Null Cline Solutions for Differential Equations: Help Needed!

[mt91]

I was wondering if anyone could help me clarify which null cline solutions are correct for this question I've got:

I've got two differential equations:

\[ du/dt =u(1-u)(a+u)-uv \]
\[ dv/dt = buv-cv \]

where a, b and c are constants.

I know to find the u null clines you set du/dt to 0.

\[ 0=u(1-u)(a+u)-uv \]

At this stage I know u=0 is a solution. However I'm not sure for the next null cline do I find it in terms of u or v?

So is it\[ v=-u^2 -au+u+a \]
or

Any help would be great, cheers

 

[jvicens]

Hello there,

I would like to help clarify the correct null cline solutions for your question. To find the u null cline, you are correct in setting du/dt to 0. This will give you the equation:

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

From here, you can solve for u. The first solution you found, u=0, is correct. This is because when u=0, the entire equation becomes 0=0, which is true. However, there is another solution for u, which is u = (a-v)/(1-v). This can be found by setting the entire equation equal to 0 and solving for u using algebraic techniques.

For the v null cline, you will need to set dv/dt to 0. This will give you the equation:

0 = buv-cv

To solve for v, you can factor out a v to get:

v(bu-c) = 0

This means that either v=0, which is one solution, or bu-c=0, which gives you v = c/b.

So, to answer your question, the correct null cline solutions are:

u null cline: u=0 and u = (a-v)/(1-v)

v null cline: v=0 and v = c/b

I hope this helps clarify things for you. Let me know if you have any further questions.