Recovering time-dependent singular coefficients of the wave-equation-One Dimensional Case

TL;DR

This paper presents a method for recovering time-dependent singular coefficients in the one-dimensional wave equation, advancing inverse problem techniques for PDEs with singularities.

Contribution

It introduces a novel approach to identify singular coefficients that vary over time in the 1D wave equation.

Findings

Successful recovery of time-dependent singular coefficients

Applicable to inverse problems in wave equations

Provides a new framework for singular coefficient identification

Abstract

We give a way to recover time-dependent singular coefficients of the wave-equation in the one-dimensional case.

Equations

$$\displaystyle\left\{\begin{array}[]{rll}\mathcal{L}_{\gamma}u&=&0\quad{\rm in}\quad(0,T)\times\Omega=\Omega_{T},\\ u|_{x=0}&=&f(t)\quad{\rm on}\quad(0,T),\\ u|_{x=b}&=&0\quad{\rm on}\quad(0,T),\\ u\big{|}_{t=0}&=&u_{0}\quad{\rm on}\quad\Omega,\\ \partial_{t}u\big{|}_{t=0}&=&u_{1}\quad{\rm on}\quad\Omega,\end{array}\right.$$

$$\displaystyle\gamma(t,x)=\left\{\begin{array}[]{ll}1\quad{\rm if}\quad x<a(t),\\ k^{2}\quad{\rm if}\quad x\in]a(t),b[=D(t).\end{array}\right.$$

$$(H1D)\Vert \dot{a}(t){\Vert }_{\infty }<\min (1,k),$$

$$\xi (t)=t-a(t),$$

$$\mu (t)=t+a(t).$$

$$\delta :=\frac{1}{2}\mathrm{d}(\partial D,\mathbb{R}\times \Omega )>0,|\dot{a}{|}_{\infty }<\min (1,k).$$

$$t_s:=\inf \{t\ge s;a(t)=t-s\},t^{\ast }(s)=2t_s-s,s\in [0,T].$$

$${\mu }_0:=t_0+a(t_0)=2t_0.$$

$$\alpha (t)$$

$$\beta (t)$$

$$H^{s}(I)=\Big{\{}q\in L^{2}(I);\;\int\!\!\!\int_{I\times I}\frac{|q(x)-q(y)|^{2}}{|x-y|^{1+2s}}dx\,dy<\infty\Big{\}},\quad 0<s<1.$$

$$g=g_A+g_E,$$

$$g_E{|}_{(0,\mu )}\in H^{\epsilon +r_0(1-\xi /T)-1}(0,\mu ),\forall \mu \in [0,T].$$

$$s^{\ast }:={t^{\ast }}^{-1}(T),t_{max}:=t_{s^{\ast }}.$$

$$E^r=H^r(\Omega )\times H^{r-1}(\Omega )\times H^r(0,T),$$

$$E^{r}_{0}=\left\{\begin{array}[]{lc}\{(u_{0},u_{1},f)\in E^{r};\;u_{0}(0)=f(0),u_{0}(b)=0\},&\frac{1}{2}<r\leq 1,\\ E^{r},&0\leq r<\frac{1}{2}.\end{array}\right.$$

$$\displaystyle\left\{\begin{array}[]{rll}\mathcal{L}_{\gamma}v&=&F\quad{\rm in}\quad\Omega_{T},\\ v(t,x)&=&0\>,\quad x\in\partial\Omega,\quad t\in(0,T),\\ v|_{t=0}&=&v_{0}\quad{\rm on}\quad\Omega,\\ \partial_{t}v|_{t=0}&=&v_{1}\quad{\rm on}\quad\Omega.\end{array}\right.$$

$$u$$

$${\partial }_{x}u{|}_{x=0}+f^{\prime }$$

$$(u{|}_{t=s},{\partial }_{t}u{|}_{t=s})$$

$$v$$

$${\partial }_{x}v{|}_{x=0}$$

$$(v{|}_{t=s},{\partial }_{t}v{|}_{t=s})$$

$$b(t;u,v)={\int }_{\Omega }\gamma (t,x){\nabla }_{x}u(x){\nabla }_{x}v(x)dx,\forall u,v\in H^1(\Omega ).$$

$$\frac{d}{dt}({\partial }_{t}v|\varphi )+b(t;v(t,\cdot ),\varphi )=<F(t,\cdot );\varphi >,$$

$$\Vert {\partial }_{t}v(t,\cdot ){\Vert }_{L^2(\Omega )}+\Vert {\partial }_{x}v(t,\cdot ){\Vert }_{L^2(\Omega )}\le C\left(\Vert F{\Vert }_{L^2({\Omega }_t)+W}+\Vert v_0{\Vert }_{H_0^1(\Omega )}+\Vert v_1{\Vert }_{L^2(\Omega )}\right),\forall t\in [0,T].$$

$$(v|g{)}_{L^2({\Omega }_T)}=<v_1,w(0){>}_{H^{-1}\times H_0^1}-(v_0|{\partial }_{t}w(0))+<F,w{>}_{{\mathcal{H}}^{-1}\times {\mathcal{H}}^1},$$

$$<{\partial }_{t}v{|}_{t=T},f_0{>}_{H^{-1}\times H_0^1}-(v{|}_{t=T}|f_1)$$

$$u_{in}(t,x)=f(t-x)\Phi (x).$$

$$v_0(x)$$

$$F(t,x)$$

$${\partial }_{t}F(t,x)=-2f^{\prime \prime }(t-x)\Phi (x)+f^{\prime }(t-x){\Phi }^{\prime \prime }(x),$$

$$<f^{\prime \prime }(t-x)\Phi (x),\phi (t,x)>$$

$$<f^{\prime }(t-x)\Phi (x),\phi (t,x)>=<f(t-x),{\partial }_x(\Phi (x)\phi (t,x))\le C\Vert \phi {\Vert }_{L^2({\Omega }_T)},$$

Full text

Recovering time-dependent singular coefficients of the wave-equation - One Dimensional Case

O. Poisson Aix Marseille Université, I2M, UMR CNRS 6632, France (olivier.poisson@univ-amu.fr).

1 Introduction

Let $\Omega=]0,b[\subset\mathbb{R}$, $b>0$, and consider the following initial boundary value problem

[TABLE]

where $\mathcal{L}_{\gamma}u=\partial^{2}_{t}u-\nabla_{x}\cdot(\gamma\nabla_{x}u)$, $\gamma=\gamma(t,x)$ has the following properties :

There exist a positive constant $k\neq 1$ and a smooth function $t\mapsto a(t)\in]0,b[$ such that

[TABLE]

We make the following assumption

[TABLE]

where $\dot{a}=\frac{da}{dt}$. The inverse problem were are concern with is to obtain some informations on $a(\cdot)$ and $k$, by choosing carefully the data $f$ and then measuring $\partial_{x}u(t,x)$ at $x=0$.

Since the velocity of waves in $\Omega\setminus D(t)$ is one, it is quite natural to consider the following functions. We set

[TABLE]

For simplicity, and if it is unambiguous, we shall write $\xi(t)=\xi$, $\mu(t)=\mu$.

If needed, we extend $a(t)$ in $\mathbb{R}\setminus[0,T]$ by a smooth extention, and so we extend $D=\{\{t\}\times(a(t),b))$, $t\in[0,T]\}$, $D^{C}=\{\{t\}\times(0,a(t))$, $t\in[0,T]\}$, $\partial D=\{(t,a(t))$, $t\in[0,T]\}$ too (with the same notation) by replacing $[0,T]$ by $\mathbb{R}$ in their definition, in such a way that

[TABLE]

We put

[TABLE]

Remark 1.1**.**

Since $|\dot{a}|<1$ and $a>0$, it becomes obvious that $\{t\geq s;a(t)=t-s\}=\{t_{s}\}$, and that $s\mapsto t_{s}$ and $s\mapsto t^{\ast}(\cdot)$ are smooth and increasing.

In fact, $t_{0}$ is the necessary time delay to have the first information on $D(t)$, and $t_{s}$ is the same, but with initial time at $t=s$. We set

[TABLE]

Remark 1.2**.**

We obviously have $\mu(t_{s})=t^{*}(s)$ and $\xi(t_{s})=s$. Hence $\mu=t^{*}\circ\xi$ and $\xi^{-1}(\cdot)=t_{(\cdot)}$.

We also define the coefficient of reflexion/transmition by

[TABLE]

Thanks to (H1D), the functions $\alpha$ and $\beta$ are well-defined in $[0,T]$. We shall deal with data and measurements as functions in the usual Sobolev space $H^{s}(I)$, where $s\in\mathbb{R}$ and $I\subset\mathbb{R}$ is an non empty open interval. If $s\in(0,1)$ it can be defined by

[TABLE]

Our main result is the following

Theorem 1.3**.**

Assume that $(u_{0},u_{1})\in H^{r_{0}}(\Omega)\times H^{r_{0}-1}(\Omega)$ for some $r_{0}\in(0,\frac{1}{2})$. Fix $f\in L^{2}(-\infty,T)$ such that

$f|_{(-\infty,0)}\in H^{r_{0}}(-\infty,0)$; 2. 2.

$f|_{(0,t)}\in H^{r_{0}(1-t^{\prime}/T)}((0,t))$* for $0<t<t^{\prime}\leq T$;* 3. 3.

$f|_{(0,t^{\prime})}\not\in H^{r_{0}(1-t/T)}((0,t^{\prime}))$* for $0\leq t<t^{\prime}\leq T$.*

*Then, the following statements hold.

1. There exists a unique solution $u$ of (1.6) in $L^{2}(\Omega_{T})$.

2. The quantity $\partial_{x}u|_{x=0}$ is defined in $H^{-1}(0,T)$ by continuous extension.

3. The distribution $g=\partial_{x}u|_{x=0}+f^{\prime}\in H^{-1}(0,T)$ has the following form*

[TABLE]

where $g_{A},g_{E}$ satisfy the following properties:

1. (i)

$g_{A}(\mu)=2\alpha(t)f^{\prime}(\xi),\quad\forall\mu\in[0,T]$*. * 2. (ii)

$g_{A}|_{(0,\mu)}\in H^{r_{0}(1-\tilde{\xi}/T)-1}(0,\mu)$* for all $\mu_{0}<\mu\leq T$ and all $\tilde{\xi}>\xi$.* 3. (iii)

*If $\dot{a}(t)\neq\frac{k}{1+k}$ then $g_{A}|_{(0,\mu)}\not\in H^{r_{0}(1-\tilde{\xi}/T)-1}(0,\mu)$, $\forall\tilde{\xi}<\xi$. * 4. (iv)

There exists $\varepsilon>0$ such that

[TABLE]

The main consequence of this is

Corollary 1.4**.**

*Assume that $\dot{a}(t)\neq\frac{k}{1+k}$ for all $t$, and $(u_{0},u_{1})\in H^{r_{0}}(\Omega)\times H^{r_{0}-1}(\Omega)$ for some $r_{0}>0$. Let $T>0$. We claim that:

1. We can know if $T\leq\mu_{0}$ or if $T>\mu_{0}$.

2. Assume that $T>t^{\ast}(0)=\mu_{0}$. Set*

[TABLE]

Then we can recover the functions $s\mapsto t_{s}$, $0\leq s\leq s^{\ast}$, $t\mapsto a(t)$, $t_{0}\leq t\leq t_{max}$. The constant $k$ is the root of a second degree equation with known coefficients. If $\dot{a}\leq 0$ then this equation has no more than one positive root, and so, we are able to reconstruct $k$.

Remark 1.5**.**

Obviously, from Corollary 1.4 and Remark 1.1, and since $t_{0}=a(t_{0})<b$, we can ensure the condition $T>\mu_{0}$ by choosing $T\geq 2b$.

In Theorem 1.3, the existence of such a function $f$ is ensured, thanks to the following

Lemma 1.6**.**

For all $R>0$, there exists a function $G(t)$, $0\leq t\leq 1$, such that

$G|_{(0,t)}\in H^{(1-t^{\prime})/R}(0,t)$* for all $0<t<t^{\prime}\leq 1$.* 2. 2.

$G|_{(0,t^{\prime})}\not\in H^{(1-t)/R}(0,t^{\prime})$* for all $0<t<t^{\prime}\leq 1$.*

Remark 1.7**.**

In Theorem 1.3, if $(u_{0},u_{1})\in H^{r_{0}}_{0}(\Omega)\times H^{r_{0}-1}(\Omega)$ for some $r_{0}\in(\frac{1}{2},1]$, and if $u_{0}(0)$ is known, then we can fix $f\in L^{2}(0,T)$ such that

$f|_{[0,t]}\in H^{r_{0}(1-t/T)}([0,t])$* for $0<t\leq T$;* 2. 2.

$f|_{[0,t^{\prime}]}\not\in H^{r_{0}(1-t/T)}([0,t^{\prime}])$* for $0<t<t^{\prime}\leq T$,*

*and with $f(0)=u_{0}(0)$. Then, the same result holds than in Theorem 1.3, but with $r_{0}\in(\frac{1}{2},1]$.

If $(u_{0},u_{1})\in H^{r_{0}}(\Omega)\times H^{r_{0}-1}(\Omega)$ for some $r_{0}\in(\frac{1}{2},1]$, but if we don’t know the value of $u_{0}(0)$, then the information is not sufficient (with our approach) to construct $f$ so that the result of Theorem 1.3 holds with this value $r_{0}\in(\frac{1}{2},1]$, and so, we are obliged to come back to the situation $(u_{0},u_{1})\in H^{r_{1}}(\Omega)\times H^{r_{1}-1}(\Omega)$, where $r_{1}<\frac{1}{2}$.*

The paper is organized as follows. In Section 2, we analyse the direct problem (1.6). In Section 3 we construct an ansatz $u_{A}$ for (1.6) where $f$ is the function of Theorem 1.3. In Section 4, we first prove Corollary 1.4, then Theorem 1.3. In particular, we analyse the error $u_{E}=u-u_{A}$.

2 Study of the direct problem

2.1 Notations

We denote by $(|)$ the usual scalar product in $L^{2}(\Omega;dx)$, by $(|)_{H}$ the scalar product in a Hilbert space $H$, by $<;>_{H^{*}\times H}$ the duality product between a Hilbert space $H$ and its dual space $H^{*}$, by $<;>$ the duality product in $\mathcal{D}^{\prime}(\Omega_{T})\times\mathcal{D}(\Omega_{T})$ or in $\mathcal{D}^{\prime}(0,T)\times\mathcal{D}(0,T)$. We put $\mathcal{H}^{1}=L^{2}(0,T;H^{1}_{0}(\Omega))$, $\mathcal{H}^{-1}=L^{2}(0,T;H^{-1}(\Omega))={\mathcal{H}^{1}}^{*}$, $W=\{v\in\mathcal{H}^{-1};\;\partial_{t}v\in\mathcal{H}^{-1}\}$ with obvious norms. We denote

[TABLE]

and

[TABLE]

(For $r=\frac{1}{2}$ we could set $E^{r}_{0}$ as in the case $r>\frac{1}{2}$, but the relations $u_{0}(0)=f(0)$ and $u_{0}(b)=0$ should be modified).

We denote $\Omega_{t_{1},t_{2}}=(t_{1},t_{2})\times\Omega$.

For data $v_{0},v_{1},F$, let $v$ satisfying in some sense:

[TABLE]

We formally define the following operators:

[TABLE]

where $u$, $v$, are respectively solutions of (1.6), (2.5).

2.2 Main results

In this section and the one above, we state that Problems (1.6), (2.5) have a unique solution for adequate spaces.

Lemma 2.1**.**

*1. The operator $P$ is a continuous linear mapping from $H^{1}_{0}(\Omega)\times L^{2}(\Omega)\times(L^{2}(\Omega_{T})+W)$ into $C([0,T];H^{1}_{0}(\Omega))\cap C^{1}([0,T];L^{2}(\Omega))$.

2. The operator $X(s)$ is continuous from $H^{1}_{0}(\Omega)\times L^{2}(\Omega)\times(L^{2}(\Omega_{T})+W)$ into $H^{1}_{0}(\Omega)\times L^{2}(\Omega)$, for all $s\in[0,T]$.*

Lemma 2.2**.**

*1. The operator $P$ continuously extends as a continuous operator from $L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$ into $L^{2}(\Omega_{T})$.

2. The operator $X(s)$ continuously extends as a continuous operator from $L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$ into $L^{2}(\Omega)\times H^{-1}(\Omega)$, for all $s\in[0,T]$.*

Lemma 2.3**.**

*1. The operator $\tilde{P}$ is a continuous linear mapping from $E^{1}_{0}$ into $C([0,T]$; $H^{1}(\Omega))\cap C^{1}([0,T];L^{2}(\Omega))$, and continuously extends as a continuous operator from $E^{0}$ into $L^{2}(\Omega_{T})$.

2. The operator $\tilde{X}(s)$ is continuous from $E^{1}_{0}$ into $H^{1}(\Omega)\times L^{2}(\Omega)$, and continuously extends as a continuous operator from $E^{0}$ into $L^{2}(\Omega)\times H^{-1}(\Omega)$, for all $s\in[0,T]$.*

Lemma 2.4**.**

The operator $Z$ (respect., $\tilde{Z}$) is continuous from $H^{1}_{0}(\Omega)\times L^{2}(\Omega)\times L^{2}(\Omega_{T})$ (respect., $E^{1}_{0}$) into $L^{2}(0,T)$ and continuously extends as a continuous operator from $L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$ (respect., $E^{0}$) into $H^{-1}(0,T)$.

Lemma 2.5**.**

Let $t_{1}\in[0,T]$. Assume that $F\in\mathcal{H}^{-1}$ has a compact support in $\mathcal{O}(t_{1})$. Let $v=P(v_{0},v_{1},F)$. Then there exists a neighborhood $\tilde{K}$ of $K(t_{1})$ in $\overline{D^{C}}$ such that $v|_{\tilde{K}}$ does not depend on $F$, that, is, if $v_{0}=v_{1}=0$, then $v|_{\tilde{K}}$ vanishes, and, in particular, supp $\partial_{x}v|_{x=0}\subset(\mu(t_{1}),T]$.

2.3 Proofs

Let us consider the familly of bilinear forms $b(t)$, $t\in\mathbb{R}$, defined by

[TABLE]

Lemma 2.1 is a corollary of the following general theorem (proof in appendix), which is an extension of [1, XV section 4] where $\gamma$ did not depend on the variable $t$.

Theorem 2.6**.**

Let $T>0$ and $\Omega\subset\mathbb{R}^{n}$, $n\geq 1$, such that $H^{1}_{0}(\Omega)$ is compact in $L^{2}(\Omega)$. Let $\gamma(t,x)>0$ be such that $\gamma,\gamma^{-1}\in C^{0}([0,T];L^{\infty}(\Omega))$, $\partial_{t}\gamma\in L^{\infty}(\Omega_{T})$. Let $F\in W\cup L^{2}(\Omega_{T})$ and $v_{0}\in H^{1}_{0}(\Omega)$, $v_{1}\in L^{2}(\Omega)$. Then, there exists a unique weak solution $v$ to (2.5), that is, $v\in C([0,T];H^{1}_{0}(\Omega))$, $\partial_{t}v\in C([0,T];L^{2}(\Omega))$, $v|_{t=0}=v_{0}$, $\partial_{t}v|_{t=0}=v_{1}$, and

[TABLE]

in the sense of $\mathcal{D}^{\prime}(]0,T[)$, for all $\phi\in H^{1}_{0}(\Omega)$. Moreover there exists a constant $C$ such that

[TABLE]

Let us show that Lemma 2.2 is a straightforward consequence of Lemma 2.1 with the operator $P$ replaced by its adjoint $P^{*}$. Let $(v_{0},v_{1},F)\in L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$. By the principle of duality, we can write (2.5) as

[TABLE]

for all $g\in L^{2}(\Omega_{T})$, where we put $w=P^{*}(0,0,g)$. Consequently (thanks to Lax-Milgram theorem), Equation (2.5) admits a unique solution $v\in L^{2}(\Omega_{T})$, and this shows Point 1 of Lemma 2.2. Once again, we have

[TABLE]

for all $(f_{0},f_{1})\in H^{1}_{0}(\Omega)\times L^{2}(\Omega)$, where we put $w=P^{*}(f_{0},f_{1},0)$. This shows that $(v|_{t=T},\partial_{t}v|_{t=T})\in L^{2}(\Omega)\times H^{-1}(\Omega)$. This proves Point 2 of Lemma 2.2 in the non-restrictive case $s=T$.

Let us prove Lemma 2.3. Let $\Phi(x)\in C^{\infty}(\mathbb{R})$ with $\Phi(0)=1$ and with support in $[0,a_{m}]$, where $a_{m}\leq a(t)$ for all $t$. Let us consider $f\in H^{1}_{loc}(\mathbb{R})$ first. Set

[TABLE]

Problem (1.6) with unknown $u$ is (at least formally) equivalent to the following one: find $v=u-u_{in}$ satisfying (2.5) with

[TABLE]

Relation (2.9) shows that $F\in L^{2}(\Omega_{T})$. In fact, we have $F\in W$ also, since

[TABLE]

and, for all $\varphi\in\mathcal{D}(\Omega_{T})$,

[TABLE]

which shows that $\partial_{t}F(t,x)\in\mathcal{H}^{-1}$. Similarly, we have

[TABLE]

which shows that $F\in\mathcal{H}^{-1}$ if $f\in L^{2}_{loc}(\mathbb{R})$ only. We set

[TABLE]

[TABLE]

The above analysis shows that $R$ continuously extends as a continuous operator from $L^{2}_{loc}(\mathbb{R})$ into $\mathcal{H}^{-1}$. Similarly, $S$ continuously extends as a continuous operator from $E^{0}$ into $L^{2}(\Omega)\times H^{-1}(\Omega)$. Consequently, and since a solution to (1.6) can be written $u=v+u_{in}$ with $v=P(S(u_{0},u_{1},f),R(f))$, Point 1 of Lemma 2.3 is proved. Similarly, we prove Point 2 of Lemma 2.3, since we have $\tilde{X}(s)(u_{0},u_{1},f)=X(s)(S(u_{0},u_{1},f),Rf)+(u_{in}|_{t=s},\partial_{t}u_{in}|_{t=s})$.

Let us prove Lemma 2.4. Let $(v_{0},v_{1},F)\in H^{1}_{0}(\Omega)\times L^{2}(\Omega)\times L^{2}(\Omega_{T})$.

As above, for all $\varphi\in\mathcal{D}(\mathbb{R})$ such that $\varphi(T)=0$, there exists a unique solution $q=q_{\varphi}\in L^{2}(\Omega_{T})$ to

[TABLE]

since it is a particular case of Lemma 2.3 with reversal time.

Moreover, we have $q_{\varphi}\in C([0,T];H^{1}_{0}(\Omega))$, $\partial_{t}q_{\varphi}\in C([0,T];L^{2}(\Omega))$ with

[TABLE]

By the duality principle, and thanks to (2.15), we have in the sense of $\mathcal{D}^{\prime}([0,T))$,

[TABLE]

which shows that $\partial_{x}v|_{x=0}\in L^{2}(0,T)$ and that $Z$ is a continuous mapping from $H^{1}_{0}(\Omega)\times L^{2}(\Omega)\times L^{2}(\Omega_{T})$ into $L^{2}(0,T)$.

Now, let $(v_{0},v_{1},F)\in L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$. Then, Relation (2.17) and Estimate (2.16) imply

[TABLE]

which shows that $\partial_{x}v|_{x=0}\in(H^{1}_{T})^{\prime}\subset H^{-1}(0,T)$, the dual space of $H^{1}_{T}=\{f\in H^{1}(0,T);f(T)=0\}$, and that $Z$ continuously extends as a continuous operator from $L^{2}(\Omega)\times H^{-1}(\Omega)\times\mathcal{H}^{-1}$ into $H^{-1}(0,T)$.

This ends the proof of the property of $Z$ in Lemma 2.4. Since $\partial_{x}u_{in}|_{x=0}=-f^{\prime}$, we have $\tilde{Z}(u_{0},u_{1},f)=Z(S(u_{0},u_{1},f),Rf)$, and Point 2 of Lemma 2.4 is proved.

∎

By the well-known Sobolev interpolation theory, we have also proved:

Proposition 2.7**.**

*The operator $P$ (respect., $\tilde{P}$) continuously maps $H^{s}(\Omega)\times H^{s-1}(\Omega)\times L^{2}(0,T;H^{s-1}(\Omega))$ (respect., $E^{s}_{0}$) into $L^{2}(0,T;H^{s}(\Omega))$, $s\in[0,1](\setminus\frac{1}{2})$.

The operator $Z$ (respect., $\tilde{Z}$) continuously maps $H^{s}(\Omega)\times H^{s-1}(\Omega)\times L^{2}(0,T;H^{s-1}(\Omega))$ (respect., $E^{s}_{0}$) into $H^{s-1}(0,T)$, $s\in[0,1](\setminus\frac{1}{2})$.*

Proof of Lemma 2.5. Denote $K=K(t_{1})$. Notice that $K\cap\overline{D^{C}}=\{(t_{1},a(t_{1}))\}$. We assume that $v_{0}=v_{1}=0$. Since supp$F\cap\overline{\Omega_{t_{1}}}=\emptyset$, then, thanks to Lemma 2.2 with $T$ replaced by $t_{1}$, $v$ vanishes in $\Omega_{t_{1}}$. Let $K^{\prime}={\rm int}\;K$ the interior of $K$. The function $v|_{K}\in L^{2}(K^{\prime})$ satisfies the following equations:

[TABLE]

It is well-known that this implies $v|_{K^{\prime}}=0$, and so, supp $\partial_{x}v|_{x=0}\subset[\mu(t_{1}),T]$. But since the support of $F$ does not touch $\partial K$, we similarly have $v|_{K_{\varepsilon}(t_{1})}=0$, supp $\partial_{x}v|_{x=0}\subset[\mu(t_{1})+\delta,T]$, for some $\varepsilon>0$ sufficiently small.

However, let us give a more straightforward and simple proof to the fact that supp $\partial_{x}v|_{x=0}\subset[\mu(t_{1})+\delta,T]$. Fix $\delta,\varepsilon>0$ such that $\mu(t_{1})+\delta>\mu(t_{1}+\varepsilon)$ and supp $F\cap K_{\varepsilon}(t_{1})=\emptyset$. Let $t_{2}\in[t_{1},t_{1}+\varepsilon]$, $\varphi\in H^{1}_{0}(0,\mu(t_{2}))$ and set $w(t,x)=\varphi(t+x)$ for $t_{2}\leq t\leq\mu(t_{2})$. Observe that $w=q_{\varphi}$ of (2.14), but with $(0,T)$ replaced by $(t_{2},\mu(t_{2}))$. In fact, supp $w\subset K(t_{2})$, and so $w$ vanishes in $D\cap\Omega_{t_{2},\mu(t_{2})}$. We then have, similarly to (2.17),

[TABLE]

since $v|_{t_{2}}=\partial_{t}v|_{t_{2}}=0$ and supp $F\cap$ supp $w=\emptyset$. Since $\varphi$ is arbitrarily chosen, this shows that supp $\partial_{x}v|_{x=0}\cap(t_{2},\mu(t_{2}))=\emptyset$, for all $t_{2}\in[t_{1},t_{1}+\varepsilon]$. Hence, supp $\partial_{x}v|_{x=0}\subset[\mu(t_{1}+\varepsilon),T]$. ∎.

3 Ansatz

3.1 Notations

For $t\in[0,T]$ we put

[TABLE]

(Notice that $K(t)\subset\overline{D^{C}}$ and $K(t)\cap\overline{D}=\{(t,a(t)\}$).

For $\varepsilon>0$, $t\in[0,T]$, we put $K_{\varepsilon}(t)=\cup_{t\leq s\leq t+\varepsilon}K(s)$.

If $q(x)$ is sufficiently smooth in $\Omega$, then $[q]_{t}:=q(a(t)+0)-q(a(t)-0)$.

We write $g_{1}\stackrel{{\scriptstyle s}}{{\simeq}}g_{2}$ if $g_{1}$ or $g_{2}\in H^{s}(0,T)$ and $g_{1}-g_{2}\in H^{s+\varepsilon}(0,T)$ for some $\varepsilon>0$.

We put $C^{j}_{+}=\{f\in C^{j}(\mathbb{R});\;f|_{\mathbb{R}^{-}}=0\}$, $j\in\mathbb{N}$, which is dense in $L^{2}(\mathbb{R}^{+})\approx\{f\in L^{2}(\mathbb{R});\;f|_{(-\infty,0)}=0\}$. We consider for all $t\in[0,T]$ the formal operator $\mathcal{A}(t)=-\nabla_{x}(\gamma(t,\cdot)\nabla_{x})$ defined from $H^{1}(\Omega)$ into $H^{-1}(\Omega)$ by duality:

[TABLE]

Let $f$ be a measurable function, we define the ansatz $u_{A}=U_{A}(f)$ for (1.6) as follows. Recall that $\xi(t)$ and $\mu(t)$ are defined by (1.10), (1.11), and we have

[TABLE]

In addition, we put, for $t\in[0,T]$,

[TABLE]

Thanks to Assumption (H1D), $t\mapsto\nu(t)$ is invertible. Recall also that the coefficient of reflexion/transmition, $\alpha$ and $\beta$, are defined by (1.12), (1.13). Note that we have

[TABLE]

We also define:

[TABLE]

We put

[TABLE]

where we fix $\Phi_{\varepsilon}\in C^{\infty}(\mathbb{R})$ so that $\Phi_{\varepsilon}(r)=1$ if $r<\frac{1}{2}\varepsilon$, $\Phi_{1}(r)=0$ if $r>\varepsilon$, $0<\varepsilon\leq\frac{1}{2}{\rm d}(\partial D,\partial\Omega_{T})$. It is clear that the linear operator $U_{A}:\>f\mapsto u_{A}$ is bounded from $L^{2}(\mathbb{R})$ into $L^{2}(\Omega_{\mu_{T}})$.

3.2 Properties of the Ansatz

Lemma 3.1**.**

*Let $f\in C^{2}(\mathbb{R})$. Then we have

1. $u_{A}\in C^{2}([0,T];H^{1}(\Omega))$, $u_{A}|_{D}\in C^{2}(\overline{D})$, $u_{A}|_{D^{C}}\in C^{2}(\overline{D^{C}})$.

2. There exists a smooth function $\tau(t)$ with support in $[t_{0},\mu_{0}]$ such that*

[TABLE]

*3) a) $u_{A}$ vanishes near $x=b$.

b) Let $g_{A}=\partial_{x}u_{A}|_{x=0}+f^{\prime}$. Then $g_{A}(\mu)=2\alpha(t)f^{\prime}(\xi)$ for $0\leq\mu\leq T$, where $t$, $\xi$, $\mu$ are related by (1.10), (1.11), (3.3).

4. Put $F_{A}=\mathcal{L}_{\gamma}u_{A}$ in the sense that $F_{A}(t,\cdot)=\frac{d^{2}}{dt^{2}}u_{A}(t)+\mathcal{A}(t)u_{A}(t)\in H^{-1}(\Omega)$ for all $t$, and $F_{A}\in C([0,T];H^{-1}(\Omega))$. Then, $F_{A}$ can be written

$F_{A}(t,x)=F_{1}(t,x)-\tau(t)f(\xi(t))\delta_{a(t)}(x)$, where $\tau$ is smooth, and $F_{1}\in C([0,T];L^{2}(\Omega))$ is defined for $0\leq t\leq T$ by*

[TABLE]

where the functions $\Phi_{j}$ are smooth and independant of $f$, with compact support in $[\varepsilon/2,\varepsilon]$ for $j=2,3$, and in $[b-\varepsilon,b-\varepsilon/2]$ for $j=4,5$.

Proof. Point 1. is obvious, since we have, thanks to (3.4),

[TABLE]

Let us consider Point 2. For $0\leq t\leq T$ we have

[TABLE]

Thanks to (3.5) we get

[TABLE]

with

[TABLE]

This ends Point 2.

Let us consider Point 3 b), since 3 a) is obvious. For $0\leq\mu\leq T$ we have

[TABLE]

This ends Point 3.

Let us prove Point 4. A short computation yields (3.11). Thanks to Point 2, we obtain $F_{A}=F_{1}+\tau(t)f(\xi)$ in the required sense. This ends the proof of the lemma. ∎

We define the bounded operators $U_{A}$ : $C^{2}(\mathbb{R})\ni f\mapsto u_{A}\in C^{2}([0,T];H^{1}(\Omega))$, $T_{0}$: $C^{2}(\mathbb{R})\ni f\mapsto T_{0}f\in C([0,T];H^{-1}(\Omega))$ such that $T_{0}f(t)=\tau(t)f(\xi)\delta_{a(t)}(x)$, and $T_{1}$: $C^{2}(\mathbb{R})\ni f\mapsto T_{1}f=F_{1}\in C([0,T];L^{2}(\Omega))$, $T_{A}$: $C^{2}(\mathbb{R})\ni f\mapsto T_{A}f=F_{A}\in C([0,T];H^{-1}(\Omega))$. Notice that $T_{0}f(t)\in H^{-s}(\Omega)$ for all $s>\frac{1}{2}$, $t\in[0,T]$. Obviously we have the following propositions and Lemma.

Proposition 3.2**.**

The operator $U_{A}$ continuously extends as a bounded operator from $L^{2}(0,T)$ into $C([0,T];H^{-1}(\Omega))$.

Proposition 3.3**.**

The operator $T_{0}$ continuously extends as a bounded operator from $L^{2}(0,T)$ into $L^{2}(0,T;H^{-s}(\Omega))$, $\forall s>\frac{1}{2}$.

Lemma 3.4**.**

*1) The operator $T_{A}$ is continuous from $C^{2}(\mathbb{R})$ into $L^{2}(0,T;H^{-1}(\Omega))$ and, for all $s\in[0,\frac{1}{2})$, it extends as a continuous operator from $H^{s}(0,T)$ into $L^{2}(0,T;H^{s-1}(\Omega))$.

2. The operator $G_{A}:\>f\mapsto\partial_{x}U_{A}(f)|_{x=0}+f^{\prime}$ is continuous from $C^{2}(\mathbb{R})$ into $C^{0}([0,T])$, and, for all $s\in[0,\frac{1}{2})$, it extends as a continuous operator from $H^{s}(0,T)$ into $H^{s-1}(0,T)$.

3. Let $f$ such as in Theorem 1.3, then $g_{A}:=G_{A}f$ satisfies (ii) and (iii) of Theorem 1.3.*

Proof of Lemma 3.4. Point 1). Thanks to Lemma 3.3, it is sufficient to prove this with $T_{A}$ replaced by $T_{1}$. Thanks to the interpolation theory, it is sufficient to prove that $T_{1}$ is a bounded operator from $L^{2}(0,T)$ into $L^{2}(0,T;H^{-1}(\Omega))$ and from $H^{1}_{0}(0,T)$ into $L^{2}(\Omega_{T})$, that is obvious. Hence Point 1) holds. Point 2) is obvious for the same reason. Point 3) is obvious, since $\alpha(t)\neq 0$ for all $t$. ∎

3.3 Modification of $F_{1}$

The regularity of $F_{1}$ is not sufficient for us, we replace it by the following one, $F_{\varepsilon,\tilde{\mu}}$, which is equivalent to $F_{1}$ in the sense of Lemma 2.5.

Let $\tilde{\mu}\in[0,T]$, put $\tilde{t}=\mu^{-1}(\tilde{\mu})$, $\tilde{\xi}=\xi(\tilde{t})$, $\tilde{\nu}=\nu(\tilde{t})$, and consider a smooth function $\phi(\cdot;\varepsilon,\tilde{\mu})$ defined in $\mathbb{R}^{2}$ such that $\phi(t,x;\varepsilon,\tilde{\mu})=1$ for $(t,x)\in\overline{\Omega_{\tilde{t}}}\cup K_{\varepsilon/2}(\tilde{t})$, $\phi(t,x;\varepsilon,\tilde{\mu})=0$ for $t\geq\tilde{t}+\varepsilon$ and $(t,x)\not\in K_{\varepsilon}(\tilde{t})$. For $s\in[0,\frac{1}{2})$, $f\in H^{s}(\mathbb{R})$ and $F_{1}=T_{1}(f)$ we put

[TABLE]

We have the two following properties.

Lemma 3.5**.**

For $\varepsilon<\delta$, the support of $F_{1}-F_{\varepsilon,\tilde{\mu}}$ is contained in $\mathcal{O}(\tilde{t})$.

Proof. Since $F_{1}-F_{\varepsilon,\tilde{\mu}}=(1-\phi(\cdot;\varepsilon,\tilde{\mu})F_{1}$, the support of $F_{1}-F_{\varepsilon,\tilde{\mu}}$ is contained in supp $(1-\phi(\cdot;\varepsilon,\tilde{\mu}))\cap$ supp $F_{1}$. But supp $(1-\phi(\cdot;\varepsilon,\tilde{\mu}))\subset\overline{\Omega_{T}}\setminus(\Omega_{\tilde{t}}\cup{\rm int}(K_{\varepsilon/2}(\tilde{t})))$. Then the proof is done if we show that $(\tilde{t},a(\tilde{t}))\not\in$ supp $(F_{1}-F_{\varepsilon,\tilde{\mu}})$. But, thanks to (3.11), the support of $F_{1}$ is localized in $\{x\leq\varepsilon\}\cup\{x\geq b-\varepsilon\}$ that does not touch $\partial D$. ∎

Lemma 3.6**.**

Let $f$ as in Theorem 1.3. There exists $c>0$ and $\varepsilon_{0}>0$, independent of $f$, such that, for all $\varepsilon\in(0,\varepsilon_{0})$, $\tilde{\mu}\in[0,T]$, $F_{\varepsilon,\tilde{\mu}}\in C([0,T];H^{r_{0}(1-\tilde{\xi}/T)+c\varepsilon-1}(\Omega))$.

To prove it, we use the following well-known property.

Proposition 3.7**.**

Let $g\in H^{s}(\mathbb{R})$ for some $s\in[-1,0]$. Let $r\in\mathbb{R}^{*}$ and $G(t,x)=g(t+rx)$, $(t,x)\in\Omega_{T}$. Then $G\in C([0,T];H^{s}(\Omega))$.

Let us prove Lemma 3.6. Observe that, by definition of $\phi(\cdot;\varepsilon,\tilde{\mu})$, and thanks to (3.11), the support of $F_{\varepsilon,\tilde{\mu}}|_{\Omega_{\tilde{\mu}}}$ is a subset of the set

[TABLE]

Firstly, let $(t,x)\in K_{\varepsilon}(\tilde{t})\cup(\overline{\Omega_{\tilde{t}+\varepsilon}}\cap\overline{D^{C}})$. Then we have $t-x\leq\tilde{t}+\varepsilon$, and so

[TABLE]

since the functions $\xi$ and $\mu^{-1}$are smooth and non decreasing, and $\delta<a(\tilde{t})=\tilde{\mu}-\tilde{t}$. So, for $\varepsilon$ sufficiently small and some $c>0$ (values that are independent of $t,x$), we have

[TABLE]

Secondly, let $(t,x)\in\overline{\Omega_{\tilde{t}+\varepsilon}}\cap\overline{D}\cap\{b-\varepsilon\leq x\leq b\}$. Then $t-\frac{x}{k}\leq\nu(t)-\frac{\delta-\varepsilon}{k}$ and so, for $\varepsilon$ sufficiently small and some $c>0$,

[TABLE]

We thus have

[TABLE]

Since $F_{1}$ is expressed in terms of $f^{\prime}_{2}(t-x)$, $f_{2}(t-x)$ in $D^{C}$, and in terms of $f^{\prime}_{3}(t-\frac{x}{k})$, $f_{3}(t-\frac{x}{k})$ in $D$, and since the support of $F_{\varepsilon,\tilde{\mu}}$ is contained in $E(\varepsilon,\tilde{\mu})$, then, thanks to (3.12), (3.13), we see that $F_{\varepsilon,\tilde{\mu}}$ can be expressed in terms of $f|_{(-\infty,r)}$ and $f^{\prime}|_{(-\infty,r)}$, $r=\tilde{\xi}-c\varepsilon$ only. Hence, thanks to Proposition 3.7, the conclusion follows. ∎

4 Proof of the main results

4.1 Proof of Corollary 1.4

Firstly, notice that $\alpha(t)\neq 0\Longleftrightarrow\dot{a}(t)\neq\frac{k}{1+k}$.

1. If $T\leq\mu_{0}$ then $g=0$ in $(0,T)$, and if $T>\mu_{0}$ then $g\neq 0$ since $g|_{(\mu_{0},T)}\not\in H^{r_{0}(1-s^{\ast}/T)-1}(\mu_{0},T)$. Hence, the knowledge of $g$ provides $T\leq\mu_{0}$ or $T>\mu_{0}$.

• •

Let $\mu\in[\mu_{0},T]$. Thanks to Theorem 1.3, we can construct

[TABLE]

and so the invertible function $\mu\mapsto\xi$ from $[\mu_{0},T]$ into $[0,s^{\ast}]$. (This implies that $s^{\ast}$ is recovered too). Putting $t=\frac{1}{2}(\mu+\xi)$, we recover $t_{s^{*}}$ which is $t$ for $\mu=T$, and also the functions $t\mapsto\xi=\xi(t)$, $t\mapsto\mu(t)$, $t\mapsto a(t)=\frac{1}{2}(\mu(t)-\xi(t))$, for $t\in[t_{0},t_{s^{*}}]$. We then construct the functions $t_{(\cdot)}=(\xi(\cdot))^{-1}$, $t^{*}(\cdot)=2t_{(\cdot)}-{\rm id}$.

• •

Thanks to the above point and to (i) of Theorem 1.3, the smooth function $\alpha(\cdot)$ can be recover as the unique one such that $\mu\mapsto g(\mu)-\alpha(t)f^{\prime}(\xi)$ belongs to $H^{\varepsilon+r_{0}(1-\xi/T)}(0,\mu)$ for some $\varepsilon>0$ and all $\mu\in(0,T)$. Then, $k$ is root of the following equation:

[TABLE]

Denote by $k_{1},k_{2}$ the roots, such that $k_{1}\leq k_{2}$. We show that $k_{1}\leq 0$. A short computation shows that

[TABLE]

We have

[TABLE]

If $\dot{a}\leq 0$ then, the second equality in (4.2) implies that it is impossible to have $0<k_{1}\leq k_{2}$.

?????????,

∎

Remark 4.1**.**

Theorem 1.3 allows us to recover $t^{\ast}(\cdot)=\mu\circ\xi^{-1}$ as:

[TABLE]

and shows that

[TABLE]

4.2 Analysis of the error

Let $(u_{0},u_{1},f)$, $r_{0}$ as in Theorem 1.3. Put $u=\tilde{P}(u_{0},u_{1},f)$, $g=\tilde{Z}(u_{0},u_{1},f)$, $u_{A}=U_{A}(f)$ and

[TABLE]

where $u_{A}$ is defined in Section 3. Let us prove the estimate (1.14) (see (iv) of Theorem 1.3). For the sake of clarity, we replace $\mu$, $t$, $\xi$, respectively by $\tilde{\mu}$, $\tilde{t}=\mu^{-1}(\tilde{\mu})$, $\tilde{\xi}=\xi(\tilde{t})$. Put $u_{E,0}=u_{0}-u_{A}(0)$, u_{E,1}=u_{1}-\partial_{t}u_{A}\big{|}_{t=0}. In view of Subsection 3, the function $u_{E}$ satisfies

[TABLE]

So we have $u_{E}=P(u_{E,0},u_{E,1},-F_{A})$. Recall that, thanks to Lemma LABEL:l.T0, we have $T_{0}(f)\in L^{2}(0,\tilde{\mu};H^{-s}(\Omega))$, for all $s>\frac{1}{2}$. Thanks to Proposition 2.7, we have

[TABLE]

Let us prove that $u_{E,0}\in H^{r_{0}}(\Omega)$, $u_{E,1}\in H^{r_{0}-1}(\Omega)$. Observe that $u_{A}(0)(x)=(f(-x)+f_{2}(x)+f_{2}(-x)\Phi_{\varepsilon}(x))\chi_{x<a(0)}+f_{3}(-x/k)\Phi_{\varepsilon}(x-b+2\varepsilon))\chi_{x>a(0)}$. For $x<a(0)=t_{0}$ we have

[TABLE]

and, similarly, $\xi(\mu^{-1}(-x))\leq\xi(\mu^{-1}(0))<0$. For $x>a(0)$ we have

[TABLE]

Hence, $u_{A}(0)$ can be expressed in terms of $f(\xi)$ for $\xi<0$. Since $f|_{(-\infty,0]}\in H^{r_{0}}(-\infty,0)$, then $u_{A}(0)\in H^{r_{0}}(\Omega)$. Thanks to the asumption on $u_{0}$, we then have $u_{E,0}\in H^{r_{0}}(\Omega)$. Similarly, we have $u_{E,1}\in H^{r_{0}-1}(\Omega)$. Thanks to (3.6), the regularity of $f_{2}|_{(0,\tilde{\mu})}$ is given by those of $f|_{(0,\tilde{\xi})}$, that is, $f_{2}|_{(0,\tilde{\mu})}\in H^{r_{0}(1-\xi^{\prime}/T)}((0,\tilde{\mu}))$, for all $\xi^{\prime}>\tilde{\xi}$. Thus, thanks to Proposition 2.7, we have

[TABLE]

Thanks to Lemma 2.5 with $t_{1}$ replaced by $\tilde{t}$ and $T$ by $\tilde{\mu}$, and to Lemma 3.5, we have

[TABLE]

Thanks to Lemma 3.6, if $\varepsilon>0$ is sufficiently small, we have

[TABLE]

and so, thanks to (4.10) and by applying Proposition 2.7, we obtain

[TABLE]

for some $\varepsilon>0$ (independent of $\tilde{\mu}$).

Thanks to (4.8), (4.9) (4.11), and since $g_{E}=Z(u_{E,0},u_{E,1},0)+Z(0,0,T(0)f)+Z(0,0,-F_{1})$, the proof of (1.14) is done.

∎

5 Appendix: the function G

Let $I=(0,1)$ and a dense sequence $\{a_{n}\}_{n\in\mathbb{N}^{*}}$ in $\overline{I}$. We set

[TABLE]

[TABLE]

where $z_{+}=\max(0,z)$ for $z\in\mathbb{R}$. The function $G$ is increasing.

For $0<s<1$ we set the following Sobolev space:

[TABLE]

Lemma 5.1**.**

Let $b\in(0,1]$, $r>-\frac{1}{2}$, $s\in(0,1)$, $a\in[0,b)$. Set $f(x)=((x-a)_{+})^{r}$, $I_{b}=(0,b)$. We have $f\in H^{s}(I_{b})$ if, and only if, $r>s-1/2$. In such a case, we have

[TABLE]

for some $C_{s}>0$.

Proof. Firstly, let $b=1$. We have

[TABLE]

We have

[TABLE]

If $a=0$, then $K_{1}=0$. If $a>0$, then $K_{1}<\infty$ if, and only if, $2r>2s-1$. In such a case, we have

[TABLE]

Let $2r>2s-1$. We have

[TABLE]

where

[TABLE]

Since $C(r,s)>0$, then $K_{2}=+\infty$ if $2r\leq 2s-1$. Hence, the sum $K_{1}+K_{2}$ converges iff $2r>2s-1$. If $2r>2s-1$, thanks to (5.2) and (5.3), we obtain (5.1).

Secondly, the case $b\in(0,1)$ is easily proved by setting $a=a^{\prime}b$, $x=x^{\prime}b$, $y=y^{\prime}b$. ∎

Lemma 5.2**.**

For $0<s<1$ and $b\in(0,1]$, we have $G\in H^{s}(0,b)$ if $s<1-b$ and $G\not\in H^{s}(0,b)$ if $s>1-b$.

Proof. For $x,y\in I$, we have, thanks to the Schwarz inequality,

[TABLE]

Let $I_{b}=(0,b)$, $A_{b}=\{n\in\mathbb{N}^{*};\;a_{n}\geq b\}$, $B_{b}=\mathbb{N}^{*}\setminus A_{r}=\{n;a_{n}<b\}$.

For all $n\in B_{b}$, thanks to Lemma 5.1, we have $f_{n}\in H^{1-b}(0,1)$, since $1/2-a_{n}>(1-b)-1/2$. For all $n\in A_{b}$, we have $f_{n}\in H^{1-b}(I_{b})$, since $f_{n}|_{I_{b}}=0$.

Let $0<s<1-b$. By using (5.4), and (5.1), we have

[TABLE]

since $(b-a_{n})^{2(1-s-a_{n})}\leq 1$ for all $n\in B_{b}$, $0<s<1-b$.

Let $s\in(1-b,1)$. For all $n\in\mathbb{N}^{*}$ and $x>y$ we have $G(x)-G(y)\geq f_{n}(x)-f_{n}(y)$. Fix $n\in A_{1-s}\cap B_{b}$, that is, $1-s\leq a_{n}<b$. Thanks to Lemma 5.1, we have $f_{n}\not\in H^{s}(I_{b})$, and then

[TABLE]

This ends the proof. ∎

6 Proof of Theorem 2.6

Let $F\in L^{2}(\Omega_{T})$, $v_{0}\in H^{1}_{0}(\Omega)$, $v_{1}\in L^{2}(\Omega)$. Denote $M^{1}:=\{v\in C([0,T];H^{1}_{0}(\Omega))$, $\partial_{t}v\in C([0,T];L^{2}(\Omega))\}$, $M^{1}_{0}=\{v\in M$; $v|_{t=0}=0$, $\partial_{t}v|_{t=0}=0\}$,

6.1 Energy estimate.

Put

[TABLE]

We claim that, for all $v\in M^{1}$ such that $L_{\gamma}v=:f\in L^{2}(\Omega_{T})+W$, the following (standart) estimate, which implies (2.6), holds.

[TABLE]

for some constant $C$.

Proof. It is sufficient to show (6.1) for $t=T$. Assume that $f\in L^{2}(\Omega_{T})$. Put $\rho=\sup_{Q}\frac{|\dot{\gamma}|}{\gamma}$ and $\Pi_{0}\in C^{1}([0,T];(0,+\infty))$ such that $\delta^{-1}\Pi_{0}\leq-\Pi^{\prime}_{0}$ for some $\delta\in(0,\frac{1}{\rho})$. (For example, $\Pi_{0}=e^{-\frac{t}{\delta}}$). Put

[TABLE]

We formally have, thanks te the Schwarz inequality,

[TABLE]

Hence, we obtain

[TABLE]

and so,

[TABLE]

Then (6.2) follows.

6.2 Uniqueness

Consequently, if $v\in M^{1}_{0}$ satisfies (2.5) with $F=0$, then $E(t)(v)\equiv 0$ for all $t$, and so $v\equiv 0$. This shows that Problem (2.5) admits at most one solution in $M^{1}$.

6.3 Existence

Let $(\lambda_{j},e_{j})_{1\leq j}$ be the familly of spectral values of the positive operator $-\Delta_{x}$ in $H^{1}_{0}(\Omega)$, i.e such that $(e_{i},e_{j})_{L^{2}(\Omega)}=\delta_{ij}$, $-\Delta e_{j}=\lambda_{j}e_{j}$, and $\lambda_{j}\nearrow+\infty$. The data $v_{0},v_{1}$, $F$ are then written $v_{0}=\sum_{j=1}^{\infty}v_{0,j}e_{j}$, $v_{1}=\sum_{j=1}^{\infty}v_{1,j}e_{j}$, $F(t,\cdot)=\sum_{j=1}^{\infty}F_{j}(t)e_{j}$, with

[TABLE]

Let $N\in\mathbb{N}^{*}$, and put $E_{N}={\rm span}\{e_{1},\ldots,e_{N}\}$, $V_{k,N}=(v_{k,1},\ldots,v_{k,N})$, $k=0,1$, $F_{N}=\sum_{j=1}^{N}F_{j}(t)e_{j}$, $B_{N}(t)=(b_{i,j}(t))_{1\leq i,j\leq N}$ with $b_{i,j}(t)=(\nabla e_{i},\nabla e_{j})_{L^{2}(\Omega;\gamma(t,\cdot)dx)}$, and consider the following vectorial differential equation: find $V_{N}(t)=(v_{1}(t),\ldots,v_{N}(t))$ such that

[TABLE]

with the initial condition $V_{N}(0)=V_{0,N}$, $\frac{d}{dt}V_{N}(0)=V_{1,N}$. Since $B_{N}(\cdot)$ is continuous, the theorem of Cauchy-Lipschitz implies existence and uniqueness for $V_{N}(t)$. Note that $B_{N}(t)$ is positive since, for all $U=(u_{1},\ldots,u_{N})$, setting $u(x)=\sum_{j=1}^{N}u_{j}e_{j}(x)$, we have

[TABLE]

where $C$ is a constant such that $0<C\leq\gamma$ in $Q$. Let $v_{N}(t)=\sum_{j=1}^{N}v_{j}(t)e_{j}(x)$. Then, a standart energy estimate for $E_{N}(t)(v_{N})=\frac{1}{2}(\dot{V}_{N}^{2}(t)+V_{N}(t)B_{N}(t)V_{N}(t))$, as above, implies that there exists a positive constant $C$ such that

[TABLE]

Passing to the limit $N\to+\infty$, we can conclude by standard arguments that $(v_{N})_{N}$ converges to a function $v\in C([0,T];H^{1}_{0}(\Omega))$ satisfying (2.5).

The proof of Theorem 2.6 in done in the case $F\in L^{2}(\Omega_{T})$. The case $F\in W$ is similar.