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Homework Help: [PDE] 2D Vibrating Plate (Unique Solution)

  1. Oct 8, 2013 #1
    We have a region [itex]Ω[/itex] in [itex]ℝ^2[/itex] with a smooth boundary. There is a plate of shape [itex]Ω[/itex] and clamped edges which is approximated by the following equation:
    $$u(x,t)=0\hspace{4ex} x\in ∂Ω$$
    $$Du(x,t)\cdot\hat{n}=0\hspace{4ex} x\in ∂Ω$$
    [itex]\hat{n}[/itex] is the outward pointing unit vector on the boundary of [itex]Ω[/itex]. Moreover, we specify the following initial conditions:
    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 [itex]u[/itex] and [itex]\tilde{u}[/itex] solve the problem. We can consturct a solution [itex]w=u-\tilde{u}[/itex] that solves the PDE with homogeneous initial data. If [itex]w\equiv 0[/itex] on [itex]Ω[/itex], 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:
    $$\frac{d}{dt}E(t)=\displaystyle\int_Ωw_tw_tt+Dw\cdot Dw_tdx$$
    I have difficulty following the next step:
    From there they go on to say that [itex]\frac{d}{dt}E(t)=0[/itex] and a chain of intuitive observations leads to the desired [itex]w\equiv 0[/itex]. I am confused by two things:

    (1)Why does [itex]Dw\cdot Dw_t=-w_tΔw[/itex]? 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
  2. jcsd
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