Mathe IIILecture 5Mathe IIILecture 5Mathe IIILecture 5Mathe IIILecture 5

2

Stability:

In the long run, the solution should be independent of the initial conditions.

The general solution of

t+2 t+1 t tx + ax + bx = c

is: *1 2t t t tx = Au + Bu + u

if :

it

tlim u = 0

The system is stable.

3

i tt i

t tlim u = lim tm = 0

where

solves

i tt i i

2

u = m m

m + am + b = 0.

i tt i

t tlim u = lim m = 0

if im < 1

1-1

2f m = m + am + b

m

2a - 4b 0, f -1 = f 1 > 0, f' -1 < 0, f' 1 > 0

The root (s) are in (-1, 1) iff:

4

1-1

2f m = m + am + b = 0

m

2a - 4b 0, f -1 = f 1 > 0, f' -1 < 0, f' 1 > 0

2a 4b,

1 - a + b > 0, 1+ a + b > 0

-2 + a < 0, 2 + a > 0 a < 2

a < 1+ b

b < 1

a < 1+ b, b < 1The system is stable iff:

5

Differential Equations

First Order Differential Equations

wheredx

= f x,t x = x tdt

first order, ordinary equation (single variable)

Differential EquationsDifferential Equations

6

x

dx= f x,t

dt

t

7

dx= x = f t

dt

x t = f t dt + C

The simplest possible equation:

x

t

8

wherex = f x,t x = x t

x t + τ - x t= f x,t

τ

An approximation:

For a given τ let: t = 0,τ,2τ,3τ, ....nτ

we obtain a difference equation , solve it and let τ 0

or graphically:

x t + τ = x t + τf x t ,t

9

x

wherex = f x,t x = x t

For t = 0, assume x(0) = x0

x0

t

0x 0 = f x ,0

τ

x1

1x τ = f x ,τ

2τ

x2

etc.

x t + τ = x t + τf x t ,t

0 0x τ = x + τf x ,0 x 2τ = x τ + τf x τ ,τ

10

x

wherex = f x,t x = x t

For t = 0, assume x(0) = x0

x0

tτ

x1

2τ

x2

Now choose a smaller τ

2ττ

As we approach a curve which solves

τ 0

x = f x,t

11

,

There exists a unique satisf

ying

0

x = x t

x = f x,t x 0 = x

12

Separable Differential Equations

x = f t g x

dx= f t g x

dt

A formal ‘trick’:

dx= f t dt

g x dxf= t dt + C

g x

13

Is this ‘trick’ valid ???

If solves the equa tio n (t) x = f t g x

then : (t) = f t g t

when then : (t)

g t 0 = f tg t

hence : (t)dt f t dt C

g t

14

hence : (t)dt f t dt C

g t

change the varia : ble : x = t dx = t dt

dxf t dt C

g x

G x = F t + C

1 G' x = , F' t = f t

g xThis defines x as an implicit function of t

15

3

6

tx =

x + 1EXAMPLE:

6 3x + 1 dx = t dt

6 3x + 1 dx = t dt C

7 41 1x + x = t + C

7 4

16

(compound interest)

r t EXAMPLE:

1

d= r t dt C

ln 1= R t C R t = r t dt

1C + R tt = e

1 R tCt = e e R t= Ce 0R0 = Ce R 0

0C =

e

R t

R 0

0t = e

e

17

r t EXAMPLE:

1 R tCt = e e R t= Ce 0R0 = Ce R 0

0C =

e

R t

R 0

0t = e

e

R t -R 0= 0 e

t

0

R t - R 0 r s ds

R t -R 0= 0 e

t

0

r s ds

t 0 e

exp t

0

0 r s ds

18

Separable Differential Equations

0 0

x t = xx = f t g x

(again)

We have seen that if solves the equation,

with then : 0 0

(t)

(t ) = x

(t)= f t

g t

0

and

0

t

t t

t(s)

ds f s dsg s

19

Separable Differential Equations (again)

0

0

t

t t

t(s)

ds f s dsg s

change the variable :

0 0= s , d = s ds, = t

0

0

t

x t

xd

f s dsg

20

0

2 x = x 0dx

= -2x tdt

EXAMPLE:

2

dx- = 2tdt

x

x t

0

2

x 0

d- = 2sds

1

0

xt2

0x

s

2= t0

1 1- =

x x

2

0

1 1= t

x x 2

0

1t +

x

1x =

Graphic description of the solution

21

0

2 x = x 0dx

= -2x tdt

EXAMPLE:

2

0

1t +

x

1x =Graphic description of the solution

denote0

1 C =

x

2t + C

1x =

0assume first 00

1 x > 0 : C =

x

22

0 0

2 x = x 0dx

= -2x tdt

EXAMPLE:

2t + C

1x =

t

x

Graphic description of the solution

x 0

23

0

2 x = x 0dx

= -2x tdt

EXAMPLE:

2t C

1x =

-

t

x

Graphic description of the solution

< 0

, 02

0

01

t +x

1 1x = C = -

x

C C

24

0

2 x = x 0dx

= -2x tdt

EXAMPLE:

2t C

1x =

-

t

x

Graphic description of the solution

< 0

C C

0 < t < Ct > C

0 > x 0

C2t C

1x =

+

25

Example

1-α α

λt0

X = AK L

K = sX

L = L e

national product

capital stock

number of workers

X t =

K t =

L t

1-α αK = sAK L α1-α λt0= sAK L e

λt0L e

1-α αAK L

sX

α 1-α λαt0

dKK = = sAL K e

dtα-1 α λαt

0K dK = sAL e dt

26

Example

α-1 α λαt0K dK = sAL e dt

0

α-1 α λατ0

K t

K 0

d = sAL e dτ

0

K tα α λατ

0K 0

1 1sAL e

α αλ

α α α λαt0 0

1 1Κ - K sAL e - 1

α αλ

27

1-α α

λt0

X = AK L

K = sX

L = L e

Example

α α α λαt0 0

1 1Κ - K sAL e - 1

α αλ

1/αα α λαt0 0

sK = K + A L e - 1

λ

This enables us to study how the evolution of capital changes with the parameters s,α, λ, A

28

1-α α

λt0

X = AK L

K = sX

L = L e

Example

1/αα α λαt0 0

sK = K + A L e - 1

λ

How does K/L behave in the long run?

α λαtα α00

α α λαt0

L e - 1KK s= + A

L L λ L e

α α α λαt0 0

sK = K + A L e - 1

λ

λt0L e

αL/

29

1-α α

λt0

X = AK L

K = sX

L = L e

Example

1/αα α λαt0 0

sK = K + A L e - 1

λ

How does K/L behave in the long run?

α λαtα α00

α α λαt0

L e - 1KK s= + A

L L λ L e 1α

-λαt0α

K s= + A e

L λ

t

sA

λ