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The motion of a thin, vibrating plate with shape $\Omega$ and clamped edges is approximated by the equation

$\frac{\partial^2 u}{\partial t^2} = -\Delta^2u$

$u(x,t) = 0$ for $x\in\partial\Omega$

$Du(x,t)\cdot\hat{n} = 0$ for $x\in\partial\Omega$

Where $\hat{n}$ is the outward pointing unit normal vector on the boundary of $\Omega$. Show that if we specify initial conditions $u(x,0)=g(x),\frac{\partial u}{\partial t}(x,0) = h(x)$, this problem has at most one solution.

My attempt: I'm using energy methods to try and solve this.

If we let $u_1,u_2$ be solutions to the equations, we let $w$ = $u_1-u_2$, and let

$E(t) = \frac{1}{2}\int_{\Omega}(\frac{\partial w}{\partial t})^2 + |\Delta w|^2dx$

be the energy functional (maybe it should be something different!)

I want the derivative to simplify to

$\dot{E}(t) = \int_{\Omega}\phi(w)(w_{tt}+\Delta^2w)dx$

I haven't figured out what $\phi(w)$ is yet, but the argument in parentheses goes to 0, showing $E$ is constant, and then with the IC will show $u_1=u_2$

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    $\begingroup$ Maybe you can factorize the operator: $\partial_t^2 + \Delta^2 = (\partial_t + i \Delta)(\partial_t - i\Delta).$ Then uniqueness amounts to proving uniqueness for the heat equation. $\endgroup$ – Kore-N Oct 10 '17 at 14:58
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$$\begin{equation} \left\{ \begin{array}{c} u_{t}+\Delta ^{2}u=0 \\ u=\partial _{n}u=0% \end{array}% \right. \label{1} \end{equation}$$

we mulltiply the equation by $u_{t}$ and integrating over $% \Omega $ we get $$\frac{d}{dt}\left( \int_{\Omega }u_{t}^{2}+\left\vert \Delta u\right\vert ^{2}\right) =\frac{d}{dt}E(t)=0$$

so $E(t)=E(0)$

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