ODEs

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Reply 20

DFranklin

Original post by ghostwalker

OK, just "seen" the solution to finding the constant.

PRSOM... {Sigh}.

Reply 21

Shadow11

Original post by ghostwalker

OK, just "seen" the solution to finding the constant.

Setting dN/dt= 0, cancel tau and move the N over to get:

N=(N_0-n_0)e^{\alpha\tau(N-N_0)}

Multiply both sides by

-\alpha\tau e^{-\alpha\tau N}

This gets all references to N over to the left, and the correct form for the Lambert function.

Thanks i now am tantalizingly close to the correct form but my final line has an ln(w(-alpha....etc)) how do i get rid of the ln?

Reply 22

ghostwalker

Original post by Shadow11

Thanks i now am tantalizingly close to the correct form but my final line has an ln(w(-alpha....etc)) how do i get rid of the ln?

Don't know how you got the "ln" there in the first place.

You may find this helpful:

Y = X e ^ X \; \Longleftrightarrow \; X = W(Y)

cribbed from wiki.

If that doesn't sort it, post your working.

Reply 23

ghostwalker

Original post by DFranklin

PRSOM... {Sigh}.

Thanks for the thought; I have the same problem with so many of your posts.

Reply 24

atsruser

Original post by ghostwalker

OK, just "seen" the solution to finding the constant.

Setting dN/dt= 0,

Isn't the problem here to show that dN/dt goes to 0? That's how I read the question. I can't see how to show that though.

Reply 25

ghostwalker

Original post by atsruser

Isn't the problem here to show that dN/dt goes to 0? That's how I read the question. I can't see how to show that though.

Neither can I, which is why I said " But assuming it approaches a constant then setting dN/dt=0 would be my approach." with the emphasis.

Hopefully the OP will enlighten us when the lecturer/tutor goes through it, or someone sees it before then.

(edited 8 years ago)

Reply 26

ghostwalker

Original post by atsruser

Isn't the problem here to show that dN/dt goes to 0? That's how I read the question.

N is a decreasing function bounded below.

Edit:
Would need a bit of analysis to then show that dN/dt goes to 0, as t goes to infinity, and I suspect that's outside the scope of what's expected in the question.

Edit2: On reflection, that, in bold, is insufficient for dN/dt to go to zero. Damn!

(edited 8 years ago)

Reply 27

DFranklin

Original post by ghostwalker

Firstly, from what's been posted, I *really* doubt you're supposed to do anything other than solve for dN/dt goes to 0.

But if you want to prove it I think you can argue as follows (details left as exercise!)

Let M be the greatest lower bound for N(t), let Y be the solution to dN/dt =0.

If M > Y, then we can find g < 0 s.t. for all large t, dN/dt < g. Then applying the MVT we can show N < M for t sufficiently large.

If M < Y, then since dN/dt is a continuous function of N, there must have been some point where dN/dt = 0.

I *think* this still leaves the question of whether dN/dt = 0 at a certain time implies the population is constant after this time. I don't think this is obvious; consider y(t) = t^3, then we can write dy/dt = 3y^{2/3} and when y = 0 this = 0 but it doesn't imply y = 0 for y > t. I think you'd need to look more closely at the DE to rule this out.

Reply 28

ghostwalker

Original post by DFranklin

But if you want to prove it I think you can argue as follows (details left as exercise!)

Let M be the greatest lower bound for N(t), let Y be the solution to dN/dt =0.

If M > Y, then we can find g < 0 s.t. for all large t, dN/dt < g.

Thanks for the reply. Perhaps I'm being dense (wouldn't be the first time), but I fail to see why the bit in bold would be true.

Reply 29

DFranklin

Original post by ghostwalker

Thanks for the reply. Perhaps I'm being dense (wouldn't be the first time), but I fail to see why the bit in bold would be true.

I have to admit I was somewhat assuming "it's not hard to show this" without going through the details. I think this fills the gaps (excepting that I assume there's a unique solution to the Lambert W-function bit):

Let's write D(N) for dN/dt. From the form we found, D(N) is cts, and and D(N_0) < 0.

I'm assuming that we showed from the rearrangement that there's a unique value Y for which D(Y) = 0 (that it's unique almost certainly gets into discussions about behaviour of the W-function but I think it was implicit that we could treat it as unique).

Then if it's always true that N>=M > Y, then since D is a cts function on [M, N_0], D is bounded and attains its bounds. Taking g = sup D, if g >=0, then D =0 somewhere on [M, N_0] (since D(N_0) < 0), which is impossible because D(Y) is the unique root and Y < M.

So g < 0 and so D(N) <=g whenever N >= M. But since this is always true, D(N) <=g for all t.

Reply 30

ghostwalker

Original post by DFranklin

...

Thanks for that. Makes sense, after reading it several times (brain's a bit slow!).

Reply 31

kate8

as the question says infect undiseased individuals at a rate alpha per diseased individual PER UNIT TIME per undressed individual why is dn/dt = an(N-n)t - n/T not correct (added in a t for time if not clear)

Reply 32

kate8

Original post by Shadow11

I've managed to get an answer for part b) but while the working I used to get there makes sense, I feel the answer I get does not, so I hope you will be able to spot my mistake, or if there isn't a mistake please explain how I am able to answer the following question :

in the question it says the final form should be in dN/dt = f(N,t). surely yours is just a function of N. Also it says at a rate of alpha per diseased individual, PER UNIT TIME etc so surely dn/dt = a*n(t)*t*(N-n) - n/T

sorry didn't mean to post twice - just really confused!

Reply 33

Shadow11

Original post by kate8

Yeh, this was pretty much one of the parts of the question that i never understood but my answer seems to be right, i mean how else could it have ultimately resulted in the Nconst b in part d without being correct? Also concerning the N,t thing i think its just possible the equation effectively has a 0t in it? But honestly i am unsure about this.

Reply 34

ghostwalker

Original post by kate8

When it was a rate of 1 per time

\tau

, you had

1/\tau

factor, to get the rate per unit time.

If it's already rate per unit time, no extra factor is required, or alternatively a factor of 1/1 = 1.

Also, note dn/dt, and dN/dt are rates per unit time.

C.f. The derivative dy/dx, which means, rate of change of y, i.e. the change in y per unit change in x.