Isoperimetric constraint (${\int }_a^{b}G(t,y,y^{\prime })dt=C$)

An isoperimetric constraint fixes the value of an integral while optimizing another integral over functions. A constant Lagrange multiplier incorporates the constraint into the variational integrand, and the resulting Euler-Lagrange equation must be supplemented by the appropriate natural boundary conditions for a free endpoint. The name includes integral constraints more general than prescribed geometric length.

For a nonnegative graph joining $(0,0)$ and $(2a,0)$ with length between $2a$ and $\pi a$, the area-maximizing smooth graph is a circular arch. The Euler-Lagrange equation integrates to the displayed circle formula. Its radius satisfies $l=2R\arcsin (a/R)$ with $R>a$. Concavity of $y-R\sqrt{1+(y^{\prime }{)}^2}$ gives global maximality on the fixed-length class.

A circular arch of chord length $2a$ described as a smooth single-valued graph has radius $R>a$ and length $2R\arcsin (a/R)$, ranging strictly between $2a$ and $\pi a$. The straight-line limit has length $2a$. At the semicircle limit the endpoint slopes are infinite. Longer circular arcs require overhangs and cannot be graphs in this class.

If the first variation of the constraint vanishes in every allowed direction, the objective coefficient in a general multiplier equation can vanish. One must then allow ${\lambda }_{0}F-{\lambda }_{1}G$ with coefficients not both zero. Normalizing ${\lambda }_0$ to one requires the regularity condition used for a normal isoperimetric extremal.

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 a regular closed planar curve, maximize signed enclosed area at fixed length using the variational integrand $\mathcal{L}=(x\dot{y}-y\dot{x})/2+\lambda \sqrt{{\dot{x}}^2+{\dot{y}}^2}$. Write the nonzero speed as $v$. The Euler-Lagrange equations integrate to $\lambda \dot{x}/v=y+c_1$ and $\lambda \dot{y}/v=-x+c_2$. A zero multiplier would force a zero velocity, so squaring and adding yields the displayed circle. The argument uses only regularity and is independent of the speed used to trace the curve. A positively oriented simple traversal of prescribed length $L$ encloses area $L^2/(4\pi )$; the planar isoperimetric inequality supplies the separate global maximality statement.

For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

For a single-valued curve below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\le \pi a$, the maximum enclosed area is attained by the lower minor circular arc. Equivalently it minimizes ${\int }_{-a}^{a}ydx$ at fixed length. Its equation and length areTo prove global minimality for $R>a$, augment the objective by $R\int \sqrt{1+y^{\prime 2}}dx$. The augmented integrand is linear in $y$ and strictly convex in $y^{\prime }$. The displayed arc satisfies its Euler-Lagrange equation, so the supporting-line inequality for the convex slope dependence, followed by integration by parts, proves it minimizes the augmented functional among curves with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The semicircle endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.

For $y(0)=y(1)=0$, write $A={\int }_0^{1}ydx$ and $w=x(1-x)$. Integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, while ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality therefore gives the displayed sharp bound, with equality exactly when $y=Kw$. A prescribed positive budget $B={\int }_0^1(y+y^{\prime 2})dx$ consequently givesThis proves the unique global optimum, rather than only stationarity from a Lagrange multiplier.

The area of a surface of revolution about the horizontal axis is $2\pi \int |y|\sqrt{1+y^{\prime 2}}dx$. For a curve below the axis with endpoints $(\pm a,0)$ and fixed arc length, its stationarity is equivalent to stationarity of $\int (y+\lambda )\sqrt{1+y^{\prime 2}}dx$. The Beltrami identity gives $(y+\lambda )/\sqrt{1+y^{\prime 2}}=C$. Choosing $C=1/k$, imposing the endpoints and integrating the length gives the displayed curve. The function $\sinh s/s$ increases strictly from $1$ to infinity on $s>0$, so every $l>2a$ determines exactly one positive $k$. Stationarity alone does not establish a global minimum.