Theoretical study of the fibrous capsule tissue growth around a disk-shaped implant

Abstract

We analyze the mathematical properties of the fibrous capsule tissue concentration around a disk-shaped implant. We establish stability estimates as well as monotonicity results that illustrate the sensitivity of this growth to the biocompatibility index parameters of the implant. In addition, we prove that the growth of the tissue increases exponentially in time toward an asymptotic regime. We also study the mathematical properties of the solution of the inverse problem consisting in the determination of the values of the biocompatibility index parameters from the knowledge of some fibrous capsule tissue measurements. We prove that this model calibration problem admits a unique solution, and establish a characterization of the index parameters. Furthermore, we demonstrate analytically that such a solution is not continuous with respect to the data, and therefore the considered inverse problem is ill-posed due to the lack of the stability requirement.

Appendices

Appendix A: Norm estimates and additional properties

We present in this section auxiliary results pertaining to (a) the equivalence of the norms employed in this paper, and (b) concentration results. The proofs are simple exercises in functional analysis, and they are included only for completeness.

The following lemma provides Poincaré–Friedrichs type inequalities (see, e.g., Brezis 1983) that will be used to prove the equivalence between the norms.

Lemma 3

Let $v \in V = \left\{ v\in H^1(\Omega )~; \quad v(L) = 0 \right\}$. Then, we have:

$$\begin{aligned} ||v||_{L^2(\Omega )}\,&\le \frac{L}{\sqrt{2}}\,||v^{\prime }||_{L^2(\Omega )}, \end{aligned}$$

(49)

$$\begin{aligned} (v(0))^2\,&\le \,L\,||v^{\prime }||^{2}_{L^2(\Omega )}, \end{aligned}$$

(50)

$$\begin{aligned} (v(0))^2\,&\le \,||v||^{2}_{H^1(\Omega )}. \end{aligned}$$

(51)

Proof

First, observe that for $v \in V$, we have $v(L) = 0$, and therefore:

$$\begin{aligned} v(x) \, = \, - \int _x^L v^{\prime }(s)\,ds~; \quad \forall x \in [0, L] \end{aligned}$$

(52)

Hence, using Cauchy–Schwarz inequality, we deduce that:

$$\begin{aligned} (v(x))^2 \,\le \, (L-x)\int _0^L (v^{\prime }(s))^2\,ds~; \quad \forall x \in [0, L] \end{aligned}$$

(53)

which proves Eq. (50). In addition, it follows from integrating Eq. (53) that:

$$\begin{aligned} \int _0^L\,(v(x))^2\,dx\,\le \, \left[-\frac{1}{2}(L-x)^2\right]_0^L\int _0^L\, (v^{\prime }(s))^2\,ds= \frac{L^2}{2}\int _0^L\,(v^{\prime }(s)^2)\,ds, \end{aligned}$$

(54)

which proves Eq. (49).

Next, we prove Eq. (51). We have:

$$\begin{aligned} (v(0))^2 \,= \,- \,\int _0^L\, ((v(x))^{2})^{\prime }\,dx =- 2\,\int _0^L\, v(x)v^{\prime }(x)\,dx. \end{aligned}$$

(55)

Hence, using Young inequality, we deduce that:

$$\begin{aligned} (v(0))^2\, \le \,\int _0^L\,\left( (v(x))^2+ (v^{\prime }(x))^2\right)\,dx \,= \,||v||^{2}_{H^1(\Omega }. \end{aligned}$$

(56)

Property A.1. Consider the following scalar product on $V \times V$:

$$\begin{aligned} (u,v)=\int _0^L\,\left(u^{\prime }(x)v^{\prime }(x)+u(0)v(0)\right)\,dx \end{aligned}$$

(57)

and

$$\begin{aligned}{}[u,v]=\int _0^L \left(e^{-c_2x^2} u^{\prime }(x)v^{\prime }(x)+u(0)v(0)\right)\,dx. \end{aligned}$$

(58)

Then, the corresponding norms denoted by $\Vert \cdot \Vert _3$ and $\Vert \cdot \Vert _4$ respectively are equivalent to $\Vert \cdot \Vert _{H^{1}(\Omega )}$, the standard $H^{1}$-norm.

Proof

First, observe that it is easy to deduce from (49) that:

$$\begin{aligned} \Vert v^{\prime }\Vert _{L^{2}(\Omega )}^2\, \le \Vert v\Vert _{H^{1}(\Omega )}^2\,\le \, \left(1+\frac{L^2}{2}\right) \ \Vert v^{\prime }\Vert _{L^{2}(\Omega )}^2~, \quad \forall \,v \in V. \end{aligned}$$

(59)

Moreover, it follows from Eq. (50) that:

$$\begin{aligned} \Vert v^{\prime }\Vert _{L^{2}(\Omega )}^2\,\le \, \Vert v\Vert _3^2 \, \le \,(1+L)\, \Vert v^{\prime }\Vert _{L^{2}(\Omega )}^2\,~, \quad \forall \,v \in V. \end{aligned}$$

(60)

Consequently, it results from Eq. (59) and Eq. (60) that the norms $\Vert \cdot \Vert _3$ and $\Vert \cdot \Vert _{H^{1}(\Omega )}$ are equivalent. Furthermore, it follows from the definition given by Eqs. (57) and (58) that:

$$\begin{aligned} \Vert v\Vert _4^2\le \Vert v\Vert _3^2\le e^{c_2L^2} \Vert v\Vert _4^2~, \quad \forall \,v \in V. \end{aligned}$$

(61)

which proves the equivalence between the norms $\Vert \cdot \Vert _3$ and $\Vert \cdot \Vert _4$ and therefore equivalent to $\Vert \cdot \Vert _{H^{1}(\Omega }$.

Lemma 4

The vector space $W \!= \!\left\{ w\in H^2(\Omega ); w(L) \!=\! 0~\mathrm {and}~ w^{\prime }(0)-c_4w(0) = 0 \right\}$ is dense in the vector space $V= \left\{ v\in H^1(\Omega ); \quad v(L) = 0 \right\}$.

Proof

For $v\in V$, we consider the sequence $\left(v_n\right)_{n \ge 1}$ satisfying the following boundary value problem:

$$\begin{aligned} \left\{ \begin{array}{l} -\frac{1}{n^2}\frac{d}{dx}(e^{-c_2x^2}\frac{dv_n}{dx}) + v_n=v\quad \text{ in}\;\; \Omega ,\\ \ \\ v_n(L)=0,\quad v^{\prime }_{n}(0) = c_4\,v_n(0). \end{array} \right. \end{aligned}$$

(62)

We also consider the following variational problem corrsponding to the boundary value problem given by (62):

$$\begin{aligned} a(v_n,\varphi ) = \ell (\varphi )~; \quad \forall \varphi \in V, \end{aligned}$$

(63)

where

$$\begin{aligned} a(v_n,\varphi )=\frac{1}{n^2}\int _0^L\,\left(e^{-c_2x^2}v^{\prime }_{n}(x){\varphi }^{\prime }(x)+v_n(x)\varphi (x)\right)dx~; \quad \forall \varphi \in V \end{aligned}$$

(64)

and

$$\begin{aligned} \ell (\varphi )=\int _0^L\, v(x)\varphi (x) \,dx~; \quad \forall \varphi \in V. \end{aligned}$$

(65)

Using the previous norm equivalent results (see Property A.1), it is then easy to check that the variational problem given by (63) admits a unique solution $v_n \in W$. Next, we prove that the sequence $(v_n)_{n\ge 1}$ converges to $v$ in $V$.

First, we substitute $\varphi = v_n$ into the variational problem (63), and then integrate by parts. We obtain:

$$\begin{aligned} \frac{1}{n^2}\int _P^L\,e^{-c_2x^2}(v^{\prime }_{n}(x))^2\,dx+\int _0^L\,(v_n(x))^2\,dx+\frac{1}{n^2}c_4(v_n(0))^2= \int _0^L\, v(x)v_n(x)\,dx.\nonumber \\ \end{aligned}$$

(66)

Second, we apply successively Cauchy–Schwarz and Young inequalities to bound the right-hand side. We obtain:

$$\begin{aligned}&\frac{1}{n^2}\int _0^L\, e^{-c_2x^2}(v^{\prime }_{n}(x))^2\,dx+\int _0^L\,(v_n(x))^2\, dx+\frac{1}{n^2}c_4(v_n(0))^2\nonumber \\&\quad \le \frac{1}{2}\left(\Vert v\Vert ^2_{L^{2}(\Omega )}+\Vert v_n\Vert ^2_{L^{2}(\Omega )}\right). \end{aligned}$$

(67)

Consequently, we deduce:

$$\begin{aligned} \left\{ \begin{array}{l} \Vert v_n\Vert _{L^{2}(\Omega )}\le \Vert v\Vert _{L^{2}(\Omega )}, \\ \ \\ \frac{\sqrt{c_4}}{n}|v_n(0)|\,\le \,\Vert v\Vert _{L^{2}(\Omega )}, \\ \ \\ \frac{1}{n}\Vert v^{\prime }_{n}\Vert ^2_{L^{2}(\Omega )}\,\le \, \Vert v\Vert ^2_{L^{2}(\Omega )}. \end{array} \right. \end{aligned}$$

(68)

Using again the variational problem (63) for $\varphi \in \mathfrak D (\Omega )$ along with the estimates given by (68), we deduce that:

$$\begin{aligned} \int _0^Lv_n(x)\varphi dx\longrightarrow \int _0^Lv(x)\varphi dx, \quad \mathrm {as} \quad n \longrightarrow \infty \quad \forall \varphi \in \mathfrak D\mathrm (\Omega ). \end{aligned}$$

Moreover, since $\mathfrak D (\Omega )$ is dense in $L^2(\Omega )$ (se, e.g., Brezis 1983; Lions and Magenes 1971; Taylor 1997), it follows that the sequence $(v_n)_{n\ge 1}$ converges weakly to $v$ into $L^2(\Omega )$. Furthermore, it follows from using the variational problem (63) with $\varphi =- (e^{-c_2x^2}v^{\prime }_{n})^{\prime }$, and then integrating by parts, that:

$$\begin{aligned} \left\{ \begin{array}{l} \frac{1}{n^2}\int _0^L\left(\left(e^{-c_2x^2}v^{\prime }_{n}(x)\right) ^{\prime } \right)^2dx+\int _0^L\,e^{-c_2x^2}(v^{\prime }_{n}(x))^2\,dx+c_4(v_n(0))^2 \\ \ \\ =\int _0^L\,e^{-c_2x^2}(v(x))^{2}(v_n(x))^{2}\,dx+c_4v(0)v_n(0). \end{array} \right. \end{aligned}$$

(69)

On the other hand, since $c_4 > 0$, the following:

$$\begin{aligned}{}[[u,v]] = c_4u(0)v(0)+\int _0^L\, e^{-c_2x^2} u^{\prime }(x)v^{\prime }(x)\,dx~; \quad \forall \, u, v \, \in \, V \end{aligned}$$

defines a scalar product on $V \times V$, and its corresponding norm, denoted by $\Vert .\Vert _5$, is equivalent to the norm $\Vert .\Vert _4$ given by Eq. (58) in Property A.1 (which is then equivalent to the standard $H^1$-norm). Consequently, it follows from Eq. (69) that:

$$\begin{aligned} \Vert v_n\Vert ^2_5\le [[v,v_n]] \le \Vert v\Vert _5\Vert v_n\Vert _5. \end{aligned}$$

Hence, the sequence $(v_n)_{n\ge 1}$ converges weakly to $v$ in $H^1(\Omega )$. Finally, using the inequality $\Vert v_n\Vert _5\le \Vert v\Vert _5$ allows to conclude that this convergence is in fact in the strong sense.

Appendix B: Properties of the solution of BVP (2)

We have already established the existence and the uniqueness of the solution $v$ of the boundary value problem BVP (2) (see Lemma 1). The goal here is to study the properties of the solution $v$ and to derive stability. These results do not appear to be standard and they may be of independent interest. These results are formulated in the following three lemmas.

Lemma 5

The solution $v$ of the boundary value problem BVP (2) satisfies the following properties:

• i. $\forall \,x\,\in [0,L[: \quad 0< v(x) \le v(0)<1.$

• ii. $||v||^2_{L^{2}(\Omega )}\,\le \,K_0; \quad {where} \quad K_0 = \min \left\{ L, \frac{c_1c_4}{4c_3}, \frac{L^2c_4}{8}e^{c_2L^2}, \frac{L^2}{2}(\frac{c_4}{4}+c_2L)\right\} .$

Proof

We set the variational problem given by Eqs. (4–6) with $\varphi = v$, that is:

$$\begin{aligned} c_1\int _0^L\,e^{-c_2x^2}v^{\prime }(x)^2\,dx+c_3\int _0^L\,(v(x))^2\, dx+c_1c_4(v(0))^2 = c_1c_4\,v(0). \end{aligned}$$

(70)

Therefore, $v(0) > 0$ (otherwise $v \equiv 0$). Moreover, by re-writing Eq. (70) as follows:

$$\begin{aligned} c_1\int _0^L\,e^{-c_2x^2}(v^{\prime }(x))^2\,dx+c_3\int _0^L\,(v(x))^2 \,dx = c_1c_4\,v(0)(1-v(0)), \end{aligned}$$

(71)

we deduce that $v(0) < 1$ (otherwise $v \equiv 0$). Next, we consider the boundary value problem BVP (2). We multiply the first equation by $v$ and integrate by parts over $[a, L[$, for $a \ge 0$. We obtain:

$$\begin{aligned} c_1\int _a^L\,e^{-c_2x^2}(v^{\prime }(x))^2\,dx+c_3\int _a^L\,(v(x))^2 \,dx =- c_1e^{-c_2a^2}v(a)v^{\prime }(a). \end{aligned}$$

(72)

Therefore, we must have $v(a)v^{\prime }(a) < 0$ for all $a \ge 0$ (otherwise $v (x) = 0$ on some interval $]a, L[$, which necessarily makes $v \equiv 0$). Consequently, we must have:

$$\begin{aligned} \forall \, x \,\in [0, L[, \quad v^{\prime }(x) \,< \,0, \end{aligned}$$

(73)

and therefore:

$$\begin{aligned} \forall \, x \,\in [0, L[, \quad 0\,< \,v(x)\,\le \,v(0)\,<\,1. \end{aligned}$$

In the following, we prove property (ii). First, observe that it follows from property (i.) that :

$$\begin{aligned} ||v||_{L^{\infty }(\Omega )}\,\le \, 1, \quad \text{ and} \text{ therefore} \quad ||v||^2_{L^{2}(\Omega )}\,\le \,L. \end{aligned}$$

In addition, using Eq. (71) and the fact that $c_1c_4\,v(0)(1-v(0))\, \le \,\frac{c_1c_4}{4}$, we deduce that:

$$\begin{aligned} \int _0^L(v^{\prime }(x))^2\,dx\le \frac{c_4}{4}e^{c_2L^2}, \quad \text{ and}\quad \int _0^L\,(v(x))^2\,dx\le \frac{c_1c_4}{4c_3}. \end{aligned}$$

(74)

Furthermore, it follows from the Friedrichs–Poincaré inequality given by Eq. (49) in Lemma 3, and Eq. (74) that:

$$\begin{aligned} \int _0^L\,(v(x))^2\,dx \le L^2\frac{c_4}{8}e^{c_2L^2}. \end{aligned}$$

(75)

Moreover, we obtain from the variational problem given by Eqs. (4–6), in which we set $\varphi = e^{c_2x^2}v$ and integrate by parts over $\Omega$, that:

$$\begin{aligned} \left\{ \begin{array}{l} c_1\int _0^L\,(v^{\prime }(x))^2\,dx+c_3\int _0^L\,e^{c_2x^2}(v(x))^2 \,dx+c_1c_4(v(0))^2\\ \ \\ =c_1c_4v(0)+ c_1c_2\int _0^L\, (v(x))^2 dx \end{array} \right. \end{aligned}$$

(76)

Since $c_1c_4v(0)\le c_1c_4(v(0))^2+\frac{c_1c_4}{4}$, and using property (i.), we deduce that:

$$\begin{aligned} c_1\int _0^L\,(v^{\prime }(x))^2\,dx+c_3\int _0^L\,e^{c_2x^2}(v(x))^2 \,dx\le \frac{c_1c_4}{4}+c_1c_2L. \end{aligned}$$

(77)

and therefore,

$$\begin{aligned} \int _0^L\,(v^{\prime }(x))^2\,dx\le \frac{c_4}{4}+c_2L. \end{aligned}$$

(78)

Using again the Friedrichs–Poincaré inequality given by Eq. (49) in Lemma 3, we deduce that:

$$\begin{aligned} \int _0^L\,(v(x))^2 \,dx \le \frac{L^2}{2}\left(\frac{c_4}{4}+c_2L\right). \end{aligned}$$

(79)

Lemma 6

The first derivative $v^{\prime }$ of the solution of the boundary value problem BVP (2) satisfies:

• i. $\forall \,x\,\in [0,L[: \quad v^{\prime }(x)<0.$

• ii. $||v^{\prime }||^2_{L^{2}(\Omega )}\le K_1$ where $K_1\,= \,\min \left\{ \frac{c_4}{4}e^{c_2L^2}, \frac{c_4}{4}+c_2L\right\} .$

• iii. $x \longrightarrow e^{-c_2x^2}v^{\prime }(x)$ is negative and increasing on $\Omega$, and satisfies $\sup _{x \in \Omega }\big |e^{-c_2x^2} v^{\prime }(x)\big | < \,c_4$.

Proof

The property (i.) (resp. property (ii.)) has already been established in the proof of Lemma B.1 [see Eq. (73), (resp. Eq. (74) and Eq. (78))].

Next, we prove property (iii.). It follows from the first equation of the boundary value problem BVP (2) that $x \longrightarrow e^{-c_2x^2}v^{\prime }(x)$ is a negative and an increasing function on $\Omega$. In addition, using the boundary condition of BVP (2), we deduce that $\sup _{x \in \Omega }\left|e^{-c_2x^2}v^{\prime }(x)\right|\,\le \,|v^{\prime }(0)| = c_4(1-v(0)) < c_4$.

Lemma 7

The second derivative of $v$, the solution of the boundary value problem BVP (2), satisfies:

$$\begin{aligned} ||v^{\prime \prime }||^2_{L^{2}(\Omega )}\,\le \,K_2e^{c_2L^2} \end{aligned}$$

where

$$\begin{aligned} K_2 = 4c_2^2\,L^2K_1\,e^{-c_2L^2}+2\frac{c_3}{c_1}\left(\frac{c_4}{4}+c_2L\right). \end{aligned}$$

Proof

It follows from the first equation of the boundary value problem BVP (2) that:

$$\begin{aligned} v^{\prime \prime }(x)\,=\,2c_2xv^{\prime }(x)+\frac{c_3}{c_1}e^{c_2x^2}v(x). \end{aligned}$$

Hence, using Young inequality, we deduce that:

$$\begin{aligned} (v^{\prime \prime }(x))^2\,\le \,4c^{2}_2x^{2}\,(v^{\prime }(x))^2+2\frac{c^{2}_3}{c^{2}_1}\left(e^{c_2x^2}v(x)\right)^{2}. \end{aligned}$$

Consequently, we have:

$$\begin{aligned} \int _0^L\, (v^{\prime \prime }(x))^2\, dx \,\le \, 4c_2^2\int _0^L x^2\,(v^{\prime }(x))^2\,dx+ 2\frac{c_3^2}{c_1^2}\int _0^L\,e^{c_2x^2}e^{c_2x^2}(v(x))^2\,dx.\qquad \end{aligned}$$

(80)

Thus,

$$\begin{aligned} \int _0^L\, (v^{\prime \prime }(x))^2\, dx \,\le \, 4c_2^2\,L^2\int _0^L \,(v^{\prime }(x))^2\,dx+ 2\frac{c_3^2}{c_1^2}\,e^{c_2L^2}\int _0^L\,e^{c_2x^2}(v(x))^2\,dx. \end{aligned}$$

(81)

Finally, using property (ii.) of Lemma 6 and Eq. (77), we obtain that:

$$\begin{aligned} \int _0^L\, (v^{\prime \prime }(x))^2\, dx \,\le \, 4c_2^2\,L^2K_1+ 2\frac{c_3^2}{c_1^2}\,e^{c_2L^2}\, \left(\frac{c_1c_4}{4c_3}+ \frac{c_1c_2}{c_3}L\right). \end{aligned}$$

(82)

Appendix C: Properties of the solution of IBVP (3)

This section is devoted to the analysis of the properties of the solution of the homogenous initial boundary value problem BVP(2). We present useful stability estimates on the solution $w$ of the initial boundary value problem IBVP(3). These results are based on the estimates established in Appendix B. They are formulated in the following three lemmas.

Lemma 8

The solution $w$ of the initial boundary value problem IBVP(3) (or (8)) satisfies the following estimate:

$$\begin{aligned} \Vert w(\cdot ,t)\Vert ^2_{L^2(\Omega )}\ \le \ K_0\, e^{-2\beta t}~; \quad \forall \, t > 0, \end{aligned}$$

(83)

where $K_0$ is defined in Lemma 5, and the positive constant $\beta$ is given by:

$$\begin{aligned} \beta =c_3+\frac{2c_1}{L^2}e^{-c_2L^2}. \end{aligned}$$

Proof

First, observe that:

$$\begin{aligned} c_1\int _0^L\,e^{-c_2x^2}\left(\frac{\partial w}{\partial x}(x,t)\right)^2\,dx \, \ge \, c_1e^{-c_2L^2}\int _0^L\,\left(\frac{\partial w}{\partial x}(x,t)\right)^2\,dx~; \quad \forall \, t \ge 0. \end{aligned}$$

Then, using the Friedrichs–Poincaré inequality given by Eq. (50), we obtain:

$$\begin{aligned} c_1\int _0^L\,e^{-c_2x^2}\Bigr (\frac{\partial w}{\partial x}(x,t)\Bigl )^2\,dx\,\ge \,2\frac{c_1}{L^2}e^{-c_2L^2}\int _0^L\,(w(x,t))^2\,dx~; \quad \forall \, t \ge 0.\qquad \end{aligned}$$

(84)

Hence, it follows from subtituting Eq. (84) into Eq. (12) (in which $w_n$ is replaced by $w$) that:

$$\begin{aligned} Y^{\prime }(t)+2\beta Y(t)\le 0~; \quad \forall \, t \ge 0. \end{aligned}$$

where the positive function $Y(t)$ is given by:

$$\begin{aligned} Y(t)=\int _0^L\,(w(x,t))^2\,dx~; \quad \forall \, t \ge 0. \end{aligned}$$

Moreover, we have:

$$\begin{aligned} \frac{d}{dt}\bigr [e^{2\beta t}Y(t)\bigl ] \,=\, e^{2\beta t}\bigr [Y^{\prime }(t)+2\beta Y(t)\bigl ]\le 0~; \quad \forall \, t \ge 0. \end{aligned}$$

Therefore, we deduce that:

$$\begin{aligned} e^{2\beta t}\,Y(t) \le Y(0)~; \quad \forall \, t \ge 0, \end{aligned}$$

which is equivalent to:

$$\begin{aligned} Y(t) \le e^{-2\beta t}\,Y(0)~; \quad \forall \, t \ge 0, \end{aligned}$$

that is:

$$\begin{aligned} \int _0^L\,(w(x,t))^2\,dx\le e^{-2\beta t}\int _0^L\,(v(x))^2\,dx~; \quad \forall \, t \ge 0. \end{aligned}$$

(85)

The use of Lemma 5 concludes the proof.

Lemma 9

The solution $w$ of the initial boundary value problem IBVP(3) (or (8)) satisfies the following estimate:

$$\begin{aligned} \left\Vert\frac{\partial w}{\partial x}(\cdot ,t)\right\Vert ^2\,<\, c_4\,e^{c_2L^2}\,e^{-2c_3t}~; \quad \forall \,t \ge 0. \end{aligned}$$

(86)

Proof

We set:

$$\begin{aligned} Z(t)=c_4(w(0,t))^2+\int _0^L\,e^{-c_2x^2}\Bigr (\frac{\partial w}{\partial x}(x,t)\Bigl )^2\,dx~; \quad \forall \, t \ge 0. \end{aligned}$$

(87)

Hence, it follows from Eq. (15) that:

$$\begin{aligned} Z^{\prime }(t)+2c_3\,Z(t)=-2c_1\int _0^L\,\left[\frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}(x,t)\right)\right]^2\,dx\le 0~; \quad \forall \, t \ge 0. \end{aligned}$$

Therefore, we have:

$$\begin{aligned} Z(t)\,\le \,e^{-2c_3t}\,Z(0)~; \quad \forall \, t \ge 0. \end{aligned}$$

On the other hand, we have:

$$\begin{aligned} Z(0) = c_4(v(0))^2+\int _0^L\,e^{-c_2x^2}\Bigr (v^{\prime }(x)\Bigl )^2\,dx. \end{aligned}$$

Hence, it follows from the variational problem given by Eqs.(4–6) that:

$$\begin{aligned} Z(0) = c_4v(0)-\frac{c_3}{c_1}\int _0^L\,(v(x))^2\,dx\,\le \,c_4v(0). \end{aligned}$$

(88)

Finally, using Lemma 5, we deduce:

$$\begin{aligned} 0 \, \le \, Z(0) \, < \, c_4. \end{aligned}$$

(89)

Consequently, we have just established the following a priori estimate:

$$\begin{aligned} 0 \, \le \, Z(t) \, = \, c_4(w(0,t))^2 + \int _0^L\,e^{-c_2x^2}\left(\frac{\partial w}{\partial x}(x,t)\right)^2\,dx \,< \,c_4e^{-2c_3t}~; \quad \forall \, t \ge 0.\nonumber \\ \end{aligned}$$

(90)

The use of $e^{-c_2x^2} \ge e^{-c_2L^2}$ for any $x \in \Omega$ allows to conclude the proof of Lemma 6.

Lemma 10

The solution $w$ of the initial boundary value problem IBVP(3) (or (8)) satisfies the following estimate:

$$\begin{aligned} \int _0^\infty \int _0^L\,\left(\frac{\partial ^2w}{\partial x^2}(x,t)\right)^2\,dx\,dt\,\le \, K_3\,e^{2c_2L^2}. \end{aligned}$$

(91)

where the positive constant $K_3$ is given by:

$$\begin{aligned} K_3 \ = \ c_4\left(\frac{1}{c_1} + 4\,c^{2}_{2}L^2\,e^{-c_2L^2}\right). \end{aligned}$$

Proof

For any positive number $T$, we integrate Eq. (15) over $]0, T[$. We then obtain:

$$\begin{aligned} Z(T)+2c_3\int _0^T Z(t)dt+2c_1\int _0^T\int _0^L\left[\frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}(x,t)\right)\right]^2dxdt = Z(0), \end{aligned}$$

where $Z(t)$ is given by Eq. (87) in the proof of Lemma 9.

Hence, for $T \longrightarrow \infty$, it follows from Eqs. (89) and (90) that:

$$\begin{aligned} \int _0^\infty \int _0^L\left[\frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}(x,t)\right)\right]^2dxdt\le \frac{c_4}{2c_1}. \end{aligned}$$

(92)

On the other hand, we also have:

$$\begin{aligned} \frac{\partial ^2w}{\partial x^2}(x,t) \, = \, e^{c_2x^2}\frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}(x,t)\right)+2c_2x\frac{\partial w}{\partial x}(x,t). \end{aligned}$$

Thus, using Young’s inequality, we obtain that:

$$\begin{aligned}&\left\Vert \frac{\partial ^2w}{\partial x^2}\right\Vert_{L^2(\Omega \times ]0, \infty [)}^{2}\\&\quad \le 2\left(\left\Vert e^{c_2x^2}\frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}\right)\right\Vert_{L^2(\Omega \times ]0, \infty [)}^{2} + 4 c_2^2 \left\Vert x\frac{\partial w}{\partial x}\right\Vert _{L^2(\Omega \times ]0, \infty [)}^{2}\right). \end{aligned}$$

Since $x \in \Omega = ]0, L[$, we deduce that:

$$\begin{aligned}&\!\!\left\Vert \frac{\partial ^2w}{\partial x^2}\right\Vert_{L^2(\Omega \times ]0, \infty [)}^{2}\nonumber \\&\!\!\quad \le 2\left( e^{2c_2L^2}\,\left\Vert \frac{\partial }{\partial x}\left(e^{-c_2x^2}\frac{\partial w}{\partial x}\right)\right\Vert_{L^2(\Omega \times ]0, \infty [)}^{2} \!+ \!4c_2^2L^2 \,\left\Vert \frac{\partial w}{\partial x}\right\Vert _{L^2(\Omega \times ]0, \infty [)}^{2}\right)\!.\qquad \qquad \end{aligned}$$

(93)

Finally, substituting Eq. (86) and Eq. (92) into Eq. (93) concludes the proof of Lemma 10.