Proof. Let $\lambda \in {\mathbb {C}}$ be fixed. For ${\boldsymbol {f}} = (f,\,g) \in {\mathcal {H}}$, we look for the unique solution to the functional equation

(4.1)\begin{equation} ({\mathbb{A}} -\lambda){\boldsymbol{u}} = {\boldsymbol{f}}. \end{equation}

In components, this reads

\[ \begin{cases} \alpha u_{xxx} + \beta \eta_{xx} -\lambda\eta= f, \\- \eta_x - \beta u_{xx} -\lambda u= g. \end{cases} \]

Setting $v=u_x$, and taking the derivative of the second equation, we get

(4.2)\begin{equation} \begin{cases} \alpha v_{xx} +\beta \eta_{xx} -\lambda\eta= f, \\ - \eta_{xx} - \beta v_{xx} -\lambda v= g_x. \end{cases} \end{equation}

We now multiply the first equation of (4.2) by $\beta$, and the second one by $\alpha$. Adding the results, we obtain

(4.3)\begin{equation} -\varkappa \eta_{xx} -\lambda(\beta\eta+\alpha v)= h, \end{equation}

where

\[ h=\beta f +\alpha g_x\in H. \]

At this point, we introduce the new variable

\[ \xi=\beta\eta+\alpha v, \]

and we consider the system made by (minus) the first equation of (4.2) and (4.3), to wit,

(4.4)\begin{equation} \begin{cases} A\xi+\lambda\eta={-}f, \\ \varkappa A\eta -\lambda\xi =h, \end{cases} \end{equation}

where $A$ is the Laplace-Dirichlet operator discussed in Subsection 2.2. Then, from the first equation above we learn that

(4.5)\begin{equation} \xi={-}\lambda A^{{-}1}\eta- A^{{-}1}f. \end{equation}

Substituting (4.5) into the second equation of (4.4), we end up with

(4.6)\begin{equation} F(A)\eta =q, \end{equation}

having set

\[ F(s)=\frac{\varkappa s^2+\lambda^2}s, \]

and

\[ q=h-\lambda A^{{-}1}f\in H. \]

By the functional calculus of $A$, we conclude that (4.6) has a unique solution given by

\[ \eta=G(A)q, \]

where

\[ G(s)=\frac{s}{\varkappa s^2+\lambda^2}, \]

if and only if $G$ is bounded on the spectrum of $A$. More explicitly, recalling (2.2),

\[ \eta=\sum_{n=1}^\infty\frac{n^2}{\varkappa n^4+\lambda^2}\langle q,w_n\rangle. \]

Thus, we have a unique solution for every $q\in H$ if and only if

\[ \lambda^2\neq{-}\varkappa n^4, \]

that is, if and only if

\[ \lambda\neq{\pm}\varrho n^2. \]

In which case, since when $s\to +\infty$

\[ G(s)\sim \frac1{\varkappa s}, \]

the solution $\eta$ belongs to the space $H^2\cap H_0^1$. By (4.5), this implies that $\xi \in H^2\cap H_0^1$. Finally, from the very definition of $\xi$, we read that $u_x \in H^1_0$ and $u \in H^3$. Summarizing, for every $\lambda \not = \pm \rho n^2$ we have found a unique solution $(\eta,\,u) \in {\mathfrak {D}}({\mathbb {A}})$ to the resolvent equation (4.1). Hence, each of these $\lambda$ belongs to the resolvent set $\rho ({\mathbb {A}})$.

Proof. For every $n = 1,\,2,\,3,\,\ldots$, let us define the functions

(4.7)\begin{equation} {\boldsymbol{u}}^{{\pm}}_n(x) = \Big(\alpha \sin nx, {\beta \pm \rho}/{n}\cos nx\Big). \end{equation}

It is apparent that

\[ {\boldsymbol{u}}^{{\pm}}_n\in {\mathfrak{D}}({\mathbb{A}}), \]

and a straightforward calculation reveals that

\[ {\mathbb{A}} {\boldsymbol{u}}^{{\pm}}_n = \lambda^\pm_n {\boldsymbol{u}}^{{\pm}}_n. \]

This tells that each $\lambda ^\pm _n$ is an eigenvalue of ${\mathbb {A}}$, and ${\boldsymbol {u}}^{\pm }_n$ is a corresponding eigenfunction. We are left to show that any other eigenfunction of $\lambda ^\pm _n$ is a multiple of ${\boldsymbol {u}}^{\pm }_n$. To this end, let $n$ be fixed, and let ${\boldsymbol {v}}=(\eta,\,u)$ be an eigenfunction of $\lambda _n^+$ (the argument for $\lambda _n^-$ is the same). Hence,

\[ {\mathbb{A}} {\boldsymbol{v}} =\lambda^+_n {\boldsymbol{v}}. \]

Exploiting the same change of variables of the previous proof, we arrive at the system

\[ \begin{cases} A\xi={-}\lambda^+_n\eta, \\ \varkappa A\eta=\lambda^+_n\xi. \end{cases} \]

Then,

\[ \xi={-}\lambda^+_nA^{{-}1}\eta, \]

and substituting this expression into the second equation above, we get

\[ A^2\eta =n^4 \eta. \]

Therefore, either $\eta = 0$, in which case $\xi = 0$, or $\eta$ is an eigenfunction of the operator $A^2$ relative to the eigenvalue $n^4$. In the latter case, since the eigenvalues of $A^2$ are all simple (a straightforward consequence of the functional calculus of $A$), $\eta$ is unique up to a multiplicative constant. More precisely, $\eta =w_n$, with $w_n$ given by (2.1). But then $\xi$ is unique as well (up to the same constant), and in turn so is $u$. In conclusion, ${\boldsymbol {v}}={\boldsymbol {u}}_n^+$, up to a multiplicative constant, meaning that $\lambda ^+_n$ is a simple eigenvalue.

Proof. Let $\lambda \in {\mathbb {C}}$ such that $\lambda \neq 0$ be fixed. We consider once again the resolvent equation

\[ ({\mathbb{A}} -\lambda){\boldsymbol{u}} = {\boldsymbol{f}}, \]

for ${\boldsymbol {f}} = (f,\,g) \in {\mathcal {H}}$. We show that for some ${\boldsymbol {f}} \in {\mathcal {H}}$ the equation above has no solution in ${\mathfrak {D}}({\mathbb {A}})$. By the same computations of the proof of theorem 4.2, we arrive at the system

(4.8)\begin{equation} \begin{cases} A\xi+\lambda\eta={-}f, \\ - \lambda\xi =h, \end{cases} \end{equation}

with

\[ h=\beta f +\alpha g_x\in H. \]

From (4.8) we immediately obtain

\[ \xi ={-}\frac{h}{\lambda}. \]

In particular, this implies that $\xi$ attains the same regularity of $h$. Thus, taking $h$ in $H$ but not more regular (e.g., $h\not \in H^2$), the resolvent equation will have no solution in ${\mathfrak {D}}({\mathbb {A}})$. On the other hand, if ${\boldsymbol {f}}=(0,\,0)$, the unique solution is clearly the null one. Accordingly, $\lambda$ is not an eigenvalue.

Instead, when $\lambda = 0$, let $u \neq 0$ be any function such that $u \in H^3$ and $u_x \in H^1_0$, and define

\[ \eta ={-}\beta u_x. \]

It is then apparent that $\eta \in H^2 \cap H^1_0$. Besides, since $\varkappa = 0$ implies $\alpha = \beta ^2$, a quick check reveals that

\[ {\mathbb{A}} {\boldsymbol{u}} = 0, \]

so that $0$ is an eigenvalue, as claimed.

Proof. In light of the linearity of the problem, it is sufficient to show that the only strong solution corresponding to null initial data is the trivial one. To this end, let $(\eta,\,u)$ be any strong solution to (1.1)–(1.3) with $(\phi,\,\psi ) = (0,\,0)$. Defining

\[ \zeta(t)=\int_0^t \eta(s)\,{\rm d}s\quad\text{and}\quad v(t)=\int_0^t u(s)\,{\rm d}s, \]

and taking into account the null initial conditions, integrating (1.1) in time we get

(5.1)\begin{equation} \begin{cases} \zeta_t - \alpha v_{xxx} - \beta \zeta_{xx} = 0, \\ v_t + \zeta_x + \beta v_{xx} = 0, \end{cases} \end{equation}

where $\zeta _t=\eta$ and $v_t=u$. Taking the derivative in space of the second equation of (5.1), after some straightforward manipulations we obtain

\[ \zeta_{xx} ={-}\frac{1}{\varkappa} \big(\beta \eta + \alpha u_{x}\big), \]

which tells that

\[ \zeta_{xx}\in H^2\cap H_0^1. \]

Let us now introduce the functional

\[ {\mathsf F}(t) = \frac12 \Big[\|\zeta_t(t)\|^2 + \varkappa \|\zeta_{xx}(t)\|^2\Big]. \]

Considering the first equation of system (1.1), written in terms of $\zeta$ and $v$, along with the first equation of (5.1), we end up with

(5.2)\begin{equation} \zeta_{tt} + \varkappa \zeta_{xxxx} = 0. \end{equation}

A simple computation yields

\[ \frac{{\rm d}}{{\rm d}t} {\mathsf F} = 0, \]

hence

(5.3)\begin{equation} {\mathsf F}(t) = {\mathsf F}(0) = 0. \end{equation}

At this point, we introduce the further functional

\[ {\mathsf G}(t) = \frac12\|\zeta(t)\|^2. \]

Then we have

\[ \frac{{\rm d}}{{\rm d}t} {\mathsf G} = \langle \zeta_t,\zeta\rangle. \]

Moreover, by (5.2) together with the fact that $\zeta _{xx}\in H^2\cap H_0^1$, and exploiting (5.3),

\[ \frac{{\rm d}^2}{{\rm d}t^2} {\mathsf G}(t) ={-}\varkappa\langle \zeta_{xxxx},\zeta \rangle + \|\zeta_t\|^2 = 2\|\zeta_t\|^2. \]

By the Cauchy–Schwarz inequality,

\[ {\mathsf G} \ddot{\mathsf G} - \dot {\mathsf G}^2 \geq 0, \]

where we used the dot to denote the time derivative. This implies that ${\mathsf G}$ is a log-convex function of time. Accordingly, for any fixed interval $[0,\,T]$, we must have that

\[ {\mathsf G}(t) \leq {\mathsf G}(0)^{1-\frac t T} {\mathsf G}(T)^{t/T}, \quad \forall\ t \in [0,T]. \]

Since ${\mathsf G}(0)= 0$, we conclude that ${\mathsf G}(t)$ is zero for every $t$, implying that $\zeta =0$. But this in turn implies that $\eta =0$. Once we know that $\eta =0$, the first equation of (1.1) simply becomes

\[ u_{xxx}=0, \]

which, as $u_{xxx}\in H^2\cap H_0^1$, readily yields $u_x=0$. Since $u\in H^1_*$, the latter equality implies that $u=0$ as well.

Proof. Let $(\eta,\,u)$ be a strong solution. Once we know from proposition 5.1 that it is unique, it is standard matter to approximate it with solutions as regular as needed. Accordingly, along this proof, we can work with functions regular enough to withstand the forthcoming calculations. Let us assume (5.4). We first define the counterpart of the functional ${\mathsf F}$ of the previous proof, that is,

\[ {\mathsf F}_0(t) = \frac12 \Big[\|\eta_t(t)\|^2 + \varkappa \|\eta_{xx}(t)\|^2\Big]. \]

Observe that ${\mathsf F}_0(t) = {\mathsf F}_0(0)$ for every $t$ and

\[ {\mathsf F}_0(0) = \frac12 \Big[\|\beta \phi_{xx} + \alpha \psi_{xxx}\|^2 + \varkappa \|\phi_{xx}\|^2 \Big] < 0. \]

Besides, let

\[ {\mathsf G}_0(t) =\frac12 \|\eta(t)\|^2 + \omega(t+t_0)^2, \]

where $\omega$ and $t_0$ are two strictly positive constants to be chosen later. By the same computations of the previous proof, just replacing $(\zeta,\,v)$ with $(\eta,\,u)$, we arrive at

\[ {\mathsf G}_0 \ddot{\mathsf G}_0 - \dot {\mathsf G}_0^2 \geq{-} 2\big[\omega + {\mathsf F}_0(0)\big]{\mathsf G}_0^2. \]

Hence, taking $\omega = - {\mathsf F}_0(0)$, we get

\[ {\mathsf G}_0 \ddot{\mathsf G}_0 - \big(\dot {\mathsf G}_0\big)^2 \geq 0, \]

and a simple calculations yields (see, e.g., [Reference Ames and Straughan1])

\[ {\mathsf G}_0(t) \geq {\mathsf G}_0(0)e^{{\dot {\mathsf G}_0(0)}/{{\mathsf G}_0(0)}t}. \]

Choosing $t_0$ such that $\dot {\mathsf G}_0(0) > 0$ we have obtained that the norm of $\eta$ (hence the norm of the solution) blows up exponentially fast. If instead (5.5) holds, in a completely similar way we can prove the exponential blow up of $\|u_x\|$.

Proof. Since $\alpha = \beta ^2$, system (1.1) becomes

(5.6)\begin{equation} \begin{cases} \eta_t - \beta^2 u_{xxx} - \beta \eta_{xx} = 0, \\ u_t + \eta_x + \beta u_{xx} = 0. \end{cases} \end{equation}

Let $(\phi,\,\psi ) \in {\mathfrak {D}}({\mathbb {A}})$. Then it is straightforward to check that the functions

(5.7)\begin{equation} \eta(t) = \phi + \beta(\phi_{xx} + \beta \psi_{xxx}) t\quad \text{and}\quad u(t) = \psi - (\phi_{x} + \beta \psi_{xx}) t \end{equation}

are (strong) solutions to (5.6) with initial conditions (1.3). To prove the uniqueness, let $(\eta,\,u)$ be a solution to (5.6) with initial conditions (1.3). Then, adding the first equation to the spatial derivative of the second equation multiplied by $\beta$, we see at once that

\[ \eta_t + \beta u_{tx} = 0. \]

This gives

\[ \eta ={-} \beta u_x + p, \]

where $p: [0,\,\pi ] \to \mathbb {R}$ is the time-independent function

\[ p(x) = \phi(x) + \beta \psi_x(x). \]

Substituting $\eta$ into the second equation, we get

\[ u_t ={-} p_x, \]

implying that

\[ u(t) = \psi - p_{x} t=\psi - (\phi_{x} + \beta \psi_{xx}) t. \]

In turn, this yields

\[ \eta(t) = \phi + \beta(\phi_{xx} + \beta \psi_{xxx}) t, \]

which establishes the desired uniqueness.

Proof. The second inequality is straightforward. Concerning the first one, notice that by the Young inequality we have

\[ 2\beta|\langle \eta, u_x \rangle | \leq \varepsilon \|\eta\|^2 + \frac{\beta^2}{\varepsilon} \|u_x\|^2. \]

Therefore,

\[ \boldsymbol{|} (\eta,u) \boldsymbol{|}_{{\mathcal{H}}}^2 \geq (1-\varepsilon)\|\eta\|^2 + \Big(\alpha-\frac{\beta^2}{\varepsilon}\Big) \|u_x\|^2. \]

Exploiting the fact that $\varkappa > 0$, we can choose $\varepsilon$ close to 1 so that

\[ 1-\varepsilon > 0 \quad \text{and} \quad \alpha-\frac{\beta^2}{\varepsilon}>0, \]

and the proof is finished.

Proof. Assume first $\boldsymbol {\phi } \in {\mathfrak {D}}({\mathbb {A}})$. Since ${\mathbb {A}}$ is the infinitesimal generator of $S(t)$, we know that ${\boldsymbol {u}}(t) \in {\mathfrak {D}}({\mathbb {A}})$ for every $t \geq 0$, and

\[ \frac{{\rm d}}{{\rm d}t}S(t)\boldsymbol{\phi} = {\mathbb{A}} S(t)\boldsymbol{\phi}. \]

Therefore, exploiting (6.3),

\[ \frac{{\rm d}}{{\rm d}t}{\mathsf E}(t) = \frac12 \frac{{\rm d}}{{\rm d}t} \boldsymbol{|} S(t)\boldsymbol{\phi} \boldsymbol{|}_{{\mathcal{H}}}^2 = ({\mathbb{A}} S(t)\boldsymbol{\phi}, S(t)\boldsymbol{\phi}) = 0. \]

Accordingly, the equality

\[ {\mathsf E}(t) = {\mathsf E}(0) \]

holds for every $t\geq 0$. By density, and due to the continuity properties of the semigroup, this equality remains valid for every $\boldsymbol {\phi } \in {\mathcal {H}}$.

Proof. By direct calculations, one can check the relations

\[ \boldsymbol{(}{\boldsymbol{w}}^+_n,{\boldsymbol{w}}^{-}_m\boldsymbol{)}_{{\mathcal{H}}}=0,\quad\forall\ n,m, \]

and

\[ \boldsymbol{(}{\boldsymbol{w}}^{{\pm}}_n,{\boldsymbol{w}}^{{\pm}}_m\boldsymbol{)}_{{\mathcal{H}}}=\delta_{nm}. \]

We are left to show that the system is complete. Let ${\boldsymbol {f}} = (f,\,g) \in {\mathcal {H}}$. The two components of ${\boldsymbol {f}}$ can be decomposed as

\[ f = \sum_{n = 1}^{\infty} a_n \sin n x \quad \text{and} \quad g = \sum_{n = 1}^{\infty} b_n \frac{\cos nx}{n}. \]

It is then sufficient to verify that each vector

\[ {\boldsymbol{f}}_n =\Big( a_n \sin nx, b_n \frac{\cos nx}{n}\Big) \]

can be expressed as a linear combination of ${\boldsymbol {w}}^+_n$ and ${\boldsymbol {w}}^-_n$. That is, we are looking for constants $c_n$ and $d_n$ such that

\[ {\boldsymbol{f}}_n=c_n{\boldsymbol{w}}^+_n +d_n{\boldsymbol{w}}^-_n, \]

which is equivalent to solve

\[ \begin{cases} \displaystyle a_n=\frac{1}{\sqrt{\alpha\varkappa\pi}}\big[\alpha c_n+\alpha d_n\big],\\ \displaystyle b_n=\frac{1}{\sqrt{\alpha\varkappa\pi}}\big[(\beta+i\sqrt{\varkappa}\,)c_n+(\beta-i\sqrt{\varkappa}\,)d_n\big]. \end{cases} \]

But for every $n$, this system has a unique solution, due to the fact that $\varkappa \neq 0$.