MHB Calculating Null Clines: Tips & Tricks

[mt91]

I need to calculate the null clines of these two equations.

I know that in order to find the null cline you set the equations to 0.

I tried to calculate the du/dt equation and got up to
\[ a+u-au-u^2 -v=0 \]
Not entirely sure where I'm supposed to go from there.

For the dv/dt equation I factorised out v to get:
\[ v(bu-c)=0 \]

giving me v=0 and bu-c = 0.

I'm not entirely sure if I'm going about this the correct way so any help would be appreciated, cheers

 

[skeeter]

$u(1-u)(a+u) - uv = 0$

$u[(1-u)(a+u) - v] = 0$

$-u[u^2+(a-1)u - (a - v)] = 0$

$u = 0$, $u = \dfrac{(1-a) \pm \sqrt{(a-1)^2 + 4(a-v)}}{2}$

I'll leave what happens from here to you.

 

[mt91]

$u(1-u)(a+u) - uv = 0$

$u[(1-u)(a+u) - v] = 0$

$-u[u^2+(a-1)u - (a - v)] = 0$

$u = 0$, $u = \dfrac{(1-a) \pm \sqrt{(a-1)^2 + 4(a-v)}}{2}$

I'll leave what happens from here to you.

Nice, so that's the u null clines?

are the v null clines then when:

\[ dv/dt =buv-cv \]
\[ 0=buv-cv \]
\[ 0=v(bu-c) \]
\[ v=0, bu=c \]