# Energy Method to show uniqueness of solution of PDE

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 $$E(t)=\frac{1}{2}\int_0^1 (u_t^2(x, t)+u_{tx}^2(x, t)+u_{x}^2(x, t))dx$$

I wanted to apply this at the following :

$$w_{tt}(x, t)-w_{xxt}(x, t)-w_{xx}(x, t)=0, 0<x<1, t>0 \\ w(x, 0)=0 \\ w_t(x, 0)=0 \\ w(0, t)=0 \\ w(1, t)=0$$

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

$$\int_0^1 w_tw_{tt}dx=\int_0^1 \frac{d}{dt}\left (\frac{1}{2}w_t^2\right )dx=\frac{d}{dt}\int_0^1 \frac{1}{2}w_t^2dx$$

$$\int_0^1 w_t w_{xxt}dx=-\int_0^1 w_{tx}\frac{d}{dx}[w_{xt}]dx=[w_t w_{xt}]_0^1-\int_0^1 w_{xt}^2dx=-\int_0^1 w_{xt}^2dx$$

$$\int_0^1 w_t w_{xx}dx=\int_0^1 w_t \frac{d}{dx}[w_x]dx=[w_t w_x]_0^1-\int_0^1 w_{tx}w_xdx=-\int_0^1 w_{tx} w_x dx=-\int_0^1 \frac{d}{dt} \left (\frac{1}{2} w_x^2\right )dx=-\frac{1}{2} \frac{d}{dt} \int_0^1 w_x^2 dx$$

How could we find of the method in this case??

EDIT1 :

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?? In this case $x \in \mathbb{R}$, do we have to take the integral on $\mathbb{R}$ ??

EDIT2 :

$$v_{tt}-v_{xt}=f$$ The characteristic curves are $$x=x_0 \text{ AND } x+t=x_0+t_0$$

$$\int_{x_0}^{x_0+t_0-t}(v_tv_{tt}-v_tv_{xt})dx=0$$

• $$\int_{x_0}^{x_0+t_0-t}v_tv_{tt}dx=\int_{x_0}^{x_0+t_0-t}\frac{\partial}{\partial{t}}\left (\frac{1}{2}v_t^2\right )dx=\frac{d}{dt}\int_{x_0}^{x_0+t_0-t}\frac{1}{2}v_t^2dx$$
• $$\int_{x_0}^{x_0+t_0-t}v_tv_{xt}dx=\int_{x_0}^{x_0+t_0-t}v_t\left [\frac{\partial}{\partial{x}}v_t\right ]dx=[v_t^2]_{x=x_0}^{x_0+t_0-t}-\int_{x_0}^{x_0+t_0-t}v_{xt}v_tdx \Rightarrow \int_{x_0}^{x_0+t_0-t}v_tv_{xt}dx=\frac{1}{2}\left (v_t^2(x_0+t_0-t, t)-v_t^2(x_0, t)\right )$$

Is this correct so far?? How could we find the energy??

In this case you have an energy inequality as opposed to an identity.

Plug in the three pieces into (1) and you will get

$$\frac{d}{dt}\int_0^1 \frac12(u_t^2+u_x^2)dx+\int_0^1u_{xt}^2dx=0$$

so

$$\frac{d}{dt}\int_0^1 \frac12(u_t^2+u_x^2)dx\le0.$$

Let $E(t)$ be the obvious energy, then you get $E(0)=0$ and $E'(0)\le0$, so $E(t)\le0$. By positivity, however, $E(t)\ge0$, so $E(t)=0$, and you are done.

P.S. Only in rare cases can you get an energy identity. Upon only the slightest perturbation of the problem, this becomes an inequality, which you can use to bound $E(t)$ in terms of $E(0)$.

• So, we have $$E(t)=\int_0^1 \frac{1}{2}(u_t^2+u_x^2)dx$$ How do we get the initial value $E(0)=0$?? It is $$E(0)=\int_0^1 \frac{1}{2}(u_t^2(x, 0)+u_x^2(x, 0))dx \overset{ u_t(x, 0)=0 }{ = } \int_0^1 \frac{1}{2} u_x^2(x, 0)dx$$ How do we get that this is equal to $0$ ?? Is it because $$u(x, 0)=0 \Rightarrow u_x(x, 0)=0$$ ?? – Mary Star Mar 21 '15 at 11:31
• I have edited my initial post... Could you take a look at it?? @FanZheng – Mary Star Mar 21 '15 at 11:53
• Yes, but did you ask a different problem in your edit? – Fan Zheng Mar 22 '15 at 0:42
• Yes, it is an other problem... @FanZheng – Mary Star Mar 22 '15 at 0:55
• @MaryStar Hint: Let $w=v_t$. Then it's a transport equation. – Fan Zheng Mar 22 '15 at 0:57