Title: VL 11 PP Folien
1 Universidad de La Habana
Lectures 5 6 Difference Equations Kurt Helmes
22nd September - 2nd October, 2008
2Part 1 Introduction Part 2 First-Order
Difference Equations Part 3 First-Order
Linear Difference Equations
3Introduction
4Part 1.1
An Example
5Dagobert- Example
6Starting Point
Given
K0
initial capital ( in Euro )
p
interest rate ( in )
r
7Objective
Find ....
- 1. The amount of capital after 1 year.
- 2. The amount of capital after 2 years.
- n. The amount of capital after n years.
8Solution
After one year the amount of capital is
How much capital do we have after 2 years?
9Solution
After one year the amount of capital is
After two years the amount of capital is
10Solution
After one year the amount of capital is
After two years the amount of capital is
11Solution
After n years the amount of capital is
12Observation
special difference equation
recursion formula
13Part 1.2
Difference Equations
14Illustration
A difference equation is a special system of
equations, with
- (countably) infinite many equations,
- (countably) infinite many unknowns.
15Hint
The solution of a difference equation is a
sequence (countably infinite many numbers).
16How do we recognize a difference equation?
17Definition Difference Equation
Explicit form
Implicit form
18First-Order Difference Equations
19Part 2.1
A Model for theHog Cycle
20Hog Cycle (Example)
21Starting Point
Given Hog-corn price ratio in Chicago in the
period 1901-1935
22Starting Point
Stylized
23Starting Point
- Find
- A (first) model, which explains /
describes the cyclical fluctuations of the
prices (ratio of prices).
24Model (Part 1) Supply and Demand
The suppply of hogs
The demand of hogs
25Model (Part 2) Supply and Price
Assumption
26Model (Part 2)
Nature of the dependance
Assumption
The supply function is linear
27Figure 1 Graphical representation of the
supply function
28Model (Part 3) Demand and Price
Assumption
For the demand we assume If the hog price
increases, the demand will decrease, thus
29Figure 2 Graphical representation of the
demand function
30Model (Part 4) Equilibrium
Postulate
Supply equals demand at any time
31Model (Part 4) Equilibrium
The equilibrium relation yields a defining
equation for the price function
32Solution (Part 4) Equilibrium
Thus we obtain the following difference equation
33Model (Part 4) Equilibrium
This difference equation is
- first-order
- linear
- inhomogeneous
34Model
(Part 5) Analysis
solution formula
35Deriving the Solution Formula
....
36Figure 2
37Model (Part 5) Analysis
Results
The equation / solution is stable.
The equation / solution is unstable.
38Figure 3 Price development for
39Figure 4 Price development for
40Figure 5 Price development for
41Summary
The given difference equation has a unique
solution it can be solved explicitly.
The price is the sum of a constant and a power
function.
42CONCLUSION
We can model and analyze dynamic processes with
difference equations.
43Part 2.2
Definitions und Concepts for First-Order
Difference Equations
44Definition
A (general) first-order nonlinear difference
equation has the form
(F is defined for all values of the variables.)
45Important Questions
- Does at least one solution exist?
- Is there a unique solution?
- How many solutions do exist?
- How does the solution change, if parameters
of the system of equations are changed
(sensitivity analysis)?
46Important Questions
- Do explicit formulae for the solution exist?
- How do we calculate the solution?
- Does the system of equations has a special
structure ?
e.g. a) linear or nonlinear,
b) one- or multidimensional ?
47Remark
If the initial value of the solution (sequence)
of a difference equation is given, i.e.
then we call our problem an
initial value problem
related to a first-order difference equation.
48Remark
The initial value problem of a first-order
difference equation has a unique solution.
49Remark
50Definition Invariant Points
invariant points.
F right-hand side.
51Invariant Points
52Invariant Points
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54Iteration rule
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62Newtons-Method (Example)
63Starting Point
Finding the roots of a nonlinear function
analytically is rarely possible. Therefore we
have to use numerical methods.
64Starting Point
For differentiable functions a numerical
root-finding algorithm exists. It goes back to
Isaac Newton (1643 1727).
65Goal
66Idea
67Idea
68Idea
Determine the intersection of the tangent with
the x-axis.
69Idea
70Idea
Repeat this operation many times.
71Figure 12 Schematic representation of
Newtons Method
72Solution
73Solution
74Solution
By the same idea we compute x2, x3, ... as
(difference equation)
75Solution
This is a (nonlinear) first-order difference
equation, and
76Numerical Example
Consider the problem of finding the root of
.
The difference equation according to Newtons
Method is
77First-Order Linear Difference Equations
78Part 3.1
First-Order Linear Difference Equations
with a Constant a-Term
79Definition
Time-dependent, inhomogeneous linear difference
equations of first order with constant a-term
take the form
Equation
80Definition
Time-dependent, inhomogeneous linear difference
equations of first order with constant a-term
take the form
Equation
Iteration Rule
81Lösungsformel
Time-dependent, inhomogeneous linear difference
equations of first order with constant a-term
have the solution
Solution formula
82Deriving the Solution Formula
....
83Special Case
For first-order linear difference equations with
constant coefficients it holds
84Example of an Exam Exercise
1
Solution
Backwards iteration yields
85Example of an Exam Exercise
1
Solution
86Example of an Exam Exercise
1
Solution
87Example of an Exam Exercise
1
Intermediate Calculation
Expanding the equation
88Example of an Exam Exercise
1
Solution
Expanding the equation
89Example of an Exam Exercise
1
Solution
General condensation of the terms
90Example of an Exam Exercise
2
91Example of an Exam Exercise
2
92Example of an Exam Exercise
2
The initial value problem can be solved either
directly by using the solution formula, i.e.
93Example of an Exam Exercise
2
94Example of an Exam Exercise
2
or by forward iteration
95Example of an Exam Exercise
2
Continuing with forward iteration
and in general
96Example of an Exam Exercise
97Example of an Exam Exercise
and it holds
98Dagobert- Example
(with deposits and payments)
99Starting Point
interest factor
100Find Formula for the account balance
Solution formula
101Formula for the account balance
The discounted capital flow is
102Summary
The discounted capital stock at time t equals the
capital stock at time t0 plus the sum of the
discounted deposits minus the sum of the
discounted payments up to time t .
103Part 3.2
First-Order Linear Difference Equations with
Variable Coefficients
104Definition
First-order linear difference equations with
variable coefficients take the form
105Solution formula
The solution of first-order linear difference
equations with variable coefficients is given by
106Dagobert- Example
with variable interest rate andproportional
deposits and payments
107Starting Point
Consider a capital model with time-dependent
interest factor
interest factor
deposits
payments
108Starting Point
Special Case Capital model with proportional
deposits and payments
109Proportional In- and Outpayments
110Numerical Example
Capital stock
1000 Euro
111Numerical Example
Capital stock
1000 Euro
Interest factor
112Numerical Example
Capital stock
1000 Euro
Interest factor
Rate of deposits
113Numerical Example
Capital stock
1000 Euro
Interest factor
Rate of deposits
Rate of payments
114Numerical Example
115Numerical Example
116Numerical Example
117Part 3.3
Stability of First-Order Linear Difference
Equations
118Definition Stability
A first-order difference equation is called
stable, if
the solution of the homogeneous equation
converges for any initial value to zero.
cf.. 1) unstable 2) chaotic
119Stability Conditions
120Stability Conditions
Remark 1
If
holds for one time point s,
121Stability Conditions
Remark 2
Stability comes along in different forms
Example
1
122Figure 13 Schematic representaion of
stability - Case A
123Stability Conditions
Remark 2
Stability comes along in different forms
Example
2
124Figure 14 Schematic representaion of
stability - Case B
125Stability Conditions
Remark 3
126Figure 15 Schematic representaion of
stability - Case C