# A NONHOMOGENEOUS BOUNDARY-VALUE PROBLEM FOR THE NONLINEAR KDV EQUATION ON STAR GRAPHS

ROBERTO DE A. CAPISTRANO-FILHO, HUGO PARADA*, AND JANDEILSON SANTOS DA SILVA

**ABSTRACT.** This paper investigates a boundary-value problem for the Korteweg-de Vries (KdV) equation on a star-graph structure. We develop a unified framework introducing the notion of *s*-compatibility, which generalizes classical compatibility conditions to star-shaped and more complex graph configurations, inspired by the works of Bona, Sun, and Zhang [14]. By combining analytical techniques with a fixed-point argument, we establish sharp global well-posedness for both the linear and nonlinear problems at the $H^s$ level. In this setting, our results extend the classical analysis for a single KdV equation [14] to star-shaped graphs composed of $N$ equations. These results provide the first comprehensive well-posedness theory for KdV equations with coupled boundary conditions on graphs. Although control issues are not treated in this article, the analytic results obtained here address several open problems, which will be addressed in a forthcoming paper.

## CONTENTS

1. Introduction 2

1.1. Dispersive systems on graph structure 2

1.2. Setting of problem and functional framework 3

1.3. Main results and structure of the article 5

2. *s*-compatibility conditions 5

2.1. Traces for $\phi_{0,j}, \phi'_{0,j}$ and $\phi''_{0,j}, j = 1, ..., N, s \in [0, 3]$ 7

2.2. Traces for $p_j, g_0$ and $g_j, j = 1, ..., N, s \in [0, 3]$ 7

2.3. Traces for $\phi_{k,j}, \phi'_{k,j}$ and $\phi''_{k,j}, j = 1, ..., N, s > 3$ 8

2.4. Traces for $p_j^{(k)}, g_0^{(k)}$ and $g_j^{(k)}, j = 1, ..., N, s > 3$ 8

3. The linear problem 9

3.1. The case $s = 0$ 10

3.2. The case $s = 3$ 20

3.3. The case $s \in (0, 3)$ 26

4. Local well-posedness of the nonlinear problem 28

5. Global well-posedness of the nonlinear problem 30

5.1. The case $s = 0$: Global a priori estimate 30

5.2. The case $s = 3$: Global a priori estimate 33

5.3. The case $s \in (0, 3)$: Global well-posedness result 37

6. Concluding Remarks 39

Acknowledgments 39

References 39

*Date*: December 17, 2025.

2020 Mathematics Subject Classification. 35Q53, 35G31, 35R02.

*Key words and phrases*. Star-Shaped Network, KdV equation, Global well-posedness, Sharp regularity.

*Corresponding author: hugo.parada@inria.fr.

Capistrano-Filho was partially supported by CAPES/COFECUB grant number 88887.879175/2023-00, CNPq grant numbers 421573/2023-6 and 301744/2025-4, and Propesqi (UFPE). Parada is supported by Agence Nationale de la Recherche under the QuBiCCS project ANR-24-CE40-3008. Da Silva acknowledges support from CNPq.
# 1. INTRODUCTION

In mathematics and physics, a quantum graph is a network composed of vertices connected by edges, where each edge carries a differential or pseudo-differential equation. When every edge is endowed with a natural metric, the structure is referred to as a metric graph. A common illustration is a power network: the transmission lines act as edges connecting transformer stations (vertices). In such a case, the governing differential equations may describe the voltage along each line, with boundary conditions at the vertices ensuring that the total current flowing into and out of each vertex sums to zero.

The concept of quantum graphs traces back to the work of Linus Pauling in the 1930s, who used them to model free electrons in organic molecules. Since then, they have emerged in a wide range of contexts, including quantum chaos, waveguide analysis, photonic crystals, and Anderson localization—the suppression of wave propagation in disordered media. Quantum graphs also arise as limit models of systems composed of thin wires and serve as important theoretical tools in mesoscopic physics and nanotechnology. A simplified formulation of quantum graphs was later proposed by Freedman et al. in [23].

Beyond solving the differential equations in specific applications, the study of quantum graphs encompasses fundamental questions such as well-posedness, controllability, and identifiability. For example, controllability concerns determining the inputs required to steer the system toward a desired configuration, such as ensuring stable power distribution across a grid. Identifiability, on the other hand, focuses on optimal measurement placement—for instance, choosing pressure gauges in a water network to detect and locate potential leaks.

## 1.1. Dispersive systems on graph structure.
In recent years, the study of nonlinear dispersive equations on metric graphs has attracted growing interest from mathematicians, physicists, chemists, and engineers (see [10, 11, 15, 25, 28] for detailed discussions and references). A fundamental setting for such investigations is the star graph $\mathcal{G}$, a metric graph consisting of $N$ half-lines $(0, \infty)$ meeting at a single common vertex $\nu = 0$. On each edge, one typically defines a nonlinear dispersive equation—most prominently, the nonlinear Schrödinger equation (see the works of Adami et al. [1, 2] and Angulo and Goloshchapova [5, 6]). The introduction of nonlinearities into dispersive models on such networks creates a rich framework for exploring soliton propagation, energy transfer, and nonlinear wave interactions. A central analytical difficulty arises at the vertex, where the graph may exhibit bifurcation or multi-bifurcation phenomena, especially in more intricate network topologies.

Other nonlinear dispersive systems on graphs have also been the subject of extensive study. For example, in the context of well-posedness theory, Cavalcante [18] established local well-posedness for the Cauchy problem associated with the Korteweg-de Vries (KdV) equation on a metric star graph composed of one negative and two positive half-lines joined at a common vertex $\nu = 0$ (the so-called $\mathcal{Y}$-junction).

Another important example is the Benjamin-Bona-Mahony (BBM) equation: Bona and Cavalcant [12] proved local well-posedness in the Sobolev space $H^1$, while Mugnolo and Rault [29] showed the existence of traveling-wave solutions on graphs. Using a different approach, Ammari and Crépeau [3] derived results concerning well-posedness and stabilization of the BBM equation in star-shaped networks with bounded edges. Moreover, Mugnolo et al. [30] provided a full characterization of boundary conditions under which the Airy-type evolution equation

$$u_t = \alpha u_{xxx} + \beta u_x, \quad \alpha \in \mathbb{R} \setminus \{0\}, \beta \in \mathbb{R},$$

generates contraction semigroups on star graphs. We also refer to the recent work on Airy dynamics on graphs with looping edges [7], which provides a complementary perspective on this setting.

Significant progress has also been achieved in the fields of control theory and inverse problems on networks. Ignat et al. [24] examined inverse problems for the heat and Schrödinger equations on tree-like structures, while Baudouin and Yamamoto [8] proposed a unified and simplified framework for coefficient identification problems. Furthermore, Ammari and Crépeau [4], Cerpa et al. [20, 21],
and Parada et al. [31, 32] established stabilization and boundary controllability results for the KdV equation on star-shaped graphs, contributing substantially to the control-theoretic understanding of dispersive models on networks. Recently, in [16], we studied the analytical structure of the entire functions associated with the KdV operator, addressing control problems involving mixed Dirichlet–Neumann boundary controls.

1.2. **Setting of problem and functional framework.** In the context of graph structure, let us consider the following: given $N \ge 2$, consider a star-shaped network $\mathcal{T}$ with $N$ edges described by intervals $I_j = (0, l_j)$, $l_j > 0$ for $j = 1, ..., N$. Denoting by $e_1, ..., e_N$ the edges of $\mathcal{T}$ one has $\mathcal{T} = \cup_{j=1}^N e_j$. Here, we study the local well-posedness for a system composed of $N$ nonlinear KdV equations posed on $\mathcal{T}$ and coupled by the boundary, namely

$$ (1.1) \quad \left\{ \begin{array}{ll} \partial_t u_j + \partial_x u_j + \partial_x^3 u_j + u_j \partial_x u_j = 0, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N \\ u_j(t, 0) = u_1(t, 0), & t \in (0, T), j = 1, \dots, N \\ \displaystyle\sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0) - \frac{N}{3} u_1^2(t, 0) + g_0(t), & t \in (0, T) \\ u_j(t, l_j) = p_j(t), \quad \partial_x u_j(t, l_j) = g_j(t), & t \in (0, T), j = 1, \dots, N \\ u_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N \end{array} \right. $$

where $u = (u_1, ..., u_N)$ stands the state of the system, $p_j$ and $g_j$ are the controls inputs and $\alpha > \frac{N}{2}$. We represent this situation in the figure (1.2).

<figure><img src="image_1.png" /><figcaption>FIGURE 1. Network with $N$ edges</figcaption></figure>

Let us introduce some notation used throughout this work. From now on, we have the following:

i. The vector $u$ is given by $u = (u_1, ..., u_N) \in L^2(\mathcal{T}) = \prod_{j=1}^N L^2(0, l_j)$ and initial data is $u^0 = (u_1^0, ..., u_N^0)$.

ii. The inner product in $L^2(\mathcal{T})$ will be given by

$$ (u, z)_{L^2(\mathcal{T})} = \sum_{j=1}^{N} \int_{0}^{l_j} u_j z_j dx, \quad u, z \in L^2(\mathcal{T}). $$

Moreover,

$$ \mathbb{H}^s(\mathcal{T}) = \prod_{j=1}^{N} H^s(0, l_j), \quad s \in \mathbb{R}. $$

iii. Define also the following spaces:

$$ \mathbb{H}_0^k(\mathcal{T}) = \prod_{j=1}^{N} H_0^k(0, l_j), \quad k \in \mathbb{N} $$
and

$$H_r^k(0, l_j) = \{v \in H^k(0, l_j), \partial_x^i v(l_j) = 0, 0 \le i \le k-1\}, k = 1, \dots, N,$$

where $v^{(m)} = \frac{dv}{dx^m}$ and the index $r$ is related to the null right boundary conditions. In addition,

$$\mathbb{H}_r^k(\mathcal{T}) = \prod_{j=1}^{N} H_r^k(0, l_j) \quad \text{and} \quad \|u\|^2_{\mathbb{H}_r^k(\mathcal{T})} = \sum_{j=1}^{n} \|u_j\|^2_{H^k(0, l_j)} \quad k \in \mathbb{N}.$$

iv. Consider the following characterization:

$$H_r^{-1}(0, l_j) = (H_r^1(0, l_j))'$$

as the dual space of $H_r(0, l_j)$ with respect to the pivot space $L^2(0, l_j)$ and $\mathbb{H}_r^{-1}$ denotes the cartesian product of $H_r^{-1}(0, l_j)$.

v. For $k \in \mathbb{N}$, the derivative $\partial_x^k u$ means the row vector

$$\partial_x^k u := (\partial_x^k u_1, \dots, \partial_x^k u_N).$$

The derivative $\partial_x^k u$ is understood in the same way.

vi. Under the above notation, we have

$$C^k([0, T]; \mathbb{H}^s(\mathcal{T})) = \prod_{j=1}^{N} C^k([0, T]; H^s(0, l_j)), \quad k = 0, 1, 2, \dots \quad \text{and} \quad s \in \mathbb{R}$$

as well as

$$L^p(0, T; \mathbb{H}^s(\mathcal{T})) = \prod_{j=1}^{N} L^p(0, T; H^s(0, l_j)), \quad p \ge 1 \quad \text{and} \quad s \in \mathbb{R}.$$

Hence, for $f \in C^k([0, T]; \mathbb{H}^s(\mathcal{T}))$ and $q \in L^p(0, T; \mathbb{H}^s(\mathcal{T}))$ we can write

$$f = (f_1, \dots, f_N) \quad \text{and} \quad q = (q_1, \dots, q_N)$$

with

$$f_j \in C^k([0, T]; H^s(0, l_j)) \quad \text{and} \quad q_j \in L^p(0, T; H^s(0, l_j)), \quad j = 1, \dots, N.$$

vii. We will consider the spaces

$$\mathbb{H}_e^k(\mathcal{T}) = \{u = (u_1, \dots, u_N) \in \mathbb{H}_r^k(\mathcal{T}); u_j(0) = u_k(0), j, k = 1, \dots, N\} \quad k \in \mathbb{N},$$

$$\mathbb{H}_c^3(\mathcal{T}) = \{u = (u_1, \dots, u_N) \in \mathbb{H}^3(\mathcal{T}); u_j(0) = u_k(0), j, k = 1, \dots, N\},$$

and

$$\mathbb{B}_{s,T} = C([0, T], \mathbb{H}^s(\mathcal{T})) \cap L^2(0, T, \mathbb{H}^{s+1}(\mathcal{T})),$$

with the norm

$$\|u\|_{\mathbb{B}_{s,T}} := \|u\|_{C([0,T], \mathbb{H}^s(\mathcal{T}))} + \|u\|_{L^2(0,T, \mathbb{H}^{s+1}(\mathcal{T}))} = \max_{t \in [0,T]} \|u\|_{\mathbb{H}^s(\mathcal{T})} + \left( \int_0^T \|u(t, \cdot)\|^2_{\mathbb{H}^{s+1} dt} \right)^{\frac{1}{2}}.$$

When $s = 0$ we write $\mathbb{B}_T$ instead $\mathbb{B}_{0,T}$.

viii. Once understand the appropriate spaces in each situation, we will use the notations $g = (g_1, \dots, g_N)$ and $p = (p_1, \dots, p_N)$.
1.3. Main results and structure of the article. With the previous framework in hand, our goal here is to prove the local and global well-posedness for the initial boundary value problem (IBVP) (1.1). The first main result establishes the local well-posedness for this system in sharp spaces.

**Theorem 1.1.** Let $T > 0$, $l_j > 0$, $j = 1, ..., N$, $s \in [0, 3]$ be given. Suppose $f \in W^{\frac{s}{3},1}(0, T; \mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0, T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T}))$ and

$$ (u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0, T) \times [H^{\frac{s}{3}}(0, T)]^N \times [H^{\frac{s+1}{3}}(0, T)]^N. $$

satisfying the s-compatibility conditions, thus there exists $T^* \in (0, T)$ such that the IBVP (1.1) admits a unique solution $u \in \mathcal{X}_{s,T^*}$, where

$$ \mathcal{X}_{s,T} = \{ v \in \mathbb{B}_{s,T}^*; \partial_x^k u_j \in L_x^\infty(0, l_j; H^{\frac{s+1-k}{3}}(0, T)), k = 0, 1, 2, j = 1, 2, ..., N \}, $$

$$ \mathbb{B}_{s,T}^* := \mathbb{B}_{s,T} \cap H^{\frac{s}{3}}(0, T; \mathbb{H}^1(\mathcal{T})). $$

The proof of the previous theorem relies on several analytical tools combined with a fixed-point argument. In addition, a new notion of s-compatibility adapted to the graph structure is introduced (see Section 2), inspired by the single-edge KdV framework in [14]. This theorem provides a sharp local well-posedness result, along with regularity properties, for the associated functions. Furthermore, we establish the existence of global solutions at the $H^s$-norm, which enables applications to control problems on star graphs and resolves several open questions that will be addressed in a forthcoming paper. The second main result of this work can be stated as follows.

**Theorem 1.2.** Let $T > 0$ and $\lambda \ge 1$ be. Consider $(u^0, g_0, g, p) \in \mathcal{X}_{s,T}^\lambda$, where $\mathcal{X}_{s,T}^\lambda$ is the set of s-compatible data

$$ u^0 \in \mathbb{H}^s(\mathcal{T}), g_0 \in H^{\frac{s+\lambda-1}{3}}(0, T), g \in [H^{\frac{s}{3}}(0, T)]^N \quad \text{and} \quad p \in [H^{\frac{s+\lambda+1}{3}}(0, T)]^N. $$

Thus, the problem (1.1) has a unique solution $u \in \mathcal{X}_{s,T}$ which satisfies

$$ \|u\|_{\mathcal{X}_{s,T}} \le C \left( \|(u^0, g_0, g, p)\|_{0,\lambda} \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s+\lambda-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} + \|p\|_{[H^{\frac{s+\lambda+1}{3}}(0,T)]^N} \right) \right) $$
where $C > 0$ is a positive constant.

Together, these two theorems provide, for the first time, a comprehensive framework for the well-posedness of general graph structures with boundary controls. Moreover, they extend the classical result for a single KdV equation [14] to star-shaped graphs composed of $N$ equations.

The paper is organized as follows. In Section 2, we introduce a general notion of s-compatibility for star graphs. Building on this framework, Section 3 addresses the well-posedness of the linear system associated with (1.1). After that, Section 4 is devoted to proving the local well-posedness result for the system (1.1), giving the proof of Theorem 1.1. The second main result, concerning global well-posedness, in the $H^s$-regime, for the system (1.1) is established in Section 5.

## 2. s-COMPATIBILITY CONDITIONS

When investigating the existence of solutions to (1.1), particular attention must be paid to the functional spaces chosen for $u^0$, $p_j$, $g_0$, and $g_j$. These spaces may allow for the existence of traces for these functions, which must be consistent with the initial and boundary conditions of (1.1), as we now make precise.

Suppose that $u \in \prod_{j=1}^N C^\infty([0, T] \times [0, l_j])$ is a solution of (1.1). Define,

$$ \phi_{0,j}(x) = u_j^0(x), \quad j = 1, ..., N $$
and observe that, according to the boundary conditions in (1.1), we have

$$
\left\{
\begin{aligned}
\phi_{0,j}(0) &= \phi_{0,1}(0), \\
\sum_{j=1}^{N} \phi_{0,j}''(0) &= -\alpha \phi_{0,1}(0) - \frac{N}{3} \phi_{0,1}^2(0) + g_0(0), \\
\phi_{0,j}(l_j) &= p_j(0), \\
\phi'_{0,j}(l_j) &= g_j(0).
\end{aligned}
\right.
$$

Now, define

$$
\phi_{1,j}(x) = -(\phi'_{0,j} + \phi'''_{0,j} + \phi_{0,j}\phi'_{0,j})(x), \quad j = 1, \dots, N,
$$

and note that

$$
\left\{
\begin{aligned}
\phi_{1,j}(0) &= \phi_{1,1}(0), \\
\sum_{j=1}^{N} \phi_{1,j}''(0) &= -\alpha \phi_{1,1}(0) - \frac{N}{3} (\phi_{0,1}(0)\phi_{1,1}(0) + \phi_{1,1}(0)\phi_{0,1}(0)) + g_0^{(1)}(0), \\
\phi_{1,j}(l_j) &= p_j^{(1)}(0), \\
\phi'_{1,j}(l_j) &= g_j^{(1)}(0).
\end{aligned}
\right.
$$

Analogously, defining

$$
\phi_{2,j}(x) = -(\phi'_{1,j} + \phi'''_{1,j} + \phi_{1,j}\phi'_{0,j} + \phi_{0,j}\phi'_{1,j})(x), \quad j = 1, \dots, N
$$

we get

$$
\left\{
\begin{aligned}
\phi_{2,j}(0) &= \phi_{2,1}(0), \\
\sum_{j=1}^{N} \phi_{2,j}''(0) &= -\alpha \phi_{2,1}(0) - \frac{N}{3} (\phi_{0,1}(0)\phi_{2,1}(0) + \phi_{1,1}(0)\phi_{1,1}(0) + \phi_{2,1}(0)\phi_{0,1}(0)) + g_0^{(2)}(0), \\
\phi_{2,j}(l_j) &= p_j^{(2)}(0), \\
\phi'_{2,j}(l_j) &= g_j^{(2)}(0).
\end{aligned}
\right.
$$

In general, we define for $j = 1, \dots, N$,

$$
\phi_{k,j} = \begin{cases} u_j^0, & k=0, \\ -\left( \phi_{k-1,j}''' + \phi_{k-1,j}' + \frac{1}{2} \sum_{l=0}^{k-1} \frac{(k-1)!}{i!(k-1-i)!} (\phi_{i,j}\phi_{k-1-i,j}')' \right), & k=1, 2, \dots \end{cases}
$$

and, by induction, one can see that, for any $k \in \mathbb{N} \cup \{0\}$,

$$
(2.1) \quad \left\{
\begin{array}{ll}
\phi_{k,j}(0) = \phi_{k,1}(0), & j=1, \dots, N, \\
\displaystyle\sum_{j=1}^{N} \phi_{k,j}''(0) = -\alpha \phi_{k,1}(0) - \frac{N}{3} \sum_{l=0}^{k} \binom{k}{l} \phi_{l,1}(0) \phi_{k-l,1}(0) + g_0^{(k)}(0), & \\
\phi_{k,j}(l_j) = p_j^{(k)}(0), & j=1, \dots, N, \\
\phi'_{k,j}(l_j) = g_j^{(k)}(0), & j=1, \dots, N.
\end{array}
\right.
$$

Here, we study the existence of solutions for (1.1) with data

$$
(u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0, T) \times [H^{\frac{s}{3}}(0, T)]^N \times [H^{\frac{s+1}{3}}(0, T)]^N.
$$

Motivated by the above, we must also require that these data satisfy the equalities in (2.1) whenever the functions and their traces are well-defined. This notion of traces follows from the Sobolev compact embedding: for $a < b$, $s \ge 0$, $k \in \mathbb{N} \cup \{0\}$, and $\sigma \in (0, 1)$ satisfying

$k + \sigma < s - \frac{1}{2},$
we have

$$H^s(a, b) \hookrightarrow C^{k,\sigma}([a, b]).$$

With this in hand, we determine which equations in (2.1) should be considered in the compatibility analysis, depending on the value of $s$. Observe that for a given $s \ge 0$, we can only define $\phi_{k,j}$ for $k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor$, since starting from $\phi_{0,j}$, each “new” $\phi_{k,j}$ is obtained by differentiating the previous one three times. Moreover,

$$\phi_{k,j} \in H^{s-3k}(0, l_j), \quad k = 0, 1, 2, \dots$$

We begin by considering $s \in [0, 3]$. In this case, $s - 3k \le 0 < \frac{1}{2}$ for every $k \ge 1$, so $H^{s-3k}(0, l_j)$ is not embedded in any space of continuous functions. Consequently, there are no traces for $\phi_{k,j}$ (and their derivatives) when $k \ge 1$. Similarly, there are no traces for $p_j^{(k)}$, $g_0^{(k)}$, and $g_j^{(k)}$ when $k \ge 1$. It thus remains to study the traces of $\phi_{0,j}$, $\phi'_{0,j}$, $\phi''_{0,j}$, $p_j$, $g_0$, and $g_j$.

2.1. Traces for $\phi_{0,j}$, $\phi'_{0,j}$ and $\phi''_{0,j}$, $j = 1, \dots, N$, $s \in [0, 3]$. We now analyze the different cases according to the value of $s$. By the same reasoning used for $k \ge 1$, there are no traces of $\phi_{0,j}$, $\phi'_{0,j}$, and $\phi''_{0,j}$ when $0 \le s \le \frac{1}{2}$. Hence, it is sufficient to consider the interval $\frac{1}{2} < s \le 3$.

i. Case $\frac{1}{2} < s \le \frac{3}{2}$. In this range, $0 < s - \frac{1}{2}$, hence

$$H^s(0, l_j) \hookrightarrow C([0, l_j]).$$

Consequently, traces of $\phi_{0,j}$ exist. On the other hand, since $s \le \frac{3}{2}$ implies $s - \frac{1}{2} \le 1$, the embedding $H^s(0, l_j) \hookrightarrow C^1([0, l_j])$ does not hold. Therefore, traces of $\phi'_{0,j}$ and $\phi''_{0,j}$ are not defined.

ii. Case $\frac{3}{2} < s \le \frac{5}{2}$. Here, $1 < s - \frac{1}{2}$, which yields

$$H^s(0, l_j) \hookrightarrow C^1([0, l_j]),$$

and hence traces of $\phi_{0,j}$ and $\phi'_{0,j}$ exist. However, since $s \le \frac{5}{2}$ implies $s - \frac{1}{2} \le 2$, the embedding $H^s(0, l_j) \hookrightarrow C^2([0, l_j])$ is not valid, and thus traces of $\phi''_{0,j}$ do not exist.

iii. Case $\frac{5}{2} < s \le 3$. In this case, since $\frac{5}{2} < s$ implies $2 < s - \frac{1}{2}$, we have

$$H^s(0, l_j) \hookrightarrow C^2([0, l_j]).$$

Therefore, traces of $\phi_{0,j}$, $\phi'_{0,j}$, and $\phi''_{0,j}$ are well-defined.

2.2. Traces for $p_j$, $g_0$ and $g_j$, $j = 1, \dots, N$, $s \in [0, 3]$. For $s \le \frac{1}{2}$, the spaces $H^{\frac{s-1}{3}}(0, T)$, $H^{\frac{s}{3}}(0, T)$, and $H^{\frac{s+1}{3}}(0, T)$ are not embedded into spaces of continuous functions. Hence, there are no traces for $p_j$, $g_0$, or $g_j$. For $\frac{1}{2} < s \le 3$, we proceed by analyzing the following cases:

i. Case $\frac{1}{2} < s \le \frac{3}{2}$. Since $\frac{1}{2} < s$ implies $0 < \frac{s+1}{3} - \frac{1}{2}$, and $s \le \frac{3}{2}$ gives $\frac{s}{3} - \frac{1}{2} \le 0$, we obtain the embedding

$$H^{\frac{s+1}{3}}(0, T) \hookrightarrow C([0, T]),$$

while neither $H^{\frac{s}{3}}(0, T) \hookrightarrow C([0, T])$ nor $H^{\frac{s-1}{3}}(0, T) \hookrightarrow C([0, T])$ holds. Therefore, traces exist for $p_j$, but not for $g_0$ or $g_j$.

ii. Case $\frac{3}{2} < s \le \frac{5}{2}$. In this range, we have the embeddings

$$H^{\frac{s+1}{3}}(0, T) \hookrightarrow C([0, T]) \quad \text{and} \quad H^{\frac{s}{3}}(0, T) \hookrightarrow C([0, T]),$$

but not $H^{\frac{s-1}{3}}(0, T) \hookrightarrow C([0, T])$. Hence, traces exist for $p_j$ and $g_j$, but not for $g_0$.

iii. Case $\frac{5}{2} < s \le 3$. In this case, all the embeddings hold:

$$H^{\frac{s-1}{3}}(0, T), \quad H^{\frac{s}{3}}(0, T), \quad H^{\frac{s+1}{3}}(0, T) \hookrightarrow C([0, T]).$$

Therefore, traces exist for $g_0$, $g_j$, and $p_j$.
In summary, we have the following relationships between values of $s \in [0,3]$ and the existence of traces:

<table><thead><tr><td>There exist traces of</td><td>When</td></tr></thead><tbody><tr><td>$\phi_{0,j}, p_j$</td><td>$1/2 < s \le 3$</td></tr><tr><td>$\phi'_{0,j}, g_j$</td><td>$3/2 < s \le 3$</td></tr><tr><td>$\phi''_{0,j}, g_0$</td><td>$5/2 < s \le 3$</td></tr></tbody></table>

Based on this, we give the following definition.

**Definition 2.1** ($s$-compatibility for $0 \le s \le 3$). Given $T > 0$ and $s \in [0,3]$ we say that

$$ (u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0, T) \times [H^{\frac{s}{3}}(0, T)]^N \times [H^{\frac{s+1}{3}}(0, T)]^N $$

is $s$-compatible when the following conditions are satisfied for $j = 1, \dots, N$:

$$ \phi_{0,j}(0) = \phi_{0,1}(0) \quad \text{and} \quad \phi_{0,j}(l_j) = p_j(0) \quad \text{if} \quad \frac{1}{2} < s \le \frac{3}{2}, $$

$$ \phi_{0,j}(0) = \phi_{0,1}(0), \quad \phi_{0,j}(l_j) = p_j(0) \quad \text{and} \quad \phi'_{0,j}(l_j) = g_j(0) \quad \text{if} \quad \frac{3}{2} < s \le \frac{5}{2} $$

or

$$ \left\{ \begin{array}{l} \phi_{0,j}(0) = \phi_{0,1}(0), \\ \displaystyle \sum_{j=1}^{N} \phi''_{0,j}(0) = -\alpha \phi_{0,1}(0) - \frac{N}{3} \phi_{0,1}^2(0) + g_0(0), \\ \phi_{0,j}(l_j) = p_j(0), \\ \phi'_{0,j}(l_j) = g_j(0). \end{array} \right. \quad \text{if } \frac{5}{2} < s \le 3. $$

2.3. Traces for $\phi_{k,j}, \phi'_{k,j}$ and $\phi''_{k,j}, j = 1, \dots, N, s > 3$. Now we will consider $s > 3$. Note that this implies $[\frac{s}{3}] - 1 \ge 0$. Since $s - 3[\frac{s}{3}] \ge 0$ we have

$$ s - 3 \left( \left[ \frac{s}{3} \right] - 1 \right) = s - 3 \left[ \frac{s}{3} \right] + 3 \ge 3 > \frac{5}{2} = 2 + \frac{1}{2} $$

and therefore $s - 3k > 2 + \frac{1}{2}$ for $k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1$, that is,

$$ 2 < (s - 3k) - \frac{1}{2}, \quad k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1. $$

Consequently, $H^{s-3k}(0, l_j) \hookrightarrow C^2([0, l_j])$ so there exist traces of $\phi_{k,j}, \phi'_{k,j}$ and $\phi''_{k,j}$ for $k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1$. We will study the existence of traces of $\phi_{[\frac{s}{3}], j}, \phi'_{[\frac{s}{3}], j}$ and $\phi''_{[\frac{s}{3}], j}$ by dividing it into cases.

2.4. Traces for $p_j^{(k)}, g_0^{(k)}$ and $g_j^{(k)}, j = 1, \dots, N, s > 3$. Observe that

$$ \frac{s-1}{3} - \frac{1}{2} = \frac{s}{3} - \frac{5}{6} > \frac{s}{3} - 1 \ge \lfloor \frac{s}{3} \rfloor - 1 \ge k, \quad k = 1, \dots, \lfloor \frac{s}{3} \rfloor - 1 $$

and consequently

$$ H^{\frac{s-1}{3}}(0, T) \hookrightarrow C^k([0, T]), \quad k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1. $$

Similarly we have

$$ H^{\frac{s}{3}}(0, T), H^{\frac{s+1}{3}}(0, T) \hookrightarrow C^k([0, T]), \quad k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1. $$

Therefore there exist traces of $p_j^{(k)}, g_0^{(k)}$ and $g_j^{(k)}$ when $k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1$. It remains to study the existence of traces for $p_j^{([\frac{s}{3}])}, g_0^{([\frac{s}{3}])}$ and $g_j^{([\frac{s}{3}])}$. As before, we do this by dividing the analysis into cases

The following table summarizes the relationship between the existence of traces and the value of $s > 3$:
<table><thead><tr><th>There exist traces of</th><th>For k equal to</th><th>When</th></tr></thead><tbody><tr><td>$\phi_{k,j}, \phi'_{k,j}, \phi''_{k,j}, p_j^{(k)}, g_0^{(k)}, g_j^{(k)}$</td><td>$0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1$</td><td>$s > 3$</td></tr><tr><td>$\phi_{k,j}, p_j^{(k)}$</td><td>$\lfloor \frac{s}{3} \rfloor$</td><td>$s > 3, 1/2 < s - 3 \lfloor \frac{s}{3} \rfloor < 3$</td></tr><tr><td>$\phi'_{k,j}, g_j^{(k)}$</td><td>$\lfloor \frac{s}{3} \rfloor$</td><td>$s > 3, 3/2 < s - 3 \lfloor \frac{s}{3} \rfloor < 3$</td></tr><tr><td>$\phi''_{k,j}, g_0^{(k)}$</td><td>$\lfloor \frac{s}{3} \rfloor$</td><td>$s > 3, 5/2 < s - 3 \lfloor \frac{s}{3} \rfloor < 3$</td></tr></tbody></table>

Hence, we arrive at the following definition for $s$ compatibility in the case $s > 3$.

**Definition 2.2** ($s$-compatibility for $s > 3$). Given $T > 0$ and $s > 3$ we say that

$$ (u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0,T) \times [H^{\frac{s}{3}}(0,T)]^N \times [H^{\frac{s+1}{3}}(0,T)]^N $$

is $s$-compatible when, for $j = 1, \dots, N$

$$ \left\{ \begin{array}{l} \phi_{k,j}(0) = \phi_{k,1}(0), \\ \alpha \sum_{j=1}^{N} \phi''_{k,j}(0) = -\alpha \phi_{k,1}(0) - \frac{N}{3} \sum_{l=0}^{k} \binom{k}{l} \phi_{l,1}(0) \phi_{k-l,1}(0) + g_0^{(k)}(0), \\ \phi_{k,j}(l_j) = p_j^{(k)}(0), \\ \phi'_{k,j}(l_j) = g_j^{(k)}(0), \end{array} \right. $$

for $k = 0, 1, \dots, \lfloor \frac{s}{3} \rfloor - 1$ and

$$ \left\{ \begin{array}{ll} \phi_{k,j}(0) = \phi_{k,1}(0) \quad \text{and} \quad \phi_{k,j}(l_j) = p_j^{(k)}(0), & \frac{1}{2} < s - 3 \lfloor \frac{s}{3} \rfloor \\ \phi'_{k,j}(l_j) = g_j^{(k)}(0), & \frac{3}{2} < s - 3 \lfloor \frac{s}{3} \rfloor \\ \sum_{j=1}^{N} \phi''_{k,j}(0) = -\alpha \phi_{k,1}(0) - \frac{N}{3} \sum_{l=0}^{k} \binom{k}{l} \phi_{l,1}(0) \phi_{k-l,1}(0) + g_0^{(k)}(0), & \frac{5}{2} < s - 3 \lfloor \frac{s}{3} \rfloor \end{array} \right. \quad j = 1, \dots, N $$

for $k = \lfloor \frac{s}{3} \rfloor$.

We conclude this part by observing that an analogous $s$-compatibility condition can be formulated for the linear system by removing the nonlinear terms. This linear compatibility will be invoked throughout the analysis, without restating its definition each time.

### 3. THE LINEAR PROBLEM

Having established all the $s$-compatibility conditions relevant to our framework, we now turn to the main objective of this section: to prove the well-posedness of the following linear nonhomogeneous initial-boundary value problem:

$$ (3.1) \quad \left\{ \begin{array}{ll} \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = f_j, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N, \\ u_j(t, 0) = u_1(t, 0), & t \in (0, T), j = 1, \dots, N, \\ \sum_{j=1}^{N} \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0) + g_0(t), & t \in (0, T), \\ u_j(t, l_j) = p_j(t), & \partial_x u_j(t, l_j) = g_j(t), \quad t \in (0, T), j = 1, \dots, N, \\ u_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N, \end{array} \right. $$

for $f \in W^{\frac{s}{3},1}(0,T; \mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T}))$ and for data

$$ (u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0,T) \times [H^{\frac{s}{3}}(0,T)]^N \times [H^{\frac{s+1}{3}}(0,T)]^N $$

satisfying the compatibility conditions, when $s \in [0, 3]$.
3.1. The case $s = 0$. We begin by considering the homogeneous boundary value problem

$$ (3.2) \quad \left\{ \begin{array}{ll} \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = f_j, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N, \\ u_j(t, 0) = u_1(t, 0), & t \in (0, T), j = 1, \dots, N, \\ \displaystyle\sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0), & t \in (0, T), \\ u_j(t, l_j) = \partial_x u_j(t, l_j) = 0, & t \in (0, T), j = 1, \dots, N, \\ u_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N. \end{array} \right. $$

Defining the operator $A : D(A) \subset L^2(\mathcal{T}) \to L^2(\mathcal{T})$ by

$$ D(A) = \left\{ u \in \mathbb{H}^3(\mathcal{T}) \cap \mathbb{H}_e^2(\mathcal{T}); \sum_{j=1}^N \partial_x^2 u_j(0) = -\alpha u_1(0) \right\} \quad \text{and} \quad Au = -\partial_x u - \partial_x^3 u. $$

The adjoint operator $A^* : D(A^*) \subset L^2(\mathcal{T}) \to L^2(\mathcal{T})$ is given by

$$ D(A^*) = \left\{ v \in \mathbb{H}^3(\mathcal{T}) \cap \mathbb{H}_e^1(\mathcal{T}); \sum_{j=1}^N \partial_x^2 v_j(0) = -\alpha v_1(0), \partial_x v_j(0) = 0, j = 1, \dots, N \right\} \quad \text{and} \quad Av = -\partial_x v - \partial_x^3 v. $$

The first result was shown in [4] and is a direct consequence of the semigroup theory.

**Proposition 3.1.** Let $T > 0$ be given. For every $f \in C^1([0, T], L^2(\mathcal{T}))$ and $u^0 \in D(A)$, the problem (3.2) has a unique classical solution $u \in C^1([0, T], L^2(\mathcal{T})) \cap C([0, T], D(A))$.

Now, let us go back to the problem (3.1). To deal with classical solutions for the nonhomogeneous problem, we recall the space $\mathbb{H}_e^3(\mathcal{T})$, consisting of functions in $\mathbb{H}^3(\mathcal{T})$ that are continuous across the junction. At this stage it is convenient to introduce the space

$$ C_l^2([0, T]) = \{ \varphi \in C^2([0, T]); \varphi(0) = 0 \}. $$

**Proposition 3.2.** For every $T > 0$, $u_0 \in D(A)$, $g_0, g_j, p_j \in C_l^2([0, T])$ and $f \in C^1([0, T]; L^2(\mathcal{T}))$ there exists a unique classical solution $u \in C^1([0, T], L^2(\mathcal{T})) \cap C([0, T], \mathbb{H}_e^3(\mathcal{T}))$ to the problem (3.1).

*Proof.* Suppose $u_0 \in D(A)$, $g_0, g_j, p_j \in \{\varphi \in C^2([0, T]); \varphi(0) = 0\}$ and $f \in C^1([0, T]; L^2(\mathcal{T}))$. Consider $\phi_j, \psi_j, \theta_j \in C^\infty([0, l_j])$ satisfying

$$ \left\{ \begin{array}{ll} \phi_j(0) = \phi_1(0), & j = 2, \dots, N, \\ \phi_j(l_j) = 0, & j = 1, \dots, N, \\ \phi_j'(l_j) = 0, & j = 1, \dots, N, \\ \displaystyle\sum_{j=1}^N \phi_j''(0) = -\alpha \phi_1(0) + 1, \\ \\ \psi_j(0) = 0, & j = 1, \dots, N, \\ \psi_j(l_j) = 1, & j = 1, \dots, N, \\ \psi_j'(l_j) = 0, & j = 1, \dots, N, \\ \psi_j''(0) = 0, & j = 1, \dots, N, \end{array} \right. \quad \left\{ \begin{array}{ll} \theta_j(0) = 0, & j = 1, \dots, N, \\ \theta_j(l_j) = 0, & j = 1, \dots, N, \\ \theta_j'(l_j) = 1, & j = 1, \dots, N, \\ \theta_j''(0) = 0, & j = 1, \dots, N. \end{array} \right. $$

For $t \in [0, T]$, define $\tilde{f}(t) := (\tilde{f}_1(t, \cdot), \dots, \tilde{f}_N(t, \cdot))$ where

$$ \tilde{f}_j := f_j - \phi_j g_{0t} - (\phi_j' + \phi_j'') g_0 - \psi_j p_{jt} - (\psi_j' + \psi_j'') p_j - \theta_j g_{jt} - (\theta_j' + \theta_j'') g_j $$
and note that $\tilde{f} \in C^1([0, T], \mathbb{L}^2(\mathcal{T}))$. Let $v \in C^1([0, T], \mathbb{L}^2(\mathcal{T})) \cap C([0, T], D(A))$ be the unique classical solution to the problem

$$ (3.3) \qquad \left\{ \begin{array}{ll} \partial_t v_j + \partial_x v_j + \partial_x^3 v_j = \tilde{f}_j, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N, \\ v_j(t, 0) = v_1(t, 0), & t \in (0, T), j = 1, \dots, N, \\ \displaystyle\sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0), & t \in (0, T), \\ v_j(t, l_j) = \partial_x v_j(t, l_j) = 0, & t \in (0, T), j = 1, \dots, N, \\ v_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N, \end{array} \right. $$

given by Proposition 3.1. For each $t \in [0, T]$, define $u(t, \cdot) = \mathbb{L}^2(\mathcal{T})$ putting

$$ (3.4) \qquad u_j(t, x) = v_j(t, x) + \phi_j(x)g_0(t) + \psi_j(x)p_j(t) + \theta_j(x)g_j(t), \quad x \in (0, l_j), \quad j = 1, \dots, N. $$

Since $\phi_j, \psi_j, \theta_j \in C^\infty([0, l_j])$, $g_0, p_j, g_j \in C^2([0, T])$ and $v \in C^1([0, T], \mathbb{L}^2(\mathcal{T})) \cap C([0, T], D(A))$, it follows that

$$ u \in C^1([0, T], \mathbb{L}^2(\mathcal{T})) \cap C([0, T], \mathbb{H}_c^3(\mathcal{T})). $$

Furthermore, we have

$$ \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = (\partial_t v_j + \partial_x v_j + \partial_x^3 v_j) + \phi_j g_{0t} + (\phi'_j + \phi''_j) g_0 \\ + \psi_j p_{jt} + (\psi'_j + \psi''_j) p_j + \theta_j g_{jt} + (\theta'_j + \theta''_j) g_j $$

so from (3.3),

$$ \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = \tilde{f} + \phi_j g_{0t} + (\phi'_j + \phi''_j) g_0 + \psi_j p_{jt} + (\psi'_j + \psi''_j) p_j \\ + \theta_j g_{jt} + (\theta'_j + \theta''_j) g_j $$

and, using the definition of $\tilde{f}$, it follows that

$$ (3.5) \qquad \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = f, \quad j = 1, \dots, N. $$

By (3.5) and by the choice of the lifting functions $\phi, \psi, \theta$, after some straightforward computations, we conclude that $u$ is a classical solution to (3.1). For the uniqueness, suppose that $\tilde{u}$ is a classical solution of (3.1). Then defining $\tilde{v} = (\tilde{v}_1, \dots, \tilde{v}_N)$, where

$$ (3.6) \qquad \tilde{v}_j(t, x) = \tilde{u}_j(t, x) - \phi_j(x)g_0(t) - \psi_j(x)p_j(t) - \theta_j(x)g_j(t), \quad x \in (0, l_j), \quad j = 1, \dots, N, $$

one can see that $\tilde{v}$ solves (3.3) and, by uniqueness guaranteed in Proposition 3.1, results $\tilde{v} = v$. But from (3.4) we have

$$ (3.7) \qquad v_j(t, x) = u_j(t, x) - \phi_j(x)g_0(t) - \psi_j(x)p_j(t) - \theta_j(x)g_j(t), \quad x \in (0, l_j), \quad j = 1, \dots, N. $$

Subtracting (3.7) from (3.6) we obtain $\tilde{u}_j = u$, finishing the proof. $\square$

Our next goal is to introduce a notion of mild solution for problem (3.1) with less regular data, namely,

$$ u^0 \in \mathbb{L}^2(\mathcal{T}), \quad g_0 \in H^{-\frac{1}{3}}(0, T), \quad g_j \in L^2(0, T), \quad p_j \in H^{\frac{1}{3}}(0, T) \text{ for } j = 1, \dots, N, $$

and

$$ f \in L^1(0, T; \mathbb{L}^2(\mathcal{T})). $$

To this end, it is convenient to recall some results concerning the well-posedness of the following single equations:

$$ (3.8) \qquad \left\{ \begin{array}{ll} \partial_t \theta + \partial_x \theta + \partial_x^3 \theta = f & t \in (0, T), x \in (0, L), \\ \theta(t, 0) = h_1(t), & \theta(t, L) = h_2(t), \\ \theta(0, x) = \phi(x) & t \in (0, T), \\ & x \in (0, L) \end{array} \right. $$
and

$$
(3.9) \qquad \left\{
\begin{array}{ll}
\partial_t \psi + \partial_x \psi + \partial_x^3 \psi = f, & t \in (0, T), x \in (0, L), \\
\partial_x^2 \psi(t, 0) + \frac{\alpha}{N} \psi(t, 0) = h_1(t), & t \in (0, T) \\
\psi(t, L) = h_2(t), \quad \partial_x \psi(t, L) = h_3(t), & t \in (0, T), \\
\psi(0, x) = \phi(x), & x \in (0, L).
\end{array}
\right.
$$

Here, it is necessary to introduce some definitions and notations. Given $T, L > 0$ and $s \ge 0$
we define the space

$$
X_{s,T} := H^s(0, L) \times H^{\frac{s+1}{3}}(0, T) \times H^{\frac{s+1}{3}}(0, T) \times H^{\frac{s}{3}}(0, T)
$$

with the norm

$$
\|(\phi, h_1, h_2, h_3)\|_{X_{s,T}} = \left( \|\phi\|_{H^s(0,L)}^2 + \|h_1\|_{H^{\frac{s+1}{3}}(0,T)}^2 + \|h_2\|_{H^{\frac{s+1}{3}}(0,T)}^2 + \|h_3\|_{H^{\frac{s}{3}}(0,T)}^2 \right)^{\frac{1}{2}},
$$

the space

$$
Y_{s,T} := H^s(0, L) \times H^{\frac{s-1}{3}}(0, T) \times H^{\frac{s+1}{3}}(0, T) \times H^{\frac{s}{3}}(0, T)
$$

with the norm

$$
\| (\phi, h_1, h_2, h_3) \|_{Y_{s,T}} = \left( \| \phi \|_{H^s(0,L)}^2 + \| h_1 \|_{H^{\frac{s-1}{3}}(0,T)}^2 + \| h_2 \|_{H^{\frac{s+1}{3}}(0,T)}^2 + \| h_3 \|_{H^{\frac{s}{3}}(0,T)}^2 \right)^{\frac{1}{2}}
$$

and the space

$$
Z_{s,T} = C([0, T], H^s(0, L)) \cap L^2(0, T, H^{s+1}(0, L))
$$

with the norm

$$
\|v\|_{Z_{s,T}} = \max_{t \in [0,T]} \|v(t, \cdot)\|_{H^s(0,L)} + \|v\|_{L^2(0,T,H^{s+1}(0,L))}.
$$

For simplicity we will use the notation $\vec{h} = (h_1, h_2, h_3)$ so $(\phi, \vec{h}) = (\phi, h_1, h_2, h_3)$. The problem (3.8) has already been addressed in [14], where studies carried out provide the following result. Hereafter, $C$ denotes a generic positive constant, possibly changing from line to line, independent of the data and of time unless otherwise specified.

**Proposition 3.3** (Bona, Sun and Zhang, [14]). *Given* $T > 0$, *for every* $(\phi, \vec{h}) \in X_{0,T}$ *and* $f \in L^1(0,T, L^2(0,L))$ *there exists a unique solution* $\theta \in Z_{0,T}$ *to the problem (3.8) which satisfies*

$$
\partial_x^k \theta \in C_b \left( [0, L]; H^{\frac{1-k}{3}}(0, T) \right), \quad k = 0, 1, 2.
$$

Moreover,

$$
\|\theta\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k \theta\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \leq C \left( \|(\phi, \vec{h})\|_{X_{0,T}} + \|f\|_{L^1(0,T;L^2(0,L))} \right)
$$

for some constant $C > 0$.

To the well-posedness for the problem (3.9), we start by considering the IBVP

$$
(3.10) \quad \left\{
\begin{array}{ll}
\partial_t v + \partial_x^3 v = f, & t \in (0, T), x \in (0, L), \\
\partial_x^2 v(t, 0) = h_1(t), \quad v(t, L) = h_2(t), & t \in (0, T), \\
v(0, x) = \phi(x), & x \in (0, L).
\end{array}
\right.
$$

This system was studied in [17], where the following result was proven.
**Proposition 3.4** (Caicedo and Zhang, [17]). Let $T > 0$ and $0 \le s \le 3$ be given. For any $f \in L^1(0, T; H^s(0, L))$ and $(\phi, \vec{h}) \in Y_{s,T}$ satisfying the s-compatibility condition

$$ \left\{ \begin{array}{ll} \phi(L) = h_2(0) & \text{if } \frac{1}{2} < s \le 3, \\ \phi'(L) = h_3(0) & \text{if } \frac{3}{2} < s \le 3, \\ \phi''(0) = h_1(0) & \text{if } \frac{5}{2} < s \le 3, \end{array} \right. $$

the IBVP (3.10) admits a unique solution $v \in Z_{s,T}$ with $\partial_x^k v \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T))$, $k = 0, 1, 2$. Moreover, there exists a positive constant $C$ such that

$$ \|v\|_{Z_{s,T}} + \sum_{k=0}^{2} \|\partial_x^k v\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \|(\phi, \vec{h})\|_{Y_{s,T}} + \|f\|_{L^1(0,T; H^s(0,L))} \right). $$

Additionally, Caicedo and Zhang in [17] showed the following lemma.

**Lemma 3.1.** Let $0 \le s \le 3$ and $T > 0$ be given. For any $g, h \in H^{\frac{s+1}{3}}(0, T)$ we have $gh \in H^{\frac{s-1}{3}}(0, T)$ and

$$ \|gh\|_{H^{\frac{s-1}{3}}(0,T)} \le C T^{\alpha} \|g\|_{H^{\frac{s+1}{3}}(0,T)} \|h\|_{H^{\frac{s+1}{3}}(0,T)} $$

where $C$ and $\alpha$ are positive constants.

Thanks to the Proposition 3.4 and the previous lemma, we can establish the well-posedness for the problem (3.9) using a fixed point argument. The result can be read as follows.

**Proposition 3.5.** Given $T > 0$. For every $f \in L^1(0, T; L^2(0, L))$ and $(\phi, \vec{h}) \in Y_{0,T}$, the IBVP (3.9) admits a unique solution $\psi \in Z_{0,T}$ which satisfies

$$ \partial_x^k \psi \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T)), \quad k = 0, 1, 2. $$

Moreover, there exists a positive constant $C$ such that

$$ \|\psi\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k \psi\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \|(\phi, \vec{h})\|_{Y_{0,T}} + \|f\|_{L^1(0,T; L^2(0,L))} \right). $$

*Proof.* For $\beta \in (0, T]$ consider the space

$$ \mathcal{Z}_{\beta} = \{ v \in Z_{0,\beta}; \partial_x^k v \in C_b([0, L]; H^{\frac{1-k}{3}}(0, \beta)), k = 0, 1, 2 \}, $$

which is a Banach space with the norm

$$ \|v\|_{\mathcal{Z}_{\beta}} = \|v\|_{Z_{0,\beta}} + \sum_{k=0}^{2} \|\partial_x^k v\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,\beta))}. $$

Let $f \in L^1(0, T; L^2(0, L))$ and $(\phi, \vec{h}) \in Y_{0,T}$ be given. Define the map $\Gamma: \mathcal{Z}_{\beta} \to \mathcal{Z}_{\beta}$ putting, for each $v \in \mathcal{Z}_{\beta}$, $\Gamma v$ as being the solution of the problem

$$ \left\{ \begin{array}{ll} \partial_t \psi + \partial_x^3 \psi = f - \partial_x v, & t \in (0, T), x \in (0, L), \\ \partial_x^2 \psi(t, 0) = h_1(t) - \frac{\alpha}{N} v(t, 0), & t \in (0, T) \\ \psi(t, L) = h_2(t), & \partial_x \psi(t, L) = h_3(t), \quad t \in (0, T), \\ \psi(0, x) = \phi(x), & x \in (0, L). \end{array} \right. $$

From Proposition 3.4 (for $s = 0$) $\Gamma$ is well defined and, for every $v \in \mathcal{Z}_{\beta}$,

$$ \|\Gamma\|_{\mathcal{Z}_{\beta}} \le C \left( \|(\phi, \vec{h})\|_{Y_{0,\beta}} + \|f\|_{L^1(0,\beta; L^2(0,L))} + \frac{\alpha}{N} \|v(\cdot, 0)\|_{H^{-\frac{1}{3}}(0,T)} + \|\partial_x v\|_{L^1(0,\beta; L^2(0,L))} \right). $$
Using the Hölder inequality, we obtain

$$ \begin{aligned} \|\partial_x v\|_{L^1(0,\beta;L^2(0,L))} &= \int_0^\beta \|\partial_x v(t, \cdot)\|_{L^2(0,L)} \le \int_0^\beta \|\partial_x v(t, \cdot)\|_{H^1(0,L)} \le \beta^{\frac{1}{2}} \|\partial_x v\|_{L^2(0,\beta;H^1(0,L))} \\ &\le \beta^{\frac{1}{2}} \|v\|_{Z_\beta}. \end{aligned} $$

On the other hand, Lemma 3.1 gives us

$$ \|v(\cdot, 0)\|_{H^{-\frac{1}{3}}(0,\beta)} \le C\beta^\alpha \|1\|_{H^{\frac{1}{3}}(0,T)} \|v(\cdot, 0)\|_{H^{\frac{1}{3}}(0,\beta)} \le C\beta^\alpha \|v\|_{Z_\beta}. $$

Consequently

$$ \begin{aligned} \|\Gamma v\|_{Z_\beta} &\le C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} + \frac{\alpha}{N} C \beta^\alpha \|v\|_{Z_\beta} + \beta^{\frac{1}{2}} \|v\|_{Z_\beta} \right) \\ &\le C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} \right) + \left( \frac{\alpha}{N} \beta^\alpha + \beta^{\frac{1}{2}} \right) C \|v\|_{Z_\beta}. \end{aligned} \quad (3.11) $$

Choosing $\beta \in (0, T]$ such that

$$ (3.12) \qquad \left( \frac{\alpha}{N} \beta^{\alpha} + \beta^{\frac{1}{2}} \right) C < \frac{1}{2} $$

and $r = 2C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} \right)$, we have $\Gamma(B_r) \subset B_r$, where $B_r = \{ v \in Z_\beta; \|v\|_{Z_\beta} \le r \}$.

Moreover $\Gamma: B_r \to B_r$ is a contraction since

$$ \|\Gamma v - \Gamma w\|_{Z_\beta} = \|\Gamma(v-w)\|_{Z_\beta} \le \left(\frac{\alpha}{N}\beta^\alpha + \beta^{\frac{1}{2}}\right) C \|v-w\|_{Z_\beta} < \frac{1}{2} \|v-w\|_{Z_\beta}. $$

From the Banach fixed point theorem it follows that $\Gamma$ has a fixed point $\psi \in B_r$, that is, $\psi$ solves (3.9) for $t \in [0, \beta]$ and data $(\phi, \vec{h}, f)$. Furthermore, by combining and (3.12) and (3.11), we obtain

$$ \begin{aligned} \|\psi\|_{Z_\beta} &\le C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} \right) + \left( \frac{\alpha}{N} \beta^\alpha + \beta^{\frac{1}{2}} \right) C \|\psi\|_{Z_\beta} \\ &\le C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} \right) + \frac{1}{2} \|\psi\|_{Z_\beta} \end{aligned} $$

hence

$$ \|\psi\|_{Z_\beta} \le 2C \left( \| (\phi, \vec{h}) \|_{Y_{0,\beta}} + \| f \|_{L^1(0,\beta;L^2(0,L))} \right). $$

Since $\beta$ does not depend on the initial data $\phi$, a standard continuation (extension) argument allows us to extend the solution $\psi$ to the entire interval $[0, T]$. Moreover, the following estimate holds:

$$ \|\psi\|_{Z_T} \le C \left( \| (\phi, \vec{h}) \|_{Y_{0,T}} + \| f \|_{L^1(0,T;L^2(0,L))} \right), $$

where $C$ is an appropriate positive constant. $\square$

We conclude this section with one final technical result, which will be instrumental in proving the main theorem of this section.

**Lemma 3.2.** Let $T > 0$ be given. For every $u_0 \in D(A)$, $g_0, g_j, p_j \in \{ \varphi \in C^2([0,T]); \varphi(0) = 0 \}$ and $f \in C^1([0,T]; L^2(\mathcal{T}))$, the corresponding solution $u$ of the problem (3.1) satisfies

$$ \partial_x^k u_j \in C([0, T] \times [0, l_j]), \quad k = 0, 1, 2. $$

*Proof.* Consider the operator

$$ P_j: H^3(0, l_j) \to L^2(0, l_j) $$

defined by $P_j w = w_x + w_{xxx}$. The graph norm in $H^3(0, l_j)$ associated to $P_j$ is

$$ \|w\|_{P_j} = \|w\|_{L^2(0,l_j)} + \|P_j w\|_{L^2(0,l_j)}. $$

From [33, Lemma A.2] there exists a constant $C > 0$ such that

$$ (3.13) \qquad \|w\|_{H^3(0,l_j)} \le C \|w\|_{P_j}, \quad \forall w \in H^3(0, l_j), \ j = 1, \dots, N. $$
On the other hand, given $v \in D(A)$ we have

$$
\begin{aligned}
\|v\|_{D(A)} &= \|v\|_{\mathbb{L}^2(\mathcal{T})} + \|v_x + v_{xxx}\|_{\mathbb{L}^2(\mathcal{T})} \\
&= \left( \sum_{j=1}^{N} \|v_j\|_{L^2(0,l_j)}^2 \right)^{\frac{1}{2}} + \left( \sum_{j=1}^{N} \|v_{jx} + v_{jxxx}\|_{L^2(0,l_j)}^2 \right)^{\frac{1}{2}} \\
&\geq \|v_j\|_{L^2(0,l_j)} + \|v_{jx} + v_{jxxx}\|_{L^2(0,l_j)} \\
&= \|v_j\|_{P_j}.
\end{aligned}
$$

Therefore, this yields

$$ (3.14) \qquad \|v_j\|_{P_j} \le \|v\|_{D(A)}, \quad j = 1, \dots, N. $$

Now, from Proposition 3.2 we have $u \in C^1([0, T], \mathbb{L}^2(\mathcal{T})) \cap C([0, T], D(A))$ and by Sobolev embedding it follows that

$$ \partial_x^k u(t, \cdot) \in H^{3-k}(0, l_j) \hookrightarrow C([0, l_j]), \quad \forall t \in [0, T], \quad k = 0, 1, 2, \quad j = 1, \dots, N. $$

Hence, fixed $x \in [0, l_j]$, we get

$$
\begin{aligned}
|\partial_x^k u_j(t, x) - \partial_x^k u_j(t_0, x)| &\le \|\partial_x^k u_j(t, \cdot) - \partial_x^k u_j(t_0, \cdot)\|_{C[0,l_j]} \\
&\le C \|\partial_x^k u_j(t, \cdot) - \partial_x^k u_j(t_0, \cdot)\|_{H^{3-k}(0,l_j)} \\
&\le C \|u_j(t, \cdot) - u_j(t_0, \cdot)\|_{H^3(0,l_j)}
\end{aligned}
$$

for any $t, t_0 \in [0, T]$. Using (3.13) and (3.14) we obtain

$$ |\partial_x^k u_j(t, x) - \partial_x^k u_j(t_0, x)| \le C \|u_j(t, \cdot) - u_j(t_0, \cdot)\|_{D(A)}, $$

from where we conclude that $\partial_x^k u_j(\cdot, x) \in C([0, T])$ showing the lemma. $\square$

We are now in a position to state the main result of this subsection, whose proof follows from a combination of Propositions 3.3 and 3.5.

**Proposition 3.6.** Given $T > 0$, there exist unique bounded linear maps

$$
\begin{aligned}
\Psi : \mathbb{L}^2(\mathcal{T}) \times H^{-\frac{1}{3}}(0, T) \times [L^2(0, T)]^N \times [H^{\frac{1}{3}}(0, T)]^N \times L^1(0, T, \mathbb{L}^2(\mathcal{T})) & \to \mathbb{B}_T \\
(u^0, g_0, g, p, f) & \mapsto \Psi(u^0, g_0, g, p, f),
\end{aligned}
$$

and, for $k = 0, 1, 2$,

$$
\begin{aligned}
\Psi_j^k : \mathbb{L}^2(\mathcal{T}) \times H^{-\frac{1}{3}}(0, T) \times [L^2(0, T)]^N \times [H^{\frac{1}{3}}(0, T)]^N \times L^1(0, T, \mathbb{L}^2(\mathcal{T})) & \to C_b([0, l_j]; H^{\frac{1-k}{3}}(0, T)) \\
(u^0, g_0, g, p, f) & \mapsto \Psi_j^k(u^0, g_0, g, p, f)
\end{aligned}
$$

such that, when $u^0 \in D(A)$, $g_0, g_j, p_j \in C_l^2([0, T])$ for $j = 1, \dots, N$ and $f \in C^1([0, T]; \mathbb{L}^2(\mathcal{T}))$ we have the following items:

i) $u := \Psi(u^0, g_0, g, p, f)$ is the unique classical solution of (3.1);

ii) $\Psi_j^k(u^0, g_0, g, p, f) = \partial_x^k u_j$.

*Proof.* We divide the proof into two cases. The constant $C$ can be adjusted for each line.

**Case 1:** $l_j = L > 0$ for $j = 1, \dots, N$.

First we define

$$
\begin{aligned}
\Psi : D(A) \times C_l^2([0, T]) \times [C_l^2([0, T])]^N \times [C_l^2([0, T])]^N \times C^1(0, T, \mathbb{L}^2(\mathcal{T})) & \to \mathbb{B} \\
(u^0, g_0, g, p, f) & \mapsto \Psi(u^0, g_0, g, p, f),
\end{aligned}
$$

and

$$
\begin{aligned}
\Psi_j^k : D(A) \times C_l^2([0, T]) \times [C_l^2([0, T])]^N \times [C_l^2([0, T])]^N \times C^1(0, T, \mathbb{L}^2(\mathcal{T})) & \to C_b([0, l_j]; H^{\frac{1-k}{3}}(0, T)) \\
(u^0, g_0, g, p, f) & \mapsto \Psi_j^k(u^0, g_0, g, p, f)
\end{aligned}
$$
for $k = 0, 1, 2$. By proving that $\Psi$ and $\Psi_j^k$ are continuous with respect to the norm of

$$ \mathbb{L}^2(\mathcal{T}) \times H^{-\frac{1}{3}}(0, T) \times [L^2(0, T)]^N \times [H^{\frac{1}{3}}(0, T)]^N \times L^1(0, T, \mathbb{L}^2(\mathcal{T})), $$

we can extend them to this space by density. Suppose $u_0 \in D(A)$, $g_0, g_j, p_j \in C_l^2([0, T])$ and $f \in C^1([0, T]; \mathbb{L}^2(\mathcal{T}))$. From Proposition 3.2, there exists a unique classical solution

$$ u \in C^1([0, T], \mathbb{L}^2(\mathcal{T})) \cap C([0, T], \mathbb{H}_c^3(\mathcal{T})) $$

for (3.1). This gives in particular $u_j \in C([0, T], L^2(0, l_j))$, $j = 1, ..., N$. Furthermore, due to Lemma 3.2, we have that

$$ \int_0^T \|u_j(t, \cdot)\|_{H^1(0, l_j)}^2 = \int_0^T \int_0^{l_j} |u_j(t, x)|^2 + \int_0^T \int_0^{l_j} |\partial_x u_j(t, x)|^2 < \infty $$

hence $u_j \in L^2(0, T, H^1(0, l_j))$, $j = 1, ..., N$. Consequently,

$$ (3.15) \qquad u_j \in C([0, T], L^2(0, l_j)) \cap L^2(0, T, H^1(0, l_j)), \quad j = 1, ..., N $$

as well as $u \in \mathbb{B} = C([0, T], \mathbb{L}^2(\mathcal{T})) \cap L^2(0, T; \mathbb{H}^1(\mathcal{T}))$. So we are motivated to define $\Psi(u^0, g_0, g, p, f) = u$. Consider $\psi, \theta$ defined by

$$ \psi := \sum_{j=1}^{N} u_j \quad \text{and} \quad \theta := Nu_1 - \psi. $$

Note that, by the regularity (3.15), we have $\theta \in Z_{0,T} = C([0, T], L^2(0, L)) \cap L^2(0, T, H^1(0, L))$. Moreover, $\theta$ is a classical solution to the IBVP

$$ \left\{
\begin{array}{ll}
\partial_t \theta + \partial_x \theta + \partial_x^3 \theta = N f_1 - \sum_{j=1}^N f_j, & t \in (0, T), x \in (0, L), j = 1, \dots, N, \\
\theta(t, 0) = 0, & t \in (0, T), j = 1, \dots, N, \\
\theta(t, L) = N p_1(t) - \sum_{j=1}^N p_j(t), & t \in (0, T), j = 1, \dots, N, \\
\partial_x \theta(t, L) = N g_1(t) - \sum_{j=1}^N g_j(t), & t \in (0, T), j = 1, \dots, N, \\
\theta(0, x) = N u_1^0(x) - \sum_{j=1}^N u_j^0(x), & x \in (0, L), j = 1, \dots, N.
\end{array}
\right. $$

Then, Proposition 3.8 ensures that $\partial_x^k \theta \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T))$ for $k = 0, 1, 2$ and

$$ \|\theta\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k \theta\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \left\| N u_1^0 - \sum_{j=1}^N u_j^0 \right\|_{L^2(0,L)} + \left\| N p_1 - \sum_{j=1}^N p_j \right\|_{H^{\frac{1}{3}}(0,T)} \right. \\ \left. + \left\| N g_1 - \sum_{j=1}^N g_j \right\|_{L^2(0,T)} + \left\| N f_1 - \sum_{j=1}^N f_j \right\|_{L^2(0,T; L^2(0,L))} \right) $$
From this inequality, we obtain

$$ \|\theta\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k \theta\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \sum_{j=1}^{N} \|u_j^0\|_{L^2(0,L)} + \sum_{j=1}^{N} \|p_j\|_{H^{\frac{1}{3}}(0,T)} \right. \\ \left. + \sum_{j=1}^{N} \|g_j\|_{L^2(0,T)} + \sum_{j=1}^{N} \|f_j\|_{L^1(0,T; L^2(0,L))} \right). $$

Analogously, (3.15) ensures that $\psi \in Z_{0,T}$. Observe that $\partial_x^2 \psi(t, 0) + \frac{\alpha}{N} \psi(t, 0) = g_0(t)$. Consequently, $\psi$ is a classical solution of

$$ \left\{ \begin{array}{ll} \partial_t \psi + \partial_x \psi + \partial_x^3 \psi = \sum_{j=1}^{N} f_j, & t \in (0, T), x \in (0, L), \\ \partial_x^2 \psi(t, 0) + (\alpha/N) \psi(t, 0) = g_0(t), & t \in (0, T), \\ \psi(t, L) = \sum_{j=1}^{N} p_j(t), \quad \partial_x \psi(t, L) = \sum_{j=1}^{N} g_j(t) & t \in (0, T), \\ \psi(0, x) = \sum_{j=1}^{N} u_j^0(x), & x \in (0, L). \end{array} \right. $$

Thus, Proposition 3.4 gives us $\partial_x^k \psi \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T))$, $k = 0, 1, 2$, and

$$ \|\psi\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k \psi\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \sum_{j=1}^{N} \|u_j^0\|_{L^2(0,L)} + \sum_{j=1}^{N} \|p_j\|_{H^{\frac{1}{3}}(0,T)} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \right. \\ \left. + \sum_{j=1}^{N} \|g_j\|_{L^2(0,T)} + \sum_{j=1}^{N} \|f_j\|_{L^1(0,T; L^2(0,L))} \right). $$

Since $u_1 = \frac{1}{N}(\theta + \psi)$ it follows that $u_1 \in Z_{0,T}$, $\partial_x^k u_1 \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T))$, for $k = 0, 1, 2$, and

$$ \|u_1\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k u_1\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \sum_{j=1}^{N} \|u_j^0\|_{L^2(0,L)} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \right. \\ \left. + \sum_{j=1}^{N} \|g_j\|_{L^2(0,T)} + \sum_{j=1}^{N} \|p_j\|_{H^{\frac{1}{3}}(0,T)} + \sum_{j=1}^{N} \|f_j\|_{L^1(0,T; L^2(0,L))} \right). \quad (3.16) $$

Hence, $u_1(\cdot, 0) \in H^{\frac{1}{3}}(0, T)$ and, for each $j = 2, \dots, N$, $u_j \in Z_{0,T}$ is the solution of the following problem

$$ \left\{ \begin{array}{ll} \partial_t u_j + \partial_x u_j + \partial_x^3 u_j = f_j, & t \in (0, T), x \in (0, L), \\ u_j(t, 0) = u_1(t, 0), \quad u_j(t, L) = p_j(t), \quad \partial_x u_j(t, L) = g_j(t), & t \in (0, T), \\ u(0, x) = u_j^0(x), & x \in (0, L). \end{array} \right. $$

By Proposition 3.3 it follows that $\partial_x^k u_j \in C_b([0, L]; H^{\frac{1-k}{3}}(0, T))$, $k = 0, 1, 2$, and

$$ \|u_j\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k u_j\|_{C_b([0,L]; H^{\frac{1-k}{3}}(0,T))} \le C \left( \|u_j^0\|_{L^2(0,L)} + \|u_1(\cdot, 0)\|_{H^{\frac{1}{3}}(0,T)} + \|p_j\|_{H^{\frac{1}{3}}(0,T)} \right. \\ \left. + \|g_j\|_{L^2(0,T)} + \|f_j\|_{L^1(0,T; L^2(0,L))} \right). $$
Using (3.16) we get

$$ (3.17) \qquad \|u_j\|_{Z_{0,T}} + \sum_{k=0}^{2} \|\partial_x^k u_j\|_{C_b([0,L];H^{-\frac{1-k}{3}}(0,T))} \le C \left( \sum_{j=1}^{N} \|u_j^0\|_{L^2(0,L)} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \right. \\ \qquad \left. + \sum_{j=1}^{N} \|g_j\|_{L^2(0,T)} + \sum_{j=1}^{N} \|p_j\|_{H^{\frac{1}{3}}(0,T)} + \sum_{j=1}^{N} \|f_j\|_{L^1(0,T;L^2(0,L))} \right), $$

for $j = 1, ..., N$. Adding for $j = 1, ..., N$ this implies, in particular

$$ (3.18) \qquad \max_{t \in [0,T]} \|u(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})} + \left( \int_0^T \|u(t, \cdot)\|_{\mathbb{H}^1}^2 dt \right)^{\frac{1}{2}} \le C \left( \sum_{j=1}^N \|u_j^0\|_{L^2(0,L)} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \right. \\ \qquad \left. + \sum_{j=1}^N \|g_j\|_{L^2(0,T)} + \sum_{j=1}^N \|p_j\|_{H^{\frac{1}{3}}(0,T)} + \sum_{j=1}^N \|f_j\|_{L^1(0,T;L^2(0,L))} \right). $$

Observe that

$$ (3.19) \qquad \sum_{j=1}^{N} \|f_j\|_{L^1(0,T;L^2(0,L))} \le N^{\frac{1}{2}} \|f\|_{L^1(0,T;L^2(\mathcal{T}))} $$

as well as

$$ (3.20) \qquad \sum_{j=1}^{N} \|u_j^0\|_{L^2(0,L)} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \sum_{j=1}^{N} \|g_j(t)\|_{L^2(0,T)} + \sum_{j=1}^{N} \|p_j(t)\|_{H^{\frac{1}{3}}(0,T)} \\ \le N^{\frac{1}{2}} \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} \right). $$

From (3.18), (3.19) and (3.20) it follows that

$$ \|u\|_{\mathbb{B}} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;\mathbb{L}^2(\mathcal{T}))} \right). $$

Therefore, due to this previous inequality, $\Psi$ is continuous. Now, combining (3.17), (3.19) and (3.20) we have

$$ \sum_{k=0}^{2} \|\partial_x^k u_j\|_{C_b([0,L];H^{-\frac{1-k}{3}}(0,T))} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;\mathbb{L}^2(\mathcal{T}))} \right). $$

Hence, defining $\Psi_j^k(u^0, g_0, g, p, f) = \partial_x^k u_j$, $k = 0, 1, 2$, we have that $\Psi_j^k$ is continuous. As mentioned before, by density, each map $\Psi$ and $\Psi_j^k$, for $k = 0, 1, 2$, has a unique continuous extension to the space $\mathbb{L}^2(\mathcal{T}) \times H^{-\frac{1}{3}}(0, T) \times [L^2(0, T)]^N \times [H^{\frac{1}{3}}(0, T)]^N \times L^1(0, T, \mathbb{L}^2(\mathcal{T}))$ which we will also denote by $\Psi$ and $\Psi_j^k$, respectively. Hence, the proposition is proved for case 1.

**Case 2:** Arbitrary $l_j > 0$, for $j = 1, ..., N$.
Let $u^0 \in L^2(\mathcal{T})$, $g_0 \in H^{-\frac{1}{3}}(0,T)$, $g \in [L^2(0,T)]^N$, $p \in [H^{\frac{1}{3}}(0,T)]^N$ and $f \in L^1(0,T, L^2(\mathcal{T}))$ be given. Define $L = \max\{l_j; j = 1, \dots, N\}$, $L^2(\tilde{\mathcal{T}}) = (L^2(0,L))^N$,

$$ \tilde{u}_j^0(x) := \begin{cases} u_j^0(x), & x \in (0, l_j), \\ 0, & x \in (0, L) \setminus (0, l_j) \end{cases} $$

and

$$ \tilde{f}_j(t, x) := \begin{cases} f_j(t, x), & x \in (0, l_j), \\ 0, & x \in (0, L) \setminus (0, l_j) \end{cases}, \quad \forall t \in [0, T]. $$

Consider $\tilde{u} = \Psi(\tilde{u}^0, g_0, g, p, \tilde{f}) \in \tilde{\mathbb{B}} := C([0, T], L^2(\tilde{\mathcal{T}})) \cap L^2(0, T, \mathbb{H}_e^1(\tilde{\mathcal{T}}))$, where $\Psi$ is the map obtained in the first case. Thus

$$ \begin{align} \|\tilde{u}\|_{\tilde{\mathbb{B}}} &\le C \left( \|\tilde{u}^0\|_{L^2(\tilde{\mathcal{T}})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|\tilde{f}\|_{L^1(0,T;L^2(\tilde{\mathcal{T}}))} \right) \tag{3.21} \\ &= C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\mathcal{T}))} \right). \end{align} $$

Now, define $u_j(t, \cdot) = \tilde{u}_j(t, \cdot)|_{[0,l_j]}$, $j = 1, \dots, N$, and $u(t, \cdot) = (u_1(t, \cdot), \dots, u_N(t, \cdot))$. For all $t \in [0, T]$,

$$ \sum_{j=1}^{N} \int_{0}^{l_j} |\partial_x^k u_j(t, x)|^2 = \sum_{j=1}^{N} \int_{0}^{l_j} |\partial_x^k \tilde{u}_j(t, x)|^2 \le \sum_{j=1}^{N} \int_{0}^{L} |\partial_x^k \tilde{u}_j(t, x)|^2, \quad k = 0, 1, $$

hence $u(t, \cdot) \in \mathbb{H}^1(\mathcal{T})$ and

$$ \|u(t, \cdot)\|_{L^2(\mathcal{T})} \le \|\tilde{u}(t, \cdot)\|_{L^2(\tilde{\mathcal{T}})}, \quad (3.22) $$

and

$$ \|u(t, \cdot)\|_{\mathbb{H}^1(\mathcal{T})}^2 \le \|\tilde{u}(t, \cdot)\|_{\mathbb{H}^1(\mathcal{T})}^2, $$

for every $t \in [0, T]$. Moreover, given $t, t_0 \in [0, T]$ we have

$$ \begin{align*} \|u(t, \cdot) - u(t_0, \cdot)\|_{L^2(\mathcal{T})}^2 &= \sum_{j=1}^{N} \int_{0}^{l_j} |u_j(t, x) - u_j(t_0, x)|^2 \\ &= \sum_{j=1}^{N} \int_{0}^{l_j} |\tilde{u}_j(t, x) - \tilde{u}_j(t_0, x)|^2 \\ &\le \sum_{j=1}^{N} \int_{0}^{L} |\tilde{u}_j(t, x) - \tilde{u}_j(t_0, x)|^2 \\ &= \|\tilde{u}(t, \cdot) - \tilde{u}(t_0, \cdot)\|_{L^2(\mathcal{T})}^2 \end{align*} $$

from where we obtain $u \in C([0, T], L^2(\mathcal{T}))$. Similarly, we get $u \in L^2(0, T; \mathbb{H}^1(\mathcal{T}))$ and

$$ \|u\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \le \|\tilde{u}\|_{L^2(0,T;\mathbb{H}^1(\tilde{\mathcal{T}}))}. \quad (3.23) $$

Combining (3.22) and (3.23) we get $u \in \mathbb{B}$ with

$$ \|u\|_{\mathbb{B}} \le \|\tilde{u}\|_{\tilde{\mathbb{B}}}. \quad (3.24) $$

Finally, define $\Psi(u_0, g_0, g, p, f) := u$ and note that, thanks to (3.21) and (3.24), we have

$$ \|\Psi(u_0, g_0, g, p, f)\|_{\mathbb{B}} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} \right. \\ \left. \qquad + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\mathcal{T}))} \right), $$
showing the continuity of $\Psi$. Moreover, by construction, for $u_0 \in D(A)$ and $g_0, g_j, p_j \in C_l^2([0, T])$, $\tilde{u}$ is the classical solution of (3.1) in $(0, T) \times (0, L)$. Hence, in this case $u$ is the classical solution of (3.1) in $(0, T) \times (0, l_j)$.

Now, for $j = 1, \dots, N$ and $k = 0, 1, 2$ define

$$ \Psi_j^k : \mathbb{L}^2(\mathcal{T}) \times H^{-\frac{1}{3}}(0,T) \times [L^2(0,T)]^N \times [H^{\frac{1}{3}}(0,T)]^N \times L^1(0,T, \mathbb{L}^2(\mathcal{T})) \to C_b([0, l_j]; H^{\frac{1-k}{3}}(0,T)) $$
putting

$$ \Psi_j^k(u_0, g_0, g, p, f)(x, \cdot) = \Psi_j^k(\tilde{u}_0, g_0, g, p, \tilde{f})(x, \cdot) \in H^{\frac{1-k}{3}}(0, T), \forall x \in [0, l_j]. $$

Clearly $\Psi_j^k$ is well defined and, for every $x \in [0, l_j]$,

$$ \begin{aligned} \| \Psi_j^k(u_0, g_0, g, p, f)(x, \cdot) \|_{H^{\frac{1-k}{3}}(0,T)} &= \| \Psi_j^k(\tilde{u}_0, g_0, g, p, \tilde{f})(x, \cdot) \|_{H^{\frac{1-k}{3}}(0,T)} \\ &\le \| \Psi_j^k(\tilde{u}_0, g_0, g, p, \tilde{f}) \|_{C_b([0,l_j]; H^{\frac{1-k}{3}}(0,T))}. \end{aligned} $$

So, we get that

$$ \| \Psi_j^k(u_0, g_0, g, p, f) \|_{C_b([0, l_j]; H^{\frac{1-k}{3}}(0, T))} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0, T)} + \|g\|_{[L^2(0, T)]^N} \right. \\ \left. \qquad + \|p\|_{[H^{\frac{1}{3}}(0, T)]^N} + \|f\|_{L^1(0, T; \mathbb{L}^2(\mathcal{T}))} \right). $$

Finally, observe that for $u_0 \in D(A)$ and $g_0, g_j, p_j \in C_l^2([0, T])$,

$$ \Psi_j^k(u_0, g_0, g, p, f)(x, \cdot) = \Psi_j^k(\tilde{u}_0, g_0, g, p, \tilde{f})(x, \cdot) = \partial_x^k \tilde{u}_j(x, \cdot) = \partial_x^k u_j(x, \cdot), \forall x \in [0, l_j], $$

that is $\Psi_j^k(u_0, g_0, g, p, f) = \partial_x^k u_j$ concluding the proof of case 2. $\square$

Proposition 3.6 motivates the introduction of the notion of mild solution for system (3.1), guaranteeing its well-posedness in this sense, including existence, uniqueness, and continuous dependence on the data.

**Definition 3.1.** Given $T > 0$, $u^0 \in \mathbb{L}^2(\mathcal{T})$, $g_0 \in H^{-\frac{1}{3}}(0, T)$, $g \in [L^2(0, T)]^N$, $p \in [H^{\frac{1}{3}}(0, T)]^N$ and $f \in L^1(0, T; \mathbb{L}^2(\mathcal{T}))$, we define the mild solution $u \in \mathbb{B}$ of (3.1) as a function $u = \Psi(u^0, g_0, g, p, f)$, where $\Psi$ is the map given in Proposition 3.6.

**Remark 3.1.** Note that, according to Proposition 3.6, if $u^0 \in D(A)$, $g_0, g_j, p_j \in C_l^2([0, T])$, for $j = 1, \dots, N$, and $f \in C^1([0, T]; \mathbb{L}^2(\mathcal{T}))$, then $u$ is actually a classical solution to the system (3.1).

### 3.2. The case $s = 3$. Given $s \ge 0$, we will consider

$$ u^0 \in \mathbb{H}^s(\mathcal{T}), \quad g_0 \in H^{\frac{s-1}{3}}(0, T), \quad g \in [H^{\frac{s}{3}}(0, T)]^N, \quad p \in [H^{\frac{s+1}{3}}(0, T)]^N $$

and

$$ f \in W^{\frac{s}{3}, 1}(0, T; \mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0, T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T})). $$

Denote by $X_{s,T}$ the set of $s$-compatible data

$$ (u^0, g_0, g, p) \in \mathbb{H}^s(\mathcal{T}) \times H^{\frac{s-1}{3}}(0, T) \times [H^{\frac{s}{3}}(0, T)]^N \times [H^{\frac{s+1}{3}}(0, T)]^N. $$
Suppose that $u$ is the solution of (3.1). Formally, we can derive the first equation of (3.1) with respect to $t$, getting

$$
\left\{
\begin{array}{ll}
\displaystyle \partial_t (\partial_t u_j) + \partial_x (\partial_t u_j) + \partial_x^3 (\partial_t u_j) = \partial_t f_j, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N, \\
\displaystyle \partial_t u_j(t, 0) = \partial_t u_1(t, 0), & t \in (0, T), j = 1, \dots, N, \\
\displaystyle \sum_{j=1}^N \partial_x^2 (\partial_t u_j) (t, 0) = -\alpha \partial_t u_1(t, 0) + g'_0(t), & t \in (0, T), \\
\displaystyle \partial_t u_j(t, l_j) = p'_j(t), \quad \partial_x (\partial_t u_j) (t, l_j) = g'_j(t), & t \in (0, T), j = 1, \dots, N.
\end{array}
\right.
$$

Moreover, $\partial_t u_j(0, x) = -\partial_x u_j^0(x) - \partial_x^3 u_j^0(x) + f_j(0, x)$ for $x \in (0, l_j)$ and $j = 1, \dots, N$. Thus, setting $v_j = \partial_t u_j$ it follows that,

$$
(3.25) \quad \left\{
\begin{array}{ll}
\displaystyle \partial_t v_j + \partial_x v_j + \partial_x^3 v_j = \partial_t f_j, & t \in (0, T), x \in (0, l_j), j = 1, \dots, N \\
v_j(t, 0) = v_1(t, 0), & t \in (0, T), j = 1, \dots, N \\
\displaystyle \sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0) + g'_0(t), & t \in (0, T) \\
v_j(t, l_j) = p'_j(t), \quad \partial_x v_j(t, l_j) = g'_j(t), & t \in (0, T), j = 1, \dots, N \\
v_j(0, x) = -\partial_x u_j^0(x) - \partial_x^3 u_j^0(x) + f_j(0, x), & x \in (0, l_j), j = 1, \dots, N
\end{array}
\right.
$$

and

$$
(3.26) \qquad u_j(t, \cdot) = u_j^0 + \int_0^t v_j(s, \cdot) ds, \quad j = 1, \dots, N.
$$

Motivated by the above, we will show that, when $s = 3$, the mild solution of (3.1) is given by (3.26)
and has the desirable regularity. Observe that, in this case,

$$
p_j \in H^{1+\frac{1}{3}}(0,T), \quad g_j \in H^1(0,T), \quad g_0 \in H^{\frac{2}{3}}(0,T) \quad \text{and} \quad f \in W^{1,1}(0,T; \mathbb{L}^2(\mathcal{T})) \cap \mathbb{L}^2(0,T; \mathbb{H}^1(\mathcal{T})).
$$

So, by results about Sobolev spaces (see for example [9]), it follows that<fnref n="1" />

$$
p'_{j} \in H^{\frac{1}{3}}(0, T), \quad g'_{j} \in L^{2}(0, T) \quad \text{and} \quad g'_{0} \in H^{-\frac{1}{3}}(0, T).
$$

Also note that

$$
-\partial_x u_j^0 - \partial_x^3 u_j^0 + f_j(0, \cdot) \in L^2(0, l_j) \quad \text{and} \quad \partial_t f \in L^2(0, T; \mathbb{L}^2(\mathcal{T}))
$$

Then, by Proposition 3.6, system (3.25) possess a unique (mild) solution $v \in \mathbb{B}_T$ satisfying

$$
\|v\|_{\mathbb{B}} \le C \left( \begin{array}{l} \| - \partial_x u^0 - \partial_x^3 u^0 + f(0, \cdot) \|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \\ \qquad + \|g'\|_{[L^2(0,T)]^N} + \|p'\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|\partial_t f\|_{L^1(0,T;\mathbb{L}^2(\mathcal{T}))} \end{array} \right).
$$

Consequently, the corresponding coordinate functions $u_j(t, \cdot)$ in (3.26) are well defined in $\mathbb{L}^2(0, l_j)$ with $j = 1, \dots, N$, as well as, the vector function $u(t, \cdot) = (u_1(t, \cdot), \dots, u_N(t, \cdot))$ is well defined in $\mathbb{L}^2(\mathcal{T})$, for any $t \in [0, T]$. This will lead us to the following result.

**Lemma 3.3.** Let $(u^0, g_0, g, p) \in \mathbb{X}_{3,T}$, $f \in W^{1,1}(0,T; \mathbb{L}^2(\mathcal{T})) \cap \mathbb{L}^2(0,T; \mathbb{H}^1(\mathcal{T}))$ and $v$ the solution of (3.25). Then for $\beta = 0, 1, 2, 3$ we have:

i. The function

$$
\begin{array}{rcl}
[0, T] & \to & \mathcal{D}'(0, l_j) \\
s & \mapsto & \partial_x^\beta v_j(s, \cdot)
\end{array}
$$

is integrable in $[0, T]$.

<footnote n="1">¹For $s > 0$ with $s - \frac{1}{2} \notin \mathbb{Z}$ and $s - |\beta| < 0$, $\partial^\beta(H^s(0, T)) \subset H^{s-|\beta|}(0, T)$ with continuous inclusion.</footnote>
ii. For any $t \in [0, T]$ we have $\int_0^t v_j(s, \cdot)ds \in H^4(0, l_j)$ and for $\beta = 0, 1, 2, 3$ we have

$$ \partial_x^\beta \int_0^t v_j(s, \cdot)ds = \int_0^t \partial_x^\beta v_j(s, \cdot)ds. $$

The previous lemma can give us the regularity for $u$.

**Proposition 3.7.** Let $(u^0, g_0, g, p) \in \mathbb{H}^3(\mathcal{T}) \times H^{\frac{3-1}{3}}(0, T) \times [H^{\frac{3}{3}}(0, T)]^N \times [H^{\frac{3+1}{3}}(0, T)]^N$ satisfying the linear compatibility conditions for $s = 3$ and $f \in W^{1,1}(0, T; \mathbb{L}^2(\mathcal{T})) \cap L^2(0, T; \mathbb{H}^1(\mathcal{T}))$ be. For $j = 1, \dots, N$ the function $t \mapsto u_j(t, \cdot)$ given by (3.26) satisfies the following items:

i. $u_j \in C^1([0, T], L^2(0, l_j)) \cap C([0, T]; H^3(0, l_j))$;

ii. $u \in \mathbb{B}_{3,T} = C([0, T]; \mathbb{H}^3(\mathcal{T})) \cap L^2(0, T; \mathbb{H}^4(\mathcal{T}))$ is the unique classical solution to the problem (3.1); and

$$ \|u\|_{\mathbb{B}_{3,T}} \le C \left( \|u^0\|_{\mathbb{H}^3(\mathcal{T})} + \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N} + \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \right) $$

for some constant $C > 0$.

iii. $u \in H^1(0, T; \mathbb{H}^1(\mathcal{T}))$ and

$$ \|u\|_{H^1(0,T;\mathbb{H}^1(\mathcal{T}))} \le C \left( \|u^0\|_{\mathbb{H}^3(\mathcal{T})} + \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N} + \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \right), $$

for some constant $C > 0$.

iv. $\partial_x^k u_j \in L_x^\infty([0, l_j]; H^{\frac{3+1-k}{3}}(0, T))$ and

$$ \|\partial_x^k u_j\|_{L_x^\infty([0,l_j];H^{\frac{3+1-k}{3}}(0,T))} \le C \left( \|u^0\|_{\mathbb{H}^3(\mathcal{T})} + \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N} + \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \right). $$

*Proof.* Consider $j \in \{1, \dots, N\}$.

i. Since $v_j \in C([0, T], L^2(0, l_j))$ it is immediate that by (3.26) that $u_j \in C^1([0, T], L^2(0, l_j))$.

ii. Let $t \in [0, T]$ be given. From (3.26) and item ii of Lemma 3.3 we have $u_j(t, \cdot) \in H^3(0, l_j)$. From (3.25) and for $r \in [0, T]$, $t \ge r$

$$ \partial_x^3 u_j(t, \cdot) - \partial_x^3 u_j(r, \cdot) = -v_j(t, \cdot) + v_j(r, \cdot) - \int_0^t \partial_x v_j(s, \cdot) ds + \int_0^r \partial_x v_j(s, \cdot) ds + f_j(t, \cdot) - f_j(r, \cdot) $$

hence

$$ \|\partial_x^3 u_j(t, \cdot) - \partial_x^3 u_j(r, \cdot)\|_{L^2(0,l_j)} \le \|v_j(t, \cdot) - v_j(r, \cdot)\|_{L^2(0,l_j)} \\ + \left\| \int_0^t \partial_x v_j(s, \cdot) ds - \int_0^r \partial_x v_j(s, \cdot) ds \right\|_{L^2(0,l_j)} \\ + \|f_j(t, \cdot) - f_j(r, \cdot)\|_{L^2(0,l_j)}. $$

Since $v \in \mathbb{B}$ and due to continuous embedding $W^{1,1}(0, T; \mathbb{L}^2(\mathcal{T})) \hookrightarrow C([0, T]; \mathbb{L}^2(\mathcal{T}))$ (see [22, Section 5.9, Theorem 2]) we have

$$ \|v_j(t, \cdot) - v_j(r, \cdot)\|_{L^2(0, l_j)}, \|f_j(t, \cdot) - f_j(r, \cdot)\|_{L^2(0, l_j)} \to 0 \text{ as } r \to t. $$
On the other hand, the Hölder inequality gives us

$$ \left\| \int_0^t \partial_x v_j(s, \cdot) ds - \int_0^r \partial_x v_j(s, \cdot) ds \right\|_{L^2(0, l_j)} \le \int_r^t \| \partial_x v_j(s, \cdot) \|_{L^2(0, l_j)} ds \\ \le |t-r|^{\frac{1}{2}} \|v\|_{\mathbb{B}}. $$

Hence

$$ (3.27) \qquad \left\| \int_0^t \partial_x v_j(s, \cdot) ds - \int_0^r \partial_x v_j(s, \cdot) ds \right\|_{L^2(0, l_j)} \to 0 \quad \text{as} \quad r \to t. $$

We conclude that

$$ \| \partial_x^3 u_j(t, \cdot) - \partial_x^3 u_j(r, \cdot) \|_{L^2(0, l_j)} \to 0 \quad \text{as} \quad r \to t $$

and, therefore, $\partial_x^3 u_j \in C([0, T]; L^2(0, l_j))$. In the same way, $\partial_x u_j \in C([0, T]; L^2(0, l_j))$ since

$$ \partial_x u_j(t, \cdot) - \partial_x u_j(r, \cdot) = \int_0^t \partial_x v_j(s, \cdot) ds - \int_0^r \partial_x v_j(s, \cdot) ds, $$

so, by (3.27), $\partial_x u_j(r, \cdot) \to \partial_x u_j(t, \cdot)$ as $r \to t$. From [33, Lemma A.2] and (3.13), we have

$$ \begin{aligned} \|u_j(t, \cdot) - u_j(r, \cdot)\|_{H^3(0, l_j)} &\le C \left( \|u_j(t, \cdot) - u_j(r, \cdot)\|_{L^2(0, l_j)} + \|P_j u_j(t, \cdot) - P_j u_j(r, \cdot)\|_{L^2(0, l_j)} \right) \\ &\le C \left( \|u_j(t, \cdot) - u_j(r, \cdot)\|_{L^2(0, l_j)} + \|\partial_x^3 u_j(t, \cdot) - \partial_x^3 u_j(r, \cdot)\|_{L^2(0, l_j)} \right) \\ &\qquad \|\partial_x u_j(t, \cdot) - \partial_x u_j(r, \cdot)\|_{L^2(0, l_j)} \right). \end{aligned} $$

Since $u_j, \partial_x u_j, \partial_x^3 u_j \in C([0, T]; L^2(0, T))$, it follows that, $u_j \in C([0, T]; H^3(0, l_j))$. Now, by (3.26) we have

$$ (3.28) \qquad \partial_t u_j(t, \cdot) + \partial_x u_j(t, \cdot) + \partial_x^3 u_j(t, \cdot) = v_j(t, \cdot) + \partial_x u_j^0 + \partial_x^3 u_j^0 + \int_0^t \partial_x v_j(s, \cdot) ds + \int_0^t \partial_x^3 v_j(s, \cdot) ds. $$

From (3.25), we have $v_j'(s, \cdot) = -\partial_x v_j(s, \cdot) - \partial_x^3 v_j(s, \cdot) + f_j'(s, \cdot)$. Using Lemma 3.3 and the hypothesis about $f$, it follows that $s \mapsto v_j'(s, \cdot)$ is integrable, where $'$ denotes the (distributional) derivative with respect to the temporal variable. Hence,

$$ \int_0^t v_j'(s, \cdot) ds = v_j(t, \cdot) - v_j(0, \cdot) $$

On the other hand,

$$ \int_0^t v_j'(s, \cdot) ds = - \int_0^t \partial_x v_j(s, \cdot) ds - \int_0^t \partial_x^3 v_j(s, \cdot) ds + \int_0^t f_j'(s, \cdot) ds $$

hence

$$ \begin{aligned} \int_0^t \partial_x v_j(s, \cdot) ds + \int_0^t \partial_x^3 v_j(s, \cdot) ds &= - \int_0^t v_j'(s, \cdot) ds + \int_0^t f_j'(s, \cdot) ds \\ &= v_j(0, \cdot) - v_j(t, \cdot) + f_j(t, \cdot) - f_j(0, \cdot). \end{aligned} $$

Using this in (3.28) we obtain

$$ \partial_t u_j(t, \cdot) + \partial_x u_j(t, \cdot) + \partial_x^3 u_j(t, \cdot) = \partial_x u_j^0 + \partial_x^3 u_j^0 + v_j(0, \cdot) + f_j(t, \cdot) - f_j(0, \cdot). $$

From this, (3.25) gives us

$$ \partial_t u_j(t, \cdot) + \partial_x u_j(t, \cdot) + \partial_x^3 u_j(t, \cdot) = f_j(t, \cdot). $$

Moreover, using (3.25) and (3.26) and the compatibility conditions, we deduce that $u$ is a classical solution to (3.1).
By the previous part we have $u \in C([0, T]; \mathbb{H}^3(\mathcal{T}))$. On the other hand, note that

$$ \partial_x^3 u_j(t, \cdot) = f_j(t, \cdot) - \partial_x u_j(t, \cdot) - \partial_t u_j(t, \cdot) = f_j(t, \cdot) - \partial_x u_j(t, \cdot) - v_j(t, \cdot) \in H^1(0, l_j) $$

so $u_j(t, \cdot) \in H^4(0, l_j)$. Moreover $\partial_x^4 u_j(t, \cdot) = \partial_x f_j(t, \cdot) - \partial_x^2 u_j(t, \cdot) - \partial_x v_j(t, \cdot)$, which gives us

$$ \|\partial_x^4 u_j(t, \cdot)\|_{L^2(0, l_j)}^2 \le C \left( \|\partial_x f_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|\partial_x^2 u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|\partial_x v_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right) $$

Thus

$$ (3.29) \qquad \|u_j(t, \cdot)\|_{H^4(0, l_j)}^2 \le C^2 \left( \|u_j(t, \cdot)\|_{H^3(0, l_j)}^2 + \|\partial_x f_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|\partial_x v_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right). $$

Integrating from 0 to T and using the regularities for f, u and v we get $u_j \in L^2(0, T; H^4(0, l_j))$ and, consequently, $u \in C([0, T]; \mathbb{H}^3(\mathcal{T})) \cap L^2(0, T; \mathbb{H}^4(\mathcal{T}))$. In addition, observe that

$$ \begin{aligned} \|u_j(t, \cdot)\|_{H^3(0, l_j)}^2 &\le C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|P_j u_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right) \\ &= C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|\partial_x u_j(t, \cdot) + \partial_x^3 u_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right) \\ &= C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|f_j(t, \cdot) - v_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right) \\ &\le C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|f_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|v_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right). \end{aligned} $$

Therefore

$$ (3.30) \qquad \|u_j(t, \cdot)\|_{H^3(0, l_j)}^2 \le C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|f_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|v_j(t, \cdot)\|_{L^2(0, l_j)}^2 \right). $$

and consequently

$$ \|u(t, \cdot)\|_{H^3(\mathcal{T})} \le C \left( \|u(t, \cdot)\|_{L^2(\mathcal{T})} + \|f(t, \cdot)\|_{L^2(\mathcal{T})} + \|v(t, \cdot)\|_{L^2(\mathcal{T})} \right). $$

Since $u, v \in C([0, T]; L^2(\mathcal{T}))$ and $f \in W^{1,1}(0, T; L^2(\mathcal{T})) \hookrightarrow C([0, T]; L^2(\mathcal{T}))$, passing to the maximum with $t \in [0, T]$, we obtain

$$ (3.31) \qquad \|u\|_{C([0,T]; \mathbb{H}^3(\mathcal{T}))} \le C \left( \|u\|_{C([0,T]; L^2(\mathcal{T}))} + \|f\|_{W^{1,1}(0,T; L^2(\mathcal{T}))} + \|v\|_{C([0,T]; L^2(\mathcal{T}))} \right). $$

Using (3.30) in (3.29) it follows that

$$ \|u_j(t, \cdot)\|_{H^4(0, l_j)}^2 \le C \left( \|u_j(t, \cdot)\|_{L^2(0, l_j)}^2 + \|f_j(t, \cdot)\|_{H^1(0, l_j)}^2 + \|v_j(t, \cdot)\|_{H^1(0, l_j)}^2 \right). $$

Taking the sum with respect to j and integrating from 0 to T we have for some $C = C(T) > 0$

$$ (3.32) \qquad \|u\|_{L^2(0, T; \mathbb{H}^4(\mathcal{T}))} \le C \left( \|u\|_{C([0, T]; L^2(\mathcal{T}))} + \|f\|_{L^2(0, T; \mathbb{H}^1(\mathcal{T}))} + \|v\|_{L^2(0, T; \mathbb{H}^1(\mathcal{T}))} \right). $$

Adding (3.32) and (3.31) it follows that

$$ (3.33) \qquad \|u\|_{\mathbb{B}_{3,T}} \le C \left( \|u\|_{\mathbb{B}_{0,T}} + \|f\|_{W^{1,1}(0,T;L^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} + \|v\|_{\mathbb{B}_{0,T}} \right). $$

Using Proposition (3.6) we have

$$ \|u\|_{\mathbb{B}_{0,T}} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\mathcal{T}))} \right) $$

and

$$ \|v\|_{\mathbb{B}_{0,T}} \le C \left( \| - \partial_x u^0 - \partial_x^3 u^0 + f(0, \cdot) \|_{L^2(\mathcal{T})} + \|g_0'\|_{H^{-\frac{1}{3}}(0,T)} \right. \\ \left. + \|g'\|_{[L^2(0,T)]^N} + \|p'\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|\partial_t f\|_{L^1(0,T;L^2(\mathcal{T}))} \right). $$

Thus, the following inequality holds

$$ (3.34) \qquad \begin{aligned} \|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|\partial_x u^0\|_{L^2(\mathcal{T})} + \|\partial_x^3 u^0\|_{L^2(\mathcal{T})} + \|f(0, \cdot)\|_{L^2(\mathcal{T})} \right. \\ \left. + \|f\|_{L^1(0, T; L^2(\mathcal{T}))} + \|\partial_t f\|_{L^1(0, T; L^2(\mathcal{T}))} + \|g_0\|_{H^{-\frac{1}{3}}(0, T)} + \|g_0'\|_{H^{-\frac{1}{3}}(0, T)} \right. \\ \left. + \|g\|_{[L^2(0, T)]^N} + \|g'\|_{[L^2(0, T)]^N} + \|p\|_{[H^{\frac{1}{3}}(0, T)]^N} + \|p'\|_{[H^{\frac{1}{3}}(0, T)]^N} \right). \end{aligned} $$
Due to embedding $W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \hookrightarrow C([0,T];\mathbb{L}^2(\mathcal{T}))$ we have

$$ (3.35) \qquad \|f(0, \cdot)\|_{\mathbb{L}^2(\mathcal{T})} \le \|f\|_{C([0,T];\mathbb{L}^2(\mathcal{T}))} \le C \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T}))}. $$

Now, due to embedding

$$ H^{\frac{3-1}{3}}(0,T) \hookrightarrow L^2(0,T) \hookrightarrow H^{-\frac{1}{3}}(0,T) \quad \text{and} \quad H^{\frac{3+1}{3}}(0,T) \hookrightarrow H^{\frac{1}{3}}(0,T) $$

the following inequalities hold

$$ (3.36) \qquad \|g_0\|_{H^{-\frac{1}{3}}(0,T)} \le C \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} \quad \text{and} \quad \|p\|_{[H^{\frac{1}{3}}(0,T)]^N} \le C \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N}. $$

Finally, since $\frac{3+1}{3} - 1 \ge 0$, $\frac{3-1}{3} > 0$, $\frac{3-1}{3} - 1 < 0$ and $\frac{3-1}{3} - \frac{1}{2} \notin \mathbb{Z}$ the operator $\frac{d}{dt}$ maps $H^{\frac{3+1}{3}}(0,T)$ (resp. $H^{\frac{3-1}{3}}(0,T)$) to $H^{\frac{1}{3}}(0,T)$ (resp. $H^{-\frac{1}{3}}(0,T)$) continuously. Thus,

$$ (3.37) \qquad \|g_0'\|_{H^{-\frac{1}{3}}(0,T)} \le c \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} \quad \text{and} \quad \|p'\|_{[H^{\frac{1}{3}}(0,T)]^N} \le c \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N}. $$

From (3.33)-(3.37) it follows that

$$ \|u\|_{\mathbb{B}_{3,T}} \le C \left( \|u^0\|_{\mathbb{H}^3(\mathcal{T})} + \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} \right. \\ \left. \qquad + \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N} + \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \right). $$

### iii. Since $u \in \mathbb{B}_{3,T}$ we have, in particular, $u \in L^2(0,T;\mathbb{H}^1(\mathcal{T}))$ and $\partial_t u = -\partial_x u - \partial_x^3 u + f \in L^2(0,T;\mathbb{H}^1(\mathcal{T}))$. Therefore, $u \in H^1(0,T;\mathbb{H}^1(\mathcal{T}))$. Furthermore,

$$ \begin{aligned} \|u\|_{H^1(0,T;\mathbb{H}^1(\mathcal{T}))}^2 &= \|u\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 + \|\partial_t u\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 \\ &= \|u\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 + \| - \partial_x u - \partial_x^3 u + f \|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 \\ &\le C \left( \|u\|_{\mathbb{B}_{3,T}}^2 + \|f\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 \right) \\ &\le C \left( \|u^0\|_{\mathbb{H}^3(\mathcal{T})} + \|g_0\|_{H^{\frac{3-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} \right. \\ & \qquad \left. + \|p\|_{[H^{\frac{3+1}{3}}(0,T)]^N} + \|f\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^2(0,T;\mathbb{H}^1(\mathcal{T}))} \right)^2. \end{aligned} $$

### iv. Since $u \in H^1(0,T;\mathbb{H}^1(\mathcal{T}))$ we have

$$ |\partial_t^k u_j(t, x)| \le \|\partial_t^k u_j(t, \cdot)\|_{L^\infty(0, l_j)} \le C \|\partial_t^k u_j(t, \cdot)\|_{H^1(0, l_j)}, \quad k = 0, 1. $$

Taking the square and integrating this inequality we see that $u_j(\cdot, x) \in H^1(0,T)$ with

$$ (3.38) \qquad \|u_j(\cdot, x)\|_{H^1(0,T)} \le C \|u\|_{H^1(0,T;\mathbb{H}^1(\mathcal{T}))}, $$

for every $x \in (0, l_j)$. Moreover

$$ (3.39) \qquad \partial_t u_j(\cdot, x) = v_j(\cdot, x) \in H^{\frac{1}{3}}(0, T), \quad \forall x \in (0, l_j). $$

From the theory of Sobolev spaces (see, for instance, [9]) it follows that $u_j \in L_x^\infty(0, T; H^{\frac{3+1-0}{3}}(0, T))$. On the other hand $u_j \in C([0, T]; H^3(0, l_j))$ implies that

$$ (3.40) \qquad |\partial_x^k u_j(t, x)| \le \|\partial_x^k u_j(t, \cdot)\|_{L^\infty(0, l_j)} \le C \|\partial_x^k u_j(t, \cdot)\|_{H^1(0, l_j)} \le C \|u_j(t, \cdot)\|_{H^3(0, l_j)}, \quad k = 1, 2, $$

that is, $\partial_x^k u_j(\cdot, x) \in L^2(0, T)$. In addition,

$$ (3.41) \qquad \partial_t \partial_x^k u_j(\cdot, x) = \partial_x v_j(\cdot, x) \in H^{\frac{1-k}{3}}(0, T), \quad k = 1, 2 $$

and therefore $\partial_x^k u_j \in L_x^\infty(0, T; H^{\frac{3+1-k}{3}}(0, T))$ for $k = 1, 2$. Finally, thanks to (3.38)-(3.41), the trace estimates hold, concluding the proof. $\square$
3.3. **The case** $s \in (0,3)$. Based on Propositions 3.6 and 3.7 and results from linear interpolation theory, we obtain the following corollary.

**Corollary 3.1.** Let $T > 0$ and $s \in (0,3)$ be. For any data

$$u^0 \in \mathbb{H}^s(\mathcal{T}), \quad g_0 \in H^{\frac{s-1}{3}}(0,T), \quad g \in [H^{\frac{s}{3}}(0,T)]^N, \quad \text{and} \quad p \in [H^{\frac{s+1}{3}}(0,T)]^N$$

satisfying the linear compatibility conditions and any $f \in W^{\frac{s}{3},1}(0,T; \mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T}))$, the problem (3.1) admits a unique solution $u \in \mathbb{B}_{s,T}$ which satisfies

$$\|u\|_{\mathbb{B}_{s,T}} \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right),$$

for some positive constant $C$. Moreover, $u \in H^{\frac{s}{3}}(0,T; \mathbb{H}^1(\mathcal{T}))$ with

$$\|u\|_{H^{\frac{s}{3}}(0,T;\mathbb{H}^1(\mathcal{T}))} \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right),$$

for some positive constant $C$.

In fact, we can prove the well-posedness for the slightly more general problem

$$(3.42) \quad \left\{ \begin{array}{ll} \displaystyle \partial_t u_j + \partial_x[(1+a_j)u_j] + \partial_x^3 u_j = f_j, & t \in (0,T), x \in (0,l_j), j=1, \dots, N, \\ u_j(t,0) = u_1(t,0), & t \in (0,T), j=1, \dots, N, \\ \displaystyle \sum_{j=1}^N \partial_x^2 u_j(t,0) = -\alpha u_1(t,0) + g_0(t), & t \in (0,T), \\ u_j(t,l_j) = p_j(t), & \partial_x u_j(t,l_j) = g_j(t), & t \in (0,T), j=1, \dots, N, \\ u_j(0,x) = u_j^0(x), & x \in (0,l_j), j=1, \dots, N. \end{array} \right. $$

To do this, it is convenient to recall the space

$$\mathbb{B}_{s,T}^* := \mathbb{B}_{s,T} \cap H^{\frac{s}{3}}(0, T; \mathbb{H}^1(\mathcal{T}))$$

with the norm of the sum. The following Lemma is a consequence of [26, Lemma 3] together with results of interpolation theory.

**Lemma 3.4.** Let $T > 0$ be given.

i. Let $s \ge 0$. There exists a constant $C > 0$ such that for any $u, v \in \mathbb{B}_{s,T}$ one have $\partial_x uv \in L^1(0, T; \mathbb{H}^s(\mathcal{T}))$ with

$$\int_0^T \|\partial_x uv\|_{\mathbb{H}^s(\mathcal{T})} dt \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{s,T}} \|v\|_{\mathbb{B}_{s,T}}.$$

ii. Let $s \in [0,3]$. There exists a constant $C > 0$ such that, for any $u, v \in \mathbb{B}_{s,T}$, we have $\partial_x uv \in L^{\frac{6}{6-s}}(0, T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T}))$ with

$$\int_0^T \|\partial_x uv\|_{\mathbb{H}^{\frac{s}{3}}(\mathcal{T})}^{\frac{6}{6-s}} \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{s,T}}^{\frac{6}{6-s}} \|v\|_{\mathbb{B}_{s,T}}^{\frac{6}{6-s}}.$$

iii. Let $s \in [0,3]$. There exists a constant $C > 0$ such that, for any $u, v \in \mathbb{B}_{s,T}^*$, we have $\partial_x uv \in W^{\frac{s}{3},1}(0, T; \mathbb{L}^2(\mathcal{T}))$ with

$$\|\partial_x uv\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T}))} \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{s,T}^*} \|v\|_{\mathbb{B}_{s,T}^*}.$$

The next result ensures the uniqueness of the problem (3.42).
**Proposition 3.8.** Let $T > 0$, $s \in [0, 3]$ and $a \in \mathbb{B}_{s,T}^*$ be. For any data

$$u^0 \in \mathbb{H}^s(\mathcal{T}), \quad g_0 \in H^{\frac{s-1}{3}}(0, T), \quad g \in [H^{\frac{s}{3}}(0, T)]^N, \quad \text{and} \quad p \in [H^{\frac{s+1}{3}}(0, T)]^N$$

satisfying the linear compatibility conditions and $f \in W^{\frac{s}{3},1}(0, T; \mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0, T; \mathbb{H}^{\frac{s}{3}}(\mathcal{T}))$, the problem (3.42) admits a unique solution $u \in \mathbb{B}_{s,T}^*$ which satisfies

$$\|u\|_{\mathbb{B}_{s,T}^*} \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right),$$

for some positive constant $C$ depending on $\|a\|_{\mathbb{B}_{s,T}^*}$.

*Proof.* Consider $\theta \in [0, T]$ and the map $\Gamma_a : \mathbb{B}_{s,\theta}^* \to \mathbb{B}_{s,\theta}^*$ where, for each $v \in \mathbb{B}_{s,\theta}^*$, $\Gamma_a v$ is the solution for

$$\left\{
\begin{array}{ll}
\partial_t u_j + \partial_x u_j + \partial_x^3 u_j = f_j - \partial_x(a_j v_j), & t \in (0, \theta), x \in (0, l_j), j = 1, \dots, N, \\
u_j(t, 0) = u_1(t, 0), & t \in (0, \theta), j = 1, \dots, N, \\
\displaystyle\sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0) + g_0(t), & t \in (0, \theta), \\
u_j(t, l_j) = p_j(t), & \partial_x u_j(t, l_j) = g_j(t), \quad t \in (0, \theta), j = 1, \dots, N, \\
u_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N.
\end{array}
\right.
$$

Thanks to Corollary 3.1 and Lemma 3.4, $\Gamma_a$ is well-defined. Moreover, defining

$$r = 2C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^1(\mathcal{T}))} \right)$$

we have

$$\begin{aligned}
\|\Gamma_a v\|_{\mathbb{B}_{s,\theta}^*} & \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,\theta)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \right. \\
& \qquad \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,\theta)]^N} + \|f - \partial_x(av)\|_{W^{\frac{s}{3},1}(0,\theta;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^1(\mathcal{T}))} \right) \\
& \le \frac{r}{2} + C \|\partial_x(av)\|_{W^{\frac{s}{3},1}(0,\theta;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^1(\mathcal{T}))} \\
& = \frac{r}{2} + C \left( \|\partial_x(av)\|_{W^{\frac{s}{3},1}(0,\theta;\mathbb{L}^2(\mathcal{T}))} + \|\partial_x(av)\|_{L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^1(\mathcal{T}))} \right) \\
& \le \frac{r}{2} + C \left( (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{s,\theta}^*} \|v\|_{\mathbb{B}_{s,\theta}^*} + (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{s,\theta}^*} \|v\|_{\mathbb{B}_{s,\theta}^*} \right) \\
& \le \frac{r}{2} + C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{s,T}^*} \cdot r.
\end{aligned}$$

Choosing $\theta \in (0, T)$ such that

$$C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{s,T}^*} < \frac{1}{2}$$

we have $\Gamma_a(B) \subset B$ where $B = \{v \in \mathbb{B}_{s,\theta}^*; \|v\|_{\mathbb{B}_{s,\theta}^*} \le r\}$. Moreover $\Gamma_a$ is a contraction on $B$ since

$$\|\Gamma_a v - \Gamma_a w\|_{\mathbb{B}_{s,\theta}^*} \le C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{s,T}^*} \|v - w\|_{\mathbb{B}_{s,\theta}^*}$$

Now, using the Banach fixed point theorem, it follows that $\Gamma_a$ has a fixed point $u \in B$, that is, $u$ solves (3.42) for $t \in [0, \theta]$ and $\|u\|_{\mathbb{B}_{s,\theta}^*} \le r$. Since $\theta$ does not depend on the data $(u^0, g_0, g, p, f)$,
using a standard continuation extension argument, the solution $u$ can be extended to the interval $[0, T]$ and the following estimate holds

$$ \|u\|_{\mathbb{B}_{s,T}^*} \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \\ + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right), $$

where the constant $C$ may depend on $\|a\|_{\mathbb{B}_{s,T}^*}$ through the number of sub-intervals of size $\theta$ needed to cover $[0, T]$. $\square$

**Remark 3.2.** By using induction, one can verify the results of this section for $s = 3k, k \in \mathbb{N}$. Indeed, assuming its validity for $s = 3k$ and using the same reasoning as in the Proposition 3.7, the same results are obtained for $s = 3(k+1)$. With this in hand, given any $s > 0$, consider $k \in \mathbb{N}$ such that $s \in [3k, 3(k+1)]$ and use the same argument of interpolation as in the Corollary 3.1 to get the results in $(3k, 3(k+1))$ and, consequently, for $s$.

### 4. LOCAL WELL-POSEDNESS OF THE NONLINEAR PROBLEM

Now, we are in a position to prove the first main result of the article, namely, the local well-posedness of the following problem

$$ (4.1) \qquad \left\{ \begin{array}{ll} \displaystyle \partial_t u_j + \partial_x[(a_j+1)u_j] + \partial_x^3 u_j + u_j \partial_x u_j = f_j, & t \in (0,T), x \in (0,l_j), \\ u_j(t,0) = u_1(t,0), & t \in (0,T), \\ \displaystyle \sum_{j=1}^N \partial_x^2 u_j(t,0) = -\alpha u_1(t,0) - \frac{N}{3}(u_1+a_1)^2(t,0) + g_0(t), & t \in (0,T) \\ u_j(t,l_j) = p_j(t), & \partial_x u_j(t,l_j) = g_j(t), & t \in (0,T), \\ u_j(0,x) = u_j^0(x), & x \in (0,l_j), & \end{array} \right. $$

where $j = 1, \dots, N$ and $a$ is a given function. Remember that: for $s \ge 0$ and $T > 0$ the space

$$ \mathcal{X}_{s,T} = \{ v \in \mathbb{B}_{s,T}^* ; \partial_x^k u_j \in L_x^\infty(0,l_j; H^{\frac{s+1-k}{3}}(0,T)), k = 0,1,2, j = 1,2,\dots,N \} $$

is a Banach space with the norm

$$ \|v\|_{\mathcal{X}_{s,T}} = \|v\|_{\mathbb{B}_{s,T}^*} + \sum_{k=0}^{2} \|\partial_x^k v\|_{\prod_{j=1}^{N} L_x^\infty(0,l_j; H^{\frac{s+1-k}{3}}(0,T))}. $$

Note that (1.1) is a particular case of (4.1) with $a = 0$.

*Proof of Theorem 1.1.* Fix $\theta \in (0, T)$ and define $\Gamma_a : \mathcal{X}_{s,\theta} \to \mathcal{X}_{s,\theta}$ putting, for each $u \in \mathcal{X}_{s,\theta}$, $\Gamma_a u$ as being the solution $v$ of

$$ \left\{ \begin{array}{ll} \displaystyle \partial_t v_j + \partial_x[(a_j+1)v_j] + \partial_x^3 v_j = -u_j \partial_x u_j + f_j, & t \in (0, \theta), x \in (0, l_j), j = 1, \dots, N, \\ v_j(t, 0) = v_1(t, 0), & t \in (0, \theta), j = 1, \dots, N, \\ \displaystyle \sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0) - \frac{N}{3}(u_1 + a_1)^2(t, 0) + g_0(t), & t \in (0, \theta), \\ v_j(t, l_j) = p_j(t), & \partial_x v_j(t, l_j) = g_j(t), & t \in (0, \theta), j = 1, \dots, N, \\ v_j(0, x) = u_j^0(x), & x \in (0, l_j), j = 1, \dots, N. & \end{array} \right. $$

From Proposition 3.8 we have

$$ \|\Gamma_a u\|_{\mathcal{X}_{s,\theta}} \le C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0 - \frac{N}{3}(u_1 + a_1)^2(\cdot, 0)\|_{H^{\frac{s-1}{3}}(0,\theta)} + \|g\|_{[H^{\frac{s}{3}}(0,\theta)]^N} \right. \\ \left. + \|p\|_{[H^{\frac{s+1}{3}}(0,\theta)]^N} + \|u \partial_x u\|_{W^{\frac{s}{3},1}(0,\theta;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} + \|f\|_{W^{\frac{s}{3},1}(0,\theta;\mathbb{L}^2(\mathcal{T})) \cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right). $$
Using Lemma 3.4 we obtain

$$ \|\partial_x uv\|_{W^{\frac{s}{3},1}(0,\theta;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \le C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{s,\theta}^*} \|u\|_{\mathbb{B}_{s,\theta}^*}. $$

Furthermore, Lemma 3.1 gives us

$$ \begin{aligned} \left\| \frac{N}{3} (u_1 + a_1)^2(\cdot, 0) \right\|_{H^{\frac{s-1}{3}}(0,\theta)} &\le C \theta^\alpha \frac{N}{3} \|(u_1 + a_1)(\cdot, 0)\|_{H^{\frac{s+1}{3}}(0,\theta)}^2 \\ &\le C \theta^\alpha \frac{N}{3} \|u + a\|_{\mathcal{X}_{s,\theta}}^2 \\ &\le \frac{4}{3} C N \theta^\alpha (\|u\|_{\mathcal{X}_{s,\theta}}^2 + \|a\|_{\mathcal{X}_{s,\theta}}^2), \end{aligned} $$

where $\alpha > 0$ is a constant. Thus,

$$ \|\Gamma_a u\|_{\mathcal{X}_{s,\theta}} \le C \left( \begin{aligned} & \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \\ & + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \\ & + C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|u\|_{\mathcal{X}_{s,\theta}}^2 + \frac{4}{3} C N \theta^\alpha \|u\|_{\mathcal{X}_{s,\theta}}^2 + \frac{4}{3} C N \theta^\alpha \|a\|_{\mathcal{X}_{s,\theta}}^2 \end{aligned} \right). $$

Defining

$$ r = 4C \left( \|u^0\|_{\mathbb{H}^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s-1}{3}}(0,T)} + \|g\|_{[H^{\frac{s}{3}}(0,T)]^N} \\ + \|p\|_{[H^{\frac{s+1}{3}}(0,T)]^N} + \|f\|_{W^{\frac{s}{3},1}(0,T;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,T;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right), $$

it follows that, for $u \in B = \{v \in \mathcal{X}_{s,\theta}; \|v\|_{\mathcal{X}_{s,\theta}} \le r\}$,

$$ (4.2) \qquad \|\Gamma_a u\|_{\mathcal{X}_{s,\theta}} \le \frac{r}{4} + C^2(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})r^2 + \frac{4}{3}C^2 N \theta^\alpha r^2 + \frac{4}{3}C^2 N \theta^\alpha \|a\|_{\mathcal{X}_{s,T}}^2. $$

On the other hand

$$ \begin{aligned} \|\Gamma_a u - \Gamma_a w\|_{\mathcal{X}_{s,\theta}} &\le C \left( \| -\frac{N}{3}(u_1 + a_1)^2(\cdot, 0) + \frac{N}{3}(w_1 + a_1)^2(\cdot, 0)\|_{H^{\frac{s-1}{3}}(0,\theta)} \right. \\ & \qquad \left. + \|u\partial_x u + w\partial_x w\|_{W^{\frac{s}{3},1}(0,\theta;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right) \\ &= C \left( \frac{N}{3} \|(u_1 + a_1)^2(\cdot, 0) - (w_1 + a_1)^2(\cdot, 0)\|_{H^{\frac{s-1}{3}}(0,\theta)} \right. \\ & \qquad \left. + \|u\partial_x u - w\partial_x w\|_{W^{\frac{s}{3},1}(0,\theta;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{s}{3}}(\mathcal{T}))} \right). \end{aligned} $$

Note that

$$ \begin{aligned} (u_1 + a_1)^2(\cdot, 0) - (w_1 + a_1)^2(\cdot, 0) &= \left[ ((u_1 + a_1) + (w_1 + a_1)) \cdot ((u_1 + a_1) - (w_1 + a_1)) \right] (\cdot, 0) \\ &= [(u_1 + w_1 + 2a_1) \cdot (u_1 - w_1)] (\cdot, 0). \end{aligned} $$

Hence, Lemma 3.1 provides

$$ \begin{aligned} & \frac{N}{3} \|(u_1 + a_1)^2(\cdot, 0) - (w_1 + a_1)^2(\cdot, 0)\|_{H^{\frac{s-1}{3}}(0,\theta)} \\ &\le \frac{N}{3} C \theta^\alpha \|(u_1 + w_1 + 2a_1)(\cdot, 0)\|_{H^{\frac{s+1}{3}}(0,\theta)} \cdot \|(u_1 - w_1)(\cdot, 0)\|_{H^{\frac{s+1}{3}}(0,\theta)} \\ &\le \frac{N}{3} C \theta^\alpha \|u + w + 2a\|_{\mathcal{X}_{s,\theta}} \|u - w\|_{\mathcal{X}_{s,\theta}} \\ &\le \frac{N}{3} C \theta^\alpha (\|u + w\|_{\mathcal{X}_{s,\theta}} + 2\|a\|_{\mathcal{X}_{s,\theta}}) \|u - w\|_{\mathcal{X}_{s,\theta}}. \end{aligned} $$
Moreover

$$u\partial_x u - w\partial_x w = (u - w)\partial_x u + w(\partial_x u - \partial_x w) = (u - w)\partial_x u + w\partial_x(u - w)$$

so, using Lemma 3.4, we have

$$\|u\partial_x u - w\partial_x w\|_{W^{\frac{6}{3},1}(0,\theta;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{6}{3}}(\mathcal{T}))} \le C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})\|u-w\|_{B_{s,\theta}^*}\|u\|_{B_{s,\theta}} \\ + C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})\|w\|_{B_{s,\theta}^*}\|u-w\|_{B_{s,\theta}^*},$$

that is,

$$\|u\partial_x u - w\partial_x w\|_{W^{\frac{6}{3},1}(0,\theta;L^2(\mathcal{T}))\cap L^{\frac{6}{6-s}}(0,\theta;\mathbb{H}^{\frac{6}{3}}(\mathcal{T}))} \le 2C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) (\|u\|_{B_{s,\theta}^*} + \|w\|_{B_{s,\theta}^*}) \|u-w\|_{B_{s,\theta}^*}.$$

Thus, we get

$$\|\Gamma_a u - \Gamma_a w\|_{X_{s,\theta}} \le \frac{1}{3}C^2 N \theta^\alpha (\|u+w\|_{X_{s,\theta}} + 2\|a\|_{X_{s,\theta}}) \|u-w\|_{X_{s,\theta}} \\ + 2C^2(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) (\|u\|_{X_{s,\theta}} + \|w\|_{X_{s,\theta}}) \|u-w\|_{X_{s,\theta}}.$$

Therefore, for $u, w \in B$ we have that

$$\|\Gamma_a u - \Gamma_a w\|_{X_{s,\theta}} \le \frac{1}{3}C^2 N \theta^\alpha (2r + 2\|a\|_{X_{s,\theta}}) \|u-w\|_{X_{s,\theta}} + 2C^2(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})2r\|u-w\|_{X_{s,\theta}} \\ \le \left(\frac{2}{3}C^2 N \theta^\alpha r + \frac{2}{3}C^2 N \theta^\alpha \|a\|_{X_{s,T}} + 4C^2(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})r\right) \|u-w\|_{X_{s,\theta}}.$$

Choosing $\theta \in (0, T)$ satisfying

$$C^2(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}})r < \frac{1}{12}, \quad \frac{4}{3}C^2 N \theta^\alpha r < \frac{1}{4}, \quad \frac{4}{3}C^2 N \theta^\alpha \|a\|_{X_{s,T}}^2 < \frac{r}{4} \quad \text{and} \quad \frac{2}{3}C^2 N \theta^\alpha \|a\|_{X_{s,T}} < \frac{1}{3},$$

it follows, from (4.2), that $\Gamma_a(B) \subset B$ and, thanks to the inequality (4.3), it follows that $\Gamma_a$ is a contraction. The Banach fixed point theorem ensures the existence of a fixed point $u \in B$ of $\Gamma_a$ which is the desired solution of (4.1) for $t \in (0, \theta)$, showing Theorem 1.1. $\square$

## 5. GLOBAL WELL-POSEDNESS OF THE NONLINEAR PROBLEM

We are now in a position to extend the previous section. Precisely, in this section, we ensure sharp global well-posedness for the system (1.1), closing the study of well-posedness in this graph structure.

### 5.1. The case $s = 0$: Global a priori estimate.
Here, we investigated energy estimates for the problem

$$\text{(5.1)} \qquad \left\{ \begin{array}{ll} \displaystyle \partial_t u_j + \partial_x u_j + \partial_x^3 u_j + u_j \partial_x u_j = f_j, & t \in (0, T), x \in (0, l_j), \\ u_j(t, 0) = u_1(t, 0), & t \in (0, T), \\ \displaystyle \sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0) - \frac{N}{3} u_1^2(t, 0) + g_0(t), & t \in (0, T) \\ u_j(t, l_j) = 0, \quad \partial_x u_j(t, l_j) = g_j(t), & t \in (0, T), \\ u_j(0, x) = u_j^0(x), & x \in (0, l_j), \end{array} \right.$$

in the case $s = 0$. Let us start with some a priori estimate that can be found in [4].

**Proposition 5.1.** [4, Proposition 2.4] Given $T > 0$, let $g_0 \in L^2(0, T)$, $g \in [L^2(0, T)]^N$, $f \in L^2(0, T; L^2(0, L))$ and $u^0 \in L^2(\mathcal{T})$. If $u \in B_{0,T}$ is a solution of (5.1) then

$$\|u\|_{B_{0,T}} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g_0\|_{L^2(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\mathcal{T}))} \right).$$

In addition, $u_1(\cdot, 0) \in L^2(0, L)$ with

$$\|u_1(\cdot, 0)\|_{L^2(0,T)} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g_0\|_{L^2(0,T)} + \|g\|_{[L^2(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\mathcal{T}))} \right).$$
Now, we study the problem

$$ (5.2) \quad \left\{
\begin{array}{ll}
\displaystyle \partial_t u_j + \partial_x u_j + \partial_x^3 u_j + u_j \partial_x u_j = -a_j \partial_x a_j - \partial_x (a_j u_j), & t \in (0, T), x \in (0, l_j), \\
u_j(t, 0) = u_1(t, 0), & t \in (0, T), \\
\displaystyle \sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha u_1(t, 0) - \frac{N}{3}(u_1 + a_1)^2(t, 0), & t \in (0, T) \\
u_j(t, l_j) = 0, \quad \partial_x u_j(t, l_j) = g_j, & t \in (0, T), \\
u_j(0, x) = u_j^0(x), & x \in (0, l_j),
\end{array}
\right. $$

where $a \in \mathcal{X}_{\lambda,T}$. The a priori estimates for this nonlinear system can be read as follows.

**Proposition 5.2.** Let $T > 0$, $\lambda > \frac{1}{2}$, $a \in \mathcal{X}_{\lambda,T}$, $u_0 \in L^2(\mathcal{T})$ and $g \in [L^2(0,T)]^N$ be given. If $u \in \mathbb{B}_{0,T}$ is a solution of (5.2) then

$$ \|u\|_{\mathbb{B}_{0,T}} \le C \left( \|u^0\|_{L^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} + \|a\|_{\mathbb{B}_{0,T}} \right). $$

*Proof.* If $u$ solves (5.2) in $(0,T)$ then, is also solution in $(0, \theta)$, for all $\theta \in (0, T)$. Observe that the central node conditions can be seen as

$$ \sum_{j=1}^{N} \partial_x^2 w_j(t, 0) = -\alpha u_1(t, 0) - \frac{N}{3} u_1^2(t, 0) - \frac{2N}{3} u_1(t, 0) a_1(t, 0) - \frac{N}{3} a_1^2(t, 0), \quad t \in (0, \theta). $$

Hence, Proposition 5.1 gives us

$$ \|u\|_{\mathbb{B}_{0,\theta}} \le C \left( \begin{array}{l} \|u^0\|_{L^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} + \| -a_j \partial_x a_j - \partial_x (a_j u_j) \|_{L^1(0,\theta;L^2(\mathcal{T}))} \\ \qquad + \| \frac{2N}{3} u_1(\cdot, 0) a_1(\cdot, 0) + \frac{N}{3} a_1^2(\cdot, 0) \|_{L^2(0,\theta)} \end{array} \right). $$

Note that by Lemma 3.4

$$ \| -a_j \partial_x a_j - \partial_x (a_j u_j) \|_{L^1(0,\theta;L^2(\mathcal{T}))} \le C(\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) (\|a\|_{\mathbb{B}_{0,T}}^2 + \|a\|_{\mathbb{B}_{0,T}} \|u\|_{\mathbb{B}_{0,\theta}}). $$

Moreover, $\lambda > \frac{1}{2}$ provides $H^{\frac{\lambda+1}{3}}(0,T) \hookrightarrow C([0,T])$, hence

$$ \begin{align*} \|u_1(\cdot, 0)a_1(\cdot, 0)\|_{L^2(0,\theta)}^2 &= \int_0^\theta |u_1(t, 0)a_1(t, 0)|^2 \le \|a_1(\cdot, 0)\|_{C([0,\theta])}^2 \int_0^\theta |u_1(t, 0)|^2 \\ &\le C \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)}^2 \int_0^\theta \|u_1(t, \cdot)\|_{L^\infty(0,l_j)}^2 \\ &\le C \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)}^2 \|u\|_{\mathbb{B}_{0,\theta}}^2. \end{align*} $$

Similarly,

$$ \|a_1^2(\cdot, 0)\|_{L^2(0,\theta)}^2 \le C \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)}^2 \|a_1\|_{L^2(0,\theta;H^1(0,l_j))}^2 $$

Joining the previous estimates

$$ \|u\|_{\mathbb{B}_{0,\theta}} \le C \left( \|u_0\|_{L^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} + (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{0,T}}^2 + (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{0,T}} \|u\|_{\mathbb{B}_{0,\theta}} \right. \\ \left. + \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)} \|u\|_{\mathbb{B}_{0,\theta}} + \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)} \|a\|_{\mathbb{B}_{0,T}} \right) $$

Choosing $\theta \in (0, T)$ such that

$$ (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{0,T}} + \|a_1(\cdot, 0)\|_{H^{\frac{\lambda+1}{3}}(0,\theta)} < \frac{1}{2} $$

we obtain

$$ \|u\|_{\mathbb{B}_{0,\theta}} \le C \left( \|u_0\|_{L^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} \right) + \frac{1}{2} \|a\|_{\mathbb{B}_{0,T}} + \frac{1}{2} \|u\|_{\mathbb{B}_{0,\theta}} $$
and, consequently,

$$ \|u\|_{\mathbb{B}_{0,\theta}} \le C \left( \|u_0\|_{\mathbb{L}^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} + \|a\|_{\mathbb{B}_{0,T}} \right), $$

Since $\theta$ does not depend on the data $u_0$, we can employ this inequality repeatedly (in a finite number of intervals of size $\theta$) to obtain the desired result in $[0, T]$. $\square$

Finally, we can state an energy estimate to problem (1.1), in the case when $s = 0$.

**Proposition 5.3.** Let $T > 0$ and $\lambda \ge 1$ be. Consider $u^0 \in \mathbb{L}^2(\mathcal{T})$, $g_0 \in H^{\frac{\lambda-1}{3}}(0,T)$, $g \in [L^2(0,T)]^N$, $p \in [H^{\frac{\lambda+1}{3}}(0,T)]^N$. If $u \in \mathbb{B}_{0,T}$ is a solution of (1.1), then

$$ \|u\|_{\mathbb{B}_{0,T}} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{\frac{\lambda-1}{3}}(0,T)} + \|g\|_{L^2(0,T)]^N} + \|p\|_{[H^{\frac{\lambda+1}{3}}(0,T)]^N} \right). $$

*Proof.* Write $u = v + w$ where $v$ and $w$ are solutions of

$$ \left\{ \begin{array}{ll} \partial_t v_j + \partial_x v_j + \partial_x^3 v_j = 0, & t \in (0, T), x \in (0, l_j), \\ v_j(t, 0) = v_1(t, 0), & t \in (0, T), \\ \displaystyle\sum_{j=1}^N \partial_x^2 u_j(t, 0) = -\alpha v_1(t, 0) + g_0(t), & t \in (0, T), \\ v_j(t, l_j) = p_j(t), & \partial_x v_j(t, l_j) = 0, \quad t \in (0, T), \\ v_j(0, x) = 0, & x \in (0, l_j), \end{array} \right. $$

and

$$ \left\{ \begin{array}{ll} \partial_t w_j + \partial_x[(v_j+1)w_j] + \partial_x^3 w_j w_j + w_j \partial_x w_j = -v_j \partial_x v_j, & t \in (0, T), x \in (0, l_j), \\ w_j(t, 0) = w_1(t, 0), & t \in (0, T), \\ \displaystyle\sum_{j=1}^N \partial_x^2 w_j(t, 0) = -\alpha w_1(t, 0) - \frac{N}{3}(w_1 + v_1)^2(t, 0), & t \in (0, T), \\ w_j(t, l_j) = 0, & \partial_x w_j(t, l_j) = g_j, \quad t \in (0, T), \\ w_j(0, x) = u_j^0, & x \in (0, l_j), \end{array} \right. $$

respectively. Propositions 3.8 and 5.2 gives us

$$ \|v\|_{\mathbb{B}_{0,T}} \le \|v\|_{\mathbb{B}_{\lambda,T}} \le C \left( \|g_0\|_{H^{\frac{\lambda-1}{3}}(0,T)} + \|p\|_{[H^{\frac{\lambda+1}{3}}(0,T)]^N} \right), $$

and

$$ \|w\|_{\mathbb{B}_{0,T}} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g\|_{[L^2(0,T)]^N} + \|v\|_{\mathbb{B}_{0,T}} \right), $$

so the result follows.

<figure><img src="image_2.png" /></figure>

We are finally in a position to establish the global well-posedness for (1.1) in the energy $L^2$-level, that is, Theorem 1.2 when $s = 0$.

**Theorem 5.1.** Let $T > 0$ and $\lambda \ge 1$ be. Given $u^0 \in \mathbb{L}^2(\mathcal{T})$, $g_0 \in H^{\frac{\lambda-1}{3}}(0,T)$, $g \in [L^2(0,T)]^N$, $p \in [H^{\frac{\lambda+1}{3}}(0,T)]^N$, the problem (1.1) has a unique solution $u \in \mathcal{X}_{0,T}$ which satisfies

$$ \|u\|_{\mathcal{X}_{0,T}} \le C \left( \|u^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{\frac{\lambda-1}{3}}(0,T)} + \|g\|_{L^2(0,T)]^N} + \|p\|_{[H^{\frac{\lambda+1}{3}}(0,T)]^N} \right), $$

for some positive constant $C > 0$ depending on $T$ and $l_1, ..., l_N$.

*Proof.* Thanks to Proposition 5.3, the radius $r$ in the proof of Theorem 1.1 can be chosen such that

$$ \|u(\theta, \cdot)\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0\|_{H^{\frac{\lambda-1}{3}}(0,T)} + \|g\|_{L^2(0,T)]^N} + \|p\|_{[H^{\frac{\lambda+1}{3}}(0,T)]^N} \le r. $$

Hence, the same radius $r$ and, consequently, the same $\theta$, can be used for data $u(\theta, \cdot)$, $g_0$, $g$ and $p$, in order to obtain a solution in $[\theta, 2\theta]$. In this way, the fixed point argument can be repeated on subintervals of length $\theta$ until the entire interval $[0, T]$ is covered. $\square$

<figure><img src="image_3.png" /></figure>
5.2. The case $s = 3$: Global a priori estimate. For simplicity we denote

$$ \|(u^0, g_0, g, p)\|_{s,\lambda} = \|u^0\|_{H^s(\mathcal{T})} + \|g_0\|_{H^{\frac{s+\lambda-1}{3}}(0,T)} + \|g\|_{H^{\frac{s}{3}}(0,T)}^N + \|p\|_{[H^{\frac{s+\lambda+1}{3}}(0,T)]^N} $$

and we reiterate that constants can be adjusted per line, with the dependencies deemed important made explicit. In addition we denote by $X_{s,T}^{\lambda}$ the set of s-compatible data

$$ u^0 \in \mathbb{H}^s(\mathcal{T}), g_0 \in H^{\frac{s+\lambda-1}{3}}(0,T), g \in [H^{\frac{s}{3}}(0,T)]^N \quad \text{and} \quad p \in [H^{\frac{s+\lambda+1}{3}}(0,T)]^N. $$

The first result in this section plays an auxiliary role, and its proof follows from a combination of a fixed-point argument and Proposition 3.6.

**Lemma 5.1.** Let $T > 0$, $v^0 \in \mathbb{L}^2(\mathcal{T})$, $\tilde{g}_0 \in H^{-\frac{1}{3}}(0,T)$, $\tilde{g} \in [L^2(0,T)]^N$, $\tilde{p} \in [H^{\frac{1}{3}}(0,T)]^N$, $f \in L^1(0,T;\mathbb{L}^2(\mathcal{T}))$ and $a \in \mathcal{X}_{0,T}$ be given. Then, the problem

$$ (5.3) \qquad \left\{ \begin{array}{ll} \displaystyle \partial_t v_j + \partial_x [(1+a_j)v_j] + \partial_x^3 v_j = f_j, & t \in (0,T), x \in (0, l_j), \\ v_j(t, 0) = v_1(t, 0), & t \in (0,T), \\ \displaystyle \sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0) - \frac{2N}{3} a_1(t, 0) v_1(t, 0) + \tilde{g}_0(t), & t \in (0,T), \\ v_j(t, l_j) = \tilde{p}_j(t), & \partial_x v_j(t, l_j) = \tilde{g}_j(t), \\ v_j(0, x) = v^0(x), & t \in (0,T), \\ x \in (0, l_j) \end{array} \right. $$

has a unique solution $v \in \mathcal{X}_{0,T}$ which satisfies

$$ \|v\|_{\mathcal{X}_{0,T}} \le C \|a\|_{\mathcal{X}_{0,T}} \left( \|v^0\|_{\mathbb{L}^2(\mathcal{T})} + \|\tilde{g}_0\|_{H^{-\frac{1}{3}}(0,T)} + \|\tilde{g}\|_{[L^2(0,T)]^N} + \|\tilde{p}\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;\mathbb{L}^2(\mathcal{T}))} \right), $$

for some positive constant $C = C(t) > 0$.

*Proof.* Consider $\theta \in [0, T]$ and the map $\Gamma_a : \mathcal{X}_{0,\theta} \to \mathcal{X}_{0,\theta}$ where, for each $w \in \mathcal{X}_{0,\theta}$, $\Gamma_a w$ is the solution of

$$ \left\{ \begin{array}{ll} \displaystyle \partial_t v_j + \partial_x v_j + \partial_x^3 v_j = f_j - \partial_x (a_j w_j), & t \in (0, \theta), x \in (0, l_j), j = 1, \dots, N, \\ v_j(t, 0) = v_1(t, 0), & t \in (0, \theta), j = 1, \dots, N, \\ \displaystyle \sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0) - \frac{2N}{3} a_1(t, 0) w_1(t, 0) + \tilde{g}_0(t), & t \in (0, \theta), \\ v_j(t, l_j) = \tilde{p}_j(t), & \partial_x v_j(t, l_j) = \tilde{g}_j(t), & t \in (0, \theta), j = 1, \dots, N, \\ v_j(0, x) = v_j^0(x), & x \in (0, l_j), j = 1, \dots, N. \end{array} \right. $$

Thanks to Corollary 3.1, Lemmas 3.1 and 3.4, we see that $\Gamma_a$ is well-defined. Moreover, defining

$$ r = 2C \left( \|v^0\|_{\mathbb{L}^2(\mathcal{T})} + \|\tilde{g}_0\|_{H^{-\frac{1}{3}}(0,T)} + \|\tilde{g}\|_{[L^2(0,T)]^N} + \|\tilde{p}\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;\mathbb{L}^2(\mathcal{T}))} \right) $$

we have

$$ \begin{aligned} \|\Gamma_a v\|_{\mathcal{X}_{0,\theta}} & \le C \left( \|v^0\|_{\mathbb{L}^2(\mathcal{T})} + \|\tilde{g}_0\|_{H^{-\frac{1}{3}}(0,\theta)} + \frac{2N}{3} \|a_1(t,0)w_1(t,0)\|_{H^{-\frac{1}{3}}(0,\theta)} + \|\tilde{g}\|_{L^2(0,T)}^N \right. \\ & \qquad \left. + \|\tilde{p}\|_{[H^{\frac{1}{3}}(0,\theta)]^N} + \|f - \partial_x(aw)\|_{L^1(0,\theta;\mathbb{L}^2(\mathcal{T}))} \right) \\ & \le \frac{r}{2} + C \left( \|\partial_x(aw)\|_{L^1(0,\theta;\mathbb{L}^2(\mathcal{T}))} + \frac{2N}{3} \|a_1(t,0)w_1(t,0)\|_{H^{-\frac{1}{3}}(0,\theta)} \right) \\ & \le \frac{r}{2} + C \left( (\theta^{\frac{1}{2}} + \theta^{\frac{1}{3}}) \|a\|_{\mathbb{B}_{0,\theta}} \|w\|_{\mathbb{B}_{0,\theta}} + \theta^{\alpha} \frac{2N}{3} \|a_1(\cdot,0)\|_{H^{\frac{1}{3}}(0,\theta)} \|w_1(\cdot,0)\|_{H^{\frac{1}{3}}(0,\theta)} \right) \\ & \le \frac{r}{2} + C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,\theta}} \|w\|_{\mathcal{X}_{0,\theta}}. \end{aligned} $$

Choosing $\theta \in (0, T)$ such that

$$ C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} < \frac{1}{2} $$
we have $\Gamma_a(B) \subset B$ where

$$B = \{v \in \mathcal{X}_{0,\theta}; \|v\|_{\mathcal{X}_{0,\theta}} \le r\}.$$

Moreover $\Gamma_a : B \to B$ is a contraction since

$$\|\Gamma_a v - \Gamma_a w\|_{\mathcal{X}_{0,\theta}} \le C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} \|v - w\|_{\mathcal{X}_{0,\theta}}.$$

Now, using the Banach fixed point theorem, it follows that $\Gamma$ has a fixed point $v \in B$, that is, $u$ solves (3.42) for $t \in [0, \beta]$. Furthermore,

$$\begin{aligned}
\|v\|_{\mathcal{X}_{0,\theta}} &= \|\Gamma_a v\|_{\mathcal{X}_{0,\theta}} \\
&\le C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} \|v\|_{\mathcal{X}_{0,\theta}} \\
&\le C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} r \\
&\le 2C^2 \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} \left( \|v^0\|_{L^2(\tau)} + \|\tilde{g}_0\|_{H^{-\frac{1}{3}}(0,T)} + \|\tilde{g}\|_{L^2(0,T)}^N \right. \\
&\qquad \left. + \|\tilde{p}\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\tau))} \right).
\end{aligned}$$

Since $\theta$ does not depend on the data $(u^0, g_0, g, p, f)$, using a standard continuation extension argument, the solution $u$ can be extended to the interval $[0, T]$ and the following estimate holds

$$\|v\|_{\mathcal{X}_{0,T}} \le C \left( \theta^{\frac{1}{2}} + \theta^{\frac{1}{3}} + \frac{2N}{3} \theta^{\alpha} \right) \|a\|_{\mathcal{X}_{0,T}} \left( \|v^0\|_{L^2(\tau)} + \|\tilde{g}_0\|_{H^{-\frac{1}{3}}(0,T)} + \|\tilde{g}\|_{L^2(0,T)}^N \right. \\
\left. + \|\tilde{p}\|_{[H^{\frac{1}{3}}(0,T)]^N} + \|f\|_{L^1(0,T;L^2(\tau))} \right),$$

for some positive constant $C := C(T)$ and $\alpha > 0$, giving the lemma. $\square$

**Proposition 5.4.** Let $T > 0$ and $\lambda \ge 1$ be given. If $u \in \mathcal{X}_{3,T}$ is a solution of (1.1) corresponding to the compatible data

$u^0 \in \mathbb{H}^3(\tau)$, $g_0 \in H^{\frac{3+\lambda-1}{3}}(0, T)$, $g \in [H^{\frac{3}{3}}(0, T)]^N$ and $p \in [H^{\frac{3+\lambda+1}{3}}(0, T)]^N$, then $u(t, x)$ is given by

$$\text{(5.4)} \qquad u_j(t, \cdot) = u_j^0 + \int_0^t v_j(s, \cdot) ds, \quad j = 1, \dots, N,$$

where $v(t, x)$ solves the problem

$$\text{(5.5)} \qquad \left\{
\begin{array}{ll}
\displaystyle \partial_t v_j + \partial_x [(1+u_j)v_j] + \partial_x^3 v_j = 0, & t \in (0, T), \quad x \in (0, l_j), \\
\displaystyle v_j(t, 0) = v_1(t, 0), & t \in (0, T), \\
\displaystyle \sum_{j=1}^N \partial_x^2 v_j(t, 0) = -\alpha v_1(t, 0) - \frac{2N}{3} u_1(t, 0) v_1(t, 0) + g_0'(t), & t \in (0, T) \\
\displaystyle v_j(t, l_j) = p_j'(t), \quad \partial_x v_j(t, l_j) = g_j'(t), & t \in (0, T), \\
\displaystyle v_j(0, x) = -\partial_x u_j^0(x) - \partial_x^3 u_j^0(x) - (u_j^0 \partial_x u_j^0)(x), & x \in (0, l_j).
\end{array}
\right.$$

Moreover

$$\text{(5.6)} \qquad \|u\|_{\mathbb{B}_{3,T}} \le C_1 \left( \|(u^0, g_0, g, p)\|_{0,\lambda} \left( \|u^0\|_{\mathbb{H}^3(\tau)} + \|g_0\|_{H^{\frac{3+\lambda-1}{3}}(0,T)} \right. \right. \\
\left. \left. + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} + \|p\|_{[H^{\frac{3+\lambda+1}{3}}(0,T)]^N} \right) \right),$$

where $C_1(r) = \alpha r^3 + \beta r^2 + \gamma r$ and $\alpha, \beta, \gamma > 0$ depends on $T$.
Proof. Suppose that $u \in \mathcal{X}_{3,T}$ is a corresponding solution of (1.1). Deriving the equalities in (1.1) with respect to $t$ we have

$$ egin{cases} 
\partial_t^2 u_j + \partial_x \partial_t u_j + \partial_x^3 \partial_t u_j + \partial_t u_j \partial_x u_j + u_j \partial_x \partial_t u_j = 0, \\
\partial_t u_j(t, 0) = \partial_t u_1(t, 0), \\
\sum_{j=1}^N \partial_x^2 \partial_t u_j(t, 0) = -\alpha \partial_t u_1(t, 0) - \frac{2N}{3} u_1(t, 0) \partial_t u_1(t, 0) + g_0'(t), \\
\partial_t u_j(t, l_j) = p_j'(t), \quad \partial_x \partial_t u_j(t, l_j) = g_j'(t). 
\end{cases} $$

Moreover

$$ egin{aligned} 
\partial_t u_j(0, x) &= -\partial_x u_j(0, x) - \partial_x^3 u_j(0, x) - u_j(0, x) \partial_x u_j(0, x) \\ 
&= -\partial_x u_j^0(x) - \partial_x^3 u_j^0(x) - u_j^0(x) \partial_x u_j^0(x). 
\end{aligned} $$

Setting $v_j = \partial_t u_j$ it follows that $v_j$ solves (5.5) and (5.4) holds. Furthermore, Lemma 5.1 provides

$$ (5.7) \qquad \|v\|_{\mathcal{X}_{0,T}} \le C \|u\|_{\mathcal{X}_{0,T}} \left( \| -\partial_x u_j^0 - \partial_x^3 u_j^0 - u_j^0 \partial_x u_j^0 \|_{\mathbb{L}^2(\mathcal{T})} \right. \\ 
\qquad \left. + \|g_0'\|_{H^{-\frac{1}{3}}(0,T)} + \|g'\|_{[L^2(0,T)]^N} + \|p'\|_{[H^{\frac{1}{3}}(0,T)]^N} \right). $$

We claim that $u \partial_x u \in C([0, T]; \mathbb{L}^2(\mathcal{T}))$ with

$$ (5.8) \qquad \|u \partial_x u\|_{C([0,T]; \mathbb{L}^2(\mathcal{T}))} \le M_1 (\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) $$

for some positive constant $M_1$ which depends on $\|(u^0, g_0, g, p)\|_{0,\lambda}$ and $T$. Indeed, observe that

$$ \partial_t (u \partial_x u) = \partial_t u \partial_x u + u \partial_x \partial_t u = v \partial_x u + u \partial_x v. $$

Then, from Lemma 3.4, we have

$$ \int_0^T \|u \partial_x u(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})} dt \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{0,T}} \|u\|_{\mathbb{B}_{0,T}} $$

and

$$ \int_0^T \| \partial_t (u \partial_x u)(t, \cdot) \|_{\mathbb{L}^2(\mathcal{T})} dt \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{0,T}} \|v\|_{\mathbb{B}_{0,T}}. $$

Consequently $u \partial_x u \in W^{1,1}(0, T; \mathbb{L}^2(\mathcal{T}))$ with

$$ \|u \partial_x u\|_{W^{1,1}(0,T;\mathbb{L}^2(\mathcal{T}))} \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|u\|_{\mathbb{B}_{0,T}} (\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}). $$

Using Theorem 5.1 and the embedding $W^{1,1}(0, T; \mathbb{L}^2(\mathcal{T})) \hookrightarrow C([0, T]; \mathbb{L}^2(\mathcal{T}))$ we have

$$ \|u \partial_x u\|_{C([0,T];\mathbb{L}^2(\mathcal{T}))} \le C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|(u^0, g_0, g, p)\|_{0,\lambda} (\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) $$

So, inequality (5.8) holds $M_1 = C(T^{\frac{1}{2}} + T^{\frac{1}{3}}) \|(u^0, g_0, g, p)\|_{0,\lambda}$, showing the claim.

Let us now prove (5.6). To do that, by [33, Lemma A.2]

$$ egin{aligned} 
\|u_j(t, \cdot)\|_{H^3(0, l_j)} &\le C \|P_j u_j(t, \cdot)\|_{L^2(0, l_j)} \\ 
&= C \|\partial_x u_j(t, \cdot) + \partial_x^3 u_j(t, \cdot)\|_{L^2(0, l_j)} \\ 
&= C \| - u_j \partial_x u_j(t, \cdot) - v_j(t, \cdot) \|_{L^2(0, l_j)} \\ 
&\le C \left( \|u_j \partial_x u_j(t, \cdot)\|_{L^2(0, l_j)} + \|v_j(t, \cdot)\|_{L^2(0, l_j)} \right). 
\end{aligned} $$

Therefore,

$$ egin{aligned} 
\|u(t, \cdot)\|_{\mathbb{H}^3(\mathcal{T})} &\le C (\|u u_x(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})} + \|v(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})}) \\ 
&\le C(1 + M_1)(\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) 
\end{aligned} $$

which gives us

$$ (5.9) \qquad \|u\|_{C([0,T];\mathbb{H}^3(\mathcal{T}))} \le C(1+M_1)(\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}). $$
On the other hand,

$$
(5.10) \qquad \partial_x^4 u(t, \cdot) = \partial_x (u \partial_x u)(t, \cdot) - \partial_x^2 u(t, \cdot) - \partial_x v(t, \cdot).
$$

However, note that

$$
\begin{align*}
\| \partial_x (u \partial_x u)(t, \cdot) \|_{\mathbb{L}^2(\mathcal{T})}^2 &= \| (\partial_x u \partial_x u + u \partial_x^2 u)(t, \cdot) \|_{\mathbb{L}^2(\mathcal{T})}^2 \\
&\le \| \partial_x u \partial_x u(t, \cdot) \|_{\mathbb{L}^2(\mathcal{T})}^2 + \| u \partial_x^2 u(t, \cdot) \|_{\mathbb{L}^2(\mathcal{T})}^2 \\
&= \sum_{j=1}^N \int_0^{l_j} |\partial_x u_j(t, x) \partial_x u_j(t, x)|^2 dx + \sum_{j=1}^N \int_0^{l_j} |u_j(t, x) \partial_x^2 u_j(t, x)|^2 dx \\
&\le \sum_{j=1}^N \| \partial_x u_j(t, \cdot) \|_{H^1(0, l_j)}^2 \| u_j(t, \cdot) \|_{H^1(0, l_j)}^2 + \sum_{j=1}^N \| u_j(t, \cdot) \|_{H^1(0, l_j)}^2 \| \partial_x^2 u_j(t, \cdot) \|_{L^2(0, l_j)}^2 \\
&\le 2 \sum_{j=1}^N \| u_j(t, \cdot) \|_{H^1(0, l_j)}^2 \| u_j(t, \cdot) \|_{H^3(0, l_j)}^2 \\
&\le 2 \| u \|_{C([0,T]; \mathbb{H}^3(\mathcal{T}))}^2 \| u(t, \cdot) \|_{\mathbb{H}^1(\mathcal{T})}^2.
\end{align*}
$$

Then $\partial_x (u \partial_x u) \in L^2(0, T; \mathbb{L}^2(\mathcal{T}))$ with

$$
\| \partial_x (u \partial_x u) \|_{L^2(0,T; \mathbb{L}^2(\mathcal{T}))} \le \sqrt{2} \| u \|_{C([0,T]; \mathbb{H}^3(\mathcal{T}))} \| u \|_{L^2(0,T; \mathbb{H}^1(\mathcal{T}))}.
$$

Using (5.9) and Theorem 5.1 it follows that

$$
(5.11) \qquad \begin{aligned}
\| \partial_x (u \partial_x u) \|_{L^2(0,T; \mathbb{L}^2(\mathcal{T}))} &\le C(1+M_1)(\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) \|u\|_{\mathbb{B}_{0,T}} \\
&\le C(1+M_1)(\|(u^0, g_0, g, p)\|_{0,\lambda}(\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}).
\end{aligned}
$$

Hence (5.10)

$$
\begin{align*}
\|u(t, \cdot)\|_{\mathbb{H}^4(\mathcal{T})}^2 &= \|u(t, \cdot)\|_{\mathbb{H}^3(\mathcal{T})}^2 + \|\partial_x^4 u(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})}^2 \\
&\le \|u(t, \cdot)\|_{\mathbb{H}^3(\mathcal{T})}^2 + C \left( \|\partial_x (u \partial_x u)(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})}^2 + \|\partial_x^2 u(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})}^2 + \|\partial_x v(t, \cdot)\|_{\mathbb{L}^2(\mathcal{T})}^2 \right)
\end{align*}
$$

and, integrating from 0 to T,

$$
\|u\|_{L^2(0,T;\mathbb{H}^4(\mathcal{T}))}^2 \le C \left( T \|u\|_{C([0,T];\mathbb{H}^3(\mathcal{T}))}^2 + \|\partial_x (u\partial_x u)\|_{L^2(0,T;\mathbb{L}^2(\mathcal{T}))}^2 + \|v\|_{L^2(0,T;\mathbb{H}^1(\mathcal{T}))}^2 \right).
$$

Due to (5.9) and (5.11) the inequality above becomes

$$
\|u\|_{L^2(0,T;\mathbb{H}^4(\mathcal{T}))}^2 \le C \left( T \left[ (1+M_1)(\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) \right]^2 \right. \\
\left. \qquad + \left[ (1+M_1)(\|u^0, g_0, g, p)\|_{0,\lambda} (\|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}}) \right]^2 + \|v\|_{\mathbb{B}_{0,T}}^2 \right)
$$

and, consequently

$$
(5.12) \quad \|u\|_{L^2(0,T;\mathbb{H}^4(\mathcal{T}))} \le C \left( (1+M_1)(T^{\frac{1}{2}} + \|(u^0, g_0, g, p)\|_{0,\lambda}) + 1 \right) \left( \|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}} \right).
$$

Thanks to (5.9) and (5.12) we have

$$
\|u\|_{\mathbb{B}_{3,T}} \le C \left( (1+M_1) \left(1+T^{\frac{1}{2}} + \|(u^0, g_0, g, p)\|_{0,\lambda}\right) + 1 \right) \left( \|u\|_{\mathbb{B}_{0,T}} + \|v\|_{\mathbb{B}_{0,T}} \right)
$$

and, from (5.7)

$$
\|u\|_{\mathbb{B}_{3,T}} \le C \left( (1+M_1) \left(1+T^{\frac{1}{2}} + \|(u^0, g_0, g, p)\|_{0,\lambda}\right) + 1 \right) \times \|u\|_{\chi_{0,T}} \left(1+\| -\partial_x u_j^0 - \partial_x^3 u_j^0 - u_j^0 \partial_x u_j^0 \|_{\mathbb{L}^2(\mathcal{T})} \right) \\
+ \|g'_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g'\|_{[L^2(0,T)]^N} + \|p'\|_{[H^{\frac{1}{3}}(0,T)]^N}.
$$
Using Theorem 5.3 and proceeding as in Proposition 3.7, we obtain

$$ \|u\|_{\mathbb{B}_{3,T}} \le C_1 \left( \|u^0\|_{\mathbb{H}^3(\tau)} + \|g_0\|_{H^{\frac{3+\lambda-1}{3}}(0,T)} + \|g\|_{[H^{\frac{3}{3}}(0,T)]^N} + \|p\|_{[H^{\frac{3+\lambda+1}{3}}(0,T)]^N} \right), $$

where

$$ C_1 = C \left( (1 + M_1) \left( 1 + T^{\frac{1}{2}} + \|(u^0, g_0, g, p)\|_{0,\lambda} \right) + 1 \right) \|(u^0, g_0, g, p)\|_{0,\lambda}. $$

Substituting $M_1 = C(T^{\frac{1}{2}} + T^{\frac{1}{3}})\|(u^0, g_0, g, p)\|_{0,\lambda}$ and rearranging the terms we see that

$$ C_1 = \alpha \|(u^0, g_0, g, p)\|_{0,\lambda}^3 + \beta \|(u^0, g_0, g, p)\|_{0,\lambda}^2 + \gamma \|(u^0, g_0, g, p)\|_{0,\lambda}, $$

where $\alpha, \beta$ and $\gamma$ depend on $T$. This gives the proof of the inequality (5.6). □

The next result ensures the global well-posedness in $H^s$, for $s = 3$.

**Theorem 5.2.** Let $T > 0$, $\lambda \ge 1$ and $(u^0, g_0, g, p) \in \mathcal{X}_{3,T}^\lambda$ be given. The problem (1.1) has a unique solution $u \in \mathcal{X}_{3,T}$ which satisfies

$$ \|u\|_{\mathcal{X}_{3,T}} \le C_1 \left( \|u^0\|_{\mathbb{H}^3(\tau)} + \|g_0\|_{H^{\frac{3+\lambda-1}{3}}(0,T)} + \|g\|_{H^{\frac{3}{3}}(0,T)]^N} + \|p\|_{[H^{\frac{3+\lambda+1}{3}}(0,T)]^N} \right), $$

where $C_1(r, u^0, g_0, g, p) = \|(u^0, g_0, g, p)\|_{0,\lambda}(\alpha r^3 + \beta r^2 + \gamma r)$ and $\alpha, \beta, \gamma > 0$ depending on $T$.

*Proof.* Thanks to Proposition 5.4, the radius $r$ in the proof of Theorem 1.1 can be chosen such that

$$ \|u(\theta, \cdot)\|_{\mathbb{H}^3(\tau)} + \|g_0\|_{H^{\frac{3+\lambda-1}{3}}(0,T)} + \|g\|_{[H^1(0,T)]^N} + \|p\|_{[H^{\frac{3+\lambda+1}{3}}(0,T)]^N} \le r. $$

Moreover, the corresponding parameter $\theta$ depends only on $\|(u^0, g_0, g, p)\|_{0,\lambda}$. Hence, the same radius $r$, and consequently the same $\theta$, can be used for the data $u(\theta, \cdot)$, $g_0$, $g$, and $p$, in order to obtain a solution on the interval $[\theta, 2\theta]$. Repeating this fixed-point argument on successive subintervals of length $\theta$ allows us to extend the solution to the entire interval $[0, T]$. □

5.3. **The case $s \in (0, 3)$: Global well-posedness result.** The results established for $s = 0$ and $s = 3$ can be extended to any $s \in (0, 3)$ by means of interpolation theory, for which we now present some preliminary considerations.

Let $B_0$ and $B_1$ be Banach spaces such that $B_1 \hookrightarrow B_0$. Given $0 < \theta < 1$ and $1 \le p \le +\infty$, we denote by

$$ B_{\theta,p} = [B_0, B_1]_{\theta,p} $$

the interpolation space obtained via the K-method (see, for instance, [34]). When $B_0$ and $B_1$ are Hilbert spaces and $p = 2$, we have

$$ [B_0, B_1]_{\theta,2} = [B_0, B_1]_{\theta}, $$

where $[B_0, B_1]_\theta$ denotes the interpolation space introduced in [27] or in [34] through the domains of self-adjoint operators.

For two pairs of indices $(\theta_1, p_1)$ and $(\theta_2, p_2)$, we write $(\theta_1, p_1) < (\theta_2, p_2)$ whenever one of the following conditions holds:

(1) $\theta_1 < \theta_2$;

(2) $\theta_1 = \theta_2$ and $p_1 > p_2$.

Whenever $(\theta_1, p_1) < (\theta_2, p_2)$, it follows that $B_{\theta_2, p_2} \hookrightarrow B_{\theta_1, p_1}$.

The following theorem, due to Bona and Scott [13], provides the first proof of global well-posedness of the initial value problem for the KdV equation posed on $\mathbb{R}$.

**Theorem 5.3.** Let $B_0^j$ and $B_1^j$ be Banach spaces with $B_1^j \hookrightarrow B_0^j$, $j = 1, 2$. Let $\sigma$ and $q$ such that $0 < \sigma < 1$ and $1 \le q \le +\infty$. Suppose $A$ is a mapping satisfying

i. $A: B_{\sigma,q}^1 \to B_0^2$ and for $f, g \in B_{\sigma,q}^1$

$$ \|Af - Ag\|_{B_0^2} \le C_0 \left( \|f\|_{B_{\sigma,q}^1} + \|g\|_{B_{\sigma,q}^1} \right) \|f - g\|_{B_0^1}; $$
ii. $A : B_1^1 \to B_1^2$ and for $h \in B_1^1$

$$ \|Ah\|_{B_1^2} \le C_1 (\|h\|_{B_{\sigma,q}^1}) \|h\|_{B_1^1}, $$

where $C_i : \mathbb{R}^+ \to \mathbb{R}^+$ are continuous non-decreasing functions, for $i = 0, 1$.

If $(\theta, p) \ge (\sigma, q)$, then $A$ maps $B_{\theta,p}^1$ into $B_{\theta,p}^2$ and for $f \in B_{\theta,p}^1$

$$ \|Af\|_{B_{\theta,p}^2} \le C (\|f\|_{B_{\sigma,q}^1}) \|f\|_{B_{\theta,p}^1}, $$

where, for $r > 0$, $C = C(r) = 4C_0(4r)^{1-\theta}C_1(3r)^{\theta}$.

With this in hand, we can obtain the global well-posedness for $s \in (0, 3)$.

*Proof of Theorem 1.2.* Given $s \in (0, 3)$, define

$$ B_0^1 = \mathbb{X}_{0,T}^{\lambda}, \quad B_1^1 = \mathbb{X}_{3,T}^{\lambda}. $$

Then, for $p = 2$ and $0 < \theta < 1$, we have

$$ B_{\theta,2}^1 = [B_0^1, B_1^1]_{\theta} = \mathbb{X}_{(1-\theta)\cdot 0+\theta\cdot 3,T}^{\lambda} = \mathbb{X}_{3\theta,T}^{\lambda}. $$

Analogously, setting $B_0^2 = \mathcal{X}_{0,T}$ and $B_1^2 = \mathcal{X}_{3,T}$, it follows that

$$ B_{\theta,2}^2 = \mathcal{X}_{3\theta,T}. $$

For $(u^0, g_0, g, p) \in \mathbb{X}_{3\theta,T}^{\lambda}$, define $A(u^0, g_0, g, p) = u$, where $u \in \mathcal{X}_{0,T}$ is the corresponding solution of (1.1). Given two elements $(u^0, g_0, g, p)$ and $(\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p}) \in \mathbb{X}_{3\theta,T}^{\lambda}$, let

$$ w = u - \tilde{u} = A(u^0, g_0, g, p) - A(\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p}). $$

Then $w$ satisfies the problem

$$ \left\{ \begin{array}{ll} \partial_t w_j + \partial_x [(1+z_j)w_j] + \partial_x^3 w_j = 0, & t \in (0,T), x \in (0, l_j), \\ w_j(t,0) = w_1(t,0), & t \in (0,T), \\ \displaystyle\sum_{j=1}^N \partial_x^2 w_j(t,0) = -\alpha w_1(t,0) - \frac{2N}{3}z_1(t,0)w_1(t,0) + g_0(t) - \tilde{g}_0(t), & t \in (0,T), \\ w_j(t, l_j) = p_j(t) - \tilde{p}_j(t), \quad \partial_x w_j(t, l_j) = g_j(t) - \tilde{g}_j(t), & t \in (0,T), \\ w_j(0,x) = u_j^0(x) - \tilde{u}_j^0(x), & x \in (0, l_j), \end{array} \right. $$

where $z = \frac{1}{2}(u + \tilde{u})$. Applying Lemma 3.1, we obtain

$$ \|w\|_{\mathcal{X}_{0,T}} \le C \|z\|_{\mathcal{X}_{0,T}} (\|u^0 - \tilde{u}^0\|_{\mathbb{L}^2(\mathcal{T})} + \|g_0 - \tilde{g}_0\|_{H^{-\frac{1}{3}}(0,T)} + \|g - \tilde{g}\|_{[L^2(0,T)]^N} + \|p - \tilde{p}\|_{[H^{\frac{1}{3}}(0,T)]^N}). $$

Hence, by Theorem 5.1,

$$ \|u - \tilde{u}\| \le C (\|(u^0, g_0, g, p)\|_{0,\lambda} + \|(\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p})\|_{0,\lambda}) \|(u^0, g_0, g, p) - (\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p})\|_{0,\lambda}, $$

where $C > 0$ is a constant. Since $\mathbb{X}_{3\theta,T}^{\lambda} \hookrightarrow \mathbb{X}_{0,T}^{\lambda}$, this inequality can be rewritten as

$$ \|u - \tilde{u}\| \le C (\|(u^0, g_0, g, p)\|_{3\theta,\lambda} + \|(\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p})\|_{3\theta,\lambda}) \|(u^0, g_0, g, p) - (\tilde{u}^0, \tilde{g}_0, \tilde{g}, \tilde{p})\|_{0,\lambda}, $$

showing that $A$ satisfies hypothesis (i) of Theorem 5.3. From Theorem 5.2, the map $A$ sends $\mathbb{X}_{3,T}^{\lambda}$ into $\mathcal{X}_{3,T}^{\lambda}$, and for every $(u^0, g_0, g, p) \in \mathbb{X}_{3\theta,T}^{\lambda}$,

$$ \begin{aligned} \|A(u^0, g_0, g, p)\|_{\mathcal{X}_{3,T}} &\le C_1 (\|(u^0, g_0, g, p)\|_{0,\lambda}) \|(u^0, g_0, g, p)\|_{3,\lambda} \\ &\le C_1 (\|(u^0, g_0, g, p)\|_{3\theta,\lambda}) \|(u^0, g_0, g, p)\|_{3,\lambda}, \end{aligned} $$

where $C_1 : \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous nondecreasing function. This shows that $A$ also satisfies hypothesis (ii) of Theorem 5.3. Therefore, $A$ maps $\mathbb{X}_{3\theta,T}^{\lambda}$ into $\mathcal{X}_{3\theta,T}^{\lambda}$, and for every $(u^0, g_0, g, p) \in \mathbb{X}_{3\theta,T}^{\lambda}$,

$$ \|A(u^0, g_0, g, p)\|_{\mathcal{X}_{3\theta,T}} \le C_1 (\|(u^0, g_0, g, p)\|_{3\theta,\lambda}) \|(u^0, g_0, g, p)\|_{3\theta,\lambda}. $$
Finally, choosing $\theta = \frac{s}{3}$ yields the desired result: for every $(u^0, g_0, g, p) \in \mathbb{X}_{s,T}^{\lambda}$, the unique solution $u \in \mathcal{X}_{0,T}$ of (1.1) actually belongs to $\mathcal{X}_{s,T}$ and satisfies

$$ \|u\|_{\mathcal{X}_{s,T}} \le C_1 (\|(u^0, g_0, g, p)\|_{s,\lambda}) \|(u^0, g_0, g, p)\|_{s,\lambda}. $$

This completes the proof. □

**Remark 5.1.** As in the linear case, an argument combining induction and interpolation ensures that the results obtained here remain valid for all $s > 3$.

## 6. CONCLUDING REMARKS

In this work, motivated by [14, 16], we established global well-posedness for the KdV equation on a star graph in the full range $s \in [0, 3]$, together with sharp compatibility conditions that guarantee the existence of classical and mild solutions. These results provide a unified functional framework that treats both linear and nonlinear models, clarifying several aspects that were previously left open in the literature. The techniques developed here, particularly the lifting procedure, energy estimates in Sobolev spaces, and the compatibility structure at the junction, pave the way for several perspectives. Among them, we mention: the study of more general graph structures, for instance, including loops [7] or other dispersive models like the Kawahara equation [19].

As we mentioned in the introduction, several control-theoretic questions naturally arise from our analysis. These topics are not treated in the present article and will be the focus of a forthcoming paper.

**Acknowledgments.** This paper was finalized during the first author's visit to several universities in France in 2025, thanks to the support of the CAPES-COFECUB program. The first author warmly thanks the Université de Lorraine, the Université de Tours, and the Université Sorbonne Paris Nord for their kind hospitality. The second author gratefully acknowledges the support of the Department of Mathematics at the Universidade Federal de Pernambuco, where this work was initiated, as well as the financial support provided by the Réseau Mathématique Franco-Brésilien during this visit.

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DEPARTAMENTO DE MATEMÁTICA, UNIVERSIDADE FEDERAL DE PERNAMBUCO, S/N CIDADE UNIVERSITÁRIA, 50740-545, RECIFE (PE), BRAZIL

Email address: roberto.capistranofilho@ufpe.br

Email address: jandeilson.santos@ufpe.br

UNIVERSITÉ DE LORRAINE, CNRS, INRIA, IECL, F-54000 NANCY, FRANCE.

Email address: hugo.parada@inria.fr