Physics: springmass system, planet motion, pendulum. Biology: ecology problem, neural conduction, epidemics


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1 Applications of nonlinear ODE systems: Physics: springmass system, planet motion, pendulum Chemistry: mixing problems, chemical reactions Biology: ecology problem, neural conduction, epidemics Economy: economic growth, competition Finance: stochastic equations 1
2 Population models with interactions: dx dt dy dt = a(x) + b(x, y) = c(y) + d(x, y) a(x): selfgrowth rate of x b(x, y): effect of interaction on x c(y): selfgrowth rate of y d(x, y): effect of interaction on y 2
3 predatorprey: a: prey, b: predator a(x) > 0, b(x, y) < 0, c(y) < 0 and d(x, y) > 0 Example: Canadian lynx and hare, fox and rabbit,... competition: a(x) > 0, b(x, y) < 0, c(y) > 0 and d(x, y) < 0 Example: tiger and lion, rabbit and sheep, USA and Japan in WWII,... cooperative (mutualism): a(x) < 0, b(x, y) > 0, c(y) < 0 and d(x, y) > 0 Example: plant and pollinator 3
4 A system of nonlinear differential equations: dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy A competition model, with x and y both grow at a logistic rate, and the competing for the same food source slows down the growth of both species Questions to study: what are the qualitative behavior of the nonlinear system? Given an initial value, what is the fate of both species? 4
5 Studies of phase portrait (1): nullclines dx = f(x, y) dt dy = g(x, y) dt The sets f(x, y) = 0 and g(x, y) = 0 are curves on the phase portrait, and these curves are called nullclines. The set f(x, y) = 0 is the xnullcline, where the vector field (f, g) is vertical. The set g(x, y) = 0 is the ynullcline, where the vector field (f, g) is horizontal. 5
6 dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy xnullcline: 2x(1 x 2 ) xy = 0 x(2 x y) = 0 so x = 0 or 2 x y = 0 ynullcline: 3y(1 y 3 ) 2xy = 0 y(3 y 2x) = 0 so y = 0 or 3 y 2x = 0 6
7 The nullclines divide the phase portraits into regions, and in each region, the direction of vector field must be one of the following: northeast, southeast, northwest and southwest (So nullclines are where the vector field is exactly east, west, north and south) In each region, we use an arrow to indicate the direction. (In 1d, we use only uparrow and downarrow in phase lines.) 7
8 dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy xnullcline: x = 0 or 2 x y = 0 On x = 0, x = 0 and y = 3y y 2 On 2 x y = 0, x = 0 and y = 3y y 2 2y(2 y) = y + y 2 ynullcline: y = 0 or 3 y 2x = 0 On y = 0, y = 0 and x = 2x x 2 On 3 y 2x = 0, y = 0 and x = 2x x 2 x(3 2x) = x + x 2 The nullclines divide the region x > 0 and y > 0 to four subregions. 8
9 Studies of phase portrait (2): equilibrium points dx = f(x, y) dt dy = g(x, y) dt Equilibrium points are points where f(x, y) = 0 and g(x, y) = 0. Equilibrium points are the intersection points of xnullcline and ynullcline. Equilibrium points are constant solutions of the system. Equilibrium points are also called steady state solutions, fixed points, etc. 9
10 dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy xnullcline: x = 0 or 2 x y = 0 ynullcline: so y = 0 or 3 y 2x = 0 Equilibrium points: (0, 0), (0, 3), (2, 0) and (1, 1) 10
11 Qualitative analysis from nullclines: Suppose that there is a solution from a point in one of the regions formed by nullclines, then there is only three possibilities for the orbit: A. tends to an equilibrium on the border of this region B. goes away to infinity C. enter other neighboring region following the arrow More information is needed for equilibrium points to further determination. 11
12 Studies of phase portrait (3): linearization at equilibrium dx = f(x, y) dt dy = g(x, y) dt Suppose that (x 0, y 0 ) is an equilibrium point. Near it, the behavior of the solutions is governed by the linearized equation dx dt = f(x 0, y 0 ) x dy dt = g(x 0, y 0 ) x Since f(x, y) f(x 0, y 0 ) + f(x 0, y 0 ) x and g(x, y) g(x 0, y 0 ) + g(x 0, y 0 ) x (x x 0 ) + f(x 0, y 0 ) (y y 0 ) y (x x 0 ) + g(x 0, y 0 ) (y y 0 ) y (x x 0 ) + f(x 0, y 0 ) (y y 0 ) y (x x 0 ) + g(x 0, y 0 ) (y y 0 ) y 12
13 Quick Review of Multivariable calculus: Linearization in 1d: f(x) f(x 0 ) + f (x 0 )(x x 0 ) Linearization in 2d: f(x, y) f(x 0, y 0 ) + f(x 0, y 0 ) (x x 0 ) + f(x 0, y 0 ) (y y 0 ) x y partial derivative: derivative of f w.r.t. x when y is fixed Notation: f(x 0, y 0 ) or f x (x 0, y 0 ) x Jacobian: ( all four partial ) derivatives of a vector field in a matrix fx (x 0, y 0 ) f y (x 0, y 0 ) g x (x 0, y 0 ) g y (x 0, y 0 ) 13
14 dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy Equilibrium points: (0, 0), (0, 3), (2, 0) and (1, 1) Jacobian J = ( 2 2x y x 2y 3 2y 2x ) J(0, 0) = ( ) (source) J(0, 3) = ( ) (sink) J(2, 0) = ( ) (sink) J(1, 1) = ( ) (saddle) 14
15 Classification of linear system: (Chap 3) Two real eigenvalues: 1. λ 1 > λ 2 > 0: source 2. λ 1 > λ 2 = 0: degenerate source 3. λ 1 > 0 > λ 2 : saddle 4. λ 1 = 0 > λ 2 : degenerate sink 5. 0 > λ 1 > λ 2 : sink Two complex eigenvalues: λ ± = a ± bi 1. a > 0: spiral source 2. a = 0: center 3. a < 0: spiral sink One real (repeated) eigenvalue: λ 1 = λ 2 = λ 1. λ > 0: star source or trying to spiral source 2. λ = 0: parallel lines or dumb system 3. λ < 0: star sink or trying to spiral sink 15
16 How to determine stability of equilibrium quickly (without solving eigenvalues)? ( a b c d ) ( x y ) = λ ( x y ), ( a λ b c d λ ) ( x y ) = ( 0 0 ), solve (a λ)(d λ) bc = 0 (characteristic equation) λ 2 (a + d)λ + (ad bc) = 0, or λ 2 T λ + D = 0 Trace=T = a + d = λ 1 + λ 2, Determinant=D = ad bc = λ 1 λ 2 (RouthHurwitz criteria for 2nd order systems): 1. Stable if T < 0 and D > 0 2. Unstable if D < 0, or T > 0 and D > 0 T 2 4D determines whether the eigenvalues are real or complex 16
17 Tracedeterminant plane 1. unstable node, source; 2. unstable focus, spiral source; 3. stable focus, spiral sink; 4. stable node, sink; 5. saddle 17
18 Phase portrait of five types: 1. unstable node, source; 2. unstable focus, spiral source; 3. stable focus, spiral sink; 4. stable node, sink; 5. saddle 18
19 Suppose that P 0 = (x 0, y 0 ) is an equilibrium point, then a solution Y(t) is a stable orbit of P 0 if lim Y(t) = P 0 ; t and a solution Y(t) is a unstable orbit of P 0 if lim t Y(t) = P 0. If all solutions nearby are stable orbits, then the equilibrium point is stable. (sink and spiral sink) If an equilibrium point not stable, then it is unstable. (saddle, source and spiral source) If the equilibrium point is a saddle point, then there is only one pair of stable orbits, and one pair of unstable orbits. These orbits are called separatrices at the saddle point. (separatrix singular) When an equilibrium point is stable, the set of all its stable orbits is called the attraction basin of this equilibrium point. 19
20 dx dt = 2x(1 x 2 ) xy dy dt = 3y(1 y 3 ) 2xy At the saddle point (1, 1), there is a pair of stable orbits and a pair of unstable orbits, and they are the separatrices of this saddle point. Indeed, the stable orbits separates the whole region into two subregions. In the subregion above this separatrix, all solutions tend to the equilibrium point (0, 3) as t ; and in the subregion below the separatrix, all solutions tend to the equilibrium point (2, 0) as t. These two subregions are the attraction basins of (0, 3) and (2, 0) respectively. 20
21 Warning: If the linearized equation at an equilibrium point is a degenerate source, degenerate sink, center, parallel lines, then the nonlinear equation may have different phase plane near the equilibrium point. Linearization Theorem in 2d: Suppose that (x 0, y 0 ) is an equilibrium point of x = f(x, y) and y = g(x, y), and the eigenvalues of Jacobian J(x 0, y 0 ) are λ 1 and λ 2. (1) λ 1 > λ 2 > 0, then the system is a source; (2) λ 1 > 0 > λ 2, then the system is a saddle; (3) 0 > λ 1 > λ 2, then the system is a sink; (4) λ 1,2 = a ± bi, a > 0, then the system is a spiral source; (5) λ 1,2 = a ± bi, a < 0, then the system is a spiral sink; (6) If the eigenvalues are other cases, then you need other information to determine the solution behavior near the equilibrium point. 21
22 Linearization Theorem in 1d: Suppose that y = y 0 is an equilibrium point of y = f(y). if f (y 0 ) < 0, then y 0 is a sink; if f (y 0 ) > 0, then y 0 is a source; if f (y 0 ) = 0, then y 0 can be any type, but in addition if f (y 0 ) > 0 or f (y 0 ) < 0, then y 0 is a node. 22
23 When the linearization cannot provide sufficient information: dx dt = y (x2 + y 2 )x dy dt = x (x2 + y 2 )y In polar coordinates: dr dt = 2r2, dθ dt = 1
24 Job List for Phase plane analysis: 1. Find nullclines, and draw arrows on nullclines and bounded regions 2. Find equilibrium points, and linearization 3. Draw possible separatrices, and sketch the phase portrait 4. Explain the phase portrait Other possibilities: Periodic solutions Some points: 1. Nullclines may not be lines. (x 2 + y 2 = 2500) 2. Separatrices usually cannot be calculated explicitly, but near saddle point, they are tangent to the eigenvector direction. 23
25 Examples: (A) x = x(2 x y), y = y(y x 2 ). (x 0, y 0) (B) A bifurcation problem (more competition models) dx dy = x( x y + 70), = y( 2x y + a), (x 0, y 0) dt dt When 0 < a < 70, three equilibrium points, (0, 0), (70, 0) and (0, a). (70, 0) is a sink, and its attraction basin is the whole region x > 0, y > 0. When 70 < a < 140, four equilibrium points, (0, 0), (70, 0), (0, a) and (a 70, 140 a). Both (70, 0) and (0, a) are stable, their attraction basin splits the region. When a > 140, three equilibrium points, (0, 0), (70, 0) and (0, a). (0, a) is a sink, and its attraction basin is the whole region x > 0, y > 0. 24
26 LotkaVolterra system of competition du 1 dτ = r 1U 1 (1 U 1 ) r 1α U 1 U 2, du 2 K 1 K 1 dτ = r 2U 2 (1 U 2 ) r 2β U 1 U 2 K 2 K 2 r 1, r 2 : growth rate K 1, K 2 : strength of the intraspecific competition α, β: strength of the interspecific competition Nondimensionalization: t = r 1 τ, u = U 1 K 1, v = U 2 K 2 New equation: du dt = u(1 u av), dv dt a = α K 2 K 1 b = β K 1 K 2 C = r 2 r 1 = cv(1 bu v) 25
27 Equilibrium points: (0, 0) (extinction), (1, 0) (u wins), (0, 1) (v wins), (u, v ) = ( 1 a 1 ab, 1 b 1 ab ) (coexistence) Case 1: a < 1 and b < 1, (u, v ) is stable Case 2: a > 1 and b > 1, (u, v ) is unstable, and (0, 1), (1, 0) are stable (bistable) Case 3: a < 1 < b, no (u, v ), (1, 0) stable, (0, 1) unstable (u is superior competitor) Case 4: a > 1 > b, no (u, v ), (0, 1) stable, (1, 0) unstable (v is superior competitor) 26
28 Principle of competition exclusion (Gause 1934) From textbook: if two species occupy the same ecological niche, then one of them will go extinct. An ecological niche is mode of existence that a species has within an ecosystem. Essentially it is the sum of all activities and relationships a species has while obtaining and using the resources needed to survive and reproduce. A species niche includes: (a). Habitat  where it lives in the ecosystem; (b). Relationships  all interactions with other species in the ecosystem; (c). Nutrition  its method of obtaining food. 27
29 Another version of Principle of competition exclusion: If the coexistence (u, v ) satisfies u 1 + v B is not possible. 1, then coexistence Evolution strategy: how to gain competition advantage r 1, r 2 : growth rate K 1, K 2 : strength of the intraspecific competition α, β: strength of the interspecific competition a = α K 2 b = β K 1 C = r 2 K 1 K 2 r 1 Kstrategy: increase your carryingcapacity rstrategy: increase your growth rate 28
30 Ecological conclusions: 1. Similar species in the same habitat will not coexist. 2. Coexistence will only occur when the competition is less intense (for competition in same niche, ab = αβ = 1) 3. This principle may be not very useful in practice, as we can always find a way in which two species differ. 4. If two competitors coexist, there is some reason for them to do so. Or to make two competitors coexist, create difference in niche. 5. Species may distributed in patchy way, so competition can be minimized; or timesharing arrangement with succession of species or seasonal variability can effect similar result. Gradual evolution of differing genetic traits (character displacement) to minimize competition, and more complex multispecies interactions in which predation mediates competition. 29
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