 Solution

Use a Lagrange multiplier $\mu$ for the norm constraint. For real variations $u+\delta \eta$ with $\eta (0)=\eta (\pi )=0$, integration by parts givesThus stationarity is equivalent to the Sturm-Liouville problemMultiplying by $u$ and integrating shows $F[u]=\mu G[u]=\mu$, so the constrained stationary values are the eigenvalues, and the normalized stationary functions are their eigenfunctions. Conversely the eigenvalue equation makes the constrained first variation vanish. The self-adjoint operator has weight one.

For $ϵ=0$, zero and negative eigenvalues yield no nonzero solution with both Dirichlet boundary conditions. Positive eigenvalues have solutions $u=A\sin (\sqrt{\lambda }x)$ after the left condition, and the right condition gives $\sqrt{\lambda }=n$, $n\ge 1$. Hence the lowest eigenpair iswhere the amplitude follows from ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$.

The Rayleigh-Ritz variational principle bounds the full lowest eigenvalue by $F[Y_1]$. Its derivative contribution is one. For the linear potential, reflection about $\pi /2$ givesThus the sine trial bound for a linear potential on an interval isFor $ϵ>0$ the bound is strict, since $-Y_1^{\prime \prime }+ϵx{Y}_1=(1+ϵx)Y_1$ is not a constant multiple of $Y_1$. Also ${\lambda }_{ϵ}\ge 1$ by the Poincaré inequality and the nonnegative potential. Equality in the upper bound holds at $ϵ=0$.