Kummer recurrence and termination

[Substitution without dividing by a solution] Put $h(r)=r^{l+1}{e}^{-kr}$ and $u=h\phi$. The multiplier $h$ is smooth and nonzero for every $r>0$, so $u\leftrightarrow \phi$ preserves $C^2$. Its derivatives satisfy $h^{\mathrm{\prime }}\mathrm{/}h=(l+1)\mathrm{/}r-k$ and $h^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}h=l(l+1)\mathrm{/}{r}^2-2k(l+1)\mathrm{/}r+k^2$. The product rule gives the identity $-u^{\mathrm{\prime }\mathrm{\prime }}+[l(l+1)\mathrm{/}{r}^2-\gamma \mathrm{/}r+k^2]u=-h[{\phi }^{\mathrm{\prime }\mathrm{\prime }}+(2(l+1)\mathrm{/}r-2k){\phi }^{\mathrm{\prime }}+(\gamma -2k(l+1))\phi \mathrm{/}r]$. Therefore the radial equation is equivalent to $r{\phi }^{\mathrm{\prime }\mathrm{\prime }}+(2l+2-2kr){\phi }^{\mathrm{\prime }}+(\gamma -2k(l+1))\phi =0$. Write $\phi (r)=\tilde{v}(2kr)$: then ${\phi }^{\mathrm{\prime }}=2k{\tilde{v}}^{\mathrm{\prime }}$ and ${\phi }^{\mathrm{\prime }\mathrm{\prime }}=(2k{)}^2{\tilde{v}}^{\mathrm{\prime }\mathrm{\prime }}$, and division by the nonzero scalar $2k$ gives $x{\tilde{v}}^{\mathrm{\prime }\mathrm{\prime }}+(2l+2-x){\tilde{v}}^{\mathrm{\prime }}+j\tilde{v}=0$. The Statement's $v(x)=e^{x\mathrm{/}2}{x}^{-(l+1)}u(x\mathrm{/}(2k))$ equals $(2k{)}^{-(l+1)}\tilde{v}(x)$; multiplication by this constant preserves the homogeneous equation. All transformations are reversible, including at zeros of $u$, and establish (a).

[Convergence with every differentiated series] Let $b=2l+2$ and $B=\max \{1,\mathrm{\mid }j\mathrm{\mid }\mathrm{/}b\}$. The recurrence gives $\mathrm{\mid }{v}_{i+1}\mathrm{\mid }\le B\mathrm{\mid }{v}_i\mathrm{\mid }\mathrm{/}(i+1)$ because $i+\mathrm{\mid }j\mathrm{\mid }\le B(i+b)$. Induction from $v_0=1$ yields $\mathrm{\mid }{v}_i\mathrm{\mid }\le B^i\mathrm{/}i!$. For any derivative order $q\ge 0$ and $R>0$, the $q$-th formal derivative on $\mathrm{\mid }x\mathrm{\mid }\le R$ has absolute sum bounded by ${\sum }_{i\ge q}{B}^i{R}^{i-q}\mathrm{/}(i-q)!=B^q{e}^{BR}$. Every differentiated series therefore converges uniformly on that compact interval. To justify differentiation explicitly, integrate the uniformly convergent first derivative series from $0$ to $x$; the integral of its finite partial sums is the corresponding original partial sum minus its value at zero, and uniform convergence allows passage to the limit. The fundamental theorem for a continuous integrand proves the derivative identity. Repeating this for successive derivative series proves $v\in C^{\mathrm{\infty }}(\mathbb{R})$ with all asserted derivatives.

[Coefficient equation] By step 1.2, substitution and collection of coefficients are legitimate. The coefficient of $x^i$ in $x{v}^{\mathrm{\prime }\mathrm{\prime }}+b{v}^{\mathrm{\prime }}-x{v}^{\mathrm{\prime }}+jv$ is $(i+1)(i+b)v_{i+1}+(j-i)v_i$ for each $i\ge 0$, including $i=0$ because the $x{v}^{\mathrm{\prime }\mathrm{\prime }}$ contribution then vanishes. The recurrence makes every coefficient zero; the equation thus holds for all real $x$. Conversely this same coefficient comparison shows that an analytic solution with constant coefficient $1$ must have exactly this recurrence. This proves (b).

[Exact termination and polynomial normalization] If $j\mathrm{\notin }{\mathbb{N}}_0$, every factor $i-j$ is nonzero, so all $v_i$ are nonzero. Such a series cannot be a polynomial: its $i$-th derivative at zero is $i!v_i$ by step 1.2. If $j=n\in {\mathbb{N}}_0$, the coefficients through $v_n$ are nonzero, $v_{n+1}=0$, and the recurrence forces all later coefficients to vanish. For the stated Laguerre polynomial, $c_i=(-1{)}^i\left(\genfrac{}{}{0px}{}{n+a}{n-i}\right)\mathrm{/}i!$ has $c_0\mathrm{\ne }0$ and $c_{i+1}\mathrm{/}{c}_i=(i-n)\mathrm{/}[(i+1)(a+i+1)]$ for $0\le i<n$, by the factorial identities in [F4]. Since $a+i+1=i+2l+2$, it has precisely the same recurrence and $v=L_n^{(2l+1)}\mathrm{/}\left(\genfrac{}{}{0px}{}{n+2l+1}{n}\right)$. Its leading coefficient is nonzero. Thus the polynomial and nonzero normalization claims in (d) follow.

[The regular branch and a zero-free initial interval] The function $u_1=u_{\mathrm{r}\mathrm{e}\mathrm{g}}=r^{l+1}{e}^{-kr}v(2kr)$ is smooth on $(0,\mathrm{\infty })$ and solves the radial equation by steps 1.1 and 2.1. Since $v(0)=1$ and $v$ is smooth, $u_1=r^{l+1}(1+O(r))$ as $r\downarrow 0$. It is real valued because the recurrence has real coefficients. Choose $\delta >0$ so that $u_1(r)\mathrm{/}{r}^{l+1}>1\mathrm{/}2$ throughout $(0,\delta )$. Hence $u_1$ has no zero there and has the required normalized endpoint limit.

[Local second solution and its exact leading asymptotic] Fix $r_{\ast }\in (0,\delta )$, put $I(r)={\int }_{r_{\ast }}^r{u}_1(s{)}^{-2}ds$, and define $u_2=u_{1}I$ only on $(0,\delta )$. Here $I^{\mathrm{\prime }}=u_1^{-2}$, so $u_2^{\mathrm{\prime }}=u_1^{\mathrm{\prime }}I+u_1^{-1}$, $u_2^{\mathrm{\prime }\mathrm{\prime }}=u_1^{\mathrm{\prime }\mathrm{\prime }}I$, and $W(u_1,u_2):=u_1{u}_2^{\mathrm{\prime }}-u_1^{\mathrm{\prime }}{u}_2=1$. Thus $u_2$ solves the same equation and is independent of $u_1$. Set $p=2l+1>0$. Since $u_1(s{)}^{-2}\mathrm{/}{s}^{-p-1}\to 1$, comparison between $(1-\eta )s^{-p-1}$ and $(1+\eta )s^{-p-1}$ below any sufficiently small fixed endpoint shows $r^p{\int }_r^{r_{\ast }}{u}_1(s{)}^{-2}ds\to 1\mathrm{/}p$: the integral above that endpoint is finite and disappears after multiplication by $r^p$, while $r^p{\int }_r^t{s}^{-p-1}ds=(1-(r\mathrm{/}t{)}^p)\mathrm{/}p$. Letting $\eta \downarrow 0$ gives $I(r)\sim -r^{-p}\mathrm{/}p$, hence $u_2(r)\sim -r^{-l}\mathrm{/}(2l+1)$. In particular $\mathrm{\mid }{u}_2\mathrm{\mid }\mathrm{/}r\to \mathrm{\infty }$ and $\mathrm{\mid }{u}_2\mathrm{\mid }\mathrm{/}{r}^{l+1}\to \mathrm{\infty }$, also when $l=0$.

[Integrability and actual form-domain approximation] Suppose $j\in {\mathbb{N}}_0$. By step 3.1, $u=u_1$ is a polynomial times $r^{l+1}{e}^{-kr}$. Near zero $u=O(r^{l+1})$, $u^{\mathrm{\prime }}=O(r^l)$, $u\mathrm{/}r=O(r^l)$; at infinity each is bounded by an exponential times a polynomial. Their squared integrals are finite: on $(0,1)$ compare with $r^{2l}$, while at infinity a polynomial times $e^{-2kr}$ is integrable, as follows by bounding each power using the exponential series or repeated integration by parts. Thus $u,u^{\mathrm{\prime }},u\mathrm{/}r\in L^2$, and $u$ has zero trace at zero. To show membership in the actual completion $V_l$, take smooth cutoffs ${\eta }_{\epsilon }$ vanishing below $\epsilon$, equal to one above $2\epsilon$, with $\mathrm{\mid }{\eta }_{\epsilon }^{\mathrm{\prime }}\mathrm{\mid }\le K\mathrm{/}\epsilon$, and ${\chi }_R$ equal to one below $R$, vanishing above $2R$, with $\mathrm{\mid }{\chi }_R^{\mathrm{\prime }}\mathrm{\mid }\le K\mathrm{/}R$. The products $u_{\epsilon ,R}={\eta }_{\epsilon }{\chi }_{R}u$ lie in $C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$. The omitted $L^2$, derivative and $u\mathrm{/}r$ tails tend to zero; the added derivative at zero obeys $\mathrm{\parallel }{\eta }_{\epsilon }^{\mathrm{\prime }}u{\mathrm{\parallel }}_2^2\le 4K^2{\int }_0^{2\epsilon }\mathrm{\mid }u\mathrm{/}r{\mathrm{\mid }}^{2}dr\to 0$, and the added derivative at infinity obeys $\mathrm{\parallel }{\chi }_R^{\mathrm{\prime }}u{\mathrm{\parallel }}_2^2\le K^2{R}^{-2}{\int }_R^{2R}\mathrm{\mid }u{\mathrm{\mid }}^{2}dr\to 0$. Hence $u_{\epsilon ,R}\to u$ in $V_l$, for instance along $\epsilon =1\mathrm{/}n$, $R=n$. This proves (d)'s $L^2$ claim and establishes $u\in V_l$.

[Global continuation across every node] Put $q(r)=l(l+1)\mathrm{/}{r}^2-\gamma \mathrm{/}r+k^2$, continuous on $(0,\mathrm{\infty })$. For $n\ge 1$ let $J_n=[r_{\ast }\mathrm{/}(n+1),(n+1)r_{\ast }]$. Apply [F2] on each $J_n$ to $u^{\mathrm{\prime }\mathrm{\prime }}=qu$ with initial data $u_2(r_{\ast })=0$, $u_2^{\mathrm{\prime }}(r_{\ast })=u_1(r_{\ast }{)}^{-1}$. The unique solutions agree on nested overlaps and with step 4.1 wherever the latter is defined, so their union is a $C^2$ solution on ${\bigcup }_n{J}_n=(0,\mathrm{\infty })$. No blowup at a finite positive node is possible because every such node lies inside one regular compact IVP interval. Differentiating the Wronskian gives $W^{\mathrm{\prime }}=u_1{u}_2^{\mathrm{\prime }\mathrm{\prime }}-u_1^{\mathrm{\prime }\mathrm{\prime }}{u}_2=0$, and its value at $r_{\ast }$ is $1$, so $W=1$ globally. If $u_1(t)=0$, then $-u_1^{\mathrm{\prime }}(t)u_2(t)=1$; this also shows that every such node is simple and the continued $u_2$ is finite and nonzero there. The reciprocal integral of step 4.1 is never used across a node.

[Form eigenidentity and the full operator domain] For $w\in C_c^{\mathrm{\infty }}(0,\mathrm{\infty })$ expand the kinetic integrand: $(u^{\mathrm{\prime }}-u\mathrm{/}r)\overline{(w^{\mathrm{\prime }}-w\mathrm{/}r)}=u^{\mathrm{\prime }}\overline{w^{\mathrm{\prime }}}-u^{\mathrm{\prime }}\bar{w}\mathrm{/}r-u\overline{w^{\mathrm{\prime }}}\mathrm{/}r+u\bar{w}\mathrm{/}{r}^2$. The last three terms are $-(u\bar{w}\mathrm{/}r{)}^{\mathrm{\prime }}$ and have zero integral because $w$ is compactly supported away from the endpoints. Therefore $a_l(u,w)=\int [u^{\mathrm{\prime }}\overline{w^{\mathrm{\prime }}}+l(l+1)u\bar{w}\mathrm{/}{r}^2-\gamma u\bar{w}\mathrm{/}r]=\int [-u^{\mathrm{\prime }\mathrm{\prime }}+(l(l+1)\mathrm{/}{r}^2-\gamma \mathrm{/}r)u]\bar{w}=-k^2⟨u,w⟩$. Compact tests are dense in $V_l$ by its definition, $a_l$ is bounded by [F1], and $w\mapsto ⟨u,w⟩$ is continuous in its norm. The identity extends to every $w\in V_l$. The defining condition in [F1] gives $u\in D(A_l)$ and $A_{l}u=-k^{2}u$. For every allowed $m$, the family with sole nonzero $(l,m)$ entry $u$ has finite graph norm; the exact block identity gives $\psi =u{Y}_{lm}\mathrm{/}r\in D(C)=H^2({\mathbb{R}}^3)$ and $C\psi =-k^2\psi$. Conversely that single-block domain identity gives the stated equivalence. This establishes (e).

[Arbitrary solutions and endpoint uniqueness] At $r_{\ast }$ the initial-data matrix with columns $(u_1,u_1^{\mathrm{\prime }})$ and $(u_2,u_2^{\mathrm{\prime }})$ has determinant $1$ by step 5.1. For any solution $u$, solve the invertible two-by-two linear system for $c_1,c_2\in \mathbb{C}$ matching $(u(r_{\ast }),u^{\mathrm{\prime }}(r_{\ast }))$. The IVP uniqueness of [F2] on the nested $J_n$ gives $u=c_1{u}_1+c_2{u}_2$ everywhere. If $c_2\mathrm{\ne }0$, the asymptotics in steps 3.2 and 4.1 give $u(r)\sim -c_2{r}^{-l}\mathrm{/}(2l+1)$, since $u_1\mathrm{/}{u}_2\to 0$. Hence $\mathrm{\mid }u\mathrm{\mid }\mathrm{/}r\to \mathrm{\infty }$ and $\mathrm{\mid }u\mathrm{\mid }\mathrm{/}{r}^{l+1}\to \mathrm{\infty }$. If $c_2=0$, the latter ratio is bounded, and every nonzero such solution is a nonzero scalar multiple of $u_1$. This proves all of (c); the zero solution is included in the general linear representation.

[Conclusion and foundational use] Step 1.1 proves (a); steps 1.2 and 2.1 prove (b); steps 3.2, 4.1, 5.1 and 6.1 prove (c); steps 3.1 and 4.2 prove (d); and step 5.2 proves (e). CC is used through the regular IVP supplier in [F2]; the compact-interval extensions in step 5.1 themselves are uniquely determined, and the cutoff approximants in step 4.2 are explicit. The actual Coulomb operator premise in [F1] carries AC in its contract; that inherited assumption is retained, without relabeling it as CC-only.