Exponential Asymptotic Stability for the Klein Gordon Equation on Non-compact Riemannian Manifolds

Abstract

The Klein Gordon equation subject to a nonlinear and locally distributed damping, posed in a complete and non compact n dimensional Riemannian manifold $(\mathcal {M}^n,\mathbf {g})$ without boundary is considered. Let us assume that the dissipative effects are effective in $(\mathcal {M}\backslash \Omega ) \cup (\Omega \backslash V)$, where $\Omega$ is an arbitrary open bounded set with smooth boundary. In the present article we introduce a new class of non compact Riemannian manifolds, namely, manifolds which admit a smooth function f, such that the Hessian of f satisfies the pinching conditions (locally in $\Omega$), for those ones, there exist a finite number of disjoint open subsets $V_k$ free of dissipative effects such that $\bigcup _k V_k \subset V$ and for all $\varepsilon >0$, $meas(V)\ge meas(\Omega )-\varepsilon$, or, in other words, the dissipative effect inside $\Omega$ possesses measure arbitrarily small. It is important to be mentioned that if the function f satisfies the pinching conditions everywhere, then it is not necessary to consider dissipative effects inside $\Omega$.

Appendix

We start this section by considering a uniqueness result due to Bur and Gérard [10] (see p. 80, Sect. 6) or Lasiecka et. al. [26]. Let us consider the wave equation posed in a compact Riemannian manifold (M, g) with boundary. If $(\omega , T_0)$ controls M, then the following observability inequality holds for ultra weak solutions to problem

$$\begin{aligned} \left\{ \begin{aligned}&v_{tt} - \Delta _g v=0 ~\hbox { in }M \times (0,T),\\&v=0 ~\hbox { on }\partial M \times (0,T),\\&v(0)=v_0\in L^2(M); v_t(0)=v_1 \in H^{-1}(M), \end{aligned} \right. \end{aligned}$$

(6.1)

holds:

$$\begin{aligned} ||v_0||_{L^2(M)}^2 + ||v_1||_{H^{-1}(M)} \le C \int _0^T \int _\omega |v(x,t)|^2\,d{\mathcal {M}}dt, \end{aligned}$$

(6.2)

for a certain constant C and for all $T\ge T_0$ and for all $\{v_0,v_1\}\in L^2(M) \times H^{-1}(M)$.

Following we will show that if the Riemannian manifold M admits a function with positive defined Hessian in some subset $U \subset \text{ int }M$, then any geodesic on U hits $\partial U$.

We consider a Riemannian manifold M Riemannian metric $G=\left< \cdot , \cdot \right>$ and Riemannian connection $\tilde{\nabla }$. We denote its Laplace–Beltrami operator by $\Delta$. Fix a coordinate system $(x_1, \ldots , x_n)$ on M, denote $G_{ij}=\left<\partial /\partial x_i,\partial /\partial x_j\right>$, let $G^{ij}$ be the inverse matrix of $G_{ij}$. The Laplace–Beltrami operator in this coordinate system is given by

$$\begin{aligned} \Delta u=\frac{1}{\sqrt{\det G_{ij}}}\sum _{i,j=1}^n \frac{\partial }{\partial x_i}\left( \sqrt{\det G_{ij}}G^{ij} \frac{\partial u}{\partial x_j} \right) , \end{aligned}$$

where $\nabla$ is the usual gradient correspondent to the Euclidean metric on the domain $(x_1, \ldots , x_n)$.

The Hessian of a smooth function $\phi :M\rightarrow \mathbb R$ is a symmetric 2-form on M defined as

$$\begin{aligned} \nabla ^2 \phi (X,Y)=XY(\phi )-\tilde{\nabla }_XY (\phi ), \end{aligned}$$

(6.3)

where X and Y are vector fields on M. Here $X(\phi )$ denotes the directional derivative of $\phi$ in the direction of the vector field X. It is well known that the value of $\nabla ^2(X,Y)(p)$ depends only on the values of X and Y on p. It means that the right-hand side of (6.3) does not depend on the smooth extension we take for X(p) and Y(p).

A curve $\gamma :(-\varepsilon ,\varepsilon ) \rightarrow M$ is a geodesic if $\tilde{\nabla }_{\gamma ^\prime (t)}\gamma ^\prime (t)\equiv 0$.

Lemma 6.1

Let M be a complete Riemannian manifold, eventually with boundary, and let $\phi :\text {int}M\rightarrow \mathbb R$ be a smooth function. Suppose that $\phi$ is bounded and $\nabla ^2 \phi (v,v)\ge c\Vert v \Vert ^2$ on an open subset $U\subset \text {int} M$, where $c>0$ is a constant. Then any geodesic on U hits $\partial U$.

Proof

Let $\gamma$ be a geodesic on U. Then

$$\begin{aligned} \frac{d^2}{dt^2}\phi (\gamma (t))= & {} \gamma ^\prime (t) \gamma ^\prime (t) \phi = \nabla ^2 \phi (\gamma ^\prime (t),\gamma ^\prime (t))+ ( \nabla _{\gamma ^\prime (t)}\gamma ^\prime (t))(\phi )\\= & {} \tilde{\nabla }^2 \phi (\gamma ^\prime (t),\gamma ^\prime (t))\ge c \Vert \gamma ^\prime (t)\Vert ^2 \end{aligned}$$

where the last equality holds because $\gamma$ is a geodesic. Observe that the last term is a positive real constant because $\Vert \gamma ^\prime (t) \Vert$ does not depend on t.

Then $\phi (\gamma (t))$ is a smooth real valued function which second derivative is bounded below by a strictly positive constant. Then it is not difficult to prove that if $\gamma$ is defined on a interval $(a,\infty )$, then $\lim _{t\rightarrow \infty } \phi (\gamma (t))=\infty$. Analogously $\lim _{t\rightarrow -\infty } \phi (\gamma (t))=\infty$ whenever $\gamma$ is defined on a interval $(-\infty ,a)$. But neither of the cases are possible if $\gamma$ remain forever in U because $\phi$ is bounded there. Therefore $\gamma$ must hit $\partial U$. $\square$

According to the construction of the function f give in [13] and mentioned in Remark 4 we have that $\nabla ^2f(v,v)\ge C\Vert v\Vert _g^2$ in V, where $V=\bigcup _{i=1}^{k} V_i$. Thus, we ensure that the geodesics find the effective dissipation region, namely $(M\backslash \overline{\Omega }) \cup M_*,$ That is, there are no geodesic “trapped” within the free sets of dissipative effects $V_i$, for all $i=1, \ldots , k$. It is worth remembering that $\Omega \subset \subset \Omega ^*$.

Now we can show that, in fact, the function v given in (4.64) is null in all $\Omega ^*$. For this we have to show that $v=0$ in $V_i$, for all $i=1, \ldots , k,$ since $(\overline{\Omega ^*} \backslash V) \subset {\mathcal {M}}_*$ and $v=0$ in ${\mathcal {M}}_*.$

By (4.64), we have

$$\begin{aligned} \left\{ \begin{array}{ll} {v^{''}-\Delta v +v =0} &{} {in\,\,\,\Omega ^*\times (0,T)} \\ {v=0} &{} {in\,\,\,\partial \Omega ^*}\times (0,T) \\ {v(0)=v_0 \in L^2(\Omega ^*);} &{} {v'(0)=v_1 \in H^{-1}(\Omega ^*),} \end{array}\right. \end{aligned}$$

(6.4)

with $v=0$ in $(\Omega ^*\backslash \Omega )\cup {\mathcal {M}}_*$.

Now, consider the problems

$$\begin{aligned} \left\{ \begin{array}{ll} {\varphi ^{''}-\Delta \varphi =0} &{} {in\,\,\,\Omega ^*\times (0,T)} \\ {\varphi =0} &{} {in\,\,\,\partial \Omega ^*}\times (0,T) \\ {\varphi (0)=v_0 \in L^2(\Omega ^*);} &{} {\varphi '(0)=v_1 \in H^{-1}(\Omega ^*),} \end{array}\right. \end{aligned}$$

(6.5)

and,

$$\begin{aligned} \left\{ \begin{array}{ll} {z^{''}-\Delta z =-v} &{} {in\,\,\,\Omega ^*\times (0,T)} \\ {z=0} &{} {in\,\,\,\partial \Omega ^*}\times (0,T) \\ {z(0)=z'(0)=0.} \end{array}\right. \end{aligned}$$

(6.6)

Defining $w= \varphi + z$, we have that w is solution of

$$\begin{aligned} \left\{ \begin{array}{ll} {w^{''}-\Delta w =-v} &{} {in\,\,\,\Omega ^*\times (0,T)} \\ {w=0} &{} {in\,\,\,\partial \Omega ^*}\times (0,T) \\ {w(0)=v_0 \in L^2(\Omega ^*);} &{} {w'(0)=v_1 \in H^{-1}(\Omega ^*),} \end{array}\right. \end{aligned}$$

(6.7)

and, if $y=v-w$ then y is solution of

$$\begin{aligned} \left\{ \begin{array}{ll} {y^{''}-\Delta y =0} &{} {in\,\,\,\Omega ^*\times (0,T)} \\ {y=0} &{} {in\,\,\,\partial \Omega ^*}\times (0,T) \\ {y(0)=y'(0)=0.} \end{array}\right. \end{aligned}$$

(6.8)

By uniqueness of solution, we conclude that $y=0$, that is, $v=w=\varphi +z.$ Note that $z''-\Delta v= -v =0$ is $(\overline{\Omega ^*}\backslash \Omega )\cup {\mathcal {M}}_*$ with $z(0)=0=z'(0)$, where it follows that $z=0$ in $(\overline{\Omega ^*}\backslash \Omega )\cup {\mathcal {M}}_*$, and consequently, $\varphi =0$ in $(\overline{\Omega ^*}\backslash \Omega )\cup {\mathcal {M}}_*$.

As $\varphi =0$ in ${\mathcal {M}}_*\supset \supset \overline{\Omega ^*}\backslash V$ we have that exists an open neighborhood $\omega _i$ of $\partial V_i$, with $\omega _i\subset V_i$, such that $\varphi =0$ in $\omega _i.$

Restricting the problem (6.5) to $V_i$, follows by (6.2) that

$$\begin{aligned} ||v_0||_{L^2(V_i)}^2 + ||v_1||_{H^{-1}(V_i)} \le C \int _0^T \int _{\omega _i} |\varphi (x,t)|^2\,d{\mathcal {M}}dt, \end{aligned}$$

(6.9)

where it follows that $v_0=v_1=0$ em $V_i,$ for all $i=1, \ldots , k,$ and therefore, by the problem (6.8), we conclude that $v=0$ in $V_i,$ for all $i=1, \ldots , k,$ as we wanted to show. $\square$