 Solution

Write $K[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. At a regular constrained extremal, the first variation of $K$ is not identically zero. Choose a variation $\zeta$ for which $\delta K[\zeta ]\ne 0$; the implicit function theorem supplies a correction in the $\zeta$ direction to each other variation so that $K$ remains one. Stationarity of $J$ on these corrected variations implies $\delta J=\lambda \delta K$ for one scalar $\lambda$.

Put $L=F-\lambda G$. Integration by parts gives, for an arbitrary twice continuously differentiable variation $\eta$,Taking variations with zero endpoint values and using the stated fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. The remaining endpoint values are free and independent, giving the natural boundary conditions:If the constraint is not regular, the general multiplier statement permits an abnormal pair ${\lambda }_{0}F-{\lambda }_{1}G$ with ${\lambda }_0=0$; the source's general normal-form derivation implicitly excludes that case. In its quadratic example, regularity follows automatically from normalization: $\delta K[y]=2\int w{y}^2=2$.

For that example $L=p{y}^{\prime 2}-q{y}^2-\lambda w{y}^2$, so the equation and endpoint conditions reduce toMultiply the equation by $y$ and integrate. The boundary term $[p{y}^{\prime }y{]}_a^b$ vanishes, givingThe last integral equals one, so $\lambda =J[y]$, the functional value at the constrained extremal.

Finally multiply the equation for $y_1$ by $y_2$ and the equation for $y_2$ by $y_1$, then subtract and integrate. The derivative terms give only the vanishing boundary term $[p(y_2{y}_1^{\prime }-y_1{y}_2^{\prime }){]}_a^b$, while the $q$ terms cancel. Therefore $({\lambda }_1-{\lambda }_2){\int }_a^{b}w{y}_1{y}_{2}dx=0$. For distinct multipliers this proves