1550078481-Ordinary_Differential_Equations__Roberts_

Applications of Systems of Equations 447

So

(4) By^2 (t) - Dx^2 (t) = By6 - Dx6 = K

where K is a constant. For K-/=- 0 the graph of equation ( 4) is a hyperbola and

for K = 0 the graph of equation ( 4) is two lines which intersect at the origin.

The trajectories in the first quadrant defined by equation (4) are sketched in

Figure 10.13. The direction (x(t), y(t)) moves along the hyperbolas (4) as t

increases, which is indicated by the arrows in Figure 10.13, was determined

by noting in system (2) that dx/dt < 0 and dy/dt < 0 for x > 0 and y > 0.

y

K = 0

ji(!B~---

x

Figure 10.13 Phase-Plane Portrait of Lanchester's Combat Model

for Two Conventional Forces with No Reinforcements

and No Operational Losses

We will say that one force "wins" the battle, if the other force vanishes first.

We see from equation ( 4) that the y-force wins if K > 0; the x-force wins if

K < 0; and there is a tie if K = 0. If K > 0, the x-force vanishes (x = 0) when

y = .JKTI3. Thus, for K > 0 the strength of the y-force at the end of the

battle is ./K]B. Likewise, if K < 0, they-force vanishes (the x-force wins the

battle) and at the end of the battle the strength of the x -force is x = J -K / D.

In order to win the battle, the y-force seeks to establish conditions under

which the inequality K = By§ - Dx§ > 0 or By§ > Dx§ holds. This can be

accomplished by increasing its combat effectiveness coefficient, B - perhaps

by using more powerful or more accurate weapons- or by increasing its initial

number of combatants, YD· Notice that doubling B causes Bya to double