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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.

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.

For real smooth coefficients with $p,w>0$, the Rayleigh quotient $\Lambda [y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ has the displayed first variation. Its stationary points solve the Sturm-Liouville equation on an admissible class whose boundary variations make $[p{y}^{\prime }\eta {]}_a^b=0$. Homogeneous Dirichlet boundary conditions do this by setting $\eta =0$ at the ends; homogeneous Neumann boundary conditions do it through $y^{\prime }=0$. More general self-adjoint boundary conditions can require a boundary contribution to the quadratic form.

For an eigenfunction, the weaker identity $[py{y}^{\prime }{]}_a^b=0$ suffices to give $\lambda =\Lambda [y]$ by integration by parts, but it does not by itself imply the condition on every variation. An explicit counterexample uses $p=q=w=1$ on $[0,1]$ and $y=\cos \pi x+\sin \pi x$. This solves $-y^{\prime \prime }+y=(1+{\pi }^2)y$ and has $[y{y}^{\prime }{]}_0^1=0$. Define a nearby familyIt keeps $y_{\epsilon }(0)=1$, $y_{\epsilon }^{\prime }(0)=\pi$, $y_{\epsilon }(1)=-1+\epsilon$, and $y_{\epsilon }^{\prime }(1)=\pi /(-1+\epsilon )$, so $[y_{\epsilon }{y}_{\epsilon }^{\prime }{]}_0^1=0$ for small $\epsilon$. Nevertheless, $\eta ={\partial }_{\epsilon }{y}_{\epsilon }{|}_0$ has $\eta (0)=0$, $\eta (1)=1$, and ${\int }_0^1{y}^{2}dx=1$, giving $\delta \Lambda =2[y^{\prime }\eta {]}_0^1=-2\pi \ne 0$. Boundary-energy cancellation alone therefore cannot replace a specified variational domain.

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.

A symmetric uniformly elliptic divergence-form energy with zeroth-order coefficient at least $\mu >0$ controls the full $H^1$ norm, including constants. Its quadratic energy minus a bounded load has a unique minimizer by the direct method, and vanishing first variation gives the natural Neumann weak equation. Unlike a pure Neumann Laplacian, this problem needs no zero-mean compatibility condition. The exact energy difference from the weak solution is half the bilinear energy of the difference.

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.

Choose Cartesian coordinates with origin at the upper pivot, $x$ to the right and $y$ upward. The two position vectors areThese expressions satisfy the rod holonomic constraints $|{\mathbf{r}}_1|=\ell$ and $|{\mathbf{r}}_2-{\mathbf{r}}_1|=\ell$ automatically. This is the practical advantage of using the two angles as coordinates: the rod tensions need not be unknowns in the motion calculation.^[ref]

Differentiate these position vectors to obtain the velocities. The dot product between their two rod contributions uses $\cos ({\theta }_1-{\theta }_2)=\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2$, givingAdding the two kinetic energies and measuring gravitational potential energy from the downward equilibrium givesThe coefficient $2$ has a geometric meaning: the upper rod raises both masses, and its motion contributes to both kinetic energies. The coupling in $T$ is present because the lower mass has a moving pivot. The zero chosen for $V$ changes $L$ by a constant and leaves the Euler-Lagrange equations unchanged.^[ref]

The principle of stationary action applies to $S=\int Ldt$. Its fixed-endpoint first variation, after integration by parts, isIndependent allowed variations therefore give one Euler-Lagrange equation for each angle. Stationary action alone does not assert a minimum.^[ref]

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.

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.

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 a fixed sufficiently regular image edge set, the Mumford–Shah functional is a strictly convex quadratic energy on the Sobolev space of its complement. A first variation gives $u-\alpha \Delta u=g$ within each region and separate one-sided Neumann boundary conditions on free edges. A jump need not vanish across an edge. The weak solution is unique by the Lax-Milgram theorem; this does not prove uniqueness when the edge set is also optimized.

For a plate occupying $x>0$, the boundary part of the first variation of its isotropic thin-plate bending energy is, up to an overall nonzero sign,At a free edge the variation and its normal derivative can be prescribed independently. Their coefficients must therefore vanish, giving the two displayed natural boundary conditions: zero bending moment and zero effective transverse shear. The mixed-derivative coefficient includes the derivative along the edge of the twisting moment; keeping only the normal derivative of $w_{xx}+\nu w_{yy}$ would miss that contribution.

For the functionalthe vanishing of its first variation under fixed endpoint data gives

For a homogeneous isotropic plate of thickness $h$, Young's modulus $E$ and Poisson's ratio $\nu$, the bending stiffness is $D=E{h}^3/[12(1-{\nu }^2)]$. Its quadratic bending energy isIt follows by integrating the strain energy density $\frac{1}{2}{\tau }_{ij}{e}_{ij}$ through the thickness: linear bending strains are proportional to $z$, and ${\int }_{-h/2}^{h/2}{z}^{2}dz=h^3/12$. Taking a first variation and integrating by parts twice gives $D{\Delta }^{2}w=0$ without applied transverse load. Clamped boundary variations have both zero value and zero normal derivative.

Lagrangian field theory specifies an action by integrating a Lagrangian density $\mathcal{L}(\varphi ,\partial \varphi ,x)$ over spacetime. Requiring its first variation to vanish for compactly supported field variations yields the field Euler-Lagrange equations. The variational Lagrangian is the integrand, whereas this field-theoretic formalism specifies the continuum fields and their dynamics.

Minimizing the log-cosh energy over zero-trace functions above an obstacle yields the displayed inequality for every feasible competitor. Convexity supplies the one-sided admissible segments. On the open noncontact region, compactly supported variations of both signs are feasible because the gap from the obstacle has a positive minimum on their support. Their first variations vanish, giving $-\Delta u+\tanh u=0$ there. This stationarity equation need not hold on the contact set.

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 $\delta y^{\prime }=0$.

For bounded BV space data and $s<t$, nested ROF level-set formulation minimizing sets cannot share an interface of positive surface measure away from data jumps. On regular graph patches, upward and downward first variations use the BV traces on a hypersurface. At contacts outside $J_f$ the forcing BV trace on a hypersurface agrees, whereas the two levels differ by $t-s$. Graph regularity and contact differentiation give the contradiction. Bounded forcing is a hypothesis of this argument; arbitrary unbounded $L^2\cap BV$ data are not covered by this statement.

The noncontact set is where the unknown function is strictly above its obstacle. For continuous functions it is open. Every smooth variation with compact support in this set remains feasible with either sign when its amplitude is small enough. The first variation therefore vanishes there, yielding the unconstrained Euler-Lagrange equation. On the complementary contact set of an obstacle problem, only the appropriate one-sided variational inequality is available globally.

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.