A note on the IBVP for wave equations with dynamic boundary conditions

Abstract

In this paper, we investigate the controllability on the IBVP for a class of wave equations with dynamic boundary conditions by the HUM method as well as the wellposedness for the related back-ward problems. After proving a new observability inequality, we establish new wellposedness and controllability theorems for the IBVP.

1 Introduction

In this paper, we consider the exact boundary controllability on the IBVP for wave equation with dynamic boundary condition as follows:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \phi''-\Delta\phi+ f(\phi)=0, &(x,t)\in Q=\Omega\times(0,T), \\ -\Delta_{T}\phi+\frac{\partial\phi}{\partial\nu}=v_{1},& \mbox{on } \Gamma_{1}\times(0,T), \\ \phi=0, & \mbox{on } \Gamma_{0}\times(0,T), \\ \phi(0,x)=\phi_{0},\qquad \phi_{t}(0,x)=\phi_{1},& x\in\Omega, \end{array}\displaystyle \right . $$

(1.1)

where $\Omega\subset\mathbb{R}^{n}$ is a bounded domain with smooth boundary $\Gamma_{0}\cup\Gamma_{1}$, $\bar{\Gamma}_{0}\cap\bar{\Gamma }_{1}=\emptyset$, and $\Delta_{T}$ is tangential Laplace operator. The boundary condition on $\Gamma_{1}$ is called the static Wentzell boundary condition and the dynamic Wentzell boundary condition is

$$ \phi''-\Delta_{T}\phi+\frac{\partial\phi}{\partial\nu}=v_{1}, \quad \mbox{on } \Gamma_{1}\times(0,T). $$

(1.2)

The system models an elastic body’s transverse vibration. For details, please see the paper of Lemrabet [1]. In [1–7] and the references therein, one can find more details as regards dynamic boundary conditions. Moreover, Heminna [3] gives the controllability for elasticity system with two controls: both tangential and normal, under the assumption of the wellposedness for the backward system, which is a key assumption for getting controllability. In this paper, we establish first of all the wellposedness theorem for back-ward systems based on the transposition method (cf. [8]) and then obtain the controllability on the IBVP for the wave equation above by using the method of HUM.

2 Boundary controllability for Wentzell systems

For simplicity, we write

$$V=H^{1}_{\Gamma_{0}}(\Omega):=\bigl\{ v\in H^{1}(\Omega): v|_{\Gamma_{1}}\in H^{1}(\Gamma_{1}), v|_{\Gamma_{0}}=0 \bigr\} ,\qquad \mathcal{H}=V\times L^{2}(\Omega), $$

with the norm

$$\begin{aligned}& \|u\|_{V}^{2}=\|\nabla u\|_{L^{2}(\Omega)}^{2}+\| \nabla_{T} u\|_{L^{2}(\Gamma _{1})}^{2}, \\& \bigl\Vert (u,v) \bigr\Vert _{\mathcal{H}}^{2}=\|u\|_{V}^{2}+\|v \|_{L^{2}(\Omega)}^{2}. \end{aligned}$$

We study the controllability under the geometric condition:

$$\exists x_{0}\in\mathbb{R}^{n}, \quad (x-x_{0}) \cdot\nu\leq0,\quad \mbox{on }\Gamma_{0}. $$

Take a look at the linear homogeneous system first,

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u''-\Delta u=0, & (x,t)\in Q=\Omega\times(0,T), \\ -\Delta_{T}u+\frac{\partial u}{\partial\nu}=0, &\mbox{on } \Gamma _{1}\times(0,T), \\ u=0, & \mbox{on } \Gamma_{0}\times(0,T), \\ u(0,x)=u_{0},\qquad u_{t}(0,x)=u_{1},& x\in\Omega. \end{array}\displaystyle \right . $$

(2.1)

The wellposedness for the problem (2.1) is not hard to see. Define an operator $\mathcal{A}: D(\mathcal{A})\rightarrow\mathcal{H}$ by

$$\mathcal{A}\left ( \textstyle\begin{array}{@{}c@{}} u\\ v \end{array}\displaystyle \right ):=\left ( \textstyle\begin{array}{@{}c@{}} v\\ \Delta u \end{array}\displaystyle \right ), $$

with

$$\begin{aligned}& D(\mathcal{A}):=\bigl\{ (u,v)\in\mathcal{H}:\Delta u\in L^{2}(\Omega),v \in V,\partial_{\nu} u-\Delta_{T} u=0\bigr\} , \\& D\bigl(\mathcal{A}^{2}\bigr)=\bigl\{ (u,v)^{T}\in D( \mathcal{A}): \mathcal{A}(u,v)^{T}\in \mathcal{H}\bigr\} . \end{aligned}$$

Write

$$E(t):=\frac{1}{2} \int_{\Omega} \bigl(|\nabla u|^{2}+\bigl\vert u'\bigr\vert ^{2} \bigr)\, dx+\frac {1}{2} \int_{\Gamma_{1}}|\nabla_{T} u|^{2} \, ds. $$

Then it is clear that $E(t)=E(0)$.

Lemma 2.1

(Observability inequality)

For $T>2R$,

$$ E(0)\leq C \int_{\Sigma_{1}} \bigl(u^{\prime 2}+u^{2}+| \nabla_{T} u|^{2}+|\Delta_{T}u|^{2} \bigr) \, ds\, dt, $$

(2.2)

where $R=\max_{x\in\bar{\Omega}}{|x-x_{0}|}$, $\Sigma_{1}=(0,T)\times\Gamma_{1}$.

Proof

Multiply the equation with the radial multiplier $(x-x_{0})\cdot\nabla u+\frac{n-1}{2}u$ and integrate by parts in Q. Then we obtain

$$\begin{aligned} \begin{aligned}[b] &\frac{1}{2} \int_{Q} \bigl(\bigl\vert u'\bigr\vert ^{2}+|\nabla u|^{2} \bigr)\,dx\,dt+\frac{1}{2} \int _{\Sigma_{1}}|\nabla_{T} u|^{2} \,ds\,dt+ \biggl\vert \biggl\langle u',(x-x_{0})\cdot\nabla u+ \frac {n-1}{2}u\biggr\rangle \biggr\vert _{0}^{T} \\ &\quad = \frac{1}{2} \int_{\Sigma_{1}}(x-x_{0})\cdot\nu\bigl\vert u'\bigr\vert ^{2} \, ds\, dt+ \int_{\Sigma _{1}}\frac{\partial u}{\partial\nu}(x-x_{0})\cdot\nabla u \, ds\, dt \\ &\qquad {}+\frac {n-1}{2} \int_{\Sigma_{1}}u\frac{\partial u}{\partial\nu} \, ds\, dt +\frac{1}{2} \int_{0}^{T} \int_{\Gamma_{0}}(x-x_{0})\cdot\nu\biggl\vert \frac{\partial u}{\partial\nu}\biggr\vert ^{2} \,ds\,dt \\ &\qquad {}+\frac{1}{2} \int_{\Sigma_{1}} \bigl(|\nabla _{T}u|^{2}-(x-x_{0}) \cdot\nu|\nabla u|^{2} \bigr) \,ds\,dt. \end{aligned} \end{aligned}$$

(2.3)

It is easy to see that

$$ \biggl\vert \biggl\langle u',(x-x_{0})\cdot\nabla u+ \frac{n-1}{2}u\biggr\rangle \biggr\vert _{0}^{T} \leq2RE(0)+c(T) \int_{\Sigma_{1}} \bigl(u^{2}+u^{\prime2} \bigr) \,ds \,dt. $$

Combining with the geometric condition $(x-x_{0})\cdot\nu\leq0$ on $\Gamma _{0}$, we deduce from (2.3) and (2.1) that

$$\begin{aligned} (T-2R)E_{0} \leq& c_{1} \int_{\Sigma_{1}}\bigl\vert u'\bigr\vert ^{2} \,ds\,dt+ \int_{\Sigma_{1}}\frac{\partial u}{\partial\nu}(x-x_{0})\cdot\nabla u\, ds\, dt \\ &{} +c(T) \int_{\Sigma_{1}}u^{2} \,ds\,dt +\frac{n-1}{2} \int_{\Sigma_{1}}u\frac{\partial u}{\partial\nu} \,ds\,dt+\frac{1}{2} \int_{\Sigma_{1}}|\nabla_{T}u|^{2} \,ds\,dt \\ \leq& c \int_{\Sigma_{1}} \bigl(\bigl\vert u'\bigr\vert ^{2}+|\Delta_{T} u|^{2}+u^{2}+|\nabla _{T}u|^{2} \bigr) \,ds\,dt. \end{aligned}$$

So, the observability inequality (2.2) holds. □

The observability inequality (2.2) enables us to define the following norm:

$$\bigl\Vert (u_{0},u_{1})\bigr\Vert _{F}^{2}:= \int_{\Sigma_{1}} \bigl(\bigl\vert u'\bigr\vert ^{2}+|\Delta_{T} u|^{2}+u^{2}+| \nabla_{T}u|^{2} \bigr) \,ds\,dt, $$

and the corresponding inner product

$$\bigl\langle (u_{0},u_{1}),(v_{0},v_{1}) \bigr\rangle _{F}:= \int_{\Sigma_{1}} \bigl(u'v'+ \Delta_{T} u\Delta_{T}v+uv+\nabla_{T}u \nabla_{T}v \bigr) \,ds\,dt, $$

where u (or v) is the solution of (2.1) with initial data $(u_{0},u_{1})$ (or $(v_{0},v_{1})$). Let

$$ F:=\overline{\bigl\{ (u_{0},u_{1}) \in C^{\infty}(\bar {\Omega})\times C^{\infty}(\bar{\Omega}): \partial_{\nu}u_{0}-\Delta_{T} u_{0}=0 \bigr\} }^{\|\cdot\|_{F}}. $$

(2.4)

Then $(F,\langle\cdot,\cdot\rangle_{F})$ is a Hilbert space.

Now we consider the wellposedness for the linear backward problem

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \phi''-\Delta\phi=0, & \mbox{in } Q, \\ \frac{\partial\phi}{\partial\nu}-\Delta_{T} \phi=v, & \mbox{on } \Gamma_{1}\times(0,T), \\ \phi=0, & \mbox{on } \Gamma_{0}\times(0,T), \end{array}\displaystyle \right . $$

(2.5)

with terminal data

$$\begin{aligned} \phi(T)=\phi_{0},\qquad \phi'(T)=\phi_{1}, \quad \mbox{in } \Omega, \end{aligned}$$

(2.6)

where

$$v(x,t)=-\partial_{t} u'+\Delta_{T}( \Delta_{T} u)-\Delta_{T} u+u $$

and $\partial_{t}$ is taken in the following sense:

$$\bigl\langle -\partial_{t} u',\psi\bigr\rangle =\bigl\langle u',\psi'\bigr\rangle ,\quad \forall \psi\in H^{1}\bigl(0,T;L^{2}(\Omega)\bigr). $$

For every

$$\bigl(\theta,\theta'\bigr)\in C \bigl((0,T+\varepsilon);D\bigl( \mathcal{A}^{2}\bigr) \bigr)\cap C^{1} \bigl((0,T+ \varepsilon);D(\mathcal{A}) \bigr)\cap C^{2} \bigl((0,T+\varepsilon); \mathcal{H} \bigr) $$

with $\theta(0)=\theta'(0)=0$, we say $\phi\in L^{\infty}(0,T;V' )$ is the solution of (2.5)-(2.6) if it satisfies the following equality:

$$\begin{aligned}& \int_{Q}\phi f \, dQ+\bigl\langle \phi'(T), \theta(T)\bigr\rangle _{F',F}-\bigl\langle \phi (T),\theta'(T) \bigr\rangle _{F',F} \\& \quad =- \int_{\Sigma_{1}} \bigl(\nabla_{T}u\nabla _{T} \theta+\Delta_{T} u\Delta_{T} \theta+u' \theta' +u\theta \bigr) \,ds\,dt, \end{aligned}$$

(2.7)

where

$$f=\theta''-\Delta\theta\in L^{1}(0,T;V). $$

It is clear that θ satisfies

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \theta''-\Delta\theta=f, & \mbox{in } Q, \\ \frac{\partial\theta}{\partial\nu}-\Delta_{T} \theta=0, & \mbox{on } \Gamma_{1}, \\ \theta=0, & \mbox{on } {\Gamma_{0}}, \\ \theta(0)=0,\qquad \theta'(0)=0, & \mbox{in } \Omega. \end{array}\displaystyle \right . $$

(2.8)

Theorem 2.2

In the sense of (2.7), the problem (2.5)-(2.6) has a unique solution ϕ satisfying

$$\phi\in L^{\infty}\bigl(0,T;V'\bigr). $$

Proof

First of all, we give the energy estimate for the nonhomogeneous system (2.8).

For the general energy (the low-order energy), since

$$\frac{1}{2}\frac{d}{dt} \biggl( \int_{\Omega}\theta^{\prime 2}+|\nabla\theta|^{2} \, dx+ \int_{\Gamma_{1}}|\nabla_{T} \theta|^{2}\, ds \biggr) = \int_{\Omega}f\theta _{t} \, dx $$

and

$$E(T)=E(t)+ \int_{t}^{T} \int_{\Omega}f\theta'\,dx\,dt, $$

we have

$$E(t)\leq C_{T} \bigl( E(T)+\|f\|_{L^{2}(0,T;L^{2}(\Omega))}^{2} \bigr), \quad \forall t\in(0,T). $$

For the high-order energy, we have

$$E_{1}(t)=\frac{1}{2} \int_{\Omega}\bigl\vert \nabla\theta'\bigr\vert ^{2}+|\Delta\theta |^{2}\, dx+\frac{1}{2} \int_{\Gamma_{1}}\bigl\vert \nabla_{T} \theta'\bigr\vert ^{2}\, ds $$

and

$$ E_{1}(T)=E_{1}(t)+ \int_{t}^{T} \int_{\Omega}f\Delta\theta'\,dx\,dt. $$

Hence,

$$\begin{aligned} E_{1}(t) =& E_{1}(T)+ \int_{t}^{T} \int_{\partial\Omega}f\frac{\partial\theta '}{\partial\nu}\, ds \, dt- \int_{t}^{T} \int_{\Omega}\nabla f\nabla\theta' \,dx\,dt \\ =& E_{1}(T)+ \int_{t}^{T} \int_{\Gamma_{1}}f\Delta_{T}\theta'\, ds\, dt- \int _{t}^{T} \int_{\Omega}\nabla f\nabla\theta' \, dx \, dt \\ \leq& E_{1}(T)+ \int_{t}^{T} \biggl( \int_{\Gamma_{1}}\Vert \nabla_{T} f\Vert ^{2} \, ds \biggr) ^{\frac{1}{2}} \biggl( \int_{\Gamma_{1}}\bigl\vert \nabla_{T}\theta' \bigr\vert ^{2}\, ds \biggr)^{\frac {1}{2}}\, dt \\ &{}+ \int_{t}^{T} \biggl( \int_{\Omega}|\nabla f|^{2}\, dx \biggr)^{\frac {1}{2}} \biggl( \int_{\Omega}\bigl\vert \nabla\theta'\bigr\vert ^{2}\, dx \biggr)^{\frac{1}{2}}\, dt \\ \leq& E_{1}(T)+\bigl\Vert E_{1}(t)\bigr\Vert _{L^{\infty}(0,T)}^{\frac{1}{2}}\Vert f\Vert _{L^{1}(0,T;V)}, \end{aligned}$$

which implies that

$$E_{1}(t)\leq C\bigl(E_{1}(T)+\|f\|_{L^{1}(0,T;V)}^{2} \bigr),\quad 0\leq t\leq T. $$

Let $\theta=\theta_{1}+\theta_{2}$, where $\theta_{1}$ satisfies

$$\left \{ \textstyle\begin{array}{l@{\quad}l} \theta_{1}''-\Delta\theta_{1}=0,&\mbox{in } Q, \\ \frac{\partial\theta_{1}}{\partial\nu}-\Delta_{T}\theta_{1}=0, &\mbox{on } \Sigma_{1}, \\ \theta_{1}=0,&\mbox{on } \Sigma_{0}, \\ \theta_{1}(T)=\theta(T),\qquad \theta_{1}'(T)=\theta'(T), &\mbox{in } \Omega, \end{array}\displaystyle \right . $$

and $\theta_{2}$ satisfies

$$\left \{ \textstyle\begin{array}{l@{\quad}l} \theta_{2}''-\Delta\theta_{2}=f,&\mbox{in } Q, \\ \frac{\partial\theta_{2}}{\partial\nu}-\Delta_{T}\theta_{2}=0, &\mbox{on } \Sigma_{1}, \\ \theta_{2}=0,&\mbox{on } \Sigma_{0}, \\ \theta_{2}(T)=0, \qquad \theta_{2}'(T)=0, &\mbox{in } \Omega. \end{array}\displaystyle \right . $$

Let

$$L\bigl(\theta(T), \theta'(T), f\bigr)= \int_{\Sigma_{1}} (\nabla_{T}u\nabla _{T}\theta+ \Delta_{T} u\Delta_{T} \theta+u_{t} \theta_{t} d+u\theta )\, ds\, dt. $$

Then we obtain

$$\begin{aligned}& L\bigl(\theta(T),\theta'(T),f\bigr) \\& \quad = \int_{\Sigma_{1}} (\nabla_{T}u\nabla _{T}\theta+ \Delta_{T} u\Delta_{T} \theta+u_{t} \theta_{t} +u\theta ) \,ds\,dt \\& \quad \leq \int_{\Sigma_{1}} \bigl(\nabla_{T} u\nabla_{T} \theta_{1}+\Delta_{T}u\Delta _{T} \theta_{1}+u_{t}\theta_{1t}+ u\theta_{1} \\& \qquad {}+ \nabla_{T} u\nabla_{T}\theta_{2}+\Delta _{T} u\Delta_{T}\theta_{2}+u' \theta_{2}'+u\theta_{2} \bigr) \,ds\,dt \\& \quad \leq C\bigl(\bigl\Vert \bigl\{ \theta(T),\theta(T)\bigr\} \bigr\Vert _{F}^{2}+\|f\|_{L^{1}(0,T;V)}^{2} \bigr)^{\frac{1}{2}}. \end{aligned}$$

Therefore, $L: F\times L^{1}(0,T;V)\rightarrow L^{\infty}(0,T;V')\times F'$ is a bounded operator. So $\exists\phi\in L^{\infty}(0,T;V')$, $(\rho _{1},-\rho_{0})\in F'$ such that

$$\begin{aligned}& \int_{Q}\phi f \,dx\,dt-\bigl\langle \rho_{1}, \theta(T)\bigr\rangle +\bigl\langle \rho_{0},\theta '(T) \bigr\rangle \\& \quad = \int_{\Sigma_{1}}\nabla_{T}u\nabla_{T}\theta+ \Delta_{T} u\Delta_{T} \theta+u' \theta'+u\theta \,ds\,dt, \end{aligned}$$

where $\int_{Q}\phi f\, dx\, dt$ means $\langle\cdot,\cdot\rangle_{L^{\infty}(0,T;V'),L^{1}(0,T;H^{1}(\Omega))}$. Next, we prove that

$$\phi(T)=\rho_{0},\qquad \phi'(T)=\rho_{1}. $$

Let λ be the eigenvalue for the Δ operator with mixed Wentzell, Dirichlet boundary conditions and m be the corresponding eigenvector. The existence of eigenvalue for the Δ operator with mixed Wentzell, Dirichlet boundary condition is based on the fact that $\Delta^{-1}:L^{2}(\Omega)\rightarrow V$ is a compact operator. That is,

$$\left \{ \textstyle\begin{array}{l@{\quad}l} -\Delta m=\lambda m, &\mbox{in } \Omega, \\ \frac{\partial m}{\partial\nu}-\Delta_{T} m=0,&\mbox{on } \Gamma_{1}, \\ m=0,&\mbox{on } \Gamma_{0}. \end{array}\displaystyle \right . $$

Set $f:=g(t)m$, where g is a smooth function in $[0,T+\varepsilon]$, and let $\theta:=h(t)m$. Then

$$ \left \{ \textstyle\begin{array}{l} h''+\lambda h=g, \\ h(0)=0,\qquad h'(0)=0. \end{array}\displaystyle \right . $$

(2.9)

Claim

$\exists g=g_{0}$ such that

$$h(T)=h'(T)=0,\qquad h''(T)\neq0. $$

If this is true, then

$$\begin{aligned}& \int_{Q}\phi g_{0}(t)m \,dx\,dt-\bigl\langle \rho_{1},h(T)m \bigr\rangle +\bigl\langle \rho_{0}, h'(T)m\bigr\rangle \\& \quad = \int_{\Sigma_{1}} \bigl(\Delta_{T}u\Delta_{T}m -u''m+\nabla _{T}u\nabla_{T}m +mu \bigr)h(t) \,ds\,dt. \end{aligned}$$

Since $h''+\lambda h=g_{0}$, we have

$$\begin{aligned}& \int_{0}^{T}\bigl\langle \phi''+ \lambda\phi, m\bigr\rangle h(t)\, dt+\bigl\langle \phi (T),m\bigr\rangle h'(T)-\bigl\langle \phi'(T),m\bigr\rangle h(T) +\langle \rho_{1},m \rangle h(T) -\langle\rho_{0},m\rangle h'(T) \\& \quad = \int_{\Sigma_{1}}\Delta_{T}u\Delta_{T}m h(t)-u''mh(t)+\nabla_{T}u \nabla_{T}m h(t)+umh(t) \,ds\,dt. \end{aligned}$$

(2.10)

Differentiate (2.10) with respect to T, we get

$$\begin{aligned}& \bigl\langle \phi''+\lambda\phi, m\bigr\rangle h(T)+ \bigl\langle \phi(T),m\bigr\rangle h''(T)+\bigl\langle \phi'(T),m\bigr\rangle h'(T)-\bigl\langle \phi''(T),m\bigr\rangle h(T) \\& \qquad {} -\bigl\langle \phi'(T),m\bigr\rangle h'(T) + \langle\rho_{1},m \rangle h'(T) -\langle \rho_{0},m\rangle h''(T) \\& \quad = \int_{\Gamma_{1}} \bigl(\Delta_{T}u\Delta_{T}m-u''m+ \nabla_{T}u\nabla_{T}m+um \bigr) \, ds \, h(T). \end{aligned}$$

Therefore

$$\bigl\langle \phi(T),m\bigr\rangle h''(T)-\langle \rho_{0},m\rangle h''(T)=0, $$

which implies that $\phi(T)=\rho_{0}$. Similarly, we obtain $\phi'(T)=\rho_{1}$.

Now we prove the claim above. Write

$$A:=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{}} 0&1\\ \lambda&0 \end{array}\displaystyle \right ), \qquad B:=\left ( \textstyle\begin{array}{@{}c@{}} 0\\ 1 \end{array}\displaystyle \right ). $$

Then, by the Kalman condition [9], we know that (2.9) is controllable. Set $X(t):=(h(t),h'(t))^{T}$. Then $\exists g_{1}(s)$, $s\in(0,\frac{T}{2})$, such that $X(\frac{T}{2})=X_{0}\neq0$. Write

$$g_{2} \biggl(s-\frac{T}{2} \biggr):=B^{T}e^{A^{T}(T-s)}w^{-1} \bigl(-e^{A\frac{T}{2}}X_{0}\bigr), $$

where $w=\int_{\frac{T}{2}}^{T}e^{A(T-s)}BB^{T}e^{A^{T}(T-s)}\, ds$. Then

$$X(t)=e^{A(t-\frac{T}{2})}X_{0}+ \int_{\frac{T}{2}}^{t}e^{A(t-s)}Bg_{2} \biggl(s-\frac {T}{2}\biggr)\, ds. $$

Clearly, $X(T)=0$, $X'(T)\neq0$. This proof is then complete.  □

The following is our exact controllability theorem.

Theorem 2.3

Let $T>2R$ and F be the Hilbert space defined in (2.4). Then for every $(\phi'(0),-\phi(0))\in F'$, there are $(u_{0},u_{1})\in F$ and a control function

$$v(x,t)=-\partial_{t} u'+\Delta_{T}( \Delta_{T} u)-\Delta_{T} u+u, $$

where u is the solution to (2.1), such that the solution $\phi (t)$ of system (2.5) with initial data $(\phi(0),\phi'(0))$ satisfies

$$\phi(T)=0, \qquad \phi'(T)=0. $$

For the nonlinear case, we assume that $f\in W^{1,\infty}_{\mathrm{loc}}(\mathbb {R})$ satisfies $f(0)=0$ and the super-linear condition (see [10]):

$$ \exists C>0, p>1\mbox{:}\quad \bigl\vert f'(s)\bigr\vert \leq C|s|^{p-1},\quad \forall s\in R \mbox{ with } p< \frac{n}{n-\frac{6}{5}+\varepsilon} \mbox{ if } n\geq2. $$

(2.11)

Proposition 2.4

Assume that f satisfies the super-linear condition (2.11). Then there exists $T_{0}>0$ such that for every $T>T_{0}$, there is a neighborhood ω of $(0,0)$ in $V\times L^{2}(\Omega)$ such that for each $(\phi_{0},\phi_{1})\in\omega$, there exists a control $v_{1}\in H^{-2}(\Gamma)$ such that the solution to (1.1) satisfies

$$\phi(T)=0,\qquad \phi'(T)=0. $$

Proof

From the results for the nonlinear system of Neumann problems (see [10]), we see that there exists a controllability $v\in L^{2}(\Gamma _{1})$ such that the solution $(\phi,\phi')$ of the following system:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} \phi''-\Delta\phi+f(\phi)=0, &\mbox{in } Q, \\ \frac{\partial\phi}{\partial\nu}=v,&\mbox{on } \Sigma_{1}, \\ \phi=0, &\mbox{on } \Sigma_{0}, \\ \phi(0)=\phi_{0},\qquad \phi'(0)=\phi_{1}, &\mbox{in } \Omega, \end{array}\displaystyle \right . $$

satisfies $(\phi(T),\phi'(T))=(0,0)$, and $\phi\in H^{\beta}(\Omega)$ where $\beta\leq\frac{3}{5}-\varepsilon$. The regularity of ϕ for Neumann problems can be found in Theorem 1.4 of [11]. Let $v_{1}=v-\Delta_{T}\phi$. Then

$$\frac{\partial\phi}{\partial\nu}-\Delta_{T}\phi=v_{1}, $$

and $v_{1}\in H^{-2}(\Gamma_{1})$ such that $\phi(T)=0$, $\phi'(T)=0$. □

Remark 2.1

For dynamic Wentzell systems with boundary condition (1.2), we can also prove the results as Theorem 2.3 and Proposition 2.4 by similar arguments.