a n (t) dn y(t) dt and equations that are not of this form are said to be nonlinear.

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1 LECTURE NOTES ON MULTI-SCALE ANALYSIS MATH 3073 WINTER 2003 UNIVERSITY OF NEW BRUNSWICK DANIEL LEMIRE, NATIONAL RESEARCH COUNCIL OF CANADA INTRODUCTION Typically, one solves parial differenial equaions PDE by ransforming hem ino ordinary differenial equaions ODE. For example, given u + u xx = 0, we se ux, = φηx we solve for φ η. The inuiion behind his approach is ha ordinary differenial equaions are easy. Are all ordinary differenial equaions easy? Well, we know how o solve linear ordinary differenial equaion, for example, pxy +qxy +rx = 0. Bu wha abou nonlinear ones such as pxy y+qxy 10 +rx = 0? The answer is ha we don know how o solve hese nonlinear equaions in general. However, in some cases, a nonlinear equaion can be nearly linear we migh be ineresed only in a coarse soluion. As we shall see, one way o proceed in such cases is o ransform he nonlinear ordinary differenial equaion ino linear parial differenaion equaions! The parial differenial equaions allow us o zoom in on he soluion. The concep of mahemaical zoom is very powerful for pracical applicaions. Waveles for example have lead o powerful image compression echniques [4]. Muli-scale mehods are rouinely used in remoe sensing many oher applicaions [5, 3]. Of course, i is also very useful o solve differenial equaions [1]. The idea of a mahemaical zoom is simple: solve he problem a successively finer scales or respecively, coarser scales. Ofen, in solving differenial equaions, we are only ineresed in coarse-scale behavior of he soluion, bu he fine scales affec his behavior so one canno simply ignore hose. Also, someimes solving for he coarse scale soluion hen laer for he finer scale soluion migh be easier. Muli-scale mehods approach he same problem on differen scales simulaneously jus like he zoom of a microscope. To do so implies using several variables even if he iniial problem only had one variable. As we shall see, his can be quie difficul, bu many of he subproblems can be hled easily wih some background in PDEs. 1. NOTATION AND REMINDERS We noe he complex conjugae of z = x + iy C by z = x iy. By Demoivre s formula, any complex number z = x + iy can be wrien as z = re iθ where r,θ R z = x 2 + y 2 = r, e iθ + e iθ = 2cosθ, e iθ e iθ = 2sinθ e iθ = cosθ + isinθ. Linear differenial equaion have he form a n dn y d n a 1 dy + a 0 y = 0 d equaions ha are no of his form are said o be nonlinear. 1

2 FIGURE 1. y 0 = cos Recall ha he binomial heorem saes ha x + a n = n k=0 n! k!n k! xk a n k. 2. A FIRST EXAMPLE THIS PART MAY NOT MAKE IT IN THE LECTURE The inuiion in wha follows is ha someimes, an equaion can be very close o a known equaion which we know how o solve, excep ha i is slighly differen. We canno ignore he difference per se, however, we can aemp o solve he problem in seps. Firs, we approximae he answer by simplifying he problem enough so ha we know how o solve i, hen we solve for he error we are making wih our simple soluion so on. Le s consider he nonlinear oscillaor equaion d 2 y d 2 + y + εy3 = 0, y0 = 1,y 0 = 0, which mos of us don know how o solve. I could be shown ha he soluion y of his equaion is bounded for all by 1 + ε/2. While you don know how o solve i, his equaion should remind you of anoher ha you know well d 2 y 0 d 2 + y 0 = 0, y 0 0 = 1,y 00 = 0. Recall ha we can solve his equaion by seing y 0 = e λ so ha we ge he characerisic equaion λ = 0 which has soluion λ { i,i}. Therefore, we have he general soluion y 0 = Acos+Bsin where A,B R mus be chosen according o he boundary condiions. We have 1 = y 0 0 = A 0 = y 0 0 = B hus he unique soluion is y 0 = cos see Fig. 1. We claim ha his is an approximae soluion o he nonlinear oscillaor equaion. To see how close i is, le s subsiue y 0 ino i... d 2 y 0 d 2 + y 0 + εy 3 0 = εy3 0 = εcos3, 2

3 so ha, as long as ε is small, i seems we have an approximaion o he real hing in ha εcos 3 0. In fac, if ε = 0 hen our soluion is exac. If y 0 is almos a soluion, hen maybe we can wrie he soluion as y 0 +somehing where his somehing is a small perurbaion. How small should he perurbaion be? How abou εsmall? Le s wrie y y 0 + εy 1. We subsiue his formula ino he full nonlinear oscillaor equaion o ge, afer cancellaions, εcos 3 + ε d2 y 1 d 2 + εy 1 + ε 3 y dividing by ε hroughou, we ge cos 3 + d2 y 1 d 2 + y 1 + ε 2 y Looking carefully a his equaion, we see ha we ve almos come full circle. Indeed, we are back o having an equaion we can solve assuming we drop one erm ε 2 y 3 1. Le s do i, le s drop ε 2 y we se 1 cos 3 + d2 y 1 d 2 + y 1 = 0. Wha abou he boundary condiions? Well, because y 0 0 = 1,y 0 0 = 0 y0 = 1,y 0 = 0, i makes sense o se y 1 0 = 0,y 1 0 = 0. How do we solve equaion 1? This could prove quie lenghy unless you use he ideniy cos 3 = 4 1 cos cos. So ha he equaion becomes cos cos + d2 y 1 d 2 + y 1 = 0. You should recall ha we find he soluion by subsiuing y 1 = Acos + Bsin +C cos3 + D cos + E sin where A cos + B sin is he homogeneous soluion whereas C cos3 + D cos is o be he inhomogeneous soluion used o mach 4 1 cos cos. We firs check he boundary condiions, 0 = y 1 0 = A +C 0 = y 1 0 = B + D. Hence, we can wrie y 1 = Acos cos3 + Bsin cos + E sin. Subsiuing his soluion ino equaion 2 gives 1 4 cos cos + d2 y 1 d 2 + y 32A + 1cos3 + 8Bsin + 8E + 3cos 1 = 4 seing his equal o zero gives A = 1/32, B = 0 E = 3/8 so ha Puing i all ogeher, we have ha y 1 = 1 32 cos cos 3 8 sin. y cos + ε 1 32 cos cos 3 8 sin. Nex, we wan o see how good a soluion y 0 + εy 1 is. Well, remember ha we said ha we were dropping he ε 2 y 3 1. Thus, we have ha our error is now of order ε2. Remember ha y 0 was accurae up o εy 3 0, hus, we have a more accurae soluion for small ε, bu we also worked harder. 3

4 1 y 0 y 0 +ε y FIGURE 2. y 0 = cos y 0 + εy 1 = cos + ε 1 32 cos cos 3 8 sin wih ε = 1/10 We ve realized our goal: we ve solved a very difficul problem incremenal by firs geing a coarse soluion hen a somewha finer one... One could prove ha if we were o coninue his process, we would evenually converge o a soluion Alernaive roue. There is a quicker way o solve he problem because we don need o solve for y 0 firs hen y 1. Firs subsiue y 0 + εy 1 for y in he differenial equaion exp ou he equaion as 0 = d2 y d 2 + y + εy3 d2 y 0 d 2 + y 0 + εy εd2 y 1 = d2 y 0 d 2 + y 0 + ε d 2 + εy 1 + ε 4 y 3 1 y d2 y 1 d 2 + y 1 + ε 4 y 3 1 Nex collec like powers of ε dropping higher powers o ge he equaions d 2 y 0 d 2 + y 0 = 0 y d2 y 1 d 2 + y 1 = 0. And finally, solve hese equaions one a a ime o ge y 0 = cos y 1 = 32 1 cos cos 3 8 sin. Exercise 1. Given d2 y d εcosy = 0 solve for y 0 give he equaion for y 1 using he approach above wih y y 0 + εy 1. The boundary condiions are no given Problems? We ve been hasy so far. We ve been collecing erms of like powers, dropping erms a will... are we sure we won have any problems? Well, we may have realized our goal, bu we ve sumbled on a problem: y 0 is bounded by 1 whereas y 1 grows wih so ha ε 2 y 3 1 migh be much larger han εy3 0. This means ha our soluion will only be good enough for small 1/ε so ha given 4

5 1.5 y 0 y 0 +ε y FIGURE 3. y 0 = cos y 0 + εy 1 = cos + ε 1 32 cos cos 3 8 sin for ε = 1/10 ε = 1/10, our soluion is only valid for 10. In Fig. 3, we see he effec because y 0 y 0 + εy 1 come apar when is of order 10 or more. 3. HERE COME THE PDES! Wha we did so far cerainly works, bu i has drawbacks. Can we improve our soluion cos wihou assuming ha is small? Now, we ll use a muli-scale analysis! Remark 1. There are lecure noes available on he web giving a similar review of his problem in differen words [2]. Our physical inuiion is ha he nonlinear oscillaor has really wo differen ime scales : τ = ε. The new variable, τ defines a long longer ime scale han because i akes a large change in o significanly change τ. Anoher way o pu i is ha τ sees he world in slow moion whereas would be he normal fas-paced ime. Thus, we wan o look a he problem boh in slow moion in normal ime... hence he word muli-scale. So, insead of wriing y y 0 + εy 1, we wrie y Y 0,τ + εy 1,τ because we have wo variables. A his poin, you migh be oally confused... Wha does Y 0,τ mean? Are τ independen variables? No, hey are no independen, τ = ε so dτ d = ε. So hen, Y 0,τ really only depends on, wha s he poin? Well, acually, you could say ha Y 0,τ depends on boh ε ha s a valuable idea as we shall see. Exercise 2. Given f, τ, prove ha d d f,τ = f + ε f 5

6 d 2 d 2 f,τ = 2 f 2 + 2ε 2 f + ε2 2 f 2. We have ha for i = 0,1, herefore dy d dy i,τ d dy 0,τ d = Y 0,τ = Y 0,τ = Y i,τ + ε dy 1,τ d + ε Y 0,τ + ε Y i,τ + ε Y 1,τ + ε 2 Y 1,τ Y0,τ + ε + Y 1,τ + ε 2 Y 1,τ bu he nonlinear oscillaor equaion has a second derivaive, so we need o differeniae again d 2 y d 2 d Y 0,τ d Y 0,τ + ε + d Y 1,τ + ε 2 d Y 1,τ d d d d = 2 Y 0,τ 2 +ε 2 2 Y 1,τ = 2 Y 0,τ 2 + ε 2 Y 0,τ + ε 2 Y 0,τ + ε + ε 3 2 Y 1,τ Y 0,τ + 2 Y 1,τ 2 + ε 2 Y 0,τ Y 1,τ 2 + ε 2 some suff. We can now subsiue his ino he nonlinear oscillaion equaion o ge 2 Y 0,τ 2 + ε 2 2 Y 0,τ + 2 Y 1,τ 2 + ε 2 some suff Nex, you can convince yourself ha +Y 0,τ + εy 1,τ + εy 0,τ + εy 1,τ ε 2 Y 1,τ Y 0,τ + εy 1,τ 3 = Y0 3,τ + εsomehing by using he binomial heorem or brue force. Therefore, subsiuing his las equaion back ino he nonlinear oscillaion gives us 2 Y 0,τ 2 +ε 2 2 Y 0,τ + 2 Y 1,τ 2 We collec like powers of ε o ge 2 Y 0,τ 2 +Y 0,τ+ε we have he corresponding equaions 3 +Y 0,τ+εY 1,τ+εY 3 0,τ+ε2 some suff 0. Y 1,τ +Y0 3,τ + Y 0,τ Y 1,τ 2 2 Y 0,τ 2 4 Y 1,τ + 2 Y 1,τ 2 +Y 0,τ = 0 +Y0 3,τ + Y 0,τ 2 2 = ε 2 some suff 0

7 We can solve equaion 3 easily by separaion of variables o ge Y 0,τ = Aτe i + Bτe i because we wan Y 0,τ o be real, we need o require ha Y 0,τ = Y 0,τ so Aτ = Bτ or equivalenly Aτ = Bτ. Thus, we wrie Y 0,τ = Aτe i + Aτe i. This is no ye a complee soluion because we don know wha Aτ is. We can hen subsiue his soluion ino equaion 4 o ge Y 1,τ + 2 Y 1,τ 2 + Aτe i + Aτe i 3 + 2iA τe i 2iA τe i = 0. We are no going o solve his equaion as i is clearly oo long difficul! However, we will ry o go some insigh ou of i as o wha Aτ should be. Again, using he binomial heorem, we see ha Aτe i + Aτe i 3 = A 3 τe 3i + A 3 τe 3i + 3AτA 2 τe i + 3A 2 τaτe i. Subsiuing his back ino our equaion gives Y 1,τ + 2 Y 1,τ 2 + A 3 τe 3i + A 3 τe 3i + 3AτA 2 τ + 2iA τ e i + 3A 2 τaτ 2iA τ e i = 0. Wriing ατ = 3AτA 2 τ + 2iA τ, we have Y 1,τ + 2 Y 1,τ 2 + A 3 τe 3i + A 3 τe 3i + ατe i + ατe i = 0. If ατ 0, hen we mus solve his differenial equaion by separaion of variables using Y 1,τ = B 1 τe i + B 2 τe i + B 3 τe 3i + B 4 τe 3i + B 5 τe i + B 6 τe i. Exercise 3. In he above, why do we need he erms B 5 τe i B 6 τe i for ατ 0? Because he erms B 5 τe i + B 6 τe i grow large for large we wan o avoid his 1, we choose ατ = 0 or 3AτA 2 τ + 2iA τ = 0. Wrie Aτ = Rτe iωτ where boh R ω are real funcions so ha Therefore, ατ is given by A τ = R τe iωτ + Rτiω τe iωτ. 3R 3 τe iωτ + 2iR τe iωτ 2Rτω τe iωτ = 0 or 3R 3 τ + 2iR τ 2Rτω τ = 0. Because z = x + iy = 0 implies x = 0 y = 0 ha if a complex number is zero, boh is real imaginary pars mus be zero as well, we have wo equaions R τ = 0 3R 3 τ 2Rτω τ = 0. 1 Avoiding large erms also called secular erms is he key ingredien in his recipe. 7

8 The firs differenial equaion ells use ha R mus be a consan le us assume R 0, he second one becomes a simple equaion 3R 2 2ω τ = 0 wih ωτ = 3R2 2 τ + K where K is some consan. Therefore Aτ = Re i 3R 2 2 τ+k while we haven solved for Y 1, we now have Y 0! Indeed, we have or Y 0,τ = Re i 3R 2 2 τ+k+ + Re i 3R 2 2 τ+k+ 3R 2 Y 0,τ = 2Rcos 2 τ + K +. All ha is lef o do is o enforce he boundary condiions 0 = dy 00,0 d = Y 00,0 Y 0 0,0 = 2RcosK = 1 + ε Y 00,0 = 2RsinK 3R 3 εsink. The las equaion gives us 2 3R 2 = 0 or sink = 0 hence we mus choose K = 0 hus, he oher equaion gives us R = 1/2. We hen have 3 y Y 0,τ = cos 8 τ + or 3ε y cos because τ = ε. Essenially, our new soluion is idenical o he previous one cos for small s bu because of a phase difference, hey evenually differ quie a bisee Fig. 4. As he nex exercise shows, cos 3ε is much beer han cos as a soluion. Exercise 4. Using y = cos 3ε show ha d 2 y 3ε d 2 + y + εy3 = εcos ε 4 εcos ε cos ε 4 εcos /4. Compare wih he resul we go wih y = cos, ha is d 2 y d 2 + y + εy3 = εcos 3. + ε 2 somehing To summarize he muli-scale mehod: subsiue Y 0,τ + εy 1,τ for y in he differenial equaion, exp collec he powers of ε solve he firs PDE, hen subsiue your parial soluion ino he second PDE, hen requiring ha here be no secular erms, finish solving for he firs order Y 0 erm. 8

9 1 0.8 cosx cos3ε/8 +1*x FIGURE 4. Comparison beween he firs order soluion y 0 = cos he muli-scale soluion y cos 3ε for ε = 1/10 4. PROBLEMS In he following problems, use he formulas dy d Y 0,τ + ε y Y 0,τ + εy 1,τ, Y0,τ + Y 1,τ + ε 2 Y 1,τ, d 2 y d 2 2 Y 0,τ 2 + ε 2 2 Y 0,τ + 2 Y 1,τ 2. 2 Exercise 5. Solve d2 y +y+ε dy d 2 d = 0, y0 = 0,y 0 = 1, using a muliscale approach. Soluion: Keeping only he firs wo powers of ε, we have 2 Y 0,τ 2 + ε 2 2 Y 0,τ + 2 Y 1,τ 2 +Y 0 + εy 1 + ε 2 Y0 0. Collecing he like powers of ε, we have firs 2 Y 0,τ +Y 2 0 = 0 hen 2 2 Y 0,τ + 2 Y 1,τ Y 1 + Y0 = 0. We solve he firs PDE o ge Y0 = Aτe i +Aτe i. Hence, he second differenial equaion becomes 2 Y 1 2 +Y 1 + 2iA τe i 2iA τe i A 2 τe 2i A 2 τe 2i + 2 Aτ 2 = 0. To avoid he secular erms erms in he soluion of Y 1 growing big, we need A τ = 0 or Aτ = Re iη hus, Y 0 = Re i+η + Re i+η = 2C cos + η = 2C cos + η where C = R+R 2 η is a consan. Because y0 = 0,y 0 = 1, we se η = π/2 C = 1/2 so ha Y 0 = sin is our soluion. 9

10 Exercise 6. Damped Oscillaor Given y + y + εy 3 = 0, find he leading order erm Y 0 using muli-scale analysis. Exercise 7. Given y + y + εy y 2 = 0, find he leading order erm Y 0 using muli-scale analysis. Exercise 8. Given y + y + εy 2 y = 0, find he leading order erm Y 0 using muli-scale analysis. Soluion: Keeping only leading erms in ε, we have 2 Y 0 2 +Y 0 + ε Y Y Y Y 2 0 Y 0 = 0. Subsiuing Y 0 = Aτe i + Aτe i ino we have Y Y Y Y 2 0 Y 0 = 0, Y Y A τie i ia τe i + iaτe i iaτe i 2 Aτe i + Aτe i = 0. Hence, we should se 2iA + A 2 A = 0 why?. Wrie A = Re iω where boh ω R are real funcions of τ, hen 2iR 2ω R + R 3 = 0 so ha R = 0 R is a consan! ω = R 2 /2 or ωτ = R 2 τ/2 + K. We have A = Re ir2 τ/2+ik herefore, Y 0 = Re ir2 τ/2+i+ik + Re ir2 τ/2 i ik = 2RcosR 2 τ/2 + + K. The parameers R K would need o be se using iniial condiions. This documen was prepared planned wih he help of professor John Sockie. REFERENCES [1] C.M. Bender S.A. Orszag, Advanced Mahemaical Mehods for Scieniss Engineers, McGraw-Hill, New York, [2] S. Bryson, Nonlinear Oscillaors, Lecure Noes, Sanford Universiy hp://sccm.sanford.edu/~bryson/ma220c_ch2.pdf. [3] R. Myers, Muli-scale Analysis Mehods for Vegeaion Classificaion, hp://ice.ucdavis.edu/~robyn/mulisclpp/. [4] D. Janssens, J2000 CODEC, hp:// [5] G.P. Pail, W.L. Myers, Muli-Scale Saisical Approach o Criical-Area Analysis Modeling of Waersheds Lscapes, hp:// 10