Local controllability of the one-dimensional nonlocal Gray–Scott model with moving controls

Abstract

In this paper, we prove the local controllability to positive constant trajectories of a nonlinear system of two coupled ODE equations, posed in the one-dimensional spatial setting, with nonlocal spatial nonlinearities, and using only one localized control with a moving support. The model we deal with is derived from the well-known nonlinear reaction–diffusion Gray–Scott model when the diffusion coefficient of the first chemical species $d_u$ tends to 0 and the diffusion coefficient of the second chemical species ${d_v}$ tends to $+ \infty$. The strategy of the proof consists in two main steps. First, we establish the local controllability of the reaction–diffusion ODE–PDE derived from the Gray–Scott model taking $d_u=0$, and uniformly with respect to the diffusion parameter ${d_v} \in (1, +\infty )$. In order to do this, we prove the (uniform) null-controllability of the linearized system thanks to an observability estimate obtained through adapted Carleman estimates for ODE–PDE. To pass to the nonlinear system, we use a precise inverse mapping argument and, secondly, we apply the shadow limit ${d_v} \rightarrow + \infty$ to reduce to the initial system.

Appendices

A Uniform parabolic Carleman estimate

In this part, we prove Lemma 2.3. The proof follows the arguments presented in [20, Appendix A] with some remarks coming grom [6]. For completeness, we give a sketch of the proof.

We start by giving some useful properties on the weight function $\alpha$ and its derivatives. By simple computations using (30), we have

$$\begin{aligned} \begin{aligned} \alpha _x&=-\lambda \xi \eta _x, \\ \alpha _{xx}&=\lambda ^2\xi (-\eta _x^2-\lambda ^{-1}\eta _{xx}), \\ |\alpha _{xx}|&\le C\lambda ^2\xi , \\ |\alpha _{xxx}|&\le C\lambda ^3\xi , \\ |\alpha _{xxxx}|&\le C\lambda ^4\xi , \\ |\alpha _t|&\le C(T+e^{2\lambda \Vert \eta \Vert _\infty })\lambda \xi ^2, \\ |\alpha _{xt}|&\le C(T+1)\lambda ^2\xi ^2, \\ |\alpha _{tt}|&\le C(T+T^2+e^{2\lambda \Vert \eta \Vert _\infty })\lambda ^2\xi ^3. \end{aligned} \end{aligned}$$

(90)

As usual, we set $w=e^{-s\alpha }\psi$ and compute

$$\begin{aligned} w_{x}=-s\alpha _x w+e^{-s\alpha }\psi _x. \end{aligned}$$

Using the boundary conditions of $\psi$, we deduce $w_x=-s\alpha _xw$ on $(0,T)\times \{0,1\}$. We introduce the parabolic operator $P=\partial _t+\frac{1}{\epsilon }\partial _{xx}$. Then, we have

$$\begin{aligned} e^{-s\alpha }P(e^{s\alpha }w)=P_e^{\epsilon }w+P_{k}^{\epsilon }w, \end{aligned}$$

(91)

where

$$\begin{aligned} P_e^{\epsilon }w&=\frac{1}{\epsilon } w_{xx}+s\alpha _t w+\frac{1}{\epsilon }s^2\alpha _x^2 w, \end{aligned}$$

(92)

$$\begin{aligned} P_k^{\epsilon }w&=w_t+\frac{2}{\epsilon }s\alpha _xw_x+\frac{1}{\epsilon }s\alpha _{xx}w. \end{aligned}$$

(93)

We take the $L^2$-norm in both sides of (91), thus obtaining

$$\begin{aligned}&\!\!\!\Vert e^{-s\alpha } P(e^{s\alpha }w)\Vert _{L^2(Q_T)}^2=\Vert P_{e}^\epsilon w\Vert ^2_{L^2(Q_T)} +\Vert P_k^\epsilon w\Vert ^2_{L^2(Q_T)}+2\left( P_e^\epsilon w,P_k^\epsilon w\right) _{L^2(Q_T)}.\nonumber \\ \end{aligned}$$

(94)

A very long, but straightforward computation gives that

$$\begin{aligned} 2\left( P_e^\epsilon w,P_k^\epsilon w\right) _{L^2(Q_T)}= BT+DT_1+DT_2, \end{aligned}$$

where

$$\begin{aligned} BT:=&\frac{1}{\epsilon }\left. \int _{0}^{T}s\alpha _{xt}w^2\text {d}t\right| _{0}^{1}+\frac{2}{\epsilon ^2}\left. \int _{0}^{T}s\alpha _{x}w_x^2\text {d}t\right| _{0}^{1}+\frac{2}{\epsilon ^2}\left. \int _{0}^{T}s\alpha _{xx}w_x w\text {d}t\right| _{0}^{1} \\&-\frac{1}{\epsilon ^2}\left. \int _{0}^{T}s\alpha _{xxx}w^2\text {d}t\right| _{0}^{1}+\frac{2}{\epsilon }\left. \int _{0}^{T}s^2\alpha _t\alpha _{x}w^2\text {d}t\right| _{0}^{1}+\frac{2}{\epsilon ^2}\left. \int _{0}^{T}s^3\alpha _{x}^3 w^2\text {d}t\right| _{0}^{1}, \\ DT_1:=&-\frac{4}{\epsilon ^2}\iint _{Q_T} s\alpha _{xx}w_x^2\,\text {d}x\text {d}t, \\ DT_2:=&-\frac{4}{\epsilon }\iint _{Q_T}s^2\alpha _{tx}\alpha _x w^2\,\text {d}x\text {d}t+\frac{1}{\epsilon ^2}\iint _{Q_T}s\alpha _{xxxx}w^2\,\text {d}x\text {d}t-\iint _{Q_T}s\alpha _{tt}w^2\,\text {d}x\text {d}t\\&-\frac{4}{\epsilon ^2}\iint _{Q_T} s^3\alpha _x^2\alpha _{xx}w^2\,\text {d}x\text {d}t. \end{aligned}$$

Using that $w_{x}=-s\alpha _x w$ in $(0,T)\times \{0,1\}$, we can obtain

$$\begin{aligned} BT=&\frac{4}{\epsilon ^2}\left. \int _{0}^T s^3\alpha _x^3 w^2 \text {d}t\right| _{0}^{1}+\frac{2}{\epsilon }\left. \int _{0}^{T}s^2\alpha _t\alpha _{x}w^2\text {d}t\right| _{0}^{1}\frac{2}{\epsilon ^2}\left. \int _{0}^{T}s^2\alpha _x\alpha _{xx}w^2\text {d}t\right| _{0}^{1} \\&+\frac{1}{\epsilon }\left. \int _{0}^{T}s\alpha _{xt}w^2\text {d}t\right| _{0}^{1}-\frac{1}{\epsilon ^2}\left. \int _{0}^{T}s\alpha _{xxx}w^2\text {d}t\right| _{0}^{1}. \end{aligned}$$

Moreover, using properties (90) together with (25)–(26), it is not difficult to see that

$$\begin{aligned} BT\ge&\frac{4C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text{ d }t\Big |_{x=1}+\frac{4C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text{ d }t\Big |_{x=0} \\ {}&-\frac{2C}{\epsilon ^2}\int _{0}^{T}s^2\lambda ^2\xi ^3(T+e^{2\lambda \Vert \eta \Vert _\infty })w^2\text{ d }t\Big |_{x=1}\\ {}&-\frac{2C}{\epsilon ^2}\int _{0}^{T}s^2\lambda ^2\xi ^3(T+e^{2\lambda \Vert \eta \Vert _\infty })w^2\text{ d }t\Big |_{x=0} \\ {}&-\frac{2}{\epsilon ^2}\int _{0}^{T}s^2\lambda ^3\xi ^2w^2\text{ d }t\Big |_{x=1}-\frac{2}{\epsilon ^2}\int _{0}^{T}s^2\lambda ^3\xi ^2w^2\text{ d }t\Big |_{x=0} \\ {}&-\frac{C}{\epsilon ^2}\int _{0}^{T}s(T+1)\lambda ^2\xi ^2w^2\text{ d }t\Big |_{x=1}-\frac{C}{\epsilon ^2}\int _{0}^{T}s(T+1)\lambda ^2\xi ^2w^2\text{ d }t\Big |_{x=0} \\ {}&-\frac{C}{\epsilon ^2}\int _{0}^{T}s\lambda ^3\xi w^2\text{ d }t\Big |_{x=1}-\frac{C}{\epsilon ^2}\int _{0}^{T}s\lambda ^3\xi w^2\text{ d }t\Big |_{x=0}, \end{aligned}$$

where we have used that $0<\epsilon \le 1$ to adjust the powers of $\epsilon$. Finally, taking $\lambda \ge C$ and $s\ge C(1+T+e^{2\lambda \Vert \eta \Vert _\infty })$, we get

$$\begin{aligned}&BT\ge \frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=1} +\frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=0}. \end{aligned}$$

(95)

Let us focus now on the distributed terms. Using (90), we can rewrite

$$\begin{aligned}&DT_1=\frac{4}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi \eta _x^2w_x^2\,\text{ d }x\text{ d }t+\frac{4}{\epsilon ^2}\iint _{Q_T}s\lambda \xi \eta _{xx}w_x^2\,\text{ d }x\text{ d }t, \end{aligned}$$

and

$$\begin{aligned} DT_2=&-\frac{4}{\epsilon }\iint _{Q_T}s^2\alpha _{xt}\alpha _x w^2\,\text {d}x\text {d}t+\frac{1}{\epsilon ^2}\iint _{Q_T}s\alpha _{xxxx} w^2\,\text {d}x\text {d}t\\&-\iint _{Q_T}s\alpha _{tt}w^2\,\text {d}x\text {d}t+\frac{4}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^3\xi ^3\eta _x^2\eta _{xx}w^2\,\text {d}x\text {d}t+\frac{4}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3\eta _x^4w^2\,\text {d}x\text {d}t. \end{aligned}$$

Using estimates (90), we can bound by below as follows

$$\begin{aligned}&DT_1\ge \frac{4}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi \eta _x^2w_x^2\,\text {d}x\text {d}t-\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda \xi w_x^2\,\text {d}x\text {d}t. \end{aligned}$$

(96)

and

$$\begin{aligned} DT_2\ge&\frac{4}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3\eta _x^4w^2\,\text {d}x\text {d}t-\frac{C(T+1)}{\epsilon }\iint _{Q_T}s^2\lambda ^3\xi ^3 w^2\,\text {d}x\text {d}t-\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda ^4\xi w^2\,\text {d}x\text {d}t\nonumber \\&-C\left( T+T^2+e^{2\lambda \Vert \eta \Vert _\infty }\right) \iint _{Q_T}s\lambda ^2\xi ^3w^2\,\text {d}x\text {d}t-\frac{C}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^3\xi ^3w^2\,\text {d}x\text {d}t. \end{aligned}$$

(97)

Collecting estimates (95)–(97) yield

$$\begin{aligned} 2\left( P_e^\epsilon w,P_k^\epsilon w\right) _{L^2(Q_T)} \ge&\frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=1} +\frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=0} \\&+\frac{4}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi \eta _x^2w_x^2\,\text {d}x\text {d}t+\frac{4}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3\eta _x^4w^2\,\text {d}x\text {d}t\\&-\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda \xi w_x^2\,\text {d}x\text {d}t-\frac{C(T+1)}{\epsilon }\iint _{Q_T}s^2\lambda ^3\xi ^3 w^2\,\text {d}x\text {d}t\\&-\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda ^4\xi w^2\,\text {d}x\text {d}t-C\left( T+T^2+e^{2\lambda \Vert \eta \Vert _\infty }\right) \\&\times \iint _{Q_T}s\lambda ^2\xi ^3w^2\,\text {d}x\text {d}t-\frac{C}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^3\xi ^3w^2\,\text {d}x\text {d}t. \end{aligned}$$

Using property (21) and taking $\lambda \ge C$ and $s\ge C(T+T^2)$, we deduce

$$\begin{aligned}&2\left( P_e^\epsilon w,P_k^\epsilon w\right) _{L^2(Q_T)} +\frac{1}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s^3\lambda ^4\xi ^3 w^2 \,\text {d}x\text {d}t+\frac{1}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s\lambda ^2\xi w_x^2 \,\text {d}x\text {d}t\nonumber \\&\quad \ge \frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=1}+\frac{C}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=0} +\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi w_x^2\,\text {d}x\text {d}t\nonumber \\&\qquad +\frac{C}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3 w^2\,\text {d}x\text {d}t. \end{aligned}$$

(98)

Combining (98) with (94), we get

$$\begin{aligned} \begin{aligned}&C\Vert e^{-s\alpha } P(e^{s\alpha }w)\Vert _{L^2(Q_T)}^2 +\frac{C}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s^3\lambda ^4\xi ^3 w^2 \,\text {d}x\text {d}t\\&\qquad +\frac{C}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s\lambda ^2\xi w_x^2 \,\text {d}x\text {d}t\\&\quad \ge \Vert P_{e}^\epsilon w\Vert ^2_{L^2(Q_T)} +\Vert P_k^\epsilon w\Vert ^2_{L^2(Q_T)}+\frac{1}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=1}\\&\qquad +\frac{1}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=0} \\&\qquad +\frac{1}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi w_x^2\,\text {d}x\text {d}t+\frac{1}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3 w^2\,\text {d}x\text {d}t. \end{aligned} \end{aligned}$$

(99)

We will add terms corresponding to $w_{xx}$ and $w_t$ to the right-hand side of (99). For this, we multiply (92) by $s^{-1/2}\xi ^{-1/2}$ and take the $L^2$-norm, that is,

$$\begin{aligned} \frac{1}{\epsilon ^2}s^{-1}\Vert \xi ^{-1/2}w_{xx}\Vert ^2_{L^2(Q_T)}&=s^{-1}\Vert \xi ^{-1/2}(P_e^{\epsilon }w-s\alpha _tw\nonumber \\ {}&\quad -\frac{1}{\epsilon }s^2\alpha _x^2 w)\Vert ^2_{L^2(Q_T)} \nonumber \le C s^{-1}\iint _{Q_T}\xi ^{-1}|P_e^{\epsilon }w|^2\,\text{ d }x\text{ d }t\nonumber \\ {}&\quad +CT^2\iint _{Q_T}s^2\lambda ^2\xi ^3w^2\,\text{ d }x\text{ d }t\nonumber +\frac{C}{\epsilon ^2}s^3\iint _{Q_T}\lambda ^4\xi ^3 w^2\,\text{ d }x\text{ d }t\nonumber \\ {}&\le C s^{-1}\!\iint _{Q_T}\xi ^{-1}|P_e^{\epsilon }w|^2\,\text{ d }x\text{ d }t\!+\!\frac{C}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3w^2\,\text{ d }x\text{ d }t, \end{aligned}$$

(100)

where we have used that $\epsilon \le 1$ and $s\ge CT^2$ in the last line. Arguing in the same way and considering (93), it is not difficult to see that

$$\begin{aligned} s^{-1}\Vert \xi ^{-1/2}w_t\Vert ^2_{L^2(Q_T)}\nonumber&\le Cs^{-1}\iint _{Q_T}\xi ^{-1}|P_k^{\epsilon }w|^2\,\text {d}x\text {d}t\nonumber \\&\quad +\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi w_{x}^2\,\text {d}x\text {d}t +\frac{C}{\epsilon ^2}\iint _{Q_T}s\lambda ^4\xi w^2\,\text {d}x\text {d}t. \end{aligned}$$

(101)

Using that $s^{-1}\xi ^{-1}\le C$ for all $(t,x)\in Q_T$, we can use (100) and (101) to estimate from below in (99) and obtain

$$\begin{aligned}&C\Vert e^{-s\alpha } P(e^{s\alpha }w)\Vert _{L^2(Q_T)}^2\nonumber +\frac{C}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s^3\lambda ^4\xi ^3 w^2 \,\text {d}x\text {d}t\nonumber \\&\qquad +\frac{C}{\epsilon ^2}\iint _{\omega _{0}(t)\times (0,T)}s\lambda ^2\xi w_x^2 \,\text {d}x\text {d}t\nonumber \\&\quad \ge \ \frac{1}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=1} +\frac{1}{\epsilon ^2}\int _0^Ts^3\lambda ^3\xi ^3w^2\text {d}t\Big |_{x=0}\nonumber \\&\qquad +\frac{1}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi w_x^2\,\text {d}x\text {d}t +\frac{1}{\epsilon ^2}\iint _{Q_T}s^3\lambda ^4\xi ^3w^2\,\text {d}x\text {d}t\nonumber \\&\qquad + \frac{1}{\epsilon ^2}\iint _{Q_T}s^{-1}\xi ^{-1}|w_{xx}|^2\,\text {d}x\text {d}t+\iint _{Q_T}s^{-1}\xi ^{-1}|w_{t}|^2\,\text {d}x\text {d}t. \end{aligned}$$

(102)

To conclude the proof, we need to eliminate the local term of $w_x$ in the above equation. For this, consider a function $\zeta \in C^\infty ([0,T]\times [0,L])$ verifying

$$\begin{aligned} {\left\{ \begin{array}{ll} 0\le \zeta \le 1 &{}\forall (t,x)\in [0,T]\times [0,L], \\ \zeta (t,x)=1 &{}\forall t\in [0,T], \;\; \forall x\in \omega _0(t),\\ \zeta (t,x)=0 &{}\forall t\in [0,T], \;\; \forall x\in [0,L]{\setminus }\overline{\omega _1(t)}. \end{array}\right. } \end{aligned}$$

We have

$$\begin{aligned} \frac{1}{\epsilon ^2}\iint _{\omega _0(t)\times (0,T)}s\lambda ^2\xi w_{x}^2\,\text {d}x\text {d}t&\le \frac{1}{\epsilon ^2}\iint _{Q_T}\zeta s\lambda ^2\xi w_{x}^2\,\text {d}x\text {d}t\\&=-\frac{1}{\epsilon ^2}\int _{0}^{T}s\lambda ^2\xi \zeta w w_{x}\text {d}t\Big |_{0}^{1} -\frac{1}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi ww_{xx}\zeta \,\text {d}x\text {d}t\\&\quad -\frac{1}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi _{x} ww_{x}\zeta \,\text {d}x\text {d}t-\frac{1}{\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi ww_{x}\zeta _{x}\,\text {d}x\text {d}t. \end{aligned}$$

Using the fact that $w_x=-s\alpha _x w$ on $(0,T)\times \{0,1\}$ together with properties (25)–(26) and Cauchy–Schwarz and Young inequalities, we get

$$\begin{aligned} \frac{C}{\epsilon ^2}\iint _{w_0(t)\times (0,T)}s\lambda ^2\xi w_x^2&\le \ \frac{1}{2\epsilon ^2}\iint _{Q_T}s^{-1}\xi ^{-1}|w_{xx}|^2\,\text {d}x\text {d}t\\&\quad +\frac{1}{2\epsilon ^2}\iint _{Q_T}s\lambda ^2\xi w_x^2\,\text {d}x\text {d}t+\frac{C}{\epsilon ^2}\iint _{\omega _1(t)\times (0,T)}s^3\lambda ^4\xi ^3w^2\,\text {d}x\text {d}t. \end{aligned}$$

Using the above estimate in (102) and recalling the change of variables $w=e^{-s\alpha }\psi$ gives the desired result. This concludes the proof.

Proof of a precise observability inequality from the Carleman estimate

The goal of this part is to prove Proposition 2.6.

Proof

The proof is by now standard and relies on well-known arguments, we follow here the presentation in [17, Lemma 4.1]. We fix $\lambda , s$ large enough such that the Carleman estimate from Proposition 2.5 holds and such that we have the following estimate

$$\begin{aligned} e^{-4s\alpha ^\star +2s\widehat{\alpha }}\le C e^{-s\alpha }\le C e^{-s\alpha ^\star } . \end{aligned}$$

(103)

All the following positive constants $C>0$ can now depend on $\lambda ,s$.

By construction, $\alpha =\beta$ and $\xi =\gamma$ in $(T/2,T)\times (0,1)$, therefore

$$\begin{aligned}&\int _{T/2}^{T} \int _0^1 \gamma |\phi |^2 e^{-2s \beta } \,\text{ d }x\text{ d }t\nonumber +\int _{T/2}^{T} \int _0^1 \gamma ^3 |\psi |^2 e^{-2s \beta } \,\text{ d }x\text{ d }t\nonumber \\ {}&\quad = \int _{T/2}^{T} \int _0^1 \xi |\phi |^2 e^{-2s \alpha } \,\text{ d }x\text{ d }t+\int _{T/2}^{T} \int _0^1 \xi ^3 |\psi |^2 e^{-2s \alpha } \,\text{ d }x\text{ d }t. \end{aligned}$$

(104)

Moreover, from the definition of $\beta ^\star$ and $\alpha ^\star$, we readily see that $e^{-s\alpha ^\star }\le e^{-s\beta ^\star }$.

From this fact and noting that $\beta$ (resp. $\alpha$) blows up exponentially as $t\rightarrow T^{-}$ (resp. $t\rightarrow 0^+$ and $t\rightarrow T^{-}$) while $\gamma$ (resp. $\xi$) blows up polynomially as $t\rightarrow T^{-}$ (resp. $t\rightarrow 0^+$ and $t\rightarrow T^{-}$), we can use (104), (103) and our Carleman inequality (34) to deduce that

$$\begin{aligned}&\int _{T/2}^{T} \int _0^1 \gamma |\phi |^2 e^{-2s \beta } \,\text {d}x\text {d}t\nonumber + \int _{T/2}^{T} \int _0^1 \gamma ^3 |\psi |^2 e^{-2s \beta } \,\text {d}x\text {d}t\nonumber \\&\quad \le \ C(s,\lambda ) \Bigg ( \iint _{Q_T}e^{-2s\beta }\gamma ^3|g_1|^2\,\text {d}x\text {d}t+ \iint _{Q_T}e^{-2s\beta } |g_2|^2\,\text {d}x\text {d}t\nonumber \\&\qquad + \iint _{\omega _2\times (0,T)} e^{-s\beta ^\star }({\gamma }^\star )^{8}|\phi |^2\,\text {d}x\text {d}t\Bigg ) . \end{aligned}$$

(105)

For the set $(0,T/2)\times (0,1)$, we will use energy estimates for the system (18). More precisely, consider the function $\nu \in C^1([0,T])$ such that

$$\begin{aligned} \nu =1 \text { in } [0,T/2], \quad \nu =0 \text { in }[3T/4,T/4], \quad |\nu ^\prime (t)|\le C/T. \end{aligned}$$

(106)

Setting $({\widetilde{\phi }},{\widetilde{\psi }})=(\nu \phi ,\nu \psi )$, it is not difficult to see that the new variables verify the system

$$\begin{aligned} {\left\{ \begin{array}{ll} -\widetilde{\phi }_t=a_{11} {\widetilde{\phi }} + a_{21}{\widetilde{\psi }} + \nu g_1-\nu ^\prime \phi &{} \text {in } Q_T, \\ -{\widetilde{\psi }}_t={d_v}{\widetilde{\psi }}_{xx} +a_{12}{\widetilde{\phi }}+a_{22} {\widetilde{\psi }} +\nu g_2 - \nu ^\prime \psi &{}\text {in } Q_T, \\ \partial _x {\widetilde{\psi }}=0 &{}\text {on } \Sigma _T, \\ ({\widetilde{\phi }},{\widetilde{\psi }})(T,\cdot )=(0,0) &{}\text {in }(0,1). \end{array}\right. } \end{aligned}$$

(107)

From standard energy estimates, we deduce that system (107) verifies

$$\begin{aligned}&\int _{\Omega }|{\widetilde{\phi }}(t,x)|^2\,\text {d}x+\int _{\Omega }|\widetilde{\psi }(t,x)|^2\,\text {d}x+{d_v} \int _{t}^{T}\!\!\!\int _{\Omega }|\widetilde{\psi }_x|^2\,\text {d}x\text {d}t\\&\quad \le \ C\left( \int _{t}^{T}\!\!\!\int _{\Omega }|{\widetilde{\phi }}|^2\,\text {d}x\text {d}t\right. +\int _{t}^{T}\!\!\!\int _{\Omega }|{\widetilde{\psi }}|^2\,\text {d}x\text {d}t+\int _{t}^{T}\!\!\!\int _{\Omega }|\eta g_1|^2\,\text {d}x\text {d}t\\&\qquad +\int _{t}^{T}\!\!\!\int _{\Omega }|\eta g_2|^2\,\text {d}x\text {d}t+ \int _{t}^{T}\!\!\!\int _{\Omega }|\eta ^\prime \phi |^2\,\text {d}x\text {d}t\left. +\int _{t}^{T}\!\!\!\int _{\Omega }|\eta ^\prime \psi |^2\,\text {d}x\text {d}t\right) , \quad \forall t\in [0,T], \end{aligned}$$

where C is a positive constant only depending on $a_{ij}$.

Dropping the third term in the left-hand side of the above expression, we use Gronwall’s inequality to deduce

$$\begin{aligned}&\Vert {\widetilde{\phi }}(0, \cdot )\Vert ^2_{L^2(\Omega )} + \Vert {\widetilde{\psi }}(0, \cdot )\Vert ^2_{L^2(\Omega )} + \iint _{Q_T}|{\widetilde{\phi }}|^2 \,\text{ d }x\text{ d }t+ \iint _{Q_T}|{\widetilde{\psi }}|^2\,\text{ d }x\text{ d }t\\ {}&\quad \le C\left( \iint _{Q_T} |\eta g_1|^2\,\text{ d }x\text{ d }t+\iint _{Q_T}|\eta g_2|^2\,\text{ d }x\text{ d }t\right. \left. + \iint _{Q_T}|\eta ^\prime \phi |^2 \,\text{ d }x\text{ d }t\right. \nonumber \\ {}&\quad \left. +\iint _{Q_T}|\eta ^\prime \psi |^2\,\text{ d }x\text{ d }t\right) , \end{aligned}$$

for some constant $C>0$ only depending on T and $a_{ij}$.

Recalling the definition of $\eta$, we obtain from the above expression

$$\begin{aligned}&\Vert \phi (0, \cdot )\Vert ^2_{L^2(\Omega )} + \Vert \psi (0, \cdot )\Vert ^2_{L^2(\Omega )} + \int _{0}^{T/2}\!\!\!\!\int _{\Omega } | \phi |^2 \,\text {d}x\text {d}t+ \int _{0}^{T/2}\!\!\!\!\int _{\Omega }| \psi |^2\,\text {d}x\text {d}t\\&\quad \le C\left( \int _{0}^{3T/4}\!\!\!\!\int _{\Omega }| g_1|^2 \,\text {d}x\text {d}t\right. \left. +\int _{0}^{3T/4}\!\!\!\!\int _{\Omega }| g_2|^2\,\text {d}x\text {d}t\right) + \frac{C}{T^2}\left( \int _{T/2}^{3T/4}\!\!\!\!\int _{\Omega } |\phi |^2 \,\text {d}x\text {d}t\right. \\&\qquad \left. +\int _{T/2}^{3T/4}\!\!\!\!\int _{\Omega }| \psi |^2\,\text {d}x\text {d}t\right) . \end{aligned}$$

Since the domain of integration in the above integrals is away from the singularity of the weight functions (48) at $t=T$ (and therefore they are bounded), we can introduce them in the above inequality as follows

$$\begin{aligned}&\Vert \phi (0, \cdot )\Vert ^2_{L^2(\Omega )} + \Vert \psi (0, \cdot )\Vert ^2_{L^2(\Omega )} + \int _{0}^{T/2}\!\!\!\!\int _{\Omega } e^{-2s\beta } \gamma \left( | \phi |^2 + | \phi _x|^2 \right) \,\text {d}x\text {d}t\\&\qquad + \int _{0}^{T/2}\!\!\!\!\int _{\Omega } \gamma ^3 | \psi |^2 e^{-2s\beta }\,\text {d}x\text {d}t\\&\quad \le C(s,\lambda ,T)\left( \int _{0}^{3T/4}\!\!\!\!\int _{\Omega } e^{-2s\beta } \gamma ^3 | g_1|^2 \,\text {d}x\text {d}t\right. \left. +\int _{0}^{3T/4}\!\!\!\!\int _{\Omega }e^{-2s\beta } | g_2|^2\,\text {d}x\text {d}t\right) \\&\qquad + C(s,\lambda ,T)\left( \int _{T/2}^{3T/4}\!\!\!\!\int _{\Omega } e^{-2s\beta }\gamma |\phi |^2 \,\text {d}x\text {d}t\right. \left. +\int _{T/2}^{3T/4}\!\!\!\!\int _{\Omega } e^{-2s\beta }\gamma ^3| \psi _2|^2\,\text {d}x\text {d}t\right) . \end{aligned}$$

Using estimate (105) to bound all the terms on the last line of the above inequality and adding up the resulting expression to (105) yields

$$\begin{aligned}&\Vert \phi (0, \cdot )\Vert ^2_{L^2(\Omega )} +\Vert \psi (0, \cdot )\Vert ^2_{L^2(\Omega )} + \iint _{Q_T} e^{-2s\beta } \gamma | \phi |^2 \,\text {d}x\text {d}t+ \iint _{Q_T} \gamma ^3 | \psi |^2 e^{-2s\beta }\,\text {d}x\text {d}t\\&\quad \le C\left( \iint _{Q_T} e^{-2s\beta } \gamma ^3 | g_1|^2 \,\text {d}x\text {d}t+\iint _{Q_T} e^{-2s\beta } | g_2|^2\,\text {d}x\text {d}t\right) \\&\qquad +C\left( \iint _{\omega _2(t)\times (0,T)} e^{-s\beta ^\star }({\gamma }^\star )^{8}|\phi |^2 \,\text {d}x\text {d}t\right) . \end{aligned}$$

To conclude, it is enough to use definitions (48) in the above inequality. This ends the proof. $\square$

Some properties of the heat semigroup

We recall in the next result some well-known facts about the heat semigroup with a diffusion parameter $d_v >0$ with homogeneous Neumann boundary conditions on the interval (0, 1), denoted by $\{e^{t{d_v}\partial _{xx}}\}_{t\ge 0}$. The proof can be found for instance in [22, Lemma A.1].

Lemma C.1

The following properties hold true.

• a. For every constant $K\in {\mathbb {R}}$, we have $e^{t{d_v} \partial _{xx}}K=K$ for all $t\ge 0$.

• b. For every $z_0\in L^2(\Omega )$, there exists a constant $C>0$ only depending on $z_0$ such that for every $t \ge 0$,

  $$\begin{aligned} \left\Vert e^{t{d_v}\partial _{xx}}\left( z_0-\int _{\Omega }z_0\,\text {d}x\right) \right\Vert _{L^2(\Omega )}\le C e^{-\lambda _1 d_v t} \left\Vert z_0\right\Vert _{L^2(\Omega )}, \end{aligned}$$

  (108)

  where $\lambda _1 >0$ is the first positive eigenvalue of the Neumann Laplacian operator $-\partial _{xx}$ on (0, 1).