Mathematics for Economists

1
Mathematics for Economists

• Differential Equations
• HLMRS 21-23

2
Difference/differential equations

• Difference equation e.g. ?(yt) yt1 - yt
  f(y,t) discrete time
• Differential equation e.g. dy/dt g(y,t)
  continuous time
• Properties of both equations are similar. Here we
  are interested in finding functions that solve
  the equations
• The general solution is sum of the solution to
  the homogeneous form plus the steady state
  solution

3
Classification

• Order whats the highest order of difference?
• Autonomous/nonautonomous does (not) depend on
  time explicitly
• Linear or nonlinear
• Solution a function that makes the difference
  equation true. Example dy/dt 3y 2
  solution y C exp(3t) - 2/3. So the
  warning multiple solutions exist (C can take any
  value in general)!

4
Contents

• Linear, First-Order Differential Equations
  autonomous and nonautonomous
• Nonlinear, First-Order Differential Equations
• Linear, Second-Order Differential Equations
• Next time Systems of Differential Equations

5
Linear First-Order Differential Equations

• Linear autonomous first-order differential
  equation dy/dt ay b
• General solution is the sum of the homogeneous
  and the particular solution y yh yp
• yh is the solution to the homogeneous equation
  dy/dt ay 0
• yh(t) C exp(-at)
• yp is the solution of dy/dt 0 yp b/a
• The process converges to yp if and only if a gt 0

6
Proof of the homogeneous solution

• dy/dt / y -a
• Integrating both the left and right-hand side
  with respect to time gives for the left-hand
  side
• ? dy/dt / y dt ? 1/y dy ln y c2
  and for the right-hand side at c1
• y exp(-at c1 - c2). Use C exp(c1 - c2)

7
Example capital accumulation

• dK/dt I - ?K. I is constant. Solve for K(t)
• Homogeneous form dK/dt ?K 0
• Solution is Kh C exp(-?t)
• Particular solution Kp I / ?
• So K(t) C exp(-?t) I / ?
• Initial conditions on K (say K0) give a
  restriction on C K0 C I / ?,
  so C (K0 I / ?)

8
Example

• Solve dy/dt 0.1y - 1, given y(0) 5
• Homogeneous form yh(t) C exp(0.1t)
• Steady state yp 10
• General solution y(t) C exp(0.1t) 10
• y(0) 5 C 10, so C -5
• y(t) -5 exp(0.1t) 10

9
Convergence to the steady state

• y(t) C exp(-at) b/a, and yp b/a
• So y(0) C exp(-at0) b/a, let t0 0
• C (y0 - b/a) exp(at0) (y0- b/a)
• y(t) (y(0) yp) exp(-at) yp
• If t goes to infinity, y(t) goes to yp for a gt 0
• In the example of the capital stock we have K(t)
  (K0 I / ?) exp(-?t) I / ?. So there is
  convergence because the depreciation rate is
  positive

10
Convergence example

• Debt grows as a percentage of GDP
  dD/dt b Y. Income grows at a rate
  dY/dt / Y g. We assume Y(0) Y0, D(0)
  D0. Interest payments are z(t) r D(t) / Y(t)
• Y(t) Y0 exp(gt), so dD/dt bY0 exp(gt)
• Integrating both sides D(t) bY0 exp(gt)/g C.
  C (D0- (b/g) Y0)
• So z(t) r (D0/Y0) exp(-gt) r (b/g)
  (1-exp(-gt)). Limit value r (b/g)

11
Nonautonomous equations

• dy/dt a(t) y b(t)
• Solution gets more complicated
• y(t) exp(-A(t)) ?0t exp(A(t)) b(t)dt C
• With A(t) ? a(t) dt

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Example

• Solve dy/dt - 2ty bt
• y(t) exp(-A(t)) ?0t exp(A(t)) b(t) dt C
• A(t) ? a(t) dt - ?2t dt -t2
• exp(A(t)) dy/dt - 2ty exp(A(t)) bt
• exp(-t2) dy/dt - 2ty exp(-t2) bt
• d(exp(-t2)y)/dt exp(-t2) bt
• Integration exp(-t2) y ? exp(-t2) bt dt
• y(t) exp(t2) ? exp(-t2)bt dt C

13
Example technological change

• y (a bk) t½
• We have dk/dt sy, so dk/dt - sbt ½k sat ½
• k(t) exp(-A(t)) ?0t exp(A(t)) b(t)dt C
• So the integrating factor is
  A(t) -?sbt ½dt -2/3 sbt3/2
• dexp(-2/3sbt 3/2)k / dt sa t ½
  exp(-2/3sbt3/2)
• So we get
  exp(-2/3sbt 3/2)k sa ? t ½
  exp(-2/3sbt3/2) dt C
• Finally we get by integration
• k(t) -a/b C exp(2/3 sbt3/2)

14
Nonlinear, First-order Differential Equations

• dy/dt g(y) with y(t0) y0
• Example dy/dt y(1 - y)
• An equilibrium point is stable if the derivative
  dy/dt / dy is negative in that point
• Take the example two equilibria y 0 and y 1.
  dy/dt / dy 1 - 2y is positive at y 0 and
  negative at y 1

15
Example Neoclassical Growth

• y f(k), y Y / L, k K / L, f(.) gt 0, f(.)
  lt 0
• dK/dt sY, s savings ratio
• dk/dt d(K/L) / dt dK/dt / L k dL/dt / L
• Let n dL/dt / L
• dk/dt sf(k) nk. Steady state y/k n/s
• Two equilibria the steady state and y 0
• dk/dt /dk sf(k) n, so d2(dk/dt)/dk2 sf(k)
  lt 0. Positive steady state is stable

16
Neoclassical Growth
dk/dt
f(k)
Slope n/s f
dk/dt sf(k) - nk
k
k
0
k
0
k
17
Linear, Second-Order Differential Equations

• d2y/dt2 a1dy/dt a2y b
• Again y yh yp, where yp is easy b/a2
• Characteristic equation r2 a1r a2 0
• yh(t) C1 exp (r1t) C2 exp(r2t) if r1?r2
• yh(t) (C1 C2t) exp(rt) if r1 r2
• Forget about the complex roots!
• Convergence needs negative real parts of the roots

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Homogeneous form

• d2y/dt2 a1dy/dt a2y 0
• Solution to first-order equations is of the form
  y(t) A exp(rt)
• dy/dt rA exp(rt) and d2y/dt2 r2A exp(rt)
• d2y/dt2 a1dy/dt a2y
  A exp(rt) (r2 a1r a2)
• Ruling out A 0 we need to find solutions to r2
  a1r a2 0, the characteristic equation

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Example

• Solve 4 d2y/dt2 - 8 dy/dt 3 y 0
• Characteristic equation r2 - 2r 3/4 0
• So r1 ½ and r2 3/2
• yh(t) C1 exp(t/2) C2 exp(3t/2)

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Three classes of solutions

• If a12 - 4a2 gt 0 two real distinct roots
  yh C1 exp(r1t) C2 exp(r2t)
• If a12 - 4a2 0 real and equal roots
  yh C1exp(rt) C2 t exp(rt)
• If a12 - 4a2 lt 0 complex roots
  yh A1exp(ht) cos vt A2
  exp(ht) sin vt
• Note that r1,2 (-a1 ?(a12 - 4a2))/2 are the
  roots of the so-called characteristic equation
  r 2 a1r a2 0. Compute h
  -a1/2. v (?4a2 - a12)/2. We do
  not discuss complex roots any further

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Particular solution example

• 4 d2y/dt2 - 8 dy/dt 3 y 9
• Steady state yp 3
• yh(t) C1 exp (t/2) C2 exp(3t/2)
• y(t) C1 exp (t/2) C2 exp(3t/2) 3
• Suppose we know that y(0) 15 and dy(0)/dt
  10. We get C1 C2 3 15. Differentiate the
  solution dy(t)/dt C1/2 exp(t/2) 3C2
  exp(3t/2), so C1/2 3C2/2 10. This gives
• y(t) 8 exp (t/2) 4 exp(3t/2) 3

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Steady State and Covergence

• In case of real roots of the characteristic
  equation, convergence of the differential
  equation requires negative real roots. This is
  intuitive from the solution
• yh(t) C1 exp (r1t) C2 exp(r2t) if r1?r2
• yh(t) (C1 C2t) exp(rt) if r1 r2

23
Linear, Second-Order Differential Equation with a
Variable Term

• d2y/dt2 a1dy/dt a2y b(t)
• Method of undetermined coefficients!
• Case 1 b(t) is a polynomial of order n pn(t),
  assume that
  yp Antn An-1 tn-1 A1 t A0
• Case 2 if b(t) is of the form exp(at) pn(t),
  assume
  yp exp(at) (Antn An-1 tn-1 A1
  t A0)

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Example (1)

• d2y/dt2 3 dy/dt - 4 y t2
• Homogeneous form characteristic equation r2 3r
  4 0 gives r1 -4, r2 1, so yh(t) C1
  exp(-4t) C2 exp(t)
• Particular solution. Guess
  yp A2t2 A1t A0. So dyp/dt 2A2t A1,
  d2yp/dt2 2A2
• 2A2 3(2A2t A1) - 4(A2t2 A1t A0) t2

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Example (2)

• 2A2 3(2A2t A1) - 4(A2t2 A1t A0) t2
• Rearranging this expression in terms
• -(4A2 1)t2 (6A2 - 4A1)t
  (2A2 3A1- 4A0) 0 has to hold for all t
• So A2 -1/4 A1 -3/8, A0 -13/32
• y(t) -1/4 t2 3/8 t -13/32 C1 exp(-4t) C2
  exp(t)