# Cannot solve recursion relation in power series solution to this ODE

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?

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

## 1 Answer

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