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This is the part of the proof for the finite speed of propagation of wave equation that i am learning. I am not understanding how the boundary term come in the first step ie while finding $\dot e(t)$ . Next doubt is after integrating by parts the first term becomes zero , why is it so ? is it because $u_{tt}-\Delta u =0$ or is it because of the conditions given ie $u_t=0$ Thanks in advance.

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2 Answers 2

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The boundary term appears because the domain depends on $t$. To see how, consider the simpler case $g(r)=\int_{B(x_0,r)}f(x)\,dx$, and write $g'(r)$ as $\lim_{h\to0}\frac1h\Bigl(\int_{B(x_0,r+h)}f(x)\,dx-\int_{B(x_0,r)}f(x)\,dx\Bigr).$ If $h>0$, the difference is an integral over a shell $\{x\colon r<\lvert x\rvert, and the limit becomes $\int_{\partial B(x_0,r)}f(x)\,dS.$ The case at hand is similar, just think of $e(t)$ as $E(t_0-t,t)$ with $E(r,t)=\int_{B(x_0,r)}(\ldots)\,dx$ and apply the multivariable chain rule.

And yes, the vanishing of the first term is indeed because $u$ satisfies the wave equation. Just pull out the common factor $2u_t$ from the the integral to see this.

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    @Herald Hanche-Olsen : Sir , yes its more convincing . I am going through it slowly2012-07-25
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For the first doubt, the basic idea is that it is an application of the "chain rule" of differentiation. To illustrate: Let $F(t,x)$ be a function, and write by $F'(t,x) = \partial_x F$. Let $x_0(t)$ and $x_1(t)$ be functions. Define

$ E(t) = \int_{x_0(t)}^{x_1(t)} F'(t,s)\mathrm{d}s = F(t,x_1(t)) - F(t,x_0(t))$

by the fundamental theorem of calculus, then you have

$ \frac{d}{dt} E(t) = \left.\partial_tF(t,\cdot)\right]_{x_0(t)}^{x_1(t)} + F'(t, x_1(t))\frac{d}{dt}x_1(t) - F'(t,x_0(t)) \frac{d}{dt}x_0(t) $

For sufficiently nice functions, we can commute the partial derivative with the integral sign and write

$ \partial_t F(t,\cdot) ]_{x_0}^{x_1} = \int_{x_0(t)}^{x_1(t)} \partial_t F'(t,s) \mathrm{d}s $

using the fundamental theorem of calculus again. Hence we have that

The time derivative of an integral whose boundary is time dependent is equal to the integral of the time derivative of the integrand plus boundary terms.

In your specific case I suggest that you compute by changing variables. First note that without loss of generality we can set $x_0, t_0$ in your expression to be both $0$, and take $t < 0$. Then

$ e(\tau) = \int_{B_0(|\tau|)} u_t^2 + |Du|^2 \mathrm{d}x = |\tau|^n \int_{B_0(1)} u_t^2(\tau, |\tau| y) + |D u(\tau,|\tau| y)|^2 \mathrm{d}y $

Now you can take derivative in $\tau$: since the domain of integration is now fixed, there is no contribution from the moving boundary, and you can take the derivative inside the integral (after using the product rule). Then it is a bit of manipulations with chain rules and integration by parts that leads you to the final expression. (Hints:

$ \frac{d}{d\tau} u_t^2(\tau,|\tau|y) = 2 u_t u_{tt} + 2 u_t (y\cdot Du_t)(\tau, |\tau| y) $

and

$ (Du_t)(\tau, |\tau| y) = \frac{1}{|\tau|} D_y( u_t(\tau,|\tau|y)) $

can be useful; the term when the derivative hits $|\tau|^n$ is used to compensate when you integrate by parts and $D$ hits on the $y$ outside.)

For the second doubt, yes, it is just applying the wave equation.

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    Yes, $C^1$ is sufficient. For the purpose of integrating by parts / Stokes theorem, you can get by with piecewise $C^1$ since one is allowed to have sets of measure 0 on which the normal is not well defined.2012-07-26