Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $u:{\mathbb{R}^ + } \times {\mathbb{R}^3} \to \mathbb{R}$ be a solution of the Cauchy problem

$\left\{ \begin{gathered} {u_{tt}} - \Delta u = 0 \\ u\left( {0,x} \right) = {u_0}\left( x \right) \\ {u_t}\left( {0,x} \right) = {u_1}\left( x \right) \\ \end{gathered} \right.$

Assuming that ${u_0}$ and ${u_1}$ are smooth and compactly supported, show that

$\left( {\exists C > 0} \right)\left( {\forall t > 0} \right)\left( {\forall x \in {\mathbb{R}^3}} \right)\left| {u\left( {t,x} \right)} \right| \leqslant \frac{C}{t}$.

Edit: Kirchoff's formula in 3 dimensions gives

$u\left( {t,x} \right) = \frac{1}{{4{t^2}\pi }}\int_{S\left( {x,t} \right)} {t{u_1}\left( y \right) + {u_0}\left( y \right) + \nabla {u_0}\left( y \right)\left( {y - x} \right)d{S_y}} $

which would directly give only $\left| {u\left( {t,x} \right)} \right| \leqslant Ct$ which isn't good enough.

Source: Evans, partial differential equations, problem 18 on page 89

share|cite|improve this question
up vote 1 down vote accepted

Since ${u_0},{u_1},\nabla {u_0}$ are compactly supported, there exists $M > 0$ such that ${u_0},{u_1},\nabla {u_0} = 0$ on $S\left( {x,t} \right)$ for all $t \geqslant M$. Since they are smooth and compactly supported, we have

$\left| {u\left( {t,x} \right)} \right| \leqslant \frac{1}{{4{t^2}\pi }}\int_{S\left( {x,t} \right)} {\left( {t{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {{u_0}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }\underbrace {\left| {y - x} \right|}_{ = t}} \right)d{S_y}} = $ $\frac{{t{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {{u_0}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }t}}{{4{t^2}\pi }}\int_{S\left( {x,t} \right)} {d{S_y}} \leqslant \frac{{\left( {t{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {{u_0}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }t} \right){M^2}}}{{{t^2}}}$

for $t \leqslant M$, and $\left| {u\left( {t,x} \right)} \right| = 0$ for $t \geqslant M$.

In case that $1 \leqslant t \leqslant M$ (if $M > 1$), we have $\left| {u\left( {t,x} \right)} \right| \leqslant \frac{{\left( {{{\left\| {{u_1}} \right\|}_\infty } + \frac{1}{t}{{\left\| {{u_0}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }} \right){M^2}}}{t} \leqslant \frac{{\left( {{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {{u_0}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }} \right){M^2}}}{t}$ and in case that $0 < t < \min \left\{ {1,M} \right\}$, we get $\left| {u\left( {t,x} \right)} \right| \leqslant \frac{{\left( {{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }} \right){M^2}}}{t} + {\left\| {{u_0}} \right\|_\infty } \leqslant \frac{{\left( {t{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }t} \right){M^2} + {{\left\| {{u_0}} \right\|}_\infty }}}{{{t^2}}}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.