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#### mt91

##### New member
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

(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.

$0=c/b(1-c/b)(a+c/b)-cv/b$
$cv/b=c/b-c^2/b^2(a+c/b)$

#### Country Boy

##### Well-known member
MHB Math Helper
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

$0=c/b(1-c/b)(a+c/b)-cv/b$
$cv/b=c/b-c^2/b^2(a+c/b)$