Green s Functions and the Heat Equation

Transcription

1 Green s Funcions and he Hea Equaion MA 436 Kur Bryan 0. Inroducion Our goal is o solve he hea equaion on he whole real line, wih given iniial daa. Specifically, we seek a funcion u(x, ) which saisfies u u x = 0, () u(x, 0) = f(x) for < x < and > 0, where f(x) is some given iniial emperaure. We ll make assumpions abou f laer. Of course we wan exisence, uniqueness, and sabiliy resuls. Recall ha he soluion o he wave equaion on he whole real line was given by he D Alember formula, which consruced he soluion via various inegrals involving he iniial daa. The soluion o he hea equaion () on he whole real line is also given by such formula, u(x, ) = e (x y) 4 f(y) dy. () π Bu of course you re wondering where in he world does ha come from? Tha s wha we should look a. 0. Mollificaion Suppose ha ϕ(x) is a reasonably smooh (say C or beer) non-negaive funcion defined for all real x which decays o zero a infiniy, wih ϕ(x) dx =. To help you visualize a specific example, we can use he funcion 0, x, 5 ϕ(x) = 6 (x ) (x + ), < x <, 0, x

2 alhough his is jus an example. Nohing ha follows depends on his paricular choice. In his case ϕ looks like x 3 Now define he funcion ψ(x, ) = ϕ(x ). You can easily check ha for any > 0 we have ψ(x, ) dx = = ϕ(x ) dx, ϕ(u) du, =, (3) where I used he subsiuion u = x/ o ge he las inegral. Wha does he funcion ψ(x, ) look like (as a funcion of x) for various choices of > 0? Well, if = hen obviously ψ(x, ) = ϕ(x), so i looks jus like he figure above. If is large hen he / facor in fron of ϕ ends o squash he funcion down oward zero, while he x/ facor inside expands he base of ϕ. Conversely, if is small hen he / facor increases he heigh of he funcion and he base shrinks (hink i hrough carefully). Here are ψ(x, ), ψ(x, 4), and ψ(x, /4) all ploed ogeher:

3 5 4 3 y x 4 Recall, however, in each case he area under he curve is exacly one. Suppose ha w(x) is a bounded and coninuous funcion on he real line, say bounded by some consan M. Consider he expression ψ(y, )w(y) dy as a funcion of. We wan o examine he behavior of his inegral as a funcion of. In paricular, wha limi if any does his inegral approach as approaches zero? Here s he informal version of he argumen: For small posiive he funcion ψ(y, ) is concenraed near y = 0. I s no hard o see ha he produc ψ(y, )w(y) is also concenraed near y = 0. Bu if y is close o zero hen w(y) w(0), wih he approximaion improving for y closer o zero, so if is really close o zero hen ψ(y, )w(y) ψ(y, )w(0). In his case since ψ(y, )w(y) dy ψ(y, ) dy =. In shor, lim 0 ψ(y, )w(0) dy = w(0) ψ(y, )w(y) dy = w(0) (4) if w is coninuous. You migh recognize ha as ges close o zero he funcion ψ(x, ) looks like a Dirac dela funcion. 3

4 Here s he hardcore version of he informal argumen (if you ve never done ϵ/δ proofs, feel free o skip his paragraph). Choose a number ϵ > 0. Since w is coninuous, for any such ϵ > 0 here is some δ > 0 (dependen on ϵ) such ha w(x) w(0) < ϵ for all x < δ. Now noe ha from equaion (3) and he fac ha w(0) is a consan we have Then w(0) = ϕ(y/)w(0) dy ϕ(y/)w(y) dy w(0) = = + + δ δ δ δ ϕ(y/)(w(y) w(0)) dy ϕ(y/)(w(y) w(0)) dy ϕ(y/)(w(y) w(0)) dy ϕ(y/)(w(y) w(0)) dy (5) (I spli he inegral ino hree ranges). Take absolue values of boh sides of equaion (5) and use he riangle inequaliy o conclude ha ϕ(y/)w(y) dy w(0) δ ϕ(y/)(w(y) w(0)) dy δ δ + ϕ(y/)(w(y) w(0)) dy + ϕ(y/)(w(y) w(0)) dy (6) δ The firs inegral on he righ above can be bounded by noing ha w(y) w(0) < ϵ for y < δ, so δ ϕ(y/)(w(y) w(0)) dy δ ϵ δ ϕ(y/) dy δ ϵ ϕ(y/) dy = ϵ. (7) (noe ϕ = ϕ). The oher wo inegrals on he righ in equaion (6) are similar o each oher. For example, since w(x) M for all x, w(y) w(0) M and so ϕ(y/)(w(y) w(0)) dy M ϕ(y/) dy. δ δ 4

5 Change variables in he las inegral (use x = y/, so y = x, dy = dx) and find ϕ(y/)(w(y) w(0)) dy M ϕ(x) dx. (8) δ δ/ A similar bound hold for he inegral from o δ. Using he bounds of equaion (7) and (8) in inequaliy (6) produces δ/ ϕ(y/)w(y) dy w(0) ϵ + M ϕ(x) dx + M ϕ(x) dx. δ/ Take he limi of boh sides above for 0 +. Boh inegrals limi o zero (hink abou why), leaving us wih lim 0 ϕ(y/)w(y) dy w(0) ϵ. Bu ϵ > 0 is arbirary, so he lef side above is zero and we have proved equaion (4). Now le s look a wha happens o he inegral ψ(y x, )w(y) dy for small. In his case he funcion ψ(y x, ) is jus ψ(y, ) ranslaed x unis o he righ, and so for small he funcion ψ(y, ) is concenraed near y = x. A simple change of variables shows ha lim ψ(y x, )w(y) dy = w(x). (9) 0 We ll call a smooh funcion ψ(x, ) wih he propery (9) a mollifier. The verb o mollify means o smooh ou, and ha s wha ψ(x, ) does. Here s an example. Le w(x) be defined as w(x) = 0, x, x +, < x 0,, 0 < x <,, x < 3, 0, x > 3 5

6 Here s wha w(x) looks like: x Figure : The Funcion w(x) Noe ha w is disconinuous. Define he funcion W (x, ) = ψ(y x, )w(y) dy (0) where ψ(y, ) = ϕ(y/) for he funcion ϕ defined earlier. We know ha lim 0 W (x, ) = w(x), bu i s insrucive o plo he funcion W (x, ) for some non-zero values of. For example, when = here is wha W (x, ) looks like compared o w(x): x Figure : The Funcion W (x, ) I s a smoohed ou version of w! If you look a he definiion of W (x, ) 6

7 you see ha he inegral simply akes he values of w(y) near y = x and averages hem, weighed by he funcion ψ(y x, ). The bigger he value of, he broader he range for he average, and he smooher W becomes. For small values of he average is over a igh range abou y = x. Since w doesn change much over such a small range, he average near any poin y = x is abou equal o w(x). Here s a plo of W (x, 4). I s prey smooh x Figure 3: The Funcion W (x, 4) And finally, here s W (x, /4). I s a much beer approximaion o w(x) x Figure 4: The Funcion W (x, /4) A noe on smoohness. Le us apply x o boh sides of equaion (0). 7

8 We obain W x = ψ (y x, )w(y) dy x where he derivaive has been slipped inside he inegral. This is permissible if w(y) and/or ψ decay fas enough a infiniy (cerainly if eiher has compac x suppor). Clearly i also requires ha ψ be differeniable in x. The poin here is ha if ψ is differeniable hen W is differeniable, even if w is no. Repeaing he process shows ha n W x n = ( ) n n ψ (y x, )w(y) dy xn again, provided ha n ψ exiss. The mollified funcion W (x, ) will be jus x n as smooh as ψ, and as approaches zero W approaches w. Mollificaion is an imporan process in PDE. Cerain hings can be done o disconinuous or non-smooh funcions (like differeniae hem). Mollificaion allows you o replace non-smooh funcions wih smooh approximaions. You hen operae on hese smooh approximaions, ake limis as goes o zero, and hope ha wha you did o he smooh approximaion makes sense for he original non-smooh funcion. Also, a funcion ψ(x, ) can be a mollifier wihou being of he form ϕ( x). For example, we can ake ψ(x, ) = ϕ( x ). () α α where ϕ is some smooh funcion and α > 0. You can easily verify ha he same argumens above apply here as ges small, ψ(x, ) ges concenraed near x = 0. The reason we re considering mollificaion is his: Look back a he funcion w(x) in Figure, and suppose ha w is he iniial emperaure daa for some bar a ime = 0. As increases we d expec ha he hea diffuses and smears ou, jus as he funcion W (x, ) does as increases in Figures o 4. In fac he diffusion of hea is a mollificaion process, if you choose he funcion ψ correcly. 8

9 0.3 Solving he Hea Equaion Suppose ha ψ(x, ) is a mollifying funcion, ha is, ψ(x, ) is defined for < x < and > 0, and lim 0 + ψ(y x, )f(y) dy = f(x) for any coninuous funcion f. To find a soluion u(x, ) o he hea equaion wih u(x, 0) = f(x) we ll define u(x, ) = ψ(y x, )f(y) dy () for > 0 for some suiable mollifier ψ. Then u(x, 0) = f(x), in he sense ha lim u(x, ) = f(x). (3) 0 + We ll someimes have o be careful in wriing u(x, 0) = f(x), since plugging = 0 ino he inegral of equaion () migh no make sense, bu he limi as 0 + will. So echnically, he iniial condiion is only saisfied in he sense of equaion (3). Now suppose ha he mollifier ψ(x, ) saisfies he hea equaion, so ψ ψ xx = 0. Then he funcion u(x, ) defined by equaion () will also saisfy he hea equaion, for ( ) u u = x ψ(y x, )f(y) dy, x = (ψ (y x, ) ψ xx (y x, ))f(y) dy, = 0. since ψ ψ xx = 0. There is also he issue of slipping he derivaives inside he inegral, bu we can do his if ψ and is derivaives decay rapidly a infiniy, or if f decays rapidly. Solving he hea equaion hus comes down o finding a mollifying funcion ψ(x, ) which saisfies he hea equaion. Equaion () provides a means for consrucing mollifiers, so le s see if we can find one ha saisfies he hea equaion. A good sar would be o ry ψ(x, ) of he form ψ(x, ) = α ϕ( x α ) 9

10 for some funcion ϕ. We have o choose ϕ appropriaely, and α also, o make his work in he hea equaion. If we plug ψ ino he hea equaion we obain ψ ψ xx = 3α ϕ (x/ α ) αx α ϕ (x/ α ) α α ϕ(x/ α ) = 0. (4) Noe ha he argumen o ϕ is x/ α. Le s ge rid of ha peculiariy. Define a new funcion η as η(x) = ϕ(x/ α ) (noe ha η depends implicily on ). Then dη dx (x) = α ϕ (x/ α ), d η dx (x) = α ϕ (x/ α ). Use hese wo equaions, and η(x) = ϕ(x/ α ), replace ϕ and is derivaives in equaion (4) wih he equivalen in η. You obain, afer a bi of simplifying algebra, η (x) + αx η (x) + α η(x) = 0. (5) This DE is acually prey easy o solve. Wrie i in he form αx η (x) + α η(x) = η (x) hen noice ha he lef and righ sides are boh exac derivaives. In fac, he lef side is α d dη (xη(x)), while he righ is obviously. Inegrae boh dx dx sides and oss in a consan c o find ha α xη(x) = η (x) + c, which is a linear firs order equaion. Wrie i as η (x) + α xη(x) = c and muliply hrough by he inegraing facor e αx. We now have d αx (e η(x)) = c e αx dx and so we can inegrae in x, add a consan of inegraion c, and do some algebra o find ha η(x) = c e αx e αx dx + c e αx. 0

11 Le s now go back o he original ϕ noaion (recall η(x) = ϕ(x/ α )) so ha ϕ( x ) = c e αx α e αx dx + c e αx. I should now be apparen wha α has o be: he righ side is a funcion of x /, or equivalenly, x/, so if his is going o work we need α = /. Also, we wan ϕ o die ou as x ges large. The erm wih he c coefficien looks like rouble. You can convince yourself ha i in fac blows up for large x. Le s se c = 0 (and remember, α = /) o obain ϕ( x ) = c e x 4. I s easy o see ha his funcion dies ou rapidly for large x. This gives us a prospecive soluion o he hea equaion ψ(x, ) = c e x 4 for any consan c. I works! You can plug ψ ino he hea equaion and check his. The las hing we need o do is adjus c o obain ψ(x, ) dx =. I s a famous fac ha e u du = π. Armed wih his fac, you can do a simple change of variable u = x/ o find ha e x 4 dx = π so we should choose c =. Thus we have π ψ(x, ) = x π e 4. (6) This funcion ψ(x, ) is of he form ϕ(x/ ) wih ϕ(u) = π e u /4. The funcion ψ(x, ) defined by equaion (6) is called he Green s Funcion, or Green s kernel, or fundamenal soluion for he hea equaion. I is indeed a mollifier as described above. Here are some plos of ψ(x, ) as a funcion of x for various. When = 0.04,, and 5 we have

12 .5 = 0.04 y = = x 0 As ges small we should ge somehing like a dela funcion, which appears o be happening. From he previous argumens, we now know ha he funcion u(x, ) defined by equaion () wih ψ(x, ) defined by equaion (6) saisfies he hea equaion wih he proper iniial condiion. This is exacly u(x, ) = π e (x y) 4 f(y) dy. I s worh noing ha he Green s funcion ψ(x, ) is in some sense he soluion o he hea equaion wih iniial condiion ψ(x, 0) = δ(x), where δ(x) is he usual Dirac dela funcion, an infinie spike a posiion zero. In he language of engineering ψ(x, ) is he impulse response of his sysem. The figure above does a good job of illusraing he naure of ψ(x, ). Consider ψ(x, ) for a very small i s very high, bu hin he hea from he dela funcion spike a x = 0 has jus begun o spread ou. As increases he hea coninues spreading ou, and ha s exacly wha you see in he figure. Exisence and Regulariy of Iniial Daa Equaion () shows ha a soluion o he hea equaion exiss. Wha do we need from he iniial daa f in order for he inegral defining u(x, ) o make sense? Cerainly if f is bounded and coninuous, he inegral is

13 perfecly well-defined; moreover, he funcion e x 4 and all of is derivaives decay so very rapidly as x, slipping and operaors under he x inegral won be an issue (so u really will saisfy he hea equaion). So we have our exisence resul: Theorem The hea equaion wih bounded coninuous iniial daa f(x) has a soluion for < x < and > 0, defined by equaion (), and u(x, ) saisfies u(x, 0) = f(x) in he sense ha lim u(x, ) = f(x). 0 + Acually, you can pu in funcions f which aren coninuous or bounded; all you really need is ha f doesn grow oo fas (faser han e x ) or be so disconinuous ha he inegral doesn make sense. Problems Show ha all x and derivaives (mixed or no) of any order of he funcion ψ(x, ) = π e x 4 are jus sums of erms of he form c k+/ x m x π e 4 for inegers k and m and some consan c. Show ha any derivaive of ψ(x, ) of any order in x and/or approaches zero as x. 0.4 Ineresing Properies of he Soluion The hea equaion has some ineresing qualiaive feaures ha are quie differen from, for example, he wave equaion. Speed of Propagaion/Causaliy Recall ha for he wave equaion informaion propagaed a a finie speed. Bu look a equaion () suppose ha he iniial condiion f(x) is zero everywhere excep on some small inerval (a, b). Because he exponenial funcion is never zero, he inegral ypically won be zero, even for small and 3

14 x arbirarily far from (a, b). More generally, he iniial condiion f affec he soluion u(x, ) for all x, no maer how small is. The hea propagaes wih infinie speed. Smoohing Properies (x y) e Firs, recall ha a funcion is said o be C if i is infiniely differeniable you can differeniae as many imes as you like and he derivaives never develop corners or disconinuiies. The funcion 4 is such a funcion, π wih respec o all variables. Also, for any > 0 his funcion decays o zero VERY rapidly away from x = y (see he las problem). As a resul, we can differeniae boh sides of equaion () wih respec o x as many imes as we like, by slipping he x derivaive inside he inegral o find n u (x, ) = xn π n (e (x y) x n 4 ) f(y) dy. The inegral always converges (if, e.g., f is bounded). In shor, no maer wha he naure of f (disconinuous, for example) he soluion u(x, ) will be C in x, for any ime > 0; he iniial daa is insanly smoohed ou! Conras his o he wave equaion imagine a square wave moving down he x axis he disconinuiy will never dissipae, i jus propagaes along a some fixed speed. 0.5 Uniqueness and Sabiliy The L Norm For he wave equaion we measured how close wo funcions f and g are on some inerval a < x < b (eiher a or b can be ± ) using he supremum norm f g = sup a<x<b f(x) g(x) (we can also alk abou he size or norm of a single funcion, as f = sup a<x<b f(x) ). There s anoher common way o measure he disance beween wo funcions on an inerval (a, b), namely ( b / f g = (f(x) g(x)) dx). a 4

15 We can also measure he norm of a funcion wih ( b / f = (f(x)) dx). a The laer quaniy is called he L norm, so he disance beween wo funcions can be measured as he L norm of heir difference. I s easy o see ha if f = 0 hen f mus be he zero funcion, and f g = 0 implies f = g. The L norm is very much like he usual Pyhagorean norm for vecors from linear algebra (or jus Calc I), which for a vecor x =< x,..., x n > is defined as / n x =. x j j= The disance beween wo vecors or poins x and y in space is hen jus x y. I urns ou ha he L norm is someimes more suiable for measuring he disance beween wo funcions. I s wha we ll use o derive sabiliy resuls (and uniqueness) for he hea equaion. Uniqueness and Sabiliy If u(x, ) saisfies he hea equaion, define he quaniy E() = u (x, ) dx, (assuming he inegral is finie) similar o he energy for he wave equaion. I s no clear if E() above is physically any kind of energy, bu I don care; i can be used o prove uniqueness and sabiliy for he hea equaion. I s easy o compue de d = u(x, )u (x, ) dx, provided u and u decay fas enough a infiniy (and are, say, coninuous) which we ll assume. Replace u in he inegral wih u xx (since u u xx = 0) o find de d = u(x, )u xx (x, ) dx. 5

16 Now inegrae by pars (ake an x derivaive off of u xx, pu i on u) o find de d = lim u x(x, )u(x, ) lim u x(x, )u(x, ) u x x x(x, ) dx. We ll assume ha u limis o zero as x in fac, we ll acually impose i as a requiremen! Assuming u x says bounded, we can drop he endpoin limis in de/d above (alernaively, we could assume u x limis o zero and u says bounded). We obain de d = u x(x, ) dx. (7) I s obvious ha de/d 0, so E() is a non-increasing (and probably decreasing) funcion of. Thus, for example, E() E(0) for > 0. Of course E() is simply he L norm of he funcion u(x, ), as a funcion of x a some fixed ime, and we ve shown ha his quaniy canno increase in size over ime. Here s a bi of noaion: we use u(, ) o mean he L norm of u as a funcion of x for a fixed value of. Wih his noaion we can formally sae he following Lemma: Lemma Suppose ha u(x, ) saisfies he hea equaion for < x < and > 0 wih iniial daa u(x, 0) = f(x). Then u(, ) f. We can wipe ou uniqueness and sabiliy a he same ime. Consider wo soluions u and u o he hea equaion, wih iniial daa f and f. Form he funcion u = u u, which has iniial daa f f. Apply Lemma o u and you immediaely obain Theorem If u and u are soluions o he hea equaion for < x < and > 0 wih iniial daa u (x, 0) = f (x) and u (x, 0) = f (x), and boh u and u decay o zero in x for all > 0, hen u (, ) u (, ) f f. To prove uniqueness ake f = f o conclude ha a all imes > 0 we have u (, ) u (, ) = 0, or (u (x, ) u (x, )) dx = 0. This forces u u = 0, so u (x, ) = u (x, ) for all x and. 6