MHB Find Steady States for Equations: u, v and Get Help with Steady State Question

[mt91]

Got a steady state question and was wondering if anyone would be able to check if I'm on the right track?

Find the steady states of these two equations:

My working out as far:

\[ 0=u*(1-u*)(a+u*)-u*v* \]
\[ 0=v*(bu*-c) \]

I looked at the 2nd equation first giving:
\[ v*=0, u*=c/b \]

subbing v*=0 into equation 1 gave:
\[ 0=u*(1-u*)(a+u*) \]
\[ u*=0, u*=1, u*=-a \]
\[ v=a+c/b - ac/b-c^2/b^2 \]

Not sure about that part of the steady state question so any help would be really helpful, cheers

Giving three steady states of
(0,0), (1,0), (-a,0)

Then I looked at subbing u*=b/c. (However this part of my working got confusing and I'm not entirely sure if it was correct.

\[ 0=c/b(1-c/b)(a+c/b)-cv/b \]
\[ cv/b=c/b-c^2/b^2(a+c/b) \]

 

[HOI]

Got a steady state question and was wondering if anyone would be able to check if I'm on the right track?

Find the steady states of these two equations:

View attachment 10548

My working out as far:

\[ 0=u*(1-u*)(a+u*)-u*v* \]
\[ 0=v*(bu*-c) \]

I looked at the 2nd equation first giving:
\[ v*=0, u*=c/b \]

Either v*= 0 or u*=c/b

subbing v*=0 into equation 1 gave:
\[ 0=u*(1-u*)(a+u*) \]
\[ u*=0, u*=1, u*=-a \]

Yes, (u*, v*)= (0, 0), (1, 0), and (-a, 0) are three steady state solutions.

\[ v=a+c/b - ac/b-c^2/b^2 \]

Yes, substituting u*= c/b into 0=u*(1-u*)(a+u*)-u*v* gives 0= (c/b)(1- c/b)(a+ c/b)- (c/b)v* .
(c/b)v*= (c/b)(1- c/b)(a+ c/b) so v*= (1- c/b)(a+ c/b)= a- ac/b+ c/b- c^2/b^2

So a fourth steady state solution is (u*, v*)= (c/b, a+ c/b- ac/b- c^2/b^2).

Not sure about that part of the steady state question so any help would be really helpful, cheers

Giving three steady states of
(0,0), (1,0), (-a,0)

Then I looked at subbing u*=b/c. (However this part of my working got confusing and I'm not entirely sure if it was correct.

\[ 0=c/b(1-c/b)(a+c/b)-cv/b \]
\[ cv/b=c/b-c^2/b^2(a+c/b) \]