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Confirm the equation for Numerov Integration method

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data
    I am given the wave eqtn: [itex] (\frac {d^2} {dr^2}+\frac{1} {r} \frac {d} {dr})\Phi(r)=−k^2\Phi(r) [/itex]
    The problems asks to 'show that the substitution $$ \Phi=r^{-\frac{1} {2}} \phi $$ gives an eqtn for which the Numerov algorithm is suitable'.

    2. Relevant equations


    3. The attempt at a solution
    I split the eqtn into 3 parts, (1)=LHS 2nd order operator, (2)=LHS 1sr order, (3)=RHS:
    (1) $$ \left(\frac{d\Phi}{dr}\right)^{\!{2}}=\left(\frac{d}{dr}\right)^{\!{2}} \left({r}^{-\frac{1}{2}}\phi\right)=\frac{3}{4}{r}^{-\frac{5}{2}}\phi - {r}^{-\frac{3}{2}}\frac{d\phi}{dr} + {r}^{-\frac{1}{2}}\left(\frac{d\phi}{dr}\right)^{\!{2}} $$
    (2)$$ \frac{1}{r}\frac{d\Phi}{dr} = \frac{1}{r}\frac{d}{dr}\left({r}^{-\frac{1}{2}}\phi\right) = -\frac{1}{2} {r}^{-\frac{5}{2}}\phi + {r}^{-\frac{3}{2}}\frac{d\phi}{dr} $$
    (3) $$ -{k}^{2}\Phi = {r}^{-\frac{1}{2}} {k}^{2}\phi $$

    then (1)+(2)=(3):
    $$ {r}^{-\frac{1}{2}} \left(\frac{d\phi}{dr}\right)^{\!{2}} + \frac{1}{4}{r}^{-\frac{5}{2}}\phi = {r}^{-\frac{1}{2}} {k}^{2}\phi $$
    Multiply by $$ {r}^{\frac{1}{2}} $$ gives:
    $$ \left(\frac{d\phi}{dr}\right)^{\!{2}} +\frac{1}{4}{r}^{-2}\phi=-{k}^{2}\phi $$
    Rearranging gives: $$ \left(\frac{d\phi}{dr}\right)^{\!{2}} =-\left({k}^{2}+\left(\frac{1}{2r}\right)^{\!{2}}\right)\phi $$
    This is almost in the correct form to use Numerov - except that I dont know what to do with the $$ \frac{1}{2r} $$ term? I was looking for something of the form $$ \left(\frac{d\phi}{dr}\right)^{\!{2}} -{k}^{2}\phi = S\left(r\right)$$ Then with S(r)=0, ki would be the eigenvalue(s)
    So I could use some help around how to proceed from here, assuming my workings above are correct?
     
  2. jcsd
  3. Apr 2, 2015 #2

    BvU

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    Hello there,

    You have dealt with the first order term and get a function of r in exchange. Why do you say that this form is 'almost' in the correct form ?
     
  4. Apr 2, 2015 #3

    DrClaude

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    Please check your signs. Also, be careful with your notation, as
    $$ \frac{d^2 \phi}{dr^2} \neq \left(\frac{d\phi}{dr}\right)^{2} $$

    As BvU pointed out, ##k^2## can have a dependence on ##r##.
     
  5. Apr 2, 2015 #4
    Hi & thanks for replying - I realized that I had the dbl diff. wrong - just couldn't code it properly in latex (new to Latex), I would appreciate seeing the correct dbl diff Latex coding (without the end tags) please?

    I thought my working might be correct and understood that k can depend on r, so a better question would have been - is it correct that the eigenvalue is $$ -(k^2 + (\frac{1}{2r})^2)^\frac{1}{2} ?$$ In other words can I treat $$ -(k^2 + (\frac{1}{2r})^2) $$ as $$ -k^2 $$ in my program(call it -l2 to avoid confusion)?

    More on the program might help. I must use a shooting method (search) and the Numerov method [S(r)=0 in this case], to find the lowest eigenvalue. The eigenvalue - $$ -(k^2 + (\frac{1}{2r})^2) $$ - varies with r, so my confusion is - should I be solving for k or for the above eigenvalue?
     
    Last edited: Apr 2, 2015
  6. Apr 3, 2015 #5

    BvU

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    ##-(k^2 + (\frac{1}{2r})^2)^\frac{1}{2}## is the second derivative, so you can treat is as such in the numerical integration. It definitely is not an eigenvalue.

    More on the program is hard to provide without a lot of giveaway. Better you get started and ask for assistance when stuck.
    You do want to think about the behaviour of ##\phi## for ##r\downarrow 0##. The singularities from ##r^{-{\tfrac 1 2}}## might need to be dealt with to make ##\Phi## behave decently.
     
  7. Apr 3, 2015 #6
    Thanks for your reply. This particular problem first gives me a simple Fortran program that calculates the eigenvalue k for the standard eqtn: $$ \frac{d^2\phi}{dr^2}=-k^2\phi $$ (note I figured out the latex). The code uses the so called shooting method to find the eigenvalue - it uses Numerov and the guesses at k to settle on the best value for k (using a tolerance).

    I am supposed to change the given code to find the eigenvalue of the wave eqtn I show the working for above - hence my questions about that as eigenvalue. In terms of operators, I think that $$ D^2\phi+k^2\phi=0 $$ makes k an eigenvalue? So to amend their program I need to understand how do I treat $$ (k^2+(\frac{1}{2r})^2) $$ instead of $$ k^2 $$
     
  8. Apr 14, 2015 #7
    Hi again, still hoping for some help with this. (I know that Φ is not defined at r=0; also k is the wave-number as well as being an eigenvalue)
    This problem comes with an existing fortran program (copy is viewable at https://drive.google.com/file/d/0B8d...ew?usp=sharing) that I must adapt to the situation above. That program outputs the eigenvalue k for the Eq. $$ D^2\phi + k^2\phi=0\:(General\: solution\:is\: k_{n}=n\pi) $$
    What I have tried is to substitute: $$ \lambda = (k^2 + (\frac{1}{2r})^2) $$
    ...so that their program gives me lambda, and then I find k by substituting back into the above eq. I also need to know r, so I adapted their sub to return r.
    They have also provided the value I should get - 2.404826. However, I get nothing close to that - see the adjusted program plus output below. If I am doing the right thing above, then the r value I am calculating is wrong, but I can't see why?

    Would really appreciate being able to finish this one, thanks
    upload_2015-4-15_12-45-9.png
     

    Attached Files:

  9. Apr 15, 2015 #8

    BvU

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    Looks as if r is a constant in your program: the upper bond of the integration.
    Instead you want to have r = ix * h (or r = ix +1 * h) inside the loop, so it really takes part in the whole range.

    Initial conditions for ##\phi## are needed anyway to make the function behave decently at r = 0 (cf here).
    So you want to be very picky with the first two (I think you call them phim and phiz, not really sure)
     
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