Chapter 23 Linear, Second-Order Differential Equations Autonomous Differential Equation Linear, second-order differential equation has following form: y¨ + a1 y˙ + a2 y = b.

(23.1)

As in Chapter 21, we divide the above problem in two parts: Steady State Part and Homogeneous Part. Steady state part is given by a2 y = b

(23.2)

b y= . a2

(23.3)

and solution is

1

Homogeneous part is given by y¨ + a1 y˙ + a2 y = 0

(23.4)

Let the solution of homogeneous part be denoted by yh , then the complete solution of (23.1) is y = yh + y.

(23.5)

Solution of Homogeneous Part In order to solve (23.4), we will use guess and verify method. Let the solution be y(t) = A exprt , A 6= 0.

(23.6)

We want to know for what values of r, (23.6) satisfies (23.4). Putting (23.6) in (23.4), we have y¨ +a1 y˙ +a2 y = A exprt (r2 +a1 r +a2 ) (23.7) 2

Then we choose r such that (r2 + a1 r + a2 ) = 0

(23.8)

(23.8) is known as characteristic equation. It is a quadratic equation and will have two roots given by p −a1 ± a21 − 4a2 r1 , r2 = . (23.9) 2 These roots are also as eigen values or characteristic roots. Three Cases (I) Real Roots & r1 6= r2 if a21 > 4a2 (II) Complex Roots & r1 6= r2 if a21 < 4a2 (III) Repeated Roots & r1 = r2 if a21 = 4a2 3

In the case of distinct roots, r1 6= r2 , we have two solutions to the homogeneous part y1 = A1 expr1 t & y2 = A2 expr2 t . (23.10) The general solution to the homogeneous part in case of distinct roots is given by yh = c1 expr1 t +c2 expr2 t

(23.11)

where c1 and c2 are two arbitrary constants (Theorem 23.2). In the case of repeated roots, r1 = r2 = r, the solutions to the homogeneous part is given by y1 = A1 exprt & y2 = tA2 exprt where r = − a21 . 4

(23.12)

The general solution to the homogeneous part in case of repeated roots is given by yh = c1 exprt +c2 t exprt

(23.13)

where c1 and c2 are two arbitrary constants (Theorem 23.2). Complex Numbers Let i be an imaginary number defined as i=

√

−1.

(23.14)

A complex number z can be expressed as z = h + iv

(23.15)

where h and v are two real numbers. Two complex numbers z1 and z2 are called conjugate complex numbers if z1 = h + iv & z2 = h − iv. 5

(23.16)

Two Properties of conjugate complex numbers 1. The sum of two conjugate complex numbers is a real-valued number, z1 + z2 = 2h. 2. The difference between two conjugate complex numbers multiplied by i yields a real valued number (z1 − z2 )i = −2v.

6

Trigonometric Functions Complex numbers can also be expressed as trigonometric or circular functions (sines and cosines). Some commonly used values of the sine and cosine functions are: Values of Sine and Cosine Functions Degree 00 900 1800 2700 3600

Radians 0

sin(θ) 0 1 0 -1 0

π 2

π

3π 2

2π

7

cos(θ) 1 0 -1 0 1

Some Important Properties of Sine and Cosine Functions 1. Both sine and cosine functions are bounded between 1 and -1. They are said to have amplitude of one. 2. Both sine and cosine functions repeat their values every 2π radian sin(θ) = sin(θ+2nπ) & cos(θ) = cos(θ+2nπ) (23.17) for n = 0, 1, 2, .... They are said to have period of 2π radians. 3. Properties 1. and 2. imply that these functions generate cyclical behavior. 4. Sine and cosine functions are symmetrical between positive and negative values of the domain. sin(−θ) = −sin(θ) & cos(−θ) = cos(θ). (23.18) 8

5. dsin(θ) dcos(θ) = cos(θ) & = −sin(θ). dθ dθ (23.19) Using the above properties, the complex number can be expressed in terms of trigonometric functions as follows h + vi = R[cos(θ) + i sin(θ)] (23.20) √ where R = h2 + v 2 , h = R cos(θ), and v = R sin(θ). Euler’s Formula

expiθ = cos(θ) + i sin(θ)

(23.21a)

exp−iθ = cos(θ) − i sin(θ)

(23.21b)

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De Moivre’s Theorem The conjugate complex numbers h ± vi, when raised to the nth power can be expressed as (h ± vi)n = Rn [cos(nθ) ± i sin(nθ)]. (23.22) Analysis of Linear Second-Order Differential Equation With Complex Roots The roots of characteristic equation are given by

r1 , r2 =

−a1 ±

p

a21 − 4a2 . 2

(23.23)

We will have complex roots if a21 < 4a2 .

10

(23.23) can be written as p −a1 ± −1 4a2 − a21 r1 , r2 = . (23.24) 2 √ 4a2 −a21 −a1 Now letting h = 2 and v = , 2 (23.24) can be written as √

r1 , r2 = h ± vi.

(23.25)

We can now express the general solution of the homogeneous part as ht

yh = exp

¡

vit

c1 exp

−vit

+c2 exp

¢

. (23.26)

Using the Euler’s formula, (23.26) can be written as yh = B1 expht cos vt + B2 expht sin vt (23.27) where B1 and B2 are two arbitrary constants. 11

Theorem 23.4: The complete solution to the linear, autonomous, second-order differential equation (23.1) is

y(t) = c1 exp

r1 t

+c2 exp

r2 t

b + , if r1 6= r2 a2

b y(t) = (c1 + tc2 ) exp + , if r1 = r2 = r a2 rt

b y(t) = B1 exp cos vt + B2 exp sin vt + , a2 ht

ht

if roots are complex.

12

Constants of Integration In solutions arbitrary constants like c1 , c2 , B1 , B2 figure. In order to pin down their values, we need some initial conditions. Since, solutions involve two constants, we need two initial conditions. Usually, initial conditions specify values of y and y˙ at time t = 0. Using these two conditions, we can find the values of constants. Convergence to Steady State We would like to derive conditions under which limt→∞ y(t) = ab2 . If limt→∞ y(t) = b a2 , then the steady state is stable equilibrium, otherwise it is unstable.

13

Theorem 23.5: The solution to the autonomous linear, second-order differential equation converges to steady state equilibrium if and only if the real parts of the characteristic roots are negative. Conditions of Convergence: Three Cases

(I) Real Roots : r1 6= r2 then r1 & r2 < 0

a1 (II) Complex : r1 6= r2 then h ≡ − < 0 2

(III) Repeated Roots : r1 = r2 = r then r < 0

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An Economic Example Inflation and Unemployment: Phillips Curve Let, p, π, m and U denote actual inflation rate, expected inflation rate, growth rate in money supply, and unemployment rate respectively. If P is the price level then p = P˙ P . Similarly, if M is the money supply then ˙ M m = M . Suppose that growth rate of money supply, m, is constant. Phillips curve posits negative association between inflation rate and unemployment rate. The basic Phillips curve assumed that there is negative association between actual inflation rate and unemployment rate: p = f (U ), f 0 (U ) < 0.

(23.27)

However, later economists modified this relationship. They posited that there is negative relationship between unexpected in15

flation and unemployment rate, where unexpected inflation is given by p − π: p − π = f (U ), f 0 (U ) < 0.

(23.28)

This later relationship is known as Expectation-Augmented Phillips Curve. Suppose that expected-augmented curve is linear p − π = −βU, β > 0.

(23.29)

Now the question is how expectation regarding inflation, π, is formed. Assume that agents use adaptive expectation i.e. π˙ = j(p − π), 0 < j < 1.

(23.30)

Finally, suppose that unemployment rate, U , evolves as follows: 16

U˙ = −k(m − p), k > 0.

(23.31)

(23.29), (23.30), and (23.31) constitute a complete macro-model in three variables, inflation rate, p, expected inflation rate, π, and unemployment rate, U . Question is: what type of time paths this model implies for these three variables? In order to find the time paths, we combine these three equations as follows. Putting (23.29) in (23.30), we get π˙ = −jβU

(23.32)

which implies that π ¨ = −jβ U˙

(23.33).

Now put (23.31) in (23.33), we have π ¨ = jβk(m − p) 17

(23.34).

We want to eliminate p from (23.34). In order to do so, rewrite (23.30) as π˙ p = + π. j

(23.35)

Putting (23.35) in (23.34), we have π ¨ + βk π˙ + jβkπ = jβkm

(23.36)

which is linear, autonomous, second-order differential equation in expected inflation rate, π. Using the method discussed earlier, we can solve it. Here, a1 = βk, a2 = jβk, b = jβkm.

(23.37)

The steady-state value, π, is given by π = m.

(23.37)

The characteristic roots of the homogeneous part is given by 18

p −βk ± β 2 k 2 − 4jβk r1 , r2 = . 2

(23.38)

(23.38) implies that we will have real and distinct roots if βk > 4j, real and repeated roots if βk = 4j, and complex and distinct roots if βk < 4j. From (23.38), we can also derive that steady-state is a stable equilibrium. In the case of repeated root, r = −βk < 0. Simi2 larly, in the case of complex root h = −βk 2 < 0. Finally, p in thepcase of real distinct roots, βk = β 2 k 2 > β 2 k 2 − 4jβk. Thus, both r1 , r2 < 0. Once, we have derived the time path of expected inflation rate, π, we can find out time path of actual inflation rate, p, by using equation (23.35). Finally, by using (23.29) we can derive time path of unemployment rate, U. 19

(23.35) implies that the steady state value of inflation rate p = π = m.

(23.39)

Thus, in the long run actual inflation rate is equal to the growth rate of money supply. Finally, (23.32) implies that the steady state value of unemployment rate is U = 0.

(23.40)

The steady state value of unemployment rate is known as natural rate of unemployment. In this particular example, it is equal to zero, but it need not always be the case. (23.39) and (23.40) suggest that long run actual inflation rate is independent of unemployment rate. This long-run relationship is captured by the notion that long run Phillips curve is vertical at U .

20

Non-Autonomous Differential Equation The general form of linear, second-order, non-autonomous differential equation is

y¨ + a1 (t)y˙ + a2 (t)y = b(t).

(23.41)

We are going to concentrate on a special form of non-autonomous equation in which a1 and a2 are constants. We will consider two special cases: Case I: b(t) = A0 + A1 t + .... + An tn Case II: b(t) = expαt [A0 + A1 t + .... + An tn ]. In both cases, one can find solution using the method of undetermined coefficients.

21

We want to find solution of following differential equation y¨ + a1 y˙ + a2 y = b(t)

(23.42)

where b(t) is as defined in above two cases. The method works as follows. First we find solution of homogeneous part y¨ + a1 y˙ + a2 y = 0

(23.43)

using the methods discussed earlier. Denote this solution by yh . Now we need to find particular solution, yp . Since, b(t) is variable there does not exist steady state. So we have to find particular solution using the method of undetermined coefficients. Case I: b(t) = A0 + A1 t + .... + An tn In this case, assume that yp = B0 + B1 t + .... + Bn tn . 22

Trick here is to find values of Bs which will satisfy differential equation (23.42). In order to find out what Bs will do, derive y˙ p and y¨p and put them in differential equation (23.42). This will provide you with series of equations in Bs. Solving these equations, you can derive particular solution, yp . Case II: b(t) = expαt [A0 + A1 t + .... + An tn ] In this case, assume that yp = expαt [B0 + B1 t + .... + Bn tn ]. and do all the steps discussed for case I. The general solution to the differential equation then is y = yh + yp .

23