Summary Difference- & Differential Equations for EOR (RUG)
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Econometrics and Operations Research
Difference- and Differential Equations
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Difference- and Differential Equations
Freeke Boerrigter
Lecture 1
First Order Ordinary Differential Equations (first-order ODE)
F ( t , x ( t ) , x ' ( t ) )=0 ,t ∈ T
Where F is a function of at most 3 variables and t ⊆T
Ordinary implies that x is only differentiated with respect to one variable.
Autonomous – a first-order ODE that does not depend on t , can be written as F ( x ( t ) , x ' ( t ) )=0
Linear – if ( y , z ) → F ( t , y , z )
Four types of first order ODE’s:
1. x ' ( t )=g ( t ) -> type I ODE
t
Solution: x (t )=∫ g ( s ) ds+ c , where c=x ( t 0 )
t0
2. x ( t )=f (t ) g ( x ( t ) ) -> separable ODE
'
t
Solution: P ( x ( t ) )=∫ f ( s ) ds+C
t0
3. x ' ( t )=f (t ) x ( t ) -> homogenous linear ODE
t ' t
x ( s)
Solution: ∫ ds=∫ f ( s )
t x (s )
0 t 0
⇒ log|x ( t )|=F ( t ) +c
⇒ x ( t ) =D e F ( t )
4. x ' ( t )=f (t ) x ( t )+ g (t ) -> in inhomogeneous linear ODE
t t
x ' ( s)
Solution: first ∫ ds=∫ f ( s )
t x (s)
0 t 0
⇒ log|x ( t )|=F ( t ) +c
¿
Then, we figure out that x ( t )=g ( t ) is a particular solution of the inhomogeneous ODE. So,
the general solution reads: x (t )=D e F (t ) + g ( t )
Method of Undetermined Coefficients g ( t )
¿
- If gis constant, find a solution x ≡a for some a ∈ R
- If g is a polynomial of degree n ≥ 1, find a solution that is an n th degree polynomial
- If g ( t )=c e p ( t ) where p is a polynomial of degree n ≥ 1, find a solution x ¿ ( t )=q ( t ) e p ( t ) for
some n th degree polynomial
¿
- If g ( t )=α sin ( rt ) + β cos ( rt ) ,find a solution x ( t )=Asin ( rt )+ Bcos ( rt ) for some A , B
Equilibrium solution – a solution of the first-order ODE of the form x ¿ ≡a . Find this by solving
F ( t , a ,0 )=0 .
Variation of constants -> replace the constant in the general solution x ' (t)=f ( t ) x (t) by a function
and then try to find the right function to obtain the general solution of x ' ( t )=f (t ) x ( t )+ g (t )
- The general solution of x ' ( t )=f (t ) x ( t ) is x (t )=c e F (t )
- Replace the constant c by an unknown function C :T → R
1
, - This results in x (t )=C ( t ) e F (t ), with C an unknown function
- Substitute x (t )=C ( t ) e F (t ) into the ODE and solve for C
Lecture 2
Consider the first order ODE
x ( t )=F ( t , x ( t ) ), where F : T × U → R with U ⊆ R
'
A solution x of (1) is called stable if for every ε > 0 there exists a δ >0 such that for every solution ~ x
~
defined on an interval [ t 0 ,t 1 ], where t 1> t 0 , with | ( 0 ) ~
x t −x ( t )| one has that x is a solution on T
≤ δ
and
|~x ( t ) −x ( t )|≤ ε , ∀ t ∈T
A stable solution x is called (locally) asymptotically stable if there exists a δ >0 such that for every
x with |~
solution ~ x ( t 0 ) −x ( t 0 )|≤ δ one has
lim (~x ( t )−x ( t ) ) =0
t→∞
A stable solution x is called globally asymptotically stable if there exists a δ >0 such that for every
x with |~
solution ~ x ( t 0 ) −x ( t 0 )|≤ δ one has
lim (~x ( t )−x ( t ) ) =0
t→∞
Globally is more powerful than locally.
Phase diagrams can be constructed as follows:
- Determine the roots of F ( x )=0. Indicate the roots with dots on the x-axis
- Determine the sign of F for each interval between such dots. If F is negative (positive) on
some interval, draw an arrow pointing to the left (right) in that interval
If both arrows point towards a dot located at a , solutions close to a converge to a : the equilibrium
solution x ≡ a is asymptotically stable.
If both arrows point away from a , solutions close to a diverge away from a : the equilibrium solution
x ≡ a is unstable.
If one arrow is pointing away from a dot located at a and the other one is pointing towards a , then
x ≡ a is unstable.
Assessing the stability of equilibrium solutions
Let F :U → R where U ⊆ R be a C 1 function. Consider this differential equation x ( t )=F ( x ( t ) ).
'
Suppose that F ( a )=0 for some a ∈ U . Then:
- If F ' ( a )< 0, then the equilibrium solution x ≡ a is asymptotically stable
- If F ' ( a )> 0, then the equilibrium solution x ≡ a is unstable
The Fundamental Theorem of Differential Equations
2
, ∂F
Let F :T × U → R where U ⊆ R ,and let A ⊂T ×U be an open and connected set. If exists and
∂x
∂F
both F and are continuous on A , then for every ( t 0 , x 0 ) ∈ A there exists a unique solution of the
∂x
differential equation x ' ( t )=F ( x ( t ) ) passing through ( t 0 , x 0 ).
Bounded Functions
A function f : S → R is bounded if there exists an M >0 such that |f ( x )|≤ M , ∀ x ∈ S .
Let B ( S , R ) be the set of all bounded real-valued functions with domain s ⊆ R
The distance d ( f , g ) between two functions f anf g in B ( S , R ) is defined as follows:
d ( f , g ) = x ∈ S|f ( x ) −g ( x )|
¿
An operator T : B ( S , R ) is a contraction mapping if there exists a β ∈ ( 0,1 ) such that:
d ( T ( f ) ,T ( g ) ) ≤ βd ( f , g ) , ∀ f , g ∈ B ( S , R )
Picard’s Method
We can obtain an approximate solution of the initial value problem
x ' ( t )=F ( t , x ( t ) ) ,t ∈ [ t 0 , t 1 ] , x ( t 0 ) =x0
Where F abides as follows:
∞
- Let y 0 ≡ x 0 and compute iteratively the sequence of functions { y n } n=1 given by
t
y n ( t ) =x 0+∫ F ( s , y n−1 ( s ) ) ds ,t ∈ [ t 0 , t 1 ] , n ≥1
t0
- Continue until d ( y n+1 , y n )
Example Picard’s Method
Consider the initial value problem
'
x ( t )=tx ( t ) ,t ≥ 0 , x ( 0 )=1
Use Picard’s Method to obtain an approximate solution of this problem
- Let y 0 ≡1. The first approximation y 1 is given by
t t
1 2
y 1 ( t )=1+∫ s y 0 ( s ) ds=¿ 1+∫ sds=1+ t ¿
0 0 2
- The second approximation y 2 is given by
t t
1 1
( 1
y 2 ( t ) =1+∫ s y 1 ( s ) ds=1+∫ s+ s3 ds=1+ t 2+ t 4
0 0 2 2 8 )
3
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