nr. 477 - 2011 - Institut for Natur, Systemer og Modeller (NSM)
nr. 477 - 2011 - Institut for Natur, Systemer og Modeller (NSM)
nr. 477 - 2011 - Institut for Natur, Systemer og Modeller (NSM)
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6.1 Fixed Points of the IM Model 55<br />
and substitute the expression <strong>for</strong> M into the two other equations. 6 So we get<br />
dMa<br />
dt<br />
dBa<br />
dt<br />
<br />
a + (k + b)Ma<br />
= f1<br />
Ba − kMa<br />
c + f1Ba<br />
<br />
a + (k + b)Ma<br />
= lMa − f1<br />
Ba − f2MaBa − dBa<br />
c + f1Ba<br />
(6.51)<br />
(6.52)<br />
Linearizing and evaluating the Jacobian Matrix of this reduced system at Ma = Ba = 0<br />
yields<br />
<br />
−k<br />
J =<br />
l<br />
f1a<br />
c<br />
− f1a<br />
<br />
c − d<br />
Which is similar to the 2 × 2 matrix shown in equation 6.24, thus the stability behavior<br />
and criteria <strong>for</strong> the healthy rest state is the same <strong>for</strong> the two dimensional system as the<br />
three dimensional system.<br />
The nullclines are found analytically by setting dMa/dt = 0 and dBa/dt = 0 in equation<br />
6.51 and 6.52 and then solving with respect to Ma (or Ba)<br />
Ma nullcline : Ma = Baf1a<br />
ck − Babf1<br />
Ba nullcline :<br />
dc + af1 + f1dBa<br />
Ma = −<br />
f2c + f1(b + k) − lf1 + f1f2Ba − lc<br />
Ba<br />
(6.53)<br />
(6.54)<br />
As it turns out, the nullcline <strong>for</strong> Ma is fairly easy to analyze whereas the nullcline <strong>for</strong><br />
Ba is a little more tricky.<br />
They both approach the point (Ba, Ma) = (0, 0). To convince oneself of this in the case<br />
of the Ba-nullcline, let Ba → 0. Then the term lc/Ba → ∞. Thus the entire fraction will<br />
approach zero as Ba → 0. Now, by differentiating the two expressions <strong>for</strong> the nullclines<br />
(equations 6.53 and 6.54) we get the slope of the Ma nullcline:<br />
and the slope of the Ba nullcline:<br />
dMa<br />
dBa<br />
dMa af1 abf<br />
=<br />
+<br />
dBa ck − bf1Ba<br />
2 1 Ba<br />
(ck − bf1Ba) 2<br />
= (cdBa + af1Ba + df1B 2 a)(f1(b + k) + cf2 − f1l + 2f1f2Ba)<br />
(f1Ba(b + k) − cl + cf2Ba − f1f2B 2 a) 2<br />
− (cd + af1 + 2df1Ba)(f1Ba(b + k) − cl + cf2Ba − f1f2B 2 a)<br />
(f1Ba(b + k) − cl + cf2Ba − f1f2B 2 a) 2<br />
(6.55)<br />
(6.56)<br />
By setting Ba = 0, the initial slope is found to be af1/ck <strong>for</strong> the Ma-nullcline and<br />
(dc+af1)/lc <strong>for</strong> the Ba-nullcline. Remember that we are working with arbitrary (positive)<br />
6 Here we are lucky enough that we are able to separate Ba from the other variables, which is not<br />
always the case. The implicit function theorem, see e.g. (Abraham et al., 1988, p.121), always tells us<br />
if such a separation is possible (in a given interval), and can be applied, if it is not apparent if it is<br />
possible to isolate one variable from the others.