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?