= Solution
Write $N[\psi]=\int_a^b(p\psi'^2+q\psi^2)dx$ and $D[\psi]=\int_a^bw\psi^2dx>0$. For a real variation, with $\Lambda=N/D$,
$$
\delta\Lambda=\frac2D\left\{[p\psi'\delta\psi]_a^b+\int_a^b[-(p\psi')'+q\psi-\Lambda w\psi]\delta\psi\,dx\right\}.
$$
Suitable endpoint conditions are homogeneous <Dirichlet boundary conditions>, homogeneous <Neumann boundary conditions> $p\psi'=0$, or one of each. They remove the variation boundary term and the product $[p\psi\psi']_a^b$. Stationarity for all interior variations is equivalent to
$$
\boxed{-(p\psi')'+q\psi=\Lambda w\psi.}
$$
Conversely, multiplication of an eigenfunction equation by $\psi$ and <integration by parts> gives $N=\lambda D$, hence $\boxed{\Lambda[\psi]=\lambda}$. General Robin data require their boundary-energy contributions unless these endpoint products happen to vanish.
Use an eigenbasis orthonormal with respect to the weight: $\int_a^bw\psi_i^*\psi_jdx=\delta_{ij}$. For complex coefficients use the Hermitian version of the quotient, with absolute squares. Self-adjointness gives
$$
\boxed{\Lambda[\widetilde\psi]=\frac{\sum_{i\geq0}|c_i|^2\lambda_i}{\sum_{i\geq0}|c_i|^2}\geq\lambda_0.}
$$
When $c_0\neq0$, dividing numerator and denominator by $|c_0|^2$ yields
$$
\boxed{\widetilde\lambda=\frac{\lambda_0+\sum_{i\geq1}|a_i|^2\lambda_i}{1+\sum_{i\geq1}|a_i|^2},\qquad a_i=c_i/c_0.}
$$
If $c_0=0$, these ratios are undefined, but the unscaled formula and upper-bound conclusion remain valid. The assumptions are a problem with a <self-adjoint operator> bounded below and a complete eigenbasis, with the trial function in its energy-form domain. The <Rayleigh-Ritz method> minimizes this quotient over an admissible trial family, obtaining an upper bound on $\lambda_0$. Higher eigenfunctions are stationary directions and need not be unrestricted extrema.
Back to article page
