Lectures Weeks 1-4

8/12/2019 Lectures Weeks 1-4

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1.4 Picard’s Theorem

Let f (x, y) and ∂f/∂y be continuous functions of x and y on a closed rectangle R withsides parallel to the axes. If (x0, y0) is any interior point of R then there exists a numberh > 0 with the property that the initial value problem

y′ = f (x, y), y(x0) = y0

has one and only one solution y = y(x) on the interval |x − x0| ≤ h.Proof . We write yn as

yn(x) = y0(x) +n

m=1

[ym − ym−1]

We therefore define

y(x) = y0(x) +∞

m=1

[ym − ym−1]

and we need to show that

1. y(x) exists

2. y(x) is a solution of the DE

3. y(x) is the only solution of the DE

We begin by proving (1) that y(x) exists, i.e. we need to show that the series aboveconverges.

We know that|f (x, y)| ≤ M

| ∂

∂yf (x, y)| ≤ K

on the interval R. The mean value theorem tell us that

|f (x, y1) − f (x, y2)| ≤ ∂

∂y f (x, y⋆) |y1 − y2|

for some number y⋆ between y1 and y2. Therefore

|f (x, y1)− f (x, y2)| ≤ K |y1 − y2|

We define h < 1/K and define R′ = {(x, y), |x−x0| ≤ h, |y−y0| ≤ Mh}. We requirethat h is sufficiently small that R ′ lies inside R.

We can show that all of the functions yn are in the interval R′.Define a = max |y1(x) − y0(x)|. Then

|f (t, y1)− f (t, y0)| ≤ K |y1(x) − y0(x)| ≤ Ka

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1.5 Lipschitz condition

If we look at the proof for the existence and uniqueness solution we see that we actuallyonly required that

|f (x, y1)− f (x, y2)| ≤ K |y1 − y2|This is a weaker condition than requiring that the function f (x, y) has a continuouspartial derivative with respect to y and the existence and uniqueness theorem can im-mediately be generalized.

Definition A function is Lipschitz continuous in an interval I if

|f (y1)− f (y0)| ≤ C |y1 − y0|

for all y1, y0

∈I for some C > 0.

What does it mean to be Lipschitz continuous? It is a stronger condition thancontinuity but it does not imply that the derivative is bounded. Consider the examples

1. f (y) = |y|

2. f (y) = y2/3

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We can extend the theorem to the case where f is only Lipshitz continuous withrespect to y. We will prove a slightly different version of the theorem

We have the following theorem: If f (y) is differentiable in I then f is Lipschitzcontinuous with constant C if and only if

supy∈I

∂f

∂y(y)

≤ C.

Proof:

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1.6 Existence and Uniqueness for the Case of Lipshitz Conti-

nuity

The existence and uniqueness proof can be generalized in many ways. Let us considerthe following theorem.

Theorem Let f (x, y) be a continuous function that satisfies a Lipshitz condition

|f (x, y1)− f (x, y2)| ≤ K |y1 − y2|

on a strip defined by a ≤ x ≤ b and −∞ < y < ∞ (note that this is not true for y ′ = y2).Then, if (x0, y0) is any point in the strip, then the initial value problem

y′ = f (x, y), y(x0) = y0.

has a unique solution on the interval a ≤ x ≤ b.Proof We define the constants

M 0 = |y0|, M 1 = max |y1(x)|, M = M 0 + M 1

It follows that |y0(x)| ≤ M and |y1(x) − y0(x)| ≤ M . If x0 ≤ x ≤ b it follows that

|y2(x)− y1(x)| =

xx0

f (t, y1)− f (t, y0) dt

≤ x

x0 |f (t, y1)− f (t, y0)| dt

≤ K

xx0

|y1(x) − y0(x)| dt

≤ KM (x− x0)

Likewise

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Therefore we obtain the following result. The difference between the true solution yand the Picard Iterate yn is

|y(x)− yn(x)| ≤ M (K (x− x0))n

n! eK (x−x0)

Note this is an upper bound and the actual iterates may be much more accurate. Thereare also bounds on the Picard iterates which can be found for the first case we considered.

Example Consider the initial value problem

y′ = x2y; y(0) = 1.

one the interval 0 ≤ x ≤ 1. Find the Picard iterates y1, y2 etc. Find an upper boundon the error for these iterates.

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2 Autonomous Systems

A system of DE’s is autonomous if the right hand side does not depend on the variableof differentiation. In two dimensions they can be written as

dx

dt = f (x, y)

dy

dt = g(x, y)

Example

dx

dt

= 0.5x

−0.4xy

dy

dt = −y + 0.2xy

2.1 Direction fields

At each point in the (x, y) plane we can determine the slope of the solution. We canthen plot these slopes to give a direction field.

So to sketch solution curves to DE,

1. plot direction field, then

2. starting at some initial point, sketch a smooth curve that follows vectors in direc-tion field.

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Example: Sketch some representative solutions for the system

dx

dt = y

dy

dt = sin(x)

The direction field is given below.

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2.2 Equilibrium solutions

The point (x0, y0) is an equilibrium point for the system

dx

dt = f (x, y)

dy

dt = g(x, y)

if f (x0, y0) = g(x0, y0) = 0.Example 1

dx

dt = 2x + y

dy

dt = 2y + x

Example 2

dx

dt = x + y

dy

dt = y (2 − x)

Behaviour of solutions near equilibriums can be observed with pplane . Note that

1. direction of vectors in direction field changes dramatically near an equilibriumpoint, and

2. solutions passing near an equilibrium go very slowly (because all components of

vector field → 0 near an equilibrium).

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Example Determine the behaviour of solutions to the system

dx

dt = −4 −2−1 −3

x.

dx/dt = − 4 x − 2 y dy/dt = − x − 3 y

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

The equilibrium point at the origin in this type of system is called a sink. If theflow is reversed then the equilibrium is a source.

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Linear systems with repeated eigenvalues

Example Consider the system

dx

dt =

2 00 2

x

dx/dt = 2 x dy/dt = 2 y

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

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Example Consider the system

dx

dt = −5 08 −5

x.

dx/dt = − 5 xdy/dt = 8 x − 5 y

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

x

y

Note that system has only one straight line solution.

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Example : Consider the one-parameter family of linear systems

dx

dt =

1 2a 0

x

where a is a parameter. Determine the type of equilibrium at the origin for all values of a. Sketch the phase portrait for representative values of a.

Eigenvalues of matrix

A =

1 2a 0

are

12 ± 1

2√ 1 + 8a

so the type of equilibrium at the origin depends on a. Also, det(A) = −2a and trace(A) =1.

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Example 1 Analyze the equilibrium points for

dx

dt = 2− x− y

dy

dt = x2 − y

dx/dt = 2 − x − y

dy/dt = x2 − y

−4 −3 −2 −1 0 1 2 3 4

−2

−1

0

1

2

3

4

5

6

x

y

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Unfortunately, linearization does not always work. In particular, if the Jacobianmatrix has a zero eigenvalue or a purely imaginary eigenvalue, then we cannot predict

the behaviour in the nonlinear system based on linearization alone.Example

dx

dt = −x3

dy

dt = −y + y2

dx/dt = − x3

dy/dt = − y + y2

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

1.5

2

x

y

Notice that in this phase portrait, (0, 0) looks like a sink and (0, 1) looks like a saddle.

These results were not predicted by the corresponding linearized systems. Linearizationdoes not work in these cases because of the zero eigenvalues of the Jacobian.

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3 Hamiltonian and Lyapunov functions

3.1 PendulumThe equation for a pendulum which is free to rotate is given by

d2θ

dt2 = −g

l sin θ

We write this (assuming g/l = 1) as

dx

dt = y

dy

dt = − sin(x)

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3.2 Hamiltonian Systems

A system is Hamiltonian if it can be written in the form

dx

dt =

∂H

∂y

dy

dt = −∂H

∂x

If we consider the equations for the pendulum we can see that H = − cos(x) + y2/2 givesthe desired equations.

One important consequence of this is that the quantity H must be conserved for oursystem. This has important consequences.

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3.3 KdV equation

The KdV (Korteweg de Vries) equations is one of the most important nonlinear partialdifferential equations. When looking for travelling wave solutions the problem is reducedto the following (assuming again all constants are one)

dx

dt = y

dy

dt = x − x2

What is the Hamiltonian for this system.

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3.5 Damped pendulum

We can modify the pendulum problem by including damping. This leads to

dx

dt = y

dy

dt = − sin(x) − y/10

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3.6 Lyapunov function

We assume that we have a dynamical system

dx

dt = f (x, y)

dy

dt = g(x, y)

We assume we have an equilibrium point at (0 , 0) (we can always transform any otherequilibrium point to the origin). A Lyapunov function is a function with the propertiesthat

1. V (x, y)

≥0

2. V (0, 0) = 0

3. dV dt ≤ 0 along solutions of the DE, in some neighbourhood of the equilibrium point.

If such a function can be found then the equilibrium point must be stable.If the function V is strictly decreasing then all solutions in the neighbourhood tend

to the equilibrium point.Note that in general it is not easy to find a Lyapunov function.

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Consider the equation for a spring with linear damping.

md2y

dt2 + cdy

dt + ky = 0

30



Consider the system

dx

dt = −2xy

dy

dt = x2 − y3

Try a function of the formV (x, y) = ax2m + by2n

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4.2 Pitchfork Bifurcation

In this bifurcation a source (or sink) is transformed into three equilibriums, two sources(sinks) and one saddle.

dx

dt = αx− x3

dy

dt = −y

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4.3 Hopf Bifurcation

In this bifurcation the real part of complex eigenvalues for the linearized system changefrom negative to positive. In this case, under fairly general conditions a closed periodicorbit is created.

dx

dt = 1 + x2y − (α + 1)x

dy

dt = αx − x2y

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Lectures Weeks 1-4