Asymptotic Solution of the Telegraph Equation (Mark A. Pinsky)

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  • 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 onfl 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).

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  • 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)lElementary calculus shows that(7) max kU exp (_ kt) 2 max k4t exp (k2t) 16kR --- e2t"From this the claim follows immediately.An Infinite IntegralProblem 93-7 (Quickie).

    Let

    I(y) fl -y(lq-2s2)(1 + 2s2)2do ds.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 of

    integration in the resulting double integral, find that

    I(y) (1- 2y) 4x/-1 e- ds + - v/-e UThe result then follows by setting y 7"

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