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Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

For a sample size $m\ge 2$ define $\overline{X}=m^{-1}{\sum }_i{X}_i$ and $S_X^2=(m-1{)}^{-1}{\sum }_i(X_i-\overline{X}{)}^2$. The normal-sample theorem givesIt also states that the mean and sample variance are independent. Combining (i) and (ii), and the independence across samples, the correctly centred pivot isThe source reverses the mean difference inside the subtraction. Its literal statistic is generally noncentral when ${\mu }_X\ne {\mu }_Y$, rather than having a central Student distribution. Under the null hypothesis both differences are zero, so this misprint does not change the requested null test.

For a two-sided level-$\alpha$ test reject when $|T|>t_{m+n-2,1-\alpha /2}$. In the numerical example the pooled variance is $0.038$, the standard error is $\sqrt{0.038(2/5)}\approx 0.12329$, and $T\approx -2.190$. There are eight degrees of freedom and the supplied critical value is $2.31$. Thus do not reject equal mean cell volume at the 5% level. This does not establish equality of the means; it states that the observed difference does not cross this test's two-sided threshold.

Age matching introduces a shared block factor: the two observations in a pair can be associated, and their age-dependent means also undermine the simple unblocked independent-samples model. A paired t-test on the within-pair drug-minus-placebo differences accounts for that design, assuming independent pairs and the corresponding normal-difference model. Shared age variation cancels in those differences; positive within-pair association reduces their variance and can increase power, although the reduction in degrees of freedom means improvement is not automatic. The paired test cannot be computed from the two marginal means and variances alone: the within-pair differences or covariance are needed.

At a regular constrained extremum, choose an admissible variation $k$ with $\delta J[k]\ne 0$. The implicit function theorem applied to $J[y+\epsilon h+sk]$ permits a compensating $s(\epsilon )$ keeping $J$ fixed. Differentiation at zero gives $\delta I[h]=\lambda \delta J[h]$ for every $h$, where $\lambda =\delta I[k]/\delta J[k]$. Integrating the first variation of $I-\lambda J$ by parts and using arbitrary compactly supported $h$ givesSmoothness alone does not exclude an abnormal constraint: if $\delta J$ vanishes identically, one needs the general multiplier form with two multipliers rather than automatically setting the coefficient of $I$ to one. The displayed normal form uses the usual regular-constraint hypothesis.

For the given quadratic functionals, the equation is $\boxed{(p{y}^{\prime }{)}^{\prime }+qy+\lambda ry=0}$. If the endpoint values of $y$ are free, the boundary variation $[p{y}^{\prime }h{]}_a^b$ gives the natural conditions $\boxed{p(a)y^{\prime }(a)=p(b)y^{\prime }(b)=0}$, interpreted by limits at singular endpoints. Multiply the equation by $y$ and integrate. These boundary terms vanish, leavingFor two eigenfunctions, multiply their equations by the other function and subtract. Integration givesThe natural conditions remove the bracket, so distinct eigenvalues imply $\boxed{{\int }_a^{b}r{y}_m{y}_{n}dx=0}$.

In the Legendre case, comparison of leading polynomial coefficients gives ${\lambda }_n=n(n+1)$, so the degrees correspond to distinct eigenvalues. A degree-zero polynomial is $y_0=C_0$. Write $y_1=Ax+B$; orthogonality to $1$ on $[-1,1]$ gives $B=0$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThusThese polynomials satisfy the natural endpoint limits because $1-x^2$ vanishes at both endpoints.

Divide the infinitesimal variation by its constant parameter $\alpha$, so the transformation generator is $(\delta {\varphi }_1,\delta {\varphi }_2)=({\varphi }_2,-{\varphi }_1)$. The variation of the kinetic term isand that of ${\varphi }_1^2+{\varphi }_2^2$ is $2{\varphi }_1{\varphi }_2-2{\varphi }_2{\varphi }_1=0$. Thus the Lagrangian density is invariant pointwise, without a nonzero boundary current. The finite transformation is a rotation of the two real scalar fields, giving an SO(2) global symmetry in field theory.

Noether's theorem supplies the parameter-independent Noether currentIts divergence is ${\partial }_{\mu }{j}^{\mu }={\varphi }_2\square {\varphi }_1-{\varphi }_1\square {\varphi }_2$, since the first-derivative terms cancel. Both fields obey the same Klein-Gordon equation, so this vanishes on solutions. With spatial boundary conditions giving zero flux at infinity, the Noether charge isA finite box with periodic boundary conditions works as well. The spatial integral is conserved because the local continuity equation converts its time derivative into a boundary flux; neither vanishing local current nor constant individual fields is required. Equivalently, $\Phi =({\varphi }_1+i{\varphi }_2)/\sqrt{2}$ transforms as $\Phi \mapsto e^{-i\alpha }\Phi$, exhibiting the equivalent global U(1) symmetry and fixing the sign of the generator.

Take a smooth variation $y_{\epsilon }=y+\epsilon \eta$. Differentiating the functional at zero gives the first variationIntegrate the second term once and the third term twice by integration by parts:Choose variations supported away from the endpoints. Stationarity and the fundamental lemma of the calculus of variations forceThis is the Euler-Lagrange equation for a second-derivative functional. With the interior equation imposed, the remaining boundary expression must vanish for every admissible variation.

At each endpoint, a freely varying value $y$ requires $\boxed{F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime }=0}$, and a freely varying slope $y^{\prime }$ requires $\boxed{F_{y^{\prime \prime }}=0}$. A fixed value or slope instead makes its corresponding variation zero and supplies no natural condition for that variable. These are the natural boundary conditions for second-derivative functionals, with the value and slope variations treated independently.

Let $g>0$ be the gravitational acceleration, and let $y$ denote positive downward depth. For a frictionless particle released from rest, conservation of energy gives $v^2=2gy$. The arc length element of $(x,-y(x))$ is $ds=\sqrt{1+y^{\prime 2}}dx$, so the travel time and length areFor a finite-time single arch, require $y>0$ in $0<x<1$, $y(0)=y(1)=0$, and interpret the endpoint integrals improperly. At a regular constrained minimizer with $S[y]=L$, the Lagrange multiplier rule for an isoperimetric constraint makes $\sqrt{2g}T+\lambda (S-L)$ stationary. Dropping the constant $-\lambda L$ givesFor $L>1$ the constraint is regular under interior variations: if its first variation vanished for every compactly supported variation, the Euler-Lagrange equation for $S$ would force $y^{\prime }$ constant, hence $y\equiv 0$ and $S=1$, a contradiction. Thus the multiplier is justified for a nondegenerate arch.

The printed case $L=1$ is degenerate. Since $\sqrt{1+y^{\prime 2}}\ge 1$, equality of the length with the horizontal separation forces $y^{\prime }=0$ and $y=0$. A particle initially at rest then has no tangential gravitational acceleration and cannot complete the trip. Its travel time is infinite. The finite-time calculation therefore concerns $L>1$, with endpoint regularity qualified below.

Use natural units and the Minkowski metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, with $A^{\mu }=(\Phi ,\mathbf{A})$ and $A_{\mu }=(\Phi ,-\mathbf{A})$. Then $F_{0i}=E_i$, $F^{i0}=E_i$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. The electromagnetic field tensor is an antisymmetric tensor, which gives givesintegration by parts in the action, with the variation vanishing on its boundary, yields $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. Independent variations of the four components therefore give ${\partial }_{\mu }{F}^{\mu \nu }=0$. The time component is $\nabla \cdot \mathbf{E}=0$; the spatial components are $\nabla \times \mathbf{B}-{\partial }_t\mathbf{E}=0$. These are the two source-free Maxwell equations obtained dynamically. The other two, $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}+\nabla \times \mathbf{E}=0$, follow identically from $F=dA$, through the Electromagnetic Bianchi identity. They are not additional Euler-Lagrange field equations.^[ref]

Hold $\varphi ,{\varphi }^{\ast }$ fixed while varying $A_{\nu }$. With the specified convention $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$, the two matter variations are $\delta (D_{\mu }\varphi )=iq\varphi \delta A_{\mu }$ and $\delta (D_{\mu }\varphi {)}^{\ast }=-iq{\varphi }^{\ast }\delta A_{\mu }$. ThusAdding the Maxwell variation from part (a) gives $\delta S=\int d^{4}x[{\partial }_{\mu }{F}^{\mu \nu }-J^{\nu }]\delta A_{\nu }$. ConsequentlyExpanding the gauge covariant derivatives supplies a useful convention check:In particular, the last sign is negative for the specified plus sign in $D_{\mu }$. Reversing that convention requires consistently changing the phase transformation and current conventions as well. This current is the matter source in scalar electrodynamics; its $A$ dependence must not be dropped.

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables during variation. With Minkowski metric $(+---)$, differentiation with respect to ${\varphi }^{\ast }$ givesThe Euler-Lagrange field equation is therefore $\square \varphi +2ie{A}^{\mu }{\partial }_{\mu }\varphi +ie({\partial }_{\mu }{A}^{\mu })\varphi +m^2\varphi =f$. The static purely temporal potential has ${\partial }_{\mu }{A}^{\mu }={\partial }_{t}V=0$, givingvariation with respect to $\varphi$ gives $(\square -2ie{A}^{\mu }{\partial }_{\mu }+m^2){\varphi }^{\ast }=f(t)$. This is the complex conjugate equation when the driving force is real, as is required for the supplied lagrangian density to be real. For a complex driving amplitude a real source term would instead require conjugate coefficients. The interaction supplied here is linear in the external potential: adding the $e^2{A}_{\mu }{A}^{\mu }|\varphi {|}^2$ term from full minimal coupling would change the problem and its equation of motion.

The finite passive dilatation is $x^{\prime }=e^{-\alpha }x$, ${\varphi }^{\prime }(x^{\prime })=e^{\alpha }\varphi (x)$. To compare at the same argument, invert the coordinate change:Consequently the fixed-argument infinitesimal variation isThis also follows from $\delta \varphi =\alpha \varphi$ and $\delta x^{\mu }=-\alpha x^{\mu }$. The plus sign on the derivative term is a consequence of this specified contraction of coordinates.

Assume four spacetime dimensions, $\Phi \ne 0$, constant $\omega$, no scalar potential and matter independent of $\Phi$. In the stated action convention, scalar variation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2-2\omega \square \Phi =0$. The trace of the metric tensor equation gives $\Phi R+\omega {\Phi }^{-1}(\nabla \Phi {)}^2+3\square \Phi =8\pi GT$. Their subtraction proves the displayed wave equation with $T=g^{ab}{T}_{ab}$. The degenerate coupling of Brans-Dicke theory needs separate treatment before dividing by its coefficient.

Adding $dF(q,t)/dt$ to a mechanical Lagrangian changes its action by $F(q_f,t_f)-F(q_i,t_i)$. Fixed endpoint variations therefore leave its bulk Euler-Lagrange equations unchanged. For fields, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by a boundary flux. The same conclusion holds for compactly supported variations or appropriate fixed boundary data. If $K$ depends on field derivatives, those boundary data must also control their variations. Such terms can still change canonical momenta, boundary observables, and representatives of Noether currents; equivalence of bulk equations does not mean that every boundary-value problem is identical.

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.

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.

Differentiating $g^{ac}{g}_{cb}={\delta }_b^a$ and multiplying by the inverse gives the displayed variation. It is the matrix inverse derivative, expressed with the covariant and contravariant components of a metric tensor.

For a family of Levi-Civita connections, the difference of affine connections is a tensor. Dividing by the family parameter and taking the limit shows that its variation is tensorial. Vary the metric compatibility equation and retain the symmetry of the two lower indices. Adding the two cyclic identities and subtracting the third gives the displayed result. In a normal coordinate frame it is the direct derivative of the Christoffel formula; its tensorial form then holds in every frame.

For a nondegenerate metric tensor whose signature is fixed during the variation, the Jacobi formula gives $\delta \det g=(\det g)g^{ab}\delta g_{ab}$. Differentiating the square root of the absolute determinant proves the displayed identity. Expressed through the variation of inverse metric, the formula is $-\sqrt{|\det g|}{g}_{ab}\delta g^{ab}/2$. Absolute values accommodate either Riemannian or Lorentzian signature.