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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

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1 Answer 1

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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 $\left| {u\left( {t,x} \right)} \right| \leqslant \frac{{\left( {{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }} \right){M^2}}}{t} + {\left\| {{u_0}} \right\|_\infty } \leqslant \frac{{\left( {t{{\left\| {{u_1}} \right\|}_\infty } + {{\left\| {\nabla {u_0}} \right\|}_\infty }t} \right){M^2} + {{\left\| {{u_0}} \right\|}_\infty }}}{{{t^2}}}$.

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