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    (Original post by ghostwalker)
    OK, just "seen" the solution to finding the constant.
    PRSOM... {Sigh}.
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    (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?
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    (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.
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    (Original post by DFranklin)
    PRSOM... {Sigh}.
    Thanks for the thought; I have the same problem with so many of your posts.
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    (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.
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    (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.
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    (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!
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    (Original post by ghostwalker)
    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!
    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.
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    (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.
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    (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.
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    (Original post by DFranklin)
    ...
    Thanks for that. Makes sense, after reading it several times (brain's a bit slow!).
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    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)
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    (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!
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    (Original post by kate8)
    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!
    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.
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    (Original post by 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)
    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.
 
 
 
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