XIII. DIFFERENCE AND DIFFERENTIAL EQUATIONS Ofen funcions, or a sysem of funcion, are paramerized in erms of some variable, usually denoed as and inerpreed as ime. The variable is wrien as a funcion of ime in one of wo formas: x = or x = x The variable belongs o some index se T, usually he nonnegaive inegers or nonnegaive reals. When we alk abou how hese sysems evolve wih ime, we have a sysem of difference equaions when he paramerizaion is discree (T = Z + ) or a sysem of differenial equaions wih coninuous paramerizaion ( T = R + ). These look like, respecively: +1 = f ( ) and: d d = g( ) Economiss ofen prefer o work wih difference equaions. We look a how variables in he economy progress from one year o he nex, which we migh describe wih he sysem of equaions: c +1 = f c,k k +1 = g c,k Nex year s consumpion can be deermined by a funcion of oday s consumpion and oday s capial sock. The same goes for nex year s capial sock. A firs order difference equaion conains only variables of one ime period on he righ-hand side; in conras, a second order difference equaion conains wo periods: c +1 = f ( c,k,c!1,k!1 ) Similarly, an n-h order difference equaion has he sae of variables for he pas n periods deermining he nex value. When I wrie diffeqs (pronounced diffy queues ) I am saing somehing abou boh difference and differenial equaions. Diffeqs are also known as laws of moion or evoluionary laws. Firs-order linear differenial equaions are he simples class of difference equaions o work wih. They look like: +1 = a + b The differenial equaion is called homogeneous when b = 0 and ime-invarian when a = a, b = b are consans. Yeserday we looked a a sysem of homogeneous, ime-invarian, firs-order linear differenial equaions. Here s a simple one: Fall 2007 mah class noes, page 99
k +1 = k! "k = ( 1! " )k Nex year s capial sock is he same as his year s, minus some consan depreciaion. Then we can forecas he capial sock wo years in he fuure: k +2 = ( 1! " )k +1 = ( 1! " ) 2 k In fac, we can say ha he capial sock n years ino he fuure will be: k +n = ( 1! " ) n k = ( 1! " ) +n k 0 where k 0 is he value of he capial sock a he beginning of ime. Firs-order diffeqs can be solved subjec o one iniial value. Tha is, provided we know wha he capial sock was a =0, we can predic i for he res of eerniy. The given value doesn necessarily have o be iniial we could have been given he value of any k and we could solve he sysem a all. Given he general firs-order difference equaion, he general soluion is:!1!1!1 = (" a s=0 s ) x 0 + # (" a s=r s )b r r =0 The soluion is much easier o express if we have consans for a and b, bu his isn always he case. Example: I sar my life wih A 0 in asses. Each year, asses grow wih a a rae of!. Each year I make invesmens (or wihdrawals, if negaive) of i, which we ake as exogenous (no depending on asses): A +1 = ( 1 +!) A + i Solve his in erms of A 0 andi. Sysems of linear differenial equaions can be expressed in marix form, in one of hese manners: 1! +1 $ # 2 & = a 11 12! a # 21 22 & # " +1 % " a a % "! x 1 ( + 1) # & = a 11 # " x 2 ( + 1) % " a 21 1 2 $ & + b 1! $ # 2 & % " b % a 12 a 22 & # % " eiher of which I will wrie compacly as: x 1 x 2 or: & + b 1 $ # & % " b 2 % +1 = A + b The soluion here is: Fall 2007 mah class noes, page 100
Example:!1 = ( A!1 A!2 A 2 A 1 A 0 )x 0 + " ( A!1 A!2 A s+2 A s+1 A )b s=0 s s The capial sock evolves according o some equaion: k +1 = ( 1+! " n)k + i Whereas populaion growh can be described as: N +1 = ( 1+ n) N +!k Solve his sysem in erms of k 0, N 0, and i. If hei = 0, wha can you say abou sabiliy? A firs-order linear difference equaion is he bes ha you can hope for. There are ricks for making hem homogeneous when hey are no. I m no going o ge ino hose. Many equaions in macro will be second-order difference equaions. In fac, you ll become very well acquained wih his one: f! " U! k ( f ( k ) # k +1 ) =! ( # k ) U f! k #1 This defines k +1 as a funcion (a leas implicily) of k and k!1. A is bes, his equaion will ake he form:!" (!k #1 # k ) =!k # k +1 $ k +1 =!( 1 + ")k #! 2 "k #1 Second-order differenial equaions are a bich o solve. Even his one, a secondorder, ime-invarian, homogenous linear difference equaion (as nice as hey ge), is ricky enough. Second-order diffeqs require wo iniial values o solve (ofen one of hese will be he acual iniial value and he oher he erminal value). We solve hem using a highly scienific mehod known as guess and verify, or more euphemisically, as he mehod of undeermined coefficiens. The idea is ha, given a difference equaion like: +1 = f (,!1 ), we can guess ha here is some funcional form g so ha and = g(!1 ). We inver he second of hese,!1 = g!1 +1 = f,!1 =?? g = f, g!1 ( ), sick i ino f, +1 = g and see wheher his guess gives us back ha +1 = g has he same funcional form. If i does, our guess is good. Le s go back o our nices possible second-order equaion: Fall 2007 mah class noes, page 101
k +1 =!( 1 + ")k #! 2 "k #1 My guess is going o be ha we can wrie each k as jus a consan imes he preceding one: k =! k "1 This means ha k!1 = 1 " k, which we subsiue back ino our difference equaion. k +1 =!( 1+ ")k #! 2 " 1 k $ % k +1 = (! ( 1 + ") #! 2 " 1 $ )k If his is rue, hen all he suff in fron of k mus also equal!.! = "( 1 + # $ "# 1! ) %! 2 " $ "#! + " 2 # = (! $ "#)(! $ ") = 0 We have deermined ha wo roos possibly saisfy his equaion; his effecively confirms ha he linear guess was correc. We should sill check ha hese are sensible soluions. This was he paricular funcional form ha solved his paricular problem. For oher difference equaions, you need o use oher guesses. If you are familiar wih he paricular funcional forms involved, you migh are more likely o guess correcly. If you re uncerain, check a book like Sydsæer, Srøm, and Berck for some recommendaions. In difference equaions, we re considering funcions ha essenially (afer some rearranging) look like: +1! = f ( )! Le s call he lef hand side of his equaion!, and define he righ hand side as some funcion g. Then a difference equaion can always be expressed as:! = g( )(!) The change in ime is! = ( + 1) " = 1, so muliplying he righ hand side by his changes nohing. This looks like a lo like a firs-order approximaion, which is exacly wha a difference equaion is. Taking he limi of his as! " 0, we can wrie his as a coninuous ime differenial equaion: d = g( )( d)! d d = g( ) Differenial equaions are always expressed a he derivaive of a variable wih respec o ime. Someimes his is denoed by d d, someimes by x!, and very ofen by!. Fall 2007 mah class noes, page 102
As wih difference equaions, he simples a firs-order, linear, ime-invarian, homogeneous equaion is he simples differenial equaion possible o solve. These equaions look like:! = a! x" = ax Wha funcion is is own derivaive? Tha s e y. Acually, i s Ce y, where C is any consan. Wha funcion imes a consan (say,! ) is is own derivaive? Well, ha s Ce! y. And ha s exacly he soluion o his differenial equaion.! = a! = Ce a = x 0 e a (You should probably remember his formula.) C is called a consan of inegraion, which equals he iniial value x 0 a =0. If you don know an iniial value, you should leave C in as an undeermined coefficien. More generally, given any linear firs order differenial equaion of he form:! = a + b where a and b are consans ha do no depend on x, he soluion o his problem should saisfy: = C exp (!" a ( s )ds 0 ) + b( r)exp! " ( " a( s)ds 0 r ) dr Higher order and nonlinear differenial equaions are messy o solve (bu generally, more solvable han difference equaions), and sysems of equaions require some ricks. There are formulas for a few soluions, bu ofen you have o employ a bunch of fun echniques o ge hem o work ou. Go look hem up if you ever need hem (and hopefully you won ). Solving diff. eq.s is difficul, bu forunaely, we re ofen no ineresed in he acual soluions. Consider an economy, he capial sock and consumpion of which can be described by he pair of differenial equaions:!k = Ak! " (# + n)k " c U $!c = c % & U ( c ) c $$ c ' ( %& A( 1"! )k! "1 " (# + n) " ) ' ( Being able o solve hese would allow us o say, given ha we know Albania s consumpion and capial sock oday, in he year 2086 hey will be such-and-such. This may be imporan in iself. However, no being able o solve hem explicily, we can sill answer many quesions abou he growh (or decline) of he variables. Growh of capial sock and sandards of living are fundamenal issues in Fall 2007 mah class noes, page 103
developmen. Growh of money socks and real money balances are cenral o moneary heory. A sock variable or sae is like he variable. In conras, he variable! is called a flow variable. Think of a pond being filled wih waer. The sock a any ime is how much waer is in he pond (gallons or cubic meers of waer); he flow is he rae a which he seam feeds ino i (measured in gallons per minue). When looking a a pair of differenial equaions in which each sock is an inpu ino he flow of each variable (as wih capial and consumpion, above), we ofen consruc phase diagrams in order o idenify regions in which each is growing or declining. There are a number of seps in his. 1. Draw a wo-dimensional graph, wih each sock variable on one axis. 2. For each variable, idenify under wha condiions here is no change in he sock. You do his by seing! = 0 and solving he equaion for one variable in erms of he oher. Plo his funcion (or correspondence). This is called an x-consan locus. 3. The x-consan locus or loci separae he graph ino several regions. In each of hese regions, figure ou wheher x is increasing or decreasing (all you need do is evaluae a one poin in each region). In he region, skech in some arrows parallel o he x-axis poining in he appropriae direcion. 4. Overlay he wo graphs. You ve now picked ou regions where x is growing and y is growing, x is growing and y is declining, and such. c Le s look a he example from he previous page. Firs we consruc he graph wih capial and consumpion on is axes. Then we pick ou capial, and se he law of moion for k! equal o zero. This solves easily for c as a funcion of k, and we skech in hese poins a which he capial sock is unchanging. [Elsewhere, we do he same hing for he law of moion for!c, and we idenify he locus along which consumpion k = 0! c = Ak " # ( $ + n)k k is unchanging.] Now we need o pick ou a poin above he k-consan locus and one below he locus o ell us he direcion of changes in he capial sock for each he Fall 2007 mah class noes, page 104
wo regions. I m going o pick one jus off he c axis and one jus off he k axis, since c k = 0 ha ll allow me o essenially se one of he variables equal o zero in each case. Evaluaing he funcion:!k = Ak! " (# + n)k " c k a c = ĉ > 0 and k =! " 0, I find ha k! =!ĉ < 0. Everywhere above he locus, capial sock is decreasing. I pencil in some lile arrows going o he lef. Then I evaluaion he funcion a c =! " 0 and k = ˆk. I find ha!k = A ˆk! " (# + n) ˆk, which looks like i should be posiive, as long as ˆk isn oo big (presumable, oo big is where i inersecs he k-axis a second ime. I skech in my arrows. In he oher picure, I do a similar hing for consumpion. Finally, I combine he wo picures, giving me somehing ha looks like his: c k = 0 c = 0 k And here you have idenified he direcions of change in he variables, so you can also race ou pahs for how hey follow, from a paricular saring poin. Knowing Albania s curren saes, we migh pu hem in he boom lef region, and hen predic ha hey follow a pah of growing consumpion and capial sock for a while bu hen hey cross over he op of he k-seady locus, and consumpion coninues o grow, bu no he capial sock (maybe hey are selling off heir asses o buy bigger and beer cars). This is he usefulness of phase diagrams. All you need is a pair of differenial equaions o skech hem. Fall 2007 mah class noes, page 105