Embed Size (px)
Citation preview
May 2018
2018 CIMPA-SENEGAL School–
Dynamical Systems Modulo
> B. Chen Charpentier, Dept. ofMathematics, University of Texas atArlington,Texas, USA.
> M. Leite, Math & Stat., University ofSouth Florida St. Petersburg, Florida USA
> D. Seack, Laboratory Mathematic ofDecision and Numerical Analysis, CheikhAnta Diop, Dakar Senegal
Dakar, Senegal
Sec 1: Introduction
Purpose:Develop some elementary yet important examples of first-orderdifferential equations. These examples illustrate some of the basicideas in the Dynamical Systems theory in the simplest possiblesetting.
> Simplest ODE but also the most important and familiar to allcalculus students.
> Logistic model and logistic model with harvesting to give ataste of certain topics (slope field,phase lines, equilibriumsolutions and their local stability, bifurcations).
May 2018
2/75
Sec 1: Introduction
The simplest example
First-order ODEdx
dt= ax (1)
> x = x(t) - unknown real-valued function of a real variable t.
> dx/dt - Derivative of x w.r.t t (also denoted by x ′ or x ′(t)).
> a - a parameter.
May 2018
3/75
Sec 1: Introduction
The simplest example
The solution of (1) is obtain from calculus and has the form
x(t) = keat k any real number (2)
To see this observe that x ′(t) = akeat = ax(t)
Furthermore, there are no other solutions. Let’s check this.
Let u(t) be any solution of (1). The derivative of u(t)e−at :
d (u(t)e−at)
dt= u′(t)e−at+u(t)
(−ae−at
)= au(t)e−at−au(t)e−at = 0
Thus, u(t)e−at = k so u(t) = keat .
May 2018
4/75
Sec 1: Introduction
The simplest example
dx
dt= ax
General solution of (1), which is a collection of all solutions of (1) isx(t) = keat , k any real number.
Note that there is a special solution of this ODE when k = 0. This isa constant solution x(t) ≡ 0.
Equilibrium (steady-state) solution or equilibrium (fixed) point areconstant solutions of an ODE.This type of solutions are often the most important solutions ofODE’s. May 2018
5/75
Sec 1: Introduction
The simplest example
Qualitative behavior of solutionsWe are generally interested in the qualitative behavior of solutions ofODE’s, which includes their geometric properties.
dx
dt= ax , a is a parameter
If a changes the ODE changes and so do the solutions.
Question: Can we describe qualitatively the way solutions change?The sign of a is critical here.
> If a > 0, limt→∞
keat equals to ∞ when k > 0 and equals to −∞when k < 0.
> If a = 0, keat = constant.> if a < 0, lim
t→∞keat = 0
May 2018
6/75
Sec 1: Introduction
The simplest example
Sketch of the graphs of solutions - dependence on a
Figure: The solution graphs (left) and phase line (right) for x ′ = ax witha > 0. Each curve represents a particular solution.
May 2018
7/75
Sec 1: Introduction
The simplest example
Geometric properties of the solutions of the ODE, a < 0
Figure: The solution graphs phase line for x ′ = ax with a < 0. Each curverepresents a particular solution.
May 2018
8/75
Sec 1: Introduction
The simplest example
Summary of Figures
> When a > 0 all nonzero solutions tend away from theequilibrium point x = 0 as t increases. Then x = 0 is call asource and is said to be unstable
> When a < 0 all solutions tend toward x = 0. Then x = 0 iscalled a sink and is said to be stable
May 2018
9/75
Sec 1: Introduction
The Logistic Model
The ODE x ′ = ax can be think of as modeling the populationgrowth if a > 0. In this case
> If x(t) population of some species at time t.
> dxdt rate of growth of population is directly proportional to thesize of the population.(Not a very realistic assumption!)
Modify the equation so that the population cannot increase withoutbound.An ODE (perhaps the simplest) that satisfies this requirement is thelogistic population growth model
May 2018
10/75
Sec 1: Introduction
The Logistic Model
x ′ = ax(
1− x
N
)(3)
> If x(t) population of some species at time t.
> a > 0 rate of population growth when x is small.
> N > 0 represents the system carry-capacity.
Without loss of generality we set N = 1 (we choose the units so thatthe carry capacity is exactly 1 unit of population). Then x(t) givesthe fraction of the ideal population present at time t. The ODE (3)reduces to:
x ′ = fa(x) = ax(1− x) (4)
May 2018
11/75
Sec 1: Introduction
The Logistic Model
Solution of the ODE (4)Use separation of variables method and integration gives
x(t) =Keat
1 + Keat(5)
K - arbitrary constant of integration.
Evaluate (5) at t = 0 and solving for K yields K =x(0)
1− x(0)
Substituting in (5) gives
x(t) =x(0)eat
1− x(0)− x(0)eat
Valid for any initial population x(0).Question: Can you tell what are the constant solutions?
May 2018
12/75
Sec 1: Introduction
The Logistic Model
x ′ = fa(x) = ax(1− x), a > 0
Equilibrium Solutions:x(t) ≡ 1 and x(t) ≡ 0
Recall: Equilibrium solutions are constant solutions of an ODE.
Remark: constant solutions occur then when dx/dt = 0.
May 2018
13/75
Sec 1: Introduction
The Logistic Model
Qualitative behavior of solutions: slope field
The right-hand side of the ODE determines the slope of the graph ofany solution at each time t. Hence we can plot little slope lines inthe tx-plane as in Figure 3 (when introduce computational methodswe will learn how to plot those). The slope of the line at (t, x) givenby the quantity ax(1− x).
Figure: The slope field of (4)
May 2018
14/75
Sec 1: Introduction
The Logistic Model
Go from Slope field to solution graphsSolutions must have graphs that are tangent to the slope field
Figure: The slope field and solution graphs of (4)
> If x(0) > 0 sol. tend to the maximum population that thesystem supports x(t) ≡ 1.
> If x(0) < 0 sol. tend to −∞, no biological meaning.
May 2018
15/75
Sec 1: Introduction
The Logistic Model
Qualitative behavior of solutions: phase lineWe then can conclude that x = 0 is a source (unstable fixed point)and x = 1 is a sink (locally stable fixed point).
Figure: The slope field, solution graphs, and phase line of (4)
May 2018
16/75
Sec 1: Introduction
The Logistic Model
Qualitative behavior from graph of fa x′ = fa(x) = ax(1 − x), a > 0
> fa crosses the x-axis at x = 0 and x = 1 so these are theequilibrium points.
> 0 < x < 1, fa(x) > 0 so slopes are positive and solutions mustincrease
> x < 0 or x > 1, fa(x) < 0 and solutions must decrease.May 2018
17/75
Sec 1: Introduction
The Logistic Model
Qualitative behavior from analytic expression of fa
x ′ = fa(x) = ax(1− x), a > 0 f ′a(x) = a− 2ax
Using “derivative tests”
> f ′a(0) = a > 0. So near x = 0 the slopes are negative for x < 0and positive for 0 < x < 1. So solutions tend away from x = 0
> f ′a(1) < 0. Hence, near x = 1 and above x = 1 the slopes arenegative so the solutions are forced to tend toward x = 1
May 2018
18/75
Sec 1: Introduction
The Logistic Model
Exercise: Applying the ideas explored so far. Consider x ′ = x − x3
with the following associated slope field
Figure: Slope field for x ′ = x − x3
Construct solutions graph and phase line.
May 2018
19/75
Sec 1: Introduction
The Logistic Model
Solution: x ′ = x − x3
Figure: Slope field, solution graphs, and phase line for x ′ = x − x3
May 2018
20/75
Sec 1: Introduction
Bifurcations
Constant harvesting in a logistic population model
Modify the logistic model to incorporate harvesting of thepopulation.
> Population obeys the logistic assumptions with a = 1.
> The population is harvested at the constant rate h.
The ODE is
x ′ = gh(x) = x(1− x)− h, h ≥ 0 (6)
Instead of solving the equation explicitly (that is, find the solutionsof (6)) we use graph of gh(x) to study the qualitative behavior of itssolutions.
May 2018
21/75
Sec 1: Introduction
Bifurcations
Qualitative behavior of the solutions of ODE (6)
Figure: Graph of the function ghx(1− x)− h in ODE (6)
May 2018
22/75
Sec 1: Introduction
Bifurcations
Summary of Figure x ′ = gh(x) = x(1− x)− h, h ≥ 0
1. 0 < h < 1/4, gh has 2 roots: x1, x2 with 0 ≤ x1 < x2. So theODE has 2 fixed points. We can see that g ′h(x1) > 0 so that x1
is a source, and g ′h(x2) < 0 so that x2 is a sink.
2. For h = 1/4 we have a bifurcation point: the 2 fixed pointsx1, x2 coalesce a h increases through 1/4 and then disappearwhen h > 1/4.
3. When h > 1/4, g ′h(x) < 0 for all x . This means that thesolutions of (6) decreases to −∞ as time increases.
May 2018
23/75
Sec 1: Introduction
Bifurcations
Bifurcation Diagram
The graphical summary given in Figure below is called bifurcationdiagram.h is the bifurcation parameter
Figure: Bifurcation diagram for gh = x(1− x)− hMay 2018
24/75
Sec 1: Introduction
Bifurcations
Biological interpretation/importance
> The bifurcation corresponds to a disaster for the species understudy. For rates of harvesting 1/4 or lower, the populationpersists, provided that initial population is sufficiently larger, i.e,x(0) ≥ x1.Very small changes in the rate of harvesting near h = 1/4 leadsto a drastic fate of the population: h > 1/4, the species becomeextinct.
> This example highlights the importance of detecting bifurcationpoints.
> The model is very simple but capture the disastrous changes inthe population that has been observed many times in realsituations.
May 2018
25/75
Sec 1: Introduction
Bifurcations
Exercise: Consider the family of ODEs
x ′ = ga(x) = x2 − ax
Construct the bifurcation diagram.
Figure: Bifurcation diagram for ga = x2 − ax May 2018
26/75
Sec 1: Introduction
Bifurcations
Hands-on activity
x ′ = fa,b(x) = ax − x3 − b
(exercise HSD, pg15)
May 2018
27/75
Sec 1: Introduction
Bifurcations
Summary for systems with 1-variable x ′ = f (x), x = x(t)
> x∗(t) is an equilibrium points or equilibrium solutions or fixedpoints or steady states solutions are solutions that satisfy
x ′ = 0
> Classification of equilibrium: sinks (locally stable solutions) iff ′(x∗) < 0 and sources (unstable solutions) if f ′(x∗) > 0
> Bifurcation: changes on solutions (number or stability) as wevary a parameter called bifurcations parameter
> Bifurcation diagram: shows in a graph how number and/orstability of equilibrium solutions changes with changes in aparameter (bifurcation parameter).
May 2018
28/75
2: Planar Linear Systems
Introduction
Planar Linear SystemsWe restrict our attention to autonomous linear systems, which arelinear systems that do not depend explicitly on time. These are thesimplest ODE- type model that can be used to describe 2-variablesystems. They have the form
x ′ = f (x , y) = ax + by
(7)
y ′ = g(x , y) = cx + dy
a, b, c, d arbitrary constants.In this section, instead of focus on finding explicitly the solutions ofthe ODE systems we search to understand the geometric properties(phase portraits) of equilibrium solutions and their stability. For thatwe use matrix theory. May 2018
29/75
2: Planar Linear Systems
Introduction
Planar linear system in matrix notation
A =
[a bc d
]X =
[xy
]X ′ =
[x ′
y ′
]Then
x ′ = f (x , y) = ax + by
y ′ = g(x , y) = cx + dy
Can be written as X ′ = AX .
May 2018
30/75
2: Planar Linear Systems
Facts from algebra
Recalling facts about systems of 2 algebraic equations, which we willoccur often
ax + by = α
cx + dy = β
We focus on the case in which at least on of a, b, c is nonzero. Usingmatrix notation
AX = V , V =
[αβ
]Determinant of A: detA = ad − bc
Fact: The system of 2-algebraic equations has one and only onesolution if and only if detA 6= 0. Otherwise, the system may or maynot have solution. However, if there is a solution in fact there areinfinitely many solutions. May 2018
31/75
2: Planar Linear Systems
Facts from algebra
Characteristic polynomial or equation of A
P(A) = P(λ) = det(A− λI2), I2 =
[1 00 1
]= det
([a− λ b
c d − λ
])= (a− λ)(d − λ)− bc
= λ2 − Tr(A)λ+ det(A)
Example: Compute the characteristic equation of A =
[1 31 −1
].
Solution: P(A) = P(λ) = λ2 − 4
May 2018
32/75
2: Planar Linear Systems
Facts from algebra
Eigenvalues and eigenvectors of A
Definition:A nonzero vector V is called an eigenvector of A if AV = λV forsome (real or complex) number λ. λ is called an eigenvalue of Aassociated to the eigenvector V .
The eigenvectors and eigenvalues can be computed used softwarebut let see how we compute them analytically.
Fact: The eigenvalues of A are the roots of P(λ). That is, tocompute eigenvalues of a matrix we solve P(λ) = 0.
May 2018
33/75
2: Planar Linear Systems
Facts from algebra
Example: Compute the eigenvalues of A =
[1 31 −1
].
Solution:P(λ) = λ2 − 4 = 0. So the eigenvalues of A are λ1 = −2 andλ2 = 2. A has two real and distinct eigenvalues.
May 2018
34/75
2: Planar Linear Systems
Facts from algebra
Compute eigenvectors V associated to an eigenvalue λ
Use the equation in the definition AV = λV :
Observe that AV = λV = λI2V because I2V = V . (multiplication ofmatrices)
Then we can write (A− λI2)V = 0 and solve for V .
Example: A as in the previous examples. Choose eigenvalue λ1 = 2and determine an eigenvector associated to it.
May 2018
35/75
2: Planar Linear Systems
Facts from algebra
Solution:
A =
[1 31 −1
]and λ = 2 V =
[v1
v2
]Then (A− λI2)V =
[1− 2 3
1 −1− 2
] [v1
v2
]= 0
This equation reduces to solving the system
−v1 + 3v2 = 0
v1 − 3v2 = 0
But these 2-equations are redundant. So we can pick only one, say
v1 − 3v2 = 0. Thus, any vector of the form
[3v2
v2
], v2 6= 0 is an
eigenvector associated to λ = 2.Compute eigenvectors for λ = −2.
May 2018
36/75
2: Planar Linear Systems
Facts from algebra
Linearly dependent and linearly independent vectors in R2.2 vectors V and W are linearly independent if V and W do not liealong the same straight line. V and W are linearly dependent ifeither V or W is zero or if both lie on the same line (that is, ifV = λW ).
Example: all vectors of the form
[3v2
v2
], v2 6= 0 computed above
are all linearly dependent.
Formal condition:
Let V =
[v1
v2
]and W =
[w1
w2
]. Then the 2 vectors are
linearly independent if and only if
det
([v1 w1
v2 w2
])6= 0
May 2018
37/75
2: Planar Linear Systems
Facts from algebra
Linearity principle & solutions of planar linear systems of ODEs
Fact: Solutions of X ′ = AX have the form
X (t) = C1eλ1tV + C2e
λ2tW
Where C1,C2 are arbitrary constants, λ1, λ2 are eigenvalues of Aassociated to linearly independent associated eigenvectors V ,Wrespectively.
Compare to x ′ = ax , a constant with solution x = keat !x ′ = ax , a linear ODE in 1-dimWhat can you say about a?
Recall:a < 0 x ≡ 0 sink (stable equilibrium)a > 0 x ≡ 0 source (unstable equilibrium) May 2018
38/75
2: Planar Linear Systems
Phase Portraits & Stab.
Stability of steady-states (equilibrium solutions)
Consider the system X ′ = AX . If detA 6= 0 the X =
[xy
]=
[00
]or (x(t), y(t)) = (0, 0) is the unique equilibrium solution, an isolatedequilibrium. The equilibrium is the origin in R2.
Stability of the equilibrium: The origin is
> asymptotically stable if the eigenvalues of A are negative orhave negative real part. In this case lim
t→∞(x(t), y(t)) = (0, 0).
> stable if the eigenvalues of A are nonpositive or havenonpositive real part.
> unstable if the eigenvalues of A are positive or have positivereal part. May 2018
39/75
2: Planar Linear Systems
Phase Portraits & Stab.
Classification of equilibria
Remark: The stability of the origin is established once theeigenvalues are known. Even without calculating the eigenvalues, thestability can he determined from the characteristic polynomial of A.(P(A) = λ2 − Tr A + detA.)- Routh-Hurwitz criteria.
Real and distinct eigenvalues [tr (A)]2 − 4 detA > 0
1. λ1 < 0 < λ2 Saddle detA < 0;
2. λ1 < λ2 < 0 Sink node detA > 0, tr A < 0;
3. 0 < λ1 < λ2 Source node detA > 0, tr A < 0.
May 2018
40/75
2: Planar Linear Systems
Phase Portraits & Stab.
Classification of equilibria (cont.)
Real and repeated eigenvalues [tr (A)]2 − 4 detA = 0
1. λ1 = λ2 < 0(> 0) 2 linearly independent eigenvectorstr A < 0(trA > 0) Stable (Unstable) proper node
2. λ1 = λ2 < 0(> 0) 1 linearly independent eigenvectortr A < 0(tr A > 0 Stable (Unstable) improper node
Complex eigenvalues (λ = a± bi) [tr (A)]2 − 4 detA < 0
1. Real(λ) = a = 0 Center tr A = 0, detA > 0
2. Real(λ) = a < 0 Spiral sink node tr A < 0, detA > 0;
3. Real(λ) > 0 Spiral Source tr A > 0, detA > 0.
May 2018
41/75
2: Planar Linear Systems
Phase Portraits & Stab.
> Given a (x0, y0) in the x,y-plane (or phase plane) there is onlyone trajectory or solution curve passing through that point.
> In Sec. 1, for an autonomous differential equationdx/dt = f (x), the direction of the fields in tx-plane wererepresented by a phase line diagram. This is a way of describingthe evolution of the variable of interest with time. That is, thedynamics of the variable is given by the phase line diagrams.
> For planar systems of ODES (2 autonomous ODEs) theevolution of the x , y variables with time (dynamics of thesystem) can be represented by phase portraits(Think of solutions of the ODE system as describing trajectoriesof particles moving in the x,y-plane) May 2018
42/75
2: Planar Linear Systems
Phase Portraits & Stab.
Phase portraits and stability summary
The trace-determinant plane, phase portraits and equilibriumsolution stability
May 2018
43/75
2: Planar Linear Systems
Phase Portraits & Stab.
A few observations from this subsection
Definition: A matrix A is hyperbolic if none of its eigenvalues hasreal part zero. The system X ′ = AX is hyperbolic and its equilibriumis said to be hyperbolic.
Remark: The equilibrium solutions are trivial periodic solutions asthey satisfy X (T + t) = X (t) for any T .
May 2018
44/75
2: Planar Linear Systems
Phase Portraits & Stab.
A few observations from this subsection (cont.)
Nontrivial periodic solutionsWhen the eigenvalues of A are purely imaginary, that is,
λ1 = bi and λ2 = −bi with b 6= 0
the solutions are periodic (non-trivial). The trajectories are circles inx,y-plane. That is, they satisfy
X (T + t) = X (t), for a fixedT > 0
T is the period of the solution.
These solutions are not equilibrium solutions (fixed points, steadystates, ...) and they can only occur for planar or higher dimensionalsystems.
May 2018
45/75
2: Planar Linear Systems
Phase Portraits & Stab.
May 2018
46/75
2: Planar Linear Systems
Phase Portraits & Stab.
May 2018
47/75
3: Nonlinear Systems
Introduction
> We continue to focus on equilibrium solutions and their localstability for nonlinear systems.
> In Sec. 1 we did this study for 1-dim autonomous ODE: thelogistic population model and the constant harvesting model.
> Here we formalize the procedure that works for autonomoussystems of ODEs (linear or nonlinear). As in Sec. 1 we willexemplify the concepts with a few well known models.
The idea is to reduce, locally, nonlinear systems to linear systemsand then apply the concepts/techniques described in Sec. 2.
May 2018
48/75
3: Nonlinear Systems
Basic Definitions
An autonomous system of ODEs has the form
dX
dt= F (X ),
where
X =
x1...xn
F (X ) =
f1(x1, ·, xn)...
fn(x1, ·, xn)
An initial value problem satisfies this system with a given initialcondition, X (t0) = X0.
Assume that a unique solution exists!
May 2018
49/75
3: Nonlinear Systems
Basic Definitions
An equilibrium solution (stead-state, fixed point, etc) of the systemof ODEs is a constant solution X such that
F (X ) = 0,
Linearized system and Jacobian matrix The system linearizedabout the equilibrium point is
dX
dt= JX
Where J is the Jacobian matrix evaluated at the equilibrium point
Now the classification of hyperbolic and non-hyperbolicequilibrium follow as described for linear systems.Additionally, the study of local stability of equilibrium reducesto the case of linear systems.
May 2018
50/75
3: Nonlinear Systems
Basic Definitions
Jacobian matrix for 1- and 2-dim systems1-dim
x = f (x), J = f ′(x) = fx
2-dim
X = F (X ) =
[f1(x , y)f2(x , y)
]= (f1(x , y), f2(x , y))
J =
∂f1∂x
∂f1∂y
∂f2∂x
∂f2∂y
=
[f1x f1yf2x f2y
]
May 2018
51/75
3: Nonlinear Systems
Examples
Recall logistic model
x ′ = fa(x) = ax(1− x), a > 0 f ′a(x) = J(x) = a− 2ax
Local Stability
> J(0) = f ′a(0) = a > 0. Thus, x = 0 is an unstable equilibriumand x = 0 is a hyperbolic equilibrium.
> J(1) = f ′a(1) < 0. Hence, x = 1 is a hyperbolic locally stableequilibrium.
May 2018
52/75
3: Nonlinear Systems
Examples
A 2-dim predator-prey model (L.S.J. Allen, 2007)
Computation/classification of steady-states, Jacobian matrix, andlocal stability.
May 2018
53/75
3: Nonlinear Systems
Examples
Discussion:
> This later equilibrium is positive if K > b/c . Therefore, apositive equilibrium exists (both prey and predator survive -interior eq.) if and only if the carry capacity of the prey K issufficiently large. Otherwise the prey population is too small tosupport the predator population.
> If K ≤ b/c , then only the prey survives (boundary eq.)May 2018
54/75
3: Nonlinear Systems
Examples
Jacobian and equilibria of the system
May 2018
55/75
3: Nonlinear Systems
Examples
May 2018
56/75
3: Nonlinear Systems
Examples
May 2018
57/75
3: Nonlinear Systems
Examples
A 3-dim competition model (L.S.J. Allen, 2007)
Jacobian, equilibria, local stability using R-H criterion.
May 2018
58/75
3: Nonlinear Systems
Examples
This model has several fixed points: the zero, 3 one-species, 3two-species.
The three-species fixed points is found by solving a 3 system ofalgebraic equations, which has unique solution if the matrix ofcoefficients is nonsingular. However, the existence of a solution doesnot guarantee that the solution is nonnegative.
May 2018
59/75
3: Nonlinear Systems
Examples
May 2018
60/75
3: Nonlinear Systems
Examples
May 2018
61/75
3: Nonlinear Systems
Nullclines
Nullclines, equilibria & infer direction of vector fields
> Consider a Lotka-Volterra predator-prey model type
> This model builds upon the one-species logistic model. In thelogistic model, intraspecies competition becomes more severe asthe population size increases, resulting in a lower per capitagrowth rate.
May 2018
62/75
3: Nonlinear Systems
Nullclines
The L-V model of competition
> The Lokta-Volterra model assumes that each individual ofspecies i experiences competition as if its own species hadpopulation size n1 + αn2. Assume that the two species can havedistinct growth rates as well as different carry capacities.
dR
dt= R − (R2 + 2RL) (8)
(9)
dL
dt= L− (L2 + 2LR) (10)
1. Compute R nullclines as well as L nullclines2. represent them in a graph & sketch the direction of the
trajectories on each distict region enclosed by nullclines
May 2018
63/75
3: Nonlinear Systems
Nullclines
Solution
> R nullclines:
dR
dt= 0 → R = 0 or L =
1
2(1− R)
> L nullclines:
dL
dt= 0 → L = 0 or L = (1− 2R)
May 2018
64/75
3: Nonlinear Systems
Nullclines
Graphical representation of the nullclines
May 2018
65/75
3: Nonlinear Systems
Nullclines
Direction of the vector filed along the nullclines
R = 0 null clineChoose a point on this line, say (0, 1/3). Substitute into
dR
dt= R − (R2 + 2RL)
dL
dt= L− (L2 + 2LR)
We obtain:dR
dt= 0 and
dL
dt= 2/9 > 0
Conclusion:arrows point up along R = 0 until the equilibrium point (0, 1) isreached.For points after this equilibrium the arrows point down.
May 2018
66/75
3: Nonlinear Systems
Nullclines
L = (1− R)/2 R nullcline
dR
dt= 0
Choose R = 1/5, then L = (1− 1/5)/2 = 2/5.
dR
dt= 2/5(1− 2/5− 2(1/5)) = 2/25 > 0
Conclusion:Arrows point up if R < 1. Then, after the equilibrium point (1, 0)the vector field changes direction. Thus after (1.0) the arrows pointdown.
May 2018
67/75
3: Nonlinear Systems
Nullclines
Nullclines & vector field direction
May 2018
68/75
3: Nonlinear Systems
Nullclines
Nullclines & vector field direction
May 2018
69/75
3: Nonlinear Systems
Periodic solutions
Classical Lotka-Volterra model & periodic solution
dx
dt= ax − bxy = x(a− by)
dy
dx= −cy + exy = y(−c + ex)
> a > 0 - intrinsic growth rate of the prey x> c > 0 - death rate of the predator.> b > 0 - per capita reduction in prey per predator> d > 0 - per capita increase in the predator per prey> x ≥ 0 - biomass of prey> y ≥ 0 - biomass of predator> bxy , dxy - referred to as functional and numerical responses.
May 2018
70/75
3: Nonlinear Systems
Periodic solutions
Classical Lotka-Volterra model & periodic solution (cont.)
dx
dt= ax − bxy = x(a− by)
dy
dx= −cy + exy = y(−c + ex)
(Procedure similar to the following is common in analysis of models)
Make the following changes of coordinates to put the model in adimensionless form
u(τ) =ex(t)
c, v(τ) =
by(t)
a, τ = at, α = c/a
The model reduces todu
dτ= u(1− v),
dv
dτ= αv(u − 1)
May 2018
71/75
3: Nonlinear Systems
Periodic solutions
Jacobian and solutionsdu
dτ= u(1− v),
dv
dτ= αv(u − 1)
The system has 2 equilibrium solutions:(0, 0) and (1, 1).
Jacobian matrix
J(u, v) =
[1− v −uαv α(u − 1)
]J at the equilibria
J(0, 0) =
[1 0α0 −α
]J(1, 1) =
[0 −1α 0
]
Eigenvalues of J(0, 0): 1, −α. The equilibrium is a Saddle.Eigenvalues of J(1, 1): ±i
√α (purely imaginary). May 2018
72/75
3: Nonlinear Systems
Periodic solutions
Lotka-Volterra model and periodic solutions
Eigenvalues of J(1, 1): ±i√α (purely imaginary). In the linear case
the equilibrium is a Center.
In the nonlinear case is ambiguous (will see more on periodicsolution in nonlinear systems latter):it could also be a spiral point (stable or unstable).
Theorem: Consider the Lotka-Volterra predator-pry system abovewith positive initial condition. Then, every solution is periodic, thatis, satisfy X (t + T ) = X (T )).
May 2018
73/75
3: Nonlinear Systems
Periodic solutions
Solutions of Lotka-Volterra model: phase portrait
May 2018
74/75
3: Nonlinear Systems
Periodic solutions
Solutions of Lotka-Volterra model: time series
May 2018
75/75
![MATH 614, Spring 2016 [3mm] Dynamical Systems …Dynamical Systems and Chaos Lecture 1: Examples of dynamical systems. A discrete dynamical system is simply a transformation f : X](https://img.dokumen.tips/doc/110x75/5fc3a613bb041d25ed5cc331/math-614-spring-2016-3mm-dynamical-systems-dynamical-systems-and-chaos-lecture.jpg)
