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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
1
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
5
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
|
7
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.
8
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.
10
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
11
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
17
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
18
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
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