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Diff. eq. with square root

  1. Nov 29, 2008 #1
    Hi all,

    I've been musing on a problem I came across whilst decanting some sloe gin (that time of the year here in the UK and all that). Essentially, I want to know given a vessel of known dimensions, open to the atmosphere at the top, with a hole in the bottom, and a pipe at the top filling with liquid at a steady rate, what height the liquid in the vessel will be at a given time. Anyway, the meat of this boils down to an equation of the form:

    [tex]\frac{dz}{dt} = a - b\sqrt{z(t)},~~~z(0)=0[/tex]

    I've tried a couple of methods to try and quell this beast, for example a simple substitution

    [tex]u(t) = \sqrt{z(t)},~~~\frac{dz}{dt} = 2u(t)\frac{du}{dt}[/tex]

    results in the equally horrible

    [tex]u(t)\left(2\frac{du}{dt} + b\right) = a[/tex]

    I've tried throwing it in Maple too, but no joy (ends up with a Lambert W function, which I don't count as an exact solution), am I missing some insight into this? A quick numerical simulation shows the equation behaves the way I would expect it to.

    (P.S. This isn't homework, I'm just amusing myself with a problem...)
  2. jcsd
  3. Nov 29, 2008 #2


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    Well, this one is separable:
    Simply set [tex]z=u^{2}\to\frac{dz}{du}=2u\to{dz}=2udu[/tex]
    Thereby, we get:
    The lefthand side may by means of:
    Therefore, we are able to find an implicit equation for z(t):
    You will probably not be able to invert this relationship into an explicit one.
  4. Nov 29, 2008 #3
    Is it possible that there is a typo here? I think it should read:
    I think there is a minus sign wrong in the solution as well. This off course if I'm not messing things up :smile:

    [Edit] The solution seems to be transformable into the Lambert W-function, as stated before. This can easily be programmed and thus a valid solution it is indeed.
  5. Nov 29, 2008 #4


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    First off:
    The typo is non-existent

    Blarrgh, you are right about the sign.

    And about the typo in the third expression.
    Hand over the bucket of water so that I can glurg myself into shame, humiliation and penitence.
    But I fear that improvement won't come after all..
  6. Nov 29, 2008 #5
    Yes, I was just plotting the implicit solution and there is a sign error. The method makes sense though - I think I had tried this method but gave up as it gave me an implicit solution. (Although I may not have, I've been toying with it on various 'napkins' whilst not doing real work, you know the way ;) ).

    Yes, it can be programmed but it is not an exact, explicit solution, relying as it does on numerical computation of the L-W function. *

    It's interesting how few DEs we can solve exactly actually...

    [edit]* Please excuse my terminology if wrong, I'm an experimental physicist not a mathematician ;)[edit]
  7. Nov 29, 2008 #6
    So, I need to check the algebra with a pencil (and that I've typed it correctly), but plotting the implicit curve shows it's not behaving the same as the numerical solution for given input parameters (which represent a 10cm x 10cm x 1m vessel, 1cm^2 hole, in flow at 0.0004m^3s^-1:



    Clearly there are many orders of magnitude differences with the exact solution, and the numerical simulation behaves in a much more physical way (i.e. swiftly asymptotic, although this may just be a scale thing - I gave up zooming out on the exact solution!).
  8. Nov 29, 2008 #7
    You should know how many times I make mistakes like these. In the years I've been doing math (I'm an engineer with a profound liking of math) I developed a sixth sence kind of feeling of whether someting is right or not. Sometimes I leave a solution lying around for a few days and start over again to find what's wrong. In the end it all comes together, but this took me years of practice. :smile: The hardest thing to do is to find a mistake in ones own calculation, it is far more easier to check someone elses. Not worth the use of a bucket. :smile:

    As stated before I'm just a simple engineer. Normally I need the use of some program to view or study the solution and this can sometimes require the need of a programming tool and thus the solution in either way is OK. Using Newton-Raphson it just happens to be an easy function to program, from there my remark. Please, use any form you wish.

    Since the problem is solved, the solution I got in the end is:
  9. Nov 29, 2008 #8
    I think it's fair to say we *all* make mistakes in calculations all the time ;)

    Oh sure, on a qualitative level that's just fine (hence my doing a quick iterative numerical simulation to explore the equation), however I find it nice to be able to make a statement, such as "This equation has no exact explicit solution in t", which seems to be the case.

    Hmm, when I plot this it does funny things, with a clear asymptote due to the behavior of
    [tex]ln(1-\sqrt{z})[/tex], although I need to double check the derivation and my typing. Will come back after supper!
  10. Nov 29, 2008 #9
    I looked a bit further on the equation and the solution and came up with the following observations.
    *) Is the following equation the correct physical one?

    \frac{dz}{dt} = \frac{\dot{V}}{A} - \frac{A}{A_0}\sqrt{2\cdot g \cdot z}
    [tex]A[/tex] the section of the vessel
    [tex]A_0[/tex] the section of the hole
    [tex]g[/tex] gravitational constant
    [tex]\dot{V}[/tex] the in flow of liquid

    Thus giving:
    b=\frac{A}{A_0}\sqrt{2\cdot g}

    In case this is correct, you get the following numerical equation (g=9.80665m/s^2)
    \frac{dz}{dt} = 0.04 - 0.04428690551393 \sqrt{z}

    *) For large time we will have a limiting height equal to:
    \frac{dz}{dt} = 0 \qquad z_0=\left(\frac{a}{b}\right)^2

    Which is thus

    *) The solution I gave is only valid for [tex]z<z_0[/tex], giving infinite time to reach this height [tex]z_0[/tex].

    *) A graphic of this function is now:

    This looks like the first one you gave.
    Hope this helps.

    Attached Files:

    Last edited: Nov 29, 2008
  11. Nov 29, 2008 #10

    Not quite, your second term is incorrect. It should read:
    \frac{dz}{dt} = \frac{\dot{V}}{A} - \frac{A_0}{A}\sqrt{2\cdot g \cdot z}

    Thus giving:
    b=\frac{A_0}{A}\sqrt{2\cdot g}

    The rest of your analysis follows nicely. I've just come back to this after eating and I must have typed something wrong when plotting, as it indeed behaves just fine within the limits [tex]0 \leq z \leq z_0[/tex].

  12. Nov 29, 2008 #11
    Indeed, I switched the ratio, it was correct on my paper. A mistake :smile:
    Glad to be of any assistance, Coomast
  13. Nov 29, 2008 #12
    Indeed, I did notice the numerical value of b was correct. A nice little problem.

    But, still safe to say there is no exact explicit solution in terms of elementary functions...

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