306 PROBLEMS AND SOLUTIONS

n+2

(P + 1)dr 2 (u- E(u))sn2udu

(- (/)e(- (//= (g-.Change y to q, with q_ 0 and q0 1:

usnuQ t + qlt2 + q2t3 +

cn udn uqn ql q2 dt

n=On + i

qo + - + -- + Q--2u du K2.

Also solved by the proposer.

Asymptotic Solution of the Telegraph Equation

Problem 92-10", by MARK A. PINSKY (Northwestern University).Let u(x; t) be a solution of the telegraph equation utt + 2ut uxx defined for

t > 0, -cx < x < cx with smooth and sufficiently rapidly decreasing initial data:u(x;0+) fl(x),ut(x;O+) f2(x). Show that there is a solution of the heat equa-tion 2vt vx defined for t > 0, -o < x < c such that u(x; t) v(x; t) O(t-1),t T oc. Specify v(x; t) in terms of its initial data v(x; 0+) and find minimal smoothnessand decay conditions on fl and f2.Solution by MICHAEL RENARDY (Virginia Tech, Blacksburg, VA).

We have v(z, 0+) f2(z)/2 + fl(z). The question of minimal assumptions on

fl and f2 obviously depends on the sense in which u v is required to be O(t-); forexample, if fl E L2(]R) and f2 E H-I(), then tllu(.,t) v(., t)ll-() is bounded as

To prove this, let z2(k, t) be the Fourier transform of u with respect to the spatialvariable, so that,, + 2, -, (, o) ]1(), ,(, o) ]().Both fi and 3 decay exponentially unless k is close to zero; so in the following discussionwe need only consider small values of k. The solution of (1) is given by

(2) z(k) a(k)exp(-(1 + V/1 k2)t) + b(k)exp(-(1 V/1 k2)t),

where

(3) a(k) + b(k) f (k), -(1 + v/1 k2)a(k) (1 y/1 k)b(k)

The first term on the right-hand side of (2) decays exponentially as t , and in thesecond term, we find

(4) b(k) 7f2(k)+ f(k) + O(k2(lf(k)l / [f2(k)l))

and

(5) 1- v/i- k2 1/2k2 + O(k4).

Dow

nloa

ded

12/1

9/14

to 1

29.4

9.17

0.18

8. R

edis

trib

utio

n su

bjec

t to

SIA

M li

cens

e or

cop

yrig

ht; s

ee h

ttp://

ww

w.s

iam

.org

/jour

nals

/ojs

a.ph

p

PROBLEMS AND SOLUTIONS 307

This yields

(6)b(k)exp(-(1-v/1-k2)t) [f2(k)+ fx(k)] exp (_ k__ft)

)+O exp --- (k2 + k4t)(Ifl(k)l + If2(k)l

Elementary calculus shows that

(7) max kU exp (_ kt) 2max k4t exp (k2t) 16kR --- e2t"

From this the claim follows immediately.

An Infinite Integral

Problem 93-7 (Quickie).Let

I(y) fl -y(lq-2s2)

(1 + 2s2)2dods.

Then,

dy2 2ds r

I(O)L1 + 2sU) 4v/,7dI fo ds

-(ol ( +1 "One can solve this simple ordinary differential equation and after changing the order ofintegration in the resulting double integral, find that

I(y) (1- 2y) 4x/-1 e- ds + - v/-e U

The result then follows by setting y 7"

Dow

nloa

ded

12/1

9/14

to 1

29.4

9.17

0.18

8. R

edis

trib

utio

n su

bjec

t to

SIA

M li

cens

e or

cop

yrig

ht; s

ee h

ttp://

ww

w.s

iam

.org

/jour

nals

/ojs

a.ph

p