Ngô Quốc Anh

March 13, 2010

Wave equations: The gradient estimates and a conservation law

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:04

Let us consider the following problem

\displaystyle u_{tt} = \Delta u, \quad x \in \mathbb R^n, t>0

with

\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x).

Assume f, g \in \mathcal S(\mathbb R^n). Note that the Schwartz space \mathcal S(\mathbb R^n) consists of all indefinitely differentiable functions f : \mathbb R^n \to \mathbb R such that

\displaystyle\mathop {\sup }\limits_{x \in {\mathbb{R}^n}} \left| {{x^\alpha }{D^\beta }f(x)} \right| < \infty

for every multi-index \alpha and \beta. We shall prove that the total energy at time t

\displaystyle E(t) = \frac{1}{2}\int_{{\mathbb{R}^n}} {\left( {u_t^2(x,t) + |{\nabla _x}u(x,t){|^2}} \right)dx}

is constant in time.

Our approach is based on the Fourier transform. Note that the Fourier transform of a Schwartz function \varphi is defined by

\displaystyle\widehat\varphi (\xi ) = \int_{{\mathbb{R}^n}} {\varphi (x){e^{ - 2\pi i\xi \cdot x}}dx} .

It is worth noticing that the Fourier transform maps \mathcal S(\mathbb R^n) to itself. Taking the Fourier transform with respect to space variable x we have

\displaystyle \widehat u_{tt}(\xi,t) = |\xi|^2 \widehat u(\xi,t)

and

\displaystyle \widehat u(\xi,0)=\widehat f(\xi), \quad \widehat u_t(\xi,0)=\widehat g(\xi).

Solving the above initial problem of the ordinary differential equation, we get

\displaystyle\widehat u(\xi ,t) = \widehat f(\xi )\cos \left( {|\xi |t} \right) + \frac{{\widehat g(\xi )}}{{|\xi |}}\sin (|\xi |t)

and therefore

\displaystyle {\widehat u_t}(\xi ,t) = - \widehat f(\xi )|\xi |\sin \left( {|\xi |t} \right) + \widehat g(\xi )\cos (|\xi |t).

As a consequence,

\displaystyle \int_{{\mathbb{R}^n}} {\left( {|{{\widehat u}_t}(\xi ,t){|^2} + |\xi {|^2}|\widehat u(\xi ,t){|^2}} \right)d\xi } = \int_{{\mathbb{R}^n}} {\left( {|\widehat g(\xi ){|^2} + |\xi {|^2}|\widehat f(\xi ){|^2}} \right)d\xi } .

We are now in a position to apply Plancherel’s Theorem to get the desired result.

Theorem (Plancherel). Suppose f \in \mathcal S(\mathbb R^n). Then

\displaystyle\varphi (x) = \int_{{\mathbb{R}^n}} {\widehat\varphi (\xi ){e^{2\pi i\xi \cdot x}}d\xi }.

Moreover

\displaystyle\int_{{\mathbb{R}^n}} {{{\left| {\widehat\varphi (\xi )} \right|}^2}d\xi } = \int_{{\mathbb{R}^n}} {{{\left| {\varphi (x)} \right|}^2}dx}.

Let us denote by E_0 this common value of the total energy. One can show that

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {u_t^2(x,t)dx} = \mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {|{\nabla _x}u(x,t){|^2}dx} = {E_0}.

The key point is to use the Riemann-Lebesgue lemma from harmonic analysis. We refer the reader to a book entitled Fourier Analysis due to Stein and Shakarchi.

3 Comments »

  1. Thanks for your post. Since FT is an isometry with L^2-norm, so it’s very nice on the conservation of total energy .

    Comment by Tuan Minh — March 24, 2010 @ 17:26

    • Actually I have lots of estimates regarding to wave and heat equations (KdV, NST, etc.), hopefully I have enough time to show.

      Comment by Ngô Quốc Anh — March 24, 2010 @ 17:33

  2. Since the total energy only depends on the initial data (Schwartz functions) so we can imply the uniqueness of the solution to the initial value problem for the wave equation in the upper half-plane. The condition that f,g are Schwartz functions is too strong. Indeed, you can see the exercise (3, 5D) in the well-known book of Folland that we only need f\in H_1(\mathbb{R}^n), g\in L^2(\mathbb{R}^n), then
    E(t)=\int_{\mathbb{R}^n} (|\nabla f|^2+|g|^2)dx

    Comment by Tuan Minh — March 7, 2012 @ 18:59


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