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|<R\}.

It is not difficult to verify that

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

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.

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