Exploring Limits of Integrals with Energy Method

[mathmari]

Hey!

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)

 

[Euge]

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

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