# the energy of scattering solution of cubic Klein-Gordon equation in $n=3$?

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?

• In principle, the equation can be solved exactly and you can answer this question. See arxiv.org/abs/1504.02299.
– Jon
Mar 5, 2020 at 13:10
• By Sobolev inequality if $u - v \to 0$ in $H^1$ we also have $u - v \in L^4$. This means that we expect $$E(v,\dot{v})(t) - \frac14 \int_{\mathbb{R}^3} |v|^4 ~dx - E(u,\dot{u})(t) \to 0$$ So it remains to show that the $L^4$ norm of a free solution decays to 0 as $t\to \infty$. This should follow by Strichartz (I don't remember the admissible exponents for KG off the top of my head, but $L^4$ in space should be admissible). Mar 5, 2020 at 14:46
• @WillieWong thank you very much! But with Strichartz estimates, which is $\lVert u \rVert_{L_t^3 L_x^6(I \times \mathbb{R}^3)} \lesssim \lVert (u_0,u_1) \rVert_{H^1 \times L^2} + \lVert F\rVert_{L_t^1L_x^2(I \times \mathbb{R}^3)}$, where $F=u^3$ is the nonlinear term, I still cannot figure out why $\lVert u(t) \rVert_{L_x^4} \to 0$?
– Tao
Mar 5, 2020 at 15:34
• You are using the wrong estimate. I'll post an answer. Mar 5, 2020 at 15:51

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, 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.

• Get it! Thanks for your help.
– Tao
Mar 5, 2020 at 16:36