Mathematical Modelling Project

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Mathematical Modelling – Group 11 – Project 3

Continuous Population Modelling - Predator Prey

Contributors

Alexandra Hunt, Paul Martin, Lorna Gollop, Andrew Haspey and Jasmine Dhaliwal

1. Introduction .

In nature all organisms are mutually dependent on each other, so a change in the population of one organism will have a knock on effect on the population of another organism. For this reason, it is important that we model the interactions between organisms. The interactions between organisms can either lead to co-existence, extinction of one or both species, or symbiosis.

In the past a variety of modelling ideas have been used. The first population modelling was carried out by Thomas Malthus in 1798 in a paper he wrote titled Essay on the Principle of Population. As shown in Figure 1, Malthus argued that populations grew logarithmically, but the populations that they depended on remained constant or only increased arithmetically. Thus, the demand for resource supply would cause the population to reach a natural level before dying off (Berryman, 1992).

Figure 1: Urbano, 2011. A graph showing Malthusian Growth, showing that the population will always over take resources.

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We decided to use the Lotka-Volterra model as it looks at multiple species interactions. The Lotka-Volterra model was developed independently by Alfred Lotka and Vito Volterra (Beals et al., 1999). The formula is the most successful attempt at modelling the relationship between prey and predator (Lotka, 1922, Volterra, 1928; cited in Bas, 2000). The Predator-Prey model plays a vital part in population dynamics because it is able to determine the biomass transferred in an association of different organisms forming a closely integrated community (Poggaile, 1998). The Lotka-Volterra predator-prey equations were built around three main assumptions:

1. The Malthus Law is applied for both models 2. The principle of mass action so the responses were assumed to be

proportional to the product rather than the ratio of population densities 3. The predator density growth is proportional to the density of eaten prey.

(Berryman, 1992; Poggaile, 1998).

If the predator was absent, the population of prey would grow exponentially, whereas in the absence of the prey, the predator population would decay exponentially due to starvation.

We decided to use the Lotka-Volterra model to investigate the predator-prey relationship between Brown Bears and Chum Salmon.

Our aim was to investigate whether the two populations could co-exist and if so we aim to investigate what impact another factor would have.

2. Mathematical Model

2.1 Assumptions • No human influence. • Constant water level and flow – a change increase in water level would make

it harder for the salmon to get up the stream, whilst a decrease in water level would make it easier for the bears to catch the salmon.

• The Bear only eats Salmon and they don’t get their nutrients from anywhere else – this is to ensure the population can be accurately modelled.

• The prey population has a constant source of food. • The rate of change of the populations are proportional to their size. • During the process, the environment does not change in favour of one

species and genetic adaptation is inconsequential. • Predators have limitless appetite.

Initially we decided to assume that there was a constant Birth and Death Rate however, we then decided not to assume this as during the period of time, the salmon do not reproduce.

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2.2 Variables

Independent Variable: Salmon Population Dependent Variable: Bear population

2.3 Model Construction and analysis

2.3.1 Sub-problem 1

This is our initial problem, with the two species interacting with each other, with no other interactions made.

The rates of change of the population’s 𝑥, 𝑦 are denoted !"!"

and !"!"

respectively, where 𝑥(𝑡) represents the Salmon population and 𝑦(𝑡) represents the Bear population. The constant positive parameters 𝑎, 𝑏, 𝑐 and 𝑑 represent net growth rate of 𝑥, the interaction of 𝑥 with 𝑦, the death rate of 𝑦 and the interaction of 𝑦 with 𝑥.

𝑑𝑥𝑑𝑡 = 𝑎𝑥 − 𝑏𝑥𝑦

𝑑𝑦𝑑𝑡 = −𝑐𝑦 + 𝑑𝑥𝑦

𝑎𝑥 − 𝑏𝑥𝑦 = 0 ⇒ 𝑥 𝑎 − 𝑏𝑦 = 0

⇒ 𝑥 = 0, 𝑦 =𝑎𝑏

−𝑐𝑦 + 𝑑𝑥𝑦 = 0 ⇒ 𝑦 −𝑐 + 𝑑𝑥 = 0

⇒ 𝑦 = 0, 𝑥 = 𝑐𝑑

Therefore the Equilibrium Points are at (0,0) and (!!, !!)

For (0,0)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 −𝑐 + 𝑑𝑥 !, !

= 𝑎 00 −𝑐

det 𝐴 − 𝜆𝐼 = 𝑎 − 𝜆 00 −𝑐 − 𝜆

= (𝑎 − 𝜆)(−𝑐 − 𝜆)

⇒ 𝜆! = 𝑎 𝜆! = −𝑐

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This implies that we have a saddle.

𝐴 − 𝜆𝐼 𝑋 = 0

⇒ 𝑎 00 −𝑐

𝑥𝑦 = 0

0

⇒ 𝑎𝑥 + 0 = 0 ⇒ 𝑎𝑥 = 0

𝑎𝑙𝑠𝑜,⇒ −𝑐𝑦 = 0 ⇒ 𝑦 = 0

For (!!, !!)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 −𝑐 + 𝑑𝑥 (𝑐𝑑,

𝑎𝑏)

=𝑎 − 𝑏

𝑎𝑏 −𝑏

𝑐𝑑

𝑑𝑎𝑏 −𝑐 + 𝑑

𝑐𝑑

det 𝐴 − 𝜆𝐼 =0 − 𝜆

−𝑏𝑐𝑑

𝑑𝑎𝑏

0 − 𝜆

= 𝜆! −−𝑏𝑐𝑑

𝑑𝑎𝑏

= 𝜆! +𝑏𝑐𝑑𝑎𝑑𝑏

⇒ 𝜆 = ±𝑎𝑏𝑐𝑑𝑑𝑏 𝑖

This is a stable centre.

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We have used Mathematica to plot the phase-plane diagram.

In Appendix 1 we used a=b=c=d=1 and plotted it on Mathematica. We can see from Figure 2, that there is a saddle point at the origin and a stable centre at (1,1).

Figure 3 shows the oscillation of the population of bear and salmon against time. 𝑥(𝑡) (red line) denotes the population of the salmon and 𝑦(𝑡) (blue line) denotes the population of the bears.

Figure 2

Figure 3

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2.3.2 Sub-problem 2


and !"!"

respectively, where 𝑥(𝑡) represents the Salmon population and 𝑦(𝑡) represents the Bear population. The constant positive parameters 𝑎, 𝑏, 𝑐, 𝑑 and 𝑟 represent net growth rate of 𝑥, the interaction of 𝑥 with 𝑦, the death rate of 𝑦, the interaction of 𝑦 with 𝑥 and carrying capacity constant in relation to 𝑎.

This is the logistic model, where the prey could grow unboundedly.

𝑑𝑥𝑑𝑡 = 𝑎𝑥 − 𝑟𝑥! − 𝑏𝑥𝑦 = 0

𝑑𝑦𝑑𝑡 = −𝑐𝑦 + 𝑑𝑥𝑦

Therefore the Equilibrium Points are at (0,0), (!!, 0) and (!

!, !!!!"

!")

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 −𝑐 + 𝑑𝑥 !, !

= 𝑎 00 −𝑐

det 𝐴 − 𝜆𝐼 = 𝑎 − 𝜆 00 −𝑐 − 𝜆

= (𝑎 − 𝜆)(−𝑐 − 𝜆)

⇒ 𝜆! = 𝑎 𝜆! = −𝑐

This implies we have a saddle.

For (!!, 0)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥0 −𝑐 + 𝑑𝑥 𝑎

!, !

=𝑎 − 2𝑎 −𝑏

𝑎𝑟

0 −𝑐 +𝑎𝑑𝑟

det 𝐴 − 𝜆𝐼 =−𝑎 − 𝜆 −

𝑏𝑎𝑟

0 −𝑐 +𝑎𝑑𝑟− 𝜆

= 0

= −𝑎 − 𝜆 −𝑐 +𝑎𝑟− 𝜆 = 0

⇒ 𝜆! = −𝑎 𝜆! = −𝑐 +𝑎𝑟

We therefore have a stable node.

For (!!, !!!!"

!")

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𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥0 −𝑐 + 𝑑𝑥 !

!, 𝑎!!!"!"

=𝑎 −

2𝑟𝑐𝑑−

𝑎𝑑 − 𝑟𝑐𝑑

−𝑏𝑐𝑑

𝑎𝑑 − 𝑟𝑐𝑏 0

det 𝐴 − 𝜆𝐼 =

−𝑟𝑐𝑑 − 𝜆 −

𝑏𝑐𝑑

𝑎𝑑 − 𝑟𝑐𝑏

0 − 𝜆= 0

= 𝜆! +𝑟𝑐𝑑𝜆 +

𝑎𝑑𝑐 − 𝑟𝑐!𝑑

⇒ 𝑇𝑟 𝐴 =−𝑟𝑐𝑑

det 𝐴 =𝑎𝑐𝑑 − 𝑟𝑐!

𝑑

4det 𝐵 =4𝑎𝑑𝑐 − 4𝑟𝑐!

𝑑

Since Tr(A)<0, both values are positive real eigenvalues.

Whenever 0<4det(B)<Tr(A)² the eigenvalues are real and distinct and we have a node that is stable if Tr(A)<0 and unstable if Tr(A)>0

Since our Tr(A) = !!"! <0 it follows we have a stable node.

Appendix 2 shows a worked example where the constant positive parameters 𝑎, 𝑏, 𝑐, 𝑑 and 𝑟 have been replaced with real values: 𝑎 = 2, 𝑟 = 𝑏 = 𝑐 = 𝑑 = 1. Figure 4 illustrates this model when plotted using Mathematica. The diagram displays a stable spiral, unstable node and a stable node. Therefore the two populations can co-exist.

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Figure 5 (below) shows a graph of each population plotted against time. As you can see, the predator starts out high and the prey low, after the interaction both populations plateau at a constant value.

Prey

Predator

Figure 4

Figure 5

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2.3.3 Sub-problem 3


and !"!"

respectively, where 𝑥(𝑡) represents the Salmon population and 𝑦(𝑡) represents the Bear population. The constant positive parameters 𝑎, 𝑏, 𝑐, 𝑑, 𝑟 and 𝑞 represent net growth rate of 𝑥, the interaction of 𝑥 with 𝑦, the death rate of 𝑦, the interaction of 𝑦 with 𝑥, carrying capacity constant in relation to 𝑎 and the carrying capacity constant in relation to 𝑐.

For our third model we have a double logistic model.

𝑑𝑥𝑑𝑡 = 𝑎𝑥 − 𝑟𝑥! − 𝑏𝑥𝑦 = 0

𝑑𝑦𝑑𝑡 = 𝑐𝑦 + 𝑞𝑦! + 𝑑𝑥𝑦 = 0

𝑎𝑥 − 𝑟𝑥2 − 𝑏𝑥𝑦 = 0 ⇒ 𝑥 𝑎 − 𝑟𝑥 − 𝑏𝑦 = 0

𝑐𝑦 − 𝑞𝑦2 + 𝑑𝑥𝑦 = 0 ⇒ 𝑦 𝑐 − 𝑞𝑦 + 𝑑𝑥 = 0

Therefore, our Equilibrium points are at 0,0 , 0, !!, !!, 0 𝑎𝑛𝑑 !"!!"

!"!!", !"!!"!"!!"

For 0,0

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 𝑐 − 2𝑞𝑦 + 𝑑𝑥 !, !

= 𝑎 00 𝑐

det 𝐴 − 𝜆𝐼 = 𝑎 − 𝜆 00 𝑐 − 𝜆

= (𝑎 − 𝜆)(𝑐 − 𝜆)

⇒ 𝜆! = 𝑎! 𝜆! = 𝑐

We therefore have an unstable node.

For 0, !!

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 𝑐 − 2𝑞𝑦 + 𝑑𝑥 !, !!

det 𝐴 − 𝜆𝐼 =𝑎 −

𝑏𝑐𝑞− 𝜆 0

𝑑𝑐𝑞

𝑐 − 2𝑐 − 𝜆

= (𝑎 −𝑏𝑐𝑞− 𝜆)(−𝑐 − 𝜆)

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⇒ 𝜆! =𝑎𝑞 − 𝑏𝑐

𝑞 𝜆! = −𝑐

This is a saddle.

For !!, 0

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 𝑎 − 2𝑟𝑥 − 𝑏𝑦 −𝑏𝑥𝑑𝑦 𝑐 − 2𝑞𝑦 + 𝑑𝑥 !

! , !

det 𝐴 − 𝜆𝐼 =𝑎 − 2𝑎 − 𝜆

−𝑏𝑑𝑟

0 𝑐 +𝑎𝑑𝑟− 𝜆

= (−𝑎 − 𝜆)𝑐𝑟 + 𝑎𝑑

𝑟− 𝜆

⇒ 𝜆! = −𝑎 𝜆! =𝑐𝑟 + 𝑎𝑑

𝑟

Due to the complexity of the 4th equilibrium point, we decided just to apply numbers straight to it, as shown in Appendix 3. We then used Mathematica to plot Figure 6. illustrates this model when plotted using Mathematica. The diagram displays a unstable node, two saddle points and a stable spiral. Therefore the two populations can co-exist.

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Figure 7 (below) shows a graph of each population plotted against time. As you can see, the predator starts out high and the prey low, after the interaction both populations plateau at a constant value. As the predator population decreases, the prey population increases. This graph is almost identical to figure 5.

Figure 6

Prey

Predator

Figure 7

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3. Discussion

As expected, all our equilibrium points are positive and in the first quadrant, since populations cannot be negative. Appendix 1 -3 show some worked examples, and the corresponding graphs also reflect what we expected the outcome to be.

Our basic model shows that the two species can co-exist at the trivial point (0,0) and then at (!

!, !!). This is what we expected, that both species can have large

populations and can exist side by side. At (0,0) the equilibrium point is a saddle, this means that it is an unstable point and a small deviation in the population will take you away from the equilibrium point. At (!

!, !!) it is a stable centre, this means that co-

existence is possible, which is what we expected.

For the logistic model, we have three equilibrium points. The trivial point of (0,0) and (!

!, 0) and (!

!, !!!!"

!"). The trivial point is a saddle which is the same as our

basic model. At (!!, 0) we have two possible outcomes, 𝜆! will always be positive, 𝜆!

can either be positive or negative. In the instant when 𝜆! is positive we will have a saddle point, however, when 𝜆! is negative we have an asymptotically stable node. This implies that we would have a stable situation. At (!

!, !"!!"

!") we have an unstable

node, this implies that the populations are unlikely to reach this point due to there being an unstable system.

For the double logistic model, we have four equilibrium points, at 0,0 , 0, !

!, !!, 0 𝑎𝑛𝑑 !"!!"

!"!!", !"!!"!"!!"

. (0,0) is an unstable node which implies that

the population will always deviate away from this point. At 0, !!

and !!, 0 we have

saddles which, again, imply that a small deviation in the populations will take you away from the equilibrium point. With regards to our 4th equilibrium point, with the figure we put in, we resulted in a stable spiral. This implies that although the populations are expected to fluctuate greatly, eventually they will reach a point where they are able to co-exist together. However, we realise that this may not always be the case for all values input.

For all our models we used Mathematica to plot the graphs. This was a very time consuming task, as only one member of the group had any knowledge of the complicated programme. The plots from Mathematica confirmed the points that we had worked out.

The largest difficulty that we had was trying to classify the eigenvalues. We got our selves stuck by doing unnecessary specific mathematics which over complicated things.

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4. Conclusion

We achieved our aim, and found that another factor impacts the model greatly. In this instance a fisherman could be the influential factor.

Due to limited time and resources we had to make a number of assumptions in order to simplify our phase-plane models. If we had more time we could have made other refinements to make our model reflect the real world more accurately. These refinements include:

-‐ Introducing another source of food for the bears and finding out what impact they have on the salmon population.

-‐ Thinking about the weather and the climate. If there was a bad storm then there could be a reduction in the number of salmon that make it back to the stream. Over longer periods of time there could be an increase in surface sea level temperatures which could cause a decrease in salmon numbers. Or, an increase in temperature could cause rivers to have more water in due to a greater rate of melting, this could make it harder for the bears to fish.

-‐ Introducing disease that could significantly reduce numbers of one or more species.

Actual Contributors

Alexandra Hunt, Paul Martin, Lorna Gollop, Andrew Haspey and Jasmine Dhaliwal

5. References

Bas, Carles (2000). Fishery science: analysis and present situation. Contributions to Science. Vol. 1, No. 4, pp. 489-510

Beals, M., Gross, L., Harrell, S. (1999). Predator-Prey Dynamics: Lotka-Volterra http://www.tiem.utk.edu/~gross/bioed/bealsmodules/predator-prey.html Accessed 02/05/15

Berryman, Alan A. (1992). The Origins and Evolution of Predator-Prey Theory. Ecology. Vol. 73, No. 5, pp. 1530-1535

Poggiale, J. C. (1998). Lotka-Volterra’s Model and Migrations: Breaking of the Well-Known Centre. Mathematical Computer Modelling. Vol. 27, No. 4, pp. 51-61

Urbano, L. (2011) Malthusian Growth. http://montessorimuddle.org/2011/06/30/malthusian-growth/ Accessed 02/05/15

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Appendix 1 – A worked example of our basic scenario

In the instant where a=b=c=d=1:

𝑑𝑥𝑑𝑡 = 𝑥 − 𝑥𝑦

𝑑𝑦𝑑𝑡 = −𝑦 + 𝑥𝑦

𝑥 − 𝑥𝑦 = 0 ⇒ 𝑥 1 − 𝑦 = 0

⇒ 𝑥 = 0, 𝑦 = 1

−𝑦 + 𝑥𝑦 = 0 ⇒ 𝑦 −1 + 𝑥 = 0

⇒ 𝑦 = 0, 𝑥 = 1

Therefore the Equilibrium Points are at (0,0) and (1,1)

For (0,0)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 1 − 𝑦 −𝑥𝑦 −1 + 𝑥 !, !

= 1 00 −1 = 𝐴

det 𝐴 − 𝜆𝐼 = 1 − 𝜆 00 −1 − 𝜆

= 1 − 𝜆 −1 − 𝜆 = 0

⇒ 𝜆!,! = ±1

This is a saddle

For (1,1)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 1 − 𝑦 −𝑥𝑦 −1 + 𝑥 !, !

= 0 −11 0 = 𝐵

det 𝐵 − 𝜆𝐼 = 0 − 𝜆 −11 0 − 𝜆

= 𝜆! + 1 = 0

⇒ 𝜆!,! = ±𝑖

This is a stable centre.

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Appendix 2-‐ A worked example of our logistic equation

In the instant where:

𝑎 = 2, 𝑟 = 𝑏 = 𝑐 = 𝑑 = 1

𝑑𝑥𝑑𝑡 = 2𝑥 − 𝑥! − 𝑥𝑦 = 0

𝑑𝑦𝑑𝑡 = −𝑦 + 𝑥𝑦 = 0

2𝑥 − 𝑥! − 𝑥𝑦 = 0 ⇒ 𝑥 2 − 𝑥 − 𝑦 = 0

⇒ 𝑥 = 0 𝑜𝑟 𝑥 = 2, 𝑦 = 1

−𝑦 + 𝑥𝑦 = 0 ⇒ 𝑦 −1 + 𝑥 = 0

⇒ 𝑦 = 0, 𝑥 = 1

Therefore the Equilibrium Points are at (0,0), (2,0) and (1,1)

For (0,0)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 2 − 2𝑥 − 𝑦 −𝑥𝑦 −1 + 𝑥 !, !

= 2 00 −1 = 𝐴

det 𝐴 − 𝜆𝐼 = 2 − 𝜆 00 −1 − 𝜆

= (2 − 𝜆)(−1 − 𝜆) = 0

⇒ 𝜆! = 2, 𝜆! = −1

This implies a saddle

For (2,0)

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 2 − 2𝑥 − 𝑦 −𝑥𝑦 −1 + 𝑥 !, !

= −2 −20 1 = 𝐵

det 𝐵 − 𝜆𝐼 = −2 − 𝜆 −20 1 − 𝜆

= (−2 − 𝜆)(1 − 𝜆) = 0

⇒ 𝜆! = −2, 𝜆! = 1

This also implies a saddle

For (1,1)

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𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 2 − 2𝑥 − 𝑦 −𝑥𝑦 −1 + 𝑥 !, !

= −1 −11 0 = 𝐶

det 𝐶 − 𝜆𝐼 = −1 − 𝜆 −11 0 − 𝜆

= 𝜆! + 𝜆 + 1 = 0

⇒ 𝜆!,! =−1 ± 3𝑖

2

This implies a stable spiral.

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Appendix 3 – A worked example of our double logistic model.

At the point a=100, r=1, b=1, c=30, q=1, d=1

𝑑𝑥𝑑𝑡 = 100𝑥 − 𝑥! − 𝑥𝑦 = 0

𝑑𝑦𝑑𝑡 = 30𝑦 + 𝑦! + 𝑥𝑦 = 0

For 0,0

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 100 − 2𝑥 − 𝑦 −𝑥𝑦 30 − 2𝑦 + 𝑥 !, !

= 100 00 30 = 𝐴

det 𝐴 − 𝜆𝐼 = 100 − 𝜆 00 30 − 𝜆

= (100 − 𝜆)(30 − 𝜆) = 0

⇒ 𝜆! = 100, 𝜆! = 30

For 0, !!

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 100 − 2𝑥 − 𝑦 −𝑥𝑦 30 − 2𝑦 + 𝑥 !, !"

= 70 030 −30

det 𝐴 − 𝜆𝐼 = 70 − 𝜆 030 −30 − 𝜆

= 70 − 𝜆 −30 − 𝜆 = 0

⇒ 𝜆! = 70, 𝜆! = −30

For !!, 0

𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 100 − 2𝑥 − 𝑦 −𝑥𝑦 30 − 2𝑦 + 𝑥 !"", !

= −100 −1000 130

det 𝐴 − 𝜆𝐼 = −100 − 𝜆 −1000 130 − 𝜆

= −100 − 𝜆 130 − 𝜆 = 0

⇒ 𝜆! = −100, 𝜆! = 130

For !"!!"!"!!"

, !"!!"!"!!"

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𝐽𝑎𝑐𝑜𝑏𝑖𝑎𝑛: 100 − 2𝑥 − 𝑦 −𝑥𝑦 30 − 2𝑦 + 𝑥 !", !"

= −35 −3565 −65

det 𝐴 − 𝜆𝐼 = −35 − 𝜆 −3565 −65 − 𝜆

= −35 − 𝜆 −65 − 𝜆 + 2275

⇒ 𝜆! + 100𝜆 + 4550

𝜆!,! =−100 ± 17200𝑖

2

Since 𝜆! and 𝜆! are complex conjugates with both real parts equalling less than zero, we therefore have a stable spiral.

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Appendix 4- Group Minutes

Meeting 1 – 2 hours – 26/03/15

Attendees: Lorna, Alexandra, Paul, Jasmine, Andrew, Saranyan

• Decided on population problem • Bears and Salmon • Will all investigate and do some reading over Easter

Meeting 2 – 1 hour – 20/04/15

Attendees: Lorna, Alexandra, Paul, Jasmine, Andrew

• Started working on the model

Meeting 3 – 1 hour – 21/04/15

Attendees: Lorna, Alexandra, Paul, Andrew, Jasmine

• Background research on both animals • Noted some factors which would affect the population rate of salmon, other

than bears. We’ve made a start on constructing our simple interaction models using Lotka-Volterra equations.

Meeting 4 – 2 hours – 22/04/15 Attendees: Lorna, Alexandra, Paul, Andrew, Jasmine

• We had started discussing the phase plane equations. Started to take into consideration all the possible scenarios i.e. the bear population decreasing and the effect this has on the salmon population and the effect a decrease in the salmon population would have on the Bears population.

Note: Andrew had turned up 25 minutes before the session ended and had no input with the discussion. Meeting 5 – 1 hour – 27/04/15 Attendees: Lorna, Alexandra, Paul, Andrew, Jasmine

• We continued work on the phase-plane diagram. • We started a rough draft of the final report.

Meeting 6 – 1hour – 28/04/15 Attendees: Lorna, Alexandra, Paul, Jasmine, Andrew

• We continued work on the phase plane equations and started plotting them on Mathematica. The report has been started and a plan made for it.

Note: Saranyan told us that he wasn’t going to be part of the group anymore

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Meeting 7 – 2.5 hours – 29/04/15 Attendees: Lorna, Alexandra, Paul, Jasmine, Andrew

• We met with Dr Naire before the lecture to clear things up.

• During the lecture Andrew and Jasmine worked on Mathematica, Alexandra and Paul worked on the maths. Lorna typed up the report so far.



• Andrew worked on Mathematica

• Alexandra and Paul worked on the maths

• Lorna had written introduction over the weekend and wrote up the maths

• Jasmine wrote about the maths and had to leave after 1 hour



• Andrew worked on Mathematica

• Alexandra and Paul worked on the maths

• Lorna worked on the report

We had a meeting with Dr Naire prior to the lecture. Paul was unable to attend due to a lecture. Lorna, Alex and Jasmine attended, Andrew wasn’t present.

Meeting 10 – 3 hours – 06/05/15 Attendees: Lorna, Andrew, Paul, Alexandra, Jasmine

• Paul and Alexandra finished up working on the third (more refined) model. • Lorna and Jasmine continued the type of the report. • Andrew created the phase-plane diagram for the third model on Mathematica.

Note: Jasmine left after 2 hours as she had another group project to work on. Andrew arrived an hour late.

Mathematical Modelling Project

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