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2 Lorenz model

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EFREI – M1 IFM Numerical analysis applied to financial issues

Practical Work #3

The aim of this session is to apply numerical methods to different differential problems and to cope with classical numerical issues.

1 A nonlinear equation (exercise 11)

We consider the ODE

(ty(tˆ ) ˆy0(t)=yˆ2(t)−1, ˆ

y(1)=2. (1)

The exact solution ˆywas calculated in exercise 11. Settn=1+(n−1)∆t, 1nN and∆t=N2−1. 1. What is the final time associated to the discretization?

2. Implement the explicit Euler scheme and the enhanced Euler scheme.

3. Show the curvest7→¡

t,ynum(t)¢

forN=100,N=200,N=500 andN=1000 as well ast7→¡ t, ˆy(t

. 4. For these four values ofN, compute max

1≤n≤N|yny(tˆ n)|. Plot these values for the two schemes and discuss their performance.

2 Lorenz model

The Lorenz model reads 





y10 =10(y2y1), y1(0)= −8, y20 =28y1y2y1y3, y2(0)=8, y30 =y1y283y3, y3(0)=27.

(2)

No exact solution is known for this system of ODEs. That is why we use numerical schemes to construct an approximation of the solution.

1. Implement the explicit Euler scheme as well as the RK4 scheme.

2. Plot the graphst7→¡

t,y1(t)¢ ,t7→¡

t,y2(t)¢

andt7→¡

t,y3(t)¢

for both schemes.

3. Plot the curvet7→¡

y1(t),y2(t)¢ .

4. Compare the numerical solutions. What can you say about the behaviour of the solution?

3 Lotka–Volterra model (exercise 12)

In the Lotka–Volterra model

(x0(t)=x(t

3−y(t

, x(0)=1, y0(t)=y(t)¡

x(t)−2¢

, y(0)=2. (3)

xandydenote respectively the rate of preys and predators in a closed area. Even if the exact solution is not explicitly known, it is proven thatxandyare periodic functions of time. Moreover, the Hamiltonian

H(x,y)=x−2 lnx+y−3 lny is preserved.

18/03/2013 1 M. Girardin – B. Meltz

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EFREI – M1 IFM Numerical analysis applied to financial issues

1. Implement the explicit Euler scheme and the Heun scheme.

2. Plot the graphst7→³ t,H¡

x(t),y(t)¢´

for both schemes as well as the theoretical value.

3. Drawt7→¡

x(t),y(t

. Is the result as expected?

4. We propose the following scheme





xn+1xn

t =xn(3−yn), yn+1yn

t =yn(xn+1−2).

Is the conclusion of Question 3. still holds?

18/03/2013 2 M. Girardin – B. Meltz

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