8.6 Second-order differential equations

General form

A second-order ordinary differential equation is a differential equation of the form
G(t, x(t), x'(t), x"(t)) = 0 for all t,
involving only t, x(t), and the first and second derivatives of x. We can write such an equation in the form
x"(t) = F (t, x(t), x'(t)).

Equations of the form x"(t) = F (t, x'(t))

An equation of the form
x"(t) = F (t, x'(t)),
in which x(t) does not appear, can be reduced to a first-order equation by making the substitution z(t) = x'(t).

Example
Let u(w) be a utility function for wealth w. The function

ρ(w) = −wu"(w)/u'(w)

is known as the Arrow-Pratt measure of relative risk aversion. (If ρu(w) > ρv(w) for two utility functions u and v then u reflects a greater degree of risk-aversion than does v.)

What utility functions have a degree of risk-aversion that is independent of the level of wealth? That is, for what utility functions u do we have

a = −wu"(w)/u'(w) for all w?

This is a second-order differential equation in which the term u(w) does not appear. (The variable is w, rather than t.) Define z(w) = u'(w). Then we have

a = −wz'(w)/z(w)

or

az(w) = −wz'(w),

a separable equation that we can write as

a·dw/w = −dz/z.

The solution is given by

a·ln w = −ln z(w) + C,

or

z(w) = Cwa.

Now, z(w) = u'(w), so to get u we need to integrate:

u(w) =
Cln w + B if a = 1
Cw1−a/(1 − a) + B if a ≠ 1

We conclude that a utility function with a constant degree of risk-aversion equal to a takes this form.

Linear second-order equations with constant coefficients

A linear second-order differential equation with constant coefficients takes the form
x"(t) + ax'(t) + bx(t) =  f (t)
for constants a and b and a function  f . Such an equation is homogeneous if  f (t) = 0 for all t.

Let x1 be a solution of the equation

x"(t) + ax'(t) + bx(t) =  f (t),
to which I refer subsequently as the "original equation". For any other solution of this equation x, define z = xx1. Then z is a solution of the homogeneous equation
x"(t) + ax'(t) + bx(t) = 0
(because z"(t) + az'(t) + bz(t) = [x"(t) + ax'(t) + bx(t)] − [x1"(t) + ax1'(t) + bx1(t)] =  f (t) −  f (t) = 0). Further, for every solution z of the homogeneous equation, x1 + z is clearly a solution of original equation.

We conclude that the set of all solutions of the original equation may be found by finding one solution of this equation and adding to it the general solution of the homogeneous equation. That is, we may use the following procedure to solve the original equation.

Procedure for finding general solution of linear second-order differential equation with constant coefficients
The general solution of the differential equation

x"(t) + ax'(t) + bx(t) =  f (t)
may be found as follows.
  1. Find the general solution of the associated homogeneous equation x"(t) + ax'(t) + bx(t) = 0.
  2. Find a single solution of the original equation x"(t) + ax'(t) + bx(t) =  f (t).
  3. Add together the solutions found in steps 1 and 2.

I now explain how to perform to the first two steps and illustrate step 3.

1. Finding the general solution of a homogeneous equation

You might guess, based on the solutions we found for first-order equations, that the homogeneous equation has a solution of the form x(t) = Aert. Let's see if it does. If x(t) = Aert then x'(t) = rAert and x"(t) = r2Aert, so that
x"(t) + ax'(t) + bx(t) = r2Aert + arAert + bAert
= Aert(r2 + ar + b).
Thus for x(t) to be a solution of the equation we need
r2 + ar + b = 0.
This equation is knows as the characteristic equation of the differential equation. It has either two distinct real roots, a single real root, or two complex roots. If a2 > 4b then it has two distinct real roots, say r and s, and we have shown that both x(t) = Aert and x(t) = Best, for any values of A and B, are solutions of the equation. Thus also x(t) = Aert + Best is a solution. It can be shown, in fact, that every solution of the equation takes this form.

The cases in which the characteristic equation has a single real root (a2 = 4b) or complex roots (a2 < 4b) require a slightly different analysis, with the following conclusions.

Proposition
Consider the homogeneous linear second-order differential equation with constant coefficients

x"(t) + ax'(t) + bx(t) = 0 for all t,
where a and b are numbers. The general solution of this equation depends on the character of the roots of the characteristic equation r2 + ar + b = 0 as follows.
Distinct real roots
If a2 > 4b, in which case the characteristic equation has distinct real roots, the general solution of the equation is
Aert + Best.

Repeated real root
If a2 = 4b, in which case the characteristic equation has a single root, the general solution of the equation is
(A + Bt)ert,
where r = −(1/2)a is the root.

Complex roots
If a2 < 4b, in which case the characteristic equation has complex roots, the general solution of the equation is
(Acos(βt) + Bsin(βt))eαt,
where α = −a/2 and β = √(ba2/4). This solution may alternatively be expressed as
Ceαt cos(βt + ω),
where the relationships between the constants C, ω, A, and B are A = C cos ω and B = −C sin ω.

Example
Consider the differential equation

x"(t) + x'(t) − 2x(t) = 0.

The characteristic equation is

r2 + r − 2 = 0

so the roots are 1 and −2. That is, the roots are real and distinct. Thus the general solution of the differential equation is

x(t) = Aet + Be−2t.

Example
Consider the differential equation

x"(t) + 6x'(t) + 9x(t) = 0.

The characteristic equation has a repeated real root, equal to −3. Thus the general solution of the differential equation is

x(t) = (A + Bt)e−3t.

Example
Consider the equation

x"(t) + 2x'(t) + 17x(t) = 0.

The characteristic roots are complex. We have a = 2 and b = 17, so α = −1 and β = 4, so the general solution of the differential equation is

[A cos(4t) + B sin(4t)]et.

2. Finding a solution of a nonhomogeneous equation

A good technique to use to find a solution of a nonhomogeneous equation is to try a linear combination of  f (t) and its first and second derivatives. If, for example,  f (t) = 3t − 6t2, then try to find values of A, B, and C such that A + Bt + Ct2 is a solution. Or if  f (t) = 2sin t + cos t, then try to find values of A and B such that  f (t) = Asin t + Bcos t is a solution. Or if  f (t) = 2eBt for some value of B, then try to find a value of A such that AeBt is a solution.

Example
Consider the differential equation

x"(t) + x'(t) − 2x(t) = t2.

The function on the right-hand side is a second-degree polynomial, so to find a solution of the equation, try a general second-degree polynomial---that is, a function of the form x(t) = C + Dt + Et2. (The important point to note is that you should not restrict to a function of the form x(t) = Et2.) For this function to be a solution,

2E + D + 2Et − 2C − 2Dt − 2Et2 = t2 for all t,

so we need (equating the coefficients of t2, t, and the constant on both sides),

2E + D − 2C = 0
2E − 2D = 0
−2E = 1.

We deduce that E = −1/2, D = −1/2, and C = −3/4. That is,

x(t) = −3/4 − t/2 − t2/2
is a solution of the differential equation.

3. Add together the solutions found in steps 1 and 2

This step is quite trivial!

Example
Consider the equation from the previous example, namely

x"(t) + x'(t) − 2x(t) = t2.

We saw above that the general solution of the associated homogeneous equation is

x(t) = Aet + Be−2t.

and that

x(t) = −3/4 − t/2 − t2/2
is a solution of the original equation.

Thus the general solution of the original equation is

x(t) = Aet + Be−2t − 3/4 − t/2 − t2/2.

Stability of solutions of homogeneous equation

Consider the homogeneous equation
x"(t) + ax'(t) + bx(t) = 0.
If b ≠ 0, this equation has a single equilibrium, namely 0. (That is, the only constant function that is a solution is equal to 0 for all t.) To consider the stability of this equilibrium, consider separately the three possible forms of the general solution of the equation.
Characteristic equation has two real roots
If the characteristic equation has two real roots, r and s, so that the general solution of the equation is Aert + Best, then the equilibrium is stable if and only if r < 0 and s < 0.
Characteristic equation has a single real root
If the characteristic equation has a single real root, then the equilibrium is stable if and only if this root is negative. (Note that if r < 0 then tor any value of k, tkert converges to 0 as t increases without bound.)
Characteristic equation has complex roots
If the characteristic equation has complex roots, the form of the solution of the equation is Aeαt cos(βt + ω), where α = −a/2, the real part of each root. Thus the equilibrium is stable if and only if the real part of each root is negative.

The real part of a real root is simply the root, so we can combine the three cases: the equilibrium is stable if and only if the real parts of both roots of the characteristic equation are negative. A bit of algebra shows that this condition is equivalent to a > 0 and b > 0.

If b = 0, then every number is an equilibrium, and none of these equilibria is stable.

In summary, we have the following result.

Proposition
An equilibrium of the homogeneous linear second-order differential equation x"(t) + ax'(t) + bx(t) = 0 is stable if and only if the real parts of both roots of the characteristic equation r2 + ar + b = 0 are negative, or, equivalently, if and only if a > 0 and b > 0.

Example
Consider the following macroeconomic model. Denote by Q aggregate supply, p the price level, and π the expected rate of inflation. Assume that aggregate demand is a linear function of p and π, equal to abp + cπ where a > 0, b > 0, and c > 0. An equilibrium condition is

Q(t) = abp(t) + cπ(t).

Denote by Q* the long-run sustainable level of output, and assume that prices adjust according to the equation

p'(t) = h(Q(t) − Q*) + π(t),

where h > 0. Finally, suppose that expectations are adaptive:

π'(t) = k(p'(t) − π(t))

for some k > 0. Is this system stable?

One way to answer this question is to reduce the system to a single second-order differential equation by differentiating the equation for p'(t) to obtain p"(t) and then substituting in for π'(t) and π(t). We obtain

p"(t) − h(kcb)p'(t) + khbp(t) = kh(aQ*).

We conclude the system is stable if and only if kc < b. (Since k > 0, h > 0, and b > 0, we have khb > 0.)

In particular, if c = 0 (i.e. expectations are ignored) then the system is stable. If expectations are taken into account, however, and respond rapidly to changes in the rate of inflation (k is large), then the system may be unstable.

Exercises


Copyright © 1997-2003 by Martin J. Osborne