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For a smooth functional $\int F(x,y,y^{\prime },y^{\prime \prime })dx$, let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. Along an extremal, the Euler-Lagrange equation for a second-derivative functional is $F_y-P^{\prime }+Q^{\prime \prime }=0$. Differentiating $H=y^{\prime \prime }Q+y^{\prime }(P-Q^{\prime })-F$ by the chain rule givesConsequently $H$ is constant whenever $F$ has no explicit dependence on $x$. This extends the Beltrami identity to a second-derivative variational problem. If $F$ is independent of $y^{\prime \prime }$, the expression reduces to $y^{\prime }{F}_{y^{\prime }}-F$.

A sufficiently regular admissible function at which the first variation of a functional vanishes for every permitted variation. An extremal is a stationary candidate; it need not be a local maximum or minimum. For a first-order integral functional with fixed endpoint values, it satisfies the Euler-Lagrange equation under the usual regularity assumptions.

A constrained extremal is normal when the constraint functional has a nonzero first variation in some allowed direction. The implicit function theorem then permits a compensating variation, yielding a multiplier equation with objective coefficient normalized to one. The Euler-Lagrange equation applies to the objective integrand minus the multiplier times the constraint integrand.

For fixed endpoint values, the Euler-Lagrange equation isIt follows by differentiating the functional along $y+\epsilon \eta$ with $\eta (0)=\eta (\pi )=0$, integrating the $F_{y^{\prime }}{\eta }^{\prime }$ term by parts and applying the fundamental lemma of the calculus of variations.

For this quadratic integrand it becomes $y^{\prime \prime }+ky=0$. If $k>0$, write $\nu =\sqrt{k}$; the condition at zero gives $y=B\sin (\nu x)$, and that at $\pi$ requires $B\sin (\nu \pi )=0$. Consequently $y=0$ is the only extremal when $\sqrt{k}$ is not an integer. If $\sqrt{k}=n$ is a positive integer, all extremals areThe integer case $n=0$ must be handled separately: $y^{\prime \prime }=0$ with both endpoints zero has only $y=0$. If negative real $k$ is allowed, the hyperbolic-function solutions also yield only zero under these endpoints. These statements concern stationary functions; no minimum is assumed.

Write $I[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. For endpoint-zero variations $\eta$, the first variations areFor a normal constrained extremal, choose a variation ${\eta }_0$ with $\delta I({\eta }_0)\ne 0$. Given any $\eta$, use the implicit function theorem to choose a small correction in the ${\eta }_0$ direction so that $y+t\eta +s(t){\eta }_0$ preserves the constraint. Differentiating it and the stationarity condition at zero givesIntegrate the derivative-of-variation term by parts. Since the variations vanish at both endpoints, the boundary term is zero. The fundamental lemma of the calculus of variations then givesThe regularity condition on the constraint is what permits the coefficient of $F$ to be normalized to 1. In a degenerate case, the general multiplier statement uses ${\lambda }_{0}F-{\lambda }_{1}G$ with the two coefficients not both zero; the source's displayed form is the normal case.

For a string described as a function graph $r(x)=(x,y(x))$, its arc-length element is $|r^{\prime }(x)|dx=\sqrt{1+y^{\prime }(x{)}^2}dx$. Inextensibility therefore givesThis also forces $l\ge 2a$, with equality only for the straight function graph.

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.

Use the chain rule along a sufficiently smooth extremal of the functional. For $E=y^{\prime }{F}_{y^{\prime }}-F$,The bracket vanishes by the Euler-Lagrange equation. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the Beltrami identity:

When an endpoint's horizontal coordinate is fixed but its height is free, the natural boundary conditions for a free endpoint require $F_{y^{\prime }}=0$ there. In this problem,Every admissible extremal is an upper circular arc with centre on $y=0$, by part (b). Its tangent is horizontal precisely above its centre. Thus imposing the natural boundary conditions for a free endpoint at $x=1$ forces $h=1$. The fixed height at $x=-1$ gives $R^2=4+4c^2$, soIts free endpoint height is $y(1)=2\sqrt{1+c^2}$. Positivity on the whole interval follows because the radicand is at least $4c^2$.

There can be no solution satisfying the natural boundary conditions for a free endpoint at both ends: an upper circular arc has only one point with a horizontal tangent, whereas the two endpoints have distinct horizontal coordinates. Therefore