Mathematical Biology- Coupled ODEs

[binbagsss]

Homework Statement

Attached

Homework Equations

Below

The Attempt at a Solution

To be honest I was going to differentiate one equation to get a 2nd order ODE and plug in the other equation, since to me $v(0)=0$ is not strong enough to do as below, am I completely mis-interpreted?

solution here:

Am I being stupid but I have no idea how we have set $v(t)=0$ for all time, de-coupling the equations trivially, and then solved for $u(t)$ , isn't this effectively claiming this is how $u$ varies with $t$ in general, given simply that $v(0)=0$, to me this is simply the behaviour at $t=0$? Or is the idea something like this holds for $t$ small only - if so, isn't this unclear in the question since it doesn't say to find how $u$ varies with $t$ small or anything like that.

Thanks

 

[Orodruin]

Since $v'$ is zero if $v$ is zero, you cannot have $v(t) \neq 0$ for any $t$ if $v(0) = 0$.

 

[binbagsss]

Since $v'$ is zero if $v$ is zero, you cannot have $v(t) \neq 0$ for any $t$ if $v(0) = 0$.

ah k thanks
why wouldn't you have to check if d^2 v /dt^2 is >0 or <0 to assess whether it is stable or not?

 

[Orodruin]

You will find that all derivatives are equal to zero. This follows directly from differentiating the differential equation. The nth derivative will have terms that are proportional to the function itself or to its lower derivatives. Since all of those are zero, the nth derivative is zero as well.

 

[binbagsss]

You will find that all derivatives are equal to zero. This follows directly from differentiating the differential equation. The nth derivative will have terms that are proportional to the function itself or to its lower derivatives. Since all of those are zero, the nth derivative is zero as well.

d^2v\dt^2 has no v dependence? well it does via u but then you're bringing du/dt in it

 

[Orodruin]

It has two terms. One is proportional to v and the other to v’. Both v and v’ are zero at t=0 and so v’’(0)=0. It does not matter what u or u’ is.