N20122 Mathematical Economics II 1 141

8/3/2019 N20122 Mathematical Economics II 1 141

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Econ 416/616 Mathematical Economics II Lau

Lecture Notes Spring 2012

Review of Integration

Indefinite integral as anti-derivative:

1.

1

11 1

( )111 Note that ( 1)

1 1

a

aa a a

xd c

x a x dx c a a x xa dx a

Example:7

76 7 1 6

33

21 3 3 12

12 2 2 2

22

31 2 1 133 3 3

( )17Note that (7)

7 7

2( )

2 2 33Note that ( )3 3 3 2

2

3( )

3 3 22Note that ( )2 2 2 3

3

xd c

x x dx c x x

dx

d x c

x x dx c x c x xdx

d x cx

x dx c x c x xdx

3

2. 11 (ln ) 1

ln Note thatd x c

x dx dx x c x dx x

3.x x

e dx e c Example:2

22 2 2

( )12Note that (2)

2 2

x

x x x x

ed c

ee dx c e e

dx

Definite integral:

( )b

a

f x dx

( )f x

a b


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Example:22 2 2 2

1 1

22

1 1

22 5 10 55

1 1

(2) (1) 4 1 3

2 2 2 2 2 2

1

ln ln 2 ln1 ln 2

5 5

x

x

xxdx

dx xx

e e ee dx

33

0 0

1ln ln 3 ln 0dx x undefined

x

0 3

Integration by parts: udv uv vdu

Proof: duv udv vdu uv duv udv vdu

Example 1 Evaluate x xe dx

Let

x x x

u x du dx

dv e dx dv e dx v e

x x x x x xe dx xe e dx xe e c

Example 2 Evaluate ln x dx

Let

1lnu x du dx

x

dv dx dv dx v x

1ln (ln )( ) ( ) ln ln x dx x x x dx x x dx x x x c

x

unbounded


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Example 3 Evaluate 2 x x e dx

Let

2 2

x x x

u x du xdx

dv e dx dv e dx v e

2 2 2

( ) ( )(2 ) 2

x x x x x

x e dx x e e xdx x e xe dx

Let

x x x

u x du dx

dv e dx dv e dx v e

2 2 2 2

2 2

2 2 ( ) ( ) 2 ( ) ( )

2 2 ( 2 2)

x x x x x x x x x

x x x x

x e dx x e xe dx x e x e e dx x e x e e c

x e xe e c e x x c

Example 4 Evaluate xe cos x dx

Let

cos cos sin

x xu e du e dx

dv x dx dv x dx v x

sin sin x x xe cos x dx e x x e dx

Let

sin sin cos

x xu e du e dx

dv x dx dv x dx v x

sin sin sin ( cos ) ( cos )

sin cos cos

2 sin cos

1sin cos

2

x x x x x x

x x x x

x x x

x x x

e cos x dx e x x e dx e x e x x e dx

e cos x dx e x e x e x dx

e cos x dx e x e x

e cos x dx e x e x c


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First Order Liner Differential Equations

Example 1: '( ) 1 or 1dy

y tdt

1 ( )dy

dy dt dy dt y t t cdt

Note: When we differentiate ( ) with respect to y t t c t , we get '( ) 1y t .

Example 2: '( ) y t y

1 1ln

1ln ( ) wheret c cy tdy dy dy

y dt dt y t c e e y t Ce C edt y y

Note: When we differentiate ( ) t y t Ce with respect to t, we get '( ) t y t Ce y .

Example 3:2'( ) y t y

2 1

2 2

1( )

dy dy dy y dt dt y t c y t

dt t cy y

Note: 2 2( )dy

t c ydt

Example 4: 1'( ) y t y 21

( ) 2( )2

dy y ydy dt ydy dt t c y t t c

dt y

Note:1 2 1 1

2 2( ) 2( )

dy

dt yt c t c

Example 5: '( ) 2 (0) 1 y t y

2 2 2 ( ) 2

(0) 2(0) 1 1

( ) 2 1

dy dy dt dy dt y t t cdt

y c c

y t t

Note: 2 (0) 2(0) 1 1dy

ydt


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Example 6: '( ) 3 (0) 1 y t y y

13

13 3 ln 3 ( ) wherectdy dy

y dt y t c y t Ce C edt y

3(0)(0) 1 y Ce C

3

( )t

y t e

Note: 3 3(0)3 3 (0) 1tdy

e y y edt

Example 7: '( ) 3 (1) 4 y t y y 3( ) t y t Ce 3(1) 3(1) 4 4 y Ce C e 3 3 3 3( ) 4 4t t y t e e e

3 3 3(1) 3Note: 4(3) 3 (1) 4 4tdy

e y y edt

Example 8: '( ) ( ) y t Py t Q

[ ( )]'( ) ( ' )

PtPt Pt Pt Pt d e y t

e y t yPe e y Py e Qdt

[ ( )] ( ) ( ) ( ) (**)Pt Pt Pt Pt Pt Pt Pt Q Q

d e y t Qe dt de y t Qe dt e y t e C y t CeP P

Note: '( ) ( ) Pt Pt dy y t C P e CPedt

'( ) ( ) [ ]Pt Pt Pt Pt Q

y t Py t CPe P Ce CPe Q CPe QP

Example 9: '( ) 3 y t y

Using (**): ( 3 ) 30

3, 0 ( ) ( )3

t tP Q y t Ce y t Ce

Example 10: '( ) 3 4 (0) 1 y t y y

Using (**): ( 3) 34 4

3, 4 ( ) ( )3 3

t tP Q y t Ce y t Ce

3(0) 34 7 7 4(0) 1 ( )3 3 3 3

t y Ce C y t e


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Example 11: '( ) ( ) ( ) ( )y t P t y t Q t ( ) ( )

( ) ( ) (***)P t dt P t dt

y t e Q t e dt c

Example 12:1

'( ) 3 y t y t t

Using (***): 1( ) , ( ) 3P t Q t t t

1 1

ln 1 2 1 3 2( ) 3 3 ( ) ( 3 ) ( )dt dt

tt tc

y t e te dt c e t t dt c t t dt c t t c t t

Example13: 22

'( ) 6 y t y t t

Using (***): 22

( ) , ( ) 6P t Q t t t

2 2 5 3

2 2 ln 2 2 2 4 2

26 6( ) 6 6 ( ) ( 6 ) ( )5 5

dt dt tt t t t c y t e t e dt c e t t dt c t t dt c t ct


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Second Order Linear Differential Equation (Homogeneous)

"( ) '( ) ( ) 0 y t Ay t By t

Let ( )rt

y t ce be the solution of the differential equation.

2'( ) , "( )rt rt y t cre y t cr e

Substitute to the differential equation, we have2 20 ( ) 0rt rt rt rt r ce Acre Bce ce r Ar B

This leads to a characteristic equation: 2 0r Ar B

The characteristic roots:2

1 2

4,

2

A A Br r

Case 1 2 4A B

1 2

1 2( )r t r t

y t c e c e

Example: " 4 0y y

0, 4 orA B characteristic equation2

2

1 2

(0) (0) 4( 4)4 0 , 2, 2

2r r r

2 2

1 2( )t t

y t c e c e

Note:2 2

1 2

2 2

1 2

'( ) 2 2

"( ) 4 4

t t

t t

y t c e c e

y t c e c e

Example: 2 " 12 ' 10 0 y y y or " 6 ' 5 0 y y y


2

1 2

(6) (6) 4(1)(5) (6) 166 5 0 , 1, 5

2 2

r r r r

5

1 2( ) t t y t c e c e

Note:5 5

1 2 1 2'( ) 5 and "( ) 25t t t t y t c e c e y t c e c e

Hence5 5 5 5

1 2 1 2 1 2 1 1 1 2 2 2( 25 ) 6( 5 ) 5( ) ( 6 5 ) (25 30 5 ) 0t t t t t t t t c e c e c e c e c e c e c c c e c c c e


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Case 2 2 4A B (complex roots)

Suppose the characteristic roots 1 2,r r a bi , then 1 2( ) ( cos sin )at

y t e k bt k bt

Example: "( ) 4 '( ) 13 ( ) 0y t y t y t


2

1 2

( 4) ( 4) 4(13)4 13 0 , 2 3 , 2 3

2r r r r i i

2

1 2 1 2( ) ( cos sin ) ( cos3 sin3 )at t y t e k bt k bt e k t k t

Note:2

1 2( ) ( cos3 sin3 )t

y t e k t k t 2 2 2

1 2 1 2 2 1 1 2

2 22 1 1 2 2 1 1 2

2

2 1 1 2

'( ) [ 3 sin 3 3 cos3 ] ( cos3 sin 3 )(2 ) [(2 3 )sin 3 (2 3 )cos3 ]

"( ) [3(2 3 )cos3 3(2 3 )sin3 ] [(2 3 )sin3 (2 3 )cos3 ]2

{[(4 6 ) (6 9 )

t t t

t t

t

y t e k t k t k t k t e e k k t k k t

y t e k k t k k t k k t k k t e

e k k k k

2 1 1 2

2

2 1 2 1

]sin3 [(6 9 ) (4 6 )]cos3 }

{( 5 12 )sin 3 (12 5 )cos3 }t

t k k k k t

e k k t k k t

2 2

2 1 2 1 2 1 1 2

2

1 2

2 2 2 2 2 2

2 1 2 1 2 2 1 1 2 1

"( ) 4 '( ) 13 ( )

{( 5 12 )sin3 (12 5 )cos3 } 4 [(2 3 )sin 3 (2 3 )cos3 ]

13 ( cos3 sin 3 )

( 5 12 ) 4 (2 3 ) 13 sin 3 (12 5 ) 4 (2 3 ) 13 cos3

t t

t

t t t t t t

y t y t y t

e k k t k k t e k k t k k t

e k t k t

e k k e k k k e t e k k e k k k e t

2 2

2 1 2 1 2 2 1 1 2 1( 5 12 8 12 13 ) sin 3 (12 5 8 12 13 ) cos3 0

t tk k k k k e t k k k k k e t

sincos

cossin

d tt

dt

d tt

dt


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Case 3 2 4A B (equal roots)

1 2( ) ( )rt

y t e c c t

Example "( ) 4 '( ) 4 ( ) 0y t y t y t


2

1 2

( 4) (4) 4(4)4 4 0 , 2

2r r r r

2

1 2 1 2( ) ( ) ( )rt ty t e c c t e c c t

Note:2

1 2

2 2 2

2 1 2 2 1 2

2 2 2

2 1 2 2 1 2 2

( ) ( )

'( ) ( ) 2 ( ) ( 2 2 )

"( ) ( 2 2 )( 2 ) ( 2 ) (4 4 4 )

t

t t t

t t t

y t e c c t

y t e c e c c t c c c t e

y t c c c t e e c c c c t e

2 2 2

1 2 2 2 1 2 1 2

2

1 2 2 2 1 2 1 2

"( ) 4 '( ) 4 ( ) (4 4 4 ) 4 ( 2 2 ) 4 ( )

[(4 4 4 ) (4 8 8 ) (4 4 )] 0

t t t

t

y t y t y t c c c t e c c c t e e c c t

e c c c t c c c t c c t

Example "( ) 8 '( ) 16 ( ) 0y t y t y t


2

1 2

( 8) (8) 4(16)8 16 0 , 4

2r r r r

41 2 1 2( ) ( ) ( )

rt ty t e c c t e c c t

Note:4

1 2

4 4 4

2 1 2 2 1 2

4 4 4

2 1 2 2 1 2 2

( ) ( )

'( ) ( ) 4 ( ) ( 4 4 )

"( ) ( 4 4 )( 4 ) ( 4 ) (16 8 16 )

t

t t t

t t t

y t e c c t

y t e c e c c t c c c t e

y t c c c t e e c c c c t e

4 4 4

1 2 2 2 1 2 1 2

4

1 2 2 2 1 2 1 2

"( ) 8 '( ) 16 ( ) (16 8 16 ) 8 ( 4 4 ) 16 ( )

[(16 8 16 ) (8 32 32 ) (16 16 )] 0

t t t

t

y t y t y t c c c t e c c c t e e c c t

e c c c t c c c t c c t


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Second Order Linear Differential Equation (Non-homogeneous)"( ) '( ) ( ) ( ) y t Ay t By t f t

Solution: ( ) H P y t y y

i) if 1 2real roots and r r , 1 21 2( )r t r t

y t c e c eB

ii) if complex roots,1 2( ) ( cos sin )

at y t e k bt k bt

B

iii) if 1 2real roots and r r , 1 2( ) ( )rt

y t e c c t

B

Example: Find the Py of "( ) '( ) ( ) y t gy t hy t t

Let ( ) '( ) "( ) 0P P P y t at b y t a y t

LHS: "( ) '( ) ( ) 0 ( ) ( ) y t gy t hy t g a h at b aht ga bh

RHS: t

Comparing coefficients, we have

2

i)

ii)

ah ah

gga h ghga bh b

h h h

2( ) ( )P

h g y t at b t

h h

Example: Find the Py of2"( ) 4 '( ) 4 ( ) 2y t y t y t t

Let 2 '( ) 2 "( ) 2P P P y at bt c y t at b y t a

LHS:2 2"( ) 4 '( ) 4 ( ) (2 ) 4(2 ) 4( ) 4 (4 8 ) (2 4 4 ) y t y t y t a at b at bt c at b a t a b c

RHS: 2 2t


4 1, 4 8 0, 2 4 4 2a b a a b c 21 1 7 7

, , ( )4 2 8 4 2 8

P

t ta b c y t

solution for the homogeneous

particular solution for the non-homogeneous equation

Solving for Py when ( ) (constant)f t : ( ) '( ) 0 "( ) 0P P P y t y t y t B

This satisfies "( ) '( ) ( ) y t Ay t By t

When ( ) is a polynomialf t of degree n , then the Py will also be a polynomial of degree n .


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Example: Find the Py of "( ) '( ) ( )qt

y t y t y t pe

Let 2( ) '( ) "( )qt qt qt P P P y t Ae y t Aqe y t Aq e

LHS:2 2"( ) '( ) ( ) ( ) ( ) ( )qt qt qt qt y t y t y t Aq e Aqe Ae e Aq Aq A

RHS: qtpe


2

2 2( )

qt

P

p pe Aq Aq A p A y t

q q q q

Note that this only works when2 0q q .

The condition2 0q q means that q is not a solution of the characteristic equation

2 0r r , that is qte is not a solution of "( ) '( ) ( ) 0y t y t y t .

If q is a simple root of

2

0q q , we look for a constant B such thatqt

Bte is a solution.

If q is a double root of2 0q q , we look for a constant C such that 2 qtCt e is a solution.

Example: Find the Py of "( ) 4 '( ) 4 ( ) 2cos2 y t y t y t t

Let ( ) sin 2 cos2 '( ) 2 cos2 2 sin 2 "( ) 4 sin 2 4 cos2P P P

y t A t B t y t A t B t y t A t B t

LHS: "( ) 4 '( ) 4 ( ) [ 4 sin 2 4 cos2 ] 4[2 cos2 2 sin 2 ] 4[ sin 2 cos2 ]y t y t y t A t B t A t B t A t B t ( 4 8 4 )sin 2 ( 4 8 4 )cos2 8 sin 2 8 cos2A B A t B A B t B t A t

RHS: 2cos2t

Comparing coefficients, we have1 1

8 0 and 8 2 0 and ( ) sin 24 4

P B A B A y t t

When ( ) sin cos f t p rt q rt , then ( ) sin cosP y t A rt B rt

When ( ) qt f t pe , then ( ) qtP y t Ae

The corresponding homogeneous second order linear differential equation is

"( ) '( ) ( ) 0y t y t y t and the characteristic equation is 2 0r r

, , ,p q are given


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Note that these techniques for obtaining particular solutions also apply if ( )f t is a sum,

difference, or product of polynomials, exponential functions, or trigonometric functions.

Example:

If2 3 2 3

( ) ( 1) sin 2 ( ) ( ) sin 2 cos2t t

P f t t e t y t At Bt C e D t E t


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Example"( ) '( ) 0 y t Ay t

Let ( )rt

y t ce be the solution of the differential equation.

2'( ) , "( )rt rt y t cre y t cr e

Substitute to the differential equation, we have2 20 ( ) 0rt rt rt r ce Acre ce r Ar

This leads to a characteristic equation: 2 0r Ar

The characteristic roots:2

1 2, or 02 2

A A A Ar r A

(0)

1 2 1 2( )At t At

y t c e c e c e c

Note:2

1 1'( ) and "( ) At At y t c Ae y t c A e

Hence2

1 1 1 2"( ) '( ) ( ) ( ) 0( ) 0 At At At y t Ay t c A e A c Ae c e c

Example

Looking for the for "( ) '( )P

y y t Ay t

Conjecture: ( )P y t C

( ) '( ) 0 and "( ) 0P P P y t C y t y t Does not work

Conjecture: ( )P y t at

( ) '( ) and "( ) 0P P P y t at y t a y t

"( ) '( ) (0) ( )

( )P

y t Ay t A a aA

y t t A

Note:

'( ) and "( ) 0P P y t y t A

"( ) '( ) (0) ( )P P y t Ay t A

A


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Example

Looking for the for "( ) '( )P y y t Ay t t

Conjecture: ( )P y t at b

( ) '( ) and "( ) 0P P P

y t at b y t a y t

"( ) '( ) (0) ( )P P y t Ay t A a t

Conjecture: 2( )P y t at bt 2( ) '( ) 2 and "( ) 2

P P P y t at bt y t at b y t a

2

2

2

"( ) '( ) (2 ) (2 )

2 (2 )

22

2

2 22

( )2

P P

P

y t Ay t a A at b t

aAt a Ab t

aA aA

a Aa Ab b

A A A A

y t t t A A A

Note:

2'( ) + and "( )P t y t y t A A A A

2"( ) '( ) ( ) +

P P

t y t Ay t A t t

A A A A A A

Does not work!!


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Example

"( ) 0y t

Conjecture: ( ) y t at b

'( ) and "( ) 0 y t a y t

Hence the solution is ( ) for any , y t at b a b

Example

Looking for the for "( )P y y t

Conjecture: ( )P y t at b

( ) '( ) and "( ) 0P P P

y t at b y t a y t Does not work.

Conjecture: 2( )P y t at bt c

2

2

( ) '( ) 2 and "( ) 22

( ) for any ,2

P P P

P

y t at bt c y t at b y t a a

y t t bt c b c

Note:

'( ) and "( )P P y t t b y t


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Steady State Equilibrium and Stable Equilibrium

Definition: Let '( ) ( )y t F y . If *y such that '( ) ( *) 0 y t F y , then *y is a steady state

equilibrium.

Definition: Let *y be a steady state equilibrium. It is a stable equilibrium if lim ( ) *t y t y .

Definition: Let ' ( ) y F y . If t does not appear explicitly on the right-hand side, it is called an

autonomous (or time-independent) differential equation.

Theorem: If ( *) 0F y and '( *) 0 *F y y is a stable equilibrium.

If ( *) 0F y and '( *) 0 *F y y is an unstable equilibrium.'y

stable equilibrium ( )F y

y

unstable equilibrium

( )'( )

dF yF y

dy


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Example:

Solows neoclassical economic growth model

Assumptions:

1. one output ( )Y t

2. linear homogenous production function ( ) [ ( ), ( )]Y t F K t L t

[ , ]

1Let [ ,1] ( ) per capita production function

Y F K L

Y KF y f k

L L L

a) marginal product of capital

1 1

'( )( ) 1

'( )

K

K

Y

y YL MP

MP f k K K L K L y f k

f kK K L

i) '( ) 0f k

ii)0 0

lim '( ) [Given , lim ]Kk K

f k L MP

iii) lim '( ) 0 [Given , lim 0]Kk K

f k L MP

iv)'( ) ''( )

''( ) 0 [ ]KMP f k f k

f kK K L

b) marginal product of labor

2

2

'( )( ) '( ) ( ) '( )

( )'( )

L L

YYL Y

y LLY f k K

L MP Y f k K MP f k kf k L L LL

y f k K f k

L L L

c) (0) 0f

d) ( )f

3. 0 1S sY s 4. 'K I 5. At equilibrium, I S

6.'

is exogenousL

n nL


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Model:

' '

( ) ( ) (1)

Y C I

Y C I

L L L

C K K C

f k f k L L L L

' ' ' 'ln ln ln

' ' '' ' (2)

K k K L K k k K L n

L k K L K

K K Kk k nk nk k nk

K L L

(1) and (2) ( ) 'C

f k k nk L

fund

' ( ) ( )

'( ) ( ) amental equation of neo-classical economic growth model

C S sY

k f k nk nk nk sf k nk L L L

k t sf k nk

nk

( )sf k

k

* steady state equilibriumk

' 0k

0 k

' 0k


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The stability of *k

@ *, * ( *)

* ( *)

k nk sf k

nk f k

s

( *) ( *) (0)

* * 0

n f k f k f

s k k

By the Mean Value Theorem,( *) (0)

(0, *) such that '( )* 0

f k f x k f x

k

Define so that (0,1)*

x

k and

( *) ( *) (0)'( *) (**)

* * 0

n f k f k f f k

s k k

.

Multiply (**) by *k * ( *) * '( *) * '( *)n

k f k k f k k f k s

* * '( *) '( *)n n

k k f k f k s s

'( *)n sf k

Note that ( ) '( ) ( )F k k t sf k nk fundamental equation

Hence '( ) '( ) and '( *) '( *) 0F k sf k n F k sf k n *k is a stable equilibrium.

Mean Value Theorem

Let ( )f x be differentiable on ( , )a b and continuous on [ , ]a b . Then there is at least one point c

where( ) ( )

'( )f b f a

f cb a

( ) y f x

y

x

a c b

(0) 0f

diminishing MPK


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System of first order ordinary differential equations

Definition

1 1 1 2

1 2

' ( , ,..., , )

...

' ( , ,..., , )

n

n n n

x f x x x t

x f x x x t

is a system of first order ordinary differential equations

Theorem (existence and uniqueness of solution)

If 1(...)

( , ... , , ) and ii n

j

f f x x t

x

are continuous, then for every point 0 0

0 1( , , ..., )

nt x x , a unique

solution ( )i ix t satisfying the system of differential equation and0

0( )

i ix t .

Definition If if does not depends (explicitly) on t, i.e.

1 1 1 2

1 2

' ( , ,..., )

...

' ( , ,..., )

n

n n n

x f x x x

x f x x x

is an autonomous system of first order ordinary differential equations

Definition Any point 1( *, ..., *)nx x is a critical point if

1 1 1 2

1 2

' ( *, *,..., *) 0

...

' ( *, *,..., *) 0

n

n n n

x f x x x

x f x x x

Definition 1( *, ..., *)nx x is an isolated critical point if a neighborhood of 1( *, ..., *)nx x

containing no other critical points.

Definition If if is linear in , 1,...,j x j n , then

1 1 1 2

1 2

' ( , ,..., )

...' ( , ,..., )

n

n n n

x f x x x

x f x x x

is a system oflinear ordinary differential equations.


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Properties of the solution of an autonomous system

For simplicity, we will consider a system of 2 equations.

' ( , )

' ( , )

x F x y

y G x y

(1) , and their derivatives are continuousF G

Lemma 1 If 1 2( ), ( ),x x t y y t t t t is a solution of (1), then for any real constant c , the

functions 1 1( ) ( ) and ( ) ( )x t x t c y t y t c are also solutions of (1).

Proof: By chain rule, we have 1 1'( ) '( ) and '( ) '( )x t x t c y t y t c .

1 1 1

1 1 1

'( ) '( ) [ ( ), ( )] [ ( ), ( )]

'( ) '( ) [ ( ), ( )] [ ( ), ( )]

x t x t c F x t c y t c F x t y t

y t y t c G x t c y t c G x t y t

1 1( ) and ( ) x t y t are solutions to (1).

Note: This lemma only applies for autonomous system.

Definition As t varies, a solution ( ), y( ) x t t of (1) describes parametrically a curve on the x y

plane. This curve is called a trajectory or a path.y

Example:

2

( )

( )

x t t

y t t

2

y x

x

Lemma 2 Through any point passes at most one path.

Remark: Note the distinction between solutions and trajectories of (1). A trajectory is a curve

that is represented by more than one solution.

Example:

2

( )

( )

x t t

y t t

and

2

4

( )

( )

x t t

y t t

are different solutions with the same trajectories.

Definition Let ( *, *)x y be an isolated critical point of (1). If [ ( ), ( )]C x t y t is a path of (1), then

we say that C APPROACHES ( *, *)x y as t if lim ( ) * and lim ( ) *t t

x t x y t y

.

Geometrically, if ( , )P x y is a point that traces out C in accordance with the

equations ( ) and ( )x x t y y t , then ( *, *) asP x y t .


23/141

23

Definition If( ) *

lim( ) *t

y t y

x t x

exists, or if it becomes either positively infinite or negatively infinite as

t , then we say that C ENTERS the critical point ( *, *x y ) as t .

Note: The quotient is the slope of the line joining ( *, *x y ) and the point ( ( ), ( ))P x t y t .

Note: This property is a property of the path C, and it does not depend on which solution isused to represent the path.

Definition A critical point is said to be STABLE if for each 0R , r R such that every path

which is inside the circle 2 2 2 0( *) ( *) @ x x y y r t t will remain inside the circle2 2 2

0( *) ( *) for x x y y R t t .

Note: Loosely speaking, a critical point is stable if all paths that get sufficiently close to the

critical point will stay close to the critical point.

Definition A critical point is ASYMPTOTICALLY STABLE if it is stable and a circle2 2 2

0( *) ( *) x x y y r such that every path which is inside this circle 0@t t

will approach the critical point as t .

Definition A critical point that is not stable is UNSTABLE.

Solving a pair of linear differential equation/ Phase diagram of the solution

1 1

2 2

'

'

x a x b y

y a x b y

(1) homogeneous system

Theorem A If the homogeneous system has 2 solutions 1

1

( )

( )

x x t

y y t

and 2

2

( )

( )

x x t

y y t

, then

1 1 2 2

1 1 2 2

( ) ( )

( ) ( )

x c x t c x t

y c y t c y t

(2) is also a solution for the system for any constant 1 2andc c .

Theorem B If the 2 solutions in Theorem A has a Wronskian1 2

1 2

( ) ( )( )

( ) ( )

x t x t W t

y t y t

that does not

vanish, then (2) is the general solution of the system.

Example:' 4

' 2

x x y

y x y

has 2 solutions

3

3

t

t

x e

y e

and

2

22

t

t

x e

y e

.

1 2

1 2

3 2

5

3 2

( ) ( )( )

( ) ( )

02

t t

t

t t

x t x t W t

y t y t

e ee

e e


24/141

24

Tryrt

rt

x Ae

y Be

,A B are constants to be determined

Substitute this into the system, 1 1

2 2

'

'

x a x b y

y a x b y

1 1

2 2

'

'

rt rt rt

rt rt rt

x rAe a Ae b Be

y rBe a Ae b Be

1 1

2 2

rA a A b B

rB a A b B

1 1

2 2

( ) 0

( ) 0

a r A b B

a A b r B

(3)

Note that 0A B are the trivial solution of the system.

Note that (3) has a non-trivial solution if 1 1

2 2

0a r b

a b r

1 1 2

1 2 1 2 1 2 1 2 2 1

2 2

0 ( )( ) 0 ( ) 0a r b

a r b r b a r a b r a b a ba b r

2

1 2 1 2 1 2 2 1( ) ( ) 4( )

2

a b a b a b a br

Solution:1

1

1 1

1 1

( )

( )

r t

r t

x t A e

y t B e

and

2

2

2 2

2 2

( )

( )

r t

r t

x t A e

y t B e

Solving the system 1 1

2 2

'

'

x a x b y

y a x b y

Note that

1

1 12

1 1 2 2

2 2

0

00 if 0

b

a r bb r

a r b a b r

a b r

characteristic equation


25/141

25

Case 1: Distinct real roots

General solution:1 2

1 2

1 1 2 2

1 1 2 2

( )

( )

r t r t

r t r t

x t c A e c A e

y t c B e c B e

Example:'

' 4 2

x x y

y x y

(1 ) 0 1 1 0

4 ( 2 ) 0 4 2 0

r A B r A

A r B r B

Characteristic equation:2(1 )( 2 ) (1)(4) 0 2 2 4 0r r r r r

2

1 26 0 ( 3)( 2) 0 3, 2r r r r r r

1 3r :[1 ( 3)] 0 4 0

4 [ 2 ( 3)] 0 4 0

A B A B

A B A B

3

1

3

1

( )

( ) 4

t

t

x t e

y t e

22r :

[1 (2)] 0 0

4 [ 2 (2)] 0 4 4 0

A B A B

A B A B

2

2

2

2

( )

( )

t

t

x t e

y t e


1 2

3 2

1 2

( )

( ) 4

t t

t t

x t c e c e

y t c e c e

Suppose (0) 1, (0) 2x y .

3(0) 2(0)

1 2 1 2

3(0) 2(0)1 2 1 2

1 (0)

2 (0) 4 4

x c e c e c c

y c e c e c c

1 2

1 6and

5 5c c

3 2

3 2

1 6( )

5 5

4 6( )5 5

t t

t t

x t e e

y t e e

A simple non-trivial solution is

1 and 4A B

A simple non-trivial solution is

1 and 1A B


26/141

26

Case 1A Real and distinct, same sign: Node Assume 1 2 0r r


1 2

1 1 2 2

1 1 2 2

( )

( )

r t r t

r t r t

x t c A e c A e

y t c B e c B e

Note: i) critical point (0,0) ii) 1 2lim lim 0

r t r t

t te e

iii) lim ( ) lim ( ) 0t t

x t y t

[i.e. ( ( ( ), ( )) (0,0) as ] x t y t t

iv) 1 2

1 2

( the roots are distinct)B B

A A

When2 2

22

2 2 2 21

2 22 2

( ) ( )0

( )( )

r t r t

r tr t

x t c A e B e By tc

x t A e Ay t c B e

When1 1

11

1 1 1 12

1 11 1

( ) ( )0

( )( )

r t r t

r tr t

x t c A e B e By tc

x t A e Ay t c B e

When

1 2

1 2 1 2

1 21 21 2

( )1 12

1 1 2 2 1 1 2 2 21 2

( )1 11 1 2 21 1 2 22

2

( ) ( )0, 0

( )( )

r r t

r t r t r t r t

r t r t r t r t r r t

c Be B

x t c A e c A e c B e c B e cy tc c

c Ax t c A e c A ey t c B e c B e e Ac

1 2

1 2

( )1 12

2 2

( )1 1 2

22

lim

r r t

t r r t

c Be B

c B

c A A

e Ac

as 1 2 0r r

Also

2 1

1 2

1 2

2 1

( )2 21

1 1 2 2 1

( )2 21 1 2 21

1

( )

( )

r r t

r t r t

r t r t r r t

c BB e

c B e c B e cy t

c Ax t c A e c A eA e

c

2 1

2 1

( )2 21

1 1

( )2 2 11

1

lim

r r t

t r r t

c BB e

c B

c A AA e

c

as 2 1 0r r

y

1

1

B

A

Node: Asymptotically stable

x

2

2

B

A

1 2andc c are arbitrary constants


27/141


28/141

28

Case 2 Equal real roots: Node 1 2r r r

General solution: 1 2 1 2

1 2 1 2

( ) ( )

( ) ( )

rt rt

rt rt

x t c Ae c A A t e

y t c Be c B B t e

Note that21 2 1 2 1 2 2 1( ) ( ) 4( )

2

a b a b a b a br

2

1 2 1 2 1 2 2 1( ) 4( ) 0r r r a b a b a b

Case 2A 1 2 2 10 and 0a b a a b 2 2 2

1 2 1 2 2 1 1 2 1 2 1 2[( ) 4( ) ( ) 4 ( ) 0]a b a b a b a b a b a b

1

2

' ( )

' ( )

rt

rt

x ax x t c e

y ay x t c e

If 0r , then lim ( ) 0 and lim ( ) 0t t

x t y t

If 0r , then lim ( ) 0 and lim ( ) 0t t

x t y t

1 2

12

( ) ( )

( )( )

rt

rt

x t c e cy t

x t cy t c e

y 2

1

c

c 0r

Node: Asymptotically stable

x

2

2

B

A

1 2andc c are arbitrary constants


29/141

29

Case 2B Any other combination: Node

General solution: 1 2 1 2

1 2 1 2

( ) ( )

( ) ( )

rt rt

rt rt

x t c Ae c A A t e

y t c Be c B B t e

If 0, lim ( ) 0 and lim ( ) 0t t

r x t y t

[Note that the term t is dominated by the term rte .]

If 0, lim ( ) 0 and lim ( ) 0t t

r x t y t .

When 12

1

( )0

( )

rt

rt

c Be y t Bc

x t c Ae A

When 2 1 2 1 212 1 2 1 2

( )( )0

( ) ( )

rt

rt

c B B t e B B t y tc

x t c A A t e A A t

12

1 2 2

11 2 22

lim limt t

BB

B B t Bt

A A A t AAt

When

1 12 2

1 2 1 21 2

1 11 2 1 22 2

( )( )( )

0, 0( ) ( )

( )

rt rt

rt rt

c B Bc B

c Be c B B t ey t t tc cc A Ax t c Ae c A A t e

c At t

1 12 2

2

1 1 22 2

( )

lim

( )t

c B Bc B

Bt tc A A A

c At t

y

0r Node: Asymptotically stable

x

B

A

2

2

B

A


30/141

30

Example:

' 3 4

'

x x y

y x y

In order to have a nontrivial solution forA

B

,3 4

0

1 1

r

r

2 2(3 )( 1 ) ( 4)(1) 0 2 1 0 ( 1) 0r r r r r real and equal roots

(3 1) 4 0 2 4 0(1)

(1 1) 0 2 0

A B A B

A B A B

A simple non-trivial solution of this system is 2, 1A B so that( ) 2

( )

t

t

x t e

y t e

A second linearly independent solution:

1 2 1 2 2 1 2 2

1 2 1 2 2 1 2 2

( ) ( ) '( ) ( ) ( )

( ) ( ) '( ) ( ) ( )

t t t t

t t t t

x t A A t e x t A A t e e A A A t A e

y t B B t e y t B B t e e B B B t B e

system 1 2 2 1 2 1 2

1 2 2 1 2 1 2

( ) 3( ) 4( )

( ) ( ) ( )

t t t

t t t

A A t A e A A t e B B t e

B B t B e A A t e B B t e

2 2 1 2 11 2 2 1 2 1 2

2 2 1 1 21 2 2 1 2 1 2

(2 4 ) (2 4 ) 0( ) 3( ) 4( )

( 2 ) ( 2 ) 0( ) ( ) ( )

t t t

t t t

A B t A A BA A t A e A A t e B B t e

A B t A B BB B t B e A A t e B B t e

2 2 1 2 1

2 2 1 1 2

2 4 0 2 4 0

2 0 2 0

A B A A B

A B A B B

The 2 equations on the left 2 22, 1A B .

The 2 equations on the right 1 1 1 1 1 12 4 2 and 2 1 1 and 0 A B A B A B

( ) (1 2 )(2)

( )

t

t

x t t e

y t te

1 2

1 2

( ) 2 (1 2 )general solution:

( )

t t

t t

x t c e c t e

y t c e c te

rt

rt

x Ae

y Be


31/141

31

Case 3 Distinct complex roots: Spirals

1 1

2 2

'

'

x a x b y

y a x b y

1 2,r r a bi

General solution: 1 1 2 2 1 2

1 1 2 2 1 2

( ) [ ( cos sin ) ( sin cos )]( ) [ ( cos sin ) ( sin cos )]

at

at x t e c A bt A bt c A bt A bt y t e c B bt B bt c B bt B bt

If 1 2 0a a b , then lim ( ) 0 and lim ( ) 0t t

x t y t

, i.e. ( ( ), ( )) (0,0) as x t y t t , i.e. (0,0) is

asymptotically stable.

However, ( ) and ( ) x t y t change sign infinitely often as t , all paths must spiral to the criticalpoint.

y

0a Spiral: Asymptotically stable

x

Case 4: Pure imaginary roots: Center

1 2,r r bi

General solution:1 1 2 2 1 2

1 1 2 2 1 2

( ) ( cos sin ) ( sin cos )

( ) ( cos sin ) ( sin cos )

x t c A bt A bt c A bt A bt

y t c B bt B bt c B bt B bt

y

Center:stable but not Asymptotically stable

x


32/141

32

Non-homogeneous system

1 1 1

2 2 2

'

'

x a x b y c

y a x b y c

Solving this system 1 1 1

2 2 2

* * 0

* * 0

a x b y c

a x b y c

will give us the critical point ( *, *)x y .

Define ( ) ( ) * and ( ) ( ) *x t x t x y t y t y [ '( ) '( ) and '( ) '( )x t x t y t y t ]

The original system can be written as1 1

2 2

'

'

x a x b y

y a x b y

The properties of the critical point of the non-homogeneous system are the same as those of the critical

point for the corresponding homogeneous system.

1 1 1 1 1 1 1 1 1 1 1

2 2 2 2 2 2 2 2 2 2 2

' ' ( *) ( *) * *

' ' ( *) ( *) * *

x a x b y x a x x b y y a x b y a x b y a x b y c

y a x b y y a x x b y y a x b y a x b y a x b y c


33/141

33

Example:

' 3

' 4 2

x x y

y x y

( 3 ) 0

4 ( 2 ) 0

r A B

A r B


B

,3 1

04 2

r

r

2 2

2

( 3 )( 2 ) (4)(1) 0 5 6 4 0 5 2 0

5 5 4(1)(2) 5 17

2 2

r r r r r r

r

1

5 17

2r

1 1 1 1

1 1 1 1

5 17 1 17( 3 ) 0 0

2 2

5 17 1 174 (2 ) 0 4 0

2 2

A B A B

A B A B

1 1 1 1

1 11 1

1

1

1 17 1 17 1 17 1 170 4 0

2 2 2 2

1 171 17 4 04 0

224 8

01 17 1 17

2

A B A B

A BA B

B

A

2

5 17

2r

2 2 2 2

2 2 2 2

5 17 1 17( 3 ) 0 0

2 2

5 17 1 174 (2 ) 0 4 0

2 2

A B A B

A B A B

2 2 2 2

2 21 2

2

2

1 17 1 17 1 17 1 170 4 02 2 2 2

1 171 174 04 0

22

4 80

1 17 1 17

2

A B A B

A BA B

B

A

rt

rt

x Ae

y Be

Both roots are real and

negative node


34/141

34

y

' 0x

1

1

8 2.561 17

BA

' 0y Node: Asymptotically stable

x

2

2

81.56

1 17

B

A


1 2

1 1 2 2

1 1 2 2

( )

( )

r t r t

r t r t

x t c A e c A e

y t c B e c B e

Also

2 1

1 2

1 22 1

( )2 21

1 1 2 2 1

( )2 21 1 2 21

1

( )

( )

r r t

r t r t

r t r t r r t

c BB e

c B e c B e cy t


c

2 1

2 1

( )2 21

1 1

( )2 2 11

1

lim

r r t

t r r t

c BB e

c B

c A AA e

c

as 2 1 0r r

Suppose (0) 1 (0)x y

1 1 2 20 0

1 1 2 2

0 01 1 2 21 1 2 2

11 (0)

8 811 (0)

1 17 1 17

c A c Ax c A e c A e

c A c Ay c B e c B e

1 2

1

1 2 2

18 8

11 17 1 17

A Ac

A A c

2

2 2

1

1 2

1 2 1

1 2

1

8 8 81 1 1

1 17 1 17 1 17

8 8 8 8

8 8 1 17 1 17 1 17 1 17

1 17 1 17

A

A A

cA A

A A A

A A


35/141

35

1

1 1

2

1 2

1 2 2

1 2

1

8 8 81 1 1

1 17 1 17 1 17

8 8 8 8

8 8 1 17 1 17 1 17 1 171 17 1 17

A

A A

cA A

A A A

A A

1 2

1

1 2

1 2

1

1

8 81 1

1 17 1 17( )

8 8 8 8

1 17 1 17 1 17 1 17

8 81 1

81 17 1 17( )

8 8 1 17

1 17 1 17

r t r t

r t

x t A e A e

A A

y t A e

A

2

2

2

8

8 8 1 17

1 17 1 17

r tA e

A

1 2

1

8 81 1

1 17 1 17( )

8 8 8 8

1 17 1 17 1 17 1 17

8 81 1

81 17 1 17

( ) 8 8 8 81 17

1 17 1 17 1 17

r t r t

r t

x t e e

y t e

28

1 17

1 17

r t

e


36/141

36

Example:

' 2 3

'

x x y

y x y

(2 ) 3 0

(1 ) 0

r A B

A r B


B

,2 3

01 1

r

r

2 2

2

(2 )(1 ) (3)(1) 0 3 2 3 0 3 1 0

( 3) ( 3) 4(1)( 1) 3 13

2 2

r r r r r r

r

1

3 13

2r

1 1 1 1

1 1 1 1

3 13 1 13(2 ) 3 0 3 0

2 2

3 13 1 13(1 ) 0 0

2 2

A B A B

A B A B

1 1 1 1

1 11 1

1

1

1 13 1 13 1 13 3 3 133 0 3 0

2 2 2 2

1 131 13 00

223 2

0.4343 03 3 13 1 13

2

A B A B

A BA B

B

A

2

3 13

2r

2 2 2 2

2 2 2 2

3 13 1 13(2 ) 3 0 3 0

2 2

3 13 1 13(1 ) 0 0

2 2

A B A B

A B A B

2 2 2 2

2 22 2

2

2

1 13 1 13 1 13 3 3 13

3 0 3 02 2 2 2

1 131 13 0022

3 20.768 0

3 3 13 1 13

2

A B A B

A BA B

B

A

The roots are real and of opposite signs saddle point

rt

rt

x Ae

y Be


37/141

37

y

1

1

20.4343

1 13

B

A

Saddle Point

x

2

2

20.768

1 13

B

A

When2

2

2 2 21

22 2

( ) ( )0 , lim ( ) 0, lim ( ) 0, lim ( ) lim ( )

( )( )

r t

r t t t t t

x t c A e By tc x t y t x t y t

x t Ay t c B e

When1

1

1 1 12

11 1

( ) ( )0 , lim ( ) lim ( ) , lim ( ) 0, lim ( ) 0

( )( )

r t

r t t t t t

x t c A e By tc x t y t x t y t

x t Ay t c B e

When1 2

1 2

1 1 2 2

1 2

1 1 2 2

( )0, 0 lim ( ) lim ( )

( )

r t r t

r t r t t t

x t c A e c A ec c x t y t

y t c B e c B e

1 2

1 2

1 21 2

( )1 12

1 1 2 2 2

( )1 11 1 2 22

2

( )( )

r r t

r t r t

r t r t r r t

c Be B

c B e c B e cy tc Ax t c A e c A e

e Ac

1 2

1 2

( )1 12

2 2

( )1 1 22

2

lim

r r t

t r r t

c Be B

c Bc A A

e Ac

as 1 2 0r r

Also

2 1

1 2

1 22 1

( )2 21

1 1 2 2 1

( )2 21 1 2 21

1

( )

( )

r r t

r t r t

r t r t r r t

c BB e

c B e c B e cy t


c

2 1

2 1

( )2 21

1 1

( )2 2 11

1

lim

r r t

t r r t

c BB e

c B

c A AA e

c

as 2 1 0r r

Suppose (0) 1 (0)x y

1 1 2 20 0

1 1 2 2

0 01 1 2 21 1 2 2

11 (0)

2 211 (0)

1 13 1 13

c A c A x c A e c A e


1 20r r


38/141

38

1 2

1

1 2 2

12 2

11 13 1 13

A Ac

A A c

2

2 2

1

1 2

1 2 1

1 2

1

2 2 21 1 11 13 1 13 1 13

2 2 2 2

2 2 1 13 1 13 1 13 1 13

1 13 1 13

A

A A

cA A

A A A

A A

1

1 1

2

1 2

1 2 2

1 2

1

2 2 21 1 1

1 13 1 13 1 13

2 2 2 2

2 2 1 13 1 13 1 13 1 13

1 13 1 13

A

A A

cA A

A A AA A

1 2

1

1 2

1 2

1

1

2 21 1

1 13 1 13( )

2 2 2 2

1 13 1 13 1 13 1 13

2 21 1

21 13 1 13( )

2 2 2 21 13

1 13 1 13 1 13 1 13

r t r t

r t

x t A e A e

A A

y t A e

A

22

2

2

1 13r tA e

A

1 2

1

2 21 1

1 13 1 13( )

2 2 2 2

1 13 1 13 1 13 1 13

2 21 1

2 21 13 1 13

( ) 2 2 2 21 13 1 13

1 13 1 13 1 13 1 13

r t r t

r t

x t e e

y t e

2r t

e


39/141

39

Suppose1 13

(0) 1, (0)6

x y

0 01 1 2 21 1 2 2

0 01 1 2 21 1 2 2

11 (0)

1 13 2 21 13 (0)6 1 13 1 136

c A c A x c A e c A e


1 2

1

1 2 2

1

2 2 1 13

1 13 1 13 6

A Ac

A A c

2

22

1

1 2

1 2 1

1 2

1

1

2 1 13 2 1 13 1 131 13 2

6 61 13 1 13 1 136 1 132 2 2 2

2 2 1 13 1 13 1 13 1 13

1 13 1 13

2(1 13) 1 13 2(1 13) 1 13

1 13 6 12 6

2 2 2 2

1 13 1 13 1 13 1 13

A

AA

cA A

A A A

A A

A A

1 1

(1 13) 1 13

6 6 02 2

1 13 1 13A

1

11

2

1 2

1 2 2

1 2

2

11 13 2 1 13 22 1 13

6 61 13 1 1361 13

2 2 2 2

2 2 1 13 1 13 1 13 1 13

1 13 1 13

( 1 13) 1 13 2 6 (1 2 13 13) 12

6 61 13 1 13 6(1 13)

2 2 2

1 13 1 13 1 1

A

AA

cA A

A A A

A A

A

2 2

2

2 13 26

6(1 13)

2 2 2

3 1 13 1 13 1 13

13 13

2 23(1 13)

1 13 1 13

A A

A


40/141

40

2 2

2

2

2

2

2

13 13 13 13( ) =

2 2 2 23(1 13) 3(1 13)

1 13 1 13 1 13 1 13

13 13 2 13 13 2

( ) 2 2 2 21 13 13(1 13) 3(1 13)

1 13 1 13 1 13 1 13

r t r t

r t

x t A e e

A

y t A eA

2

13

r

e

Note that as , ( ) 0 and ( ) 0t x t y t .

In general, so long as the given point fulfill the condition( ) 1 13

0.768( ) 6

y t

x t

, then the path will

converge to the critical point.

The critical point is a SADDLE POINT.


41/141

41

Example:

' 5 2

' 17 5

x x y

y x y

(5 ) 2 0

17 ( 5 ) 0

r A B

A r B


B

,5 2

017 5

r

r

2 2(5 )( 5 ) (2)( 17) 0 25 34 0 9 0

3

r r r r

r i

General solution:

1 1 2 2 1 2

1 1 2 2 1 2

( ) ( cos3 sin 3 ) ( sin 3 cos3 )

( ) ( cos3 sin3 ) ( sin 3 cos3 )

x t c A t A t c A t A t

y t c B t B t c B t B t

The roots are complex roots center

rt

rt

x Ae

y Be


42/141

42

Taylor Expansion2"( *)( *) ( )( *)

( ) ( *) '( *)( *) ... ( , *)2! !

n n f x x x f a x x

f x f x f x x x a x xn

2 2

( , ) ( *, *) ( *, *)( *) ( *, *)( *)

( *, *)( *) 2 ( *, *)( *)( *) ( *, *)( *)...

2!

x y

xx xy yy

f x y f x y f x y x x f x y y y

f x y x x f x y x x y y f x y y y

Example: Expand ( ) around 0x f x e x

( ) (0) 1

'( ) '(0) 1

''( ) ''(0) 1

...

x

x

x

f x e f

f x e f

f x e f

2 3

2 3 4

''(0) '''(0)( ) (0) '(0)( 0) ( 0) ( 0) ...2! 3!

1 ...2! 3! 4!

f f f x f f x x x

x x xx

Example: Expand ( ) ln(1 ) around 0 f x x x .

1

2

( ) ln(1 ) (0) 0

'( ) (1 ) '(0) 1

''( ) 1(1 )

f x x f

f x x f

f x x

3 3

4 4 4

''(0) 1

'''( ) ( 2)(1 ) =(2!)(1 ) '''(0) 2!

( ) 2( 3)(1 ) = (3!)(1 )

f

f x x x f

f x x x

4

5 5 5 5

(0) 3!

( ) (2)(3)( 4)(1 ) (4!)(1 ) (0) 4!

...

f

f x x x f

2 3

2 3 4 5 2 3 4 5 6

''(0) '''(0)( ) (0) '(0)( 0) ( 0) ( 0) ...

2! 3!

( 1) (2!) ( 3!) (4!)( ) 0 ... ...

2! 3! 4! 5! 2 3 4 5 6

f f f x f f x x x

x x x x x x x x x f x x x


43/141

43

Phase diagram

Shell Model of Inventive Activity and Capital Accumulation

( ) [ ( ), ( ), ( )]Y t F A t K t L t

( ) : stock of technical knowledge

( ) : capital stock

( ) : labor

A t

K t

L t

: output allocated for inventive activity

(1 )(1 ) : output allocated for consumption

(1 ) : output allocated for capital accumulation

Y

s Y

sY

success rate depreciation

'( ) ( ) ( )

'( ) (1 ) ( ) ( )

A t Y t A t

K t s Y t K t

Let ( , ) where ( , ) is homogenous to degree 1Y AF K L F K L ( ) ( ) 0, ' 0, " 0, (0) 0, ( ) , '(0) , '( ) 0 y Af k Y ALf k f f f f f f f

Let' ' ' ' ' '

1 so that ( ); ; ' 'Y k K L K K K

L Y Af k y Y k k K L k K L K K L

' ( ) (1)

' (1 ) (1 ) ( ) (2)

A Y A Af k A

k s y k s Af k k

Constructing the ' 0A curve

1' ( ) 0 ( ) 0 ( ) ( )S A Af k A f k f k k f

A

' 0A

I J

1( )

Sk f

k

Consider the points ( , ) and ( , ) I k A J k A I I J J

.

Since

i) ( ) ( ) f k f k I J

ii) ( ) 0Jf k

' ( ) 0I A f k

Since andI Jare arbitrary ' 0A on the left of the' 0A curve.


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44

Constructing the ' 0k curve

' (1 ) ( ) 0(1 ) ( )

kk s Af k k A

s f k

Differentiating

2 2

' 0

[ (1 ) ( )] ( )[ (1 ) '( )] [ ( ) '( )]with respect to 0[ (1 ) ( )] (1 )[ ( )]k

dA s f k k s f k f k kf k A k dk s f k s f k

The ' 0k is upward sloping.

A ' 0k

I

J

k

A

' 0A ' 0k

SA saddle point

k

Sk

Consider the points ( , ) and ( , ) I k A J k A I I J J

.

Since and I J I J A A k k

(1 ) ( ) (1 ) ( ) 0 I I I J J J s A f k k s A f k k

Since andI Jare arbitrary ' 0k for the pointswhich lie above the ' 0k curve.

Similarly, ' 0k all the points which lie below the' 0k curve.


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45

Stability

Take a Taylor series expansion around ( , )S Sk A .

' ' ' '

' '( , ) ( ) ( ) ( ) ( )

' ' ' '' '( , ) ( ) ( ) ( ) ( )

S S S S

S S S S

S S S S

S S S S

S S S S S S A A A A A A A Ak k k k k k k k

S S S S S S A A A A A A A Ak k k k k k k k

A A A A

A A k A A A k k A A k k A k A k

k k k k k k k A A A k k A A k k

A k A k

1

'( ) 0

S

S

SA Ak k

A f k a

A

1

''( ) 0

S

S

S SA Ak k

A A f k b

k

2' (1 ) ( ) 0S

S

SA Ak k

k s f k aA

2

'(1 ) '( ) (sign unknown)

S

S

S SA Ak k

ks A f k b

k

2 2

1 2 1 2 1 2 2 1 2 2 2 1( ) ( ) 4( ) 4,

1 2 2 2

a b a b a b a b b b a br r

2 2 2 2 22 2 2 1 2 2 2 1

1 2 2 1

( 4 ) ( 4 )0

4 4

b b a b b b a br r a b

1 2,r r are real and opposite sign

saddle point


46/141

46

Phase Diagram of'( ) '( ')

" , convex functions, " 0, " 0"( ')

g x rf x x f g f g

f x

We will construct a phase diagram with and 'x x as the 2 variables.

Constructing ' 0x 'x

' 0x x

Constructing " 0x " 0 '( ) '( ') 0 x g x rf x

Total differential "( ) "( ') ' 0g x dx rf x dx

' 0

' "( )0

"( ')x

dx g x

dx rf x

'x

B

x

A

" 0x

Phase diagram'x

x

x

" 0x " 0x

0x Tx

Consider points A and B.

' ' , B A B A x x x x

'( ' ) '( ' ) ( "( ') 0)B A

rf x rf x f x

'( ) '( ' ) '( ) '( ' ) 0 B B A Ag x rf x g x rf x

Hence any point lying above the " 0x

curve has the property that " 0 'x x

The solution to the differential equation must

be either single-peaked, monotonic, or

single-troughed in ( )x t , '( )x t changes sign at

most once.

There is only 1 path going from0

x to Tx and takes exactly T .


47/141

47

Example

1 2' { ( ) [ ( ), ] } 0, 0, [0, ] 0, [ ( ),0] ,

' [ ( ), ] ( ) '(.) 0, (0) 0

k s f k A f k r nk A A A A f k

A f k n f f

:k capital-labor ratios : saving rate : debt per capita

A : aid per capita

: retirement of principal

r: interest rate

n : labor growth rate

' 0 curve

' [ ( ), ] ( ) 0 A f k n

1 2

1 2

11 2

' 0 2

'( ) ( ) 0'( ) ( ) 0

'( )0 0, 0

A f k dk A d n d A f k dk A n d

A f k dA A

dk A n

' 0

I J

k

0 0k

, I J J I k k

'

'

[ ( ), ] ( )

[ ( ), ] ( ) 0

J J J J

I I I I

A f k n

A f k n


48/141

48

' 0 curvek

1 2

1 2

11

' 0 2

' { ( ) [ ( ), ] } 0

{ '( ) [ '( ) ] } 0

{ [ '( ) '( )] } [ ( ) ] 0

[ '( ) '( )]0 iff '( )(1 ) 0

( )k

k s f k A f k r nk

s f k dk A f k dk A d rd d ndk

s f k A f k n dk s A r d

s f k A f k ndsf k A n

dk s r A

I

J

' 0k

k

' 0

' 0k

k

When k is small, '( )f k is large.

, I J I J

k k

'

'

{ ( ) [ ( ), ] }

{ ( ) [ ( ), ] } 0

J J J J J J J

I I I I I I I

k s f k A f k r nk

s f k A f k r nk k


49/141

49

Stability

Take a Taylor series expansion around ( , )S Sk .

' ' ' '

' '( , ) ( ) ( ) ( ) ( )

' ' ' '' '( , ) ( ) ( ) ( ) ( )

S S S S

S S S S

S S S S

S S S S

S S S S S Sk k k k k k k k

S S S S S Sk k k k k k k k

k k k k

k k k k k k k k k

k k k k k k k

1 1 1

'{ '( ) '( )} '( )(1 ) 0

S

S

S S Sk k

ks f k A f k n sf k A n a

k

2 1

'( ) 0

S

S

k k

ks A r b

1 2' '( ) 0S

S

k k

A f k ak

2 2

'( )


50/141

50

Calculus of Variation

Motivation

Find the shortest distance in the plane between the points, ( , ) and ( , )a A b B .

ds dx

21

2 2 2 22[( ) ( ) ] ( ) [1 ] 1 '( )dx

ds dt dx dt x t dt dt

dt1

22min [1 '( ) ] s.t. ( ) , ( )

b

a

x t dt x a A x b B

General problem1

0

0 0 1 1( ) [ , ( ), '( )] s.t. ( ) , ( )

t

x tt

optimize F t x t x t dt x t x x t x

Necessary condition for an optimum (Eulers Equation): '[ , *( ), *'( )]

[ , *( ), *'( )] xxdF t x t x t

F t x t x t dt

Example1

22

( )min [1 '( ) ] s.t. ( ) , ( )

b

x ta

x t dt x a A x b B 1

2 2

12 2

' 1

2 2

[ , ( ), '( )] [1 '( ) ]

1 '0 and [1 '( ) ] [2 '( )]2

[1 ' ]

x x

F t x t x t x t

F xF F x t x t x

x

(Eulers Equation): '[ , *( ), * '( )]

[ , *( ), * '( )] xx

dF t x t x t F t x t x t

dt

1

2 122 2 2 2 2 2 2 22

1

2 2

2 22 2 2 2

2 2

'

[1 ( ') ] '0 ' [1 ( ') ] ( ') [1 ( ') ] ( ') ( ')

[1 ( ') ]

(1 )( ') ( ') '( ) ( ) , are constants1 1

xd

x xc x c x x c x x c c x

dtx

c c dxc c x x x t k k x t kt h k h

c c dt

( ) 1,

( ) 1

x a ka h A a k A A B aB Abk h

x b kb h B b h B a b a b


51/141

51

Optimal control theory

Problem:1

0

( )[ , ( ), ( )] (1)

t

u tt

Max f t x t u t dt

s.t. '( ) [ , ( ), ( )] (2)x t g t x t u t dynamic equation/ state equation/ transition equation

0 1

0 0

1

, fixed

( ) fixed (3)

( ) free

t t

x t x

x t

[ , ( ), ( )]f t x t u t

( ) :x t state variable

( ) :u t control variable (piecewise continuous)

,f g continuously differentiable

0t 1t

Let ( )t be any continuously differentiable function on 0 1[ , ]t t .

For any ( ), ( )x t u t satisfying (2) and (3), we have

1 1

0 0

1

0

[ , ( ), ( )] [ , ( ), ( )] ( )[ ( , ( ), ( )) '( )]

[ , ( ), ( )] ( ) [ , ( ), ( )] ( ) '( ) (4)

t t

t t

t

t

f t x t u t dt f t x t u t t g t x t u t x t dt

f t x t u t t g t x t u t t x t dt

Integrating by part the last term of (4),

1 1 1

1

0

0 0 0

1

0

1 1 0 0

1 1 0 0

( ) '( ) ( ) ( ) ( ) '( ) ( ) ( ) ( ) ( ) ( ) '( )

( ) ( ) ( ) ( ) ( ) '( )

t t tt

t

t t t

t

t

t x t dt t x t x t t dt t x t t x t x t t dt

t x t t x t x t t dt

(5)

1

0

1

0

1 1 0 0

(5) (4) [ , ( ), ( )]

[ , ( ), ( )] ( ) [ , ( ), ( )] ( ) '( ) ( ) ( ) ( ) ( ) (6)

t

t

t

t

f t x t u t dt

f t x t u t t g t x t u t x t t dt t x t t x t

Let *( )u t be the optimal control which maximizes the integral1

0

[ , ( ), ( )]

t

t

f t x t u t dt .

Let ( ) *( ) ( )u t u t ah t where ( ) is an arbitrary "fixed" function and is a parameterh t a .

Clearly, 0 ( ) *( )a u t u t .

Find a path

which maximize

the area/integral.


52/141

52

Let 0 1( , ),y t a t t t denotes the state variable generated by (2) and (3) with control ( )u t .

This ( , ) y t a satisfies the following conditions: 0 0( , ) and ( ,0) *( )y t a x y t x t .

Let

1

0

( ) [ , ( , ), *( ) ( )]

t

t

J a f t y t a u t ah t dt

Using (6), 1

0

( ) [ , ( , ), *( ) ( )] ( ) [ , ( , ), *( ) ( )] ( , ) '( )

t

t

J a f t y t a u t ah t t g t y t a u t ah t y t a t dt

1 1 0 0( ) ( , ) ( ) ( , )t y t a t y t a

Since *( )u t is the optimal control, ( )J a assumes its maximum at 0 '(0) 0a J .

1

0

11 0

'( ) [ , ( , ), *( ) ( )] ( , ) [ , ( , ), *( ) ( )] ( )

( ) [ , ( , ), *( ) ( )] ( , ) ( ) [ , ( , ), *( ) ( )] ( )

(( , )( , ) '( ) ( ) ( )

t

y a u

t

y a u

a

J a f t y t a u t ah t y t a f t y t a u t ah t h t

t g t y t a u t ah t y t a t g t y t a u t ah t h t

dy tdy t a y t a t dt t t

da

0 , )a

da

As 00 0

( , )( , ) (initial condition) 0

dy t ay t a x

da

1

0

1 1

'( ) [ , ( , ), *( ) ( )] ( , ) [ , ( , ), *( ) ( )] ( )

( ) [ , ( , ), *( ) ( )] ( , ) ( ) [ , ( , ), *( ) ( )] ( )

( , ) '( ) ( ) ( , )

t

y a u

t

y a u

a a



y t a t dt t y t a

1

0

1 1

'(0) [ , ( ,0), *( )] ( ,0) [ , ( ,0), *( )] ( ) ( ) [ , ( ,0), *( )] ( ,0)

( ) [ , ( ,0), *( )] ( ) ( ,0) '( ) ( ) ( ,0)

= [ , ( ), *( )

t

y a u y a

t

u a a

x

J f t y t u t y t f t y t u t h t t g t y t u t y t

t g t y t u t h t y t t dt t y t

f t x t u t

1

0

1 1

] ( ) [ , ( ), *( )] '( ) ( ,0)

[ , ( ), *( )] ( ) [ , ( ), *( )] ( ) ( ) ( ,0) 0

t

x a

t

u u a

t g t x t u t t y t

f t x t u t t g t x t u t h t dt t y t

Since ( , 0), ( )a y t h t can be of any form, therefore the necessary conditions for '(0) 0J are

i) ' 0x xf g

ii) 0u uf g

iii) 1( ) 0t


53/141

53

Sum up:

If the functions *( ), *( )u t x t maximizes (1) subject to (2) and (3), then there is a continuously

differentiable ( )t such that

i) state equation 0 0'( ) ( , ( ), ( )), ( )x t g t x t u t x t x

ii) co-state equation 1'( ) [ , ( ), ( )] ( ) [ , ( ), ( )] , ( ) 0x xt f t x t u t t g t x t u t t

iii) optimality condition 0 1[ , ( ), ( )] ( ) [ , ( ), ( )] 0 foru uf t x t u t t g t x t u t t t t

Short-cut formulas:

Problem:1

0( )

0 0

( , , )

. . ' ( , , )

( )

t

u tt

Optimize f t x u dt

s t x g t x u

x t x

Set up a Hamiltonian function: [ , , , ] ( , , ) ( , , )H t x u f t x u g t x u

Necessary conditions:

i) 0 [ 0]u u uH

H f gu

ii) ' [ ' ]x xH f gx

iii) 1( ) 0t transversality condition

iv) ' ( , , )H

H x g t x u


54/141

54

Example1

2

0

max ( ) . . ' 1 , (0) 1, (1) freeu

x u dt s t x u x x

2( , , , ) (1 )H t x u x u u


2

1 2 0 (1)

1 ' (2)

' 1 (3)

(1) 0

uH u

H u x

H

x

(4)

(3): '( ) 1 ( )t t t c Using (4) ( ) 1 (5)t t

2

12

1 1 1(1) : ( ) ( )

2 2(1 ) 2(1 )

1 1(2) : '( ) 1 1 ( )

4(1 ) 4(1 )

u t u t t t

x t u x t t ct t

1 1

1 5 1 5Using (0) 1 1 0 ( )

4(1 0) 4 4(1 ) 4 x c c x t t

t


55/141

55

Example10

2

0

max 5 s.t. ' , (0) 2, (10) freeu

x dt x x u x x

2( , , , ) 5 ( )H t x u x x u


2

2 0 (1)

' (2)

' (5 ) (3)

(10) 0

uH u

H x u x

H

x

(4)

(1)

10 10 10 10

0

5(3) : ' 5 ( ) 5

1

Using (10) 0 0 (10) 5 5 5 ( ) 5 5

(1) ( ) 0

(2) ' ( )

Using (0) 2 2 (0) ( ) 2

t t

t

t

t

t ce ce

ce ce c e t e

u t

x x x t ke

x x ke k x t e

10

2

0

max ( 5 ) . . ' , (0) 2, (10) freeu

u x dt s t x x u x x

2( , , , ) 5 ( )H t x u u x x u


2

1 2 0 (1)

' (2)

' (5 ) (3)

(10) 0 (4)

u

HH u

u

H x H x u

H

x

10 10 10 10

5(3) ( ) 5

1

(4) 0 (10) 5 5 ( ) 5 (5 ) ( ) 5 5 (5)

t t

t t

t ce ce

ce c e t e e t e

10 10

1 1 1(5) (1) ( )

2 ( ) 2( 5 5 ) 10 10t tu t

t e e

' ( ) PtQ

y Py Q y t ceP


56/141

56

2 2

10 10

1 1(2) ' '

10 10 10 10t t

x x x xe e

2 21 1

10 10

10

10

1 1( ) ( )

10 10 10 10

1( )

10 10 10

dt dt t t

t t

t

t

x t e e dt k x t e e dt k

e e

e x t e k

e

Using (0) 2x , we have10 10

0

10 0 10

1 12 (0) 2

10 1010 10 10 10

e e x e k k

e e

10 10

10 10

1 1( ) 2

10 10 10 10 10 10

t

t

e e x t e

e e


57/141

57

5

2 2

1

max ( )

. . ' , (1) 2

uux u x dt

s t x x u x

2 2

( ) H ux u x x u


2 0 (1)

' ( 2 ) 2

u

X

H x u

H u x u x

(2)

(5) 0 (3)

' x x u

(4)

(1) 2 (5)x

(1) 2u x (6)

(4) ' (4')u x x

(4') (6) 2( ' ) 2 ' 3 x x x x x (7)

Differentiate (7) with respect to ' 2 " 3 't x x (8)

(6) (2) ' 2 (2 ) 3 3u x u x x u (9)

(8), (4) and (9) 2 " 3 ' 3 3( ' ) 2 " 3 ' 6 3 ' 2 " 6 0 " 3 0x x x x x x x x x x x x x

Characteristic equation:2 3 0 3r r

3 3

1 2( ) t t x t c e c e (10)

Differentiating (10) with respect to 3 31 2' 3 3t t

t x c e c e

(4): 3 3 3 31 2 1 2

' ( 3 3 ) ( )t t t t u x x c e c e c e c e 3 3

1 2( 3 1) ( 3 1)t t

u c e c e (11)

(6) and (11)3 3 3 3

1 2 1 22 2 ( 3 1) ( 3 1)t t t t u x c e c e c e c e

3 3 3 3

1 2 1 2( ) (2 3 2 1) (2 3 2 1) (2 3 3) (2 3 3)t t t t t c e c e c e c e

Using (3), we have 5 3 5 31 2(5) (2 3 3) (2 3 3) 0c e c e (12)


58/141

58

(5) and (10) 3 31 2

(1) 2 x c e c e (13)

(12) and (13)

5 3 5 3

1

3 32

(2 3 3) (2 3 3) 0

2

ce e

ce e

5 3

3 5 3 5 3

1 4 3 4 3 4 3 4 35 3 5 3

3 3

5 3

3 5 3 5 3

2 4 3 4 3 4 3 4 35 3 5 3

3 3

0 (2 3 3)2 2(2 3 3) 2

(2 3 3) (2 3 3) (7 4 3)(2 3 3) (2 3 3)

(2 3 3) 0

2 2(2 3 3) 2

(2 3 3) (2 3 3) (7 4 3)(2 3 3) (2 3 3)

ee e e

ce e e ee e

e e

e

e e ec

e e e ee e

e e

Solution:3 3

1 2( )t t

x t c e c e

3 3

1 2( ) ( 3 1) ( 3 1)t tu t c e c e

3 3

1 2( ) (2 3 3) (2 3 3)t tt c e c e


59/141

59

Example: Find the shortest distance between a point ( , )At A and the line bt t .

A x ds dx

dt

t

A Bt t

Problem:

2

2

min min (1 ( ') s.t. ( )

min (1 )

s.t. '

( )

( ) free

B B

A A

B

A

t t

A A

t t

t

ut

A A

B

ds x dt x t X

u dt

x u

x t X

x t

Hamiltonian:21 H u u


20 (1)

(1 )

' (2)

' 0 (3)

( ) 0 (4)

u

X

B

uH

u

x u

H

t

(3) ( )t c (5)

(4) and (5) ( ) 0t

Substitute2

( ) 0 (1) 0 01

ut u

u

0 (2) ' 0 ( )u x x t k ( )x t is a horizontal line.

2 2 2 2 2( ) ( ) [1 ( ) ]( ) 1 ( ')dx

ds dt dx dt x dt dt


60/141

60

Example: Optimal price path

( )P t : price @ t

( )Q t : cumulative sales @ t

( )C Q : unit production cost '( ) 0C Q

( ) ( )f P g Q : the rate that a new product can be sold1

'( ) 0; "( ) 0

'( ) 0 for

f P f P

g Q Q Q

Problem:

T

( )0

max [ ( )] ( ) ( )P t

P C Q f P g Q dt

s.t. ' ( ) ( )Q f P g Q

0(0) 0Q Q

( , , , ) ( , , , ) [ ( )] ( ) ( ) ( ) ( ) ( ) ( )[ ( ) ]H t x u H t Q P P C Q f P g Q f P g Q f P g Q P C Q

Optimal conditions:

( ) ( ) [ ( ) ] '( ) 0PH g Q f P P C Q f P (1)

' ( ) [ ( ) ] '( ) ( ) '( )

( ) ( ) '( ) [ ( ) ] '( )

QH f P P C Q g Q g Q C Q

f P g Q C Q P C Q g Q

(2)

( ) 0T (3)

' ( ) ( )Q f P g Q (4)

(1): ( )( ) [ ( ) ] '( ) 0 ( )'( )

f Pf P P C Q f P P C Qf P

( )( ) ( )

'( )

f Pt C Q P

f P (5)

Differentiate (5) with respect to t:

2

2

2 2

'( ) '( ) '( ) ( ) "( ) '( )'( ) '( ) '( ) '( )

'( )

' '( ) ( ) "( ) ' ( ) "( )' '( ) ' ' '( ) ' ' ''( ) '( )

f P f P P t f P f P P t t C Q Q t P t

f P

P f P f P f P P f P f PC Q Q P C Q Q P P f P f P

2

( ) "( )' '( ) ' ' 2

'( )

f P f PC Q Q P

f P

(6)

: control variable

: state variable

P

Q


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61

(5) (2) ' ( ) ( ) '( ) [ ( ) ] '( )

( )( ) ( ) '( ) ( ) ( ) '( )

'( )

( )( ) ( ) '( ) '( )'( )

f P g Q C Q P C Q g Q

f Pf P g Q C Q P C Q C Q P g Q

f P

f P

f P g Q C Q g Qf P

(2')

(2) and (6)

2

( ) "( ) ( )'( ) ' ' 2 ( ) ( ) '( ) '( )

'( )'( )

f P f P f PC Q Q P f P g Q C Q g Q

f Pf P

2

2

2

( ) "( ) ( )' ( ) ( ) '( ) ( ) ( ) ' 2 ( ) ( ) '( ) '( )

'( )'( )

( ) ( ) (+)

( ) "( ) '( )' 2 ( )

'( )'( )

f P f P f PQ f P g Q C Q f P g Q P f P g Q C Q g Q

f Pf P

f P f P g QP f P

f Pf P

( )

'( ) '( )sign P t sign g Q

When the market is expanding (i.e. 1Q Q ), ( )P t is increasing.

When the market is shrinking (i.e. 1Q Q ), ( )P t is decreasing.


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62

4 Interpretation of ( )t

1

0

0 0

0 0

Define ( , ) max ( , , )

s.t. '( ) ( , , )

( )

t

t

V x t f t x u dt

x t g t x u

x t x

(1)

Let *( )u t be the optimal control and *( )x t be the optimal path of ( )x t .

1 1

0 0

1

0

0 0( , ) ( , *, *) ( , *, *) [ ( , *, *) ']

= ( , *, *) ( , *, *) '] (2)

t t

t t

t

t

V x t f t x u dt f t x u g t x u x dt

f t x u g t x u x dt

where ( )t is the corresponding continuously differentiable multiplier function

Integrate the last term of (2) by parts:

1

0

1

0

0 0

1 1 0 0

( , ) ( , *, *) ( , *, *) ']

( , *, *) ( , *, *) * '] ( ) * ( ) ( ) ( ) (3)

t

t

t

t

V x t f t x u g t x u x dt

f t x u g t x u x dt t x t t x t

Now, using the same ( )t , define

1

0

0 0

0 0

( , ) max ( , , )

s.t. '( ) ( , , )

( )

t

t

V x a t f t x u dt

x t g t x u

x t x a

1 1 1

1

0

0 0 0

1 1 0 0

Integrate by parts:

Let ( ) ( ) '( ) , ' ( ) ' ( )

' ( ) ( ) ' ( ) ( ) ( ) ( ) '

t t tt

t

t t t

udv uv vdu duv udv vdu uv duv udv vdu

u t t du t dt dv x dt v t dv x dt x t

x dt t x t x dt t x t t x t x dt

0 0*( ) ( ) x t x t


63/141

63

1 1

0 0

1

0

0 0

1 1 0 0

( , ) ( , , ) ( , , ) ( , , ) ']

( , , ) ( , , ) '] ( ) ( ) ( ) ( ) (4)

t t

t t

t

t

V x a t f t x u dt f t x u g t x u x dt

f t x u g t x u x dt t x t t x t a

where ( ) and ( ) are optimal for the new problem. x t u t

1

0

1

0

0 0 0 0 1 1 0 0

1 1 0 0

0 1 1

( , ) ( , ) ( , , ) ( , , ) '] ( ) ( ) ( ) ( )

( , *, *) ( , *, *) * '] ( ) *( ) ( ) ( )

[ ' * * ' *] ( ) ( )[ ( )

t

t

t

t

V x a t V x t f t x u g t x u x dt t x t t x t a

f t x u g t x u x dt t x t t x t

f g x f g x dt t a t x t

1

0

1*( )] (5)

t

tx t

Expand ( , , ) ( , , ) around ( *, *)f t x u g t x u x u

( , , ) ( , , ) ( , *, *) ( , *, *) ( , *, *) ( , *, *) ( *)

* * ( *) ...

x x

u u

f t x u g t x u f t x u g t x u f t x u g t x u x x

f g u u

1

0

0 0 0 0

0 1 1 1

(5) ( , ) ( , ) [ * * ']( *) [ * *]( *)

( ) ( )[ ( ) *( )] (6)

t

x x u u

t

V x a t V x t f g x x f g u u dt

t a t x t x t

Note that

1

' ( * *)

*, *, and satisfy * * 0

( ) 0

x x

u u

f g

x u f g

t

0 0 0 0 0

0 0 0 00

0 0 0 0 0 00 0 0

00

(6) ( , ) ( , ) ( )

( , ) ( , )( )

( , ) ( , ) ( , )( ) lim ( , )

xa

V x a t V x t t a

V x a t V x t t

aV x a t V x t V x t

t V x t a x


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64

Principle of optimality: If *( ) and *( )u t x t are optimal paths, then any sub-paths are also optimal.

*( )x t

#x

t#

0 1t t t

Suppose we follow#*( ), * ( ) tox t u t t and stop. We then try to solve the following problem:

1

#

# #max ( , , ) s.t. ' ( , , ) and ( )

t

t

f t x u dt x g t x u x t x

Let ( )x t be the solution to this problem.

Claim; ( ) *( ) x t x t

Suppose not.

That is1

#

# #( ) maximizes ( , , ) s.t. ' ( , , ) and ( )

t

t

x t f t x u dt x g t x u x t x so that

1 1

# #

( , , ) ( , *, *)

t t

t t

f t x u dt f t x u dt

Now define# #

0 0

# #

1 1

*( ) for [ , ] *( ) for [ , ]( ) and ( )

( ) for ( , ] ( ) for ( , ]

x t t t t u t t t t x t u t

x t t t t u t t t t

Clearly

# #1 1 1

# #0 0 0

( , , ) ( , *, *) ( , , ) ( , *, *) ( , *, *)

t t tt t

t t tt t

f t x u dt f t x u dt f t x u dt f t x u dt f t x u dt

This implies that *( )x t cannot be the optimal path of the original problem. Contradiction.

Note that

#1 1

#0 0

( , , ) ( , , ) ( , , )t tt

t t t

f t x u dt f t x u dt f t x u dt

In general, # # #[ ( ), ] ( )xV

V x t t t x


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65

Example:

x : productive capital [which decays at a constant proportionate rate 0b ]( )P x : profit rate that can be earned with the stock of productive capital '(0) 0 and " 0P P

u : gross investment rate

( )C x : cost of investment '(0) 0 and " 0C C

Problem:

0( )

0

max [ ( ) ( )] s.t. ' , (0) 0, 0

T

rt

u te P x C u dt x u bx x x u

[Note that as 0u and 0 0 ( ) 0 [0, ] x x t t T ].

( , , , ) [ ( ) ( )] ( )rtH t x u e P x C u u bx


'( ) 0 (1)

' '( ) (2)

( ) 0 (3)

'

rt

urt

x

H e C u

H e P x b

T

x u bx

(4)

(1) '( ) ( )rtC u e t

(2): ' '( ) ' '( )rt rt e P x b b e P x

( ) ( )

( ) ( )

( ) ( ) (

( ' ) '( ) [ '( ) ( )] '( ( ))

( ) '( ( )) ( ) ( ) '( ( ))

( ) '( ( )) ( )

T T

bt r b t bs r b s

t t

T TT

bs r b s bT bt r b s

tt t

T

bt r b s r b t bt r

t

e b e P x e s b s ds e P x s ds

e s e P x s ds e T e t e P x s ds

e t e P x s ds e e t e

) ( )

( )( )

'( ( ))

( ) '( ( ))

T

b t r b s

t

T

rt r b s t

t

e P x s ds

e t e P x s ds

Finally, we have ( )( )'( ( )) '( ( ))T

r b s t

t

C u t e P x s ds

marginal cost

marginal cost of investment current value of the marginal value of capital

Value @ t of a marginal unit of capital the discount stream of future marginal profits it generates

The calculation reflects the fact that capital decays, and therefore at each time s t a unit of capitalcontributes only a fraction bse of what it contributed originally @ t.


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66

Sufficiency

Problem:1

0

0 0

max ( , , )

. . ' ( , , ), ( )

t

ut

f t x u dt

s t x g t x u x t x

Theorem: Suppose that

i) ( , , )f t x u and ( , , )g t x u are both differentiable concave functions of ( , )x u ,

ii) *( ), *( ), ( ) x t u t t satisfy the necessary conditions,

iii) ( ) and ( ) x t t are continuous with ( ) 0t for all t if ( , , )g t x u is nonlinear in orx u , or both,

then *( ), *( ) x t u t solve the problem.

Note:

i) If the function ( , , )g t x u is linear in ( , )x u , then ( )t may assume any sign.

ii) If ( , , )f t x u is concave while ( , , )g t x u is convex and ( ) 0t , then the necessary conditionswill also be sufficient for optimality.


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67

6 Fixed endpoint problem

1

0

( )

0 0

1 1

max ( , , )

. . ' ( , , )( )

( )

t

u tt

f t x u dt

s t x g t x u x t x

x t x

We need to modify the equations on p. 46-47 in the notes.

1

0

( ) [ , ( , ), *( ) ( )] ( ) [ , ( , ), *( ) ( ) ( , ) '( )

t

t

J a f t y t a u t ah t t g t y t a u t ah t y t a t dt

1 1 0 0( ) ( , ) ( ) ( , )t y t a t y t a

1

0

11 0

'( ) [ , ( , ), *( ) ( )] ( , ) [ , ( , ), *( ) ( )] ( )

( ) [ , ( , ), *( ) ( )] ( , ) ( ) [ , ( , ), *( ) ( )] ( )

(( , )( , ) '( ) ( ) ( )

t

y a u

t

y a u

a



dy tdy t a y t a t dt t t

da

0, )a

da

As 00 0

( , )( , ) (initial condition) 0

dy t ay t a x

da

Also1

1 1

( , )

( , ) (endpoint condition) 0

dy t a

y t a x da

1

0

'(0) = [ , ( ), *( )] ( ) [ , ( ), *( )] '( ) ( ,0)

[ , ( ), *( )] ( ) [ , ( ), *( )] ( ) 0

t

x x a

t

u u

J f t x t u t t g t x t u t t y t

f t x t u t t g t x t u t h t dt

Since ( , 0), ( )a y t h t can be of any form, therefore the necessary conditions for '(0) 0J are

i) ' 0x xf g

ii) 0u uf g

No transversality condition.


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68

Production planning problem

Objective: produce B units of goods in a period of T in a way such that the sum of production

cost and storage cost is minimized.

Problem:2

1 2( )

0

min ( ) s.t. '( ) , (0) 0, ( ) , 0

T

u tc u c x dt x t u x x T B u

2

1 2( , , , )H t x u c u c x u


1

2

0 2 0 (1)

' (2)

' (3)

u

x

H c u

H c

H x u

(2) 2 2 1'( ) ( )t c t c t k

2 1 2 1

1 1 1 1

( )( )(1) ( )

2 2 2 2

c t k c t k tu t

c c c c

22 1 2 12

1 1 1 1

(3) '( ) ( )2 2 4 2

c k c k x t u t x t t t k

c c c c

Using the initial condition and endpoint conditions,

22 12 2 2

1 1

(0) 0 (0) (0) 04 2

c k x k k k

c c

2 2

22 1 2 1 1 2 2 12 1

1 1 1 1 1 1

2( ) ( ) ( )

4 2 4 2 2 4 2

c k c T k T k T c T c T Bc x T B T T k B B k

c c c c c c T

2 12 1 2

2( )

2

c T Bct c t k c t

T

2 1

2 1 2 2 2 2

1 1 1 1 1 1 1

2

(2 )2( )2 2 2 2 2 4 4

c T Bc

c t k c t c t c T c t T B BTu tc c c c c c T c T

2 1

2 2 22 1 2 2 2 2

2

1 1 1 1 1 1 1

2

( )2( ) (0)

4 2 4 2 4 4 4

c T Bc

c k c c c T c t t T T B Bt x t t t k t t t t t

c c c c c c T c T

Note that we assume( ) 0, [0, ]u t t T


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69

Example:

Two factors, capital ( )K t and extractive resource ( )R t are used to produce a good Q according to the

production function1 a aQ AK R

, where 0 1a .

The output can be consumed, yielding utility ( ) lnU C C , or it may be invested.

Problem:

( ), ( )0

0

1

0

max ln

. . ' , (0) , ( ) 0

' , (0) , ( ) 0

0, 0

T

C t R t

a a

C dt

s t x R x x x T

K AK R C K K K T

C R

2 state variables: ,x K

2 control variables: ,C R

Define( )

( )( )

R ty t

K t resource-capital ratio [ ( ) ( ) ( )]R t y t K t

Problem:

( ), y( )0

0

0

max ln

. . ' , (0) , ( ) 0

' , (0) , ( ) 00, 0

T

C t t

a

C dt

s t x Ky x x x T

K AKy C K K K T C y

1 2ln ( )a H C Ky AKy C


2

1

1 2

1

2 1 2

10 (1)

0 (2)

' 0 (3)'

C

a

y

x

a

K

HC

H K a AKy

H H y Ay

(4)

' (5)

' (6)a

x KY

K AKy C

Since 0 ( ) 0 MU C t t

Since marginal product ofR as 0 0 ( ) 0 R R y t t


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70

(2): 11 2 ( 0) (2')a

a Ay K

1

1 12 1

1

1 1 1

2 1

(7)

[(1 ) '] ' (1 ) '' [ ' 0 by (3)]

a

a

a a a

y

aAy aA

a y y y a y y

aA aA

Using (7)

1

2

1

12

(1 ) '

' (1 ) '(8)

a

a

a y y

a yaA

y y

aA

1 22 2 22

'(2 ') (4) ' (1 ) (9)a a a a aa Ay y Ay aAy Ay a Ay

1 1

1

' '(8) and (9) (1 ) (1 )

1 1 (10)

a

a a

a

a

a a

y y dya a Ay A A dt

y y y

dy y A dt At k aAt ak y y a y aAt ak

1

( ) 1( )

( )

aR ty t

K t aAt ak

Also, the capital-output ratio

( ) 1is linear in , i.e. it is growing linearly at the rate

( ) a aK t K aAt ak ak

at t aQ t AKy Ay A A

2

2 1

' 1 (1 )(10) (9) (1 ) (1 )a aa Ay aak at k

atA

2

2 2

22 1 2

2 1

1

2 1

1 1

1 1

(1 ) (1 )ln ( ) ln( )

( ) ( )

1(1) ( ) ( ) ( )

( )

ak

N20122 Mathematical Economics II 1 141

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