MHB Confirm equation for Numerov method

[ognik]

Thanks for reading.
I am given the wave eqtn $ {[(\d{}{r}})^{2} +\frac{1}{r}\d{}{r}]\Phi\left(r\right)=-{k}^{2}\Phi $
The problems asks to 'show that the substitution $ \Phi={r}^{-\frac{1}{2}}\phi $ gives an eqtn for which the Numerov algorithm is suitable'.
I get $ {(\d{\phi}{r}})^{2}=-\left({k}^{2}+\left(\frac{1}{2r}\right)^{\!{2}}\right)\phi $
Its close, but not quite what I expected, I have checked it a couple of times. I'd appreciate if someone would check if I have it right?
Then the eiganvalues would be $ {\left({k}^{2}+\left(\frac{1}{2r}\right)^{\!{2}}\right)\ }^{-\frac{1}{2}} $?

 

[ognik]

I should show my working I think... I wasn't sure how to show the 2nd order DEs in latex, hope I've got them right...

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 = -{k}^{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 don't 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)$
So I could use some help around how to proceed from here, assuming my workings above are correct?
Thanks

 

[I like Serena]

Hi ognik! :)

Just a quick observation:
$$\left(\!\d {}r\!\right)^{\!2} \ne \frac{d^2}{dr^2}$$

 

[I like Serena]

The wikipedia article about Numerov's method says that the equation should be of the form:
$$\left(\!\frac{d^2}{dx^2}+a(x)\!\right)y(x) = 0$$
Or equivalently, applied to your problem statement:
$$\frac{d^2}{dr^2}\phi(r) = -a(r)\phi(r)$$This suggests that your problem statement has a typo.
Can it be that it should be $$\frac{d^2}{dr^2}$$ instead of $$\left(\!\frac{d}{dr}\!\right)^{\!2}$$?
If that is the case, your derivation is correct.
Moreover, it has the proper form with $$a(r) = k^2+\left(\!\frac{1}{2r}\!\right)^{\!2}$$. (Wasntme)

 

[ognik]

Sorry, mea culpa with the dbl deriv, I just got the latex wrong at first. So I'm happy that my derivation is correct and I can see that it is in the correct form, but I am blank on what to do next. They have given me a Fortan program that uses a 'shooting method' (ie successive corrected guesses) to find the first eigenvalue k, where a(r)=k². The program calls a sub-routine which used Numerov to check the guessed k. It stops when k is close enough.

My task is to modify that program to do the same thing with what I have derived. So far sounds straight forward, I have no problem with coding if I understand what is going on.

But, I just can't see how the a(k) from their program is related to my a(k, r)? (Please excuse the bad notation)
If I had (k +1/2r)² I could manage that, but not the sum of products as I have. Probably I just don't understand the eigenvalue thing, if k is the eigenvalue in their fortran program, what is the eigenvalue in my derivation? Sorry, probably not explaining well, because I probably don't know what I don't know :-)

 

[I like Serena]

Without having seen your fortran program I'll make an educated guess.

In your problem statement r is the variable and k is some constant. I expect that in your fortran program k is the variable and there is no particular constant.

You can probably use the program if you rewrite your problem as:
$$\phi''(k)=-\left(\!c^2+\left(\!\frac 1 {2k}\!\right)^{\!\!2}\right)\phi(k)$$

 

[ognik]

Hi - I've gone through the section again and have gleaned some more information. The section is about finding eigenvalues analytically, their program (copy is viewable at https://drive.google.com/file/d/0B8djt9QSh5rAWldCekRoOGJKVkk/view?usp=sharing) finds the lowest eigenvalue of a stretched string - so its the wave-number I am sure. The (un-normalised) eigenvalues and eigenvectors are known as k_n = n$$\pi$$, $$\phi$$_n = sin n$$\pi$$x

The logic can be summarised as:
guess a trial eigenvalue and generate a solution by integrating the DE as an initial value problem. Iterate until a trial value is found for which the boundary conditions are satisfied within a given tolerance. For each trial value of k, we integrate forward from x=0 to x=1 with the initial conditions $$\phi$$(x=0) = 0, $$\phi$$^'(x=0) = $$\delta$$ (arbitrary). Finding a value of k for which $$\phi$$(1) vanishes is a root finding problem ...

The program I have uploaded to the URL above is probably clearer than all this :-)

I follow it all quite clearly, its just reconciling the constant k in the program with the more complex sum-of-squares function a(k, r) I must work with. Hope that makes it clear enough, if not please ask for anything else you might need to help me out here?

 

[ognik]

Hi, still hoping for some help with this, here's what I've done since the above.

Without that much confidence, I resorted to arithmetic and substituted $$\lambda=k^2 + \frac{1}{2r}^2$$
Then the given program already calculates my 'new' eigenvalue - $$\lambda$$ and I can easily substitute back to find k - a trivial change to the program. (While that sounds sensible, I can't shake the feeling I am missing something critical...).

Anyway, I also need to know the value of r for that, so I adapted their subroutine to return r as well.
But instead of getting the expected value for k (k BTW is just the wave number) which should be 2.404826 (solution to problem according to the text), I get k just smaller than pi (pi is the eigenvalue their original program returns, which is correct - $$k_{n}=v\pi$$)...

So to try and figure this out, I calculated manually what r should have been, and it should have been 0.2473. But I have no idea why my r (1.01) is wrong? or even if I am on the right track. My program and output is attached, be really grateful to get this finished, it is chapters of the text behind where I am busy now ...