= Solution
{c}
Write $K[y]=\int_a^bG(x,y,y')dx$. At a regular constrained extremal, the <first variation> of $K$ is not identically zero. Choose a variation $\zeta$ for which $\delta K[\zeta]\neq0$; 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 $\eta$,
$$
0=\delta(J-\lambda K)[\eta]
=\int_a^b\left(L_y-\frac d{dx}L_{y'}\right)\eta\,dx
+[L_{y'}\eta]_a^b.
$$
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:
$$
\boxed{F_y-\lambda G_y-\frac d{dx}(F_{y'}-\lambda G_{y'})=0,\qquad
(F_{y'}-\lambda G_{y'})(a)=(F_{y'}-\lambda G_{y'})(b)=0.}
$$
If the constraint is not regular, the general multiplier statement permits an abnormal pair $\lambda_0F-\lambda_1G$ with $\lambda_0=0$; the source's general normal-form derivation implicitly excludes that case. In its quadratic example, regularity follows automatically from normalization: $\delta K[y]=2\int wy^2=2$.
For that example $L=py'^2-qy^2-\lambda wy^2$, so the equation and endpoint conditions reduce to
$$
(py')'+qy+\lambda wy=0,\qquad py'(a)=py'(b)=0.
$$
Multiply the equation by $y$ and integrate. The boundary term $[py'y]_a^b$ vanishes, giving
$$
-\int_a^bpy'^2dx+\int_a^bqy^2dx+\lambda\int_a^bwy^2dx=0.
$$
The last integral equals one, so \b[$\lambda=J[y]$], 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 $[p(y_2y_1'-y_1y_2')]_a^b$, while the $q$ terms cancel. Therefore $(\lambda_1-\lambda_2)\int_a^bwy_1y_2dx=0$. For distinct multipliers this proves
$$
\boxed{\int_a^bwy_1y_2dx=0.}
$$
Back to article page
