13 Differetial Equatios the aswer to your questio ca best be epressed as a series of partial differetial equatios... Castiel, Superatural, Seaso 6, Episode 18 A differetial equatio is a mathematical equatio for a ukow fuctio of oe (ODE-Ordial DE) or several (PDE-Partial DE) variables that relates the values of the fuctio itself ad its derivatives of various orders. Differetial equatios play a promiet role i egieerig, physics, ecoomics, biology, ad other disciplies. May real life processes ca be described as differetial equatios. Oe of reasos is a way we lear about the process: we ivestigate a process by observig its chage ad a rate of the chage. I this course we preset a quick survey of idea of differetial equatios i a utshell. 13.1 Modelig DE (7.1) 13.1.1 Model of populatio growth Cosider a populatio problem i ideal coditio (ulimited eviromet ad resources, o disease or predator). I such situatio we may assume that the populatio growth is proportioal to the populatio size. Sice the rate of populatio growth is the first derivative of the populatio size this ca be epressed as dp dt = kp, where k is the proportioality costat. If k>0 ad iitial populatio is >0 we have P' ( t) > 0, i.e. the populatio is always icreasig, furthermore the P' ( t) icreases as the populatio icreases. The solutio to P' ( t) = k( Ce kt ) = kp t dp dt = kp is epoetial fuctio, i.e. P ( t ) =, ote that Cekt ( ). The costat C is positive because the ature of the problem (egative populatio make o sese). We also cosider t=0 as iitial time ad iterested i t>0. Note that P 0 ( ) = Ce k 0 = C, so C is the iitial populatio. Cosider ow more complicated model, where the populatio caot grow above specific size M (due to limited resources, for eample). I this case we look for P' ( t) > 0 while P<M ad P' ( t) < 0 whe P>M. Thus we got dp dt = kp 1 P M. This
equatio is kow as logistic equatio, we will lear to solve it if we get to sectio 7.5. Logistic DE have equilibrium solutios P=0 ad P=M, from which it ever chage. Whe P>M P' ( t) < 0 for 0<P<M P' ( t) > 0 as epected. 13.1. Model of the motio of a sprig By Hook s Law, whe we release a sprig beig stretched it eerts force =-k, from the other side, by Newto s Secod Law, the force is mass times acceleratio. Thus = k. The secod derivative if is proportioal to with opposite m sig, therefore ca be either si t, cos t or their combiatio. The presece of trigoometric fuctios i the solutio is t surprisig, sice the sprig is supposed to oscillate about its equilibrium positio. m d dt = k d dt 13.1.3 Geeral DE I DE we cosider y (y()) a depedet variable ad idepedet variable (sice the equatio should hold for ay value of ). I may physical problems idepedet variable is time, therefore i DE s it ofte amed t, e.g. y (t) = 3y(t). The order of DE is the highest derivative that occurs i equatio, thus it ca be formed as followig: First order DE: y =f(,y) (read y ()=f(,y()). Secod order DE: y =f(,y,y ) ( ( ) ) N th order DE: y ( ) = f, y, y',..., y 1 Eamples: -liear DE y ( 3) = ay'' ( ) + by' ( ) + cy( ) + d - o liear y'' = y ( ) Whe we solve DE s we ofte have a family of solutio rather tha a sigle solutio, for eample E 1. y' = cos have ifiite umber of solutios si + C oe per a costat C.
I may physical (or so) problems we ofte eed a particular solutio that satisfies a iitial coditio y( t 0 ) = y 0 or a set of IC y ( ) ( t 0 ) = y,0. The umber of iitial coditio is equal to order of DE. Such problem is called a iitial value problem. E. Solve y'' = 0 with respect to iitial coditios y( 1) =, y' ( 1) = 1. From y'' = 0 we lear that y is a lie, i.e. y = a + b, applyig ICs oe get y( 1) = a 1+ b = a + b = ad y' ( 1) = b = 1 a = 1, thus y = +1(verify). Aother useful sort of problems called boudary value problems, i this case some of coditios are give at the ed poit of iterval of iterest t 1. The total umber of coditios is still equal to the order of DE. E 3. Solve y'' = 0 with respect to iitial coditios y( 0) = 1, y( 1) = 1. We apply BC to y = a + b to get y( 0) = a 0 + b = 1 b = 1 ad y( 1) = a 1+1 = 1 a =, thus y = +1(verify). 13. Directio Fields ad Euler Method (7.) Ufortuately, oly certai DE s ca be solved i sese of obtaiig eplicit formula for the solutio. Still we ca lear a lot about the solutio. 13..1 Directio Fields DF is a method to sketch the solutio to equatio y' = f (, y), without solve it. Such, graph provide importat iformatio about solutio s behavior. Note, that we do ot look at iitial or boudary coditio yet, meas we lookig at the family of solutio to DE. The idea is to defie a grid ad to draw a o crossig little lies or arrows with a slope m = f (, y) at each grid poit. The solutio to particular problem y' = f (, y) with IC ( ) = y 0 is a thread o a DF graph. y 0 E 4. y' = y
E 5. y' = ( y y ) ( 1 y) = ( y ) ( y +1) ( 1 y) y < 1 y' > 0 (e.g. y = y' = 36 ) 1 < y < 1 y' < 0 (e.g. y = 0 y' = ) 1 < y < y' < 0 (e.g. y = 1.5 y' = 0.31) y > y' > 0 (e.g. y = 3 y' = 16 ) 0 15 10 5 - -1 1 3 13.. Euler Method Cosider problem y' = f ( t, y) with respect to IC y( t 0 ) = y 0. Let h = Δt, deote t = t 0 + h, ad y( t ) = y, the we itegrate both sides of the equatio to get y( ) y( ) = f t, y( t) y' ( t)dt = f ( t, y( t) )dt. Evaluatig the itegral of the left side oe get ( )dt or y k+1 = y k + f ( t, y( t) )dt rectagular rule to get y k+1 = y k + ( ) f, y( ) the last itegral oe solve usig ( ) = y k + hf, y k ( ) which is called Euler scheme or method. Differet approimatios of the itegral lead to differet schemes, but this is out of the scope of this course. E 6. Let solve y' = y y 0 ( ) = 1 (the eact solutio is e ) The Euler method gives us y y hf ( y ) y hy y ( h) = + + 1, = + = 1 +
y 0 = 1 1 0 1 ( 1 ) y = y + h ( 1 ) ( 1 ) y = y + h = + h Thus y ( 1 h) = +, chage h h y = 1+ h = 1+ h e = to get ( ) h ( ) 1/ h 0 E 7. Use Euler method to approimate y ( 1) usig h = 1,1 / for y( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) h= 1: y = y + ht 1 y = y = 1 1 0 0 0 0 y = y = y + h = y 0+ 1 = 1 1 1 0 h= y = y + ht y = + h = h = = 1/: 1 1 1 1 1 1 1 1 1/4 3/4 y = y = y + h = y 1 = 3 / 4 0.75 0 1 the aalytical solutio is ( 1 t + e ) y' + ty= t 0 = 1, so 1 y( ) ( e) eample that i order to get precise result oe eed very small h. Further more, whe we lookig for y( t 1 ) where t 1 >> t 0 we will eed may iteratios to get good approimatio. 1 = 1+ 1/ 0.684. Oe lear from the