Mathematics for Economists

1
Mathematics for Economists

• Systems of Differential and Difference Equations
• HLMRS 24

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Contents

• Two methods to solve Systems of Linear
  Differential Equation (1) the Substitution
  Method for 2-dimensional cases, and (2) the
  Direct Method for general cases
• Stability Analysis and Linear Phase Diagrams
• Systems of Linear Difference Equations

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Systems of Differential Equations General set-up

• Equation 1 dy1/dt a11 y1 a12 y2 b1
• Equation 2 dy2/dt a21 y1 a22 y2 b2
• Two autonomous differential equations
• Complete solution is again the sum of the
  homogeneous and the particular solutions
• Trick for a 2-dimensional system substitute to
  get one second-order differential equation of the
  homogeneous form

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Trick Substitution

• d2y1/dt2 a11 dy1/dt a12 dy2/dt
• Use dy2/dt a21 y1 a22 y2
• d2y1/dt2 a11 dy1/dt (a21 y1 a22 y2)
• Use a transformation of the first equation y2
  (dy1/dt - a11 y1) / a12
• d2y1/dt2 - (a11 a22) dy1/dt
  (a11a22 - a12 a21) y1 0,
  which we can solve as a normal
  second-order differential equation

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Example for two distinct real roots

• Equation 1 dy1/dt y1 - 3 y2
• Equation 2 dy2/dt 1/4 y1 3 y2
• Trick d2y1/dt2 - 4 dy1/dt 15/4 y1 0
• r2 - 4r 15/4 0 so r1 3/2, r2 5/2
• y1(t) C1 exp(3/2 t) C2 exp(5/2 t)
• y2(t) (-dy1/dt y1)/3, so
• y2(t) -1/6 C1 exp(3/2 t) ½ C2 exp(5/2 t)

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Particular Solutions

• a11 y1p a12 y2p b1 0
• a21 y1p a22 y2p b2 0
• So y1p (a12b2 - a22b1)/(a11a22 - a12a21)
• and y2p (a21b1 - a11b2)/(a11a22 - a12a21)
• Combine the homogeneous and particular solution
  to get the general solution

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Example

• Equation 1 dy1/dt y1 - 3 y2 - 5
• Equation 2 dy2/dt 1/4 y1 3 y2 - 5
• Homogeneous solution
• y1(t) C1 exp(3/2 t) C2 exp(5/2 t)
• y2(t) -1/6 C1 exp(3/2 t) ½ C2 exp(5/2 t)
• Particular solution y1p - 3 y2p 5 0 and
  1/4 y1p 3 y2p 5 0, so y1p 8 and
  y2p 1
• y1(t) C1 exp(3/2 t) C2 exp(5/2 t) 8
• y2(t) -1/6 C1 exp(3/2 t) ½ C2 exp(5/2 t) 1

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Initial conditions

• Can we compute C1 and C2? Yes, if we have
  initial conditions. In the previous example
  y1(0) 1 and y2(0) 3 gives
• y1(0) C1 C2 8 1
• y2(0) -1/6 C1 -1/2 C2 1 3
• C1 -9/2 and C2 -5/2
• So the full solution is
• y1(t) - 9/2 exp(3/2 t) 5/2 exp(5/2 t) 8
• y2(t) 3/4 exp(3/2 t) 5/4 exp(5/2 t) 1

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Direct Method Matrix Theory

• dy/dt A y b
• y and b are n-vectors, A is a (n,n)-matrix
• Guess the homogeneous solution
  y k exp(rt), so r k exp(rt) A k exp(rt),
  or A r I k 0. So the
  determinant of A r I must
  be equal to 0 A r I 0. A vector ki that
  satisfies A ri I ki 0 is the eigenvector
  corresponding to root ri
• Solution is yi(t) Ci ki exp(rit). Total
  solution is a linear summation of i individual
  solutions
• The particular solution is yp - A-1 b

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Example
dy/dt
y
A-rI (4-r)2 -4 r2 -8r -12 0, so r1 2 and
r2 6
0, which gives k1 1, k2 2
For r1 2 we solve
For r2 6 we solve
0, which gives k1 1, k2 -2
exp(2t) C2
exp(6t)
The final solution is y(t) C1
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Stability Analysis

• The system is asymptotically stable if the
  solution converges to the steady state
• This is the case if the characteristic roots are
  negative (like for single equations)
• For a (2,2)-case the equilibrium is called a
  saddle-point equilibrium of one root is negative
  and the other positive
• The saddle path is then
  y2 (r1 - a11) / a12 (y1
  y1p) y2p

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Linear Phase Diagrams

• Simple example. Stable system
• Equation 1 dy1/dt - 2 y1 2
• Equation 2 dy2/dt -3 y2 6
• Equations are independent, steady state y1 1
  and y2 2
• Solutiony1(t) C1 exp(-2t) 1 and
  y2(t) C2 exp(-3t) 2
• The roots are negative stable solution

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Phase Diagram (1)
Phase plane
dy1/dt 0
y2
Laws of motion
dy2/dt 0
2
trajectory
Isosector
Isosector
Isokline
0
y1
1
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Phase Diagram (2)
dy1/dt 0

• dy1/dt 2 y1 - 2
• dy2/dt 3 y2 - 6

y2
dy2/dt 0
2
Positive real roots Unstable node
0
y1
1
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Phase Diagram (3)
y2
dy2/dt 0

• dy1/dt y2 - 2
• dy2/dt y1/4 - 1/2

dy1/dt 0
2
One positive real root r1 1/2 One negative r2
-1/2
Saddle path y2 3 -1/2 y1
0
y1
2
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Determining stability from A

• For a (2,2)-case A a11 a22 a12 a21 is the
  determinant of A. Assume A ? 0
• If A lt 0, the roots are real and of opposite
  sign saddle-point stability
• If A gt 0 and a11 a22 lt 0 stability.
• If A gt 0 and a11 a22 gt 0 real parts of both
  roots are positive unstable!

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Example Dornbusch Overshooting

• Three markets assets, money, and goods. Two
  equilibrium variables the exchange rate and the
  price level. How do the markets respond to an
  increase in money supply?
• Real money demand mD -ar by r is the
  interest rate and y is fixed output (variables in
  logs). Money supply is m - p
• r r Ede/dt e domestic price of foreign
  currency (an increase is a depreciation)
  r foreign interest rate
• Perfect foresight Ede/dt de/dt

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Dornbusch Overshooting (2)

• Solve for de/dt p/a (by - m) / a r
• Domestic prices adjust slowly to changes in
  demand yD dp/dt ? (yD y)
• Demand is driven by the relative price of
  domestic goods yD u v (e - p)
• So dp/dt - ?vp ?ve ? (u - y)
• Two equations in dp/dt and de/dt with a11
  - ?v a12 ?v, a21 1/a and a22 0

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Overshooting Solution

• Steady state pss ar - by m
  ess pss (u - y)/v
• p(t) C1 exp(r1t) C2 exp(r2t) pss
• e(t) (r1 ?v) / ?v C1 exp(r1t)
  (r2 ?v) / ?v C2 exp(r2t) ess
• A - ?v lt 0, so saddle-point stable

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Overshooting Phase Diagram
p
p e (u - y) / v is the dp/dt-line
dp/dt 0
pss
de/dt 0
0
e
ess
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Overshooting
p
p e (u - y) / v is the dp/dt-line
New saddlepath
dp/dt 0
p2ss
de/dt 0
p1ss
0
e
e1ss
e2ss
e2
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Linearization around the steady state

• dy1/dt F(y1, y2)
• dy2/dt G(y1, y2)
• Let y1p and y2p be the steady state values. a11
  ?F/?y1, a12 ?F/?y2, etc.
• dy1/dt a11 (y1 - y1p) a12 (y2 y2p)
• dy2/dt a21 (y1 - y1p) a22 (y2 y2p)
• So we can analyze the properties of nonlinear
  systems from the linearization

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Systems of Difference Equations

• yt1 a11 yt a12 xt b1
• xt1 a21 yt a22 xt b2
• Homogeneous part (1) substitution or (2)
  the direct method
• Particular solution z y x (I - A)-1 b
• We illustrate by means of examples

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Substitution method

• yt1 6 yt 8 xt
• xt1 yt xt
• yt2 6 yt1 8 xt1 6 yt1 8 (yt xt)
• xt (-6 yt yt1)/8, so
• yt2 6 yt1 8 yt (-6 yt yt1)
• Characteristic equation r2 -7r -2
• yt C1 r1t C2 r2t
• xt -6/8 (C1 r1t C2 r2t) 1/8 (C1 r1t1 C2
  r2t1)

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Direct method
yt1
yt
A - rI r2 - 6r 8 0, so r1 2 and r2 4
0, which gives k1 1, k2 1/2
For r1 2 we solve
For r2 4 we solve
0, which gives k1 1, k2 1/4
2t C2
4t
The final solution is y(t) C1
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Full solution

• yt1 6 yt - 8 xt 10
• xt1 yt 1
• Same homogeneous solution as the previous example
• Steady state y 2 and x 3
• yt C1 2t C2 4t 2
• xt C1/2 2t C2/4 4t 3

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Stability

• A system of two linear difference equations with
  constant parameters is asymptotically stable if
  and only if the absolute values of the
  characteristic roots are less than unity
• yt C1 2t C2 4t 2
• xt C1/2 2t C2/4 4t 3
• This system is typically unstable