Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

1 Answer

From comments of Yimin, more-or-less:

2) Integrate by parts. See Appendix C, theorem 3, part (ii) in Evans. Note that the boundary term vanishes due to the assumptions.

1) Energy, in PDE, often means (the integral of) a quantity you can minimise in order to solve the equation in question, by which I mean that the function that minimises the energy also solves the PDE. This often corresponds to energy as it is used in physics.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.