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


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.


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.

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.


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

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Li, C., Xiao, T. A note on the IBVP for wave equations with dynamic boundary conditions. Bound Value Probl 2016, 34 (2016).

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  • Wentzell boundary condition
  • wave equation
  • wellposedness
  • controllability