1
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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

1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Get it! Thanks for your help. $\endgroup$
    – Tao
    Mar 5, 2020 at 16:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.