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Non-Markovian demography dynamics

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  1. Oct 9, 2014 #1
    1. The problem statement, all variables and given/known data
    Let P(t) be a population at time t, b(t) is the birth rate at time t and d(t) is the death rate at time t.
    • Define the relative mortality μ(t) as the probability to die at age t.
    • How is it normalized?
    • Set up an equation of the absolute number of deaths at time t as a convolution of b(t) and μ(t).
    • Suppose the relative mortality decays exponentially as a function of age [tex]\mu(t)=\frac{e^{-t/T}}{T}[/tex] Is it correctly normalized?

    2. Relevant equations
    [tex]P(t)=P_0 +b(t)\cdot t-d(t)\cdot t[/tex]
    [tex]\frac{d}{dt}P(t)=b(t)-d(t)[/tex]
    3. The attempt at a solution
    • The first part was confusing because of the ambiguity age=time but I figured out that if you take into acount all ages you can define an absolute mortality at a time t [tex]μ(t)=d(t)/P(t)[/tex] but I can be wrong.
    • This part was confusing as well. The only help I could get was: "The mortality is normalized taking into acount that everyone must die at a certain age" so my attempt was [tex]P_0=\Sigma\mu(t)P(t)[/tex] because the total population at a time is the sum of the people who dies from that moment (b=0, of course) but I'm not so sure about this. Also I'd want to know if the sums can be converted into an integral so I can plug it into one of the equations I got.
    • So it basically asks me to do something like this: [tex]d(t)=(μ*b)(t)=
      \int_{0}^{\infty}\mu(t-t')b(t')dt'[/tex] and then replace into equation in 2. and solve the equation using Laplace transforms... but I can't see why the appliance of convolution is valid in this case.
    • Ignoring the fact that it doesn't make sense that mortality decays with time, I know that [tex]\int\frac{e^{-t/T}}{T}dt=e^{-t/T}[/tex] but I think it is not correctly normalized because it should be equal to 1 but it does not seems correct divide the thing by ##e^{-t/T}## am I getting something wrong?
     
    Last edited: Oct 9, 2014
  2. jcsd
  3. Oct 14, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Oct 14, 2014 #3

    Ray Vickson

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    I can say one thing for sure: if
    [tex] \frac{d}{dt}P(t)=b(t)-\delta(t) [/tex]
    is correct, and if ##b(t), \delta(t)## are truly non-constant functions of ##t##, then
    [tex] P(t)=P_0 +b(t)\cdot t-\delta(t)\cdot t \: \longleftarrow \: \text{ this is FALSE} [/tex]
    I have used ##\delta(t)## instead of ##d(t)## because it is a bad idea to use ##d## both as a function and as a derivative operator.

    I really could not understand the rest of the post, or at least, could not understand the nature or the questions. I concur that the OP should re-write it or seek clarification from whoever set the problem.
     
  5. Oct 15, 2014 #4

    BvU

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    P(t) as a population at time t is a number. It does not say anything about the age distribution.$$\frac{d}{dt}P(t)=b(t)-d(t)$$is about the only thing that you can say about it.

    in contrast, is a probability density distribution: ##\int_0^\infty \mu(\tau)\; d\tau = 1## means: you are certain to die (at some age). It says nothing at all about d(t) as it is described in the OP, because that doesn't tell you how many people have an age between ##\tau## and ##\tau + d\tau##.

    Like Ray, I'd really like to see the original problem tekst.

    (don't have demographic experience, but do know something about particle size distributions as a function of time. .)
     
  6. Oct 31, 2014 #5
    Thanks to Ray and BvU for their replies. Although I couldn't manage to get clarifications from my teacher, and despite the fact the deadline was a week ago, I ought at least to answer you with the complete homework instructions (besides I'd like to know the answer ):

    1. The problem statement, all variables and given/known data

    Consider the time-dependency of a population that is characterized by the following parameters:
    ##P(t)##: total population at time ##t##
    ##b(t)##: birth rate at time ##t##
    ##d(t)##: death rate at time ##t##
    1. Construct a simple first-order differential equation for ##P(t)## as a function of birth and death rates.

    2. Define the relative mortality ##\mu(t)## as the fraction of people of age ##t## who die at this age; that is, as the probability to die at age ##t##. How is it normalized? How can you calculate the absolute number of deaths at time ##t## as a convolution of ##b(t)## and ##\mu(t)##? Substituting this relationship into the differential equation for ##P(t)##, you obtain an integro-differential equation for the population

    3. To be specific, make the following assumptions for these quantities:
    • There is a constant influx of population ##\dot{P_0}##
    • The birth rate is directly proportional to the total population
    • The relative mortality decays exponentially as a function of age ##\mu(t)=T^{-1}e^{-t/T}## Is it correctly normalized?
    • The population is in equilibrium, that is, it neither grows nor shrinks ##dP(t)/dt=0##
    With these assumptions, find the equilibrium population ##P(t)=\text{constant}=P_0##

    4. For the equilibrium situation, calculate the age profile ##p(t)## of the population-the fraction of the population that has age ##t##- and the average age.

    2. Relevant equations
    The convolution of two functions μ(t) and b(t) is [tex](μ*b)(t)= \int_{0}^{\infty}\mu(t-t')b(t')dt'[/tex] The Laplace transform of a convolution is [tex]\mathcal{L}[(\mu*b)(t)]=M(s)B(s)[/tex] where [tex]M(s)=\mathcal{L}[\mu(t)]\qquad B(s)=\mathcal{L}[b(t)][/tex](I think we're going to need this at some point)

    3. The attempt at a solution
    1. I think we all agree that [tex]\frac{d}{dt}P(t)=b(t)−d(t)[/tex]The other equation is valid for a timestep, say, ##\Delta t##[tex]P(t+\Delta t)=P_0(t)+b(\Delta t)\cdot \Delta t-d(\Delta t)\cdot \Delta t[/tex]
    2. The confusion starts: if ##t## is supposed to be the time why does it say that ##\mu(t)## is the fraction of people of age t who die at age ##t##? In an effort to not complicate this problem I proposed[tex]\mu(t)=\frac{\text{Number of persons who die at a time } t}{\text{Total of persons at a time } t}=\frac{d(t)}{P(t)}[/tex]The normalization proposed by BvU seems right, I answered that, setting b(t)=0, you can say that [tex]P_0=\sum\limits_{t=0}^\infty \mu(t)P(t)[/tex]If the absolute number of deaths ##d(t)## must be a convolution of ##\mu(t)## and ##b(t)## then [tex]d(t)=(\mu*b)(t)=\int_0^\infty \mu(t-t')b(t')dt'[/tex] and substituting back into the agreed equation gives [tex]\frac{d}{dt}P(t)=b(t)-\int_0^\infty \mu(t-t')b(t')dt'[/tex] which in essence is an integro-differential equation.
    At this point I must admit that I tried to figure out the reason of the appliance of the convolution but couldn't came up with it, so if you could explain me I would be grateful.

    3.If there is an influx ##\dot{P_0}## then [tex]\frac{d}{dt}P(t)=\dot{P_0}+b(t)-\int_0^\infty \mu(t-t')b(t')dt'[/tex]
    We are told that ##b(t)=\beta P(t)## so the equation becomes [tex]\frac{d}{dt}P(t)=\dot{P_0}+\beta P(t)-\int_0^\infty \mu(t-t')\beta P(t')dt'[/tex] Now, because the mortality decays with time (I still don't get it) the equation would be [tex]\frac{d}{dt}P(t)=\dot{P_0}+\beta P(t)-\int_0^\infty \frac{e^{-(t-t')/T}}{T}\beta P(t')dt'[/tex] Finally, if the net poblational change is null then
    [tex]0=\dot{P_0}+\beta P(t)-\int_0^\infty \frac{e^{-(t-t')/T}}{T}\beta P(t')dt'\\ \Rightarrow\dot{P_0}+\beta P(t)=\int_0^\infty \frac{e^{-(t-t')/T}}{T}\beta P(t')dt'[/tex]Then resolving the last equation by the method of the Laplace transform[tex]
    \mathcal{L}[\dot{P_0}]=\frac{\dot{P_0}}{s}[/tex] [tex]\mathcal{L}[\beta P(t)]=\beta \Psi(t)[/tex] [tex]\mathcal{L}[(\mu*b)(t)]=M(s)B(s)[/tex]Now this part may be wrong (haven't done Laplace transforms in a while)[tex]M(s)=\mathcal{L}\left[\frac{e^{-t/t}}{T}\right]=\frac{T}{s-T}[/tex][tex]B(s)=\mathcal{L}[b(t)]=\mathcal{L}[\beta P(t)]=\beta \Psi(s)[/tex]So the main equation becomes [tex]\frac{\dot{P_0}}{s}+\beta \Psi(s)=\beta \Psi(s)\left(\frac{T}{s-T}\right)[/tex][tex]\Rightarrow \frac{\dot{P_0}}{s}=\left(\frac{2T-s}{s-T}\right)\beta\Psi(s)[/tex]Finally, by the appliance of the inverse Laplace transform (again: this could be wrong),[tex]P(t)=A\beta\cos\left(\frac{e^{-t/T}}{\dot{P_0}}\right)[/tex] where ##A## is a constant.

    4. This part is definitely wrong... my answer was that the age profile is [tex]\varpi=\int_{t_0}^{t_f}A\beta\cos\left(\frac{e^{-t/T}}{\dot{P_0}}\right)dt[/tex] So the average age is [tex]p(t)=\frac{P(t)}{\varpi}[/tex]Anything you can correct or add I'd be grateful :)
     
  7. Nov 1, 2014 #6

    BvU

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    You did an impressive amount of work to post the whole thing !
    I am pleased I stumbled on the answer to 1) and we agree. For 2) I prepared a little bit by introducing ##\tau## for age. That way you can work towards the convolution they hint at: after all, the number of people that have age ##\tau## at time t is ##b(t-\tau)##; a fraction ##\mu(\tau)## of them die, so the total death rate at time t is ##\displaystyle \int_{-\infty}^0 b(t-\tau)\mu(\tau) \;d\tau##.

    [edit] oh, sorry, you had this worked out already with t' to distinguish from t.

    You ask whence the convolution thing: it comes in more or less naturally and gets rid of d . It replaces d by a model for d (generally ##\mu(\tau)## is less dependent on t than d and here it's considered a time-independent function of ##\tau## only).
    In the same way ##b(t)## is modelled further down as ##\beta p(t)## with ##\beta## considered as a constant.

    You must understand that I strongly object to $$\mu(t) = {d(t)\over P(t)}$$ left hand side is a function of age, right hand side is a function of time.

    Equally $$P_0=\sum\limits_{t=0}^\infty \mu(t)P(t)$$makes no sense: you can't multiply ##\mu(\tau)## and ##P(t)##. And ##P_0## is a number yet to be found (from the steady state solution of the equation that we have and that will be extended with ##\dot P_0##.


    I get a little lost in 3) for lack of time. But it's a very interesting exercise, so if no one else chips in I'll certainly be back later :)
     
  8. Nov 3, 2014 #7
    This makes sense.
    That is exactly were I was confused at first! I opted to discard any dependence of ##\mu## with respect to age and instead make it a time-dependent mortality... the reason is that the main equation would depend on two variables and it seemed to over-complicate the problem.
    My fault: at the time I was thinking (like before) in a timestep, i.e., $$P_0=\sum\limits_{\Delta t=0}^\infty \mu(\Delta t)P(\Delta t)$$ My thought process was: how to determine the initial population ##P_0##?, if nobody borns, then you can sum up the people who dies from that moment...
    No hurry ;)
     
  9. Nov 3, 2014 #8

    BvU

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    I don't think there is a need for P(0): if you wait long enough P(t) will become stationary (what they call P0) independent of the value of P(0).

    Your solution for 3) can't be right: it goes to P = 0

    The D.E. to solve is like the one you wrote, but with a different mortality expression: the one I propose in #6.
     
  10. Nov 6, 2014 #9
    Hmm... let me check:[tex]P(t)=A\beta\cos\left(\frac{e^{-t/T}}{\dot{P_0}}\right)[/tex]When the time is long enough (##t\rightarrow\infty##) my answer gives[tex] \lim_{t \to \infty}\frac{e^{-t/T}}{\dot{P_0}}=0\qquad\Rightarrow\qquad \lim_{t \to \infty}P(t)=A\beta\cos(0)=A\beta[/tex]which in essence is a constant.

    That way my answer says that the population tends to oscillate around a constant and is "dampening" until it becomes that stable solution.
     
  11. Nov 7, 2014 #10

    BvU

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    Oops, o:), oops, oops. My mistake. Been reading too fast.

    Did the change in the convolution result in a change in the result ?
    Does P(t) satisfy the D.E. ?
    I half expected A to be found too: something with ##\dot P_0## and T in it. Have to think this over.

    By the way:
    Doesn't make sense at all. Should have been ##\displaystyle \int_0^{\infty} b(t-\tau)\mu(\tau) \;d\tau##
     
    Last edited: Nov 7, 2014
  12. Nov 7, 2014 #11

    BvU

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    Intriguing exercise ...

    -----------------------------

    Intermezzo:
    There was another question that wasn't answered: is ##\mu(t)=\frac{e^{-t/T}}{T}## correctly normalized ?

    You were unhappy
    But I'm happy, because ##\displaystyle \int_0^\infty\frac{e^{-\tau/T}}{T}d\tau=-e^{-\tau/T}\ \Big |_0^\infty = 1\;##.

    -------------------------------

    Coming back to the original D.E. I'm less happy.

    Let ##P_0## be the steady state solution (for which you found ##P_0 = A\beta\ ##, which I don't believe -- but if you don't specify A, then it can of course be written in this form ).

    In fact, I'm beginning to consider the possibility that there is no single steady state solution for P(t): any number can be stable. What could gradually go to a steady distribution is the age distribution.

    Clearly ##{dP(t)\over dt} = 0## then, and also ##P(t-\tau) = P(t) = P_0 ##.
    That may in fact enable us to actually do the convolution and find d(t)

    (it's unfortunate they use the symbol ##\dot P_0## for the influx. We have to distinguish that from ##{dP(t)\over dt}\ ##)

    Furthermore, ##b(t) = \beta P(t) =\beta P_0 ##.

    Then the full D.E. , which reads (I think):

    $${dP(t)\over dt} = \dot{P_0} + \beta P(t) - \int_0^\infty \frac {e^{-\tau/T} } {T} \beta P(t-\tau)\; d\tau$$ simplifies to
    $$0 = \dot{P_0} + \beta P_0 - \int_0^\infty \frac {e^{-\tau/T} } {T} \beta P_0 \; d\tau\\ $$ leading to ##0 =\dot P_0##, cleary unsatisfactory.

    So this time I'm the one to ask: what's going wrong ?

    My own answer: we have the wrong death rate. The influx isn't immortal. But we don't know their age distribution. What do you think ?

    Another thing: For the steady state solution, there's no need to do all these Laplace transforms!

    I am really curious to hear from you if teacher found a decent way to work out this exercise . . .
     
  13. Nov 23, 2014 #12
    First: your normalization is correct.

    Second: My answer is not correct. The cosine function has to have negative values somewhere in its domain unless you have very specific conditions over ##T## and ##\dot{P}_0## -which were not given-. The negative values mean a negative population: impossible.
    Agree. I think that my teacher made up that exercise without actually doing it, or maybe copied it from somewhere in a sloppy way because at the beggining it took into account actual age of the population but never mentioned it again.
    He didn't actually do it. When asked he said that almost everyone solved it with a different method and most of them were correct, even when the answers were not the same... When I presented the homework he said my answer was ok (I was not satisfied though).

    To BvU: Thank you for your help but I think I'm going to leave this problem as it is, partially because it does not provide the information necessary to solve it and partially because I have a lot to do right now. Maybe next year I'll have another look at it.
     
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