# Rapid stabilization for a wave equation with boundary disturbance

Patricio Guzmán, Agustin Huerta, Hugo Parada

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Patricio Guzmán, Agustin Huerta, Hugo Parada. Rapid stabilization for a wave equation with boundary disturbance. 2025. hal-05296068

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# Rapid stabilization for a wave equation with boundary disturbance

Patricio Guzmán,*, Agustín Huerta* and Hugo Parada†

## Abstract

In this paper we study the rapid stabilization of an unstable wave equation, in which an unknown disturbance is located at the boundary condition. We address two different boundary conditions: Dirichlet-Dirichlet and Dirichlet-Neumann. In both cases, we design a feedback law, located at the same place as the unknown disturbance, that forces the exponential decay of the energy for any desired decay rate while suppressing the effects of the unknown disturbance. For the feedback design we employ the backstepping method, Lyapunov techniques and the sign multivalued operator. The well-posedness of the closed-loop system, which is a differential inclusion, is shown with the maximal monotone operator theory.

**2020 Mathematics Subject Classification:** 35B40, 34G20, 35L05, 74K05, 93D15, 93D23, 93D09.

**Keywords:** Wave equation, boundary disturbance, feedback stabilization, Backstepping, Lyapunov techniques, exponential stability.

## 1 Introduction

**Some general comments.** Partial differential equations have been widely and successfully used to derive mathematical models for diverse phenomena, such as the temperature on solids or fluids [7, 17, 47], the propagation of waves in a medium [16, 20, 52], the deflection of strings or beams [35, 45], to name a few remarkable examples. Once a mathematical model is established one relevant task in control theory is to design feedback laws to stabilize the state of the system to their equilibria or to another state of interest [14, 40, 32, 39]. In the literature we can frequently find that mathematical models are analyzed under ideal assumptions in which disturbances are neglected for the sake of simplicity, but disturbances are always present and indeed may correspond to an additional source of instability. Accordingly, it is relevant to include disturbances in the study of the stabilization problem.

**A wave equation with boundary disturbance.** Let $L \in (0, \infty)$. Let $\lambda, \alpha$ and $\beta$ be time independent coefficients. Let $p = p(t)$ represent an unknown disturbance and let $U = U(t)$ represent a control to be designed. Let $(u_0, u_1)$ be an initial condition. Then, as in [49], let us consider the following two cases.

<footnote>*Departamento de Matemática, Universidad Técnica Federico Santa María, Valparaíso, Chile.
E-mail: patricio.guzmanm@usm.cl, huertasotoagustin@gmail.com (corresponding author)
†Université de Lorraine, CNRS, Inria, IECL, F-54000 Nancy, France.
E-mail: hugo.parada@inria.fr</footnote>
• Dirichlet-Dirichlet case:

$$ egin{cases} u_{tt} = u_{xx} + 2\lambda(x)u_t + \alpha(x)u_x + \beta(x)u, & (t, x) \in (0, \infty) \times (0, L), \\ u(t, 0) = 0, \quad u(t, L) = p(t) + U(t), & t \in (0, \infty), \\ u(0, x) = u_0(x), \quad u_t(0, x) = u_1(x), & x \in (0, L). \end{cases} \tag{1.1} $$

• Dirichlet-Neumann case:

$$ egin{cases} u_{tt} = u_{xx} + 2\lambda(x)u_t + \alpha(x)u_x + \beta(x)u, & (t, x) \in (0, \infty) \times (0, L), \\ u(t, 0) = 0, \quad u_x(t, L) = p(t) + U(t), & t \in (0, \infty), \\ u(0, x) = u_0(x), \quad u_t(0, x) = u_1(x), & x \in (0, L). \end{cases} \tag{1.2} $$

**Rapid stabilization.** For a regular enough state $(u, u_t)$ of (1.1) or (1.2), we define its energy by

$$ E(t) = \frac{1}{2} \int_{0}^{L} (|u_t|^2 + |u_x|^2) dx, \quad t \in [0, \infty). \tag{1.3} $$

The purpose of this paper is to exponentially stabilize the regular enough state $(u, u_t)$ to the rest position $(0, 0)$ by means of a control $U$ that suppresses the effects of an unknown disturbance $p$. Being precise, we aim to prove the rapid stabilization, which in this paper is understood as follows:

$$ egin{cases} 
\text{For any desired decay rate } d \in (0, \infty) \\
\text{there exists a control } U(t) \\
\text{and there exists a constant } C \in [1, \infty) \\
\text{such that } E(t) \le Ce^{-2dt}E(0) \text{ for all } t \in [0, \infty). 
\end{cases} \tag{1.4} $$

In virtue of [38, 50, 49, 8, 11, 12, 46, 31, 19, 6] and also of many other works, as can be consulted in [34, 51] for instance, we have that the backstepping method has been successfully applied to prove the rapid stabilization of several partial differential equations in the undisturbed case, which in this paper is when $p = 0$.

**Remark 1.1** In [49] it has been shown (1.4) for (1.1) and for (1.2) when $p = 0$. In this paper we address the disturbed case. Then, if we denote by $U_*(t)$ the control designed in [49], unfortunately we cannot ask $U(t) = U_*(t) - p(t)$ to solve the disturbed case, since we assume that $p(t)$ is unknown.

In this paper we employ the backstepping transformation constructed in [49] and apply the idea behind the feedback design in [27] to show (1.4) for (1.1) and for (1.2). Accordingly, although the disturbance is assumed unknown, we ask it to satisfy some assumptions, that shall be introduced next.

**Stabilization in the disturbed case.** The disturbed case has been object of recent interest, as can be consulted in Table 1. There we present some of the concerned literature in the stabilization of partial differential equations subjected to unknown disturbances, acting either in the domain or at the boundary.
<table><caption>Table 1: Stabilization of partial differential equations subjected to unknown disturbances. Assumptions (A1) and (A2) for the unknown disturbance considered in this paper are based on the assumptions asked in the referenced works.</caption><thead><tr><th>Equation</th><th>Distributed disturbance</th><th>Boundary disturbance</th></tr></thead><tbody><tr><td>Heat</td><td>[28]</td><td>[30, 37, 26, 27]</td></tr><tr><td>Wave</td><td>[18]</td><td>[22, 21, 42]</td></tr><tr><td>Beam</td><td>[1]</td><td>[23, 29, 36]</td></tr><tr><td>Schrödinger</td><td>—</td><td>[24, 31]</td></tr></tbody></table>

Given the works in Table 1, it follows that the commonly accepted assumptions for the unknown disturbance involves knowing a uniform-in-time upper bound for its norm and also knowing its regularity. In this paper these assumptions are the following:

(A1) There exists a constant $P \in (0, \infty)$ such that $|p(t)| \le P$ for every $t \in [0, \infty)$.

(A2)DD $p \in W^{2,1}(0, \infty)$ and $p(0) = 0$.

(A2)DN $p \in W^{3,1}(0, \infty)$ and $p(0) = p'(0) = 0$.

Some comments on these assumptions are the following.

The works in Table 1 use the knowledge of the uniform-in-time upper bound for the norm of the unknown disturbance for the design of controls as feedback laws. In that design it is commonly used the sign multivalued operator sign : $\mathbb{R} \to 2^{\mathbb{R}}$ ($2^{\mathbb{R}}$ denotes the power set of $\mathbb{R}$) given by

$$ \text{sign}(f) = \begin{cases} \frac{f}{|f|} & \text{if } f \neq 0, \\ [-1, 1] & \text{if } f = 0, \end{cases} \qquad (1.5) $$

to suppress the instability that may produce the unknown disturbance. To that end it is adequately used the property $|f| = \text{sign}(f)f$. These feedback laws also contain other (univalued) terms, but are not involved in the suppression.

The works in Table 1 use the knowledge of the regularity of the unknown disturbance to satisfy the required hypotheses to apply the well-posedness results for the corresponding closed-loop systems, which are differential inclusions due to (1.5). Regarding the well-posedness, only [30, 23, 29, 31] consider the notion of Filippov solution, requiring the introduction of a sliding surface, while the rest of the works consider the notion of classical solution, as understood in the maximal monotone operator theory. In this paper we consider the latter notion, beign the choice only due to mathematical convenience.

**Contribution of this paper.** Except by [31] (Schrödinger equation) and by [28, 27] (Heat equation), all the works in Table 1 prove the exponential decay of some energy for a certain decay rate that results from the proof. Accordingly, the obtained decay rate in those works is unchosen. Therefore, most of the results are not rapid in the sense described in (1.4), being the case addressed in this paper (Wave equation) an open problem. To fill that gap, in this paper we employ the backstepping transformation constructed in [49] and apply the idea behind the feedback design in [27] to show the rapid stabilization of an unstable wave equation with boundary disturbance, meaning (1.4) for (1.1) and for (1.2). The well-posedness results for the corresponding closed-loop systems are obtained thanks to the maximal monotone operator theory by adapting an optimization argument in [9].
For the feedback design (see [33, 40, 10, 34, 39, 41, 25] for instance) we take inspiration from [27]. Indeed, in (1.1) or (1.2) we consider $U(t) = U_1(t) + U_2(t)$, where $U_1(t)$ is used to achieve the desired decay rate and $U_2(t)$ is used to reject the effects of the unknown disturbance. Being precise, it is in the choice of $U_1(t)$ where the backstepping transformation constructed in [49] is employed and it is in the design of $U_2(t)$ where Lyapunov techniques and (1.5) are adequately applied. Since the final expression for the feedback law $U(t)$ is multivalued because of (1.5), the corresponding closed-loop system is a differential inclusion, with the inclusion being located at the boundary condition. We prove the well-posedness result with the aid of the maximal monotone operator theory (see [3, 48, 2] and/or [5, 13, 15, 44] for instance), in which we adapt an optimization argument in [9] that uses the Moreau regularization to treat the terms at the boundary condition. Our main results are given by Theorem 2.2 (Dirichlet-Dirichlet case) and by Theorem 3.1 (Dirichlet-Neumann case).

**Organization of this paper.** The rest of this paper is organized as follows. The feedback design for the Dirichlet-Dirichlet case (respectively: Dirichlet-Neumann case) is done in Section 2.1 (respectively: Section 3.1), while the well-posedness of the corresponding closed-loop system is shown in Section 2.2 (respectively: Section 3.2). The main result gathering the conclusions of both the feedback design and the well-posedness parts is stated in Section 2.3 (respectively: Section 3.3). Finally, in Section 4 we provide closing comments.

## 2 Dirichlet-Dirichlet case

In this section we present all the required analysis to prove Theorem 2.2, which is one of our main results and corresponds to the obtention of (1.4) for (1.1). We start with a comment.

**Remark 2.1** As stated in [49, Section 2], we can transform (1.1) into another wave equation without first-order spatial derivative term by applying the transformation $v(t, x) = (e^{\frac{1}{2} \int_{0}^{x} \alpha(\tau) d\tau})u(t, x)$, thus obtaining

$$ egin{cases} v_{tt} = v_{xx} + 2\lambda(x)v_t + \left[-\frac{1}{2}\alpha'(x) - \frac{1}{4}\alpha(x)^2 + \beta(x)\right]v, & (t, x) \in (0, \infty) \times (0, L), \\ v(t, 0) = 0, \quad v(t, L) = \left(e^{\frac{1}{2}\int_{0}^{L} \alpha(\tau) d\tau}\right)p(t) + \left(e^{\frac{1}{2}\int_{0}^{L} \alpha(\tau) d\tau}\right)U(t), & t \in (0, \infty), \\ v(0, x) = \left(e^{\frac{1}{2}\int_{0}^{x} \alpha(\tau) d\tau}\right)u_0(x), \quad v_t(0, x) = \left(e^{\frac{1}{2}\int_{0}^{x} \alpha(\tau) d\tau}\right)u_1(x), & x \in (0, L). \end{cases} \quad (2.1) $$

Thus, in this section we consider

$$ egin{cases} u_{tt} = u_{xx} + 2\lambda(x)u_t + \beta(x)u, & (t, x) \in (0, \infty) \times (0, L), \\ u(t, 0) = 0, \quad u(t, L) = p(t) + U(t), & t \in (0, \infty), \\ u(0, x) = u_0(x), \quad u_t(0, x) = u_1(x), & x \in (0, L). \end{cases} \quad (2.2) $$

### 2.1 Feedback design

In this section we prove the feedback design part of Theorem 2.2. To that end, we start by introducing all the required elements of the backstepping transformation constructed in [49, Section 3].

Let $d \in (0, \infty)$ be the desired decay rate of the exponential decay of the energy (1.3). We proceed to introduce some elements from [49, Pages 4017, 4018 and 4019], with $c(x) = d^2$ and $d(x) = d$ in their notation. For $x \in [0, L]$ let us consider
$$ egin{cases} a(x) = \lambda(x) + d, \\ h(x) = \cosh \left( \int_0^x a(\tau) d\tau \right), \\ m(x) = \frac{h'(x)}{2a(x)} \left( 2\lambda(x) + a(x) + a(0) \right) + \frac{h(x)}{2} \int_0^x \left( d^2 - \lambda(y)^2 - \beta(y) - d^2 \right) dy. \end{cases} \quad (2.3) $$

Let $\mathcal{T} = \{(x, y) \in \mathbb{R}^2 / 0 \le y \le x \le L\}$. For $(x, y) \in \mathcal{T}$ let us consider

$$ egin{cases} \rho_1(x, y) = 2\lambda(y) + 2d, \\ \rho_2(x, y) = d^2 + \beta(y), \\ \rho_3(x, y) = 2\lambda(y)\beta(y) + 2\lambda''(y) + 2d\beta(y), \\ \rho_4(x, y) = 4\lambda'(y), \\ \rho_5(x, y) = 4\lambda(y)^2 + 4d\lambda(y) + d^2 + \beta(y). \end{cases} \quad (2.4) $$

Finally, we recall the following result from [49], which states the existence and uniqueness of regular enough kernel functions involved in the backstepping transformation.

**Theorem 2.1 (Theorem 3.1 in [49])** Let $\lambda \in C^2([0, L])$ and $\beta \in C([0, L])$. Then, there exist a unique $k = k(x, y)$ in $C^2(\mathcal{T})$ and a unique $s = s(x, y)$ in $C^2(\mathcal{T})$ such that

$$ egin{cases} k_{xx} - k_{yy} = \rho_1 s_{yy} + \rho_2 k + \rho_3 s + \rho_4 s_y, & (x, y) \in \mathcal{T}, \\ k(x, x) = m(x), & x \in [0, L], \\ k(x, 0) = 0, & x \in [0, L]. \end{cases} \quad (2.5) $$

$$ egin{cases} s_{xx} - s_{yy} = \rho_1 k + \rho_5 s, & (x, y) \in \mathcal{T}, \\ s(x, x) = -\sinh \left( \int_0^x a(\tau) d\tau \right), & x \in [0, L], \\ s(x, 0) = 0, & x \in [0, L]. \end{cases} \quad (2.6) $$

We have introduced all the required elements. We continue with the feedback design. In (2.2) we consider $U(t) = U_1(t) + U_2(t)$. Then, in virtue of the computations done in [49, Section 3.1], we employ the backstepping transformation

$$ w(t, x) = h(x)u(t, x) - \int_{0}^{x} k(x, y)u(t, y) dy - \int_{0}^{x} s(x, y)u_{t}(t, y) dy \quad (2.7) $$

and choose

$$ U_1(t) = \frac{1}{h(L)} \left( \int_{0}^{L} k(L, y)u(t, y) dy + \int_{0}^{L} s(L, y)u_{t}(t, y) dy \right) \quad (2.8) $$

to transform (2.2) into the target system

$$ egin{cases} w_{tt} = w_{xx} - 2dw_t - d^2w, & (t, x) \in (0, \infty) \times (0, L), \\ w(t, 0) = 0, \quad w(t, L) = h(L)p(t) + h(L)U_2(t), & t \in (0, \infty), \\ w(0, x) = w_0(x), \quad w_t(0, x) = w_1(x), & x \in (0, L), \end{cases} \quad (2.9) $$

where the transformed initial condition $(w_0, w_1)$, as stated in [49, Section 3.3], is given by
$$
\left\{
\begin{aligned}
w_0(x) &= h(x)u_0(x) - \int_0^x k(x, y)u_0(y) \, dy - \int_0^x s(x, y)u_1(y) \, dy, \\
w_1(x) &= h(x)u_1(x) + s_y(x, x)u_0(x) - s(x, x)u'_0(x) \\
&\quad - \int_0^x [2\lambda(y)s(x, y) + k(x, y)] u_1(y) \, dy - \int_0^x [\beta(y)s(x, y) + s_{yy}(x, y)] u_0(y) \, dy.
\end{aligned}
\right.
\qquad (2.10)
$$

As an intermediate step, we introduce a modified energy as follows: for a regular enough state $(w, w_t)$
of the target system (2.9), we define its modified energy by

$$
V(t) = \frac{1}{2} \int_{0}^{L} (w_t + dw)^2 dx + \frac{1}{2} \int_{0}^{L} |w_x|^2 dx, \ t \in [0, \infty). \qquad (2.11)
$$

We proceed to perform formal computations, which later are justified in view of the analysis done in
Section 2.2. Performing integration by parts when needed, we have

$$
\begin{align*}
V'(t) &= \int_0^L (w_t + dw)(w_{tt} + dw_t) \, dx + \int_0^L w_x w_{xt} \, dx \\
&= \int_0^L \left( w_t(w_{xx} - 2dw_t - d^2w) + d|w_t|^2 + dw(w_{xx} - 2dw_t - d^2w) + d^2ww_t \right) \, dx \\
&\quad - \int_0^L w_{xx}w_t \, dx + \int_0^L w_x(t, L)w_t(t, L) \, dx \\
&= \int_0^L \left( -d|w_t|^2 + dww_{xx} - 2d^2ww_t - d^3|w|^2 \right) \, dx + w_x(t, L)w_t(t, L) \\
&= -2d \int_0^L \frac{1}{2} \left( |w_t|^2 + 2dww_t + d^2|w|^2 + |w_x|^2 \right) \, dx + dw(t, L)w_x(t, L) + w_x(t, L)w_t(t, L).
\end{align*}
$$

Accordingly, we arrive at

$$
V'(t) = -2dV(t) + w_x(t, L)(w_t(t, L) + dw(t, L)). \tag{2.12}
$$

From the previous expression we shall design $U_2(t)$ as a feedback law. Let us assume (A1). Then,
as $w(t, L) = h(L)p(t) + h(L)U_2(t)$ it follows

$$
V'(t) + 2dV(t) = w_x(t, L)w_t(t, L) + w_x(t, L)dh(L)p(t) + w_x(t, L)dh(L)U_2(t). \quad (2.13)
$$

Then, with the choice of $U_2(t)$ so that

$$
U_2(t) = -\frac{1}{dh(L)} w_t(t, L) - P \operatorname{sign}(w_x(t, L)), \qquad (2.14)
$$

the right-hand side of (2.13) becomes non-positive. Indeed, thanks to the property $\theta f = |f|$ for every
$\theta \in \operatorname{sign}(f)$, we have

$$
\begin{align*}
& w_x(t, L)w_t(t, L) + w_x(t, L)dh(L)p(t) + w_x(t, L)dh(L)U_2(t) \\
& \le |w_x(t, L)|dh(L)P + w_x(t, L)dh(L)\theta = |w_x(t, L)|dh(L)P - |w_x(t, L)|dh(L)P \\
& = 0 \quad \text{for every } \theta \in -P \operatorname{sign}(w_x(t, L)). \tag{2.15}
\end{align*}
$$
Therefore, in virtue of (2.13) and (2.15) we conclude that the solutions of the corresponding closed-loop system, obtained by considering the target system (2.9) together with (2.14), satisfy

$$V(t) \le e^{-2dt}V(0), \ t \in [0, \infty). \tag{2.16}$$

The closed-loop system is the following one.

$$\left\{
\begin{array}{ll}
w_{tt} = w_{xx} - 2dw_t - d^2w, & (t, x) \in (0, \infty) \times (0, L), \\
w(t, 0) = 0, & t \in (0, \infty), \\
w(t, L) + \frac{1}{d}w_t(t, L) \in h(L)p(t) - h(L)P \operatorname{sign}(w_x(t, L)), & t \in (0, \infty), \\
w(0, x) = w_0(x), \ w_t(0, x) = w_1(x), & x \in (0, L).
\end{array}
\right.
\tag{2.17}$$

The next result states an equivalence between the modified energy and the energy.

**Lemma 2.1** Let $(w, w_t)$ be such that $w(t, \cdot) \in H^1(0, L)$, $w(t, 0) = 0$ and $w_t(t, \cdot) \in L^2(0, L)$ for every $t \in [0, \infty)$. Then,

$$\begin{aligned}
& \min \left\{ \frac{1}{8L^2d^2 + 1}, \frac{1}{2} \right\} \int_0^L (|w_t|^2 + |w_x|^2) dx \le 2V(t) \\
& \le \max \{8L^2d^2 + 1, 2\} \int_0^L (|w_t|^2 + |w_x|^2) dx, \quad t \in [0, \infty).
\end{aligned}
\tag{2.18}$$

**Proof** By hypotheses it is available Poincaré inequality $\int_0^L |w(t, x)|^2 dx \le 4L^2 \int_0^L |w_x(t, x)|^2 dx$ for every $t \in [0, \infty)$. We prove the first inequality only (from left to right), since the proof of the second one is similar. Thanks to Poincaré and Cauchy inequalities, for every $\delta > 0$ we have

$$\begin{aligned}
2V(t) &= \int_0^L (|w_t|^2 + 2w_t dw + d^2|w|^2) \, dx + \int_0^L |w_x|^2 \left( \frac{1}{2} + \frac{1}{2} \right) \, dx \\
&\ge \int_0^L |w_t|^2 (1 - d\delta) \, dx + \int_0^L |w_x|^2 \frac{1}{2} \, dx + \int_0^L |w|^2 \left( d^2 - \frac{d}{\delta} + \frac{1}{8L^2} \right) \, dx.
\end{aligned}
\tag{2.19}$$

By choosing $\delta > 0$ so that $d^2 - \frac{d}{\delta} + \frac{1}{8L^2} = 0$, which is when $\delta = \frac{8L^2d}{8L^2d^2+1}$, it follows that $1 - d\delta = \frac{1}{8L^2d^2+1}$. Accordingly, from (2.19) we get the first inequality because

$$2V(t) \ge \frac{1}{8L^2d^2 + 1} \int_0^L |w_t|^2 dx + \frac{1}{2} \int_0^L |w_x|^2 dx.$$

The second inequality also follows by applying Poincaré and Cauchy inequalities. ■

To conclude the feedback design part of Theorem 2.2, let us obtain (1.4) for (2.2). To that end, we use that both the backstepping operator and its inverse are linear and bounded operators. Let us introduce the space

$$H = \{h \in H^1(0, L) / h(0) = 0\} \quad \text{and} \quad X = H \times L^2(0, L), \tag{2.20}$$

with norm $\|(a, b)\|_X^2 = \|b\|_{L^2(0, L)}^2 + \|a'\|_{L^2(0, L)}^2$. From [49, Section 3.3] we have that the backstepping operator $\Pi: X \to X$ given by $\Pi(u_0, u_1) = (w_0, w_1)$, with $(w_0, w_1)$ defined in (2.10), has a linear and bounded inverse $\Pi^{-1}: X \to X$.
**Remark 2.2** As stated in [49, Section 3.3], the backstepping operator Π maps the state $(u, u_t)$ of (2.2) with (2.8) into the state $(w, w_t)$ of the target system (2.9).

Let us set $C_1 = \|\Pi^{-1}\|_{\mathcal{L}(X)}$ and $C_2 = \|\Pi\|_{\mathcal{L}(X)}$, which satisfy $C_1C_2 \ge 1$. As the energy satisfies $2E(t) = \|(u, u_t)(t, \cdot)\|_X^2$ in virtue of (1.3), it follows from the continuity of $\Pi^{-1}$, (2.18), (2.16) and the continuity of Π that

$$ \begin{aligned} \|(u, u_t)(t, \cdot)\|_X^2 & \le C_1^2 \|(w, w_t)(t, \cdot)\|_X^2 \\ & \le C_1^2 \frac{1}{\min\left\{\frac{1}{8L^2d^2+1}, \frac{1}{2}\right\}} 2V(t) \\ & \le C_1^2 \frac{1}{\min\left\{\frac{1}{8L^2d^2+1}, \frac{1}{2}\right\}} e^{-2dt} 2V(0) \\ & \le C_1^2 \frac{\max\{8L^2d^2+1, 2\}}{\min\left\{\frac{1}{8L^2d^2+1}, \frac{1}{2}\right\}} e^{-2dt} \|(w_0, w_1)\|_X^2 \\ & \le C_1^2 C_2^2 \frac{\max\{8L^2d^2+1, 2\}}{\min\left\{\frac{1}{8L^2d^2+1}, \frac{1}{2}\right\}} e^{-2dt} \|(u_0, u_1)\|_X^2 \quad \text{for all } t \in [0, \infty). \end{aligned} \quad (2.21) $$

Accordingly, we have shown (1.4) for (2.2). In other words, we have shown the feedback design part of Theorem 2.2 and (2.44) with $C = C_1C_2\sqrt{K}$, where

$$ K = \max\{8L^2d^2 + 1, 2\} / \min\left\{\frac{1}{8L^2d^2 + 1}, \frac{1}{2}\right\}. \quad (2.22) $$

**Remark 2.3** As stated in [51, Section 2.6], let us note that the inverse backstepping transformation is not used in the feedback design. Nevertheless, its existence and properties are indeed used for the proof of the closed-loop stability of (2.43), as we have just exposed for the obtention of (2.21).

## 2.2 Well-posedness

In this section we apply the maximal monotone operator theory to prove the well-posedness of the closed-loop system (2.17), which is a differential inclusion, with the inclusion being located at the boundary condition.

Let us recall (2.20). Also, on the Hilbert space $X$ let us consider the inner product

$$ \langle (q_1, l_1), (q_2, l_2) \rangle_X = \int_0^L q'_1(x)q'_2(x) \, dx + \int_0^L (l_1(x) + dq_1(x))(l_2(x) + dq_2(x)) \, dx. \quad (2.23) $$

Let us introduce

$$ y(t, x) = w(t, x) - \psi(t)\phi(x) \quad \text{with} \quad \psi(t) = dh(L) \left( e^{-dt} \int_0^t e^{ds} p(s) \, ds \right) \quad \text{and} \quad \phi(x) = \frac{x}{L} \left( 2 - \frac{x}{L} \right). \quad (2.24) $$

Since $\phi(0) = 0$, $\phi(L) = 1$ and $\phi'(L) = 0$ we get that $y(t, 0) = w(t, 0)$, $y(t, L) + \frac{1}{d}y_t(t, L) = w(t, L) + \frac{1}{d}w_t(t, L) - h(L)p(t)$ and $y_x(t, L) = w_x(t, L)$. Let us assume $(\mathbf{A}2)_{\text{DD}}$, thus obtaining that $y(0, x) = w(0, x)$ and $y_t(0, x) = w_t(0, x)$. Then, it follows from (2.17) that $y = y(t, x)$ satisfies
$$ egin{cases} y_{tt} = y_{xx} - 2dy_t - d^2y + f, & (t, x) \in (0, \infty) \times (0, L), \\ y(t, 0) = 0, & t \in (0, \infty), \\ y(t, L) + \frac{1}{d}y_t(t, L) \in -h(L)P \operatorname{sign}(y_x(t, L)), & t \in (0, \infty), \\ y(0, x) = w_0(x), y_t(0, x) = w_1(x), & x \in (0, L), \end{cases} \qquad (2.25) $$

where $f(x, t) = (-\partial_{tt} + \partial_{xx} - 2d\partial_t - d^2)(\psi(t)\phi(x))$. Finally, let us consider the operator

$$ egin{cases} \mathcal{A} : D(\mathcal{A}) \subset X \to X, \\ \mathcal{A}(q, l) = (-l, -q'' + 2dl + d^2q), \\ D(\mathcal{A}) = \{(q, l) \in X \mid \mathcal{A}(q, l) \in X, dq(L) + l(L) \in -dh(L)P \operatorname{sign}(q'(L))\}. \end{cases} \qquad (2.26) $$

**Remark 2.4** We have that $D(\mathcal{A})$ is a non-empty set because $H_0^2(0, L) \times H_0^1(0, L) \subset D(\mathcal{A})$ due to the true statement $0 \in -dh(L)P[-1, 1]$. Because of the sign multivalued operator (1.5) it follows that $D(\mathcal{A})$ is not a linear subspace, and hence, $\mathcal{A}$ is not a linear operator. Also, $D(\mathcal{A})$ is dense in $X$. Finally,

$$ D(\mathcal{A}) = \{(q, l) \in (H^2(0, L) \cap H) \times H \mid dq(L) + l(L) \in -dh(L)P \operatorname{sign}(q'(L))\}. $$

Thanks to (2.26) we can write (2.25) in operator form as follows:

$$ egin{cases} \frac{d}{dt}(y, y_t) + \mathcal{A}(y, y_t) = (0, f), & t \in [0, \infty), \\ (y, y_t)(0) = (w_0, w_1). \end{cases} \qquad (2.27) $$

We proceed to prove that (2.27) is well-posed by applying the maximal monotone operator theory. In that direction we have the following two results: the first one (Proposition 2.1) states that the operator $\mathcal{A}$ is monotone, while the second one (Proposition 2.2) states that the operator $I + \mathcal{A}$ is surjective.

**Proposition 2.1** The operator defined by (2.26) is monotone.

**Proof** Let $(q_1, l_1)$ and $(q_2, l_2)$ be in $D(\mathcal{A})$. For brevity in notation and only in this proof, let us set $q = q_1 - q_2$ and $l = l_1 - l_2$. Then, in virtue of the inner product (2.23), we have, after two integrations by parts,

$$ egin{align*} & \langle \mathcal{A}(q_1, l_1) - \mathcal{A}(q_2, l_2), (q_1, l_1) - (q_2, l_2) \rangle_X \\ &= \int_0^L (-l)'q' dx + \int_0^L ((-q'' + 2dl + d^2q) + d(-l)) (l + dq) dx \\ &= d\|q'\|_{L^2(0,L)}^2 + d^3\|q\|_{L^2(0,L)}^2 + d\|l\|_{L^2(0,L)}^2 + 2d^2 \int_0^L ql dx + (-l(L)q'(L) - dq(L)q'(L)) \\ &= I_1 + I_2. \end{align*} $$

On the one hand, since $|2d^2 \int_0^L ql dx| \le d^3\|q\|_{L^2(0,L)}^2 + d\|l\|_{L^2(0,L)}^2$ we get that $I_1 \ge d\|q'\|_{L^2(0,L)}^2 \ge 0$. On the other hand, as $dq_j(L) + l_j(L) \in -dh(L)P \operatorname{sign}(q'_j(L))$ for $j \in \{1, 2\}$, there exists $\theta_j \in \operatorname{sign}(q'_j(L))$ such that $dq_j(L) + l_j(L) = -dh(L)P\theta_j$. Therefore,
$$
\begin{align*}
I_2 &= -l(L)q'(L) - dq(L)q'(L) \\
&= -((l_1(L) + dq_1(L)) - (l_2(L) + dq_2(L))) (q'_1(L) - q'_2(L)) \\
&= dh(L)P(\theta_1 - \theta_2)(q'_1(L) - q'_2(L)).
\end{align*}
$$

We conclude that $I_2 \ge 0$ because the sign multivalued operator (1.5) is monotone. Accordingly, the operator $\mathcal{A}$ is monotone as $I_1 + I_2 \ge 0$. ■

**Proposition 2.2** The operator defined by (2.26) satisfies $R(I + \mathcal{A}) = X$.

**Proof** Given $(m, n) \in X$ we need to prove the existence of $(q, l) \in D(\mathcal{A})$, domain being defined in (2.26), such that $(q, l) + \mathcal{A}(q, l) = (m, n)$. To that end, let us introduce the problem

$$
\left\{
\begin{array}{l}
-q'' + q(1 + 2d + d^2) = n + m(1 + 2d), \\
q(0) = 0, \quad dq(L) + (q(L) - m(L)) \in -dh(L)P \operatorname{sign}(q'(L)).
\end{array}
\right.
\qquad (2.28)
$$

By proving that (2.28) possesses a unique solution $q \in H^2(0, L)$ and by putting $l = q - m$, then we get that $(q, l) \in D(\mathcal{A})$ and that $(q, l) + \mathcal{A}(q, l) = (m, n)$. For the proof we proceed as in [9, Section 2] and analyze an optimization problem.

**Remark 2.5** The analysis in [9, Section 2] has also been adapted for the proof of [26, Proposition 3.1] and the proof of [27, Proposition 3.2].

Let us introduce the functional $J : H^2(0, L) \cap H \to \mathbb{R}$ by

$$
J(q) = \frac{1}{2} \int_{0}^{L} \left[ (q'')^2 + (q')^2(1 + 2d + d^2) + 2q''(n + m(1 + 2d)) \right] dx \\
\qquad + (1 + d) \left[ -m(L)q'(L) + dh(L)P \varphi_{\sigma}(q'(L)) \right], \qquad (2.29)
$$

where $\varphi_{\sigma}: \mathbb{R} \to \mathbb{R}$ is the Moreau regularization of the convex and continuous function $\varphi: \mathbb{R} \to \mathbb{R}$ given by $\varphi(x) = |x|$. Thanks to [48, Chapter IV, Proposition 1.8] (see also [2, Theorem 2.9]) we have

$$
\varphi_{\sigma}(x) = \frac{\sigma}{2} |\alpha_{\sigma}(x)|^2 + \varphi(J_{\sigma}(x)), \quad (\sigma, x) \in (0, \infty) \times \mathbb{R},
$$

where $\alpha_{\sigma} = \sigma^{-1}(I - J_{\sigma}) : \mathbb{R} \to \mathbb{R}$ is the Yosida approximation of the maximal monotone operator $\alpha : \mathbb{R} \to 2^{\mathbb{R}}$ given by $\alpha(x) = (\partial\varphi)(x) = \operatorname{sign}(x)$ and $J_{\sigma} = (I + \sigma\alpha)^{-1} : \mathbb{R} \to \mathbb{R}$ is the resolvent of $\alpha$. Furthermore, also from [48, Chapter IV, Proposition 1.8] (see also [2, Theorem 2.9]) we have that $\varphi_{\sigma}$ is a convex and differentiable function satisfying that $\varphi'_{\sigma}(x) = \alpha_{\sigma}(x)$.

Since $q \in H^2(0, L) \cap H$ we have the following inequalities.

$$
\|q\|_{L^2(0,L)}^2 \le 4L^2 \|q'\|_{L^2(0,L)}^2, \quad |q(L)| \le 2L^{1/2} \|q'\|_{L^2(0,L)} \quad \text{and} \quad |q'(L)| \le (1+L^{-1})^{1/2} \|q\|_{H^2(0,L)} \cdot \quad (2.30)
$$

The functional $J$ is convex and continuous, the latter being a consequence of the continuous injection of $H^2(0, L)$ into $C^1([0, L])$. Also, considering the non-negativity of $\varphi_{\sigma}$ in (2.29) and applying (2.30), there exist two positive constants, $C_1 = C_1(L)$ and $C_2 = C_2(L, d, m, n)$, such that

$$
J(q) \geq C_1 \|q\|^2_{H^2(0,L)} - C_2 \|q\|_{H^2(0,L)},
$$
from which it follows that $J(q) \to \infty$ as $\|q\|_{H^2(0,L)} \to \infty$, concluding that the functional $J$ is coercive. Therefore, [43, Theorem 2.19] yields the existence of a minimizer $q \in H^2(0, L) \cap H$ for the functional $J$. Then, in virtue of [43, Proposition 3.20] and [43, Theorem 3.24], we have that the Gâteaux derivative of the functional $J$ at such minimizer must vanish for each direction $\rho \in H^2(0, L) \cap H$. Accordingly, as $\varphi'_\sigma(x) = \alpha_\sigma(x)$ we obtain

$$ J'(q; \rho) = \int_0^L \left[ q''\rho'' + q'\rho'(1 + 2d + d^2) + \rho''(n + m(1 + 2d)) \right] dx \\ + (1 + d) \left[ -m(L) + dh(L)P\alpha_\sigma(q'(L)) \right] \rho'(L) = 0 \quad \forall \rho \in H^2(0, L) \cap H, $$

and after one integration,

$$ J'(q; \rho) = \int_0^L \left[ q'' - q(1 + 2d + d^2) + (n + m(1 + 2d)) \right] \rho'' dx \\ + (1 + d) \left[ dq(L) + (q(L) - m(L)) + dh(L)P\alpha_\sigma(q'(L)) \right] \rho'(L) = 0 \quad \forall \rho \in H^2(0, L) \cap H. \quad (2.31) $$

By considering $C_0^\infty(0, L) \subset H^2(0, L) \cap H$ in (2.31), we can deduce thanks to [4, Lemma 8.1] the existence of two constants, $C_3 \in \mathbb{R}$ and $C_4 \in \mathbb{R}$, so that $q'' - q(1 + 2d + d^2) + (n + m(1 + 2d)) = C_3x + C_4$ for almost every $x \in (0, L)$. Then, from (2.31) we get

$$ -C_3\rho(L) + C_3L\rho'(L) + C_4(\rho'(L) - \rho'(0)) \\ + (1 + d) [dq(L) + (q(L) - m(L)) + dh(L)P\alpha_\sigma(q'(L))] \rho'(L) = 0 \quad \forall \rho \in H^2(0, L) \cap H. \quad (2.32) $$

From (2.32) we can deduce that $C_3 = 0$ by choosing a third degree polynomial satisfying $\rho(L) = 1$ and $\rho(0) = \rho'(0) = \rho'(L) = 0$. Similarly, we can deduce that $C_4 = 0$ by choosing a second degree polynomial satisfying $\rho'(0) = 1$ and $\rho(0) = \rho'(L) = 0$.

**Remark 2.6** Those two polynomials do exist. Indeed, the conclusion $C_3 = 0$ can be obtained with $\rho(x) = -2L^{-3}x^3 + 3L^{-2}x^2$, and $C_4 = 0$, with $\rho(x) = -(2L)^{-1}x^2 + x$. Both polynomials satisfy the requirements.

Finally, we can also deduce that

$$ dq(L) + (q(L) - m(L)) + dh(L)P\alpha_\sigma(q'(L)) = 0. $$

Accordingly, so far we have shown:

**Lemma 2.2** For any $\sigma \in (0, \infty)$ there exists $q_\sigma \in H^2(0, L)$ such that $-q_\sigma'' + q_\sigma(1+2d+d^2) = n+m(1+2d)$ for almost every $x \in (0, L)$, $q_\sigma(0) = 0$ and $dq_\sigma(L) + (q_\sigma(L) - m(L)) = -dh(L)P\alpha_\sigma(q_\sigma'(L))$.

In view of Lemma 2.2 it seems that we are close to prove that (2.28) possesses a unique solution $q \in H^2(0, L)$. To that end, let $q_\sigma$ be given by Lemma 2.2 and let us see what happens to it when $\sigma \to 0^+$. We proceed to obtain upper bounds, independent of $\sigma$, for $\|q_\sigma\|_{H^2(0,L)}$ and $|\alpha_\sigma(q_\sigma'(L))|$.
• Upper bound for $||q_{\sigma}||_{H^2(0,L)}$ independent of $\sigma$

In virtue of Lemma 2.2 we can get

$$||q'_{\sigma}||_{L^2(0,L)}^2 - q'_{\sigma}(L)q_{\sigma}(L) + \frac{1}{2}||q_{\sigma}||_{L^2(0,L)}^2 \le \frac{1}{2}||n + m(1+2d)||_{L^2(0,L)}^2.$$

In order to treat the term $-q'_{\sigma}(L)q_{\sigma}(L)$, let us note that from $0 \in \alpha(0)$ we conclude that $J_{\sigma}(0) = 0$, and consequently, $\alpha_{\sigma}(0) = 0$. Then, by Lemma 2.2 and as $\alpha_{\sigma}$ is a monotone operator, we infer

$$-q'_{\sigma}(L)q_{\sigma}(L) = -\frac{1}{1+d}m(L)q'_{\sigma}(L) + \frac{d}{1+d}h(L)P\alpha_{\sigma}(q'_{\sigma}(L))q'_{\sigma}(L) \ge -\frac{1}{1+d}m(L)q'_{\sigma}(L).$$

Therefore, applying (2.30) and Cauchy inequality, for every $\delta > 0$ we have

$$||q'_{\sigma}||_{L^2(0,L)}^2 + \frac{1}{2}||q_{\sigma}||_{L^2(0,L)}^2 \le \frac{1}{2}||n + m(1+2d)||_{L^2(0,L)}^2 + (1+L^{-1})\frac{|m(L)|^2}{(1+d)^2}\frac{\delta}{2} + \frac{1}{2\delta}||q_{\sigma}||_{H^2(0,L)}^2. \quad (2.33)$$

Thanks to Lemma 2.2 we can get

$$||q''_{\sigma}||_{L^2(0,L)}^2 \le 2(1+d)^4||q_{\sigma}||_{L^2(0,L)}^2 + 2||n+m(1+2d)||_{L^2(0,L)}^2. \quad (2.34)$$

Finally, from the combination of (2.33) together with (2.34) and a suitable choice for $\delta > 0$, we can obtain the existence of two positive constants, $C_5 = C_5(d)$ and $C_6 = C_6(L, d)$, such that

$$||q_{\sigma}||_{H^2(0,L)}^2 \le C_5||n + m(1+2d)||_{L^2(0,L)}^2 + C_6|m(L)|^2. \quad (2.35)$$

• Upper bound for $|\alpha_{\sigma}(q'_{\sigma}(L))|$ independent of $\sigma$

By Lemma 2.2 it follows that

$$|\alpha_{\sigma}(q'_{\sigma}(L))| \le \frac{(d+1)}{dh(L)P}|q(L)| + \frac{1}{dh(L)P}|m(L)|.$$

Then, in view of (2.30) and (2.35), we see that there exist two positive constants, $C_7 = C_7(L, d)$ and $C_8 = C_8(L, d)$, such that

$$|\alpha_{\sigma}(q'_{\sigma}(L))| \le \frac{C_7}{h(L)P}||n + m(1+2d)||_{L^2(0,L)} + \frac{C_8}{h(L)P}|m(L)|. \quad (2.36)$$

Therefore, in virtue of (2.35) and (2.36) we conclude that the sequences $\{q_{\sigma}\}_{\sigma>0} \subset H^2(0, L)$ and $\{\alpha_{\sigma}(q'_{\sigma}(L))\}_{\sigma>0} \subset \mathbb{R}$ are bounded. Then, there exist $(q, b) \in H^2(0, L) \times \mathbb{R}$ and subsequences, which we denote by the same symbols, such that $q_{\sigma} \to q$ in $H^2(0, L)$ (weak convergence) and $\alpha_{\sigma}(q'_{\sigma}(L)) \to b$ in $\mathbb{R}$ as $\sigma \to 0^+$. Moreover, we infer that $q_{\sigma} \to q$ in $C^1([0, L])$ as $\sigma \to 0^+$ since the injection of $H^2(0, L)$ into $C^1([0, L])$ is compact, implying that $q_{\sigma}(x) \to q(x)$ and $q'_{\sigma}(x) \to q'(x)$ in $\mathbb{R}$ for all $x \in [0, L]$ as $\sigma \to 0^+$.

Accordingly, Lemma 2.2 and the previous arguments yield:

**Lemma 2.3** There exists $q \in H^2(0, L)$ such that $-q'' + q(1+2d+d^2) = n+m(1+2d)$ for almost every $x \in (0, L)$, $q(0) = 0$ and $dq(L) + (q(L) - m(L)) = -dh(L)Pb$.
In view of Lemma 2.3 and (2.28) we see that in order to complete the proof of $R(I + \mathcal{A}) = X$ we just need to prove that $b \in \alpha(q'(L)) = \text{sign}(q'(L))$. The required arguments can be found at the end of the proof of [26, Proposition 3.1] or at the end of the proof of [27, Proposition 3.2]. The proof of Proposition 2.2 is complete. ■

Let us return to (2.27). Due to Proposition 2.1 and Proposition 2.2 we can apply [48, Chapter IV, Lemma 1.3] to conclude that the operator $\mathcal{A}$ defined in (2.26) is maximal monotone. Regarding the data $(0, f)$ and $(w_0, w_1)$ in (2.27): recalling that $f(x, t) = (-\partial_{tt} + \partial_{xx} - 2d\partial_t - d^2)(\psi(t)\phi(x))$, by (A2)$_{DD}$ we get that $f \in W^{1,1}(0, \infty; C^\infty(\mathbb{R}))$; also, let us assume that $(w_0, w_1) \in D(\mathcal{A})$. Then, we can apply [48, Chapter IV, Theorem 4.1] to conclude the existence of a unique $(y, y_t) \in W^{1,1}(0, \infty; X)$ such that

$$ egin{cases} 
\displaystyle \frac{d}{dt}(y, y_t) + \mathcal{A}(y, y_t) = (0, f) & \text{for almost every } t > 0, \\[1.5ex] 
(y, y_t) \in D(\mathcal{A}) & \text{for every } t \ge 0, \\[1.5ex] 
(y, y_t)(0) = (w_0, w_1). 
\end{cases} 
\tag{2.37} $$

With (2.17) (the closed-loop system), (2.24) and (2.25) in mind, due to (2.37) we conclude that $(w, w_t) \in W^{1,1}(0, \infty; X)$ uniquely solves

$$ egin{cases} 
w_{tt} = w_{xx} - 2dw_t - d^2w & \text{for almost every } (t, x) \in (0, \infty) \times (0, L), \\ w(t, 0) = 0 & \text{for every } t \in [0, \infty), \\ w(t, L) + \frac{1}{d}w_t(t, L) \in h(L)p(t) - h(L)P \text{ sign}(w_x(t, L)) & \text{for every } t \in [0, \infty), \\ w(0, x) = w_0(x) \text{ and } w_t(0, x) = w_1(x) & \text{for every } x \in (0, L). 
\end{cases} 
\tag{2.38} $$

We have shown the well-posedness part of Theorem 2.2.

## 2.3 Conclusion

Here we gather the conclusions of both the feedback design (see Section 2.1) and the well-posedness (see Section 2.2) parts to establish our main result regarding the Dirichlet-Dirichlet case.

In Section 2.1 we mentioned that from [49, Section 3.3] it follows that the backstepping operator $\Pi: X \to X$ given by $\Pi(u_0, u_1) = (w_0, w_1)$, with $(w_0, w_1)$ defined in (2.10), has a linear and bounded inverse $\Pi^{-1}: X \to X$. We also mentioned Remark 2.2. Similarly to Remark 2.2, we also have that the inverse backstepping operator $\Pi^{-1}$ maps the state $(w, w_t)$ of (2.38) into the state $(u, u_t)$ of (2.2) with $U(t) = U_1(t) + U_2(t)$, where $U_1(t)$ is given by (2.8) and $U_2(t)$ is given by (2.14). Therefore, $(u, u_t) \in W^{1,1}(0, \infty; X)$.

Let us note that $U_1(t)$ in (2.8) is written in terms of $(u, u_t)$ but $U_2(t)$ in (2.14) is written in terms of $(w, w_t)$. However, we can rewrite $U_2(t)$ in (2.14) in terms of $(u, u_t)$ thanks to (2.10). Then, in order to present the final expression of the designed feedback law, for a regular enough function $f = f(t, x)$ let us introduce

$$ U_1(t, f) = \frac{1}{h(L)} \left( \int_0^L k(L, y)f(t, y) dy + \int_0^L s(L, y)f_t(t, y) dy \right), \tag{2.39} $$
$$U_2(t, f) = -\frac{1}{dh(L)}U_3(t, f) - P \operatorname{sign}(U_4(t, f)), \qquad (2.40)$$

$$U_3(t, f) = h(L)f_t(t, L) + s_y(L, L)f(t, L) - s(L, L)f_x(t, L) \\ - \int_0^L (2\lambda(y)s(L, y) + k(L, y))f_t(t, y) dy - \int_0^L (\beta(y)s(L, y) + s_{yy}(L, y))f(t, y) dy, \qquad (2.41)$$

$$U_4(t, f) = h'(L)f(t, L) + h(L)f_x(t, L) - s(L, L)f_t(t, L) - k(L, L)f(t, L) \\ - \int_0^L s_x(L, y)f_t(t, y) dy - \int_0^L k_x(L, y)f(t, y) dy. \qquad (2.42)$$

Accordingly, in virtue of (2.10) it follows from (2.41) that $w_t(t, L) = U_3(t, u)$, Also, in view of (2.7) we see from (2.42) that $w_x(t, L) = U_4(t, u)$. Then, $U_1(t, u) = U_1(t)$ and $U_2(t, u) = U_2(t)$, where $U_1(t)$ and $U_2(t)$ are respectively given by (2.8) and (2.14).

Finally, the main result regarding the Dirichlet-Dirichlet case is the following one.

**Theorem 2.2** Let $\lambda \in C^2([0, L])$ and $\beta \in C([0, L])$. Let $d \in (0, \infty)$ be the desired decay rate. Let $k = k(x, y)$ and $s = s(x, y)$ be the kernel functions respectively obtained from (2.5) and (2.6). Let us assume (A1) and (A2)DD. Let $(u_0, u_1) \in \Pi^{-1}(\mathcal{I}_{DD}) = \{\Pi^{-1}(y_0, y_1) / (y_0, y_1) \in \mathcal{I}_{DD}\}$ be an initial condition, where $\mathcal{I}_{DD} = D(\mathcal{A})$ and $D(\mathcal{A})$ is given by (2.26). Then, there exists a unique $u = u(t, x)$ with $(u, u_t) \in W^{1,1}(0, \infty; X)$, where $X$ is given by (2.20), such that

$$\left\{
\begin{array}{l}
u_{tt} = u_{xx} + 2\lambda(x)u_t + \beta(x)u \quad \text{for almost every } (t, x) \in (0, \infty) \times (0, L), \\
u(t, 0) = 0 \quad \text{for every } t \in [0, \infty), \\
u(t, L) \in p(t) + U_1(t, u) + U_2(t, u) \quad \text{for every } t \in [0, \infty), \\
u(0, x) = u_0(x) \quad \text{and } u_t(0, x) = u_1(x) \quad \text{for every } x \in (0, L).
\end{array}
\right.
\qquad (2.43)$$

Moreover, there exists a constant $C \in [1, \infty)$ such that

$$\|u_t(t, \cdot)\|_{L^2(0,L)}^2 + \|u_x(t, \cdot)\|_{L^2(0,L)}^2 \le C e^{-2dt} (\|u_1\|_{L^2(0,L)}^2 + \|u_0'\|_{L^2(0,L)}^2) \quad \text{for all } t \in [0, \infty). \qquad (2.44)$$

**Remark 2.7** Theorem 2.2 can be applied to obtain (1.4) for (1.1). To that end, it is necessary to take into account Remark 2.1 and make the necessary changes. Then, in view of Theorem 2.1 (existence and uniqueness of regular enough kernel functions) we see that the additional assumption $\alpha \in C^1([0, L])$ is needed.

# 3 Dirichlet-Neumann case

In this section we prove Theorem 3.1, which is one of our main results and corresponds to the obtention of (1.4) for (1.2). Most of the arguments in this section are mild variations to the ones presented previously for the Dirichlet-Dirichlet case.
**Remark 3.1** We can transform (1.2) into another wave equation without first-order spatial derivative term by applying the transformation $v(t, x) = (e^{\frac{1}{2} \int_{0}^{x} \alpha(\tau) d\tau})u(t, x)$. Indeed, we get (2.1) but with

$$v_x(t, L) = (\alpha(L)/2)v(t, L) + (e^{\frac{1}{2} \int_{0}^{L} \alpha(\tau) d\tau})p(t) + (e^{\frac{1}{2} \int_{0}^{L} \alpha(\tau) d\tau})U(t)$$

instead of

$$v(t, L) = (e^{\frac{1}{2} \int_{0}^{L} \alpha(\tau) d\tau})p(t) + (e^{\frac{1}{2} \int_{0}^{L} \alpha(\tau) d\tau})U(t).$$

Thus, for $\gamma \in \mathbb{R}$, in this section we consider

$$\left\{
\begin{array}{ll}
u_{tt} = u_{xx} + 2\lambda(x)u_t + \beta(x)u, & (t, x) \in (0, \infty) \times (0, L), \\
u(t, 0) = 0, \quad u_x(t, L) = \gamma u(t, L) + p(t) + U(t), & t \in (0, \infty), \\
u(0, x) = u_0(x), \quad u_t(0, x) = u_1(x), & x \in (0, L).
\end{array}
\right.
\qquad (3.1)$$

## 3.1 Feedback design

In this section we prove the feedback design part of Theorem 3.1. To that end, let us recall the required elements of the backstepping transformation constructed in [49, Section 3], which are: (2.3), (2.4), Theorem 2.1 (existence and uniqueness of regular enough kernel functions) and (2.10).

As in the Dirichlet-Dirichlet case, in (3.1) we consider $U(t) = U_1(t) + U_2(t)$. In order to choose $U_1(t)$ we first employ the backstepping transformation (2.7) to get

$$\begin{aligned}
w_x(t, x) = h'(x)u(t, x) + h(x)u_x(t, x) & \\
& - k(x, x)u(t, x) - \int_0^x k_x(x, y)u(t, y) dy - s(x, x)u_t(t, x) - \int_0^x s_x(x, y)u_t(t, y) dy.
\end{aligned}$$

Then, in virtue of the computations done in [49, Section 3.1] we choose

$$\begin{aligned}
U_1(t) = & -\gamma u(t, L) \\
& + \frac{1}{h(L)} \left( -h'(L)u(t, L) + k(L, L)u(t, L) + \int_0^L k_x(L, y)u(t, y) dy \right. \\
& \left. + s(L, L)u_t(t, L) + \int_0^L s_x(L, y)u_t(t, y) dy \right),
\end{aligned}
\qquad (3.2)$$

to transform (3.1) into the target system

$$\left\{
\begin{array}{ll}
w_{tt} = w_{xx} - 2dw_t - d^2w, & (t, x) \in (0, \infty) \times (0, L), \\
w(t, 0) = 0, \quad w_x(t, L) = h(L)p(t) + h(L)U_2(t), & t \in (0, \infty), \\
w(0, x) = w_0(x), \quad w_t(0, x) = w_1(x), & x \in (0, L),
\end{array}
\right.
\qquad (3.3)$$

where the transformed initial condition $(w_0, w_1)$ is given by (2.10).

Let us note that the formal computations performed in Section 2.1 (Feedback design) for the Dirichlet-Dirichlet case are also valid for the Dirichlet-Neumann case. In particular, by keeping the definition of the modified energy (2.11), but now for a regular enough state $(w, w_t)$ of the target system (3.3), we find that both (2.12) and Lemma 2.1 are still valid.
As in the Dirichlet-Dirichlet case, we shall design $U_2(t)$ as a feedback law from (2.12). Let us assume (A1). Then, as $w_x(t, L) = h(L)p(t) + h(L)U_2(t)$, it follows

$$ 
\begin{aligned}
V'(t) + 2dV(t) &= U_2(t)h(L)(w_t(t, L) + dw(t, L)) + p(t)h(L)(w_t(t, L) + dw(t, L)) \\
&\leq U_2(t)h(L)(w_t(t, L) + dw(t, L)) + Ph(L)|w_t(t, L) + dw(t, L)|.
\end{aligned} 
\quad (3.4) $$

Then, with the choice of $U_2(t)$ so that

$$ U_2(t) = -P \operatorname{sign}(w_t(t, L) + dw(t, L)), \quad (3.5) $$

the right-hand side of (3.4) becomes non-positive. Indeed, thanks to the property $\theta f = |f|$ for every $\theta \in \operatorname{sign}(f)$, we have

$$ 
\begin{aligned}
& U_2(t)h(L)(w_t(t, L) + dw(t, L)) + Ph(L)|w_t(t, L) + dw(t, L)| \\
&= \theta h(L)(w_t(t, L) + dw(t, L)) + Ph(L)|w_t(t, L) + dw(t, L)| \\
&= -Ph(L)|w_t(t, L) + dw(t, L)| + Ph(L)|w_t(t, L) + dw(t, L)| \\
&= 0 \quad \text{for every } \theta \in -P \operatorname{sign}(w_t(t, L) + dw(t, L)).
\end{aligned} 
\quad (3.6) $$

Therefore, in virtue of (3.4) and (3.6) we conclude that the solutions of the corresponding closed-loop system, obtained by considering the target system (3.3) together with (3.5), satisfy $V(t) \leq e^{-2dt}V(0)$ for all $t \in [0, \infty)$. The closed-loop system this time is the following one.

$$ 
\left\{
\begin{array}{ll}
w_{tt} = w_{xx} - 2dw_t - d^2w, & (t, x) \in (0, \infty) \times (0, L), \\
w(t, 0) = 0, & t \in (0, \infty), \\
w_x(t, L) \in h(L)p(t) - h(L)P \operatorname{sign}(w_t(t, L) + dw(t, L)), & t \in (0, \infty), \\
w(0, x) = w_0(x), \quad w_t(0, x) = w_1(x), & x \in (0, L).
\end{array}
\right. 
\quad (3.7) $$

To conclude, let us note that the arguments presented at the end of Section 2.1 (Feedback design) for the Dirichlet-Dirichlet case regarding the backstepping operator $\Pi: X \to X$ and its inverse are also valid for the Dirichlet-Neumann case. Also:

**Remark 3.2** The backstepping operator $\Pi$ maps the state $(u, u_t)$ of (3.1) with (3.2) into the state $(w, w_t)$ of the target system (3.3).

Accordingly, (2.21) is still valid, thus concluding (1.4) for (3.1). We have shown the feedback design part of Theorem 3.1 and (3.28) with $C = C_1C_2\sqrt{K}$, where $K$ is defined in (2.22).

## 3.2 Well-posedness

In this section we apply the maximal monotone operator theory to prove the well-posedness of the closed-loop system (3.7), which is a differential inclusion, with the inclusion being located at the boundary condition. To that end, let us recall the Hilbert space (2.20) together with its inner product (2.23).
Let us introduce

$$y(t, x) = w(t, x) - \psi(t)\phi(x) \text{ with } \psi(t) = h(L)p(t) \text{ and } \phi(x) = \frac{x}{L}(x - L). \quad (3.8)$$

Since $\phi(0) = 0$, $\phi(L) = 0$ and $\phi'(L) = 1$ we get that $y(t, 0) = w(t, 0)$, $y(t, L) = w(t, L)$, $y_t(t, L) = w_t(t, L)$ and $y_x(t, L) = w_x(t, L) - h(L)p(t)$. Let us assume $(A2)_{DN}$, thus obtaining that $y(0, x) = w(0, x)$ and $y_t(0, x) = w_t(0, x)$. Then, it follows from (3.7) that $y = y(t, x)$ satisfies

$$\left\{
\begin{array}{ll}
y_{tt} = y_{xx} - 2dy_t - d^2y + f, & (t, x) \in (0, \infty) \times (0, L), \\
y(t, 0) = 0, & t \in (0, \infty), \\
y_x(t, L) \in -h(L)P \operatorname{sign}(y_t(t, L) + dy(t, L)), & t \in (0, \infty), \\
y(0, x) = w_0(x), y_t(0, x) = w_1(x), & x \in (0, L),
\end{array}
\right.
\quad (3.9)$$

where $f(x, t) = (-\partial_{tt} + \partial_{xx} - 2d\partial_t - d^2)(\psi(t)\phi(x))$. Finally, let us consider the operator

$$\left\{
\begin{array}{l}
\mathcal{A} : D(\mathcal{A}) \subset X \to X, \\
\mathcal{A}(q, l) = (-l, -q'' + 2dl + d^2q), \\
D(\mathcal{A}) = \{(q, l) \in X / \mathcal{A}(q, l) \in X, q'(L) \in -h(L)P \operatorname{sign}(l(L) + dq(L))\}.
\end{array}
\right.
\quad (3.10)$$

**Remark 3.3** As in Remark 2.4 we have that $D(\mathcal{A})$ is a non-empty set. Also, $\mathcal{A}$ is not a linear operator and $D(\mathcal{A})$ is dense in $X$. Finally,

$$D(\mathcal{A}) = \{(q, l) \in (H^2(0, L) \cap H) \times H / q'(L) \in -h(L)P \operatorname{sign}(l(L) + dq(L))\}.$$

Thanks to (3.10) we can write (3.9) in operator form as follows:

$$\left\{
\begin{array}{ll}
\displaystyle \frac{d}{dt}(y, y_t) + \mathcal{A}(y, y_t) = (0, f), & t \in [0, \infty), \\
\\
\displaystyle (y, y_t)(0) = (w_0, w_1).
\end{array}
\right.
\quad (3.11)$$

We proceed to prove that (3.11) is well-posed by applying the maximal monotone operator theory. In that direction we have the following two results: the first one (Proposition 3.1) states that the operator $\mathcal{A}$ is monotone, while the second one (Proposition 3.2) states that the operator $I + \mathcal{A}$ is surjective.

**Proposition 3.1** The operator defined by (3.10) is monotone.

**Proof** The proof is similar to that of Proposition 2.1 (Dirichlet-Dirichlet case). Let us employ the notation of that result. This time we also get that $\langle \mathcal{A}(q_1, l_1) - \mathcal{A}(q_2, l_2), (q_1, l_1) - (q_2, l_2) \rangle_X = I_1 + I_2$ and conclude $I_1 \ge 0$ by the same means.

Let us handle $I_2 = -l(L)q'(L) - dq(L)q'(L)$. As $q'_j(L) \in -h(L)P \operatorname{sign}(l_j(L) + dq_j(L))$ for $j \in \{1, 2\}$, there exists $\theta_j \in \operatorname{sign}(l_j(L) + dq_j(L))$ such that $q'_j(L) = -h(L)P\theta_j$. Therefore,

$$\begin{aligned}
I_2 &= -l(L)q'(L) - dq(L)q'(L) \\
&= -(q'_1(L) - q'_2(L))((l_1(L) + dq_1(L)) - (l_2(L) + dq_2(L))) \\
&= h(L)P(\theta_1 - \theta_2)((l_1(L) + dq_1(L)) - (l_2(L) + dq_2(L))).
\end{aligned}$$
We conclude that $I_2 \ge 0$ because the sign multivalued operator (1.5) is monotone. Accordingly, the operator $\mathcal{A}$ is monotone as $I_1 + I_2 \ge 0$.

**Proposition 3.2** The operator defined by (3.10) satisfies $R(I + \mathcal{A}) = X$.

**Proof** The proof is similar to that of Proposition 2.2 (Dirichlet-Dirichlet case). Given $(m, n) \in X$ we need to prove the existence of $(q, l) \in D(\mathcal{A})$, domain being defined in (3.10), such that $(q, l) + \mathcal{A}(q, l) = (m, n)$. To that end, let us introduce the problem

$$ \begin{cases} -q'' + q(1 + 2d + d^2) = n + m(1 + 2d), \\ q(0) = 0, \quad q'(L) \in -h(L)P \operatorname{sign}((1 + d)q(L) - m(L)). \end{cases} \qquad (3.12) $$

By proving that (3.12) possesses a unique solution $q \in H^2(0, L)$ and by putting $l = q - m$, then we get that $(q, l) \in D(\mathcal{A})$ and that $(q, l) + \mathcal{A}(q, l) = (m, n)$. As in the proof of Proposition 2.2 (Dirichlet-Dirichlet case), we analyze an optimization problem. Thus, let us introduce the functional $J: H \to \mathbb{R}$ by

$$ J(q) = \frac{1}{2} \int_{0}^{L} [(q')^2 + q^2(1 + 2d + d^2) - 2q(n + m(1 + 2d))] dx \\ + \frac{1}{1+d} h(L) P \varphi_{\sigma}((1 + d)q(L) - m(L)), \qquad (3.13) $$

where $\varphi_{\sigma}: \mathbb{R} \to \mathbb{R}$ is the Moreau regularization of the convex and continuous function $\varphi: \mathbb{R} \to \mathbb{R}$ given by $\varphi(x) = |x|$.

As in the proof of Proposition 2.2 (Dirichlet-Dirichlet case), we can argue the existence of a minimizer $q \in H$ for the functional $J$. Also, the Gâteaux derivative of the functional $J$ at such minimizer must vanish for each direction $\rho \in H$. Accordingly, as $\varphi'_{\sigma}(x) = \alpha_{\sigma}(x)$ we obtain

$$ J'(q; \rho) = \int_{0}^{L} [q'\rho' + q\rho(1 + 2d + d^2) - \rho(n + m(1 + 2d))] dx \\ + h(L)P \alpha_{\sigma}((1 + d)q(L) - m(L)) \rho(L) = 0 \quad \forall \rho \in H. \qquad (3.14) $$

As $C_{0}^{\infty}(0, L) \subset H$, we can argue that $q \in H^2(0, L)$ and then that $-q'' + q(1 + 2d + d^2) = n + m(1 + 2d)$ for almost every $x \in (0, L)$. Moreover, after one integration by parts in (3.14) we get

$$ [q'(L) + h(L)P \alpha_{\sigma}((1 + d)q(L) - m(L))] \rho(L) = 0 \quad \forall \rho \in H. $$

Accordingly, so far we have shown:

**Lemma 3.1** For any $\sigma \in (0, \infty)$ there exists $q_{\sigma} \in H^2(0, L)$ such that $-q_{\sigma}'' + q_{\sigma}(1+2d+d^2) = n+m(1+2d)$ for almost every $x \in (0, L)$, $q_{\sigma}(0) = 0$ and $q'_{\sigma}(L) = -h(L)P \alpha_{\sigma}((1+d)q_{\sigma}(L) - m(L))$.

As in the proof of Proposition 2.2 (Dirichlet-Dirichlet case), let $q_{\sigma}$ be given by Lemma 3.1. We proceed to obtain upper bounds, independent of $\sigma$, for $\|q_{\sigma}\|_{H^2(0,L)}$ and $|\alpha_{\sigma}((1+d)q_{\sigma}(L) - m(L))|$.

* Upper bound for $\|q_{\sigma}\|_{H^2(0,L)}$ independent of $\sigma$

By Lemma 3.1 we have
$$ \int_{0}^{L} [-q''_{\sigma} + q_{\sigma}(1 + 2d + d^{2})] [(1 + d)q_{\sigma} - m] dx = \int_{0}^{L} [n + m(1 + 2d)] [(1 + d)q_{\sigma} - m] dx. $$

Let us recall that $m \in H$, and hence, $m \in H^1(0, L)$ with $m(0) = 0$. Then, after one integration by parts and applying Cauchy inequality we get

$$ \begin{aligned} & (1 + d) \int_{0}^{L} |q'_{\sigma}|^2 dx - q'_{\sigma}(L) [(1 + d)q_{\sigma}(L) - m(L)] - \int_{0}^{L} q'_{\sigma} m' dx \\ & \quad + (1 + d)(1 + d)^2 \int_{0}^{L} |q_{\sigma}|^2 dx - (1 + d)^2 \int_{0}^{L} q_{\sigma} m dx \\ & \leq \frac{1}{4}(1 + d)^2 \int_{0}^{L} |q_{\sigma}|^2 dx + \|n + m(1 + 2d)\|_{L^2(0,L)}^2 + \|n + m(1 + 2d)\|_{L^2(0,L)} \|m\|_{L^2(0,L)}. \end{aligned} \quad (3.15) $$

We shall use (3.15) later. Once again applying Cauchy inequality we arrive at

$$ \int_{0}^{L} q'_{\sigma} m' dx \leq \frac{1}{2} \int_{0}^{L} |q'_{\sigma}|^2 dx + \frac{1}{2} \|m'\|_{L^2(0,L)}^2, \quad (3.16) $$

$$ (1 + d)^2 \int_{0}^{L} q_{\sigma} m dx \leq \frac{1}{2}(1 + d)^2 \int_{0}^{L} |q_{\sigma}|^2 dx + \frac{1}{2}(1 + d)^2 \|m\|_{L^2(0,L)}^2. \quad (3.17) $$

In order to treat the term $-q'_{\sigma}(L) [(1 + d)q_{\sigma}(L) - m(L)]$ we use that $\alpha_{\sigma}(0) = 0$. Then, by Lemma 3.1 and as $\alpha_{\sigma}$ is a monotone operator, we infer

$$ \begin{aligned} & -q'_{\sigma}(L) [(1 + d)q_{\sigma}(L) - m(L)] \\ &= h(L)P[\alpha_{\sigma}((1 + d)q_{\sigma}(L) - m(L))] [(1 + d)q_{\sigma}(L) - m(L)] \\ &= h(L)P[\alpha_{\sigma}((1 + d)q_{\sigma}(L) - m(L)) - \alpha_{\sigma}(0)] [((1 + d)q_{\sigma}(L) - m(L)) - 0] \geq 0. \end{aligned} \quad (3.18) $$

Accordingly, from the combination of (3.15), (3.16), (3.17) and (3.18) we can get an upper bound for $\|q_{\sigma}\|_{H^1(0,L)}$, independent of $\sigma$, in terms of $\|m\|_{H^1(0,L)}$ and $\|n\|_{L^2(0,L)}$. Furthermore, thanks to Lemma 3.1 we can get the same for $\|q_{\sigma}\|_{H^2(0,L)}$. The conclusion is: there exists a positive constant $C_1 = C_1(d)$, independent of $\sigma$, such that

$$ \|q_{\sigma}\|_{H^2(0,L)} \leq C_1 \|m\|_{H^1(0,L)} + C_1 \|n\|_{L^2(0,L)}. \quad (3.19) $$

• Upper bound for $|\alpha_{\sigma}((1 + d)q_{\sigma}(L) - m(L))|$ independent of $\sigma$

There exists a positive constant $C_2 = C_2(L)$ such that $\|q\|_{C^1([0,L])} \leq C_2 \|q\|_{H^2(0,L)}$ for any $q \in H^2(0, L)$ because the injection of $H^2(0, L)$ into $C^1([0, L])$ is continuous. Then, by Lemma 3.1 and (3.19) we obtain

$$ |\alpha_{\sigma}((1 + d)q_{\sigma}(L) - m(L))| \leq \frac{1}{h(L)P}|q'_{\sigma}(L)| \leq \frac{C_2 C_1}{h(L)P} \|m\|_{H^1(0,L)} + \frac{C_2 C_1}{h(L)P} \|n\|_{L^2(0,L)}. \quad (3.20) $$

Therefore, in view of (3.19) and (3.20) we see that the sequences

$$ \{q_{\sigma}\}_{\sigma>0} \subset H^2(0, L) \text{ and } \{\alpha_{\sigma}((1 + d)q_{\sigma}(L) - m(L))\}_{\sigma>0} \subset \mathbb{R} $$
are bounded. Then, there exist $(q, b) \in H^2(0, L) \times \mathbb{R}$ and subsequences, which we denote by the same symbols, such that $q_\sigma \rightharpoonup q$ in $H^2(0, L)$ (weak convergence) and $\alpha_\sigma ((1+d)q_\sigma(L) - m(L)) \to b$ in $\mathbb{R}$ as $\sigma \to 0^+$. Moreover, we infer that $q_\sigma \to q$ in $C^1([0, L])$ as $\sigma \to 0^+$ since the injection of $H^2(0, L)$ into $C^1([0, L])$ is compact, implying that $q_\sigma(x) \to q(x)$ and $q'_\sigma(x) \to q'(x)$ in $\mathbb{R}$ for all $x \in [0, L]$ as $\sigma \to 0^+$.

Accordingly, Lemma 3.1 and the previous arguments yield:

**Lemma 3.2** There exists $q \in H^2(0, L)$ such that $-q'' + q(1+2d+d^2) = n+m(1+2d)$ for almost every $x \in (0, L)$, $q(0) = 0$ and $q'(L) = -h(L)Pb$.

In view of Lemma 3.2 and (3.12) we see that in order to complete the proof of $R(I+A) = X$ we just need to prove that $b \in \alpha((1+d)q(L) - m(L)) = \text{sign}((1+d)q(L) - m(L))$. The required arguments can be found at the end of the proof of [26, Proposition 3.1] or at the end of the proof of [27, Proposition 3.2]. The proof of Proposition 3.2 is complete. ■

Let us return to (3.11). We argue as we did at the end of Section 2.2 (Well-posedness) for the Dirichlet-Dirichlet case. We have that the operator $\mathcal{A}$ defined in (3.10) is maximal monotone. Let us recall (A2)$_{DN}$ and let us assume $(w_0, w_1) \in D(\mathcal{A})$. Then, we conclude the existence of a unique $(y, y_t) \in W^{1,1}(0, \infty; X)$ such that

$$ egin{cases} 
\displaystyle \frac{d}{dt}(y, y_t) + \mathcal{A}(y, y_t) = (0, f) & \text{for almost every } t > 0, \\[1.5ex] 
(y, y_t) \in D(\mathcal{A}) & \text{for every } t \ge 0, \\[1.5ex] 
(y, y_t)(0) = (w_0, w_1). 
\end{cases} 
\quad (3.21) $$

With (3.7) (the closed-loop system), (3.8) and (3.9) in mind, due to (3.21) we conclude that $(w, w_t) \in W^{1,1}(0, \infty; X)$ uniquely solves

$$ egin{cases} 
w_{tt} = w_{xx} - 2dw_t - d^2w & \text{for almost every } (t, x) \in (0, \infty) \times (0, L), \\ w(t, 0) = 0 & \text{for every } t \in [0, \infty), \\ w_x(t, L) \in h(L)p(t) - h(L)P \operatorname{sign}(w_t(t, L) + dw(t, L)) & \text{for every } t \in [0, \infty), \\ w(0, x) = w_0(x) \text{ and } w_t(0, x) = w_1(x) & \text{for every } x \in (0, L). 
\end{cases} 
\quad (3.22) $$

We have shown the well-posedness part of Theorem 3.1.

## 3.3 Conclusion

Here we gather the conclusions of both the feedback design (see Section 3.1) and the well-posedness (see Section 3.2) parts to establish our main result regarding the Dirichlet-Neumann case. The arguments are similar to that of Section 2.3 (Dirichlet-Dirichlet case).

From [49, Section 3.3] it follows that the backstepping operator $\Pi: X \to X$ given by $\Pi(u_0, u_1) = (w_0, w_1)$, with $(w_0, w_1)$ defined in (2.10), has a linear and bounded inverse $\Pi^{-1}: X \to X$. We also mentioned Remark 3.2. Similarly to Remark 3.2, we also have that the inverse backstepping operator $\Pi^{-1}$ maps the state $(w, w_t)$ of (3.22) into the state $(u, u_t)$ of (3.1) with $U(t) = U_1(t) + U_2(t)$, where $U_1(t)$ is given by (3.2) and $U_2(t)$ is given by (3.5). Therefore, $(u, u_t) \in W^{1,1}(0, \infty; X)$.
Let us note that $U_1(t)$ in (3.2) is written in terms of $(u, u_t)$ but $U_2(t)$ in (3.5) is written in terms of $(w, w_t)$. However, we can rewrite $U_2(t)$ in (3.5) in terms of $(u, u_t)$ thanks to (2.10). Then, in order to present the final expression of the designed feedback law, for a regular enough function $f = f(t, x)$ let us introduce

$$ U_1(t, f) = -\gamma f(t, L) + \frac{1}{h(L)} \left( -h'(L)f(t, L) + k(L, L)f(t, L) + \int_0^L k_x(L, y)f(t, y) dy \right. \tag{3.23} \\ \left. +s(L, L)f_t(t, L) + \int_0^L s_x(L, y)f_t(t, y) dy \right), $$

$$ U_2(t, f) = -P \operatorname{sign}(U_3(t, f) + dU_4(t, f)), \tag{3.24} $$

$$ U_3(t, f) = h(L)f_t(t, L) + s_y(L, L)f(t, L) - s(L, L)f_x(t, L) \\ - \int_0^L (2\lambda(y)s(L, y) + k(L, y))f_t(t, y) dy - \int_0^L (\beta(y)s(L, y) + s_{yy}(L, y))f(t, y) dy, \tag{3.25} $$

$$ U_4(t, f) = h(L)f(t, L) - \int_0^L k(L, y)f(t, y) dy - \int_0^L s(L, y)f_t(t, y) dy. \tag{3.26} $$

Accordingly, in virtue of (2.10) it follows from (3.25) that $w_t(t, L) = U_3(t, u)$, Also, in view of (2.7) we see from (3.26) that $w(t, L) = U_4(t, u)$. Then, $U_1(t, u) = U_1(t)$ and $U_2(t, u) = U_2(t)$, where $U_1(t)$ and $U_2(t)$ are respectively given by (3.2) and (3.5).

Finally, the main result regarding the Dirichlet-Neumann case is the following one.

**Theorem 3.1** Let $\lambda \in C^2([0, L])$, $\beta \in C([0, L])$ and $\gamma \in \mathbb{R}$. Let $d \in (0, \infty)$ be the desired decay rate. Let $k = k(x, y)$ and $s = s(x, y)$ be the kernel functions respectively obtained from (2.5) and (2.6). Let us assume (A1) and (A2)DN. Let $(u_0, u_1) \in \Pi^{-1}(\mathcal{I}_{DN}) = \{\Pi^{-1}(y_0, y_1) / (y_0, y_1) \in \mathcal{I}_{DN}\}$ be an initial condition, where $\mathcal{I}_{DN} = D(\mathcal{A})$ and $D(\mathcal{A})$ is given by (3.10). Then, there exists a unique $u = u(t, x)$ with $(u, u_t) \in W^{1,1}(0, \infty; X)$, where $X$ is given by (2.20), such that

$$ \left\{ \begin{array}{l} u_{tt} = u_{xx} + 2\lambda(x)u_t + \beta(x)u \text{ for almost every } (t, x) \in (0, \infty) \times (0, L), \\ u(t, 0) = 0 \text{ for every } t \in [0, \infty), \\ u_x(t, L) \in \gamma u(t, L) + p(t) + U_1(t, u) + U_2(t, u) \text{ for every } t \in [0, \infty), \\ u(0, x) = u_0(x) \text{ and } u_t(0, x) = u_1(x) \text{ for every } x \in (0, L). \end{array} \right. \tag{3.27} $$

Moreover, there exists a constant $C \in [1, \infty)$ such that

$$ \|u_t(t, \cdot)\|_{L^2(0,L)}^2 + \|u_x(t, \cdot)\|_{L^2(0,L)}^2 \le C e^{-2dt} \left( \|u_1\|_{L^2(0,L)}^2 + \|u'_0\|_{L^2(0,L)}^2 \right) \text{ for all } t \in [0, \infty). \tag{3.28} $$

**Remark 3.4** Theorem 3.1 can be applied to obtain (1.4) for (1.2). To that end, it is necessary to take into account Remark 3.1, put $\gamma = \alpha(L)/2$ and make the necessary changes. Then, in view of Theorem 2.1 (existence and uniqueness of regular enough kernel functions) we see that the additional assumption $\alpha \in C^1([0, L])$ is needed.
# 4 Concluding remarks

In this paper we have shown the rapid stabilization (1.4) for the unstable wave equations (1.1) and (1.2), in which there is an unknown disturbance acting at the boundary condition. We have employed the backstepping method, Lyapunov techniques and the sign multivalued operator (1.5) for the design of suitable feedback laws. The assumptions made on the unknown disturbance, namely (A1) and (A2), are the standard ones that can be found in the literature. The well-posedness of the corresponding closed-loop system has been shown with the maximal monotone operator theory. The main limitation of the approach followed in this paper is that it is needed a backstepping transformation to map the original plant into a target system so that Lyapunov techniques can be applied.

# 5 Acknowledgments

Patricio Guzmán and Agustín Huerta received financial support from FONDECYT 11240290. Hugo Parada was partially supported by the Agence Nationale de la Recherche through the Labex CIMI (ANR-11-LABX-0040). He is currently supported by the QuBiCCS project (ANR-24-CE40-3008).

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