Remarks: (a) The Dirac delta is the function zero on the domain R {0}.
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1 Sectio Objective(s): The Dirc s Delt. Mi Properties. Applictios. The Impulse Respose Fuctio The Dirc Delt Geerlized Sources Defiitio The Dirc delt geerlized fuctio is the limit δ(t) = lim δ (t), for every fixed t R of the sequece fuctios {δ } =,! δ (t) = u(t) u " t # $. δ Remrk: The sequece of bump fuctios itroduced bove c be rewritte s follows, 0, t < 0 δ (t) =, 0! t < 0, t ". We the obti the equivlet expressio, ) 0 for t = 0, δ(t) = for t = δ 3 (t) δ 2 (t) δ (t) Remrk: There re ifiitely my sequeces {δ } of fuctios with the Dirc delt s their limit s. 0 /3 /2 t Remrks: () The Dirc delt is the fuctio zero o the domi R {0}. (b) The Dirc delt is ot fuctio o R. Theorem. Every fuctio i the sequece {δ } bove stisfies * c+ c δ (t c) dt =.
2 Mi Properties. Remrk: We use limits to defie opertios o Dirc s delts. Defiitio We itroduce the followig opertios o the Dirc delt:! f(t) δ(t c) + g(t) δ(t c) = lim f(t) δ (t c) + g(t) δ (t c) " #,. $ b δ(t c) dt = lim $ b δ (t c) dt, L[δ(t c)] = lim L[δ (t c)]. Theorem For every c R d ϵ > 0 holds, $ c+ϵ c ϵ δ(t c) dt =. Proof of Theorem 4.4.3: The itegrl of Dirc s delt geerlized fuctio is computed s limit of itegrls, $ c+ϵ c ϵ $ c+ϵ δ(t c) dt = lim δ (t c) dt c ϵ $ c+ = lim dt, for c % = lim c + & c < ϵ, = lim =. This estblishes the Theorem.
3 3 Theorem If f is cotiuous o (, b) d c (, b), the f(t) δ(t c) dt = f(c). Proof of Theorem 4.4.4: We gi compute the itegrl of Dirc s delt s limit of sequece of itegrls, δ(t c) f(t) dt = lim = lim = lim! c+ c δ (t c) f(t) dt " u(t c) u # t c $ % f(t) dt f(t) dt, < (b c). To get the lst lie we used tht c [, b]. Let F be y primitive of f, so F (t) = & f(t) dt. The we c write, δ(t c) f(t) dt = lim ' F # c + $ ( F (c) = lim # $ ' F # c + $ ( F (c) = F (c) = f(c). This estblishes the Theorem.
4 4 Theorem For ll s R holds e cs for c! 0, L[δ(t c)] = 0 for c < 0. Proof of Theorem 4.4.5: We use the previous Theorem o the itegrl tht defies Lplce trsform, L[δ(t c)] = This estblishes the Theorem. % 0 e cs for c! 0, e st δ(t c) dt = 0 for c < 0,
5 Applictios of the Dirc Delt. Remrks: () Dirc s delt geerlized fuctio is useful to describe impulsive forces. (b) A impulsive force trsfers fiite mometum i ifiitely short time. (c) For exmple, pedulum t rest tht is hit by hmmer.
6 6 Exmple 4.4.3: Use Newto s equtio of motio d Dirc s delt to describe the chge of mometum whe prticle is hit by hmmer. Solutio: A poit prticle with mss m, movig o oe spce directio, x, with force F ctig o it is described by m = F mx (t) = F (t, x(t)), We use the prticle mometum, p = mv, to write the Newto s equtio, mx = mv = (mv) = F p = F. So the force F chges the mometum, P. If we itegrte o itervl [t, t 2 ] we get p = p(t 2 ) p(t ) =! t2 t F (t, x(t)) dt. Suppose tht impulsive force is ctig o prticle t t 0 trsmittig fiite mometum, sy p 0. This is where the Dirc delt is uselful for, becuse we c write the force s F (t) = p 0 δ(t t 0 ), the F = 0 o R {t 0 } d the mometum trsferred to the prticle by the force is p =! t0 + t t 0 t p 0 δ(t t 0 ) dt = p 0. The mometum trferred is p = p 0, but the force is ideticlly zero o R {t 0 }. We hve trsferred fiite mometum to the prticle by iterctio t sigle time t 0.
7 The Impulse Respose Fuctio. Defiitio The impulse respose fuctio t the poit c! 0 of the lier opertor L(y) = y + y + 0 y, with, 0 costts, is the solutio y δ of L(y δ ) = δ(t c), y δ (0) = 0, y δ(0) = 0. Theorem The fuctio y δ is the impulse respose fuctio t c! 0 of the costt coefficiets opertor L(y) = y + y + 0 y iff holds y δ = L! e cs ". p(s) where p is the chrcteristic polyomil of L. Remrk: The impulse respose fuctio y δ t c = 0 stifies y δ = L! " p(s). Proof of Theorem 4.4.7: Compute the Lplce trsform of the differetil equtio for for the impulse respose fuctio y δ, L[y ] + L[y ] + 0 L[y] = L[δ(t c)] = e cs. Sice the iitil dt for y δ is trivil, we get (s 2 + s + 0 ) L[y] = e cs. Sice p(s) = s 2 + s + 0 is the chrcteristic polyomil of L, we get L[y] = e cs p(s) y(t) = L! e cs p(s) ". We otice tht ll the steps i this clcultio re if d oly ifs. This estblishes the Theorem.
8 8 Exmple Similr to 4.4.6: Fid the solutio y to the iitil vlue problem y y = δ(t 3), y(0) = 0, y (0) = 0. Solutio: The source is geerlized fuctio, so we eed to solve this problem usig the Lpce Trsform. So we compute the Lplce Trsform of the differetil equtio, L[y ] L[y] = L[δ(t 3)] (s 2 ) L[y] = e 3s, where i the secod equtio we hve lredy itroduced the iitil coditios y(0) = 0, y (0) = 0. We rrive to the equtio Recllig the trsltio idetity L[y] = e 3s (s 2 ) e cs L[f(t)] = L[u(t c) f(t c)], we get tht L[y] = L[u(t 3) sih(t 3)], which leds to the solutio y(t) = u(t 3) sih(t 3).
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