4
$\begingroup$

I'm trying to solve the differential equation $$\frac{d^2u}{dr^2} - \left[V_0(r-1)^2 + \frac{\ell(\ell+1)}{r^2} \right]u = -\lambda u$$ where $r\geq0$ is the radial component in spherical coordinates, $\ell\in \mathbb N$ and $V_0 >0$, and $\lambda>0$ are the discrete eigenvalues of the system. I am looking for solutions which behave nice asymptotically, i.e. $u \to 0$ as $r \to 0$ and $r \to \infty$. Using asymptotic analysis, we have $$\frac{d^2u}{dr^2} \approx V_0(r-1)^2u \implies u \approx e^{-\sqrt{V_0}(r-1)^2/2}$$ as $r \to \infty$ and $$\frac{d^2u}{dr^2} \approx \frac{\ell(\ell+1)}{r^2}u \implies u \approx r^{\ell+1}$$ as $r \to 0$. I then guess a solution of the form $$u = e^{-\sqrt{V_0}(r-1)^2/2} r^{\ell+1} \sum_{k=0}^\infty a_k r^k$$ for some power series coefficients $a_k$. Plugging this into the differential equation however, I get recursion equations which depend on 3 coefficients, so I can't solve them. How should one solve this equation using asymptotic analysis and a series solution?

$\endgroup$
  • $\begingroup$ @user10354138 No I am solving a 2nd order equation... what makes you say that? $\endgroup$ – gene Nov 4 '18 at 4:44
  • $\begingroup$ @user10354138 Oh yes, that was just a typo on my part. I've corrected that now. $\endgroup$ – gene Nov 4 '18 at 4:51
  • $\begingroup$ Could you show the recursion relation you obtained? $\endgroup$ – Yuriy S Nov 11 '18 at 12:11
  • $\begingroup$ What happens if you try a solution with $r^{-l}$ instead of $r^{l+1}$? $\endgroup$ – DisintegratingByParts Nov 14 '18 at 19:34
1
$\begingroup$

Hint:

$\dfrac{d^2u}{dr^2}-\left(V_0(r-1)^2+\dfrac{\ell(\ell+1)}{r^2}\right)u=-\lambda u$

$\dfrac{d^2u}{dr^2}-\left(V_0r^2-2V_0r+V_0-\lambda+\dfrac{\ell(\ell+1)}{r^2}\right)u=0$

Let $u=e^{ar^2}v$ ,

Then $\dfrac{du}{dr}=e^{ar^2}\dfrac{dv}{dr}+2are^{ar^2}v$

$\dfrac{d^2u}{dr^2}=e^{ar^2}\dfrac{d^2v}{dr^2}+2are^{ar^2}\dfrac{dv}{dr}+2are^{ar^2}\dfrac{dv}{dr}+(4a^2r^2+2a)e^{ar^2}v=e^{ar^2}\dfrac{d^2v}{dr^2}+4are^{ar^2}\dfrac{dv}{dr}+(4a^2r^2+2a)e^{ar^2}v$

$\therefore e^{ar^2}\dfrac{d^2v}{dr^2}+4are^{ar^2}\dfrac{dv}{dr}+(4a^2r^2+2a)e^{ar^2}v-\left(V_0r^2-2V_0r+V_0-\lambda+\dfrac{\ell(\ell+1)}{r^2}\right)e^{ar^2}v=0$

$\dfrac{d^2v}{dr^2}+4ar\dfrac{dv}{dr}+\left((4a^2-V_0)r^2+2V_0r+2a-V_0+\lambda-\dfrac{\ell(\ell+1)}{r^2}\right)v=0$

Choose $4a^2-V_0=0$ , i.e. $a=\pm\dfrac{\sqrt{V_0}}{2}$ , the ODE becomes

$\dfrac{d^2v}{dr^2}\pm2\sqrt{V_0}r\dfrac{dv}{dr}+\left(2V_0r\pm\sqrt{V_0}-V_0+\lambda-\dfrac{\ell(\ell+1)}{r^2}\right)v=0$

Let $v=e^{\mp\sqrt{V_0}r}w$ ,

Then $\dfrac{dv}{dr}=e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}\mp\sqrt{V_0}e^{\mp\sqrt{V_0}r}w$

$\dfrac{d^2v}{dr^2}=e^{\mp\sqrt{V_0}r}\dfrac{d^2w}{dr^2}\mp\sqrt{V_0}e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}\mp\sqrt{V_0}e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}+V_0e^{\mp\sqrt{V_0}r}w=e^{\mp\sqrt{V_0}r}\dfrac{d^2w}{dr^2}\mp2\sqrt{V_0}e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}+V_0e^{\mp\sqrt{V_0}r}w$

$\therefore e^{\mp\sqrt{V_0}r}\dfrac{d^2w}{dr^2}\mp2\sqrt{V_0}e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}+V_0e^{\mp\sqrt{V_0}r}w\pm2\sqrt{V_0}r\left(e^{\mp\sqrt{V_0}r}\dfrac{dw}{dr}\mp\sqrt{V_0}e^{\mp\sqrt{V_0}r}w\right)+\left(2V_0r\pm\sqrt{V_0}-V_0+\lambda-\dfrac{\ell(\ell+1)}{r^2}\right)e^{\mp\sqrt{V_0}r}w=0$

$\dfrac{d^2w}{dr^2}\pm2\sqrt{V_0}(r-1)\dfrac{dw}{dr}+\left(\lambda\pm\sqrt{V_0}-\dfrac{\ell(\ell+1)}{r^2}\right)w=0$

Let $w=r^bz$ ,

Then $\dfrac{dw}{dr}=r^b\dfrac{dz}{dr}+br^{b-1}z$

$\dfrac{d^2w}{dr^2}=r^b\dfrac{d^2z}{dr^2}+br^{b-1}\dfrac{dz}{dr}+br^{b-1}\dfrac{dz}{dr}+b(b-1)r^{b-2}z=r^b\dfrac{d^2z}{dr^2}+2br^{b-1}\dfrac{dz}{dr}+b(b-1)r^{b-2}z$

$\therefore r^b\dfrac{d^2z}{dr^2}+2br^{b-1}\dfrac{dz}{dr}+b(b-1)r^{b-2}z\pm2\sqrt{V_0}(r-1)\left(r^b\dfrac{dz}{dr}+br^{b-1}z\right)+\left(\lambda\pm\sqrt{V_0}-\dfrac{\ell(\ell+1)}{r^2}\right)r^bz=0$

$\dfrac{d^2z}{dr^2}\pm2\left(\sqrt{V_0}(r-1)-\dfrac{b}{r}\right)\dfrac{dz}{dr}+\left(\lambda\pm\sqrt{V_0}\pm2\sqrt{V_0}\left(b-\dfrac{b}{r}\right)+\dfrac{b(b-1)-\ell(\ell+1)}{r^2}\right)z=0$

Which can converts to Heun's Biconfluent Equation similar to Can one explicitly solve the shifted harmonic oscillator

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.