Stabilization of the Critical Semilinear Klein–Gordon Equation in Compact Space

Abstract

We study the internal stabilization and the boundary stabilization of the critical semilinear Klein–Gordon equation in compact space. The observability inequalities are proved by the Morawetz estimates on Riemannian manifold, and then the compactness–uniqueness arguments are used to prove the main stabilization results.

A Appendix for Unique Continuation

Proposition A.1

Assume that

$$\begin{aligned}&DH(X,X)\ge \delta |X|_g^2,\quad X\in {\mathbb {R}}^3_x, x\in {\overline{\Omega }},\end{aligned}$$

(6.1)

$$\begin{aligned}&{\,\mathrm div\,}_g H =\delta _1\end{aligned}$$

(6.2)

and

$$\begin{aligned} \langle H,\nu \rangle \ge \theta ,\quad x\in \Gamma ,\end{aligned}$$

(6.3)

where $\delta ,\delta _1,\theta$ are positive constants. Then there exists $T_0\ge 0$ such that for any $T>T_0$, the only solution

$$\begin{aligned} (u,u_t)\in C\left( [0,T],H^1(\Omega )\times L^2(\Omega ) \right) \end{aligned}$$

(6.4)

to the system

$$\begin{aligned} {\left\{ \begin{array}{ll}u_{tt}-\Delta _g u+u+u^{5}=0\qquad (x,t)\in \Omega \times (0,T),\\ \displaystyle \frac{\partial u}{\partial \nu }=u_t=0\qquad (x,t)\in \Gamma \times (0,T),\end{array}\right. } \end{aligned}$$

(6.5)

is the trivial one $u\equiv 0$.

Proof

It follows from (3.1) that

$$\begin{aligned} E(t)=E(0),\quad t>0. \end{aligned}$$

(6.6)

Let ${\mathcal {H}}=H$. Note that

$$\begin{aligned} {\,\mathrm div\,}_g\left( u^2 {\mathcal {H}}\right) =2u{\mathcal {H}}(u)+u^2{\,\mathrm div\,}_g {\mathcal {H}}.\end{aligned}$$

(6.7)

Then

$$\begin{aligned} \int _{\Gamma } u^2 \langle {\mathcal {H}},\nu \rangle _g \text {d}\Gamma _g = \int _{\Omega } \left( 2u{\mathcal {H}}(u)+u^2{\,\mathrm div\,}_g {\mathcal {H}}\right) \text {d}x_g. \end{aligned}$$

(6.8)

Therefore,

$$\begin{aligned} \int _{\Omega } u^2 \text {d}x_g \le C \int _{\Omega } \left| \nabla _g u\right| _g ^2 \text {d}x_g+ C \int _{\Gamma } u^2 \text {d}\Gamma _g. \end{aligned}$$

(6.9)

Note that

$$\begin{aligned}&\int _0^T\int _{\Gamma }\displaystyle \frac{\partial u}{\partial \nu }{\mathcal {H}}(u) \text {d}\Gamma _g \text {d}t+\frac{1}{2}\int _0^T\int _{\Gamma } \left( u_t^2-\left| \nabla _g u\right| _g ^2-u^2-\frac{1}{3}u^{6}\right) \langle {\mathcal {H}}, \nu \rangle _g \text {d}\Gamma _g \text {d}t\nonumber \\&\quad \le -\frac{\theta }{2}\int _0^T\int _{\Gamma } \left( \left| \nabla _g u\right| _g^2+u^2+\frac{1}{3}u^{6}\right) \text {d}\Gamma _g \text {d}t. \end{aligned}$$

(6.10)

It follows from (3.4) that

$$\begin{aligned}&0\ge \frac{\theta }{2}\int _0^T\int _{\Gamma } u^2\text {d}\Gamma _g \text {d}t+\int _{\Omega }u_tH(u)\text {d}x_g\Big |^T_0 +\int _0^T\int _{\Omega }\delta \left| \nabla _g u\right| _g^2 \text {d}x_g \text {d}t\nonumber \\&\quad +\frac{\delta _1}{3}\int _0^T\int _{\Omega } u^{6} \text {d}x_g \text {d}t+\frac{\delta _1}{2}\int _0^T\int _{\Omega }\left( u_t^2-\left| \nabla _g u\right| _g ^2-u^{2}-u^{6}\right) \text {d}x_g \text {d}t.\qquad \end{aligned}$$

(6.11)

With (6.9), for sufficiently small positive constant $\delta _c$, we obtain

$$\begin{aligned}&0\ge \int _{\Omega }u_tH(u)\text {d}x_g\Big |^T_0 +\frac{\delta _c}{2}\int _0^T\int _{\Omega }\left( u_t^2+\left| \nabla _g u\right| _g ^2+u^{2}+u^{6}\right) \text {d}x_g \text {d}t\nonumber \\&\quad +\frac{\delta _1-\delta _c}{2}\int _0^T\int _{\Omega }\left( u_t^2-\left| \nabla _g u\right| _g ^2-u^{2}-u^{6}\right) \text {d}x_g \text {d}t. \end{aligned}$$

(6.12)

Let $P=\frac{\delta _1-\delta _c}{2}$. Substituting (3.9) into (6.12), we obtain

$$\begin{aligned}&0\ge \int _{\Omega }u_t\left( {\mathcal {H}}(u)+Pu\right) \text {d}x_g\Big |^T_0\nonumber \\&\quad +\frac{\delta _c}{2}\int _0^T\int _{\Omega }\left( u_t^2+ |\nabla _g u|_g^2 +u^{2}+u^{6}\right) \text {d}x_g \text {d}t. \end{aligned}$$

(6.13)

Therefore,

$$\begin{aligned} \int _0^T E(t)\text {d}t \le CE(0), \end{aligned}$$

(6.14)

which implies

$$\begin{aligned} (T-C)E(0)\le 0.\end{aligned}$$

(6.15)

The assertion holds true. $\square$

The following lemma is given by Lemma 3.2 in [30].

Lemma A.1

Let $H(x)=x$, then

$$\begin{aligned} DH(X,X)=\left\langle \left( G(x)+\frac{ r(x)}{2}\frac{\partial G(x)}{\partial r}\right) X,X\right\rangle ,\quad \text {for}\ \ X\in {\mathbb {R}}^3_x, x \in {\mathbb {R}}^3. \end{aligned}$$

(6.16)

The assumptions (6.1), (6.2) and (6.3) can be checked in view of the metric g as follows.

Example A.1

Let $G(x)=diag\{ \alpha _1(x),\alpha _2(x),\alpha _3(x)\}$, where $\alpha _i(x)(i=1,2,3)$ is a smooth positive function defined on ${\mathbb {R}}^3$. Assume that for $i=1,2,3$

$$\begin{aligned}&\frac{1}{\alpha _i(x) }\times \frac{ r}{2 }\frac{\partial \alpha _i(x)}{\partial r} \ge \delta -1 \quad for \quad x\in \Omega , \end{aligned}$$

(6.17)

$$\begin{aligned}&\sum _{i=1}^3\frac{1}{\alpha _i(x) }\times \frac{ r}{2 }\frac{\partial \alpha _i(x)}{\partial r}=\delta _1-3 \quad for \quad x\in \Omega , \end{aligned}$$

(6.18)

A concise example to check (6.17) and (6.18) is $\alpha _i(x)=r^{c_i}$, where $c_i\ge 2(\delta -1)$ and $\sum _{i=1}^3c_i=2(\delta _1-3)$.

Let $H=x$ and $X=(X_1,X_2,X_3)^T \in {\mathbb {R}}^3$ . Then it follows from Lemma A.1 that

$$\begin{aligned} DH(X,X)= & {} \left\langle \left( G(x)+\frac{ r}{2}\frac{\partial G(x)}{\partial r}\right) X,X \right\rangle \nonumber \\= & {} \sum _{i=1}^3 \left( \alpha _i(x) + \frac{ r}{2 }\frac{\partial \alpha _i(x)}{\partial r} \right) X_i^2\nonumber \\\ge & {} \delta \sum _{i=1}^3 \alpha _i(x) X_i^2 \nonumber \\= & {} \delta |X|_g^2 \quad for \quad X\in {\mathbb {R}}^3_x,\ \ x\in \Omega , \end{aligned}$$

(6.19)

$$\begin{aligned} {\,\mathrm div\,}_g H= & {} \frac{1}{\sqrt{\hbox {det }(G(x))}} {\,\mathrm div\,}\left( \sqrt{\hbox {det }(G(x))} H\right) \nonumber \\= & {} 3+\frac{r}{2} \frac{\partial \ln (\hbox {det }(G(x)) ) }{\partial r}\nonumber \\= & {} 3+ \sum _{i=1}^3\frac{1}{\alpha _i(x) }\cdot \frac{ r}{2 }\frac{\partial \alpha _i(x)}{\partial r} . \nonumber \\= & {} \delta _1 \quad for \quad x\in \Omega . \end{aligned}$$

(6.20)