1Intro tomathmodelingIntroductiontoregularperturbationtheoryVery often, a mathematical problem cannot be solved exactly or, if the exact solution isavailable, it exhibits such an intricate dependency in the parameters that it is hard to useas such.It may be the case, however, that a parameter can be identified, say e, such thatthe solution is available and reasonably simple for e = 0. Then, one may wonder how thissolution is altered for non-zero but small e. Perturbation theory gives a systematic answerto this question.Perturbation theory for algebraic equations.Consider the quadratic equationx-1=Ex.(1)The two roots of this equation areXi=e/2+V1+e2/4, X2 =e/2 -/1+e2/4.(2)For small e,these roots are well approximated by the firstfew terms of their Taylor seriesexpansion (seefigure 1)(3)X =1+e/2+e2/8+0(e3), x2 =-1+=/2 -e2/8+0(e3)Can we obtain (3)without priorknowledge of the exact solutions of (1)? Yes, using regularperturbation theory. The technique involves four steps.STEP A. Assume that the solution(s) of (1) can be Taylor expanded in . Then we have(4)X= Xo +eXI +e?X2+ O(e3),forXo, Xi,X2 tobedetermined.STEP B. Substitute (4) into (1) written as x? -1 - Ex = 0, and expand the left hand sideof theresultingequation inpower series ofe.UsingX = X + 2eXoX +e2(X+ 2XX2) + O(e3),(5)Ex=eXo+e2Xi + O(e3),'a(e) = O(b(e) as → 0, (a(e) is big-oh of b(e)")if there exists a positive constant M such that |a(e)/ ≤ M|b(e)wheneveris sufficientlycloseto0
Intro to math modeling 1 Introduction to regular perturbation theory Very often, a mathematical problem cannot be solved exactly or, if the exact solution is available, it exhibits such an intricate dependency in the parameters that it is hard to use as such. It may be the case, however, that a parameter can be identified, say ε, such that the solution is available and reasonably simple for ε = 0. Then, one may wonder how this solution is altered for non-zero but small ε. Perturbation theory gives a systematic answer to this question. Perturbation theory for algebraic equations. Consider the quadratic equation x2 − 1 = εx. (1) The two roots of this equation are x1 = ε/2 + q 1 + ε2/4, x2 = ε/2 − q 1 + ε2/4. (2) For small ε, these roots are well approximated by the first few terms of their Taylor series expansion (see figure 1)1 x1 = 1 + ε/2 + ε2 /8 + O(ε3 ), x2 = −1 + ε/2 − ε2 /8 + O(ε3 ). (3) Can we obtain (3) without prior knowledge of the exact solutions of (1)? Yes, using regular perturbation theory. The technique involves four steps. STEP A. Assume that the solution(s) of (1) can be Taylor expanded in ε. Then we have x = X0 + εX1 + ε2X2 + O(ε3 ), (4) for X0, X1, X2 to be determined. STEP B. Substitute (4) into (1) written as x2 − 1 − εx = 0, and expand the left hand side of the resulting equation in power series of ε. Using x2 = X2 0 + 2εX0X1 + ε2 (X2 1 + 2X0X2) + O(ε3 ), εx = εX0 + ε2X1 + O(ε3 ), (5) 1a() = O(b()) as ε → 0, (“a(ε) is big-oh of b()”) if there exists a positive constant M such that |a(ε)| ≤ M|b(ε)| whenever ε is sufficiently close to 0
2Regular perturbation theory2.5x1261.50.51.50EFigure 1:The root Xi plotted as a function of (solid line), compared with theapproximations by truncation of the Taylor series at O(e2), xi = 1 + /2 (dotted line), andO(e3), xi =1 + e/2+?/8 (dashed line). Notice that even though the approximations area priori valid in the range e< 1 only, the approximation Xi =1 + e/2 +X /8 is fairly goodeven up to = = 2.this givesX - 1 + E(2XgX1 - X) +e2(X +2XoX2 - X)) + O(e3) = 0(6)STEP C.Equate to zero the successive terms of the series in the left hand side of (6):0(e0) :X -1= 0,O(el) :2 XoX - Xo = 0,(7)0(e) :X + 2XoX2 - X1 = 00(e3) :STEP D. Successively solve the sequence of equations obtained in (7). Since X? -1 = 0has two roots, Xo =±1, oneobtainsXo = 1,Xi = 1/2,X2 = 1/8,(8)Xo = -1.X = 1/2,X2= -1/8.Itcanbechecked that substituting (8)into (4)onerecovers (3)From the previous example it might not be clear what the advantage of regularperturbation theoryis,since one canobtain (3)moredirectlybyTaylorexpansion oftherootsin (2).To see the strengthof regular perturbation theory,considerthefollowing equation(9)x-1=eer
2 Regular perturbation theory 0 0.5 1 1.5 2 1 1.5 2 2.5 ε x1 Figure 1: The root x1 plotted as a function of ε (solid line), compared with the approximations by truncation of the Taylor series at O(ε2), x1 = 1 + ε/2 (dotted line), and O(ε3), x1 = 1 + ε/2 + ε2/8 (dashed line). Notice that even though the approximations are a priori valid in the range ε 1 only, the approximation x1 = 1 + ε/2 + x2/8 is fairly good even up to ε = 2. this gives X2 0 − 1 + ε(2X0X1 − X0) + ε2 (X2 1 + 2X0X2 − X1) + O(ε3 ) = 0. (6) STEP C. Equate to zero the successive terms of the series in the left hand side of (6): O(ε0) : X2 0 − 1 = 0, O(ε1): 2X0X1 − X0 = 0, O(ε2) : X2 1 + 2X0X2 − X1 = 0, O(ε3) : ··· (7) STEP D. Successively solve the sequence of equations obtained in (7). Since X2 0 − 1 = 0 has two roots, X0 = ±1, one obtains X0 = 1, X1 = 1/2, X2 = 1/8, X0 = −1, X1 = 1/2, X2 = −1/8. (8) It can be checked that substituting (8) into (4) one recovers (3). From the previous example it might not be clear what the advantage of regular perturbation theory is, since one can obtain (3) more directly by Taylor expansion of the roots in (2). To see the strength of regular perturbation theory, consider the following equation x2 − 1 = εex. (9)
3Intro to math modelingx12.521.50.10.20.30.40.51&Figure 2:The solid line is thegraph of two (why two?)of the three solutions of (9)obtained numerically and plotted as a function of (solid line).Also plotted are theapproximations by truncation of the Taylor series at O(2), xi = 1 + ee/2 (dotted line), andO(3), Xi = 1 + Ee/2 + 2e2 /8 (dashed line) (see (14).The solutions of this equation are not available; therefore the direct method is inapplicablehere.However, the Taylor series expansion of these solutions can be obtained byperturbation theory. We introduce the expansion (4) as in Step A. In Step B, we use (recallthat e? = 1 + z + z /2 + O(2)e' = ee0+eXi+e*x+(e) = ee*0eXi+-x+(e) = ee0 +e2Xie%+ O(e).(10)Substituting this expression in (9)written as x?_1-eer=0 and using (5),weobtainX -1+e (2XoXi -e%) +e2 (X +2XoX2 - Xie0) + 0(e3) = 0(11)Thus,the sequence ofequations obtained in Step C is0(e) :X -1 = 0,0(el) :2XoX1 -eXo = 0,(12)0(e2) :X + 2XoX2 - Xiexo = 0,0(e3) :fromwhichweobtain (stepD)Xo = 1,Xf = e/2,X2 = e2 /8,(13)X2 = -1/(8e2),Xo = -1,Xi = -1/(2e)
Intro to math modeling 3 0 0.1 0.2 0.3 0.4 0.5 1 1.5 2 2.5 3 ε x1 Figure 2: The solid line is the graph of two (why two?) of the three solutions of (9) obtained numerically and plotted as a function of ε (solid line). Also plotted are the approximations by truncation of the Taylor series at O(ε2), x1 = 1 + εe/2 (dotted line), and O(ε3), x1 = 1 + εe/2 + ε2e2/8 (dashed line) (see (14)). The solutions of this equation are not available; therefore the direct method is inapplicable here. However, the Taylor series expansion of these solutions can be obtained by perturbation theory. We introduce the expansion (4) as in Step A. In Step B, we use (recall that ez = 1 + z + z2/2 + O(z3)) εex = εeX0+εX1+ε2X2+O(ε3) = εeX0 e εX1+ε2X2+O(ε3) = εeX0 + ε2X1eX0 + O(ε3 ). (10) Substituting this expression in (9) written as x2 − 1 − εex = 0 and using (5), we obtain X2 0 − 1 + ε 2X0X1 − eX0 + ε2 X2 1 + 2X0X2 − X1eX0 + O(ε3) = 0. (11) Thus, the sequence of equations obtained in Step C is O(ε0) : X2 0 − 1 = 0, O(ε1): 2X0X1 − eX0 = 0, O(ε2) : X2 1 + 2X0X2 − X1eX0 = 0, O(ε3) : ··· (12) from which we obtain (step D) X0 = 1, X1 = e/2, X2 = e2/8, X0 = −1, X1 = −1/(2e), X2 = −1/(8e2), (13)
4Regularperturbation theory0.5y(t)0.2500.511.50Figure 3:The exact solution (17) (solid) line is compared with the approximations bytruncation of the Taylor series (see (18)at O(e) (dotted line), O(e2) (dashed line), and O(3)(indistinguishable from solid line).or,equivalentlyX1 =1+Ee/2+e?e/8+0(e3)(14)X2 = -1 - E/(2e) - e2 /(8e) + O(e3)The expression for Xi is compared to the numerical solution of (9) on figure 2.Remark:Infact (9) has three solutionsfor 0E1.The solution which exists for all > 0 is the one with expansion given in x2 in (14); thesolution with the expansion given in xi in (14) disappears for > Ei; and the third solution(see figure 2: the solid line is the graph of a two-valued function) cannot be obtained byregular perturbation.Exercise 1. Solve by perturbationx2- 4=eln(x).(15)Notice that, as → 0, (15) formally reduces to the equation x2 - 4 = 0, with two rootsXi,2 = ±2. (15) also has two solutions (why?). How are they related to X1,2 = ±2? Canboth solutions of (15) be obtained by perturbation?Perturbationtheoryfordifferential equations.Consider%(0) =1.dydy-1,y(0) = 0,(16)d-2drdt
4 Regular perturbation theory 0 0.5 1 1.5 2 0 0.25 0.5 τ y( )τ Figure 3: The exact solution (17) (solid) line is compared with the approximations by truncation of the Taylor series (see (18)) at O(ε) (dotted line), O(ε2) (dashed line), and O(ε3) (indistinguishable from solid line). or, equivalently, x1 = 1 + εe/2 + ε2e2/8 + O(ε3) x2 = −1 − ε/(2e) − ε2/(8e2) + O(ε3). (14) The expression for x1 is compared to the numerical solution of (9) on figure 2. Remark: In fact (9) has three solutions for 0 ε1. The solution which exists for all ε > 0 is the one with expansion given in x2 in (14); the solution with the expansion given in x1 in (14) disappears for ε>ε1; and the third solution (see figure 2: the solid line is the graph of a two-valued function) cannot be obtained by regular perturbation. Exercise 1. Solve by perturbation x2 − 4 = ε ln(x). (15) Notice that, as ε → 0, (15) formally reduces to the equation x2 − 4 = 0, with two roots x1,2 = ±2. (15) also has two solutions (why?). How are they related to x1,2 = ±2? Can both solutions of (15) be obtained by perturbation? Perturbation theory for differential equations. Consider d2 y dτ 2 = −ε dy dτ − 1, y(0) = 0, dy dτ (0) = 1. (16)
5Intro to math modelingRecall that this equation governs the dynamics of a projectile thrown vertically into the airif air friction is taken into account.Here e=kv/(mg), where V is the initial velocity ofthe projectile, m is its mass, k is the friction constant ([K] = MT-1) and g is the accelerationfrom gravity. In (16), the altitude is measured in units of l = V /g, and the time in units ofTi = V/g.Theexactsolution of (16)is (canyou showthis?)(1 +) (1 - e-eT) -J(r)= (17)E2For o < 1, which corresponds to a situation where air friction is small, we canapproximate y(t) by the first few terms of its Taylor series expansion in (see figure 3).Using 1 - e-z = z -z /2 +2 /3! + O(z), we obtain(18)y(T) = T - T2 /2 +E(-T2 /2 + +3 /6) + ε2(-3 /6 - T4 /24) + O(e3).This expansion is uniformly valid in T in the range 0 < T < 1/e, which is the range ofphysical interest since the projectile hits the ground well before 1/e if e< 1 (see below)Wewishtoobtainthe expansion in (18)withoutprior knowledge of the exactsolution (17), using regular perturbation theory.We proceed similarly as for algebraicequations.STEP A. Introduce the expansiony(T) = yo(T) +Ey1(T) +? y2(T) + O(e3)(19)where yo(+), yi(+), y2(+) are functions of to be determinedSTEPB.Substitute (19)into (16)(differential equation and initial conditions)written asy+e+1=0,%(0) -1 = 0,y(0) = 0,dr2+drdrand expand the resulting equations in power series of e. This givesdyo(dyi+dyo)d'yzdyi+ 0(e3) = 01+8d-2d-2+"drd-2R(20)yo(0) + Eyi(0) + 2 y2 (0) + O(e3) = 0,(0) -1+() +些(0) + 0(e)=0.dTdrdr
Intro to math modeling 5 Recall that this equation governs the dynamics of a projectile thrown vertically into the air if air friction is taken into account. Here ε = kV/(mg), where V is the initial velocity of the projectile, m is its mass, k is the friction constant ([k] = MT−1) and g is the acceleration from gravity. In (16), the altitude is measured in units of `1 = V2/g, and the time in units of τ1 = V/g. The exact solution of (16) is (can you show this?) y(τ ) = (1 + ε) ε2 1 − e −ετ − τ ε . (17) For 0 < ε 1, which corresponds to a situation where air friction is small, we can approximate y(τ ) by the first few terms of its Taylor series expansion in ε (see figure 3). Using 1 − e−z = z − z2/2 + z3/3! + O(z4), we obtain y(τ ) = τ − τ 2 /2 + ε(−τ 2 /2 + τ 3 /6) + ε2 (τ 3/6 − τ 4/24) + O(ε3 ). (18) This expansion is uniformly valid in τ in the range 0 < τ 1/ε, which is the range of physical interest since the projectile hits the ground well before 1/ε if ε 1 (see below). We wish to obtain the expansion in (18) without prior knowledge of the exact solution (17), using regular perturbation theory. We proceed similarly as for algebraic equations. STEP A. Introduce the expansion y(τ ) = y0(τ ) + εy1(τ ) + ε2 y2(τ ) + O(ε3 ), (19) where y0(τ ), y1(τ ), y2(τ ) are functions of τ to be determined. STEP B. Substitute (19) into (16) (differential equation and initial conditions) written as d2 y dτ 2 + ε dy dτ + 1 = 0, y(0) = 0, dy dτ (0) − 1 = 0, and expand the resulting equations in power series of ε. This gives d2 y0 dτ 2 + 1 + ε d2 y1 dτ 2 + dy0 dτ + ε2 d2 y2 dτ 2 + dy1 dτ + O(ε3 ) = 0, y0(0) + εy1(0) + ε2 y2(0) + O(ε3 ) = 0, dy0 dτ (0) − 1 + ε dy1 dτ (0) + ε2 dy2 dτ (0) + O(ε3 ) = 0. (20)
6RegularperturbationtheorySTEP C. Equate to zero the successive terms of the series in the left hand side of (6):dyodyo (0)O(e0) :+ 1 = 0,yo(0) = 0,) -1= 0,dr2drdyi,dyo(0) =0,O(el) := 0.yi(0) = 0,dr2 + drdr(21)d y2+ dyid2 (0) = 0,O(e2) :y2(0) = 0,d-2drdr0(e3) :STEPD.Successively solve the sequence of equations obtained in (21)yo(T) = T - T2 /2. y1(T) = --? /2 +T3 /6,y2(T) = T3 /6 - T4 /24.(22)Substituting (22) into (19) gives (18).Exercise 2. Show that yn(↑) =(-1)"(-n+1/(n + 1)! - +n+2 /(n + 2))Exercise 3. The maximum altitude reached by the projectile is y+ := y(t+), where T+ is thetime such that dy(↑+)/dt = 0 (why?). From (17) we havedy (1 +a)e-st1dreEHenceIn(1 + )(23)T+ :Eandy,= y(r.) = 1_ In(1 +e)(24)=2For small e, T+ and y+ can be approximated by (recall that In(1 + 2) = z - 2 /2 + z /3 +0(2),T+=1-E/2+e2/2+0(e3),y+=1/2-=/3+e2/4+0(3)(25)Re-obtaintheseexpansionsfrom (18)byperturbationtheoryExercise 4.Check that (24)is consistent with the result we obtained for h usingdimensional analysis.Exercise 5. Recall that in the high friction limit, the appropriate equation for the projectileisdz=-些-8,些(0) =1,(26)z(0) = 0,ds2= dsdswhere d :=1/e=mg/(kV). Here the altitude is measuredin units of l2=Vm/k, and the timein units ofT2=m/k.Solve this equationby regular perturbation technique when 0<<1
6 Regular perturbation theory STEP C. Equate to zero the successive terms of the series in the left hand side of (6): O(ε0 ) : d2 y0 dτ 2 + 1 = 0, y0(0) = 0, dy0 dτ (0) − 1 = 0, O(ε1 ) : d2 y1 dτ 2 + dy0 dτ = 0, y1(0) = 0, dy1 dτ (0) = 0, O(ε2) : d2 y2 dτ 2 + dy1 dτ , y2(0) = 0, dy2 dτ (0) = 0, O(ε3) : ··· (21) STEP D. Successively solve the sequence of equations obtained in (21): y0(τ ) = τ − τ 2 /2, y1(τ ) = −τ 2 /2 + τ 3 /6, y2(τ ) = τ 3 /6 − τ 4 /24. (22) Substituting (22) into (19) gives (18). Exercise 2. Show that yn(τ ) = (−1)n(τ n+1/(n + 1)! − τ n+2/(n + 2)!). Exercise 3. The maximum altitude reached by the projectile is y? := y(τ?), where τ? is the time such that dy(τ?)/dτ = 0 (why?). From (17) we have dy dτ = (1 + ε)e−ετ ε − 1 ε . Hence τ? = ln(1 + ε) ε , (23) and y? = y(τ?) = 1 ε − ln(1 + ε) ε2 . (24) For small ε, τ? and y? can be approximated by (recall that ln(1 + z) = z − z2/2 + z3/3 + O(z4)), τ? = 1 − ε/2 + ε2/2 + O(ε3 ), y? = 1/2 − ε/3 + ε2/4 + O(ε3 ). (25) Re-obtain these expansions from (18) by perturbation theory. Exercise 4. Check that (24) is consistent with the result we obtained for h using dimensional analysis. Exercise 5. Recall that in the high friction limit, the appropriate equation for the projectile is d2z ds2 = −dz ds − δ, z(0) = 0, dz ds(0) = 1, (26) where δ := 1/ε = mg/(kV). Here the altitude is measured in units of `2 = Vm/k, and the time in units of τ2 = m/k. Solve this equation by regular perturbation technique when 0 < δ 1
7Intro to math modelingWhat isthemaximalaltitudereachedbytheprojectile?What isthetime of theflight?Whatis the ratio between the ascent and the descent times of the projectile?Exercise6.Thedimensionlessformof theequationforaperfectpendulumof lengthlisdodo() =0,22= -sin(), 0(0) = 由,where the time is measured in units of Ve/g. Solve this equation by regular perturbationwhen0<1.Isyourapproximationuniformlyvalidintime?
Intro to math modeling 7 What is the maximal altitude reached by the projectile? What is the time of the flight? What is the ratio between the ascent and the descent times of the projectile? Exercise 6. The dimensionless form of the equation for a perfect pendulum of length ` is d2θ dτ 2 = −sin(θ), θ(0) = φ, dθ dτ (0) = 0, where the time is measured in units of p`/g. Solve this equation by regular perturbation when 0 < φ 1. Is your approximation uniformly valid in time?