Euler-Lagrange equation

For $I[y]={\int }_a^{b}F(x,y,y^{\prime })dx$, stationarity under fixed-endpoint variations gives

For an autonomous scalar integral $\int f(u^{\prime })$, the first variation gives the integrated Euler-Lagrange equation $f^{\prime }(u^{\prime })=c$. If $u$ is $C^1$ and each fiber of $f^{\prime }$ is finite, the connected image of the interval under the continuous derivative $u^{\prime }$ lies in a finite set and is therefore a singleton. The fixed endpoints then determine exactly one affine solution. If $f$ is strictly convex, the strict Jensen's inequality applied to $u^{\prime }$ additionally makes this solution the unique global minimizer among fixed-endpoint Lipschitz functions. Without convexity the same unique smooth stationary solution can be a maximizer, for example at zero for $f(v)=1/(1+v^{2008})$ and zero endpoint values.

If a smooth functional has integrand $f(x,v)$ independent of the function value, a weak local minimizer satisfies the displayed first integral almost everywhere. Its first variation gives $\int f_v(x,u^{\prime }){\phi }^{\prime }=0$ for every compactly supported test function. Thus the distributional derivative of $f_v(x,u^{\prime })$ vanishes, making it constant on the interval. Equivalently, any smooth compactly supported function with integral zero is the derivative of such a test function, so the first-variation identity says that $f_v(x,u^{\prime })$ agrees with a single constant against all tests. This reduces the Euler-Lagrange equation from a second-order equation to an algebraic relation in $u^{\prime }$.

For positive twice differentiable functions $y$, the Euler-Lagrange equation for this functional reduces to $y{y}^{\prime \prime }+1+y^{\prime 2}=0$. Thus $(y{y}^{\prime }{)}^{\prime }=-1$ and integration gives $(x-h{)}^2+y^2=R^2$. The upper circular arc is a graph wherever its radicand is positive. A fixed horizontal endpoint with free height imposes $F_{y^{\prime }}=y^{\prime }/(y\sqrt{1+y^{\prime 2}})=0$, so that endpoint must lie above the centre. Distinct horizontal endpoints cannot both satisfy this natural boundary conditions for a free endpoint. The Beltrami identity alone also admits constant functions, which are excluded by the full Euler-Lagrange equation.

Adding a total derivative $dg(q(t))/dt$ to a Lagrangian changes its action only by $g(q(T))-g(q(0))$. Under fixed-endpoint variations, this boundary term is constant, so the Euler-Lagrange equations and their minimizing paths are unchanged. Componentwise, $d[\partial (dg/dt)/\partial {\dot{q}}_i]/dt={\sum }_j{g}_{ij}{\dot{q}}_j=\partial (dg/dt)/\partial q_i$ by equality of mixed partial derivatives. Endpoint freedom would require reconsidering the resulting natural boundary conditions for a free endpoint.

Motion on a sphere with speed proportional to $\sin \theta$ has travel-time line element proportional to $\sqrt{d{\theta }^2/{\sin }^2\theta +d{\varphi }^2}$. With $z=\mathrm{arsinh}(\cot \theta )$, this becomes the flat cylinder metric $\sqrt{d{z}^2+d{\varphi }^2}$. Shortest lifted paths are straight lines. Nonvertical lines give $\cot \theta =\sinh (A\varphi +B)$; vertical lines are meridians of constant $\varphi$. The poles lie at infinite travel-time distance.

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.

When a variational integrand $f(z,z^{\prime })$ has no explicit dependence on the independent variable, every extremal satisfies

If the integrand $L(q_1,\dots ,q_m,q_1^{\prime },\dots ,q_m^{\prime })$ has no explicit dependence on its independent variable $z$, its Euler-Lagrange equations imply the displayed first integral. Indeed, differentiate the left side and cancel the second-derivative terms to obtain ${\sum }_j{q}_j^{\prime }({\partial }_{q_j}L-d({\partial }_{q_j^{\prime }}L)/dz)=0$. This is the several-variable form of the Beltrami identity. Applied to Fermat principle, it expresses conservation of the component of the optical momentum of a ray along a translation-invariant direction.

A catenary is a curve of the form $y=a\cosh ((x-x_0)/a)+y_0$. It is the equilibrium shape of a uniform flexible chain in a uniform gravitational field.

Among curves of fixed arc length $L>2a$ joining $(-a,0)$ to $(a,0)$, the hanging catenary globally minimizes $\int yds$. Parametrize a competitor by arc length $s\in [-L/2,L/2]$, so $x^{\prime 2}+y^{\prime 2}=1$. Integration by parts and the fixed horizontal endpoint separation give, for every $b>0$,The inequality is the Cauchy-Schwarz inequality. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$. Integration gives the catenaryThe last equation has a unique positive $b$, since $\sinh q/q$ increases strictly for $q>0$. The equality curve is a single-valued graph below the chord, so the same global bound proves the result under those additional restrictions.

When the values of $y$ are free at both fixed endpoint locations, integration by parts leaves the boundary term $[F_{y^{\prime }}\eta {]}_a^b$. Since the endpoint variations are arbitrary, stationarity requires

For a functional $J[y]=\int F(x,y,y^{\prime },y^{\prime \prime })dx$, two applications of integration by parts leave the boundary termIf the endpoint displacement is fixed but its slope is free, $\delta y=0$ while $\delta y^{\prime }$ is arbitrary. The first variation therefore forces $F_{y^{\prime \prime }}=0$ at that endpoint. For the beam energy density $F=\beta (y^{\prime \prime }{)}^2/2-\alpha (y^{\prime }{)}^2/2$ with $\beta >0$, this gives $y^{\prime \prime }=0$, the zero-bending moment condition. A clamped endpoint instead prescribes the slope and has $\delta y^{\prime }=0$.

A pointwise constraint on the initial derivative need not be closed in the norm controlling the value and first-derivative square integrals. For example, minimize $b{\int }_0^T{\dot{V}}^{2}dt$ with $b,L,T>0$, $V(0)=0$ and ${\int }_0^{T}Vdt=L$. The attained minimum is $3b{L}^2/T^3$, uniquely achieved by $V_{\ast }=3Lt(2T-t)/(2T^3)$, with natural condition ${\dot{V}}_{\ast }(T)=0$. Adding $\dot{V}(0)=0$ excludes this minimizer but leaves the same infimum. Multiplying $V_{\ast }$ by $1-e^{-t/\epsilon }$ and rescaling to restore the integral gives smooth admissible approximations converging in energy. The formal quadratic from the interior Euler-Lagrange equation and two initial conditions alone is therefore not a true minimizer.

Forwith free terminal values, the Euler-Lagrange equations and natural conditions give a nonzero stationary family exactly when $x_0=(k+\frac{1}{2})\pi$. It is

Forindependent variations and integration by parts give