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Zusammenfassung

Summary Dynamical Systems

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Vorschau 3 aus 20 Seiten

Summary for Dynamical systems.

Inhaltsvorschau

Dynamical Systems

1 First-Order Equations
1.1 The simplest example
If x′ = ax, the solution is x = keat . We call the collection of all solutions of a differential
equation the general solution of the equation. A specific solution can be found using an initial
condition.
A solution when x′ = 0 is called an equilibruim point for the equation.
We say that the equilibrium point is a source when nearby solutions tend away from it. The
equilibrium point is a sink when nearby solutions tend toward it.
Phase line: we view x(t) as a particle moving along the real line.

1.2 The logistic population model
logistic population growth model:  x
x′ = ax 1 −
N
a gives the rate of population growth when x is small, while N represents a sort of “ideal”
population or “carrying capacity.”
When N = 1, the solution (found by sep of variables) is given by

Keat x(0)
x(t) = , K=
1 + Keat 1 − x(0)

You can sketch a slope field using the right hand side of the equation.
We have that for the equation x′ = g(x), when g ′ (x) > 0 then x is a source and if g ′ (x) < 0 it
is a sink.

1.3 Constant harvesting and bifurcations
Harvesting equation: x′ = x(1 − x) − h with h ≥ 0
In a bifurcation diagram, we plot the parameter h horizontally. Over each h-value we plot the
corresponding phase line. The curve in this picture represents the equilibrium points for each
value of h. This gives another view of the sink and source merging into a single equilibrium
point and then disappearing.
Saddle-node bifurcation: two equilibria of opposite stability collide and cease to exist.

1.4 Periodic harvesting and periodic solutions
A model for periodic harvesting:

x′ = ax(1 − x) − h(1 + sin(2πt))

(depends on time, so is nonautonomous).

1

,If a solution is periodic and you know things about one period, you know something about the
entire function.
Poincaré map: p(x0 ) = x(1). Applying iteratively: pn (x0 ) = x(n)
If we have that p(x0 ) = x0 , then x is periodic (i.e. a solution is periodic if the poincaré map
has a fixed point).

1.5 Computing the poincaré map
We define the flow: ϕ : R × R → R : t 7→ ϕ(t, x0 ) (just an alternative expression for the solution
of the differential equation satisfying the initial condition x0 )
If a solution is periodic with period 1, we have that p(x0 ) = ϕ(1, x0 ).
The poincaré map is an increasing function. It can have either 0, 1 or 2 fixed points.



2 Planar linear systems
2.2 Planar systems
A planar system is of the form
x′ = f (x, y)
y ′ = g(x, y)

2.4 Planar linear systems
We restrict our attention now to planar linear systems, that is
x′ = ax + by
y ′ = cxd y
Which we can write in matrix form
Å ′ã Å ãÅ ã
′ x a b x
X = AX, =
y′ c d y
Proposition The planar linear system X ′ = AX has
1. A unique equilibrium point (0, 0) if det A ̸= 0
2. A straight line of equilibrium points if det A = 0 and A is not the 0 matrix

2.5 Eigenvalues and Eigenvectors
Suppose that V0 is an eigenvector for the matrix A with associated eigenvalue λ. Then the
function X(t) = eλt V0 is a solution of the system X ′ = AX

2.6 Solving Linear Systems
Theorem Suppose A as a pair of real eigenvalues λ1 ̸= λ2 and associated eigenvectors V1 and
V2 . Then the general solution of the linear system X ′ = AX is given by
X(t) = αeλ1 t V1 + βeλ2 t V2


2

, 3 Phase portraits for Planar systems
3.1 Real distinct eigenvalues
Å ã
λ1 0
Case 1: λ1 < 0 < λ2 (saddle) Let A = We then have that
0 λ2
Å ã Å ã
λ1 t 1 λ2 t 0
αe + αe
0 1

Note that as t → ∞, αeλ1 t (1, 0) = (0, 0) (stable line). As t → ∞, βeλ2 t (1, 0) goes away from
(0, 0). The equilibrium point (0, 0) is called a saddle.




Case 2: λ1 < λ2 < 0 When λ1 < λ2 < 0 everything tends to (0, 0). How do they approach
the origin? We compute dy/dx = (dy/dt)/(dx/dt) = λλ12 αβ e(λ2 −λ1 )t . Since λ2 − λ1 > 0, these
slopes approach ±∞. Thus these solutoons tend to the original tangentially to the y-axis. Since
λ1 < λ2 , we call λ1 the stronger eigenvalue and λ2 the weaker eigenvalue.
Case 3: 0 < λ2 < λ1 (Source) Basially the negative from case 2, all solutions tend away from
(0, 0) instead of towards it but they follow the same paths




3

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