dynamical systems ii - cheikh anta diop...

Dynamical systems

Dynamical systems II

Benito Chen-Charpentier

Department of Mathematics, University of Texas at Arlington, Arlington, TX76019, USA

[email protected]

African Mathematical School on Insight from MathematicalModeling

into Problems in Conservation, Ecology, and EpidemiologyUniversity of Cheikh Anta Diop, Dakar Senegal

May 2018

mailto:[email protected]

Dynamical systems

Equilibria in nonlinear systems


It is often impossible to solve nonlinear differential equationsexactly. Most of the time we can only find equilibrium solutionswhen the time derivatives are zero. Then we only need to solve thealgebraic system of the right hand sides equal zero. Equilibriumsolutions are very important and by perturbating around them agood idea of the behavior of the dynamical system.

Dynamical systems


Some examples

Consider the system

x ′ = x + y2

y ′ =−y ,

which has a single equilibrium point, x = 0,y = 0. Near the originthe system of equations looks like the linearized system

x ′ = x

y ′ =−y .

Dynamical systems


The solution is

x(t) = x0et , y(t) = y0e

−t .

So there is a saddle at the origin for the linearized system, stablealong the y axis and unstable along the x axis.For the nonlinear system, we have the same solution for y and thex equation is

x ′ = x + y20 e−2t ,

which can be solved by finding and integrating factor or by usingundetermined coefficients to find the particular solution. Thesolution is

x(t) = (x0 +1

3y20 )et − 1

3y20 e

−2t

y(t) = y0e−t

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Figure : Phase plane of nonlinear system.

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The trajectory with y0 = 0 moves away from the origin for any x0.The trajectory through x0 =−1

3y20 with arbitrary y0 moves to the

origin.In this example, a saddle point of the linearized system correspondsto a saddle point of the nonlinear system.

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In this example a change of variables u = x + 13y

2,v = y changesthe system to

u′ = u, v ′ =−v .

So the conservation of the saddle is obvious.But what in general?

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Consider now the system

x ′ =1

2x−y − 1

2(x3 + y2x)

y ′ = x +1

2y − 1

2(y3 + x2x).

The linearized system is

X ′ =

(12 −11 1

2

)X ,

which has eigenvalues λ = 12 ± i . So all solution spiral away from

the origin.

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The system is easier to analyze in polar coordinates

r ′ = r(1− r2)/2

θ′ = 1.

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Since θ = t + c , with c a constant, the solutions spiral in thecounterclockwise direction and the distance to the origin increasesif r < 1, decreases if r > 1 and is constant for r = 1.

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The linearized system in polar coordinates is

r ′ =r

2θ′ = 1.

We clearly see all solutions spiral toward ∞.

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Figure : Phase plane of nonlinear system

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Figure : Solution starting at (2,1)

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Figure : Solution starting at (.1,.1)

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The next example is

x ′ =−y + εx(x2 + y2)

y ′ = x + εy(x2 + y2),

with ε an arbitrary constant. The linearized system is

x ′ =−yy ′ = x ,

with eigenvalues ±i . So the origin is a center with trajectoriescircles moving in the counterclockwise direction.

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The nonlinear system in polar coordinates is

r ′ = εr3

θ′ = 1.

For ε > 0 all solutions spiral away from the origin, and if ε < 0spiral towards the origin, both in the counterclockwise direction.

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Figure : Phase plane linearized system (ε = 0)

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Figure : Phase plane nonlinear system (ε = 1)

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Figure : Phase plane nonlinear system (ε =−1)

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A final example is

x ′ = x2

y ′ =−y .

The only equilibrium point is the origin. The linearized system is

x ′ = 0

y ′ =−y ,

which has as equilibrium points all the x−axis. The eigenvalues atany equilibrium point are 0 and −1. The zero eigenvaluecompletely changes the behavior.

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Figure : Phase plane linearized system

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Figure : Phase plane nonlinear system

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Local stability of nonlinear systems


The previous examples illustrate that the behavior of the solutionof a nonlinear system near an equilibrium point resembles that ofthe solution of the linear system near the same point only whenthe real part of all the eigenvalues of the Jacobian matrix isdifferent from zero, that is the equilibrium point is hyperbolic.The linearized system about X0 is

X ′ = DF (X0)X ,

where DF is the Jacobian matrix of F .The following results are valid in Rn but we will only prove them inR2.Let X ′ = F (X ) and suppose F (X0) = 0. If X0 6= 0 we can doU = X −X0 and the new system will have the equilibrium at theorigin. So we will assume that X0 = 0.

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An equilibrium point is said to be stable if nearby solutions staynearby for all future time. Only stable equilibrium points arephysically meaningful.More generally, A solution ψ of a system

X ′ = F (t,X ), t ≥ 0

is said to be stable if, given any ε > 0, there exists a δ > 0 suchthat any solution φ of the system satisfying

|φ(0)−ψ(0)|< δ

satisfies|φ(t)−ψ(t)|< ε, t ≥ 0.

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The solution is asymptotically stable if, in addition to being stable,

|φ(t)−ψ(t)| → 0 as t→ ∞.

If the equilibrium is not stable it is said to be unstable.

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Consider the planar system

x ′ = f (x ,y)

y ′ = g(x ,y),

with f (0,0) = g(0,0) = 0. The Routh-Hurwitz criteria states thatthe eigenvalues of a 2x2 matrix J have negative real parts if thecoefficients of characteristic polynomial are both positive. Thecharacteristic polynomial of the Jacobian matrix is

λ2−λTr(DF (X0)) +det(DF (X0)),

then the eigenvalues have negative real part iff Tr(DF (X0)) < anddet(DF (X0)) > 0. So we have the next theorem for the stability.

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TheoremAssume the first-order partial derivatives of f and g are continuousin some open set containing the equilibrium (x0,y0) of the systemx ′ = f (x ,y),y ′ = g(x ,y). Then the equilibrium is locallyasymptotically stable if

Tr(DF (X0)) < 0 and det(DF (X0)) > 0.

In addition, the equilibrium is unstable if either Tr(DF (X0)) > 0 ordet(F (X0)) < 0).

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The same classification of equilibrium points that was used forlinear systems can be used for nonlinear systems. However sincethe linearization is only an approximation to the nonlinear system,the nonlinear system may behave differently from the linearizedsystem in three cases:

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1. det(DF ) = 0 . There is at least one zero eigenvalue. In thelinear system the equilibrium points are not isolated. In thenonlinear system, this may also be true. If there is an isolatedequilibrium, it may be a node, spiral or saddle.

2. Tr(DF ) = 0 and det(DF ) > 0. The eigenvalues are purelyimaginary. The equilibrium is a center of the linearized systembut may be a center or spiral or the nonlinear system.

3. Tr(DF )2 = 4det(DF ). This represents the borderline betweencomplex and real eigenvalues so the equilibrium may be anode or a spiral.

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Nodes

A node can be a sink or a source.Consider the system

x ′ =−λx +h1(x ,y)

y ′ =−µy +h2(x ,y),

where h1 and h2 contain terms that are quadratic or higher in xand y . Assume −λ <−µ < 0. The linearized system is

x ′ =−λx

y ′ =−µy ,

for which the origin is a sink.

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Saddles

Now consider

x ′ = λx +h1(x ,y)

y ′ =−µy +h2(x ,y),

where h1 and h2 contain terms that are quadratic or higher in xand y . Assume <−µ < 0 < λ . The linearized system is

x ′ = λx

y ′ =−µy ,

which has a saddle at the origin. The y -axis is the stable line andthe x-axis the unstable line. The nonlinear system also has asaddle at the origin but the stable and unstable lines change.

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Theorem (The Stable Curve Theorem)

Suppose the system

x ′ = λx +h1(x ,y)

y ′ =−µy +h2(x ,y),

satisfies µ < 0 < λ and hj(x ,y)/r → 0 as r → 0. Then there is anε > 0 and a curve x = hs(y) that is defined for |y |< ε and satisfieshs(0) = 0. Furthermore:

1. All solutions whose initial conditions lie on this curve remainon this curve for all t ≥ 0 and tend to the origin as t→ ∞;

2. The curve x = hs(y) passes through the origin tangent to they -axis;

3. All other solutions whose initial conditions lie in the disk ofradius ε centered at the origin leave this disk as timeincreases.

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As an example consider

x ′ =−x + (x2 + y2)/2

y ′ = y + (x2 + y2)/2,

which has the origin as a saddle just as the linearized system.

x ′ =−xy ′ = y .

But the nonlinear system also has a center at (1,−1).

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Figure : Phase plane linearized system (ε = 0). Origin is a saddle

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Figure : Phase plane of nonlinear system (ε = .5). A saddle at the originand a center at (1,-1)

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Bifurcations

BifurcationsConsider a family of differential systems

X ′ = Fa(X ),

where a is a real parameter. Fa has continuous derivatives withrespect to a of all orders. Informally, a bifurcation occurs whenthere is a significant change in the solutions as a varies.Example:

x ′ = fa(x) = x2 +a,

which has a single equilibrium point x = 0 if a = 0, and f ′0(0) = 0but f ′′0 (0) 6= 0. For a> 0 this equation has no equilibrium pointssince fa(x) > 0 for all x , but for a< 0 this equation has a pair ofequilibria. Thus a bifurcation occurs as the parameter passesthrough a = 0. This kind of bifurcation is called a saddle-nodebifurcation.

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Bifurcations

Figure : Saddle-node bifurcation

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Bifurcations

Theorem (Saddle-Node Bifurcation)

Suppose x ′ = fa(x) is a first-order differential equation for which

1. fa0(x0) = 0;

2. f ′a0(x0) = 0;

3. f ′′a0(x0) 6= 0;

4.∂ fa0∂a (x0) 6= 0.

Then this differential equation undergoes a saddle-node bifurcationat a = a0.

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Bifurcations

Example (Pitchfork bifurcation): Consider:

x ′ = x3−ax .

There are three equilibria for this equation, at x = 0 and =±√a

when a> 0. When a≤ 0,x = 0 is the only equilibrium point.

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Bifurcations

Figure : Pitchfork bifurcation

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Bifurcations

Example (Transcritical bifurcation):

x ′ = ax + x2.

For a 6= 0 there are two solutions x = 0 and x =−a. At a = 0 thereis only one solution, x = 0. Both branches of solutions intersect ata = 0, and there is an exchange of stability.

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Bifurcations

Figure : Transcritical bifurcation

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Bifurcations

I In a saddle node bifurcation, as the bifurcation parameterpasses through the bifurcation point, two equilibria disappear,so that there are no equilibria afterward. One of the twoequilibria is stable and the other one is unstable, before theydisappear.

I In a pitchfork bifurcation, there are two stable equilibriaseparated by an unstable equilibrium. When the bifurcationpoint is passed, there is only one stable equilibrium. Or thereare two unstable equilibria separated by a stable equilibrium,until the bifurcation point is passed. Then there is only oneunstable equilibrium.

I In a transcritical bifurcation, there are two equilibria, onestable and one unstable. When the bifurcation point is passed,there is an exchange of stability; the unstable equilibriumbecomes stable and the stable one becomes unstable.

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Bifurcations

These types of bifurcations also exist for systems of two or moreequations. An easy way to obtain the bifurcations for twodifferential equations is to add the second equation y ′ =−y to thex equation. For example

x ′ = x2 +a

y ′ =−y ,

has a saddle node bifurcation.

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Bifurcations

For systems there is a fourth type of bifurcation. a Hopfbifurcation. Consider the linear system:

x ′ = ax−y

y ′ = x +ax .

The origin is an equilibrium. The trace and determinant of theJacobian matrix evaluated at the origin are 2a and a2 + 1,respectively. Since the discriminant of the Jacobian matrix isnegative, (2a)2−4a2−4 =−4, the eigenvalues are a± i . If a< 0,the origin is a stable spiral. If a = 0, the origin is a center, and ifa> 0, it is an unstable spiral. The bifurcation value is at a = 0.

Dynamical systems


Bifurcations

As the bifurcation parameter a increases through the bifurcationvalue a = 0, the equilibrium (0, 0) changes from a stable spiral to aneutral center to an unstable spiral. There are infinitely manyperiodic solutions at the bifurcation value a = 0. There is a Hopfbifurcation at the origin.

Dynamical systems


Bifurcations

Consider the system

x ′ = ax−y −x(x2 + y2)

y ′ = x +ay −y(x2 + y2).

The origin is an equilibrium point. The linearized system is

X ′ =

(a −11 a

)X ,

with eigenvalues a± i .

Dynamical systems


Bifurcations

There is a bifurcation for a = 0. To study the behavior, change topolar coordinates

r ′ = ar − r3

θ′ = 1.

For a< 0 the origin is a sink (stable spiral). For a> 0 it is a source(unstable spiral). But also r ′ = 0 if r =

√a so there is a periodic

solution. r ′ > 0 if 0 < r <√a and r ′ < 0 if r >

√a, so all nonzero

solutions spiral toward the periodic solution as t→ ∞. A periodicsolution appears as a increases through zero. We have a Hopfbifurcation.

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Bifurcations

Figure : Phase plane, a =−1, stable spiral

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Bifurcations

Figure : Phase plane a = 0, stable spiral.

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Bifurcations

Figure : Phase plane a = 1, limit cycle.

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Bifurcations

In general, at a bifurcation, as a passes through the bifurcationvalue a0 with the eigenvalues pure imaginary, there are threepossible dynamics that may occur.

1. At the bifurcation value a0 infinitely many neutrally stableconcentric closed orbits encircle the equilibrium.

2. A stable spiral changes to a stable limit cycle for values of theparameter close to a0 (supercritical hifurcation).

3. A stable spiral and unstable limit cycle change to an unstablespiral for values of the parameter close to a0 (subcriticalbifurcation).

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Bifurcations

Theorem (Hopf Bifurcation Theorem in two dimensions)

Consider the system

X ′ = A(a)X +F (X ,a).

Let F have continuous third-order derivatives in x and y . Assumethat the origin is an equilibrium point and that the Jacobian matrixA(a) is valid for all |a| sufficiently small. In addition, assume thatthe eigenvalues of matrix A(a) are α(r)± iβ (a) with α(0) = 0 andβ (0) 6= 0 such that the eigenvalues cross the imaginary axis withnonzero speed (transversal),

dα

da|a=0 6= 0.

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Bifurcations

Theorem (Continuation)

Then, in any open set U containing the origin in R2 and for anya0 > 0, there exists a value a∗,a∗ < a0 such that the system ofdifferential equations has a periodic solution for a = a∗ in U (withapproximate period T = 2π/β (0).

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Periodic solutions

Periodic solutions

Equilibrium solutions of systems of differential equations areundoubtedly among the most important solutions, but there areother types of solutions that are important as well. Anotherimportant type of solution is the periodic solution or closed orbit.Like equilibrium points that are asymptotically stable, periodicsolutions may also attract other solutions. That is, solutions mayapproach periodic solutions just as they can approach equilibria.In the plane, the limiting behavior of solutions is essentiallyrestricted to equilibria and closed orbits, although there are a fewexceptional cases. The most important result is known asPoincare-Bendixson theorem. In higher dimensions there are fewresults.

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Periodic solutions

The notation Γ(X0, t) denotes a solution trajectory as a function oftime t beginning at the initial point X0 = (x(t0),y(t0)) = (x0,y0),In addition, Γ+(X0, t) is the part of the solution trajectory wheret ≥ t0, a positive orbit, and Γ−(X0, t) is the part of the solutionwhere t ≤ t0, a negative orbit.

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Periodic solutions

If solutions are bounded, then their negative and positive orbitsapproach limiting sets as t→−∞ or as t→ ∞. The α-limit set,denoted α(X0), is the set of points in the plane that areapproached by the negative orbit as t→−∞. The ω-limit set,denoted ω(X0), is the set of points in the plane that areapproached by the positive orbit as t→ ∞

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Periodic solutions

Theorem (Poincare-Bendixson Theorem)

. Let Γ+(X0, t) be a positive orbit that remains in a closed andbounded region of the plane. Suppose the ω-limit set does notcontain any equilibria. Then either

1. Γ+(X0, t) is a periodic orbit (Γ+(X0, t) = ω(X0)) or

2. the ω-limit set, ω(X0), is a periodic orbit.

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Periodic solutions

Theorem (Poincare-Bendixson dichotomy)

. Let Γ+(X0, t) be a positive orbit that remains in a closed andbounded region B of the plane. Suppose B contains only a finitenumber of equilibria. Then the ω-limit set takes one of thefollowing three forms;

1. ω(X0) is an equilibrium.

2. ω(X0) is a periodic orbit.

3. ω(X0) contains a finite number of equilibria and a set oftrajectories Γi whose α- and ω-limit sets consist of one ofthese equilibria for each trajectory Γi .

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Periodic solutions

An important assumption in both of these theorems is thatsolutions are bounded. In Case 2, if Γ+(X0, t) 6= ω(X0) butapproaches the periodic orbit, then the periodic orbit may be alimit cycle. In Case 3, the limiting set is referred to as a cyclegraph. The cycle graph may consist of either an equilibrium and ahomoclinic orbit (connecting an equilibrium to itself) or severalequilibria and heteroclinic orbits (connecting two differentequilibria).

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Periodic solutions

Consider the example:

x ′ = 8x−y2

y ′ =−y + x2.

It has two equilibrium points, (0,0) and (2,4). The Jacobianmatrix is

DF =

(8 −2y

2x −1

).

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Periodic solutions

At (0,0)

DF =

(8 00 −1

).

Because one eigenvalue is positive (λ1 = 8) and one is negative(λ2 =−1), the origin is a saddle point. Solutions move away fromthe origin along the x-axis (unstable manifold) and move towardthe origin along the y -axis (stable manifold), This behavior can beseen by finding the eigenvectors.

Dynamical systems


Periodic solutions

At the second equilibrium (2,4)

DF =

(8 −84 −1

),

The characteristic equation is λ 2−7λ + 24 = 0 orλ1,2 = 7/2± i

√47/2. The equilibrium (2,4) is an unstable spiral

point.

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Periodic solutions

Figure : Phase plane. Two equilibria, (0,0) saddle, (2,4) unstable spiral

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Periodic solutions

Two results results give sufficient conditions that rule out thepossibility of periodic solutions. They are Bendixson’s criterion andDulac’s criterion.

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Periodic solutions

Theorem (Bendixson’s Criterion)

Suppose D is a simply connected open subset of R2. If theexpressiondiv(f ,g) = ∂ f /∂x + ∂g/∂y is not identically zero anddoes not change sign in D, then there are no periodic orbits of theautonomous system in D.

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Periodic solutions

Theorem (Dulac’s Criterion)

Suppose D is a simply connected open subset of R2 and B(x ,y) isa real-valued C 1 function in D. If the expression

div(Bf ,Bg) =∂ (Bf )

∂x+

∂ (Bg)

∂y

is not identically zero and does not change sign in D, then thereare no periodic solutions of the autonomous system in D.

Dynamical systems


Periodic solutions

Example: Consider the predator prey system with logistic growthfor the prey x , and constant recruitment for the predator y :

x ′ = x(1−ax−by)

y ′ = y(1 + cx−dy),

with a,b,c ,d > 0. Let B(x ,y) = 1/(xy), D = {(x ,y)|x > 0,y > 0}.div(Bx(1−ax−by),By(1 + cx−dy)) =−a/y −d/x < 0 in D.Therefore Dulac’s criterion implies that the system has no periodicsolutions in D.

Dynamical systems


Periodic solutions

For a = 1,b = 1/2,c = 1 and d = 1 he system has four equilibriumpoints: (0,0),(1,0),(0,1) and (1/3,4/3). The origin is an unstablenode, (1,0) and (0,1) are saddle points, and (1/3,4/3) is a stablenode.

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Periodic solutions

Figure : Phase plane, four equilibria. Only (1/3, 4/3) is stable (node)

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Periodic solutions

This notes are based mainly on [4] and [1]. Other good referencesare [2] and [3].

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Periodic solutions

Linda JS Allen.An Introduction to Mathematical Biology.Pearson, 2007.

Gerda De Vries, Thomas Hillen, Mark Lewis, Birgitt

SchOnfisch, et al.A course in mathematical biology: quantitative modeling withmathematical and computational methods, volume 12.Siam, 2006.

Leah Edelstein-Keshet.Mathematical models in biology, volume 46.Siam, 1988.

Dynamical systems


Periodic solutions

Morris W Hirsch, Stephen Smale, and Robert L Devaney.Differential equations, dynamical systems, and an introductionto chaos.Academic press, 2012.

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