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Tricky little integral

  1. Mar 9, 2008 #1
    let me apologize in advance for not using latex

    one of my students (I'm an english teacher in japan) mentioned one of the DE's he had tried to solve way back in the day when he was a chemical engineer. He often had trouble and went to his math friend for help. Being a typical mathematician, his friend would ponder the problem for a few moments and then say confidently that there was indeed a solution. Of course he never actually wrote one out.

    now I find myself in a similar situation...

    To start, a simple DE:
    dB/dt = k1*(B +k2)^0.6 * B^0.8 [B is the concentration of something, k1, k2 constants]

    next part is easy:

    (k1*(B +k2)^0.6 * B^0.8)^-1 dB = dt and we just integrate. easy, right?

    int[(B +k)^-0.6 * B^-0.8] dB = t <I dropped the k1> but how can we actually perform this integration?

    I banged my head a couple times, but couldn't think of any nice substitution.
    I'm thinking that maybe there is a pretty way to do this using contour integration, but I haven't come up with anything yet...
    Maple is no help.

    Any thoughts?
     
    Last edited: Mar 9, 2008
  2. jcsd
  3. Mar 9, 2008 #2

    HallsofIvy

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    I doubt there is any elementary anti-derivative. Do you have reason to believe there is one?
     
  4. Mar 9, 2008 #3
    no I have no reason to believe so, but it would be nice.
    Out of curiosity, why do you doubt the existence of an elementary anti-derivative?


    by integrating by parts and iterating the process, it gets big and ugly pretty quickly

    my study was mostly pure math, so I have no ideas as to a numerical solution. Anyone else?
     
  5. Mar 9, 2008 #4

    HallsofIvy

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    Because "almost all" "integrable" functions do not have elementary anti-derivatives. Given an arbitrary (integrable) function, it would be very surprising if it did have an elementary anti-derivative.
     
  6. Mar 9, 2008 #5
    Last edited: Mar 9, 2008
  7. Mar 10, 2008 #6
    I hadn't thought of it. I'll give it a try.
    As for k2, I have no idea. Arbitrary I guess.

    thanks John
     
  8. Mar 10, 2008 #7
    John,

    just tried the expansion. it works fine i guess.
    when I actually perform the integral in Maple I get:
    >>
    > sum(int(binomial(-.6, k)*x^k/x^.8, x), k = 0 .. infinity);
    5 x^(1/5) hypergeom([0.2000000000, 0.6000000000], [1.200000000], -1. x)
    >>

    now I'll research hypergeometric functions...

    can any of you tell me what these mean physically?
     
  9. Mar 27, 2008 #8

    epenguin

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    Firstly this looks to be a very non-fundamental empirical engineer's formula. I think it is being instructive for me as it made me realise something I didn't previously. Hope this is right :shy: . At first I thought ah dB/dt increases with B, so as long as you start with B>0 you will have an accelerating reaction. dB/dt = kB gives you an exponential, this should start a bit slower than an exponential but then when B has built up beyond a certain point it should be faster than exponential. A very dim light was telling me there is supposed to be nothing that grows faster than exponential whatever that means. Then I thought if it grows then for most of the time x will have become much greater than k2 so then it will be sufficiently well approximated by dB/dt = kB^1.4. So solve that and if I am right :surprised it explodes! That is it goes to infinity in a finite time. Try also dB/dt = x^2 or dB/dt = B^n, for any n>1 - it explodes. I graphed some of these on a calculator and confirmed it, also for the original equation. I am also comforted by finding here some similar phenomenon for d.e.s of this general type which you can see doing a Search for 'blow up'. So now I realise what was meant by 'nothing increases faster than exponential' (or slower than logarithm). Nothing that extends over the t axis to infinity anyway; maybe I am not the only one who hadn't grasped this?:shy:


    So I guess there are no real physical laws dB/dt = B^n, for any n>1, over an extended range of B, which would explain why I hadn't met something like this before. So either this was a part of a study of how to avoid explosions in reactors, or maybe you remembered the formula wrong? k1*(B +k2)^-0.6 * B^0.8 sounds nearer a plausible reactor kinetics.


    For actually solving it, I tried the variable B = x^5 ; this gives something apparently simpler like dx/dt = (x^5 + k2)^0.6, but that is still an irrational exponent and seems to get you nowhere much. In fact I found a formula for that and also for the original form in http://integrals.wolfram.com/index.jsp the first time I ever used that too. There is no advantage in my variable change. I know nothing about hypergeometic functions and it is not even evident to me you could see the explosion phenomenon in those formulae. For most people I imagine there is no virtue in not looking it up, if fact there seems to be little point in solving this at all :biggrin: but as I was surprised and unsure I would appreciate comments on the qualitative considerations above.
     
  10. Mar 28, 2008 #9
    epenguin,
    I think your ideas are well founded.

    I'm not sure if you're interested, but I think the engineer in this case is of the mind that B is decreasing.

    I'm not sure exactly how it would work as am not an engineer or chemist, but I'll make a guess...

    The equation might be describing the concentration of a compound/element in a solution which is decreasing over time as it is reacting with something else.

    That's a total guess
     
  11. Mar 28, 2008 #10

    epenguin

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    Thank you grmnsplx. There had to be a minus sign in it somewhere I think, e.g. before the whole expression.
     
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