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?

  • $\begingroup$ In principle, the equation can be solved exactly and you can answer this question. See arxiv.org/abs/1504.02299. $\endgroup$
    – Jon
    Mar 5, 2020 at 13:10
  • $\begingroup$ 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). $\endgroup$ Mar 5, 2020 at 14:46
  • $\begingroup$ @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 $\lVert u(t) \rVert_{L_x^4} \to 0$? $\endgroup$
    – Tao
    Mar 5, 2020 at 15:34
  • $\begingroup$ You are using the wrong estimate. I'll post an answer. $\endgroup$ Mar 5, 2020 at 15:51

1 Answer 1


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.

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

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