# Degenerate/singular beam-type equations: a controllability result

MOHAMMAD AKIL$^a$, GENNI FRAGNELLI$^b$, AMINE SBAI$^c$

$^a$INSA Hauts-de-France, CERAMATHS-Laboratoire,
Univ. Polytechnique Hauts-de-France,
Valenciennes F-59313, France
email: mohammad.akil@uphf.fr

$^b$Dipartimento di Ingegneria dell'Informazione e Scienze Matematiche,
Università degli Studi di Siena,
via Roma 56, 53100 Siena, Italy,
email: genni.fragnelli@unitsi.it

$^c$Hassan first University of Settat,
Faculty of Sciences and Technology, MISI Laboratory,
B.P. 577, Settat 26000, Morocco.
email: a.sbai@uhp.ac.ma

## Abstract

The paper deals with the controllability of a degenerate/singular beam-type equation in divergence or in non-divergence form. In particular, we assume that the degeneracy and the singularity are at the same boundary point and we consider a suitable control $f$ localized on the non degenerate boundary point. As a first step, we prove the existence of a solution for the homogeneous problems, then we prove some estimates on their energies. Thanks to them, we prove two observability inequalities, and using the notion of solution by transposition, we prove that the initial problems are null controllable.

Keywords: Boundary controllability, degenerate beam equations, observability, singular potentials.
MSC 2020: 35L80, 35L81, 93B05, 93B07.

## 1 Introduction

We consider two boundary controllability problems for a system modeling the bending vibrations of a degenerate beam with singular potential of length $L = 1$. Denote by $u$ the deflection of the beam and assume that the left end of the beam is fixed, while a suitable shear force $f$ acts on the right end of the beam. In this case, the motion describing beam bending is given by the following problems

$$ egin{cases} u_{tt}(t,x) + A_i u(t,x) - \frac{\lambda}{d(x)} u(t,x) = 0, & (t,x) \in Q, \\ B_i(t,0) = 0 = u(t,1) = 0, u_x(t,1) = f(t), & t > 0, \\ u(0,x) = u_0(x), & x \in (0,1), \\ u_t(0,x) = u_1(x), & x \in (0,1), \end{cases} \quad (1.1) $$

$i = 1, 2$, where $Q := (0, +\infty) \times (0, 1)$, $\lambda \in \mathbb{R}$,

$$ A_i u := \begin{cases} a(x)u_{xxxx}(t,x), & i = 1, \\ (au_{xx})_{xx}(t,x), & i = 2, \end{cases} $$
and $B_i u(t, 0)$ are suitable boundary conditions related to the degeneracy of the function $a$. In particular, we consider two types of degeneracy for the function $a$ according to the following definitions.

**Definition 1.1.** A function $g: [0, 1] \to \mathbb{R}$ is weakly degenerate at 0, (WD) for short, if $g \in C^0[0, 1] \cap C^1(0, 1]$ is such that $g(0) = 0, g > 0$ on $(0, 1]$ and, if

$$ \sup_{x \in (0,1]} \frac{x|g'(x)|}{g(x)} := K_g, \qquad (1.2) $$

then $K_g \in (0, 1)$.

**Definition 1.2.** A function $g: [0, 1] \to \mathbb{R}$ is strongly degenerate at 0, (SD) for short, if $g \in C^1[0, 1]$ is such that $g(0) = 0, g > 0$ on $(0, 1]$ and in (1.2) we have $K_g \in [1, 2)$.

Roughly speaking, when $g(x) \sim x^K$, it is (WD) if $K \in (0, 1)$ and (SD) if $K \in [1, 2)$. According to the previous definitions, we have

$$ B_i u(t, 0) := \begin{cases} u(t, 0) = u_x(t, 0) = 0, & i = 1, \\ u(t, 0) = \begin{cases} u_x(t, 0) = 0, & \text{if } a \text{ is (WD)}, \\ (au_{xx})(t, 0) = 0, & \text{if } a \text{ is (SD)}, \end{cases} & i = 2. \end{cases} $$

We underline that, contrary to degenerate wave equations for which null controllability fails if $K_a \ge 2$, for degenerate beam equations null controllability is an open problem if $K_a \ge 2$; indeed, in this case the assumption $K_a < 2$ is essential just for technical reasons. Clearly, when the problem is degenerate and singular, then not only $K_a$ plays an important role, but also $K_d$, as we will see.

Equation similar to the one of (1.1) can be found, for example, in models that describe the vibrations of a bridge. Indeed, a suspension bridge may be seen as a beam of given length $L$, with hinged ends and whose downward deflection is measured by a function $u(t, x)$ subject to three forces. These forces can be summarized as the stays holding the bridge up as nonlinear springs with spring constant $k$, the constant weight per unit length of the bridge $W$ pushing it down, and the external forcing term $f(t, x)$. This leads to the equation

$$ u_{tt} + \gamma u_{xxxx} = -ku^{+} + W + f(t, x). $$

For more details on these problems we refer to [6] and [9].

Boundary exact controllability on linear beam problems has been studied for many years by a lot of authors (see, for example, [4], [14] - [20] and the references therein). In all the previous papers, the equation is always non-degenerate. The first results on boundary controllability for degenerate problems can be found in [1] (see also the arxiv version of 2015), where the authors consider the equation

$$ u_{tt} - (a(x)u_x)_x = 0, $$

and later in [3] where the equation is in non-divergence form.

Clearly, the presence of the degenerate operators $A_i u$ leads us to introduce different spaces with respect to the ones used for the degenerate wave equations as in [1] or in [3]. Actually, the spaces considered in this paper are the same introduced in [6] and in [9] where the problems (1.1) with $\lambda = 0$ are studied. In these new spaces we prove some estimates on the energy associated to the homogeneous problems. Thanks to these inequalities and to the notion of the solutions by transposition, we are able to prove the null controllability for the original systems (1.1). Thus the novelty of this paper is that for the first time we consider the boundary controllability for a degenerate and a singular beam-type equation.

The paper is organized in the following way: in Section 2, we study the well-posedness of the homogeneous problems associated to (1.1); in Section 3, we consider the energy associated to the homogeneous problems and we prove some estimates on the energies from below and from above. Thanks to these estimates, in Section 4 we prove that the original problems have a unique solution by transposition and these solutions are null controllable. Every section is divided into two subsections: the first one for the non-divergence problem and the second one for the divergence one. The paper ends with some open problems.
## 2 Well-posedness for the problems with homogeneous Dirichlet boundary conditions

In this section we will study the well-posedness of the problem in non-divergence or in divergence form with homogeneous boundary conditions.

In both cases, we assume the following hypothesis.

**Hypothesis 1.** The functions $a, d \in C^0[0, 1]$ are such that

1. $a(0) = d(0) = 0$, $a, d > 0$ on $(0, 1]$,

2. there exist $K_a, K_d \in (0, 2)$ such that the functions

$$x \mapsto \frac{x^{K_a}}{a(x)} \qquad (2.1)$$

and

$$x \mapsto \frac{x^{K_d}}{d(x)} \qquad (2.2)$$

are nondecreasing in a right neighborhood of $x = 0$.

It is clear that, if Hypothesis 1 holds, then

$$\lim_{x \to 0} \frac{x^{\gamma}}{a(x)} = 0 \qquad (2.3)$$

for all $\gamma > K_a$ and

$$\lim_{x \to 0} \frac{x^{\gamma}}{d(x)} = 0 \qquad (2.4)$$

for all $\gamma > K_d$.

Let us remark that if $a$ is (WD) or (SD), then (1.2) implies that (2.1) holds on the whole domain $(0, 1]$. Analogously for $d$.

### 2.1 The non-divergence case

In this subsection we consider the following degenerate/singular beam-type problem with Dirichlet boundary conditions

$$\left\{
\begin{array}{l}
y_{tt}(t,x) + a y_{xxxx}(t,x) - \frac{\lambda}{d} y(t,x) = 0, \quad (t,x) \in (0, +\infty) \times (0, 1), \\
y(t, 0) = y(t, 1) = 0, \qquad t \in (0, +\infty), \\
y_x(t, 0) = y_x(t, 1) = 0, \qquad t \in (0, +\infty), \\
y(0, x) = y_T^0(x), \qquad x \in (0, 1), \\
y_t(0, x) = y_T^1(x), \qquad x \in (0, 1).
\end{array}
\right.
\qquad (2.5)$$

In order to treat (2.5), let us introduce the following Hilbert spaces with the related inner products and norms given by:

$$L_{\frac{1}{a}}^2(0, 1) := \{u \in L^2(0, 1) \mid \|u\|_{\frac{1}{a}} < \infty\},$$

$$\langle u, v \rangle_{\frac{1}{a}} := \int_{0}^{1} uv \frac{1}{a} dx, \quad \|u\|_{\frac{1}{a}}^{2} = \int_{0}^{1} \frac{u^{2}}{a} dx,$$

for all $u, v \in L_{\frac{1}{a}}^{2}(0, 1)$;

$$H_{\frac{1}{a}}^{i}(0, 1) := L_{\frac{1}{a}}^{2}(0, 1) \cap H_{0}^{i}(0, 1), \quad i = 1, 2,$$

$$\langle u, v \rangle_{i, \frac{1}{a}} := \langle u, v \rangle_{\frac{1}{a}} + \sum_{k=1}^{i} \int_{0}^{1} u^{(k)}(x)v^{(k)}(x)dx$$
and

$$ \|u\|_{H_{\frac{1}{a}}^{i}(0,1)}^{2} := \|u\|_{\frac{1}{a}}^{2} + \sum_{k=1}^{i} \|u^{(k)}\|_{L^{2}(0,1)}^{2}, $$

For all $u, v \in H_{\frac{1}{a}}^{i}(0,1)$, $i = 1, 2$. We recall that $H_{0}^{i}(0,1) := \{u \in H^{i}(0,1) : u^{(k)}(j) = 0, j = 0, 1, k = 0, \dots, i-1\}$, with $u^{(0)} = u$ and $i = 1, 2$. Observe that it is easy to prove the equivalence in $H_{\frac{1}{a}}^{i}(0,1)$, $i = 1, 2$ of the two norms $\|u\|_{H_{\frac{1}{a}}^{i}(0,1)}^{2}$ and

$$ \|u\|_{i}^{2} := \|u\|_{\frac{1}{a}}^{2} + \|u^{(i)}\|_{L^{2}(0,1)}^{2}, $$

which is induced by the inner product

$$ \langle u, v \rangle_{i} := \langle u, v \rangle_{\frac{1}{a}} + \langle u^{(i)}, v^{(i)} \rangle_{L^{2}(0,1)}, $$

For all $u, v \in H_{\frac{1}{a}}^{i}(0,1)$, $i = 1, 2$. Indeed, if $i = 1$ the previous assertion is clearly true; if $i = 2$, using the fact that $u^{(k)}(j) = 0$ for all $k = 0, 1$ and $j = 0, 1$ one has immediately the equivalence. Moreover, the next result holds.

**Proposition 2.1.** (see [6, Proposition 2.1]) Assume Hypothesis 1. Setting $\|u\|_{i,o} := \|u^{(i)}\|_{L^{2}(0,1)}$ for all $u \in H_{\frac{1}{a}}^{i}(0,1)$, the norms $\| \cdot \|_{H_{\frac{1}{a}}^{i}(0,1)}$, $\| \cdot \|_{i}$ and $\| \cdot \|_{i,o}$ are equivalent in $H_{\frac{1}{a}}^{i}(0,1)$.

Observe that $\| \cdot \|_{i,o}$ is induced by the inner product

$$ \langle u, v \rangle_{i,o} = \int_{0}^{1} u^{(i)}(x)v^{(i)}(x)dx, $$

for all $u, v \in H_{\frac{1}{a}}^{i}(0,1)$.

Using the previous spaces, we can define the operator $(A_1, D(A_1))$ by

$$ A_1 u := au'''' \text{ for all } u \in D(A_1) := \{u \in H_{\frac{1}{a}}^{2}(0,1) : au'''' \in L_{\frac{1}{a}}^{2}(0,1)\}. $$

Moreover,

$$ \langle A_1 u, v \rangle_{L_{\frac{1}{a}}^2(0,1)} = \int_0^1 u''v''dx $$

i.e.

$$ \int_{0}^{1} u'''' v dx = \int_{0}^{1} u'' v'' dx \qquad (2.6) $$

for all $(u, v) \in D(A_1) \times H_{\frac{1}{a}}^{2}(0,1)$, see [5, Proposition 2.1].

An immediate consequence of the previous Gauss Green formula is the next corollary:

**Corollary 2.1.** Assume Hypothesis 1. The space $C_{c}^{\infty}(0,1)$ is dense in $H_{\frac{1}{a}}^{2}(0,1)$.

*Proof.* It is sufficient to prove that if $u \in H_{\frac{1}{a}}^{2}(0,1)$ is orthogonal to $C_{c}^{\infty}(0,1)$, then $u = 0$. To this aim, consider the scalar product associated to $\| \cdot \|_{2}$. Hence, for all $\varphi \in C_{c}^{\infty}(0,1)$ one has

$$ \langle u, v \rangle_2 = 0 \iff \int_0^1 u \varphi \frac{1}{a} dx + \int_0^1 u'' \varphi'' dx = 0 \qquad (2.7) $$

$$ \iff \int_{0}^{1} u \varphi \frac{1}{a} dx = - \int_{0}^{1} u'' \varphi'' dx = - \int_{0}^{1} u'''' \varphi dx. \qquad (2.8) $$

Hence $au'''' = -u$ in the distributional sense and, in particular, $au'''' \in L_{\frac{1}{a}}^{2}(0,1)$. Consequently, $u \in D(A_1)$ and applying (2.6), we have

$$ \langle u, v \rangle_2 = \int_0^1 u \varphi \frac{1}{a} dx + \int_0^1 u'' \varphi'' dx = - \int_0^1 u'''' \varphi dx + \int_0^1 u'' v'' dx = 0 $$

for all $v \in H_{\frac{1}{a}}^{2}(0,1)$. $\square$
In the rest of the paper, the next result will be crucial.

**Proposition 2.2.** Assume Hypothesis 1 and take $K_a, K_d$ such that $K_a + K_d \le 2$. If $u \in H_{\frac{1}{a}}^2(0,1)$, then $\frac{u}{\sqrt{ad}} \in L^2(0,1)$ and there exists a positive constant $C > 0$ such that

$$ \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx \le C \int_0^1 (u''(x))^2 dx. \qquad (2.9) $$

*Proof.* Taking $u \in H_{\frac{1}{a}}^2(0,1)$ and setting $z(x) := u'(x)$, one has that $z(0) = z(1) = 0$ and

$$ \begin{aligned} \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx &= \int_0^1 \frac{x^{K_a+K_d}}{a(x)d(x)} \frac{u^2(x)}{x^{K_a+K_d}} dx \\ &\le \frac{1}{a(1)d(1)} \int_0^1 \frac{u^2(x)}{x^{K_a+K_d}} dx \le \frac{1}{a(1)d(1)} \int_0^1 \frac{u^2(x)}{x^2} dx \\ &\le \frac{4}{a(1)d(1)} \int_0^1 (u'(x))^2 dx = \frac{4}{a(1)d(1)} \int_0^1 (z(x))^2 dx \\ &\le \frac{16}{a(1)d(1)} \int_0^1 (z'(x))^2 dx = \frac{16}{a(1)d(1)} \int_0^1 (u''(x))^2 dx \end{aligned} $$

and the thesis follows.

□

Let

$$ C_{HP} \text{ be the best constant of (2.9).} \qquad (2.10) $$

As in [10], [11] or in [12, Chapter V], we assume the next hypothesis:

**Hypothesis 2.** The constant $\lambda \in \mathbb{R}$ is such that $\lambda \ne 0$ and

$$ \lambda < \frac{1}{C_{HP}}. \qquad (2.11) $$

Observe that the case $\lambda = 0$ is already considered in [6]. Thus it is not restrictive to assume $\lambda \ne 0$. Moreover, if $\lambda \in (0, \frac{1}{C_{HP}})$, we can take $\epsilon \in (0,1)$ such that

$$ \lambda = \frac{1 - \epsilon}{C_{HP}} > 0. \qquad (2.12) $$

Hence, as a consequence of Proposition 2.2, one has the next estimate.

**Proposition 2.3.** Assume Hypothesis 1 and $\lambda \in (0, \frac{1}{C_{HP}})$. If $u \in H_{\frac{1}{a}}^2(0,1)$, then

$$ \int_0^1 (u''(x))^2 dx - \lambda \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx \ge \epsilon \int_0^1 (u''(x))^2 dx. $$

*Proof.* By Proposition 2.2 one has

$$ \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx \le C_{HP} \int_0^1 (u''(x))^2 dx, $$

hence

$$ \lambda \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx \le (1 - \epsilon) \int_0^1 (u''(x))^2 dx, $$

$$ \int_0^1 (u''(x))^2 dx - \lambda \int_0^1 \frac{u^2(x)}{a(x)d(x)} dx \ge \epsilon \int_0^1 (u''(x))^2 dx. $$

□
Under Hypotheses 1 and 2, one can consider in $H_{\frac{1}{a}}^2(0,1)$ also the product

$$ \langle u, v \rangle_{2, \sim} := \langle u, v \rangle_{2, \circ} - \lambda \int_{0}^{1} \frac{uv}{ad} dx, $$

which induces the norm

$$ \|u\|_{2, \sim}^2 = \|u\|_{2, \circ}^2 - \lambda \int_{0}^{1} \frac{u^2}{ad} dx. $$

By Propositions 2.1 and 2.2, one can prove the following equivalence.

**Corollary 2.2.** Assume Hypotheses 1 and 2 and $K_a + K_d \le 2$. Then the norms $\| \cdot \|_{H_{\frac{1}{a}}^2(0,1)}$, $\| \cdot \|_2$, $\| \cdot \|_{2, \circ}$ and $\| \cdot \|_{2, \sim}$ are equivalent in $H_{\frac{1}{a}}^2(0,1)$.

In order to study the well-posedness of (2.5), we introduce the operator

$$ A_{1,\lambda}u := A_1u - \frac{\lambda}{d}u, \quad \forall u \in D(A_{1,\lambda}), $$

where

$$ D(A_{1,\lambda}) := \{u \in H_{\frac{1}{a}}^2(0,1) \mid A_{1,\lambda}u \in L_{\frac{1}{a}}^2(0,1)\}. \qquad (2.13) $$

Observe that if $u \in H_{\frac{1}{a}}^2(0,1)$ and $K_a + 2K_d \le 2$, reasoning as in Proposition 2.2, one has that $\frac{u}{d} \in L_{\frac{1}{a}}^2(0,1)$; hence $u \in D(A_{1,\lambda})$ if and only if $u \in D(A_1)$, i.e. $D(A_{1,\lambda}) = D(A_1)$.

If $K_a + 2K_d > 2$, the definition of $D(A_{1,\lambda})$ is not sufficient to ensure that the members are smooth enough to justify (2.6) (for more details we refer to [13]). For this reason in the following we assume the next assumption.

**Hypothesis 3.** Assume Hypothesis 1 and $K_a + 2K_d \le 2$.

Under this assumption, it is clear that $d$ cannot be (SD). On the other hand, $a$ can be (SD), but in this case $K_d$ has to be very small.

Now, let us consider the matrix operator $A_1 : D(A_1) \subset \mathcal{H}_0 \to \mathcal{H}_0$ given by

$$ A_1 := \begin{pmatrix} 0 & Id \\ A_{1,\lambda} & 0 \end{pmatrix}, \quad D(A_1) := D(A_{1,\lambda}) \times H_{\frac{1}{a}}^2(0,1), \qquad (2.14) $$

where

$$ \mathcal{H}_0 := H_{\frac{1}{a}}^2(0,1) \times L_{\frac{1}{a}}^2(0,1) \qquad (2.15) $$

is endowed with the scalar product and norm given by

$$ \langle \begin{pmatrix} u \\ v \end{pmatrix}, \begin{pmatrix} w \\ z \end{pmatrix} \rangle_{\mathcal{H}_0} = \langle u, w \rangle_{2, \circ} + \langle v, z \rangle_{\frac{1}{a}} $$

and

$$ \|(u, v)\|_{\mathcal{H}_0}^2 := \|u''\|_{L^2(0,1)}^2 + \|v\|_{\frac{1}{a}}^2, $$

for all $(u, v)$, $(w, z) \in \mathcal{H}_0$. In this way, we can rewrite (2.5) as the Cauchy problem

$$ \begin{cases} \dot{\mathcal{Y}}(t) = A_1 \mathcal{Y}(t), & t \ge 0, \\ \mathcal{Y}(0) = \mathcal{Y}_0, \end{cases} \qquad (2.16) $$

with

$$ \mathcal{Y}(t) := \begin{pmatrix} y \\ y_t \end{pmatrix} \quad \text{and} \quad \mathcal{Y}_0 := \begin{pmatrix} y_T^0 \\ y_T^1 \end{pmatrix}. $$

**Theorem 2.1.** Assume Hypotheses 2 and 3. Then the operator $(A_1, D(A_1))$ is non positive with dense domain and generates a contraction semigroup $(T(t))_{t \ge 0}$.
Since the proof of the previous result is standard we refer for example to [6]. Anyway, we can underline that, as usual in semigroup theory, the mild solution of (2.16) given by Theorem 2.1 can be more regular: if $\mathcal{Y}_0 \in D(\mathcal{A}_1)$, then the solution is classical, in the sense that $\mathcal{Y} \in C^1([0, +\infty); \mathcal{H}_0) \cap C([0, +\infty); D(\mathcal{A}_1))$ and the equation in (2.5) holds for all $t \ge 0$. Hence, as in [1, Corollary 4.2] or in [2, Proposition 3.15], one can deduce the next result.

**Theorem 2.2.** Hypotheses 2 and 3 hold. If $(y_T^0, y_T^1) \in H_{\frac{1}{a}}^2(0, 1) \times L_{\frac{1}{a}}^2(0, 1)$, then there exists a unique mild solution

$$y \in C^1([0, +\infty); L_{\frac{1}{a}}^2(0, 1)) \cap C([0, +\infty); H_{\frac{1}{a}}^2(0, 1))$$

of (2.5) which depends continuously on the initial data. Moreover, there exists $C > 0$ such that

$$\int_{0}^{1} \frac{y_t(t, x)^2}{a} dx + \int_{0}^{1} y_{xx}^2(t, x) dx \le C \left( \int_{0}^{1} \frac{(y_T^1)^2}{a} dx + \int_{0}^{1} (y_T^0)_{xx}^2 dx \right) \quad (2.17)$$

for every $t > 0$. In addition, if $(y_T^0, y_T^1) \in D(A_{1,\lambda}) \times H_{\frac{1}{a}}^2(0, 1)$, then the solution $y$ is classical, in the sense that

$$y \in C^2([0, +\infty); L_{\frac{1}{a}}^2(0, 1)) \cap C^1([0, +\infty); H_{\frac{1}{a}}^2(0, 1)) \cap C([0, +\infty); D(A_{1,\lambda}))$$

and the equation of (2.5) holds for all $t \ge 0$.

*Proof.* Since the proof is standard, we only prove (2.17). First, assume that $y$ is a classical solution for which (2.6) holds, for instance when $(y_T^0, y_T^1) \in C_c^\infty(0, 1) \times C_c^\infty(0, 1) \subset D(A_{1,\lambda}) \times H_{\frac{1}{a}}^2(0, 1)$. Then, taken any $t > 0$, multiplying the equation by $\frac{y_t}{a}$, integrating over $[0, 1] \times [0, t]$ and using (2.6), we obtain

$$\begin{aligned} & \frac{1}{2} \int_{0}^{t} \int_{0}^{1} \frac{1}{a} \frac{d}{d\tau} (y_{\tau}(\tau, x))^2 dx d\tau + \frac{1}{2} \int_{0}^{t} \int_{0}^{1} \frac{d}{d\tau} (y_{xx}(\tau, x)^2) dx d\tau \\ & = \frac{\lambda}{2} \int_{0}^{t} \int_{0}^{1} \frac{1}{ad} \frac{d}{d\tau} y^2(\tau, x) dx d\tau. \end{aligned}$$

As a consequence,

$$\begin{aligned} \int_{0}^{1} \frac{y_t^2(t, x)}{a} dx + \int_{0}^{1} y_{xx}^2(t, x) dx = & \int_{0}^{1} \frac{(y_T^1)^2}{a} dx + \int_{0}^{1} (y_T^0)_{xx}^2 dx \\ & + \lambda \int_{0}^{1} \frac{y^2(t, x)}{ad} dx - \lambda \int_{0}^{1} \frac{(y_T^0)^2}{ad} dx. \end{aligned} \quad (2.18)$$

Now, if $\lambda \le 0$, the previous inequality, together with Proposition 2.2 applied to $y_T^0$, immediately give (2.17). On the other hand, if $\lambda \in (0, \frac{1}{C_{HP}})$, by (2.12) one has that $\lambda C_{HP} = 1 - \epsilon$ and, by Proposition 2.2,

$$\lambda \int_{0}^{1} \frac{y^2(t, x)}{ad} dx \le (1 - \epsilon) \int_{0}^{1} y_{xx}^2(t, x) dx,$$

which, together with (2.18), gives

$$\begin{aligned} \int_{0}^{1} \frac{y_t^2(t, x)}{a} dx + \int_{0}^{1} y_{xx}^2(t, x) dx \le & \int_{0}^{1} \frac{(y_T^1)^2}{a} dx + \int_{0}^{1} (y_T^0)_{xx}^2 dx \\ & + (1 - \epsilon) \int_{0}^{1} y_{xx}^2(t, x) dx. \end{aligned}$$

Hence

$$\epsilon \int_{0}^{1} \frac{y_t^2(t, x)}{a} dx + \epsilon \int_{0}^{1} y_{xx}^2(t, x) dx \le \int_{0}^{1} \frac{(y_T^1)^2}{a} dx + \int_{0}^{1} (y_T^0)_{xx}^2 dx$$

and the desired statement holds.
Now, let $(y_T^0, y_T^1) \in H_{\frac{1}{a}}^2(0,1) \times L_{\frac{1}{a}}^2(0,1)$ and take $(y_{T,n}^0, y_{T,n}^1) \in C_c^\infty(0,1) \times C_c^\infty(0,1)$ be such that $(y_{T,n}^0, y_{T,n}^1) \to (y_T^0, y_T^1)$ in $H_{\frac{1}{a}}^2(0,1) \times L_{\frac{1}{a}}^2(0,1)$ - recall that $C_c^\infty(0,1)$ is dense in $H_{\frac{1}{a}}^2(0,1)$, and so in $L_{\frac{1}{a}}^2(0,1)$, see Corollary 2.1. In this way, the mild solution $y_n$ to problem (2.5) with $y_T^0$ replaced by $y_{T,n}^0$ and $y_T^1$ replaced by $y_{T,n}^1$ is also classical, and thus (2.17) holds for $y_n$. Moreover, by linearity of the equation, it is readily seen that (2.17) implies that $(\frac{y_n}{\sqrt{a}})_t + (y_n)_{xx}$ is a Cauchy sequence in $L^2(0,1)$. This permits to pass to the limit and show that (2.17) holds also for mild solutions. $\square$

**Remark 1.** Due to the reversibility in time of the equation, solutions exist with the same regularity for $t < 0$. We will use this fact in the proof of the controllability result, by considering a backward problem whose final time data will be transformed in initial ones: this is the reason for the notation of the initial data in problem (2.5).

## 2.2 The divergence case

Let us now move to the divergence case considering the following degenerate/singular beam problem with Dirichlet boundary conditions

$$ egin{cases} y_{tt}(t,x) + (ay_{xx})_{xx}(t,x) - \frac{\lambda}{d}y(t,x) = 0, & (t,x) \in (0, +\infty) \times (0,1), \\ y(t,0) = \begin{cases} y_x(t,0) = 0, & \text{if } a \text{ is (WD)}, \\ (ay_{xx})(t,0) = 0, & \text{if } a \text{ is (SD)}, \end{cases} & t \in (0, +\infty), \\ y(t,1) = y_x(t,1) = 0, & t \in (0, +\infty), \\ y(0,x) = y_T^0(x), & x \in (0,1), \\ y_t(0,x) = y_T^1(x), & x \in (0,1). \end{cases} \quad (2.19) $$

As a first step, we consider the next spaces introduced in [8] which are different according to the kind of degeneracy of $a$.

* If $a$ is (WD):

$$ V_a^2(0,1) := \left\{ \begin{array}{l} u \in H^1(0,1) : u' \text{ is absolutely continuous in } [0,1], \\ \sqrt{a}u'' \in L^2(0,1) \end{array} \right\} $$

and

$$ egin{aligned} H_a^2(0,1) &:= \{u \in V_a^2(0,1) : u(0) = u(1) = 0\} \\ &= \{u \in H_0^1(0,1) : u' \text{ is absolutely continuous in } [0,1], \\ & \qquad \sqrt{a}u'' \in L^2(0,1)\} \end{aligned} $$

* If $a$ is (SD):

$$ V_a^2(0,1) := \left\{ \begin{array}{l} u \in H^1(0,1) : u' \text{ is locally absolutely continuous in } (0,1], \\ \sqrt{a}u'' \in L^2(0,1) \end{array} \right\} $$

and

$$ egin{aligned} H_a^2(0,1) &:= \{u \in V_a^2(0,1) : u(0) = u(1) = 0\} \\ &= \{u \in H_0^1(0,1) : u' \text{ is locally absolutely continuous in } (0,1], \\ & \qquad \sqrt{a}u'' \in L^2(0,1)\}. \end{aligned} $$

In both cases we consider on $V_a^2(0,1)$ the norm

$$ \|u\|_{2,a}^2 := \|u\|_{L^2(0,1)}^2 + \|u'\|_{L^2(0,1)}^2 + \|\sqrt{a}u''\|_{L^2(0,1)}^2, \quad \forall u \in V_a^2(0,1), \quad (2.20) $$
which is equivalent to the following one

$$ \|u\|_{2, \bullet}^2 := \|u\|_{L^2(0,1)}^2 + \|\sqrt{a}u''\|_{L^2(0,1)}^2, \quad \forall u \in V_a^2(0,1) \qquad (2.21) $$

(see [8, Propositions 2.1 and 2.2]). The previous norms are induced, respectively, by the inner products

$$ \langle u, v \rangle_{2,a} := \langle u, v \rangle_{L^2(0,1)} + \langle u', v' \rangle_{L^2(0,1)} + \langle \sqrt{a}u'', \sqrt{a}v'' \rangle_{L^2(0,1)} $$

and

$$ \langle u, v \rangle_{2, \bullet} := \langle u, v \rangle_{L^2(0,1)} + \langle \sqrt{a}u'', \sqrt{a}v'' \rangle_{L^2(0,1)} $$

for all $u, v \in V_a^2(0,1)$.

Clearly, also on $H_a^2(0,1)$ we consider indifferently the norms $\| \cdot \|_{2,a}$ or $\| \cdot \|_{2, \bullet}$. Using the previous spaces, it is possible to prove the following Gauss-Green formula: setting

$$ \mathcal{Q}(0,1) := \{u \in V_a^2(0,1) : au'' \in H^2(0,1)\}, \qquad (2.22) $$

one has

$$ \int_0^1 (au'')'' vdx = [(au'')' v]_{x=0}^{x=1} - [au''v']_{x=0}^{x=1} + \int_0^1 au''v''dx \qquad (2.23) $$

for all $(u, v) \in \mathcal{Q}(0,1) \times V_a^2(0,1)$ (see [7, Lemmas 2.1 and 2.3]). Moreover, setting

$$ \mathcal{Z}(0,1) := \{u \in H_a^2(0,1) : au'' \in H^2(0,1)\}, $$

(2.23) becomes

$$ \int_0^1 (au'')'' vdx = -[au''v']_{x=0}^{x=1} + \int_0^1 au''v''dx \qquad (2.24) $$

for all $(u, v) \in \mathcal{Z}(0,1) \times H_a^2(0,1)$. Now, we define the following space

• If a is (WD):

$$ \begin{aligned} H_{a,0}^2(0,1) &:= \{u \in H_a^2(0,1) : u'(0) = u'(1) = 0\} \\ &= \{u \in H_0^1(0,1) : u' \text{ is absolutely continuous in } [0,1], \\ & \qquad u'(0) = u'(1) = 0, \sqrt{a}u'' \in L^2(0,1)\}. \end{aligned} $$

• If a is (SD):

$$ \begin{aligned} H_{a,0}^2(0,1) &:= \{u \in H_a^2(0,1) : u'(1) = 0\} \\ &= \{u \in H_0^1(0,1) : u' \text{ is locally absolutely continuous in } (0,1], \\ & \qquad u'(1) = 0, \sqrt{a}u'' \in L^2(0,1)\}. \end{aligned} $$

On $H_{a,0}^2(0,1)$ we consider the same norms of $H_a^2(0,1)$ and the following one

$$ \|u\|_{2,\star} := \|\sqrt{a}u''\|_{L^2(0,1)}, \quad \forall u \in H_{a,0}^2(0,1), $$

induced by the inner product

$$ \langle u, v \rangle_{2, \star} := \int_0^1 a(x)u''(x)v''(x)dx, $$

for all $u, v \in H_{a,0}^2(0,1)$.

**Proposition 2.4.** [9, Proposition 3.1] Assume Hypothesis 1. Then the norms $\| \cdot \|_{2,a}$, $\| \cdot \|_{2, \bullet}$ and $\| \cdot \|_{2, \star}$ are equivalent in $H_{a,0}^2(0,1)$. In particular,

$$ \|u\|_{L^2(0,1)}^2 \le \|u'\|_{L^2(0,1)}^2 \le \frac{1}{a(1)(2-K_a)} \|\sqrt{a}u''\|_{L^2(0,1)}^2, \qquad (2.25) $$

for all $u \in H_{a,0}^2(0,1)$.
As a consequence, one has the following result.

**Proposition 2.5.** Assume Hypothesis 1. If $u \in H_{a,0}^2(0,1)$, then $\frac{u}{\sqrt{d}} \in L^2(0,1)$ and there exists a positive constant $C > 0$ such that

$$ \int_0^1 \frac{u^2(x)}{d(x)} dx \le C \int_0^1 a(x)(u''(x))^2 dx. \qquad (2.26) $$

*Proof.* Take $u \in H_{a,0}^2(0,1)$. Then, by (2.25),

$$ \begin{aligned} \int_0^1 \frac{u^2(x)}{d(x)} dx &= \int_0^1 \frac{x^{K_d} u^2(x)}{d(x) x^{K_d}} dx \le \frac{1}{d(1)} \int_0^1 \frac{u^2(x)}{x^{K_d}} dx \\ &\le \frac{1}{d(1)} \int_0^1 \frac{u^2(x)}{x^2} dx \le \frac{4}{d(1)} \int_0^1 (u'(x))^2 dx \\ &\le \frac{4}{a(1)(2-K_a)d(1)} \int_0^1 a(x)(u''(x))^2 dx \end{aligned} $$

and the thesis follows.

□

Let

$$ \tilde{C}_{HP} \text{ be the best constant of (2.26).} \qquad (2.27) $$

As in the previous subsection, assume the next hypothesis:

**Hypothesis 4.** The constant $\lambda \in \mathbb{R}$ is such that $\lambda \neq 0$ and

$$ \lambda < \frac{1}{\tilde{C}_{HP}}. \qquad (2.28) $$

Also in this case, it is not restrictive to assume $\lambda \neq 0$, since the case $\lambda = 0$ is considered in [9].

Moreover, if $\lambda \in (0, \frac{1}{\tilde{C}_{HP}})$, we can take $\tilde{\epsilon} \in (0,1)$ such that

$$ \lambda = \frac{1 - \tilde{\epsilon}}{\tilde{C}_{HP}} > 0. \qquad (2.29) $$

Thanks to (2.29) and Proposition 2.5, one has the next estimate.

**Proposition 2.6.** Assume Hypothesis 1 and $\lambda \in (0, \frac{1}{\tilde{C}_{HP}})$. If $u \in H_{a,0}^2(0,1)$, then

$$ \int_0^1 a(x)(u''(x))^2 dx - \lambda \int_0^1 \frac{u^2(x)}{d(x)} dx \ge \tilde{\epsilon} \int_0^1 a(x)(u''(x))^2 dx. \qquad (2.30) $$

*Proof.* By Proposition 2.5 one has

$$ \int_0^1 \frac{u^2(x)}{d(x)} dx \le \tilde{C}_{HP} \int_0^1 a(x)(u''(x))^2 dx, $$

hence

$$ \lambda \int_0^1 \frac{u^2(x)}{d(x)} dx \le (1 - \tilde{\epsilon}) \int_0^1 a(x)(u''(x))^2 dx, $$

and, in particular,

$$ \int_0^1 a(x)(u''(x))^2 dx - \lambda \int_0^1 \frac{u^2(x)}{d(x)} dx \ge \tilde{\epsilon} \int_0^1 a(x)(u''(x))^2 dx. $$

□
Under Hypotheses 1 and 4, one can consider on $H_{a,0}^2(0, 1)$ also the inner product

$$ \langle u, v \rangle_{2, \diamond} := \langle u, v \rangle_{2, \star} - \lambda \int_0^1 \frac{uv}{d} dx $$

which induces the norm

$$ \|u\|_{2, \diamond}^2 = \|u\|_{2, \star}^2 - \lambda \int_0^1 \frac{u^2}{d} dx, $$

for all $u, v \in H_{a,0}^2(0, 1)$. Thanks to Proposition 2.5, the following equivalence can be proved.

**Corollary 2.3.** *Assume Hypotheses 1 and 4. Then the norms $\| \cdot \|_{2,a}$, $\| \cdot \|_{2,\bullet}$, $\| \cdot \|_{2,\star}$ and $\| \cdot \|_{2,\diamond}$ are equivalent in $H_{a,0}^2(0, 1)$.*

Now, define the operator $(A_2, D(A_2))$ by

$$ A_2 u := (au'')'', $$

for all $u \in D(A_2)$, where

$$ \begin{aligned} D(A_2) &:= \{u \in \mathcal{Z}(0, 1) : u'(0) = u'(1) = 0, \text{ if } a \text{ is (WD), or} \\ & \qquad (au'')'(0) = u'(1) = 0, \text{ if } a \text{ is (SD)} \} \\ &= \{u \in H_a^2(0, 1) : au'' \in H^2(0, 1) \text{ and } u'(0) = u'(1) = 0, \text{ if } a \text{ is (WD), or} \\ & \qquad (au'')'(0) = u'(1) = 0, \text{ if } a \text{ is (SD)} \}. \end{aligned} $$

As in [9], one can prove that (2.24) becomes

$$ \int_0^1 (au'')'' vdx = \int_0^1 au''v''dx, \qquad (2.31) $$

for all $(u, v) \in D(A_2) \times H_{a,0}^2(0, 1)$. In order to study the well-posedness of (2.19), we need to introduce the operator

$$ A_{2,\lambda} u := A_2 u - \frac{\lambda}{d} u, \quad \forall u \in D(A_{2,\lambda}), $$

where

$$ D(A_{2,\lambda}) := \{u \in H_{a,0}^2(0, 1) \mid A_{2,\lambda}u \in L^2(0, 1)\}. \qquad (2.32) $$

Observe that if $u \in H_{a,0}^2(0, 1)$ and $K_d \le 1$, reasoning as in Proposition 2.5, one has that $\frac{u}{d} \in L^2(0, 1)$; hence $u \in D(A_{2,\lambda})$ if and only if $u \in D(A_2)$, i.e. $D(A_{2,\lambda}) = D(A_2)$.

**Hypothesis 5.** Assume Hypothesis 1 and $K_d \le 1$.

Under this assumption, it is clear that $d$ can be (SD), but in this case we can consider only the case $K_d = 1$.

Now, let us consider the matrix operator $A_2 : D(A_2) \subset K_0 \to K_0$ given by

$$ A_2 := \begin{pmatrix} 0 & Id \\ A_{2,\lambda} & 0 \end{pmatrix}, \quad D(A_2) := D(A_{2,\lambda}) \times H_{a,0}^2(0, 1), \qquad (2.33) $$

where

$$ K_0 := H_{a,0}^2(0, 1) \times L^2(0, 1) \qquad (2.34) $$

is endowed with the scalar product and norm given by

$$ \langle \begin{pmatrix} u \\ v \end{pmatrix}, \begin{pmatrix} w \\ z \end{pmatrix} \rangle_{K_0} := \langle u, w \rangle_{2, \star} + \langle v, z \rangle_{L^2(0, 1)} $$

and

$$ \|(u, v)\|_{K_0}^2 := \|\sqrt{a}u''\|_{L^2(0, 1)}^2 + \|v\|_{L^2(0, 1)}^2, $$
for all $(u, v)$, $(w, z) \in \mathcal{K}_0$. In this way, we can rewrite (2.19) as the Cauchy problem

$$ \begin{cases} \dot{\mathcal{Y}}(t) = \mathcal{A}_2\mathcal{Y}(t), & t \ge 0, \\ \mathcal{Y}(0) = \mathcal{Y}_0, \end{cases} \qquad (2.35) $$

with

$$ \mathcal{Y}(t) := \begin{pmatrix} y \\ y_t \end{pmatrix} \quad \text{and} \quad \mathcal{Y}_0 := \begin{pmatrix} y_T^0 \\ y_T^1 \end{pmatrix}. $$

**Theorem 2.3.** Assume Hypotheses 4 and 5. The operator $(\mathcal{A}_2, D(\mathcal{A}_2))$ is non positive with dense domain and generates a contraction semigroup $(S(t))_{t \ge 0}$.

For the proof of Theorem 2.3 we refer to [9]. The analogue of Theorem 2.2 becomes:

**Theorem 2.4.** Assume Hypotheses 4 and 5. If $(y_T^0, y_T^1) \in H_{a,0}^2(0,1) \times L^2(0,1)$, then there exists a unique mild solution

$$ y \in C^1([0, +\infty); L^2(0,1)) \cap C([0, +\infty); H_{a,0}^2(0,1)) $$

of (2.19) which depends continuously on the initial data. Moreover, there exists $C > 0$ such that

$$ \int_0^1 y_t(t,x)^2 dx + \int_0^1 ay_{xx}^2(t,x) dx \le C \left( \int_0^1 (y_T^1)^2 dx + \int_0^1 a(y_T^0)_{xx}^2 dx \right) \qquad (2.36) $$

for every $t > 0$. In addition, if $(y_T^0, y_T^1) \in D(A_{2,\lambda}) \times H_{a,0}^2(0,1)$, then the solution $y$ is classical, in the sense that

$$ y \in C^2([0, +\infty); L^2(0,1)) \cap C^1([0, +\infty); H_{a,0}^2(0,1)) \cap C([0, +\infty); D(A_{2,\lambda})) $$

and the equation of (2.19) holds for all $t \ge 0$.

*Proof.* Since the proof is standard, we prove only (2.36). Thus, assume that $y$ is a classical solution for which (2.31) holds. Then, taken any $t > 0$, multiplying the equation in (2.19) by $y_t$, integrating over $[0, 1] \times [0, t]$ and using (2.31), we obtain

$$ \begin{aligned} & \frac{1}{2} \int_0^t \int_0^1 \frac{d}{d\tau} (y_\tau(\tau, x))^2 dx d\tau + \frac{1}{2} \int_0^t \int_0^1 a \frac{d}{d\tau} (y_{xx}(\tau, x)^2) dx d\tau \\ & = \frac{\lambda}{2} \int_0^t \int_0^1 \frac{1}{d} \frac{d}{d\tau} y^2(\tau, x) dx d\tau. \end{aligned} $$

As a consequence,

$$ \int_0^1 y_t^2(t,x)dx + \int_0^1 ay_{xx}^2(t,x)dx = \int_0^1 (y_T^1)^2 dx + \int_0^1 a(y_T^0)_{xx}^2 dx \\ + \lambda \int_0^1 \frac{y^2(t,x)}{d} dx - \lambda \int_0^1 \frac{(y_T^0)^2}{d} dx. \qquad (2.37) $$

Now, if $\lambda \le 0$, the previous inequality, together with Proposition 2.5 applied to $y_T^0$, immediately give (2.36). On the other hand, if $\lambda \in (0, \frac{1}{C_{HP}})$, by (2.29) one has $\lambda \tilde{C}_{HP} = 1 - \tilde{\epsilon}$. In this way, by Proposition 2.5 we find

$$ \lambda \int_0^1 \frac{y^2(t,x)}{d} dx \le (1 - \tilde{\epsilon}) \int_0^1 ay_{xx}^2(t,x) dx, $$

which, together with (2.37), gives

$$ \int_0^1 y_t^2(t,x)dx + \int_0^1 ay_{xx}^2(t,x)dx \le \int_0^1 (y_T^1)^2 dx + \int_0^1 a(y_T^0)_{xx}^2 dx \\ + (1 - \tilde{\epsilon}) \int_0^1 ay_{xx}^2(t,x)dx. $$
Hence

$$ \tilde{\epsilon} \int_{0}^{1} y_{t}^{2}(t, x) dx + \tilde{\epsilon} \int_{0}^{1} ay_{xx}^{2}(t, x) dx \leq \int_{0}^{1} (y_{T}^{1})^{2} dx + \int_{0}^{1} a(y_{T}^{0})^{2}_{xx} dx $$

and the desired statement holds. The rest of the proof is similar to the one of Theorem 2.2 so we omit it. □

Remark 1 holds also for the divergence case.

# 3 Energy estimates

In this section we prove some estimates of the energy associated to the solutions of (2.5) and (2.19). These estimates will be crucial to obtain the observability inequalities and thus the null controllability for the original problems.

## 3.1 The non-divergence case

To begin with, we give the next definition.

**Definition 3.1.** For a mild solution $y$ of (2.5) we define its energy as the continuous function

$$ E_y(t) = \frac{1}{2} \int_0^1 \left[ \frac{1}{a} y_t^2(t, x) + y_{xx}^2(t, x) - \frac{\lambda}{ad} y^2(t, x) \right] dx, \quad \forall t \geq 0. \qquad (3.1) $$

The above definition guarantees that the classical conservation of the energy still holds true also in the degenerate/singular situation. Indeed, proceeding in a standard way (see for example [6] or [13]), one can prove the next result.

**Theorem 3.1.** *Assume Hypothesis 1 and let $y$ be a mild solution of (2.5). Then*

$$ E_y(t) = E_y(0), \quad \forall t \geq 0. \qquad (3.2) $$

Let us now introduce the following lemma.

**Lemma 3.1.** [see [6, Lemma 3.1] and [13, Lemma 3.1]] Assume Hypothesis 1.

1. If $u \in H_{a}^{1}(0, 1)$, then $\lim_{x \to 0} \frac{x}{a}u^2(x) = 0$.

2. If $u \in H_{a}^{1}(0, 1)$ and $K_a + K_d < 2$, then $\lim_{x \to 0} \frac{x}{ad}u^2(x) = 0$.

3. If $u \in H_{a}^{1}(0, 1)$, then $\lim_{x \to 0} \frac{x^2}{ad}u^2(x) = 0$.

4. Assume $a$ (SD). If $y \in D(A_1)$, then $y'' \in W^{1,1}(0, 1)$.

Thanks to the previous lemma, we can prove the next equality.

**Theorem 3.2.** Assume $a$ and $d$ (WD) or (SD) and $K_a + 2K_2 \leq 2$. Fixed $T > 0$, if $y$ is a classical solution of (2.5), then

$$ \frac{1}{2} \int_{0}^{T} y_{xx}^{2}(t, 1) dt = \int_{0}^{1} \left[ \frac{x^2}{a} y_{t} y_{x} \right]_{t=0}^{t=T} dx + \frac{1}{2} \int_{Q_T} \frac{x}{a} y_{t}^{2} \left( 2 - \frac{xa'}{a} \right) dxdt \\ + 3 \int_{Q_T} xy_{xx}^{2} dxdt + \frac{\lambda}{2} \int_{Q_T} \left( 2 - \left( \frac{a'}{a} + \frac{d'}{d} \right) x \right) \frac{xy^2}{ad} dxdt. \qquad (3.3) $$

Here $Q_T := (0, T) \times (0, 1)$.
Proof. Since some computations are similar to the ones of [6, Theorem 3.2], we sketch them. Multiplying the equation in (2.5) by $\frac{x^2 y_x}{a}$ and integrating by parts over $Q_T$, we obtain

$$0 = \int_0^1 \left[ y_t \frac{x^2 y_x}{a} \right]_{t=0}^{t=T} dx - \frac{1}{2} \int_0^T \left[ \frac{x^2}{a} y_t^2 \right]_{x=0}^{x=1} dt + \frac{1}{2} \int_{Q_T} \left( \frac{x^2}{a} \right)' y_t^2 dxdt \tag{3.4} \\ + \int_{Q_T} x^2 y_x y_{xxxx} dxdt - \lambda \int_{Q_T} \frac{x^2}{ad} y y_x dxdt.$$

Now consider the last integral. Clearly

$$\int_{Q_T} \frac{x^2}{ad} y y_x dxdt \in \mathbb{R};$$

indeed

$$\left| \int_{Q_T} \frac{x^2}{ad} y y_x dxdt \right| \le \int_{Q_T} \frac{|y|}{\sqrt{ad}} |y_x| \frac{x^{K_a+K_d}}{\sqrt{ad}} dxdt \le \frac{1}{\sqrt{a(1)d(1)}} \int_{Q_T} \frac{|y|}{\sqrt{ad}} |y_x| dxdt.$$

By Proposition 2.2 and the Hölder inequality, one has that this integral is well defined; moreover, it results:

$$-\lambda \int_{Q_T} \frac{x^2}{ad} y y_x dxdt = -\frac{\lambda}{2} \int_{Q_T} \frac{x^2}{ad} (y^2)_x dxdt \\ = -\frac{\lambda}{2} \int_0^T \left[ \frac{x^2 y^2}{ad} \right]_{x=0}^{x=1} + \frac{\lambda}{2} \int_{Q_T} \left( \frac{x^2}{ad} \right)' y^2 dxdt.$$

Taking into account the fact that

$$\left(\frac{x^2}{a}\right)' = \frac{x}{a}\left(2 - \frac{xa'}{a}\right) \quad \text{and} \quad \left(\frac{x^2}{ad}\right)' = \frac{x}{ad}\left(2 - \frac{xa'}{a} - \frac{xd'}{d}\right),$$

(3.4) becomes

$$0 = \int_0^1 \left[ y_t \frac{x^2 y_x}{a} \right]_{t=0}^{t=T} dx - \frac{1}{2} \int_0^T \left[ \frac{x^2}{a} y_t^2 \right]_{x=0}^{x=1} dt + \frac{1}{2} \int_{Q_T} \left( 2 - \frac{xa'}{a} \right) \frac{xy_t^2}{a} dxdt \tag{3.5} \\ + \int_{Q_T} x^2 y_x y_{xxxx} dxdt - \frac{\lambda}{2} \int_0^T \left[ \frac{x^2 y^2}{ad} \right]_{x=0}^{x=1} + \frac{\lambda}{2} \int_{Q_T} \left( 2 - \frac{xa'}{a} - \frac{xd'}{d} \right) \frac{xy^2}{ad} dxdt.$$

By Lemma 3.1 and by the regularity of the solution (recall that $y_t \in H_{\frac{1}{a}}^2(0,1) \subset H_{\frac{1}{a}}^1(0,1)$), one has

$$\frac{1}{a(1)} y_t^2(t, 1) = 0, \lim_{x \to 0} \frac{x^2}{a(x)} y_t^2(t, x) = 0$$

and

$$\lambda \int_0^T \left[ \frac{x^2 y^2}{ad} \right]_{x=0}^{x=1} = 0.$$

It remains to consider the term $\int_{Q_T} x^2 y_x y_{xxxx} dxdt$ which is well-defined (see [6, Theorem 3.2]). Now, let $\delta > 0$ and write

$$\int_{Q_T} x^2 y_x y_{xxxx} dxdt = \int_0^T \int_0^\delta x^2 y_x y_{xxxx} dxdt + \int_0^T \int_\delta^1 x^2 y_x y_{xxxx} dxdt.$$

Obviously, by the absolute continuity of the integral,

$$\lim_{\delta \to 0} \int_0^T \int_0^\delta x^2 y_x y_{xxxx} dxdt = 0,$$

and proceeding as in [6, Theorem 3.2] (thanks also to Lemma 3.1), we obtain

$$\lim_{\delta \to 0} \int_0^T \int_\delta^1 x^2 y_x y_{xxxx} dxdt = -\frac{1}{2} \int_0^T y_{xx}^2 dt + 3 \int_0^T \int_0^1 xy_{xx}^2 dxdt.$$

Hence, (3.3) follows. □
Thanks to (3.3), one can obtain the following estimate from below on the energy.
The next assumption collects all the needed requirements to prove all the estimates for the energy.

**Hypothesis 6.** Hypothesis 2 holds. Moreover, the functions $a$ and $d$ are (WD) or (SD) with $K_a+2K_d \le 2$.

**Theorem 3.3.** Assume Hypothesis 6. Fixed $T > 0$, if $y$ is a mild solution of (2.5), then

$$ \int_{0}^{T} y_{xx}^{2}(t, 1)dt \leq C_{1}E_{y}(0) + C_{2}TE_{y}(0), \qquad (3.6) $$

where

$$ C_1 = 2 \max \left\{ C_{\epsilon} \frac{4}{a(1)}, 1 \right\}, \qquad (3.7) $$

$$ C_2 = \max \{C_{\epsilon}(6 + |\lambda|C_{HP}(2 + K_a + K_d)), 2 + K_a\}, \qquad (3.8) $$

and

$$ C_{\epsilon} := \begin{cases} 1, & \lambda < 0, \\ \frac{1}{\epsilon}, & \lambda > 0. \end{cases} \qquad (3.9) $$

Here $\epsilon$ is as in (2.12).

*Proof.* As a first step, assume that $y$ is a classical solution of (2.5); thus, (3.3) holds for a fixed $T > 0$. Now, set $z(t, x) := y_x(t, x)$; since $y_x(t, 0) = 0$, by the classical Hardy inequality, we obtain

$$ \int_{0}^{1} y_{x}^{2} dx = \int_{0}^{1} x^{2} \frac{z^{2}}{x^{2}} dx \leq 4 \int_{0}^{1} z_{x}^{2} dx = 4 \int_{0}^{1} y_{xx}^{2} dx. \qquad (3.10) $$

Case $\lambda < 0$. In this case

$$ \int_{0}^{1} \left[ \frac{x^{2}y_{x}(\tau, x)y_{t}(\tau, x)}{a(x)} \right]_{\tau=0}^{\tau=T} dx \leq 2 \max \left\{ \frac{4}{a(1)}, 1 \right\} E_{y}(0) $$

(see [6, Theorem 3.3]),

$$ \left| \int_{Q_T} \frac{x}{a} y_t^2 \left( 2 - \frac{xa'}{a} \right) dxdt \right| \leq (2 + K_a) \int_{Q_T} \frac{y_t^2}{a} dxdt, $$

and

$$ \int_{Q_T} xy_{xx}^2 dxdt \leq \int_{Q_T} y_{xx}^2 dxdt. $$

Moreover, by Proposition 2.2, we can obtain

$$ \left| \frac{\lambda}{2} \int_{Q_T} \frac{y^2 x}{ad} \left( 2 - \frac{xa'}{a} - \frac{xd'}{d} \right) dxdt \right| \leq \frac{|\lambda|}{2} (2 + K_a + K_d) \int_{Q_T} \frac{y^2}{ad} dxdt \\ \leq \frac{|\lambda|}{2} C_{HP} (2 + K_a + K_d) \int_{Q_T} y_{xx}^2 dxdt. $$

Thus, by (3.3), the previous estimates and Theorem 3.1, one has

$$ \begin{aligned} \frac{1}{2} \int_{0}^{T} y_{xx}^{2}(t, 1)dt \leq 2 \max \left\{ \frac{4}{a(1)}, 1 \right\} E_{y}(0) + \frac{1}{2} (2 + K_{a}) \int_{Q_T} \frac{y_{t}^{2}}{a} dxdt \\ + 3 \int_{Q_T} y_{xx}^{2} dxdt + \frac{|\lambda|}{2} C_{HP} (2 + K_{a} + K_{d}) \int_{Q_T} y_{xx}^{2} dxdt \\ \leq C_{1}E_{y}(0) + C_{2}TE_{y}(0), \end{aligned} $$

where $C_1$ and $C_2$ are as in (3.7) and (3.8), respectively, and the thesis follows if $y$ is a classical solution.
Case $\lambda \in (0, \frac{1}{C_{HP}})$. By (3.10) and Proposition 2.3

$$
\begin{aligned}
& \int_0^1 \left[ \frac{x^2 y_x(\tau, x) y_t(\tau, x)}{a(x)} \right]_{\tau=0}^{\tau=T} dx \le \frac{2}{a(1)} \int_0^1 y_{xx}^2(T, x) dx + \frac{1}{2} \int_0^1 \frac{y_t^2(T, x)}{a(x)} dx \\
& + \frac{2}{a(1)} \int_0^1 y_{xx}^2(0, x) dx + \frac{1}{2} \int_0^1 \frac{y_t^2(0, x)}{a(x)} dx \\
& \le \frac{2}{\epsilon a(1)} \left( \int_0^1 y_{xx}^2(T, x) dx - \lambda \int_0^1 \frac{y^2(T, x)}{a(x) d(x)} dx \right) \\
& + \frac{2}{\epsilon a(1)} \left( \int_0^1 y_{xx}^2(0, x) dx - \lambda \int_0^1 \frac{y^2(0, x)}{a(x) d(x)} dx \right) + \frac{1}{2} \int_0^1 \frac{y_t^2(T, x)}{a(x)} dx + \frac{1}{2} \int_0^1 \frac{y_t^2(0, x)}{a(x)} dx \\
& \le \max \left\{ \frac{4}{\epsilon a(1)}, 1 \right\} (E_y(T) + E_y(0)) = 2 \max \left\{ \frac{4}{\epsilon a(1)}, 1 \right\} E_y(0).
\end{aligned}
$$

Moreover, as in [6, Theorem 3.3], one can prove

$$
\left| \int_{Q_T} \frac{x}{a} y_t^2 \left( 2 - \frac{xa'}{a} \right) dxdt \right| \le (2 + K_a) \int_{Q_T} \frac{y_t^2}{a} dxdt,
$$

and

$$
\int_{Q_T} xy_{xx}^2 dxdt \le \int_{Q_T} y_{xx}^2 dxdt.
$$

Finally, by Proposition 2.2, we have

$$
\begin{aligned}
\left| \frac{\lambda}{2} \int_{Q_T} \frac{y^2 x}{ad} \left( 2 - \frac{xa'}{a} - \frac{xd'}{d} \right) dxdt \right| &\le \frac{\lambda}{2} (2 + K_a + K_d) \int_{Q_T} \frac{y^2}{ad} dxdt \\
&\le \frac{\lambda}{2} C_{HP} (2 + K_a + K_d) \int_{Q_T} y_{xx}^2 dxdt.
\end{aligned}
$$

Thus, by (3.3), the previous estimates, Theorem 3.1 and Proposition 2.3 one has

$$
\begin{aligned}
& \frac{1}{2} \int_0^T y_{xx}^2(t, 1) dt \le 2 \max \left\{ \frac{4}{\epsilon a(1)}, 1 \right\} E_y(0) + \frac{1}{2} (2 + K_a) \int_{Q_T} \frac{y_t^2}{a} dxdt \\
& \qquad + (6 + \lambda C_{HP} (2 + K_a + K_d)) \frac{1}{2} \int_{Q_T} y_{xx}^2 dxdt \\
& \le 2 \max \left\{ \frac{4}{\epsilon a(1)}, 1 \right\} E_y(0) + \frac{1}{2} (2 + K_a) \int_{Q_T} \frac{y_t^2}{a} dxdt \\
& \qquad + \frac{1}{2\epsilon} (6 + \lambda C_{HP} (2 + K_a + K_d)) \left( \int_{Q_T} y_{xx}^2 dxdt - \lambda \int_{Q_T} \frac{y^2(t, x)}{a(x) d(x)} dt \right) \\
& \le C_1 E_y(0) + C_2 T E_y(0),
\end{aligned}
$$

where $C_1$ and $C_2$ are as in (3.7) and (3.8), respectively. Hence, the thesis is proved if $y$ is a classical solution. Now, let $y$ be the mild solution associated to the initial data $(y_T^0, y_T^1) \in \mathcal{H}_0$. Then, consider a sequence $\{(y_{T,n}^0, y_{T,n}^1)\}_{n \in \mathbb{N}} \subset D(A_1)$ that approximate $(y_T^0, y_T^1)$ and let $y_n$ be the classical solution of (2.5) associated to $(y_{T,n}^0, y_{T,n}^1)$. Clearly $y_n$ satisfies (3.6) and thanks to Theorem 2.2 we can pass to the limit obtaining the thesis. $\square$

In the following we shall also need an estimate from above for the energy. To this aim we need an
equality different from (3.3).
**Theorem 3.4.** Assume a and d (WD) or (SD) and $K_a + 2K_d \le 2$. Fixed $T > 0$, if y is a classical solution of (2.5), then

$$ \frac{1}{2} \int_{0}^{T} y_{xx}^{2}(t, 1) dt = \int_{0}^{1} \left[ \frac{xy_{t}y_{x}}{a} \right]_{t=0}^{t=T} dx + \frac{1}{2} \int_{Q_{T}} \frac{y_{t}^{2}}{a} \left( 1 - \frac{xa'}{a} \right) dxdt + \frac{3}{2} \int_{Q_{T}} y_{xx}^{2} dxdt \\ + \frac{\lambda}{2} \int_{Q_{T}} \frac{y^{2}}{ad} \left[ 1 - x \left( \frac{a'}{a} + \frac{d'}{d} \right) \right] dxdt. \quad (3.11) $$

*Proof.* Multiplying the equation in (2.5) by $\frac{xy_x}{a}$ and integrating over $Q_T$, we obtain

$$ 0 = \int_{0}^{1} \left[ \frac{xy_{x}y_{t}}{a} \right]_{t=0}^{t=T} dx - \frac{1}{2} \int_{0}^{T} \left[ \frac{x}{a} y_{t}^{2} \right]_{x=0}^{x=1} dt + \frac{1}{2} \int_{Q_{T}} \left( \frac{x}{a} \right)' y_{t}^{2} dxdt \\ + \int_{Q_{T}} xy_{x}y_{xxxx} dxdt - \frac{\lambda}{2} \int_{0}^{T} \left[ \frac{x}{ad} y^{2} \right]_{x=0}^{x=1} dt + \frac{\lambda}{2} \int_{Q_{T}} \left( \frac{x}{ad} \right)' y^{2} dxdt. $$

Using the fact that

$$ \left(\frac{x}{a}\right)' = \frac{1}{a}\left(1 - \frac{xa'}{a}\right) \quad \text{and} \quad \left(\frac{x}{ad}\right)' = \frac{1}{ad}\left[1 - \left(\frac{a'}{a} + \frac{d'}{d}\right)x\right], $$

one has

$$ 0 = \int_{0}^{1} \left[ \frac{xy_{x}y_{t}}{a} \right]_{t=0}^{t=T} dx - \frac{1}{2} \int_{0}^{T} \left[ \frac{x}{a} y_{t}^{2} \right]_{x=0}^{x=1} dt + \frac{1}{2} \int_{Q_{T}} \frac{y_{t}^{2}}{a} \left( 1 - \frac{xa'}{a} \right) dxdt \\ + \int_{Q_{T}} xy_{x}y_{xxxx} dxdt - \frac{\lambda}{2} \int_{0}^{T} \left[ \frac{x}{ad} y^{2} \right]_{x=0}^{x=1} dt + \frac{\lambda}{2} \int_{Q_{T}} \left[ 1 - \left( \frac{a'}{a} + \frac{d'}{d} \right) x \right] \frac{y^{2}}{ad} dxdt. \quad (3.12) $$

Thanks to Lemma 3.1, we have

$$ \lim_{x \to 0} \frac{x}{a(x)} y_t^2(t, x) = 0 \quad \text{and} \quad \lim_{x \to 0} \frac{x}{a(x)} y(t, x) = 0; $$

thus

$$ \int_{0}^{T} \left[ \frac{x}{a} y_{t}^{2} \right]_{x=0}^{x=1} dt = 0 \quad \text{and} \quad \int_{0}^{T} \left[ \frac{x}{ad} y^{2} \right]_{x=0}^{x=1} dt = 0. $$

Moreover, the term $\int_{Q_T} xy_x y_{xxxx} dxdt$ is well-defined and, as in [6, Theorem 3.4], one has

$$ \int_{Q_T} xy_x y_{xxxx} dxdt = -\frac{1}{2} \int_0^T y_{xx}^2(t, 1) dt + \frac{3}{2} \int_0^T \int_0^1 y_{xx}^2 dxdt. $$

Coming back to (3.12), the thesis follows. □

Thanks to (3.11), we have the next estimate from above on the energy.

**Theorem 3.5.** Assume Hypothesis 6. Fixed $T > 0$, if y is a mild solution of (2.5), then

$$ \int_{0}^{T} y_{xx}^{2}(t, 1) dt \geq C_{3} T E_{y}(0) - C_{4} E_{y}(0), $$

where

$$ C_3 := \begin{cases} 2 \left\{ \left(1 - \frac{K_a}{2}\right) + \lambda C_{HP} \left(2 + K_a + K_d\right) \right\}, & \lambda < 0, \\ 2 \left(1 - \frac{K_a}{2}\right), & \lambda > 0, \end{cases} \quad (3.13) $$

$$ C_4 := C_{\epsilon} \left( 4 \max \left\{ \frac{4}{a(1)}, 1 \right\} + K_a \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} \right) \quad (3.14) $$

and $C_{\epsilon}$ is as in (3.9).
Proof. As a first step, assume that $y$ is a classical solution of (2.5). Multiplying the equation in (2.5) by $\frac{-K_a y}{2a}$ and integrating by parts on $Q_T$, we have

$$0 = -\frac{K_a}{2} \int_0^1 \left[ \frac{yy_t}{a} \right]_{t=0}^{t=T} dx + \frac{K_a}{2} \int_{Q_T} \frac{y_t^2}{a} dxdt - \frac{K_a}{2} \int_{Q_T} y_{xx}^2 dxdt + \frac{\lambda K_a}{2} \int_{Q_T} \frac{y^2}{ad} dxdt.$$

Summing the previous equality to (3.11) multiplied by 2 and using the degeneracy condition (1.2), we have

$$\begin{aligned} \int_0^T y_{xx}^2(t, 1)dt = 2 \int_0^1 \left[ \frac{xy_x y_t}{a} \right]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 \left[ \frac{yy_t}{a} \right]_{t=0}^{t=T} dx + \frac{K_a}{2} \int_{Q_T} \frac{y_t^2}{a} dxdt \\ - \frac{K_a}{2} \int_{Q_T} y_{xx}^2 dxdt + \int_{Q_T} \frac{y_t^2}{a} dxdt + 3 \int_{Q_T} y_{xx}^2 dxdt - \int_{Q_T} \frac{xa'}{a} \frac{y_t^2}{a} dxdt \\ + \lambda \int_{Q_T} \frac{y^2}{ad} \left( 1 - x \left( \frac{d'}{d} + \frac{a'}{a} \right) \right) + \frac{\lambda K_a}{2} \int_{Q_T} \frac{y^2}{ad} dxdt \\ \geq 2 \int_0^1 \left[ \frac{xy_x y_t}{a} \right]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 \left[ \frac{yy_t}{a} \right]_{t=0}^{t=T} dx + \int_{Q_T} \left( 1 - \frac{K_a}{2} \right) \frac{y_t^2}{a} dxdt \\ + \int_{Q_T} \left( 1 - \frac{K_a}{2} \right) y_{xx}^2 dxdt + \lambda \int_{Q_T} \frac{y^2}{ad} \left( \frac{K_a}{2} + 1 - x \left( \frac{d'}{d} + \frac{a'}{a} \right) \right) dxdt. \end{aligned}$$

Hence,

$$\begin{aligned} \int_0^T y_{xx}^2(t, 1)dt \geq 2 \int_0^1 \left[ \frac{xy_x y_t}{a} \right]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 \left[ \frac{yy_t}{a} \right]_{t=0}^{t=T} dx \\ + \left(1 - \frac{K_a}{2}\right) \int_{Q_T} \left(\frac{y_t^2}{a} + y_{xx}^2\right) dxdt + \lambda \int_{Q_T} \frac{y^2}{ad} \left(\frac{K_a}{2} + 1 - x\left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt. \end{aligned} \quad (3.15)$$

Now, we consider the boundary terms. As in [6, Theorem 3.3], one has that for all $\tau \in [0, T]$,

$$\begin{aligned} \left| \int_0^1 \frac{xy_x y_t}{a}(\tau, x)dx \right| &\le \frac{1}{2} \int_0^1 \frac{x^2}{a} y_x^2(\tau, x)dx + \frac{1}{2} \int_0^1 \frac{y_t^2}{a}(\tau, x)dx \\ &\le \frac{1}{2a(1)} 4 \int_0^1 y_{xx}^2(\tau, x)dx + \frac{1}{2} \int_0^1 \frac{y_t^2}{a}(\tau, x)dx \quad (3.16) \\ &\le \max \left\{ \frac{4}{a(1)}, 1 \right\} \frac{1}{2} \left( \int_0^1 y_{xx}^2(\tau, x)dx + \int_0^1 \frac{y_t^2}{a}(\tau, x)dx \right), \end{aligned}$$

by (3.10). Moreover, for all $\tau \in [0, T]$,

$$\begin{aligned} \left| \int_0^1 \left( \frac{yy_t}{a} \right) (\tau, x) dx \right| &\le \frac{1}{2} \int_0^1 \frac{y_t^2}{a} (\tau, x) dx + \frac{1}{2} \int_0^1 \frac{y^2}{a} (\tau, x) dx \\ &\le \frac{1}{2} \int_0^1 \frac{y_t^2}{a} (\tau, x) dx + \frac{\max_{x \in [0,1]} d(x)}{2} \int_0^1 \frac{y^2}{ad}(\tau, x) dx \\ &\le \frac{1}{2} \int_0^1 \frac{y_t^2}{a} (\tau, x) dx + \frac{\max_{x \in [0,1]} d(x)}{2} C_{HP} \int_0^1 y_{xx}^2(\tau, x) dx, \end{aligned}$$

by (2.9). Hence

$$\begin{aligned} & \frac{K_a}{2} \left| \int_0^1 \frac{yy_t}{\sigma}(\tau, x) dx \right| \\ &\le \frac{K_a}{4} \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} \left( \int_0^1 y_{xx}^2(\tau, x) dx + \int_0^1 \frac{y_t^2(\tau, x)}{a} dx \right) \end{aligned} \quad (3.17)$$

for all $\tau \in [0, T]$.
Now, we proceed in two different ways, according to the sign of $\lambda$.
Case $\lambda \in (0, \frac{1}{C_{HP}})$. In this case, by (2.12) and by (2.9),

$$ \int_{0}^{1} y_{xx}^{2} dx - \lambda \int_{0}^{1} \frac{y_{xx}^{2}}{\sigma d} dx \geq \int_{0}^{1} y_{xx}^{2} dx - (1 - \epsilon) \int_{0}^{1} y_{xx}^{2} dx = \epsilon \int_{0}^{1} y_{xx}^{2} dx. $$

Thus, by Theorem 3.1, for all $\tau \in [0, T]$

$$ \begin{aligned} 2E_y(0) = 2E_y(\tau) &\geq \epsilon \int_0^1 (y_{xx}^2)(\tau, x) dx + \int_0^1 \frac{y_t^2(\tau, x)}{a} dx \\ &\geq \epsilon \left( \int_0^1 (y_{xx}^2)(\tau, x) dx + \int_0^1 \frac{y_t^2(\tau, x)}{a} dx \right). \end{aligned} $$

Thus,

$$ \int_{0}^{1} (y_{xx}^{2})(\tau, x) dx + \int_{0}^{1} \frac{y_{t}^{2}(\tau, x)}{a} dx \leq \frac{2}{\epsilon} E_{y}(0) \quad (3.18) $$

for all $\tau \in [0, T]$. Applying this inequality to (3.16) and (3.17), we obtain

$$ 2 \left| \int_{0}^{1} \left[ \frac{xy_x y_t}{a} \right]_{t=0}^{t=T} dx \right| \leq 2 \max \left\{ \frac{4}{a(1)}, 1 \right\} \frac{2}{\epsilon} E_y(0). $$

and

$$ \frac{K_a}{2} \left| \int_{0}^{1} \left[ \frac{yy_t}{\sigma} (\tau, x) dx \right]_{t=0}^{t=T} \right| \leq \frac{K_a}{\epsilon} \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} E_y(0). $$

On the other hand, since $\lambda > 0$, we have

$$ \begin{aligned} & \left(1 - \frac{K_a}{2}\right) \int_{Q_T} \left(\frac{y_t^2}{a} + y_{xx}^2\right) dxdt + \lambda \int_{Q_T} \frac{y^2}{ad} \left(\frac{K_a}{2} + 1 - x \left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt \\ &= \left(1 - \frac{K_a}{2}\right) \int_{Q_T} \left(\frac{y_t^2}{a} + y_{xx}^2 - \lambda \frac{y^2}{ad}\right) dxdt \\ &\quad + \lambda \int_{Q_T} \frac{y^2}{ad} \left(1 - \frac{K_a}{2} + \frac{K_a}{2} + 1 - x \left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt \\ &= \left(1 - \frac{K_a}{2}\right) 2TE_y(0) + \lambda \int_{Q_T} \frac{y^2}{ad} \left(2 - x \left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt \\ &\geq \left(1 - \frac{K_a}{2}\right) 2TE_y(0) + \lambda \left(2 - K_a - K_d\right) \int_{Q_T} \frac{y^2}{ad} dxdt \geq \left(1 - \frac{K_a}{2}\right) 2TE_y(0). \end{aligned} $$

Hence, by (3.15) and the previous estimates, we find

$$ \int_{0}^{T} y_{xx}^{2}(t, 1) dt \geq 2T \left(1 - \frac{K_a}{2}\right) E_{y}(0) - \left(4 \max \left\{\frac{4}{a(1)}, 1\right\} + K_a \max \left\{1, C_{HP} \max_{x \in [0,1]} d(x)\right\}\right) \frac{1}{\epsilon} E_{y}(0) $$

and the thesis holds if $y$ is a classical solution.

Case $\lambda < 0$. Proceeding as before, by (3.16), (3.17) and using the fact that $\lambda < 0$, one has

$$ 2 \left| \int_{0}^{1} \left[ \frac{xy_x y_t}{a} \right]_{t=0}^{t=T} dx \right| \leq 4 \max \left\{ \frac{4}{a(1)}, 1 \right\} E_y(0) $$

and

$$ \begin{aligned} \frac{K_a}{2} \left| \int_{0}^{1} \left[ \frac{yy_t}{a} \right]_{t=0}^{t=T} dx \right| &\leq 2 \frac{K_a}{4} \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} \\ &\quad \left( \int_{0}^{1} y_{xx}^{2}(\tau, x) dx + \int_{0}^{1} \frac{y_{t}^{2}(\tau, x)}{a} dx \right) \\ &\leq K_a \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} E_y(0). \end{aligned} $$
In this case, since $\lambda < 0$, by (2.9), we have

$$ \lambda \int_{Q_T} \frac{y^2}{ad} dxdt \geq \lambda C_{HP} \int_{Q_T} y_{xx}^2 dxdt \geq 2\lambda C_{HP} T E_y(0), $$

and, as before,

$$ \begin{aligned} & \left(1 - \frac{K_a}{2}\right) \int_{Q_T} \left(\frac{y_t^2}{a} + y_{xx}^2\right) dxdt + \lambda \int_{Q_T} \frac{y^2}{ad} \left(\frac{K_a}{2} + 1 - x \left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt \\ &= \left(1 - \frac{K_a}{2}\right) 2TE_y(0) + \lambda \int_{Q_T} \frac{y^2}{ad} \left(2 - x \left(\frac{d'}{d} + \frac{a'}{a}\right)\right) dxdt \\ &\geq \left(1 - \frac{K_a}{2}\right) 2TE_y(0) + \lambda(2 + K_a + K_d) \int_{Q_T} \frac{y^2}{ad} dxdt \\ &\geq \left(1 - \frac{K_a}{2}\right) 2TE_y(0) - 2|\lambda|(2 + K_a + K_d)C_{HP}TE_y(0). \end{aligned} $$

By (3.15) and the previous estimates, we have

$$ \int_0^T y_{xx}^2(t, 1) dt \geq 2T \left\{ \left(1 - \frac{K_a}{2}\right) - |\lambda| C_{HP} (2 + K_a + K_d) \right\} E_y(0) \\ - \left( 4 \max \left\{ \frac{4}{a(1)}, 1 \right\} + K_a \max \left\{ 1, C_{HP} \max_{x \in [0,1]} d(x) \right\} \right) E_y(0) $$

and the thesis follows if $y$ is a classical solution.

Now, let $y$ be the mild solution associated to the initial data $(y_T^0, y_T^1) \in \mathcal{H}_0$. Proceeding as in Theorem 3.3, we obtain the desired inequalities. $\square$

## 3.2 The divergence case

Let us now treat the energy estimates related to the divergence case. First of all we give the following definition.

**Definition 3.2.** For a mild solution $y$ of (2.19) we define its energy as the continuous function

$$ E_y(t) = \frac{1}{2} \int_0^1 \left[ y_t^2(t, x) + a y_{xx}^2(t, x) - \frac{\lambda}{d} y^2(t, x) \right] dx, \quad \forall t \geq 0. \quad (3.19) $$

As for the non-divergence case, the definition above guarantees that the classical conservation of the energy still holds true also in this case. Indeed, as in [9], one can prove the following result.

**Theorem 3.6.** Assume Hypothesis 1 and let $y$ be a mild solution of (2.19). Then

$$ E_y(t) = E_y(0), \quad \forall t \geq 0. \quad (3.20) $$

For the following the next lemmas will be crucial.

**Lemma 3.2.** (see [9, Lemma 2.1]) Assume $a(WD)$ and $y \in Q(0, 1)$. The following assertions are true:

1. If $y'(0) = 0$ or $(ay'')'(0) = 0$, then $\lim_{\delta \to 0^+} y'(\delta)(ay'')'(\delta) = 0$.

2. If $y'(0) = 0$ or $(ay'')'(0) = 0$, then $\lim_{\delta \to 0^+} y'(\delta)(ay'')'(\delta) = 0$.

3. $\lim_{\delta \to 0^+} a(\delta)(y')^2(\delta) = 0$.

4. $\lim_{\delta \to 0^+} \delta a(\delta)(y'')^2(\delta) = 0$.

**Lemma 3.3.** (see [9, Lemma 2.2]) Assume $a(SD)$ and $y \in Q(0, 1)$. The following assertions are true:

1. $\lim_{\delta \to 0^+} \delta y'(\delta) = 0$.
2. If $(ay'') (0) = 0$, then $\lim_{\delta \to 0^+} y'(\delta) (ay'') (\delta) = 0$.

3. If $(ay'') (0) = 0$, then $\lim_{\delta \to 0^+} a(\delta) (y')^2 (\delta) = 0$.

4. If $(ay'') (0) = 0$, then $\lim_{\delta \to 0^+} \delta a(\delta) (y'')^2 (\delta) = 0$.

The analogue of Theorem 3.2 is the following result.

**Theorem 3.7.** Assume a and d (WD) or (SD) and $K_d \le 1$. Fixed $T > 0$, if y is a classical solution of (2.19), then

$$ \frac{1}{2}a(1) \int_0^T y_{xx}^2(t, 1)dt = \int_0^1 [y_t x^2 y_x]_{t=0}^{t=T} dx + \int_{Q_T} xy_t^2 dxdt - \int_{Q_T} a' y_x^2 dxdt \\ + \int_{Q_T} \left(3xa - \frac{x^2}{2}a'\right) y_{xx}^2 dxdt + \frac{\lambda}{2} \int_{Q_T} \frac{x}{d} \left(2 - \frac{xd'}{d}\right) y^2 dxdt. \quad (3.21) $$

*Proof.* Multiplying the equation in (2.19) by $x^2 y_x$ and integrating over $Q_T$, we obtain

$$ \begin{aligned} 0 &= \int_{Q_T} y_{tt} x^2 y_x dxdt + \int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt - \lambda \int_{Q_T} \frac{x^2}{d} yy_x dxdt \\ &= \int_0^1 [y_t x^2 y_x]_{t=0}^{t=T} dx - \frac{1}{2} \int_0^T [x^2 y_t^2]_{x=0}^{x=1} dt + \int_{Q_T} xy_t^2 dxdt + \int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt \\ &\quad - \frac{\lambda}{2} \int_0^T \left[ \frac{x^2 y^2}{d} \right]_{x=0}^{x=1} + \frac{\lambda}{2} \int_{Q_T} \frac{x}{d} \left( 2 - \frac{xd'}{d} \right) y^2 dxdt. \end{aligned} $$

Then using the boundary conditions of y, we have

$$ \begin{aligned} 0 = \int_{0}^{1} [y_t x^2 y_x]_{t=0}^{t=T} dx &+ \int_{Q_T} xy_t^2 dxdt + \int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt - \frac{\lambda}{2} \int_{0}^{T} \left[ \frac{x^2 y^2}{d} \right]_{x=0}^{x=1} \\ &+ \frac{\lambda}{2} \int_{Q_T} \frac{x}{d} \left( 2 - \frac{xd'}{d} \right) y^2 dxdt. \end{aligned} \quad (3.22) $$

It is easy to check that $\lim_{x \to 0} y^2(t, x) \frac{x^2}{d} = 0$. Indeed $y(0) = 0$, thus

$$ |y(x)| \le \int_0^x |y'(t)| dt \le \sqrt{x} \|y'\|_{L^2(0,1)}. $$

Hence

$$ \frac{x}{d(x)} y^2(x) \le \frac{x^2}{d(x)} \|y'\|_{L^2(0,1)}^2 \quad (3.23) $$

and the statement follows by (2.4). As a consequence, one has

$$ \int_0^T \left[ \frac{x^2}{d} y^2 \right]_{x=0}^{x=1} dt = 0. $$

Now, consider the term $\int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt$, which is clearly well defined, and let $\delta > 0$. Then

$$ \int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt = \int_0^T \int_0^\delta x^2 y_x (ay_{xx})_{xx} dxdt + \int_0^T \int_\delta^1 x^2 y_x (ay_{xx})_{xx} dxdt. $$

By the Hölder inequality, $x^2 y_x (ay_{xx})_{xx} \in L^1(0, 1)$, hence the absolute continuity of the integral implies

$$ \lim_{\delta \to 0} \int_0^T \int_0^\delta x^2 y_x (ay_{xx})_{xx} dxdt = 0. $$
Let us now estimate the integral $\int_0^T \int_\delta^1 x^2 y_x (ay_{xx})_{xx} dxdt$. As in [9, Theorem 3.7], using Lemmas 3.2 and 3.3, one can prove the following equality

$$ \begin{aligned} \int_{Q_T} x^2 y_x (ay_{xx})_{xx} dxdt &= \lim_{\delta \to 0} \int_0^T \int_\delta^1 x^2 y_x (ay_{xx})_{xx} dxdt \\ &= -\frac{1}{2}a(1) \int_0^T y_{xx}^2(t, 1)dt - \int_{Q_T} a'y_x^2 dxdt \\ &\quad + \int_{Q_T} \left(3xa - \frac{x^2}{2}a'\right) y_{xx}^2 dxdt \end{aligned} $$

and the thesis follows.

<figure><img src="image_1.png" /></figure>

Thanks to (3.21), one can obtain the following estimate from below on the energy.

The next assumption collects all the needed requirements to prove the estimates on the energy.

**Hypothesis 7.** Hypothesis 4 holds. Moreover, the functions $a$ and $d$ are (WD) or (SD) with $K_d \le 1$.

**Theorem 3.8.** Assume Hypothesis 7. Fixed $T > 0$, if $y$ is a mild solution of (2.19), then

$$ a(1) \int_0^T y_{xx}^2(t, 1)dt \le \tilde{C}_1 E_y(0) + \tilde{C}_2 T E_y(0), \quad (3.24) $$

where

$$ \tilde{C}_1 = 4 \max \left\{ \tilde{C}_\epsilon \frac{1}{a(1)(2-K_a)}, 1 \right\}, \quad (3.25) $$

$$ \tilde{C}_2 = 4 \max \left\{ 1, \tilde{C}_\epsilon \left( \tilde{C}_{2,1} + \frac{6+K_a}{2} + \frac{|\lambda|}{2}(2+K_d)\tilde{C}_{HP} \right) \right\}, \quad (3.26) $$

$$ \tilde{C}_{2,1} := \begin{cases} \frac{\max_{x \in [0,1]} |a'(x)|}{a(1)(2-K_a)}, & a \text{ is } (SD), \\ \frac{4K_a}{(1-K_a)^2}, & a \text{ is } (WD) \end{cases} \quad (3.27) $$

and

$$ \tilde{C}_{\epsilon} := \begin{cases} 1, & \lambda < 0, \\ \frac{1}{\tilde{\epsilon}}, & \lambda > 0. \end{cases} \quad (3.28) $$

Here $\tilde{\epsilon}$ is as in (2.29).

**Proof.** As a first step we estimate the distributed terms given in (3.21). As in [9], one has

$$ \begin{aligned} \left| \int_{Q_T} \left( 3xa - \frac{x^2 a'}{2} \right) y_{xx}^2 dxdt \right| &\le \frac{6+K_a}{2} \int_{Q_T} ay_{xx}^2 dxdt, \\ \int_{Q_T} xy_t^2 dxdt &\le \int_{Q_T} y_t^2 dxdt, \end{aligned} $$

and

$$ \left| \int_{Q_T} a' y_x^2 dxdt \right| \le \tilde{C}_{2,1} \int_{Q_T} ay_{xx}^2 dxdt, $$

where $\tilde{C}_{2,1}$ is as in (3.27). Moreover, by Proposition 2.5

$$ \begin{aligned} \left| \frac{\lambda}{2} \int_{Q_T} \frac{x}{d} \left( 2 - \frac{xd'}{d} \right) y^2 dxdt \right| &\le \frac{|\lambda|}{2} (2+K_d) \int_{Q_T} \frac{y^2}{d} dxdt \\ &\le \frac{|\lambda|}{2} (2+K_d) \tilde{C}_{HP} \int_{Q_T} ay_{xx}^2 dxdt. \end{aligned} \quad (3.29) $$
Now, we will estimate the boundary term in (3.21). By (2.25), one has

$$
\begin{aligned}
\left| \int_0^1 x^2 y_x(\tau, x) y_t(\tau, x) dx \right| &\le \frac{1}{2} \int_0^1 x^4 y_x^2(\tau, x) dx + \frac{1}{2} \int_0^1 y_t^2(\tau, x) dx \\
&\le \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x) y_{xx}^2(\tau, x) dx dt + \frac{1}{2} \int_0^1 y_t^2(\tau, x) dx,
\end{aligned}
$$

for all $\tau \in [0, T]$; hence, we get

$$
\begin{aligned}
\left| \int_0^1 \left[ x^2 y_x(\tau, x) y_t(\tau, x) \right]_{\tau=0}^{\tau=T} dx \right| &\le \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x) y_{xx}^2(T, x) dx + \frac{1}{2} \int_0^1 y_t^2(T, x) dx \\
&\quad + \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x) y_{xx}^2(0, x) dx + \frac{1}{2} \int_0^1 y_t^2(0, x) dx.
\end{aligned}
$$

Case $\lambda < 0$. By Theorem 3.6, one has

$$
\left| \int_0^1 \left[ x^2 y_x(\tau, x) y_t(\tau, x) \right]_{\tau=0}^{\tau=T} dx \right| \le 2 \max \left\{ \frac{1}{a(1)(2-K_a)}, 1 \right\} E_y(0).
$$

Hence, by (3.21)

$$
\begin{aligned}
a(1) \int_0^T y_{xx}^2(t, 1) dt &\le 4 \max \left\{ \frac{1}{a(1)(2-K_a)}, 1 \right\} E_y(0) + 2 \int_{Q_T} y_t^2 dx dt \\
&\quad + 2 \left( \tilde{C}_{2,1} + \frac{6+K_a}{2} + \frac{|\lambda|}{2} (2+K_d) \tilde{C}_{HP} \right) \int_{Q_T} a y_{xx}^2 dx dt \\
&\le \tilde{C}_1 E_y(0) + T \tilde{C}_2 E_y(0),
\end{aligned}
$$

where $\tilde{C}_1$ and $\tilde{C}_2$ are as in (3.25) and (3.26), respectively.

Case $\lambda \in (0, \frac{1}{\tilde{C}_{HP}})$. By Proposition 2.6

$$
\begin{aligned}
\left| \int_0^1 \left[ x^2 y_x(\tau, x) y_t(\tau, x) \right]_{\tau=0}^{\tau=T} dx \right| &\le \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x) y_{xx}^2(T, x) dx + \frac{1}{2} \int_0^1 y_t^2(T, x) dx \\
&\quad + \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x) y_{xx}^2(0, x) dx + \frac{1}{2} \int_0^1 y_t^2(0, x) dx \\
&\le \frac{1}{2} \frac{1}{\tilde{c}a(1)(2-K_a)} \left( \int_0^1 a(x) y_{xx}^2(T, x) dx - \lambda \int_0^1 \frac{y^2(T, x)}{d(x)} dx \right) + \frac{1}{2} \int_0^1 y_t^2(T, x) dx \\
&\quad + \frac{1}{2} \frac{1}{\tilde{c}a(1)(2-K_a)} \left( \int_0^1 a(x) y_{xx}^2(0, x) dx - \lambda \int_0^1 \frac{y^2(0, x)}{d(x)} dx \right) + \frac{1}{2} \int_0^1 y_t^2(0, x) dx \\
&\le 2 \max \left\{ \frac{1}{\tilde{c}a(1)(2-K_a)}, 1 \right\} E_y(0).
\end{aligned}
$$

Hence, by (3.21) and Proposition 2.6, one has

$$
\begin{aligned}
a(1) \int_0^T y_{xx}^2(t, 1) dt &\le 4 \max \left\{ \frac{1}{\tilde{c}a(1)(2-K_a)}, 1 \right\} E_y(0) + 2 \int_{Q_T} y_t^2 dx dt \\
&\quad + 2 \left( \tilde{C}_{2,1} + \frac{6+K_a}{2} + \frac{|\lambda|}{2}(2+K_d)\tilde{C}_{HP} \right) \int_{Q_T} a y_{xx}^2 dx dt \\
&\le 4 \max \left\{ \frac{1}{\tilde{c}a(1)(2-K_a)}, 1 \right\} E_y(0) + 2 \int_{Q_T} y_t^2 dx dt \\
&\quad + \frac{2}{\tilde{c}} \left( \tilde{C}_{2,1} + \frac{6+K_a}{2} + \frac{|\lambda|}{2}(2+K_d)\tilde{C}_{HP} \right) \left( \int_{Q_T} a y_{xx}^2 dx dt - \lambda \int_0^1 \frac{y^2(t, x)}{d(x)} dx \right) \\
&\le \tilde{C}_1 E_y(0) + T \tilde{C}_2 E_y(0),
\end{aligned}
$$

and the thesis follows if $y$ is a classical solution. If $y$ is a mild solution then we can proceed as in Theorem 3.3. $\square$
In order to obtain an estimate from above on the energy, the next preliminary result is useful.

**Theorem 3.9.** Assume a and d (WD) or (SD) and $K_d \le 1$. Fixed $T > 0$, if y is a classical solution of (2.19), then the following identity holds:

$$ \frac{1}{2}a(1) \int_0^T y_{xx}^2(t, 1)dt = \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx + \frac{1}{2} \int_{Q_T} y_t^2 dxdt + \frac{1}{2} \int_{Q_T} (3a - xa')y_{xx}^2 dxdt \\ + \frac{\lambda}{2} \int_{Q_T} \frac{y^2}{d} \left[1 - \frac{xd'}{d}\right] dxdt. \qquad (3.30) $$

*Proof.* Multiplying the equation in (2.19) by $xy_x$ and integrating over $Q_T$, one has

$$ 0 = \int_0^1 [xy_t y_t]_{t=0}^{t=T} dx + \frac{1}{2} \int_{Q_T} y_t^2 dxdt + \int_{Q_T} xy_x (ay_{xx})_{xx} dxdt - \frac{\lambda}{2} \int_0^T \left[ \frac{x}{d} y^2 \right]_{x=0}^{x=1} dxdt \\ + \frac{\lambda}{2} \int_{Q_T} \frac{1}{d} \left(1 - \frac{xd'}{d}\right) y^2 dxdt. \qquad (3.31) $$

As in the proof of Theorem 3.7 (see (3.23)), one has that $\lim_{x \to 0} y^2(t, x) \frac{x}{d} = 0$ and hence

$$ \int_0^T \left[ \frac{x}{d} y^2 \right]_{x=0}^{x=1} dt = 0. $$

Now, let $\delta > 0$ and rewrite the well defined integral $\int_{Q_T} xy_x (ay_{xx})_{xx} dxdt$ as

$$ \int_{Q_T} xy_x (ay_{xx})_{xx} dxdt = \int_0^T \int_0^\delta xy_x (ay_{xx})_{xx} dxdt + \int_0^T \int_\delta^1 xy_x (ay_{xx})_{xx} dxdt. $$

Due to the absolute continuity of the integral we get

$$ \lim_{\delta \to 0} \int_0^T \int_0^\delta xy_x (ay_{xx})_{xx} dxdt = 0. $$

Moreover, as in [9]

$$ \int_0^T \int_\delta^1 xy_x (ay_{xx})_{xx} dxdt = - \int_0^T \delta y_x(t, \delta) (ay_{xx})_x (t, \delta) dt + \int_0^T a(\delta) y_x(t, \delta) y_{xx}(t, \delta) dt \\ - \frac{1}{2} \int_0^T [axy_{xx}^2]_{x=\delta}^{x=1} dt + \frac{1}{2} \int_0^T \int_\delta^1 (3a - xa') y_{xx}^2 dxdt. \qquad (3.32) $$

As in the proof of Theorem 3.7, one has

$$ \lim_{\delta \to 0} \int_0^T \int_\delta^1 (3a - xa') y_{xx}^2 dxdt = \int_{Q_T} (3a - xa') y_{xx}^2 dxdt $$

and, thanks to Lemma 3.2 or 3.3, (3.32) becomes

$$ \int_0^T \int_\delta^1 xy_x (ay_{xx})_{xx} dxdt = -\frac{1}{2}a(1)y_{xx}^2(t, 1) + \frac{1}{2} \int_{Q_T} (3a - xa') y_{xx}^2 dxdt, $$

and the thesis follows.

Thanks to the equality stated in the previous theorem, one can prove an estimate from above of the energy. This estimate will be crucial to obtain the observability result.

**Theorem 3.10.** Assume Hypothesis 7. Fixed $T > 0$, if y is a mild solution of (2.19), then

$$ \int_0^1 y_{xx}^2 dx \ge \tilde{C}_3 T E_y(0) - \tilde{C}_4 E_y(0), $$
where

$$ \tilde{C}_3 := \begin{cases} \frac{1}{a(1)}(m_a + (2 + K_a - 2K_d + m_a)\lambda\tilde{C}_{HP}), & \lambda < 0, \\ \frac{1}{a(1)}m_a, & \lambda > 0, \end{cases} \quad (3.33) $$

and

$$ \tilde{C}_4 := \frac{\tilde{C}_\epsilon}{a(1)} \left( 4 \max \left\{ 1, \frac{1}{a(1)(2-K_a)} \right\} + K_a \max \left\{ 1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x) \right\} \right), \quad (3.34) $$

where $m_a = \min\{6 - 3K_a, K_a + 2\}$ and $\tilde{C}_\epsilon$ is as in (3.28).

*Proof.* As a first step, assume that $y$ is a classical solution of (2.19). Then, multiplying the equation in (2.19) by $\frac{-K_a y}{2}$, integrating by parts over $Q_T$ and applying (2.31), we have

$$ 0 = -\frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx + \frac{K_a}{2} \int_{Q_T} y_t^2 dxdt - \frac{K_a}{2} \int_{Q_T} ay_{xx}^2 dxdt + \frac{K_a}{2} \int_{Q_T} \lambda \frac{y^2}{d} dxdt. $$

Summing the previous equality to (3.30) multiplied by 2, using (1.2) and Theorem 3.6, we have

$$ \begin{aligned} a(1) \int_0^T y_{xx}^2(t, 1) dt &= 2 \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx + \left(\frac{K_a}{2} + 1\right) \int_{Q_T} y_t^2 dxdt \\ &\quad + \int_{Q_T} \left(3 - \frac{xa'}{a} - \frac{K_a}{2}\right) ay_{xx}^2 dxdt + \int_{Q_T} \left(1 + \frac{K_a}{2} - \frac{xd'}{d}\right) \lambda \frac{y^2}{d} dxdt \\ &\geq 2 \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx + (K_a + 2) \frac{1}{2} \int_{Q_T} y_t^2 dxdt \\ &\quad + (6 - 3K_a) \frac{1}{2} \int_{Q_T} ay_{xx}^2 dxdt - (2K_d - 2 - K_a) \frac{1}{2} \int_{Q_T} \lambda \frac{y^2}{d} dxdt \\ &\geq 2 \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx \\ &\quad + m_a T E_y(0) + \frac{1}{2} (2 + K_a - 2K_d + m_a) \int_{Q_T} \lambda \frac{y^2}{d} dxdt. \end{aligned} \quad (3.35) $$

Observe that $2 + K_a - 2K_d + m_a > 0$. Indeed, if $K_a \ge 1$, then $\min\{6 - 3K_a, K_a + 2\} = 6 - 3K_a$ and, using the fact that $K_d \le 1$ by assumption, we have $K_a - K_d \le K_a + K_d < 4$. Thus, as a consequence,

$$ 2 + K_a - 2K_d + 6 - 3K_a = 8 - 2K_a + 2K_d > 0. $$

On the other hand, if $K_a > 1$, then $\min\{6 - 3K_a, K_a + 2\} = K_a + 2$, and

$$ 2 + K_a - 2K_d + K_a + 2 = 4 + 2K_a - 2K_d > 0 $$

using the fact that $2 + K_a > K_d$.

Now, we proceed in two different ways, according to the sign of $\lambda$.

Case $\lambda \in (0, \frac{1}{C_{HP}})$. In this case, by Proposition 2.6

$$ \int_0^1 ay_{xx}^2 dx - \lambda \int_0^1 \frac{y^2}{d} dx \ge \tilde{\epsilon} \int_0^1 ay_{xx}^2 dx. $$

Thus, by Theorem 3.6, for all $\tau \in [0, T]$

$$ \begin{aligned} 2E_y(0) = 2E_y(\tau) &\ge \tilde{\epsilon} \int_0^1 ay_{xx}^2 dx + \int_0^1 y_t^2(\tau, x) dx \\ &\ge \tilde{\epsilon} \left( \int_0^1 ay_{xx}^2 dx + \int_0^1 y_t^2(\tau, x) dx \right), \end{aligned} $$

so that

$$ \int_0^1 ay_{xx}^2 dx + \int_0^1 y_t^2(\tau, x) dx \le \frac{2}{\tilde{\epsilon}} E_y(0), \quad (3.36) $$
for all $\tau \in [0, T]$. Let us now consider the boundary terms that appear in (3.35). In particular, for all $\tau \in [0, T]$,

$$ \begin{aligned} \left| \int_0^1 xy_x(\tau, x)y_t(\tau, x)dx \right| & \le \frac{1}{2} \int_0^1 x^2 y_x^2(\tau, x)dx + \frac{1}{2} \int_0^1 y_t^2(\tau, x)dx \\ & \le \frac{1}{2} \frac{1}{a(1)(2-K_a)} \int_0^1 a(x)y_{xx}^2(\tau, x)dxdt + \frac{1}{2} \int_0^1 y_t^2(\tau, x)dx. \end{aligned} \quad (3.37) $$

Hence, by (3.36), one has

$$ \left| \int_0^1 xy_x(\tau, x)y_t(\tau, x)dx \right| \le \max \left\{ 1, \frac{1}{a(1)(2-K_a)} \right\} \frac{1}{\tilde{\epsilon}} E_y(0), $$

for all $\tau \in [0, T]$, so that

$$ 2 \left| \int_0^1 [xy_x y_t]_{t=0}^{t=T} dx \right| \le 4 \max \left\{ 1, \frac{1}{a(1)(2-K_a)} \right\} \frac{1}{\tilde{\epsilon}} E_y(0). $$

Moreover, by Proposition 2.5, one has

$$ \begin{aligned} \left| \int_0^1 yy_t(\tau, x)dx \right| & \le \frac{1}{2} \int_0^1 y_t^2(\tau, x)dx + \frac{1}{2} \int_0^1 y^2(\tau, x)dx \\ & \le \frac{1}{2} \int_0^1 y_t^2(\tau, x)dx + \frac{\max_{x \in [0,1]} d(x)}{2} \int_0^1 \frac{y^2}{d}(\tau, x)dx \quad (3.38) \\ & \le \frac{1}{2} \int_0^1 y_t^2(\tau, x)dx + \frac{\max_{x \in [0,1]} d(x)}{2} \tilde{C}_{HP} \int_0^1 ay_{xx}^2(\tau, x)dx \end{aligned} $$

for all $\tau \in [0, T]$. Hence,

$$ \left| \int_0^1 yy_t(\tau, x)dx \right| \le \max \left\{ 1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x) \right\} \frac{1}{\tilde{\epsilon}} E_y(0), $$

for all $\tau \in [0, T]$ so that

$$ \frac{K_a}{2} \left| \int_0^1 [yy_t(t, x)]_{t=0}^{t=T} dx \right| \le \frac{K_a}{\tilde{\epsilon}} \max \left\{ 1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x) \right\} E_y(0). $$

Hence, by (3.35) and the previous estimates, we find

$$ \begin{aligned} a(1) \int_0^T y_{xx}^2(t, 1)dt & \ge 2 \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx \\ & + m_a T E_y(0) + \frac{1}{2}(2+K_a-2K_d+m_a) \int_{Q_T} \lambda \frac{y^2}{d} dxdt \\ & \ge -\left(4 \max \left\{1, \frac{1}{a(1)(2-K_a)}\right\} + K_a \max \left\{1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x)\right\}\right) \frac{1}{\tilde{\epsilon}} E_y(0) \\ & + m_a T E_y(0). \end{aligned} $$

Case $\lambda < 0$. Let us consider again the boundary terms that appear in (3.35). In particular, by (3.37) and (3.38) which hold for all considered $\lambda$,

$$ 2 \left| \int_0^1 [xy_x(t, x)y_t(t, x)]_{t=0}^{t=T} dx \right| \le 4 \max \left\{ 1, \frac{1}{a(1)(2-K_a)} \right\} E_y(0) $$

and

$$ \frac{K_a}{2} \left| \int_0^1 [yy_t(t, x)]_{t=0}^{t=T} dx \right| \le K_a \max \left\{ 1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x) \right\} E_y(0). $$
Moreover, by Proposition 2.5,

$$ \lambda \int_{Q_T} \frac{y^2}{d} dxdt \geq \lambda \tilde{C}_{HP} \int_{Q_T} ay_{xx}^2 dxdt \geq 2\lambda \tilde{C}_{HP} T E_y(0). $$

Thus, by the previous estimates and (3.35), we obtain

$$ \begin{aligned} a(1) \int_0^T y_{xx}^2(t, 1) dt &\geq 2 \int_0^1 [xy_t y_x]_{t=0}^{t=T} dx - \frac{K_a}{2} \int_0^1 [yy_t]_{t=0}^{t=T} dx \\ &\quad + m_a T E_y(0) + (2 + K_a - 2K_d + m_a) \frac{1}{2} \int_{Q_T} \lambda \frac{y^2}{d} dxdt \tag{3.39} \\ &\geq - \left( 4 \max \left\{ 1, \frac{1}{a(1)(2-K_a)} \right\} + K_a \max \left\{ 1, \tilde{C}_{HP} \max_{x \in [0,1]} d(x) \right\} \right) E_y(0) \\ &\quad + m_a T E_y(0) + (2 + K_a - 2K_d + m_a) \lambda \tilde{C}_{HP} T E_y(0) \end{aligned} $$

and the thesis holds if $y$ is a classical solution.

Now, let $y$ be the mild solution associated to the initial data $(y_T^0, y_T^1) \in \mathcal{K}_0$. Then, by approximation as in Theorem 3.8, we obtain the desired inequalities. $\square$

# 4 Boundary observability and null controllability

According to [1], [3] and [6], we state the following definition:

**Definition 4.1.** Problem (2.5) (resp., (2.19)) is said to be observable in time $T > 0$ via the second derivative at $x = 1$ if there exists a constant $C > 0$ such that for any $(y_T^0, y_T^1) \in \mathcal{H}_0$ (resp., $(y_T^0, y_T^1) \in \mathcal{K}_0$) , the classical solution $y$ of (2.5) (resp., (2.19)) satisfies

$$ CE_y(0) \leq \int_0^T y_{xx}^2(t, 1) dt. \qquad (4.1) $$

Moreover, any constant satisfying (4.1) is called observability constant for (2.5) (resp., (2.19)) in time $T$. Setting

$$ C_T := \sup\{C > 0 : C \text{ satisfies (4.1)}\}, $$

we have that problem (2.5) (resp., (2.19)) is observable if and only if

$$ C_T = \inf_{(y_T^0, y_T^1) \neq (0,0)} \frac{\int_0^T y_{xx}^2(t, 1) dt}{E_y(0)} > 0. $$

The inverse of $C_T$, $c_t := \frac{1}{C_T}$, is called the cost of observability (or the cost of control) in time $T$.

## 4.1 The non-divergence case

Theorem 3.5 admits the following straightforward corollaries.

**Corollary 4.1.** Assume the functions $a$ and $d$ (WD) or (SD) with $K_a + 2K_d \leq 2$ and

$$ -\frac{2 - K_a}{2C_{HP}(2 + K_a + K_d)} < \lambda < 0. $$

If $T > \frac{C_4}{C_3}$, then (2.5) is observable in time $T$; moreover

$$ C_3T - C_4 \leq C_T. $$

Here $C_3$ and $C_4$ are as in (3.13) and (3.14), respectively.
**Corollary 4.2.** Assume the functions a and d (WD) or (SD) with $K_a + 2K_d \le 2$ and

$$ \lambda \in \left(0, \frac{1}{C_{HP}}\right). $$

If $T > \frac{C_4}{C_3}$, then the same statement of Corollary 4.1 holds.

Now, we are ready to prove the null controllability for (1.1) when $i = 1$. More precisely, given $(u_0, u_1) \in L_{\frac{1}{a}}^2(0, 1) \times (H_{\frac{1}{a}}^2(0, 1))'$, where $(H_{\frac{1}{a}}^2(0, 1))'$ is the dual space of $H_{\frac{1}{a}}^2(0, 1)$, we look for a control $f \in L^2(0, T)$ such that the solution of (1.1) satisfies

$$ u(T, x) = u_t(T, x) = 0, \quad \text{for all } x \in (0, 1). \tag{4.2} $$

For this last purpose, following [6], we give the definition of a solution for (1.1) by transposition, which permits low regularity on the notion of solution itself. To be more precise, we have the following:

**Definition 4.2.** Let $f \in L_{loc}^2[0, +\infty)$ and $(u_0, u_1) \in L_{\frac{1}{a}}^2(0, 1) \times (H_{\frac{1}{a}}^2(0, 1))'$. We say that $u$ is a solution by transposition of (1.1) if

$$ u \in \mathcal{U} := C^1 \left( [0, +\infty); (H_{\frac{1}{a}}^2(0, 1))' \right) \cap C \left( [0, +\infty); L_{\frac{1}{a}}^2(0, 1) \right) $$

and for all $T > 0$

$$ \langle u_t(T), v_T^0 \rangle (H_{\frac{1}{a}}^2(0,1))', H_{\frac{1}{a}}^2(0,1) - \int_0^1 \frac{1}{a} u(T) v_T^1 dx = \langle u_1, v(0) \rangle (H_{\frac{1}{a}}^2(0,1))', H_{\frac{1}{a}}^2(0,1) \\ - \int_0^1 \frac{1}{a} u_0 v_t(0,x) dx + \int_0^T f(t) v_{xx}(t,1) dt \tag{4.3} $$

for all $(v_T^0, v_T^1) \in \mathcal{H}_0$, where $v$ solves the backward problem

$$ \left\{ \begin{array}{ll} v_{tt}(t,x) + a(x)v_{xxxx}(t,x) - \frac{\lambda}{d}v(t,x) = 0, & (t,x) \in (0, +\infty) \times (0, 1), \\ v(t,0) = v_x(t,0) = 0, & t \in (0, +\infty) \\ v(t,1) = v_x(t,1) = 0, & t \in (0, +\infty) \\ v(T,x) = v_T^0(x), \quad v_T(x) = v_T^1(x), & x \in (0, 1). \end{array} \right. \tag{4.4} $$

Here $\mathcal{H}_0$ is the space defined in (2.15).

Observe that, setting $y(t,x) := v(T - t, x)$, one has that $y$ satisfies (2.5) with $y_T^0(x) = v_T^0(x)$ and $y_T^1(x) = -v_T^1(x)$. Hence, thanks to Theorem 2.2 and Remark 1, one has that there exists a unique mild solution $y$ of (2.5) in $\mathbb{R}$. In particular, there exists a unique mild solution

$$ v \in C^1 \left( [0, +\infty); L_{\frac{1}{a}}^2(0, 1) \right) \cap C \left( [0, +\infty); H_{\frac{1}{a}}^2(0, 1) \right) $$

of (4.4) which depends continuously on the initial data $V_T := (v_T^0, v_T^1) \in \mathcal{H}_0$.

By Theorem 3.1, the energy is preserved in our setting, as well, so that the method of transposition done in [1] continues to hold thanks to (3.6), and so there exists a unique solution by transposition $u \in \mathcal{U}$ of (1.1) when $i = 1$, namely a solution of (4.3). For more details we refer to [6].

Now, we are ready to pass to null controllability, recalling that by linearity and reversibility of equation (1.1), null controllability for any initial data $(u_0, u_1)$ is equivalent to exact controllability, see [1]. In order to prove that (1.1) is null controllable, let us start with

**Hypothesis 8.** Assume $T > \frac{C_4}{C_3}$, a and d (WD) or (SD) with $K_a + 2K_d \le 2$, and

$$ -\frac{2 - K_a}{2C_{HP}(2 + K_a + K_d)} < \lambda < 0 $$

or

$$ \lambda \in \left(0, \frac{1}{C_{HP}}\right). $$
Now, consider the bilinear form $\Lambda$ on the product space $\mathcal{H}_0 \times \mathcal{H}_0$ as follows

$$ \Lambda(V_T, W_T) := \int_0^T v_{xx}(t, 1)w_{xx}(t, 1) $$

where $v$ and $w$ are the solutions of (4.4) associated to the final data $V_T := (v_T^0, v_T^1)$ and $W_T := (w_T^0, w_T^1)$, respectively. The bilinear form $\Lambda$ is continuous and coercive, and thanks to these properties one can prove the null controllability for the original problem (1.1). To this aim, define $T_0$ as the lower bound found in Corollaries 4.1 and 4.2, i.e.

$$ T_0 := \frac{C_4}{C_3}, $$

where $C_3$, and $C_4$ are the constants defined in (3.13) and (3.14), respectively. Then, as in [6], one can prove the next result.

**Theorem 4.1.** Assume Hypothesis 8. Then, for all $T > T_0$ and for every $(u_0, u_1)$ in $L^2_{\frac{1}{a}}(0, 1) \times (H^2_{\frac{1}{a}}(0, 1))'$ there exists a control $f \in L^2(0, T)$ such that the solution of (1.1) satisfies (4.2).

## 4.2 The divergence case

Theorem 3.10 admits the following straightforward corollaries, which are the analogue of Corollaries 4.1 and 4.2.

**Corollary 4.3.** Assume the functions $a$ and $d$ (WD) or (SD) with $K_d \le 1$ and

$$ \frac{m_a}{(2 + K_a - 2K_d + m_a) \tilde{C}_{HP}} < \lambda < 0 $$

If $T > \frac{\tilde{C}_4}{C_3}$, then (2.19) is observable in time $T$; moreover

$$ \tilde{C}_3 T - \tilde{C}_4 \le C_T. $$

**Corollary 4.4.** Assume the functions $a$ and $d$ (WD) or (SD) with $K_d \le 1$ and

$$ \lambda \in \left(0, \frac{1}{\tilde{C}_{HP}}\right). $$

If $T > \frac{\tilde{C}_4}{C_3}$, then the same statement of Corollary 4.3 holds.

Now, we give the definition of solution by transposition for the problem (1.1) when $i = 2$.

**Definition 4.3.** Let $f \in L^2_{loc}[0, +\infty)$ and $(u_0, u_1) \in L^2(0, 1) \times (H^2_{a,0}(0, 1))'$. We say that $u$ is a solution by transposition of (1.1) if

$$ u \in \mathcal{V} := C^1([0, +\infty); (H^2_{a,0}(0, 1))' \cap C([0, +\infty); L^2(0, 1)) $$

and

$$ \langle u_t(T), v_T^0 \rangle_{(H^2_{a,0}(0,1))', H^2_{a,0}(0,1)} - \int_0^1 u(T,x)v_T^1 dx \\ = \langle u_1, v(0) \rangle_{(H^2_{a,0}(0,1))', H^2_{a,0}(0,1)} - \int_0^1 u_0(x)v_t(0,x)dx - \int_0^T a(1)v_{xx}(t,1)f(t)dt, $$

for all $T > 0$ and for all $(v_T^0, v_T^1) \in \mathcal{K}_0$, where $v$ solves the backward problem

$$ \left\{ \begin{array}{ll} v_{tt}(t,x) + (av_{xx})_{xx}(t,x) - \frac{\lambda}{d}v(t,x) = 0, & (t,x) \in (0, +\infty) \times (0, 1), \\ v(t, 0) = 0, v(t, 1) = 0, & t \in (0, +\infty), \\ \begin{cases} v_x(t, 0) = 0, & \text{if } a \text{ is (WD)}, \\ (av_{xx})(t, 0) = 0, & \text{if } a \text{ is (SD)}, \end{cases} & t \in (0, +\infty), \\ v_x(t, 1) = 0, & t \in (0, +\infty), \\ v(T,x) = v_T^0(x), v_t(T,x) = v_T^1(x), & x \in (0, 1). \end{array} \right. \quad (4.5) $$

Here $\mathcal{K}_0$ is the space defined in (2.34).
Also for the divergence case, one can prove that there exists a unique solution by transposition $u \in \mathcal{V}$ of (1.1) when $i = 2$, namely a solution of (4.3). For more details we refer to [6].

**Hypothesis 9.** Assume $T > \frac{\tilde{C}_4}{C_3}$, $a$ and $d$ (WD) or (SD) with $K_d \le 1$, and

$$-\frac{m_a}{(2 + K_a - 2K_d + m_a) \tilde{C}_{HP}} < \lambda < 0$$

or

$$\lambda \in \left(0, \frac{1}{\tilde{C}_{HP}}\right).$$

Recall that $2 + K_a - 2K_d + m_a \tilde{C}_{HP} > 0$ (see the proof of Theorem 3.10).

In order to study the null controllability for (1.1), we proceed as before considering the bilinear form $\tilde{\Lambda} : \mathcal{K}_0 \times \mathcal{K}_0 \to \mathbb{R}$ defined as

$$\tilde{\Lambda}(V_T, W_T) := \int_0^T a(1)v_{xx}(t, 1)w_{xx}(t, 1)dt,$$

where $v$ and $w$ are the solutions of (4.5) associated to the data $V_T := (v_T^0, v_T^1)$ and $W_T := (w_T^0, w_T^1)$, respectively, $\tilde{T}_0 := \frac{\tilde{C}_4}{C_3}$ and $\tilde{C}_3, \tilde{C}_4$ are as in (3.33) and (3.34), respectively. Then, as in [9], one has the next result.

**Theorem 4.2.** Assume Hypothesis 9. Then, for all $T > \tilde{T}_0$ and for every $(u_0, u_1) \in L^2(0, 1) \times (H_{a,0}^2(0, 1))'$, there exists a control $f \in L^2(0, T)$ such that the solution of (1.1) satisfies (4.2).

# 5 Open problems

There are some open problems related to the boundary controllability studied in this paper.

For example, if the problem is in non-divergence form we have seen that the assumption $K_a+2K_d \le 2$ is crucial to justify (2.6). Clearly this is a technical assumption, thus it would be interesting to study the problem when $K_a+2K_d > 2$. This assumption is also required in [13] and so this is an open problem also for degenerate and singular wave systems. We underline that in [13] also a drift term is considered. Thus, one can think to extend the results of this paper to (1.1) in presence of a drift term. Indeed there are no results on boundary controllability for (1.1) with a drift term even if $\lambda = 0$.

If the problem is in divergence form then the previous assumption $K_a + 2K_d \le 2$ is replaced by $K_d \le 1$. Hence, what happens if $K_d > 1$?

Another interesting open problem is the following: according to the techniques used in the previous proofs we need an estimate from below if $\lambda$ is negative. Hence, the problem is to prove the boundary null controllability without this additional condition.

# Acknowledgements

G. Fragnelli is a member of: Gruppo Nazionale per l'Analisi Matematica, la Probabilità e le loro Applicazioni (GNAMPA) of the Istituto Nazionale di Alta Matematica (INdAM), UMI "Modellistica Socio-Epidemiologica (MSE)" and UMI "ClinMath". She is partially supported by INdAM GNAMPA Project Mathematical Methods for Climate Impacts: Control, Game Theory, and Inverse Problems for EBCM (CUP E5324001950001).

# 6 Declarations, Conflict of interest

The authors have no competing interests to declare that are relevant to the content of this article.
References

[1] F. Alabau-Boussouira, P. Cannarsa, Günter Leugering, *Control and stabilization of degenerate wave equations*, SIAM J. Control Optim., **55** (2017), 2052–2087.

[2] A. Bensoussan, G. Da Prato, M. C. Delfour, S. K. Mitter, *Representation and Control of Infinite Dimensional Systems, Second edition*. Birkhäuser Boston, Inc., Boston, MA, 2007.

[3] I. Boutaayamou, G. Fragnelli, D. Mugnai, *Boundary controllability for a degenerate wave equation in non divergence form with drift*, SIAM J. Control Optim., **61** (2023), 1934–1954.

[4] I. F. Bugariu, S. Micu, I. Roventa, *Approximation of the controls for the beam equation with vanishing viscosity*, Electron. J. Qual. Theory Differ. Equ., **85** (2016), 2259–2303.

[5] A. Camasta, G. Fragnelli, *A degenerate operator in non divergence form*, Recent advances in mathematical analysis. Trends in mathematics, Birkhäuser, Cham, 2023, 209–235.

[6] A. Camasta, G. Fragnelli, *Boundary controllability for a degenerate beam equation*, Math. Meth. Appl. Sci., **47** (2024), 907–927.

[7] A. Camasta, G. Fragnelli, *Degenerate fourth order parabolic equations with Neumann boundary conditions*, in corso di pubblicazione su Analysis and Numerics of Design, Control and Inverse Problems, volume INdAM-Springer. arXiv:2203.02739.

[8] A. Camasta, G. Fragnelli, *Fourth-order differential operators with interior degeneracy and generalized Wentzell boundary conditions*, Electron. J. Differential Equations, **2022** (2022), paper No. 87, 1-22.

[9] A. Camasta, G. Fragnelli, *New results on controllability and stability for degenerate Euler-Bernoulli type equations*, Discrete and Continuous Dynamical Systems, **44** (2024), 2193-2231.

[10] G. Fragnelli, *Interior degenerate/singular parabolic equations in nondivergence form: well-posedness and Carleman estimates*, J. Differential Equations, **260** (2016), 1314–1371.

[11] G. Fragnelli, D. Mugnai, *Carleman estimates for singular parabolic equations with interior degeneracy and non smooth coefficients*, Adv. Nonlinear Anal., **6** (2017), 61–84.

[12] G. Fragnelli, F. Mugnai, *Control of Degenerate and Singular Parabolic Equations. Carleman Estimates and Observability*, SpringerBriefs in Mathematics: Springer International Publishing, 2021.

[13] G. Fragnelli, D. Mugnai, A. Sbai, *Boundary controllability for degenerate/singular hyperbolic equations in nondivergence form with drift*, submitted for publication; arXiv:2402.18247.

[14] W. Krabs, G. Leugering, T. I. Seidman, *On boundary controllability of a vibrating plate*, Appl. Math. Optim., **13** (1985), 205–229.

[15] J. E. Lagnese, *Recent progress in exact boundary controllability and uniform, stabilizability of thin beams and plates. Distributed parameter control systems*, New Trends Appl. Lect. Notes Pure Appl. Math., **128** (1990), 61–112.

[16] J. E. Lagnese, J. L. Lions, *Modelling analysis and control of thin plates*, 1998.

[17] I. Lasiecka, R. Triggiani, *Exact controllability of the Euler-Bernoulli equation with controls in the Dirichlet and Neumann boundary conditions: a nonconservative case*, SIAM J. Control Optim., **27** (1989), 330–373.

[18] L. Léon and E. Zuazua, *Boundary controllability of the finite-difference space semidiscretizations of the beam equation*, ESAIM Control Optim. Calc. Var., **8** (2002), 827–862.

[19] J. L. Lions, *Exact controllability, stabilization and perturbations for distributed system*, SIAM Rev., **30** (1988), 1–68.

[20] B. P. Rao, *Exact boundary controllability of a hybrid system of elasticity by the HUM method*, ESAIM: Control, Optim. Calc. Var., **6** (2001), 183–199.