[PDE] 2D Vibrating Plate (Unique Solution)

[Arkuski]

We have a region $Ω$ in $ℝ^2$ with a smooth boundary. There is a plate of shape $Ω$ and clamped edges which is approximated by the following equation:
$$\frac{∂^2u}{∂t^2}=-Δ^2u$$
$$u(x,t)=0\hspace{4ex} x\in ∂Ω$$
$$Du(x,t)\cdot\hat{n}=0\hspace{4ex} x\in ∂Ω$$
$\hat{n}$ is the outward pointing unit vector on the boundary of $Ω$. Moreover, we specify the following initial conditions:
$$u(x,0)=g(x)$$
$$u_t(x,0)=h(x)$$
Given all of this, we wish to show our problem has at most one solution.

So the way I went about this was to let $u$ and $\tilde{u}$ solve the problem. We can consturct a solution $w=u-\tilde{u}$ that solves the PDE with homogeneous initial data. If $w\equiv 0$ on $Ω$, then our solution is unique.

I am using the second edition of Lawrence Evans' Partial Differential Equations, and they use an energy method to prove uniqueness of a solution of the wave equation with given boundary/initial data. They define energy and its derivative with respect to time to be the following:
$$E(t):=\frac{1}{2}\displaystyle\int_Ωw^2_t(x,t)+|Dw(x,t)|^2dx$$
$$\frac{d}{dt}E(t)=\displaystyle\int_Ωw_tw_tt+Dw\cdot Dw_tdx$$
I have difficulty following the next step:
$$\frac{d}{dt}E(t)=\displaystyle\int_Ωw_t(w_{tt}-Δw)dx$$
From there they go on to say that $\frac{d}{dt}E(t)=0$ and a chain of intuitive observations leads to the desired $w\equiv 0$. I am confused by two things:

(1)Why does $Dw\cdot Dw_t=-w_tΔw$? Where does the negative come from?

(2)If I were to replicate this for the higher order problem I posted, would I have to find more derivatives of energy to prove uniqueness?

Thanks a bunch

 

[lippyka]

!For (1), the negative comes from the fact that Δw is a scalar and Dw_t is a vector, which means that their dot product will involve a negative sign. For (2), yes, you would need to find more derivatives of energy in order to prove uniqueness for the higher order problem.