}^{1}\theta '(x)w(x)dx -\frac{bk}{\lambda } \int _{0}^{1}\theta (x)\theta ''(x)dx } \\{} & {} {+b\int _{0}^{1}\theta (x)w'(x)dx +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} +\frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} } \\= & {} -c^{2} w(1)v'(1)-\int _{0}^{1}w(x)v(x)dx +(\mu +1)\left\| w\right\| _{2}^{2} \\{} & {} {+\frac{bk}{\lambda } \left\| \theta '\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} +\frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} } \\\ge & {} ac^{2} w^{2} (1)-\left\| w\right\| _{2} \left\| v\right\| _{2} +(\mu +1)\left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\left\| v\right\| _{2}^{2} +\frac{bk}{\lambda } \left\| \theta '\right\| _{2}^{2} +\frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} \\\ge & {} ac^{2} w^{2} (1)+\left( \mu +\frac{1}{2} \right) \left\| w\right\| _{2}^{2} +c^{2} \left\| v'\right\| _{2}^{2} +\frac{1}{2} \left\| v\right\| _{2}^{2} +\frac{bk}{\lambda } \left\| \theta '\right\| _{2}^{2} +\frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} \ge 0 \end{aligned}$$

In the above we have used (A12), (A14), (A15), three integrations by parts, the Cauchy-Schwarz inequality and the fact that $\left\| w\right\| _{2} \left\| v\right\| _{2} \le \frac{1}{2} \left\| w\right\| _{2}^{2} +\frac{1}{2} \left\| v\right\| _{2}^{2}$. Moreover, for every ${\bar{F}}=\left( {\bar{F}}_{1},{\bar{F}}_{2},{\bar{F}}_{3} \right) \in X$ the equation $(A+2I)U={\bar{F}}\in X$ has a unique solution $U=(v,w,\theta )\in D(A)$. Indeed, the equation $(A+2I)U={\bar{F}}\in X$ gives $w=2v-{\bar{F}}_{1}$ and $v(x)={\tilde{v}}(x)+\frac{a{\bar{F}}_{1} (1)}{2a+1} x$ for $x\in [0,1]$, where $({\tilde{v}},\theta )\in H^{2} (0,1)\times H^{2} (0,1)$ is the unique solution of the boundary-value problem

$$\begin{aligned}{} & {} {-c^{2} {\tilde{v}}''(x)+2(\mu +2){\tilde{v}}(x)+b\theta '(x)=\varphi _{1} (x)} \nonumber \\{} & {} \quad {-k\theta ''(x)+2\lambda {\tilde{v}}'(x)+2\theta (x)=\varphi _{2} (x)} \nonumber \\{} & {} \quad { \ for \ x\in (0,1) \ a.e. \ with ~{\tilde{v}}(0)=\theta (0)=0 } \nonumber \\{} & {} \quad {\ and \ {\tilde{v}}'(1)+2a{\tilde{v}}(1)=\theta (1)=0 } \end{aligned}$$

(A16)

with $\varphi _{1} (x)={\bar{F}}_{2} (x)+(\mu +2){\bar{F}}_{1} (x)-2(\mu +2)\frac{a{\bar{F}}_{1} (1)}{2a+1} x$ and $\varphi _{2} (x)={\bar{F}}_{3} (x)+\lambda {\bar{F}}'_{1} (x)-\frac{2\lambda a{\bar{F}}_{1} (1)}{2a+1}$. Let Y be the Hilbert space

$$\begin{aligned} Y=\left\{ \, ({\tilde{v}},\theta )\in H^{1} (0,1)\times H^{1} (0,1)\,:\, {\tilde{v}}(0)=\theta (0)=\theta (1)=0\, \right\} \end{aligned}$$

(A17)

with the usual scalar product

$$\begin{aligned} \left( ({\tilde{v}},\theta ),(r,p)\right)= & {} \int _{0}^{1}{\tilde{v}}'(x)r'(x)dx +\int _{0}^{1}{\tilde{v}}(x)r(x)dx \nonumber \\{} & {} {+\int _{0}^{1}\theta '(x)p'(x)dx +\int _{0}^{1}\theta (x)p(x)dx, } \nonumber \\{} & {} \qquad \ for \ ({\tilde{v}},\theta )\in Y, \ (r,p)\in Y \end{aligned}$$

(A18)

The fact that the boundary-value problem (A16) has a unique solution $({\tilde{v}},\theta )\in H^{2} (0,1)\times H^{2} (0,1)$ for every $(\varphi _{1},\varphi _{2} )\in L^{2} (0,1)\times L^{2} (0,1)$ is a direct consequence of the methodology described on pages 221-229 in [5] and the Lax-Milgram Theorem (Corollary 5.8 on page 140 in [5]) applied to the continuous, coercive, bilinear form $\alpha$ on Y defined by the formula

$$\begin{aligned} \alpha \left( ({\tilde{v}},\theta ),(r,p)\right)= & {} 2ac^{2} {\tilde{v}}(1)r(1)+c^{2} \int _{0}^{1}{\tilde{v}}'(x)r'(x)dx \\{} & {} {+2(\mu +2)\int _{0}^{1}{\tilde{v}}(x)r(x)dx +b\int _{0}^{1}\theta '(x)r(x)dx } \\{} & {} {+\frac{bk}{2\lambda } \int _{0}^{1}\theta '(x)p'(x)dx +b\int _{0}^{1}{\tilde{v}}'(x)p(x)dx +\frac{b}{\lambda } \int _{0}^{1}\theta (x)p(x)dx } \\{} & {} \qquad { \ for \ all \ ({\tilde{v}},\theta )\in Y, \ (r,p)\in Y } \end{aligned}$$

Thus $A+I$ is a maximal monotone operator and consequently (using the Hille-Yosida Theorem) A is the generator of a continuous semigroup of contractions $S(t):X\rightarrow X$. It follows from Theorem 7.10 on page 198 in [5] that for every $F\in C^{1} \left( {\mathbb {R}} _{+};X\right)$ and for every $U_{0} \in D(A)$ there exists a unique solution $U\in C^{1} \left( {\mathbb {R}} _{+};X\right) \cap C^{0} \left( {\mathbb {R}} _{+};D(A)\right)$ of the initial-value problem (A2) with initial condition $U(0)=U_{0}$.

Going back to the original state variables (and using (A1)), we conclude that for every $d\in C^{2} \left( {\mathbb {R}} _{+} \right)$, $f\in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$ and for every $u_{0} \in H^{2} (0,1)$, $w_{0} \in H^{1} (0,1)$, $\theta _{0} \in H^{2} (0,1)$ with $u_{0} (0)=w_{0} (0)=\theta (0)=\theta (1)=0$, $u'_{0} (1)=-aw_{0} (1) +d(0)$ there exists a unique solution $u\in C^{1} \left( {\mathbb {R}} _{+};H^{1} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$, $\theta \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$ with $u_{t} \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{1} (0,1)\right)$ of the initial-boundary value problem (2), (6), (7), (8), (9) with $u[0]=u_{0}$, $u_{t} [0]=w_{0}$, $\theta [0]=\theta _{0}$ that additionally satisfies $u_{t} (t,0)=0$ for all $t\ge 0$.

(3) Well-posedness analysis of Model (D).

Model (C) consists of equations (2), (8), (9), (10), (11). Applying the transformation

$$\begin{aligned}{} & {} {w(t,x)=u_{t} (t,x)-\sigma u_{xx} (t,x)+gu(t,x)}\nonumber \\{} & {} {v(t,x)=u_{t} (t,x)} \nonumber \\{} & {} {y(t)=u_{t} (t,1)} \nonumber \\{} & {} \quad { for \ t\ge 0, \ x\in [0,1] } \end{aligned}$$

(A19)

where

$$\begin{aligned} g:=\mu -\sigma ^{-1} c^{2}, \ \kappa :=\sigma ^{-1} c^{2} \end{aligned}$$

(A20)

we obtain the first-order in time model (A2), where

$$\begin{aligned}{} & {} U=(u,v,\theta ,w,y) \end{aligned}$$

(A21)

$$\begin{aligned}{} & {} F(t,x)=\left( 0,f(t,x),0,f(t,x),0\right) , \ for \ t\ge 0, \ x\in [0,1] \end{aligned}$$

(A22)

and $A:D(A)\rightarrow X$ is the linear unbounded operator defined by

$$\begin{aligned}{} & {} AU={\left( -\sigma u''-h,-\sigma v''+gv+\kappa h+b\theta ',-k\theta ''+\lambda v',\kappa h+b\theta ',a^{-1} v'(1)\right) } \nonumber \\{} & {} \quad { \ for \ U=(u,v,\theta ,w,y)\in D(A), \ where \ h=w-gu } \end{aligned}$$

(A23)

with X being the Hilbert space

$$\begin{aligned} X=\left( L^{2} (0,1)\right) ^{4} \times {{\mathbb {R}}} \end{aligned}$$

(A24)

with scalar product

$$\begin{aligned}{} & {} {\left( U,{\bar{U}}\right) =\int _{0}^{1}u(x){\bar{u}}(x)dx +\int _{0}^{1}v(x){\bar{v}}(x)dx } \nonumber \\{} & {} {\quad +\frac{b}{\lambda } \int _{0}^{1}\theta (x){\bar{\theta }}(x)dx +\int _{0}^{1}w(x){\bar{w}}(x)dx +a\sigma y{\bar{y}}, } \nonumber \\{} & {} {\qquad for \ U=(u,v,\theta ,w,y)\in X, \ {\bar{U}}=({\bar{u}},{\bar{v}},{\bar{\theta }},{\bar{w}},{\bar{y}})\in X } \end{aligned}$$

(A25)

and $D(A)\subset X$ being the following linear subspace of X

$$\begin{aligned}{} & {} {D(A)= \left\{ \, (u,v,\theta ,w,y)\in Z\,:\, \begin{array}{c} {u(0)=\theta (0)=v(0)=0} \\ {v(1)-y=u'(1)+ay=\theta (1)=0} \\ { where \ Z=\left( H^{2} (0,1)\right) ^{3} \times L^{2} (0,1)\times {{\mathbb {R}}} } \end{array}\; \right\} }\nonumber \\ \end{aligned}$$

(A26)

Let $I:X\rightarrow X$ be the identity operator. We next show that $A+\beta I$ is a maximal monotone operator (see [5]) for a sufficiently large constant $\beta >0$. Indeed, we have for all $U=(u,v,\theta ,w,y)\in D(A)$:

$$\begin{aligned}{} & {} {\left( (A+\beta I)U,U\right) =\int _{0}^{1}u(x)\left( -\sigma u''(x)+gu(x)-w(x)\right) dx } \\{} & {} \qquad {+\int _{0}^{1}v(x)\left( -\sigma v''(x)+gv(x)+\kappa (w(x)-gu(x))+b\theta '(x)\right) dx } \\{} & {} \qquad {+\frac{b}{\lambda } \int _{0}^{1}\theta (x)\left( -k\theta ''(x)+\lambda v'(x)\right) dx +\beta \left\| u\right\| _{2}^{2} +\beta \left\| v\right\| _{2}^{2} } \\{} & {} \qquad { +\beta \frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} +\beta \left\| w\right\| _{2}^{2} +a\sigma \beta y^{2} } \\{} & {} \qquad {+\int _{0}^{1}w(x)\left( \kappa (w(x)-gu(x))+b\theta '(x)\right) dx +\sigma yv'(1)} \\{} & {} \quad {=a\sigma yu(1)+\sigma \left\| u'\right\| _{2}^{2} -\left( 1+\kappa g\right) \int _{0}^{1}u(x)w(x)dx +b\int _{0}^{1}w(x)\theta '(x)dx } \\{} & {} \qquad {+\sigma \left\| v'\right\| _{2}^{2} +\sigma ^{-1} c^{2} \int _{0}^{1}v(x)w(x)dx -\kappa g\int _{0}^{1}v(x)u(x)dx +\frac{bk}{\lambda } \left\| \theta '\right\| _{2}^{2} } \\{} & {} \qquad {+\left( \beta +g\right) \left\| u\right\| _{2}^{2} +\left( \beta +g\right) \left\| v\right\| _{2}^{2} +\beta \frac{b}{\lambda } \left\| \theta \right\| _{2}^{2} +\left( \beta +\sigma ^{-1} c^{2} \right) \left\| w\right\| _{2}^{2} +a\sigma \beta y^{2} } \end{aligned}$$

In the above we have used (A23), (A25), (A26) and integration by parts. Using the fact that $yu(1) \ge -\beta y^{2} -\frac{u^{2} (1)}{4\beta }$ and the Cauchy-Schwarz inequality we obtain:

$$\begin{aligned} \left( (A+\beta I)U,U\right)\ge & {} -\frac{a\sigma }{4\beta } u^{2} (1)+\sigma \left\| u'\right\| _{2}^{2} -\left| 1+\kappa g\right| \left\| u\right\| _{2} \left\| w\right\| _{2} -b\left\| w\right\| _{2} \left\| \theta '\right\| _{2}\\{} & {} {+\sigma \left\| v'\right\| _{2}^{2} -\kappa \left\| v\right\| _{2} \left\| w\right\| _{2} -\kappa \left| g\right| \left\| u\right\| _{2} \left\| v\right\| _{2} +\frac{bk}{\lambda } \left\| \theta '\right\| _{2}^{2} } \\{} & {} {+\left( \beta +g\right) \left\| u\right\| _{2}^{2} +\left( \beta +g\right) \left\| v\right\| _{2}^{2} +\frac{\beta b}{\lambda } \left\| \theta \right\| _{2}^{2} +\left( \beta +\kappa \right) \left\| w\right\| _{2}^{2} } \end{aligned}$$

Since $u\in H^{2} (0,1)$ with $u(0)=0$ we get $u^{2} (1) =2\int _{0}^{1}u(x)u'(x)dx \le 2\left\| u\right\| _{2} \left\| u'\right\| _{2}$. Using this inequality and the inequalities $\left\| u\right\| _{2} \left\| w\right\| _{2} \le \left\| u\right\| _{2}^{2} +\left\| w\right\| _{2}^{2}$, $\left\| v\right\| _{2} \left\| w\right\| _{2} \le \left\| v\right\| _{2}^{2} +\left\| w\right\| _{2}^{2}$, $\left\| u\right\| _{2} \left\| v\right\| _{2} \le \left\| u\right\| _{2}^{2} +\left\| v\right\| _{2}^{2}$, $\left\| \theta '\right\| _{2} \left\| w\right\| _{2} \le \frac{k}{\lambda } \left\| \theta '\right\| _{2}^{2} +\frac{\lambda }{4k} \left\| w\right\| _{2}^{2}$, we get:

$$\begin{aligned} \left( (A+\beta I)U,U\right)\ge & {} -\frac{a\sigma }{2\beta } \left\| u\right\| _{2} \left\| u'\right\| _{2} +\sigma \left\| u'\right\| _{2}^{2} +\sigma \left\| v'\right\| _{2}^{2} \\{} & {} {+\left( \beta -\left| g\right| -1-2\kappa \left| g\right| \right) \left\| u\right\| _{2}^{2} +\left( \beta -\left| g\right| -\kappa -\kappa \left| g\right| \right) \left\| v\right\| _{2}^{2} } \\{} & {} {+\left( \beta -\frac{b\lambda }{4k} -1-\kappa \left| g\right| \right) \left\| w\right\| _{2}^{2} } \end{aligned}$$

Finally, using the inequality $\frac{a\sigma }{2\beta } \left\| u\right\| _{2} \left\| u'\right\| _{2} \le \sigma \left\| u'\right\| _{2}^{2} +\frac{a^{2} \sigma }{16\beta ^{2} } \left\| u\right\| _{2}^{2}$ we get:

$$\begin{aligned}{} & {} {\left( (A+\beta I)U,U\right) \ge \left( \beta -\left| g\right| -1-2\kappa \left| g\right| -\frac{a^{2} \sigma }{16\beta ^{2} } \right) \left\| u\right\| _{2}^{2} } \\{} & {} \quad {+\left( \beta -\left| g\right| -\kappa -\kappa \left| g\right| \right) \left\| v\right\| _{2}^{2} +\left( \beta -\frac{b\lambda }{4k} -1-\kappa \left| g\right| \right) \left\| w\right\| _{2}^{2} } \end{aligned}$$

The above inequality shows that for $\beta \ge 1+\kappa +\left( 2\kappa +1\right) \left| g\right| +\frac{a^{2} \sigma }{16\beta ^{2} } +\frac{b\lambda }{4k}$ we have $\left( (A+\beta I)U,U\right) \ge 0$ for all $U=(u,v,\theta ,w,y)\in D(A)$.

We next show that for sufficiently large $\beta >0$, the range of the linear operator $(A+(\beta +1)I)$ is X. We show that for every ${\bar{F}}=\left( {\bar{F}}_{1},{\bar{F}}_{2},{\bar{F}}_{3},{\bar{F}}_{4},{\bar{F}}_{5} \right) \in X$ the equation $(A+(\beta +1)I)U={\bar{F}}\in X$ has a unique solution $U=(u,v,\theta ,w,y)\in D(A)$ provided that $\beta >0$ is sufficiently large. The equation $(A+(\beta +1)I)U={\bar{F}}\in X$ gives $y=\frac{a{\bar{F}}_{5} }{1+a(\beta +1)} -\frac{{\tilde{v}}'(1)}{a(\beta +1)}$, $u(x)={\tilde{u}}(x)-a\left( {\tilde{v}}(1)+\frac{a{\bar{F}}_{5} }{1+a(\beta +1)} \right) x$, $w(x)=\frac{{\bar{F}}_{4} (x)-b\theta '(x)}{\beta +1+\kappa } +\frac{\kappa g}{\beta +1+\kappa } \left( {\tilde{u}}(x)-a\left( {\tilde{v}}(1)+\frac{a{\bar{F}}_{5} }{1+a(\beta +1)} \right) x\right)$ and $v(x)={\tilde{v}}(x)+\frac{a{\bar{F}}_{5} x}{1+a(\beta +1)}$ for $x\in [0,1]$, where $({\tilde{u}},{\tilde{v}},\theta )\in \left( H^{2} (0,1)\right) ^{3}$ is a solution of the boundary-value problem

$$\begin{aligned}{} & {} {-\sigma {\tilde{u}}''(x)+(\beta +1)\left( 1+\frac{g}{\beta +1+\kappa } \right) {\tilde{u}}(x) } \nonumber \\{} & {} \quad {-a(\beta +1)\left( 1+\frac{g}{\beta +1+\kappa } \right) {\tilde{v}}(1)x+\frac{b}{\beta +1+\kappa } \theta '(x)=\varphi _{1} (x)} \nonumber \\{} & {} \quad {-\sigma {\tilde{v}}''(x)+(\beta +1+g){\tilde{v}}(x)+\frac{b\left( \beta +\kappa \right) }{\beta +1+\kappa } \theta '(x) } \nonumber \\{} & {} \quad {-\frac{\kappa g\left( \beta +\kappa \right) }{\beta +1+\kappa } {\tilde{u}}(x)+\frac{a\kappa g\left( \beta +\kappa \right) }{\beta +1+\kappa } {\tilde{v}}(1)x=\varphi _{2} (x)} \nonumber \\{} & {} \quad {-k\theta ''(x)+\lambda {\tilde{v}}'(x)+(\beta +1)\theta (x)=\varphi _{3} (x)} \nonumber \\{} & {} \qquad {\ for \ x\in (0,1) \ a.e. \ with \ {\tilde{u}}(0)={\tilde{v}}(0)=\theta (0)=0} \nonumber \\{} & {} \qquad { and \ {\tilde{u}}'(1)={\tilde{v}}'(1)+a(\beta +1){\tilde{v}}(1)=\theta (1)=0 } \end{aligned}$$

(A27)

with $\varphi _{3} (x)={\bar{F}}_{3} (x)-\frac{a\lambda {\bar{F}}_{5} }{1+a(\beta +1)}$, $\varphi _{1} (x)={\bar{F}}_{1} (x)+\frac{1}{\beta +1+\kappa } {\bar{F}}_{4} (x)+\frac{a^{2} (\beta +1){\bar{F}}_{5} }{1+a(\beta +1)} \left( 1+\frac{g}{\beta +1+\kappa } \right) x$ and $\varphi _{2} (x)={\bar{F}}_{2} (x)-\frac{1}{\beta +1+\kappa } {\bar{F}}_{4} (x)-\left( \beta +1+g+\frac{a\kappa g\left( \beta +\kappa \right) }{\beta +1+\kappa } \right) \frac{a{\bar{F}}_{5} x}{1+a(\beta +1)}$. Let Y be the Hilbert space

$$\begin{aligned} Y=\left\{ \, ({\tilde{u}},{\tilde{v}},\theta )\in \left( H^{1} (0,1)\right) ^{3} \,:\, {\tilde{u}}(0)={\tilde{v}}(0)=\theta (0)=\theta (1)=0\, \right\} \end{aligned}$$

(A28)

with the usual scalar product

$$\begin{aligned}{} & {} \left( ({\tilde{u}},{\tilde{v}},\theta ),(q,r,p)\right) =\int _{0}^{1}{\tilde{u}}'(x)q'(x)dx +\int _{0}^{1}{\tilde{u}}(x)q(x)dx \nonumber \\{} & {} \quad {+\int _{0}^{1}{\tilde{v}}'(x)r'(x)dx +\int _{0}^{1}{\tilde{v}}(x)r(x)dx +\int _{0}^{1}\theta '(x)p'(x)dx +\int _{0}^{1}\theta (x)p(x)dx } \nonumber \\{} & {} \qquad {for \ ({\tilde{u}},{\tilde{v}},\theta )\in Y, \ (q,r,p)\in Y } \end{aligned}$$

(A29)

We show that the boundary-value problem (A27) has a unique solution $({\tilde{u}},{\tilde{v}},\theta )\in \left( H^{2} (0,1)\right) ^{3}$ for every $(\varphi _{1},\varphi _{2},\varphi _{3} )\in \left( L^{2} (0,1)\right) ^{3}$ provided that $\beta >0$ is sufficiently large. Let $R>0$ be an arbitrary constant and consider the continuous, bilinear form $\alpha$ on Y defined by the formula

$$\begin{aligned} \alpha \left( ({\tilde{u}},{\tilde{v}},\theta ),(q,r,p)\right)= & {} \frac{\sigma \sqrt{15\sigma } }{a^{2} \sqrt{\beta +1} } \int _{0}^{1}{\tilde{u}}'(x)q'(x)dx \nonumber \\{} & {} {+\frac{\sqrt{15\sigma \left( \beta +1\right) } }{a^{2} } \left( 1+\frac{g}{\beta +1+\kappa } \right) \int _{0}^{1}{\tilde{u}}(x)q(x)dx } \nonumber \\{} & {} {-\frac{\sqrt{15\sigma \left( \beta +1\right) } }{a} \left( 1+\frac{g}{\beta +1+\kappa } \right) {\tilde{v}}(1)\int _{0}^{1}xq(x)dx } \nonumber \\{} & {} { +\frac{b\sqrt{15\sigma } }{a^{2} \left( \beta +1+\kappa \right) \sqrt{\beta +1} } \int _{0}^{1}\theta '(x)q(x)dx }\nonumber \\{} & {} {+\sigma a(\beta +1){\tilde{v}}(1)r(1)+\sigma \int _{0}^{1}{\tilde{v}}'(x)r'(x)dx }\nonumber \\{} & {} {+(\beta +1+g)\int _{0}^{1}{\tilde{v}}(x)r(x)dx +\frac{b\left( \beta +\kappa \right) }{\beta +1+\kappa } \int _{0}^{1}\theta '(x)r(x)dx }\nonumber \\{} & {} {-\frac{\kappa g\left( \beta +\kappa \right) }{\beta +1+\kappa } \int _{0}^{1}{\tilde{u}}(x)r(x)dx +\frac{a\kappa g\left( \beta +\kappa \right) }{\beta +1+\kappa } {\tilde{v}}(1)\int _{0}^{1}xr(x)dx } \nonumber \\{} & {} {+\frac{bk\left( \beta +\kappa \right) }{\lambda (\beta +1+\kappa )} \int _{0}^{1}\theta '(x)p'(x)dx +\frac{b\left( \beta +\kappa \right) }{(\beta +1+\kappa )} \int _{0}^{1}{\tilde{v}}'(x)p(x)dx } \nonumber \\{} & {} { +\frac{b(\beta +1)\left( \beta +\kappa \right) }{\lambda (\beta +1+\kappa )} \int _{0}^{1}\theta (x)p(x)dx }\nonumber \\{} & {} {for \ all \ ({\tilde{u}},{\tilde{v}},\theta )\in Y, \ (q,r,p)\in Y } \end{aligned}$$

(A30)

We notice that continuity of the bilinear form $\alpha$ on Y defined by (A30) is established by means of the inequality $\left| {\tilde{v}}(1)\right| \le \sqrt{2\left\| {\tilde{v}}\right\| _{2} \left\| {\tilde{v}}'\right\| _{2} } \le \left\| {\tilde{v}}\right\| _{2} +\left\| {\tilde{v}}'\right\| _{2}$, which holds since ${\tilde{v}}\in H^{1} (0,1)$ with ${\tilde{v}}(0)=0$ (recall (A28)). Completing the squares and using the Cauchy-Schwarz inequality and the fact that $\left| {\tilde{v}}(1)\right| \le \sqrt{2\left\| {\tilde{v}}\right\| _{2} \left\| {\tilde{v}}'\right\| _{2} }$, we are in a position to establish the following inequality for all $({\tilde{u}},{\tilde{v}},\theta )\in Y$:

$$\begin{aligned}{} & {} {\alpha \left( ({\tilde{u}},{\tilde{v}},\theta ),({\tilde{u}},{\tilde{v}},\theta )\right) \ge \frac{\sigma \sqrt{15\sigma } }{a^{2} \sqrt{\beta +1} } \left\| {\tilde{u}}'\right\| _{2}^{2} +\frac{\sigma a(\beta +1)}{4} {\tilde{v}}^{2} (1)+\frac{\sigma }{2} \left\| {\tilde{v}}'\right\| _{2}^{2} } \nonumber \\{} & {} \quad {+\frac{\left\| {\tilde{u}}\right\| _{2}^{2} }{2a^{4} k\sqrt{\beta +1} } (\beta +1)a^{2} k\sqrt{15\sigma } } \nonumber \\{} & {} \quad {-\frac{\left\| {\tilde{u}}\right\| _{2}^{2} }{2a^{4} k\sqrt{\beta +1} } \left( \left( 2\sqrt{15\sigma } +a^{2} \kappa ^{2} \right) a^{2} k\left| g\right| +k\kappa a^{4} \left| g\right| \sqrt{\beta +1} +15\sigma b\lambda +10ak\left| g\right| ^{2} \right) } \nonumber \\{} & {} \quad {+\frac{1}{6\sigma } \left( \sigma \left( \beta +1\right) -\left( 6\sigma +3\sigma \kappa +a\kappa ^{2} \left| g\right| \right) \left| g\right| \right) \left\| {\tilde{v}}\right\| _{2}^{2} } \nonumber \\{} & {} \quad {+\frac{bk\left( \beta +\kappa \right) }{2\lambda (\beta +1+\kappa )} \left\| \theta '\right\| _{2}^{2} +\frac{b(\beta +1)\left( \beta +\kappa \right) }{\lambda (\beta +1+\kappa )} \left\| \theta \right\| _{2}^{2} } \end{aligned}$$

(A31)

It follows from (A31) that the bilinear form $\alpha$ on Y defined by (A30) is coercive when

$$\begin{aligned}{} & {} {\beta +1-\frac{\kappa a^{2} \left| g\right| }{\sqrt{15\sigma } } \sqrt{\beta +1}>\frac{\left( 2\sqrt{15\sigma } +a^{2} \kappa ^{2} \right) a^{2} k\left| g\right| +15\sigma b\lambda +10ak\left| g\right| ^{2} }{a^{2} k\sqrt{15\sigma } } } \\{} & {} \quad {\beta +1>\left( 6+3\kappa +\sigma ^{-1} a\kappa ^{2} \left| g\right| \right) \left| g\right| } \end{aligned}$$

The above inequalities are satisfied when $\beta >0$ is sufficiently large (notice that left hand sides of the above inequalities tend to $+\infty$ as $\beta \rightarrow +\infty$). Following the methodology described on pages 221-229 in [5] and using the Lax-Milgram Theorem (Corollary 5.8 on page 140 in [5]) applied to the continuous, coercive, bilinear form $\alpha$ on Y defined by (A30) we conclude that the boundary-value problem (A27) has a unique solution $({\tilde{u}},{\tilde{v}},\theta )\in \left( H^{2} (0,1)\right) ^{3}$ for every $(\varphi _{1},\varphi _{2},\varphi _{3} )\in \left( L^{2} (0,1)\right) ^{3}$ provided that $\beta >0$ is sufficiently large.

Thus $A+\beta I$ is a maximal monotone operator when $\beta >0$ is sufficiently large and consequently (using the Hille-Yosida Theorem) A is the generator of a continuous semigroup of contractions $S(t):X\rightarrow X$. It follows from Theorem 7.10 on page 198 in [5] that for every $F\in C^{1} \left( {\mathbb {R}} _{+};X\right)$ and for every $U_{0} \in D(A)$ there exists a unique solution $U\in C^{1} \left( {\mathbb {R}} _{+};X\right) \cap C^{0} \left( {\mathbb {R}} _{+};D(A)\right)$ of the initial-value problem (A2) with initial condition $U(0)=U_{0}$.

Going back to the original state variables (and using (A19)), we conclude that for every $f\in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$ and for every $u_{0} \in H^{2} (0,1)$, $v_{0} \in H^{2} (0,1)$, $\theta _{0} \in H^{2} (0,1)$ with $u_{0} (0)=v_{0} (0)=\theta (0)=\theta (1)=0$, $u'_{0} (1)=-av_{0} (1)$ there exists a unique solution $u\in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$, $\theta \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$ with $u_{t} \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right) \cap C^{0} \left( {\mathbb {R}} _{+};H^{2} (0,1)\right)$, $u_{xx} \in C^{1} \left( {\mathbb {R}} _{+};L^{2} (0,1)\right)$, $u_{t} (\, \cdot \,,1)\in C^{1} \left( {\mathbb {R}} _{+} \right)$ of the initial-boundary value problem (2), (8), (9), (10), (11) with $u[0]=u_{0}$, $u_{t} [0]=v_{0}$, $\theta [0]=\theta _{0}$ that additionally satisfies $u_{t} (t,0)=0$, $u_{xt} (t,1)=-a\frac{d}{d\, t} \left( u_{t} (t,1)\right)$ for all $t\ge 0$.