3 First order nonlinear equations

Transcription

1 Separable equations 3 First order nonlinear equations d f dx = where f(x,) is not (in general) a linear function of. ( x, ) The equation is autonomous if there is no explicit dependence on the independent variable, i.e. f(x,) = f(). 3.1 Separable equations These are often straight forward to solve. Suppose we can write f ( x, ) d dx = ( ) ( ) g x = h Then the ode ma be written as h( ) g( x) h( ) d = ( ) g x dx H() = G(x) + const h = dh/d; g = dg/dx d sin x = dx 1+ Rearrange: d d = + = sin x dx dx 1+ 1 Integrate d ln sin xdx cos x A + = + = + Arbitrar constant of integration A determined b boundar condition () = 1 + ½ = 1 + ½ A A = ln + = cos x ln = ½ ln ln + = 3 cos x (*) End of Lecture 7 Cannot give explicit solution for, but clearl periodic with period π. For some quantit u, the value of u + ln u is a monotonicall increasing function of u. The minimum value of the right-hand-side of (*) is 3 = 1, at x = nπ, and the maximum value is 3+ = 5 at x = (n+1)π. + ln = 1 = 1 is the minimum value. The maximum value is around = 1.9. Note: if is a solution, then is also a solution to this equation. 56

2 Separable equations FAMILIES OF SOLUTIONS In the above example we considered a specific solution, selecting A = 3 in order to satisf the boundar condition () = 1. This is one of a famil of solutions that ma be obtained using other values of A. 4 Increasing A 3 1 A cos x ln + = ; red A=3; green A=4, blue A=5, black A=6; magenta A=.5, can A=.1, dark gre A=.1. [See S3_1_FamilOfSolutionsForSeparableEquation.nb] As A, and dominated b ~ A amplitude of oscillation becomes small. As A, and dominated b ln ~ A amplitude of oscillation becomes small. Isoclines: contours of constant d/dx. Here vertical lines at x = nπ (where d/dx vanishes) are isoclines Can be helpful for sketching solution. Solution with greatest amplitude for ln + = A cos x. max() = t + ln = A+ t t min() = b + ln = A t b + ln( t / b ) = 4 Let Y ½ ( t + b ), B ½ ( b b ) t = Y + B, b = Y B Y + B 4BY + ln = 4 Y B b b 57

3 Separable equations Look for extremum db/da = 1 1+ B 1 B BY + B+ = Y + B Y B ( ) 1Y B Y + B B 1 B+ = B = B 1 = Y B Y B Y B Y = B + 1 Y = ± (B + 1) Substitute back B B 1 B + 1+ B + 1+ ln 1= B + 1 B Which requires numerical solution, giving B =.48, Y = 1.11, t = 1.59, b = FLOW MAP d = [Note: separable, but we won t separate] dt Consider t( 1 ) Here we wish to map out the solution, without directl solving the ode. Construct the flow, i.e. arrows showing the slope. d/dt = t at =. For large, then d/dt large and negative. For < 1, then d/dt >. 58

4 Separable equations d/dt t [Fields produced b Flow.dfc] Arrows are vectors showing direction of solution; the flow. Arrows often referred to as flow vectors. Note that since f(t,) is single valued then solutions cannot cross. 59

5 Separable equations t Can draw on solution easil using arrows. There is an analog between the flow vectors and velocit vectors in a fluid flow, and between the solutions and streamlines. Note: solutions drawn with uniform spacing for (), but onl up to () = t The green lines are orthogonal to the red ones. These might represent, for example, height contours above a plane, or contours of constant potential (see IB Fluids next ear). Note potential contours drawn with uniform spacing for t at =. 6

6 Separable equations To construct this map of the solution we did not require the equation to be separable. However, since it is, we can determine the analtical solution. d t( 1 ) dt = 1 d d = + = t 1 dt 1 1+ dt Integrating ln 1+ ln 1 = ln = ( t + c) 1+ =± Ae =± e 1 t t + c t t t + c ± Ae 1 Ae 1 e 1 = = = t t t + c ± Ae + 1 Ae ± 1 e ± 1 = tanh ½(t +c) or = coth ½(t +c) Note: We could have approximated the behaviour for large without solving the exact equation b noting that 1 d so t dt 1 d t dt 1 1 t + const for t t some arbitrar t. Exploring the behaviour c > () > c < () < = tanh ½(t +c) has < < 1; 1 as t = coth ½(t +c) has > 1; 1 as t = tanh ½(t +c) has 1 < < 1; 1 as t = coth ½(t +c) has < 1; as t EQUILIBRIUM SOLUTIONS A stead or equilibrium solution is one for which d/dt = = const. For the previous example, d/dt = on = ± 1 = const is a solution if = ± 1. The equation converges towards = 1 and diverges awa from = 1. = 1 is said to be a stable equilibrium. = 1 is said to be an unstable equilibrium ISOCLINES These are contours of constant slope, i.e. f(t,) = const. For the previous example, f(t,) = t(1 ), 61

7 Separable equations. 1.5 c < c > c < t Note: All curves asmptote to t = and = ±1. This emphasises that approach to stable equilibrium (and departure from unstable equilibrium) is faster at high t. d sin x sin x α In the earlier example =, we can set f = =, sa. The isoclines are therefore dx given b (/α) sin x + 1 = sin x sin x = 1 α ± α for α 1 For solution to exist, require sin x α >. When sin x = α, then = 1 and location of isoclines is single-valued so the must be vertical at =. As d/dx = when sin x =, then x = nπ must have vertical isoclines with α =. Upper branch of isocline has x x x = + as α becomes small. sin sin sin 1 α α α 6

8 Stabilit Lower branch of isolcine has 4. sin x sin x sin x α = 1= 1 1 sin α α α x 4 sin x 1 α 1 α = ! sin 8sin 4 α x x 1 α α = cosec x sinx Yellow line: sin x Red line: cosec x. End of Lecture 8 3. Stabilit t α > α > α < The technique we shall explore here is not reall necessar for first order odes, but is much more useful for higher order odes. 63

9 Stabilit 3..1 PERTURBATION Suppose that = is the equilibrium solution, i.e. f(t, ) =. Consider the behaviour of = + ε u where < ε $ 1 and u = u(t) = O(1) For the example used in 3.1. t( 1 ) d dt =, = ± 1 and consider = ±1 + εu. ( ) d u t u dt Substitute into the ode: ( + ε ) = 1 ( + ε ) ( 1 ) ( ) du ε = t + ε u+ ε u = tεu + εu dt ( ) and linearise (discard terms in ε and higher) du tu dt Solve with u() = u = 1 u u e t u u e t as t Solutions converge towards = 1 = 1 is stable. = 1 t u u e ± as t Solutions diverge from = 1 = 1 is unstable. As we shall see, a solution is stable if f/ < and unstable if f/ >, but what if f/ =? 3.. SEMI-STABLE d 1 dt = Consider ( ) f ( ) Equilibrium solution requires f() = = 1. Since f is independent of t, all solutions have the same shape. 64

10 Stabilit t t Solutions for < 1 converge on equilibrium = 1, but for > 1 solutions diverge. 1 1 The equation is separable: ( 1 ) d = = t t 1 = 1 t t Stabilit For t > t, ~ 1 1/t 1 from below as t. For t < t, ~ 1/(t t) + as t t from below, so rises from - as t increases from t. Stabilit analsis: let = 1 + εu d dt du 1+ u = = 1 1 u = u dt ( ) ( ) ε ε ε ε du dt = εu 65

11 Stabilit Marginall stable GENERIC STABILITY PROBLEM d dt = Consider f ( t, ) 1. Find fixed points (equilibrium solutions) b solving f(,t) = to find solutions with = const t. [It is not sufficient for d/dt to vanish for onl some period of time.]. Expand f about an equilibrium solution =, sa. Note: Each equilibrium solution must be investigated separatel. a. = + u, sa, and expand f in Talor series about = : f 1 f f t = f t + u + u +! b. (, ) (, ) 3. Substitute into equation = = a. du f 1 f u u dt = = = + +! Solutions: f If = α () t, then du αu for sufficientl small u. dt = u u e α if α > unstable if α < stable. If α(t) changes sign as t varies, then must stud with greater care; ma be stable or unstable. Fortunatel, there are a large number of problems in which f is independent of t, so α is constant. dt If α = t, then du dt βu with β () t 1 = f = 1 1 du = = u u 1 u =, β dt β dt and sstem can be semi-stable. For β = const, have u = 1 β ( t t ) as t increases towards t from below.. If t > t, then u as t increases. If t < t, then u sign(β) 66

12 Stabilit d f = [For previous example, ( 1 ) dt = ( 1) u = 1 ( t t ) and semi-stable., f =, so at = 1 equilibrium, If β =, need to look at further terms of Talor series in order to determine stabilit PHASE PORTRAITS Further insight ma be gained b considering the phase portrait of the differential function. The phase portrait is a plot of d/dt against (effectivel f(,t) against t). Consider d 1 dt =. This clearl has equilibrium solutions or fixed points at = ±1. d/dt Recall that f/ indicates the stabilit of the fixed points. When f/ < the solution is stable as d/dt will become negative if a perturbation moves above the fixed point, thus causing it to move back towards the fixed point. Conversel, f/ > leads to d/dt increasing as we move awa from the fixed point, giving an unstable solution. A stable fixed point is an attractor, while an unstable fixed point is a repellor. d The semi-stable equation = ( 1 ) explored in 3.. has the attractor and repellor merging: dt 67

13 Exact equations d/dt End of Lecture Exact equations Recall that we can convert a linear equation of the form + q = f into an exact equation of the d form ( I ) = If using the integrating factor I = e qdx. Here we shall tr to do a similar thing for dx nonlinear equations. Note: There is no simple guaranteed procedure for this it is not alwas possible. It will be introduced b example. d 3 Consider 3x x dx + + =. This equation is neither linear nor separable. Can we solve it? 3 3 B inspection we can see 3 d d x + + x = ( x + x ) =, and the equation is exact. dx dx 3 Hence the solution is (, ) ψ x x + x = c (c = const) dψ is an exact differential. ψ is sometimes called a potential. It is a conserved quantit. c x 3 =. x FINDING AN EXACT EQUATION d dψ, +, =, which we wish to write as = for some ψ(x,). dx dx Consider f ( x ) g( x ) Recall d ψ ψ d ψ = + dx dx x In the previous example, ψ, so we need = f ( x, ) ψ x and = g( x, ). 68

14 Exact equations ψ = 3x ψ = x 3 + A(x), ψ 3 = + x ψ = x 3 + x + B(), x where A(x) and B() are the constants of integration. Now the two expressions for ψ must be identical, so A(x) = x + c and B() = c, thus ψ = x 3 + x + c. However, not all equations are exact, so this procedure is not alwas possible. ψ ψ Since = x x, then we must have f g = ; can therefore test to see if an exact differential x is possible before tring to find it! d 3 3x dx + = 1 d Note: separable, so it can be written as dx = 3x, but we look to make it exact. f = 3x f/ x = 3 g = 3 g/ = 6 f/ x equation is not exact. But the equation can be made exact b multipling the equation b x: d 3 3x x dx + = f/ x = 6x = g/. Integrating f = ψ/ ψ = x 3 + A(x) Integrating g = ψ/ x ψ = x 3 + B(x). Equating ψ = x 3 + const, hence = cx /3. Here we have converted the equation into an exact equation b multipling it b an integrating factor (here x) INTEGRATING FACTOR FOR NONLINEAR EQUATIONS The idea of an integrating factor was introduced in.3. for linear equations. The process of finding the integrating factor for a nonlinear equation is more complex. d Multipl the general equation f + g = b the function µ(x,), so dx d µ f + µ g =. dx For this new equation to be exact, we must have x ( µ f ) = ( µ g ) (*) 69

15 and ψ = µ f Examples ψ ( x, ) = µ f d+ A( x) ψ = µ g ψ ( x, ) = µ g dx+ B( ) x Unfortunatel, determining µ b solving (*) ma be as difficult (or impossible) as solving the original differential equation. [Of course, equations on the Examples Sheet ma be possible ] [Finding appropriate A(x) and B() is tpicall more straight forward.] Sometimes the solution of (*) ma be simplified b proceeding on the assumption that µ is a function of x or alone. d 3 3x + = [Our previous example] dx Multipl b integrating factor, assuming µ = µ() [We alread know µ = x works!] f µ g µ 3 ( µ f ) = µ = 3µ = ( µ g) = g+ µ = + 6µ x x 1 dµ 3 = µ ~ 3/ µ d So ψ = µ f = 3x 1/ ψ = x 3/ + A(x) ψ 3/ and = µ g = ψ = x 3/ + B(). x Hence ψ = x 3/ + c and = Cx /3 as before! Note that the integrating factor is therefore not unique. Since the solution has ~ x /3, then µ ~ 3/ ~ x, the previous integrating factor. Moreover, one would expect an combination of x and that is equal to the integrating factor x would also be an integrating factor, e.g. µ = x Examples CHEMICAL KINETICS Consider the chemical reaction X + Y U + V e.g. NaOH + HCl H O + NaCl Suppose the concentration of X is x(t), of Y is (t) and U is u(t) (the reaction gives equal quantities of U and V, so v(t) = u(t)). If x() = x, () = and u() =, then the conservation relations give x + u = x and + u =. Suppose the reaction rate r is λx, where λ = λ(t) [Note that if X + Y, then λ x, etc.] du Then = r = λx = λ( x u)( u) f ( u). dt Fixed points (equilibria) are where f(u) = u = x and u = (i.e. all of one of the species is converted). 7

16 Examples Need to keep in mind the phsical meaning of x and : clearl cannot have a negative concentration! We shall assume x < (the converse ma be treated similarl). The phase portrait ma then be drawn: f x u As u() =, we have the reaction rate f(u) decreases monotonicall towards the equilibrium u = x. u x t Of course, here we can determine the exact solution as equation is separable: 1 du du = = λ, x u u dt x x u u dt ( )( ) 1 x ln ( x u) ln ( u) = λ ( t t ) x u = e = Ae u λ( t t )( x ) λt( x ) λt( x) λt( x) ( 1 ) = u Ae x Ae 71

17 Examples Noting that u() = gives A = /x so x Ae u = Ae e e λt λxt λt λxt e e λt λxt u = x e λt xt xe λ Note: Problems such as this including the derivation of the ode(s) are fair game for examiners. Maths is not onl about solving the equations, but also about deriving them and drawing conclusions from the solutions. In Part IA the examiners would help ou with an derivation required in an exam POPULATION DYNAMICS Constant birth and death rates Suppose birth rate is proportional to population, birth rate = α [For man species, the number of females matter, but not the number of males, so α might not be constant. Moreover, for some species, e.g. rabbits, overcrowding causes the birth rate to decline.] Suppose also that the death rate is proportional to the population, death rate = β [This ma be true for death b natural causes and hunting b another species. It will tpicall not be true for epidemics, combat between members, etc.] d = = r dt ( α β) = e population grows or decreases exponentiall, depending on sign of α β. End of Lecture 1 Fighting for limited resources If α > β then the population will grow too large for the available resources (e.g. food suppl). Scarcit of food ma lead to fighting (& death) between individuals who happen across the same food at the same time. The probabilit of one individual being at a given location at a given time is proportional to the population,, so the probabilit of two individuals meeting at a given location is proportional to, hence where s is a constant. Thus death rate due to fighting = s d = = dt Y rt. ( ) r s r 1 f with Y = r/s. Note that this should hold even if α < β. This is the logistic differential equation. The equation is autonomous as it has no explicit t dependence. 7

18 Examples The phase portrait shows critical points (fixed points or equilibria) at =, Y. f Y Y/ The = fixed point has f/d > so is unstable, whereas at = Y f/d < so the fixed point is stable. 73

19 Examples positive curvature negative curvature t Clearl, = Y is a stable equilibrium, whereas = is unstable. Logistics equation is separable and can be solved explicitl: 1 d 1 1 d = r + = dt Y dt 1 Y ln rt c Y = + Suppose = at t = t, then A = e rt /(Y ) and Ae Y = rt 74

20 Examples ( t ) rt Ye = Y + e which gives Y as t. But can population reall rise from just above zero? With careful management it is possible (e.g. Old Blue see later), but in general Mating opportunities Individuals need to find mates to procreate. This can become difficult when population densities are ver low. The probabilit of finding a mate is similar to the probabilit of fighting over food and is again proportional to. Of course, fighting or procreating depends on who meets! For low population densities (where fighting is not an issue) the birth rate is λ, while the death rate remains β, so ( t ) d = 1 f ( ) dt β + λ = β X This equation has the same structure as previousl, with critical points at = and = X. 3. rt t Now, = X is unstable, so if population falls below = X, then it will fall to zero and extinction! 75

21 Examples Passenger pigeon The North American passenger pigeon is one of the most famous recent examples of extinction. In mid 19 th centur, ~ 1 9, the most numerous bird on earth. Widel hunted for sport and food in late 19 th centur; as man as 3, shot at a time! B 1896 onl 5, remaining Extinct b 19. Extinction was ultimatel caused not b hunters killing the remaining birds, but b their inabilit to find mates. Their extinction was one of the earl triggers for conservation. [Lots of information on the web.] Coming back from the brink Introduction of mammals (cats, rats, ferrets, opossum, etc.) to New Zealand has devastated the populations of man species of birds. Some have become extinct, but others have brushed with the edge of extinction onl to recover through careful management, with human intervention ( divine intervention?) effectivel bpassing the problem of finding a mate. 76

22 Examples The New Zealand black robin is an extreme example: in 198 there was onl a single breading female and four males. Now there are around 5 and the population is growing. Logistic growth with a threshold Populations above the critical threshold = X cannot reall grow without limit, so some new limiting factor must come into pla when is large enough. Could model this as d = r 1 1 f dt Y Z For Y < Z the phase portrait is f Y Z Clearl equilibria are =, Y, Z, with two stable and one unstable fixed points. 77

23 Comparison with discrete equations t Stable equilibria at =, Z, and unstable at = Y. 3.5 Comparison with discrete equations Consider the logistics equation d = r 1 dt Y r > ; Y >. [ = Y stable, = unstable] The Euler finite difference approximation (see..3) to this ma be written as ( ρ ) n+1 n = r dt (1 n /Y) n = + Y ρ n = ρ Y + n = λ 1 n ξ n n+ 1 n ρ 1 n where ρ = r dt, λ = (1 + ρ) and ξ = Y(1+ρ)/ρ. Let u n = n /ξ u n+1 = λ (1 u n ) u n This normalised nonlinear first order difference equation has onl one important parameter, λ, whereas in the original we had r and Y. n 78

24 Comparison with discrete equations EQUILIBRIUM SOLUTIONS Put u n+1 = u n : u n = λ (1 u n ) u n (λ 1 λ u n )u n = u n = or u n = (λ 1)/λ U. [c.f. =, Y] 3.5. STABILITY Similar to what we did for the differential equations. Near u n =, have u n $ u n u n+1 λ u n u n λ n u Hence unstable if λ > 1. Now since λ = 1 + ρ, and ρ = r dt >, then u n = unstable equilibrium. Near u n = U = (λ 1)/λ, suppose u n = (λ 1)/λ + v n v n $ U λ 1 λ 1 λ 1 1 λ 1 + vn+ 1 = λ 1 vn + vn = λ vn + vn λ λ λ λ λ λ 1 = + ( ) vn λvn λ λ 1 = + ( λ) vn λvn λ 1 ( λ 1) v ( λ ) n+ 1 λ n ( ) Stable if λ < 1 1 < λ < 1 1 < λ < 3 v n v v n If we are using the difference equation as an approximation to the logistic differential equation, then ρ =r dt will be small, so λ < 3 as required for stabilit. However, for real populations, the population ma be better approximated b the difference equation than the differential equation. For example, most births occur at almost the same time in the breeding season, so ρ ma represent a whole ear, or even a whole generation, thus the meaning of λ ma be different and ma not be close to 1. Following the same arguments as for the stabilit of differential equations ( 3..3), we can relate the stabilit to the gradient in the difference formula. Let u = U + v u = U is equilibrium solution Then u = U + v = f ( u ) = f ( U + v ) f ( U) + v f ( U) n+ 1 n+ 1 n n n 79

25 Comparison with discrete equations n so we require + f ( U) v v for stabilit (v n as n ). For the current equation, we therefore require n 1 < 1 f d = u du λ = λ ( ( 1 uu ) ) ( 1 u) is less than unit when u =U = (λ 1)/λ (the equilibrium solution). Substituting λ 1 + = λ < 1, λ which gives the same result as before BEHAVIOUR CLOSE TO LIMIT OF STABILITY How does u n approach U as n for different values of λ? 8

26 Comparison with discrete equations 1..8 Ke lambda =.8 U = -.5 =. =.4 =.6 =.8 = Ke lambda = 1.5 U =.33 =. =.4 =.6 =.8 = u u n n 1..8 Ke lambda =.8 U =.64 =. =.4 =.6 =.8 = Ke lambda = 3. U =.69 =. =.4 =.6 =.8 = u u n n The λ = 1.5 and λ =.8 solutions are stable and approach the equilibrium solution u = U. The convergence can be monotonic, or oscillator. Wh? End of Lecture GRAPHICAL APPROACH TO DIFFERENCE EQUATION Plotting u n+1 against u n provides a straight forward and instructive method of graphicall solving the difference equation. If u n+1 = f(u n ) then begin b plotting f(u n ) and the line u n+1 = u n : For λ =.8: 81

27 Comparison with discrete equations u n+1 u n Clearl u n as n regardless of starting point. For λ = 1.5: u n+1 u n Convergence is monotonic close to equilibrium solution. For λ =.8: 8

28 Comparison with discrete equations u n+1 u n Initial monotonic convergence changes to oscillator convergence. For λ = 3.: u n+1 u n Initial convergence changes to a stead oscillation with period two. The appearance of this new oscillator solution is called a bifurcation. In this case the period two oscillation is stable; the original solution still exists but it is now unstable. The phenomenon is called period-doubling. At λ = there is a second period-doubling bifurcation in which the period two solution becomes unstable and a stable period four oscillation appears. 83

29 Comparison with discrete equations u n Ke lambda = 3.5 U =.71 =. =.4 =.6 =.8 = 1..6 u.4. u n n For λ slightl larger get another period doubling to an oscillation with period eight, then to sixteen, etc., with values of λ converging to λ c = LOGISTIC MAP Valuable insight into the behaviour of the difference equation ma be gained b plotting its asmptotic solutions (as n ) as a function of the control parameter(s). For our logistic difference equation u n+1 = λ (1 u n ) u n, this means plotting u n against λ for large values of n. 84

30 Comparison with discrete equations u lambda When λ < 1, the difference equation converges on u =. In the range < λ < 3 a stead solution is obtained with u = (λ 1)/λ. As noted before, we have a bifurcation at λ = 3 to a period two oscillator solution (hence two values of u are observed). From λ.3499, four solutions are observed following a period-doubling bifurcation. Increasing λ further results in additional perioddoubling bifurcations until λ = λ c = Beond this point all periodic solutions are unstable and we have deterministic chaos (the nearl solid black bands in the figure). Tpicall two solutions that ma be close together at one n diverge rapidl as n increases. 85

31 Comparison with discrete equations BANDS u.6 P8 P16.5 P lambda For λ > λ c there are intervals of values of λ in which stable periodic orbits exist. These attract almost all initial conditions and so show up as clear bands in the diagram. If we look hard enough it turns out that there is an interval containing a stable periodic orbit of ever integer period. 86

32 Comparison with discrete equations u lambda For λ > λ c there are intervals of values of λ in which stable periodic orbits exist. These attract almost all initial conditions and so show up as clear bands in the diagram. If we look hard enough it turns out that there is an interval containing a stable periodic orbit of ever integer period. For λ > 4, tpical initial conditions are mapped out of the interval [,1] and of to LOCATION OF BIFURCATIONS Recall that when we were looking for equilibrium solutions we set u n+1 = u n and solving u n = λ (1 u n ) u n For the period two orbit, we can do a similar thing, but set u n+ = u n. Now u n = u n+ = λ (1 u n+1 ) u n+1 = λ (1 λ (1 u n ) u n ) λ (1 u n ) u n (*) which ields a quartic in u n. However, we alread know two of the roots: since u n+1 =, U satisf both u n+ = u n+1 and u n+1 = u n, then the are also solutions to (*). Factorising, u n (1 λ + λu n )[1 + λ λ(1+λ)u n + λ u n ] =. 87

33 Comparison with discrete equations The term in square brackets provides additional roots 1+ λ± λ λ 3 u n = (**) λ provided λ λ 3 > λ < 1 or λ > 3. The new roots exist when λ 3 and correspond to the new period two orbit. From a perturbation analsis we can show that the are stable for λ close to 3; in this case the period-doubling bifurcation is termed supercritical. Note that at λ = 3, the new root is a double root with u n = /3, whereas the equilibrium root also has u n = /3. In 3.5. we saw that f/ u < 1 for stabilit of the equilibrium solution. We can use the same ideas here for the oscillator solution, where we have u n = f(f(u n )). The stabilit boundar therefore requires ( f ( u) ) f = λ ( 1 u)( 1 λu+ λu ) =± 1 u Substituting in either of the oscillator solutions (**) and simplifing gives 4 + λ λ = ± 1, with roots λ = 1 6, λ = 1, λ = 3 and λ = Onl the last two of these are relevant, and show that the period two solution is stable for 3 < λ < We ma repeat this analsis to find where the period four values of u n become possible b setting u n+4 = u n. This leads to a sixteenth order polnomial for u n, sharing the four roots of the period two orbits. Plotting the difference equation, looking for potential orbits up to period eight, shows how the structure becomes more complex as λ increases, with each successive bifurcation adding a new periodic solution branch on top of a previousl stable branch of periodic points that is now unstable. 88

34 Comparison with discrete equations u(n+?) u(n+?) lambda=.9 u(n+1) u(n+) u(n+3) u(n+4) u(n+5) u(n+6) u(n+7) u(n+8) u(n).. lambda= 3.1 u(n+1) u(n+) u(n+3) u(n+4) u(n+5) u(n+6) u(n+7) u(n+8) u(n) u(n+?) u(n+?) lambda= 3.5 u(n+1) u(n+) u(n+3) u(n+4) u(n+5) u(n+6) u(n+7) u(n+8) u(n).. lambda= 3.57 u(n+1) u(n+) u(n+3) u(n+4) u(n+5) u(n+6) u(n+7) u(n+8) u(n) 89