On the Controllability of a Free-Boundary Problem for 1D Heat Equation with Local and Nonlocal Nonlinearities

Abstract

This paper deals with the analysis of the internal control of a free-boundary problem for the 1D heat equation with local and nonlocal nonlinearities. We prove a local null controllability result with distributed controls, locally suported in space. The proof is based on Schauder’s fixed point theorem combined with some appropriate specific estimates.

1 Introduction

Let $T > 0$, $0< \kappa _1< \kappa _2< L_{*}< L_0 < B$ and $y_0 \in C^{2+\frac{1}{2}}([0,L_0])$ be given. For any function $L \in C^{1+\frac{1}{4}}([0,T])$ with $0< L_{*} \le L(t) \le B,\,\,t \in (0,T)$ we will set

$$\begin{aligned}Q_L := \{(x,t) : x \in (0, L(t)) \text{ and } t \in (0,T) \}.\end{aligned}$$

In this paper, we will investigate the null controllability properties of a free-boundary problem for the nonlinear 1D parabolic equation of the form

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_t - \beta \left( \int _{0}^{L(t)}y dx \right) y_{xx} + g(y,y_x) = v\mathrm {1}_{\omega }, &{} (x,t) \in Q_L,\\ y(0,t) = 0; y(L(t),t) = 0, &{} t \in (0,T),\\ y(x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. \end{aligned}$$

(1.1)

with the additional boundary condition

$$\begin{aligned} L(t)&= L_0 - \int _{0}^{t}\left[ \beta \left( \int _{0}^{L(s)} y(x,s)dx\right) y_x(L(s),s)\right] ds, \text{ therefore, } \nonumber \\ -L'(t)&= \beta \left( \int _{0}^{L(t)}y(x,t)dx\right) y_x(L(t),t), \end{aligned}$$

(1.2)

with $t \in (0,T)$ and $L_0 = L(0)$. Here $y=y(x,t)$ is the state, $v = v(x,t)$ is a control function that acts on the system at any time through a nonempty open set $\omega = (\kappa _1,\kappa _2)$, and $\mathrm {1}_{\omega }$ denotes the characteristic function of the $\omega$. Regarding the functions $\beta$ and g, we make the following assumptions:

• (A1) $\beta : \mathbb {R} \rightarrow \mathbb {R}$ is a $C^1$ function that possesses bounded derivatives and satisfies

  $$\begin{aligned} 0< \beta _0<\beta (r)<\beta _1< +\infty ,\,\,\forall r \in \mathbb {R}. \end{aligned}$$

• (A2) $g : \mathbb {R} \times \mathbb {R} \rightarrow \mathbb {R}$ is a $C^2$ function, with bounded derivatives, such that $g(0,0)=0$.

The main purpose of this paper is to prove the local null controllability result of (1.1). To accomplish this goal, let us recall the following classical controllability concept:

Definition 1.1

It will be said that (1.1) is null controllable at time T if there exist a control $v \in L^2(\omega \times (0,T))$, a function $L \in C^{1+\frac{1}{4}}([0,T])$ and an associated solution $y = y(x,t)$ satisfying (1.1), (1.2), and

$$\begin{aligned} y(x,T) = 0, \,x \in (0, L(T)), \end{aligned}$$

(1.3)

for each $y_0 \in C^{2+\frac{1}{2}}([0,L_0])$.

Definition 1.2

It will be said that (1.1) is approximately controllable at time T if there exist a control $v \in L^2(\omega \times (0,T))$, a function $L \in C^{1+\frac{1}{4}}([0,T])$ and an associated state $y = y(x,t)$ satisfying (1.1), (1.2) and

$$\begin{aligned} \Vert y(\cdot ,T) \Vert _{L^{2}(0,L(T))} \le \varepsilon , \end{aligned}$$

(1.4)

for any $y_{0} \in C^{2+\frac{1}{2}}([0,L_0])$ and $\varepsilon > 0$.

As we have already mentioned, we are interested in the local null controllability of (1.1), that is, in other words, the system (1.1) is said to be locally null controllable at any time $T>0$ if, there exists $\delta > 0$ such that, if $\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_0])}\le \delta$, there exists a triplet (L, v, y) with

$$\begin{aligned} \left\{ \begin{array}{ll} L \in C^{1+\frac{1}{4}}([0,T]), &{} L_{*} \le L(t) \le B, \\ v \in L^2(\omega \times (0,T)), \end{array}\right. \end{aligned}$$

(1.5)

satisfying (1.1), (1.2), and (1.3).

In mathematics, the expression free-boundary problem (FBP) refers to a problem in which one or several variables must be determined in different domains in space or in space-time. In a brief definition, we can say that a FBP is a boundary value problem defined in a domain that is not given a priori, therefore, a part of the unknown. If the domains are known, the problem reduces to solve equations, usually partial differential equations or ordinary differential equations. Free-boundary problems arise in various mathematical models that encompass applications ranging from Physics to Economics, Finances and biological phenomena where there is an extra effect of the environment. This effect in general deals with a qualitative change of the environment and an appearance of a phase transition; for example, ice to water, liquid to crystal, purchases to sales (assets), active to inactive (biology).

Free-boundary problems similar to (1.1)–(1.2) are connected to several interesting applications. We mention the following works:

• Tumor growth and other phenomena from mathematical biology; see Friedman [23, 24].

• Fluid-solid interaction; see Doubova and Fernández-Cara [12], Vázquez and Zuazua [36] and Liu et al. [31].

• Gas flow through porous media; see Aronson [2], Fasano [15] and Vázquez [35].

• Solidification and related Stefan problems; see Friedman [22].

• The analysis and computation of free surfaces flows; see Hermans [26, 27], Stoker [33, 34] and Wrobel and Brebbia [38].

In the last years, there are many works addressing controllability problems of linear and semilinear PDE’s. In particular, let us mention Fursikov and Imanuvilov [25], Barbu [3], Fernández-Cara and Zuazua [20], Doubova et al. [13] and Xu Liu and Xu Zhang [30] and the references therein in the context of bounded domains. In the context of the linear and semilinear PDE’s, we also mention the following articles [11, 14, 16, 17, 21, 29].

For parabolic free-boundary problems, controllability questions have been considered only in a few papers; see for instance Fernández-Cara et al. in [18] and Fernández-Cara and de Sousa in [19]. In both cited papers, the common point is that the main operator is linear and the free-boundary condition is given by

$$\begin{aligned} -L'(t)=y_x (L(t),t),\, t\in (0,T). \end{aligned}$$

(1.6)

In the present paper, with an extension in mind for another more realistic and interesting problems, we have considered a nonlocal term in the main part of the partial derivative operator. In this way, the free-boundary condition (1.2) becomes more general than condition (1.6). This is the main novelty in this work.

In addition, in [37] the authors studied the null controllability of a free-boundary problem for the quasi-linear 1D parabolic equation.

The nonlocal term in (1.1) appears naturally in some physical models. For example, they can arise in heat conduction in materials with memory, nuclear reactors, and population dynamics, for instance the bacteria in a container, the diffusion coefficients may depend on the total amount of individuals; see for instance [9, 39]. We also mention that, in the context of elasticity theory, terms in the form

$$\begin{aligned}&\displaystyle {\beta \Bigl ( \int _{0}^{L(t)}|y(x,t)|^{2} dx \Bigr )} \quad \text{ and } \quad \\&\displaystyle {\beta \Bigl ( \int _{0}^{L(t)}|y_{x}(x,t)|^{2} dx \Bigr )} \end{aligned}$$

appear, respectively, in Carrier and Kirchhoff equations. These equations arise in nonlinear vibration theory; see for instance [32].

Our main result is the following:

Theorem 1.1

Assume that $T > 0$ and $0< \kappa _1< \kappa _2< L_{*}< L_0 < B$. Under the previous assumptions on $\beta$ and g, the nonlinear system (1.1) is locally null-controllable.

For the proof of this theorem, we will first fix $\varepsilon >0$ and prove the existence of triplets $(L_{\varepsilon },y_{\varepsilon },v_{\varepsilon })$ that are uniformly bounded in an appropriate space and satisfy (1.1), (1.2) and (1.4). To this end, we will introduce a fixed point reformulation relying suitable linearized problems and we check that, if the initial data $y_0$ is sufficiently small, than the Schauder’s Fixed Point Theorem can be applied. Finally, we take limits as $\varepsilon \rightarrow 0$ and we see that, at least for a subsequence, we get convergence to a solution of (1.1), (1.2) and (1.3).

Throughout this paper, we denote by C a generic positive constant; for example: $C_1$, $C_2$, etc. are other positive (specific) constants; when it makes sense.

The paper is organized as follows: Sect. 2 is devoted to recall some known results and prove the approximate controllability of the linearized system (2.1). The Sect. 3 deals with the proof of Theorem 1.1. We present in Sect. 4 some open questions. In Appendix A, we sketch the proof of a Carleman estimate and in Appendix B we prove some relevant lemmas.

2 Analysis of the Controllability of the Linearized System in a Non-cylindrical Domain and Regularity Property

Given $L_0 > 0$, $T > 0$, and $0< \kappa _1< \kappa _2< L_{*}< L_0 < B$, and fixing $y_0 \in L^{2}(\Omega )$, assume that $L \in C^{1+\frac{1}{4}}([0,T])$ is a prescribed function satisfying

$$\begin{aligned}0< L_{*} \le L(t) \le B,\,\,t \in (0,T)\,.\end{aligned}$$

In this section we will prove that, for any $(\overline{y}, \overline{L})\in C^{2,1}_{x,t}(\overline{Q_{\overline{L}}}) \times C^{1+\frac{1}{4}}([0,T])$ , the linear system

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_t - \beta \left( \int _{0}^{\overline{L}(t)}\overline{y}(x,t) dx \right) y_{xx} + a(\overline{y},\overline{y}_x)y + b(\overline{y},\overline{y}_x)y_x = v\widetilde{1}_{\omega }, &{} (x,t) \in Q_{\overline{L}},\\ y(0,t) = 0; y(\overline{L}(t),t) = 0, &{} t \in (0,T),\\ y(x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. . \end{aligned}$$

(2.1)

is approximate controllable and also, that we can find a pair $(v_\varepsilon ,y_\varepsilon )$ satisfying $\Vert y_\varepsilon (\cdot ,T)\Vert _{L^2(0,\overline{L}(T))}\le \varepsilon$

for $\varepsilon >0$. In (2.1) we consider $L_*< \overline{L}(t) < B$, with $\overline{L}(0) = L_0$, $\widetilde{1}_{\omega } \in C^{\infty }_{0}(\omega )$, with $\widetilde{1}_{\omega }=1$ in $\omega _{1} \subset \subset \omega$ and $Q_{\overline{L}} = \{(x,t) : x \in (0, \overline{L}(t)) \text{ and } t \in (0,T) \}$.

We can verify that, for every $v \in L^{2}(\omega \times (0,T))$ and every $y_0 \in L^{2}(0,L_0)$, there exists a unique solution y to (2.1), with $y \in L^{2}(0,T;H^{1}_{0}(0,\overline{L}(t)))$ and $y_{t} \in L^{2}(0,T;H^{-1}(0,\overline{L}(t)))$. Consequently,

$$\begin{aligned} y \in C^{0}([0,T];L^{2}(0,\overline{L}(t)))\,. \end{aligned}$$

To this end, an appropriate change of variable $w(\zeta ,t):= y(x,t)$ allows us to rewrite (2.1) as a similar problem for a parabolic PDE of the form

$$\begin{aligned} \displaystyle w_{t} - \frac{\beta (t)}{L^2(t)}w_{\zeta \zeta } + \widetilde{a}(\zeta , t)w + \widetilde{b}(\zeta , t)w_{\zeta } = G(\zeta ,t), \end{aligned}$$

in a cylindrical domain, with bounded coefficients $\widetilde{a}$, $\widetilde{b} \in C^{1/2,1/4}_{\xi ,t}(\overline{Q})$ and square-integrable right hand side $G \in L^{2}(Q)$, where $Q:= (0,1)\times (0,T)$ (see Sect. 2.4 for the definition of $\zeta$ in Control Regularity).

As usual, the controllability of (2.1) is closely related to the properties of the solutions to the associated adjoint state. In this case, the adjoint system is

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\varphi _t - \beta \left( t\right) \varphi _{xx} + a(x,t)\varphi + b(x,t)\varphi _x = F(x,t), &{} (x,t) \in Q_L,\\ \varphi (0,t) = 0; \varphi (\overline{L}(t),t) = 0, &{} t \in (0,T),\\ \varphi (x,T) = \varphi ^T(x), &{} x \in (0,\overline{L}(T)), \end{array}\right. \end{aligned}$$

(2.2)

where $F \in L^{2}(Q_{\overline{L}})$ and $\varphi ^T \in L^2(0,\overline{L}(T))$.

Next, we sketch the points used in the proof of the null controllability of the linearized system using an observability estimate. First, we use a global Carleman estimate satisfied by the solutions of (2.2). Second, this estimate allows us to establish an observability estimate. Third, we prove the approximate controllability of (2.1) by using the observability estimate. Finaly, we establish a regularity property for the pair control-state in a certain Hölder space.

2.1 A Carleman Estimate for the Solutions to (2.2)

In this section, we will recall some Carleman estimates satisfied by the solutions to (2.2).

Denote by $\Sigma _{\overline{L}} \!:=\! \{(x,t) : x \!=\! 0 \text{ or } x \!=\! \overline{L}(t) \text{ with } 0< t < T\}$ the lateral boundary of $Q_{\overline{L}}$.

The following technical result, due to Fursikov and Imanuvilov [25], is fundamental:

Lemma 2.1

Let $\omega _0$ be an arbitrary non-empty open subdomain with $\overline{\omega }_0 \subset \omega$. There exists a function $\alpha _0 \in C^{1}\left( \overline{Q}_{\overline{L}}\right)$ with $\alpha _{0,xx} \in C^{0}\left( \overline{Q}_{\overline{L}}\right)$ such that

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle \alpha _0(x,t) = 0 &{} \forall (x,t) \in \Sigma _{\overline{L}},\\ |\alpha _{0,x}| > 0 &{} \text{ in } \overline{Q}_{\overline{L}} \backslash (\omega _0 \times (0,T)), \\ \displaystyle \alpha _0(x,t) = 1 - \frac{x- b}{\overline{L}(t) - b} &{} \forall x \in (b,\overline{L}(t)) \text{ and } \forall t \in [0,T]. \end{array}\right. \end{aligned}$$

The proof of this lemma can be found in [18] (see Lemma 2.1).

Let $\alpha _1$ and $\gamma$ be real functions, and let $\xi$ and $\alpha$ be weights defined by

$$\begin{aligned} \displaystyle \alpha _1(x,t):= & {} \alpha _0(x,t) + 1, \qquad \gamma (t) := t^k(T - t)^k,\nonumber \\ \displaystyle \xi (x,t):= & {} \frac{e^{\lambda \alpha _1(x,t)}}{\gamma (t)}, \qquad \alpha (x,t) := \displaystyle \frac{e^{2\lambda \parallel \alpha _1\parallel _{\infty } } - e^{\lambda \alpha _1(x,t)}}{\gamma (t)}, \end{aligned}$$

(2.3)

where $\lambda > 0$ and $k \ge 2$ are real numbers.

The next result is a Carleman estimate for the solutions to adjoint system (2.2).

Theorem 2.1

Let $\alpha _0$ and $\gamma$, $\xi$, $\alpha$ be as defined as in Lemma 2.1 and (2.3), respectively. There exist $\lambda _0$, $s_0$, and $C_0$, positive constants, only depending on $a, a' , L_*,\omega$, and T, such that, for any $s \ge s_0$, $\lambda \ge \lambda _0$, any $F \in L^2(Q_{\overline{L}})$, and any $\varphi ^T \in L^2(0,\overline{L}(T))$, one has the following inequality:

$$\begin{aligned}&\displaystyle \iint _{Q_{\overline{L}}} e^{-2s\alpha }\left[ (s\xi )^{-1}(|\varphi _t|^2 + |\varphi _{xx}|^2 ) + (s\xi )\lambda ^2|\varphi _x|^2 + (s\xi )^3\lambda ^4|\varphi |^2 \right] dxdt \nonumber \\&\qquad + s\lambda \int _{0}^T \left[ e^{-2s\alpha (\overline{L}(t),t)}\xi (\overline{L}(t),t)|\varphi _x(\overline{L}(t),t)|^2 + e^{-2s\alpha (0,t)}\xi (0,t)|\varphi _x(0,t)|^2 \right] dt \nonumber \\&\qquad \quad \le C\left( \iint _{Q_{\overline{L}}} e^{-2s\alpha } |F|^2 dxdt + \iint _{\omega \times (0,T)} e^{-2s\alpha }(s\xi )^3\lambda ^4|\varphi |^2dxdt \right) , \end{aligned}$$

(2.4)

where $\varphi$ is the corresponding solution to (2.2).

The proof is given in Appendix.

2.2 An Observability Inequality

Consider the homogeneous adjoint system:

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\varphi _t - \beta \left( t\right) \varphi _{xx} + a(x,t)\varphi + b(x,t)\varphi _x = 0, &{} (x,t) \in Q_{\overline{L}},\\ \varphi (0,t) = 0; \varphi (\overline{L}(t),t) = 0, &{} t \in (0,T),\\ \varphi (x,T) = \varphi ^T(x), &{} x \in (0,\overline{L}(T)), \end{array}\right. \end{aligned}$$

(2.5)

where $\varphi ^T \in L^2(0,\overline{L}(T))$.

Now, we will prove the observability inequality for weak solutions of the adjoint system (2.5). Observe that it is a consequence of the previous Carleman inequality.

Proposition 2.1

There exists $C > 0$, only depending on $\Vert a \Vert _{L^\infty (Q_{\overline{L}}) }$, $\Vert b \Vert _{L^\infty (Q_{\overline{L}}) }$,$\Vert \overline{L}'\Vert _{\infty }$,$L_{*}, B, \omega$, and T, such that, for any $\varphi ^T \in L^2(0,\overline{L}(T))$, the associated solution to (2.5) satisfies

$$\begin{aligned} \int _{0}^{L_0}|\varphi (x,0)|^2dx \le C\iint _{\omega \times (0,T)} e^{-2s_0\alpha }\xi ^3|\varphi |^2dxdt. \end{aligned}$$

(2.6)

Proof

Let us take $\lambda = \lambda _0$ and $s = s_0$ in (2.4). Then

$$\begin{aligned} \int \int _{Q_{\overline{L}}} e^{-2s_0\alpha } \xi ^3 |\varphi |^2 \,dx\,dt\, \le C \int \int _{\omega \times (0,T)} e^{-2s_0\alpha } \xi ^3 |\varphi |^2 \,dx\,dt \end{aligned}$$

and, consequently,

$$\begin{aligned}&\int _{T/4}^{3T/4} \int _0^{\overline{L}(t)} |\varphi |^2 \,dx\,dt\,\nonumber \\&\quad \le C \int _{T/4}^{3T/4} \int _0^{\overline{L}(t)} e^{-2s_0\alpha } \xi ^3 |\varphi |^2\,dx\,dt\,\nonumber \\&\quad \le C \int \int _{\omega \times (0,T)} e^{-2s_0\alpha } \xi ^3|\varphi |^2 \,dx\,dt. \end{aligned}$$

(2.7)

On the other hand, multiplying the PDE in (2.5) by $\varphi$ and integrating in $(0, \overline{L}(t))$, we get the following identity

$$\begin{aligned}&\displaystyle -\frac{1}{2}\frac{d}{dt}\left( \int _0^{\overline{L}(t)} |\varphi |^2 \,dx \right) \\&\qquad \quad + \frac{1}{2}\overline{L}'(t) \, |\varphi (\overline{L}(t), t)|^2 + \int _0^{\overline{L}(t)} \beta (t)|\varphi _x|^2 \, dx \\&\qquad \displaystyle = - \int _0^{\overline{L}(t)} a|\varphi |^2 \,dx - \int _0^{\overline{L}(t)} b\varphi \varphi _x \,dx, \quad \quad \forall t \in (0,T). \end{aligned}$$

Since $\varphi (\overline{L}(t), t) \equiv 0$, we deduce that, for a small $\epsilon > 0$,

$$\begin{aligned}&\displaystyle -\frac{1}{2}\frac{d}{dt}\left( \int _0^{\overline{L}(t)} |\varphi (x,t)|^2 \,dx \right) + \int _0^{\overline{L}(t)} \beta (t)|\varphi _x|^2 \, dx \\&\quad \displaystyle \le c_a\int _0^{\overline{L}(t)} |\varphi |^2 \,dx + \frac{c_b}{4\epsilon } \int _0^{\overline{L}(t)} |\varphi |^2 \,dx + \epsilon \int _0^{\overline{L}(t)} |\varphi _x|^2 \,dx, \end{aligned}$$

and this implies that

$$\begin{aligned} -\frac{d}{dt}\left( \int _0^{\overline{L}(t)} |\varphi (x,t)|^2 \,dx \right) \le 2M\int _0^{\overline{L}(t)} |\varphi (x,t)|^2 \,dx . \end{aligned}$$

(2.8)

Integrating (2.8) in time, we have

$$\begin{aligned} \Vert \varphi (\cdot ,0)\Vert ^2_{L^2(0,L(0))} \le e^{2MT} \Vert \varphi (\cdot ,t)\Vert ^2_{L^2(0,\overline{L}(t))}, \quad \forall t \in (0,T) \end{aligned}$$

and

$$\begin{aligned} \frac{T}{2}\int _0^{L(0)} |\varphi (x,0)|^2 \,dx \le \int _{T/4}^{3T/4} \int _0^{\overline{L}(t)} e^{2MT}|\varphi (x,t)|^2 \,dx\,dt. \end{aligned}$$

(2.9)

From (2.7) and (2.9), we find (2.6) and the proof is done. $\square$

2.3 Approximate Controllability of the Linearized System

The approximate controllability of system (2.1) is obtained as a consequence of the observability inequality seen in Proposition 2.1.

Theorem 2.2

For any $y_0 \in L^{2}(0,L_0)$ and $\varepsilon >0$, there exists pairs $(v_\epsilon ,y_\epsilon )$, with $y_\epsilon \in C^0([0,T];L^{2}(0,\overline{L}(t)))$ and $v_\epsilon \in L^2(\omega \times (0,T))$, satisfying (2.1) and

$$\begin{aligned} \Vert y_\epsilon (\cdot ,T)\Vert _{L^2(0,\overline{L}(T))}\le \varepsilon \,. \end{aligned}$$

(2.10)

Furthermore, $v_\epsilon$ can be found such that

$$\begin{aligned} \Vert v_\epsilon \Vert _{L^2(\omega \times (0,T))} \le C_1\Vert y_0 \Vert _{L^{2}(0,L_0)}, \end{aligned}$$

(2.11)

where $C_1$ depends on $\Vert a \Vert _{L^\infty (Q_{\overline{L}}) }$, $\Vert b \Vert _{L^\infty (Q_{\overline{L}}) }$, $\Vert \overline{L}'\Vert _{\infty }$, $L_*$, $\omega$, and T.

Proof

Thus, let $y_0 \in L^2(0,L_0)$ and $\varepsilon > 0$ be given and let us introduce the functional $J_\varepsilon (\cdot ; a,b,L)$ with

$$\begin{aligned}&J_\varepsilon (\varphi ^T ; a,b,\overline{L}) = \frac{1}{2}\iint _{\omega \times (0,T)} e^{-2s_0\alpha } \xi ^3|\varphi |^2dxdt\\&\quad + \varepsilon \Vert \varphi ^T\Vert _{L^2(0,\overline{L}(T))} + \left( \varphi (\cdot ,0),y_0 \right) _{L^2(0,L_0)}, \end{aligned}$$

for all $\varphi ^T \in L^2(0,\overline{L}(T))$.

Here, $\varphi$ is the solution of (2.2) associated to $\varphi ^{T}$. Using (2.6), it is relatively easy to check that $J_\varepsilon (\cdot ;a,b,\overline{L})$ is strictly convex, continuous, and coercive in $L^2(0,\overline{L}(T))$, so it possesses a unique minimum $\widehat{\varphi }^{T}_\varepsilon \in L^2(0,\overline{L}(T))$, whose associated solution is denoted by $\widehat{\varphi }_\varepsilon$.

Let us now introduce the control $v_\varepsilon = e^{-2s_0\alpha } \xi ^3 \widehat{\varphi }_\varepsilon \widetilde{1}_{\omega }$ and denote by $y_\varepsilon$ the solution to (2.1) associated to $v_\varepsilon$, that is

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_{\varepsilon , t} - \beta \left( \int _{0}^{\overline{L}(t)}\overline{y}(x,t) dx \right) y_{\varepsilon , xx} + a(\overline{y},\overline{y}_x)y_\varepsilon + b(\overline{y},\overline{y}_x)y_{_\varepsilon , x} = v_\varepsilon \widetilde{1}_{\omega }, &{} (x,t) \in Q_{\overline{L}},\\ y_\varepsilon (x,t) = 0; y_\varepsilon (\overline{L}(t),t) = 0, &{} t \in (0,T),\\ y_\varepsilon (x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. . \nonumber \\ \end{aligned}$$

(2.12)

Then, either $\widehat{\varphi }^T_\varepsilon = 0$ or we can differentiate the functional at $\widehat{\varphi }^T_\varepsilon$ and obtain a necessary condition to reach a minimum at $\widehat{\varphi }^T_\varepsilon$:

$$\begin{aligned} \left\{ \begin{array}{l} \displaystyle \iint _{\omega \times (0,T)}e^{-2s_0\alpha } \xi ^3\widehat{\varphi }_\varepsilon \varphi dxdt+ \varepsilon \left( \frac{\widehat{\varphi }^T_\varepsilon }{\Vert \widehat{\varphi }^T_\varepsilon \Vert _{L^2(0,L_0)}},\varphi ^T \right) _{L^2(0,L(T))} +\left( \widehat{\varphi }_{\varepsilon }(\cdot ,0 ),y_0 \right) _{L^2(0,L_0)} = 0 \\ \\ \forall \varphi ^T \in L^2(0,\overline{L}(T)) \end{array}\right. . \nonumber \\ \end{aligned}$$

(2.13)

Furthermore, from the inequality $J_\varepsilon (\widehat{\varphi }_\varepsilon ^T) \le J_\varepsilon (0) = 0$ and (2.6), we deduce that

$$\begin{aligned}&\displaystyle \frac{1 }{2}\iint _{\omega \times (0,T)}e^{-2s_0\alpha }\xi ^3|\widehat{\varphi }_\varepsilon |^2dxdt +\varepsilon \Vert \widehat{\varphi }^T_\varepsilon \Vert _{L^2(0,\overline{L}(T))} \le -\left( \widehat{\varphi }_\varepsilon (0),y_0 \right) _{L^2(0, \overline{L}(T))}\\&\quad \displaystyle \le \frac{1}{4}\iint _{\omega \times (0,T)}e^{-2s_0\alpha }\xi ^3|\widehat{\varphi }_\varepsilon |^2dxdt + \Vert y_0 \Vert ^2_{L^2(0,L_0)} \end{aligned}$$

and, consequently,

$$\begin{aligned} \Vert e^{s_0\alpha }\xi ^{-3/2}v_\varepsilon \Vert ^2_{L^2(\omega \times (0,T))}\le & {} \iint _{\omega \times (0,T)}e^{-2s_0\alpha }\xi ^3|\widehat{\varphi }_\varepsilon |^2dxdt\nonumber \\&+\varepsilon \Vert \widehat{\varphi }^T_\varepsilon \Vert ^2_{L^2(0,L(T))} \le 4C\Vert y_0 \Vert ^2_{L^2(0,L_0)}. \end{aligned}$$

(2.14)

So,

$$\begin{aligned} \Vert e^{s_0\alpha }\xi ^{-3/2}v_\varepsilon \Vert _{L^2(\omega \times (0,T))} \le 2C\Vert y_0 \Vert _{L^2(0,L_0)}. \end{aligned}$$

(2.15)

It is not difficulty to check that $v_{\varepsilon }=e^{-2s_0\alpha } \xi ^3 \widehat{\varphi }_\varepsilon \widetilde{1}_{\omega }$ is a solution of

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\varphi _{\varepsilon , t} - \beta \left( t\right) \varphi _{\varepsilon , xx} + a(x,t)\varphi _\varepsilon + b(x,t)\varphi _{\varepsilon , x} = 0, &{} (x,t) \in Q_L,\\ \varphi _\varepsilon (0,t) = 0; \varphi _\varepsilon (L(t),t) = 0, &{} t \in (0,T),\\ \varphi _\varepsilon (x,T) = -\frac{1}{\varepsilon }y_\varepsilon (x,T), &{} x \in (0,L(T)). \end{array}\right. ; \end{aligned}$$

(2.16)

Multiplying both sides of the first equation of (2.12) by $\widehat{\varphi }_{\varepsilon }$ and integrating it on $Q_{\overline{L}}$, we obtain

$$\begin{aligned} \frac{1}{\varepsilon } \int _{0}^{\overline{L}(T)} |y_{\varepsilon }(T)|^{2}dx +\iint _{\omega \times (0,T)}e^{-2s_0 \alpha } \xi ^3 |\widehat{\varphi }_{\varepsilon }|^{2} dxdt =-(\widehat{\varphi }_{\varepsilon }(\cdot ,0), y_0)_{L^{2}(0,L_0)}\,, \end{aligned}$$

which implies that

$$\begin{aligned} \frac{1}{\varepsilon } \int _{0}^{\overline{L}(T)} |y_{\varepsilon }(T)|^{2}dx +\iint _{\omega \times (0,T)}e^{-2s_0 \alpha } \xi ^3 |\widehat{\varphi }_{\varepsilon }|^{2} dxdt\le C\Vert y_0 \Vert ^{2}_{L^{2}(0,L_0)} \end{aligned}$$

(2.17)

and therefore

$$\begin{aligned} \Vert y_{\varepsilon }\Vert _{L^{2}(0,\overline{L}(T))} \le \varepsilon \,C\Vert y_0 \Vert _{L^{2}(0,L_0)}\,. \end{aligned}$$

(2.18)

As $\varepsilon \rightarrow 0$, $y_{\varepsilon }(T) \rightarrow 0$ in $L^{2}(0,\overline{L}(T))$ and therefore $v_{\varepsilon }$ is an approximated null-control for (2.1).

The proof of Theorem 2.2 is completed. $\square$

An immediate consequence of Theorem 2.2 is the following null controllability:

Corollary 2.1

For any $y_0 \in L^{2}(0,L_0)$, there exists pairs (v, y), with $y \in C^0([0,T];L^{2}(0,\overline{L}(t)))$ and $v \in L^2(\omega \times (0,T))$, satisfying (2.1) and $y(x,T)= 0$, $\forall x \in (0,\overline{L}(T))$. Furthermore, v can be found such that

$$\begin{aligned} \Vert v \Vert _{L^2(\omega \times (0,T))} \le C_2\Vert y_0 \Vert _{L^{2}(0,L_0)}, \end{aligned}$$

(2.19)

where $C_2$ depends on $\Vert a \Vert _{L^\infty (Q_{\overline{L}}) }$, $\Vert b \Vert _{L^\infty (Q_{\overline{L}}) }$, $\Vert \overline{L}'\Vert _{\infty }$, $L_*$, $\omega$, and T.

2.4 A Regularity Property

Let (v, y) be a control-state pair furnished by Corollary 2.1. We will see in this section that, for some $\theta \in [0,1)$ only depending on $\Vert \overline{L}'\Vert _{\infty }$, $L_*$, $L_*$, $\omega$ and T, one has

$$\begin{aligned} y \in C^{2+\theta ,1+\theta /2}_{x,t}\left( \overline{Q_{\overline{L}}}\right) \quad \text{ and } \quad v \in C^{\theta ,\frac{\theta }{2}}_{x,t}(\overline{Q_{\overline{L}}}) \end{aligned}$$

(2.20)

where $C^{m+\theta ,(m+\theta )/2}_{x,t}\left( \overline{Q_{\overline{L}}}\right)$ is the space of functions $u:\overline{Q_{\overline{L}}}\rightarrow \mathbb {R}$ such that $D_{t}^{r}D_{x}^{s}u(x,t)$ is continuous in $\overline{Q_{\overline{L}}}$ for $2r+s < m +\theta$, with m a non-negative integer, and the norm is given by

$$\begin{aligned}&\displaystyle \Vert u \Vert _{C_{x,t}^{m+\theta ,(m+\theta )/2}\left( \overline{Q_{\overline{L}}}\right) } \displaystyle = \sum _{2r+s\le m} \Vert D_{t}^{r}D_{x}^{s}u(x,t) \Vert _{\infty }\\&\quad \displaystyle + \sum _{2r+s = m}\left( \sup _{(x,t),(x',t')\in \overline{Q_{\overline{L}}}}\frac{|D_{t}^{r}D_{x}^{s}u(x,t) - D_{t}^{r}D_{x}^{s}u(x',t')|}{|x - x'|^\theta + |t - t'|^{\theta /2}}\right) <+\infty .\\ \end{aligned}$$

In the sequel, we recall some relevant Lemmas (their proofs will be given in Appendix B):

Lemma 2.2

Assume that $\beta \in C_b^1(\mathbb {R})$, $g\in C_b^2(\mathbb {R}^2)$, $\overline{L} \in C^{1+\frac{1}{4}}([0,T])$, and $\overline{y}\in C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})$.

• (i) If $\displaystyle { \beta (t) = \beta \left( \int _0^{\overline{L}(t)} \overline{y}(x,t) \ dx \right) }$, then $\beta \in C^{\theta }([0,T])$, for all $0 \le \theta < 1$;

• (ii) If $\displaystyle { a(x,t) = \int _0^1 \frac{\partial {g}}{\partial {s}}( \lambda \overline{y}(x,t), \lambda \overline{y}_x(x,t)) \ d\lambda }$ and $\displaystyle { b(x,t) = \int _0^1 \frac{\partial {g}}{\partial {p}}( \lambda \overline{y}(x,t),} \lambda \overline{y}_x(x,t)) \ d\lambda$, then one has $a, b \in C_{x,t}^{\theta , \theta /2}(\overline{Q_{\overline{L}}})$, for all $0 \le \theta < 1$;

• (iii) If $\widetilde{a}(\xi ,t) = a(L(t)\xi ,t)$, then $\widetilde{a} \in C_{\xi ,t}^{\theta , \theta /2}(\overline{Q}),$ for all $0 \le \theta < 1$. If we consider the function $\displaystyle { \widetilde{b}(\xi ,t) = b(L(t)\xi ,t) - \frac{\xi {L}'}{{L}} }$, with $L\in C^{1+ \gamma /2}([0,T])$, then $\widetilde{b} \in C_{\xi ,t}^{\gamma , \gamma /2}(\overline{Q})$, for all $0 \le \gamma < 1$.

Lemma 2.3

If $\overline{L} \in C^{1+\frac{1}{4}}([0,T])$, $\overline{y}\in C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})$ and $\beta \in C_b^1(\mathbb {R})$, then the function $L: [0,T] \rightarrow \mathbb {R}$ given by

$$\begin{aligned} L(t) = L_0 - \int _{0}^{t}\left[ \beta \left( \int _{0}^{\overline{L}(s)} y(x,s)dx\right) y_x(\overline{L}(s),s) \right] ds \end{aligned}$$

is such that $L\in C^{1+\theta }([0,T])$, with $0< \theta < 1$.

Now, we will apply a standard technique that leads to the construction of a control-state with the required regularity (similar ideas were used in [30]). In this way, let us detail the following steps:

Step 1 Control regularity

From Sect. 2.3, $v_{\varepsilon } := e^{-2s_0\alpha }\xi ^{3}\widehat{\varphi }_{\varepsilon }\widetilde{1}_{\omega }$, where $\widehat{\varphi }_{\varepsilon }$ is solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\widehat{\varphi }_{\varepsilon , t} - \beta (t)\widehat{\varphi }_{\varepsilon , xx} + a(x,t)\widehat{\varphi }_{\varepsilon } + b(x,t)\widehat{\varphi }_{\varepsilon , x} = 0, &{} (x,t) \in Q_L,\\ \widehat{\varphi }_{\varepsilon }(0,t) = 0; \widehat{\varphi }_{\varepsilon }(L(t),t) = 0, &{} t \in (0,T),\\ \widehat{\varphi }_{\varepsilon }(x,T) = \widehat{\varphi }^{T}_{\varepsilon }(x), &{} x \in (0,L_0), \end{array}\right. . \end{aligned}$$

(2.21)

Let us introduce:

$$\begin{aligned} \displaystyle \widetilde{\alpha }(x,t) = \min _{{\mathop {0< t< T}\limits ^{x \in (0,L(t))}}}\left\{ e^{2\lambda _0\Vert \alpha _1 \Vert _{\infty }} - e^{\lambda _0\alpha _1(x,t)} \right\} \, \text{ and } \, \, \overline{\alpha }(x,t) = \max _{{\mathop {0< t < T}\limits ^{x \in (0,L(t))}}}\left\{ e^{2\lambda _0\Vert \alpha _1 \Vert _{\infty }} - e^{\lambda _0\alpha _1(x,t)} \right\} , \end{aligned}$$

where $\alpha _1$ was given in (2.3). Let $\delta$ be a real number such that $0 < \delta \le 1/4$ with $2\widetilde{\alpha }- (1+\delta )\overline{\alpha } > 0$.

Thus, one has

$$\begin{aligned} \alpha \le \frac{\overline{\alpha }}{\gamma } \le (1 + \delta )\frac{\overline{\alpha }}{\gamma }. \end{aligned}$$

Let us consider $\displaystyle z_{\varepsilon } = e^{-s_0(1 + \delta )\frac{\overline{\alpha }}{\gamma }}\frac{1}{\gamma ^3}\widehat{\varphi }_{\varepsilon }$, such that $z_{\varepsilon }$ satisfies:

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -z_{\varepsilon , t} - \beta (t)z_{\varepsilon , xx} + a(x,t)z_{\varepsilon } + b(x,t)z_{\varepsilon , x} = F_{\varepsilon }, &{} (x,t) \in Q_L,\\ z_{\varepsilon }(0,t) = 0; z_{\varepsilon }(L(t),t) = 0, &{} t \in (0,T),\\ z_{\varepsilon }(x,T) = 0, &{} x \in (0,L_0), \end{array}\right. , \end{aligned}$$

(2.22)

where $F_{\varepsilon } = \left( e^{-s_0(1 + \delta )\frac{\overline{\alpha }}{\gamma }}\frac{1}{\gamma ^3}\right) _{t}\widehat{\varphi }_{\varepsilon }$. Observe that

$$\begin{aligned} \displaystyle |F_{\varepsilon }|^2 = \left| \left( e^{-s_0(1 + \delta )\frac{\overline{\alpha }}{\gamma }}\frac{1}{\gamma ^3}\right) _{t}\widehat{\varphi }_{\varepsilon } \right| ^2 = \displaystyle \left| e^{-s_0(1 + \delta ) \frac{\overline{\alpha }}{\gamma }} \left( \frac{\gamma '}{\gamma ^2}\frac{1}{\gamma ^3} - \frac{3\gamma '}{\gamma ^4} \right) \widehat{\varphi }_{\varepsilon }\right| ^2 , \end{aligned}$$

Then, we obtain

$$\begin{aligned} \begin{array}{lcl} \displaystyle |F_{\varepsilon }|^2 &{} \le &{} \displaystyle C\left| e^{-2s_0\frac{\overline{\alpha }}{\gamma }} \left( \frac{1}{\gamma ^5} e^{-s_0\delta \frac{\overline{\alpha }}{\gamma }}\right) \widehat{\varphi }_{\varepsilon }\right| ^2 \\ &{} \le &{} \displaystyle C\left| e^{-2s_0\alpha }\xi ^3 \left( e^{-s_0\delta \frac{\overline{\alpha }}{\gamma }} \frac{1}{\gamma ^2}\right) \widehat{\varphi }_{\varepsilon } \right| ^2 \\ &{} \le &{} \displaystyle C_1 e^{-2s_0\alpha }\xi ^3|\widehat{\varphi }_{\varepsilon }|^2 , \end{array} \end{aligned}$$

whence

$$\begin{aligned} \iint _{Q_L} |F_{\varepsilon }|^2dxdt \le C_1\iint _{Q_L} e^{-2s_0\alpha }\xi ^3|\widehat{\varphi }_{\varepsilon }|^2dxdt. \end{aligned}$$

From Carleman inequality and (2.15), we have

$$\begin{aligned} \iint _{Q_L} |F_{\varepsilon }|^2dxdt \le C_2\iint _{\omega \times (0,T)} e^{-2s_0\alpha }\xi ^3|^2\widehat{\varphi }_{\varepsilon }|^2dxdt \le C \Vert y_0 \Vert ^2_{L^2(0,L_0)}. \end{aligned}$$

Let us now consider the following change of variables:

$$\begin{aligned} \displaystyle \zeta= & {} \frac{x}{L(t)}, \quad \widetilde{a}(\zeta ,t) = a(L(t)\zeta ,t), \quad \widetilde{b}(\zeta ,t) = b(L(t)\zeta ,t) - \zeta \frac{L'(t)}{L(t)},\\ \displaystyle \quad \eta _{\varepsilon }(\zeta ,t)= & {} z_{\varepsilon }(x,t) \quad \text{ and } \quad G_{\varepsilon }(\zeta ,t) = F_{\varepsilon }(x,t). \end{aligned}$$

We have $\eta _{\varepsilon }(\zeta ,t)$ is well defined in $Q:= (0,1)\times (0,T)$ and, moreover,

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle -\eta _{\varepsilon , t} - \frac{\beta (t)}{L^2(t)}\eta _{\varepsilon , \zeta \zeta } + \widetilde{a}(\zeta ,t)\eta _{\varepsilon } + \widetilde{b}(\zeta ,t)\eta _{\varepsilon , \zeta } = G_{\varepsilon }(\zeta ,t), &{} (\zeta ,t) \in Q,\\ \eta _{\varepsilon }(0,t) = 0; \eta _{\varepsilon }(1,t) = 0, &{} t \in (0,T),\\ \eta _{\varepsilon }(\zeta ,T) = 0, &{} \zeta \in (0,1), \end{array}\right. . \end{aligned}$$

(2.23)

Since $\displaystyle G_{\varepsilon }(\zeta ,t)\in L^2\left( (0,1)\times (0,T) \right)$, then $\displaystyle \eta _{\varepsilon }(\zeta ,t)\in W^{2,1}_{2}\left( (0,1)\times (0,T) \right)$, and

$$\begin{aligned} \displaystyle \Vert \eta _{\varepsilon }\Vert _{W^{2,1}_{2}\left( Q \right) } \le C\Vert G_{\varepsilon } \Vert _{L^2(Q)} \le C\Vert F_{\varepsilon } \Vert _{L^2(Q_L)}.\end{aligned}$$

Since $W^{2,1}_{2}\left( (0,1)\times (0,T) \right) \hookrightarrow C_{\zeta ,t}^{1/2,1/4}\left( [0,1]\times [0,T] \right)$ is a continuous embedding, then

$$\begin{aligned} \displaystyle \Vert \eta _{\varepsilon }\Vert _{C_{\zeta ,t}^{1/2,1/4}\left( \overline{Q} \right) } \le C\Vert F_{\varepsilon } \Vert _{L^2(Q_L)} \le C\Vert y_0 \Vert _{L^2(0,L_0)}. \end{aligned}$$

Therefore, from the change of variables, $\displaystyle z_{\varepsilon } \in C_{x,t}^{1/2,1/4}\left( \overline{Q}\right)$, and $\displaystyle \Vert z_{\varepsilon }\Vert _{C_{x,t}^{1/2,1/4}\left( \overline{Q}\right) }\le C\Vert y_0\Vert _{L^2(0,L_0)}$, thus, $v_{\varepsilon } = e^{-2s_0\alpha }\xi ^{3}\gamma ^3e^{s_0(1+\delta )\frac{\overline{\alpha }}{\gamma }}z_{\varepsilon }\widetilde{1}_{\omega } \in C_{x,t}^{1/2,1/4}\left( \overline{Q_{\overline{L}}}\right)$. Observe that:

$$\begin{aligned} e^{-2s_0\alpha }\xi ^3\gamma ^3e^{s_0(1+\delta )\frac{\overline{\alpha }}{\gamma }}\le \xi ^3\gamma ^3e^{\frac{-s_0}{\gamma }\left( 2\widetilde{\alpha }-(1+\delta )\overline{\alpha }\right) }, \end{aligned}$$

where the right-hand side of the last inequality is bounded, and consequently,

$$\begin{aligned} \displaystyle \Vert v_{\varepsilon }\Vert _{C_{x,t}^{1/2,1/4}\left( \overline{Q_{\overline{L}}} \right) } \le C\Vert y_0 \Vert _{L^2(0,L_0)}. \end{aligned}$$

Step 2 State regularity

If $\beta \in C^1(\mathbb {R})$, $L \in C^{1+\frac{1}{4}}\left( [0,T]\right)$, and $\overline{y} \in C_{x,t}^{2,1}\left( \overline{Q_{\overline{L}}}\right)$, then, from Lemma 2.2, the functions

$$\begin{aligned} \displaystyle \beta (t) = \beta \left( \int _{0}^{L(t)}\overline{y}(s,t) ds \right) , \ \ \ \ \displaystyle a(x,t) = \int _0^1 \frac{\partial g}{\partial s}\left( \lambda \overline{y}(x,t),\lambda \overline{y}_x(x,t)\right) d\lambda , \end{aligned}$$

and

$$\begin{aligned} \displaystyle b(x,t) = \int _0^1 \frac{\partial g}{\partial p}\left( \lambda \overline{y}(x,t),\lambda \overline{y}_x(x,t)\right) d\lambda \end{aligned}$$

satisfy

$$\begin{aligned} \begin{array}{l} \beta (t) \in C^{\theta }([0,T]), \text{ and } a(x,t), b(x,t) \in C_{x,t}^{\theta ,\theta /2}\left( \overline{Q_{\overline{L}}}\right) , \text{ for } \text{ all } 0 \le \theta < 1, \end{array} \nonumber \\ \end{aligned}$$

(2.24)

therefore, $y_{\varepsilon }$ is solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_{\varepsilon , t} - \beta \left( t\right) y_{\varepsilon , xx} + a(x,t)y_{\varepsilon } + b(x,t)y_{\varepsilon , x} = v_{\varepsilon }\mathrm {\widetilde{1}}_{\omega }, &{} (x,t) \in Q_L,\\ y_{\varepsilon }(0,t) = 0; y_{\varepsilon }(L(t),t) = 0, &{} t \in (0,T),\\ y_{\varepsilon }(x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. \end{aligned}$$

(2.25)

Since $L \in C^{1+1/4}([0,T])$, from (2.24) and change of variables $\displaystyle \zeta = \frac{x}{L(t)}$, $\widetilde{a}(\zeta ,t)= a(L(t)\zeta ,t)$, and $\displaystyle \widetilde{b}(\zeta ,t)= b(L(t)\zeta ,t) - \zeta \frac{L'(t)}{L(t)}$, we have

$$\begin{aligned} \displaystyle v_{\varepsilon } \in C_{x,t}^{1/2,1/4} \left( \overline{Q_{\overline{L}}} \right) , \ \ \ \ \widetilde{a}, \widetilde{b} \in C_{\zeta ,t}^{1/2,1/4}\left( \overline{Q}\right) , \text{ and } \ \ \ \ \displaystyle \frac{\beta (t)}{\left( L(t)\right) ^2} \in C^{1/4}([0,T]), \end{aligned}$$

thus, $w_{\varepsilon }$ is solution to

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle w_{\varepsilon , t} - \frac{\beta (t)}{L^2(t)}w_{\varepsilon , \zeta \zeta } + \widetilde{a}(\zeta ,t)w_{\varepsilon } + \widetilde{b}(\zeta ,t)w_{\varepsilon , \zeta } = \widetilde{v}_{\varepsilon }(\zeta ,t), &{} (\zeta ,t) \in Q,\\ w_{\varepsilon }(0,t) = 0; w_{\varepsilon }(1,t) = 0, &{} t \in (0,T),\\ w_{\varepsilon }(\zeta ,0) = y_0(L(0)\zeta ,0), &{} \zeta \in (0,1), \end{array}\right. . \end{aligned}$$

From Theorem 5.2, Chap. IV, p. 320 [28], the function $w_{\varepsilon } \in C_{\zeta ,t}^{2+1/2,1+1/4}\left( \overline{Q}\right)$ and

$$\begin{aligned} \Vert w_{\varepsilon }\Vert _{C_{\zeta ,t}^{2+1/2,1+1/4}\left( \overline{Q}\right) } \le C \left( \Vert \tilde{v}_{\varepsilon }\Vert _{C_{\zeta , t}^{1/2,1/4}\left( \overline{Q}\right) } + \Vert y_{0}\Vert _{C^{2+1/2}\left( 0,1\right) } \right) \end{aligned}$$

Therefore, for $\theta = 1/2$ fixed, we have $y_{\varepsilon } \in C_{x,t}^{2+1/2,1+1/4}\left( \overline{Q_{\overline{L}}}\right)$ and

$$\begin{aligned} \begin{array}{l} \Vert y_{\varepsilon }\Vert _{C_{x,t}^{2+1/2,1+1/4}\left( \overline{Q_{\overline{L}}}\right) } \le C \left( \Vert y_0\Vert _{L^{2}(0,L_0)} + \Vert y_{0}\Vert _{C^{2+1/2}\left( 0,L_0\right) } \right) \end{array} \end{aligned}$$

(2.26)

Fig. 1

a Domain ${Q_{L}}$; b Domain Q

3 Main Result

This section is devoted to prove the main result, namely, TheoremReferences [11, 14, 16, 17, 21 and 29] were provided in the reference list; however, these were not mentioned or cited in the manuscript. As a rule, if a citation is present in the text, then it should be present in the list. Please provide the location of where to insert the reference citation in the main body text. Kindly ensure that all references are cited in ascending numerical order. 1.1. It will be a consequence of Theorem 2.2 and a fixed-point argument (Fig. 1).

For this purpose, let $(\overline{y}, \overline{L}) \in C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})\times C^{1+\frac{1}{4}}([0,T])$ be given, with $L_*\le \overline{L}(t) \le B$, $\overline{L}(0) = L_0$ and $y_0 \in C^{2+\frac{1}{2}}([0,L_0])$.

Next, we recall the problem given by (2.12) with an additional condition

$$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_{\varepsilon , t} - \beta \left( \int _{0}^{\overline{L}(t)}\overline{y}(x,t) dx \right) y_{\varepsilon ,xx} + a(\overline{y},\overline{y}_x)y_{\varepsilon } + b(\overline{y},\overline{y}_x)y_{\varepsilon ,x} = v_{\varepsilon }\mathrm {\widetilde{1}}_{\omega }, &{} (x,t) \in Q_{\overline{L}},\\ y_{\varepsilon }(0,t) = 0; y_{\varepsilon }(\overline{L}(t),t) = 0, &{} t \in (0,T),\\ y_{\varepsilon }(x,0) = y_0(x), &{} x \in (0,L_0),\\ \Vert y_{\varepsilon }(\cdot ,T)\Vert _{L^2(0,\overline{L}(T))}\le \varepsilon \end{array}\right. . \nonumber \\ \end{aligned}$$

(3.1)

Let us introduce the sets

$$\begin{aligned} \mathsf {Y}&= \left\{ \overline{y} \in C_{x,t}^{2,1}\big (\overline{Q_{\overline{L}}}\big ) : \Vert \overline{y}\Vert _{C_{x,t}^{2,1}\big (\overline{Q_{\overline{L}}}\big )} \le R \right\} \quad \text{ and } \\ \mathsf {Z}&= \left\{ \overline{L} \in C^{1+\frac{1}{4}}([0,T]) : L_*\le \overline{L}(t) \le B,\, \overline{L}(0) = L_0 ,\, \Vert \overline{L}\Vert _{C^{1+\frac{1}{4}}([0,T])} \le R \right\} \,. \end{aligned}$$

where the constant $R>0$ will be determined later.

Now, we set the mapping

$$\begin{aligned} \begin{array}{rccc} \mathsf {H}_\varepsilon : &{} \mathsf {Y} \times \mathsf {Z} &{} \longrightarrow &{} C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})\times C^{1+\frac{1}{4}}([0,T]) \\ &{} (\overline{y},\overline{L}) &{} \longmapsto &{} (y_\varepsilon ,L_\varepsilon ) \end{array}, \end{aligned}$$

where $y_\varepsilon$ satisfies (3.1) for $v_{\varepsilon } = e^{-2 s_{0} \alpha } \xi ^{3}\widehat{\varphi }_\varepsilon \widetilde{1}_{\omega }$, $\widehat{\varphi }_\varepsilon$ is the unique minimum of $J_\varepsilon (\cdot ,g(\overline{y},\overline{y}_{x}),\overline{L})$ and

$$\begin{aligned} \displaystyle L_\varepsilon (t) = L_0 - \int _{0}^{t}\left[ \beta \left( \int _{0}^{\overline{L}(s)} \overline{y}(x,s)\,\,dx\right) y_{\varepsilon ,x }(\overline{L}(s),s)\right] ds. \end{aligned}$$

(3.2)

Our goal is to prove that $\mathsf {H}_\varepsilon$ satisfies the hypothesis of the Schauder’s Fixed Point Theorem. We can verify from the results in Sect. 2 that the mapping $\mathsf {H}_\varepsilon$ is well defined.

For $\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])}$ sufficiently small, from (2.26) we have that

$$\begin{aligned} \Vert y_{\varepsilon } \Vert _{C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})} \le C \left( \Vert y_0\Vert _{L^{2}(0,L_0)} + \Vert y_{0}\Vert _{C^{2+\frac{1}{2}}\left( [0,L_0]\right) } \right) \le C_1 \Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])} \le R_1\,. \end{aligned}$$

Furthermore, we have the following estimate:

$$\begin{aligned} \Biggl |\int ^{\overline{L}(s)}_{0} \overline{y}(x,s) \,\,dx\Biggr |&\le \int ^{\overline{L}(s)}_{0} \Vert \overline{y}\Vert _{C^{0}(\overline{Q_{\overline{L}}})} \,\,dx \\&\le \Vert \overline{y}\Vert _{C^{0}(\overline{Q_{\overline{L}}})} \Vert \overline{L}\Vert _{C^{0}([0,T])}\\&\le C\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])}. \end{aligned}$$

From the last estimate, one has

$$\begin{aligned} |L_{\varepsilon }(t)-L_{0}|&\le \int ^{t}_{0} |\beta | \Biggl | \int _{0}^{\overline{L}(s)} \overline{y}(x,s)\,\,dx \Biggr | |y_{\varepsilon ,x}(\overline{L}(s),s)| \,\,ds\\&\le C_{2}\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])}\,T. \end{aligned}$$

Moreover

$$\begin{aligned} |L_{\varepsilon }'(t)|&= \Biggl | \beta \left( \int _{0}^{\overline{L}(s)} \overline{y}(x,s)\,\,dx\right) y_{\varepsilon ,x}(\overline{L}(t),t)\Biggr |\\&\le C_{3}\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])}. \end{aligned}$$

In view of the Lemma 2.3, we obtain

$$\begin{aligned} \Vert L_{\varepsilon } \Vert _{C^{1+\frac{1}{4}}([0,T])} \le C_{4}\Vert y_0\Vert _{C^{2+\frac{1}{2}}([0,L_0])}. \end{aligned}$$

Now, we take $R=\min \left\{ \frac{R_1}{C_{1}}, \frac{L_{0}-L_{*}}{C_{2}T}, \frac{B-L_{0}}{C_{2}T}, \frac{R_1}{C_{4}} \right\}$

Therefore, for $\Vert y_{0} \Vert _{C^{2+\frac{1}{2}}([0,L_{0}])} \le R$ one has

$$\begin{aligned} \Vert y_\varepsilon \Vert _{C^{2,1}_{x,t}(\overline{Q_{\overline{L}}})} \le R, \quad \Vert L_{\varepsilon }\Vert _{C^{1+\frac{1}{4}}([0,T])} \le R \quad \text{ and } \quad L_{*} \le L_{\varepsilon }(t) \le B. \end{aligned}$$

Thus, we verify that $\mathsf {H}_{\varepsilon }$ maps $\mathsf {Y} \times \mathsf {Z}$ into itself, that is

$$\begin{aligned}\mathsf {H}_\varepsilon (\mathsf {Y} \times \mathsf {Z}) \subset \mathsf {Y} \times \mathsf {Z}\,.\end{aligned}$$

Furthermore,

$$\begin{aligned} y_{\varepsilon } \in C^{2+\frac{1}{2},1+\frac{1}{4}}_{x,t}(\overline{Q_{\overline{L}}})\hookrightarrow C_{x,t}^{2,1}(\overline{Q_{\overline{L}}}) \quad \text{ compactly } \text{ embedded }. \end{aligned}$$

From Lemma 2.3, we have that

$$\begin{aligned} L_{\varepsilon } \in C^{1+\sigma }([0,T]) \hookrightarrow C^{1+\frac{1}{4}}([0,T])\, \text{ compactly } \text{ embedded, } \text{ for } \, \frac{1}{4}< \sigma < 1\,. \end{aligned}$$

Therefore $\mathsf {H}_{\varepsilon }$ maps $\mathsf {Y} \times \mathsf {Z}$ into a compact set of $C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})\times C^{1+\frac{1}{4}}([0,T])$.

In view of the previous properties of $\mathsf {H}_{\varepsilon }$, there exists $\delta >0$ (independent of $\varepsilon$) such that, if $\Vert y_0 \Vert _{C^{2+\frac{1}{2}}([0,L_0])} \le \delta$, we can apply Schauder’s Fixed Point Theorem to the mapping $\mathsf {H}_\varepsilon : \mathsf {Y} \times \mathsf {Z} \mapsto C_{x,t}^{2,1}(\overline{Q_{\overline{L}}})\times C^{1+\frac{1}{4}}([0,T])$.

Let $(y_{\varepsilon },L_{\varepsilon })$ be a fixed point of $\mathsf {H}_{\varepsilon }$ for each $\varepsilon >0$. Then, it is clear that $(y_{\varepsilon },L_{\varepsilon })$, together with $v_{\varepsilon }$, satisfies (1.1),(1.2), (2.10) and (2.11).

So, we can extract subsequences indexed by $\varepsilon$ satisfying

$$\begin{aligned} \left\{ \begin{array}{lll} y_{\varepsilon } \rightarrow y\,\,&{}\text{ in }&{}\,\,C_{x,t}^{2,1}(\overline{Q_{\overline{L}}}) \\ L_{\varepsilon } \rightarrow L\,\,&{}\text{ in }&{}\,\,C^{1}([0,T]) \\ v_{\varepsilon }\rightharpoonup v\,\,&{}\text{ in }&{}\,\,L^{2}(\omega \times (0,T)). \\ \end{array} \right. \end{aligned}$$

(3.3)

From (3.3), we can take limits in system (3.1) and deduce that y is the state associated to control v and (1.1)–(1.2) is locally null controllable.

Hence, Theorem 1.1 is proved.

4 Open Questions

As a first comment, an interest question concerns the global null controllability to (1.1)–(1.2), which does not seem to be simple. To prove a global result, we would have to use a global inverse mapping theorem, but this requires much more complicated estimates, which do not seem to be accesible.

Other important topics arise from our current research:

• In the system (1.1), we can replace the nonlocal nonlinearity $\beta \left( \int _{0}^{L(t)}y dx \right)$ by $\beta \left( \int _{0}^{L(t)}y_x dx \right)$, in order to analyze whether it is possible to prove results about null controllability.

• An interesting case deals with the null controllability of the degenerate system

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_t - \left( \beta \left( x,\int _{0}^{L(t)}y dx \right) y_{x}\right) _x + g(y,y_x) = v\mathrm {1}_{\omega }, &{} (x,t) \in Q_L,\\ y(0,t) = 0; y(L(t),t) = 0, &{} t \in (0,T),\\ y(x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. \end{aligned}$$

  (4.1)

  with the additional boundary condition

  $$\begin{aligned} -L'(t) = \beta \left( x,\int _{0}^{L(t)}y(x,t)dx\right) y_x(L(t),t), \end{aligned}$$

  where $\beta$ is a separated variables function given by $\beta (x,r)=\ell (r)a(x)$ and $\beta$ defines an operator which degenerates at $x=0$ and has a nonlocal term. More precisely, the function a behaves $x^{\alpha }$, with $\alpha \in (0,1)$.

  On the controllability of degenerate parabolic equations, for an instance, we mention the following works: Cannarsa et al. [4,5,6,7,8], Alabau-Boussouira et al. [1] and Demarque et al. [10].

• Another interesting case is found when the control function acts on the free boundary, as we can see in the system below:

  $$\begin{aligned} \left\{ \begin{array}{ll} \displaystyle y_t - \beta \left( \int _{0}^{L(t)}y dx \right) y_{xx} + g(y,y_x) = 0, &{} (x,t) \in Q_L,\\ y(0,t) = 0; y(L(t),t) = v(t), &{} t \in (0,T),\\ y(x,0) = y_0(x), &{} x \in (0,L_0), \end{array}\right. \end{aligned}$$

  (4.2)

  together with (1.2) and (1.3).

  However, this control problem needs a deeper analysis.

Appendices

Appendix A: Proof of Theorem 2.1

Let $\psi$ be new variable defined by $\varphi (x,t) := e^{s\alpha (x,t)}\psi (x,t)$. It yields:

$$\begin{aligned} \displaystyle \varphi _t= & {} s\alpha _te^{s\alpha }\psi + e^{s\alpha }\psi _t, \quad \varphi _x = s\alpha _xe^{s\alpha } \psi + e^{s\alpha }\psi _x \,\, \text{ and } \\ \displaystyle \varphi _{xx}= & {} e^{s\alpha }\left( \psi _{xx} + s^2\alpha ^2_x\psi + 2s\alpha _x\psi _x + s\alpha _{xx}\psi \right) . \end{aligned}$$

The functions $\alpha$ and $\xi$ yield:

$$\begin{aligned} \displaystyle \alpha _x= & {} -\xi _x = -\lambda \alpha _{0,x}\xi , \quad \alpha _{xx} = -\lambda ^2\alpha _{0,x}^2\xi - \lambda \alpha _{0,xx}\xi , \\ \displaystyle \alpha _t= & {} -k\frac{T-2t}{(T-t)^{k+1} t^{k+1}}\left( e^{2\lambda \parallel \alpha _1\parallel _{\infty } } - e^{\lambda \alpha _1(x,t)} \right) - \lambda \alpha _{0,t}\xi , \\ \displaystyle \alpha _{x,t}= & {} k \lambda \xi \left( \frac{(T-2t)}{t(T-t)}\alpha _{0,x} -\frac{\alpha _{0,tx}}{k}-\lambda \frac{\alpha _{0,t} \alpha _{0,x}}{k}\right) \end{aligned}$$

and, for $\lambda$ sufficiently large, one has:

$$\begin{aligned} \displaystyle |\alpha _x|\le & {} C\lambda \xi , \quad |\alpha _{xx}| \le C\lambda ^2\xi ,\\ \displaystyle |\alpha _t|\le & {} C\xi ^{1+\frac{1}{k}} + C\lambda \xi \le C \lambda \xi ^{\frac{3}{2}} \quad \text{ and } \\ |\alpha _{x,t}|\le & {} C \lambda \xi ^2. \end{aligned}$$

On the other hand, we have that $\psi (x,0) \equiv 0$ in (0, L(0)) and $\psi (x,T) \equiv 0$ in $(0,\overline{L}(T))$.

Moreover, replacing $\varphi$ by $e^{s\alpha }\psi$ in the PDE (2.2), it yields:

$$\begin{aligned}&e^{-2s\alpha }(s\alpha _t\psi + \psi _t) + e^{-2s\alpha }\Big [\beta \left( t\right) \psi _{xx} + \beta \left( t\right) s^2\alpha _{x}^2\psi + 2\beta \left( t\right) s\alpha _x\psi _x\\&\quad + \beta \left( t\right) s\alpha _{xx}\psi \Big ] + e^{s\alpha }(a\psi ) + e^{s\alpha }\Big [s\alpha _x(b\psi ) + (b\psi _x) \Big ] = F(x,t), \end{aligned}$$

therefore,

$$\begin{aligned}&\Big [\psi _t + 2\beta \left( t\right) s\alpha _x\psi _x \Big ] + \Big [\beta \left( t\right) \psi _{xx} + \beta \left( t \right) s^2\alpha _{x}^2\psi \Big ] \\&\quad = - e^{-s\alpha } F(x,t) - s\alpha _t\psi - \beta \left( t\right) s\alpha _{xx}\psi - a\psi - s\alpha _xb\psi - b\psi _x. \end{aligned}$$

Considering the following notation

$$\begin{aligned} \left\{ \begin{array}{l} U\psi := \psi _t + 2\beta \left( t\right) s\alpha _x\psi _x = (U\psi )_1 + (U\psi )_2\\ V\psi := \beta \left( t\right) \psi _{xx} + \beta \left( t\right) s^2\alpha _{x}^2\psi = (V\psi )_1 + (V\psi )_2\\ \end{array}\right. , \end{aligned}$$

it yields:

$$\begin{aligned} U\psi + V\psi = G(x,t), \end{aligned}$$

where

$$\begin{aligned} G(x,t)= & {} - e^{-s\alpha (x,t)} F(x,t) - s\alpha _t(x,t)\psi (x,t) - \beta (t)s\alpha _{xx}(x,t)\psi (x,t) \\&- a(x,t)\psi (x,t) - s\alpha _x(x,t)b(x,t)\psi (x,t) - b(x,t)\psi _x(x,t). \end{aligned}$$

Therefore,

$$\begin{aligned}&\displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})} \nonumber \\&\quad \displaystyle = \Vert e^{-s\alpha } F + s\alpha _t\psi + \beta s\alpha _{xx}\psi + a\psi + s\alpha _xb\psi + b\psi _x \Vert _{L^2(Q_{\overline{L}})}^2 \nonumber \\&\quad \displaystyle \le C\Big (\Vert e^{-2s\alpha }F\Vert _{L^2(Q_{\overline{L}})}^2 + \Vert s\alpha _t\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert \beta s\alpha _{xx}\psi \Vert ^2_{L^2(Q_{\overline{L}})}\Big .\nonumber \\&\qquad + \Big .\Vert a\psi \Vert ^2_{L^2(Q_{\overline{L}})} + \Vert s\alpha _xb\psi \Vert ^2_{L^2(Q_{\overline{L}})} + \Vert b\psi _x \Vert _{L^2(Q_{\overline{L}})}^2 \Big ) \nonumber \\&\quad \displaystyle \le C\Bigg (\iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2|\alpha _t|^2|\psi |^2dxdt \nonumber \\&\qquad + \iint _{Q_{\overline{L}}}\beta s^2|\alpha _{xx}|^2|\psi |^2dxdt \Big . \displaystyle + \Bigg . \iint _{Q_{\overline{L}}} |a|_{L^{\infty }(Q_{\overline{L}})}^2|\psi |^2dxdt \nonumber \\&\qquad + \iint _{Q_{\overline{L}}} |b|_{L^{\infty }(Q_{\overline{L}})}^2 s^2|\alpha _x|^2|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |b|_{L^{\infty }(Q_{\overline{L}})}^2|\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \displaystyle \le C\Bigg (\iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2\widetilde{C}\lambda ^2\xi ^3|\psi |^2dxdt \nonumber \\&\qquad + C_{\beta }\iint _{Q_{\overline{L}}} s^2\widetilde{C}\lambda ^4\xi ^2 |\psi |^2dxdt \Big . \displaystyle + \Bigg . |a|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} |\psi |^2dxdt \nonumber \\&\qquad + |b|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} s^2\lambda ^2|\alpha _{0,x}|^2\xi ^2|\psi |^2dxdt + |b|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \displaystyle \le C\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2\lambda ^4\xi ^3|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg ). \nonumber \\ \end{aligned}$$

(4.3)

Let us compute $(U\psi ,V\psi )_{L^2(Q_{\overline{L}})}$ and use the fact that $\displaystyle \iint _{Q_{\overline{L}}} = \int _0^T\int _0^{{\overline{L}}(t)}$.

$$\begin{aligned} I_1= & {} \left( (U\psi )_1,(V\psi )_1\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}\beta \psi _{t}\psi _{xx}dxdt \\= & {} \displaystyle \int _0^T\beta \left( \psi _t\psi _x \Big \vert _{0}^{{\overline{L}}(t)} - \int _{0}^{{\overline{L}}(t)}\psi _x\psi _{xt}ds \right) dt \\= & {} \displaystyle -\frac{1}{2}\int _0^T\beta \int _0^{{\overline{L}}(t)}\frac{d}{dt}[\psi _{x}]^2dxdt \\= & {} \displaystyle -\frac{1}{2}\int _0^T\left[ \beta \frac{d}{dt}\int _0^{{\overline{L}}(t)}(\psi _{x})^2dx - \beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)\right] dt\\= & {} \displaystyle \frac{1}{2}\int _0^T\int _{0}^{{\overline{L}}(t)}\beta '(\psi _{x})^2dxdt + \frac{1}{2}\int _{0}^{T}\beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)dt;\\ I_2= & {} \left( (U\psi )_1,(V\psi )_2\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}s^2\beta \psi _{t}\alpha _x^2\psi dxdt \\= & {} \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}s^2\beta \psi _{t}\alpha _x^2\psi dxdt \\= & {} \displaystyle \frac{s^2}{2}\int _0^T\beta \int _0^{{\overline{L}}(t)}\frac{d}{dt}\left( \psi \right) ^2\alpha _x^2 dxdt \\= & {} \displaystyle -\frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}\left( \beta '\alpha ^2_x + 2\beta \alpha _x\alpha _{xt} \right) \psi ^2dxdt;\\ I_3= & {} \left( (U\psi )_2,(V\psi )_1\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} 2s\beta ^2\alpha _x\psi _x\psi _{xx} dxdt \\= & {} \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _x\frac{d}{dx}\left( \psi _x\right) ^2 dxdt \\= & {} \displaystyle -\int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _{xx}\psi _x^2 dxdt \displaystyle + \int _{0}^{T}s\psi _x^2({\overline{L}}(t),t)\beta ^2\alpha _x({\overline{L}}(t),t)dt\\&\displaystyle - \int _{0}^{T}s\psi _x^2(0,t)\beta ^2\alpha _x(0,t)dt;\\ I_4= & {} \left( (U\psi )_2,(V\psi )_2\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} 2s^3\beta ^2\alpha _x^3\psi \psi _x dxdt \\= & {} \displaystyle s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _x^3\frac{d}{dx}(\psi )^2 dxdt = \displaystyle -3s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _{x}^2\alpha _{xx}\psi ^2 dxdt. \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle (U\psi ,V\psi )_{L^2(Q_{\overline{L}})}= & {} \displaystyle -\int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _{xx}\psi _x^2 dxdt -3s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _{x}^2\alpha _{xx}\psi ^2 dxdt\\&\displaystyle -\frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}2\beta \alpha _x\alpha _{xt}\psi ^2dxdt + \int _{0}^{T}s\psi _x^2({\overline{L}}(t),t)\beta ^2\alpha _x(L(t),t)dt\\&\displaystyle + \frac{1}{2}\int _{0}^{T}\beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)dt - \int _{0}^{T}s\psi _x^2(0,t)\beta ^2\alpha _x(0,t)dt\\&\displaystyle + \frac{1}{2}\int _0^T\int _{0}^{{\overline{L}}(t)}\beta '(\psi _{x})^2dxdt - \frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}\beta '\alpha ^2_x\psi ^2dxdt. \end{aligned}$$

Now, computing $\displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})}$, we have:

$$\begin{aligned} (U\psi ,V\psi )_{L^2(Q_{\overline{L}})} \displaystyle= & {} I_1 + I_2 + I_3 + I_4 \nonumber \\ \displaystyle\ge & {} C\Bigg (\frac{1}{2}\int _0^T \beta \psi ^2_x({\overline{L}}(t),t){\overline{L}}'(t) dt + \frac{1}{2}\int _0^T\beta '\int _0^{{\overline{L}}(t)}\psi ^2_xdxdt\Bigg ) \nonumber \\&\displaystyle + \Bigg ( \frac{-s^2}{2}\iint _{Q_{\overline{L}}} \beta '\alpha _x^2\psi ^2dxdt - C_2s^2\lambda ^2\iint _{Q_{\overline{L}}} \xi ^3|\psi |^2dxdt \Bigg )\nonumber \\&\displaystyle + \Bigg (\frac{sC_3}{2}\iint _{Q_{\overline{L}}} \lambda ^2\xi |\psi _x|^2 dxdt \nonumber \\&- s\iint _{\omega _0\times (0,T)} \lambda ^2C_3\xi |\psi _x|^2dxdt + s\lambda C_3\int _0^T \left[ \xi ({\overline{L}}(t),t)\psi _x^2({\overline{L}}(t),t) \right. \nonumber \\&\displaystyle \left. +\xi (0,t)\psi _x^2(0,t) \right] dt \Bigg ) + \Bigg ( C_4\iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3\psi ^2dxdt\nonumber \\&- \overline{C}\iint _{\omega _0\times (0,T)} s^3\lambda ^4\xi ^3\psi ^2dxdt \Bigg ); \end{aligned}$$

(4.4)

$$\begin{aligned} \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 \displaystyle= & {} \iint _{Q_{\overline{L}}} |\psi _t + 2\beta s\alpha _x\psi _x|^2dxdt \displaystyle = \iint _{Q_{\overline{L}}} \big |\psi _t + 2\beta s(\lambda \alpha _{0,x}\xi )\psi _x\big |^2dxdt \nonumber \\ \displaystyle\ge & {} C\Bigg ( \iint _{Q_{\overline{L}}} |\psi _t|^2dxdt - \iint _{Q_{\overline{L}}}s^2\lambda ^2\xi ^2|\psi _x|^2dxdt\Bigg ) \nonumber \\ \displaystyle\ge & {} \widetilde{C} \Bigg ( \iint _{Q_{\overline{L}}} (s\xi )^{-1} |\psi _t|^2dxdt - \iint _{Q_{\overline{L}}}(s\xi )\lambda ^2|\psi _x|^2dxdt\Bigg ); \end{aligned}$$

(4.5)

$$\begin{aligned} \displaystyle \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 \displaystyle= & {} \iint _{Q_{\overline{L}}} |\beta \psi _{xx} + \beta s^2\alpha _x^2\psi |^2dxdt\nonumber \\ \displaystyle= & {} \iint _{Q_{\overline{L}}} \big |\beta \psi _{xx} - (-\beta s^2\lambda ^2 \alpha _{0,x}^2\xi ^2\psi )\big |^2dxdt \nonumber \\ \displaystyle\ge & {} C\Bigg ( \iint _{Q_{\overline{L}}} |\psi _{xx}|^2dxdt - \iint _{Q_{\overline{L}}} s^4\lambda ^4\xi ^4|\psi |^2dxdt\Bigg ) \nonumber \\ \displaystyle\ge & {} \widetilde{C} \Bigg ( \iint _{Q_{\overline{L}}} (s\xi )^{-1} |\psi _{xx}|^2dxdt - \iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3|\psi |^2dxdt\Bigg ). \end{aligned}$$

(4.6)

Combining the estimates (4.3) and (4.4), they yield:

$$\begin{aligned}&\displaystyle C\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt \Bigg . +\displaystyle \Bigg . \iint _{Q_{\overline{L}}} s^2\lambda ^4\xi ^3|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \ge \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})}\nonumber \\&\qquad \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \bigg ( \int _0^T \beta \psi ^2_x({\overline{L}}(t),t){\overline{L}}'(t) dt\bigg .\nonumber \\&\qquad + \displaystyle \int _0^T\beta '\int _0^{{\overline{L}}(t)}\psi ^2_xdxdt - s^2\iint _{Q_{\overline{L}}} \beta '\alpha _x^2\psi ^2dxdt\nonumber \\&\qquad -\displaystyle s^2\lambda ^2\iint _{Q_{\overline{L}}} \xi ^3|\psi |^2dxdt + s\iint _{Q_{\overline{L}}} \lambda ^2\xi |\psi _x|^2 dxdt \nonumber \\&\qquad + \displaystyle s\iint _{\omega _0\times (0,T)} \lambda ^2\xi |\psi _x|^2dxdt + s\lambda \int _0^T \left[ \xi ({\overline{L}}(t),t)\psi _x^2({\overline{L}}(t),t)\right. \nonumber \\&\qquad - \Bigg . \displaystyle \left. \xi (0,t)\psi _x^2(0,t) \right] dt + \iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3\psi ^2dxdt - \iint _{\omega _0\times (0,T)} s^3\lambda ^4\xi ^3\psi ^2dxdt\Bigg ). \nonumber \\ \end{aligned}$$

(4.7)

Also, given $\varepsilon > 0$, we have:

$$\begin{aligned} \iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2|\psi _x|^2dxdt&\le \varepsilon \iint _{Q_{\overline{L}}}(s\xi )^{-1}|\psi _{xx}|^2 dxdt \nonumber \\&\quad + C_{\varepsilon }\iint _{\omega _0\times (0,T)}(s\xi )^3\lambda ^4|\psi |^2dxdt \end{aligned}$$

(4.8)

In fact,

$$\begin{aligned}&\displaystyle \Bigg |\iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2\psi _x^2dxdt\Bigg | = \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2(\xi \psi _x)_x\psi dxdt\Bigg |\\ \\&\quad \le \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2\xi _x\psi _x\psi dxdt\Bigg | + \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2\xi \psi _{xx}\psi dxdt\Bigg |\\ \\&\quad \le \displaystyle C\Bigg |\iint _{\omega _0\times (0,T)} (s\xi )(\lambda \psi _x)(\lambda ^2\psi )dxdt\Bigg | + \Bigg |\iint _{\omega _0\times (0,T)} (s\xi )\psi _{xx}(\lambda ^2\psi )dxdt\Bigg |\\ \\&\quad \le \displaystyle C\Bigg (\varepsilon _1 \iint _{\omega _0\times (0,T)} (s\xi )|\lambda \psi _x|^2dxdt + \frac{1}{4\varepsilon _1}\iint _{\omega _0\times (0,T)} (s\xi )|\lambda ^2\psi |^2dxdt \Bigg )\\ \\&\qquad + \displaystyle \Bigg (\varepsilon _2 \iint _{\omega _0\times (0,T)} (s\xi )^{-1}|\psi _{xx}|^2dxdt + \frac{1}{4\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}|\lambda ^2\psi |^2dxdt \Bigg ). \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle \iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2|\psi _x|^2dxdt\le & {} \displaystyle \frac{\varepsilon _2 }{1-C\varepsilon _1} \iint _{\omega _0\times (0,T)} (s\xi )^{-1}|\psi _{xx}|^2dxdt \\&+ C_{\varepsilon _1\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}\lambda ^4|\psi |^2dxdt \\\le & {} \displaystyle \frac{\varepsilon _2 }{1-C\varepsilon _1} \iint _{Q_{\overline{L}}} (s\xi )^{-1}|\psi _{xx}|^2dxdt \\&+ C_{\varepsilon _1\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}\lambda ^4|\psi |^2dxdt . \end{aligned}$$

Combining the inequalities (4.5)–(4.8), they yield:

$$\begin{aligned}&\displaystyle \iint _{Q_{\overline{L}}} (s\xi )^{-1} \left[ |\psi _t|^2 + |\psi _{xx}|^2\right] dxdt + \iint _{Q_{\overline{L}}} \left[ (s\xi )\lambda ^2|\psi _x|^2 + (s\xi )^3\lambda ^4|\psi |^2\right] dxdt \\&\qquad +\displaystyle (s\lambda ) \int _{0}^{T}\left[ \xi ({\overline{L}}(t),t)|\psi _x({\overline{L}}(t),t)|^2 + \xi (0,t)|\psi _x(0,t)|^2 \right] dt \\&\qquad + \int _0^T\beta '\int _0^{{\overline{L}}(t)}|\psi _x|^2dxdt \\&\quad \le \displaystyle C \,\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{\omega \times (0,T)} s^3\lambda ^4\xi ^3|\psi |^2dxdt\\&\qquad + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt + \iint _{Q_{\overline{L}}} \beta '(s\xi )^2\lambda ^2|\psi |^2dxdt \Bigg ). \end{aligned}$$

Therefore,

$$\begin{aligned}&\displaystyle \iint _{Q_{\overline{L}}} \left[ (s\xi )^{-1}(|\psi _t|^2 + |\psi _{xx}|^2 ) + (s\xi )\lambda ^2|\psi _x|^2 + (s\xi )^3\lambda ^4|\psi |^2 \right] dxdt \\&\qquad + s\lambda \int _{0}^T \left[ \xi ({\overline{L}}(t),t)|\psi _x({\overline{L}}(t),t)|^2 + \xi (0,t)|\psi _x(0,t)|^2 \right] dt \\&\quad \le C \left( \iint _{Q_{\overline{L}}} e^{-2s\alpha } |F|^2 dxdt + \iint _{\omega \times (0,T)} (s\xi )^3\lambda ^4|\psi |^2dxdt \right) . \end{aligned}$$

Coming back to the original variable $\varphi$, we have (2.4).

Appendix B: Proof of Lemma 2.2 and 2.3

Proof of Lemma 2.2

(i) For all $0 \le \theta < 1$, we have:

$$\begin{aligned} \frac{| \beta (t_1) - \beta (t_2) |}{ |t_1 - t_2|^{\theta } }= & {} \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \Big | \beta \left( \int _0^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx \right) - \beta \left( \int _0^{\overline{L}(t_2)} \overline{y}(x,t_2) \ dx \right) \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _0^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx - \int _0^{\overline{L}(t_2)} \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _{\overline{L}(t_2)}^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx + \int _0^{\overline{L}(t_2)} \overline{y}(x,t_1) - \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot ( | \overline{L}(t_1) - \overline{L}(t_2) | + |t_1 - t_2| ) \\\le & {} C\frac{|t_1 - t_2|}{ |t_1 - t_2|^{\theta } } < \infty . \end{aligned}$$

Then, $\beta \in C^{\theta }([0,T])$, for all $0 \le \theta < 1$.

(ii) For all $0 \le \theta < 1$, one has:

$$\begin{aligned}&\frac{| a(x_1,t_1) - a(x_2,t_2) |}{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le \frac{ \displaystyle { \int _0^1 \Big | \frac{\partial {g}}{\partial {s}}( \lambda \overline{y}(x_1,t_1), \lambda \overline{y}_x(x_1,t_1)) - \frac{\partial {g}}{\partial {s}}( \lambda \overline{y}(x_2,t_2), \lambda \overline{y}_x(x_2,t_2)) \Big | \ d\lambda } }{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le C \cdot \frac{( |\overline{y}(x_1,t_1) - \overline{y}(x_2,t_2)| + |\overline{y}_x(x_1,t_1) - \overline{y}_x(x_2,t_2)| )}{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le C \cdot \frac{ ( |x_1-x_2| + |t_1 - t_2| ) }{ ( |x_1-x_2| + |t_1 - t_2|^{1/2} )^{\theta } } < \infty . \end{aligned}$$

Then, $a \in C_{x,t}^{\theta , \theta /2}(\overline{Q}_L)$, for all $0 \le \theta < 1$. The proof is analogous to b(x, t).

(iii) For all $0 \le \theta <1$, one has:

$$\begin{aligned} \frac{| \widetilde{a}(\xi _1, t_1) - \widetilde{a}(\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta }}= & {} \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }\\= & {} \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2)| }{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \cdot \\&\frac{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } , \end{aligned}$$

On the other hand, from (ii), we have

$$\begin{aligned} \displaystyle { \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2)| }{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } } \le C. \end{aligned}$$

Then, one has,

$$\begin{aligned} \frac{| \widetilde{a}(\xi _1, t_1) - \widetilde{a}(\xi _2, t_2) |}{(|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }\le & {} C\cdot \frac{ (L(t_1)| \xi _1 - \xi _2 | + |\xi _2 | .|L(t_1) - L(t_2)| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\le C_1\cdot \frac{ (| \xi _1 - \xi _2 | + |L(t_1) - L(t_2)| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \\&< \infty , \quad \forall \theta \in [0,1). \end{aligned}$$

Then, $\widetilde{a} \in C_{\xi ,t}^{\theta , \theta /2}(\overline{Q})$, for all $0 \le \theta < 1$.

In a similar way, we obtain $\widehat{b}(\xi ,t) = b(L(t)\xi ,t) \in C_{\xi ,t}^{\theta , \theta /2}(\overline{Q})$. Let us analyse the function $\displaystyle { \overline{b}(\xi ,t) = \xi \frac{L'(t)}{L(t)} }$, for $L\in C^{1+ \gamma /2}([0,T])$.

For all $0 \le \gamma < 1$,

$$\begin{aligned}&\frac{| \overline{b}(\xi _1, t_1) - \overline{b}(\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } }\\&\quad = \frac{1}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } \cdot \Big | \xi _1\frac{L'(t_1)}{L(t_1)} - \xi _2\frac{L'(t_2)}{L(t_2)} \Big | \\&\quad \le \frac{C}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } \cdot \left( | \xi _1 - \xi _2| \cdot \Big | \frac{L'(t_1)}{L(t_1)} \Big | + |\xi _2| \cdot \Big | \frac{L'(t_1)}{L(t_1)} - \frac{L'(t_2)}{L(t_2)} \Big | \right) \\&\quad \le C_1 \cdot \frac{\left( \displaystyle { | \xi _1 - \xi _2| + |L'(t_1) - L'(t_2)|\frac{1}{|L(t_1)|} + |L'(t_2)| \cdot \frac{|L(t_2)-L(t_1)|}{|L(t_1)L(t_2)|} } \right) }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } }\\&\quad \le C_2 \cdot \frac{\left( \displaystyle { | \xi _1 - \xi _2| + |L'(t_1) - L'(t_2)| + |L(t_2)-L(t_1)| } \right) }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } < \infty , \end{aligned}$$

since $L\in C^{1+ \gamma /2}([0,T])$.

Then, $\widetilde{b} \in C_{\xi ,t}^{\gamma , \gamma /2}(\overline{Q})$, for all $0 \le \gamma < 1$. $\square$

Proof of Lemma 2.3

For $0 \le \theta < 1$, one has:

$$\begin{aligned}&\frac{| L'(t_1) - L'(t_2) |}{ |t_1 - t_2|^{\theta } } \\&\quad = \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \left[ \displaystyle \beta \left( \int _{0}^{\overline{L}(t_1)} y(x,t_1)dx\right) y_x(\overline{L}(t_1),t_1) \right. \\&\qquad \left. - \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) y_x(\overline{L}(t_2),t_2) \right] \\&\quad =\frac{1}{ |t_1 - t_2|^{\theta } } \cdot {\displaystyle { \left[ \displaystyle { \beta \left( \int _{0}^{\overline{L}(t_1)} y(x,t_1)dx\right) - \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) } \right] y_x(\overline{L}(t_1),t_1) }} \\&\qquad + \displaystyle { \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \displaystyle { \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) \cdot \left[ y_x(\overline{L}(t_1),t_1) - y_x(\overline{L}(t_2),t_2) \right] } } \\&\quad = I_1 + I_2 \end{aligned}$$

For $0 \le \theta < 1$, we obtain the following estimates for $I_1$ and $I_2$:

$$\begin{aligned} |I_1|\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \displaystyle { {\displaystyle { \displaystyle { \int _{0}^{\overline{L}(t_1)} y(x,t_1) \ dx - \int _{0}^{\overline{L}(t_2)} y(x,t_2) \ dx } }} } \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _{\overline{L}(t_2)}^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx + \int _0^{\overline{L}(t_2)} \overline{y}(x,t_1) - \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C_1}{ |t_1 - t_2|^{\theta } } \cdot |\overline{L}(t_1) - \overline{L}(t_2)| + c\int _0^{\overline{L}(t_2)} \frac{ |\overline{y}(x,t_1) - \overline{y}(x,t_2)| }{|t_1 - t_2|^{\theta }} \ dx \\\le & {} C_2 \left( \sup _{t_1, t_2\in [0,T]} \frac{|\overline{L}(t_1) - \overline{L}(t_2)|}{|t_1 - t_2|^{\theta }} + \sup _{t_1, t_2\in [0,T]} \frac{|\overline{y}(x,t_1) - \overline{y}(x,t_2)|}{|t_1 - t_2|^{\theta }} \right) < \infty , \end{aligned}$$

and

$$\begin{aligned} |I_2|\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot |y_x(\overline{L}(t_1),t_1) - y_x(\overline{L}(t_2),t_2)| \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot ( |\overline{L}(t_1) - \overline{L}(t_2)| + |t_1 - t_2| ) < \infty . \end{aligned}$$

$\square$