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 Figure 1 in which y = f ( y ˙ ) is described by a mirror-imaged N-shaped curve in the ( y ˙ , y ) plane.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

¹Mathematica 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 Figure 1 are

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 Figure 2. A = B holds at the upper kinked point of the blue curve and C = D at the lower kinked point. The red trajectory from 0 to t 1 describes the movement of y ˙ ( t ) from point S to A. At time t 1 when the left-most red curve defined on interval [ 0, t 1 ] arrives at the upper horizontal dotted line, the red curve jumps straightly down to the starting point of the lower red curve defined on interval [ t 1 , t 2 ] , the point of which correspond to point B. At time t 2 , the red curve crosses the lower dotted line from below and the intersection corresponds to point C at which the red curve jumps straightly

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. Figure 2 illustrates the same dynamics of Figure 1 from a different view point.

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 Figure 1. Solutions in (7) describe the movement from point S to point A. At t = t 1 , the dynamic system (4) is switched to Equation (5) with d = − n . Equations in (8) with i = 2 presents explicit forms of the solution and its time derivative,

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 Figure 2. Since y ˙ 2 ( t 1 ) < − n / ν and y ˙ 2 ( t ) increases in t, solving y ˙ 2 ( t ) = − n / ν gives a necessary time to arrive at − n / ν ,

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 [1] and [2] . Needless to say, these particular values of the parameters are selected only for analytical simplicity and do not affect qualitative aspects of the results to be obtained.

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

In particular, Figure 1 and Figure 2 are illustrated under Assumption 1 and the initial values of y ( 0 ) = 2 and y ˙ ( 0 ) = 8 / 15 , where the pair ( y ˙ ( 0 ) , y ( 0 ) ) corresponds to point S in Figure 2. The critical times at which the system switching occurs are given

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 .

²in [4] , it is assumed that θ = 1 is one year delay. Under the specified values of the parameters, t i + 2 − t i ≃ 5.218 , implying the length is considered to be 5.218 years.

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.

³Goodwin 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)

[6] derive explicit conditions for stability/instability of the n-th order linear scalar neutral delay differential equation with a single delay. Since (13) is a special case of the n-th order equation, applying their result (i.e., Theorem 2.1) leads to the following: the real parts of the solutions of Equation (16) are positive for all θ > 0 if a > 1 . The first result on the fixed delay model is summarized as follows:

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. If the stability switches at θ = θ ¯ , then (16) must have a pair of pure conjugate imaginary roots with θ = θ ¯ . Thus to find the critical value of θ ¯ , we assume that λ = i ω , with ω > 0 , is a root of (16) for θ = θ ¯ > 0 . Substituting λ = i ω into (16), we have

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 Figure 3. For given value of y 0 , the dynamic system runs for 0 ≤ t ≤ T = 500 . The solution for t ≤ 450 are discarded to eliminate the initial disturbances and the maximum and minimum values of the resultant solutions for 450 ≤ t ≤ 500 are plotted against y 0 . The bifurcation parameter y 0 increases from −10 to 6 with increment of 0.01 and for each value of y 0 , the same calculation procedure is repeated. As is seen in Figure 3 and already pointed out by [5] , the delay dynamic system with the constant initial function has the two threshold initial values y 0 m a x and y 0 m i n such that the sawtooth oscillations arise for y 0 m i n ≤ y 0 ≤ y 0 m a x and so do the Goodwinian oscillations otherwise. These values depend on the length of the delay and are numerically determined as y 0 m a x ≃ 2.39 and y 0 m i n ≃ − 6.28 under θ = 1 . In the following, we first set y 0 = y 0 G = 4.5 and consider Goodwinian oscillations in Section 4 and then examine sawtooth oscillations with y 0 = y 0 S = − 2 in Section 5.

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

⁴The explicit forms are given in Appendix I.

where K I ≃ 21.19 . At, the red trajectory y ˙ I ( t ) crosses the lower horizontal dotted line at − n / ν and the crossing point is denoted by the left most green dot in Figure 4. The boundary values of this interval are

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 Figure 4, the blue trajectory is kinked and the red curve jumps downward at t = t S . This discontinuity is shown as follows. Let y 0 ( t ) and y ˙ 0 ( t ) be a solution and its derivative in interval [ 0, t S ] .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 Figure 4, the red curve y ˙ I I I ( t ) crosses the upper horizontal dotted line from below at t b ≃ 5.19 and y ˙ I I I ( t ) < 3 n / ν for t I I < t < t b . This crossing point is also denoted by the green dot.

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 Figure 4 to avoid the notational congestion. Due to the continuity of the blue and red trajectories at t = t I I I , y I I I ( t I I I ) = y I V ( t I I I ) and y ˙ I I I ( t I I I ) = y ˙ I V ( t I I I ) hold. The boundary values for the right end point of interval I I V are calculated as

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 Figure 5.

( 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

⁵With the same parameter values but different initial values, [1] analytically obtained a 9 years cycle and [4] numerically got an 8.12 year cycle.

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 [1] aims to obtain. So we summarize this results as follows:

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.

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 Figure 7. The red curves are the locus of y ( t ) = f [ y ˙ ( t − θ ) ] and the green parallelogram is a sawtooth limit cycle. Black dotted curve connects the boundary values. The following points are shown in Figure 7:

( 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 Figure 6. The move reaches point (7) at the end of I 3 and jumps to point (8) at the beginning of interval I 4 in which we see signs of sawtooth oscillations. Two intersections obtained in interval I 3 causes two changes of the dynamic system; the linear Equation (5) with d = 3 n governs the movement from point (8) to point (9) and the delay Equation (13) controls the movement from point (9) to point (10) and then the system is changed to the linear Equation (5) with d = − n managing the movement along the lower downward line from point (10) to point (11). At the beginning of interval I 5 , a jump from point (11) to point (12) occurs and the further movement along the upper red line to point (13) is controlled by the linear equation with d = 3 n , point (13) to point (14) by the delay equation and point (14) to point (15) by the linear equation with d = − n . Point (15) jumps to a point on the upper red line and the dynamic system change occurs as well in I k for integer k ≥ 6 as in   I 5 . By doing so, the trajectory gradually approaches to the green sawtooth limit cycle as time goes on. It is noticed that a jump occurs at the local maximum or minimum point in the non-delay model whereas even at the middle of these boundary values in the delay model.