Differential Equations II MATC46H3S. Lisa Jeffrey. Paul Selick

Transcription

1 Differentil Equtions II MATC46H3S Lis Jeffrey Pul Selick E-mil ddress, Lis Jeffrey: E-mil ddress, Pul Selick: Lis Jeffrey Bhen Centre, room BA6211, 4 St. George Street, Toronto, Ontrio, M5S 2E4 Pul Selick Bhen Centre, room BA626, 4 St. George Street, Toronto, Ontrio, M5S 2E4 URL, Lis Jeffrey: URL, Pul Selick:

2 Typeset by Brin Wu

3 Contents Chpter 1. Lplce Trnsforms 1 1. Definitions 1 2. Lplce Trnsforms of Derivtives 2 3. The Gmm Function 7 4. Convolutions 9 5. Lplce Trnsforms of Some Discontinuous Functions Step Functions Impulse Functions 15 Chpter 2. Phse Portrits: Qulittive nd Pictoril Descriptions of Solutions of Two-Dimensionl Systems Introduction Phse Portrits of Liner Systems Rel Distinct Eigenvlues Complex Eigenvlues Repeted Rel Roots Phse Portrits of Non-Liner Systems Applictions The Pendulum The Dmped Pendulum Predtor-Prey Equtions Lipunov s Second Method Periodic Solutions Index Theory 5 Chpter 3. Boundry Vlue Problems Boundry Vlue Problems Smple Appliction Leding to BVPs Homogeneous Boundry Vlue Problems Introduction Eigenvlue Problems Sturm-Liouville Nonhomogeneous Boundry Vlue Problems Determinnt Cses The Cse When The Cse When 82 iii

4 iv CONTENTS 3.2. Green s Functions Prtil Differentil Equtions Vibrting String The Lplce Eqution The Het Eqution The Schrödinger Eqution Zeros of Solutions of Second Order Liner Differentil Equtions Proof of the Properties of Sturm-Liouville Problems 99 Chpter 4. Midterm Review Lplce Trnsforms Phse Portrits Liner Systems Nonliner Systems Index Theory 15 Chpter 5. Review 17 Appendix A. The Grm-Schmidt Process 113

5 CHAPTER 1 Lplce Trnsforms 1. Definitions Definition 1.1 Lplce trnsform. Let f : [, R. The Lplce trnsform of f, denoted Lf or ˆf, is the function ˆfs : e st ft dt. The domin of ˆf is the set of s for which the integrl converges. Theorem 1.2. If ˆfs converges, then ˆfs converges for ll s > s. Proof. Suppose tht s > s. We wish to show tht the til of the integrl is smll, i.e., show tht, given n ɛ >, there exists n such tht b e st ft dt < ɛk for ll b, where k is constnt, i.e., independent of nd b. Let βx x e st ft dt, where x. The integrl exists since ˆfs converges. Note tht βx is lso differentible fundmentl theorem of clculus with β x e sx fx. Therefore, lim x βx. Choose n such tht x βx < ɛ. Then Choosing b e st ft dt b b e st e st e st ft dt e s st β t dt e s stβt b + b s s e s st βt dt. u e s st, du s s e s st dt, dv β t dt, v βt, we hve b e st ft dt e s sb βb e s s β + s s b e s st βt dt. Since s s nd b, we hve e s sb 1 nd e s s 1. Therefore, b e st ft dt b βb + β + s s e s st βt dt 1

6 2 1. LAPLACE TRANSFORMS Therefore, letting k 2 + s s 2, we hve b ɛ + ɛ + s s e s st ɛ dt 2ɛ + ɛ s s 2 s s 2 b 2ɛ + ɛ s s 2 e s s e s sb 2ɛ + ɛ s s 2 e s s 2ɛ + ɛ s s 2 ɛ 2 + s s 2. b e st ft dt < ɛk for ll b. Recll from MATA3/36/37 tht 1 if gx hx, then hx dx converges implies tht gx dx converges. 2 gx dx converges implies tht gx dx converges. Also note tht s > s e st e st. Therefore, if e st ft dt converges, then Theorem 1.2 follows immeditely. The generl cse requires more creful nlysis. Exmple 1.3. Compute Lf for fx e x. Solution. We hve ˆfs lim b where the domin is s >, i.e.,,. e st e t dt e sb s 1 s e st dt lim b e st s 1 s 1 s, b Definition 1.4 Exponentil order. Given constnt α, continuous function f : [, R is sid to hve exponentil order α if there exists constnt C such tht fx Ce αx for sufficiently lrge x. More precisely, f hs exponentil order if there exist constnts C nd b such tht fx Ce αx for x > b. We write f ξ α to men tht f hs exponentil order α. Theorem 1.5 Comprison theorem. If f ξ α, then ˆfs is defined for ll s > α. From now on, we will ssume tht f ξ α for some α. 2. Lplce Trnsforms of Derivtives To tke into ccount the Lplce trnsform of derivtives, note tht, from the definition of the Lplce trnsform, we hve Lf e st f t dt lim b b e st f t dt.

7 2. LAPLACE TRANSFORMS OF DERIVATIVES 3 Letting u e st, du se st dt, dv f t dt, v ft, we hve Lf lim e st ft b + s b b e st ft dt Our interest now lies in lim b e sb fb. Note tht e sb fb Ce } sb {{ e αb }. lim e sb fb f + slf. b But s b. So by squeezing, we hve lim b e st fb. But then limb e sb fb while so we conclude tht lim b e sb fb. Hence, e sb fb e sb fb e sb fb, Lf f + slf slf f. Exmple 1.6. Solve y 4y e x with y 1. Solution. Tking the Lplce trnsform of the entire eqution, we hve Ly 4y Le x, Ly 4Ly 1 s 1, sly y 4Ly 1 s 1, s 4 Ly 1 1 s 1, s 4 Ly 1 s 1 + 1, s 4 Ly s s 1, Ly s s 1 s 4. Rerrnging the expression to revel terms with esily identifible inverse Lplce trnsforms, we hve Tking the inverse Lplce trnsform gives Ly s s 1. y 4 3 e4x 1 3 ex. Property 1.7 Lplce trnsform. 1 Lf + bg Lf + blg

8 4 1. LAPLACE TRANSFORMS 2 Lf slf f, Lf slf f s 2 Lf sf f,.. L f n s n Lf s n 1 f s n 2 f f n 1 3 If f nd g re continuous on [, nd Lf Lg, then f g. 4 L e x fx ˆfs 5 ˆf is differentible nd L x n fx 1 n ˆf n s. b ˆf is integrble nd L fx/x s ˆfu du. 6 lim s ˆfs Proof of 1. This is trivil. Letting Proof of 2. Note tht L f x e sx f x dx. u e sx, du se sx dx, dv f x dx, v fx, we hve L f x e sx fx + s 1fx + slf e sx fx dx slf f. Proof of 3. Suppose tht Lf Lg nd let h f g. Then Lh. To show tht h s well, we use the corollry to the Weierstrss Approximtion Theorem MATC37. If f is continuous on [, 1] nd 1 tn ft dt for ll n, 1, 2,..., then f. Suppose tht ˆfs for ll s s. Consider s s + n. Then ĥs + n e s+nt ht dt e nt e st ht dt e nt v t dt. Let vx x e st ht dt so tht v x e sx hx. Letting u e nt, dv v t dt, du ne nx dx, v v,

9 2. LAPLACE TRANSFORMS OF DERIVATIVES 5 we hve ĥs + n e nt vt lim b e nb b + n ĥs v + n + n n e nt vt dt e st ht v + n e nt vt dt. e nt vt dt e nt vt dt e nt vt dt Therefore, e nt vt dt for ll n. Now let z e t so tht t ln1/z lnz nd dt 1/z dz. Then t z nd lim t e t. Therefore, 1 1 e nt vt dt z n v lnz 1 z z n 1 v lnz dz for ll n. Therefore, v lnz. As z runs through [, 1], lnz runs through [,, i.e., vt for ll t [,. Therefore, vt nd v t e st ht for ll t [,. Therefore, ht s well since e st. dz Proof of 4. We hve L e x fx e sx e x fx dx e s x fx dx ˆfs. nd Proof of 5. We hve ˆfs lim h ˆfs + h ˆfs h L xfx e st tft dt. We must show tht for ll ɛ >, we hve ˆfs + h ˆfs I h < ɛ for sufficiently smll h. We hve ˆfs + h ˆfs I h e s+ht ft dt h e st ft dt e st tft dt

10 6 1. LAPLACE TRANSFORMS Note tht e ht 1 h e st e ht 1 ft dt + h e st e ht 1 h + t ft dt. e st tft dt + t 1 ht + h 2 t 2 2! t h ht2 2 + h2 t ht 2 2 th 3! + t2 h 2 + 4! < ht th 2! + t2 h 2 + 3! ht 2 e th. Therefore, ˆfs + h ˆfs I h If h is sufficiently smll, then h s α <. So for h sufficiently smll. So we cn mke for h sufficiently smll. Therefore, h h t 2 e h s αt dt < t 2 e h s αt dt < ɛ ht 2 e th e st ft dt t 2 e h s αt dt. We cn differentite gin by the sme procedure. ˆfs + h ˆfs ˆf s lim I L xfx. h h Proof of 6. For lrge x, we hve fx Ce αx for some C nd α. So ˆfs }{{} R e sx fx dx + b b Therefore, by the Squeeze Theorem, we hve e sx fx dx e sx fx dx C e s αx dx b b + e sx fx dx. }{{} C s α +R b fx e sx dx b lim s lim ˆfs, s e sx fx dx,

11 b 3. THE GAMMA FUNCTION 7 lim e sx fx dx. s Tble 1.1 shows prtil list of Lplce trnsforms. Note tht s/ s is defined for ll s, but it Tble 1.1: Lplce trnsforms f Lf x n Γn + 1 cosx sinx x cosx x sinx s n+1 s s 2 + 2, s > s s 2 2 s s s equls e sx cosx dx only when s > it does not converge for s. 3. The Gmm Function Definition 1.8 Gmm function. The Gmm function is defined s Γn : x n 1 e x dx, n >. Property 1.9 Gmm function. 1 Γn n 1 Γn 1 2 Γn n 1! if n is positive integer 3 Γ 1 2 π Proof of 1. From the definition of the Gmm function, we hve Considering integrtion by prts, we hve Γn x n 1 e x dx. u x n 1, dv e x dx, du n 1 x n 2 dx, v e x. Therefore, Γn x n 1 e x + n n 1 Γn 1 x n 2 e x dx See [?, p. 34] for more comprehensive list.

12 8 1. LAPLACE TRANSFORMS n 1 Γn 1. Proof of 2. First note tht Γ1 e x dx e x 1. So by induction in prt 1, Γn n 1! for ny positive integer n. Proof of 3. From the definition of the Gmm function, we hve 1 e x Γ dx. 2 x Considering the substitution u x, it follows tht x 2 u nd dx 2u du. Therefore, 1 e u2 Γ 2u du 2 e u2 du. 2 u Let I e u2 du. Then I 2 π/2 π 2 Therefore, I π/2. So Γ1/2 2I π. e r2 2 e u2 +v 2 dv du e r2 r dr dθ π 4. With the Gmm function nd its properties estblished, we cn now prove the entries in Tble 1.1. Proof of Tble We hve If n is n integer, then nd so on. Let t sx. Then 2 We hve Let b i. Then Lx n Lx n e sx x n dx. L1 L e x 1 s 1 s, Lx d 1 1 ds s s 2, L x 2 d 1 ds s 2 2! s 3,... t tn dt e s n s 1 s n+1 L e bx 1 s b. e t t n dt L e ix 1 s i s + i s Γn + 1 s n+1.

13 4. CONVOLUTIONS 9 Therefore, L cosx Re L e ix s s It follows immeditely from 2 tht L sinx Im L e ix s We hve L xe bx 1 s b 2. Therefore, L xe ix 2 1 s + i 2 s i s s is s So it follows tht L x cosx s2 2 s It follows immeditely from 4 tht L x sinx 2s s Convolutions Let Lf ˆf nd Lg ĝ. Then we wish to find n h such tht Lh ˆfĝ. To do so, we hve Lhs ˆfsĝs e sx fx dx e sx+y fxgy dx dy. e sy gy dy Let u x + y nd t y. Figure 1.1 shows grphiclly this substitution. Then y t x u The xy-plne. b The hlf-plne u > t. Figure 1.1: The xy-plne nd hlf-plne below t u.

14 1 1. LAPLACE TRANSFORMS nd J 1. Thus, So Lhs J u [ ] e su fu tgt dt du e su u e sx x x L fx tgt dt. hx clled the convolution of f nd g, written f g. x fu tgt dt du fx tgt dt dx fx tgt dt, Property 1.1 Convolution. 1 f g g f 2 f g h f g h 3 f g + h f g + f h 4 λf g λ f g, where λ is constnt Exmple Solve y + y fx, where y nd y. In ddition, consider the cse where fx tnx. Solution. Tking the Lplce trnsform of every term, we hve Therefore, so With fx tnx, we hve y x x x tnt sinx t dt y L 1 1 Ly sly y s 2 ŷ, Ly sŷ y sŷ. s ŷ ˆf ŷ 1 s ˆf, s ˆf sinx f x tnt sinx cos t + cosx sin t dt tnt sinx cost cosx sint dt 1 L 1 s 2 L 1 ˆf + 1 ft sinx t dt.

15 x x 5. LAPLACE TRANSFORMS OF SOME DISCONTINUOUS FUNCTIONS 11 sinx sint cosx sin2 t cost sinx sint dt x x x dt 1 cos 2 t cosx cost x dt sinx sint dt cosx sect dt + cosx cost dt sinx cost cosx ln sect + tnt x + cosx sint x sinx cosx + 1 secx cosx ln + tnx x ln cosx sinx secx sinx cosx + sinx cosx ln + tnx + cosx sinx sinx cosx ln secx + tnx Step Functions. Let 5. Lplce Trnsforms of Some Discontinuous Functions 1, x, ux :, x <, clled the step function. Suppose x nd let gx : ux x. Then 1, x x, gx, x < x. We hve Therefore, Lĝ s e sx gx dx x e L 1 sx ux x. s e sx dx e sx. s Recll tht L1 1/s. This is the cse where x. More generlly, we hve the following theorem. Theorem We hve In prticulr, f shifted to the right by x. Figure 1.2 illustrtes the ide of these shifts. Proof 1. We hve L 1 sx e ˆfs L ux x fx x sx e ˆfs. L 1 sx e ˆfs L 1 e Lf + f sx s e L 1 sx e L 1 f + L 1 sx f s s ux x f x + fux x

16 12 1. LAPLACE TRANSFORMS fx b ux x fx x Figure 1.2: A plot showing fx nd fx shifted to the right by x. Proof 2. We hve x f x tut x dt + fux x x x f x t dt, x x, + fux x, x < x x ux x f x t dt + f x ux x fx t x + f x ux x f + fx x + f ux x fx x. L ux x fx x Considering substitution, let v t x so tht dv dt. Then L ux x fx x x e st ft x dt. e sv x fv dv e sx e sv fv dv e sx ˆfs. Also note tht L ux x fx L ux x gx x e sx Lg e sx L fx x, where gx x fx nd gx fx + x.

17 5. LAPLACE TRANSFORMS OF SOME DISCONTINUOUS FUNCTIONS 13 Exmple Solve 3y + 7y + 2y fx with y nd y, where x, x 2, fx 1, x < 2. Solution. Tking the Lplce trnsform, we hve Ly sŷ y sŷ, Ly s sŷ y s 2 ŷ. Therefore, our eqution becomes 3s 2 + 7s + 2 ŷ L ux 2x + 1 L1, s + 2 3s + 1 ŷ e 2s Lx + 3 L1 1 e 2s s s s 3s + 1 e 2s s 2 1 s. Solving for ŷ gives ŷ e 2s 1 1 s 2 s + 2 s s + 2 3s + 1. }{{} We now must consider prtil frctions. First considering, note tht Compring coefficients, we see tht nd 1 s 2 s + 2 As + B s 2 + C s + 2 As + B s + 2 Cs2 s 2. s + 2 s 2 1 4C C 1 4, s 2B 1 B 1 2, 3 A + B 3 + C 1, A , 12A , 12A 3, A 1 4. Therefore, 1 s 2 s /4 s + 1 /2 s /4 s + 2

18 14 1. LAPLACE TRANSFORMS nd we tenttively hve Now considering, note tht Compring coefficients gives us ŷ e 2s 1 /4 s + 1 /2 s /4 s s s + 2 3s + 1 A s + B s C 3s s s + 2 3s + 1 }{{} A s + 2 3s Bs 3s Cs s + 2. s s + 2 3s + 1 s 1 2A A 1 2, s B 1 1, s C 9 5. Therefore, nd we finlly hve Therefore, 1 s s + 2 3s /2 s + 1 /1 s /5 3s + 1 ŷ e 2s 1 /4 s + 1 /2 s /4 1 /2 s + 2 s 1 /1 s /5 s + 1/3. y ux x e 2x e 2x e x/3. Exmple 1.14 Trick. Solve xy + 2x + 3 y + x + 3 y 3e x with y. Solution. First note tht the initil condition y lone specifies the solution s it implies the vlue of y. More precisely, Tking the Lplce trnsform, we hve d ds + 3y + 3y 3 y 1. Lxy + 2Lxy + 3Ly + Lxy + 3Ly 3L e x, s2ŷ sy y 2 d ds sŷ y + 3 sŷ y d dsŷ + 3ŷ 3 s + 1, 2sŷ s 2 dŷ dŷ 2ŷ 2sdŷ + 3sŷ ds ds ds + 3ŷ 3 s + 1, s 2 + 2s + 1 dŷ ds + sŷ + ŷ 3 s + 1, s dŷ ds s + 1 ŷ 3 s + 1, dŷ ds 1 s + 1ŷ 3 s

19 5. LAPLACE TRANSFORMS OF SOME DISCONTINUOUS FUNCTIONS 15 At this point it is pproprite to introduce the integrting fctor I e R Multiplying both sides by 1/ s + 1 gives us Note tht lim s ŷ C. Therefore, nd ds s+1 e lns+1 1 s dŷ s + 1 ds 1 s ŷ 3 s + 1 4, ŷ s s C, ŷs 1 ŷ 2 + C s + 1. s s y e x L 1 1 s 2 xe x. Note tht if ny x 2 hd ppered in the originl eqution, the resulting differentil eqution for ŷ would hve hd order 2. So this trick hs limited pplicbility. b 5.2. Impulse Functions. A force F t cting between t nd t b produces momentum ρ F dt. An instntneous trnsfer of momentum ρ t time cn be thought of s the limit s ɛ of the result of force of size ρ/ɛ cting over time ɛ from to + ɛ. Consider the step function shown in Figure 1.3. We hve 1 ² Ft ² t Figure 1.3: The step function with width ɛ nd height 1/ɛ, bounded by the y-xis, encloses region with re 1. where f ɛ x dx ɛ 1 ɛ 1, f ɛ 1 ɛ 1 ux ɛ.

20 16 1. LAPLACE TRANSFORMS Thus, Tking the limit, we hve Lf ɛ 1 ɛ 1 s e ɛs 1 e ɛs. s ɛs lim Lf 1 e ɛs se ɛs ɛ lim lim 1. ɛ ɛ }{{ ɛs ɛ } s l Hôpitl s Rule Exmple A block of wood of mss 8 g is motionless t the end of spring with spring constnt 1 g /sec 2. At time t, it is hit by bullet weighing 1 g nd trveling upwrd t 1 m /sec. Find the eqution of motion of the block of wood ssuming tht there is no resistnce. Solution. Let xt be the distnce bove the strting position t time t. Then we hve 9 d2 x + 1x F t dt2 with initil conditions x nd x, where F t is the impulse function with momentum Tking the Lplce trnsform, we hve 1 g 1 m /sec 1 g m /sec. Therefore, s 2ˆx + 1ˆx 1, ˆx 1 9s s s /9 1 3 x 1 3 sin t 3. 1/3 s /9. Figure 1.4 shows the plot of the eqution. t p 2 p 3 p 4 p 5 p 6 p x Figure 1.4: The plot of the eqution of the motion of the block of wood in Exmple 1.15.

21 CHAPTER 2 Phse Portrits: Qulittive nd Pictoril Descriptions of Solutions of Two-Dimensionl Systems 1. Introduction Let V : R 2 R 2 be vector field. Imgine, for exmple, tht Vx, y represents the velocity of river t the point x, y. We wish to get the description of the pth tht lef dropped in the river t the point x, y will follow. For exmple, Figure 2.1 shows the vector field of Vx, y y, x 2. Let The vector field plot of `y, x b The trjectories of `y, x 2 Figure 2.1: The vector field plot of y, x 2 nd its trjectories. γt xt, yt be such pth. At ny time, the lef will go in the direction tht the river is flowing t the point t which it is presently locted, i.e., for ll t, we hve γ t V xt, yt. If V F, G, then dx dt F x, y, }{{} y dy dt Gx, y. }{{} x 2 In generl, it will be impossible to solve this system exctly, but we wnt to be ble to get the overll shpe of the solution curves, e.g., we cn see tht in Figure 2.1, no mtter where the lef is dropped, it will hed towrds, s t. We re ssuming here tht V depends only on the position x, y nd not lso on time t. 17

22 18 2. PHASE PORTRAITS 2. Phse Portrits of Liner Systems Before considering the generl cse, let us look t the liner cse where we cn solve it exctly, i.e., V x + by, cx + dy with or x Ax, where dx dt x x + by, dy [ x y ] dt, A cx + dy, [ b Recll the existence nd uniqueness theorem for ODE s from MATB44: if ll the entries of A re continuous, then for ny point x, y, there is unique solution of x Ax stisfying xt x nd yt y. In other words, there exists unique solution through ech point; in prticulr, the solution curves do not cross. The bove cse cn be solved explicitly, where x e At [ x is solution pssing through x, y t time t. We will consider only cses where deta Rel Distinct Eigenvlues. Let λ 1 nd λ 2 be distinct eigenvlues of A nd let v nd w be their corresponding eigenvectors. Let P [v, w]. Then [ ] P 1 λ1 AP. λ 2 }{{} D Therefore, At P Dt P 1 nd we hve [ ] [ ] [ ] [ ] x e At x Pe Dt P 1 x e λ 1t C1 [v, w] y y e λ2t C 2 [ ] C1 e λ1t [v, w] C C 2 e λ2t 1 e λ1t v + C 2 e λ2t w. Different C 1 nd C 2 vlues give vrious solution curves. Note tht C 1 1 nd C 2 implies tht x e λ1t v. If λ 1 <, then the rrows point towrd the origin, s shown in Figure 2.2 which contins stble node. Note tht x C 1 e λ1t v + C 2 e λ2t w e λ2t C 1 e λ1 λ2t v + C 2 w. The coefficient of v goes to s t, i.e., s t, x,, pproching long curve whose tngent is w. As t, x e λ1t C 1 v + C 2 e λ2 λ1t w, i.e., the curves get closer nd closer to being prllel to v s t. We hve the degenerte cse when λ 1 < λ 2, in which cse x C 1 e λ1t v + C 2 w. The cse when λ 1 < < λ 2 gives us the phse portrit shown in Figure 2.2b which contins sddle point. This occurs when deta <. The cse when < λ 1 < λ 2 gives us the phse portrit shown in Figure 2.2c which contins n unstble node. We hve x C 1 e λ1t v + C 2 e λ2t w e λ1t C 1 v + C 2 e λ2 λ1t w. y ] c d ].

23 2. PHASE PORTRAITS OF LINEAR SYSTEMS 19 λ 1 < λ 2 < b λ 1 < < λ 2 c < λ 1 < λ 2 Figure 2.2: The cses for λ 1 nd λ 2. Therefore, s t, x [, ], pproching v s tngent; s t, x pproches prllel to w symptoticlly. Note tht in ll cses, the origin itself is fixed point, i.e., t the origin, x nd y, so nything dropped t the origin stys there. Such point is clled n equilibrium point; in stble node, if it is disturbed, it will come bck; in n unstble node, if perturbed slightly, it will leve the vicinity of the origin Complex Eigenvlues. Complex eigenvlues come in the form λ α ± βi, where β. In such cse, we hve x C 1 Re e λt v + C 2 Im e λt v, where v p + iq is n eigenvector for λ. Then e λt v e αt e βit p + iq e αt cosβt + i sinβt p + iq

24 2 2. PHASE PORTRAITS e αt cosβtp sinβtq + i cosβtq + i sinβtp. Therefore, x e αt[ C1 cosβtp C 1 sinβtq ] + C 1 cosβtq + C 2 sinβtp [ ] e αt k1 cosβt + k 2 sinβt k 3 cosβt + k 4 sinβt ] ] e αt cosβt [ k1 k 3 + sinβt [ k2 k 4. Note tht tra 2α. So α tra, α > tra <, α < tra >. Consider first α. To consider the xes of the ellipses, first note tht [ ] [ ] [ ] p1 q 1 C1 C 2 cosβt x. p 2 q 2 C 2 C 1 sinβt }{{}}{{} P C Except in the degenerte cse, where p nd q re linerly dependent, we hve [ ] cosβt C 1 P 1 x. sinβt Therefore, cosβt + sinβt [ cosβt sinβt ] x t P 1 t C 1 t C 1 P 1 x, cos 2 βt + sin 2 βt x t P 1 t C 1 t C 1 P 1 x, 1 x t P 1 t C 1 t C 1 P 1 x. Note tht C C t, so Therefore, [ ] [ ] CC t C1 C 2 C1 C 2, C 2 C 1 C 2 C 1 [ ] C C C2 2 C 2 C1 2 + C C2 2 I. C 1 C C1 2 +, C2 2 C 1 t C 1 C C1 2 +, C2 2 Recll tht the trce of mtrix is the sum of the elements on the min digonl.

25 Therefore, 2. PHASE PORTRAITS OF LINEAR SYSTEMS 21 C 1 t C 1 C 1 2 C C2 2 I C1 2 + I C2 2 2 C C2 2 1 x t P 1 t C 1 t C 1 P 1 x 1 x t P 1 t P 1 x. C1 2 + C2 2 Let T P 1 t P 1. Then x t Tx C C 2 2 nd T T t T is symmetric. Therefore, the eigenvectors of T re mutully orthogonl nd form the xes of the ellipses. Figure 2.3 shows stble spirl nd n unstble spirl. α < b α > Figure 2.3: The cses for α, where we hve stble spirl when α > nd n unstble spirl when α > Repeted Rel Roots. We hve N A λi, where N 2 nd A N + λi. So [ ] e e At e Nt+λtI e Nt e λti λt I + Nt. e λt Therefore, x e At [ C1 C 2 ] I + Nt [ e λt e λt ] [ C1 Note tht N 2 detn 2 detn. Therefore, [ ] n1 n 2 N. αn 1 αn 2 C 2 ] e λt I + Nt [ C1 C 2 ]. Also, N 2 trn n 1 + αn 2. Let [ 1 v α ]. Then Nv [ n 1 + αn 2 α n 1 + αn 2 ] [ ].

26 22 2. PHASE PORTRAITS So Av N + λi v λv, i.e., v is n eigenvector for λ. Therefore, [ ] [ x e λt C1 I + Nt e λt C1 + n 1 C 1 + n 2 C 2 t C 2 C 2 + α n 1 + n 2 C 2 t [ ] [ ] e λt C1 1 + n 1 C 1 + n 2 C 2 t C 2 α [ ] e λt C1 + n 1 C 1 + n 2 C 2 tv. C 2 ] If λ <, we hve s t. If λ >, then [ x y [ x y ] ] s t. Wht is the limit of the slope? In other words, wht line is pproched symptoticlly? We hve y lim t x lim t [ [ ] ] C 2 + n 1 C 1 + n 2 C 2 tv 2 C 1 + n 1 C 1 + n 2 C 2 tv 1 v 2 v 1, i.e., it pproches v. Similrly, y lim t x v 2, v 1 i.e., it lso pproches v s t. Figure 2.4 illustrtes the sitution. λ < b λ > Figure 2.4: The cses for λ, where we hve stble node when λ < nd n unstble node when λ >. We encounter the degenerte cse when N. This does not work, but then A λi, so [ ] [ ] [ ] [ ] x e At C1 e λt C1 e λt C1, C 2 e λt C 2 C 2

27 3. PHASE PORTRAITS OF NON-LINEAR SYSTEMS 23 which is just stright line through Figure 2.5 illustrtes this sitution. [ C1 C 2 ]. λ < b λ > Figure 2.5: The degenerte cses for λ when N, where we hve stble node when λ < nd n unstble node when λ >. Returning to the generl cse, we hve 3. Phse Portrits of Non-Liner Systems dx F x, y, dt dy Gx, y. dt Definition 2.1 Equilibrium point. A point where dx/dy nd dy/dt is clled n equilibrium point or singulr point or criticl point. We cn get n pproximtion to the behviour in the vicinity of ech equilibrium point by determining the behviour of the liner pproximtion. Let p, q be n equilibrium point. Since F p, q nd Gp, q, the Tylor expnsion of F x, y nd Gx, y round p, q re F x, y F x x p + F p,q y y p +, p,q Gx, y G x x p + G p,q y y p +. p,q Let x x p nd ỹ y q. So the behviour ner p, q is pproximted by tht of dx/dt Ax, where [ ] [ ] F F x x p x p,q y x, A p,q ỹ y q G G. x p,q y p,q

28 24 2. PHASE PORTRAITS Definition 2.2 Stble equilibrium point. An equilibrium point p is clled stble if for ll ɛ >, there exists δ > such tht ny solution which comes within δ of p never gets frther thn ɛ from p t ny lter time. Definition 2.3 Asymptoticlly stble equilibrium point. A stble equilibrium point p is clled symptoticlly stble if, in ddition to the properties of stble equilibrium point, there exists n r such tht every solution which comes within r of p pproches p s t. Figure 2.6 illustrtes this sitution. y y x x Stble but not symptoticlly stble. b Asymptoticlly stble. Figure 2.6: Stbility nd symptotic stbility. From 2.1, liner systems in which both eigenvlues hve negtive rel prts re stble, while, if t lest one eigenvlue hs positive rel prt, it is unstble. The follow theorem ties these ides together. Theorem 2.4. An equilibrium point is stble if the rel prts of both eigenvlues of the corresponding liner system re negtive. It is unstble if the rel prt of t lest one eigenvlue is positive. In these cses, stbility is determined by behviour of the corresponding liner system. In other words, e.g., no eigenvlue with positive rel prt but t lest one eigenvlue with no rel prt we would require the need to nlyze higher order terms not just liner terms in the Tylor expnsion to determine its behviour. Exmple 2.5. Find nd clssify the equilibrium points of F x, y 3x 3y x 2 + xy, Gx, y 3y + x 2 4xy. Solution. To find the equilibrium points, we set 3x 3y x 2 + xy, 3y + x 2 4xy. See 5, p. 32.

29 3. PHASE PORTRAITS OF NON-LINEAR SYSTEMS 25 Eqution implies tht 3 x y x x y 3 x x y x 3 or x y. If x 3, then 3y y, 9 9y, so y 1 nd 3, 1 is n equilibrium point. If x y, then 3y + y 2 4y 2, 3y 3y 2, nd y y 2 implies tht y or y 1. Therefore, two more equilibrium points re, nd 1, 1. So in summry, the equilibrium points re Note tht A F x p,q G x p,q,, 1, 1, 3, 1. F y p,q G y p,q 3 2p + q 2p 4q 3 + p 3 4p where p, q is n equilibrium point, i.e., the mtrix A is obtined by evluting its entries t the equilibrium points. At,, we hve A [ ] [ which gives us double root λ 3, which is indictive of n unstble equilibrium point. Note tht [ ] [ ] [ ] 3 b. b 3b ],, The eigenvector is [1, ]. At 1, 1, we hve [ ] [ A Finding eigenvlues, we hve [ ] 2 2 λi 2 1, 2 λ 1 λ 4, ]. 2 λ + λ 2 4, λ 2 λ 6, λ 3 λ + 2.

30 26 2. PHASE PORTRAITS Therefore, λ { 2, 3}, which is indictive of n unstble equilibrium point. For λ 3, we hve [ ] [ ] [ ] 1 2 2b 2b, 2 4 b 2 4b which gives us the eigenvector [ 2, 1]. For λ 2, we hve [ ] [ ] [ b 2 1 b 2 + b ] b 2, which gives us the eigenvector [1, 2]. At 3, 1, we hve [ ] [ A Finding eigenvlues, we hve [ ] 2 λi 2 9, λ λ + 18, ]. λ + 9 λ + 2. Therefore, λ { 9, 2}, which is indictive of stble equilibrium point. For λ 2, we hve [ 2 7 ] [ b ] [ 2 7b ], so 2 7b nd the eigenvector is [7, 2]. For λ 9, we hve [ ] [ ] [ ] 7 7, 2 b 2 so nd the eigenvector is [, 1]. Figure 2.7 shows the phse portrit. 4. Applictions 4.1. The Pendulum. Consider the pendulum in Figure 2.8. Let x, y denote the ccelertion nd let norml denote its component in the norml direction. Then we know tht Since x l sinθ, we hve Therefore, g sinθ norml x cosθ y sinθ. x d2 x dt 2 Similrly, since y l cosθ, we hve dx dt l cosθdθ dt. l sinθ 2 dθ + l cosθ d2 θ dt dt 2. dy dt l sinθdθ dt,

31 4. APPLICATIONS ,1 3,1-1, x -1 Figure 2.7: The phse portrit of F, G. y θ ` m x mg sin θ mg cos θ mg Figure 2.8: A pendulum with length l nd mss m. nd it follows tht 2 y d2 y dθ dt 2 l cosθ + l sinθ d2 θ dt dt 2. Therefore, we hve g sinθ 2 dθ l sinθ cosθ l cos 2 θ d2 θ dt dt 2 2 dθ l cosθ sinθ l cos 2 θ d2 θ dt dt 2 which finlly results in l d2 θ dt 2, d 2 θ dt 2 + g sinθ, 2.1 l

32 28 2. PHASE PORTRAITS where θ is the ngle, t is time, g is the ccelertion due to grvity, nd l is the length of the pendulum. Chnge the nottion: use x to represent the ngle. Then Eqution 2.1 becomes Let y dx/dt. Then x [ x y d 2 x dt 2 + g sinx. l ], [ dx dt y g l sinx Let F x, y y nd Gx, y g/l sinx. Then the equilibrium points re nπ, for n Z. It then follows tht nd For n even, we hve F x, F y 1, G x g l cosx, g x g nπ l 1n 1 n+1 g l. A [ 1 g l where λ ±i g/l is centre. For n odd, we hve [ ] 1 A, g l ], ]. G y, where λ ± g/l is sddle point. Figure 2.9 shows the phse portrit. The ctul solution curves re y p -2 p -p p 2 p 3 p x -2-4 Figure 2.9: The phse portrit of pendulum. given by Reducing its order gives us which finlly gives us x x + g l sinxx. x 2 g cosx C, 2 l y 2 2 g cosx + C. l

33 4. APPLICATIONS 29 Note tht closed loop in phse portrit, e.g., the ones surrounding the centres in Figure 2.9, indictes periodic solution The Dmped Pendulum. Adding n ir resistnce term rl dx dt ml d2 x dt 2 d 2 x dt 2 + r dx m dt + g sinx. l mg sinx rldx dt, to Eqution 2.1, we hve Letting y dx/dt, we hve dy dt r m y g l sinx. Let F y nd G g/l sinx r/m y. Then the equilibrium points re y nd x nπ, where n Z. Note tht nd For n even, we hve which gives us F x, F y 1, G x g l cosx, G x 1 n+1 g nπ l. A [ g l 1 r m ] λ 2 + r m λ + g l. Solving for λ gives us λ r m ± i 4g l r2 m 2. 2 Assuming tht r < 2m g/l, this gives stble spirl. For n odd, we hve which gives us A [ 1 g l r m ] λ 2 + r m λ g l.,, G y r m, Solving for λ gives us λ r m ± 4g l + r2 m 2. 2 Assuming once gin tht r < 2m g/l, this gives sddle. Figure 2.1 shows the phse portrit of dmped pendulum Predtor-Prey Equtions. Exmple 2.6 Predtor-Prey. Consider lnd populted by foxes nd rbbits, where the foxes prey upon the rbbits. Let xt nd yt be the number of rbbits nd foxes, respectively, t time t. In the bsence of predtors, t ny time, the number of rbbits would grow t rte proportionl to the number of rbbits t tht time. However, the presence of predtors lso cuses the number of rbbits to decline in proportion to the number of encounters between fox nd rbbit, which is proportionl to the product

34 3 2. PHASE PORTRAITS y p -3 p -2 p -p p 2 p 3 p 4 p -2 x Figure 2.1: The phse portrit of dmped pendulum. xtyt. Therefore, dx/dt Ax Bxy for some positive constnts nd b. For the foxes, the presence of other foxes represents competition for food, so the number declines proportionlly to the number of foxes but grows proportionlly to the number of encounters. Therefore dy/dt Cy + Dxy for some positive constnts c nd d. The system is our mthemticl model. dx Ax Bxy, dt dy Cy + Dxy dt If we wnt to find the function yx, which gives the wy tht the number of foxes nd rbbits re relted, we begin by dividing to get the differentil eqution dy Cy + Dxy dx Ax Bxy with A, B, C, D, xt, yt positive. In this cse, we cn solve explicitly s dy y C + Dx dx x A By, A By C + Dx dy dx, y A y B dy x Cx + D dx, A ln y By C ln x + Dx + C, y A e By kx C e Dx 2.2 for some constnt k. We cn use the method of implicit differentition to verify tht it is indeed solution of the eqution for ny k. MATA3.

35 4. APPLICATIONS 31 Explicitly, if yx is the function defined implicitly by Eqution 2.2, then Replcing k from Eqution 2.2 gives Ay A 1 y e By + y A B e By y k C x C 1 e Dx + kx C De Dx. Ay A 1 y e By + y A B e By y Cx C 1 e Dx ya e By x C e Dx + x C De Dx ya e By x C e Dx Dividing by y A 1 e By gives nd so solving for y gives s desired. CyA e By x + Dy A e By. Ay + y By Cy x + Dy, y Cy + Dxy Ax Bxy, The grph of typicl solution is shown in Figure y D C B 2 A x Figure 2.11: A typicl solution of the Predtor-Prey model with 9.4, b 1.58, c 6.84, d 1.3, nd k Beginning t point such s A, where there re few rbbits nd few foxes, the fox popultion does not initilly increse much due to the lck of food, but with so few predtors, the number of rbbits multiplies rpidly. After while, the point B is reched, t which time the lrge food supply cuses the rpid increse in the number of foxes, which in turn curtils the growth of the rbbits. By the time point C is reched, the lrge number of predtors cuses the number of rbbits to decrese. Eventully, point D is reched, where the number of rbbits hs declined to the point where the lck of food cuses the fox popultion to decrese, eventully returning the sitution to point A.

36 32 2. PHASE PORTRAITS To find the equilibrium points, we know tht we either must hve x or A By nd y or C Dx. Therefore, the equilibrium points re so we hve At,, we hve A,, C D, A, B [ A By Bx Dy [ A C giving us sddle point. At C/D, A/B, we hve [ BC D AD B C + Dx the determinnt of which is AC, so λ ±i AC, giving us centre point. Figure 2.12 shows the phse portrit. ], ], ] C D,A B 1, x Figure 2.12: The phse portrit of Exmple 2.6, showing sddle point t the origin nd centre point t C D, A B. 5. Lipunov s Second Method We hve been exmining linerized systems bout ech equilibrium point to get n ide of how the originl system behves. But to wht extent is it possible to conclude tht the properties of linerized systems ccurtely reflect properties of the ctul system?

37 5. LIAPUNOV S SECOND METHOD 33 Theorem 2.8. Consider x F x, y, 2.3 y Gx, y. Let V F, G nd let be n equilibrium point of System 2.3. Suppose there exists function E with the following properties: 1 Ex, y > for x, y, nd E,. 2 E is differentible. 3 For ny solution xt, yt of System 2.3, there exists n r > such tht E V whenever x 2 + y 2 < r. Then is stble equilibrium point of System 2.3. Proof. The ide of the theorem is this. Consider contour line E C, s shown in Figure Intuitively, the hypothesis tht E V sys tht V points inwrds, so tht once solution enters the E E C V Figure 2.13: Some contour line E C. region surrounded by E C it cn never leve. More precisely, if xt, yt is solution, then d dt E xt, yt E dx x dt + E dy y dt E x F + E y G E V }{{} So if p 1 xt 1, yt 1 nd p 2 xt 2, yt 2 re points on solution curve with t 2 > t 1, then t2 de dt, t 1 dt Ex, y t2, t 1 Ep 2 Ep 1. Therefore, Ep 2 Ep 1, i.e., E decreses with t, so once it enters region bounded by contour line of E, it cn never leve. We cn lwys move our point to the origin by trnsltion. Such function is clled Lipunov s Function for the system.

38 34 2. PHASE PORTRAITS Recll the definition of stbility p. 24 tht sttes tht given n ɛ >, there exists δ > such tht ny solution coming within δ of p never therefter gets frther thn ɛ of p. So given n ɛ >, let m be the minimum vlue of R on x 2 + y 2 ɛ. Such vlue exists becuse E is continuous nd the locus of x 2 + y 2 ɛ is compct. Furthermore, m > becuse E > on x 2 + y 2 ɛ. Then the contour line Ex, y m/2 lies entirely inside x 2 + y 2 ɛ, s illustrted in Figure Since E is continuous nd E,, there 2 2 x + y ² Exy, m 2 Figure 2.14: The contour line Ex, y m/2 lies entirely inside x 2 + y 2 ɛ. They cn t touch becuse there is no point on x 2 + y 2 ɛ where Ex, y m/2. exists δ > such tht Ex, y < m/2 whenever x 2 + y 2 < δ. Once solution enters x 2 + y 2 δ, then Ex, y < m/2, so it cn never therefter leve E m/2 nd thus cn never leve x 2 + y 2 ɛ. to Theorem 2.8b. Assume the hypotheses of Theorem 2.8 hold except tht condition 3 is strengthened 3 There exists n r > nd α > such tht E V αe whenever < x 2 + y 2 < r 2. Then we get the stronger conclusion tht is symptoticlly stble. Proof. Suppose tht for ny solution xt, yt of System 2.3, there exists n r > nd α > such tht E V αe whenever < x 2 + y 2 < r 2. Then Therefore, nd so Furthermore, de dt E V αe. de dt + αe αt de e dt + eαt E. e αt E C E Ce αt lim t E xt, yt, tht is, on ech solution curve, E, so x, y. Theorem 2.8c. Assume the conditions of Theorem 2.8 but conditions 2 nd 3 re strengthened to 2 E is continuously differentible s Boyce nd Di Prim ssume

39 5. LIAPUNOV S SECOND METHOD 35 stble. 3 There exists n r > such tht E V < whenever < x 2 + y 2 < r. Then is symptoticlly Proof. As in the proof of Theorem 2.8, E is decresing function long ech solution curve. We lredy showed stbility. Therefore, suppose k hs the property tht solution entering x 2 + y 2 k never leves. We wnt to show tht lim t E xt for ny solution curve xt. Suppose xt is solution curve which does not hve this property. Then there exists c > such tht E xt c for ll t. Therefore, the solution xt voids the open set E 1 [, c nd so there exists rdius R such tht the solution never enters the bll x < R. Thus, for some t, the solution lies in the nnulus R x r for ll t t. Since de/dt E V is continuous nd negtive, it ttins mximum M where M > on the compct set R x r for ll t t. Therefore, for ll t > t, This implies tht t d t dt E xt } {{ } E xt E xt t t M dt M t t. E xt E xt + Mt }{{}}{{} Mt. constnt for ll t. This is contrdiction s E xt >. Therefore, E xt eventully gets less thn ny c, i.e., lim E xt xt. t Theorem 2.9. Let be n equilibrium point of System 2.3. Suppose there exists function E with the following properties: 1 Ex, y > for some x, y in every neighbourhood of the origin nd E,. 2 E is differentible. 3 For ny solution xt, yt of System 2.3, there exists n r such tht E V > whenever < x 2 + y 2 < r. Then is n unstble equilibrium point of System 2.3. Proof. Using ides similr to previous proofs, one cn show tht this is true. Corollry 2.1. An equilibrium point is symptoticlly stble if the rel prts of both eigenvlues of the corresponding linerized system re negtive. It is unstble if the rel prt of t lest one eigenvlue is positive. Note tht d dt E`xt, yt E dx x dt + E dy dx y dt E dt, dy «E V. dt {z } V There is no conclusion if λ 1 while λ 2, e.g., if the linerized system hs centre t p, p my or my not be stble. If there exists n E > with E, such tht E V >, then the origin is not stble.

40 36 2. PHASE PORTRAITS Proof sketch. Suppose tht the rel prts of both eigenvlues re negtive. Let x [x, y]. Write dx F x, y Ax + By + fx, y, dt dy Gx, y Cx + Dy + gx, y, dt where f nd g re continuous with f, g,, nd there exist constnts k 1 nd k 2 such tht fx, y k 1 x nd gx, y k 2 x whenever x is sufficiently smll. Let [ ] [ ] [ ] A B F f A, V Ax +. C D G g Let p tra A+D nd q deta AD BC. The chrcteristic eqution then becomes λ 2 pλ+q. If the roots re rel, then, by the hypothesis, they re negtive, so their sum p is negtive nd their product q is positive. If the roots re complex, sy u ± iv, then, by the hypothesis, u < nd so gin p 2u is negtive nd q u 2 + v 2 is positive. Let Q Ax + By 2 + Cx + Dy 2 Ax Ax. nd set E Q + q x 2 + y 2. Clerly, E > for x, nd E,. Why is E V < for smll x? To see this, note tht E Q + q x 2 + y 2 Q + 2q x, y Q + 2qx nd Therefore, [ 2 Ax + By A + 2 Cx + Dy C Q 2 Ax + By B + 2 Cx + Dy D [ ] [ ] A C Ax + By 2 B D Cx + Dy 2A t Ax. E V 2A t Ax + 2qx Ax + [f, g] ] Note tht by the Cyley-Hmilton Theorem, we hve 2 X t A t A t Ax + qx t Ax + 2 A t Ax + qx [f, g]. A 2 traa + detai. Tht is, Tking the trnspose gives A 2 pa + qi. A t 2 pa t + qi, A t 2 + qi pa t.

41 5. LIAPUNOV S SECOND METHOD 37 Therefore, Eqution now becomes E V 2 x t A t 2 + qi Ax + 2 A t Ax + qx [f, g] 2px t A t Ax + 2 A t A + qx [f, g] 2pAx Ax + 2 A t A + qx [f, g] 2pQ +2 A t A + qx [f, g]. }{{} < We hve 2pQ < since p < nd Q >. Using the fct tht [f, g] k k2 2 x, we cn show tht the second term is less thn or equl to p Q for smll x. Therefore, it is not big enough to ffect the sign of E V, i.e., E V < for smll nonzero x. Exmple Consider V 2xy, x 2 y 3. Is the origin stble? Solution. First note tht the only equilibrium point is x, y,. Suppose we try Ex, y x 2 + by 2 for suitble, b >. Then E V [2x, 2by] [ 2xy, x 2 y 3] 4x 2 y + 2bx 2 y 2by 4. Choose 1 nd b 2 so tht the x 2 y term will cncel. Then E V 4y 4. Therefore, by Lipunov, the origin is stble. Figure 2.15 shows the phse portrit of V. 1 y - 1 x -1 Figure 2.15: The phse portrit of 2xy, x 2 y 3 of Exmple 2.11, showing tht the origin is stble.

42 38 2. PHASE PORTRAITS 6. Periodic Solutions Theorem 2.12 Poincré-Bendixson. Let R be closed bounded region in R 2. Suppose dx F x, y, dt dy Gx, y dt hs solution xt, yt which lies in R for ll t t. If System 2.4 hs no equilibrium points in R, then either 1 xt, yt is periodic solution i.e., closed curve or loop, s shown in Figure 2.16 or 2 xt, yt spirls towrds periodic solution, s shown in Figure 2.16b. 2.4 A periodic solution. b Spirls towrds periodic solution. Figure 2.16: Proof ide. Let C xt, yt be our given solution curve. Let p n xt n + n, yt + n. Unless C is periodic solution, the points {p n } re distinct, so by the Bolzno-Weierstrss Theorem, there exists n ccumultion point p of {p n } lying in R since R is compct. Let C be the solution curve pssing through p. Note tht since we ssumed no equilibrium points in R, p is not n equilibrium point, so C is curve, not just the point p. Intuitively, since the solution curves cnnot cross nd C hs points on it tht pproch p s limit, C must spirl towrds C. More precisely, we hve the following. Lemm Let C xt, yt be solution curve to System 2.4, let p n xt n + n, yt + n, let p be n ccumultion point of {p n } lying in R, nd let C be the solution curve pssing through p. Then there exists short line segment l through p hving the following properties: 1 The curves C nd C cross l infinitely often in every neighbourhood of p. 2 Every solution crossing l does so in the sme direction. Proof. Proof is omitted, but it uses continuity nd the Jordn Curve Theorem Theorem Let q be the next point t which C crosses l. We show tht q p so tht C is periodic solution. The curve C crosses l ner p sy, t p, so by continuity, it must cross gin ner q sy, t q. This is illustrted in Figure But then every subsequent crossing of l by C must be frther wy from p thn

43 6. PERIODIC SOLUTIONS 39 ` C q p p q Figure 2.17: The curve C crossing l. from q, since C cnnot cross itself nd it cnnot cross l in the wrong direction. But this contrdicts C crossing l infinitely often in every neighbourhood of p. Therefore, p q, so C is closed curve. The point is this. If p is frther from p thn from q, then the next crossing would be ever frther wy. But if p q, then q cn be closer to p thn p ws, so everything is oky. Thus, q is closer to p thn p ws nd subsequent crossings re even closer. Applying this rgument now to other points on C nd other lines, we cn see tht C must be pproching C. Theorem 2.14 Jordn Curve Theorem. Let C be closed curve in R 2 which does not cross itself. Then C divides R 2 into two disjoint non-empty connected open subsets, hving C s their common boundry, nmely, R 2 \ C I O. One of these open sets is bounded nd the other is unbounded. Exmple Consider Let x x y x y x + y y F x, y x y x Gx, y x + y y x y2, x y2. x y2, x y2. Find equilibrium points. Setting F gives x 1 x 2 32 y2 y nd setting G gives y 1 x 2 12 y2 x. Therefore, 1 x y2 x 2 12 y2 y x x 1 y They re clled, respectively, the inside nd outside of C.

44 4 2. PHASE PORTRAITS unless x or y. If x, then Eqution implies tht y. If y, then Eqution implies tht x. Therefore,, is one solution. Let 1 x 2 1/2 y 2. Then y 2 1. Therefore, one fctor is positive nd one is negtive, which implies tht > nd y 2 <. 2 y 2 1, 2 y To hve rel solutions for, we need y 4 4 y 2 2. But then > x y2 < 1 y 2 < 2, which is contrdiction. Therefore, no solution exists other thn,. Let V F, G. Consider the behviour of V on circles x 2 + y 2 c 2, s shown in Figure y n V x Figure 2.18: The vector V on circle x 2 + y 2 c 2. To determine if V points into the circle or out of the circle, we look t V n. Then V n > implies tht V is pointing out. V n implies tht V is tngent to the circle. V n < implies tht V is pointing in. To find out which condition it stisfies, we compute V n F, G x, y F x + Gy x 2 xy x 2 x y2 + xy + y 2 y 2 x y2 x 2 x x2 y 2 + y 2 x 2 y y4 x 2 + y 2 x y4 5 2 x2 y 2 r 2 x 4 2x 2 y 2 y y4 1 2 x2 y 2

45 6. PERIODIC SOLUTIONS 41 r 2 x 2 + y y2 y 2 x 2 r 2 r y2 x 2 y 2 r 2 r r2 sin 2 θ r 2 cos 2 θ r 2 sin 2 θ r 2 r r4 sin 2 θ cos2θ r 2 r sin2 θ cos2θ r 2 1 r sin2 θ cos2θ. Note tht sin2 θ cos2θ sin2 θ cos2θ 3 2. If r 2, then r 2 4, so r sin2 θ cos2θ 2 1 r sin2 θ cos2θ < V n <. If r 1/2, then r 2 1/4, so r sin2 θ cos2θ 38 1 r sin2 θ cos2θ > V n >. This sitution is illustrted in Figure So once solution comes within r 2, it stys within r 2, but solution outside r 1/2 stys outside r 1/2. Therefore, let R be the region between r 1/2 nd r 2. This region contins no equilibrium points, but ny solution which enters it stys within it. So pplying the Poincré-Bendixson Theorem Theorem 2.12, p. 38 shows tht ny solution within R spirls towrds periodic solution within R, s shown in Figure Figure 2.2 shows the phse portrit y R V x V 1 r 2 r 2 Figure 2.19: If r 2, then V points out. If r 1/2, then it points in.

46 42 2. PHASE PORTRAITS of the system. 2 y x -1-2 Figure 2.2: The phse portrit of Exmple 2.15, showing tht the origin is the only equilibrium point. Exmple Consider x x y + 1 x 2 + y 2, x2 + y 2 y y x + 1 x 2 + y 2. x2 + y 2 solve Immeditely, note tht x, y, s we must enforce x 2 + y 2. To find equilibrium points, It follows tht y + x + y 2 }{{} Eq. x x2 + y 2 1 x 2 + y 2, y x2 + y 2 1 x 2 + y 2. xy x2 + y 2 1 x 2 + y 2 }{{} Eq. Therefore, there is no solution in the domin of V, which is R 2 \ {}. Consider V n on the circle x 2 + y 2 c 2. Then V n F x + Gy xy + x 2. x 2 x2 + y 2 1 x 2 + y 2

47 6. PERIODIC SOLUTIONS 43 + xy + y 2 1 x 2 + y 2 x2 + y 2 x 2 + y 2 1 x 2 + y 2 r 1 r 2. Therefore, if r > 1, then V n <, while if r < 1, then V n >. So the solutions entering the nnulus 1/2 r 3/2 sty there. Since there re no equilibrium points in this nnulus, by the Poincré-Bendixson Theorem Theorem 2.12, p. 38, it hs periodic solution. In fct, let r 2 x 2 + y 2. Then where r. Now, we hve 2rr 2xx + 2yy, rr xx + yy xf + yg r 1 r 2, r 1 r 2, dr dt 1 r2, dr 1 r 2 dt, 1/2 1 r + 1 /2 dr dt, 1 + r 1 1 r + 1 dr 2 dt, 1 + r 1 + r ln 1 r 2t + C, 1 + r 1 r ke2t, r ke2t 1 ke 2t + 1. Therefore, lim t r k/k 1, i.e., ll solutions spirl towrds r 1 s t, s Figure 2.21 shows. Exmple 2.17 vn der Pol Eqution. Consider x + µ x 2 1 x + x, µ >. 2.5 Let To find equilibrium points, we let x y F, y x µ x 2 1 y x G. y,

48 44 2. PHASE PORTRAITS x -1-2 Figure 2.21: The phse portrit of Exmple 2.16, showing tht ll solutions spirl towrds r 1 s t. µ x 2 1 y x. Since y, immeditely x. Therefore,, is the only equilibrium point. Consider the linerized system { x y, Then the mtrix of the system is [ y x + µy. 1 1 µ The chrcteristic eqution is λ 2 µλ + 1. The solutions of this re Note tht ] λ µ ± µ µ > 2 gives us n unstble node distinct rel roots. µ 2 gives us n unstble node repeted rel positive roots. µ < 2 gives us n unstble spirl complex roots. Does it hve ny periodic solutions? We cn ttempt to find out with V n F x + Gx xy µ x 2 1 y 2 xy µ x 2 1 y 2. But this indictes nothing. Thus, we need to try different-looking region R not n nnulus. To crry on with this solution, we need new tool: Liénrd s Theorem..

49 6. PERIODIC SOLUTIONS 45 Theorem 2.18 Liénrd s Theorem. Let f, h : R R nd let gx x ft dt. Suppose tht 1 f is continuous. 2 f is even. 3 there exists n > such tht g < for < x <. gx > for x >. fx > for x >. 4 lim x gx. 5 h is odd nd hx > for x >. Then x + fxx + hx hs unique periodic solution nd every other solution spirls towrds it. Proof. We convert x + fxx + hx to system. Let y x + gx. Therefore, So the system is y x + dg dx x fxx hx + fxx hx. { x y gx, y hx. Note tht f is even implies tht g is odd. Therefore, replcing x by x nd y by y leves the equtions unchnged, so the solutions re symmetric bout the origin. Hence, if we know the solutions for x, we cn get those with x by reflection bout the origin. So ssume tht x. Let γ be the grph of y gx with x. Let x, y lie on γ nd let C x be the solution which psses through x, y t t. y A 2 C x γ x x y, A 1 Figure 2.22: The solution C x reching the y-xis t both ends. Lemm As t increses from, x decreses nd y decreses until eventully the y-xis is reched. As t decreses from, x decreses nd y increses until eventully the y-xis is reched. Proof. Since y hx <, y lwys decreses s t increses. On γ, V, hx, so C x γ heding stright down, nd fter pssing x, y cn never get bove γ. leves

50 46 2. PHASE PORTRAITS So x y gx < when t. Therefore, x decreses s t increses from. Similrly, C x enters γ heding stright down, so it ws never below γ for t. So x y gx > when t. Therefore, x decreses s t decreses from. It remins to be shown tht C x ctully reches the y-xis on both ends. From wht we hve shown so fr, it might look like the sitution illustrted in Figure y C x γ x y, x Figure 2.23: Let kx 2 so tht kx for x. Given constnt b, let Then differentiting with respect to t gives Therefore, r b r b hx b gx. x hx dx r 2 b kx + y b 2. 2r b r b dk dx x + 2 y b y 2hx y gx + 2 y b hx 2hx b gx. Since g is continuous nd [, x ] is compct, g hs both minimum m nd mximum M on [, x ]. Choosing b m gives gx b for ll x [, x ], so r m r m <. Therefore, r m <, so r m decreses with t. But r m y m 2, so y m 2 does not go to s in the digrm we wish to rule out. Similrly, choosing b M gives gx b, so r m increses with incresing t. Equivlently, r m decreses with decresing t, so the distnce from C x to, M decreses s t, nd in prticulr does not go to. Thus, the y-xis is reched on this side lso. Given x, let y 1 x nd y 2 x be the vlues of y where C x crosses the y-xis. Reflection bout the origin gives nother section of the solution curve C x s shown in Figure We wish to show tht there is vlue of x for which y 2 x y 1 x so tht the two hlves piece together to give periodic solution.

51 6. PERIODIC SOLUTIONS 47 A 1 y C x A, y x 2 2 x A 2 x y, A, y x 1 1 Figure 2.24: The solution curve C x crosses y t y 1 x nd y 2 x. Lemm 2.2. There exists unique x such tht Proof. Recll tht nd tht given constnt b, x < x y 2 x < y 1 x, x x y 2 x y 1 x, x > x y 2 x > y 1 x. kx : 2 We choose b to obtin r 2 kx + y 2. Therefore, x hx dx r 2 b : kx + y b2. rr hxgx gx dy dt holds on ny solution curve. Let ω gx dy, first order differentil form. Let Ix : with C x directed bckwrds from A 1 to A 2. Let t 1 nd t 2 be the vlues of t t A 1 nd A 2, respectively. Then We now need to show the following: C x ω Ix ω C x gx dy C x t2 t 1 r dr dt dt r2 2 y2 2 + k y 2 1 k 2 t2 t 1 gx dy dt dt tt 2 1 rt2 2 rt 1 2 tt 1 2 y2 2 y1 2. 2

52 48 2. PHASE PORTRAITS 1 Ix < if x <. 2 Ix strictly increses with incresing x for x >. 3 lim x Ix. To show 1, if x <, then gx < for ll x on C x, nd dy/dt is incresing in the direction of C x re following. So Ix gx dy < C x for x <. To show 2, consider < x < x. Let we C x C 1 C 2 C 3, C x C 1 C 2 C 3 s shown in Figure Then y A 2 C 3 A 2 C 3 D 2 y g x end of τ τ b x y, bg, b x, y x C 2 σ A 1 A 1 C 1 C 1 C 2 D 1 Figure 2.25: ω gx dy C 1 C 1 gx hx y gx dx gx dy dx dx gx dy /dt dx/dt dx hx gx gx y dx.

53 6. PERIODIC SOLUTIONS 49 On C 1, gx y is lrger thn it is on C 1, so gx y 1 is smller. But gx when x [, ], so gxhx gx y 1 is lrger less negtive on C1 thn it is on C 1. Therefore ω ω. C 1 C 1 Similrly, C 3 ω gx dy dx dx gx hx y gx dx hx gx y gx dx. On C 3, y gx is lrger thn it is on C 3, so y gx 1 is smller. But gx, so gxhx y gx 1 is lrger less negtive on C 3 thn on C 3. Therefore, ω < C 3 Finlly, let σ be the portion of C 2 between D 1 nd D 2. Since fx dg/dx > when x >, ech point on σ hs lrger vlue of gx thn the corresponding point the one with the sme y-coordinte on C 2. Therefore, ω < C 2 σ C 3 ω. ω < where the second inequlity comes from the fct tht since gx > on x >, the integrl over C 2 σ is positive. Therefore ω + ω + ω < C 1 C 2 C }{{ 3 } Ix i.e., Ix strictly increses with incresing x when x >. C 2 ω, ω + C 1 ω + C 2 C 3 ω } {{ } I x To show 3, select b so tht < b nd b is less thn the x-coordinte of the point where C x crosses the x-xis. Let τ be the verticl line segment through b s shown in Figure Then ω < τ C 2 ω by the rgument we used to show C 2 ω < ω. Note tht C2 ω gx dy gb τ τ gb length of τ gby gbgx. Since lim x gx, we hve C 2 ω s x. Therefore, lim Ix. x It follows from 1, 2, 3, nd from continuity tht there exists unique x such tht Ix, Ix <, x < x, Ix >, x > x. So C x pieces together with its reflection bout the origin to form periodic solution, s shown in Figure Also, x < x y 1 > y 2. Therefore, the solutions inside C spirl out to C. Similrly, solutions outside τ dy,