# Ngô Quốc Anh

## February 28, 2010

### Wave equations: The L^2-norm estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:06

Now we talk about $L^2$-norm of solution to homogeneous wave equations. Let $u(x,t)$ be a solution of the Cauchy problem

$\displaystyle u_{tt}- \Delta u =0, \quad x \in \mathbb R^3, t>0$

with the following conditions

$\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x), \quad x\in \mathbb R^3$.

Assume that $f, g \in C_0^\infty(\mathbb R^3)$. If

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx} = 0$

then

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {g{{(x)}^2}dx} = 0$.

Proof. Let $R>0$ such that

$\displaystyle {\rm supp}(f),{\rm supp}(g) \subset B_R = \{x \in \mathbb R^3, |x|.

It is not difficult to verify that

$\displaystyle {\rm supp}(u(\cdot,t)) \subset \{ x\in \mathbb R^3, t-R<|x|

for any $t>R$.

By using the Green formula, we get the following equality from the equation

$\displaystyle\int_{{\mathbb{R}^3}} {{u_{tt}}(x,t)dx} = \int_{{\mathbb{R}^3}} {\Delta u(x,t)dx} = 0$.

Therefore, we have

$\displaystyle\frac{d}{{dt}}\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = 0$,

and

$\displaystyle\int_{{\mathbb{R}^3}} {{u_t}(x,t)dx} = \int_{{\mathbb{R}^3}} {g(x)dx}$,

and

$\displaystyle\int_{{\mathbb{R}^3}} {u(x,t)dx} = \int_{{\mathbb{R}^3}} {f(x)dx} + t\int_{{\mathbb{R}^3}} {g(x)dx}$

by the initial conditions. In the other side, by Cauchy inequality, it follows that

$\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u(x,t)dx} \leqslant {\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}}{\left( {\int_{t - R < |x| < t + R} {dx} } \right)^{\frac{1}{2}}} \hfill \\ \qquad= {\left( {\frac{8}{3}\pi R} \right)^{\frac{1}{2}}}{\left( {3{t^2} + {R^2}} \right)^{\frac{1}{2}}}{\left( {\int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} } \right)^{\frac{1}{2}}} \hfill \\ \end{gathered}$

for any $t>R$. This is a contradiction provided that

$\displaystyle\int_{{\mathbb{R}^3}} {g(x)dx} \ne 0$.

We now try to find the limit

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx}$

in general case. For simplicity, we assume $f \equiv 0$. We will show that there exists a non-negative constant $c$ such that

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{\mathbb R^3} {u{{(x,t)}^2}dx} = c$.

Proof. Let $\widehat u$ denote the Fourier transform of $u$ with respect to the variable $x$. By taking the Fourier transform of the wave equation and the initial conditions we can get

$\displaystyle\widehat u(\xi ) = \frac{{\left| {\widehat g(\xi )} \right|}}{{\left| \xi \right|}}\sin (\left| \xi \right|t)$.

Therefore

$\displaystyle\begin{gathered} \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {\widehat u{{(\xi ,t)}^2}d\xi } \hfill \\ \qquad= \frac{1}{{{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {|\widehat g(\xi ){|^2}\frac{{{{\sin }^2}(\left| \xi \right|t)}}{{{{\left| \xi \right|}^2}}}d\xi } \hfill \\ \qquad= \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}d\xi } - \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g(\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } . \hfill \\ \end{gathered}$

Noting

$\displaystyle {\left| \xi \right|^{ - 2}}|\widehat g(\xi ){|^2} \in {L^1}({\mathbb{R}^3})$

we have

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi ){|^2}\cos (2\left| \xi \right|t)d\xi } = 0$.

We then get

$\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^3}} {u{{(x,t)}^2}dx} = \underbrace {\frac{1}{{2{{(2\pi )}^3}}}\int_{{\mathbb{R}^3}} {{{\left| \xi \right|}^{ - 2}}|\widehat g (\xi )|^2d\xi } }_c$.

Note that this constant $c$ is positive if $g \not\equiv 0$.