21.06.2014 Views

Numerical Methods Contents - SAM

Numerical Methods Contents - SAM

Numerical Methods Contents - SAM

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

Model:<br />

autonomous Lotka-Volterra ODE:<br />

˙u = (α − βv)u<br />

˙v = (δu − γ)v<br />

population sizes:<br />

u(t) → no. of prey at time t,<br />

v(t) → no. of predators at time t<br />

↔ ẏ = f(y) with y =<br />

( u<br />

v)<br />

, f(y) =<br />

( )<br />

(α − βv)u<br />

. (11.1.3)<br />

(δu − γ)v<br />

State of heart described by quantities:<br />

with parameters:<br />

Phenomenological model:<br />

l = l(t) ˆ= length of muscle fiber<br />

p = p(t) ˆ= electro-chemical potential<br />

˙l = −(l 3 − αl + p) ,<br />

ṗ = βl ,<br />

α ˆ= pre-tension of muscle fiber<br />

β ˆ= (phenomenological) feedback parameter<br />

(11.1.4)<br />

This is the so-called Zeeman model: it is a phenomenological model entirely based on macroscopic<br />

observations without relying on knowledge about the underlying molecular mechanisms.<br />

v<br />

vector field f for Lotka-Volterra ODE<br />

✄<br />

α/β<br />

Vector fields and solutions for different choices of parameters:<br />

Solution curves are trajectories of particles carried<br />

along by velocity field f.<br />

γ/δ<br />

u<br />

Fig. 123<br />

Parameter values for Fig. 123: α = 2,β = 1, δ = 1,γ = 1<br />

Ôº¾ ½½º½<br />

Ôº¾ ½½º½<br />

6<br />

5<br />

u = y 1<br />

6<br />

5<br />

2.5<br />

2<br />

1.5<br />

Phase flow for Zeeman model (α = 3,β=1.000000e−01)<br />

3<br />

2<br />

Heartbeat according to Zeeman model (α = 3,β=1.000000e−01)<br />

l(t)<br />

p(t)<br />

4<br />

4<br />

1<br />

1<br />

0.5<br />

y<br />

3<br />

v = y 2<br />

3<br />

p<br />

0<br />

l/p<br />

0<br />

2<br />

v = y 2<br />

Fig. 124<br />

2<br />

−0.5<br />

−1<br />

−1<br />

1<br />

1<br />

−1.5<br />

−2<br />

−2<br />

0<br />

1 2 3 4 5 6 7 8 9 10<br />

t<br />

Solution ( u(t)<br />

v(t)<br />

)<br />

for y0 := ( u(0) ) (<br />

v(0) = 42 )<br />

0<br />

0 0.5 1 1.5 2 2.5 3 3.5 4<br />

u = y 1<br />

Fig. 125<br />

Solution curves for (11.1.3)<br />

−2.5<br />

0<br />

l<br />

2 2.5<br />

Fig. 126<br />

−2.5 −2 −1.5 −1 −0.5 0.5 1 1.5<br />

−3<br />

0 10 20 30 40 50 60 70 80 90 100<br />

time t<br />

Fig. 127<br />

Parameter values for Figs. 125, 124: α = 1, β = 1, δ = 1,γ = 2<br />

Example 11.1.3 (Heartbeat model). → [10, p. 655]<br />

stationary point<br />

✸<br />

Ôº¾ ½½º½<br />

Ôº¾ ½½º½

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!