^{1}

^{2}

^{*}

^{3}

It is well-known that Goodwin ’ s nonlinear delay accelerator model can generate diverse oscillations (i.e., smooth and sawtooth oscillations). It is, however, less-known what conditions are needed for these dynamics to emerge. In this study, using a piecewise linear investment function, we solve the governing delay differential equation and obtain the explicit forms of the time trajectories. In doing so, we detect conditions for persistent oscillations and also conditions for the birth of such cyclic dynamics.

It has been well-known that Goodwin’s business cycle model with a delayed nonlinear accelerator [

[

This paper reconstructs the third version having a piecewise linear investment function with fixed time delay. It is a complement to [

The paper is organized as follows. In Section 2, the Goodwin model without delay is considered to see how nonlinearities of the model contribute emergence of cyclic dynamics. In Section 3, an investment delay is introduced to construct. Effects of investment delay on the smooth oscillations are considered in Section 4 and those on the sawtooth oscillations are done in Section 5. Section 6 contains some concluding remarks.

To find out how nonlinearity works to generate endogenous cycles, we review the second version of Goodwin’s model, which we call the basic model,

{ ε y ˙ ( t ) = k ˙ ( t ) − ( 1 − α ) y ( t ) , k ˙ ( t ) = φ [ y ˙ ( t ) ] . (1)

Here k is the capital stock, y the national income, a the marginal propensity to consume, which is positive and less than unity, and the reciprocal of e is a positive adjustment coefficient. Since the dot over variables means time differentiation, k ˙ ( t ) and y ˙ ( t ) are the rates of change in capital (i.e., investment) and national income. The first equation of (1) defines an adjustment process of the national income in which national income rises or falls according to whether investment is larger or smaller than savings. The second equation determines the induced investment based on the acceleration principle with which investment depends on the rate of changes in the national income in the following way,

φ [ y ˙ ( t ) ] = { 3 n if y ˙ ( t ) ∈ I U = ( 3 n ν , ∞ ) , ν y ˙ ( t ) if y ˙ ( t ) ∈ I M = [ − n ν , 3 n ν ] , − n if y ˙ ( t ) ∈ I L = ( − ∞ , − n ν ) (2)

where ν > 0 and n > 0 . This function is piecewise linear and has three distinct regions. Accordingly, there are two threshold values of y ˙ ( t ) denoted as 3 n / ν and − n / ν and the investment is proportional to the change in the national income in the middle region, I M , but becomes perfectly inflexible (i.e., inelastic) in the upper region I U or the lower region I L . These values are thought to be “ceiling” and “floor” of investment where the ceiling is assumed to be three time higher than the floor as it was the case in Goodwin’s model.

Inserting the second equation of (1) into the first one and moving the terms on the right hand side to the left give an implicit form of the dynamic equation for national income y,

ε y ˙ ( t ) − φ ( y ˙ ( t ) ) + ( 1 − α ) y ( t ) = 0 (3)

where the stationary point of the basic model is y ( t ) = y ˙ ( t ) = 0 for all t. With Equation (2) it is reduced to either

( ε − ν ) y ˙ ( t ) = − ( 1 − α ) y ( t ) (4)

if y ˙ ( t ) is in the middle region or

ε y ˙ ( t ) = − ( 1 − α ) y ( t ) + d (5)

where d = 3 n if y ˙ ( t ) is in the upper region and d = − n if in the lower region. Equations (4) and (5) are linear and thus solvable. We see graphically and then analytically how dynamics proceeds.

Solving (4) and (5) for y ( t ) presents an alternative expression of dynamic equation y ( t ) = f [ y ˙ ( t ) ] where

f [ y ˙ ( t ) ] = { 3 n − ε y ˙ ( t ) 1 − α if y ˙ ( t ) ∈ I U , ( ν − ε ) y ˙ ( t ) 1 − α if y ˙ ( t ) ∈ I M , − n + ε y ˙ ( t ) 1 − α if y ˙ ( t ) ∈ I L . (6)

Once the initial value is given, the whole evolution of national income is determined. The phase diagram with ε < ν is shown in ^{1} The stationary point is at the origin denoted by E. The locus of y = f ( y ˙ ) is the positive sloping line in the middle region while it is the negative sloping upper or lower line in the upper or lower region. For each value of y ˙ , there is a unique corresponding y value determined to make a point ( y ˙ , y ) satisfy Equation (6) and it is also determined whether y is increasing or decreasing at that point. So the direction of the trajectory is given in all points of the phase diagram. The directions are shown by arrows. Let A denote the local maximum point of the curve with positive y and y ˙ always, and let C be the local minimum point with negative y and y ˙ . Point B and D have the same y values as at points A and C, respectively. Notice that the direction of the dynamic evolution goes from − ∞ to C, from the origin to C, from the origin to A and also from + ∞ to A. Selecting the initial point denoted as S on the positive-sloping line, the evolution starts at this point and moves upward until point A as indicated by arrow. Then it cannot continue on the continuous curve after point A since the direction of evolution changes. Therefore it jumps to point B and continues along the same direction until point C, where the same problem occurs, so another jump occurs to point D and evolution continues until point A, at which the next round repeats itself. Thus the differential Equation (3) with the piecewise linear investment function (2) can give rise to a closed orbit ABCD constituting a self-sustaining slow-rapid oscillation. The stationary point is unstable if ε < ν , however, the oscillation is stable in a sense that the locus ( y ˙ , y ) sooner or later converges to the same oscillation regardless of a

^{1}Mathematica version 11 is used to perform simulations and illustrate this and the following figures. The color versions of the figures are found in DP278 at http://www.chuo-u.ac.jp/research/institutes/economic/publication/disccusion.

selection of the initial point. This is a simple exhibition of emerging a stable endogenous cycle of national income.

The vertex of the closed oscillation in

A = ( 3 n ν , y max ) , B = ( y ˙ m , y max ) , C = ( − n ν , y min ) and D = ( y ˙ M , y min )

where the maximum and minimum values of y ˙ along the cycle are

y ˙ M = n ( 4 ν − ε ) ε ν and y ˙ m = − n ( 4 ν − 3 ε ) ε ν

while the maximum and minimum values of y along the cycle are

y max = 3 n ( ν − ε ) ν ( 1 − α ) and y min = − n ( ν − ε ) ν ( 1 − α ) .

It should be noticed that the instability and the nonlinearity is crucial sources for the birth of persistent oscillations since the instability of the stationary point prevents trajectories from converging and the nonlinearities such as the ceiling and floor prevent trajectories from diverging.

Jumping behavior leads to the kinked time trajectory of y ( t ) and the discontinuous time trajectory of y ˙ ( t ) that are shown as the blue and red curves in

up to the downward-sloping red curve defined on interval [ t 2 , t 3 ] , that then crosses the upper horizontal dotted line and the cross-point correspond to point A.

Selecting an initial point, we can determine an explicit form of the corresponding time trajectory and its rate of change. In particular, we take an initial point on the positive sloping part of the y ( 0 ) = f [ y ˙ ( 0 ) ] curve such as

0 < y ( 0 ) = constant and y ˙ ( 0 ) = 1 − α ν − ε y ( 0 ) < 3 n ν

where y min < y ( 0 ) < y max . If y ˙ ( t ) is in the middle region, Equation (4) yields explicit forms of the solution y ( t ) and its time derivative,

y 1 ( t ) = e 1 − α ν − ε t K 1 and y ˙ 1 ( t ) = e 1 − α ν − ε t ( 1 − α ν − ε K 1 ) with K 1 = y ( 0 ) (7)

and if y ˙ ( t ) enters the upper or lower region, Equation (5) yields the following forms of the solution y ( t ) and its time derivative,

y i ( t ) = e − 1 − α ε t K i + d i 1 − α and y ˙ i ( t ) = e − 1 − α ε t ( − 1 − α ε K i ) for i = 2 , 3 ,< /mo> ⋯ (8)

where d i = − n if i is even and d i = 3 n if i is odd.

Since y ˙ 1 ( 0 ) < 3 n / ν and y ˙ 1 ( t ) is increasing in t, solving y ˙ 1 ( t ) = 3 n / ν for t presents an arrival time t = t 1 ,

t 1 = ν − ε 1 − α log [ 3 n ( ν − ε ) ν ( 1 − α ) K 1 ] .

Substituting t 1 into y 1 ( t ) and y ˙ 1 ( t ) yields

y 1 ( t 1 ) = y max and y ˙ 1 ( t ) = 3 n ν

implying that point ( y ˙ 1 ( t 1 ) , y 1 ( t 1 ) ) corresponds to point A in

y 2 ( t ) = e − 1 − α ε t K 2 − n 1 − α and y ˙ 2 ( t ) = e − 1 − α ε t ( − 1 − α ε K 2 ) . (9)

Since the time trajectory y ( t ) is continuous in t, solving y 1 ( t 1 ) = y 2 ( t 1 ) gives the value of K 2 ,

K 2 = n ( 4 ν − ω ) ν ( 1 − α ) e − 1 − α ε t 1 .

With this K 2 , we have

y ˙ 2 ( t 1 ) = y ˙ m and y 2 ( t 1 ) = y max .

Thus point ( y ˙ 2 ( t 1 ) , y 2 ( t 1 ) ) corresponds to point B to which point A jumps. This rapid change is described by the vertical movement of the red curve along the vertical line at t = t 1 in

t 2 = t 1 + ω 1 − α log [ 4 ν − 3 ε ε ] .

In the same way above, it is possible to show that

( y ˙ 2 ( t 2 ) , y 2 ( t 2 ) ) = ( − n ν , y min )

which is point C. The movement from point B to C is described by solutions in (9). At time t = t 2 , the dynamic system with d = − n is switched to the dynamic system with d = 3 n and (8) with i = 3 presents explicit forms of the solutions

y 3 ( t ) = e − 1 − α ε t K 3 + 3 n 1 − α and y 3 ( t ) = e − 1 − α ε t ( − 1 − α ε K 3 ) . (10)

Solving y 2 ( t 2 ) = y 3 ( t 2 ) presents the value of K 3 ,

K 3 = − n ( 4 ν − ε ) ν ( 1 − α ) e − 1 − α ε t 2 .

It is also able to be shown that point ( y ˙ 3 ( t 2 ) , y 3 ( t 2 ) ) is identical with point D, implying a jump to point D from point C. Solving y ˙ 3 ( t ) = 3 n / ν presents a time when y ˙ 3 ( t ) arrives at 3 n / ν ,

t 3 = t 2 + ε 1 − α log [ 4 ν − ε 3 ε ] .

At t = t 3 , we can confine that the trajectory comes back to point A at which the following holds,

( y ˙ 3 ( t 3 ) , y 3 ( t 3 ) ) = ( 3 n ν , y M ) .

A new round starts as time goes further and the same procedure is applied to obtain explicit forms of the solutions for t ≥ t 3 . Time segments of y ( t ) and y ˙ ( t ) that constitute one cycle of national income are now given by (9) and (10). Since the length of one cycle is measured by the time period between one upper (or lower) kinked point and next upper (or lower) kinked point, it is given by

t 3 − t 1 = ε 1 − α log [ ( 4 ν − ε ) ( 4 ν − 3 ε ) 3 ε 2 ] .

Further the length of the recession period along segment BC in which national income is decreasing is

t 2 − t 1 = ω 1 − α log [ 4 ν − 3 ε ε ]

while the length of the recovery period along segment DA in which national income is increasing is

t 3 − t 2 = ε 1 − α log [ 4 ν − ε 3 ε ] .

In what follows, we will perform numerical simulations with the set of the parameter values given below which are the same parameter values used in [

Assumption 1 α = 0.6 , ε = 0.5 , θ = 1 , ν = 2 and n = 3

In particular,

t 1 ≃ 7.998, t 2 ≃ 11.204, t 3 ≃ 13.216, t 4 ≃ 16.422, t 5 ≃ 7.998 and t 6 ≃ 21.640

by solving

y ˙ i ( t ) = 3 n ν if i is odd or y ˙ i ( t ) = − n ν if i is even .

^{2}in [

The length of one cycle given by t i + 2 − t i is about 5.218 years.^{2} In the same way, the recession period from one peak to trough of the cycle is given by t i − t i − 1 ≃ 3.206 years for i being even while the recovery period from one trough to peak by t i − t i − 1 ≃ 2.012 years for i being odd. The constant K i solves

y i ( t i ) = y i + 1 ( t i )

with K 1 = 2 and the numerical results are as follows,

K 2 ≃ 14641.52, K 3 ≃ − 219622.71, K 4 ≃ 951698.40, K 5 ≃ − 14275475.98 and K 6 ≃ 61860395.91.

We now investigate how the investment delay affects time paths of national income. Observing the fact that, in real economy, plans and their realizations need time to take effects, [

φ [ y ˙ ( t − θ ) ] = { 3 n if y ˙ ( t − θ ) in I U , ν y ˙ ( t − θ ) if y ˙ ( t − θ ) in I M , − n if y ˙ ( t − θ ) in I M . (11)

With this modification, the dynamic Equation (3) turns to be

ε y ˙ ( t ) − φ ( y ˙ ( t − θ ) ) + ( 1 − α ) y ( t ) = 0 (12)

that we call the delay model.^{3} Equation (12) is reduced to a linear delay differential equation of neutral type if the delayed rate of change in national income stays in the middle region

ε y ˙ ( t ) − ν y ˙ ( t − θ ) + ( 1 − α ) y ( t ) = 0 (13)

and it remains to be a linear ordinary differential Equation (5) if the delayed rate is in the upper or lower region. To solve the delay equation, we need an initial function that determines behavior of y prior to time zero,

y ( t ) = Φ ( t ) for − θ ≤ t ≤ 0.

Although [

It has been examined by [

Assumption 2 Φ ( t ) = y 0 ≠ 0 and Φ ˙ ( t ) = 0 for − θ ≤ t ≤ 0. (14)

^{3}Goodwin assumes a smooth form of the nonlinear investment function in his third version.

It is well known that if the characteristic polynomial of a linear neutral equation has roots only with negative real parts, then the stationary point is locally asymptotically stable. The normal procedure for solving this equation is to try an exponential form of the solution. Substituting y ( t ) = y 0 e λ t into (13) and rearranging terms, we obtain the corresponding characteristic equation:

ε λ − ν λ e − λ θ + ( 1 − α ) = 0.

To check stability, we determine conditions under which all roots of this characteristic equation lie in the left or right half of the complex plane. Dividing both sides of the characteristic equation by e and introducing the new variables

a = ν ε > 0 and b = 1 − α ε > 0 , (15)

we rewrite the characteristic equation as

λ − a λ e − λ θ + b = 0. (16)

[

Lemma 1 If ν > ε , then the zero solution of the fixed delay model (13) is locally unstable for all θ > 0 .

On the other hand, if v ≤ ε or a ≤ 1 , characteristic Equation (16) has at most finitely many eigenvalues with positive real parts. The eigenvalue is real and negative when θ = 0 . The roots of the characteristic equation are functions of the delay. Although it is expected that all roots have negative real parts for small values of θ , the real parts of some roots may change their signs to positive from negative as the lengths of the delay increases. The stability of the zero solution may change. Such phenomena are often referred to as stability switches. We will next prove that stability switching, however, cannot take place in the delayed model.

Lemma 2 If v ≤ ε , then the zero solution of the fixed delay model (13) is locally stable for all θ > 0 .

Proof. 1) It can be checked that λ = 0 is not a solution of (16) because substituting λ = 0 yields b = 0 that contradicts

b − a ω sin ω θ = 0 ,

and

ω − a ω cos ω θ = 0.

Moving b and w to the right hand sides and adding the squares of the resultant equations, we obtain

b 2 + ( 1 − a 2 ) ω 2 = 0.

Since b > 0 and 1 − a 2 > 0 as a < 1 is assumed, there is no w that satisfies the last equation. In other words, there are no roots of (16) crossing the imaginary axis when q increases. No stability switch occurs and thus the zero solution is locally asymptotically stable for any θ > 0 .

2) In case of ε = ν in which a = 1 , the characteristic equation becomes

λ ( 1 − e − λ θ ) + b = 0. (17)

It is clear that λ = 0 is not a solution of (17) since b > 0 . Thus we can assume that a root of (17) has non-negative real part, λ = u + i v with u ≥ 0 for some θ > 0 . From (17), we have

( u + b ) 2 + v 2 = e − 2 u θ ( u 2 + v 2 ) ≤ ( u 2 + v 2 ) ,

where the last inequality is due to e − 2 u θ ≤ 1 for u ≥ 0 and θ > 0 . Hence

2 u b + b 2 ≤ 0,

where the direction of inequality contradicts the assumption that u ≥ 0 and b > 0 . Hence it is impossible for the characteristic equation to have roots with nonnegative real parts. Accordingly, all roots of (17) must have negative real parts for all θ > 0 .

Lemmas 1 and 2 imply the following theorem concerning local stability of the delay model (13).

Theorem 3 For any θ > 0 , the zero solution of the delay model (13) is locally asymptotically stable if ν ≤ ε and unstable if ν > ε .

We call y 0 in Assumption 2 an initial value for convenience. Fixing the length of delay at θ = 1 , we illustrate a bifurcation diagram with respect to the initial value in

Given Assumptions 1 and 2 with y 0 = y 0 G , the time trajectories of y ( t ) and y ˙ ( t ) are illustrated by the blue and red curves, respectively, in

which we can see that delay time trajectories show sharp differences from non-delay time trajectories depicted in

We omit consideration in interval [ 0, t S ] with t S = θ as behavior there strongly depends on a choice of initial point. In the first interval I I = [ t S , t I ] where t I = t a + θ ≃ 4.03 and t a ≃ 3.03 solves the equation y ˙ I ( t ) = − n / ν , the blue and red trajectories are controlled by Equation (5) with d = − n ,

( G-I ) { y I ( t ) = e − 1 − α ε t K I − n 1 − α y ˙ I ( t ) = e − 1 − α ε t ( − 1 − α ε K I ) for t ∈ I I

^{4}The explicit forms are given in Appendix I.

where K I ≃ 21.19 . At

y I ( t S ) ≃ 2.02, y ˙ I ( t S ) ≃ − 7.62 and y I ( t I ) ≃ − 6.66, y ˙ I ( t I ) ≃ 0.67.

As seen in ^{4} Then, constant K I is determined so as to satisfy y 0 ( t S ) = y I ( t S ) , that is, the end point of y 0 ( t S ) is coincided with the starting point of y I ( t S ) . Hence it first implies the continuity of y ( t ) at t = t S . Secondly, the solutions of y 0 ( t ) and y I ( t ) should satisfy the following dynamic equations respectively,

ε y ˙ 0 ( t S ) + ( 1 − α ) y 0 ( t S ) = 0, ε y ˙ I ( t S ) + ( 1 − α ) y I ( t S ) = − n .

Subtracting the second equation from the first presents

y ˙ 0 ( t S ) − y ˙ I ( t S ) = n ε > 0 or y ˙ 0 ( t S ) > y ˙ I ( t S )

where the last inequality implies discontinuity of the derivative at t = t S .

In the second interval I I I = [ t I , t I I ] with t I I = t I + θ ≃ 5.03 , applying successive integration for dynamic equation ε y ˙ ( t ) + ( 1 − α ) y ( t ) = ν y ˙ I ( t − θ ) gives explicit forms of the solutions,

( G-II ) { y I I ( t ) = e − 1 − α ε t ( α 1 I I t + α 0 I I ) y ˙ I I ( t ) = e − 1 − α ε t ( β 1 I I t + β 0 I I ) for t ∈ I I I

where

α 1 I I = − ν ( 1 − α ) ε 2 e 1 − α ε θ K I ≃ − 150.92, β 1 I I = − 1 − α ε α 1 I I ≃ 120.74, α 0 I I ≃ 440.94, β 0 I I = α 1 I I − 1 − α ε α 0 I I ≃ − 503.67.

The integral constant α 0 I I is obtained by solving y I ( t I ) = y I I ( t I ) . The boundary values for the end points of interval I I I are

y I I ( t I ) ≃ − 6.66, y ˙ I I ( t I ) ≃ 0.67 and y I I ( t I I ) ≃ − 5.69, y ˙ I I ( t I I ) ≃ 1.85.

It can be numerically as well as graphically checked that y I I ( t I ) = y I ( t I ) and y ˙ I I ( t I ) = y ˙ I ( t I ) . In consequence y I ( t ) and y I I ( t ) are smoothly connected (i.e., continuous and differentiable). On the other hand, y ˙ ( t ) ≥ − n / ν for t ≥ t a induces system change at t = t I leading to that y ˙ I ( t I ) and y ˙ I I ( t I ) are connected with kink (i.e., continuous and non-differentiable).

For t in the third interval I I I I = [ t I I , t I I I ] with t I I I = t I I + θ ≃ 6.03 , we have 3 n / ν > y ˙ I I ( t − θ ) > − n / ν . Hence, successive integration implies that Equation (13) with Φ [ y ˙ I I ( t − θ ) ] = ν y ˙ I I ( t − θ ) yields the trajectories described by

( G-III ) { y I I I ( t ) = e − 1 − α ε t ( α 2 I I I t 2 + α 1 I I I t + α 0 I I I ) y ˙ I I I ( t ) = e − 1 − α ε t ( β 2 I I I t 2 + β 1 I I I t + β 0 I I I ) for t ∈ I I I I

where

α 2 I I I = ν ε e 1 − α ε θ β 1 I I 2 ⋅ 537.41, β 2 I I I = − 1 − α ε α 2 I I I ≃ − 429,93, α 1 I I I = ν ε e 1 − α ε θ ( β 0 I I − θ β 1 I I ) ≃ 5521.58, β 1 I I I = 2 α 2 I I I − 1 − α ε α 1 I I I ≃ 5521.68, α 0 I I I ≃ 14044.6, β 0 I I I = α 1 I I I − 1 − α ε α 0 I I I ≃ − 16794.2.

Since y I I ( t I I ) = y I I I ( t I I ) and y ˙ I I ( t I I ) = y ˙ I I I ( t I I ) hold, the blue and red trajectories are continuous at t = t I I . The boundary values for the right endpoint of interval I I I I are

y I I I ( t I I I ) ≃ 0.55 and y ˙ I I I ( t I I I ) ≃ 6.98.

As is seen in

In the fourth interval I I V = [ t I I I , t I V ] with t I V = t b + θ ≃ 6.19 , Equation (13) with Φ [ y ˙ I I I ( t − θ ) ] = ν y ˙ I I I ( t − θ ) determines the trajectories,

( G-IV ) { y I V ( t ) = e − 1 − α ε t ( α 3 I V t 3 + α 2 I V t 2 + α 1 I V t + α 0 I V ) y ˙ I V ( t ) = e − 1 − α ε t ( β 3 I V t 3 + β 2 I V t 2 + β 1 I V t + β 0 I V ) for t ∈ I I V

where

α 3 I V = ν ε e 1 − α ε θ β 1 I I I 3 ≃ − 1275.76, β 3 I V = − 1 − α ε α 3 I V ≃ 1020.61, α 2 I V = ν ε e 1 − α ε θ β 1 I I I − θ β 2 I I I 2 ≃ 28404.7, β 2 I V = 3 α 3 I V − 1 − α ε α 2 I V ≃ − 26551, α 1 I V = β 0 I I I − θ β 1 I I I + θ 2 β 2 I I I ≃ − 202487, β 1 I V = 2 α 2 I V − 1 − α ε α 1 I V ≃ 218799, α 0 I V ≃ − 467961, β 0 I V = α 1 I V − 1 − α ε α 0 I V ≃ − 576852.

Since interval I I V is very narrow, the right end point of I I V is labelled at the upper part of

y I V ( t I V ) ≃ 2.49 and y ˙ I V ( t I V ) ≃ 16.01.

Since y ˙ I V ( t − θ ) > 3 n / ν holds in the fifth interval I V = [ t I V , t V ] where t V = t c + θ ≃ 8.78 and t c ≃ 7.78 solves y ˙ V ( t ) = 3 n / ν , Equation (13) with Φ [ y ˙ I V ( t − θ ) ] = 3 n gives the following forms of the solution and its time derivative,

( G-V ) { y V ( t ) = e − 1 − α ε t K V + 3 n 1 − α y ˙ V ( t ) = e − 1 − α ε t ( − 1 − α ε K V ) for t ∈ I V

where K V ≃ − 2839.05 . The crossing point of the red curve with the upper horizontal dotted line at t c is denoted by the green dot. The boundary values for the right end point of this interval are

y V ( t V ) ≃ 19.97 and y ˙ V ( t V ) ≃ 2.02.

Since y V ( t − θ ) < 3 n / ν holds for t of the sixth interval I V I = [ t V , t V I ] with t V I = t V + θ ≃ 9.78, dynamic Equation (13) with φ [ y ˙ V ( t − θ ) ] = ν y ˙ V ( t − θ ) determines the following evolution of y ( t ) and y ˙ ( t ) :

( G-VI ) { y V I ( t ) = e − 1 − α ε t ( α 1 V I t + α 0 V I ) y ˙ V I ( t ) = e − 1 − α ε t ( β 1 V I t + β 0 V I ) for t ∈ I V I

where

α 1 V I = − ν ( 1 − α ) ε 2 e 1 − α ε θ K V ≃ 20129, β 1 V I = − 1 − α ε α 1 V I ≃ − 16175.2, α 0 V I ≃ − 155088, β 0 V I = α 1 V I − 1 − α ε α 0 V I ≃ 144289.

The boundary values for the right end point of interval I V I are

y V I ( t V I ) ≃ 17.06 and y ˙ V I ( t V I ) ≃ − 5.56.

It is seen that the red curve crosses the lower horizontal dotted line from above at t d ≃ 9.05 .

In the seventh interval I V I I = [ t V I , t V I I ] with t V I I = t d + θ ≃ 10.05 , applying successive integration to dynamic Equation (13) with φ [ y ˙ V I ( t − θ ) ] = ν y ˙ V I ( t − θ ) yields the forms of the solutions,

( G-VII ) { y V I I ( t ) = e − 1 − α ε t ( α 2 V I I t 2 + α 1 V I I t + α 0 V I I ) y ˙ V I I ( t ) = e − 1 − α ε t ( β 2 V I I t 2 + β 1 V I I t + β 0 V I I ) for t ∈ I V I I

where

α 2 V I I = ν ε e 1 − α ε θ β 1 V I 2 ≃ − 71997, β 2 V I I = − 1 − α ε α 2 V I I ≃ 51597.63, α 1 V I I = ν ε e 1 − α ε θ ( β 0 V I − θ β 1 B I ) ≃ 142848.2, β 1 V I I = 2 α 2 V I I − 1 − α ε α 1 V I I ≃ − 128678, α 0 V I I ≃ − 704149.86, β 0 V I I = α 1 I V I − 1 − α ε α 0 V I I ≃ 706168.

The boundary values for the right end point of interval I V I I are

y V I I ( t V I I ) ≃ 13.82 and y ˙ V I I ( t V I I ) ≃ − 17.06.

Finally, in the eighth interval I V I I I = [ t V I I , t E ] with t E = t V I I + θ ≃ 11.05 , y ˙ V I I ( t ) < n / ν for t ∈ I V I I implies dynamic Equation (5) with φ [ y ˙ V I I ( t − θ ) ] = − n controls the trajectories,

( G-VIII ) { y V I I I ( t ) = e − 1 − α ε t K V I I I − n 1 − α y ˙ V I I I ( t ) = e − 1 − α ε t ( − 1 − α ε K V I I I ) for t ∈ I V I I I

where K V I I I ≃ 66133.90 . Notice that y V I I I ( t E ) = y I ( t S ) and y ˙ V I I I ( t E ) = y ˙ I ( t S ) hold. The length of the period is about 10 years.^{5} Very roughly speaking, the recovery period could be approximately 4.7 years from t I to t V and then the recession period is 5.3 years. The same cycle repeats itself for t > t E .

Calculating the boundary values of each interval I i , we have the following set of points ( y ˙ ( t ) , y ( t ) ) in the phase diagram of

( S ) = ( − 7 . 6 2 , 2 . 0 2 ) ( 2 ) = ( − 0.67, − 6.66 ) ( 3 ) = ( 1.85, − 5.69 ) ( 4 ) = ( 7.00,0.55 ) ( 5 ) = ( 16.01,2.49 ) ( 6 ) = ( 2.02,19.97 ) ( 7 ) = ( − 5.56,17.06 ) ( 8 ) = ( − 17.06,13.82 ) ( E ) = ( − 7 . 6 2 , 2 . 0 2 )

Point denoted by (S) and (E) is the starting point and the ending point of the cyclic oscillation, both of which are identical. Equation (5) with d = − n governs decreasing movements from (S) to (2) and from (8) to (E) along the lower red line whereas Equation (5) with d = 3 n controls upward movements

^{5}With the same parameter values but different initial values, [

from (5) to (6) along the upper red line. On the other hand, movements from (2) to (5) and from (6) to (8) along the dotted curves between these two lines are described by Equation (13). The switching of dynamic equations occurs at the following points:

Point (2) at which Equation (5) with d = − n is changed to Equation (13);

Point (5) at which Equation (13) to Equation (5) with d = 3 n ;

Point (6) at which Equation (5) with d = 3 n to Equation (13);

Point (8) at which Equation (13) to Equation (5) with d = − n ;

Points (3), (4) and (7) at which Equations (13) have at different forms of Φ [ y ˙ ( t − θ ) ]

We can verify the following.

Theorem 4 The cyclic time trajectories of y ( t ) and y ˙ ( t ) are continuous at these switching points.

Proof. 1) At point (2) with t = t I , y I ( t ) is connected to y I I ( t ) and so is y ˙ I ( t ) to y ˙ I I ( t ) . Integral constant α 0 I I of y I I ( t ) is determined so as to solve y I ( t I ) = y I I ( t I ) . Further y I ( t ) and y I I ( t ) should satisfy the dynamic equations at t = t I ,

ε y ˙ I ( t I ) + ( 1 − α ) y I ( t I ) = − n , ε y ˙ I I ( t I ) + ( 1 − α ) y I I ( t I ) = ν y ˙ I ( t I − θ ) .

y ˙ I ( t I − θ ) = y ˙ I ( t a ) holds at t I = t a + θ and y ˙ I ( t a ) = − n / ν by definition of t a , both of which lead to ν y ˙ I ( t I − θ ) = − n . Therefore the above two dynamic equations are identical and thus two solutions of these dynamic equations take the same values at t = t I , namely, y I ( t I ) = y I I ( t I ) and y ˙ I ( t I ) = y ˙ I I ( t I ) .

2) At point (5) with t = t I V , y I V ( t ) is connected to y V ( t ) and so is y ˙ I V ( t ) to y ˙ V ( t ) . Integral constant K V of y I V ( t ) is determined so as to solve y I V ( t I V ) = y V ( t I V ) . Further, y I V ( t ) and y V ( t ) should satisfy the dynamic equations at t = t I V ,

ε y ˙ I V ( t I V ) + ( 1 − α ) y I V ( t I V ) = θ y ˙ I I I ( t I V − θ ) , ε y ˙ V ( t I ) + ( 1 − α ) y I V ( t I ) = 3 n .

y ˙ I I I ( t I V − θ ) = y ˙ I ( t a ) holds at t I V = t b + θ and y ˙ I I I ( t b ) = 3 n / ν by definition of t b , both of which lead to θ y ˙ I I I ( t I V − θ ) = 3 n . Therefore the above two dynamic equations are identical and thus two solutions of these dynamic equations take the same values at t = t I V , namely, y I V ( t I V ) = y V ( t I V ) and y ˙ I V ( t I V ) = y ˙ V ( t I V ) . The same procedure applies for points (6) and (8).

3) At point (3) with t = t I I , y I I ( t ) and y I I I ( t ) satisfy the dynamic equations

ε y ˙ I I ( t I I ) + ( 1 − α ) y I I ( t I I ) = θ y ˙ I ( t I I − θ ) , ε y ˙ I I I ( t I I ) + ( 1 − α ) y I I I ( t I I ) = θ y ˙ I I ( t I I − θ ) .

y ˙ I ( t I I − θ ) = y ˙ I ( t I ) and y ˙ I I ( t I I − θ ) = y ˙ I I ( t I ) as t I I = t I + θ . From (i), we already have y ˙ I ( t I ) = y ˙ I I ( t I ) . Further the integral constant α 1 I I I of y I I I ( t ) solves y I I ( t I I ) = y I I ( t I I ) . Then substituting the second equation from the first equation presents y ˙ I I ( t I I ) = y ˙ I I I ( t I I ) . The same procedure applies for Points (4) and (7).

This theorem confirms no jumps of the derivatives at the switching points of the dynamic system, implying the smooth time trajectory of national income just like observed business cycle. This is what [

Theorem 5 If the initial value y 0 of the initial function and the length of delay θ are selected such as y 0 < y 0 min ( θ ) or y 0 > y 0 max ( θ ) , then the delay model can generate smooth oscillations of national income.

Under Assumptions 1 and 2 with y o S = − 2 ,

It has been pointed out by [^{6} Our main aim of this section is to analytically reproduce these numerical results to understand why a trajectory y ( t ) has kinks and its derivative y ˙ ( t ) makes jumps. To this end, we start to divide the interval I = [ 0 , 5 ] into five subintervals with respect to the length of delay θ ,

I i = [ t i − 1 , t i ] for i = 1,2,3,4,5

where t i − 1 = i − 1 and t i = t i − 1 + θ with θ = 1 . Detailed derivations of the forms of y ( t ) and y ˙ ( t ) in each interval are presented in Appendix II.

^{6}More precisely, [

The constant initial function Φ ( t − θ ) = − 2 for t ∈ I 1 = [ t 0 , t 1 ] is selected. The dynamic Equation (13) with φ [ Φ ˙ ( t − θ ) ] = 0 yields the following forms of the solution and its derivative

( S-I ) { y 1 ( t ) = e − 1 − α ε t K 1 with K 1 = − 2, y ˙ 1 ( t ) = e − 1 − α ε t ( − 1 − α ε K 1 ) .

The boundary values for the end points of interval I 1 are

y 1 ( t 0 ) ≃ − 2, y ˙ 1 ( t 0 ) ≃ 1.6 and y 1 ( t 1 ) ≃ − 0.897, y ˙ 1 ( t 1 ) ≃ 0.719.

In the second interval I 2 = [ t 1 , t 2 ] , solving (13) with φ [ y ˙ 1 ( t − θ ) ] = ν y ˙ 1 ( t − θ ) by successive integration yields the following forms

( S-II ) { y 2 ( t ) = e − 1 − α ε t ( α 1 I I t + α 0 I I ) y ˙ 2 ( t ) = e − 1 − α ε t ( β 1 I I t + β 0 I I )

where

α 1 I I = − k ( 1 − α ) ε 2 K 1 e 1 − α ε θ ≃ 14.244 , β 1 I I = − k ( 1 − α ) 2 ε 3 e − 1 − α ε θ K 1 ≃ − 11.395 , α 0 I I ≃ − 16.244 , β 0 I I = − 1 − α ε ( k ε e 1 − α ε θ K 1 + K 2 ) ≃ 27.238.

Integral constant α 0 I I is determined so as to satisfy y 2 ( t 1 ) = y 1 ( t 1 ) , implying the continuity of the blue curve at t 1 . The discontinuity of the red curve at that point can be shown in the same way as in the case of Goodwin cycle. The solutions y 1 ( t 1 ) and y 2 ( t 1 ) satisfy the corresponding dynamic equations at t = t 1 ,

ε y ˙ 1 ( t 1 ) + ( 1 − α ) y 1 ( t 1 ) = 0 , ε y ˙ 2 ( t 1 ) + ( 1 − α ) y 2 ( t 1 ) = ν y ˙ 1 ( t 1 − θ )

where t 1 = t 0 + θ and t 0 = 0 implying that y ˙ 1 ( t 1 − θ ) = y ˙ 1 ( 0 ) > 0 . Subtracting the first equation from the second equation presents

y ˙ 2 ( t 1 ) − y ˙ 1 ( t 1 ) = ν ε y ˙ 1 ( 0 ) > 0 or y ˙ 2 ( t 1 ) > y ˙ 1 ( t 1 ) .

The last inequality confirms the discontinuity of the red curve at t = t 1 . The boundary values for the end points of interval I 2 are

y 2 ( t 1 ) ≃ − 0.897, y ˙ 2 ( t 1 ) ≃ 7.119 and y 2 ( t 2 ) ≃ 2.472, y ˙ 2 ( t 2 ) ≃ 0.898.

It is then numerically confirmed that

y 1 ( t 1 ) = y 2 ( t 1 ) ⇒ continuityofthebluecurveat t = t 1 , y ˙ 1 ( t 0 ) ≠ y ˙ 2 ( t 1 ) ⇒ discontinuityoftheredcurveat t = t 1 .

As shown in the Appendix II, t a is the value at which the red curve crosses the upper horizontal dotted line once from above and divides the interval I 3 = [ t 2 , t 3 ] into two subintervals, I 3 a = [ t 2 , t a θ ] and I 3 b = [ t a θ , t 3 ] where t a θ = t a + θ . So we derive a solution of the differential equation in each interval. In interval I 3 a Equation (5) with d = 3 n presents the forms of y ( t ) and y ˙ ( t ) :

( S-IIIa ) { y 3 a ( t ) = e − 1 − α ε t K 3 + 3 n 1 − α , y ˙ 3 a ( t ) = − 1 − α ε K 3 e − 1 − α ε t ,

where solving y 3 a ( t 2 ) = y 2 ( t 2 ) gives constant value K 3 ≃ − 99.20 . The boundary values for the end points of interval I 3 a are

y 3 a ( t 2 ) ≃ 2.472, y ˙ 3 a ( t 2 ) ≃ 16.023 and y 3 a ( t a θ ) ≃ 6.565, y ˙ 3 a ( t a θ ) ≃ 12.748

where the continuity of the blue curve and the discontinuity of the red curve at t = t 2 are also numerically confirmed,

y 2 ( t 2 ) = y 3 a ( t 2 ) and y ˙ 2 ( t 2 ) ≠ y ˙ 3 a ( t 2 ) .

On the other hand, in interval I 3 b , Equation (13) with φ [ y ˙ 2 ( t − θ ) ] = ν y ˙ 2 ( t − θ ) yields the solution and its derivative,

( S-IIIb ) { y 3 b ( t ) = e − 1 − α ε t ( α 2 I I I t 2 + α 1 I I I t + α 0 I I I ) y ˙ 3 b ( t ) = e − 1 − α ε t ( β 2 I I I t 2 + β 1 I I I t + β 0 I I I )

where

α 2 I I I = k ε e 1 − α ε θ β 1 I I 2 ≃ − 50.719 , β 2 I I I = − 1 − α ε α 2 I I I ≃ − 22.79 , α 1 I I I = k ε e 1 − α ε θ ( β 0 I I − θ β 1 I I ) ≃ 343.918 , β 1 I I I = 2 α 2 I I I − 1 − α ε α 1 I I I ≃ − 255.97 α 0 I I I = − 480.253 , β 0 I I I = α 1 I I I − 1 − α ε α 0 I I I ≃ 559.71.

The boundary values for the end points of interval I 3 b are

y 3 b ( t a θ ) ≃ 6.565, y ˙ 3 b ( t a θ ) ≃ 12.748 and y 3 b ( t 3 ) ≃ 8.621, y ˙ 3 b ( t 3 ) ≃ − 3.309

where the blue and red curves are confirmed to be continuous at t = t a θ ,

y 3 a ( t a θ ) = y 3 b ( t a θ ) and y ˙ 3 a ( t a θ ) = y ˙ 3 b ( t a θ ) .

Due to the values of t b and t c obtained in the Appendix II, interval I 4 = [ t 3 , t 4 ] is divided into three subintervals I 4 a = [ t 3 , t b θ ] , I 4 b = [ t b θ , t c θ ] and I 4 c = [ t c θ , t 4 ] by t b θ = t b + θ and t c θ = t c + θ . Since φ [ y ˙ 3 ( t − θ ) ] = φ [ y ˙ 4 ( t − θ ) ] = 3 n for t ∈ I 4 a , Equation (5) with d = 3 n yields the solution of the differential equation and its derivative

( S-VIa ) { y 4 a ( t ) = e − 1 − α ε t K 4 a + 3 n 1 − α with K 4 a ≃ − 152.994, y ˙ 4 a ( t ) = e − 1 − α ε t ( − 1 − α ε K 4 a ) with − 1 − α ε K 4 a ≃ 122.395.

The boundary values for the end points of interval I 4 a are

y 4 a ( t 3 ) ≃ 8.621, y ˙ 4 a ( t 3 ) ≃ 11.103 and y 4 a ( t b θ ) ≃ 13.456, y ˙ 4 a ( t b θ ) ≃ 7.236

where the blue curve is continuous and the red curve jumps at t = t 3 ,

y 3 b ( t 3 ) = y 4 a ( t 3 ) and y ˙ 3 b ( t 3 ) ≠ y ˙ 4 a ( t 3 ) .

In I 4 b , Equation (13) with φ [ y ˙ 4 a ( t − θ ) ] = ν y ˙ 4 a ( t − θ ) gives the solution and its derivative

( S-IVb ) { y 4 b ( t ) = e − 1 − α ε t { α 3 I V t 3 + α 2 I V t 2 + α 1 I V t + α 0 I V } y ˙ 4 b ( t ) = e − 1 − α ε t { β 3 I V t 3 + β 2 I V t 2 + β 1 I V t + β 0 I V }

where

α 3 I V = k ε e 1 − α ε θ β 2 I I I 3 ≃ 180.604 , β 3 I V = − 1 − α ε α 3 I V ≃ − 96.322 , α 2 I V = k ε e 1 − α ε θ β 1 I I I − 2 θ β 2 I I I 2 ≃ − 2037.36 , β 2 I V = 3 α 3 I V − 1 − α ε α 2 I V ≃ 1991.1 , α 1 I V = k ε e 1 − α ε θ ( β 0 I I I − θ β 1 I I I + θ 2 β 2 I I I ) ≃ 10195.4 , β 1 I V = 2 α 2 I V − 1 − α ε α 1 I V ≃ − 12231 , α 0 I V ≃ − 15672.428 , β 0 I V = α 1 I V − 1 − α ε α 0 I V ≃ 22733.3.

The boundary values for the end points of interval I 4 b are

y 4 b ( t b θ ) ≃ 13.456, y ˙ 4 b ( t b θ ) ≃ 7.236 and y 4 b ( t c θ ) ≃ 11.677, y ˙ 4 b ( t c θ ) ≃ − 15.342

where the blue and red curves are continuous at t = t b θ ,

y 4 a ( t b θ ) = y 4 a ( t b θ ) and y ˙ 4 a ( t b θ ) = y ˙ 4 a ( t b θ ) .

Since φ [ y ˙ 4 b ( t − θ ) ] = − n for t ∈ I 4 c Equation (5) with d = − n yields the solution and its derivative

( S-IVc ) { y 4 c ( t ) = e − 1 − α ε t K 7 − n 1 − α with K 7 ≃ 415.092 y ˙ 4 c ( t ) = e − 1 − α ε t ( − 1 − α ε K 7 ) with − 1 − α ε K 7 ≃ − 340.074

The boundary values for the end points of interval I 4 c are

where the blue and red curves are continuous at t = t c θ

y 4 b ( t c θ ) = y 4 c ( t c θ ) and y ˙ 4 b ( t c θ ) = y ˙ 4 c ( t c θ ) .

Due to the crossing values t d and t e obtained in the Appendix II, interval I 5 = [ t 4 , t 5 ] is divided into three subintervals I 5 a = [ t 4 , t d θ ] and I 5 b = [ t d θ , t e θ ] and I 5 c = [ t e θ , t 5 ] by t d θ = t d + θ and t e θ = t e + θ . Since φ [ y ˙ 4 b ( t − θ ) ] = φ [ y ˙ 4 b ( t − θ ) ] = 3 n for t ∈ I 5 a , Equation (5) with d = 3 n implies the solution and its derivative,

( S-Va ) { y 5 a ( t ) = e − 1 − α ε t K 8 + 3 n 1 − α with K 8 ≃ − 320.88 y ˙ 5 a ( t ) = e − 1 − α ε t ( − 1 − α ε K 8 ) with − 1 − α ε K 8 ≃ 256.704

The boundary values for the end points of interval I 5 a are

y 5 a ( t 4 ) ≃ 9.420, y ˙ 5 a ( t 4 ) ≃ 10.464 and y 5 a ( t d θ ) ≃ 14.150, y ˙ 5 a ( t d θ ) ≃ 6.680

where the blue curve of y ( t ) is continuous but kinked at t = t 4 and accordingly, the red curve of y ˙ ( t ) jumps at t = t 4 ,

y 4 c ( t 4 ) = y 5 a ( t 4 ) and y ˙ 4 c ( t 4 ) ≠ y ˙ 5 a ( t 4 )

In I 5 b , Equation (13) with φ [ y ˙ 4 a ( t − θ ) ] = ν y ˙ 4 a ( t − θ ) yields the solution and its derivative

( S-Vb ) { y 5 b ( t ) = e − 1 − α ε t { α 4 V t 4 + α 3 V t 3 + α 2 V t 2 + α 1 V t + α 0 V } y ˙ 5 b ( t ) = e − 1 − α ε t { β 4 V t 4 + β 3 V t 3 + β 2 V t 2 + β 1 V t + β 0 V }

where

α 4 V = k ε e 1 − α ε θ β 3 I V 3 ≃ − 214.369 , β 4 V = − 1 − α ε α 4 I V ≃ 171.495 , α 3 V = k ε e 1 − α ε θ β 2 I V − 3 θ β 3 I V 3 ≃ 6765.83 , β 3 V = 4 α 4 V − 1 − α ε α 3 V ≃ − 9270.135 ,

α 2 V = k ε e 1 − α ε θ β 1 I V − 2 θ β 2 I V + 3 θ 2 β 3 I V 2 ≃ − 73452.5 , β 2 V = 3 α 3 V − 1 − α ε α 2 V ≃ 79059.451 , α 1 V = k ε e 1 − α ε θ ( β 0 I V − θ β 1 I V + θ 2 β 2 I V − θ 3 β 3 I V ) ≃ 329841 , β 1 V = 2 α 2 V − 1 − α ε α 1 V ≃ − 410777.286 ,

α 0 V = K 9 ≃ − 525028.550 , β 0 V = α 1 V − 1 − α ε α 0 V ≃ 749863.287.

The boundary values for the end points of interval I 5 b are

y 5 b ( t d θ ) ≃ 14.150, y ˙ 5 b ( t d θ ) ≃ 6.680 and y 5 b ( t e θ ) ≃ 13.793, y ˙ 5 b ( t e θ ) ≃ − 17.034.

Since φ [ f ˙ 4 c ( t − θ ) ] = − n for t ∈ I 5 c , Equation (5) with d = − n yields the solution and its derivative

( S-Vc ) { y 5 c ( t ) = e − 1 − α ε t K 10 − n 1 − α with K 10 ≃ 860.508 y ˙ 5 c ( t ) = e − 1 − α ε t ( − 1 − α ε K 10 ) with − 1 − α ε K 10 ≃ − 688.406

The boundary values for the end points of interval I 5 c are

y 5 c ( t e θ ) ≃ 13.793, y ˙ 5 c ( t e θ ) ≃ − 17.034 and y 5 c ( t 5 ) ≃ 8.267, y ˙ 5 c ( t 5 ) ≃ − 12.614

where the blue and red curves are continuous at t = t e θ ,

y 5 b ( t e θ ) = y 5 c ( t e θ ) and y ˙ 5 b ( t e θ ) = y ˙ 5 b ( t e θ ) .

Notice that the red curves in intervals, I 3 , I 4 and I 5 intersect the upper and lower horizontal dotted curves and a difference between t-values is getting smaller,

t c − t b ≃ 0.308, t d − t e ≃ 0.063 and t f − t g ≃ 0.015.

As seen above, the delay differential Equation (13) describes dynamic behavior of y ( t ) for t ∈ I k b for k = 3 , 4 , 5 while the linear ordinary Equation (5) determines the form of y ( t ) for t ∈ I k a or I k c . For k ≥ 6 , the same types of the solutions are obtained and the size of I k b shrinks, implying that as k increases, the resultant shape of the solution form of y ( t ) approaches the sawtooth shape at whose vertices y ˙ ( t ) jumps.

We now turn attention to the phase diagram in the ( y ˙ ( t ) , y ( t ) ) plane. The boundary values of each trajectory that have been obtained are summarized in the following table and plotted in

( 1 ) = ( 1.6 , − 2 ) ( 2 ) = ( 0.72 , − 0.90 ) ( 3 ) = ( 7.12 , − 0.90 ) ( 4 ) = ( 0.90 , 2.47 ) ( 5 ) = ( 16.02 , 2.4 ) ( 6 ) = ( 12.75 , 6 , 57 ) ( 7 ) = ( − 3.31 , 8.62 ) ( 8 ) = ( 11.10 , 8.62 ) ( 9 ) = ( 7.24 , 13.46 ) ( 10 ) = ( − 15.34 , 11.68 ) ( 11 ) = ( − 13.54 , 9.42 ) ( 12 ) = ( 10.46 , 9.42 ) ( 13 ) = ( 6.68 , 14.15 ) ( 14 ) = ( − 17.03 , 13.79 ) ( 15 ) = ( − 12.61 , 8.27 )

Point (1) is the starting point of interval I 1 at t = 0 and the delay Equation (13) transports it to point (2) at t = 1 . At the beginning of the second interval I 2 , y ˙ ( t ) jumps, which makes the horizontal move of point (2) to point (3) that arrives at point (4) at the end of I 2 . At the beginning of the third interval I 3 , y ˙ ( t ) jumps again and point (4) horizontally shifts to point (5) at which the dynamic system is changed to the linear Equation (5) with d = 3 n making a move along the upper downward line to point (6) as t proceeds from t = t 2 to

t = t a θ . The dynamic system is changed back to the delay equation at t = t a θ and then the dotted trajectory leaves the upper red curve heading to point (7). This is because for t ∈ [ t a θ , t 3 ] , the delay equation with y ˙ ( t − θ ) ∈ I M controls dynamic behavior. On the way, the y ˙ ( t ) curve crosses the upper and lower horizontal dotted curves as seen in

This paper presented Goodwin’s nonlinear accelerator model augmented with investment delay in continuous time scales. Assuming a piecewise linear investment function and specifying the values of the model’s parameters, explicit forms of sawtooth oscillations were derived when the initial value of the constant initial function was selected in the neighborhood of the steady state. Otherwise the same was done for Goodwin oscillation. With these numerical results, the paper exhibited valuable insights into the macro dynamics of market economies: the delay nonlinear accelerator-multiplier mechanism can be a source of various types of business cycles; economies starting in the neighborhood of the steady state could achieve regular ups and downs while economies starting away from the steady state presented persistent and irregular cycles.

Matsumoto, A., Nakayama, K. and Szidarovszky, F. (2018) Goodwin Accelerator Model Revisited with Piecewise Linear Delay Investment. Advances in Pure Mathematics, 8, 178-217. https://doi.org/10.4236/apm.2018.82010

In this Appendix, we provide mathematical underpinnings for Goodwin oscillations. Since the investment delay could make y ( t ) kinked and y ˙ ( t ) discontinuous at t n = n θ for integer n, the time interval for t ≥ 0 is reconstructed as the union of unit intervals I n = [ t n , t n + 1 ] for n and then a dynamic equation defined over interval I n is solved to obtain explicit forms of time trajectory and its derivative. Dynamic equation is solved with successive integration in which an initial point or function is the solution of dynamic equation defined in the proceeding subinterval.

Interval 0: I 0 = [ t 0 , t 1 ] where t 0 = 0 and t 1 = 1 .

The initial function Φ ( t − θ ) determines dynamics for t ≤ 0 . Since φ [ Φ ˙ ( t − θ ) ] = 0 by Assumption 2, solving ε y ˙ ( t ) + ( 1 − α ) y ( t ) = 0 presents explicit forms of the solution and its derivative

y 0 ( t ) = e − 1 − α ε t K 0 and y ˙ 0 ( t ) = e − 1 − α ε t ( − 1 − α ε K 0 ) with K 0 = 3 n ν .

and as can be seen in

y ˙ 0 ( t ) < − n ν for t ∈ I 0 . (A-1)

Derivation of (G-I)

Interval 1: I 1 = [ t 1 , t 2 ] where t 2 = 2 .

(A-1) implies φ [ y ˙ 0 ( t − θ ) ] = − n for t ∈ I 1 and then (8) with d = − n are

y 1 ( t ) = e − 1 − α ε t K 1 + ( − n 1 − α ) and y ˙ 1 ( t ) = e − 1 − α ε t ( − 1 − α ε K 1 )

where solving y 0 ( t 1 ) = y 1 ( t 1 ) gives

K 1 = K 0 + n 1 − α e − 1 − α ε t 1 ≃ 21.19.

and the following holds,

y ˙ 1 ( t ) < − n ν for t ∈ I 1 . (A-2)

Intervals 2 and 3: I i = [ t i , t i + 1 ] where t i + 1 = t i + θ for i = 2 , 3 .

In the same way as in interval I 1 , (8) with d = − n leads to the identical forms of y i ( t ) and y ˙ i ( t ) for t ∈ I i as the ones defined in I 1 ,

y i ( t ) = y 1 ( t ) and y ˙ i ( t ) = y ˙ 1 ( t ) with K i = K 1 .

Notice that the red curve crosses the lower dotted horizontal line from below in interval I 3 . Solving y ˙ 3 ( t ) = − n / ν for t gives

t a ≃ 3.03

with which the following inequalities hold

y ˙ 3 ( t ) < − n ν for t < t a and − n ν < y ˙ 3 ( t ) < 3 n ν for t a < t ≤ t 4 . (A-3)

Interval 4: I 4 = [ t 4 , t 5 ] where t 4 = 4 and t 5 = 5 .

Due to (A-3), t I = t a + θ ≃ 4.03 divides interval I 4 into two subintervals, I 4 a = [ t 4 , t I ] and I 4 b = [ t I , t 5 ] . First, (2) with φ [ y ˙ 3 ( t − θ ) ] = − n for t ∈ I 4 a presents the same forms

y 4 a ( t ) = y 1 ( t ) and y ˙ 4 a ( t ) = y ˙ 1 ( t ) with K 4 a = K 1 .

So far we have seen that in I I = I 1 ∪ I 2 ∪ I 3 ∪ I 4 a = [ t S , t I ] with t S = t 1 , a time trajectory of y ( t ) is described by

y I ( t ) = e − 1 − α ε t K I + ( − n 1 − α )

where y I ( t ) = y 1 ( t ) and K I = K 1 . A form of y ˙ I ( t ) is obtained by time-differentiating y I ( t ) and is identical with y ˙ 1 ( t ) . y I ( t ) and y ˙ I ( t ) construct system (G-I) defined in Section 4.1.

Derivation of (G-II)

On the other hand for t ∈ I 4 b , the investment is delayed and (12) with φ [ y ˙ 3 ( t − θ ) ] = ν y ˙ 3 ( t − θ ) is written as

y ˙ ( t ) + 1 − α ε y ( t ) = ν ε e − 1 − α ε ( t − θ ) ( − 1 − α ε K 3 ) .

Multiplying both sides by the term e 1 − α ε t and arranging the terms present

d d t [ y ( t ) e 1 − α ε t ] = e − 1 − α ε t ( − ν ( 1 − α ) ε 2 K 3 e − 1 − α ε θ e 1 − α ε t ) .

Integrating both sides yields

y ( t ) e 1 − α ε t = ( − ν ( 1 − α ) ε 2 K 3 e − 1 − α ε θ ) t + K 4 b .

Thus the form of the solution is

y 4 b ( t ) = e − 1 − α ε t ( α 1 i v t + α 0 i v )

where

α 1 i v = − ν ( 1 − α ) ε 2 K 3 e − 1 − α ε θ and α 0 i v = K 4 b .

The integral constant K 4 b is obtained by solving y 4 a ( t I ) = y 4 b ( t I ) ,

K 4 b ≃ 440.938.

The derivative of y ( t ) is

y ˙ 4 b ( t ) = e − 1 − α ε t ( β 1 i v t + β 0 i v )

where

β 1 i v = − 1 − α ε α 1 i v and β 0 i v = α 1 i v − 1 − α ε α 1 i v .

It can be checked that

− n ν < y ˙ 4 ( t ) < 3 n ν for t ∈ I 4 (A-4)

where

y ˙ 4 ( t ) = y ˙ 4 a ( t ) for t ∈ I 4 a and y ˙ 4 ( t ) = y ˙ 4 b ( t ) for t ∈ I 4 b .

Interval 5: I 5 = [ t 5 , t 6 ] where t 6 = 6 .

As in interval I 4 , the threshold value t I I = t I + θ ≃ 5.03 divides interval I 5 into two subintervals, I 5 a = [ t 5 , t I I ] and I 5 b = [ t I I , t 6 ] . For t ∈ I 5 a , Equation (12) with φ [ y ˙ 4 a ( t − θ ) ] = ν y ˙ 4 a ( t − θ ) leads to the solution and its derivative that are the same as the ones obtained in I 4 b ,

y 5 a ( t ) = y 4 b ( t ) and y ˙ 5 a ( t ) = y 4 b ( t ) with K 5 a = K 4 b .

Therefore time trajectories for t ∈ I I I = I 4 b ∪ I 5 a = [ t I , t I I ] are described by

y I I ( t ) = y 4 b ( t ) and y ˙ I I ( t ) = y ˙ 4 b ( t )

both of which form (G-II) defined in Section 4.1 where α j i v and β j i v for j = 0 , 1 are written as α j I I and β j I I .

Derivation of (G-III)

Dynamic Equation (12) with φ [ y ˙ 4 b ( t − θ ) ] = ν y ˙ 4 b ( t − θ ) for t ∈ I 5 b has a solution

y 5 b ( t ) = e − 1 − α ε t ( α 2 v t 2 + α 1 v t + α 0 v )

where

α 2 v = ν ε e 1 − α ε θ β 1 i v 2 α 1 v = ν ε e 1 − α ε θ ( β 0 i v − θ β 1 i v ) α 0 v = K 5 b

Solving y 5 a ( t I I ) = y 5 b ( t I I ) presents

K 5 b ≃ 140446.6.

Differentiating y 5 b ( t ) gives

y ˙ 5 b ( t ) = e − 1 − α ε t ( β 2 v t 2 + β 1 v t + β 0 v )

where

β 2 v = − 1 − α ε α 2 v , β 1 v = 2 α 2 V − 1 − α ε α 1 v , β 0 v = α 1 V − 1 − α ε α 0 v .

Since the y ˙ 5 b ( t ) curve crosses the upper horizontal line from below at

t b ≃ 5.19395,

we then have

− n ν < y ˙ 5 a ( t ) < 3 n ν for t 5 ≤ t ≤ t I I , − n ν < y ˙ 5 b ( t ) < 3 n ν for t I I ≤ t ≤ t b , y ˙ 5 b ( t ) > 3 n ν for t b ≤ t ≤ t 6 . (A-5)

Interval 6: I 6 = [ t 6 , t 7 ] where t 7 = 7 .

Due to the two values, t I I and t b , we define two threshold values t I I I = t I I + θ ≃ 6.03 and t I V = t b + θ ≃ 6.19 , both of which then divide interval I 6 into three subintervals, I 6 a = [ t 6 , t I I I ] , I 6 b = [ t I I I , t I V ] and I 6 c = [ t I V , t 7 ] . Accordingly conditions in (A-5) determine the induced investment as

φ [ y ˙ ( t − θ ) ] = { φ [ y ˙ 5 a ( t − θ ) ] = ν y ˙ 5 a ( t − θ ) for t ∈ I 6 a , φ [ y ˙ 5 b ( t − θ ) ] = ν y ˙ 5 b ( t − θ ) for t ∈ I 6 b , φ [ y ˙ 5 b ( t − θ ) ] = 3 n for t ∈ I 6 c .

In consequence, the form of the solution and its derivative in I 6 a are

y 6 a ( t ) = y 5 b ( t ) and y ˙ 6 a ( t ) = y ˙ 5 b ( t ) .

Therefore blue and red trajectories for t ∈ I I I I = I 5 b ∪ I 6 a = [ t I I , t I I I ] are described by

y I I I ( t ) = y 5 b ( t ) and y ˙ I I I ( t ) = y ˙ 5 b ( t )

both of which form (G-III) defined in Section 4.1 where α j v and β j v for j = 0 , 1 , 2 are written as α j I I I and β j I I I .

Derivation of (G-IV)

For t ∈ I 6 b , successive integral leads to the solution

y 6 b ( t ) = e − 1 − α ε t ( α 3 v i t 3 + α 2 v i t 2 + α 1 v i t + α 0 v i )

where

α 3 v i = ν ε e 1 − α ε θ β 2 v 2 , α 2 v i = ν ε e 1 − α ε θ β 1 v − 2 θ β 2 v 2 , α 1 v i = β 0 v − θ β 1 v + θ 2 β 2 v , α 0 v i = K 6 b .

Solving y 6 a ( t I I I ) = y 6 b ( t I I I ) yields

K 6 b ≃ − 155088.

Differentiating y 6 b ( t ) with respect to t is, after arranging the terms, it can be written as

y ˙ 6 b ( t ) = e − 1 − α ε t ( β 3 v i t 3 + β 2 v i t 2 + β 1 v i t + β 0 v i )

where

β 3 v i = − 1 − α ε α 3 v i , β 2 v i = 3 α 3 V − 1 − α ε α 2 v i , β 1 v i = 2 α 2 V − 1 − α ε α 1 v i , β 0 v i = α 1 V − 1 − α ε α 0 v i .

Therefore blue and red trajectories in I I V = I 6 b are described by

y I V ( t ) = y 6 b ( t ) and y ˙ I V ( t ) = y ˙ 6 b ( t )

both of which form (G-IV) defined in Section 4.1 where α j v i and β j v i for j = 0 , 1 , 2 , 3 are written as α j I V and β j I V .

Derivation of (G-V)

For t ∈ I 6 c , Equation (9) implies a form of the solution,

y 6 c ( t ) = e − 1 − α ε t K 6 c + 3 n 1 − α y ˙ 6 c ( t ) = e − 1 − α ε t ( − 1 − α ε K 6 c )

where solving y 6 b ( t I V ) = y 6 c ( t I V ) gives

K 6 c ≃ − 2839.05.

Interval 7: I 7 = [ t 7 , t 8 ] where t 8 = 8 .

Since Equation (2) implies φ [ y ˙ 6 n ( t − θ ) ] = 3 n for n = a , b , c and t ∈ I 7 , Equation (9) implies that

y 7 ( t ) = y 6 c ( t ) , y ˙ 7 ( t ) = y ˙ 6 c ( t ) .

Notice that the y ˙ 7 ( t ) curve intersects the horizontal dotted line at 3 n / ν from above at the following point,

t c ≃ 7.78001.

Thereby,

y ˙ 7 ( t ) > 3 n ν for t 7 ≤ t < t c , − n ν < y ˙ 7 ( t ) < 3 n ν for t c < t ≤ t 8 . (A-7)

Interval 8: I 8 = [ t 8 , t 9 ] where t 9 = 9 .

The threshold value in interval I 7 defines a new threshold value t V = t c + θ ≃ 8.78 in interval I 8 that divides interval I 8 into two subintervals, I 8 a = [ t 8 , t V ] and I 8 b = [ t V , t 9 ] . Since φ [ y ˙ 7 ( t − θ ) ] = 3 n for t ∈ I 8 a , (9) implies

y 8 a ( t ) = y 7 ( t ) and y ˙ 8 a ( t ) = y ˙ 7 ( t )

Therefore trajectories in I V = I 6 c ∪ I 7 ∪ I 8 a = [ t I V , t V ] are described by

y V ( t ) = y 6 c ( t ) = y 7 ( t ) = y 8 a ( t ) and y ˙ V ( t ) = y ˙ 6 c ( t ) = y ˙ 7 ( t ) = y ˙ 8 a ( t ) .

y V ( t ) and y ˙ V ( t ) construct (G-V) defined in Section 4.1 where K 6 c is replaced with K V .

Derivation of (G-VI)

On the other hand, φ [ y ˙ 7 ( t − θ ) ] = ν y ˙ 7 ( t − θ ) for t ∈ I 8 b , successive integration implies that the solution of y ( t ) has the form,

y 8 b ( t ) = e − 1 − α ε t ( α 1 v i i i t + α 0 v i i i )

where

α 1 v i i i = − ν ( 1 − α ) ε 2 K 8 e 1 − α ε θ , α 0 v i i i = K 8 b ≃ − 155088.

Time differentiation of y 8 b ( t ) is

y ˙ 8 b ( t ) = e − 1 − α ε t ( β 1 v i i i t + β 0 v i i i )

where

β 1 v i i i = − 1 − α ε α 1 v i i i , β 0 v i i i = α 1 v i i i − 1 − α ε α 0 v i i i .

It can be checked that

− n ν < y ˙ 8 a ( t ) < 3 n ν for t ∈ I 8 a − n ν < y ˙ 8 b ( t ) < 3 n ν for t ∈ I 8 b . (A-8)

Interval 9: I 9 = [ t 9 , t 10 ] where t 10 = t 9 + θ .

The threshold value t V I = t V + θ divides interval I 9 into two subintervals, I 9 a = [ t 9 , t V I ] and I 9 b = [ t V I , t 9 ] . Since the first equation of (A-8) implies φ [ y ˙ 8 ( t − θ ) ] = ν y ˙ 8 a ( t − θ ) for t ∈ I 9 a , the solution of (12) is

y 9 a ( t ) = y 8 b ( t ) and y ˙ 9 a ( t ) = y ˙ 8 b ( t )

Therefore trajectories in I V I = I 8 b ∪ I 9 a = [ t V , t V I ] are described by

y V I ( t ) = y 8 b ( t ) = y 9 a ( t ) and y ˙ V I ( t ) = y ˙ 8 b ( t ) = y ˙ 9 a ( t )

both of which form (G-VI) defined in Section 4.1where α j v i i i and β j v i i i for j = 0 , 1 , are written as α j V I and β j V I .

Derivation of (G-VII)

On the other hand, φ [ y ˙ 8 ( t − θ ) ] = ν y ˙ 8 b ( t − θ ) for t ∈ I 9 b implies the following form of the solution,

y 9 b ( t ) = e − 1 − α ε t ( α 2 i x t 2 + α 1 i x t + α 0 i x )

where

α 2 i x = ν ε e 1 − α ε θ β 1 v i i i 2 , α 1 i x = ν ε e 1 − α ε θ ( β 0 v i i i − θ β 1 v i i i ) , α 0 i x = K 9 b ≃ − 7.0415 × 10 6

and

y ˙ 9 b ( t ) = e − 1 − α ε t ( β 2 i x t 2 + β 1 i x t + β 0 i x )

where

β 2 i x = − 1 − α ε α 3 i x , β 1 i x = 2 α 2 I X − 1 − α ε α 1 i x , β 0 i x = α 1 I X − 1 − α ε α 0 i x .

It is to be noticed that the y ˙ 6 a ( t ) curve intersects the horizontal line at − n / ν from above at

t d ≃ 9.04967

with which

y ˙ 9 a ( t ) > − n ν for t 9 ≤ t < t d

and

( y ˙ 9 a ( t ) , y ˙ 9 b ( t ) ) < − n ν for t d < t ≤ t 9 .

Interval 10: I 10 = [ t 10 , t E ] where t E ≃ 11.05729 .

The threshold value t V I I = t d + θ ≃ 10.04 divides interval I 10 into two subintervals, I 10 a = [ t 10 , t V I I ] and I 10 b = [ t V I I , t E ] . Since φ [ y ˙ 9 a ( t − θ ) ] = ν y ˙ 9 a ( t − θ ) for t ∈ I 10 a , we have

y 10 a ( t ) = y 9 b ( t ) and y ˙ 10 a ( t ) = y ˙ 9 b ( t )

Therefore time trajectories in I V I I = I 9 b ∪ I 10 a = [ t V I , t V I I ] are described by

y V I I ( t ) = y 9 b ( t ) = y 10 a ( t ) and y ˙ V I I ( t ) = y ˙ 9 b ( t ) = y ˙ 10 a ( t )

both of which form (G-VII) defined in Section 4.1 where α j i x and β j i x for j = 0 , 1 are written as α j V I I and β j V I I .

Derivation of (G-VIII)

On the other hand φ [ y ˙ 9 a ( t − θ ) ] = φ [ y ˙ 9 b ( t − θ ) ] = − n for t ∈ I 10 b implies that

y 10 b ( t ) = e − 1 − α ε t K 10 b + ( − n 1 − α ) with K 10 b ≃ 66133.895

and

y ˙ 10 b ( t ) = e − 1 − α ε t ( − 1 − α ε K 10 b ) .

The end point t ˜ e is obtained by solving y 10 b ( t ) = y 1 ( t s ) for t. Therefore time trajectories in I V I I I = I 10 b are given by

y V I I I ( t ) = y 10 b ( t ) and y ˙ V I I I ( t ) = y ˙ 10 b ( t )

y V I I I ( t ) and y ˙ V I I I ( t ) form (G-VIII) defined in Section 4.1 where K 10 b is denoted by K V I I I . A cycle starts at t = t S y I ( t S ) ≃ 2.02 and finishes at t = t E with y V I I I ( t E ) ≃ 2.02 . The length of this cycle is equal to t E − t S that is about 10.06.

Our main aim of this appendix is to analytically reproduce these numerical results of sawtooth oscillations to understand why a trajectory y ( t ) has kinks (alternatively, its derivative y ˙ ( t ) makes jumps). To this end, we start to divide the whole interval I = [ 0 , 5 ] into five subintervals with respect to the length of delay θ = 1 ,

I i = [ t i − 1 , t i ] for i = 1,2,3,4,5

with t i = i and t i = t i − 1 + θ .

Derivative of (S-I)

Interval I: I 1 = [ t 0 , t 1 ] where t 0 = 0 and t 1 = 1 .

Equation (13) with the constant initial function y 0 ( t − θ ) = Φ ( t − θ ) yields the solution of the form

y 1 ( t ) = e − 1 − α ε t ⋅ K 1 with K 1 = − 2

and differentiation gives its derivative form

y ˙ 1 ( t ) = e − 1 − α ε t ⋅ ( − 1 − α ε K 1 ) .

Both of which form (S-I) defined in Section 5.1. Since y ˙ 1 ( t 0 ) = 1.6 , y ˙ 1 ( t 1 ) ≃ 0.719 and y ˙ 1 ( t ) > 0 for t ∈ I 1 , y ˙ 1 ( t ) stays in the middle region,

− n k ≤ y ˙ 1 ( t ) ≤ 3 n k for t ∈ I 1 . (A-9)

Derivative of (S-II)

Interval II: I 2 = [ t 1 , t 2 ] where t 2 = t 1 + θ .

Due to (2) and (A-9),

φ [ y ˙ 1 ( t − θ ) ] = ν ⋅ y ˙ 1 ( t − θ )

which is substituted into (13) to obtain,

y ˙ ( t ) + 1 − α ε y ( t ) = Q ( t ) with Q ( t ) = − k ( 1 − α ) ε 2 K 1 e − 1 − α ε θ e − 1 − α ε t .

Since this equation can be written as

d d t ( y ( t ) e 1 − α ε t ) = Q ( t ) e 1 − α ε t

Integrating both sides and arranging the terms present the solution of y ( t ) , denoted as y 2 ( t ) ,

y 2 ( t ) = e − 1 − α ε t [ ( − k ( 1 − α ) ε 2 K 1 e 1 − α ε θ ) t + K 2 ] .

Since the trajectory of y 2 ( t ) is piecewise continuous, solving y 1 ( t 1 ) = y 2 ( t 1 ) presents

K 2 ≃ − 16.244.

The form of y 2 ( t ) is rewritten as

y 2 ( t ) = e − 1 − α ε t ( α 1 I I t + α 0 I I )

with

α 1 I I = − k ( 1 − α ) ε 2 K 1 e 1 − α ε θ ≃ 14.244 and α 0 I I = K 2

A time derivative of y 2 ( t ) is

y ˙ 2 ( t ) = e − 1 − α ε t ( β 1 I I t + β 0 I I )

with

β 1 I I = − k ( 1 − α ) 2 ε 3 e − 1 − α ε θ K 1 ≃ − 11.395 and β 0 I I = − 1 − α ε ( k ε e 1 − α ε θ K 1 + K 2 ) ≃ 27.238.

These y 2 ( t ) and y ˙ 2 ( t ) form (S-II) defined in Section 5.1. Under Assumption 1, we calculate the boundary values of interval I 2 ,

y 2 ( t 1 ) ≃ − 0.899 , y ˙ 2 ( t 1 ) ≃ 7.119 and y 2 ( t 2 ) ≃ 2.472 and y ˙ 2 ( t 2 ) ≃ 0.898.

The red curve crosses the upper horizontal dotted line once from above at point

t a ≃ 1.286

with which the following inequalities hold, as is seen in I 2 in

y ˙ 2 ( t ) > 3 n ν for t 1 ≤ t < t a − n ν < y ˙ 2 ( t ) < 3 n ν for t a < t ≤ t 2 . (A-10)

Derivations of (G-IIIa), (G-IIIb) and (G-IIIc)

Interval III: I 3 = [ t 2 , t 3 ] where t 3 = t 2 + θ .

Due to the value of t a , the interval I 3 is divided into two subintervals, I 3 a = [ t 2 θ , t a ) and I 3 b = ( t a θ , t 3 ] where t a θ = t a + θ . Delay investment is differently determined according to conditions in (A-10),

φ [ y ˙ 2 ( t − θ ) ] = { 3 n if t ∈ I 3 a k y ˙ 2 ( t − θ ) if t ∈ I 3 b . (18)

different dynamic systems are defined on different subintervals. So we derive the solution of the differential equation in each subinterval.

Interval III-1: t 2 ≤ t < t a θ .

In this subinterval, Equation (8) with the first equation of (A-11) yields the solution

y 3 ( t ) = e − 1 − α ε t K 3 + 3 n 1 − α

and its derivative

y ˙ 3 ( t ) = − 1 − α ε K 3 e − 1 − α ε t

where solving y 3 ( t 2 ) = y 2 ( t 2 ) gives

K 3 ≃ − 99.2.

These two functions form (S-IIIa) where y 3 ( t ) and y ˙ 3 ( t ) are denoted by y 3 a ( t ) and y ˙ 3 a ( t ) . Boundary values of this interval are

y 3 a ( t 2 ) ≃ 2.472, y ˙ 3 a ( t 2 ) ≃ 16.02 and y 3 a ( t a θ ) ≃ 6.565, y ˙ 3 a ( t a θ ) ≃ 12.748.

Interval III-2: t a θ < t ≤ t 3 .

In this interval, we have Equation (13) with the second equation of (A-11) that is rewritten as

y ˙ ( t ) + 1 − α ε y ( t ) = Q ( t )

where

Q ( t ) = k ε y ˙ 2 ( t − θ ) = k ε 2 e − 1 − α ε ( t − θ ) { β 1 I I ( t − θ ) + β 0 I I } .

Rewriting the dynamic equation as

d d t ( y ( t ) e 1 − α ε t ) = Q ( t ) e 1 − α ε t

and integrating both sides give the following form of a solution,

y 4 ( t ) = e − 1 − α ε t { k ε e 1 − α ε θ [ β 1 I I 2 t 2 + ( β 0 I I − θ β 1 I I ) t ] + K 4 }

where solving y 3 ( t a θ ) = y 4 ( t a θ ) gives

K 4 ≃ − 480.253.

Then the form of y 4 ( t ) is rewritten as

y 4 ( t ) = e − 1 − α ε t ( α 2 I I I t 2 + α 1 I I I t + α 0 I I I )

where

α 2 I I I = k ε e 1 − α ε θ β 1 I I 2 ≃ − 50.719

α 1 I I I = k ε e 1 − α ε θ ( β 0 I I − θ β 1 I I ) ≃ 343.918 ,

and

α 0 I I I = K 4 .

A derivative of y 4 ( t ) is

y ˙ 4 ( t ) = e − 1 − α ε t ( β 2 I I I t 2 + β 1 I I I t + β 0 I I I )

where

β 2 I I I = − 1 − α ε α 2 I I I ≃ − 22.79 ,

β 1 I I I = 2 α 2 I I I − 1 − α ε α 1 I I I ≃ − 255.97

and

β 0 I I I = α 1 I I I − 1 − α ε α 0 I I I ≃ 559.71.

Boundary values are

y 4 ( t a θ ) ≃ 6.565, y ˙ 4 ( t a θ ) ≃ 12.748

and

y 4 ( t 3 ) ≃ 8.621, y ˙ 4 ( t 3 ) ≃ − 3.309.

These y 4 ( t ) and y ˙ 4 ( t ) form (S-IIIb) in which y 4 ( t ) and y ˙ 4 ( t ) are denoted as y 3 b ( t ) and y ˙ 3 b ( t ) . As is seen in

t b ≃ 2.535 with y ˙ 3 b ( t b ) = 3 n k

t c ≃ 2.843 with y ˙ 3 b ( t c ) = − n k .

It is apparent from

y ˙ 3 a ( t ) > 3 n k for t 3 ≤ t < t a θ , y ˙ 3 b ( t ) > 3 n k for t a θ ≤ t < t b , 3 n k > y ˙ 3 b ( t ) > − n k for t b < t < t c , − n k > y ˙ 3 c ( t ) for t c < t ≤ t 3 . (A-12)

Derivation of (S-IVa), (S-IVb) and (S-IVc)

Interval IV: I 4 = [ t 3 , t 4 ] where t 4 = t 3 + θ .

Due to the properties described in (A-12), interval I 4 is divided into three subintervals by t b θ = t b + θ and t c θ = t c + θ in which

φ [ y ˙ 3 a ( t − θ ) ] = φ [ y ˙ 3 b ( t − θ ) ] = 3 n for t 3 ≤ t < t b θ (A-13)

φ [ y ˙ 3 b ( t − θ ) ] = k y ˙ 3 b ( t − θ ) for t b θ < t < t c θ (A-14)

and

φ [ y ˙ 3 b ( t − θ ) ] = − n for t c θ < t ≤ t 4 . (A-15)

Interval IV-1: t 3 ≤ t < t b θ

Equation (2) with (A-13) implies that the solution of the differential equation

ε y ˙ ( t ) + ( 1 − α ) y ( t ) = 3 n

is given by

y 5 ( t ) = e − 1 − α ε t K 5 + 3 n 1 − α with K 5 ≃ − 152.994

where K 5 solves

y 4 ( t 3 ) = y 5 ( t 3 ) .

A derivative of y 5 ( t ) is

y ˙ 5 ( t ) = e − 1 − α ε t ( − 1 − α ε K 5 ) with − 1 − α ε K 5 ≃ 122.395.

These y 5 ( t ) and y ˙ 5 ( t ) form (S-IVa) in which y 5 ( t ) and y ˙ 5 ( t ) are denoted as y 4 a ( t ) and y ˙ 4 a ( t ) We then have the boundary values of y 4 a ( t ) of interval [ t 3 , t b θ ] ,

y 4 a ( t 3 ) ≃ 8.621, y ˙ 4 a ( t 3 ) ≃ 11.103 and y 4 a ( t b θ ) ≃ 13.456, y ˙ 4 a ( t b θ ) ≃ 7.236.

Interval IV-2: t b θ < t < t c θ .

Rewriting the delay differential Equation (2) with (A-14) as

y ˙ ( t ) + 1 − α ε y ( t ) = Q ( t )

where

Q ( t ) = k ε y ˙ 4 ( t − θ ) = k ε e − 1 − α ε ( t − θ ) { β 2 I I I ( t − θ ) 2 + β 1 I I I ( t − θ ) + β 0 I I I } .

Successive integration yields the solution,

y 6 ( t ) = e − 1 − α ε t { ∫ k ε e 1 − α ε θ [ β 2 I I I t 2 + ( β 1 I I I − 2 θ β 2 I I I ) t + ( β 0 I I I − θ β 1 I I I + θ 2 β 2 I I I ) ] d t + K 6 }

or

y 6 ( t ) = e − 1 − α ε t { α 3 I V t 3 + α 2 I V t 2 + α 1 I V t + α 0 I V }

with

α 3 I V = k ε e 1 − α ε θ β 2 I I I 3 ≃ 180.604 , α 2 I V = k ε e 1 − α ε θ β 1 I I I − 2 θ β 2 I I I 2 ≃ − 2037.36 , α 1 I V = k ε e 1 − α ε θ ( β 0 I I I − θ β 1 I I I + θ 2 β 2 I I I ) ≃ 10195.4 , α 0 I V = K 6 ≃ − 15672.428

where K 6 solves

f 5 ( t b θ ) = f 6 ( t b θ ) .

A derivative of y 6 ( t ) is

y ˙ 6 ( t ) = e − 1 − α ε t { β 3 I V t 3 + β 2 I V t 2 + β 1 I V t + β 0 I V }

with

β 3 I V = − 1 − α ε α 3 I V ≃ − 96.322 , β 2 I V = 3 α 3 I V − 1 − α ε α 2 I V ≃ 1991.1 , β 1 I V = 2 α 2 I V − 1 − α ε α 1 I V ≃ − 12231 , β 0 I V = α 1 I V − 1 − α ε α 0 I V ≃ 22733.3

These y 6 ( t ) and y ˙ 6 ( t ) form (S-IVb) in which y 6 ( t ) and y ˙ 6 ( t ) are denoted as y 4 b ( t ) and y ˙ 4 b ( t ) . The boundary values of the interval are

y 4 b ( t b θ ) ≃ 13.455, y ˙ 4 b ( t b θ ) ≃ 7.236 and y 4 b ( t c θ ) ≃ 11.677, y ˙ 4 b ( t c θ ) ≃ − 15.342.

Further the downward-sloping curve of y ˙ 4 b ( t ) intersects each of the two dotted horizontal lines at 3 n / k and − n / k at the points

t d ≃ 3.561 with y ˙ 4 b ( t d ) = 3 n k

and

t e ≃ 3.624 with y ˙ 4 b ( t e ) = − n k .

Interval IV-3: t c θ < t ≤ t 4

Equation (A-15) implies that the dynamic Equation (2) has the following forms for the solution y 7 ( t ) and its derivative y ˙ 7 ( t )

y 7 ( t ) = e − 1 − α ε t K 7 − n 1 − α with K 7 ≃ 415.092

an

y ˙ 7 ( t ) = e − 1 − α ε t ( − 1 − α ε K 7 ) with − 1 − α ε K 7 ≃ − 340.074

where K 7 solves

y 6 ( t c θ ) = y 7 ( t c θ ) .

These y 7 ( t ) and y ˙ 7 ( t ) form (S-IVc) in which y 7 ( t ) and y ˙ 7 ( t ) are denoted as y 4 c ( t ) and y ˙ 4 c ( t ) . The boundary values of the interval are

y 4 c ( t c θ ) ≃ 11.671, y ˙ 4 c ( t c θ ) ≃ − 15.342 and y 4 c ( t 4 ) ≃ 9.420, y ˙ 4 c ( t 4 ) ≃ − 13.536

It is seen that

y ˙ 4 a ( t ) > 3 n k for t 3 ≤ t < t b θ , y ˙ 4 b ( t ) > 3 n k for t b θ ≤ t < t d , 3 n k > y ˙ 4 b ( t ) > − n k for t d < t < t e , − n k > y ˙ 4 b ( t ) for t e < t ≤ t 4 .

Derivation of (S-V)

Interval V: I 5 = [ t 4 , t 5 ]

Due to the threshold values, t A I V and t B I V , interval I 5 is divided into three subintervals by t A V = t A I V + θ and t B V = t B I V + θ in which

φ [ f ˙ 5 ( t − θ ) ] = φ [ f ˙ 6 ( t − θ ) ] = 3 n for t 4 ≤ t < t A V (A-16)

φ [ f ˙ 6 ( t − θ ) ] = k f ˙ 6 ( t − θ ) for t A V < t < t B V (A-17)

and

φ [ f ˙ 7 ( t − θ ) ] = − n for t B V < t ≤ t 5 . (A-18)

Repeating the same procedure done just above, we can derive the explicit forms of a solution of the delay dynamic equation.

Interval V-1: t 4 ≤ t < t A V .

Equation (8) with (A-16) presents the following forms of the solutions,

f 8 ( t ) = e − 1 − α ε t K 8 + 3 n 1 − α with K 8 ≃ − 320.88

and

f ˙ 8 ( t ) = e − 1 − α ε t ( − 1 − α ε K 8 ) with − 1 − α ε K 8 ≃ 256.704

where K 8 solves f 7 ( t 4 ) = f 8 ( t 4 ) . These f 8 ( t ) and f ˙ 8 ( t ) form (S-Va) in which f 8 ( t ) and f ˙ 8 ( t ) are replaced with y 5 a ( t ) and y ˙ 5 a ( t ) .

Interval V-2: t A V < t < t B V .

Applying successive integration to Equation (2) with (A-17) yields the following forms of the solutions,

f 9 ( t ) = e − 1 − α ε t { α 4 V t 4 + α 3 V t 3 + α 2 V t 2 + α 1 V t + α 0 V }

with

α 4 V = k ε e 1 − α ε θ β 3 I V 3 ≃ − 214.369 α 3 V = k ε e 1 − α ε θ β 2 I V − 3 θ β 3 I V 3 ≃ 6765.83 , α 2 V = k ε e 1 − α ε θ β 1 I V − 2 θ β 2 I V + 3 θ 2 β 3 I V 2 ≃ − 73452.5 , α 1 V = k ε e 1 − α ε θ ( β 0 I V − θ β 1 I V + θ 2 β 2 I V − θ 3 β 3 I V ) ≃ 329841 , α 0 V = K 9 ≃ − 525029

where K 9 solves f 8 ( t A I V ) = f 9 ( t A I V ) . A derivative of f 6 ( t ) is

f ˙ 9 ( t ) = e − 1 − α ε t { β 4 V t 4 + β 3 V t 3 + β 2 V t 2 + β 1 V t + β 0 V }

with

β 4 V = − 1 − α ε α 4 I V ≃ 171.495 , β 3 V = 4 α 4 V − 1 − α ε α 3 V ≃ − 9270.135 , β 2 V = 3 α 3 V − 1 − α ε α 2 V ≃ 79059.451 , β 1 V = 2 α 2 V − 1 − α ε α 1 V ≃ − 410777.286 , β 0 V = α 1 V − 1 − α ε α 0 V ≃ 749863.287

These f 9 ( t ) and f ˙ 9 ( t ) form (S-Vb) in which f 9 ( t ) and f ˙ 9 ( t ) are replaced with y 5 b ( t ) and y ˙ 5 b ( t ) . As is seen in

t f ≃ 4.566 with y ˙ 5 b ( t f ) = 3 n k

t g ≃ 4.581 with y ˙ 5 b ( t g ) = − n k .

To avoid notational congestion in

Interval lV-3: t B V < t ≤ t 5

Equation (8) with (A-18) presents the following forms of the solutions,

f 10 ( t ) = e − 1 − α ε t K 10 − n 1 − α with K 10 ≃ 860.508

and

f ˙ 10 ( t ) = e − 1 − α ε t ( − 1 − α ε K 10 ) with − 1 − α ε K 10 ≃ − 688.406

where K 10 solves f 9 ( t B V ) = f 10 ( t B V ) . These f 10 ( t ) and f ˙ 10 ( t ) form (S-Vc) in which f 10 ( t ) and f ˙ 10 ( t ) are replaced with y 5 c ( t ) and y ˙ 5 c ( t ) .

It is seen that

y ˙ 5 a ( t ) > 3 n k for t 4 ≤ t < t d θ , y ˙ 5 b ( t ) > 3 n k for t d θ ≤ t < t f , 3 n k > y ˙ 5 b ( t ) > − n k for t f < t < t g , − n k > y ˙ 5 c ( t ) for t g < t ≤ t 5 .