8.6.1
Homogeneous (stationary) equations
Consider the equation:
α¨x + β ˙x + γx = 0. (8.4) Recall ˙x = αx ⇒ x = ceαt. Differentiating an exponential function yields an
exponential, but differentiating the exponential twice still yields the exponential! So, conjecture a solution x = ert. The would-be identity is:
αr2ert+ βrert+ γert = 0 ert(αr2+ βr + γ) = 0.
Since ert> 0, for the identity to hold, r must be a root of the polynomial
L(r) = αr2+ βr + γ. We call this the characteristic polynomial. Its roots are called characteristic values. For a characteristic value r, exp(rt) is indeed a solution to (8.4). Notice that roots may be complex numbers if the determinant of the polynomial is negative. We will not discuss this case (save for a mention at the end of the section) as it is relatively rare and exotic in economic applications, and focus instead on the real roots, which may be different or equal.
Real unequal roots
For r1 6= r2, the solution is given by:
x(t) = k1er1t+ k2er2t (8.5)
for arbitrary k1, k2. Perform a quick verification:
˙x = k1r1er1t+ k2r2er2t ˙x = k1r21er1t+ k2r22er2t. α¨x + β ˙x + γx = αk1r21e r1t+ αk 2r22e r2t+ βk 1r21e r1t+ βk 2r22e r2t +γk1er1t+ γk2er2t = (αr12+ βr1+ γ)k1er1t+ (αr22+ βr2+ γ)k2er2t.
8.6. Linear second order equations
The family of solutions has two degrees of freedom. In many applications one has to find a unique solution to satisfy certain conditions, for example, initial conditions for the function itself and its derivative: x(t0) = x0, x0(t0) = z0. These conditions exhaust
the two degrees of freedom and lead to a unique solution of k1, k2. Alternatively, one
may have to deal with a Cauchy (boundary values) problem: x(t0) = x), x(t1) = x1.
Remark See also example 24.13 in Simon and Blume, Mathematics for Economists (W.W. Norton, 1994).
Real equal roots
Suppose r1 = r2, then the family of solutions is given by:
x(t) = k1er1t+ k2ter1t. (8.6)
Notice the term t premultiplying the second term. We know k1er1t is a valid solution to
the ODE, so it suffices to verify k2ter1t:
˙x = k2r1ter1t+ k2er1t ¨ x = k2r12e r1t+ k 2r1er1t+ k2r1er1t = k2r12te r1t+ 2k 2r1er1t. α¨x + β ˙x + γx = αk2r12te r1t+ 2αk 2r1er1t+ βk2r1ter1t+ βk2er1t+ γk2ter1t = αk2r12te r1t+ βk 2r1ter1t+ γk2ter1t+ 2αk2r1er1t+ βk2er1t = k2ter1t(αr21+ βr1+ γ) + k2er1t(2αr1 + β) = k2er1t(2αr1+ β) = 0.
The last equality follows from the fact that for a ‘double’ root:
r = −β ± (β
2− 4αγ)1/2
2α = − β 2α.
Two complex roots
Purely for future reference, in case the roots of the charcateristic polynomial are complex values, r = α ± βi, the solutions are given by:
x(t) = eαt(C1cos βt + C2sin βt).
8.6.2
Non-homogeneous second order linear equations
We have already mentioned non-homogeneous equations. In the context of second order equations, these are written with a non-zero function of t on the RHS:
α¨x + β ˙x + γx = g(t). (8.7)
To characterise the family of solutions suppose that at least one solution is known, xp(t). This will serve as a particular solution.
The solution to the homogenised equation:
α¨x + β ˙x + γx = 0
is called the general solution of (8.7) and is given by (8.5) or (8.6).
We claim that the entire family of solutions to (8.7) is given by xp(t) + xg(t). To verify
this, substitute into (8.7):
α(xp+ xg)··+ β(xp+ xg)·+ γ(xp+ xg)
= (α¨x + βxp+ γxp) + (α¨g + βxg+ γxg)
= g(t) + 0 = g(t).
How does one go about finding xp(t)? Sometimes this is purely a matter of luck with
the method of undetermined coefficients. In general, one conjectures a sufficiently flexible parameterised functional form resembling the function on the RHS of (8.7): for example for a polynomial g = Λk(t) try x
p(t) = ¯Λk(t), a polynomial of the same degree.
For an exponential function g = eαt try x
p(t) = keαt, for a logarithmic g = log t a
generously parameterised guess is xp = t2log t + c1t log t + c2log t + c3t + c4. Notice that
overparameterisation is not a problem for the method, as the coefficients that should not have been there will come out as zeros upon imposing the identity (think: why?).
Activity 8.2 Find the entire family of solutions to ODEs: (a) y00− y0− 6y = 0 (b) 2y00+ 3y0− 2y = 0 (c) y00− 3y0+ 2y = 0 (d) y + 14 ˙¨ y + 13y = t (e) y00− 3y0+ 2y = e−x (f) y00− 3y0+ 2y = (x + 1)2.
Activity 8.3 Solve for a particular solution subject to initial conditions: (a) y00− y0− 30y = 0, y(0) = 1, y0(0) = −4
(b) y00− 2y0− 15y = 0, y(0) = 2, y0(0) = 0.
8.6.3
Phase portraits
So far we focused on solving ODEs in closed form. This, however, is often not possible. Instead we could learn a lot about the behaviour of a model described by an ODE in purely qualitative terms. Consider a non-linear equation:
8.6. Linear second order equations
Figure 8.1: Family of solutions to ˙x = (1 − x)x(x + 1).
The exact solution (with one degree of freedom) is given by:
x(t) = ±√ 1
1 + ce−2t (where well-defined) and x(t) ≡ 0 (verify this).
Figure 8.1 plots representative solutions for different values of c.
Notice that the solutions as functions of time look essentially the same, indeed, they are replicas of each other obtained by a parallel horizontal shift. This should not be
surprising as (8.8) is stationary: if you alter a solution x(t) by defining ˜x(t) = x(t + s) for any real number s, you will have obtained another valid solution to (8.8):
x(t + s) = ±√ 1 1 + ce−2(t+s) = ± 1 p1 + (ce−2s)e−2t = ± 1 √ 1 + ˜ce−2t.
In the applied context a researcher might be interested in the behaviour of the system when ‘restarted’ at a particular point.
To ‘restart system’ at y0 at time t0 means to analyse the solution with an initial
value condition y(t0) = y0.
Definition 37
If x denoted, say, national debt level, it would be meaningful to ask what would happen to a country that initially had neither debt nor foreign assets. The diagram depicts the constant solution x ≡ 0, thus we would expect that foreign asset position to prevail. However, if the country at any given time starts with a small but positive level of debt, it would asymptotically converge to the level given by 1 (unspecified unit). If the debt level is negative (the country is a net creditor to the rest of the world), it is expected to fall to −1; if the debt level is above 1 in absolute value, it is expected to
Figure 8.2: Graph of the RHS of ˙x = (1 − x)x(x + 1).
Figure 8.3: Phase portrait of ˙x = (1 − x)x(x + 1).
regress back to either positive or negative steady state ±1 (the above statements do not follow from any actual model of foreign debt and are purely for illustrating the concept of qualitative analysis). Clearly the diagram helps uncovering important qualitative properties of the model, but it is also quite redundant in doing so. It would be easier to observe the sign of the function on the RHS, f (x) = (1 − x)x(x + 1) to achieve the same end, as made evident by Figure 8.2. The arrows show the direction in which the system should evolve according to the differential ‘laws of motion’ (this is the synonym for the ODE).
To make the same information even more succinct, omit the graph of the function f (·), leaving only the indication of its sign (embodied in the arrows). We have Figure 8.3. This figure is the phase diagram of the stationary equation (8.8). The space of the values taken by the endogenous variable (the unknown function of time) is called the phase space.
8.6.4
The notion of stability
Figures 8.2 and 8.3 above highlighted a special nature of certain points in the phase space. At points 0 and 2 the system described by (8.8) is expected to rest forever. Recall that we call the constant solutions steady states, and we extend the use of the term to the actual constant values these solutions take. In many applications it is of interest whether the system is expected to return to a steady state if it accidentally leaves it (in the formal language of the theory of ODEs, we are interested in the behaviour of the system restarted at various points away from the steady state, to see