Lebeau–Robbiano Inequality for Heat Equation with Dynamic Boundary Conditions and Optimal Null Controllability

Abstract

The paper deals with time and norm optimal control problems for the heat equation with dynamic boundary conditions in the context of the null controllability. More precisely, we answer an open question left in our previous paper (Boutaayamou et al., in Math Methods Appl Sci 45:1359–1376, 2021). To do so, we first prove a new Lebeau–Robbiano spectral inequality using a logarithmic convexity inequality, then an observability inequality on any set of positive measure. This plays a relevant role in proving the existence and uniqueness of optimal null controls. Finally, the connection between time and norm optimal null controls is presented.

Appendix

Lemma 5.3

Assume that there are two positive constants $D_1$ and $D_2$, such that, for any $\left( a_i\right) _{i \ge 1} \in l^2(\mathbb {R})$ and any $r > 0$,

$$\begin{aligned} \sqrt{\sum _{\lambda _{k} \le r}\mid a_{k}\mid ^{2}} \le D_{1} e^{D_{2} \sqrt{r}} \left\| \sum _{\lambda _{k} \le r} a_{k} \psi _{k} \right\| _{L^{2}(\omega )}. \end{aligned}$$

(15)

Then, for any $T>0$, there exist $\beta \in (0,1)$ such that

$$\begin{aligned} \Vert \Psi (\cdot , T)\Vert \le \left( \mu \text {e}^{\frac{K}{T}}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}\right) ^{\beta }\Vert \Psi (\cdot , 0)\Vert ^{1-\beta }, \end{aligned}$$

Proof

Let $\displaystyle \Psi _{0}=\sum _{k \ge 1} a_{k} \Phi _{k} \in \mathbb {L}^2.$ First, we have

$$\begin{aligned} \left\| \Psi (\cdot , T)\right\| ^{2} = \left\| \sum _{k \ge 1}a_{k} e^{-\lambda _{k} T} \Psi _{k}\right\| ^{2} = \sum _{\lambda _{k} \le r}\mid a_{k} e^{-\lambda _{k} T} \mid ^{2}+\sum _{\lambda _{k} > r}\mid a_{k} e^{-\lambda _{k} T} \mid ^{2}. \end{aligned}$$

By the inequality (15), we obtain

$$\begin{aligned} \sqrt{\sum _{\lambda _{k} \le r}\mid a_{k}e^{- \lambda _{k} T} \mid ^{2}} \le D_{1} e^{D_{2} \sqrt{r}} \left\| \sum _{\lambda _{k} \le r} a_{k}e^{-\lambda _{k} T}\psi _k \right\| _{L^{2}(\omega )}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \Vert \Psi (\cdot , T)\Vert&\le \sqrt{\sum _{\lambda _{k} \le r}\mid a_{k} e^{-\lambda _{k} T} \mid ^{2}}+\sqrt{\sum _{\lambda _{k}> r}\mid a_{k} e^{-\lambda _{k} T} \mid ^{2}}\\&\le D_{1} e^{D_{2} \sqrt{r}}\left\| \sum _{\lambda _{k} \le r} a_{k} e^{-\lambda _{k} T} \psi _{k}\right\| _{L^{2}(\omega )}+\sqrt{\sum _{\lambda _{k}>r}\mid a_{k} e^{-\lambda _{k} T}\mid ^{2}} \\&\le D_{1} e^{D_{2} \sqrt{r}}\left\| \sum _{\lambda _{k} \le r} a_{k} e^{-\lambda _{k} T} \psi _{k}+\sum _{\lambda _{k}>r} a_{k} e^{-\lambda _{k} T}\psi _{k}\right\| _{L^{2}(\omega )} \\&\quad +D_{1} e^{D_{2} \sqrt{r}}\left\| \sum _{\lambda _{k}>r} a_{k} e^{-\lambda _{k} T} \psi _{k}\right\| _{L^{2}(\omega )}+\sqrt{\sum _{\lambda _{k}>r}\mid a_{k} e^{-\lambda _{k} T}\mid ^{2}} \\&\le D_{1} e^{D_{2} \sqrt{r}}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}+\left( 1+ D_{1}\right) e^{D_{2} \sqrt{r}} e^{-r T} \sqrt{\sum _{\lambda _{k}>r}\mid a_{k}\mid ^{2}}\\&\le D_{1} e^{D_{2} \sqrt{r}}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}+\left( 1+ D_{1}\right) e^{D_{2} \sqrt{r}} e^{-r T} \left\| \Psi \left( \cdot ,0 \right) \right\| .\\ \end{aligned} \end{aligned}$$

Using Young inequality, for any $\varepsilon >0$, we obtain

$$\begin{aligned} D_{2} \sqrt{r}=\frac{D_{2}}{\sqrt{\varepsilon T}} \sqrt{\varepsilon r T} \le \varepsilon r T+\left( \frac{D_{2}}{\sqrt{\varepsilon T}}\right) ^{2}. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \Vert \Psi (\cdot , T)\Vert&\le D_{1} e^{D_{2} \sqrt{r}}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}+\left( 1+ D_{1}\right) e^{D_{2} \sqrt{r}} e^{-r T} \left\| \Psi \left( \cdot ,0 \right) \right\| \\&\le D_{1} e^{\frac{D_{2}^2}{\varepsilon T}} e^{\varepsilon r T}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}+\left( 1+ D_{1}\right) e^{\frac{D_{2}^2}{\varepsilon T}} e^{(\varepsilon -1)r T} \left\| \Psi \left( \cdot ,0 \right) \right\| \\&\le (1+ D_{1}) e^{\frac{D_{2}^2}{\varepsilon T}} \left( e^{\varepsilon r T}\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}+ e^{(\varepsilon -1)r T} \left\| \Psi \left( \cdot ,0 \right) \right\| \right) . \end{aligned} \end{aligned}$$

By taking

$$\begin{aligned} rT=\ln \left( \frac{\left\| \Psi (\cdot , 0)\right\| }{\left\| \psi (\cdot , T)\right\| _{L^{2}(\omega )}}\right) , \end{aligned}$$

we deduce

$$\begin{aligned} \begin{aligned} \Vert \Psi (\cdot , T)\Vert&\le (1+ D_{1}) e^{\frac{D_{2}^2}{\varepsilon T}} \left[ \left( \frac{\Vert \Psi (\cdot , 0)\Vert }{\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}} \right) ^{\varepsilon } \Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}\right. \\&\quad \left. + \left( \frac{\Vert \Psi (\cdot , 0)\Vert }{\Vert \psi (\cdot , T)\Vert _{L^{2}(\omega )}} \right) ^{\varepsilon -1} \left\| \Psi \left( \cdot ,0 \right) \right\| \right] . \end{aligned} \end{aligned}$$

Setting $\beta = 1-\varepsilon$, yields to the desired formula

$$\begin{aligned} \begin{aligned} \quad \Vert \Psi (\cdot , T)\Vert&\le 2 (1+ D_{1}) e^{\frac{D_{2}^2}{\varepsilon T}} \left\| \Psi (\cdot , 0)\right\| ^{1-\beta } \left\| \psi (\cdot , T)\right\| _{L^{2}(\omega )}^{\beta }. \end{aligned} \end{aligned}$$

$\square$