How Does the Energy Method Prove Uniqueness in Differential Equations?

[mathmari]

Hey!

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

We want to show that the problem $$w_{tt}(x, t)-w_{xxtt}(x, t)-w_{xx}(x, t)=f(x, t), 0<x<1, t>0$$
$$w(x, 0)=\phi(x) \\ w_t(x, 0)=\psi(x) \\ w_x(0, t)=h(t), t>0 \\ w_x(1, t)=g(t)$$
has an unique solution.

We suppose that $w_1, w_2$ are two distinct solutions. Then $u=w_1-w_2$ solves the problem :
$$u_{tt}(x, t)-u_{xxtt}(x, t)-u_{xx}(x, t)=0, \, 0<x<1, \, t>0$$
$$u(x, 0)=0 \\ u_t(x, 0)=0 \\ u_x(0, t)=0 \\ u_x(1, t)=0$$

$$$$

To find the energy we do the following:

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

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

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

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

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

The energy of the system is $$E(t)=\frac{1}{2}\int_0^1 (u_t^2(x, t)+u_{tx}^2(x, t)+u_{x}^2(x, t))dx$$

$$\Rightarrow E'(t)=\int_0^1 (u_t u_{tt}+u_{tx}u_{xtt}+u_xu_{xy})dx= \dots =0$$

(The energy is always positive.)

$$\Rightarrow 0 \leq E(t) = E(0)=0 \Rightarrow u_t=0 \Rightarrow u(x, t)=u(x, 0)=0 \text{ Contradiction}$$

So, the initial problem has an inuque solution.

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I wanted to apply this at the following :

$$v_{tt}(x, t)-v_{xt}(x, t)=f(x, t), x \in \mathbb{R}, t>0 \\ v(x, 0)=g(x), x \in \mathbb{R} \\ v_t(x, 0)=h(x), x \in \mathbb{R}$$

and I have done the following:

We suppose that $v_1, v_2$ are two distinct solutions. Then $u=v_1-v_2$ solve the problem:
$$u_{tt}-u_{xt}=0, x \in \mathbb{R}, t>0 \\ v(x, 0)=0, x \in \mathbb{R} \\ v_t=0, x \in \mathbb{R}$$
In this case since $x\in \mathbb{R}$ we are looking for the characteristic curves to use them as the limits of the integral.

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

$$\int_{x_0}^{x_0+t_0-t}(u_tu_{tt}-u_tu_{xt})dx=0 \tag 2$$

- $$\int_{x_0}^{x_0+t_0-t}u_tu_{tt}dx=\int_{x_0}^{x_0+t_0-t}\frac{\partial}{\partial{t}}\left (\frac{1}{2}u_t^2\right )dx=\frac{d}{dt}\int_{x_0}^{x_0+t_0-t}\frac{1}{2}u_t^2dx$$

-
\begin{align}
\int_{x_0}^{x_0+t_0-t}u_tu_{xt}dx &= \int_{x_0}^{x_0+t_0-t}u_t\left [\frac{\partial}{\partial{x}}u_t\right ]dx=[u_t^2]_{x=x_0}^{x_0+t_0-t}-\int_{x_0}^{x_0+t_0-t}u_{xt}u_tdx \\ &\Rightarrow \int_{x_0}^{x_0+t_0-t}u_tu_{xt}dx=\frac{1}{2}\left (u_t^2(x_0+t_0-t, t)-u_t^2(x_0, t)\right )
\end{align}

$$(2) \Rightarrow \frac{d}{dt}\int_{x_0}^{x_0+t_0-t}\frac{1}{2}u_t^2dx-\frac{1}{2}\left (u_t^2(x_0+t_0-t, t)-u_t^2(x_0, t)\right )=0 \\ \Rightarrow \frac{d}{dt}\int_{x_0}^{x_0+t_0-t}\frac{1}{2}u_t^2dx=\frac{1}{2}\left (u_t^2(x_0+t_0-t, t)-u_t^2(x_0, t)\right )$$

The energy of the system is $$E(t)=\int_{x_0}^{x_0+t_0-t}\frac{1}{2}u_t^2dx$$

$$E'(t)=\frac{d}{dt}\int_{x_0}^{x_0+t_0-t}\frac{1}{2}u_t^2dx \Rightarrow E'(t)=\frac{1}{2}\left (u_t^2(x_0+t_0-t, t)-u_t^2(x_0, t)\right )$$

We have also that $E(0)=0, E'(0)=0$, right?? (Wondering)

How could we continue to show that $u=0$ ?? (Wondering) We could show that if we would know that $E'(t) \leq 0$, but how could we get this inequality?? (Wondering)

 

[jvicens]

Hello,

Thank you for sharing your example and asking for further clarification. To continue showing that $u=0$, we can use the energy method to prove that $E'(t) \leq 0$. This can be done by showing that $\frac{d}{dt} \int_{x_0}^{x_0+t_0-t} \frac{1}{2}u_t^2dx \leq 0$.

To do this, we can use the Cauchy-Schwarz inequality, which states that for any two functions $f$ and $g$ on a given interval,

$$\left(\int_a^b f(x)g(x)dx\right)^2 \leq \int_a^b f(x)^2dx \int_a^b g(x)^2dx$$

Applying this to our problem, we have

$$\frac{d}{dt} \int_{x_0}^{x_0+t_0-t} \frac{1}{2}u_t^2dx = \int_{x_0}^{x_0+t_0-t} u_t u_{tt}dx \leq \sqrt{\int_{x_0}^{x_0+t_0-t} u_t^2dx \int_{x_0}^{x_0+t_0-t} u_{tt}^2dx}$$

We can then use the fact that $u_{tt} = f+u_{xt}$ and the fact that $u_t = 0$ on the characteristic curves, to get

$$\frac{d}{dt} \int_{x_0}^{x_0+t_0-t} \frac{1}{2}u_t^2dx \leq \sqrt{\int_{x_0}^{x_0+t_0-t} u_t^2dx \int_{x_0}^{x_0+t_0-t} (f+u_{xt})^2dx}$$

We can then use the fact that $u_t = 0$ and $f$ is bounded, to show that

$$\frac{d}{dt} \int_{x_0}^{x_0+t_0-t} \frac{1}{2}u_t^2dx \leq \sqrt{\int_{x_0}^{x_