Let $n=3$ and $u$ be the solution to Klein-Gordon equation 
\begin{equation}
\begin{cases}\ddot{u}-\Delta u +u=u^3 \\
 u(0)=u\_0, \partial\_t u(0)=u\_1, 
\end{cases}
\end{equation}
where $(u\_0,u\_1) \in H^1 \times L^2$. If we assume that $u$ exists globally and scatters to a solution $v$ of a free Klein-Gordon equation (the nonlinear term=0) as $t \to \infty$ with initial data $(v\_0,v\_1)$, then the energy $E(u,\dot{u})$ would equal to the energy of free Klein-Gordon $E(v,\dot{v})$?


where the energies are defined as below:
\begin{equation}
E(u,\dot{u})(t)=\int\_{\mathbb{R}^3} \frac{1}{2} \left( |u|^2+|\nabla u|^2 +|\dot{u}|^2 \right) -\frac{1}{4}|u|^4 dx, \\
E(v,\dot{v})(t)= \int\_{\mathbb{R}^3} \frac{1}{2} \left( |v|^2+|\nabla v|^2 +|\dot{v}|^2 \right) dx.
\end{equation}
And $u$ scatters to $v$ means that $\lVert u(t)-v(t) \rVert\_{H^1}+\lVert \dot u(t)-\dot v(t) \rVert\_{L^2} \to 0$ as $ t\to \infty$.


With some classical results of conservation of energy, we can see that $E(u,\dot{u})(t)$ doesn't depend on time, and the same as $v$. But my question is: does $E(u,\dot{ u})$ equal to $E(v,\dot{v})$? And how can I check it precisely?



the energy of scattering solution of cubic Klein-Gordon equation in $n=3$?
Step 1: assuming scattering, there exists a solution $v$ to the linear Klein-Gordon equation such that $u-v \to 0$ in $H^1(\mathbb{R}^3)$ as $t \to \infty$. By Sobolev embedding this means that for any $p\in [2,6]$ you also have $u-v \to 0$ in $L^p(\mathbb{R}^3)$. 


In particular, this means that 
$$ E(u,\dot{u})(t) - E(v,\dot{v})(t) + \frac14 \int\_{\mathbb{R}^3} |v(t)|^4 ~\mathrm{d}x \to 0 $$
as $t \to \infty$. 
Since the first two quantity is constant. It suffices to show that 
$$ \liminf \|v\|\_{L^4\_x} \to 0 $$


Step 2: Since $v$ solves the linear Klein-Gordon equation, by the [granddaddy of Strichartz estimates](https://projecteuclid.org/euclid.dmj/1077312392), you see that in three dimensions (see final page of the linked article), for any $q\in [10/3,\infty)$ the space-time integral estimate


$$ \|v\|\_{L^q\_{t,x}} \lesssim E(v, \dot{v}). $$


In particular you can apply this to $q = 4 > 10/3$ and have that $\|v(t)\|\_{L^4\_x}^4$ is integrable in time, and hence has vanishing liminf. 


