3. Calculus with distributions

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Dwight Parsons
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1 6 RODICA D. COSTIN 3.1. Limits of istributios. 3. Calculus with istributios Defiitio 4. A sequece of istributios {u } coverges to the istributio u (all efie o the same space of test fuctios) if (φ, u ) (φ, u) for all test fuctios φ. The poit is that istributios ca be obtaie as limits of fuctios Fuametal sequeces for δ. Ituitively, oe may thik of δ() as a fuctio which is zero at all poits ecept for = 0, where it is ifiite, a such that δ() = 1. Of course, there are o such fuctios, but δ() is a limit (i istributio sese) of fuctios which are zero outsie itervals arou 0), of legths shrikig to zero, while the total itegral of these fuctios remais oe. Of course, i this case the maimum of these fuctios must go to ifiity. There are may sequeces of fuctios which coverge, i the istributio sese, to Dirac s δ fuctio. They are calle fuametal sequeces for δ A sequece of step fuctios. Cosier a arrower a arrower step, cetere at 0, of total area 1: 2 if 1, 1 () = 0 otherwise The lim = δ. Iee, for ay test fuctio φ, (φ, )= φ() = φ(c ) for some c 1, 1 (by the mea value theorem) which coverges to φ(0) = (φ, δ) Other sequeces. We coul smooth out the corers of our step fuctios to make them eve C. I fact, oe ca fi may fuametal sequeces for δ by rescalig, as follows. Pick a fuctio f which is C o R which is zero outsie some iterval [,] a has total itegral oe: f() = 1. For each >0 use rescalig to shrik [,] to[, ] while ialtig f so that the total area remais oe: efie f () = 1 f The f () = 0 outsie [, ] a f () =1(check!). Wehave lim 0 f = δ sice (φ, f )= f ()φ() = = f(y)φ(y)y = {f(y)[φ(y) φ(0)] + f(y)φ(0)} y f(y)φ(y)y

2 DISTRIBUTIONS 7 φ(y) φ(0) = φ(0) f(y) y + yf(y) y y a usig the mea value theorem for the fuctio φ(y) = φ(0) + yf(y)φ (c) for some c (0,) hece the last term i the sum goes to 0 as 0 a the limit is φ(0). 4. Gree s fuctio 4.1. No-homogeeous coitios ca be reuce to homogeeous coitios, but for a o-homogeeous equatio. If the iitial a bouary coitios are ot zero, the problem is first ture ito oe with homogeeous coitios (but with a o-homogeeous term) by subtractig from the ukow fuctio a fuctio witch has the prescribe iitial a/or bouary coitios. This is easiest to see whe illustrate o a few eamples First orer equatios. Cosier the simplest eample (5) y + a()y = g(), y( 0)=y 0, for 0 We fi a fuctio h() so that h( 0 )=y 0,thesubstitutey = h + u. For eample, we ca take h() =y 0. Problem (5) becomes u + a()u = g() a()y 0, u( 0 )=0, for 0 which is the type (11) stuie before Seco orer equatios. A equtio with o-homogeeous bouary coitios, say: P ()y + Q()y + R()y =0, y(0) = A, y(1) = B ca be trasforme ito a seco-orer o-homogeeous equatio, butwith homogeeous bouary coitios: just fi a fuctio g() so that g(0) = A a g(1) = B, thesubstitutey = g + u. (For eample, oe coul take g() =(B A) + A). The problem becomes P ()y + Q()y + R()y = f(), u(0) = A, u(1) = B where f() = (P ()g + Q()g + R()g). So from ow o, whe ealig with o-homogeeous equatios, we ca restrict our cosieratios to homogeeous coitios.

3 8 RODICA D. COSTIN 4.2. Superpositio. Suppose we have a liear, o-homogeeous ifferetial equatio, Lu = f where f() is a give fuctio (the forcig term, or the o-homogeeity), a L = L(, ) is a liear ifferetial operator, for eample, (6) L = u + α for equatio + αu = f() (7) L = u + a() for equatio + a()u = f() (8) L = 1 w() (p() )+q() for equatio 1 w() u (p() )+q()u = f() (9) L = t 2 for equatio u 2 t u = f a so o. The: if u 1 solves Lu 1 = f 1 (), a u 2 solves Lu 2 = f 2 (), the a liear combiatio u = c 1 u 1 + c 2 u 2 solves Lu = c 1 f 1 + c 2 f 2 (because L is liear). Similarly, we ca superimpose ay umber of solutios: if u k solves Lu k = f k () for k =1,..., the u = k c ku k solves Lu = k c kf k. The same is true if we also have some homogeeous coitios, iitial (IC) or bouary (BC). A coitio is calle homogeeous whe a liear fuctioal of u is require to be zero. For eample, we coul cosier the problem Lu = f with -L as i (7) a the (IC) u( 0 )=0 -L as i (8) a the (IC) u( 0 )=0, u ( 0 )=0 -L as i (8) a the (BC) u( 0 )=0, u( 1 ) 3u ( 1 )=0 -L as i (9) a the (IC) u(t 0,) = 0 a (BC) u( 0 )=0, u( 1 )=0 a so o. For this o-homogeeous problems with homogeeous coitios, agai, a superpositio of forcig terms is solve by a superpositio of solutios. Why ot cosier a cotiuous superpositio? writte as a superpositio of impulses : f() = f(t)δ( t) t After all, f ca be The solve, for each parameter t, the equatio LG = δ( t), eote its solutio by G = G(, t), a the superimpose: (10) u() = f(t)g(, t) t is epecte to solve Lu = f. We ca similarly solve homogeeous (IC) or (BC) problems. G(, t) is calle the Gree s fuctio of the problem. The tasks ahea are the: to fi ways of calculatig the Gree s fuctio of give problems, to show that the superpositio (10) is iee a solutio,

4 DISTRIBUTIONS 9 a to see what iformatio about solutios ca be rea irectly i the Gree s fuctio First eamples The simplest eample, C case. Cosier L as i (7) a the homogeeous iitial value problem (11) u + a()u = f(), u( 0)=0, for 0 Let us fi its Gree s fuctio: solve (12) G + a()g = δ( t), G( 0,t) = 0 for all t The itegratig factor is ep( a() ). Assume a C. The so is the itegratig factor, a the we ca multiply the equatio by it (sice δ ca be multiplie by C fuctios) a we get e a(s) s 0 G = e a(s) s 0 δ( t) hece (13) e a(s) s t 0 G = e a(s) s 0 δ( t) Recallig that H = δ(), itegratig (13) 2 therefore a similarly H( t) =δ( t) e 0 a(s) s G(, t) =e t 0 a(s) s H( t)+c(t) G(, t) =e t a(s) s H( t) + e a(s) s 0 C(t) particular sol. geeral sol. homog. eq. Imposig the iitial coitio we obtai C(t) = e 0 t a(s) s H( 0 t) hece (14) G(, t) =e t a(s) s χ [0,](t) where χ [0,](t) =H( t) H( 0 t) = is a step fuctio. The solutio of (11) is u() = 0 f(t)g(, t) t = 1ift [0,] 0 otherwise 0 f(t)e t a t 2 We ee to use here the fact that the oly istributios whose erivatives are 0 are the costats, which requires a proof (ot iclue here).

5 10 RODICA D. COSTIN The simplest eample, o-smooth coefficiets case. Cosier agai the problem (11), oly ow with a() ot C (for eample, we coul have a() = ). The we o ot kow that we ca multiply (meaigfully) the istributio δ by the itegratig factor. The the equatio is solve as follows: we solve for <t, a separately for >ta we match the two pieces. There are two mai facts here. Fact 1. We ca rely o the ituitio that δ() = 0 for <0 a also δ() = 0 for >0. (This ca be rigorously state, but we will ot o that.) Therefore δ( t) = 0 for <ta also for >t. Fact 2. Eamiig (12) we see that G (as a fuctio of ) has a jump iscotiuity of magitue 1 at = t. This coitio will match the two solutios o the two itervals. Now let us solve (12). For <tthe equatio is G + a()g =0with the geeral solutio G(, t) =A(t)e a 0. Imposig the iitial coitio we must have A(t) = 0, hece G(, t) = 0 for <t. For > t we have agai G + a()g = 0, with the geeral solutio G(, t) =B(t)e t a. There are o other coitios o this iterval, a B(t) is etermie from the jump coitio at = t: wemusthave G(, t) =t+ G(, t) =t =1 hece B(t) = 1, givig for G the formula (14).

Chapter 4. Fourier Series

Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,