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A Problems with identities involving Legendre polynomials

  1. Aug 8, 2017 #1

    hunt_mat

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    I am studying the linear oscillation of the spherical droplet of water with azimuthal symmetry. I have written the surface of the droplet as [tex]F=r-R-f(t,\theta)\equiv 0.[/tex]

    I have boiled the problem down to a Laplace equation for the perturbed pressure, [itex]p_{1}(t,r,\theta)[/itex]. I have also reasoned that the function [tex]f(t,\theta)=\sum_{n=0}^{\infty}\beta_{n}(t)P_{n}(\cos\theta).[/tex]

    The pressure is given by the following:
    [tex]p_{1}=\sum_{n=0}^{\infty}\frac{\ddot{\beta}_{n}(t)}{(n+2)r^{n+1}}P_{n}(\cos\theta)[/tex]

    There is a boundary condition to determine an equation for [itex]\beta_{n}(t)[/itex] which is:
    [tex]p_{1}=-\cot\theta\frac{\partial f}{\partial\theta}-\frac{\partial^{2}f}{\partial\theta^{2}}[/tex]

    I've tried expanding and using a few identities but have gotten bogged down, I can't seem to eliminate the derivatives of [itex]P_{n}(\cos\theta)[/itex] but get bogged down and I am unsure where to go from here.

    Any suggestions?

    I have done the following:
    [tex]
    \begin{eqnarray}
    p_{1}\Big|_{r=1} & = & -\cot\theta\frac{\partial f}{\partial\theta}-\frac{\partial^{2}f}{\partial\theta^{2}} \\
    & = & -\cot\theta\sum_{n=0}^{\infty}\beta_{n}(t)\frac{\partial}{\partial\theta}P_{n}(\cos\theta)-\sum_{n=0}^{\infty}\beta_{n}(t)\frac{\partial^{2}}{\partial\theta^{2}}P_{n}(\cos\theta) \\
    & = & \cot\theta\sum_{n=0}^{\infty}\beta_{n}(t)P_{n}'(\cos\theta)\sin\theta+\sum_{n=0}^{\infty}\beta_{n}(t)\frac{\partial}{\partial\theta}\left[P_{n}'(\cos\theta)\sin\theta\right] \\
    & = & \sum_{n=0}^{\infty}\beta_{n}(t)P_{n}'(t)\cos\theta+\sum_{n=0}^{\infty}\beta_{n}(t)\left[P_{n}'(\cos\theta)\cos\theta-P_{n}''(\cos\theta)\sin^{2}\theta\right] \\
    & = & \sum_{n=0}^{\infty}\beta_{n}(t)\left[2\cos\theta P_{n}'(\cos\theta)-P_{n}''(\cos\theta)\sin^{2}\theta\right]
    \end{eqnarray}
    [/tex]
     
    Last edited: Aug 8, 2017
  2. jcsd
  3. Aug 8, 2017 #2

    MathematicalPhysicist

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    I think there's something wrong here (or that I misunderstood something here), the LHS of the last equation depends on ##(r,t,\theta)## and the RHS depends on ##(\theta,t)## only, that necessarily means that the LHS should vanish for all ##n\ge 0##.

    Obviously it should also be ##\ddot{\alpha_n}##, in your notation you forgot the index ##n##.
     
  4. Aug 8, 2017 #3

    hunt_mat

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    I think I have corrected all the errors you pointed out.
     
  5. Aug 8, 2017 #4

    MathematicalPhysicist

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    Or that ##r## is some constant which doesn't change.

    Anyway, I would check for Rodrigues identity, maybe there are more identities.
     
  6. Aug 8, 2017 #5

    hunt_mat

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    I can use the governing equation to get rid of the double derivative but that leaves the single derivative tern which still needs to be gotten rid of. I was then thinking of setting [itex]\theta=0[/itex] for simplicity.
     
  7. Aug 8, 2017 #6

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    Yes, you can check for ##\theta =0 ##, and then equate both sides for the same index ##n##.

    This way you'll receive a recursion-differential equation for ##\beta_n##.

    I am not sure how to solve this problem, I know of a book that talks about these sort of equations by Bellman and Cooke, never found the time to read it though.
     
  8. Aug 8, 2017 #7

    hunt_mat

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    I was thinking that I could rearrange the equation to get the equation, I think it won't be a recurrence relation though.
     
  9. Aug 8, 2017 #8

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    Why not recursion?, on the LHS you have: ##\ddot{\beta}_n(t)P_n(1)/(n+2) ## and on the right it's ##\beta_n(t)\cdot 2 P_n^'(1)##.

    You have dependence on both ##t## and on ##n##, by the Rodrigues formula for P'.

    Edit: you are right there's no recursion, then it's simpler.
     
  10. Aug 8, 2017 #9

    hunt_mat

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    This problem has been done but I am redoing it by a different perspective to make sure I understand the answer.
     
  11. Aug 8, 2017 #10

    hunt_mat

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    Using the governing equation for Legendre polynomials I have the following:
    [tex]-\sin^{2}\theta P_{n}''(\cos\theta)=\cos\theta\sin\theta P_{n}'(\cos\theta)+n(n+1)P_{n}(\cos\theta)\sin^{2}\theta[/tex]

    Not sure what to do from here.

    I suppose that I could make my job easier by assuming a form of [tex]\beta_{n}(t)=\cos(k(n)t)[/tex] and find values for [itex]k(n)[/itex] but I want to do more later on so that won't help me.
     
  12. Aug 8, 2017 #11

    MathematicalPhysicist

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    If you want to find ##\beta(t)## then in your first post in equation (5) you are basically done, just plug ##\theta=0##, so the term with ##\sin(\theta)## vanishes, and you have a simple ODE for beta.
     
  13. Aug 8, 2017 #12

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    You need to be clearer as to what you want to find, I know that research is a bit unknown territory, but without knowing what you want to get I am not sure how can I or anyone else help you here.
     
  14. Aug 8, 2017 #13

    hunt_mat

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    I have explained what I want, I want an ODE for [itex]\beta_{n}(t)[/itex], my problem is finding out what the derivatives for the Legendre polynomial is.

    Your suggestion for plugging in [itex]\theta=0[/itex] is a tad naive as you don't know what the derivatives are.
     
  15. Aug 8, 2017 #14

    MathematicalPhysicist

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    The derivatives of Legendre polynomial are given by the Rodrigues identity, for example:

    Yes, you are correct as to the fact that you need to solve for each ##n##, but there's no other way around it.

    So we get: ##\ddot{\beta}_n(t)=\beta_n(t)(n+2)(2P'_n(1)/P_n(1))##, that's in case ##P_n(1)\ne 0 ##.
    For each value of ##n##, you calculate both ##P_n(1)## and ##P_n'(1)##, well you can use: ##P_n'(x)=2(n+1)P_{n+1}(x)##;
    but other than that I don't see what can you do.

    BTW, the solution of the ODE depends of course on the sign of the coefficient that multiplies ##\beta##, anyway we guess a solution of the form ##e^{\alpha_n t}##, of course the problem is we don't know the sign of the coefficient, but according to wikipedia there's a recursion relation to ##P_n(x)## so you can find out that this coefficient is indeed positive.
     
  16. Aug 10, 2017 #15

    hunt_mat

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    You have sums to consider which is why I don't thing substituting values is necessarily the right thing to do here. I think that you have to take inner products and as the inner product is an integral, there are some functions which are floating around so it's going to require some work I think.
     
  17. Aug 10, 2017 #16

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    Yes, you are right, then take the inner product, you have the relation between the derivatives of Legendre polynomial and the polynomial itself so it shouldn't be that hard, maybe tedious, but in maths you are accustomed to long and tedious work, aren't you? :-)
     
  18. Aug 10, 2017 #17

    hunt_mat

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    With stuff like this it is very easy to get lost in messy calculation unless you have experience in doing it.
     
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