# Wave Equation, Energy methods.

I am reading the book of Evans, Partial differential Equations ... wave equation section 2.4; subsection 2.4.3: Energy methods. Arriving at the theorem:

Theorem 5 (Uniqueness for wave equation). There exists at most one function $u \in C^{2}(\overline{U}_{T})$ solving

$u_{tt} -\Delta u=f$ in $U_{T}$

$u=g$ on $\Gamma_{T}$

$u_{t}=h$ on $U \times \{t=0\}.$

Proof. If $\tilde{u}$ is another such solution, then $w:=u-\tilde{u}$ solves

$w_{tt} -\Delta w=0$ in $U_{T}$

$w=0$ on $\Gamma_{T}$

$w_{t}=0$ on $U \times \{t=0\}.$

Define the "energy"

$e(t):=\frac{1}{2} \int_{U} w^{2}_{t}(x,t)+ \mid Dw(x,t)\mid ^{2} dx (0\leq t \leq T).$

We compute

$\dot{e}(t)=\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx (\cdot = \frac{d}{dt})$

$=\int_{U}w_{t}(w_{tt} - \Delta w)dx=0$.

There is no boundary term since $w=0$, and hence $w_{t}=0$, on $\partial U \times [0,T].$ Thus for all $0\leq t \leq T, e(t)= e(0)=0$, and so $w_{t}, Dw \equiv 0$ within $U_{T}$. Since $w \equiv 0$ on$U \times \{t=0\}$, we conclude $w=u-\tilde {u}\equiv 0$ in $U_{T}$.

I have two questions:

1) What is the motivation for the definition of $e(t)$

2)$\int_{U} w_{t}w_{tt}+ Dw \cdot Dw_{t}dx$

$=\int_{U}w_{t}(w_{tt} - \Delta w)dx$. How to justify this equality?

Thank very much.

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I think you might take the energy as a sum of potential and kinetic energy. –  Yimin Feb 16 '13 at 19:53
The second you may use integral by part to justify it. Or use Green's formula. –  Yimin Feb 16 '13 at 19:59
If Yimin could put those answers in the answer section, and Manoel could accept it, we could mark this question as closed. –  Ray Yang Feb 17 '13 at 15:58