 Solution

At a regular constrained extremum, choose an admissible variation $k$ with $\delta J[k]\ne 0$. The implicit function theorem applied to $J[y+\epsilon h+sk]$ permits a compensating $s(\epsilon )$ keeping $J$ fixed. Differentiation at zero gives $\delta I[h]=\lambda \delta J[h]$ for every $h$, where $\lambda =\delta I[k]/\delta J[k]$. Integrating the first variation of $I-\lambda J$ by parts and using arbitrary compactly supported $h$ givesSmoothness alone does not exclude an abnormal constraint: if $\delta J$ vanishes identically, one needs the general multiplier form with two multipliers rather than automatically setting the coefficient of $I$ to one. The displayed normal form uses the usual regular-constraint hypothesis.

For the given quadratic functionals, the equation is $\boxed{(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0}$. If the endpoint values of $y$ are free, the boundary variation $[p{y}^{\prime }h{]}_a^b$ gives the natural conditions $\boxed{p(a)y^{\prime }(a)=p(b)y^{\prime }(b)=0}$, interpreted by limits at singular endpoints. Multiply the equation by $y$ and integrate. These boundary terms vanish, leavingFor two eigenfunctions, multiply their equations by the other function and subtract. Integration givesThe natural conditions remove the bracket, so distinct eigenvalues imply $\boxed{{\int }_a^{b}r{y}_m{y}_{n}dx=0}$.

In the Legendre case, comparison of leading polynomial coefficients gives ${\lambda }_n=n(n+1)$, so the degrees correspond to distinct eigenvalues. A degree-zero polynomial is $y_0=C_0$. Write $y_1=Ax+B$; orthogonality to $1$ on $[-1,1]$ gives $B=0$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThusThese polynomials satisfy the natural endpoint limits because $1-x^2$ vanishes at both endpoints.