Past exam of the mathematical physiology course of the University of Oxford / 2011 / 1 / c / ii / Solution

On the attracting right branch, the leading fast balance is

w = 1 - v

, so

\overset{w}{˙} = v = 1 - w

. On the left branch it is

w = - v

, so

\overset{w}{˙} = - w

. The slow travel times are therefore

T_{R} = \int_{w_{L}}^{w_{R}} \frac{d w}{1 - w} = lo g \frac{1 + α}{β}, T_{L} = \int_{w_{R}}^{w_{L}} \frac{d w}{- w} = lo g \frac{1 - β}{- α} .

Both are positive because

w_{R} > w_{L}

. The fast jumps contribute vanishing time as

ϵ \to 0

, giving the piecewise linear relaxation-oscillator period

T \sim lo g \frac{1 + α}{β} + lo g \frac{1 - β}{- α} = lo g \frac{( 1 + α ) ( 1 - β )}{- α β} .

For the illustrated

α = - 0.25

,

β = 0.4

, the leading period is

1.5041

in the given nondimensional time units.

 Ancestors (13)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version