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Exercise on Stein's/Sharkachi book chapter 6:

Let $u(x, t)$ be a smooth solution of the wave equation and let $E(t)$ denote the energy of this wave $$E(t) = \int_{\mathbb R^d} \bigg|\frac{\partial u(x,t)}{\partial t}\bigg|^2+\sum_{j=1}^d \int_{\mathbb R^d}\bigg | \frac{\partial u(x,t)}{\partial x_j}\bigg |^2 dx.$$ We have seen that $E(t)$ is constant using Plancherel’s formula. One can give an alternate proof of this fact by differentiating the integral with respect to $t$ and showing that $\frac{dE}{dt} = 0.$

I remember that when we were in the 1-dimensional case, to prove a similar formula (which didn't involved modules back then), we had to multiply $u_{tt}= u_{xx}$ by $u_t$, integrate with respect to $x$ and then realize integration by parts. Noticing that $1/2(u_t^2)_t = u_{tt}u_t $ and doing similarly for $u_{xx}u_t$, we would arrive that: $\frac{d}{dt}(1/2\int u_t^2dx+1/2\int u_x^2dx ) =0 $ which implies the conservation of energy.

So the hint given for the problem in $\mathbb R^d$ is the same: use integration by parts. But I dunno exactly what is the d-dimensional analog of integration by parts. How can I solve this problem?

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This is a typical energy method approach to use this to show the wave equation have but one solution. Rewrite your integral as

$$E(t) = \int_{\mathbb{R}^d} u_{t}^2 + |\nabla u|^2 dx$$

Then

$$E'(t) = \int_{\mathbb{R}^d} 2u_{t}u_{tt} + 2(\nabla u \cdot \nabla u_t) dx$$

Here is the step where IBP is performed. Since you're curious, in higher dimension, the formula is

$$\int_U u_{x_{i}}v dx = -\int_U uv_{x_{i}}dx + \int_{\partial U}uv\nu^{i} dS$$

for $i = 1,2,...n$ and $\nu^i$ is the $ith$ component of the outward pointing unit normal vector of $U$, an open set bounded in $\mathbb{R}^n$. Applying it here we get

$$\int_{\mathbb{R}^d}\nabla u \cdot \nabla u_t \,dx = -\int_{\mathbb{R}^d}u_t \Delta u \, dx$$

since the boundary of $\Bbb{R}^n$ is empty. Hence

$$E'(t) = 2\int_{\mathbb{R}^d} u_{t}(u_{tt} - \Delta u) dx = 0$$

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  • $\begingroup$ Why the last integral equals 0? $u$ being solution of the wave equation only implies that $u_{tt}= \Delta u$ $\endgroup$ – math.h Nov 20 '18 at 11:42
  • $\begingroup$ Ah, I made an error, I will edit. $\endgroup$ – DaveNine Nov 20 '18 at 12:55

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