Calculus of variations

Calculus of variations finds stationary points of functionals, often among functions or regular curves.

The inequalities $0\le \log \cosh t\le t^2/2$ make this energy finite on $H^1$. Since the second derivative of its scalar potential is ${\mathrm{sech}}^{2}t\in (0,1]$, its Fréchet derivative is $DI(v)[h]=\int (\nabla v\cdot \nabla h+\tanh (v)h)$, with Taylor remainder between zero and half the squared $H^1$ norm of $h$. The derivative operator is strictly monotone because the hyperbolic tangent is strictly increasing.

This excess measures the height of the slope-dependent integrand above its tangent line at the reference slope $p$, evaluated at a trial slope $q$. A convex function of slope has nonnegative excess. At a smooth strong local minimizer, localized needle variations force nonnegative excess along the minimizing curve for every trial slope. It detects fast-gradient competitors that the second variation around a fixed derivative can miss.

At a strong local minimizer of a variational integral, almost every reference slope $u^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,u(x),v)$. Equivalently its Weierstrass excess function is nonnegative for every trial slope. Uniformly small variations can have large derivatives, which is why this condition tests all trial slopes rather than only nearby ones. It is a necessary condition along the minimizing curve; the strengthened Weierstrass condition additionally controls a neighborhood of that curve and its slopes.

Suppose the Weierstrass necessary condition applies to a strong local minimum of this fixed-endpoint integral. It requires the tangent to $g$ at $u^{\prime }(x)$ to lie below the entire graph for almost every $x$. A line of nonzero slope is unbounded above in one direction, while $g$ is bounded; hence this supporting line has slope zero. Its contact value is then a global minimum of $g$. The resulting function minimizes the integral globally, since its integrand is almost everywhere equal to the global lower bound. Thus for a bounded differentiable slope-only integrand a strong local minimum can exist only at global energy level. Weak local minima can behave differently because their competitors have nearby derivatives.

For a $C^1$ curve $u$, some $\delta >0$ must make the Weierstrass excess function nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ satisfying $|y-u(x)|<\delta$, $|p-u^{\prime }(x)|<\delta$. Thus the supporting-line property persists in a tube of nearby heights and reference slopes. The term “strengthened” here refers to this neighborhood requirement; strict excess for $q\ne p$ is a further condition.

Under the usual $C^2$ integrand regularity, an admissible $C^1$ curve satisfying the weak Euler-Lagrange equation, a coercively positive second variation, and the strengthened Weierstrass condition is a strong local minimizer of a variational integral. Strict excess away from the reference slope gives strict minimality. Coercive positivity means ${\delta }^{2}I(u)[\eta ,\eta ]\ge c\Vert \eta {\Vert }_{W^{1,2}}^2$ for every zero-trace variation and some $c>0$.

The second variation provides a neighboring field of extremals with slope $p(x,y)$. The associated endpoint-invariant Hilbert integral yields $I(v)-I(u)=\int E(x,v,p(x,v),v^{\prime })dx$ for uniformly close admissible $v$. The tube condition makes this integral nonnegative even when $v^{\prime }$ is large, explaining the passage from weak to strong minimality.

For a fixed-endpoint variational integral, an admissible function $u$ is a strong local minimizer if $I(v)\ge I(u)$ for all admissible $v$ whose uniform norm distance from $u$ is sufficiently small. Derivatives can be far apart. Hence strong local minimality implies weak local minimality, while a global minimizer implies both. Small narrow bumps and rapid low-amplitude oscillations explain why the reverse implication can fail.

With fixed endpoint values on a finite interval, the displayed derivative coercivity makes a finite-energy strong local minimizer of a variational integral also a local minimizer in $L^1$. Otherwise choose better competitors converging in $L^1$. The upper energy bound controls their derivatives in $L^p$, and a common endpoint controls their function norms. Compact Sobolev embedding gives a uniformly convergent subsequence; its limit is the $L^1$ limit. This contradicts strong local minimality. Convexity of the integrand is unnecessary for this neighborhood-transfer argument.

For a fixed-endpoint integral on $W^{1,\infty }(a,b)$, an admissible function $u$ is a weak local minimizer if $I(v)\ge I(u)$ whenever the Sobolev norm $\Vert v-u{\Vert }_{W^{1,\infty }}$ is sufficiently small. Both function and derivative must be close. “Weak” here labels the variational neighborhood convention, not the weak topology of a Banach space.

For $0<\delta <1$, choose the zero-trace piecewise affine curve with slope $\delta$ up to $1-\delta /4$ and slope $\delta -4$ thereafter. Periodicity makes $g^{\prime }$ equal to the same nonzero constant at both slopes, and $g^{\prime \prime }>0$ on small neighborhoods of both. For sufficiently small Lipschitz perturbations, local strict convexity bounds the energy difference below by $c\int {\varphi }^{\prime 2}$, since the common first-variation coefficient multiplies $\int {\varphi }^{\prime }=0$. This proves a strict weak minimum. The global supporting-tangent condition fails because its nonzero slope cannot stay below bounded $g$, so it is not a strong minimum. By contrast, tents of slopes $4n,-4n$ attain $g=-1$ everywhere and provide infinitely many distinct global minimizers.

For an autonomous smooth Lagrangian homogeneous of degree two in velocities, the Beltrami identity gives $\dot{q}\cdot L_{\dot{q}}-L=L$ constant on its Euler-Lagrange extremals. On a regular segment with this constant positive, the Euler-Lagrange expression for $\sqrt{L}$ equals that for $L$ divided by $2\sqrt{L}$, since the extra term is proportional to $\dot{L}$. Hence an energy extremal has the same path as a length extremal. Length is invariant under regular reparametrization: in two coordinates, on a graph $v=v(u)$, its density is $\sqrt{E+2F{v}_u+G{v}_u^2}|du|$. If a nonnegative semidefinite energy vanishes along a curve, the curve instead minimizes length at zero; division by $\sqrt{L}$ is not permitted there. For a positive definite metric, regular nonconstant curves have positive energy.

For a sufficiently smooth functional $I[y]={\int }_a^{b}F(x,y,y^{\prime },y^{\prime \prime })dx$, the first variation in direction $\eta$ is $\int (F_y\eta +F_{y^{\prime }}{\eta }^{\prime }+F_{y^{\prime \prime }}{\eta }^{\prime \prime })dx$. Twice applying integration by parts gives the interior coefficient $F_y-(F_{y^{\prime }}{)}^{\prime }+(F_{y^{\prime \prime }}{)}^{\prime \prime }$ and boundary expression $[(F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime })\eta +F_{y^{\prime \prime }}{\eta }^{\prime }{]}_a^b$. Variations supported in the interior and the fundamental lemma of the calculus of variations force the displayed equation. The remaining boundary terms give the natural boundary conditions for second-derivative functionals.

With values $y(0)=y(\pi /2)=y^{\prime }(\pi /2)=0$ and $y^{\prime }(0)=1$, the higher-order Euler-Lagrange equation is $y^{(4)}-y=0$. Write $c=\cosh (\pi /2)$ and $s=\sinh (\pi /2)$. Solving the four linear boundary equations gives $y=[c(\cos x-\cosh x)+(1-s)\sin x+(1+s)\sinh x]/2$. The added function $x^2$ does not affect the first variation.

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 $y^{\prime }{F}_{y^{\prime }}-F$.

At an endpoint of a second-derivative variational problem, a free value $y$ gives the first displayed condition, while a free slope $y^{\prime }$ gives the second. Each condition is required only when its corresponding endpoint variation is free; a prescribed value or slope instead forces that variation to vanish. For $F=K(y^{\prime \prime }{)}^2/2+qy$, the two conditions become $y^{\prime \prime \prime }=0$ and $y^{\prime \prime }=0$, respectively. A clamped beam has both value and slope fixed, a pinned beam fixes value but leaves slope free, and a free end leaves both free.

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 variational principle specifies a function, path or field by requiring a functional to be stationary under an allowed class of variations. The admissible class includes the prescribed boundary conditions and constraints. A stationary value need not be a minimum: the Euler-Lagrange equation expresses the first-order stationarity condition, while a minimum requires additional analysis. Fermat principle applies this idea to optical path length, and the Rayleigh-Ritz method approximates a variational problem in a chosen family of trial functions.

Let $H$ be a lower-bounded self-adjoint Hamiltonian operator and $I$ a commuting self-adjoint operator with $I^2=1$. For a normalized trial vector $\varphi$, define $a=⟨H{⟩}_{\varphi }$, $b=⟨I{⟩}_{\varphi }$, $c=⟨HI{⟩}_{\varphi }$. The Rayleigh quotient on $\varphi +tI\varphi$ is $[(1+t^2)a+2tc]/[1+t^2+2tb]$, with derivative $2(c-ab)(1-t^2)/(1+t^2+2tb{)}^2$. Unless $c=ab$, its extrema are the normalized nonzero vectors $\varphi \pm I\varphi$, with energies $(a\pm c)/(1\pm b)$. Their difference is $2(c-ab)/(1-b^2)$, identifying the maximum and minimum.

The orthogonal projections $(1\pm I)/2$ split any trial energy into a weighted mean of the two sector energies, so one nonzero sector projection never raises the trial energy. If $|b|=1$, one projected vector is zero and its quotient is undefined; the remaining family has constant energy. Exchange symmetry makes these sectors bosonic and fermionic spatial wavefunctions. Stationarity within this two-dimensional trial space does not generally make a vector an exact eigenvector of $H$.

For constant $K$, this quadratic combination of two-by-two gradient minors is a divergence. Indeed,because the term containing $u_{j,\beta \alpha }$ vanishes by antisymmetry of the Levi-Civita symbol. Its volume integral vanishes for homogeneous displacement boundary data. It also vanishes pointwise on rank-one matrices, since ${\nu }_{\alpha }{\nu }_{\beta }$ is symmetric. Consequently one can strengthen an elastic energy density by such a term without changing fixed-boundary incremental energies or acoustic tensors.

The brachistochrone problem asks for the curve along which a particle released from rest reaches a specified target fastest under a uniform gravitational field, without friction. With downward depth $u$, conservation of energy gives speed $\sqrt{2gu}$ and travel time $\int \sqrt{(1+u^{\prime 2})/(2gu)}dx$. The Beltrami identity makes $u(1+u^{\prime 2})$ constant, leading to a cycloid. If only the horizontal endpoint coordinate $x_0>0$ is fixed, the natural boundary conditions for a free endpoint select the half arch ending at its deepest point, with depth $2x_0/\pi$ and minimum time $\sqrt{\pi x_0/g}$.

For horizontal separation one, let $y>0$ be downward depth and constrain the arc length to $L$. Conservation of energy makes the travel time proportional to $\int \sqrt{1+y^{\prime 2}}/\sqrt{y}dx$. An isoperimetric constraint adds a Lagrange multiplier $\lambda$, and the Beltrami identity gives $(y^{-1/2}+\lambda )/\sqrt{1+y^{\prime 2}}=H>0$. With $b=\lambda /H$, $c=H^2$ and $s=\sin (\varphi /2)$, the stationary arch is $y=s^2/[c(1-bs{)}^2]$, $dx/d\varphi =s^2/[c(1-bs{)}^3]$, for $0\le \varphi \le 2\pi$. The endpoint and length integrals fix $b,c$ uniquely for $L>1$; symmetry gives $x(2\pi -\varphi )=1-x(\varphi )$. At $L=1$ the only curve is horizontal and a particle released from rest never completes the journey. The finite-time stationary arches have vertical endpoint tangents, so the graph is smooth only on the open interval. These first-variation equations alone do not establish global minimality in a larger admissible class.

Measure polar angle $\theta$ from the downward vertical on the lower hemisphere of radius $R$. A particle released from rest at the rim and sliding without energy loss has speed $\sqrt{2gR\cos \theta }$. The spherical arc length along $\varphi =\varphi (\theta )$ is $R\sqrt{1+{\sin }^2\theta {\varphi }^{\prime 2}}|d\theta |$. Thus its travel time has integrand proportional to $\sqrt{(1+{\sin }^2\theta {\varphi }^{\prime 2})/\cos \theta }$. Since the azimuth is a cyclic coordinate, the Euler-Lagrange equation givesSolving for ${\varphi }^{\prime 2}$ gives the displayed first integral on a segment where $\theta$ is a valid path parameter. For finite $C$, approaching the rim $\theta \uparrow \pi /2$ gives ${\varphi }^{\prime }\to 0$. Thus a regular stationary path begins along the meridian toward the bottom. This is an initial direction condition; a path joining other rim points can turn before reaching the bottom. Monotone segments use positive $|d\theta |$ in the time integral, and a turning point requires another path parameter or a split into segments.

Consider a frictionless track from rest between equal-height endpoints separated horizontally by $l>0$. Let depth $y$ be positive downward and let the maximum depth be $h>0$. Conservation of energy gives speed $\sqrt{2gy}$. The Cauchy-Schwarz inequality, applied to the tangent displacement, givesThe squared norm of the coefficient vector is $1/h+(h-y)/(hy)=1/y$. The track goes from depth zero to $h$ and returns to zero, so the second term integrates to at least $2{\int }_0^h\sqrt{(h-y)/(hy)}dy=\pi \sqrt{h}$. Integrating the first term gives $l/\sqrt{h}$, even if the track backtracks horizontally. Division by $\sqrt{2g}$ proves the first bound. The arithmetic-geometric mean inequality proves the second.

Equality requires $h=l/\pi$ and tangent slope $|dy/dx|=\sqrt{(h-y)/y}$, on the descending and ascending halves respectively. Substitution $y=h{\sin }^2\theta$ integrates to the cycloid $x=h(\theta -\frac{1}{2}\sin 2\theta )$, $0\le \theta \le \pi$. It attains the bound, proving global minimality of the single-arch brachistochrone problem between these endpoints. Endpoint singularities are handled as limits of the integrals; the cycloid has finite travel time.

Measure $\theta \in [-\pi /2,\pi /2]$ from the vertically downward direction on the lower half of a cylinder of radius $R$, and let $z$ be axial distance. From rest at an edge, conservation of energy gives speed $\sqrt{2gR\cos \theta }$. The travel time is $T=(2gR{)}^{-1/2}\int \sqrt{R^2+z^{\prime 2}}/\sqrt{\cos \theta }d\theta$. Since $z$ is a cyclic coordinate, its conjugate quantity $z^{\prime }/[\sqrt{\cos \theta }\sqrt{R^2+z^{\prime 2}}]=K$ is constant, giving the displayed formula. For finite positive axial separation, $0<K<1$ is fixed uniquely by integrating $z^{\prime }$ between the edges: the separation grows strictly from zero to infinity as $K$ increases from zero to one. The time integrand is strictly convex in $z^{\prime }$, so its supporting-line inequality proves global minimality for the fixed endpoints. As the starting edge is approached, $z^{\prime }\to 0$: the initial direction is directly down the curved surface, with no axial component. This differs from the constant-pitch shortest geodesic on the unrolled cylinder.

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.

A sequence of functionals Gamma-converges in a specified topology when every convergent $u_n\to u$ satisfies $F(u)\le {\mathrm{liminf}}_n{F}_n(u_n)$, and each $u$ has a recovery sequence $u_n\to u$ with $F_n(u_n)\to F(u)$. With appropriate compactness of energy sublevels, limits of approximate global minimizers minimize $F$, and the minimum values converge. This is suited to variational approximation such as the Ambrosio–Tortorelli approximation; it does not describe arbitrary local minima.

A variational problem seeks a stationary point or minimizer of a functional on a specified function space. For a symmetric coercive bilinear form $a$ and bounded linear functional $\ell$ on a real Hilbert space, minimizing $J(v)=a(v,v)/2-\ell (v)$ is equivalent to the weak formulation $a(u,v)=\ell (v)$ for every test vector. The Lax-Milgram theorem gives a unique solution, and $J(u+w)-J(u)=a(w,w)/2$ proves unique minimization. Specifying the admissible space and its boundary constraints is part of the problem.

The direct method takes a minimizing sequence, extracts a subsequence using compactness, and uses lower semicontinuity to show that its limit attains the infimum.

A Lavrentiev gap is a strict change in an infimum when the admissible regularity class is restricted, despite retaining the same endpoint values. For example $I(u)={\int }_0^1(u^3-x{)}^2{u}^{\prime 6}dx$ with $u(0)=0,u(1)=1$ has value zero at the absolutely continuous function $u=x^{1/3}$. Every Lipschitz competitor has a uniform positive lower energy bound. Indeed its ratio to $x^{1/3}$ starts at zero, so let $x_0$ be its first crossing of $u(x)=x^{1/3}/2$. Before that point, $|u^3-x|\ge 7x/8$. Holder inequality givesThus $I(u)\ge (49/4096)(3/5{)}^5/x_0\ge (49/4096)(3/5{)}^5>0$. This proves a gap directly. The example's integrand lacks a uniform superlinear lower bound along its zero set; it illustrates the admissible-class issue rather than by itself furnishing a counterexample under stronger coercivity assumptions.

For $I_{\epsilon }(u)={\int }_0^1[(u^4-x{)}^2+\epsilon |u^{\prime }{|}^{4/3}]dx$ with endpoint values zero and one, the zero-penalty minimizer $x^{1/4}$ has infinite derivative penalty. Use $u=x^{1/4+\delta }$ instead. The mismatch is at most $(32/27){\delta }^2$, while the derivative penalty is $(3\epsilon /(4\delta ))(1/4+\delta {)}^{4/3}$. Balancing ${\delta }^2$ and $\epsilon /\delta$ by $\delta ={\epsilon }^{1/3}$ gives an upper energy bound of order ${\epsilon }^{2/3}$. Nonnegativity then proves convergence of minimum energies to zero. This is a recovery construction; no claim of a matching lower asymptotic constant is needed.

For a scalar fixed-endpoint problem on a bounded interval, suppose the continuous integrand $f(x,y,p)$ is convex in $p$ and satisfies $f(x,y,p)\ge \Theta (|p|)-C$, where $\Theta$ is nonnegative convex and $\Theta (r)/r\to \infty$. If there is one admissible absolutely continuous function of finite energy, a global minimizer exists in $W^{1,1}$.

The lower bound gives uniform integrability of derivatives in a minimizing sequence. Fixed endpoint values then give boundedness and equicontinuity of the functions. Extract uniform convergence of the functions and weak $L^1$ convergence of their derivatives; the limit retains the endpoint values. Convexity in the derivative gives weak lower semicontinuity of convex gradient energies, so the limit attains the infimum. Strict convexity of the full integrand on a convex admissible class additionally gives uniqueness.

Suppose the domain has finite measure and $F\in C^1$ is convex, with $|F(p)|\le C(1+|p{|}^2)$ and $|DF(p)|\le C(1+|p|)$. The supporting inequality $F(p)\ge F(q)+DF(q)\cdot (p-q)$, integrated with $q=Du$, gives $\int F(D{u}_k)\ge \int F(Du)+\int DF(Du)\cdot (D{u}_k-Du)$. The last term tends to zero by weak convergence of the gradients and the fixed $L^2$ test field $DF(Du)$. This proves weak lower semicontinuity without requiring pointwise convergence of gradients.

For a bounded domain, bounded symmetric uniformly elliptic $A$, $q\le 0$, and $f\in L^2$, minimizing $J$ on $\psi +H_0^1$ solves $\mathrm{div}(A\nabla u)+qu=f$. The nonnegative quadratic terms are weakly lower semicontinuous, and Poincaré inequality makes the gradient norm coercive on the zero-boundary part. The first variation gives the weak equation with the forcing sign displayed above. If $q$ is positive, coercivity may fail at a Dirichlet eigenvalue; orthogonality to the homogeneous kernel is then needed for solvability.

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.

Take $a=3\pi /2$, $L=10$ and $u(x)=a\exp (-|x{|}^2/L^2)$ on ${\mathbb{R}}^3$. Set $f=-\Delta u+u+\sin u$. Both functions are smooth and belong to L2 space, and $u\in H^1$. The strictly increasing reaction $h(s)=s+\sin s$ has $0\le h(u)\le h(a)=a-1$, whileConsequently $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert u{\Vert }_{\infty }$. This $u$ is a minimizer of the screened sine-Gordon energy: it solves the Euler-Lagrange equation and that energy is convex. Thus even the minimizing and smooth hypotheses do not justify the stronger source-size bound. The appropriate general estimate is maximum bound for a monotone reaction term.

A minimizing sequence satisfies $E(u_n)\to \inf E$. Coercivity bounds an appropriate sublevel set, Compactness supplies a convergent subsequence, and sequential lower semicontinuity turns its limit into a global minimizer.

A minimizing sequence can converge uniformly while its derivatives keep oscillating between distinct energy wells. For the functional $I(u)={\int }_0^1[u^4+(u^{\prime 4}-16{)}^4]$, with zero endpoint values and Lipschitz admissible functions, the triangular waves displayed above have slopes $\pm 2$ almost everywhere and height $1/N$. Direct integration gives $I(u_N)=1/(5N^4)$. Nonnegativity therefore makes the infimum zero. Attainment would require both $u=0$ and $|u^{\prime }|=2$ almost everywhere, an impossibility. The uniform limit is zero but its energy is $16^4$; thus the energy fails to be lower semicontinuous for this convergence. The example explains why control of function values alone cannot replace suitable convexity in the derivative variable in the direct method in the calculus of variations.

Boundary data $g$ satisfy the bounded slope condition with constant $K$ when every boundary point admits affine lower and upper supporting functions of Lipschitz constant at most $K$. Comparison with these barriers gives a global Lipschitz bound for convex variational problems.

For suitable convex integral functionals, ordered boundary values of two minimizers imply the same ordering in the domain. Affine minimizers can therefore serve as upper and lower barriers.

A functional is a function whose inputs are themselves functions, curves, or other elements of a function space.

A quadratic functional on a real vector space combines a quadratic form $B(v,v)$, a linear functional $\ell (v)$ and a constant. If $B$ is a symmetric bounded bilinear form, its first variation is $DI(v)[w]=2B(v,w)-2\ell (w)$. For a strictly positive $B$, an existing stationary point is the unique minimizer by the quadratic variational principle for a symmetric positive operator.

The Euler-Lagrange equation is $y^{\prime \prime }+ky=0$. For positive $k$ a nonzero stationary function exists exactly when $\sqrt{k}$ is a positive integer $n$, and all such functions are $B\sin (nx)$. These values are the Dirichlet eigenvalues of $-d^2/d{x}^2$ on the interval. At $k=0$ the solutions are affine and the endpoint conditions leave only zero; negative $k$ likewise leaves only zero. Stationarity alone does not assert minimization of the quadratic functional.

Impose zero Dirichlet boundary conditions on an interval. The isoperimetric constraint $I[y]=1$ is regular because $D{I}_y[y]=2I[y]=2\ne 0$. The Lagrange multiplier rule applied to $J-\lambda (I-1)$ gives the Sturm-Liouville equationMultiplying by $y$ and using integration by parts gives $J[y]=\lambda I[y]=\lambda$. Thus the multiplier is the stationary value of the original functional, or its Rayleigh quotient. This calculation does not require $w$ to have one sign; positive-weight spectral conclusions require additional hypotheses.

If two stationary functions have different multipliers, multiplying their equations crosswise and subtracting gives $({\lambda }_n-{\lambda }_m)\int w{y}_n{y}_m=0$. The endpoint terms vanish. This proves the weighted orthogonality of Sturm-Liouville eigenfunctions. Higher stationary values need not be local minima of the constrained functional.

The epigraph is $\{(u,t):E(u)\le t\}$. It is a convex set exactly when $E$ is a convex function, and is a sequentially closed set in the product topology exactly when $E$ is sequentially lower semicontinuous.

The functional derivative is defined by the first variation

The first variation of a differentiable action functional gives its change under a small field deformation. With $E=\delta S/\delta \varphi$, the change $\varphi \mapsto \varphi -\eta f$ generates $-\eta \int fE$ to first order. This is the basic cancellation behind the equivalence theorem for field redefinitions; the functional Jacobian and transformed operator insertions must also be treated consistently.

Along a differentiable path of fields $\varphi (x,t)$, a differentiable functional obeysIntegrating in time expresses the functional change as the line integral of its functional derivative along the path.

An energy functional assigns an integral energy to a function. Minimization of a suitable energy is the basis of the Dirichlet principle. For the modified Helmholtz equation, a natural example is

A field configuration is critical for an energy functional $E$ if its first variation vanishes for every admissible infinitesimal field variation. For $E[\varphi ]=\int (\frac{1}{2}|\nabla \varphi {|}^2+U(\varphi ))$, integration by parts against compactly supported variations gives $\delta E=\int (-\Delta \varphi +U^{\prime }(\varphi ))\delta \varphi$, so stationary field configurations satisfy the Euler-Lagrange equation $\Delta \varphi =U^{\prime }(\varphi )$. A constrained target instead allows only tangent variations, changing that field equation. Criticality is stationarity, not necessarily energetic stability or a local minimum.

The p-energy of a Sobolev function isIts stationary points under fixed boundary data are weak solutions of the p-Laplacian equation.

A Lagrangian is a function or density whose stationary integral encodes an extremization problem. In Lagrangian mechanics its action gives the equations of motion; in geometry it describes geodesics and minimal surfaces.

In Lagrangian, the action integrates such a density over spacetime.

A finite-dimensional optimization Lagrangian combines an objective function with Lagrange multiplier terms for its constraints.

A variation of a function $y$ is a one-parameter family $y+\epsilon \eta$, where $\eta$ satisfies the required boundary conditions.

The first variation of a functional $L$ at $y$ in the direction $\eta$ is

If a continuous function $f$ satisfiesfor every smooth compactly supported test function $\eta$, then $f=0$ on $(a,b)$.

A point is stationary for a differentiable function or functional when its first derivative or first variation vanishes in every admissible direction.

The gradient vanishes only when $y=0$ and $\tan x=-2x$. On each tangent branch, $\tan x+2x$ increases strictly from negative to positive infinity, so it has exactly one root $x_s$. With $z_s=e^{-x_s^2}\cos x_s$, the Hessian matrix is diagonal with entries $-(4x_s^2+3)z_s$ and $-2z_s$. Its determinant is positive, so every stationary point is a strict maximum when $z_s>0$ or a strict minimum when $z_s<0$. Signs alternate between successive tangent branches. This gives infinitely many nondegenerate extrema under a decaying Gaussian function envelope.

Where $q(x)\ne 0$, the product rule and chain rule give the displayed real derivative. Its stationary points therefore satisfy $3q+(x-a)q^{\prime }=0$. The denominator uses the square of the real cube root, which stays positive for negative $q$. Points with $q=0$ must be checked in the original function: a simple zero of $q$ with $x\ne a$ creates a vertical tangent, not a stationary point. If the prefactor also vanishes, a cancellation can change differentiability and must be analyzed separately.

Suppose $h$ is differentiable and $F^{\prime }>0$ on its range. The multivariable chain rule shows that $F\circ h$ has exactly the same stationary points as $h$. Since $F$ is strictly increasing, the type of every local maximum, local minimum, and saddle point is preserved. If both functions are twice differentiable, then at a stationary pointso the Hessian matrix signature is unchanged. The hypothesis $F^{\prime }>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero derivative and create new stationary points in a composition.

Among functions with fixed boundary values, the solution of $\nabla \cdot (\kappa \nabla \varphi )=0$ with $\kappa >0$ uniquely minimizes the weighted energy

Let $V$ be a closed linear subspace of a real Hilbert space, and let $Q(v)=a(v,v)$ for a symmetric bounded bilinear form that is nonnegative on the ambient space and coercive on $V$. Then $Q$ has a unique minimizer on every closed affine subspace $g+V$. If $Q(v_j)\to m={\inf }_{g+V}Q$, the midpoint belongs to $g+V$ and the parallelogram law gives $Q(v_j-v_k)\le 2Q(v_j)+2Q(v_k)-4m\to 0$. Coercivity on $V$ makes the minimizing sequence Cauchy; completeness and continuity give a minimizer. The vanishing first variation is $a(u,v)=0$ for all $v\in V$. Two minimizers differ by $w\in V$ with $Q(w)=0$, so coercivity gives uniqueness. This proves the symmetric existence theorem directly, without invoking the Lax-Milgram theorem.

For $g(1)=1$ and finite displayed energy, the angular mode $g(r)\cos \theta$ has Dirichlet energy equal to $\pi$ times that integral. The harmonic function $r\cos \theta$ minimizes the energy with the same boundary data by the Dirichlet principle, giving the bound. Finite energy forces $g(r)\to 0$ as $r\to 0$: Cauchy–Schwarz bounds changes of $g^2$ by twice the product of the two energy tails, so $g^2$ has a limit, and a nonzero limit would make $\int g^2/r$ divergent. Consequently completing the square gives the exact identity ${\int }_0^1[r{g}^{\prime 2}+g^2/r]=1+{\int }_0^{1}r(g^{\prime }-g/r{)}^2$. Equality occurs exactly for $g(r)=r$. This identity also covers finite-energy functions whose polar expression is not classically smooth at the centre.

Fermat-type optical paths in cylindrical coordinates extremize refractive index times Euclidean arclength.

A helical extremal has constant radius and a polar angle linear in axial position.

Noether’s theorem associates a conserved quantity to each continuous variational symmetry.

For independent coordinates with $L={\sum }_i{m}_i({\dot{x}}_i^2-{\omega }^2{x}_i^2)/2$, the variables $X_i=\sqrt{m_i}{x}_i$ give an isotropic oscillator even at unequal masses. Rotations of the $X_i$ are continuous symmetries. Their conserved Noether charges $X_i{P}_j-X_j{P}_i$ therefore predate any equality of physical masses. Equal masses alone cannot create new invariances of arbitrary additional interactions.

Invariance under a shift of the time coordinate, with other required background structures held fixed. A continuous time-translation symmetry gives an energy conservation law through Noether theorem. A generic expanding cosmological background is not stationary and need not supply a corresponding globally conserved total energy.

Suppose an action is invariant under a local transformation $\delta {\Phi }_a=R_a\alpha +R_a^{\mu }{\partial }_{\mu }\alpha$ for every compactly supported function $\alpha$. Variation and integration by parts give the off-shell identityThus arbitrary local symmetry parameters imply dependencies among the field equations. In gauge theory, these identities accompany constraints rather than independent physical charges for each function. This complements the conserved-current statement of Noether theorem.