# Homework Help: Lotka-Volterra Equation Assistance

1. Jun 25, 2012

### Modulus85

1. The problem statement, all variables and given/known data
I'm working a problem that involves a predator-prey relationship scenario. There will be a total of 600 "steps" and I need to calculate the population count for each step.

At this point, I'm just trying to calculate the prey population count for step one.

It's been a long time since I had much math practice and I am confused/frustrated.

a = 0.165
b = 0.0006
x = 250 (prey count)
y = 10

2. Relevant equations
dx/dt = ax - bxy

3. The attempt at a solution
My attempt was to just plug in numbers for the right hand side of the equation, like so:
dx/dt = (0.165 * 250) - (0.0006 * 250 * 10). I came up with 39.75, but considering the count started at 250, 39.75 seems very wrong.

I believe dx/dt is change in population over change time, however I am not sure what to do with that part of the equation.

Any help would be appreciated.

2. Jun 25, 2012

### HallsofIvy

You are leaving a great deal unsaid but I think you mean that you are to use Euler's numerical method to solve the equation. Unfortunately, you don't seem to have any idea what that method is! And you don't appear to know preciselywhat the question is. In Euler's method, we approximate the differential "dt" by some small time step so that we can write dx/dt= f(x,t) as dx= f(x,t)dt. If you are to use "600 steps" to go from t= 0 to t= T, the dt= T/600- but you don't give any final time for the "T" value.

Also, if you are going to have a "predator-prey" problem, you will need both a "predator" and a "prey"! Your equation for dx/dt has both x and y in it but you have no equation for dy/dx.

Last edited by a moderator: Jun 28, 2012
3. Jun 25, 2012

### jackmell

May I add something guys?

Modulus, I agree with everything Hall said. Now, how do we resolve this problem and I don't mean the DE? Well, you've got to step back from this and solve numerically, a simple single DE equation first. Say:

$$\frac{dy}{dt}=2y-1,\quad y(0)=1$$

That's right out of "Differential Equations" by Blanchard, Devaney, and Hall and they solve it numerically via Euler. Get it straight first with just one equation then try and move up to using Euler for a system say:

$$\frac{dx}{dt}=2x-xy$$
$$\frac{dy}{dt}=-y+xy$$

That one is solved numerically in the text too. So worst-case scenario: get that text and study what they did.

4. Jun 27, 2012

### mickeysmiles

I think Modulus85's problem is simpler than everyone is making it out. Modulus: It appears you are confusing dx/dt (the CHANGE) with the resulting population total. Your calculation is correct, and it's as simple as that, however, the equation is for the CHANGE in population, not the population itself. Simply add dx/dt to the the initial population to get the current population (i.e., 39.75+250 = 289.75, which would obviously round down to 289).

5. Jun 28, 2012

### HallsofIvy

It certainly is NOT "that simple" as long as half the problem is not given. We are given that dx/dt depends on y but have not equation telling how y changes. The "Lotke- Volterra" stystem must have two equations.

6. Jun 28, 2012

### mickeysmiles

You are correct that two equations are needed and the problem is not simple IF the calculation was to go 600 steps. However, the OP is only trying to get the calculation correct for step "one." Since the initial y value is given, you don't need dy/dt to solve step one for dx/dt. You just plug in the values. You will obviously need the other equation to go any further, which I'm sure he can easily figure out how to solve dy/dt now that he knows how to correctly solve dx/dt.

7. Jun 29, 2012

### jackmell

You do need a $\Delta t$ though to compute the value of x after step one as it's given by the expression:

$$x_{k+1}=x_k+(a x_k-b x_k y_k)\Delta t$$

so say the time step is one, then:

$$x_{1}=x_0+(a x_0-b x_0 y_0)=250+(0.165 250-0.0006 250 10)=289.75$$

Now, if I had the equation for dy/dt, I could compute $y_1$. Get that one, then I can compute $x_2$ and $y_2$ and so forth.

Frankly though I think me and Hall ran him off. I was just tryin' to help.