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Math Help - Forest growth equation

  1. #1
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    Forest growth equation

    Good morning !

    I have an equation for the growth of a tree: let B be the amount of wood in my tree, and t the time since I planted it.
    The growth is as follow:

    dW/dt = a.(W^0.5) - b.W

    This is the normal growth of my tree, when nothing special happens. Parameters a and b are such that my tree starts growing slowly, then reaches a maximum growth, and progressively goes down asymptotically towards zero growth.

    The accumulation of vegetation in my forest, after integration of that function, is:

    W(t) = Q.[1-exp(-Kt)]^2

    (Q = (a/b)^0.5 and K=b/2)

    Up to now, everything is fine. The tree nicely accumulates vegetation following a sigmoid curve towards its maximum, which is equal to Q. Now I plant a whole forest, and I use the same equations to represent the average amount of wood per tree in my forest.

    Now I want to have some disturbances in my forest. Every day, there is a probability p for a tree to burn. When a tree burns, at t+1, it starts regrowing, but may burn again anytime with the same probability p.

    And now I want to know the new asymptotic value of wood in my forest after it reaches equilibrium under the new fire probability (let's call it Wfinal).
    I've built a model of my forest to do that, which is fine, but I need to be able to predict Wfinal mathematically. Basically I need the new function W(t) as a function of a, b and p.

    I modified the growth function to have the fires going on (dB/dt = a.(W^0.5) - b.W - p.W), but that's wrong, because the growth of my trees is not linear through time.

    I gave it long thoughts and many many scribbles on paper but couldn't make it...

    Many thanks for your help !
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  2. #2
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    Re: Forest growth equation

    You didn't define B(t). I assume this is just your modified W(t), ie the size of a single tree allowing for randomly distributed burns.

    Is a differential equation the right modelling approach? It seems that B(t) is not differentiable at the point the tree burns.


    PS Im not an expert on this topic, so dont treat this as gospel. im only replying as no one else has.
    Last edited by SpringFan25; June 29th 2011 at 01:51 PM.
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  3. #3
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    Re: Forest growth equation

    Hi ! Thanks for your reply !
    Sorry B(t) should have been B(t).

    I've made some progress on that but still dont have a straightforward solution. It's probably not the right topic to post in, but I'll just post the answer here anyways for anyone who might be interested.

    The idea is that I just want to know the final average biomass in my forest. That is, I take a forest that is big enough so that the probabilistic model of fire occurence wont make my average biomass varying up and down when t is high (i.e. when I've reached the equilibrium between forest growth and forest disturbance). I should say here if it wasn't clear that at each timestep, only a patch of my forest will burn (i.e. I dont burn my whole forest with probability p). I burn my trees inside the forest with probability p. So basically if my forest is big enough, almost exactly p% of it will burn every year.

    Based on that, I can say that at t1, after I planted my forest, my forest biomass is:

    W = (1-p) * Q.[1-exp(-K*t1)]^2 + p * Q.[1-exp(-K*t0]^2;
    First element is the biomass of 90% of my forest that wasn't burned, second element are the 10% that burned down (thus having to, back to initial growth).

    And at t2:

    W = (1-p) - (1-p)*p * Q.[1-exp(-K*t2)]^2 + (p - p*p) * Q.[1-exp(-K*t1]^2 + p * Q.[1-exp(-K*t0]^2;
    First element is the biomass of the 81% (90% - 10% of 90%) of my forest that still didnt burn since the beginning, second element is the biomass of my 9% of the forest that burned once at t1, and third element are the newly burned trees (10% : 9% from the first element at t1, and 1% of unlucky trees that burned twice). I dont consider any fuel threshold for a tree to burn, i.e. it can burn even if it burned the year before already.

    I take it that in my case, at t400, my forest will be at equilibrium, so I could write a very long function by expanding the t1 and t2 steps up to t400... but that's not convenient. I did some simplification of that to come up with a way to compute W at t400 with a loop on t:

    for t=1 to t=400
    W = W + (p) * exp(-p * (400-t)) * (Q*(1-exp(-k*(400-t))^2;
    end

    But I'd really like to have a way to compute Wt400 (or Wfinal) without looping... it's part of a big model and I need to save running time as much as I can.

    Hope that wasn't too confusing !

    Thank you !
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  4. #4
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    Re: Forest growth equation

    The idea is that I just want to know the final average biomass
    For the avoidance of ambiguity i take "average" to mean "expectation" in the statistical sense. By assuming that the "burn" events follow a poisson process with average rate \lambda per period I obtained the following:


     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left(     \frac{2(e^{-t(k + \lambda)}-1)}{k + \lambda} - \frac{e^{-t(2K +   \lambda)}-1}{2K + \lambda} -\frac{e^{-\lambda t}-1}{\lambda}   \right)

    and hence the asymtotic result is:
    \lim_{t \to \infty} E(B_t) =  -Q \lambda \left(   \frac{2}{K + \lambda} - \frac{1}{2K + \lambda} -\frac{1}{\lambda}  \right)

    Where E(B_t) is the expected size of a tree at time t. Multiply by the number of trees in the forest to get the expected size of the forest (this requires the assumption that all trees burn independently of one another - highly unlikely! perhaps you could model the forest as a big tree instead? just a thought)

    I made these assumptions:

    • When a tree burns it's mass is immediately drops to zero. The tree grows normally afterwards like any other new tree.
    • Tree burns follow a poisson process and occur at an average rate of \lambda per period.
    • Mass at time 0 is 0.
    • The formula for W(t) in your first post is correct.



    Definitions:
    Q and K are as defined in your first post.

    Deriviation (you need to understand undergrad level statistics to follow this):
    Spoiler:

    You have already worked out the mass of a tree if it has been growing continuously for the whole period t:
    W(t) = Q(1-e^{kt})^2

    If burns follow a poisson process it is known that the time between burns follows an exp(\lambda) distribution.

    Define X_t as the amount of time that the tree has been continuously growing at time t. This is the time since the last burn (if there has been a burn), and t otherwise. I believe (but wont prove) that X_t follows the following mixed distribution:

     X_t = r \text{ with probability density } \lambda e^{-\lambda r} ~for~ {0 \leq r \leq t
     X_t = t \text{ with probability mass } e^{-\lambda t}

    it immediately follows that the Mass of the tree (Bt) is also a mixed distribution:

     B_t = W(r) \text{ with probability density } \lambda e^{-\lambda r} ~for~ {0 \leq r \leq t
     B_t = W(t)  \text{ with probability mass } e^{-\lambda t}


    Now evaluate E(B_t) in the normal way:

     E(B_t) = w(t)e^{-\lambda t} +  \int_{0}^{t} w(r) \lambda e^{-\lambda r} dr

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  \int_{0}^{t} \left(Q(1-e^{-Kr})^2 \right) \lambda e^{-\lambda r} dr

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \int_{0}^{t} (1-e^{-Kr})(1-e^{-Kr})  e^{-\lambda r} dr

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \int_{0}^{t} \left( 1 - 2e^{-Kr} + e^{-2Kr} \right) e^{-\lambda r} dr

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \int_{0}^{t} e^{-\lambda r} - 2e^{-r(k + \lambda)} + e^{-r(2K + \lambda)}   dr

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left[  \frac{e^{-\lambda r}}{-\lambda} - \frac{2e^{-r(K + \lambda)}}{-K - \lambda} + \frac{e^{-r(2K + \lambda)}}{-2K - \lambda}  \right]_{0}^{t}

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left[   \frac{2e^{-r(K + \lambda)}}{K + \lambda} - \frac{e^{-r(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda r}}{\lambda}  \right]_{0}^{t}

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left(   \frac{2e^{-t(K + \lambda)}}{K + \lambda} - \frac{e^{-t(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda t}}{\lambda}  \right) - Q \lambda \left(   \frac{2e^{-0(K + \lambda)}}{K + \lambda} - \frac{e^{-0(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda 0}}{\lambda}  \right)

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left(   \frac{2e^{-t(k + \lambda)}}{k + \lambda} - \frac{e^{-t(2K + \lambda)}}{2K + \lambda} -\frac{e^{-\lambda t}}{\lambda}  \right) - Q \lambda \left(   \frac{2}{K + \lambda} - \frac{1}{2K + \lambda} -\frac{1}{\lambda}  \right)

     E(B_t) = Q(1-e^{-K t})^2 e^{-\lambda t} +  Q \lambda \left(    \frac{2(e^{-t(k + \lambda)}-1)}{k + \lambda} - \frac{e^{-t(2K +  \lambda)}-1}{2K + \lambda} -\frac{e^{-\lambda t}-1}{\lambda}  \right)




    READ MY SIGNITURE CAREFULLY. I am not an expert on this topic and am not particularly confident any of the above is correct.
    Last edited by SpringFan25; July 1st 2011 at 03:50 PM.
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  5. #5
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    Re: Forest growth equation

    update:

    i plotted E(B_t) for the arbitrary parameter values: Q=\lambda = K = 1

    i also plotted your original W(t) for comparison (purple line).

    LINK TO GRAPH


    the results look fairly sensible, the "with burns" function has a similar shape but grows slower and reaches a lower asymptote than your original W(t). This does NOT prove that my deriviation was correct, but is obviously a good sign.

    One caveat, i did notice that wofram shows a discontinuity when you plot my function on its own. I hope this is a flaw in wolfram rather than my function!

    http://www.wolframalpha.com/input/?i=plot+%281-e^%28-t%29%29^2+e^%28-t%29+%2B+%28%282%28e^%28-2t%29-1%29%29%2F%282%29+-+%28e^%28-3t%29-1%29%2F%283%29+-%28e^%28-t%29-1%29%2F%281%29%29%2Cfor+t%3D0+to+10

    given that it shows no discontinuity in the first link, i suspect this is a problem with wolfram alpha, but cant guaranteee it.
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  6. #6
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    Re: Forest growth equation

    Great ! That's exactly what I needed ! It fits the spatial model I implemented.

    Thank you so much !
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