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

Boundary Value Problems20162016:34

Received: 5 November 2015

Accepted: 27 January 2016

Published: 5 February 2016


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.


Wentzell boundary conditionwave equationwellposednesscontrollability

1 Introduction

In this paper, we consider the exact boundary controllability on the IBVP for wave equation with dynamic boundary condition as follows:
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \phi''-\Delta\phi+ f(\phi)=0, &(x,t)\in Q=\Omega\times(0,T), \\ -\Delta_{T}\phi+\frac{\partial\phi}{\partial\nu}=v_{1},& \mbox{on } \Gamma_{1}\times(0,T), \\ \phi=0, & \mbox{on } \Gamma_{0}\times(0,T), \\ \phi(0,x)=\phi_{0},\qquad \phi_{t}(0,x)=\phi_{1},& x\in\Omega, \end{array}\displaystyle \right . $$
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
$$ \phi''-\Delta_{T}\phi+\frac{\partial\phi}{\partial\nu}=v_{1}, \quad \mbox{on } \Gamma_{1}\times(0,T). $$
The system models an elastic body’s transverse vibration. For details, please see the paper of Lemrabet [1]. In [17] 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,
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} u''-\Delta u=0, & (x,t)\in Q=\Omega\times(0,T), \\ -\Delta_{T}u+\frac{\partial u}{\partial\nu}=0, &\mbox{on } \Gamma _{1}\times(0,T), \\ u=0, & \mbox{on } \Gamma_{0}\times(0,T), \\ u(0,x)=u_{0},\qquad u_{t}(0,x)=u_{1},& x\in\Omega. \end{array}\displaystyle \right . $$
The wellposedness for the problem (2.1) is not hard to see. Define an operator \(\mathcal{A}: D(\mathcal{A})\rightarrow\mathcal{H}\) by
$$\mathcal{A}\left ( \textstyle\begin{array}{@{}c@{}} u\\ v \end{array}\displaystyle \right ):=\left ( \textstyle\begin{array}{@{}c@{}} v\\ \Delta u \end{array}\displaystyle \right ), $$
$$\begin{aligned}& D(\mathcal{A}):=\bigl\{ (u,v)\in\mathcal{H}:\Delta u\in L^{2}(\Omega),v \in V,\partial_{\nu} u-\Delta_{T} u=0\bigr\} , \\& D\bigl(\mathcal{A}^{2}\bigr)=\bigl\{ (u,v)^{T}\in D( \mathcal{A}): \mathcal{A}(u,v)^{T}\in \mathcal{H}\bigr\} . \end{aligned}$$
$$E(t):=\frac{1}{2} \int_{\Omega} \bigl(|\nabla u|^{2}+\bigl\vert u'\bigr\vert ^{2} \bigr)\, dx+\frac {1}{2} \int_{\Gamma_{1}}|\nabla_{T} u|^{2} \, ds. $$
Then it is clear that \(E(t)=E(0)\).

Lemma 2.1

(Observability inequality)

For \(T>2R\),
$$ E(0)\leq C \int_{\Sigma_{1}} \bigl(u^{\prime 2}+u^{2}+| \nabla_{T} u|^{2}+|\Delta_{T}u|^{2} \bigr) \, ds\, dt, $$
where \(R=\max_{x\in\bar{\Omega}}{|x-x_{0}|}\), \(\Sigma_{1}=(0,T)\times\Gamma_{1}\).


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
$$\begin{aligned} \begin{aligned}[b] &\frac{1}{2} \int_{Q} \bigl(\bigl\vert u'\bigr\vert ^{2}+|\nabla u|^{2} \bigr)\,dx\,dt+\frac{1}{2} \int _{\Sigma_{1}}|\nabla_{T} u|^{2} \,ds\,dt+ \biggl\vert \biggl\langle u',(x-x_{0})\cdot\nabla u+ \frac {n-1}{2}u\biggr\rangle \biggr\vert _{0}^{T} \\ &\quad = \frac{1}{2} \int_{\Sigma_{1}}(x-x_{0})\cdot\nu\bigl\vert u'\bigr\vert ^{2} \, ds\, dt+ \int_{\Sigma _{1}}\frac{\partial u}{\partial\nu}(x-x_{0})\cdot\nabla u \, ds\, dt \\ &\qquad {}+\frac {n-1}{2} \int_{\Sigma_{1}}u\frac{\partial u}{\partial\nu} \, ds\, dt +\frac{1}{2} \int_{0}^{T} \int_{\Gamma_{0}}(x-x_{0})\cdot\nu\biggl\vert \frac{\partial u}{\partial\nu}\biggr\vert ^{2} \,ds\,dt \\ &\qquad {}+\frac{1}{2} \int_{\Sigma_{1}} \bigl(|\nabla _{T}u|^{2}-(x-x_{0}) \cdot\nu|\nabla u|^{2} \bigr) \,ds\,dt. \end{aligned} \end{aligned}$$
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
$$ F:=\overline{\bigl\{ (u_{0},u_{1}) \in C^{\infty}(\bar {\Omega})\times C^{\infty}(\bar{\Omega}): \partial_{\nu}u_{0}-\Delta_{T} u_{0}=0 \bigr\} }^{\|\cdot\|_{F}}. $$
Then \((F,\langle\cdot,\cdot\rangle_{F})\) is a Hilbert space.
Now we consider the wellposedness for the linear backward problem
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \phi''-\Delta\phi=0, & \mbox{in } Q, \\ \frac{\partial\phi}{\partial\nu}-\Delta_{T} \phi=v, & \mbox{on } \Gamma_{1}\times(0,T), \\ \phi=0, & \mbox{on } \Gamma_{0}\times(0,T), \end{array}\displaystyle \right . $$
with terminal data
$$\begin{aligned} \phi(T)=\phi_{0},\qquad \phi'(T)=\phi_{1}, \quad \mbox{in } \Omega, \end{aligned}$$
$$v(x,t)=-\partial_{t} u'+\Delta_{T}( \Delta_{T} u)-\Delta_{T} u+u $$
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:
$$\begin{aligned}& \int_{Q}\phi f \, dQ+\bigl\langle \phi'(T), \theta(T)\bigr\rangle _{F',F}-\bigl\langle \phi (T),\theta'(T) \bigr\rangle _{F',F} \\& \quad =- \int_{\Sigma_{1}} \bigl(\nabla_{T}u\nabla _{T} \theta+\Delta_{T} u\Delta_{T} \theta+u' \theta' +u\theta \bigr) \,ds\,dt, \end{aligned}$$
$$f=\theta''-\Delta\theta\in L^{1}(0,T;V). $$
It is clear that θ satisfies
$$ \left \{ \textstyle\begin{array}{l@{\quad}l} \theta''-\Delta\theta=f, & \mbox{in } Q, \\ \frac{\partial\theta}{\partial\nu}-\Delta_{T} \theta=0, & \mbox{on } \Gamma_{1}, \\ \theta=0, & \mbox{on } {\Gamma_{0}}, \\ \theta(0)=0,\qquad \theta'(0)=0, & \mbox{in } \Omega. \end{array}\displaystyle \right . $$

Theorem 2.2

In the sense of (2.7), the problem (2.5)-(2.6) has a unique solution ϕ satisfying
$$\phi\in L^{\infty}\bigl(0,T;V'\bigr). $$


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

For the general energy (the low-order energy), since
$$\frac{1}{2}\frac{d}{dt} \biggl( \int_{\Omega}\theta^{\prime 2}+|\nabla\theta|^{2} \, dx+ \int_{\Gamma_{1}}|\nabla_{T} \theta|^{2}\, ds \biggr) = \int_{\Omega}f\theta _{t} \, dx $$
$$E(T)=E(t)+ \int_{t}^{T} \int_{\Omega}f\theta'\,dx\,dt, $$
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
$$E_{1}(t)=\frac{1}{2} \int_{\Omega}\bigl\vert \nabla\theta'\bigr\vert ^{2}+|\Delta\theta |^{2}\, dx+\frac{1}{2} \int_{\Gamma_{1}}\bigl\vert \nabla_{T} \theta'\bigr\vert ^{2}\, ds $$
$$ E_{1}(T)=E_{1}(t)+ \int_{t}^{T} \int_{\Omega}f\Delta\theta'\,dx\,dt. $$
$$\begin{aligned} E_{1}(t) =& E_{1}(T)+ \int_{t}^{T} \int_{\partial\Omega}f\frac{\partial\theta '}{\partial\nu}\, ds \, dt- \int_{t}^{T} \int_{\Omega}\nabla f\nabla\theta' \,dx\,dt \\ =& E_{1}(T)+ \int_{t}^{T} \int_{\Gamma_{1}}f\Delta_{T}\theta'\, ds\, dt- \int _{t}^{T} \int_{\Omega}\nabla f\nabla\theta' \, dx \, dt \\ \leq& E_{1}(T)+ \int_{t}^{T} \biggl( \int_{\Gamma_{1}}\Vert \nabla_{T} f\Vert ^{2} \, ds \biggr) ^{\frac{1}{2}} \biggl( \int_{\Gamma_{1}}\bigl\vert \nabla_{T}\theta' \bigr\vert ^{2}\, ds \biggr)^{\frac {1}{2}}\, dt \\ &{}+ \int_{t}^{T} \biggl( \int_{\Omega}|\nabla f|^{2}\, dx \biggr)^{\frac {1}{2}} \biggl( \int_{\Omega}\bigl\vert \nabla\theta'\bigr\vert ^{2}\, dx \biggr)^{\frac{1}{2}}\, dt \\ \leq& E_{1}(T)+\bigl\Vert E_{1}(t)\bigr\Vert _{L^{\infty}(0,T)}^{\frac{1}{2}}\Vert f\Vert _{L^{1}(0,T;V)}, \end{aligned}$$
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
$$\left \{ \textstyle\begin{array}{l@{\quad}l} \theta_{2}''-\Delta\theta_{2}=f,&\mbox{in } Q, \\ \frac{\partial\theta_{2}}{\partial\nu}-\Delta_{T}\theta_{2}=0, &\mbox{on } \Sigma_{1}, \\ \theta_{2}=0,&\mbox{on } \Sigma_{0}, \\ \theta_{2}(T)=0, \qquad \theta_{2}'(T)=0, &\mbox{in } \Omega. \end{array}\displaystyle \right . $$
$$L\bigl(\theta(T), \theta'(T), f\bigr)= \int_{\Sigma_{1}} (\nabla_{T}u\nabla _{T}\theta+ \Delta_{T} u\Delta_{T} \theta+u_{t} \theta_{t} d+u\theta )\, ds\, dt. $$
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
$$ \left \{ \textstyle\begin{array}{l} h''+\lambda h=g, \\ h(0)=0,\qquad h'(0)=0. \end{array}\displaystyle \right . $$


\(\exists g=g_{0}\) 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
$$\begin{aligned}& \int_{0}^{T}\bigl\langle \phi''+ \lambda\phi, m\bigr\rangle h(t)\, dt+\bigl\langle \phi (T),m\bigr\rangle h'(T)-\bigl\langle \phi'(T),m\bigr\rangle h(T) +\langle \rho_{1},m \rangle h(T) -\langle\rho_{0},m\rangle h'(T) \\& \quad = \int_{\Sigma_{1}}\Delta_{T}u\Delta_{T}m h(t)-u''mh(t)+\nabla_{T}u \nabla_{T}m h(t)+umh(t) \,ds\,dt. \end{aligned}$$
Differentiate (2.10) with respect to T, we get
$$\begin{aligned}& \bigl\langle \phi''+\lambda\phi, m\bigr\rangle h(T)+ \bigl\langle \phi(T),m\bigr\rangle h''(T)+\bigl\langle \phi'(T),m\bigr\rangle h'(T)-\bigl\langle \phi''(T),m\bigr\rangle h(T) \\& \qquad {} -\bigl\langle \phi'(T),m\bigr\rangle h'(T) + \langle\rho_{1},m \rangle h'(T) -\langle \rho_{0},m\rangle h''(T) \\& \quad = \int_{\Gamma_{1}} \bigl(\Delta_{T}u\Delta_{T}m-u''m+ \nabla_{T}u\nabla_{T}m+um \bigr) \, ds \, h(T). \end{aligned}$$
$$\bigl\langle \phi(T),m\bigr\rangle h''(T)-\langle \rho_{0},m\rangle h''(T)=0, $$
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]):
$$ \exists C>0, p>1\mbox{:}\quad \bigl\vert f'(s)\bigr\vert \leq C|s|^{p-1},\quad \forall s\in R \mbox{ with } p< \frac{n}{n-\frac{6}{5}+\varepsilon} \mbox{ if } n\geq2. $$

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


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.



Ti-Jun Xiao acknowledges support from NSFC (Nos. 11271082, 11371095).

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Authors’ Affiliations

Shanghai Key Laboratory for Contemporary Applied Mathematics, School of Mathematical Sciences, Fudan University, Shanghai, P.R. China


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