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Mathematical Foundations -1- Difference equations

© John Riley August 14, 2012

Systems of difference equations

Life cycle model 2

Phase diagram 4

Eigenvalue and eigenvector 5

The general two variable model 9

Complex eigenvalues 15

Stable and unstable systems 24



Review: Life-cycle model:

the evolution of the system is defined by two linear equations.

1 (1 )t tc r cα+ = + , FOC

1 (1 )( )t t tW r W c+ = + − . wealth equation

For the baseline case we make the following assumptions.

Assumptions: (1 ) 1r α+ > and 1α <

Consumption sequence: If 0tc > consumption is growing larger at a constant rate.

If 0tc < consumption is growing more negative at a constant rate.



More analysis:

( 1) (1 ) ( )c t r c tα+ = + ( 1) (1 )( ( ) ( ))W t r W t C t+ = + − .

Constant growth paths. ( 1) ( )x t x tλ+ =

Consumption and wealth grow at the same rate if ( 1) (1 ) ( )W t r W tα+ = + .

Substituting, ( 1) (1 )( ( ) ( )) (1 ) ( )W t r W t C t r W tα+ = + − = + . Then ( ) ( ) ( )W t C t W tα− = and so

( ) (1 ) ( )C t W tα= − .

Summarizing, if ( ) (1 ) ( )C t W tα= − both consumption and wealth grow at the rate 1 (1 )rλ α= + .

Also if ( ) 0C t = , then ( 1) 0C t + = and ( 1) (1 ) ( )W t r W t+ = + so both grow at the same rate 1 r+



Phase Diagram

We can express the FOC and wealth equation

in matrix form as follows.

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

W t r r W tc t r c tα

+ + − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

(1)

This is a linear difference equation system.



1. Solution with a constant growth rate

( 1) ( )( 1) ( )

W t W tc t c t

λ+⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦.

Any λ for which this holds is called an eigenvalue (or characteristic root.) Any vector

1

2

( )( )

vW tvc t⎡ ⎤⎡ ⎤

= ⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

for which this holds is called an eigenvector.

Appealing to (1)

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

W t r r W t W t W t W tc t r c t c t c t c t

λλ λ

α λ+ + − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

= = = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦I

1 (1 ) ( ) 0 ( )

0 (1 ) ( ) 0 ( )r r W t W t

r c t c tλ

α λ+ − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥+⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦



Solving for the eigenvalues (constant growth rates)

If you write out the equation system it will be clear that this equation can be rewritten as follows.

1 (1 ) ( ) 00 (1 ) ( ) 0r r W t

r c tλ

α λ+ − − +⎡ ⎤ ⎡ ⎤ ⎡ ⎤

=⎢ ⎥ ⎢ ⎥ ⎢ ⎥+ −⎣ ⎦ ⎣ ⎦ ⎣ ⎦. (2)

It follows that the determinant of this matrix is zero.

1 (1 )0

0 (1 )r r

rλ

α λ+ − − +

=+ −

that is ((1 ) )(1 ) 0r rα λ λ+ − + − =

Thus the eigenvalues are 1 (1 )rλ α= + and 2 1 rλ = + .



Solving for the eigenvectors

Let 1 1 11 2( , )v v v= be an eigenvector associated with the first eigenvalue 1λ . From (2),

11 1

12

1 (1 ) 00 (1 ) 0r r v

r vλ

α λ+ − − + ⎡ ⎤⎡ ⎤ ⎡ ⎤

=⎢ ⎥⎢ ⎥ ⎢ ⎥+ −⎣ ⎦ ⎣ ⎦⎣ ⎦.

Substituting for 1λ .

1112

(1 )(1 ) (1 ) 00 0 0

r r vv

α+ − − + ⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦.

That is, 1112

(1 )(1 ) (1 ) 00 0 0

r r vv

α+ − − + ⎡ ⎤⎡ ⎤ ⎡ ⎤=⎢ ⎥⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎣ ⎦.

Solving, 1 11 2(1 )v vα− = . Thus 1 (1,1 )v α= − is an eigenvector.



1 (1,1 )v α= − is an eigenvector. Note

that this vector is in the positive quadrant or

negative quadrant if and only if 1α < . This

is the case depicted.

Exercise: Consider the dynamics if 1α > .

Similarly, substituting for 2λ ,

2 (1,0)v = is an eigenvector.

If 1(1)x v= , then 11( 1) tx t vλ+ =

If 2(1) ,x v= then 22( 1) tx t vλ+ = .

The eigenvectors and constant growth paths are depicted in a “phase diagram”



1 (1 )t tc r cα+ = +

1 (1 )( )t t tW r W c+ = + − .

Hence 1 ((1 )( )1t t t t

rW W r W cr+ − = + −

+



Consider any other (1)x . Since 1v and 2v are linearly independent we can write (1)x as some linear

combination 1 21 2(1)x v vα α= +

Then 1 2 1 2 1 21 2 1 2 1 1 2 2(2) (1) ( )x x v v v v v vα α α α α λ α λ= = + = + = +A A A A .

Similarly 1 2 1 2 2 1 2 21 1 2 2 1 1 2 2 1 1 2 2(3) (2) ( ) ( ) ( )x x v v v v v vα λ α λ α λ α λ α λ α λ= = + = + = +A A A A

Repeating this.....

1 21 1 2 2( 1) t tx t v vα λ α λ+ = + .

Suppose 2 1λ λ> Then 1 2 212 1 2 2 2

2

( 1) ( ( ) )t t tx t v v vλλ α α λ αλ

+ = + →

Thus the solution converges to the constant growth path with the higher eigenvector.



SYSTEMS OF LINEAR DIFFERENCE EQUATIONS

The general two variable model

( 1) ( ) ( ) ( )x t x t x t x tλ λ+ = = =A I .

Rearranging,

( 1) ( ) ( ) ( ) ( ) ( ) 0x t x t x t x t x tλ λ λ+ − = − = − =A I A I .

For a stationary state this must hold with 1λ = , that is ( ) ( ) 0x t− =A I . We assume that 0− ≠A I , so

that the matrix A - I is invertible. Then the unique solution to the equation system ( ) ( ) 0x t− =A I is

( ) 0x t = .



Next consider 1λ ≠ . For a constant growth path the equation system ( ) ( ) 0x tλ− =A I must have a

non-zero solution. Arguing as above, this is not possible if λ−A I is invertible. Hence for constant

growth,

11 12 211 22

21 22

( ) 0a a

a aa a

λλ λ λ

λ−

− = = − + + =−

A I A . (1)

This equation is known as the characteristic equation of the matrix A. The two roots 1 2,λ λ of the

quadratic equation are known as the characteristic roots or eigenvalues. Associated with each, is an

initial value of the state vector 1v and 2v . These are known as the eigenvectors. As long as the two

eigenvalues differ, the two initial state vectors are independent.



Real eigenvectors: We have seen that 211 22( ) 0a aλ λ− + + =A (1)

We can also write the characteristic equation as follows.

21 2 1 2 1 2( )( ) ( ) 0λ λ λ λ λ λ λ λ λ λ− − = − + − = . (2)

Comparing (1) and (2), the sum of the roots is the sum of the diagonal elements of A and the product

of the roots is the determinant of A. Solving the quadratic characteristic equation yields the two roots

21 1

1 11 22 11 222 4

21 12 11 22 11 222 4

( ) ( )

( ) ( )

a a a a A

a a a a A

λ

λ

= + + + −

= + − + −, (3)

Thus there are two distinct real roots if and only if

2 211 22 11 22 21 12( ) 4 ( ) 4 0a a A a a a a+ − = − + > .

Thus a sufficient condition for two distinct real roots is that 12 21 0a a > .



Suppose this condition holds. Arguing exactly as in the example, any initial state vector (1)x can be

expressed as a linear combination of 1v and 2v , that is

1 21 2(1)x v vα α= + .

Since , 1,2i iiv v iλ= =A , it follows that , 1,2t i t i

iv v iλ= =A . Then

1 21 1 2 2( 1) t tx t v vα λ α λ+ = + .

Thus the long run dynamics are determined by the eigenvalue with the larger absolute value.



Complex eigenvalues

11 12 211 22 1 2

21 22

( ) ( )( ) 0a a

a aa a

λλ λ λ λ λ λ λ

λ−

− = = − + + = − − =−

A I A . (4)

Consider the solution to the characteristic equation.

21 11 11 22 11 222 4

21 12 11 22 11 222 4

( ) ( )

( ) ( )

a a a a A

a a a a A

λ

λ

= + + + −

= + − + −, (5)

Suppose that the expression under the square root in (5) is negative. Then define 111 222 ( )a aα = + and

2 2111 224 ( )A a aβ = − + . Then (5) can be rewritten as follows.

21

22

λ α β

λ α β

= + −

= − −



21

22

λ α β

λ α β

= + −

= − −

It follows that there is no solution in terms of real numbers and so there can be no constant growth

paths. We employ a remarkable mathematical sleight of hand and introduce complex numbers. We

define 1i = − . Then

1 iλ α β= + and 2 iλ α β= −



Example: Suppose that 12

12

11−⎡ ⎤

= ⎢ ⎥⎣ ⎦

A .

The solution for a particular starting value

is depicted.

2 14

1 1/ 2(1 ) 0

1/ 2 1λ

λ λλ

− −− = = − + =

−A I .

While there is no real root we note that

2 1i = − and so 2 214(1 ) iλ− = .

Taking the square root,

121 iλ = ± █



As the first step in solving for the general solution, note that the eigenvalue 1λ and associated

eigenvector v must satisfy the constant growth condition:

1( ) 0vλ− =A I ,

That is 11 1 1 12 2( ) 0a v a vλ− + = . Without loss of generality we may choose 1 1v = then

2 11 1 12( ) /v a aλ= − − . Since 1λ is a complex number, so is 2v . We will write this more succinctly as

2 1 2v k ik= + .



It proves extremely useful to express the vector of parameters ( , )α β in polar coordinates. See Fig.

B.4-4 below. Note that 2 2r α β= + = A and tan /θ β α= .

1 iλ α β= + and 2 iλ α β= −

Then we can rewrite the eigenvalue as

1 (cos sin )r iλ θ θ= + , 2 (cos sin )r iλ θ θ= −

Fig. B.4-4: Polar coordinates



Mathematical aside: For any z, cos sin izz i z e+ = .

To see this consider the Taylor expansion around 0z = .

2 3 42 3 4

2 3 4

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

df d f d f d ff z f z z z zdz dz dz dz

= + + + + +

2 312

1sin sin 0 cos(0) sin(0) cos(0) ...3!

z z z z= + − − + so 31sin ...3!

z z z= − +

2 3 41 1 1cos cos(0) sin(0) cos(0) sin(0) sin(0) ...2! 3! 4!

z z z z z= − − + + + so 2 4

cos 1 ...2! 4!z zz = − + +

31sin ...3!

z z z= − +

Summing: 2 3 4

cos sin 1 ...1! 2! 3! 4!iz z iz zz i z+ = + − − + +

Finally Taylor’s Expansion of ize .

0 0 2 0 2 3 0 3 4 0 41 1 12! 3! 4!

iz i i i i ie e ie z i e z i e z i e z= + + + + so 2 3 4

1 ...1! 2! 3! 4!

iz iz z iz ze = + − − + +



cos sin cos sinii e iθθ θ θ θ+ = = +

Then (cos sin ) cos sint i ti e t i tθθ θ θ θ+ = = +

And so if 1 2

1(1)x v

k ik⎡ ⎤

= = ⎢ ⎥+⎣ ⎦, and (cos sin )r iλ θ θ= +

( 1) ( ) t tx t x t v vλ+ = = =A A

1 2

1t i tr ek ik

θ ⎡ ⎤= ⎢ ⎥+⎣ ⎦

1 2

1(cos sin )tr t i t

k ikθ θ

⎡ ⎤= + ⎢ ⎥+⎣ ⎦



Collecting real and complex terms,

1 2 1 2

cos sin( 1)

cos sin sin cost tt t

x t r irk t k t k t k t

θ θθ θ θ θ

⎡ ⎤ ⎡ ⎤+ = +⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦

. (6)

Define

1 2

1 2 1 2

cos sin( ) and ( )

cos sin sin cost tt t

v t r v t rk t k t k t k t

θ θθ θ θ θ

⎡ ⎤ ⎡ ⎤= =⎢ ⎥ ⎢ ⎥− +⎣ ⎦ ⎣ ⎦

.

We now argue that both 1( )v t and 2 ( )v t are solutions to the difference equation system. Since 1 2(( ) ( )v t iv t+ is a solution,

1 2 1 2( 1) ( 1) ( 1) ( ) ( ) ( )x t v t iv t x t v t i v t+ = + + + = = +A A A .



Collecting real and complex terms,

1 1 2 2( 1) ( ) and ( 1) ( )v t v t v t v t+ = + =A A .

Choose 1α and 2α so that 1 21 2(1) (1) (1)v v xα α+ = . Since 1( )v t and 2 ( )v t are solutions, so is any linear

combination. Then

1 21 2( ) ( ) ( )x t v t v tα α= +

is the general solution, given the initial state (1)x .



Stable and unstable systems

From (3)

21 11 11 22 11 222 4

21 12 11 22 11 222 4

( ) ( )

( ) ( )

i a a a a A

i a a a a A

λ α β

λ α β

= + = + + + −

= − = + − + −

From Fig. B.4-4, we can rewrite these eigenvalues in polar coordinates as follows.

1 (cos sin )r iλ θ θ= + , 2 (cos sin )r iλ θ θ= −

where 2 2r α β= + and 21 111 22 11 222 4( , ) ( ( ), ( ) )a a A a aα β = + − + .

Substituting for α and β , r A= .

From (1) the amplitude of the oscillation is tr at time t. Thus the amplitude of the oscillations is

increasing if 1r > and decreasing if 1r < . Therefore the cycles of an oscillating system with complex

eigenvalues are damped if and only if 1A < .



Example (continued)

If 12

12

11−⎡ ⎤

= ⎢ ⎥⎣ ⎦

A , we have seen that the eigenvalues

are 11 21 iλ = + and 1

2 21 iλ = − .

Transforming the eigenvalues into polar coordinates,

15 (cos sin )

2t i tλ θ θ= + and

25 (cos sin )

2t i tλ θ θ= − , where 1

2tanθ = .

Note that, since 1r = >A the amplitude of the oscillations is increasing. Thus the dynamic system is

unstable.

Fig. B.4-5: Polar coordinates

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