= Solution
For example, impose homogeneous <Dirichlet boundary conditions> $y(a)=y(b)=0$, and use variations vanishing there. Homogeneous <Neumann boundary conditions> $py'=0$ are another choice removing the boundary terms. Put $N[y]=\int_a^b(py'^2+qy^2)dx$, $D[y]=\int_a^bwy^2dx>0$, and $I=N/D$. At a stationary function let $\lambda=I[y]$. The quotient rule gives $\delta I=(\delta N-\lambda\delta D)/D$. After <integration by parts>, stationarity under every allowed variation gives
$$
0=\int_a^b[-(py')'+qy-\lambda wy]\eta\,dx,
$$
so the <fundamental lemma of the calculus of variations> yields the stated <Sturm-Liouville problem>. Conversely an eigenfunction with those boundary conditions has $N=\lambda D$ and makes the quotient stationary. The lowest eigenvalue minimizes the <Rayleigh quotient> under the usual self-adjoint, lower-bounded hypotheses; higher eigenvalues are stationary values with orthogonality constraints when minimizing successively.
The hydrogen equation becomes $-(r^2\psi')'-2r\psi=2Er^2\psi$. Its radial energy quotient is therefore
$$
E[\psi]=\frac{\int_0^\infty[r^2\psi'^2-2r\psi^2]dr}{2\int_0^\infty r^2\psi^2dr}.
$$
Regularity at zero and decay at infinity remove the boundary term $r^2\psi\psi'$. The condition $\psi(0)=1$ fixes an arbitrary amplitude for these trial functions; it is not their three-dimensional probability normalization, which cancels from the quotient.
For $\psi=e^{-\alpha r}$, elementary exponential integrals give $\int r^2\psi^2dr=1/(4\alpha^3)$, $\int r^2\psi'^2dr=1/(4\alpha)$, and $\int r\psi^2dr=1/(4\alpha^2)$. Hence
$$
E(\alpha)=\frac{\alpha^2}{2}-\alpha=\frac12(\alpha-1)^2-\frac12,\qquad\boxed{\alpha=1,\quad E_{\exp}=-\frac12.}
$$
For the <compactly supported linear trial for hydrogen>, the norm, gradient and Coulomb integrals are $\beta^3/30$, $\beta/3$, and $\beta^2/12$, respectively. Thus
$$
E(\beta)=\frac5{\beta^2}-\frac5{2\beta},\qquad E'(\beta)=\frac{5(\beta-4)}{2\beta^3},\qquad\boxed{\beta=4,\quad E_{\mathrm{linear}}=-\frac5{16}.}
$$
The derivative changes from negative to positive at four, so this is the minimum over positive $\beta$. The <quantum variational principle> gives upper bounds on the <ground-state energy>; \b[the exponential trial is better], and in fact $e^{-r}$ satisfies the exact differential equation at $E=-1/2$. It is the nodeless hydrogen ground state in the given units. The linear trial is admissible for the energy quadratic form despite its derivative jump at $r=\beta$; it need not possess a classical second derivative at that point.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1-hydrogen-variational-trials.png]
{title=Variational energy curves for exponential and compact linear hydrogen trials}
{description=Original energy curves in the equation's atomic units. The exponential family reaches the exact ground-state energy, while the best compact linear trial remains above it.}
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