- #1
ognik
- 643
- 2
Homework Statement
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'.
Homework Equations
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 don't know what to do with the $$ \frac{1}{2r} $$ term? I was looking for something of the form $$ \left(\frac{d\phi}{dr}\right)^{\!{2}} -{k}^{2}\phi = S\left(r\right)$$ 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?