Matthew Lesko-Krleza

Vanier College

ESP Research Paper

L. F. Richardson’s Theory of Conflict and Lanchester’s Combat Models

Matthew Lesko-Krleza

Differential Equations 201-HTL-VA

Ivan Ivanov

May 27, 2015

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L. F. Richardson’s Theory of Conflict and Lanchester’s Combat Models

Abstract

This research paper will examine several mathematical theories of war. The first is L. F.

Richardson’s theory of conflict and the second is Lanchester’s combat model. Our first problem

at hand is the question: “how can one define a mathematical model that expresses the relation

of two nations who are determined to defend themselves from a possible attack by the other.”

Essentially, the goal is to construct a model that will allow one to know if two nations will

progress into a state of war or of disarmament. The first model is based off the work of Lewis

Fry Richardson. This model is limited modern era warfare. A linear system of differential

equations is the basis for Richardson's model and applications in Differential Equations and

Eigenvalue/vector analysis are used to assess the system and find a solution. Our second

problem is to define a mathematical model to show which out of two forces would win in a

conventional combat scenario and which one would win in a guerilla combat scenario. This

model is based off of Lanchester’s work and it has similar constrictions to that of Richardson’s

model.

Theory behind Richardson’s Model

The models to be presented are not an attempt to determine the date at which a war

will break out nor are they capable to make scientific statements about foreign politics. The

solutions to these equations are not inevitable events. They describe what would occur if a

nation felt compelled to follow mechanical impulses.

For simplicity, we will call the first nation Jedesland and the second one Andersland. We

will let the war potential or armaments of Jedesland be denoted by 𝑥 = 𝑥(𝑡). Its rate of

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change will depend on the readiness/defence of Andersland which we will denote as 𝑘, a

positive constant. Jedesland’s feelings towards Andersland (denoted as 𝑔) and its cost of

armaments (denoted as 𝛼) will also affect the rate of change of 𝑥(𝑡).

The rate of change of armaments for Jedesland takes into account 4 terms:

𝑥 = 𝑥(𝑡) ∶ The war potential or armaments of Jedesland;

𝑘 ∶ War readiness of Andersland;

𝑔 ∶ Feelings/Grievances Jedesland has towards Andersland;

𝛼 ∶ Cost of Jedesland’s armaments.

𝑑𝑥/𝑑𝑡 = 𝑘𝑦 − 𝛼𝑥 + 𝑔

The rate of change of armaments for Andersland takes into account 4 terms:

𝑦 = 𝑦(𝑡) ∶ The war potential or armaments of Andersland;

𝑙 ∶ War readiness of Jedesland;

ℎ ∶ Feelings/Grievances Andersland has towards Jedesland;

𝛽 ∶ Cost of Anderson's armaments.

𝑑𝑦/𝑑𝑡 = 𝑙𝑥 − 𝛽𝑦 + ℎ

The linear system of these two differential equations is used for the first model

(1) 𝑑𝑥

𝑑𝑡 = 𝑘𝑦 − 𝛼𝑥 + 𝑔

𝑑𝑦

𝑑𝑡 = 𝑙𝑥 − 𝛽𝑦 + ℎ

The terms 𝑘, 𝑔 and 𝑙, ℎ cause the variables 𝑥 and 𝑦 to increase respectively, thus they

are positive constants. While as the terms 𝛼 and 𝛽 cause 𝑥 and 𝑦 to decrease respectively,

which is why they subtract the other terms in model (1).

Over millennia there has been constant debate about what causes war. Sir Edward Grey,

the British Foreign Secretary during World War I believes that “the increase of armaments [...]

produces a consciousness of the strength of other nations and a sense of fear. The enormous

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growth of armaments in Europe, the sense of insecurity and fear caused by them - it was these

that made war inevitable. This is the real and final account of the origin of the Great War,” (4.5

Mathematical theories of war page 399). While as L. S. Amery of Britain’s 1930 parliament who

does not agree with this statement, believes that it were the “insoluble conflicts of ambitions

[between European countries] and not in the armaments themselves, that the cause of the War

lay,” (4.5 Mathematical theories of war page 39-400).

Model (1) takes both of these conflicting theories into account. Sir Edward Grey would

take 𝑔, ℎ small compared to 𝑘, 𝑙. L. S. Amery would take 𝑘, 𝑙 small compared to 𝑔, ℎ.

Equilibrium Solution

Whether a nation will run into an arms race or disarmament, one has to know the

equilibrium solutions for Model (1) in order to interpret how changes in certain variables can

affect the rate of change of armaments.

An equilibrium solution is attained when

𝑑𝑥

𝑑𝑡 = 𝑥′ = 0 ;

𝑑𝑦

𝑑𝑡 = 𝑦′ = 0

To find this single equilibrium solution, we will let

�⃑� = |𝑥𝑦| ; 𝑓 = |

𝑔ℎ

| ; 𝑥′⃑⃑⃑ ⃑ = |𝑥′

𝑦′| = 𝐴�⃑� + 𝑓 ; 𝐴 = | – 𝛼 𝑘

𝑙 −𝛽|

To solve for �⃑� , we will set

𝑥′⃑⃑⃑ ⃑ = 0⃑⃑ = |00

|

|00

| = | – 𝛼 𝑘

𝑙 −𝛽| |

𝑥𝑦| + |

𝑔ℎ

| → |−𝑔−ℎ

| = | – 𝛼 𝑘

𝑙 −𝛽| |

𝑥𝑦|

To isolate �⃑�, 𝐴−1 has to calculated

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𝐴−1 =1

(– 𝛼)(−𝛽) − (𝑘)(𝑙)|−𝛽 −𝑘−𝑙 – 𝛼

| = 1

𝛼𝛽 − 𝑘𝑙|−𝛽 −𝑘−𝑙 – 𝛼

|

Multiply both left sides by 𝐴−1 and isolate �⃑�

(1

𝛼𝛽 − 𝑘𝑙|−𝛽 −𝑘−𝑙 – 𝛼

|) |−𝑔−ℎ

| = (1

𝛼𝛽 − 𝑘𝑙|−𝛽 −𝑘−𝑙 – 𝛼

|) | – 𝛼 𝑘

𝑙 −𝛽| |

𝑥𝑦|

|𝑥𝑦| =

1

𝛼𝛽 − 𝑘𝑙|−𝛽 −𝑘−𝑙 – 𝛼

| |−𝑔−ℎ

| → |𝑥𝑦| =

1

𝛼𝛽 − 𝑘𝑙|𝛽𝑔 + 𝑘ℎlg + 𝛼ℎ

|

Our Equilibrium Solutions are:

𝑥𝑜 = 𝛽𝑔 + 𝑘ℎ

𝛼𝛽 − 𝑘𝑙 ; 𝑦𝑜 =

𝑙𝑔 + 𝛼ℎ

𝛼𝛽 − 𝑘𝑙 ; 𝛼𝛽 − 𝑘𝑙 ≠ 0

Roots of Characteristic Polynomial

To know if a nation will run into an arms race or disarmament, we have to know what

will cause the equilibrium solution to become stable or unstable. To do so, the system of

differential equations has to be solved, but only partially. It is the eigenvalues that demonstrate

whether a solution is stable or unstable.

The computation of the eigenvalues of A is as follows

𝑥′ = 𝑘𝑦 – 𝛼𝑥 + 𝑔 𝑦′ = 𝑙𝑥 − 𝛽𝑦 + ℎ

𝑥′⃑⃑⃑ ⃑ = |𝑥′

𝑦′| = 𝐴�⃑� ; 𝐴 = | – 𝛼 𝑘

𝑙 −𝛽|

𝑝(𝜆) = det (|−𝛼 − 𝜆 𝑘

𝑙 −𝛽 − 𝜆|) = (−𝛼 − 𝜆)(−𝛽 − 𝜆) − 𝑘𝑙

= 𝜆2 + (𝛼 + 𝛽)𝜆 + (𝛼𝛽 − 𝑘𝑙) = 0

𝜆1,2 =−(𝛼 + 𝛽) ± √(𝛼 + 𝛽)2 − 4(𝛼𝛽 − 𝑘𝑙)

2

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Let eigenvectors of 𝜆1 𝑎𝑛𝑑 𝜆2 be 𝑣1⃑⃑⃑⃑⃑ 𝑎𝑛𝑑 𝑣2⃑⃑⃑⃑⃑ . The solution is

𝑦′⃑⃑ ⃑⃑ = 𝑐1𝑦1 + 𝑐2𝑦2 + 𝑗(𝑡) = 𝑐1𝑒𝜆1𝑡�⃑�1 + 𝑐2𝑒𝜆2𝑡�⃑�2 + 𝑗(𝑡)

The particular solution of the system (1) is denoted as 𝑗(𝑡).

Stable Solution: 𝜆1 , 𝜆2 < 0 ; 𝑦1 , 𝑦2 will head to their equilibrium solutions 𝑥𝑜 , 𝑦𝑜 as 𝑡 → ∞.

The equilibrium solution 𝑥(𝑡) = 𝑥𝑜(𝑡) ; 𝑦(𝑡) = 𝑦𝑜(𝑡) is stable if the roots of the

characteristic polynomial are negative. For the eigenvalues to be negative:

𝛼𝛽 − 𝑘𝑙 > 0

When that section is positive, (𝛼 + 𝛽)2 will be subtracted, thus

(𝛼 + 𝛽) > √(𝛼 + 𝛽)2 − 4(𝛼𝛽 − 𝑘𝑙)

𝜆1,2 =−(𝛼 + 𝛽) ± √(𝛼 + 𝛽)2 − 4(𝛼𝛽 − 𝑘𝑙)

2< 0

A stable solution signifies that the nations will go into disarmament.

If 𝑔, 𝑥(𝑡), 𝑦(𝑡) 𝑎𝑛𝑑 ℎ were made zero simultaneously, this would set an ideal scenario

for permanent peace by disarmament and satisfaction.

𝑑𝑥

𝑑𝑡= 0 ;

𝑑𝑦

𝑑𝑡= 0

If 𝑥(𝑡) 𝑎𝑛𝑑 𝑦(𝑡) would simultaneously hit zero at a certain time 𝑡, but the constants

𝑔 𝑎𝑛𝑑 ℎ did not vanish, this would imply Mutual Disarmament. 𝑥(𝑡) 𝑎𝑛𝑑 𝑦(𝑡) will not remain

zero and both nations will rearm because of dissatisfaction.

𝑑𝑥

𝑑𝑡= 𝑔 ;

𝑑𝑦

𝑑𝑡= ℎ

Unilateral Disarmament implies that at a certain instant of time one country is disarmed

but will not remain so.

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At a certain instant of time

𝐼𝑓 𝑥 = 0 ; 𝑑𝑥

𝑑𝑡= 𝑘𝑦 + 𝑔 𝑜𝑟 𝐼𝑓 𝑦 = 0 ;

𝑑𝑦

𝑑𝑡= 𝑙𝑥 + ℎ

Unstable Solution: 𝜆1 𝑜𝑟 𝜆2 > 0 ; 𝑦1 𝑜𝑟 𝑦2 will respectively head to ∞ as 𝑡 → ∞.

The equilibrium solution 𝑥(𝑡) = 𝑥𝑜(𝑡) ; 𝑦(𝑡) = 𝑦𝑜(𝑡) is unstable if one of the roots of the

characteristic polynomial is positive. For an eigenvalue to be positive:

𝛼𝛽 − 𝑘𝑙 < 0

When that section is negative, because of the minus in front, (𝛼 + 𝛽)2 will be added, thus

(𝛼 + 𝛽) < √(𝛼 + 𝛽)2 − 4(𝛼𝛽 − 𝑘𝑙)

𝜆1 =−(𝛼 + 𝛽) + √(𝛼 + 𝛽)2 − 4(𝛼𝛽 − 𝑘𝑙)

2> 0

An unstable solution signifies that the two nations will run into a runaway arms race and

potentially war.

Richardson’s Second Model

If the War readiness/ “defense” terms predominate in model (1). Then our system of

equations would look like so

𝑑𝑥

𝑑𝑡= 𝑘𝑦 ;

𝑑𝑦

𝑑𝑡= 𝑙𝑥

This system has an equilibrium solution at 𝑥 = 𝑥𝑜 = 0, 𝑦 = 𝑦𝑜 = 0

Finding a solution to these equations is simpler, but at the cost of not being to be as a precise

model as model (1).

To find a general solution, we let

𝑥′⃑⃑⃑ ⃑ = |𝑥′𝑦′

| ; �⃑� = |𝑥𝑦| ; 𝑥′⃑⃑⃑ ⃑ = 𝐴�⃑� ; 𝐴 = |

0 𝑘𝑙 0

|

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The eigenvalues of matrix A are computed

𝑝(𝜆) = det (|−𝜆 𝑘

𝑙 −𝜆|) = 𝜆2 − 𝑘𝑙 = 0

𝜆1 = √𝑘𝑙 ; 𝜆2 = −√𝑘𝑙

The eigenvectors of matrix A are solved

𝜆1 = √𝑘𝑙, 𝑣1⃑⃑⃑⃑⃑ ∶ 𝐴 − 𝜆𝐼 = 0 = |−√𝑘𝑙 𝑘

𝑙 −√𝑘𝑙 |

00

| → 𝑅𝑅𝐸𝐹 → |1 −√𝑘

𝑙

0 0

|00

|

𝑣1⃑⃑⃑⃑⃑ = |√𝑘

𝑙

1

|

𝜆2 = −√𝑘𝑙, 𝑣2⃑⃑⃑⃑⃑ ∶ 𝐴 − 𝜆𝐼 = 0 = |√𝑘𝑙 𝑘

𝑙 √𝑘𝑙 |

00

| → 𝑅𝑅𝐸𝐹 → |1 √𝑘

𝑙

0 0

|00

|

𝑣2⃑⃑⃑⃑⃑ = |−√𝑘

𝑙

1

|

The general solution is

𝑦(𝑡) = 𝑐1𝑒𝜆1𝑡�⃑�1 + 𝑐2𝑒𝜆2𝑡�⃑�2 = 𝑐1𝑒√𝑘𝑙𝑡 |√𝑘

𝑙

1

| + 𝑐2𝑒−√𝑘𝑙𝑡 |−√𝑘

𝑙

1

|

Stable Solution: The equilibrium solution is stable when 𝜆 = ±√𝑘𝑙 = 0, 𝑘, 𝑙 = 0 implying that

the nations have disarmed and will remain so.

Unstable Solution: The equilibrium solution is unstable when 𝜆 = √𝑘𝑙 > 0, 𝑘, 𝑙 > 0. This

means that the nations will run into an arms race and potentially war.

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Numerical Example

An application consisting of Fundamental Matrices and eigenvalue/vector analysis will be used

𝑥′ = 3𝑦 − 4𝑥 + 6 𝑥(0) = 0

𝑦′ = 𝑥 − 2𝑦 + 1 𝑦(0) = 0

𝑥′⃑⃑⃑⃑ = |𝑥′

𝑦′| = 𝐴�⃑� + 𝑓 ; �⃑� = |𝑥𝑦| ; 𝐴 = |

−4 31 −2

| ; 𝑓 = |61

|

The eigenvalues and eigenvectors of matrix A are calculated to establish the complementary

solution

𝑝(𝜆) = det(𝐴 − 𝜆𝐼) = det (|−4 − 𝜆 3

1 −2 − 𝜆|) = (−4 − 𝜆)(−2 − 𝜆) − 3

= 𝜆2 + 6𝜆 + 5 = (𝜆 + 5)(𝜆 + 1) = 0

𝜆1 = −1 ∶ 𝑣1⃑⃑⃑⃑⃑ ∶ 𝐴 − 𝜆𝐼 = |−3 31 −3

|00

| → 𝑅𝑅𝐸𝐹 → |1 −10 0

|00

| 𝑣1⃑⃑⃑⃑⃑ = |11

|

𝜆2 = −5 ∶ 𝑣2⃑⃑⃑⃑⃑ ∶ 𝐴 − 𝜆𝐼 = |1 31 3

|00

| → 𝑅𝑅𝐸𝐹 → |1 30 0

|00

| 𝑣2⃑⃑⃑⃑⃑ = |−31

|

𝑥𝑐⃑⃑ ⃑⃑ = 𝑐1𝑦1 + 𝑐2𝑦2 = 𝑐1𝑒−𝑡 |11

| + 𝑐2𝑒−5𝑡 |−31

|

An application involving a Fundamental Matrix is required to solve for the particular solution

𝑥𝑝⃑⃑⃑⃑⃑ = 𝜙 ∫ 𝜙−1 𝑓𝑑𝑡

𝜙 = |𝑦1 𝑦2| 𝜙 = |𝑒−𝑡 −3𝑒−5𝑡

𝑒−𝑡 𝑒−5𝑡 | 𝜙−1 =1

(𝑒−𝑡)(𝑒−5𝑡)−(−3𝑒−5𝑡)( 𝑒−5𝑡) | 𝑒−5𝑡 3𝑒−5𝑡

−𝑒−𝑡 𝑒−𝑡 |

𝜙−1 =1

4𝑒−6𝑡 | 𝑒−5𝑡 3𝑒−5𝑡

−𝑒−𝑡 𝑒−𝑡 | = 1

4 | 𝑒𝑡 3𝑒𝑡

−𝑒5𝑡 𝑒5𝑡 |

𝜙−1𝑓 =1

4 | 𝑒𝑡 3𝑒𝑡

−𝑒5𝑡 𝑒5𝑡 | |61

| = |

9

4𝑒𝑡

−5

4𝑒5𝑡

|

9

∫ 𝜙−1𝑓𝑑𝑡 = ||∫

9

4𝑒𝑡𝑑𝑡

∫ −5

4𝑒5𝑡𝑑𝑡

|| = |

9

4𝑒𝑡

−1

4𝑒5𝑡

|

𝑥𝑝⃑⃑⃑⃑⃑ = 𝜙 ∫ 𝜙−1 𝑓𝑑𝑡 = |𝑒−𝑡 −3𝑒−5𝑡

𝑒−𝑡 𝑒−5𝑡 | |

9

4𝑒𝑡

−1

4𝑒5𝑡

| = |32

|

𝑥(𝑡)⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑⃑ = 𝑥𝑐⃑⃑ ⃑⃑ + 𝑥𝑝⃑⃑⃑⃑⃑ = 𝑐1𝑒−𝑡 |11

| + 𝑐2𝑒−5𝑡 |−31

| + |32

|

IVP

𝑥(0) = 0 = 𝑐1 − 3𝑐2 + 3

𝑦(0) = 0 = 𝑐1 + 𝑐2 + 2

|1 −31 1

|−3−2

| → 𝑅𝐸𝐹 → |1 −30 4

|−31

| 𝑐1 = −9

4 𝑐2 =

1

4

𝑥(𝑡)⃑⃑ ⃑⃑ ⃑⃑ ⃑⃑⃑ = −9

4𝑒−𝑡 |

11

| + 1

4 𝑒−5𝑡 |

−31

| + |32

|

As 𝑡 → ∞, 𝑥 = 3, 𝑦 = 2. Nation x will have a cache of armaments 1.5 times greater than that

of nation y. Both nations will reach Mutual Disarmament, because the driving force 𝑓 does not

vanish.

Conclusion

To conclude, L. F. Richardson’s theory of conflict is an interesting theory that can

demonstrate whether two nations will head into an arms race or disarmament. It is a rather

accurate model. Historical battles such as those in World War 2 have been successfully

modelled using Richardson’s theory.

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Lanchester’s Combat Models

During the First World War, F. W. Lanchester constructed mathematical models that can

determine the results of a combat engagement between two enemy forces. This section will

present two of the models from his work. The first model will account for the combat of a

conventional force versus a conventional force. The second model will account for the combat

of a conventional force versus a guerilla force.

Theory and Construction of the Models

We will call the two forces that are in combat with one another “x-force” and “y-force,”

where the strengths of these two forces is determined by their number of combatants. We will

denote 𝑥(𝑡) and 𝑦(𝑡) as the number of combatants for the two forces, where 𝑡 is the number

of days since the start of combat.

The rate of change of 𝑥(𝑡) and 𝑦(𝑡) in Lanchester’s model depends on three factors:

reinforcement rate, operational loss rate, and combat loss rate.

Operational loss rate is the loss of combat force due to non-combat mishaps such as

desertion and disease. Since these mishaps depend on factors that cannot be quantified such as

the psychology of a soldier, the engagements that will be looked upon will consider the

operational loss rate as negligible.

The combat loss rate is the loss of combat force due to the other force’s combat

effectiveness. Suppose that x-force is a conventional combat force in which every combatant is

in “kill-range” of their enemy y-force. It is assumed that as x-force suffers losses, y-force will

concentrate fire on the remaining combatants. Under these conditions, the combat loss rate of

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force x depends on the combat effectiveness and strength of force y. We will denote the

combat effectiveness of force y as 𝑎, a positive constant.

The situation is different if x-force is a guerilla force which use a form of irregular

military warfare. Guerilla forces are smaller compared to their opponent and consist of non-

standard military combatants who use ambush, hit-and-run, and sabotage tactics as opposed to

full-frontal assault. Guerilla combat tactics focus heavily on mobility and invisibility, implying

that when a guerilla force is engaged with the enemy, the opponent does not know when they

make a kill or not. We will consider the x-force to be a guerilla force that occupies a region R. Y-

force fires into R but cannot know when a shot is a hit or miss. It is correct to say that x-force’s

combat loss rate is proportional to its own strength 𝑥(𝑡), as the larger 𝑥(𝑡) is, the greater the

probability that an opponent’s shot will kill a guerilla combatant. The combat loss rate of 𝑥(𝑡) is

also proportional to 𝑦(𝑡) and y-force’s combat effectiveness, which we will denote as 𝑐, as the

more combatants there are, the greater number of x-casualties.

The rate at which new soldiers enter or withdraw from battle is called the

reinforcement rate. We will denote these rates for x and y forces by 𝑓(𝑡) and 𝑔(𝑡) respectively.

The construction of the conventional combat model is as follows:

𝑥 = 𝑥(𝑡) ∶ The strength of force x;

𝑎 ∶ Combat effectiveness of force y;

𝑓(𝑡) ∶ Reinforcement rate of force x.

𝑦 = 𝑦(𝑡) ∶ The strength of force y;

𝑏 ∶ Combat effectiveness of force x;

𝑔(𝑡) ∶ Reinforcement rate of force y.

𝑑𝑥

𝑑𝑡 = −𝑎𝑦 + 𝑓(𝑡)

𝑑𝑦

𝑑𝑡 = −𝑏𝑥 + 𝑔(𝑡) (1a)

The construction of the conventional-guerilla combat model is as follows:

𝑥 = 𝑥(𝑡) ∶ The strength of force x;

𝑐 ∶ Combat effectiveness of force y;

𝑓(𝑡) ∶ Reinforcement rate of force x.

𝑦 = 𝑦(𝑡) ∶ The strength of force y;

𝑣 ∶ Combat effectiveness of force x;

𝑔(𝑡) ∶ Reinforcement rate of force y.

𝑑𝑥

𝑑𝑡 = −𝑐𝑥𝑦 + 𝑓(𝑡)

𝑑𝑦

𝑑𝑡 = −𝑣𝑥 + 𝑔(𝑡) (1b)

The systems of equations (1a) is a linear system and can be solved explicitly. While as the

system of equations (1b) is non-linear and the aid of a computer would be necessary for its

solution.

Isolated Scenario

If forces x and y were engaged against one another in an isolated environment, we

would take their reinforcement rates to be zero. Under these conditions, the systems (1a) and

(1b) reduce to the following:

𝑑𝑥

𝑑𝑡 = −𝑎𝑦

𝑑𝑦

𝑑𝑡 = −𝑏𝑥 (2a)

𝑑𝑥

𝑑𝑡 = −𝑐𝑥𝑦

𝑑𝑦

𝑑𝑡 = −𝑣𝑥 (2b)

Conventional Combat: The Square Law

The orbits of the system (2a) are the solution curves of the equation

𝑑𝑦

𝑑𝑡∗

𝑑𝑡

𝑑𝑥 =

𝑑𝑦

𝑑𝑥=

𝑏𝑥

𝑎𝑦

Integrating this equation gives:

∫ 𝑎𝑦 𝑑𝑦 = ∫ 𝑏𝑥 𝑑𝑥 → 𝑎𝑦2

2+ 𝑐1 =

𝑏𝑦2

2+ 𝑐2 → 𝑎𝑦2 − 𝑏𝑦2 = 𝐾

𝑎𝑦𝑜2 − 𝑏𝑥𝑜

2 = 𝐾 (3)

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Equation (3) is in a closed environment, thus we will represent the initial strength of both forces

as 𝑦𝑜 and 𝑥𝑜

. This equation is called the “Lanchester’s Square Law,” the strengths of the forces

appear quadratically. System (2a) is typically referred to as the square law model.

Equation (3) can essentially determine which force will win, based off their combat

effectiveness and strength.

For some constant K, equation (3) defines a set of hyperbolas in the x-y plane that are indicated

in the following graph.

Figure (A)

Y-force seeks to establish a setting in which 𝐾 > 0. This would mean that y-force wants

to establish and hold the inequality

𝑎𝑦𝑜2 > 𝑏𝑥𝑜

2

This inequality demonstrates that y-force has a greater combat effectiveness and

strength than x-force. In order to achieve this inequality, y-force can either increase 𝑎, by

means of using better weapons, or by increasing their initial strength 𝑦𝑜 . It is more beneficial to

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increase 𝑦𝑜 , as a doubling of 𝑎 results in the doubling of 𝑎𝑦𝑜

2, while as the doubling of 𝑦𝑜 results

in the fourfold increase of 𝑎𝑦𝑜2.

As this inequality is held, equation (3) will yield a positive constant, expressing that y-

force will win the battle when 𝑦 reaches a strength of √𝐾

𝑎.

With the inequality still held, 𝑦 decreases to a strength of√𝐾

𝑎 → 𝑎𝑦2 = 𝐾. Plugging

this into equation (3), and isolating 𝑏𝑥𝑜2, we notice that 𝑏𝑥𝑜

2 = 0, expressing that x-force is

eliminated. Similarly, x wins if 𝐾 < 0. If 𝐾 = 0, it is a tie as both forces have the same strength.

This is the basis of Lanchester’s Square Law of conventional combat.

Conventional-Guerilla Combat

The orbits of the system (2b) are the solution curves of the equation

𝑑𝑦

𝑑𝑡∗

𝑑𝑡

𝑑𝑥 =

𝑑𝑦

𝑑𝑥=

𝑣𝑥

𝑐𝑥𝑦 =

𝑣

𝑐𝑦

𝑑𝑦

𝑑𝑥=

𝑣

𝑐𝑦

Multiplying both sides of this equation by 𝑐𝑦 then integrating gives:

𝑐𝑦2 − 2𝑣𝑥 = 𝑀

𝑐𝑦𝑜2 − 2𝑣𝑥𝑜 = 𝑀 (4)

For some constant M, equation (4) defines a set of hyperbolas in the x-y plane that are

indicated in the figure (B). Y-force desires to establish the inequality

𝑐𝑦𝑜2 > 2𝑣𝑥𝑜

This would mean that y-force has a greater strength than x-force and we notice that as this

inequality is held, M will be a positive constant.

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Figure (B)

Y-force wins if 𝑀 > 0, because this would mean that x-force has been eliminated by the time

𝑦(𝑡) decreased to √𝑀

𝑐. Plugging √

𝑀

𝑐 as 𝑦 in equation (4) we notice that 𝑐𝑦2 = 𝑀 and isolating

for 2𝑣𝑥𝑜 will yield 2𝑣𝑥𝑜 = 0, expressing that x-force is eliminated. Similarly, x-force wins if

𝑀 < 0.

The use of a computer is necessary for the solution of the nonlinear system (2b). The following

examples will demonstrate how a change in the combat effectiveness coefficients and initial

forces can change the hyperbolas in the x-y plane in figure (B) and how it can change the

outcome of a battle.

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Guerilla Combat Examples

EX1: Let us consider the scenario where the conventional army has a much greater initial

strength than the guerilla army, but the guerilla combatants are extremely combat effective.

In this graph, 𝑐 = 1, 𝑣 = 250, 𝑦𝑜 = 200, 𝑥𝑜 = 40.

Figure (C)

We can see that in this situation, y-force will win the engagement as 𝑐𝑦𝑜2 > 2𝑣𝑥𝑜, meaning that

𝑀 > 0. By analyzing figure (C), we do notice that the hyperbolas bend, showing that the

guerilla forces do have an effect on the conventional army, even if they are 5 times smaller. By

looking at hyperbola (H1) we notice that when the conventional army has an initial strength of

200, it will lose more than a quarter of its strength by the time the guerilla force is eliminated.

To summarize, y-force will win at the cost of losing more than 50 of its initial strength.

H1

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EX2: Let us consider the scenario where both x-force and y-force have similar initial

strengths. The guerilla force is still smaller than the conventional force but is still extremely

combat effective.

In this graph, 𝑐 = 1, 𝑣 = 150, 𝑦𝑜 = 200 𝑥𝑜 = 180.

Figure (D)

X-force shall win the engagement as 𝑐𝑦𝑜2 < 2𝑣𝑥𝑜 , meaning that 𝑀 < 0. By analyzing figure (D),

we can see that the hyperbolas do bend, showing that both armies do have an effect on one

another. By looking at hyperbola (H2) we notice that when the guerilla army starts with an

initial strength of 180, by the time they win, they will remain with a strength of roughly 45.

They will have lost a considerable amount of their army. The only way for the conventional

H2

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army to win this scenario is if their initial force is of at least around 225. To summarize, x-force

will win the combat but at a great of loss of roughly 125 of its initial strength.

EX3: Let us consider the scenario where both x-force and y-force have identical initial

strengths. The guerilla force is much more combat effective than the conventional army.

In this graph, 𝑐 = 1, 𝑣 = 250, 𝑦𝑜 = 200 𝑥𝑜 = 200.

Figure (D)

In this scenario, the guerilla combatants will win as 𝑐𝑦𝑜2 < 2𝑣𝑥𝑜 , meaning that 𝑀 < 0. The

hyperbola (H3) shows us that they will be at a loss of 80 of its initial strength. Starting with the

same strength, it is correct to say that the winning factor for the guerilla force is their combat

effectiveness. While as the conventional army holds strength in larger numbers of soldiers.

H3

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Conclusion

To conclude, the conventional army typically gains the upper hand when its initial

strength is larger than that of the guerilla force. The guerilla force is still very combat capable,

thus in most cases where the gap between the initial forces is not immensely large, the guerilla

force will put up a fight and affect the conventional army’s final strength. If the initial strengths

are very close, the guerilla army has a strong chance of winning the engagement due to their

combat capabilities. Lanchester’s work has shown to be precise in historic examples such as the

battle of Iwo Jima.

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References

Differential Equations & Their Applications: An Introduction to Applied Mathematics by Martin

Braun. Springer; 4th edition (December 5, 1992), 978-0387978949