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Analytic solution of this, Advice.

  1. Oct 21, 2006 #1

    Clausius2

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    I am looking for the analytic solution of this ODE (if it were one):

    [tex]s^2G''+sG'-(1+s^2+s\text{coth}(s))G=-4s^2e^{-s}[/tex]

    I have solved this equation numerically, it only gives one physically realizable configuration rejecting conveniently one of the homogenous solutions. I don't have those solutions BUT I have the asymptotic behavior of [tex]G(s)[/tex], which turns out to be [tex]G\sim As^{\sqrt{2}}[/tex] and [tex]G(s)\sim se^{-s}}[/tex] for small and large [tex]s[/tex] respectively, where A is a coefficient that I have worked out by means of a linear shooting.

    When writting it on Maple in OdeAdvisor, it says to me that it is a Linear ODE (easy thing to know) and with Linear Symmetries. Does this last thing have something to do with Lie Symmetries?. May this ODE be solvable employing that theory, I don't have a clue.

    Thanks.
     
  2. jcsd
  3. Oct 23, 2006 #2
    For small s we can start with the equation

    [tex]
    u^{\prime \prime} +\lambda u^{\prime}
    +\lambda^{\prime}u=\gamma e^{\int \eta ds}
    [/tex]

    and make the substitution

    [tex]
    u = G(s) e^{\int \eta ds}
    [/tex]

    This will give us the equation

    [tex]
    G^{\prime \prime}(s) + (2 \eta + \lambda) G^{\prime}(s)
    + (\eta^{\prime} + \eta^2 + \lambda \eta + \lambda^{\prime}) G(s) = \gamma
    [/tex]


    which if we set

    [tex]
    \lambda = \frac{1}{s} - 2 \eta
    [/tex]

    and

    [tex]
    \gamma = -4 e^{-s}
    [/tex]

    gives us

    [tex]
    G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s}
    + ( \frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2}) G(s) = -4e^{-s}
    [/tex]

    which we can see is similar to your equation once we divide it by [tex]s^2[/tex].

    [tex]
    G^{\prime \prime}(s) + \frac{G^{\prime}(s)}{s}
    - ( \frac{1}{s^2} + 1 + \frac{\coth{(s)}}{s}) G(s) = -4e^{-s}
    [/tex]


    Now the next bit is to match the [tex]G(s)[/tex] term with your equation. So,

    [tex]
    \frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2}
    = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}
    [/tex]

    Hence we need to choose an [tex]$\eta$[/tex] that fits. If we try substituting

    [tex]
    \eta = \coth{(s)} \pm \frac{\sqrt{2}}{s}
    [/tex]

    we get, with some fiddling,

    [tex]
    \frac{2(1 - \sqrt{2})}{s} \coth{(s)} =
    \frac{2(1 - \sqrt{2})}{s^2}
    [/tex]

    or

    [tex]
    \frac{2(1 + \sqrt{2})}{s} \coth{(s)} =
    \frac{2(1 + \sqrt{2})}{s^2}
    [/tex]

    So, for small s, where we can make the assumption that

    [tex]
    \coth{(s)} \simeq \frac{1}{s}
    [/tex]

    we therefore have a solution.
    The only thing left is to solve the very first equation.
    By inspection we can see that it can be integrated directly to yield

    [tex]
    u^{\prime} + \lambda u = \int \gamma e^{\int \eta ds} ds
    [/tex]

    which is straightforward to solve for u, and hence G.

    For large, positive s we can use the same method, except in our choice of [tex]\eta[/tex]. If we choose

    [tex]\eta = \coth{(s)} - 2 [/tex]

    and plug this into

    [tex]
    \frac{\eta}{s} - \eta^{\prime} - \eta^2 - \frac{1}{s^2}
    = -\frac{1}{s^2} -1 - \frac{\coth{(s)}}{s}
    [/tex]

    and rearrange, we get

    [tex]
    (\frac{2}{s} + 4) \coth{(s)} = \frac{2}{s} + 4 [/tex]

    which as s tends to (positive) infinity, coth(s) tends to 1, and this becomes true.

    You can then solve for u and G as above.
     
    Last edited: Oct 23, 2006
  4. Oct 23, 2006 #3
    will get back to this later...
     
    Last edited: Oct 23, 2006
  5. Oct 23, 2006 #4

    Clausius2

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    Mathew! THANKS A LOT MAN!!. Keep on posting, I will keep track of every line. At first sight it seems you don't find a closed form though....it's an asymptotic behavior isn't it?.
     
  6. Oct 23, 2006 #5

    Clausius2

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    Gold Member

    Hey man. I re-did all the steps, and it pretty much works. The only problem is that finding an exact solution of:

    [tex]\eta^2+\eta'-\eta/s=1+coth(s)/s[/tex]

    seems a pretty difficult task. It's a Riccatti equation. It would be nice to have it for closing the problem. I will mess around with some books to look for a solution (Maple doesn't say anything as always), and also I am gonna see how good is your asymptotic approximation compared with the one (coming from Frobenius at s=0 and approximating [tex]scoth(s)\sim s[/tex] for large s) that I had before. Where did you find this stuff?. I mean, you have the book where this stuff can be found?.

    Thanks.
     
  7. Oct 25, 2006 #6
    Sorry, no book - I've just been fiddling around with it, seeing what comes up.

    Yeah - for the general case I can't find a solution (one that doesn't involve an approximation), but I'll keep looking.
     
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