Exact boundary controllability of 1D semilinear wave equations through a constructive approach

Abstract

The exact controllability of the semilinear wave equation $y_{tt}-y_{xx}+ f(y)=0$, $x\in (0,1)$ assuming that f is locally Lipschitz continuous and satisfies the growth condition $\limsup _{\vert r\vert \rightarrow \infty } \vert f(r)\vert /(\vert r\vert \ln ^{p}\vert r\vert )\leqslant \beta$ for some $\beta$ small enough and $p=2$ has been obtained by Zuazua (Ann Inst H Poincaré Anal Non Linéaire 10(1):109–129, 1993). The proof based on a non-constructive fixed point arguments makes use of precise estimates of the observability constant for a linearized wave equation. Under the above asymptotic assumption with $p=3/2$, by introducing a different fixed point application, we present a simpler proof of the exact boundary controllability which is not based on the cost of observability of the wave equation with respect to potentials. Then, assuming that f is locally Lipschitz continuous and satisfies the growth condition $\limsup _{\vert r\vert \rightarrow \infty } \vert f^\prime (r)\vert /\ln ^{3/2}\vert r\vert \leqslant \beta$ for some $\beta$ small enough, we show that the above fixed point application is contracting yielding a constructive method to approximate the controls for the semilinear equation. Numerical experiments illustrate the results. The results can be extended to the multi-dimensional case and for nonlinearities involving the gradient of the solution.

Appendix: Proof of Theorem 7

In what follows, in order to simplify the notations, we shall just write $\rho =\rho (t)$ and $\rho _1=\rho _1(t)$ instead of $\rho =\rho (s;x,t)$ and $\rho _1=\rho _1(s;t)$.

Preliminary to the proof and following [18], for all $f\in \mathcal {C}^0(\mathbb R;E)$ (where E is a Banach space) and any $\tau >0$, we define $\delta _\tau f:= f\left( t+\frac{\tau }{2} \right) - f\left( t-\frac{\tau }{2} \right)$ and

$$\begin{aligned} \mathcal {T}_\tau f := \frac{1}{\tau }\delta _\tau \left( \frac{\delta _\tau f}{\tau }\right) = \frac{f(t+\tau )-2f(t)+f(t-\tau )}{\tau ^2}. \end{aligned}$$

Let now $w_s\in P_s$ and the solution $y_s \in L^2(Q_T)$ be given by 6. Then, z defined by $z=\mathcal {T}_\tau w_s$ belongs to $P_s$, where $w_s$ as well as $y_s$ can be extended uniquely on $(-\infty ,0)$ and $(T,+\infty )$. Indeed, in the interval $(-\infty , 0)$ the solution $y_s$ satisfies the following set of equations

(53)

where the source term $B\in L^2(Q_T)$ is assumed to be extendable by 0 outside (0, T). Recall that the boundary condition $y_s(1,t) =0$ holds outside (0, T) since $\eta =0$ (appearing in the formula of $v_s$) vanishes outside $(\delta , T-\delta )$.

Similarly, in $(T, + \infty )$ we can define the solution $y_s$ uniquely, and $y_s(t)=0$ for all $t\geqslant T$. It follows that the solution $y_s$ satisfies $y_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))\cap \mathcal {C}^1(\mathbb R;H^{-1}(\Omega ))$ and $y_s\in \mathcal {C}^0((-\infty ,\delta ];H^1(\Omega ))\cap \mathcal {C}^1((-\infty ,\delta ];L^2(\Omega ))$ and $y_s\in \mathcal {C}^0([T-\delta , +\infty );H^1(\Omega ))\cap \mathcal {C}^1([T-\delta ,+ \infty );L^2(\Omega ))$ (see [26]). We extend as well the weights $\rho$ and $\rho _1$ in $\Omega \times \mathbb {R}$ so that it preserves smoothness and positivity properties.

This ensures the extension of the solution $w_s$ which satisfies the following set of equations in $\mathbb R$

(54)

Moreover, it can be seen that $Lw_s=0$ in $[T, + \infty )$, since $y_s$ is a controlled solution to (53).

We now proceed to the proof of Theorem 7, done in three steps.

Step 1 : We suppose first that $u_0\in H^1_0(\Omega )\cap H^2(\Omega )$, $u_1\in H_0^1(\Omega )$ and $B\in \mathcal {D}(0,T;L^2(\Omega ))$ and prove that $v_s\in H^1(0,T)$ and $(y_s)_t\in L^2(Q_T)$.

We start by considering the variational formulation (12) by choosing $z=\mathcal {T}_\tau w_s$ as test function. Since $w_s \in \mathcal {C}^0(\mathbb R;H^1_0(\Omega ))\cap \mathcal {C}^1(\mathbb R;L^2(\Omega ))$ solves (54), it is clear that $\mathcal {T}_\tau w_s \in \mathcal {C}^0([0,T]; H^1_0(\Omega )) \cap \mathcal {C}^1([0,T]; L^2(\Omega ))$, $(\mathcal {T}_\tau w_s)_x\in L^2(0,T)$. With this z, the formulation reads

$$\begin{aligned} \int _{Q_T}\rho ^{-2} L w_s L \mathcal {T}_\tau w_s \,\text {d}x\text {d}t+ s \int _0^T \eta ^2(t) \rho ^{-2}_1 (w_s)_x(1,t) \mathcal {T}_\tau (w_s)_x(1,t) \text {d}t\nonumber \\ = \int _\Omega u_1(x) \mathcal {T}_\tau w_s(x,0) \,\text {d}x- \int _\Omega u_0(x) \mathcal {T}_\tau (w_s)_t(x,0) \,\text {d}x+ \int _{Q_T}B \mathcal {T}_\tau w_s \,\text {d}x\text {d}t. \end{aligned}$$

(55)

Sub-step 1. Let us start with the first integral in the left hand side of (55). We have

$$\begin{aligned} \begin{aligned}&\int _{Q_T}\rho ^{-2}(t)Lw_s(t) L \mathcal {T}_\tau w_s(t) \,\text {d}x\text {d}t\\&=\frac{1}{\tau }\int _{Q_T} \rho ^{-2}(t) Lw_s(t)\frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\,\text {d}x\text {d}t\\&-\frac{1}{\tau }\int _{Q_T} \rho ^{-2}(t) Lw_s(t)\frac{L w_s(t)-Lw_s(t-\tau )}{\tau }\,\text {d}x\text {d}t\\&=\frac{1}{\tau }\int _{Q_T} \rho ^{-2}(t) Lw_s(t)\frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\,\text {d}x\text {d}t\\&-\frac{1}{\tau }\int _{-\tau }^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\,\text {d}x\text {d}t\\&= \int _{Q_T} \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\\&-\frac{1}{\tau }\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\\&{-\frac{1}{\tau }\int _T^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t.} \end{aligned} \end{aligned}$$

(56)

Now, observe that

$$\begin{aligned} \begin{aligned}&\int _{Q_T} \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\\&= \int _{Q_T} \rho ^2(t)\left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \rho ^{-2}(t)\\&\times \left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\\&= - \int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\\& + \int _{Q_T} \rho ^2(t) \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \\&\times \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau ) \,\text {d}x\text {d}t. \end{aligned} \end{aligned}$$

(57)

The equality (56) then reads

$$\begin{aligned}&\int _{Q_T}\rho ^{-2} Lw_s(t) L \mathcal {T}_\tau w_s(t) \,\text {d}x\text {d}t\nonumber \\ =&- \int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\nonumber \\&+ \int _{Q_T} \rho ^2(t) \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \nonumber \\&\times \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau ) \,\text {d}x\text {d}t\nonumber \\&~~~~~~~~~~~~~~ -\frac{1}{\tau }\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\nonumber \\&{-\frac{1}{\tau }\int _T^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t.} \end{aligned}$$

(58)

Next, we shall look into the second term in the left hand side of (55). First, recall the smooth function $\eta$ given by (9) satisfies $\eta =0$ in $(-\infty , \delta ]\cup [T-\delta , +\infty )$ (with $\delta >0$ given in (9)). Then, in a similar way that have lead to (56), we have assuming $|\tau |\leqslant \delta$ :

$$\begin{aligned}&\int _0^T \eta ^2(t)\rho _1^{-2}(t)(w_s)_x(1,t) \mathcal {T}_\tau (w_s)_x(1,t) \text {d}t\nonumber \\&\quad = \int _0^T \left( \frac{\eta ^2(t)\rho _1^{-2}(t)(w_s)_x(1,t)-\eta ^2(t+\tau )\rho _1^{-2}(t+\tau ) (w_s)_x(1,t+\tau )}{\tau }\right) \nonumber \\&\qquad \times \left( \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right) \text {d}t. \end{aligned}$$

(59)

Then, using the identity

$$\begin{aligned} ad- bc = \frac{(a-c)(b+d)-(a+c)(b-d)}{2}, \qquad \forall (a,b,c,d) \in \mathbb R^4 \end{aligned}$$

(60)

with $a = \eta ^2(t) \rho ^{-2}_1(t)$, $b = (w_s)_x(1,t+\tau )$, $c=\eta ^2(t+\tau ) \rho ^{-2}_1(t+\tau )$ and $d = (w_s)_x(1,t)$, we obtain from (59)

$$\begin{aligned}&\int _0^T \eta ^2(t) \rho _1^{-2}(t)(w_s)_x(1,t) \mathcal {T}_\tau (w_s)_x(1,t) \text {d}t\nonumber \\ =&\int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t) - \eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) \left( (w_s)_x(1,t)+(w_s)_x(1,t+\tau ) \right) }{2\tau } \nonumber \\&\times \left( \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right) \text {d}t\nonumber \\&\qquad \quad -\int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau ) \right) }{2}\left| \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right| ^2 \text {d}t. \end{aligned}$$

(61)

Now, using (58) and (61) in the formulation (55), we have

$$\begin{aligned}&\int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\nonumber \\&+s\int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau ) \right) }{2}\left| \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right| ^2 \text {d}t\nonumber \\ =&\int _{Q_T} \rho ^2(t) \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \nonumber \\&\qquad \times \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau ) \,\text {d}x\text {d}t\nonumber \\&~~~~~~~~~~~~~~ -\frac{1}{\tau }\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\nonumber \\&{-\frac{1}{\tau }\int _T^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t}\nonumber \\&+s \int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t) - \eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) \left( (w_s)_x(1,t)+(w_s)_x(1,t+\tau ) \right) }{2\tau } \nonumber \\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \times \left( \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right) \text {d}t\nonumber \\&- \int _{Q_T}B \mathcal {T}_\tau w_s \,\text {d}x\text {d}t-\int _\Omega u_1 \mathcal {T}_\tau w_s(\cdot ,0) \,\text {d}x+ \int _\Omega u_0 \mathcal {T}_\tau (w_s)_t(\cdot ,0) \,\text {d}x\nonumber \\ :=&I_1 +I_2 + I_3 +I_4 +I_5 +I_6+I_7. \end{aligned}$$

(62)

Sub-step 2. In this step, we obtain precise estimates for the terms $I_1$ and ${I_4}$ and then an estimate of the left hand side of (62).

(i) Estimate of ${I_1}$. Young’s inequality leads to

$$\begin{aligned} |I_1| \leqslant&\frac{1}{2} \int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\nonumber \\&\quad + \frac{1}{2} \int _{Q_T} \rho ^2(t)\left| \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau )\right| ^2 \,\text {d}x\text {d}t. \end{aligned}$$

(63)

(ii) Estimate of ${I_4}$. We have

$$\begin{aligned} |{I_4}|&=s\left| \int _0^T \frac{\left( \eta (t)\rho _1^{-1}(t)-\eta (t+\tau )\rho _1^{-1}(t+\tau )\right) \left( (w_s)_x(1,t)+ (w_s)_x(1,t+\tau )\right) }{2\tau } \right. \nonumber \\&\left. \times \frac{\left( \eta (t)\rho _1^{-1}(t)+\eta (t+\tau )\rho _1^{-1}(t+\tau ) \right) \left( (w_s)_x(1,t+\tau )- (w_s)_x(1,t)\right) }{\tau } \text {d}t\right| \nonumber \\&\leqslant 2s\int _0^T \left| \frac{\left( \eta (t)\rho _1^{-1}(t)-\eta (t+\tau )\rho _1^{-1}(t+\tau )\right) \left( (w_s)_x(1,t)+ (w_s)_x(1,t+\tau )\right) }{2\tau } \right| ^2 \text {d}t\nonumber \\&\ \ +\frac{s}{8}\int _0^T \left| \eta (t)\rho _1^{-1}(t)+\eta (t+\tau )\rho _1^{-1}(t+\tau ) \right| ^2\left| \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right| ^2 \text {d}t\nonumber \\&\leqslant 2s\int _0^T \left| \frac{\left( \eta (t)\rho _1^{-1}(t)-\eta (t+\tau )\rho _1^{-1}(t+\tau )\right) \left( (w_s)_x(1,t)+ (w_s)_x(1,t+\tau )\right) }{2\tau } \right| ^2 \text {d}t\nonumber \\&\ \ +\frac{s}{2}\int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau ) \right) }{2}\left| \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right| ^2 \text {d}t. \end{aligned}$$

(64)

(iii) A first estimate of the left hand side of (62). The previous estimates and (62) give

$$\begin{aligned}&\frac{1}{2}\int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\nonumber \\&+ \frac{s}{2}\int _0^T\frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) }{2}\left| \frac{(w_s)_x(1,t)- (w_s)_x(1,t+\tau )}{\tau }\right| ^2 \text {d}t\nonumber \\&\leqslant \frac{1}{2} \int _{Q_T} \rho ^2(t)\left| \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau )\right| ^2 \,\text {d}x\text {d}t\nonumber \\&+ \left| {\frac{1}{\tau }}\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\right| \nonumber \\&+{\left| \frac{1}{\tau }\int _T^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\right| }\nonumber \\&+2s\int _0^T\left| \frac{\left( \eta (t)\rho _1^{-1}(t)-\eta (t+\tau )\rho _1^{-1}(t+\tau )\right) \left( (w_s)_x(1,t)+(w_s)_x(1,t+\tau ) \right) }{2\tau }\right| ^2 \text {d}t\nonumber \\&+\left| \int _{Q_T}B \mathcal {T}_\tau w_s \,\text {d}x\text {d}t\right| +\left| \int _\Omega u_0 \mathcal {T}_\tau (w_s)_t(\cdot ,0) \,\text {d}x\right| +\left| \int _\Omega u_1 \mathcal {T}_\tau w_s(\cdot ,0) \,\text {d}x\right| \nonumber \\&:=J_1+J_2+J_3+J_4+J_5+J_6+ J_7. \end{aligned}$$

(65)

Sub-step 3 : We prove that the left hand side of (65) is bounded uniformly with respect to ${|\tau |}\in [0,\delta ]$.

(i) $J_1$ is bounded. Since $\rho ^{-2}\in {\mathcal {C}^\infty (\mathbb {R}\times \overline{\Omega })}$, $(\rho ^{-1})_t=-2s\lambda \beta (t-\frac{T}{2}) \phi \rho ^{-1}$ and $Lw_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))$:

$$\begin{aligned}&\int _{Q_T} \rho ^2(t)\left| \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau )\right| ^2 \,\text {d}x\text {d}t\\&\quad \rightarrow 4s^2\lambda ^2\beta ^2\int _{Q_T} \biggl (t-\frac{T}{2}\biggr )^2 \phi ^2(t)\rho ^{2}(t)y_s^2 \,\text {d}x\text {d}t\end{aligned}$$

as $\tau \rightarrow 0$ and thus $J_1$ is bounded.

(ii) $J_2$ is bounded. Since $\rho ^{-2}Lw_s=y_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))$, $\rho ^{-2}\in {\mathcal {C}^\infty (\mathbb {R}\times \overline{\Omega })}$ and $L w_s=\rho ^2y_s\in \mathcal {C}^1((-\infty ,\delta ];L^2(\Omega ))$ we have, as $\tau \rightarrow 0$

$$\begin{aligned}&\left| {\frac{1}{\tau }}\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\right| \\&\quad \rightarrow \left| \int _\Omega y_s(0)(\rho ^2y_s)_t(0)\right| \end{aligned}$$

and thus $J_2$ is bounded.

(iii) $J_3$ is bounded. Since $\rho ^{-2}Lw_s=y_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))$, $\rho ^{-2}\in \mathcal {C}^\infty (\mathbb {R}\times \overline{\Omega })$ and $L w_s=\rho ^2y_s\in \mathcal {C}^1([T-\delta , +\infty );L^2(\Omega ))$ we have, as $\tau \rightarrow 0$

$$\begin{aligned}&\left| \frac{1}{\tau }\int _T^{T-\tau }\!\!\!\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\right| \\&\quad \rightarrow \left| \int _\Omega y_s(T)(\rho ^2y_s)_t(T)\right| =0 \end{aligned}$$

and thus $J_3$ is bounded.

(iv) $J_4$ is bounded. Since $(w_s)_x(1,\cdot )\in L^2(0,T)$ and $\eta \rho ^{-1}_1\in \mathcal {C}^1(\mathbb R)$ we have

$$\begin{aligned} \begin{aligned}&2s\int _0^T\left| \frac{\left( \eta (t)\rho _1^{-1}(t)-\eta (t+\tau )\rho _1^{-1}(t+\tau )\right) \left( (w_s)_x(1,t)+(w_s)_x(1,t+\tau ) \right) }{2\tau }\right| ^2 \text {d}t\\&\rightarrow 2s\int _0^T\left| (\eta \rho _1^{-1})_t(t) (w_s)_x(1,t)\right| ^2 \text {d}t\end{aligned} \end{aligned}$$

as $\tau \rightarrow 0$ and thus ${J_4}$ is bounded.

(v) $J_5$ is bounded. For $\tau$ small enough, since $B\in \mathcal {D}(\mathbb R;L^2(\Omega ))$ and $w_s\in \mathcal {C}{{}^0}(\mathbb R; L^2(\Omega ))$, we have

$$\begin{aligned} \left| \int _{Q_T}B \mathcal {T}_\tau w_s \,\text {d}x\text {d}t\right| = \left| \int _{-\tau }^{T+\tau }\!\!\!\int _\Omega \mathcal {T}_\tau B w_s\,\text {d}x\text {d}t\right| \rightarrow \left| \int _{Q_T}B_{tt} w_s\,\text {d}x\text {d}t\right| \end{aligned}$$

(66)

as $\tau \rightarrow 0$ and thus ${ J_5}$ is bounded.

(vi) $J_6$ is bounded. We have $L w_s=\rho ^2 y_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))$ and $w_s\in \mathcal {C}^0(\mathbb R;H^1_0(\Omega ))$, thus $(w_s)_{tt}= Lw_s + (w_s)_{xx} \in \mathcal {C}(\mathbb R;H^{-1}(\Omega ))$. We then have, for all $t \in \mathbb R$:

$$\begin{aligned} (w_s)_t(t)- (w_s)_t(0)=\int _0^t Lw_s(\xi ) \,\text {d}\xi + \int _0^t (w_s)_{xx}(\xi )\,\text {d}\xi . \end{aligned}$$

This yields

$$\begin{aligned} \mathcal {T}_\tau (w_s)_t(0)&=\frac{(w_s)_t(\tau )-2 (w_s)_t(0)+ (w_s)_t(-\tau )}{\tau ^2}\\&=\frac{1}{\tau ^2}\bigg (\int _0^{\tau } L w_s(\xi )\,\text {d}\xi +\int _0^{-\tau } L w_s(\xi )\,\text {d}\xi \\&\quad +\int _0^{\tau } (w_s)_{xx}(\xi )\,\text {d}\xi +\int _0^{-\tau } (w_s)_{xx} (\xi )\,\text {d}\xi \bigg ) \\&=\frac{2}{\tau } \int _0^{\tau }\frac{\xi }{\tau } \bigg (\frac{L w_s(\xi )-L w_s(-\xi )}{2\xi }\bigg )\,\text {d}\xi \\&\quad + \frac{2}{\tau } \int _0^{\tau }\frac{\xi }{\tau } \bigg (\frac{ (w_s)_{xx}(\xi )- (w_s)_{xx} (-\xi )}{2\xi }\bigg ) \,\text {d}\xi . \end{aligned}$$

Now, since $L w_s=\rho ^2y_s\in \mathcal {C}^1((-\infty ,\delta ];L^2(\Omega ))$, we write that

$$\begin{aligned}&\int _\Omega u_0\frac{2}{\tau } \int _0^{\tau } \frac{\xi }{\tau } \bigg (\frac{L w_s(\xi )-L w_s(-\xi )}{2\xi } \bigg )\,\text {d}\xi \,\text {d}x\nonumber \\&\rightarrow \int _\Omega u_0(\rho ^2y_s)_t(0)\,\text {d}x=-2s\lambda \beta T\int _\Omega \phi (0)\rho ^2(0)u_0^2\,\text {d}x+ \int _\Omega \rho ^2(0) u_0u_1 \,\text {d}x\ \ \text {as } \tau \rightarrow 0. \end{aligned}$$

(67)

On the other hand, since $u_0\in H^2(\Omega )\cap H^1_0(\Omega )$ and $w_s\in \mathcal {C}^0(\mathbb R; H^1_0(\Omega ))$:

$$\begin{aligned}&\frac{2}{\tau } \int _0^\tau \left\langle \frac{\xi }{\tau } \bigg (\frac{ (w_s)_{xx}(\xi )- (w_s)_{xx} (-\xi )}{2\xi }\bigg ) , u_0 \right\rangle _{H^{-1}, H^1_0} \,\text {d}\xi \nonumber \\&= \frac{2}{\tau } \int _0^\tau \frac{\xi }{\tau } \int _\Omega (u_0)_{xx} \frac{(w_s)(\xi )-(w_s)(-\xi )}{2\xi } \,\text {d}x \,\text {d}\xi \nonumber \\&\rightarrow \int _\Omega (u_0)_{xx} (w_s)_t(0) \,\text {d}x\quad \text {as}\, \tau \rightarrow 0 \end{aligned}$$

(68)

since moreover $w_s\in \mathcal {C}^1(\mathbb {R};L^2(\Omega ))$. Thus

$$\begin{aligned} \int _\Omega u_0 \mathcal {T}_\tau (w_s)_t(\cdot ,0) \,\text {d}x&\rightarrow -2s\lambda \beta T\int _\Omega \phi (0)\rho ^2(0)u_0^2 + \int _\Omega \rho ^2(0) u_0u_1\nonumber \\&\quad +\int _\Omega (u_0)_{xx} (w_s)_t(0) \,\text {d}x \end{aligned}$$

(69)

as $\tau \rightarrow 0$ and thus ${J_6}$ is bounded.

(vii) $J_7$ is bounded. We have $L w_s=\rho ^2 y_s\in \mathcal {C}^0(\mathbb R;L^2(\Omega ))$ and $w_s\in \mathcal {C}^1(\mathbb R;H^1_0(\Omega ))$, thus $(w_s)_{tt}= \rho ^2 y_s + (w_s)_{xx} \in \mathcal {C}^0(\mathbb R;H^{-1}(\Omega ))$. Therefore

$$ \mathcal {T}_\tau w(0)=\frac{w(\tau )-2w(0)+w(-\tau )}{\tau ^2}\rightarrow (w_s)_{tt}(0)=(w_s)_{xx}(0)+\rho ^2(0)u_0\hbox { in } H^{-1}(\Omega )$$

as $\tau \rightarrow 0$ and thus

$$\begin{aligned} \int _\Omega u_1 \mathcal {T}_\tau w (\cdot ,0) \,\text {d}x\rightarrow&\langle (w_s)_{tt}(0),u_1\rangle _{H^{-1}(\Omega )\times H^1_0(\Omega )}\nonumber \\&=\langle (w_s)_{xx}(0),u_1\rangle _{H^{-1}(\Omega )\times H^1_0(\Omega )}+\int _\Omega \rho ^2(0) u_1 u_0\,\text {d}x\\&= -\int _\Omega (w_s)_{x}(\cdot ,0)(u_1)_x\,\text {d}x+\int _\Omega \rho ^2(0) u_1 u_0 \nonumber \,\text {d}x. \end{aligned}$$

(70)

as $\tau \rightarrow 0$.

${J_7}= \left| \int _\Omega u_1 \mathcal {T}_\tau w (\cdot ,0) \,\text {d}x\right|$ is therefore bounded.

(viii) Then we can conclude, from (65), that the terms

$$\begin{aligned} \int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\end{aligned}$$

and

$$\begin{aligned} \int _0^T\frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) }{2}\left| \frac{(w_s)_x(1,t)- (w_s)_x(1,t+\tau )}{\tau }\right| ^2 \text {d}t\end{aligned}$$

are bounded. Remark that this implies that the two terms $\int _{Q_T}\rho ^{-2}(t) |L (\frac{\delta _\tau w_s}{\tau })|^2 \,\text {d}x\text {d}t$ and $\int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(\frac{\delta _\tau w_s}{\tau })_x(1,t)|^2 \text {d}t$ are bounded; indeed,

$$\begin{aligned}&\int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(\frac{\delta _\tau w_s}{\tau })_x(1,t)|^2 \text {d}t\\&\leqslant 2\int _0^T\frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) }{2}\left| \frac{(w_s)_x(1,t)- (w_s)_x(1,t+\tau )}{\tau }\right| ^2 \text {d}t. \end{aligned}$$

We also have

$$\begin{aligned} \begin{aligned}&\int _{Q_T}\rho ^{-2}(t) |L (\frac{\delta _\tau w_s}{\tau })|^2 \,\text {d}x\text {d}t\\&\quad \leqslant 2\int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\\&\qquad +2\int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t)-\rho ^{-2}(t+\tau )}{\tau }Lw_s(t+\tau )\right| ^2 \,\text {d}x\text {d}t\\ \end{aligned} \end{aligned}$$

and each term of the right hand side is bounded.

Sub-step 4. In this step, we prove that $v_s\in H^1(0,T)$ and $y_s\in \mathcal {C}^0([0,T];H^1(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$.

Since $\frac{\delta _\tau w_s}{\tau }\in \mathcal {C}^0([0,T];H_0^1(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$ and satisfies $(\frac{\delta _\tau w_s}{\tau })_x(1,\cdot )\in L^2(0,T)$ then the Carleman estimates (10) gives

$$\begin{aligned}&s \int _{Q_T}\rho ^{-2}(t) \biggl (| (\frac{\delta _\tau w_s}{\tau })_{t}|^2 + | (\frac{\delta _\tau w_s}{\tau })_x|^2\biggr ) \,\text {d}x\text {d}t\\&\quad + s^3 \int _{Q_T}\rho ^{-2}(t) | \frac{\delta _\tau w_s}{\tau }|^2 \,\text {d}x\text {d}t\\&\quad + s \int _\Omega \rho ^{-2}(0) \biggl (|(\frac{\delta _\tau w_s}{\tau })_t(x,0)|^2 + |(\frac{\delta _\tau w_s}{\tau })_x(x,0)|^2\biggr ) \,\text {d}x\\&\quad + s^3 \int _\Omega \rho ^{-2}(0) |\frac{\delta _\tau w_s}{\tau }(x,0)|^2 \,\text {d}x\\&\quad \leqslant C \int _{Q_T}\rho ^{-2}(t) |L (\frac{\delta _\tau w_s}{\tau })|^2 \,\text {d}x\text {d}t+ C s \int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(\frac{\delta _\tau w_s}{\tau })_x(1,t)|^2 \text {d}t. \end{aligned}$$

Therefore, since the right hand side is bounded, $(\frac{\delta _\tau w_s}{\tau })_{t}$ and $(\frac{\delta _\tau w_s}{\tau })_x$ are bounded in $L^2(Q_T)$ and thus $(w_s)_{tt}\in L^2(Q_T)$ and $(w_s)_{t}\in L^2(0,T;H^1_0(\Omega ))$. Moreover, $\frac{\delta _\tau w_s}{\tau }(\cdot ,0)$ is bounded in $H^1_0(\Omega )$ thus $(w_s)_t(\cdot ,0)\in H^1_0(\Omega )$. We also have $L (\frac{\delta _\tau w_s}{\tau })$ bounded in $L^2(Q_T)$ so $L( w_s)_t\in L^2(Q_T)$. Thus $(w_s)_t$ satisfies

$$\begin{aligned} \left\{ \begin{aligned}&L (w_s)_t \in L^2(Q_T), \\&(w_s)_t(0,t)=(w_s)_t(1,t) =0, \quad t\in (0,T) \\&((w_s)_t (0),(w_s)_{tt}(0))\in H_0^1(\Omega )\times L^2(\Omega ) \end{aligned} \right. \end{aligned}$$

(71)

and thus $(w_s)_t\in \mathcal {C}^0([0,T];H^1_0(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$ and $(w_s)_{tx}(1,\cdot )\in L^2(0,T)$. Therefore from the definition of $v_s$, $v_s\in H^1(0,T)$ while from the equation satisfied by $(y_s,v_s)$ (see (17)), $y_s\in \mathcal {C}^0([0,T];H^1(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$.

Remark 1 We then have $w_s\in \mathcal {C}^1([0,T];H^1_0(\Omega ))\cap \mathcal {C}^2([0,T];L^2(\Omega ))$ and from the equation satisfied by $w_s$, since $Lw_s\in \mathcal {C}^1([0,T];L^2(\Omega ))$ we deduce that $(w_s)_{xx}=(w_s)_{tt}-L w_s \in \mathcal {C}^0([0,T];L^2(\Omega ))$ and thus that $(w_s)_{xx}(\cdot ,0)\in L^2(\Omega )$.

Step 2 : In this step, we give estimates on $(v_s)_t$ and $(y_s)_t$.

First of all, since $(w_s)_t\in \mathcal {C}^0([0;T];H^1_0(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$, $L (w_s)_t\in L^2(Q_T)$ and $(w_s)_{tx}(1,\cdot )\in L^2(0,T)$, we can write the Carleman estimate (10) for $(w_s)_t$ leading to

$$\begin{aligned}&s \int _{Q_T}\rho ^{-2}(t) (| (w_s)_{tt}|^2 + | (w_s)_{tx}|^2) \,\text {d}x\text {d}t+ s^3 \int _{Q_T}\rho ^{-2}(t) | (w_s)_t|^2 \,\text {d}x\text {d}t\nonumber \\&\quad + s \int _\Omega \rho ^{-2}(0) (|(w_s)_{tt}(x,0)|^2 + (w_s)_{tx}(x,0)|^2) \,\text {d}x+ s^3 \int _\Omega \rho ^{-2}(0) |(w_s)_t(x,0)|^2 \,\text {d}x\nonumber \\&\quad \leqslant C \int _{Q_T}\rho ^{-2}(t) |(L w_s)_t|^2 \,\text {d}x\text {d}t+ C s \int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(w_s)_{tx}(1,t)|^2 \text {d}t. \end{aligned}$$

(72)

Sub-step 1 : In this step, we pass to the limit when $\tau \rightarrow 0$ in equation (62). We have, since $y_s=\rho ^{-2} Lw_s\in \mathcal {C}^1(\mathbb {R};L^2(\Omega ))$ :

$$\begin{aligned} \int _{Q_T} \rho ^2(t)\left| \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right| ^2 \,\text {d}x\text {d}t\rightarrow \int _{Q_T} \rho ^2(t)\left| (y_s)_t\right| ^2 \,\text {d}x\text {d}t\end{aligned}$$

as $\tau \rightarrow 0$ and since $(w_s)_{tx}(1,\cdot )\in L^2(-\delta ,T+\delta )$ and $\eta \rho _1^{-1}\in \mathcal {C}(\mathbb {R})$ :

$$\begin{aligned} \begin{aligned} \int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t)+\eta ^2(t+\tau )\rho _1^{-2}(t+\tau ) \right) }{2}&\left| \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right| ^2 \text {d}t\\&\rightarrow \int _0^T \eta ^2(t)\rho _1^{-2}(t) \left| (w_s)_{tx}(1,t)\right| ^2 \text {d}t\end{aligned} \end{aligned}$$

as $\tau \rightarrow 0$. Since $y_s=\rho ^{-2} Lw_s\in \mathcal {C}^1(\mathbb R;L^2(\Omega ))$, $Lw_s\in \mathcal {C}^1(\mathbb R;L^2(\Omega ))$ and $(\rho ^{-1})_t=-2s\lambda \beta (t-\frac{T}{2}) \phi \rho ^{-1}$ in $Q_T$, we infer that

$$\begin{aligned}&\int _{Q_T} \rho ^2(t) \left( \frac{\rho ^{-2}(t) Lw_s(t)-\rho ^{-2}(t+\tau )Lw_s(t+\tau )}{\tau }\right) \\&\quad \times \left( \frac{\rho ^{-2}(t) -\rho ^{-2}(t+\tau )}{\tau }\right) Lw_s(t+\tau ) \,\text {d}x\text {d}t\\&\quad \rightarrow -2s\lambda \beta \int _{Q_T}(t-\frac{T}{2}) \phi (t) \rho ^2(t) (y_s)_ty_s\,\text {d}x\text {d}t, \end{aligned}$$

$$\begin{aligned}&\frac{1}{\tau }\int _{-\tau }^{0}\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\\&\quad \rightarrow \int _\Omega y_s(0)(\rho ^{2}y_s)_t(0)\,\text {d}x\end{aligned}$$

and

$$\begin{aligned} \frac{1}{\tau }\int _T^{T-\tau }\int _\Omega \rho ^{-2}(t+\tau )Lw_s(t+\tau )\left( \frac{L w_s(t+\tau )-Lw_s(t)}{\tau }\right) \,\text {d}x\text {d}t\rightarrow 0 \end{aligned}$$

as $\tau \rightarrow 0$.

Similarly, since $w_s\in \mathcal {C}^2(\mathbb R;L^2(\Omega ))$ and $(w_s)_{tx}(1,\cdot )\in L^2(-\delta ,T+\delta )$,

$$\begin{aligned}&\int _0^T \frac{\left( \eta ^2(t)\rho _1^{-2}(t) - \eta ^2(t+\tau )\rho _1^{-2}(t+\tau )\right) \left( (w_s)_x(1,t)+(w_s)_x(1,t+\tau ) \right) }{2\tau } \nonumber \\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \times \left( \frac{(w_s)_x(1,t+\tau )-(w_s)_x(1,t)}{\tau }\right) \text {d}t\\&\rightarrow \int _0^T (\eta ^2\rho _1^{-2})_t(w_s)_x(1,t)(w_s)_{tx}(1,t)\text {d}t\end{aligned}$$

and

$$\begin{aligned} \int _{Q_T}B \mathcal {T}_\tau w_s \,\text {d}x\text {d}t\rightarrow \int _{Q_T}B (w_s)_{tt} \,\text {d}x\text {d}t\end{aligned}$$

as $\tau \rightarrow 0$. Since $(w_s)_{tt}(\cdot ,0)\in L^2(\Omega )$, the convergence (70) reads

$$\begin{aligned} \int _\Omega u_1 \mathcal {T}_\tau w_s(\cdot ,0) \,\text {d}x\rightarrow \int _\Omega (w_s)_{tt}(\cdot ,0)u_1\,\text {d}x. \end{aligned}$$

Similarly, since $(w_s)_t(\cdot ,0)\in H^1_0(\Omega )$, (69) reads

$$\begin{aligned}&\int _\Omega u_0 \mathcal {T}_\tau (w_s)_t(\cdot ,0) \,\text {d}x\rightarrow -2s\lambda \beta T\int _\Omega \phi (0)\rho ^2(0)u_0^2\,\text {d}x+ \int _\Omega \rho ^2(0) u_0u_1\,\text {d}x\\&\quad -\int _\Omega (u_0)_{x} (w_s)_{tx}(\cdot ,0) \,\text {d}x. \end{aligned}$$

We conclude that the limit with respect to $\tau \rightarrow 0$ in (62) leads to the following equality

$$\begin{aligned}&\int _{Q_T} \rho ^2(t)\left| (y_s)_t\right| ^2 \,\text {d}x\text {d}t+s\int _0^T \eta ^2(t)\rho _1^{-2}(t) \left| (w_s)_{tx}(1,t)\right| ^2 \text {d}t\nonumber \\&\quad =-2s\lambda \beta \int _{Q_T}(t-\frac{T}{2}) \phi (t) \rho ^2(t) (y_s)_ty_s\,\text {d}x\text {d}t- \int _\Omega y_s(0)( \rho ^2y_s)_t(0)\nonumber \\&\quad +s\int _0^T (\eta ^2\rho _1^{-2})_t(t)(w_s)_x(1,t)(w_s)_{tx}(1,t)\text {d}t\nonumber \\&\quad - \int _{Q_T}B (w_s)_{tt} \,\text {d}x\text {d}t-\int _\Omega (w_s)_{tt}(\cdot ,0)u_1\,\text {d}x\nonumber \\&\quad -2s\lambda \beta T\int _\Omega \phi (0)\rho ^2(0)u_0^2\,\text {d}x+ \int _\Omega \rho ^2(0) u_1 u_0\,\text {d}x\nonumber \\&\quad -\int _\Omega (u_0)_{x} (w_s)_{tx}(\cdot ,0) \,\text {d}x\nonumber \\&\quad :=K_1+K_2+K_3+K_4+K_5+K_6+K_7+K_8. \end{aligned}$$

(73)

Sub-step 2 : In this step, we estimate each term $K_i, i=1,\cdots ,8$.

(i) We get that, there exists $C>0$ only depending on T such that

$$|K_1|\leqslant \frac{1}{8}\int _{Q_T} \rho ^2(t) |(y_s)_t|^2\,\text {d}x\text {d}t+Cs^2\int _{Q_T} \rho ^2(t) |y_s|^2\,\text {d}x\text {d}t.$$

(ii) Similarly, recalling that $y_s(\cdot ,0)=u_0$ and $(y_s)_t(\cdot ,0)=u_1$, there exists $C>0$ such that

$$|K_2|\leqslant Cs \Vert \rho (0)u_0\Vert ^2_{L^2(\Omega )}+ \Vert \rho (0)u_0\Vert _{L^2(\Omega )}\Vert \rho (0)u_1\Vert _{L^2(\Omega )}.$$

(iii) Using that $(\rho _1^{-1})_t=-2s\lambda \beta (t-\frac{T}{2}) \phi \rho _1^{-1}$, we obtain

$$\begin{aligned}&|K_3|=\left| s\int _0^T (\eta ^2\rho _1^{-2})_t(w_s)_x(1,t)(w_s)_{tx}(1,t)\text {d}t\right| \\&\leqslant 2\left( s\int _0^T |(\eta \rho _1^{-1})_t|^2|(w_s)_x(1,t)|^2 \text {d}t\right) ^{1/2}\left( s\int _0^T \eta ^2(t)\rho _1^{-2}(t)| (w_s)_{tx}(1,t)|^2\text {d}t\right) ^{1/2}\\&\leqslant C\left( s^3\int _0^T \eta ^2(t)\rho _1^{-2}(t)|(w_s)_x(1,t)|^2 \text {d}t+s\int _0^T\rho _1^{-2}(t)|(w_s)_x(1,t)|^2 \text {d}t\right) ^{1/2}\\&\times \left( s\int _0^T \eta ^2(t)\rho _1^{-2}(t)| (w_s)_{tx}(1,t)|^2\text {d}t\right) ^{1/2}\\&\leqslant C\left( s\int _\delta ^{T-\delta } \frac{\rho _1^2(t)}{\eta ^2(t)}v_s^2 \text {d}t+s\int _0^T\rho _1^{-2}(t)|(w_s)_x(1,t)|^2 \text {d}t\right) \\&\qquad +\frac{s}{8}\int _0^T \eta ^2(t)\rho _1^{-2} | (w_s)_{tx}(1,t)|^2\text {d}t. \end{aligned}$$

We now estimate the term $\int _0^T\rho _1^{-2}|(w_s)_x(1,t)|^2 \text {d}t$ appearing in the previous inequality: proceeding as in [23, Lemma 3.7] with $q(x,t)=x\rho ^{-2}(x,t)$ such that $q(0,t)=0$ and $q(1,t)=\rho _1^{-2}(t)$, we get the equality

$$\begin{aligned}&\frac{1}{2}\int _0^T \rho _1^{-2}(t)|(w_s)_x(1,t)|^2\\&\quad =2\int _{Q_T} x \rho ^{-1}\rho _t^{-1} w_x w_t +\frac{1}{2}\int _{Q_T} (\rho ^{-2}-2x\rho ^{-1}\rho _x^{-1})(w_x^2+w_t^2)\\&\qquad +\int _\Omega [x\rho ^{-2} w_tw_x]_0^T. \end{aligned}$$

Writing that $\vert \rho ^{-1}\rho _x^{-1}\vert \leqslant Cs \rho ^{-2}$ and $\vert \rho ^{-1}\rho _t^{-1}\vert \leqslant Cs \rho ^{-2}$, we obtain (since $s\geqslant 1$)

$$\begin{aligned}&\frac{1}{2}\int _0^T \rho _1^{-2}(t)|(w_s)_x(1,t)|^2\\&\quad \leqslant Cs\int _{Q_T} \rho ^{-2}(w_x^2+w_t^2) +Cs\int _\Omega \biggl (\rho ^{-2}(0)(w_t^2+w_x^2)(0)\\&\qquad +\rho ^{-2}(T)(w_t^2+w_x^2)(T)\biggr ) \end{aligned}$$

leading, using the Carleman estimate (10), to

$$\begin{aligned} s\int _0^T\rho _1^{-2}|(w_s)_x(1,t)|^2 \text {d}t\leqslant Cs\left( \Vert \rho y_s\Vert _{L^2(Q_T)}^2+s^{-1}\Vert \frac{\rho _1}{\eta }v_s\Vert _{L^2(\delta ,T-\delta )}^2\right) . \end{aligned}$$

(74)

Thus,

$$|K_3|\leqslant Cs\left( \Vert \rho y_s\Vert _{L^2(Q_T)}^2+ \Vert \frac{\rho _1}{\eta }v_s\Vert _{L^2(\delta ,T-\delta )}^2\right) +\frac{s}{8}\int _0^T \eta ^2(t)\rho _1^{-2} (w_s)_{tx}(1,t)\text {d}t.$$

(iv) Using the Carleman estimate (72) we have

$$\begin{aligned}&|K_4|=\left| \int _{Q_T}B (w_s)_{tt} \,\text {d}x\text {d}t\right| \leqslant \left( s^{-1}\int _{Q_T}\rho ^2B^2\right) ^{1/2} \left( s\int _{Q_T}\rho ^{-2}|(w_s)_{tt}|^2\right) ^{1/2}\\&\leqslant C\left( s^{-1}\int _{Q_T}\rho ^2B^2\right) ^{1/2} \left( \int _{Q_T}\rho ^{-2} |(\rho ^2 y_s)_t |^2 \,\text {d}x\text {d}t+ s \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant C\left( s^{-1}\int _{Q_T}\rho ^2B^2\right) ^{1/2} \left( s^2 \int _{Q_T}\rho ^{2} | y_s |^2 \,\text {d}x\text {d}t+\int _{Q_T}\rho ^{2}|( y_s)_t |^2 \,\text {d}x\text {d}t\right. \\&\quad \left. + s \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant C\left( s^{-1}\int _{Q_T}\rho ^2B^2 + s^2 \int _{Q_T}\rho ^{2} | y_s |^2 \,\text {d}x\text {d}t\right) +\frac{1}{8}\int _{Q_T}\rho ^{2} |( y_s)_t |^2 \,\text {d}x\text {d}t\\&\quad + \frac{s}{8} \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t. \end{aligned}$$

(v) Similarly, using again the Carleman estimate (72) we have

$$\begin{aligned} |K_5|&=\left| \int _\Omega (w_s)_{tt}(\cdot ,0)u_1\,\text {d}x\right| \leqslant \Vert \rho (0)u_1\Vert _{L^2(\Omega )} \Vert \rho ^{-1}(0)(w_s)_{tt}(\cdot ,0)\Vert _{L^2(\Omega )}\\&\leqslant C s^{-1/2}\Vert \rho (0) u_1\Vert _{L^2(\Omega )} \left( \int _{Q_T}\rho ^{-2}(t) |(\rho ^2 y_s)_t |^2 \,\text {d}x\text {d}t\right. \\&\quad \left. + s \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant C s^{-1/2}\Vert \rho (0) u_1\Vert _{L^2(\Omega )} \left( s^2 \int _{Q_T}\rho ^{2} | y_s |^2 \,\text {d}x\text {d}t+\int _{Q_T}\rho ^{2} |( y_s)_t |^2 \,\text {d}x\text {d}t\right. \\&\left. + s \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant C\left( s^{-1}\Vert \rho (0) u_1\Vert ^2_{L^2(\Omega )} + s^2 \int _{Q_T}\rho ^{2} | y_s |^2 \,\text {d}x\text {d}t\right) +\frac{1}{8} \int _{Q_T}\rho ^{2} |( y_s)_t |^2 \,\text {d}x\text {d}t\\&\quad + \frac{s}{8} \int _0^T \eta ^2(t)\rho ^{-2}_1 |(w_s)_{tx}(1,t)|^2 \text {d}t. \end{aligned}$$

(vi) Simpler, we get

$$|K_6|\leqslant Cs\Vert \rho (0)u_0\Vert ^2_{L^2(\Omega )} $$

(vii) and

$$ |K_7|\leqslant \Vert \rho (0)u_0\Vert _{L^2(\Omega )}\Vert \rho (0)u_1\Vert _{L^2(\Omega )}. $$

(viii) Eventually, (72) leads to

$$\begin{aligned} |K_8|&=\left| \int _\Omega (u_0)_{x} (w_s)_{tx}(\cdot ,0)\right| \leqslant \Vert \rho (0)(u_0)_x\Vert _{L^2(\Omega )} \Vert \rho ^{-1}(0)(w_s)_{tx}(\cdot ,0)\Vert _{L^2(\Omega )}\\&\leqslant Cs^{-1/2} \Vert \rho (0) (u_0)_{x}\Vert _{L^2(\Omega )} \left( \int _{Q_T}\rho ^{-2}(t) |(\rho ^2 y_s)_t |^2 \,\text {d}x\text {d}t\right. \\&\quad \left. + C s \int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant Cs^{-1/2} \Vert \rho (0) (u_0)_{x}\Vert _{L^2(\Omega )} \left( s^2 \int _{Q_T}\rho ^{2}(t) | y_s |^2 \,\text {d}x\text {d}t+\int _{Q_T}\rho ^{2}(t) |( y_s)_t |^2 \,\text {d}x\text {d}t\right. \\&\quad + \left. s \int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(w_s)_{tx}(1,t)|^2 \text {d}t\right) ^{1/2}\\&\leqslant C \left( s^{-1} \Vert \rho (0) (u_0)_{x}\Vert ^2_{L^2(\Omega )} + s^2 \int _{Q_T}\rho ^{2}(t) | y_s |^2 \,\text {d}x\text {d}t\right) \\&\quad +\frac{1}{8}\int _{Q_T}\rho ^{2}(t) |( y_s)_t |^2 \,\text {d}x\text {d}t\\&\quad + \frac{s}{8} \int _0^T \eta ^2(t)\rho ^{-2}_1(t) |(w_s)_{tx}(1,t)|^2 \text {d}t. \end{aligned}$$

Sub-step 3 : In this step, we give estimates on $(v_s)_t$ and $(y_s)_t$. Collecting the previous estimates, we get from (73)

$$\begin{aligned} \begin{aligned} \int _{Q_T}&\rho ^2(t)\left| (y_s)_t\right| ^2 \,\text {d}x\text {d}t+s\int _0^T \eta ^2(t)\rho _1^{-2}(t) \left| (w_s)_{tx}(1,t)\right| ^2 \text {d}t\\&\leqslant C\left( s^2\Vert \rho y_s \Vert ^2_{L^2(Q_T)}+s \Vert \frac{\rho _1}{\eta }v\Vert _{L^2(\delta ,T-\delta )}^2 + s^{-1}\int _{Q_T}\rho ^2B^2 \right. \\&\qquad \left. +s\Vert \rho (0)u_0\Vert ^2_{L^2(\Omega )} +s^{-1}\Vert \rho (0)(u_0)_x\Vert _{L^2(\Omega )} ^2+s^{-1}\Vert \rho (0) u_1\Vert _{L^2(\Omega )}^2\right) . \end{aligned} \end{aligned}$$

(75)

We have

$$\begin{aligned} \begin{aligned}&s^{-1}\int _0^T \rho _1^2(t) \left| (v_s)_t\right| ^2dt\\&\leqslant C\left( s\int _0^T \eta ^2(t)\rho _1^{-2}(t) \left| (w_s)_{tx}(1,t)\right| ^2+s\int _\delta ^{T-\delta } \frac{\rho _1^2(t)}{\eta ^2(t)}v_s^2(t) \text {d}t\right. \\&\quad \left. +s\int _0^T \rho _1^{-2}(t) \left| (w_s)_{x}(1,t)\right| ^2\right) \end{aligned} \end{aligned}$$

(76)

thus using the estimates (74) and (14), (75) implies for $s\geqslant s_0\geqslant 1$ that

$$\begin{aligned} \begin{aligned}&\int _{Q_T} \rho ^2(t)\left| (y_s)_t\right| ^2 \,\text {d}x\text {d}t+s^{-1}\int _0^T \rho _1^2(t) \left| (v_s)_t\right| ^2dt\\&\leqslant C \left( s^{-1} \Vert \rho B\Vert ^2_{L^2(Q_T)} +se^{-2s}\Vert u_0\Vert ^2_{L^2(\Omega )} +s^{-1}e^{-2s}\Vert (u_0)_x\Vert _{L^2(\Omega )} ^2\right. \\&\quad \left. +s^{-1}e^{-2s}\Vert u_1\Vert _{L^2(\Omega )}^2\right) \\ \end{aligned} \end{aligned}$$

(77)

which gives the announced estimate (19) in the case of regular data.

Step 3 : Case where $B\in L^2(Q_T)$ and $(u_0,u_1)\in H^1_0(\Omega )\times L^2(\Omega )$. We proceed by density: there exist $(u_0^n)_{n\in \mathbb N}\in H^2(\Omega )\cap H^1_0(\Omega )$, $(u_1^n)_{n\in \mathbb N}\in H^1_0(\Omega )$ and $(B^n)_{n\in \mathbb N}\in \mathcal {D}(0,T;L^2(\Omega ))$ such that $u_0^n\rightarrow u_0$ in $H^1_0(\Omega )$, $u_1^n\rightarrow u_1$ in $L^2(\Omega )$ and $B^n\rightarrow B$ in $L^2(Q_T)$ as $n\rightarrow \infty$.

Let $(y^n_s,v^n_s)$ be the solution of (6) given in Theorem 6 associated to $(u_0^n, u_1^n,B^n)$. Then, by linearity, we have for all $(n,m)\in \mathbb N^2$, from (14)

$$\begin{aligned}&\Vert \rho (y^n_s-y^m_s)\Vert _{L^2(Q_T)} + s^{-1/2}\Vert \rho _1( v^n_s-v^m_s)\Vert _{L^2(0,T)} \\&\quad \leqslant \Vert \rho (y^n_s-y^m_s)\Vert _{L^2(Q_T)} + s^{-1/2}\left\| \frac{\rho _1}{\eta } (v^n_s-v^m_s)\right\| _{L^2(\delta ,T-\delta )}\\&\quad \leqslant C \left( s^{-3/2} \Vert \rho (B^n-B^m)\Vert _{L^2(Q_T)} + s^{-1/2} e^{-s} \Vert u_0^n-u_0^m\Vert _{L^2(\Omega )} \right. \\&\quad \left. + s^{-3/2} e^{-s} \Vert u_1^n-u_1^m\Vert _{L^{2}(\Omega )} \right) , \end{aligned}$$

while from (19)

$$\begin{aligned} \Vert \rho (y^n_s-y^m_s)_t \Vert _{L^2(Q_T)}&+ s^{-1/2} \Vert \rho _1 (v^n_s-v^m_s)_t \Vert _{L^2(0,T)} \\ \leqslant&\left( s^{-1/2} \Vert \rho (B^n-B^m)\Vert _{L^2(Q_T)} + s^{1/2} e^{-s} \Vert u_0^n-u_0^m\Vert _{L^2(\Omega )} \right. \\&\left. +s^{-1/2}e^{-s} \Vert (u_0^n-u_0^m)_x\Vert _{L^2(\Omega )} + s^{-1/2} e^{-s} \Vert u_1^n-u_1^m\Vert _{L^2(\Omega )}\right) \end{aligned}$$

and from (21).

$$\begin{aligned}&\Vert (y^n_s-y^m_s)_t \Vert _{L^\infty (0,T;L^2(\Omega ))}+\Vert (y^n_s-y^m_s)_x \Vert _{L^\infty (0,T;L^2(\Omega ))}\\&\quad \leqslant C \left( \Vert B^n-B^m\Vert _{L^2(Q_T)} +\Vert u_0^n-u_0^m\Vert _{L^2(\Omega )}\right. \\&\quad \left. + \Vert (u_1^n-u_1^m)_x\Vert _{L^2(\Omega )}+\Vert v^n_s-v^m_s \Vert _{H^1(0,T)}\right) . \end{aligned}$$

Therefore $v^n_s\rightarrow v_s$ in $H^1(0,T)$ and $y^n_s\rightarrow y_s\in \mathcal {C}^0([0,T];H^1(\Omega ))\cap \mathcal {C}^1([0,T];L^2(\Omega ))$ and, passing to the limit in the equation (6) satisfied by $(y^n,v^n)$, we obtain that $(y_s,v_s)$ solves (6). Moreover, passing to the limit in the estimate (19) satisfied by $(y^n_s,v^n_s)$, we deduce that $(y_s,v_s)$ also satisfies (19). Using (10), we easily check that $(y_s,v_s)$ satisfies $v_s = s\eta ^2\rho _1^{-2}(w_s)_x(1,\cdot )$ and $y_s=\rho ^{-2}L w_s$ where $w_s\in P_s$ is the unique solution of (12). The proof of Theorem 7 is complete.