Null Clines - II

[ol98]

I think you are misunderstanding what "cline" means here. It is NOT a specific value of u or v. A "cline" is a line or curve. "u= 0" is not the number 0 but is the straight line in the uv plane of all points (0, v), the v-axis. The other "cline" is the parabola v= -u^2+ (1- a)u+ a.

I've got a new null cline question, I was wondering if you could confirm I have the correct clines:

The equations are:



For the null clines I got:



Would be great if you could confirm whether this is correct or not.

 

[HallsofIvy]

I think you are still not understanding what a "null cline" is.
There are two "null clines"

At a "null cline" we must have
\(\displaystyle \frac{dn}{d\tau}= \frac{n}{n+ 1}- \beta nc= 0\)
and
\(\displaystyle \frac{dc}{d\tau}= \alpha- \mu c= 0\).

Since the second equation does not involve n, we can solve it directly:
\(\displaystyle c= \frac{\alpha}{\mu}\).

Now replace c in the first equation with that:
\(\displaystyle \frac{n}{n+ 1}- \frac{\alpha\beta}{\mu}n= 0\)
\(\displaystyle \frac{n}{n+ 1}= \frac{\alpha\beta}{\mu}n\)

Obviously n= 0 is a solution. if n is not 0 we can divide both sides by it to get \(\displaystyle \frac{1}{n+ 1}= \frac{\alpha\beta}{\mu}\) so that
\(\displaystyle n+ 1= \frac{\mu}{\alpha\beta}\)
\(\displaystyle n= \frac{\mu}{\alpha\beta}- 1= \frac{\mu- \alpha\beta}{\alpha\beta}\).

The two null clines are
a) \(\displaystyle n= 0\), \(\displaystyle c= \frac{\alpha}{\mu}\). and
b) \(\displaystyle n= \frac{\mu- \alpha\beta}{\alpha\beta}\), \(\displaystyle c= \frac{\alpha}{\mu}\).