What are the limits of integrals in the energy method for wave equations?

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SUMMARY

The discussion focuses on the application of the energy method to wave equations, specifically analyzing the limits of integrals in the context of energy functionals. The example provided illustrates the energy conservation in a bounded domain, leading to the conclusion that for unbounded domains, such as when \(x \in \mathbb{R}\), the limits of the integral for the energy functional are indeed \(-\infty\) and \(\infty\). The energy functional is defined as \(E(t) = \frac{1}{2}\int_{-\infty}^\infty (v_t^2(x,t) + v_x^2(x,t))\, dx\).

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mathmari
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Hey! :o

In my notes there is the following example about the energy method.

$$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, 0<x<1, t>0 \\ u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$

$$\int_0^1(u_tu_{tt}-u_tu_{xxtt}-u_tu_{xx})dx=0 \tag 1$$

$$\int_0^1 u_tu_{tt}dx=\int_0^1\frac{1}{2}(u_t^2)_tdx=\frac{d}{dt}\int_0^1 \frac{1}{2}u_t^2dx$$

$$\int_0^1 u_t u_{xxtt}dx=-\int_0^1 u_{tx}u_{xtt}dx+[u_t u_{xtt}]_0^1=-\int_0^1\frac{1}{2}(u_{tx}^2)_tdx$$

$$\int_0^1 u_t u_{xx}dx=-\int_0^1 u_{tx}u_x dx+[u_t u_x]_0^1=-\frac{1}{2} \frac{d}{dt} \int_0^1 u_x^2dx$$

$$(1) \Rightarrow \frac{d}{dt}\int_0^1 \frac{1}{2}u_t^2dx+\frac{d}{dt}\frac{1}{2}\int_0^1 u_{tx}^2dx+\frac{d}{dt} \frac{1}{2} \int_0^1 u_x^2dx=0$$

The energy of the system is $$E(t)=\frac{1}{2}\int_0^1 (u_t^2(x, t)+u_{tx}^2(x, t)+u_{x}^2(x, t))dx$$
When we have the problem $$v_{tt}(x, t)-v_{xt}(x, t)=0, x \in \mathbb{R}, t>0 \\ v(x, 0)=0, x \in \mathbb{R} \\ v_t(x, 0)=0, x \in \mathbb{R}$$

which are the limits of the integral?? (Wondering) In this case $x \in \mathbb{R}$, do we have to take the integral on $\mathbb{R}$ ?? (Wondering) Or do we use the characteristic curves?? (Wondering)
 
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You've probably solved this by now but for the sake of completion, the limits of the energy functional are $-\infty$ and $\infty$. In fact, the energy is

$$E(t) = \frac{1}{2}\int_{-\infty}^\infty (v_t^2(x,t) + v_x^2(x,t))\, dx$$
 

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